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2300	https://classwork.com/i-do-we-do-you-do-strategy-gradual-release-strategy/	Skip to content Are you looking for the Classworks Special Education program from TouchMath? Click here to go to their site. I Do, We Do, You Do Strategy: How to use the gradual release strategy [PLUS Examples!] You’ve heard about the I Do, We Do, You Do strategy as a way to scaffold new information in your classroom. But chances are, you’re already doing it – teachers are naturals at the technique. We’ll break down the strategy, give examples of using it, and show you how to do Gradual Release using technology. What is I Do, We Do, You Do? “I do, we do, you do” is a teaching strategy that involves a gradual release of responsibility from the teacher to the students. The three phases are: I do: In this phase, the teacher models how to complete a task or solve a problem. The teacher may use think-alouds, demonstrations, or other methods to show the students how to do the task. We do: In this phase, the teacher and the students work together to complete the same task or solve the same problem. The teacher provides support and guidance as needed, but the students actively participate in the task. You do: In this phase, the students work independently to complete a similar task or solve a similar problem. The teacher provides feedback and support as needed, but the students are responsible for completing the task independently. The goal of the “I do, we do, you do” strategy is to gradually shift the responsibility for learning from the teacher to the students. Students can build their skills and confidence over time by starting with explicit instruction and modeling, moving to guided practice, and finally to independent practice. 10 I Do, We Do, You Do Examples To Jumpstart Your Lesson Plans Here are a few examples of the Gradual Release strategy at work, so you can understand how to apply I Do, We Do, You Do in your class. Math problem-solving: The teacher models how to solve a math problem (I do), then the class solves similar problems together (we do), and finally, students solve similar problems independently (you do). Close reading: The teacher reads a passage aloud and models how to annotate it (I do), then the class annotates a similar selection together (we do), and finally, students annotate a new passage independently (you do). Writing: The teacher models how to write a paragraph (I do), then the class writes a similar paragraph together (we do), and finally, students write a new paragraph independently (you do). Science experiments: The teacher demonstrates how to conduct an experiment (I do), then the class conducts a similar experiment together (we do), and finally, students conduct a new experiment independently (you do). Language learning: The teacher models how to use a new vocabulary word in a sentence (I do), then the class practices using the word together (we do), and finally, students use the word in a new sentence independently (you do). Art projects: The teacher demonstrates how to create a specific type of artwork (I do), then the class creates similar artwork together (we do), and finally, students create a new piece of artwork independently (you do). Debate: The teacher models how to make an argument (I do), then the class debates a similar topic together (we do), and finally, students debate a new topic independently (you do). Social studies research: The teacher models how to conduct research on a specific topic (I do), then the class conducts similar research together (we do), and finally, students conduct research on a new topic independently (you do). Music performance: The teacher demonstrates how to play a piece of music (I do), then the class plays the piece together (we do), and finally, students perform a new piece of music independently (you do). Physical education: The teacher models how to perform a specific exercise (I do), then the class performs the exercise together (we do), and finally, students perform a new exercise independently (you do). Incorporate Gradual Release into your Google Slides and PowerPoint presentations during instruction. One way to increase student engagement is to incorporate I Do, We Do, You Do into direct instruction. You can do this effortlessly with TeacherMade’s Slide View. Slide View can integrate questions and activities into your Google Slides or PowerPoints. So you can demonstrate a new skill or concept, work a question together as a class using our 20+ question types, and then let your students try the concept on their own. When students submit their answers through Slide View, you get instant access to the results. So you can address common issues instantly. That instant connection to feedback makes your lesson more engaging and effective. With TeacherMade Slide View you can: Upload whole slide fecks from Google Slides or PowerPoint Merge Slideshows with interactive activities Create interactive presentation slides Embed video Get Instant Feedback and data on your students with a cost-effective Nearpod alternative. Design integrated classroom experiences where teaching, practice, and assessment happen and students learn more. Try TeacherMade 2.0 Slide View today with our risk-free 30-day trial.
2301	https://zhuanlan.zhihu.com/p/585479186	初一数学一元一次方程市场经济打折销售问题专题，配详细解题分析 - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册初一数学一元一次方程市场经济打折销售问题专题，配详细解题分析首发于初中数学提升切换模式初一数学一元一次方程市场经济打折销售问题专题，配详细解题分析 Harleytang小课堂有关教育的知识分享和学习方法探讨收录于 · 初中数学提升 3 人赞同了该文章初一数学一元一次方程市场经济打折销售问题专题，配详细解题分析初一数学上学期有关一元一次方程这一章节内容的学习过程当中，除了对方程的解法以及表示的实际意义有充分地了解以外，对于一元一次方程在实际问题中的应用这一学期学习的重点内容。那么利用一元一次方程来解决实际问题的过程当中，有关于市场经济打折销售的问题，是同学们在学习中的重点和难点。本身同学们对于市场活动在了解就不多的情况下，要结合实际解决问题，相对来说比较困难，但是其实质还是利用方程在方法通过对所给题目当中的数量关系来建立方程。达到解决实际问题的效果。很多同学在学习这部分内容时总觉得很吃力，或者是对题目的意图所表达的意思理解不透彻，主要是对于其中涉及的市场经济的计算关系存在不熟悉或者是运用不熟练的情况。所以同学们在学习当中想要搞定一元一次方程，解决实际问题中市场经济和打折售的问题，那么要从以下几个方面着手。首先，对于一元一次方程的理解一定要透彻。其实一元一次方程是小学数学学习过程当中的重点内容，如果基础掌握牢固的情况下，进入初一的一元一次方程解题过程当中对于方程的理解应当是不太困难的，而且在解方程的过程当中，利用小学所学的知识内容也能够轻松解决，只是初中解决一元一次方程的过程当中，要注意解题过程的明确，毕竟按照考试评分的标准来说，过程也是要给分的，所以我们只需要注重对方程解的过程当中顾城的详细标准化即可。其次，对于解方程中如何列方程成为大家解题过程中的难点。而列方程最主要的内容在于如何从题目中提取重要的数量关系这一信息点结合实际的数据来建立方程。其中最关键的就是对于市场经济打折问题所涉及的几个重要公式。一定要理解透彻，然后学会熟练的将公式进行变形。比如利润等于销售价格减去进价或成本价。而利润率则是销售价格减去进价比上成本价。往往在提醒中出现时会出现比较复杂的价格变动关系，大家只需要顺着题目的意思进行提价或打折，确定其销售价和成本价的情况下，利用公式建立方程，即可达到这一步骤的目的。第三，对于一元一次方程的市场经济问题，需要同学们对销售价，进价，成本价等相关的概念有清晰的了解。这一过程当中所涉及的这些内容，他们之间的联系是如何进行的？对于不同的商品其成本价，销售价或标价等关系是如何界定的对于列出方程来说至关重要。比如有的题型当中会出现成本价，而有的题型当中则出现的是进货价不同的市场情况所涉及的这些内容所对应的关系只要搞清楚那么对于列方程来说也就非常简单了。在解决市场经济问题过程当中，对于一元一次方程的应用更为复杂的问题就是对于方案的选择，这时候建立数量关系求出其进价或标价的同时，需要对其商品的利润进行计算，只需要按照各自的方案进行计算之后进行比较就可以得出最佳的方案选择。就是在学习和掌握一元一次方程应用。带基础之上的进一步学习，所以其前提还是对市场经济打折销售问题的明确要求。并且能够熟练使用常用的几个公式进行列方程，才能对方案选择问题有更高效的学习。唐老师为大家准备的这一份。一元一次方程解决实际问题中有关市场经济打折销售问题的专题训练，有助于帮助大家对这类问题进行集中解决，并且建立对市场经济的初步了解，其中最关键的为如何提取题目当中涉及的数量关系，通过常用的几个公式来列方程，解决这一难题之后，其他的问题将会迎刃而解，当然，如果对解一元一次方程的步骤和过程还不太清晰的同学，一定要加强训练，否则问题的叠加将会使得对解一元一次方程的应用难上加难。另外在进行专项训练的过程当中，尽量进行集中的训练，等自己完全训练完以后，再参照最后的答案解析，进行解题思路的分析和过程比对，确定解题思路和数量关系的提取是正确的，那么其他的内容也就迎刃而解。相信通过这一类专题训练的练习，同学们对市场经济打折销售问题将会有更加深入的了解，并且对于涉及的常用公式也会有更加熟练地使用。那么解决这一类问题也就非常轻松。写在最后，初一数学上学期有关一元一次方程的实际应用中市场经济打折销售问题，是同学们在学习当中比较困难的一部分主要是整个计算的过程比较复杂而导致同学们在计算时出现思维混乱，只要搞清楚题目中所涉及的数量关系，利用常用的几个公式进行列方程，那么其他问题也就得到解决，所以在专题训练过程当中一定要注意信息的提取以及如何列方程，抓住这一重点，相信同学们的这一问题将会得到解决。发布于 2022-11-21 12:46・云南数学市场经济专题赞同 3添加评论分享喜欢收藏申请转载写下你的评论... 还没有评论，发表第一个评论吧关于作者 Harleytang小课堂有关教育的知识分享和学习方法探讨回答 553文章 735关注者 5,132 关注他发私信推荐阅读一元一次方程是整个初中数学的基础，掌握好这13种应用题型，考试稳得高分！初一学生必看 ========================================== 一元一次方程应用考试题型大全 1、工程问题列方程解应用题是初中数学的重要内容之一，其核心思想就是将等量关系从情景中剥离出来，把实际问题转化成方程或方程组，从而解决问题。列方程… 志愿填报邹老师一元一次方程应用专题，十大题型（包括数轴上动点问题） ========================== 轻松中考数...发表于初中数学各...教研课题--实际问题与一元一次方程 ================= 实际问题与一元一次方程题型大致分为配套问题，工程问题，折扣问题，销售盈亏问题，积分问题等，每种问题都对应的有解题方法和技巧，每种题型掌握方法之后就会迎刃而解。一、关于配套问题… 秦晶晶一元一次方程9大题型超全解析，很重要，一定要看！ ======================== 白白白想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
2302	https://faculty.uml.edu/jpropp/491/P7.pdf	Math 491, Problem Set #7 (due 10/2/03) 1. One basis for the space of polynomials of degree less than d is the monomial basis 1, t, t2, ..., td−1. Another is the shifted monomial basis 1, (t + 1), (t + 1)2, ..., (t + 1)d−1. Call these bases u1, ..., ud and v1, ..., vd respectively. (a) Derive a formula for the entries of the change-of-basis matrix M expressing the ui’s as linear combinations of the vj’s. (b) Derive a formula for the entries of the change-of-basis matrix N expressing the vj’s as linear combinations of the ui’s. (c) From the description of M and N as basis-change matrices, we know that MN = NM = I. Forgetting for the moment what M and N mean, rewrite the assertions MN = NM = I as bino-mial coeﬃcient identities, and prove them either algebraically or bijectively.
2303	https://cstheory.stackexchange.com/questions/138/many-one-reductions-vs-turing-reductions-to-define-npc	cc.complexity theory - Many-one reductions vs. Turing reductions to define NPC - Theoretical Computer Science Stack Exchange Join Theoretical Computer Science By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Many-one reductions vs. Turing reductions to define NPC Ask Question Asked 15 years, 1 month ago Modified10 years, 11 months ago Viewed 7k times This question shows research effort; it is useful and clear 44 Save this question. Show activity on this post. Why do most people prefer to use many-one reductions to define NP-completeness instead of, for instance, Turing reductions? cc.complexity-theory np-hardness reductions Share Share a link to this question Copy linkCC BY-SA 2.5 Cite Improve this question Follow Follow this question to receive notifications asked Aug 17, 2010 at 12:58 MatthiasMatthias 1,708 1 1 gold badge 14 14 silver badges 22 22 bronze badges Add a comment\| 7 Answers 7 Sorted by: Reset to default This answer is useful 39 Save this answer. Show activity on this post. Two reasons: (1) just a matter of minimality: being NPC under many-one reductions is a formally stronger statement and if you get the stronger statement (as Karp did and as you almost always do) then why not say so? (2) Talking about many-one reductions gives rise to a richer, more delicate, hierarchy. For example the distinction NP vs co-NP disappears under Turing reductions. This is similar in spirit to why often one uses Logspace-reductions rather than polytime ones. Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Aug 17, 2010 at 17:00 NoamNoam 9,439 49 49 silver badges 59 59 bronze badges 1 20 While (2) is certainly true, I can use (1) to argue that we should use one-one reductions. Since most many-one reductions we build are in fact one-one reductions, why don't we study those when they are formally stronger and we get them most of the time anyways? I think because it's simpler not to have to bother proving injectivity, even though we usually have it. In that sense, maybe many-one reductions are sort of the "Goldilocks reductions" -- just the right power, just the right simplicity of proof.Joshua Grochow –Joshua Grochow 2010-08-18 14:54:51 +00:00 Commented Aug 18, 2010 at 14:54 Add a comment\| This answer is useful 21 Save this answer. Show activity on this post. I don't know whether there is a preference, but they are conjectured to be distinct notions. That is, Turing reducibility is conjectured to be a stronger notion. (There exist A and B such that A is T-reducible to B, but not m-o reducible to B.) One paper that discusses this is this one by Lutz and Mayordomo. They propose a strengthening of the statement P != NP; roughly, that NP includes a non-negligible amount of EXPTIME. This assumption allows them to show that the two notions of reducibility are distinct. Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Aug 17, 2010 at 13:08 Aaron SterlingAaron Sterling 7,090 7 7 gold badges 44 44 silver badges 75 75 bronze badges 0 Add a comment\| This answer is useful 19 Save this answer. Show activity on this post. I think the reason people prefer (to start with) many-one reductions is pedagogical -- a many-one reduction from A to B is actually a function on strings, whereas a Turing reduction requires the introduction of oracles. Note that Cook reduction (polynomial-time Turing) and Karp-Levin reduction (polynomial-time many-one) are known to be distinct on E unconditionally, by Ko and Moore, and separately by Watanabe (as referenced in the Lutz and Mayordomo paper in Aaron Sterling's response). Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Aug 17, 2010 at 13:37 Joshua GrochowJoshua Grochow 39.2k 4 4 gold badges 139 139 silver badges 244 244 bronze badges Add a comment\| This answer is useful 7 Save this answer. Show activity on this post. Turing reductions are more powerful than many-one mapping reductions in this regard: Turing reductions let you map a language to its complement. As a result it can kind of obscure the difference between (for example) NP and coNP. In Cook's original paper he didn't look at this distinction (iirc Cook actually used DNF formulas instead of CNF), but it probably became clear very quickly that this was an important separation, and many-one reductions made it easier to deal with this. Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Aug 17, 2010 at 15:07 KurtKurt 849 10 10 silver badges 13 13 bronze badges 1 12 Stephen Cook pointed out during his keynote at FLoC 2010 that his 1971 paper actually claims to prove that SAT is complete for P^NP under Turing reductions... Of course, the usual formulation follows from the same proof, so this is a situation of someone claiming less than they proved! See 4mhz.de/cook.html for a re-typeset version of the paper. Also, the sentence "We have not been able to add either {primes} or {isomorphic graphpairs} to [the list of 4 NP-complete problems]" always makes me smile!András Salamon –András Salamon 2010-08-18 00:44:38 +00:00 Commented Aug 18, 2010 at 0:44 Add a comment\| This answer is useful 5 Save this answer. Show activity on this post. to jump off somewhat on other angle/ answer here by AS, this is an open question (also here) at the frontiers of TCS whether Cook ("Turing") reductions are different than Karp-Levin ("many-one") reductions, possibly equivalent to (major? key?) open questions of complexity class separations. here is a new result along these lines Separating Cook Completeness from Karp-Levin Completeness under a Worst-Case Hardness Hypothesis / Debasis Mandal, A. Pavan, Rajeswari Venugopalan (ECCC TR14-126) We show that there is a language that is Turing complete for NP but not many-one complete for NP, under a worst case hardness hypothesis. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:32 CommunityBot 1 answered Oct 10, 2014 at 16:02 vznvzn 11.3k 2 2 gold badges 32 32 silver badges 68 68 bronze badges Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. The definitions of efficient reducibility are motivated in part by an analogy with recursion theory. In recursion theory, the m-reductions are closely connected to the arithmetical hierarchy. (m-reductions preserve arithmetical degree). Arithmetical classifications are important beyond mere computability. For example, one can say that true Σ 1 Σ 1 statements are provable in Robinson's Q Q. In complexity theory, there is also a notion of "polynomial hierarchy", though unlike the arithmetical hierarchy it is only conjectured to exist. This leads to classifications that are more subtle than "Is this problem as hard to solve as NP?" 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 Nov 23, 2011 at 12:23 David HarrisDavid Harris 3,558 24 24 silver badges 27 27 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Generally, Many-one (Karp) reduction is easier to design because it is a restricted form of reduction that makes one call and the main task involves transforming the input into different encoding. Turing reduction may involve complex logic. Existence of a set that is complete for NP under Turing reduction but not under many-one reduction implies that P!=NP. For instance, Unsatisfiability is complete for NP under Cook reduction but it is not known to be complete for NP under Karp reduction. So, if you prove that there is no Karp reduction from SAT to UNSAT (eqivalently from UNSAT to SAT) then you would prove that NP != CoNP and hence P!=NP. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Nov 23, 2011 at 11:53 answered Aug 25, 2010 at 14:32 Mohammad Al-TurkistanyMohammad Al-Turkistany 21.4k 5 5 gold badges 67 67 silver badges 157 157 bronze badges 2 can you give a reference to your last sentence or explain it?Tayfun Pay –Tayfun Pay 2011-11-23 02:20:56 +00:00 Commented Nov 23, 2011 at 2:20 3 I explained my last sentence.Mohammad Al-Turkistany –Mohammad Al-Turkistany 2011-11-23 11:54:35 +00:00 Commented Nov 23, 2011 at 11:54 Add a comment\| Your Answer Thanks for contributing an answer to Theoretical Computer Science Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions cc.complexity-theory np-hardness reductions 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 Report this ad Linked 5Contained in NP and Turing-reduction from an NP-complete problem ⇒⇒ NP-complete under Karp reductions? 237Major unsolved problems in theoretical computer science? 11Do many-one reductions and Turing reductions define the same class NPC 12Completeness under injective Karp reductions 19The motivation for using Karp-reductions in the theory of N P N P-completeness 10Are the notions of #P-completeness via Turing reductions and polynomial many-one counting reductions equivalent? 9Examples of NP-complete languages without Levin reductions? 6Oracle complexity classes and hardness under different notions of reduction Related 11Do many-one reductions and Turing reductions define the same class NPC 11Instance of FPT-reductions that is not a polynomial-time reduction 3Exercises of polynomial and turing reductions 8Reference for Turing to many-one reductions 12Completeness under injective Karp reductions 13Slowest many-one reduction? 7Reductions weaker than polynomial-time for ∃R∃R 9Examples of NP-complete languages without Levin reductions? 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2304	https://courses.lumenlearning.com/calculus2/chapter/combining-and-multiplying-power-series/	Combining and Multiplying Power Series \| Calculus II Skip to main content Calculus II Module 6: Power Series Search for: Combining and Multiplying Power Series Learning Outcomes Combine power series by addition or subtraction Create a new power series by multiplication by a power of the variable or a constant, or by substitution Multiply two power series together Combining Power Series If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of x or evaluate a power series at x m x m for a positive integer m to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for f(x)=1 1−x f(x)=1 1−x, we can find power series representations for related functions, such as y=3 x 1−x 2 and y=1(x−1)(x−3)y=3 x 1−x 2 and y=1(x−1)(x−3). In Combining Power Series we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at x=0 x=0. Similar results hold for power series centered at x=a x=a. theorem: Combining Power Series Suppose that the two power series ∞∑n=0 c n x n∑n=0∞c n x n and ∞∑n=0 d n x n∑n=0∞d n x n converge to the functions f and g, respectively, on a common interval I. The power series ∞∑n=0(c n x n±d n x n)∑n=0∞(c n x n±d n x n) converges to f±g f±g on I. For any integer m≥0 m≥0 and any real number b, the power series ∞∑n=0 b x m c n x n∑n=0∞b x m c n x n converges to b x m f(x)b x m f(x) on I. For any integer m≥0 m≥0 and any real number b, the series ∞∑n=0 c n(b x m)n∑n=0∞c n(b x m)n converges to f(b x m)f(b x m) for all x such that b x m b x m is in I. Proof We prove i. in the case of the series ∞∑n=0(c n x n+d n x n)∑n=0∞(c n x n+d n x n). Suppose that ∞∑n=0 c n x n∑n=0∞c n x n and ∞∑n=0 d n x n∑n=0∞d n x n converge to the functions f and g, respectively, on the interval I. Let x be a point in I and let S N(x)S N(x) and T N(x)T N(x) denote the N th partial sums of the series ∞∑n=0 c n x n∑n=0∞c n x n and ∞∑n=0 d n x n∑n=0∞d n x n, respectively. Then the sequence {S N(x)}{S N(x)} converges to f(x)f(x) and the sequence {T N(x)}{T N(x)} converges to g(x)g(x). Furthermore, the N th partial sum of ∞∑n=0(c n x n+d n x n)∑n=0∞(c n x n+d n x n) is N∑n=0(c n x n+d n x n)=N∑n=0 c n x n+N∑n=0 d n x n=S N(x)+T N(x).∑n=0 N(c n x n+d n x n)=∑n=0 N c n x n+∑n=0 N d n x n=S N(x)+T N(x). Because lim N→∞(S N(x)+T N(x))=lim N→∞S N(x)+lim N→∞T N(x)=f(x)+g(x),lim N→∞(S N(x)+T N(x))=lim N→∞S N(x)+lim N→∞T N(x)=f(x)+g(x), we conclude that the series ∞∑n=0(c n x n+d n x n)∑n=0∞(c n x n+d n x n) converges to f(x)+g(x)f(x)+g(x). ■◼ We examine products of power series in a later theorem. First, we show several applications of Combining Power Series and how to find the interval of convergence of a power series given the interval of convergence of a related power series. Example: Combining Power Series Suppose that ∞∑n=0 a n x n∑n=0∞a n x n is a power series whose interval of convergence is (−1,1)(−1,1), and suppose that ∞∑n=0 b n x n∑n=0∞b n x n is a power series whose interval of convergence is (−2,2)(−2,2). Find the interval of convergence of the series ∞∑n=0(a n x n+b n x n)∑n=0∞(a n x n+b n x n). Find the interval of convergence of the series ∞∑n=0 a n 3 n x n∑n=0∞a n 3 n x n. Show Solution Since the interval (−1,1)(−1,1) is a common interval of convergence of the series ∞∑n=0 a n x n∑n=0∞a n x n and ∞∑n=0 b n x n∑n=0∞b n x n, the interval of convergence of the series ∞∑n=0(a n x n+b n x n)∑n=0∞(a n x n+b n x n) is (−1,1)(−1,1). Since ∞∑n=0 a n x n∑n=0∞a n x n is a power series centered at zero with radius of convergence 1, it converges for all x in the interval (−1,1)(−1,1). By Combining Power Series, the series ∞∑n=0 a n 3 n x n=∞∑n=0 a n(3 x)n∑n=0∞a n 3 n x n=∑n=0∞a n(3 x)n converges if 3 x is in the interval (−1,1)(−1,1). Therefore, the series converges for all x in the interval (−1 3,1 3)(−1 3,1 3). Watch the following video to see the worked solution to Example: Combining Power Series. You can view the transcript for “6.2.1” here (opens in new window). try it Suppose that ∞∑n=0 a n x n∑n=0∞a n x n has an interval of convergence of (−1,1)(−1,1). Find the interval of convergence of ∞∑n=0 a n(x 2)n∑n=0∞a n(x 2)n. Hint Find the values of x such that x 2 x 2 is in the interval (−1,1)(−1,1). Show Solution Interval of convergence is (−2,2)(−2,2). In the next example, we show how to use Combining Power Series and the power series for a function f to construct power series for functions related to f. Specifically, we consider functions related to the function f(x)=1 1−x f(x)=1 1−x and we use the fact that 1 1−x=∞∑n=0 x n=1+x+x 2+x 3+⋯1 1−x=∑n=0∞x n=1+x+x 2+x 3+⋯ for \|x\|<1\|x\|<1. Example: Constructing Power Series from Known Power Series Use the power series representation for f(x)=1 1−x f(x)=1 1−x combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series. f(x)=3 x 1+x 2 f(x)=3 x 1+x 2 f(x)=1(x−1)(x−3)f(x)=1(x−1)(x−3) Show Solution First write f(x)f(x) as f(x)=3 x(1 1−(-x 2))f(x)=3 x(1 1−(-x 2)). Using the power series representation for f(x)=1 1−x f(x)=1 1−x and parts ii. and iii. of Combining Power Series, we find that a power series representation for f is given by ∞∑n=0 3 x(-x 2)n=∞∑n=0 3(−1)n x 2 n+1∑n=0∞3 x(-x 2)n=∑n=0∞3(−1)n x 2 n+1. Since the interval of convergence of the series for 1 1−x 1 1−x is (−1,1)(−1,1), the interval of convergence for this new series is the set of real numbers x such that \|x 2\|<1\|x 2\|<1. Therefore, the interval of convergence is (−1,1)(−1,1). To find the power series representation, use partial fractions to write f(x)=1(1−x)(x−3)f(x)=1(1−x)(x−3) as the sum of two fractions. We have 1(x−1)(x−3)=-1 2 x−1+1 2 x−3=1 2 1−x−1 2 3−x=1 2 1−x−1 6 1−x 3.1(x−1)(x−3)=-1 2 x−1+1 2 x−3=1 2 1−x−1 2 3−x=1 2 1−x−1 6 1−x 3. First, using part ii. of Combining Power Series, we obtain 1 2 1−x=∞∑n=0 1 2 x n for\|x\|<1 1 2 1−x=∑n=0∞1 2 x n for\|x\|<1. Then, using parts ii. and iii. of Combining Power Series, we have 1 6 1−x 3=∞∑n=0 1 6(x 3)n for\|x\|<3 1 6 1−x 3=∑n=0∞1 6(x 3)n for\|x\|<3. Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we have 1(x−1)(x−3)=∞∑n=0(1 2−1 6⋅3 n)x n 1(x−1)(x−3)=∑n=0∞(1 2−1 6⋅3 n)x n where the interval of convergence is (−1,1)(−1,1). try it Use the series for f(x)=1 1−x f(x)=1 1−x on \|x\|<1\|x\|<1 to construct a series for 1(1−x)(x−2)1(1−x)(x−2). Determine the interval of convergence. Hint Use partial fractions to rewrite 1(1−x)(x−2)1(1−x)(x−2) as the difference of two fractions. Show Solution ∞∑n=0(−1+1 2 n+1)x n∑n=0∞(−1+1 2 n+1)x n. The interval of convergence is (−1,1)(−1,1). In the previous example, we showed how to find power series for certain functions. In the next example we show how to do the opposite: given a power series, determine which function it represents. Example: Finding the Function Represented by a Given Power Series Consider the power series ∞∑n=0 2 n x n∑n=0∞2 n x n. Find the function f represented by this series. Determine the interval of convergence of the series. Show Solution Writing the given series as ∞∑n=0 2 n x n=∞∑n=0(2 x)n∑n=0∞2 n x n=∑n=0∞(2 x)n, we can recognize this series as the power series for f(x)=1 1−2 x f(x)=1 1−2 x. Since this is a geometric series, the series converges if and only if \|2 x\|<1\|2 x\|<1. Therefore, the interval of convergence is (−1 2,1 2)(−1 2,1 2). try it Find the function represented by the power series ∞∑n=0 1 3 n x n∑n=0∞1 3 n x n. Determine its interval of convergence. Hint Write 1 3 n x n=(x 3)n 1 3 n x n=(x 3)n. Show Solution f(x)=3 3−x f(x)=3 3−x. The interval of convergence is (−3,3)(−3,3). Try It Multiplication of Power Series We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions. The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply ∞∑n=0 c n x n=c 0+c 1 x+c 2 x 2+⋯∑n=0∞c n x n=c 0+c 1 x+c 2 x 2+⋯ and ∞∑n=0 d n x n=d 0+d 1 x+d 2 x 2+⋯∑n=0∞d n x n=d 0+d 1 x+d 2 x 2+⋯. It appears that the product should satisfy (∞∑n=0 c n x n)(∞∑n=−0 d n x n)=(c 0+c 1 x+c 2 x 2+⋯)⋅(d 0+d 1 x+d 2 x 2+⋯)=c 0 d 0+(c 1 d 0+c 0 d 1)x+(c 2 d 0+c 1 d 1+c 0 d 2)x 2+⋯.(∑n=0∞c n x n)(∑n=−0∞d n x n)=(c 0+c 1 x+c 2 x 2+⋯)⋅(d 0+d 1 x+d 2 x 2+⋯)=c 0 d 0+(c 1 d 0+c 0 d 1)x+(c 2 d 0+c 1 d 1+c 0 d 2)x 2+⋯. In Multiplying Power Series, we state the main result regarding multiplying power series, showing that if ∞∑n=0 c n x n∑n=0∞c n x n and ∞∑n=0 d n x n∑n=0∞d n x n converge on a common interval I, then we can multiply the series in this way, and the resulting series also converges on the interval I. theorem: Multiplying Power Series Suppose that the power series ∞∑n=0 c n x n∑n=0∞c n x n and ∞∑n=0 d n x n∑n=0∞d n x n converge to f and g, respectively, on a common interval I. Let e n=c 0 d n+c 1 d n−1+c 2 d n−2+⋯+c n−1 d 1+c n d 0=n∑k=0 c k d n−k.e n=c 0 d n+c 1 d n−1+c 2 d n−2+⋯+c n−1 d 1+c n d 0=∑k=0 n c k d n−k. Then (∞∑n=0 c n x n)(∞∑n=0 d n x n)=∞∑n=0 e n x n(∑n=0∞c n x n)(∑n=0∞d n x n)=∑n=0∞e n x n and ∞∑n=0 e n x n converges to f(x)⋅g(x)on I∑n=0∞e n x n converges to f(x)⋅g(x)on I. The series ∞∑n=0 e n x n∑n=0∞e n x n is known as the Cauchy product of the series ∞∑n=0 c n x n∑n=0∞c n x n and ∞∑n=0 d n x n∑n=0∞d n x n. We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for f(x)=1(1−x)(1−x 2)f(x)=1(1−x)(1−x 2) using the power series representations for y=1 1−x and y=1 1−x 2 y=1 1−x and y=1 1−x 2. Example: Multiplying Power Series Multiply the power series representation 1 1−x=∞∑n=0 x n=1+x+x 2+x 3+⋯1 1−x=∑n=0∞x n=1+x+x 2+x 3+⋯ for \|x\|<1\|x\|<1 with the power series representation 1 1−x 2=∞∑n=0(x 2)n=1+x 2+x 4+x 6+⋯1 1−x 2=∑n=0∞(x 2)n=1+x 2+x 4+x 6+⋯ for \|x\|<1\|x\|<1 to construct a power series for f(x)=1(1−x)(1−x 2)f(x)=1(1−x)(1−x 2) on the interval (−1,1)(−1,1). Show Solution We need to multiply (1+x+x 2+x 3+⋯)(1+x 2+x 4+x 6+⋯)(1+x+x 2+x 3+⋯)(1+x 2+x 4+x 6+⋯). Writing out the first several terms, we see that the product is given by (1+x 2+x 4+x 6+⋯)+(x+x 3+x 5+x 7+⋯)+(x 2+x 4+x 6+x 8+⋯)+(x 3+x 5+x 7+x 9+⋯)=1+x+(1+1)x 2+(1+1)x 3+(1+1+1)x 4+(1+1+1)x 5+⋯=1+x+2 x 2+2 x 3+3 x 4+3 x 5+⋯.(1+x 2+x 4+x 6+⋯)+(x+x 3+x 5+x 7+⋯)+(x 2+x 4+x 6+x 8+⋯)+(x 3+x 5+x 7+x 9+⋯)=1+x+(1+1)x 2+(1+1)x 3+(1+1+1)x 4+(1+1+1)x 5+⋯=1+x+2 x 2+2 x 3+3 x 4+3 x 5+⋯. Since the series for y=1 1−x y=1 1−x and y=1 1−x 2 y=1 1−x 2 both converge on the interval (−1,1)(−1,1), the series for the product also converges on the interval (−1,1)(−1,1). Watch the following video to see the worked solution to Example: Multiplying Power Series. For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip of “6.2.3” here (opens in new window). try it Multiply the series 1 1−x=∞∑n=0 x n 1 1−x=∑n=0∞x n by itself to construct a series for 1(1−x)(1−x)1(1−x)(1−x). Hint Multiply the first few terms of (1+x+x 2+x 3+⋯)(1+x+x 2+x 3+⋯)(1+x+x 2+x 3+⋯)(1+x+x 2+x 3+⋯). Show Solution 1+2 x+3 x 2+4 x 3+⋯1+2 x+3 x 2+4 x 3+⋯ Try It Candela Citations CC licensed content, Original 6.2.1. Authored by: Ryan Melton. License: CC BY: Attribution 6.2.3. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 2. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at Licenses and Attributions CC licensed content, Original 6.2.1. Authored by: Ryan Melton. License: CC BY: Attribution 6.2.3. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 2. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at PreviousNext Privacy Policy
2305	https://www.quora.com/What-is-the-sum-of-numbers-from-1-to-9	Something went wrong. Wait a moment and try again. Summation of Series Mathematics Homework Ques... Elementary Mathematics Ed... Arithmetical Calculations Basic Math 3 Number Series Mathematics Simple Mathematics Number and Arithmetic 5 What is the sum of numbers from 1 to 9? · The sum of the numbers from 1 to 9 can be calculated using the formula for the sum of an arithmetic series: Sn=n2×(a+l) where: - n is the number of terms, - a is the first term, - l is the last term. In this case: - n=9, - a=1, - l=9. Plugging in the values: S9=92×(1+9)=92×10=9×5=45 So, the sum of the numbers from 1 to 9 is 45. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. 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If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions What is the sum of 1 and 9? What is the sum of integers from 1-10? What is the sum of the numbers in the number 9? What is the answer to 1! +2! +…29!? What will be the sum of whole number from 1 to 40? Md. Ifrail Ajam Knows Bengali · 5y Solve, Let me write the numbers… 1, 2, 3, 4, 5, 6, 7, 8, 9 Now make pair using first and last numbers….. 1+9=10 2+8=10 3+7=10 4+6=10 Here, the number doesn’t make any pair due to the lack of the number… Here, we have found 4 pairs. Each pair makes 10… So, the sum of 4 pairs is, 10+10+10+10=40 ………We can write (104=40) So, the sum of the numbers is, = 40+5 =45…. We can also solve it using a formula… n(n+1)/2 Here, n= The last number Let’s apply this formula now, n(n+1)/2 = 9(9+1)/2 = 910/2 = 90/2 => 45 The sum of the number is, 45 (Ans.) Solve, Let me write the numbers… 1, 2, 3, 4, 5, 6, 7, 8, 9 Now make pair using first and last numbers….. 1+9=10 2+8=10 3+7=10 4+6=10 Here, the number doesn’t make any pair due to the lack of the number… Here, we have found 4 pairs. Each pair makes 10… So, the sum of 4 pairs is, 10+10+10+10=40 ………We can write (104=40) So, the sum of the numbers is, = 40+5 =45…. We can also solve it using a formula… n(n+1)/2 Here, n= The last number Let’s apply this formula now, n(n+1)/2 = 9(9+1)/2 = 910/2 = 90/2 => 45 The sum of the number is, 45 (Ans.) Paul Holloway Upvoted by Rhys Thomas , MSc Mathematics & Physics, De Montfort University (2002) · Author has 681 answers and 1.3M answer views · 7y We could just whip out a calculator but there’s a cleaver trick we can use to add them up really quickly in our heads. 1 + 9 = 10 2 + 8 = 10 3 + 7 = 10 4 + 6 = 10 which leaves us with just 5 add them up and we get 45. We can generalise this to get the sum of all counting numbers between 1 and n. We’re basically multiplying n+1 by half of n. (n+1) (n/2) = n(n+1)/2 Knowing this adding up all the numbers between 1 and 100 should be child’s play and you can do it in your head far faster than you could do it on a calculator. 100(101/2) = 10100/2 = 5050. If you’re asked to do a party trick you can remember this We could just whip out a calculator but there’s a cleaver trick we can use to add them up really quickly in our heads. 1 + 9 = 10 2 + 8 = 10 3 + 7 = 10 4 + 6 = 10 which leaves us with just 5 add them up and we get 45. We can generalise this to get the sum of all counting numbers between 1 and n. We’re basically multiplying n+1 by half of n. (n+1) (n/2) = n(n+1)/2 Knowing this adding up all the numbers between 1 and 100 should be child’s play and you can do it in your head far faster than you could do it on a calculator. 100(101/2) = 10100/2 = 5050. If you’re asked to do a party trick you can remember this simple technique and look like a genius as you ask people to get out the calculator app on their smart phones and get them to add up all the numbers from 1 to some arbitrarily large number and you do it in your head or on the back of an envelope faster than they can do it on their smart phones. Ronnie Elli 5y The long way to solve it is add 1+2+3+4+5+6+7+8+9=45 However there is a short solution: add first plus last numbers in the sequence and divide the result by two. Then multiple the result by last number in the sequence. 1+9 = 10 10/2 = 5 95 = 45 Here’s another example: the sum of numbers from 1 to 50 1+50 = 51 51/2=25.50 5025.50= 1,275 How do we solve the problem when the first number is not 1? Let’s say we want the sum of 6 to 50. Stage 1 sum of 1 to 50 We already know it’s 1,275 From the result deduct the sum of 1 to 5 1+5= 6 6/2 = 3 35= 15 1,275 -15 = 1,260 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Related questions What is the sum of the following arithmetic: 1, 2, 3, 99, 100? What is the sum of numbers that start from 1 to 100? What is the sum of 1+11+111+1111+11111+111111…up to 100 digits of 1? How do I solve 1 + 3 + 5 + 7 + 9 + . . . + 97 + 99 ? What is the sum of the counting numbers from 1 to 60? Soni .S. Melathil Studied Mathematics (Graduated 1994) · Author has 54 answers and 23.9K answer views · 2y here we have to use the formula is S = n x (n+1)/2 S - sum of the series, n - last term in the series, here n = 9, therefore S = 9 x (9+1)/2 S = 9 x 10/2 S = 45 Therefore, the sum of numbers from 1 to 9 is 45. Ans is 45 OR we have to find normal method The sum of numbers from 1 to 9 can be found by adding each of the numbers together. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 Therefore, Ans is 45 Subham Kumar Gupta Former Systems Engineer at Tata Consultancy Services (company) (2020–2024) · Author has 402 answers and 573.5K answer views · 2y Originally Answered: What is the sum of the numbers from 1 to 10? · Sponsored by CDW Corporation What would an evergreen data storage architecture do for you? Get uninterrupted access to data that is update-proof when you invest in storage from CDW & Pure Storage. Doctor Sachidanand Das PhD,Former Scientific Officer-G, Government of India · Author has 11.3K answers and 16M answer views · 7y Sum: 45 Calculation: The formula for finding the sum of a sequence of natural numbers ,which is an A.P. , is n(n+1)/2 where n is the last number . In our problem, n = 9 and therefore Sum = 9.10/2 = 9.5 = 45 Ritika Satapathy Lives in Bhubaneswar, Odisha, India · 5y We have a Arithmetic formula And according to that we need to add 1,2,3….till ‘n’ ’n’ can be any integer but in this case as per our question n=9 Then the formula is {n×(n+1)}/2(the sum of the first ’n’ natural number) So the sum is {n×(n+1)}/2={9×(9+1)}/2=45 The sum of the number Is,45 (ans) Risha C Shaji MA ECONOMICS from Union Christian College, Aluva (Graduated 2024) · Author has 64 answers and 9K answer views · 1y To find the sum of numbers the equation is n/2(first term + last term) Here the first term is 1 and last term is 9 By Substituting the values into equation we get 4.5×10 = 45 Related questions What is the sum of 1 and 9? What is the sum of integers from 1-10? What is the sum of the numbers in the number 9? What is the answer to 1! +2! +…29!? What will be the sum of whole number from 1 to 40? What is the sum of the following arithmetic: 1, 2, 3, 99, 100? What is the sum of numbers that start from 1 to 100? What is the sum of 1+11+111+1111+11111+111111…up to 100 digits of 1? How do I solve 1 + 3 + 5 + 7 + 9 + . . . + 97 + 99 ? What is the sum of the counting numbers from 1 to 60? What is the sum of the even numbers between 1 and 9? What's 100+99+98+97+96…? And how do I figure out the sums of other numbers like this? What is the next number in the series 5, 15, 10, 20, 15, 25? Here are the numbers in some kind of sequence. 3, 7, 14, 23, 36, 49, 66. What are the next three numbers? What number comes next in the sequence: 1, 6, 3, 8, 5? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2306	https://www.ixl.com/math/grade-5/compare-sums-of-unit-fractions	IXL \| Compare sums of unit fractions \| 5th grade math SKIP TO CONTENT [x] - [x] IXL Learning Sign in- [x] Remember Sign in nowJoin now IXL Learning Learning Math Skills Lessons Videos Games Fluency Zone New! Language arts Skills Videos Games Science Social studies Spanish Recommendations Recommendations wall Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Student awards Assessment Analytics Takeoff Inspiration Learning All Learning Math Language arts Science Social studies Spanish Recommendations Skill plans Learning Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Assessment Analytics Takeoff Inspiration Membership Sign in Math Math Language arts Language arts Science Science Social studies Social studies Spanish Spanish Recommendations Recommendations Skill plans Skill plans IXL plans South Carolina state standards Textbooks Test prep Awards Awards Everyday Mathematics - 5th grade Compare sums of unit fractions ZX8 Share skill Copy the link to this skill share to facebook share to twitter Time to get in the zone! Your teacher would like you to focus on skills in . Let's pick a skill from these categories. Let's go! 1 Compare sums of unit fractions ZX8 Share video Copy the link to this video share to facebook share to twitter You are watching a video preview. Become a member to get full access! You've reached the end of this video preview, but the learning doesn't have to stop! Join IXL today! Become a memberSign in Incomplete answer You did not finish the question. Do you want to go back to the question? Go back Submit Learn with an example or Watch a video Without adding, find which sum is greater. 1 8 + 1 6 1 8 + 1 9 neither; they are equal Submit Back to practice ref_doc_title. Back to practice Learn with an example or Watch a video Learn with an example question Without adding, find which sum is less. 1 9 + 1 5 1 9 + 1 8 neither; they are equal key idea A unit fraction has a numerator of 1. A unit fraction with a smaller denominator has a larger value. A unit fraction with a larger denominator has a smaller value. 1 3 1 5 1 3 has a smaller denominator than 1 5 , so 1 3 is greater than 1 5 . solution Each fraction has a numerator of 1, so they are all unit fractions. Look at their denominators to compare them. Compare the first fractions. 1 9 + 1 5 1 9 + 1 8 The first fractions are equal. Compare the second fractions. 1 9 + 1 5 1 9 + 1 8 The unit fraction with the smaller denominator has a greater value. So, 1 5 is greater than 1 8 . 1 5 1 8 Compare the sums. The first fractions are equal. The second sum has a smaller second fraction. So, the second sum is less than the first. 1 9 + 1 5 1 9 + 1 8 Looking for more? Try these: Video: Compare sums of unit fractions Lesson: Comparing fractions Learn how to compare fractions! Comparing fractions with like denominators Comparing fractions with like numerators Comparing fractions using benchmark fractions Comparing fractions by making equivalent fractions Lesson: Comparing fractions Learn with an example Excellent! You got that right! Continue Learn with an example or Watch a video Jumping to level 1 of 1 Excellent! Now entering the Challenge Zone—are you ready? Questions answered Questions 0 Time elapsed Time 00 00 03 hr min sec SmartScore out of 100 IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more 0 You met your goal! Teacher tools Group Jam Live Classroom Leaderboards Work it out Not feeling ready yet? These can help:K.3 Compare fractions and mixed numbersK.3 Compare fractions and mixed numbers - Fifth grade U8JW.6 Solve inequalities using addition and subtraction shortcutsW.6 Solve inequalities using addition and subtraction shortcuts - Second grade 87QLesson: Comparing fractions Learn how to compare fractions! Comparing fractions with like denominators Comparing fractions with like numerators Comparing fractions using benchmark fractions Comparing fractions by making equivalent fractions Lesson: Comparing fractions Company \| Membership \| Blog \| Help center \| User guides \| Tell us what you think \| Testimonials \| Careers \| Contact us \| Terms of service \| Privacy policy © 2025 IXL Learning. All rights reserved. Follow us First time here? 1 in 4 students uses IXL for academic help and enrichment. Pre-K through 12th grade Sign up nowKeep exploring
2307	https://pmc.ncbi.nlm.nih.gov/articles/PMC5764449/	Testicular ptosis as a sign of L2 radiculopathy - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Neurol Clin Pract . 2015 Apr;5(2):178–181. doi: 10.1212/CPJ.0000000000000121 Search in PMC Search in PubMed View in NLM Catalog Add to search Testicular ptosis as a sign of L2 radiculopathy JD Bartleson JD Bartleson, MD 1 Department of Neurology (JDB), Department of Radiology (TM), and Department of Neurologic Surgery (WRM), Mayo Clinic, Rochester, MN. Find articles by JD Bartleson 1, Timothy Maus Timothy Maus, MD 1 Department of Neurology (JDB), Department of Radiology (TM), and Department of Neurologic Surgery (WRM), Mayo Clinic, Rochester, MN. Find articles by Timothy Maus 1, W Richard Marsh W Richard Marsh, MD 1 Department of Neurology (JDB), Department of Radiology (TM), and Department of Neurologic Surgery (WRM), Mayo Clinic, Rochester, MN. Find articles by W Richard Marsh 1 Author information Copyright and License information 1 Department of Neurology (JDB), Department of Radiology (TM), and Department of Neurologic Surgery (WRM), Mayo Clinic, Rochester, MN. © 2015 American Academy of Neurology PMC Copyright notice PMCID: PMC5764449 PMID: 29443205 Practical Implications In men with low back and unilateral proximal lower limb pain, if a markedly lower testicle is observed on the same side as the pain, consider an upper lumbar radiculopathy most likely affecting the L2 spinal nerve. A 71-year-old man was referred to the Mayo Comprehensive Spine Center for possible surgery on a left L2–3 disk extrusion. Superimposed on a past history of episodic low back pain, 6 months prior, the patient developed major low back and left flank pain after moving some rocks. Over the course of a few days, the pain became so severe that he visited his local emergency department and received temporary benefit from an injection of morphine and a brief course of oral prednisone. He continued to have severe pain in his lower back and left flank and developed pain in his left anterior iliac crest and proximal left medial, anterior, and lateral thigh, with numbness, sensitivity, and brief neuralgic pains over his left anterolateral thigh. His lower limb reflexes were reportedly normal. Within a month of symptom onset, he underwent lumbar spine MRI and subsequent CT myelography, both of which showed a left L2–3 disk extrusion with superior migration and L2 nerve root impingement (see figure 1). An EMG was reported to show a left upper lumbar radiculopathy. He had a left L2–3 transforaminal epidural steroid injection that helped him briefly. He did not notice any weakness or bowel or bladder difficulty. He had noted that ever since his pain began, his left testis was hanging considerably lower than the right. He estimated that the left testicle had hung 6 cm lower than the right at its nadir. His pain and testicular ptosis had improved over time, albeit incompletely, when we saw him. He was otherwise in good health except for a history of right rotator cuff disease, and he was taking only ibuprofen for pain. Open in a new tab Imaging Figure 1. (A) Sagittal T2 (left) and T1 (right) MRIs demonstrate an L2 disk extrusion with cranial migration of disk material behind the body of L2 (arrows). The high signal at the margin of the disk extrusion on the T1 image may be an associated hemorrhage. (B) Coronal (left) and sagittal (right) CT reconstructions from CT myelography again demonstrate the cranial migration of the extruded material into the axilla of the L2 dural sleeve (arrows). (C) Contiguous axial T2-weighted MRI from the L2 pedicle to the L2 disk. The disk extrusion (single arrows) effaces the L2 dural sleeve but has little effect on the traversing L3 nerve. Note that the disk extrusion has an extraforaminal component (double arrows). (D) Contiguous axial CT myelography images from the L2 pedicle to the L2 disk also demonstrate the extrusion (arrows). Note that the right L2 dural sleeve fills with intrathecal contrast while the left L2 dural sleeve is effaced and does not fill. (E) Radiographic image from the CT myelogram also shows the filling defect in the axilla of the left L2 dural sleeve (arrow). The imaging studies support the presence of a left L2 radiculopathy. Neurologic examination was normal, including normal left lower limb strength and intact symmetric knee reflexes. Cremasteric reflexes were absent bilaterally. His left testicle hung about 4 cm lower than the right (see figure 2). The patient reported that his left testicular ptosis had been at least 50% lower when his pain was at its peak. Open in a new tab Patient's testes Figure 2. The top of the patient's left testicle is at the level of the bottom of his right testicle, a distance of approximately 4 cm. The patient reported that at the height of his L2 radiculopathy the left testicle hung about 6 cm lower than the right. He was seen in neurosurgical consultation. We all agreed that because his symptoms were improving and he had no motor deficit, we should continue conservative treatment. Over the next 6 months, his pain continued to improve, with very mild residual low back pain and left anterolateral thigh dysesthesias. His left testicle still hung about 4 cm lower than the right. DISCUSSION The cremaster muscle is supplied by the genital branch of the genitofemoral nerve and the L1, and especially L2, spinal nerves.1–3 The cremaster muscle lacks voluntary control and pulls the testis upward toward the superficial inguinal ring. The muscle is thought to play a key role in testicular thermoregulation.1,2 The left testicle commonly hangs lower than the right, but only by about 1 cm or less.2 Our patient's asymmetry was far in excess of 1 cm and he had observed that this was a new and notable finding. The cremasteric reflex (stroking of the medial thigh provokes reflex contraction of the cremaster muscle and elevation of the testicle) is mediated by the L1 and L2 spinal nerves. Although his cremasteric reflexes were absent on both sides, this finding is not uncommon in older men.3 The patient's pain was consistent with an L2 distribution on the left, and his imaging studies clearly documented severe L2 nerve root compression. Except for the cremaster muscle weakness, he did not have any focal neurologic signs pointing to a lumbar radiculopathy. As a result, his pain could have prompted extensive investigation of the abdomen, pelvis, and left lower limb looking for the cause of his symptoms. The patient's observation of drooping of his left testicle alerted us to the connection between cremaster muscle weakness and the discogenic L2 nerve root impingement. The patient's gradual, spontaneous improvement also supports an extruded disk as the mechanism of his symptoms and physical findings. L2 radiculopathy can cause back, groin, and anterior thigh pain with anterolateral thigh sensory loss and/or paresthesias.4 Deep tendon reflexes are typically normal and hip flexion weakness is absent or mild. The cremasteric reflex is a superficial reflex that can be absent in upper and lower motor neuron lesions, including genitofemoral nerve injury. It can be due to local processes affecting the groin or testes, occur as a result of aging, or be missing for no apparent reason. As a result, loss of the cremasteric reflex is frequently unreliable as a localizing finding. Loss of cremaster muscle strength with testicular ptosis is rarely mentioned in the literature. To our knowledge, the new onset of cremaster muscle weakness secondary to L2 radiculopathy has not been previously described. Upper lumbar radiculopathies are rare and more difficult to diagnose than the much more common L4, L5, and S1 radiculopathies. Upper lumbar radiculopathies do not cause sciatica and often lack characteristic motor, reflex, and/or sensory deficits. Upper lumbar radiculopathies are less likely to be caused by disk compression than their caudal counterparts. For example, Spangfort reviewed 15,235 lumbar disk operations in 49 publications and noted that 0.1% occurred at L1–2, 0.4% at L2–3, 2.8% at L3–4, 49.8% at L4–5, and 46.9% at L5–S1.5 Spangfort also reported on 2,157 positive lumbar disk operations performed in 2 hospitals in Sweden from 1951 to 1966 and found that 0.05% occurred at L1–2, 0.2% at L2–3, 1.8% at L3–4, 47.4% at L4–5, and 50.5% at L5–S1.5 Disk operations were more common in men than women at all ages, and upper lumbar disk herniations were somewhat more likely with age. Because upper lumbar disk disease is rare, symptoms of an upper lumbar radiculopathy can suggest another cause such as tumor, infection, diabetes, or an alternative spondylotic mechanism. Spine imaging aids in diagnosis. In cases of low back and unilateral proximal lower limb pain in men, one should consider inspecting the patient's genitalia to look for marked asymmetry of testicular height. If his much lower testicle is on the same side as the pain, one should consider an upper lumbar radiculopathy most likely affecting the L2 spinal nerve. The patient can also be asked whether he has observed this finding in temporal association with the onset of the pain syndrome. Men who wear boxer shorts (as does our patient) may be more likely to notice a drooping testicle than men who wear jockey shorts. Unfortunately, there is no similar physical finding for women with L2 radiculopathy. STUDY FUNDING No targeted funding reported. DISCLOSURES J.D. Bartleson receives publishing royalties for Spine Disorders: Medical and Surgical Management (Cambridge University Press, 2009). T. Maus has received travel compensation as a member of the Executive Board of the International Spine Intervention Society and practices interventional pain management (80% clinical effort) at Mayo Clinic. W.R. Marsh reports no disclosures. Full disclosure form information provided by the authors is available with the full text of this article atNeurology.org/cp. Correspondence to:bartleson.john@mayo.edu Funding information and disclosures are provided at the end of the article. Full disclosure form information provided by the authors is available with the full text of this article atNeurology.org/cp. Footnotes Correspondence to:bartleson.john@mayo.edu Funding information and disclosures are provided at the end of the article. Full disclosure form information provided by the authors is available with the full text of this article atNeurology.org/cp. REFERENCES 1.Standring S, editor. Gray's Anatomy: The Anatomical Basis of Clinical Practice, 40th ed. Spain: Elsevier; 2008:1064, 1262. 2.Williams PL, editor. Gray's Anatomy, 38th ed. New York, NY: Churchill Livingstone; 1995:824, 825, 1848. 3.Campbell WW, editor. DeJong's The Neurologic Examination, 7th ed. Philadelphia, PA: Lippincott Williams & Wilkins; 2013:580. 4.Tarulli AW, Raynor EM. Lumbosacral radiculopathy. Neurol Clin. 2007;25:387–405. doi: 10.1016/j.ncl.2007.01.008. [DOI] [PubMed] [Google Scholar] 5.Spangfort EV. The lumbar disk herniation. A computer-aided analysis of 2,504 operations. Acta Orthop Scand Suppl. 1972;142:1–95. doi: 10.3109/ort.1972.43.suppl-142.01. [DOI] [PubMed] [Google Scholar] Articles from Neurology: Clinical Practice are provided here courtesy of American Academy of Neurology ACTIONS View on publisher site PDF (601.9 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page DISCUSSION STUDY FUNDING DISCLOSURES Footnotes REFERENCES Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
2308	https://www.khanacademy.org/math/basic-geo/x7fa91416:decomposing-to-find-area/x7fa91416:area-of-triangles/v/example-finding-area-of-triangle	How to find area of triangle (formula walkthrough) (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content Basic geometry and measurement Course: Basic geometry and measurement>Unit 8 Lesson 2: Area of triangles Area of a triangle Finding area of triangles Area of triangles Find base and height on a triangle Area of triangles Triangle missing side example Math> Basic geometry and measurement> Decomposing to find area> Area of triangles © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Finding area of triangles Google Classroom Microsoft Teams About About this video Transcript The area of a triangle is found by multiplying one half of the base by the height. In our example, the base is 18 and the height is 6. So, half of 18 is 9, and 9 times 6 equals 54 square units. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Paarth Bamb 7 years ago Posted 7 years ago. Direct link to Paarth Bamb's post “In a case when the height...” more In a case when the height isn't given, how do you find out the height? Answer Button navigates to signup page •9 comments Comment on Paarth Bamb's post “In a case when the height...” (38 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Brooks Wang 7 years ago Posted 7 years ago. Direct link to Brooks Wang's post “In that case, you would u...” more In that case, you would use the Pythagorean theorem. That's a squared plus b squared equals c squared. In this specific scenario, the hypotenuse is already given and you would need to know one of the sides or legs in the equation. That means you would square the hypotenuse (the side not connected to a right angle), subtract the square of the side/leg you already know, then take the square root of that final number. That would be the height of the triangle 15 comments Comment on Brooks Wang's post “In that case, you would u...” (48 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Goezonme 6 years ago Posted 6 years ago. Direct link to Goezonme's post “Can I also write the solu...” more Can I also write the solution like this:bh/2? Answer Button navigates to signup page •3 comments Comment on Goezonme's post “Can I also write the solu...” (24 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Aishwarya P. 2 years ago Posted 2 years ago. Direct link to Aishwarya P.'s post “Answer: OFC!! To find the...” more Answer: OFC!! To find the answer you would, base x height and then divide by 2 or multiply by 1/2! :) Hope this helps! Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Nicky the smartest a year ago Posted a year ago. Direct link to Nicky the smartest's post “why is this a lot more fu...” more why is this a lot more fun then acual tests and conversations with a teacher. :) Answer Button navigates to signup page •2 comments Comment on Nicky the smartest's post “why is this a lot more fu...” (13 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer drishtibhatia06 2 months ago Posted 2 months ago. Direct link to drishtibhatia06's post “For real though, It's wa...” more For real though, It's way easier to understand when a person knows your strengths and weaknesses Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more unknown..... 2 years ago Posted 2 years ago. Direct link to unknown.....'s post “what if your base or heig...” more what if your base or height is a fraction? PLS HELP ME! Answer Button navigates to signup page •3 comments Comment on unknown.....'s post “what if your base or heig...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer juliet 2 years ago Posted 2 years ago. Direct link to juliet's post “If the base or height is ...” more If the base or height is a fraction, then do the formula as normal. If one is a fraction, and the other isn't, make the whole number a fraction, for example, 3= 3/1. Hope this helps! 1 comment Comment on juliet's post “If the base or height is ...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... egg 5 years ago Posted 5 years ago. Direct link to egg's post “OK, is it just base times...” more OK, is it just base times height and then divide by two? i am kinda confused. can some one plz help me? Answer Button navigates to signup page •Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer ccostello30 𓃵 𓃵 𓃵 𓃵 a year ago Posted a year ago. Direct link to ccostello30 𓃵 𓃵 𓃵 𓃵's post “Bruv you have a good char...” more Bruv you have a good character so you shouldn’t be confused Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Luna Lopez 6 months ago Posted 6 months ago. Direct link to Luna Lopez's post “Wait, shouldn't he have m...” more Wait, shouldn't he have multiplied 6 by one half too? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer louisaandgreta 5 years ago Posted 5 years ago. Direct link to louisaandgreta's post “This is perhaps a silly q...” more This is perhaps a silly question, if so sorry. (Or perhaps covered later?) Since the length of a triangle’s sides is correlated to the triangle’s angles, shouldn’t it follow that you could calculate the area of a triangle with only the measure of it’s angles? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Bryant 7 years ago Posted 7 years ago. Direct link to Bryant's post “how do you find the heigh...” more how do you find the height without it giving it Answer Button navigates to signup page •1 comment Comment on Bryant's post “how do you find the heigh...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kirsten Morales 2 years ago Posted 2 years ago. Direct link to Kirsten Morales's post “Can a triangle have more ...” more Can a triangle have more than one base? Answer Button navigates to signup page •2 comments Comment on Kirsten Morales's post “Can a triangle have more ...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Loflin Sakran 2 years ago Posted 2 years ago. Direct link to Loflin Sakran's post “no then it's not a triang...” more no then it's not a triangle its a square or rectangle. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... LaurenV 2 years ago Posted 2 years ago. Direct link to LaurenV's post “can I also just do the fo...” more can I also just do the formula H(B / 1/2)? Answer Button navigates to signup page •1 comment Comment on LaurenV's post “can I also just do the fo...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript [Instructor] What is the area of the triangle? And they're talking about this triangle here. And like always, pause this video and see if you can figure it out on your own. All right, now let's work through it together. So we know that the area of a triangle is going to be equal to 1/2 times our base times our height. Let me do the height in a different color. Times, times our height. So, what is the length of our base in this scenario? Well, the base is this 18 right over here. Let me highlight it. This length right over here is our base. So the base is 18. And what is the height? Well, the height, we see, is six. They give it to us. They don't always give it to you, but in this example they do. So the height is, the height... The height is, I'm having trouble with my pen, is six. So now we just have to compute what 1/2 times 18 times six is. Well, let's think about this a little bit. Well, 1/2 times 18 is nine. Nine times six is equal to what? Nine times six is equal to 54. And we are done. This is 54 square units. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: article 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. 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2309	https://math.stackexchange.com/questions/258876/proving-square-root-of-a-square-is-the-same-as-absolute-value	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 Proving square root of a square is the same as absolute value Ask Question Asked Modified 9 years, 1 month ago Viewed 52k times 21 $\begingroup$ Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them: $\begin{align} f(x) &= \sqrt {x^2} \ f(-1) &= \sqrt {(-1)^2} \ f(-1) &= \sqrt {1} \ f(-1) &= 1 \end{align}$ $\begin{align} g(x) &= x \ g(-1) &= -1 \end{align}$ thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions: $y = \sqrt {x^2}$: $y = x$: $y = \mid x \mid$: From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$? radicals absolute-value Share edited Aug 27, 2016 at 16:02 Antonio Vargas 25.6k22 gold badges7171 silver badges161161 bronze badges asked Dec 14, 2012 at 19:56 Cole TobinCole Tobin 1,43344 gold badges1515 silver badges2424 bronze badges $\endgroup$ 2 $\begingroup$ How is $y=\sqrt{x^2}$ a "2d line?" As for how you prove this, first you have to specify how you define things. Specifically, how do you define $\|z\|$ and $\sqrt{z}$? (I'm not entirely sure why you branch off into infinity. Infinity is not a number, and treating it as one is almost always an error. In a few circumstances, you can, but most of the time it is not worth it.) $\endgroup$ Thomas Andrews – Thomas Andrews 2012-12-14 20:03:36 +00:00 Commented Dec 14, 2012 at 20:03 4 $\begingroup$ Can't you break into two cases, $x>=0$ and $x<0$, and check that for each one it's the same as $\|x\|$ ? $\endgroup$ Ricbit – Ricbit 2012-12-14 20:04:09 +00:00 Commented Dec 14, 2012 at 20:04 Add a comment \| 3 Answers 3 Reset to default 31 $\begingroup$ It is the definition of square root of a number. The square root is defined in the sense that $s(x^2) = \sqrt{ x^2 } = \|x\|$ for all real $x$. Thus, the domain is the real numbers and the codomain is the non-negative real numbers. The reason it is defined this way is to make sure that $s$ is a function. Assume for a minute that $s(x^2) = x$, then: $$\sqrt{(-5)^2} = -5, \qquad \sqrt{5^2} = 5$$ But we know that $\sqrt{(-5)^2} = \sqrt{25} = \sqrt{5^2}$. Thus we see that $s(25) = -5, 5$. And thus $s$ is not a function. To keep it as a function, we have to 'sacrifice' and say that $s(x^2) \neq x$. Rather, $s(x^2) = \|x\|$. This will be consistent with the definition of a function. Since it is a definition, it cannot be proven. The problem is that many think that $\sqrt {x^2} = x$ because we study positive numbers before we study negative numbers, which is understandable, because I used to make that mistake all the time. Share answered Dec 14, 2012 at 22:38 HusainHusain 79455 silver badges1313 bronze badges $\endgroup$ 1 6 $\begingroup$ Don't you mean principal square root? $\endgroup$ Kyle Delaney – Kyle Delaney 2015-12-11 01:58:48 +00:00 Commented Dec 11, 2015 at 1:58 Add a comment \| 13 $\begingroup$ Given a non-negative real number $\alpha$, the number $\sqrt\alpha$ is defined to be the unique non-negative real number $\beta$ such that $\beta^2=\alpha$. Since $\sqrt{\alpha}\geq 0$ for all $\alpha\geq 0$, then for any real $\gamma$, it follows that $$\sqrt{\gamma^2}=\begin{cases}\gamma & \gamma\geq 0\-\gamma & \gamma<0,\end{cases}$$ that is, $$\sqrt{\gamma^2}=\|\gamma\|.$$ Share answered Dec 14, 2012 at 20:05 Cameron BuieCameron Buie 105k1010 gold badges106106 silver badges245245 bronze badges $\endgroup$ Add a comment \| 4 $\begingroup$ When $x < 0$, $\|x\| = -x > 0$. $-x \ne x$ (unless $x=0$) and $(-x)^2 = x^2$. There are two "square roots" of any positive number $y$, i.e. numbers whose square is $y$, and the positive one is called $\sqrt{y}$. So $\sqrt{x^2} = -x = \|x\|$ when $x < 0$, and $\sqrt{x^2} = x = \|x\|$ when $x \ge 0$. Share answered Dec 14, 2012 at 20:04 Robert IsraelRobert Israel 472k2828 gold badges376376 silver badges714714 bronze badges $\endgroup$ 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 radicals absolute-value 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 2 Why is $\sqrt{x^2} = \|x\|$? 2 Why is $\sqrt{x^2}= \|x\|$ rather than $\pm x$? 1 Tiny doubt with square root 0 what is the reasoning behind $\sqrt{x^2}=\lvert x\rvert$ 1 Understanding why If $\|a\|+1>0$ then $\sqrt{\|a\|+1}>0$ 10 Why does $\sqrt{x^2}$ seem to equal $x$ and not $\|x\|$ when you multiply the exponents? 3 Confused About Squareroots & Absolute Values 2 Solving for the derivative of absolute value, gone wrong 2 Question regarding the square root of a squared number. 1 Another square root question... 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2310	https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2021/1WLUcGY6_kxyR6eTmnGkcS2nR9ATRXFkE_transcript.pdf	MITOCW \| 18102-sp21-lecture-16.mp4 [SQUEAKING] [RUSTLING] [CLICKING] CASEY RODRIGUEZ: OK, so let's continue with our discussion of Fourier series from last time. For a given function in L2, we define the Fourier coefficient f hat of n to be the integral of 1 over 2 pi minus pi to pi f of t e to the minus int dt, which up to a factor of 1 over root t, is equal to the inner product of f with e to the int over square root of 2 pi in the Hilbert space L2, OK? And the question that we had-- so let me-- and we also had that the n-th partial sum for the Fourier series associated to f was given by sum from n equals minus n to n f hat of n e to the inx. And the question which we're trying to resolve is do we have for all f and L2 limit as n goes to infinity of f minus SN of f 2 equals 0, OK? All right? In other words, is f equal to its Fourier series, at least when we interpret equals as in this sense here? Now, based on what we've done for Hilbert spaces, this question is equivalent to the following statement. If f is an L2, and the Fourier coefficients are all 0, does this imply f equals 0, right? So this question is-- by what we've done for Hilbert spaces, big L2 is a Hilbert space. This question here is equivalent to this statement, which is that the collection of orthonormal vectors and big L2 consisting of the exponentials divided by square root of 2-- is this a maximal orthonormal subset? Or as we were using the terminology we had from last time, does that form an orthonormal basis? So this is the statement that we're going to prove this class. And we're going to proceed via Fejer's method, if you like to give it a name, where what we did last time was-- when we recall, we had the Cesaro Fourier mean we defined to be the average of the first n partial sums with the hope that somehow this behaves a little bit better than the partial sums, because that's the thing we're trying to study. And that's a hard question. And typically, means of sequences might behave better than the sequences themselves. But if the original sequence converges, then the means converge. So we should expect this to converge to f but hopefully faster or have better, more recognizable properties than just studying the partial sums directly. And we'll get to-- in the next statement, it's a little bit clearer why the Cesaro Fourier means converge to f. And so what our goal-- what we're going to show is that if f is an L2, then the Cesaro Fourier means converts to 0, I mean converts to f as n goes to infinity. OK? And so once we've proven that, then that gives us what we want in the yellow box, right? That proves what's in the yellow box, because let's take f and L2 with 48 coefficients all 0. Then, all of the partial sums will be 0. Then, all of the means will be 0. And since the means converge to f, that proves f is 0. And we get what's in the yellow box. And therefore, the partial the Fourier sums converts to f as capital N goes to infinity in L2, all right? And then, once we prove that, I'll make a couple of comments about other types of questions one can ask and what you can do, or a brief comment. So this is our goal for this lecture. And we should be able to get through it. So let me first rewrite the Cesaro Fourier means slightly differently. How we did in the previous lecture for the partial Fourier sums-- we wrote them as what's called a convolution. I haven't defined convolution-- but an integral of a function depending on x and t times f of t dt. And we're going to do the same now for the Cesaro means. And we'll see here in what-- it's a little bit more clear, although I didn't talk so much about the Dirichlet kernel that appears for these guys-- but why the Cesaro means converge to f. OK, all right, so the statement is for all f in L2 minus pi to pi, we have that the n-th Cesaro mean of f, which is our Fourier mean, I can write as the integral from minus pi to pi of a function kn of x minus t times f of t dt. So remember, for the partial sums s sub n, we could write it as d sub n, where d denoted a Dirichlet kernel, where kn of x-- this is equal to n plus 1 over 2 pi, and then 1 over 2 pi times n plus 1 times sine n plus 1 over 2x over sine x over 2 squared. And this holds that x equals 0. This one holds at x0 equal to 0. OK? And this thing, we call Fejer's or the Fejer kernel. So all right? And now, let me just list off a few properties that we'll get from this. Moreover, we have the following properties. 1, kn is non-negative. kn of x is equal to kn of minus x. It's even. And kn is 2 pi periodic. The second is that the integral of k sub n of x from minus pi to pi-- dx, or let's make this t-- dt equals 1. And the third is if delta is a positive number less than pi, less than or equal to pi, then for all x with absolute value bigger than or equal to delta and less than or equal to pi, we have that kn of x, which is equal-- I don't need the absolute values, because it's non-negative-- is less than or equal to 1 over 2 pi over n plus 1 times sine squared delta over 2, OK? So OK. So let's prove this theorem. And then, I'm going to say a few comments about-- well, since I have these properties right here, let me go ahead and make a few comments before we prove it. What does that mean kn looks like? Let me draw 0 pi minus pi. So kn is non-negative. It's even. And away from a small neighborhood, it's quite small if capital N is very big. So what it's looking like is maybe the first one-- and it's large at the origin. OK, so maybe that's n equals let's say 1. And then, let's say this is delta, and then minus delta. If I were to now look at, let's say, n equals-- I don't know-- a billion, it looks more like something that's very concentrated at the origin, but in such a way that the area underneath the graph-- so the integral-- the area equals 1, OK? And the same with what I drew in white, because white was supposed to be n equals 1. Yellow was supposed to be n equals-- I don't know-- 1,000. The area is always 1, OK? So this is telling you that if I look at sigma n of f-- so this is just some remarks. This is not to be taken completely literally. This is just the intuition on why we believe that the Cesaro means converge to f. And I'll say how this picture differs from if we looked at just SN. So this means that sigma n of f is, in fact-- so remember, we're going to get, in the end, that this is equal to kn of x minus t f of t dt. Now, kn is very concentrated near where t equals x, OK? So based on the picture, as n gets very large, this thing is getting more and more concentrated near where n equals x, OK? Now, and therefore, at least for let's say very nice f, if this thing is concentrated near where t equals x, then f of t will be approximately f of x. So f of x comes out of the integral because this is an integral dt. So since this thing is concentrated at-- and because the area underneath the curve is always 1, this integral is always equal to the integral of kn over any. So kn is 2 pi periodic. This integral is equal to the same integral over any 2 pi periodic interval, which means I could put here-- I could add an x to both top and bottom, and therefore change variables to get this is kn of t dt, which equals-- because the integral is 1, I would get something like f of x, OK? So this is a heuristic reason on why one should expect the Cesaro means to converge to f. OK? If you look back at the kernel that we had for the partial sums, it had some of the same-ish properties. It was 2 pi periodic, and also even. The integral was 1. And it did decay away from 0. However, it's non-negative. I'm talking about the Dirichlet kernel dn, which if you look back in your notes, was sine of plus n plus 1/2 times x over sine x over 2 with a constant out in front. And that little difference, the fact that this kernel is non-negative-- and the Dirichlet kernel is not-- makes a big difference. So although this heuristic argument-- maybe you don't see it there-- in the actual proof itself, that oscillation--and what I mean by isolation is the fact that dn actually does oscillate between negative and positive values--this bit of oscillation is actually what you can use to build up a continuous function whose partial sums do not converge to that continuous function at a point, OK? But as we'll see for the Cesaro means, the Cesaro Fourier means, basically, pick a space. And the Cesaro sums or Cesaro means converge to the function in whatever space you're talking about. And I'll say a little bit more about that in a minute. But OK. So let's prove the theorem that the Cesaro means are written in this way, and the kernel has these three properties. So let me recall that we have SN of x-- or let's put a k there-- this is equal to, as we wrote last time, minus pi to pi DK of x minus t f of t dt, where DK of, say, t was, from last time, equal to 2n plus 1 over 2 pi at t equals 0 and sine n plus 1/2 t over sine t over 2, and then with 1 over 2 pi out in front, I believe. Let me make sure I got the right exponent. Right. For t not equal to 0. OK? Oh, and this should be k. OK, so using this, we have that the Cesaro sum of x-- this is equal to 1 over n plus 1, sum from k equals 0 to n, the mean of the first and partial sums. And this is equal to--now, sk f of x is equal to this. So I can write this as integral from minus pi to pi of 1 over n plus 1 sum from k equals 0 to n of DK x minus t f of t dt. And so this here is kn of x minus t. All right? So now, I'm just going to verify that kn of x takes that form that we had before. And kn of x-- this is equal to 1 over n plus 1 sum from k equals 0 to n of DK of x. And let's go to the next half board. So I can write this as 1 over 2 pi n plus 1 and times-- so I will look at the case that x is non-0. x equals 0 is, you'll get what you get. But let's look at x not equal to 0. So then, I plug in this formula here and pull out a sine t over 2-- or sine x over 2 squared on the bottom. And then, I get k equals 0 to n of sine x over 2 times sine n plus 1/2 x, OK? And because I feel like it, let me put a 2 here and a 2 here. Why do I feel like it? Well, it's because if I have 2 times sine of a sine b, I can write that as using my angle sum formulas from trigonometry. You wondered why those would be useful. Well, here they are appearing in the advanced MIT class. You can write this as sum from k equals 0 to n of cosine n x minus cosine n plus 1x. Let me make sure I got that right. Or this should be k. I'm sorry. That should have been k. k, k, all right? Now this, is a telescoping sum, right? I have a sum of cosine kx. I have a cosine k plus 1x. So this is equal to-- so let's just write this out. And let me just indicate why this is a telescoping sum. We get cosine 0x minus cosine 1x plus cosine 1x minus cosine 2x dot dot dot plus the last one, which is cosine nx minus cosine n plus 1x. And OK, so this telescopes. That cancels with this. That will cancel with so on. And that last one will cancel. So all that we're left with is this one minus this one divided by this 2 that I have right there. And I get 1 over 2 pi n plus 1 times 1 over sine squared x over 2 times 1 minus cosine n plus 1 x over 2. And again, using a trig formula-- 1 minus cosine 2a equals sine squared-- divided by 2 is equal to sine squared a. So I get this is equal to 1 over 2 pi n plus 1 times sine squared n plus 1 over 2x divided by sine squared x over 2, OK? So that verifies the formula for the Fejer kernel. What about the properties that we have there? These properties-- at least the first two-- follow directly from this formula and the definition. So 1, follows immediately. This is clearly non-negative. It's even, taking x to minus x does not change this, because we have squares. And also because of the squares, it's 2 pi periodic rather than 4 pi periodic, OK? OK, so that's 1. For 2, we note that if we take the integral for minus pi to pi of the Dirichlet kernel, this is-- OK, we had a formula for the Dirichlet kernel, but remember, this is nothing but-- this was defined to be the sum from n equals minus k to k of e to the int dt, OK? Now, e to the int when n is not equal to 0 is 2 pi periodic. And when I integrate it from minus pi to pi, the integral from minus pi to pi of this 2 pi periodic thing-- you can just check. It's the integral of sine, nt, and cosine nt over its period. That's going to give me 0. So all I pick up is when n equals 0, right? And so that's equal to just the n equals 0 term. So that gives me 1, OK? So since the integral of each kernel is 1, then the integral of the Fejer kernel-- which remember, this is equal to the average of the Dirichlet kernels. And each of these is 1 sum from k equals 0 to n 1. I get n plus 1 divided by n plus 1. I get 1, OK? So that gives me 2. And for the third property, we have-- what do we have? Then, the function sine squared x over 2-- what does it look like? This is increasing. Or I should say it's even and increasing on 0 to pi. So what it looks like is sine squared x over 2. So there's pi minus pi sine squared. Looks like it goes up to 1. So if I'm looking at all x outside of-- so in that shaded region-- then, if x is outside of this delta region, then I get that sine squared x over 2 is going to be bigger than or equal to whatever, so it sits above the value that I get here, which is sine squared delta over 2. And therefore, I get that kn of x, which is equal to its absolute value, is less than or equal to 1 over 2 pi n plus 1 sine squared n plus 1 over 2x over-- I had sine squared x over 2, but since sine squared x over 2 is bigger than or equal to sine squared delta over 2, taking 1 over reverses the inequalities. And I get sine squared delta over 2 here. Sine of anything is always bounded above by 1. So I get this is less than or equal to 1 over 2 pi n plus 1 sine squared delta over 2, OK? So for the moment, let me put this absolute value there. I'm not doing it because I think it looks better. It's because I'm going to make a comment in a minute. OK, let me just make a small comment. Well, let me prove the next theorem. And then, I'll make the comment. OK, so we have these properties of the Fejer kernel. And now, what we're going to do is on our way to proving that we have convergence of the Cesaro means to a function in L2, we're first going to do it for continuous function. So you proved in the assignments that in L2 minus pi to pi, the continuous functions vanishing at the two endpoints are dense in the space big L2, OK? Now, if a function's continuous and equals 0 at both of the endpoints, it's 2 pi periodic in the sense that it has the same value at both endpoints. And therefore, the subspace of continuous functions that are 2 pi periodic is dense in L2. So if we're going to be able to show that the Cesaro means converge to a function in L2 for arbitrary L2 function, maybe it makes sense to try and do it first for continuous functions. And it's there that this argument that I just--this heuristic argument I gave here will be more math-like. OK, so we have a following theorem due to Fejer, which is the following. If f is continuous and 2 pi periodic, meaning f of pi is equal to f minus pi, then not only do we have the Cesaro means converging to f in L2, we actually have it in the best sense that you could for a continuous function. Then, sigma n of f converges to f uniformly in minus pi to pi, all right? So before, we were looking at Fourier series in L2. So convergence in L2 was the way one makes sense of infinite series or something converging to something else, all right? If we're looking at continuous functions, then we already a different norm there if we want to just consider a complete space containing continuous functions. We have the uniform norm, or the infinity norm. And so what this says is that even in this smaller space and in this stronger norm, we have convergence of the Cesaro means to the function f. But again, this doesn't imply that the Fourier series converges to f uniformly. Like I said, one can, in fact, use this oscillatory behavior of the Dirichlet kernel to prove there exist continuous functions whose Fourier series diverges at a point. And therefore, it doesn't converge uniformly to the function. But this is true for the Cesaro means because of these properties of the Fejer kernel, because it has this shape where it's non-negative. It's peaking near the origin. And it has total mass 1, and total integral 1. In some sense, you should think of, as n goes to infinity, sigma n is looking more and more like the Dirac delta function at 0, which maybe you encountered in physics. If that doesn't mean anything, don't worry about it. Just skip to the next part of the talk, which is supposed to have this magical property that it's 0 away from 0, which these are looking like, as integral 1. And when you integrate it against a function, you get f evaluated at the origin, which is like what we're saying here, OK? So again, that's some more heuristics. But linear operators depending on a parameter that appear like this, where it's a function of this form times f of t integrated dt, pop up all the time in harmonic analysis, OK? And having these properties, in fact, pops up also in harmonic analysis, OK? Harmonic analysis being a fancy name for Fourier analysis and other stuff. So let's prove this. So the first thing that I want to do is-- so f is a continuous function on minus pi to pi. That's 2 pi periodic. So I can extend f to all of R by periodicity. In other words, so we extend to all of R, meaning I have-- so there's pi minus pi pi. Here's a 3 pi. Here's minus 3 pi. So supposedly, I have this continuous function, which is 2 pi periodic. Now, I take that continuous function and just extend it by how it is here and so on, OK? I'm not saying I extend it by 0 outside. I'm saying I extend it periodically, OK? OK, now, I can write down a formula for exactly how you do that. But just trust me. You can do that. And also the following simple properties, then-- f, now referring to it as a function defined on all of R that's 2 pi periodic, this is also continuous, is 2 pi periodic, which implies that f is uniformly continuous and bounded, i.e. If I look at the infinity norm of f first off, because by periodicity, this is just equal to sup xn minus pi to pi, and because f is continuous, this thing is finite, OK? All right. Now, it's not difficult to believe that f is-- if I extend it by periodicity, it's going to be continuous. But using that and the fact that it's 2 pi periodic, you can then also conclude that it's uniformly continuous, meaning-- let's just quickly review what uniformly continuous means. This means for all epsilon positive, there exists a delta positive such that if y minus z is less than delta, then f of y minus f of z is less than epsilon, meaning I can choose a delta independent of y and of the point, right? Continuous at a point means I fix x. Then, for all epsilon, there exists a delta. Uniformly continuous means the delta doesn't depend on x, the point that I'm looking at. All right, so we have basic observation that we're going to make there. And maybe I'll just leave this up for now. So we want to prove the sigma n's converge to f uniformly on minus pi to pi. So that means we should be able to find, for every epsilon, a capital M such that for all n bigger than or equal to M sigma int and for all x in minus pi to pi sigma n of f minus f is less than epsilon in absolute value. All right, so let epsilon be positive. Since f is uniformly continuous, as I stated-- recalling the definition-- that implies that there exists a delta positive such that if y minus z is less than delta, then f of y minus f of z is less than-- and let me get this right so it comes out pretty in the end-- is less than epsilon over 2, OK? So now, what we're going to go through is make that argument which I just erased actually precise, all right? So here, we're saying if f is very close to-- if any two points are sufficiently close, f is going to be close in value. OK, now choose M natural number so that for all n bigger than or equal to M, the quantity twice times the L infinity norm over n plus 1 times sine squared delta over 2 is less than epsilon over 2, OK? So n plus 1-- that's the thing that's changing. So I have these fixed numbers here now. I've fixed delta. I have the L infinity norm of f. So I have this number here. And I'm just saying, choose a capital M so that for all n bigger than or equal to M, this number of times-- and I'll even put it here-- times 1 over n plus 1 is small, is less than epsilon over 2. And I can do that because this, as capital N goes to infinity, converges to 0, right? OK. Now, since f and k sub n, the Dirichlet kernel, are 2 pi periodic, I can write the Cesaro mean, which is given by minus pi to pi kn of x minus t f of t dt. I can make a change of variables, set tau equal to x minus t. And then, this will be equal to-- what is it going to be equal to? x minus pi x plus pi kn of tau f of x minus tau d tau, OK? All of this change of variable stuff is fine, because I'm dealing with continuous functions. I'm integrating continuous functions. So that's the Riemann integral. We have a change of variables for the Riemann integral. So that's completely fine. OK, now this is the product of 2 pi periodic functions. And if I take the integral of that quantity of a 2 pi periodic function, the integral of that is equal to the integral over any interval of length 2 pi, all right? So we're integrating over an interval of length 2 pi, right? We're going from x minus pi to x plus pi. That is equal to the integral of the same quantity over any interval of length 2 pi. So it's also equal to the integral over minus pi to pi. OK? So all I'm saying is I can change variables and move the x minus t. And let me even go back to t instead of tau here. Because of periodicity, I can switch this x minus t over here to f. All right, now we're going to start seeing some magic happen. And this is where that heuristic argument that I gave earlier actually starts to make sense. So then, I have that for all n bigger than or equal to M-- so I have that condition that quantity was less than epsilon over 2. And for all x and minus pi to pi, I have that sigma n f of x minus f of x-- so this is equal to minus pi to pi. And again, I'm going to write this now as kn of t f of x minus t dt minus-- now, here's the trick. The Fejer kernel has integral 1. So I can actually write f of x as minus pi to pi integral kn of t times f of x dt. I'm integrating dt, right? Then this just pops out. I get f of x times the integral of the Fejer kernel, which is 1, OK? And this equals minus pi pi. So just combining things-- kn of t f of x minus t minus f of x, which is good, because we have a continuous function. And now, we have something inside that looks like I'm subtracting f of some argument minus f of the argument minus something, OK? Now, I'm going to split this integral into two parts, and then use the triangle inequality and bring the triangle inequality inside. In fact, I'm going to go ahead and do that here. This is less than or equal to if I combine terms like I did and then bring the absolute value inside. OK? And now, I'm going to split this integral up into two parts. This is equal to the integral over t less than delta. And because kn is non-negative, this is just kn of t f of x minus t minus f of x dt plus the other term. OK? kn of t f of x minus t minus f of t dt, all right? Now, what do we know? If the absolute value of t is less than delta, then x minus t minus x is equal to minus t, which is an absolute value less than delta. So note that x minus t minus x equals t is less than delta here, right? And therefore, this quantity here is less than epsilon over 2 by how we chose delta. So this is less than epsilon over 2 times the integral over this region of kn of t. OK? But I can make this region larger and just go back to-- so let me just leave it here as it is. Plus now what do I do with this piece? I have this. I bound by twice the L infinity norm of f. The absolute value of this is less than or equal to by the triangle inequality, the sum of the absolute values, which is less than or equal to the sup of this plus the sup of that and x, which is equal to twice the infinity norm. So I get 2 times the infinity norm of f popping out from this term, and kn of t-- oh, I'm away from t less than delta. And this is where I use that third property that I have from before, that it's less than 1 over 2 pi. So let me leave this here. Sum 1 over 2 pi n plus 1, sine squared delta over 2 dt, OK? And now, this, I can say, is less than or equal to the whole integral over minus pi to pi, which is equal to 1. Plus again, making this an integral over the entire region, I get 2 pi times-- or divided by 2 pi gives me 1. So I get twice infinity over n plus 1 sine squared delta over 2. And we chose n plus 1 so that this second quantity here is less than epsilon over 2. OK? And therefore, uniformly, we prove that for all capital N bigger than or equal to M for all x in minus pi to pi, the difference in sigma nf in f is less than epsilon, proving uniform convergence, OK? So here's the remark I was going to make, is that the same proof can be modified if instead of kn of x being bigger than or equal to 0-- let me make sure I'm saying the right thing. So if instead of this property, which we had for the Fejer kernel, we have that sup over n of the integral from minus pi to pi of kn of x is finite, meaning if I have a function or if I have a sequence of functions, kn's, and I have the corresponding operators that look like that-- maybe they're not associated to any questions about Fourier analysis, but I'm just saying-- and it satisfies the three properties I had before with the exception of being non-negative, but instead of that, it satisfies this property, then I can do redo the same proof and show that those things converge to f uniformly, OK? Why am I saying that? Because maybe you would like to then try your hand at replacing kn with dn, the Dirichlet kernel, OK? The Dirichlet kernel satisfies all of the other properties we had up there. The integral is 1. In absolute value, it decays away from x is less than delta. And it's even in 2 pi periodic, OK? But it doesn't satisfy this. And if I look at minus pi to pi of the Dirichlet kernel, what one can prove is that this is something like log n for large enough n, OK? All right? So that was just a tiny remark I wanted to say on why, if you thought about maybe redoing this proof using the Dirichlet kernel, which satisfies almost all the same properties with the exception of being non-negative, you could, if the Dirichlet kernel had satisfied this bound. But it doesn't. It satisfies this bound. It's like log n. And therefore, if I take the sup, I don't get something finite, OK? All right, so we've proven that the Cesaro means of a continuous function convert uniformly to a continuous function. So we're almost to the point where we can say that the Cesaro means of an L2 to function converge to an L2 function and conclude that the subset of exponentials divided by square root of 2 pi form a maximal orthonormal subset of L2, and therefore is in orthonormal basis so that the partial Fourier sums converge back to the function in L2. We just need one more bit of information. So we have the following theorem. For all f in L2 of minus pi to pi, if I look at sigma n of f-- so first off, this is just a finite linear combination of exponentials, right? So this is clearly an L2. It's a continuous function. But if I take the L2 norm of that, it could depend on n. But in fact, it's less than or equal to the L2 norm of f. OK? So how we'll prove this is we'll first-- and this bound is what allows us to go from the 2 pi periodic continuous functions to general L2 functions by a density argument, OK? So first, we'll do this for 2 pi periodic continuous functions, and then by density, conclude it for L2 functions. So suppose first that f is 2 pi periodic. And then of course, extend it to R by periodicity like we did before. Then, as before, we had that the Cesaro mean of f is equal to the integral from minus pi to pi of f of x minus t kn of t dt. And so if I compute the integral sigma n f of x squared dx, this is equal to-- so each one of these is equal to an integral over minus pi to pi. So I'm going to have three integrals. And f of x minus s times the complex conjugate fx minus t times kn of s and kn of t, And Then ds dt dx, OK? Now, all these functions are continuous. So we have a Fubini's theorem, which says we can reverse the order of integration however we please. So I can write this as the integral for minus pi to pi, minus pi to pi, of now integrating first with respect to x-- kn t. And now, integrating first with respect to x. dx, let's say ds, dt, OK? Now I do Cauchy-Schwarz on this. And so this is less than or equal to minus pi to pi minus pi to pi pi kn of s kn of t times-- I'm using Cauchy-Schwarz in x now-- so times the L2 norm of the function minus s. So I'm taking the L2 norm in this variable-- 2 times 2 ds dt, OK? What I mean by this is I'm taking the L2 norm of this function depending on s, But In the first variable-- in this x variable, OK? So just write it out to see what I mean. Now, this is the integral of a function over an interval of length 2 pi. That's 2 pi periodic. That's equal to the integral of that function over any 2 pi periodic interval or any interval of length 2 pi. So I can, in fact, remove this s and remove this t, and just pick up the L2 norm of f in both places. So this is, in fact, equal to-- and because these two things no longer depend on s and t, they come all the way out of the integral. And I get L2 norm squared times minus pi to pi kn of s ds times the integral from minus pi to pi kn of t dt. Both of these integrals equal 1. So I get norm squared. And I started off with the L2 norm squared, or the L2 norm squared of the Cesaro mean of f. So I get this for all 2 pi periodic continuous functions, OK? Now, how do we then get the bound for general f? We use the density argument. So by what you've done in the assignments, there exists a sequence of 2 pi-- so let me start over real quick. Now, let's take a general element in L2. OK, now we start. By assignments, you know that there exists a sequence of 2 pi periodic continuous functions converging to f in L2-- fn a of 2 pi periodic continuous functions such that the fn's converge to f in L2. And one can verify simply from the definition of each of the Cesaro means that then-- so this is as little n goes to infinity--that then the Cesaro means also converge as little n goes to infinity. So capital N here is fixed, OK? Just using the definition of what the Cesaro mean is and Cauchy-Schwarz, basically, OK? And the fact that fn's converge to f in L2. Thus we get that the L2 norm of the Cesaro mean is equal to the limit, as n goes to infinity, of-- so this is little n-- of the L2 norm of the Cesaro means of these continuous 2 pi periodic functions, which as we've proven already-- these are all less than or equal to the L2 norm of fn, because they're 2 pi periodic. And again, because f is converging to fn, the norms converge. And I get the result I wanted for general L2 functions. OK? So now, we're almost there. What we have is this bound. And we have that the Cesaro remains converge to-- so if I take the Cesaro means of a continuous function, those converge to the continuous function uniformly on the interval. We're going to use that, this bound, and the density, again, of the 2 pi periodic continuous functions in L2 to conclude the following theorem. For all f in L2, the Cesaro means converge to f as capital N goes to infinity. In particular, we get, as an immediate corollary, if all of the Fourier coefficients are 0, then f is 0, right? Because if I've proven this and all the Fourier coefficients are 0, then the Cesaro means are all 0. And therefore, since this is 0 converging to f, f must be 0. OK? And therefore, the set of exponentials--normalized, of course-- form a maximal orthonormal subset of L2, i.e. that they're an orthonormal basis for big L2, which answers the question we had about Fourier series converging to a function in L2, OK? All right. So we'll do this just as a standard epsilon n argument. Let epsilon-- so let f be in L2. Let epsilon be positive. So we know that the continuous 2 pi periodic functions are dense in L2, because we did this in the assignment that for any f in L2 over an interval, I can find a continuous function that vanishes at the endpoints and therefore is periodic, which is close to f in L2. So there exists a g that's continuous 2 pi periodic such that f minus g in L2 norm is less than or epsilon over 3. So since sigma N g converges to g uniformly on minus pi to pi, there exists a natural number M such that for all N bigger than or equal to M, for all x minus pi to pi, I have that sigma N g of x minus g of x is less than epsilon over 3 square root of 2 pi. OK? Now, we go about the part where we replace f with g, OK? Then, for all N bigger than or equal to M, if I look at the L2 norm of sigma N of f minus f in L2, and I apply the triangle-- I add and subtract terms and apply the triangle inequality-- I get that this is less than or equal to sigma N of f minus g 2 plus sigma N of g minus g in L2 plus g minus f in L2, OK? So sigma N of f minus g is equal to sigma N of f minus sigma N of g. So I use that there without explicitly stating that. So let me say sigma N of f minus sigma N of g. Just from the definition, you can check this is equal to sigma N of f minus g, OK? Now, by the bound I just proved, the L2 norm of the Cesaro mean is less than or equal to the L2 norm of the function here. So this is less than or equal to f minus g2. And then, I also have this L2 norm of f minus G there. So I'll put a 2 there plus-- and I'll actually write out what this is-- sigma N g of x minus g of x squared dx 1/2, OK? Now, f minus g is less than epsilon over 3 in L2 norm. So this is less than twice epsilon over 3. Sigma N g minus g is less than epsilon over 3 square root of 2 pi here. So I get epsilon over 3-- that pulls all the way out-- minus pi pi 1 over 2 pi dx. And I just get epsilon in the end, OK? OK. So that concludes what I wanted to do for Fourier series, at least for now, which applies what we've done for Lebesgue integration, these big LP spaces, and also some of this general machinery we've built up for Hilbert spaces to actually answer a more concrete question rather than just trying to prove general statements. General statements are very, very useful. I'm not saying they're not. But I'm just saying so that you can see a concrete problem why one would want and use functional analysis in the first place. Now, coming back to what we've done so far, so let me just make a couple of remarks about what we haven't shown. It's a very deep theorem due to Carleson. So what we've shown is that the partial sums-- so we showed the set of exponentials normalized, or a maximal orthonormal set-- I mean that they're orthonormal basis. So the partial sums converge to f in L2. So this is what we've shown. For all f in L2, the partial sums converge to f in L2, all right? But this does not translate into a point-wise statement. This does not say that the partial sums converge to f almost everywhere. OK? There is a general theorem one can say that is covered in more advanced measure theory classes where one can say that there exists a subsequence converging to f almost everywhere. But that's not very good, or at least very clean. Now, for a long time, it was not necessarily believed that the partial sums converged to f almost everywhere. But a theorem due to Carleson shows that for all f in L2, partial sums do converge to the function almost everywhere, OK? This is, in fact-- maybe this is true. Maybe this is not. I heard this from my advisor. Carleson spent a few decades trying to prove the negation of the statement, trying to come up with an example of a function whose partial sums converge don't converge almost everywhere back to the function. And then, he came up with the bright idea that, well, maybe that's not true. Let me spend some time trying to put myself in the other shoes. And within a year or a couple of years, he was able to prove this theorem, OK? So this is Carleson's theorem that we do have convergence almost everywhere. Now, you can also ask, what about convergence? So this convergence in L2 of the partial sums. We have other LP spaces, right? What about in those LP spaces? Can I replace this 2 with p? The Fourier coefficients and partial sums-- these all make sense for any big LP space. So what is known is that also-- and now, the name is escaping me, but I'll just state it. For all p between 1 and infinity, the partial sums converge to the function in lp. When p equals 1, this is false, OK? And when p equals infinity, this is also false, because the partial sums-- these are a finite linear combination of exponentials, and therefore continuous function, OK? So you can't have, for an arbitrary function in L infinity--which can be discontinuous, just has to be bounded-- these converging to L in such a function. Because then, the limit would have to be continuous, OK? The uniform limit of continuous functions, which L infinity kind of is, has to be continuous, OK? So that's why you wouldn't expect it for L infinity. And for what one would call duality, because infinity is the dual of L1, you also don't get p equals 1. But in fact, things are worse there. You can come up with an L1 function. So that the Fourier series-- I don't think I'm lying when I say this, but-- diverges almost everywhere, I want to say, OK? I don't think I'm lying. But if p equals 1, one can come up with an example where the partial sums diverge point-wise almost everywhere, OK? OK. But to prove this flavor of statements requires deeper harmonic analysis, harmonic analysis being the umbrella that Fourier analysis sits in, and requires a knowledge of, or at least working with certain operators, which are called singular integral operators, which were developed back in the last century, middle of the last century at the University of Chicago by my mathematical grandfather and great grandfather, which gives you some beautiful results about, again, convergence of Fourier series, but also some applications to PDEs, which were why they were originally created in the first place and so on. But perhaps you'll encounter that if you take a class in harmonic analysis or Fourier series. I haven't taught the Fourier series class, so I don't know what it's about. But that kind of material will not be covered in this class. And this will be as far as we go as far as these types of questions, all right? So next time, we'll move on to minimizers over closed convex sets and consequences of that, one being that we can identify-- which is the most important application-- we can identify the dual of a Hilbert space with the Hilbert space in a canonical way. You can already prove that if you wish using the fact that every Hilbert space is asymmetrically isomorphic to little l2. You know that the dual of little l2 is 1 over q, is lq, where 1 over 2 plus 1 over q equals 1. And therefore, q equals 2. So little l2 is a dual of itself. But we'll prove it for general Hilbert spaces, which has some very important and interesting consequences when it comes to now studying, solving equations in Hilbert spaces, meaning you have linear operators. When can you solve equations involving these linear operators, and so on? All right, so we'll stop there.
2311	https://www.albert.io/blog/derivative-notation-and-defining-the-derivative-ap-calculus-ab-bc-review/	Skip to content ➜ AP® Calculus AB-BC Derivative Notation and Defining the Derivative: AP® Calculus AB-BC Review The Albert Team Last Updated On: Everyday life is packed with rapid changes—car speedometers, stock growth rates, and profit margins adjust in real time. Therefore, calculus steps in to measure those changes precisely. Understanding the derivative, its notation, and its link to the slope of the tangent line builds the foundation for nearly every AP® Calculus topic. Start practicing AP® Calculus AB-BC on Albert now! What We Review Building the Core Idea: What Is a Derivative? Intuitive Picture The derivative is the instantaneous rate of change of a function. Meanwhile, the average rate of change uses two points on a curve, forming a secant line. However, the derivative zooms in to a single point, giving the slope of the tangent line instead. Mini-Analogy: Zooming In Imagine pinching-to-zoom on a curved graph until one tiny piece looks straight. At that micro-level, the curve behaves like a straight line. The slope of that “almost straight” segment is the derivative. The Limit Definition of a Derivative The formal definition is \displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. Worked Example #1 Find f'(x) for f(x)=x^{2}-3x. Solution Write the difference quotient: \displaystyle \frac{(x+h)^{2}-3(x+h)-\bigl(x^{2}-3x\bigr)}{h} Expand numerator: \displaystyle \frac{x^{2}+2xh+h^{2}-3x-3h-x^{2}+3x}{h} Simplify: \displaystyle \frac{2xh+h^{2}-3h}{h} Factor out h: \displaystyle \frac{h(2x+h-3)}{h}=2x+h-3 Take the limit as h\to 0: \displaystyle f'(x)=2x-3. Key Takeaways Therefore, if the limit fails to exist or is infinite, the derivative at that point does not exist. Derivative Notation Common Symbols f'(x) y' \dfrac{dy}{dx} \dfrac{d}{dx}[f(x)] Reading Aloud “f prime of x” “y prime” “dee-y dee-x” “dee by dee-x of f of x” Worked Example #2 Given y=\sqrt{x}, the derivative is y'=\dfrac{1}{2\sqrt{x}}. The same result may be written as f'(x)=\dfrac{1}{2\sqrt{x}} or \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}. Pro Tip Choose derivative notation that matches the question format. For graphs, f'(a) is efficient. However, tables often favor \dfrac{dy}{dx}. Four Representations of a Derivative (CHA-2.B.4) Graphical Representation Picture a parabola y=x^{2}. Draw a tangent at x=1. The slope visually appears about 2. (See Example #3.) Ready to boost your AP® scores? Explore our plans and pricing here! Numerical (Tabular) Representation Make a small-h table for f(x)=x^{2} at x=1. \| \| \| --- \| \| h \| Average rate \dfrac{f(1+h)-f(1)}{h} \| \| 0.1 \| 2.1 \| \| 0.01 \| 2.01 \| \| 0.001 \| 2.001 \| Therefore, the numbers approach 2, confirming the derivative. Analytical (Symbolic) Representation After mastering limits, rules like the Power Rule speed things up. For f(x)=x^{2}, directly write f'(x)=2x. Verbal Representation “Temperature is rising at 2 °C per minute” is a plain-English derivative. Derivative at a Point = Slope of the Tangent Line (CHA-2.C.1) Geometric Meaning Zooming again, the tangent line touches the curve at one point and mimics its direction there. Its slope equals the derivative value. Worked Example #4 For g(x)=\ln x at x=e: Use the known derivative rule: g'(x)=\dfrac{1}{x}. Therefore, g'(e)=\dfrac{1}{e}. The tangent line formula: \displaystyle y-g(e)=g'(e)(x-e) which simplifies to \displaystyle y-1=\dfrac{1}{e}(x-e). Common Pitfalls Students sometimes confuse the secant slope between two points with the slope of the tangent line at a single point. Always let h\to 0. Putting It All Together: Mixed-Format AP-Style Example A piecewise velocity graph shows a straight segment rising from 0 m/s to 10 m/s over 5 s, then a flat segment at 10 m/s. Tasks Estimate the derivative at t=2\text{ s} from the graph. State units. Decide where the derivative fails to exist. Solution The line through (0, 0) and (5, 10) has slope 2, so v'(2)\approx 2\text{ m/s}^2. Therefore, the units are meters per second squared (acceleration). At t=5\text{ s} the graph has a sharp corner; thus, the derivative (acceleration) does not exist there. This example required switching among graphical, numerical (slope), and verbal interpretations—exactly the AP® style. Quick Reference Chart: Essential Vocabulary \| \| \| --- \| \| Term \| Definition or Key Feature \| \| Derivative \| The instantaneous rate of change; slope of the tangent line \| \| Limit \| The value a function approaches as the input nears some point \| \| Tangent line \| Straight line that touches a curve at one point and matches its slope there \| \| \dfrac{dy}{dx} \| Leibniz notation for the derivative of y with respect to x \| \| Instantaneous rate of change \| Value of the derivative at a single point \| \| Secant line \| Line through two points on a curve; gives an average rate of change \| \| Power Rule \| If f(x)=x^{n}, then f'(x)=nx^{n-1} \| Conclusion: Key Takeaways Derivatives describe how quantities change at a specific instant. Moreover, mastering the limit definition and derivative notation unlocks deeper skills: graph analysis, motion problems, and optimization. Therefore, practice translating among graphical, numerical, analytical, and verbal views. With consistent effort, the slope of the tangent line will feel as natural as the slope of a straight line. Sharpen Your Skills for AP® Calculus AB-BC Are you preparing for the AP® Calculus exam? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now! 2.1 Defining Average and Instantaneous Rates of Change at a Point 2.3 Estimating Derivatives of a Function at a Point Need help preparing for your AP® Calculus AB-BC exam? Albert has hundreds of AP® Calculus AB-BC practice questions, free responses, and an AP® Calculus AB-BC practice test to try out. Start practicing AP® Calculus AB-BC on Albert now! Interested in a school license? 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2312	https://van.physics.illinois.edu/ask/listing/19313	Photon Number, Energy, and Wavelength \| Physics Van \| Illinois University of Illinois Urbana-Champaign The Grainger College of Engineering Physics Van Search Menu Search Home Request a Show Frequently Asked Questions Ask the Van Volunteers Contact Us Home Ask the Van Photon Number, Energy, and Wavelength Category Subcategory Search Most recent answer: 05/06/2012 Q: Why the number of the photons decreases when its energy increases ? Gehad (age 18) Egypt A: Hello Gehad, Your premise about the number of photons and energy is not quite correct. The relation that we now believe to be correct is that the wavelength of the photon decreases as the energy of the photon increases. They are related by λ = hc/ E photon where h is Plank's constant, c is the velocity of light and λ is the wavelength of the photon. In a given packet of photons the total energy is proportional to the number of photons. LeeH (published on 05/06/2012) Follow-Up #1: More on energy and photon number. Q: Firstly, Thanks for answering me so fast. The total energy of photons is what I meant. I read in a book the following paragraph and there was no explanation: "The emission is a great flood of photons, their energy increases when their frequency increases. But their number decreases when the energy increases" Is it incorrect ? Gehad (age 18) Egypt A: Hi Gehad, The first statement is correct. The only way the second statement is correct is if the writer implied that "For a constant amount of total energy, the number of photons would decrease as the frequency increases". Otherwise it's wrong as explained in the previous post. LeeH (published on 05/07/2012) Follow-Up #2: why is photon energy hf? Q: why number of photon is greater in red light than in blue light,assuming their energies equal?please tell me theoritical reason .thanks hajra naeem (age 18) hadali A: The number of photons is just the energy divided by the energy per photon.So why is the energy E of a photon proportional to its frequency, f,: E=hf? That's a deep question. It leads straight to a deeper question: the energy of anything is just hf where h is Planck's constant and f is the frequency at which the quantum state changes phase. Like some very deep questions, the answer is almost trivial. If E=hf,always, then E and f are just the same thing, expressed in different units. Planck's constant is just the conversion factor from conventional frequency units to conventional energy units. Now we get to the real physics. Why is the quantum f from that universal definition, E-hf, the same as the frequency you get from measuring the oscillation of the electric field in a wave made of photons? This is subtle. The electric field for any single photon has an expected value of zero. Electric field oscillations come from beats between states with different numbers of photons, differing by one. So the beat frequencies are just f, the single-photon quantum frequency. Really, we don't expect that this short last argument should be very clear. Usually it isn't learned until one takes a graduate quantum mechanics course. Mike W. (published on 02/12/2015) Related Questions Can you use light to attract or repel an item? Absorption of short light pulses light from Hiroshima light dependent switches refraction and reflection light reflection from glass light from old sources Seeing reflected and emitted light Speed of light in various directions The optics of sunlight Still Curious? Expore Q&As in related categories Properties of Light Physics Van Twitter Instagram Facebook LinkedIn YouTube Urbana, IL 61801 Phone: Fax: Email: The Grainger College of Engineering This program is supported in part by the National Science Foundation (DMR 21-44256) and by the Department of Physics. Any opinions, findings, and conclusions or recommendations expressed in this website are those of the author(s) and do not necessarily reflect the views of the National Science Foundation. 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2313	https://www.mathsisfun.com/geometry/symbols.html	Symbols in Geometry Symbols in Algebra Symbols in Mathematics Show Ads Hide Ads \| About Ads We may use Cookies OK Home Algebra Data Geometry Physics Dictionary Games Puzzles [x] Algebra Calculus Data Geometry Money Numbers Physics Activities Dictionary Games Puzzles Worksheets Hide Ads Show Ads About Ads Donate Login Close Symbols in Geometry Common Symbols Used in Geometry Symbols save time and space when writing. Here are the most common geometrical symbols (also see Symbols in Algebra): \| Symbol \| Meaning \| Example \| In Words \| --- --- \| \| △ \| Triangle \| △ABC has 3 equal sides \| Triangle ABC has three equal sides \| \| ∠ \| Angle \| ∠ABC is 45° \| The angle formed by ABC is 45 degrees. \| \| ⊥ \| Perpendicular \| AB⊥CD \| The line AB is perpendicular to line CD \| \| ∥ \| Parallel \| EF∥GH \| The line EF is parallel to line GH \| \| ° \| Degrees \| 360° \| 360 degrees (a full rotation!) \| \| ∟ \| Right Angle (90°) \| ∟ is 90° \| A right angle is 90 degrees \| \| \| Line Segment "AB" \| AB \| The line segment between A and B \| \| \| Line "AB" \| \| The infinite line that includes A and B \| \| \| Ray "AB" \| \| The line that starts at A, goes through B and continues on \| \| ≅ \| Congruent (same shape and size) \| △ABC ≅ △DEF \| Triangle ABC is congruent to triangle DEF \| \| ∼ \| Similar (same shape, different size) \| △DEF∼△MNO \| Triangle DEF is similar to triangle MNO \| \| ∴ \| Therefore \| a=b b=a \| a equals b, therefore b equals a \| Example: In △ABC, ∠BAC is ∟ Is really saying: "In triangle ABC, the angle BAC is a right angle" Naming Angles For angles the central letter is where the angle is. Example: ∠ABC is 45° The point "B" is where the angle is. Mathopolis:Q1)Q2)Q3)Q4)Q5)Q6)Q7)Q8)Q9)Q10) Symbols in AlgebraSymbols in MathematicsGreek LettersGeometry Index Donate ○ Search ○ Index ○ About ○ Contact ○ Cite This Page ○ Privacy Copyright © 2025 Rod Pierce
2314	https://www.scribd.com/document/385893542/Volumetric-or-Cubical-Expansion-Coefficients-of-Liquids	Volumetric or Cubical Expansion Coefficients of Liquids \| PDF \| Thermal Expansion \| Methanol 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) 1K views 1 page Volumetric or Cubical Expansion Coefficients of Liquids This document lists the volumetric or cubical expansion coefficients of various common liquids. It provides the coefficient of expansion in units of 1/K and 1/°C for liquids such as water, e… Full description Uploaded by Immer AI-enhanced description Go to previous items Go to next items Download Save Save Volumetric or Cubical Expansion Coefficients of Li... For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Volumetric or Cubical Expansion Coefficients of Li... For Later You are on page 1/ 1 Search Fullscreen Search the most efficient way to navigate the Engineering ToolBox! Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! V olumetric or Cubical Expansion Coefficients of Liquids Volum etric expansion coefficients of some common liquids Sponsored Links Coefficients of cubical or volumetric thermal expansion of some common liquids: Liquid Volumetric Coefficient of Expansion (1/K, 1/ o C)(1/ o F) Acetic acid 0.001 10 0.00061 Acetone 0.00143 0.00079 Alcohol, ethyl (ethanol)0.00109 0.00061 Alcohol, meth yl (methanol,wo od alcohol, wo od naphtha,wood spirits, CH 3 OH)0.0 0 1 4 9 0.0 0 0 8 3 Ammonia 0.00245 0.00136 Aniline 0.00085 0.00047 B e n z e n e 0.0 0 1 2 5 0.0 0 0 6 9 B r o m i n e 0.0 0 1 1 0 0.0 0 0 6 1 C a l c i u m C h l o r i d e, 5.8% s o l u t i o n 0.0 0 0 2 5 C a l c i u m C h l o r i d e, 4 0.9% s o l u t i o n 0.0 0 0 4 6 C a r b o n d i s u l f i d e 0.0 0 1 1 9 0.0 0 0 6 6 C a r b o n t e t r a c h l o r i d e 0.0 0 1 2 2 0.0 0 0 6 8 C h l o r o f o r m 0.0 0 1 2 7 0.0 0 0 7 1 E t h e r 0.0 0 1 6 0 0.0 0 0 8 9 E t h y l a c e t a t e 0.0 0 1 3 8 0.0 0 0 7 7 E t h y l e n e g l y c o l 0.0 0 0 5 7 0.0 0 0 3 2 D i c h l o r o d i f l u o r o m e t h a n e r e f r i g e r a n t R-1 2 0.0 0 2 6 0.0 0 1 4 4 n-H e p t a n e 0.0 0 1 2 4 0.0 0 0 6 9 H y d r o c h l o r i c a c i d, 3 3.2% s o l u t i o n 0.0 0 0 4 6 I s o b u t y l a l c o h o l 0.0 0 0 9 4 0.0 0 0 5 2 G a s o l i n e 0.0 0 0 9 5 0.0 0 0 5 3 G l y c e r i n e (g l y c e r o l)0.0 0 0 5 0 0.0 0 0 2 8 K e r o s e n e, j e t f u e l 0.0 0 0 9 9 0.0 0 0 5 5 M e r c u r y 0.0 0 0 1 8 0.0 0 0 1 0 M e t h y l i o d i d e 0.0 0 1 2 0.0 0 0 6 7 n-O c t a n e 0.0 0 1 1 4 0.0 0 0 6 3 O i l (u n u s e d e n g i n e o i l)0.0 0 0 7 0 0.0 0 0 3 9 O l i v e o i l 0.0 0 0 7 0 P a r a f f i n o i l 0.0 0 0 7 6 4 0.0 0 0 4 2 P e t r o l e u m 0.0 0 1 0 0.0 0 0 5 6 n-P e n t a n e 0.0 0 1 5 8 0.0 0 0 8 8 P h e n o l 0.0 0 0 9 0.0 0 0 5 0 P o t a s s i u m c h l o r i d e, 2 4.3% s o l u t i u o n 0.0 0 0 3 5 S o d i u m c h l o r i d e, 2 0.6% s o l u t i o n 0.0 0 0 4 1 S o d i u m s u l f a t e, 2 4% s o l u t i o n 0.0 0 0 4 1 S u l f u r i c a c i d, c o n c e n t r a t e d 0.0 0 0 5 5 0.0 0 0 3 1 T o l u e n e 0.0 0 1 0 8 0.0 0 0 6 0 T r i c h l o r o e t h y l e n e 0.0 0 1 1 7 0 0.0 0 0 6 5 T u r p e n t i n e 0.0 0 1 0 0 0 0.0 0 0 5 6 Water 1) 0.0 0 0 2 1 4 0.0 0 0 1 2 1) Volume tric expansion coefficients for water at different temperatures Example and Calculator - Volumetric Expansio n Sponsored Links V ol u me tr ic or Cu bi ca l E xp an s io n C oe f f ic ie nt s o f L i qu i ds ht tp s://ww w.e ng in ee r in gt oo l bo x.co m/cu bi ca l-ex pa ns io n-co ef fi ci e nt s-d_...1 d e 6 1 0/0 1/2 0 1 8 9:0 0 a. m. 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 Vessel Sizing - Knockout Drums Design Guide No ratings yet Vessel Sizing - Knockout Drums Design Guide 2 pages UNIT 2 Pollution PDF No ratings yet UNIT 2 Pollution PDF 5 pages Sellsheet - Silkflo New Aug07 2 No ratings yet Sellsheet - Silkflo New Aug07 2 2 pages LixTRA™ Novel Copper Leach Technology No ratings yet LixTRA™ Novel Copper Leach Technology 6 pages Biological Stains - Handbook On The Nature and Uses of The Dyes Employed in The Biological Laboratory No ratings yet Biological Stains - 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2315	https://courses.lumenlearning.com/calculus3/chapter/three-dimensional-coordinate-systems/	Module 2: Vectors in Space Three-Dimensional Coordinate Systems Learning Objectives Describe three-dimensional space mathematically. Locate points in space using coordinates. As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontal [latex]x[/latex]-axis and the vertical [latex]y[/latex]-axis. We can add a third dimension, the [latex]z[/latex]-axis, which is perpendicular to both the [latex]x[/latex]-axis and the [latex]y[/latex]-axis. We call this system the three-dimensional rectangular coordinate system. It represents the three dimensions we encounter in real life. definition The three-dimensional rectangular coordinate system consists of three perpendicular axes: the [latex]x[/latex]-axis, the [latex]y[/latex]-axis, and the [latex]z[/latex]-axis. Because each axis is a number line representing all real numbers in [latex]\mathbb{R}[/latex] the three-dimensional system is often denoted by [latex]\mathbb{R}^3[/latex]. In Figure 1(a), the positive [latex]z[/latex]-axis is shown above the plane containing the [latex]x[/latex]– and [latex]y[/latex]-axes. The positive [latex]x[/latex]-axis appears to the left and the positive [latex]y[/latex]-axis is to the right. A natural question to ask is: How was arrangement determined? The system displayed follows theright-hand rule. If we take our right hand and align the fingers with the positive [latex]x[/latex]-axis, then curl the fingers so they point in the direction of the positive [latex]y[/latex]-axis, our thumb points in the direction of the positive [latex]z[/latex]-axis. In this text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow a left-hand rule, but the right-hand rule is considered the standard representation. Figure 1. (a) We can extend the two-dimensional rectangular coordinate system by adding a third axis, the [latex]z[/latex]-axis, that is perpendicular to both the [latex]x[/latex]-axis and the [latex]y[/latex]-axis. (b) The right-hand rule is used to determine the placement of the coordinate axes in the standard Cartesian plane. In two dimensions, we describe a point in the plane with the coordinates latex[/latex]. Each coordinate describes how the point aligns with the corresponding axis. In three dimensions, a new coordinate, [latex]z[/latex], is appended to indicate alignment with the [latex]z[/latex]-axis: latex[/latex]. A point in space is identified by all three coordinates (Figure 2). To plot the point latex[/latex], go [latex]x[/latex] units along the [latex]x[/latex]-axis, then [latex]y[/latex] units in the direction of the [latex]y[/latex]-axis, then [latex]z[/latex] units in the direction of the [latex]z[/latex]-axis. Figure 2. To plot the point latex [/latex] go [latex] x [/latex] units along the [latex]x[/latex]-axis, then [latex]y[/latex] units in the direction of the [latex]y[/latex]-axis, then [latex] z[/latex] units in the direction of the z-axis. Example: locating points in space Sketch the point latex[/latex] in three-dimensional space. Show Solution To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in the positive [latex]x[/latex] direction, 2 units in the negative [latex]y[/latex] direction, and 3 units in the positive [latex]z[/latex] direction. Complete the prism to plot the point (Figure 3). Figure 3. Sketching the point latex[/latex]. try it Sketch the point latex[/latex] in three-dimensional space. Show Solution Figure 4. Sketch of point latex[/latex] In two-dimensional space, the coordinate plane is defined by a pair of perpendicular axes. These axes allow us to name any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just as in two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms a coordinate plane: the [latex]xy[/latex]-plane, the [latex]xz[/latex]-plane, and the [latex]yz[/latex]-plane (Figure 5). We define the [latex]xy[/latex]-plane formally as the following set: [latex]{(x,y,0) :x,y\in\mathbb{R}}[/latex]. Similarly, the [latex]xz[/latex]-plane and the [latex]yz[/latex]-plane are defined as [latex]{(x,0,z) :x,z\in\mathbb{R}}[/latex] and [latex]{(0,y,z) :y,z\in\mathbb{R}}[/latex], respectively. To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished. (Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finished walls meet, facing out into the room, the floor is the [latex]xy[/latex]-plane, the wall to your right is the [latex]xz[/latex]-plane, and the wall to your left is the [latex]yz[/latex]-plane. Figure 5. The plane containing the [latex]x[/latex]– and [latex]y [/latex]-axes is called the [latex]xy [/latex]-plane. The plane containing the [latex]x [/latex]– and [latex]z [/latex]-axes is called the [latex]xz [/latex]-plane, and the [latex]y [/latex]– and [latex]z [/latex]-axes define the [latex]yz [/latex]-plane. In two dimensions, the coordinate axes partition the plane into four quadrants. Similarly, the coordinate planes divide space between them into eight regions about the origin, called octants. The octants fill [latex]\mathbb{R}^3[/latex] in the same way that quadrants fill [latex]\mathbb{R}^2[/latex], as shown in Figure 6. Figure 6. Points that lie in octants have three nonzero coordinates. Most work in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. In this section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to three dimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space. If two points lie in the same coordinate plane, then it is straightforward to calculate the distance between them. We that the distance [latex]d[/latex] between two points latex[/latex] and latex[/latex] in the [latex]xy[/latex]-coordinate plane is given by the formula [latex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/latex]. The formula for the distance between two points in space is a natural extension of this formula. theorem: the distance between two points in space The distance [latex]d[/latex] between points latex[/latex] and latex[/latex] is given by the formula [latex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}[/latex]. The proof of this theorem is left as an exercise. (Hint: First find the distance [latex]d_1[/latex] between the points latex[/latex] and latex[/latex] as shown in Figure 7.) Figure 7. The distance between [latex]P_1[/latex] and [latex]P_2[/latex] is the length of the diagonal of the rectangular prism having [latex]P_1[/latex] and [latex]P_2[/latex] as opposite corners. Example: distance in space Find the distance between points [latex]P_1=(3, -1, 5)[/latex] and [latex]P_2=(2, 1, -1)[/latex]. Figure 8. Find the distance between the two points. Show Solution Substitute values directly into the distance formula: [latex]\begin{align} d(P1,P2) &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\ &= \sqrt{(2 - 3)^2 + (1 - (-1))^2 + (-1 - 5)^2} \ &= \sqrt{(-1)^2 + 2^2 + (-6)^2} \ &= \sqrt{41} \ \end{align}[/latex] try it Find the distance between points [latex]P_1=(1, -5, 4)[/latex] and [latex]P_2=(4, -1, -1)[/latex]. Show Solution [latex]5\sqrt{2}[/latex] Watch the following video to see the worked solution to the above Try It. You can view the transcript for “CP 2.12” here (opens in new window) Before moving on to the next section, let’s get a feel for how [latex]\mathbb{R}^3[/latex] differs from [latex]\mathbb{R}^2[/latex]. For example, in [latex]\mathbb{R}^2[/latex], lines that are not parallel must always intersect. This is not the case in [latex]\mathbb{R}^3[/latex].For example, consider the line shown in Figure 9. These two lines are not parallel, nor do they intersect. Figure 9. These two lines are not parallel, but still do not intersect. You can also have circles that are interconnected but have no points in common, as in Figure 10. Figure 10. These circles are interconnected, but have no points in common. We have a lot more flexibility working in three dimensions than we do if we stuck with only two dimensions. Candela Citations CC licensed content, Original CP 2.12. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at Licenses and Attributions CC licensed content, Original CP 2.12. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 3. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at Privacy Policy
2316	https://www.mathisfunforum.com/viewtopic.php?pid=79369	Index Laws Practice (Page 1) / Exercises / Math Is Fun Forum Math Is Fun Forum Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ ° Index User list Rules Search Register Login You are not logged in. Topics: Active \| Unanswered Index »Exercises »Index Laws Practice Pages: 1 12007-05-07 02:19:38 ToastReal MemberRegistered: 2006-10-08 Posts: 1,321 Index Laws Practice To complete these questions it will be necessary to know the Index Laws: Law 1: Law 2: Law 3: Law 4: Law 5: And the rational index laws: Also, we define , where To prove this: , but 1: Simplify with positive indices: a) b) c) d) e) 2: Simplify with positive indices: a) b) c) d) e) 3: Simplify with positive indices: a) b) c) d) e) 4: Express as a power of x: a) b) c) d) e) Last edited by Toast (2007-05-07 02:22:15) Offline 22007-09-24 00:57:56 nova_angelMemberRegistered: 2007-09-24 Posts: 1 Re: Index Laws Practice Hey..thats cool...may you post some level 9 or secondary 3(asia) practice to the forum... i need more practise...anyway thank you~~this is kinda cool... Offline 32007-12-28 18:58:29 kmlb123MemberRegistered: 2007-12-28 Posts: 1 Re: Index Laws Practice ahhh...finally something i can understand ty for this stuff and could u upload some other stuff like medium level coordinate geometry (i drag at it) TY4 THIS Offline 42007-12-28 20:25:44 Jai GaneshAdministratorRegistered: 2005-06-28 Posts: 51,883 Re: Index Laws Practice Hi kmlb123! Happy to learn that you found this Exercise useful! I shall post more exercises at Middle School/High School level; certainly I shall try to post an exercise exclusively on Coordinate Geometry. It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel. Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking. Offline 52013-01-07 17:38:53 Ash OkasMemberRegistered: 2013-01-07 Posts: 1 Re: Index Laws Practice Exercise online, good see the stuff, helpful for kids and all, It inspired me to join the lovely forum, evoked my forgotten love for maths Offline 62013-01-07 20:27:10 BobAdministratorRegistered: 2010-06-20 Posts: 10,795 Re: Index Laws Practice hi Some people discover this forum but don't realise there is a huge wealth of excellent math teaching material including exercises at Try it! Bob Children are not defined by school ...........The Fonz You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei Sometimes I deliberately make mistakes, just to test you! …………….Bob Offline 72013-01-07 20:46:43 bobbymbumpkinFrom: Bumpkinland Registered: 2009-04-12 Posts: 109,606 Re: Index Laws Practice Hi Ash Okas; Welcome to the forum. Tell us about yourself in "Introductions." In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline 82013-01-08 07:16:45 noelevansMemberRegistered: 2012-07-20 Posts: 236 Re: Index Laws Practice Hi Toast! Nice set of problems! Here is a variation of law 2 that is easy to use and always ends up with a non-negative exponent. It takes care of all three cases: m>n, m=n, m<n. It is especially nice when the problem involves negative exponents. (p is the opposite of the smaller of m and n) For example recalling that a^0 = 1 we get ( Whether this is true for a=0 has been thoroughly explored in other threads of this forum.) m n p a^(m+p)/a^(n+p) 2 5-2 a^0 / a^3 = 1/(a^3) 3-2 2 a^5 / a^0 = a^5 -5-3 5 a^0 / a^2 = 1/(a^2) 4-7 7 a^11 / a^0 = a^11 3 3-3 a^0 / a^0 = 1 -4-4 4 a^0 / a^0 = 1 Of course the last two cases here would obviously be 1 from the start. This law takes care of all cases for positive, negative or zero exponents m and n, leaving the answer with a non-negative exponent for a in the numerator or denominator as most books require for the answer. If p is anything else but -min(m,n) the equality is still true, but it will not be in "simplified" form. Play around with it a bit and I think you will find it is quite handy and easy to use. Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional). LaTex is like painting on many strips of paper and then stacking them to see what picture they make. Offline Pages: 1 Index »Exercises »Index Laws Practice Board footer Jump to Atom topic feed Powered by FluxBB
2317	https://forum.wordreference.com/threads/handy-vs-easy-to-use.2009879/	handy vs easy-to-use \| WordReference Forums WordReference.comLanguage Forums ForumsRules/Help/FAQHelp/FAQ MembersCurrent visitors Interface Language Dictionary search: Log inRegister What's newSearch Search [x] Search titles and first posts only [x] Search titles only By: Search Advanced search… Rules/Help/FAQHelp/FAQ MembersCurrent visitors Interface Language Menu Log in Register Install the app Install How to install the app on iOS Follow along with the video below to see how to install our site as a web app on your home screen. Note: This feature may not be available in some browsers. English Only English Only handy vs easy-to-use Thread starterNavyBlue Start dateDec 15, 2010 NavyBlue Senior Member Spain/Spanish Dec 15, 2010 #1 Is it possible to use handy meaning easy-to-use? Thank you. F > Senior Member United Kingdom British English Dec 15, 2010 #2 No, "handy" means that something is useful, nothing close to an "easy-to-use" meaning. pickarooney Senior Member Provence, France English (Ireland) Dec 15, 2010 #3 It can be used to mean 'simple' and I suppose it could mean easy-to-use as well. Definition 2 here would suggest you're right: P Parla Member Emeritus New York City English - US Dec 16, 2010 #4 In the US, we generally use "handy" to describe something that's useful (for a particular purpose) and/or convenient (in the sense of near at hand, easily accessible). Dictionaries are handy for learning the meanings of words, and I keep several handy (right next to my desk). Neither would imply that the thing is necessarily easy to use. EStjarn Senior Member Spanish Dec 16, 2010 #5 According to the American Heritage Dictionary, handy does mean 'easy to use': handy adj.handier, handiest 1.Skillful in using one's hands; manually adroit. See Synonyms at dexterous. 2.Readily accessible: found a handy spot for the hammer. 3.Useful; convenient: a handy tool. 4.Easy to use or handle: a handy reference book. Click to expand... ~ I think 'easy to use' (or 'relatively easy to use') is an inherent quality of something that is handy. We wouldn't call something handy unless we found the operation of it at least relatively easy to perform. Last edited: Dec 17, 2010 Myridon Senior Member Texas English - US Dec 16, 2010 #6 EStjarn said: ~ I think 'easy to use' (or 'relatively easy to use') is an inherent quality of something that is handy. Click to expand... You seem to be treating those definitions as if the word always means all four things at the same time and/or that the fourth one is just as common as the first. It also happens that I disagree with the example given. I think a "handy reference book" is either accessible/convenient or useful. I also find it odd that "convenient" is part of definition 3 when it seems to go with definition 2. EStjarn Senior Member Spanish Dec 16, 2010 #7 Myridon said: You seem to be treating those definitions as if the word always means all four things at the same time and/or that the fourth one is just as common as the first. Click to expand... That wasn't my intention. I was referring to the third sense: 'handy' meaning 'useful'. Thank you for pointing it out. Going back to the original question - Is it possible to use handy meaning easy-to-use? Fish says it is not, pickarooney suggest that it is, Parla hints that it is not common to do so, AHD says it's possible, Myridon seems to agree with Parla. Personally I don't have an opinion about it. I only sense that when we describe something as 'handy', meaning that it is useful, there is an implication that this something, apart from being useful, is relatively easy for us to use. I think a "handy reference book" is either accessible/convenient or useful. Click to expand... The word 'accessible' has multiple meanings. In the AHD entry for 'handy', the word seems to refer to what Parla calls 'near at hand', which I interpret as 'easy to reach'. But it seems you're using the word 'accessible' in the sense of 'easy to understand', which roughly translates to 'easy to use'. Myridon Senior Member Texas English - US Dec 16, 2010 #8 EStjarn said: That wasn't my intention. I was referring to the third sense: 'handy' meaning 'useful'. Thank you for pointing it out. Click to expand... So you used the exact words of the fourth definition to refer to third definition? No wonder we're not getting anywhere. "Easy to use" You push one button and nothing happens. It's not useful. "Useful" You push a million buttons, stand on your head, then do six backhand springs and it does exactly what you want. It's not easy to use. Somethings are both in varying degrees of both qualities. Myridon seems to agree with Parla. Click to expand... I don't think I do. That fourth definition seems spurious since it comes with an example that doesn't seem to support it. EStjarn Senior Member Spanish Dec 16, 2010 #9 Myridon said: So you used the exact words of the fourth definition to refer to third definition? Click to expand... Not really. I was focusing on the thread subject which deals with the term 'easy-to-use' in relation to the term 'handy'. If it might help ro clear up the confusion, I'll make a second attempt at clarifying myself: Post #5 a) states that 'handy' - according to AHD - indeed means 'easy to use'; b) comments that the quality of being 'easy to use' seems inherent in the quality of being 'useful' when we use the word 'handy' to describe something which we find useful. If this something wouldn't be at least relatively easy to use, we wouldn't choose the word 'handy' to describe it. panjandrum Senior Member Belfast, Ireland English-Ireland (top end) Dec 16, 2010 #10 Here are the OED meanings. They are quite distinct, but could, of course, apply to the same thing. Ready to hand; near at hand; conveniently accessible or ready for use. Convenient to handle or hold in the hand; easy to be manipulated, managed, or directed. They could also be quite separate. I always keep my floppledinger handy when I'm peeling tomatoes. It's a real horror to use, but it is so fast I wouldn't do without it. You must log in or register to reply here. Share: BlueskyLinkedInWhatsAppEmailShareLink English Only English Only WR styleSystemLightDark English (EN-us) Contact us Terms and rules Privacy policy Help RSS Community platform by XenForo®© 2010-2025 XenForo Ltd. Back TopBottom
2318	https://en.wikipedia.org/wiki/Prince_Rupert%27s_cube	Jump to content Prince Rupert's cube Español Français Magyar 日本語 Português Русский Српски / srpski Svenska Türkçe Українська Edit links From Wikipedia, the free encyclopedia Cube that fits through hole in smaller cube In geometry, Prince Rupert's cube is the largest cube that can pass through a hole cut through a unit cube without splitting it into separate pieces. Its side length is approximately 1.06, 6% larger than the side length 1 of the unit cube through which it passes. The problem of finding the largest square that lies entirely within a unit cube is closely related, and has the same solution. Prince Rupert's cube is named after Prince Rupert of the Rhine, who asked whether a cube could be passed through a hole made in another cube of the same size without splitting the cube into two pieces. A positive answer was given by John Wallis. Approximately 100 years later, Pieter Nieuwland found the largest possible cube that can pass through a hole in a unit cube. Many other convex polyhedra, including all five Platonic solids, have been shown to have the Rupert property: a copy of the polyhedron, of the same or larger shape, can be passed through a hole in the polyhedron. It was unknown whether this is true for all convex polyhedra; an August 2025 preprint claims the answer is no. Solution [edit] Place two points on two adjacent edges of a unit cube, each at a distance of 3/4 from the point where the two edges meet, and two more points symmetrically on the opposite face of the cube. Then these four points form a square with side length One way to see this is to first observe that these four points form a rectangle, by the symmetries of their construction. The lengths of all four sides of this rectangle equal , by the Pythagorean theorem or (equivalently) the formula for Euclidean distance in three dimensions. For instance, the first two points, together with the third point where their two edges meet, form an isosceles right triangle with legs of length , and the distance between the first two points is the hypotenuse of the triangle. As a rectangle with four equal sides, the shape formed by these four points is a square. Extruding the square in both directions perpendicularly to itself forms the hole through which a cube larger than the original one, up to side length , may pass. The parts of the unit cube that remain, after emptying this hole, form two triangular prisms and two irregular tetrahedra, connected by thin bridges at the four vertices of the square. Each prism has as its six vertices two adjacent vertices of the cube, and four points along the edges of the cube at distance 1/4 from these cube vertices. Each tetrahedron has as its four vertices one vertex of the cube, two points at distance 3/4 from it on two of the adjacent edges, and one point at distance 3/16 from the cube vertex along the third adjacent edge. History [edit] Prince Rupert's cube is named after Prince Rupert of the Rhine. According to a story recounted in 1693 by English mathematician John Wallis, Prince Rupert wagered that a hole could be cut through a cube, large enough to let another cube of the same size pass through it. Wallis showed that in fact such a hole was possible (with some errors that were not corrected until much later), and Prince Rupert won his wager. Wallis assumed that the hole would be parallel to a space diagonal of the cube. The projection of the cube onto a plane perpendicular to this diagonal is a regular hexagon, and the best hole parallel to the diagonal can be found by drawing the largest possible square that can be inscribed into this hexagon. Calculating the size of this square shows that a cube with side length : , slightly larger than one, is capable of passing through the hole. Approximately 100 years later, Dutch mathematician Pieter Nieuwland found that a better solution may be achieved by using a hole with a different angle than the space diagonal. In fact, Nieuwland's solution is optimal. Nieuwland died in 1794, a year after taking a position as a professor at the University of Leiden, and his solution was published posthumously in 1816 by Nieuwland's mentor, Jean Henri van Swinden. Since then, the problem has been repeated in many books on recreational mathematics, in some cases with Wallis' suboptimal solution instead of the optimal solution. Models [edit] The construction of a physical model of Prince Rupert's cube is made challenging by the accuracy with which such a model needs to be measured, and the thinness of the connections between the remaining parts of the unit cube after the hole is cut through it. For the maximally sized inner cube with length ≈1.06 relative to the length 1 outer cube, constructing a model is "mathematically possible but practically impossible". On the other hand, using the orientation of the maximal cube but making a smaller hole, big enough only for a unit cube, leaves additional thickness that allows for structural integrity. For the example using two cubes of the same size, as originally proposed by Prince Rupert, model construction is possible. In a 1950 survey of the problem, D. J. E. Schrek published photographs of a model of a cube passing through a hole in another cube. Martin Raynsford has designed a template for constructing paper models of a cube with another cube passing through it; however, to account for the tolerances of paper construction and not tear the paper at the narrow joints between parts of the punctured cube, the hole in Raynsford's model only lets cubes through that are slightly smaller than the outer cube. Since the advent of 3D printing, construction of a Prince Rupert cube of the full 1:1 ratio has become easy. Generalizations [edit] Unsolved problem in mathematics Do all convex polyhedra have the Rupert property? More unsolved problems in mathematics A polyhedron is said to have the Rupert property if a polyhedron of the same or larger size and the same shape as can pass through a hole in . All five Platonic solids—the cube, regular tetrahedron, regular octahedron, regular dodecahedron, and regular icosahedron—have the Rupert property. Of the 13 Archimedean solids, it is known that at least these ten have the Rupert property: the cuboctahedron, truncated octahedron, truncated cube, rhombicuboctahedron, icosidodecahedron, truncated cuboctahedron, truncated icosahedron, truncated dodecahedron, and the truncated tetrahedron, as well as the truncated icosidodecahedron. It has been conjectured that all 3-dimensional convex polyhedra have this property, but also, to the contrary, that the rhombicosidodecahedron does not have Rupert's property. More recently, a Steininger & Yurkevich (2025) preprint describes a counterexample, which they call the Noperthedron: an explicit convex polyhedron with 90 vertices, which is not Rupert. Cubes and all rectangular solids have Rupert passages in every direction that is not parallel to any of their faces. Another way to express the same problem is to ask for the largest square that lies within a unit cube. More generally, Jerrard & Wetzel (2004) show how to find the largest rectangle of a given aspect ratio that lies within a unit cube. As they observe, the optimal rectangle must always be centered at the center of the cube, with its vertices on the edges of the cube. Depending on its aspect ratio, the ratio between its long and short sides, there are two cases for how it can be placed within the cube. For an aspect ratio of or more, the optimal rectangle lies within the rectangle connecting two opposite edges of the cube, which has aspect ratio exactly . For aspect ratios closer to 1 (including aspect ratio 1 for the square of Prince Rupert's cube), two of the four vertices of an optimal rectangle are equidistant from a vertex of the cube, along two of the three edges touching that vertex. The other two rectangle vertices are the reflections of the first two across the center of the cube. If the aspect ratio is not constrained, the rectangle with the largest area that fits within a cube is the one of aspect ratio that has two opposite edges of the cube as two of its sides, and two face diagonals as the other two sides. Eleven of the 13 Catalan solids and 87 of the 92 Johnson solids—all but gyrate rhombicosidodecahedron , parabigyrate rhombicosidodecahedron , metabigyrate rhombicosidodecahedron , trigyrate rhombicosidodecahedron , and paragyrate diminished rhombicosidodecahedron —are known to have the Rupert property. The Catalan solids for which the Rupert property is not known are the deltoidal hexecontahedron and pentagonal hexecontahedron. For all , the -dimensional hypercube also has the Rupert property. Moreover, one may ask for the largest -dimensional hypercube that may be drawn within an -dimensional unit hypercube. The answer is always an algebraic number. For instance, the problem for asks for the largest (three-dimensional) cube within a four-dimensional hypercube. After Martin Gardner posed this question in Scientific American, Kay R. Pechenick DeVicci and several other readers showed that the answer for the (3,4) case is the square root of the smaller of two real roots of the polynomial , which works out to approximately 1.007435. For , the optimal side length of the largest square in an -dimensional hypercube is either or , depending on whether is even or odd respectively. References [edit] ^ Jump up to: a b Steininger, Jakob; Yurkevich, Sergey (2025), A convex polyhedron without Rupert's property, arXiv:2508.18475 ^ Jump up to: a b c Gardner, Martin (2001), The Colossal Book of Mathematics: Classic Puzzles, Paradoxes, and Problems : Number Theory, Algebra, Geometry, Probability, Topology, Game Theory, Infinity, and Other Topics of Recreational Mathematics, W. W. Norton & Company, pp. 172–173, ISBN 9780393020236 ^ Jump up to: a b Wells, David (1997), The Penguin Dictionary of Curious and Interesting Numbers (3rd ed.), Penguin, p. 16, ISBN 9780140261493 ^ Jump up to: a b c Rickey, V. Frederick (2005), Dürer's Magic Square, Cardano's Rings, Prince Rupert's Cube, and Other Neat Things (PDF), archived from the original (PDF) on 2010-07-05; notes for “Recreational Mathematics: A Short Course in Honor of the 300th Birthday of Benjamin Franklin,” Mathematical Association of America, Albuquerque, NM, August 2–3, 2005 ^ Jump up to: a b c Jerrard, Richard P.; Wetzel, John E. (2004), "Prince Rupert's rectangles", The American Mathematical Monthly, 111 (1): 22–31, doi:10.2307/4145012, JSTOR 4145012, MR 2026310 ^ Swinden, J. H. Van (1816), Grondbeginsels der Meetkunde (in Dutch) (2nd ed.), Amsterdam: P. den Hengst en zoon, pp. 512–513 ^ Ozanam, Jacques (1803), Montucla, Jean Étienne; Hutton, Charles (eds.), Recreations in Mathematics and Natural Philosophy: Containing Amusing Dissertations and Enquiries Concerning a Variety of Subjects the Most Remarkable and Proper to Excite Curiosity and Attention to the Whole Range of the Mathematical and Philosophical Sciences, G. Kearsley, pp. 315–316 ^ Dudeney, Henry Ernest (1936), Modern puzzles and how to solve them, p. 149 ^ Ogilvy, C. Stanley (1956), Through the Mathescope, Oxford University Press, pp. 54–55. Reprinted as Ogilvy, C. Stanley (1994), Excursions in mathematics, New York: Dover Publications Inc., ISBN 0-486-28283-X, MR 1313725 ^ Ehrenfeucht, Aniela (1964), The Cube Made Interesting, translated by Zawadowski, Wacław, New York: The Macmillan Co., p. 77, MR 0170242 ^ Stewart, Ian (2001), Flatterland: Like Flatland Only More So, Macmillan, pp. 49–50, ISBN 9780333783122 ^ Darling, David (2004), The Universal Book of Mathematics: From Abracadabra to Zeno's Paradoxes, John Wiley & Sons, p. 255, ISBN 9780471667001 ^ Pickover, Clifford A. (2009), The Math Book: From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publishing Company, Inc., p. 214, ISBN 9781402757969 ^ Sriraman, Bharath (2009), "Mathematics and literature (the sequel): imagination as a pathway to advanced mathematical ideas and philosophy", in Sriraman, Bharath; Freiman, Viktor; Lirette-Pitre, Nicole (eds.), Interdisciplinarity, Creativity, and Learning: Mathematics With Literature, Paradoxes, History, Technology, and Modeling, The Montana Mathematics Enthusiast: Monograph Series in Mathematics Education, vol. 7, Information Age Publishing, Inc., pp. 41–54, ISBN 9781607521013 ^ Parker, Matt (2015), Things to Make and Do in the Fourth Dimension: A Mathematician's Journey Through Narcissistic Numbers, Optimal Dating Algorithms, at Least Two Kinds of Infinity, and More, New York: Farrar, Straus and Giroux, p. 98, ISBN 978-0-374-53563-6, MR 3753642 ^ Schrek, D. J. E. (1950), "Prince Rupert's problem and its extension by Pieter Nieuwland", Scripta Mathematica, 16: 73–80 and 261–267; as cited by Rickey (2005) and Jerrard & Wetzel (2004) ^ Hart, George W. (January 30, 2012), Math Monday: Passing a Cube Through Another Cube, Museum of Mathematics; originally published in Make Online ^ 3geek14, Prince Rupert's Cube, Shapeways, retrieved 2017-02-06{{citation}}: CS1 maint: numeric names: authors list (link) ^ Jump up to: a b Jerrard, Richard P.; Wetzel, John E.; Yuan, Liping (April 2017), "Platonic passages", Mathematics Magazine, 90 (2), Washington, DC: Mathematical Association of America: 87–98, doi:10.4169/math.mag.90.2.87, S2CID 218542147 ^ Scriba, Christoph J. (1968), "Das Problem des Prinzen Ruprecht von der Pfalz", Praxis der Mathematik (in German), 10 (9): 241–246, MR 0497615 ^ Chai, Ying; Yuan, Liping; Zamfirescu, Tudor (June–July 2018), "Rupert Property of Archimedean Solids", The American Mathematical Monthly, 125 (6): 497–504, doi:10.1080/00029890.2018.1449505, S2CID 125508192 ^ Hoffmann, Balazs (2019), "Rupert properties of polyhedra and the generalized Nieuwland constant", Journal for Geometry and Graphics, 23 (1): 29–35 ^ Lavau, Gérard (December 2019), "The Truncated Tetrahedron is Rupert", The American Mathematical Monthly, 126 (10): 929–932, doi:10.1080/00029890.2019.1656958, S2CID 213502432 ^ Jump up to: a b Steininger, Jakob; Yurkevich, Sergey (2022), "Extended Abstract for: Solving Rupert's Problem Algorithmically" (PDF), ACM Commun. Comput. Algebra, 56 (2): 32–35, doi:10.1145/3572867.3572870, S2CID 253802715 ^ Jump up to: a b Steininger, Jakob; Yurkevich, Sergey (2023), "An algorithmic approach to Rupert's problem", Mathematics of Computation, 92 (342): 1905–1929, arXiv:2112.13754, doi:10.1090/mcom/3831, MR 4570346 ^ Bezdek, András; Guan, Zhenyue; Hujter, Mihály; Joós, Antal (2021), "Cubes and boxes have Rupert's passages in every nontrivial direction", The American Mathematical Monthly, 128 (6): 534–542, arXiv:2111.03817, doi:10.1080/00029890.2021.1901461, MR 4265479, S2CID 235234134 ^ Thompson, Silvanus P.; Gardner, Martin (1998), Calculus Made Easy (3rd ed.), Macmillan, p. 315, ISBN 9780312185480 ^ Fredriksson, Albin (2024), "Optimizing for the Rupert property", The American Mathematical Monthly, 131 (3): 255–261, arXiv:2210.00601, doi:10.1080/00029890.2023.2285200 ^ Huber, Greg; Shultz, Kay Pechenick; Wetzel, John E. (June–July 2018), "The n-cube is Rupert", The American Mathematical Monthly, 125 (6): 505–512, doi:10.1080/00029890.2018.1448197, S2CID 51841349 ^ Guy, Richard K.; Nowakowski, Richard J. (1997), "Unsolved Problems: Monthly Unsolved Problems, 1969-1997", The American Mathematical Monthly, 104 (10): 967–973, doi:10.2307/2974481, JSTOR 2974481, MR 1543116 ^ Weisstein, Eric W., "Cube Square Inscribing", MathWorld External links [edit] Weisstein, Eric W., "Prince Rupert's Cube", MathWorld Retrieved from " Category: Cubes Hidden categories: CS1 Dutch-language sources (nl) CS1 maint: numeric names: authors list CS1 German-language sources (de) Good articles Articles with short description Short description is different from Wikidata Commons category link from Wikidata
2319	https://www.sciencedirect.com/topics/neuroscience/ventral-posterolateral-nucleus	Skip to main content Sign in Generated by AI, this Topic Page draws on reliable ScienceDirect content and is continuously improved. This Topic Page was generated using a generative AI system developed by Elsevier. It summarizes foundational concepts using structured information from ScienceDirect’s trusted sources, including books and journal articles. The content was produced by an AI model and has not been manually authored or reviewed. While the system is designed to reflect scientific consensus and is continuously improved by our data science team, it may still contain errors or incomplete information. To support quality improvement, some Topic Pages are reviewed by subject matter experts (SMEs) as part of ongoing evaluation. This expert input informs system refinement but does not extend to individual quality assurance for each page. We are also running A/B tests to evaluate whether these pages provide greater value than our previous versions, across engagement, discoverability, and usefulness. We encourage you to explore the cited sources for in-depth reading, and if something feels unclear or inaccurate, please let us know. Your feedback helps us improve this AI-powered experience for all users. Outline 1. Introduction to the Ventral Posterolateral Nucleus 2. Anatomical Structure and Connectivity 3. Functional Role in Somatosensory Processing 4. Neurophysiological Mechanisms and Signal Integration 5. Clinical Significance and Neurological Disorders 6. Conclusion Topic summaryAI 1. Introduction to the Ventral Posterolateral Nucleus The ventral posterolateral nucleus (VPL) is a distinct thalamic nucleus characterized by the presence of large stellate or polygonal cells, alongside medium and small cell types. It is histologically separated from the ventral posteromedial nucleus (VPM), which processes sensory information from the head and face. The VPL serves as the primary relay station for somatosensory information from the trunk and extremities, receiving afferent inputs from the spinal cord and dorsal column nuclei, and transmitting these signals to the primary somatosensory cortex located in the postcentral gyrus. This nucleus is somatotopically organized, with the contralateral lower and upper limbs represented in specific regions within the VPL. Its main function is to relay sensory signals—including pain, temperature, touch, and proprioception—from peripheral receptors to the cortex, excluding inputs from the head and face, which are processed by the VPM. 2. Anatomical Structure and Connectivity The VPL is characterized by fairly large stellate or polygonal cells, which serve as its main distinctive feature, alongside medium and small cell types. It is subdivided into anterior, posterior, medial, and lateral regions, with these divisions based on cell size, density, distribution, molecular properties, or responses to cutaneous stimuli. Immunohistochemical analysis reveals a very intense acetylcholinesterase (AChE) reaction in the VPL, distinguishing it from the ventral lateral nucleus (VLp), which exhibits low AChE activity. The distinction between VLp and the anterior part of the VPL (VPLa) can be made by the transition from large neurons in VLp to a mixture of large and small-sized neurons in the VPL. Anatomically, in monkeys, the VPL and the VPM are separated by a narrow cell-poor septum known as the lamella arcuata. Afferent inputs to the VPL include fibers from the medial lemniscus, which convey discriminative touch and proprioceptive information from the dorsal column nuclei (cuneate and gracile nuclei). The spinothalamic tract also terminates in the VPL. Fibers from proprioceptive receptors of the trunk, extremities, and head project to an area within the rostral VPL, though this area has not been fully delineated in humans. The VPL receives sparse serotonergic and noradrenergic innervation, with noradrenergic terminals present in slightly higher levels; these terminals form axo-axonic and axo-dendritic contacts, some on dendritic spines. Efferent projections from the VPL are directed primarily to the postcentral gyrus, encompassing the primary somatosensory cortex (Brodmann areas 3, 1, and 2). 3. Functional Role in Somatosensory Processing The VPL serves as a primary relay for somatosensory information, receiving afferents from the spinal cord and dorsal column nuclei and projecting to the primary somatosensory cortex via thalamocortical fibers. It encodes and transmits multiple sensory modalities, including discriminative touch, pressure, vibration, proprioception, pain, and temperature. Both protopathic fibers (noxious stimuli, temperature, touch) and epicritic fibers (fine touch, proprioception) from the trunk and extremities terminate in this nucleus. Somatotopic mapping within the VPL preserves spatial information, with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively. Rostral zones of the VPL process proprioceptive inputs. Electrophysiological studies have identified neurons in the VPL with firm receptive fields capable of encoding noxious stimuli at varying intensities, as well as low-threshold mechanoreceptive and wide dynamic range neurons. Microstimulation of VPL neurons in humans elicits discrete sensations of pain or cooling. Somatosensory projections from the VPL are organized to maintain topographic fidelity from the periphery to the cortex. 4. Neurophysiological Mechanisms and Signal Integration Thalamocortical relay neurons in the VPL are activated predominantly by driver inputs, which utilize ionotropic glutamate receptors such as α-amino-3-hydroxy-5-methyl-4-isoxazolepropionic acid (AMPA) and N-methyl-D-aspartate (NMDA). These driver inputs are characterized by thick axons and large terminals on proximal dendrites. Modulatory inputs, including corticothalamic feedback, activate both ionotropic and metabotropic glutamate receptors, possess thin axons, and have small terminals on proximal dendrites. Within the VPL, there is a somatotopic organization, and the nucleus receives convergent sensory inputs from the spinothalamic and medial lemniscal pathways, which terminate in distinct regions and contribute to the integration of pain, temperature, touch, and proprioceptive signals. Neurons in the VPL encode information evoked by noxious stimuli at different intensities, and peripheral nociceptive stimuli can excite VPL neurons, which play an important role in relaying pain-related sensory-discriminative information to the cortex. Noxious stimulation inhibits neurons in the thalamic reticular nucleus, which projects to the VPL and ventral basal nucleus group; the thalamic reticular nucleus is proposed to play a role in gating transthalamic information flow. The VPL sends its output to the posterior limb of the internal capsule and corona radiata en route to the primary somatosensory cortex. 5. Clinical Significance and Neurological Disorders Thalamic stroke involving the VPL can result in sensory deficits such as numbness, paresthesia, and central post-stroke pain (CPSP), also known as thalamic pain syndrome or Dejerine–Roussy syndrome. Patients may experience spontaneous pain, cold allodynia, and sensory loss in body regions corresponding to the affected brain territory, with symptoms including increased thermal and mechanical sensitivity, and attacks of dysesthesia and allodynia. Clinical presentations include abnormal sensitivity to painful stimuli (hyperpathia), dysesthesia, and persistent deep sensory deficits. Electrophysiological recordings in patients with stroke- or spinal cord injury-induced chronic pain have demonstrated plastic changes in thalamic neuron properties, including increased spontaneous firing and decreased excitation thresholds in the VPL. Animal models of thalamic syndrome induced by collagenase injection into the VPL have shown hyperesthesia and increased thermal sensitivity localized to the contralateral limb. The VPL is a target for neurosurgical interventions such as deep brain stimulation (DBS) for intractable neuropathic pain, including post-stroke pain and phantom limb pain. High-frequency stimulation (HFS, 100–150 Hz) within the VPL has been shown to decrease neuronal hyperexcitability and thermal hyperalgesia in animal models, while low-frequency stimulation (LFS, 20–40 Hz) had no effect on neuronal firing. Clinical studies have reported that DBS of the VPL thalamus and periaqueductal gray (PAG) achieved an overall reduction of 48.8% in pain in 80% of patients with poststroke neuropathic pain. Longitudinal studies have shown that 59% of patients experienced significant acute pain relief after VPL DBS, with 31% maintaining pain relief after approximately 80 months of follow-up. The mechanisms of analgesia are not fully understood but may involve modulation of ascending pain pathways and activation of suprasegmental descending endogenous pain inhibition systems. Limitations of DBS for pain control include inconsistent target localization, heterogeneity of pain diagnoses, and lack of large randomized trials. 6. Conclusion The ventral posterolateral nucleus of the thalamus is characterized by large stellate or polygonal cells, with additional medium and small cell types, and is divided into anterior, posterior, medial, and lateral subdivisions based on cellular size, density, distribution, molecular properties, and responses to cutaneous stimuli. It serves as a primary relay nucleus for somatosensory information from the trunk and extremities, receiving inputs from the spinal cord via the spinothalamic tract and dorsal column-medial lemniscus system, and transmitting these signals to the primary somatosensory cortex. Somatotopic organization within the VPL ensures precise representation of contralateral lower and upper limbs. The VPL integrates diverse sensory modalities, including pain, temperature, touch, and proprioception, with protopathic and epicritic fibers from both the anterolateral and dorsal column-medial lemniscus systems terminating in this nucleus. Neurophysiological mechanisms in the VPL enable encoding of noxious stimuli at varying intensities, contributing to sensory-discriminative processing. Spinothalamic tract projections to the VPL are involved in sensory-discriminative aspects of pain, and microstimulation of VPL neurons in humans elicits discrete sensations of pain or cooling. Clinically, lesions in the VPL are strongly associated with central post-stroke pain, with the posterolateral and inferior parts of the VPL most frequently implicated. The VPL is a target for deep brain stimulation in pain control, reflecting its importance in sensory disorders and therapeutic interventions. Reference 1 Book Chapter Thalamus Jürgen K. Mai, F. Forutan The Human Nervous System , 2012 pp 618-677 View PDFView chapter Related quote(s)1 / 5 "... Ventral Posterolateral Nucleus (VPL) VPL stands out against its surrounding by fairly large stellate or polygonal cells. This property represents the main distinctive feature of the nucleus ( Figure 19.17 D). In addition to the large cells there are also medium to small cell types found. VPL has been divided into several subdivisions (anterior, posterior, medial, lateral) on the basis of size, density, and distribution of cells, molecular properties, and by their responses to cutaneous stimuli . The distinction between VL (VLp) and the anterior part of the ventral posterolateral nucleus (VPLa) can be made by the transition from the large neurons in VL to the mixed large and small-sized neurons in the VPL. Immunohistochemically, there is a difference in the AChE-reaction: low in VLp, very intense in VPL. Ventral Posteromedial Nucleus (VPM) VPM corresponds to the area of the Ncl. arcuatus, Ncl. semilunaris, and Ncl. posterointernus of earlier authors . VPM receives the ascending secondary trigeminal afferents via the trigeminal lemniscus from the head, face, and intraoral structures (afferents from proprioceptors and nociceptors from the stomatognathic system). In monkeys VPL and VPM are separated by a narrow cell-poor septum (lamella arcuata) that is well seen in sections stained for PV and CO . In humans, this lamella can be distinguished during fetal development as distinct (CD15-negative) lamina that surrounds the entire VPM. Against the centromedian nucleus VPM is delimited by a fibrous capsule, the lamella centralis . Medially and ventrally VPM borders the ventral posteromedial nucleus, parvicellular part (VPPC) that is considered by some researchers as part of VPM. Both nuclei are, however, well distinguishable parts ( Figure 19.24 ). ..." Related quote(s)2 / 5 "... Ventral Posterolateral Nucleus (VPL) VPL stands out against its surrounding by fairly large stellate or polygonal cells. This property represents the main distinctive feature of the nucleus ( Figure 19.17 D). In addition to the large cells there are also medium to small cell types found. VPL has been divided into several subdivisions (anterior, posterior, medial, lateral) on the basis of size, density, and distribution of cells, molecular properties, and by their responses to cutaneous stimuli . The distinction between VL (VLp) and the anterior part of the ventral posterolateral nucleus (VPLa) can be made by the transition from the large neurons in VL to the mixed large and small-sized neurons in the VPL. Immunohistochemically, there is a difference in the AChE-reaction: low in VLp, very intense in VPL. ..." Related quote(s)3 / 5 "... Sensory Thalamus The sensory thalamus can be subdivided into a somatosensory unit with several nuclei (ventral posterior nucleus,VP; ventroposterior inferior nucleus,VPI; ventroposterior superior nucleus, VPS; ventromedial nucleus, VM; ventral medial nucleus, posterior part, VPMpc; anterior pulvinar, APul) (see Chapter 30 ) and the viscerosensory nunit (ventral posteromedial nucleus, parvicell. part, VPPC). Each nucleus has its individual set of afferents and distinctive projections to specific cortical areas. Ventroposterior Nucleus or Complex (Ncl. ventralis posterior, VP) The somatosensory unit represents the primary relay nucleus of the thalamus where all fiber tracts for the transmission of surface and deep sensitivity of the trunk, extremities, and head of (mainly) the opposite side of the body terminate . Because of the convergence of the many afferents, the different nuclear components of the unit are grouped together as the ventrobasal complex ( Table 19.1 ; see Chapter 30 ). VP represents a functional unit that is histologically separated in two main subnuclei: the ventral posterolateral nucleus, VPL, and the ventral posteromedial nucleus, VPM, . The lateral portion provides the receptive area for the spinal and the lemniscal fibers; the medial portion is the representation of the trigeminal fibers. The fibers of the protective (protopathic) sensitivity (from neurons that respond to noxious stimuli, temperature, and touch) from trunk and extremities are carried by the anterolateral system (spino-thalamic tract); the discriminative (epicritic) sensitivity and proprioception are mediated by the dorsal column-medial lemniscus system. Protopathic and epicritic fibers of both systems terminate in the ventral posterolateral nucleus (VPL). ..." Related quote(s)4 / 5 "... The corresponding fibers from the head (from the trigeminal complex and from the remaining branchial nerves), which mediate impulses from the cutaneous receptors, are carried to the ventral posteromedial nucleus (VPM). The fibers from the proprioceptive receptors from trunk, extremities and the head (temporomandibular joint) project to an area within the rostral VPL which has not been delineated with complete certainty in the human. In the monkey the territory for the proprioceptive or kinesthetic fibers has been located within the VP “shell” or within a separate nucleus, the ventroposterior superior nucleus, VPS (VPO) ( Kaas, 2008 ; Chapter 30 ; see Chapter 31 ). These fibers are joined by those from the vestibular nuclei and taste pathway. Among the afferents to the lateral thalamic nuclei, the so-called secondary trigeminal pathway has created some misunderstandings. In the human the connection between the spinal trigeminal nucleus (but also from the main sensory nucleus) and VPM was described as a complex, dorsally ascending bundle that was later named as Wallenberg’s bundle, compromised in the Wallenberg syndrome . Re-examination showed that the secondary trigeminal pathway and the supposed Wallenberg bundle represent structurally related but functionally different trajectories. The Wallenberg bundle is complex with multiple sites of origin, related to the reticular activating system and ends in different parts of the thalamus, preferentially within the intralaminar nuclei (but also in VPM) . Ventral Posterolateral Nucleus (VPL) VPL stands out against its surrounding by fairly large stellate or polygonal cells. This property represents the main distinctive feature of the nucleus ( Figure 19.17 D). In addition to the large cells there are also medium to small cell types found. VPL has been divided into several subdivisions (anterior, posterior, medial, lateral) on the basis of size, density, and distribution of cells, molecular properties, and by their responses to cutaneous stimuli . 1 ..." Related quote(s)5 / 5 "... The fibers of the protective (protopathic) sensitivity (from neurons that respond to noxious stimuli, temperature, and touch) from trunk and extremities are carried by the anterolateral system (spino-thalamic tract); the discriminative (epicritic) sensitivity and proprioception are mediated by the dorsal column-medial lemniscus system. Protopathic and epicritic fibers of both systems terminate in the ventral posterolateral nucleus (VPL). The corresponding fibers from the head (from the trigeminal complex and from the remaining branchial nerves), which mediate impulses from the cutaneous receptors, are carried to the ventral posteromedial nucleus (VPM). The fibers from the proprioceptive receptors from trunk, extremities and the head (temporomandibular joint) project to an area within the rostral VPL which has not been delineated with complete certainty in the human. In the monkey the territory for the proprioceptive or kinesthetic fibers has been located within the VP “shell” or within a separate nucleus, the ventroposterior superior nucleus, VPS (VPO) ( Kaas, 2008 ; Chapter 30 ; see Chapter 31 ). These fibers are joined by those from the vestibular nuclei and taste pathway. Among the afferents to the lateral thalamic nuclei, the so-called secondary trigeminal pathway has created some misunderstandings. In the human the connection between the spinal trigeminal nucleus (but also from the main sensory nucleus) and VPM was described as a complex, dorsally ascending bundle that was later named as Wallenberg’s bundle, compromised in the Wallenberg syndrome . Re-examination showed that the secondary trigeminal pathway and the supposed Wallenberg bundle represent structurally related but functionally different trajectories. The Wallenberg bundle is complex with multiple sites of origin, related to the reticular activating system and ends in different parts of the thalamus, preferentially within the intralaminar nuclei (but also in VPM) . 1 ..." Reference 2 Review article The neural basis of somatosensory temporal discrimination threshold as a paradigm for time processing in the sub-second range: An updated review Ordas C.M., Alonso-Frech F. Neuroscience & Biobehavioral Reviews , 2024 pp 105486 View PDFView article Related quote(s)1 / 1 "... Thalamus The somatosensory input from all modalities and all regions of the body converge within the ventroposterior (VP) region of the thalamus. Since the classical descriptions by Hassler , nomenclature of the thalamus has been diverse and even controversial. However, efforts have been made to reach a common terminology. Following a recent revision by Mai and Majtanik, the VP nucleus has been histologically separated into the lateral ventroposterior nucleus (ventral posterolateral nucleus, VPL,), the medial ventroposterior nucleus (ventral posteromedial nucleus, VPM) and the ventrocaudal nucleus (V.c.pc) . The latter has been further divided into two smaller divisions named the external and the internal divisions, equivalent to the ventral posterior inferior nucleus (VPI) and the medial ventroposterior nucleus (ventral posteromedial nucleus, parvocellular part, VPMpc) . VPL has been divided into several subdivisions (anterior, posterior, medial, lateral). The spinal, lemniscal and trigeminal afferents to the thalamus are difficult to separate in humans. Nevertheless, VPM is thought to receive the ascending secondary trigeminal afferents via the trigeminal lemniscus from the head and face. On the other hand, proprioceptive fibers project to an area anterior and dorsal to VPM and VPL, which has been termed “superior ventroposterior” (VPS), oral part (VPO) or “shell region” . Projections to somatosensory cortex are mainly directed to areas 3b and 1 although they are highly convergent and divergent . ..." Reference 3 Book Chapter Diencephalon Reha Erzurumlu, Gulgun Sengul, Emel Ulupinar Human Neuroanatomy , 2024 pp 57-78 View PDFView chapter Related quote(s)1 / 2 "... Thalamic nuclei The thalamic nuclei are named according to their locations and abbreviations as follows: anterior nuclei (AN), ventral anterior nucleus (VA), ventral lateral nucleus (VL), ventral posteromedial nucleus (VPM), ventral posterolateral nucleus (VPL), mediodorsal (MD), median (M) as the median nucleus group, lateral dorsal nucleus (LD), lateral posterior nucleus (LP), intralaminar nuclei (IN), pulvinar nuclei (P), and the nuclei that make up the metathalamus, mediale geniculate body (MGB), and lateral geniculate body (LGB). These nuclei are further classified as specific, primary sensory (VPL, VPM, LGB, MGB), motor (VA, VL), association (P, AN, MD, LD, LP), and nonspecific (or nonspecific) nuclei (M, IN) based on their functions and specific connections with cortical areas ( Table 5.2 , Fig. 5.4 ). ..." Related quote(s)2 / 2 "... Reciprocally, axons from lamina VI in the same region also project to LGB ( Fig. 5.4 ). Clinical concepts related to the thalamus Due to the different perforating branches that supply the posterior cerebral artery, various clinical findings can occur in cases of vascular disorders such as bleeding or blockage involving these segments. Lateral thalamic infarction: It occurs when the thalamogeniculate artery, which supplies the ventral lateral thalamus including VPL and VPM, CM, and rostro-lateral part of the pulvinar, is blocked. Lateral thalamic infarction can start with paresthesia and numbness on one side of the body and can also affect all sensory modalities. If the infarction also affects the poster limb of the internal capsule adjacent to VPL, motor defects can occur in addition to sensory deficits. In the more common form of lateral thalamic infarction known as Dejerine–Roussy syndrome (“thalamic pain” syndrome), initially mild contralateral hemiparesis and persistent deep sensory deficits are observed. Involuntary movements resembling dystonia, choreoathetosis, and tremor can occur in the contralateral arm and leg. The dystonic posture deficit that develops in the hand is called “thalamic hand.” In the late stage, severe, persistent, and unresponsive pain can be observed on the side where numbness and paresthesia occurred before. The initial symptoms observed in thalamic syndrome are usually abnormal sensitivity to painful stimuli (hyperpathia) or abnormal perception of touch sensation as discomfort (dysesthesia) on the face, arms, and/or legs. Allodynia refers to the perception of pain in response to a light touch that would not normally be painful, or even in the absence of a stimulus. Pain associated with thalamic syndrome can increase due to exposure to heat or cold and emotional distress. 1 ..." Reference 4 Book Chapter Thalamus Groenewegen H.J., Witter M.P. The Rat Nervous System , 2004 pp 407-453 View PDFView chapter Related quote(s)1 / 1 "... Afferent and efferent connections The ventral posterolateral and ventral posteromedial nuclei receive their main ascending inputs from somatosensory afferents originating in the spinal cord, dorsal column nuclei, and trigeminal complex (see also Tracey, Chapters 7 and 25 , and Waite, Chapter 26 , this volume). These somatosensory projections are somatotopically organized such that afferents from the trunk and limbs terminate in the VPL and those from the head terminate in the VPM. Spinal and trigeminal fibers not only reach the VPL and VPM but also distribute over much wider areas of the thalamus, including the posterior nucleus (see page 417 ) and the intralaminar nuclei (see section “Midline and Intralaminar Thalamic Nuclei”). Spinothalamic fibers, transferring among others nociceptive signals, originate from different laminae of the dorsal horn as well as the central gray of the spinal cord and terminate as large boutons in the ventral posterolateral nucleus . Nociceptive information may reach the VPL also indirectly via the caudal medullary reticular formation (medullary dorsal reticular nucleus) in addition to the direct spinothalamic pathway . Dorsal column nuclear afferents likewise terminate with large boutons in the VPL . Spinal and lemniscal fibers, in part, converge on individual thalamocortical neurons, the lemniscal fibers on the soma, and more proximal parts of dendrites of the neurons than the spinothalamic fibers . Lemniscal, but also spinothalamic, fibers probably use glutamate as neurotransmitter . Spinothalamic fibers have also been shown to contain substance P . The spinal and principal trigeminal nuclei both project to the ventral posteromedial nucleus as well as, less densely, to the posterior complex (cf. also Waite, Chapter 26 , this volume). Projections to the VPM are more focussed, those to the medial part of posterior nucleus more diffuse . ..." Reference 5 Book Chapter The Diencephalon Srikant S. Chakravarthi, Alejandro Monroy-Sosa, Kost Elisevich From Anatomy to Function of the Central Nervous System , 2025 pp 255-284 View PDFView chapter Related quote(s)1 / 4 "... There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. Following its inputs from the spinothalamic axons, the VPL sends its output to the posterior limb of the internal capsule and corona radiata en route to the primary somatosensory area of the cortex. In addition, apart from the contribution of the VPL, the posterior and intralaminar groups of the medial dorsal nucleus also receive inputs from the spinal lemniscus . The Motor Thalamus The motor thalamus (MThal) represents an important and strategic location within the ventral thalamus and shares a strong interconnection with both motor cortices, as well as the cerebellum (dentate and interposed nuclei) and basal ganglia (substantia nigra pars reticulata and internal segment of the globus pallidus), to regulate both cognitive and proprioceptive control of movement. The current system shares many of the features of the “driver-modulator” concept, as proposed by Sherman and Guillery for the sensory thalamus. As described by Bosch-Bouju (2013) , the MThal acts as a “super-integrator” of motor information from two functional driver-input subdivisionst: the basal ganglia and cerebral cortex and the cerebellum and cerebral cortex. ..." Related quote(s)2 / 4 "... The drivers differ from modulators in that they predominantly activate ionotropic glutamate receptors (AMPA and NMDA), contain thick axons, and have large terminals on proximal dendrites. Modulators have activating ionotropic and metabotropic glutamate receptors, contain thin axons, and have small terminals on proximal dendrites . It is also important to note that for the thalamus, drivers are much less numerous than modulators, as in the example provided, where layer 6 inputs have many more synapses than the retinal (driver) inputs . Sensory Relay The sensory relay system of the thalamus is perhaps the most commonly known and robust of all relay circuits. There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. Following its inputs from the spinothalamic axons, the VPL sends its output to the posterior limb of the internal capsule and corona radiata en route to the primary somatosensory area of the cortex. In addition, apart from the contribution of the VPL, the posterior and intralaminar groups of the medial dorsal nucleus also receive inputs from the spinal lemniscus . ..." Related quote(s)3 / 4 "... Sensory Relay The sensory relay system of the thalamus is perhaps the most commonly known and robust of all relay circuits. There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. Following its inputs from the spinothalamic axons, the VPL sends its output to the posterior limb of the internal capsule and corona radiata en route to the primary somatosensory area of the cortex. In addition, apart from the contribution of the VPL, the posterior and intralaminar groups of the medial dorsal nucleus also receive inputs from the spinal lemniscus . 1 ..." Related quote(s)4 / 4 "... In an effort to control the various stimulatory and inhibitory inputs and outputs contained within every thalamic circuit, there are critical drivers and modulators which control the degree of signal traveling through these circuits. Again, an apt and highly studied example is the communication between retinal receptive field inputs and the visual layer 6 of the cortex that includes the lateral geniculate nucleus. The retinal afferent drivers and cortical input modulators work together to control signals sent to the lateral geniculate nucleus for processing. The drivers differ from modulators in that they predominantly activate ionotropic glutamate receptors (AMPA and NMDA), contain thick axons, and have large terminals on proximal dendrites. Modulators have activating ionotropic and metabotropic glutamate receptors, contain thin axons, and have small terminals on proximal dendrites . It is also important to note that for the thalamus, drivers are much less numerous than modulators, as in the example provided, where layer 6 inputs have many more synapses than the retinal (driver) inputs . Sensory Relay The sensory relay system of the thalamus is perhaps the most commonly known and robust of all relay circuits. There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. 1 ..." Reference 6 Review article The structural and functional importance of the thalamus in migraine processes with and without aura. A literature review Valenzuela-Fuenzalida J.J., Suazo-Santibanez A., Semmler M.G., Cariseo-Avila C., Santana-Machuca E., Orellana-Donoso M. Translational Research in Anatomy , 2021 pp 100130 View PDFView article Related quote(s)1 / 1 "... The lateral thalamic nuclei group are comprised by lateral posterior nucleus (LP, lateral dorsal group), ventral anterior nucleus or group (VA), ventral lateral nucleus or group (VL), ventral medial nucleus or group (VM), ventral posterior nucleus or group (VP), ventral posterolateral nucleus (VPL), ventral posteromedial nucleus (VPM), ventral posterior inferior nucleus (VPI). VPL and VPM nuclei are part of the somatosensory system, VPL relays medial lemniscal and spinothalamic input and relays the inferior portion of the postcentral gyrus. VL receives input mainly from the deep cerebellar dentate nucleus, and a smaller contribution from the basal nuclei to the rostral part of the VL. Projections from the VL, are the primary motor area, area 4, of the precentral gyrus and also has a smaller projection to premotor areas. Thus, involving the VL to motor feedback from the cerebellum and basal nuclei to the cerebral cortex [ 17 ]. The basal nuclei, especially the medial globus pallidus and the parts reticulata of the substantia nigra, send inputs to the VA, which in turn projects to premotor cortex including the supplementary motor area of the frontal lobes and is involved in planning and initiating movements [ 17 ]. As for the LP, it has been demonstrated that plays an important integrative function for the limbic spatial learning systems, since it projects to the hippocampal formation directional information that it is not just a reflection of sensory inputs [ 18 ]. The VPI is the thalamic relay center in the vestibulo-cortical pathway, since its projections are mainly directed to the primary vestibular cortex [ 19 ]. The largest portion of the thalamic posterior nuclei is the pulvinar, and it's also the largest association nuclei found in the dorsal part of the thalamus. It receives inputs from the superior colliculus and from the association cortex. ..." Reference 7 Book Chapter Human perception and neurocognitive development across the lifespan Shu-Chen Li, Evelyn Muschter, Jakub Limanowski, Adamantini Hatzipanayioti Tactile Internet , 2021 pp 199-221 View PDFView chapter Related quote(s)1 / 1 "... Sensory signals from different modalities are processed in different subregions within the thalamus, specifically, the optic nerve fibers project to the Lateral Geniculate Nucleus (LGN), the auditory nerve fibers project to the Medial Geniculate Complex (MGC), whereas the somatosensory nerve fibers project to the Ventral Posterolateral Nucleus (VPL) and the Ventral Posteromedia Nucleus (VPM). Another brain area that is also involved in early sensory processing is the superior colliculus, which is a small structure within the midbrain that sits right below the thalamus. The initial sensory representations formed within the different regions of the thalamus and in the superior colliculus are then relayed to other brain areas, which include regions in the midbrain (e.g., the basal ganglia) and in the cerebral cortex (e.g., visual, auditory and the somatosensory cortices), as well as regions in the parietal and frontal lobes that are important for attentional and cognitive control processes (see for review). These later regions are engaged in downstream processes that ( i ) compare and ( i i ) bind sensory representations acquired from the different modalities, as well as ( i i i ) interpret the current subjective sensory experiences as guided by attention and memories of past experiences. These processes together eventually give rise to the perception of an integral whole of an object, such as a tiny ball that is flying across the court as in the case of tennis playing. Most sensory signals are attained or, at least, modulated by motor sampling processes; thus perception inherently also involves sensorimotor processes. 9.2.2 Principles of multisensory integration In navigating through the manifold of sensory signals in the environment, the brain's sensory and perceptual processes need to distinguish between signals that belong to the same event from those that do not. When sensory information of an object or event is provided by more than one input modality, three general processing principles are involved in multisensory integration. ..." Reference 8 Book Chapter Somatosensory Evoked Potentials Daniel Dumitru, Lawrence R. Robinson, Machiel J. Zwarts Electrodiagnostic Medicine , 2002 pp 357-413 View PDFView chapter Related quote(s)1 / 1 "... 22, 134, 318 This tract terminates by synapsing in a specific region of the thalamus, the ventral posterolateral nucleus or VPL ( Fig. 9-1 ). Third-order nerve fibers originating in the VPL, thalamocortical fibers , project to the postcentral gyrus of the parietal lobe, known as the somatosensory cortex ( Fig. 9-1 ). 123 A characteristic topographic arrangement of sensory regions is discretely arranged along the cortex corresponding to lemniscal fibers from specific portions of the body, forming the so-called homunculus ( Fig. 9-2 ). 134 The somatosensory region representing the upper limb is located on the cortex superior to the face area. The lower limb somatosensory representation, however, is along the medial aspect of the cerebral hemisphere, separated from its contralateral body region in the opposite hemisphere by the interhemispheric fissure. With respect to the lower limb somatosensory information, a second fiber route has been proposed, i.e., the spinomedullothalamic tract . 318 This view suggests that the dorsal aspect of the lateral funiculus may convey a significant portion of the type IA and II afferent information to the somatosensory cortex. 55, 197, 229, 254 The proposed pathway is a continuation of primary afferents that synapse at the level of Clark's column to then ascend in the dorsal spinocerebellar tract and subsequently in the nucleus Z at the level of the medullopontine junction (see Fig. 9-1 ). 214 Nerve fibers from nucleus Z then travel to the medial lemniscus and ascend to the rostral region of the VPL. The traditional pathway from the VPL to the somatosensory cortex then serves as the final pathway for the fibers from the lower limb. 1 ..." Reference 9 Book series Chapter Noradrenergic innervation of somatosensory thalamus and spinal cord Westlund K.N., Zhang D., Carlton S.M., Sorkin L.S., Willis W.D. Progress in Brain Research, 1991 pp 77-88 View PDFView chapter Related quote(s)1 / 2 "... Monoamine terminals in the somatosensory thalamus More recent studies have examined specific anatomical connectivities and have required the development of techniques with better resolution. Light and electron microscopic observations have confirmed the presence of a sparse innervation by serotonergic and noradrenergic terminals of the ventral posterolateral nucleus of the macaque monkey ( Fig. 1 ). The ventral posterolateral nucleus of the thalamus receives sensory information from the spinal cord. A sparse distribution was noted for both kinds of aminergic terminals, with noradrenergic terminals present in slightly higher levels. En passant endings and intervaricose segments were evident at the ultrastructural level. Profiles immunocytochemically labeled for serotonin ( Fig. 1A,C ) or for the noradrenergic marker, DβH ( Fig. 1B,D ), were either axo-axonic or axo-dendritic. Some of the noradrenergic terminals appeared to contact dendritic spines. 1 ..." Related quote(s)2 / 2 "... Conclusions These studies have sought to reveal a more definitive anatomical basis for the mechanisms proposed for the modulation of nociception by norepinephrine by carefully mapping the locations of catecholamine terminals in relation to STT cells in the spinal cord and their targets in the thalamus. Thus, although descending pathways which could form the anatomical basis for monoaminergic interactions with the spinothalamic tract at the spinal cord are well known, continued analysis of chemically defined anatomical connectivities must be done before we can begin to understand the role of monoamines in complex functions such as somatosensory processing. These studies have been extended into the ventral posterolateral nucleus of the thalamus to determine the sources of noradrenergic innervation to this sensory relay nucleus. It is likely that these connectivities contribute to what is known as “stimulation-produced analgesia” and for the clinical effectiveness of some pharmaceuticals used in pain control. Extension of these studies to other regions is necessary to understand the role of norepinephrine in motor and autonomic control. 1 ..." Reference 10 Book Chapter Anatomy and Physiology of Pain Bajwa Z.H., Wilsey B.L., Fishman S.M. Office Practice of Neurology , 2003 pp 1383-1389 View PDFView chapter Related quote(s)1 / 1 "... SUPRASPINAL STRUCTURES The supraspinal system comprises the reticular formation, thalamus, hypothalamus, limbic system, and cerebral cortex. Within this system are extensive communicating projections for ascending algesic and descending analgesic pathways. The reticular formation extends through the entire length of the brainstem. The reticular formation receives input mainly from the spinoreticular tract but also from other structures in the supraspinal system. At the reticular formation level, receptive fields usually are extremely large, arising from both ipsilateral and contralateral parts of the body. The reticular formation controls the state of arousal and is important in autonomic reflex responses (fast defense reactions), inducing powerful analgesia in crisis situations. The thalamus consists of multiple nuclei and acts as the major relay station for incoming nociceptive stimuli. The thalamus is subdivided phylogenetically into paleothalamus and neothalamus or by nuclei location. The paleothalamus or medial thalamus has input mainly from the spinothalamic tract and the reticular formation. Its receptive fields are large, and it is involved primarily with the motivational-affective aspect of pain. The paleothalamus has extensive connections with the cerebral cortex. The neothalamus or lateral thalamus rests at the ventrobasal portion of the thalamus. The neothalamus, unlike the paleothalamus, is organized somatotopically and subdivided into the ventral posterolateral nucleus and the ventral posteromedial nucleus. The ventral posterolateral nucleus receives input mainly from the spinothalamic tract but also from the dorsal column system and the somatosensory cortex. The ventral posteromedial nucleus receives input mainly from the trigeminothalamic tract, which carries sensory input from the head and face and projects to the somatosensory cortex, involved with craniofacial pain. Although most of the neothalamic neurons respond to mechanoreceptive input, some are nociceptive-specific and are wide-dynamic range neurons. 1 ..." Reference 11 Book Chapter Development of the Nervous System Grow W.A. Fundamental Neuroscience for Basic and Clinical Applications , 2018 pp 72-90.e1 View PDFView chapter Related quote(s)1 / 1 "... In this way, the map of visual space on the retina is maintained in the lateral geniculate nucleus and ultimately in the visual cortex. Similar maps are formed in the medial geniculate nucleus (tonotopic mapping) and ventral posterolateral nucleus (somatotopic mapping). 1 ..." Reference 12 Book Chapter Proprioception: a sense to facilitate action Kyle P. Blum, Christopher Versteeg, Joseph Sombeck, Raeed H. Chowdhury, Lee E. Miller Somatosensory Feedback for Neuroprosthetics , 2021 pp 41-76 View PDFView chapter Related quote(s)1 / 1 "... Thalamic proprioceptive encoding As the medial lemniscus leaves the dorsal column nuclei, it decussates before synapsing in the ventroposterior complex of the thalamus. The nomenclature used to delineate proprioceptive regions of the thalamus is inconsistent across species and even research groups [see Bota et al. (2019) for a review]. In rats, a complex proprioceptive somatotopy exists in the most rostral regions of the ventral posterolateral nucleus (rVPL) . In monkeys, classic studies describe a region rostral and dorsal within the thalamus, termed the ventroposterior superior nucleus or sometimes ventroposterior oralis that projects to cortical area 3a . Other thalamic “motor” nuclei (ventrolateral; VL and ventral intermediate nucleus; Vim) also receive inputs from the cerebellum, carrying signals partially derived from proprioceptive mossy fiber inputs [for a review, see Shadmehr (2020 )]. To our knowledge, no recordings have been made of the primate proprioceptive thalamus during reaching. Proprioceptive recordings from monkey VPL during head rotations have been reported, containing a mixture of proprioceptive and vestibular modalities . 1 ..." Reference 13 Review article Thalamus: The ‘promoter’ of endogenous modulation of pain and potential therapeutic target in pathological pain You H.-J., Lei J., Pertovaara A. Neuroscience & Biobehavioral Reviews , 2022 pp 104745 View PDFView article Related quote(s)1 / 1 "... 2.2 Thalamus: function as ‘discriminator’ in control of nociception In the ventral posterolateral nucleus (VPL) of the thalamus, there are neurons that have firm receptive fields and that are capable of encoding information evoked by noxious stimuli at different intensities applied not only to a somatic area, but also to a visceral area (such as intraperitoneal injection of bradykinin or the dilatation of the uterus) . Peripheral nociceptive stimuli can also cause excitement of other thalamic nuclei, such as the intralaminar nuclei, submedius (SM) nucleus, the PO nucleus, and the VM nucleus . In contrast to the excitation of nociceptive thalamic neurons in the VPL that is considered to play an important role in relaying pain-related sensory-discriminative information to the cortex, noxious stimulation inhibits neurons in the thalamic reticular nucleus . This finding is in line with evidence that the thalamic reticular nucleus has projections to the thalamic VPL and the ventral basal nucleus group, and with the proposal that the thalamic reticular nucleus plays a role in gating of transthalamic information flow . Electrophysiological recordings in patients with stroke- or spinal cord injury-induced chronic pain indicate that the functional properties of thalamic neurons exhibit plastic changes in pathophysiological conditions . In patients with limb amputation, the thalamic representation of the missing limb remains functional and its stimulation can produce phantom limb pain in the affected limb . Animal studies in rats exposed to peripheral nerve injury or hind limb inflammatory nociception showed that the ventral basal nucleus of the thalamus, including the VPL and VPM nuclei, exhibits significantly decreased neuronal excitation thresholds . In line with this, increased spontaneous firing and increased responses to noxious stimulation were reported in the mediodorsal (MD) thalamic neurons in animals with chronic pain induced by spinal cord injury . 1 ..." Reference 14 Book Chapter Spinal Cord Gulgun Sengul, Charles Watson The Human Nervous System , 2012 pp 233-258 View PDFView chapter Related quote(s)1 / 1 "... The lateral spinothalamic tract continues as the spinal lemniscus. The spinothalamic tracts terminate mainly in the ventral posterolateral nucleus, the ventral posteromedial nucleus, the central lateral nucleus of the intralaminar nuclei, and the posterior nuclear group of the thalamus. There are both nociceptive and thermoreceptive-specific neurons in the ventral posterolateral and posteromedial nuclei . In humans, microstimulation of these neurons has been shown to elicit discrete sensations of pain or cooling . Studies in primates have shown that the dorsal and ventral spinothalamic terminations largely overlap in the lateral and medial thalamus ( Apkarian and Hodge, 1989c ). Spinothalamic tract projections to the central lateral nucleus of the thalamus play a part in motivational-affective responses to pain. The projection to the lateral thalamus (the ventrobasal complex) is involved in sensory-discriminative aspects of pain. Animal studies have shown that spinothalamic tract neurons send collateral branches to the medullary reticular formation, parabrachial area, periaqueductal gray, and nucleus accumbens. These projections distribute information to multiple brainstem sites, which might in turn activate autonomic or affective responses, or descending pain-modulatory mechanisms . The spinothalamic sytem reaches multiple cortical areas in the contralateral hemisphere. Using anterograde transneuronal transport of herpes simplex virus in the monkey, Dum et al. (2009) found that major targets for the spinothalamic input are the granular insular cortex, the secondary somatosensory cortex, and several cortical areas in the cingulate sulcus. Comparable cortical regions in humans have also been shown to display activation when subjects are exposed to painful stimuli. Using diffusion tensor tractography, Hong et al. (2010) observed that the spinothalamic tract and related thalamocortical fibers in the human brain originate from the ventral posterolateral nucleus of the thalamus. 1 ..." Reference 15 Review article Neuropathic pain modeling: Focus on synaptic and ion channel mechanisms Bouali-Benazzouz R., Landry M., Benazzouz A., Fossat P. Progress in Neurobiology , 2021 pp 102030 View PDFView article Related quote(s)1 / 1 "... Thalamic syndrome Stroke is the leading cause of disability in the industrialized world and 8% of stroke victims suffer from post-stroke central pain (PSCP) . Thalamic syndrome is one of PSCP occurring after a cerebrovascular accident in the thalamus. In patients, this syndrome is characterized by spontaneous pain, cold allodynia and sensory loss in the body parts that correspond to the brain territory injured . In the rat, stereotaxic injection of collagenase or blood in the ventral postero-lateral nucleus of the thalamus (VPL) resulted in small hemorrhagic stroke lesions, mimicking the human thalamic syndrome . The authors observed an increased thermal (hot) and mechanical sensitivity in the contralateral limb after thalamic hemorrhage over the 3-week testing period without hypersensitivity to cold. These results in rats, do not reproduce the cold allodynia phenotype seen in patients in whom localized lesions in the ventro-caudal thalamus are sufficient to cause deficiencies in cold sensation without heat pain perception . However, recently two different mice models have been developed by unilateral stereotaxic injection of kainate or collagenase in the VPL . In these models, tactile and cold hypersensitivity are observed with a delay of three to five days after injury. These mice models develop symptoms similar to the sensory abnormalities observed in patients suffering from CPSP, such spontaneous pain, attacks of dysesthesia and allodynia. 1 ..." Reference 16 Book Chapter Microangiopathies (Lacunes) Marti-Vilalta J.L., Arboix A.G., Mohr J.P. Stroke , 2011 pp 485-515 View PDFView chapter Related quote(s)1 / 1 "... The small vessels, individually occluded, may cause an infarct that cuts across the functional anatomic fields of somatosensory projections, causing clusters of symptoms and signs from lesions that are at variance with the normal organization. Nature of the Sensory Complaints The patients complain of striking alterations in spontaneous sensations. 136,145 The parts feel stretched, hot, and sunburned, as if being stuck by pins, larger, smaller, or heavier. Contacts with the skin from eyeglasses, bedclothes, rings, watches, and sheets feel heavier on the affected side and may transiently aggravate the sensory disturbance. The stimulus seems to persist for a few seconds after its removal. In the patients with severe disturbances, the occurrence of a stimulus is better reported than its exact location. In a series of 21 cases, impairment of all sensory types (touch, pinprick, vibration, and position sense) was usually associated with large lacunes in the lateral thalamus; restricted sensory complaints suggest small lacunes at any level of the sensory pathway. 146 The Dejerine-Roussy syndrome 147 is an uncommon accompaniment of lacunar infarction of the thalamus, although dysesthetic accompaniments are common in pure sensory stroke, as described previously. The full Dejerine-Roussy syndrome was originally described as the effect of occlusion of the thalamogeniculate branch of the posterior cerebral artery, with infarction of the ventral posterolateral and ventral posteromedial nuclei, largely sparing the remaining nuclei of the thalamus. Cases documented only by CT have shown a lesion small enough to qualify for a clinical diagnosis of lacunar infarction. 148 The initial deficit usually consists of hemiparesis and a hemisensory syndrome. The pain, which is an inconstant feature in cases with such infarcts, may begin at the onset of the syndrome or appear only later; delays of up to several months are common. The pains are intermittent or constant, appear spontaneously, or at other times are provoked by contact with the affected parts. 1 ..." Reference 17 Book Chapter Central Pain and Complex Regional Pain Syndromes Richard Wilson, Preeti Raghavan, Andrew K. Treister, Eric Y. Chang Stroke Rehabilitation , 2019 pp 105-114 View PDFView chapter Related quote(s)1 / 1 "... The spinothalamic tract The most studied tract associated with pain is the spinothalamic tract, which transmits the modalities of pain, temperature, and deep touch from the body. The spinothalamic tract courses from the lateral portion of the spinal cord, through the lateral medulla and pons, to the ventral posterolateral nucleus (VPL) of the thalamus, terminating in the post-central gyrus ( Fig. 7.1A ). Lesions or injury to any part of this tract can potentially result in CPSP; however, some structures are more highly associated with this syndrome than others. CPSP was originally described as a thalamic pain syndrome, and the thalamus continues to be the most commonly documented and studied neural structure associated with CPSP. 5,9 Modern studies have shown that specific areas within the thalamus are more correlated to the development of CPSP than others. Studies have shown that CPSP patients have lesions within the VPL and/or the ventral posteromedial (VPM) of the thalamus ( Fig. 7.1B ). 10–14 A more recent study using MRI and digital radiographic atlases in thalamic stroke patients with and without CPSP found that the CPSP group had lesions largely involving the VPL, with some also involving the VPM nucleus. 15,16 Specifically, lesions in the posterolateral and inferior parts of the VPL were most associated with CPSP. A few of the CPSP patients in this study did have lesions confined to the pulvinar nucleus as well, an area that processes visual input. The development of CPSP in these patients was thought to be due to the shared vascular supply and close proximity to the VPL 15 ( Fig. 7.1B ); again implying the strong association of the VPL and CPSP. Indeed, another study found that thalamic lesions involving the area where the ventral posterior nuclei and the pulvinar meet were 81 times more likely to lead to CPSP than other thalamic lesions, confirming this area as high risk for CPSP, and opening the door for potential preemptive treatments against CPSP as a future avenue of research. 16 1 ..." Reference 18 Book Chapter Vertebrobasilar Disease J.P. Mohr, Louis R. Caplan Stroke , 2004 pp 207-274 View PDFView chapter Related quote(s)1 / 1 "... 387 The lesion responsible for ataxic hemiparesis is usually in the more rostral pons, interrupting crossing fibers as well as the pyramidal fibers in the pontine base. We have also seen a patient with this syndrome whose postmortem lesion involved the brachium conjunctivum and cerebral peduncle at the midbrain level. Slight ataxia of the other (normal) leg has been a clue to the pontine location of this syndrome. At times, horizontal or vertical nystagmus and dysarthria accompany the limb ataxia and hemiparesis. 137 Pure Sensory Stroke. A lacune in the somatosensory nuclei of the thalamus and ventral posterolateral and ventral posteromedial nuclei produces sensory symptoms on the opposite side of the body and face without motor, cerebellar, or higher cortical function abnormalities. 375 388 389 390 Paresthesias are usually characterized as tingling, prickling, or a sleepy, cold, hard, numb, or dead feeling. Usually at least two parts, such as face and arm or arm and leg, are involved. 388 The limbs are involved more than the face, but often the whole hemicorpus is affected, including the abdomen, chest, and face (as well as the eye, ear, and inside of the mouth). Cortical representation in the postcentral gyrus for the ear, eye, and trunk is quite small. Involvement of these structures usually signifies a lesion in a tract or a thalamic nucleus rather than a parietal cortical lesion. When the hand is affected, usually all digits are affected. The symptoms of sensory dysfunction commonly far exceed the objective signs; in fact, objective parameters of sensory function may be completely normal. In one reported patient with a right lateral thalamic lacune, the sensory loss consisted only of proprioceptive loss, the temperature and pain sensations both being intact. 391 Helgason and Wilbur 392 described 10 patients with pontine infarcts identified on MRI in whom sensory symptoms and signs were the most prominent clinical findings. 1 ..." Reference 19 Review article Murine models of human neuropathic pain Colleoni M., Sacerdote P. Biochimica et Biophysica Acta (BBA) - Molecular Basis of Disease , 2010 pp 924-933 View PDFView article Related quote(s)1 / 1 "... Thalamic syndrome is a form of central pain that typically results from stroke in the thalamus and is characterized by spontaneous pain, attacks of allodynia, and dysesthesia. The lack of suitable thalamic syndrome models has hampered research into the dysregulated perception of pain resulting from stroke in the thalamus. Therefore, the development and characterization of a rodent model of thalamic syndrome was the first step to discover the underlying mechanisms of this disease and possible therapeutics. Very recently a rat model of this syndrome was developed based on a small hemorrhagic stroke lesion induced by collagenase injection in the ventral posterolateral nucleus of the rat thalamus . Animals displayed hyperesthesia in response to mechanical pinch stimulation, with sensitivity localized in the hind limb and increased thermal sensitivity. This novel model has not been developed in mice yet. 1 ..." Reference 20 Book Chapter Deep Brain Stimulation for Pain Senatus P., Condit D. Schmidek and Sweet Operative Neurosurgical Techniques , 2012 pp 1433-1442 View PDFView chapter Related quote(s)1 / 1 "... Central neuropathic pain originates from damage to the central nervous system, that is, the brain and spinal cord. Specific types of central pain include spinal cord injury, poststroke pain, postherpetic neuralgia, and other forms of neuralgia. 29 The ST has been the principal target region for DBS treatment of neuropathic pain. The thalamus, a large, paired mass of gray matter, resides in the diencephalon. It is located in the center of the brain, along the midline, and sits right above the brain stem. The thalamus’s functions include motor control and transmission of sensory stimuli. 30 Stimulation of the lateral nuclei, specifically the ventral posterolateral (VPL) nucleus and the ventral posteromedial (VPM) nucleus, has been shown to reduce pain symptoms of neuropathic origin. CH is a severe primary neurovascular headache disorder causing 10% to 20% of patients with the disorder to experience debilitating headaches. 31 Pain arising from this syndrome is indicated by attacks of intense unilateral periorbital pain, often jointly experienced with ipsilateral cranial autonomic disturbance. Through utilization of PET studies, increased blood flow to the posterior hypothalamic region was observed in patients amid an acute CH attack. The hypothalamus, forming the ventral aspect of the diencephalon and located beneath the thalamus, is linked with endocrine and pituitary function. 30 DBS of the posterior hypothalamic region was introduced to treat pain arising from acute CH attacks. 23 1 ..." Reference 21 Review article A Review of Medical and Surgical Options for the Treatment of Facial Pain Penn M.C., Choi W., Brasfield K., Wu K., Briggs R.G., Dallapiazza R., Russin J.J., Giannotta S.L., Lee D.J. Otolaryngologic Clinics of North America , 2022 pp 607-632 View PDFView article Related quote(s)1 / 2 "... Deep Brain Stimulation DBS is an approved treatment option for a variety of neurologic and psychiatric diseases, including essential tremor, Parkinson disease, idiopathic dystonia, and obsessive-compulsive disorder. 116 , 117 It has been explored since the 1970s and 1980s for its potential to treat chronic pain syndromes but has yet to be Food and Drug Administration approved. Evidence for DBS in the treatment of specific pain syndromes and disorders has been mixed, and after the failure of clinical trials for DBS for pain in 2001, 118 less attention has been given to the study of DBS to manage and treat chronic facial pain. 119 Common DBS targets for chronic pain syndromes include the periaqueductal gray (PAG) region and the ventral posterolateral nucleus of the thalamus (VPL) ( Fig. 6 ). 116 , 120 The exact process by which pain relief occurs from DBS targeting of these regions is not known, but some theories suggest it may lead to the release of endogenous opioids or to modulation of ascending pain pathways and neural circuitry. 119 DBS of these regions has demonstrated functional changes in regional cerebral blood flow, which may lead to long-term changes in pain perception. 121 As DBS is designed to target neural pathways centrally, its use seems especially relevant for patients with demyelinating pain syndromes refractory to medical management. 117 Despite the failure of previous trials, recent evidence for DBS in the treatment of chronic facial pain syndromes seems more promising. A prospective study of 56 patients with chronic pain syndromes, including six with trigeminal neuropathic pain, treated with DBS to the PAG and lateral somatosensory thalamus demonstrated mixed results across different syndromes, with the most success seen in patients with chronic low back or leg pain. However, 50% of patients in this study with trigeminal neuropathic pain and/or dysesthesia dolorosa achieved at least a 50% reduction in their pain. 120 A recent trial of 15 patients with poststroke neuropathic pain who underwent DBS of the VPL thalamus and PAG achieved an overall reduction of 48.8% in 80% of patients. 1 ..." Related quote(s)2 / 2 "... Evidence for DBS in the treatment of specific pain syndromes and disorders has been mixed, and after the failure of clinical trials for DBS for pain in 2001, 118 less attention has been given to the study of DBS to manage and treat chronic facial pain. 119 Common DBS targets for chronic pain syndromes include the periaqueductal gray (PAG) region and the ventral posterolateral nucleus of the thalamus (VPL) ( Fig. 6 ). 116 , 120 The exact process by which pain relief occurs from DBS targeting of these regions is not known, but some theories suggest it may lead to the release of endogenous opioids or to modulation of ascending pain pathways and neural circuitry. 119 DBS of these regions has demonstrated functional changes in regional cerebral blood flow, which may lead to long-term changes in pain perception. 121 As DBS is designed to target neural pathways centrally, its use seems especially relevant for patients with demyelinating pain syndromes refractory to medical management. 117 Despite the failure of previous trials, recent evidence for DBS in the treatment of chronic facial pain syndromes seems more promising. A prospective study of 56 patients with chronic pain syndromes, including six with trigeminal neuropathic pain, treated with DBS to the PAG and lateral somatosensory thalamus demonstrated mixed results across different syndromes, with the most success seen in patients with chronic low back or leg pain. However, 50% of patients in this study with trigeminal neuropathic pain and/or dysesthesia dolorosa achieved at least a 50% reduction in their pain. 120 A recent trial of 15 patients with poststroke neuropathic pain who underwent DBS of the VPL thalamus and PAG achieved an overall reduction of 48.8% in 80% of patients. 122 For those who continued to require analgesic medications after DBS therapy, all were able to switch to nonopiate pain medications. Similarly, a small case series of 7 patients with unilateral, intractable facial pain found pain scores were significantly decreased at 1-year follow-up after DBS implantation. 1 ..." Reference 22 Review article Advances and Future Directions of Neuromodulation in Neurologic Disorders Burns M.R., Chiu S.Y., Patel B., Mitropanopoulos S.G., Wong J.K., Ramirez-Zamora A. Neurologic Clinics , 2021 pp 71-85 View PDFView article Related quote(s)1 / 1 "... Pain DBS has been applied to a variety of pain syndromes ranging from poststroke pain, spinal cord injury, brachial plexus injury, and headache. 78 There are 3 commonly targeted structures for neuromodulation of pain: (1) the thalamus, specifically the ventral posterolateral nucleus and ventral posteromedial nucleus (VPL/VPM), (2) the periventricular and periaqueductal gray (PVG/PAG), and (3) the anterior cingulate cortex (ACC). Neuromodulation of the VPL/VPM and PVG/PAG have been the most well established in the literature. 79–81 One longitudinal study of VPL/VPM and PVG/PAG DBS for various pain conditions revealed 59% of patients experienced significant acute pain relief. 81 After approximately 80 months of follow-up, 31% of patients continued to experience pain relief. In another study with up to 15 years of follow-up after VPL/VPM or PVG/PAG DBS for various pain syndromes, 62% of patients continued to experience adequate pain relief. 82 A meta-analysis of DBS for pain reported that modulation of the PVG/PAG alone or PVG/PAG with VPL/VPM or internal capsule was more effective than VPL/VPM alone. In addition, overall 58% of patients achieved pain relief. DBS was most effective in treating intractable low back pain and least successful in treating central thalamic pain/poststroke pain. Although central thalamic pain syndrome has historically been difficult to treat, Franzini and colleagues 83 recently investigated DBS of the posterior limb of the internal capsule in 4 patients. Three of four patients achieved long-term pain relief post-DBS. After a mean follow-up of 5.88 years, the average reduction in pain was 38% based on the 10-point visual analog scale. Inspired by lesional therapies for cancer-related pain, ACC DBS has emerged as the newest potential target for pain. 84 , 85 Spooner and colleagues 86 published the first case report of ACC DBS in 2007 in a patient with medically refractory neuropathic pain from a complete C4 spinal cord injury. 1 ..." Reference 23 Book Chapter Discussion Saab C.Y. Chronic Pain and Brain Abnormalities , 2014 pp 127-141 View PDFView chapter Related quote(s)1 / 2 "... Neuromodulation by Electricity Deep brain stimulation (DBS) is considered primarily a neurosurgical intervention intended to manage severe intractable pain. Although the safety and efficacy of DBS were established firmly for movement disorders, DBS for pain remains “off-label” and its mechanisms remain largely unknown. In general, the primary disadvantage of using electrodes is the requirement for surgical procedures, with the potential to trigger inflammation, cell death and gliosis cascades. 19 The main targets of DBS for pain include thalamic and brain stem structures. Current indications include pain secondary to stroke, amputation (phantom pain), failed back surgeries and cancer, thus mainly of neuropathic origin. Interestingly, the frequencies used during DBS for pain range typically from 50–70 Hz (see discussion below regarding low- versus high-frequency stimulation). Today, conventional wisdom equates “electrical stimulation” with “neuronal excitation,” but it is becoming increasingly clear that different electrical stimulation parameters may result in gain or loss of function in a neuronal population exposed to a stimulating probe. For example, though the mechanisms of DBS remain speculative, 20 studies related to DBS for motor disorders suggest that high-frequency stimulation (HFS, >100 Hz) can mimic the functional effects of ablation, 21 also referred to as “jamming” of neuronal circuitry. 20 In animal models, two studies have recently tested the effects of deep brain stimulation (DBS) on sensitization of single-units in the brain, as well as on pain behavior. 22,23 Aiming to reverse neuronal sensitization in the sensory ventral posterolateral (VPL) nucleus of the thalamus, HFS (100–150 Hz) within the VPL was shown to effectively decrease neuronal hyperexcitability and thermal hyperalgesia in an animal model of peripheral pain, whereas stimulation at a low-frequency (LFS, 20–40 Hz) had no effect on neuronal firing. 22 Interestingly, LFS (50 Hz) in the VPL is known to produce little or no analgesia in rats. 1 ..." Related quote(s)2 / 2 "... Current indications include pain secondary to stroke, amputation (phantom pain), failed back surgeries and cancer, thus mainly of neuropathic origin. Interestingly, the frequencies used during DBS for pain range typically from 50–70 Hz (see discussion below regarding low- versus high-frequency stimulation). Today, conventional wisdom equates “electrical stimulation” with “neuronal excitation,” but it is becoming increasingly clear that different electrical stimulation parameters may result in gain or loss of function in a neuronal population exposed to a stimulating probe. For example, though the mechanisms of DBS remain speculative, 20 studies related to DBS for motor disorders suggest that high-frequency stimulation (HFS, >100 Hz) can mimic the functional effects of ablation, 21 also referred to as “jamming” of neuronal circuitry. 20 In animal models, two studies have recently tested the effects of deep brain stimulation (DBS) on sensitization of single-units in the brain, as well as on pain behavior. 22,23 Aiming to reverse neuronal sensitization in the sensory ventral posterolateral (VPL) nucleus of the thalamus, HFS (100–150 Hz) within the VPL was shown to effectively decrease neuronal hyperexcitability and thermal hyperalgesia in an animal model of peripheral pain, whereas stimulation at a low-frequency (LFS, 20–40 Hz) had no effect on neuronal firing. 22 Interestingly, LFS (50 Hz) in the VPL is known to produce little or no analgesia in rats. 24 In another animal model of central pain, LFS (50 Hz) within the motor cortex reduced mechanical allodynia and thermal hyperalgesia. 23 These effects were shown to be mimicked by DBS in the zona incerta of the subthalamus and blocked by reversible inactivation of this region (see Chapter 6 in this book by Keller and Masri). This suggests that the potential analgesic effects of motor cortex stimulation may be due to disinhibition of the zona incerta, thus, causing analgesia by restoring inhibition in the thalamus. 1 ..." Reference 24 Book Chapter Neurosurgical Approaches to Pain Management Raslan A.M., Burchiel K.J. Practical Management of Pain , 2014 pp 328-334.e2 View PDFView chapter Related quote(s)1 / 1 "... The use of DBS for pain control has failed to gain much acceptance in the neurosurgical community, and the use of DBS electrodes as pain control implants has never achieved U.S. Federal Drug Administration (FDA) approval. The lack of data to support the procedure is due, in part, to the small number of patients treated, inconsistent target localization, heterogeneity of the pain diagnoses treated, and failure to mount a prospective randomized trial that was sufficiently powered to answer the question of efficacy. The mechanism of pain relief by DBS is poorly understood but appears to be dependent on the site. The thalamus and PVG / PAG were the most commonly 65 targeted sites for DBS implants for pain. Hosobuchi and colleagues 58 suggested that the pain-relieving effect of PVG and PAG stimulation might involve endogenous opioid receptors based on their studies in which it was found that the pain-relieving effect of DBS could be reversed by naloxone. Evidence to support this mechanism of action of PAG/PVG DBS is inconsistent. Some investigators supported the concept whereas others disagreed. Currently, it is postulated that the pain-relieving effect of PAG / PVG DBS is due to activation of multiple supraspinal descending pain modulatory systems, both opioid and nonopioid. 66 Pain relief resulting from stimulation of the ventral posterolateral (VPL) nucleus and ventral posteromedial (VPM) nucleus (Vc nucleus in the European Hassler terminology), the major sensory nuclei of the thalamus, is poorly understood. Inhibition of spinothalamic tract neurons 67 and activation of dopaminergic mechanisms have both been proposed. 68 The most accepted hypothesis is that thalamic stimulation activates the nucleus raphe magnus of the rostroventral medulla, which results in activation of a suprasegmental descending endogenous pain inhibition system. 1 ..." Reference 25 Review article Surgical Options for Atypical Facial Pain Syndromes Rahimpour S., Lad S.P. Neurosurgery Clinics of North America , 2016 pp 365-370 View PDFView article Related quote(s)1 / 1 "... Since its inception in the middle of the twentieth century, deep brain stimulation of the thalamus and caudate has also been used in the treatment of chronic pain. 31 However, results of 2 multicenter trails were unfortunately inconclusive. 32 Several stimulation targets have been used, including PAG, zona incerta, and the ventral posterolateral nucleus of the thalamus. In a recent double-blinded study by Rasche and colleagues 33 of 56 patients with neuropathic pain, 6 patients had trigeminal neuropathic pain. Of these patients, half, at 30 months of mean follow-up, had an unsatisfactory response to the treatment. Deep brain stimulation, however, remains a poorly understood intervention for neuropathic pain; much remains to be learned about its role in the facial pain treatment algorithm. 1 ..." Reference 26 Review article Deep brain stimulation for chronic pain: Intracranial targets, clinical outcomes, and trial design considerations Keifer O.P., Riley J.P., Boulis N.M. Neurosurgery Clinics of North America , 2014 pp 671-692 View PDFView article Related quote(s)1 / 1 "... From there, DBS targets for pain control would expand to include the internal capsule (IC), the ventral posterolateral nucleus (VPLP) and the ventral posteromedial nucleus (VPM) of the sensory thalamus (STH), the centro-median parafasicular region (CM-Pf) of the thalamus, the periaqueductal/paraventricular gray (PAG/PVG), the posterior hypothalamus (PH), the motor cortex, the nucleus accumbens (NAcc), and the anterior cingulate cortex. The following sections highlight the past, present, and future DBS targets used to treat various types of pain. 2 ..." References (26) Reference 1 Book Chapter Thalamus Jürgen K. Mai, F. Forutan The Human Nervous System , 2012 pp 618-677 View PDFView chapter Related quote(s)1 / 5 "... Ventral Posterolateral Nucleus (VPL) VPL stands out against its surrounding by fairly large stellate or polygonal cells. This property represents the main distinctive feature of the nucleus ( Figure 19.17 D). In addition to the large cells there are also medium to small cell types found. VPL has been divided into several subdivisions (anterior, posterior, medial, lateral) on the basis of size, density, and distribution of cells, molecular properties, and by their responses to cutaneous stimuli . The distinction between VL (VLp) and the anterior part of the ventral posterolateral nucleus (VPLa) can be made by the transition from the large neurons in VL to the mixed large and small-sized neurons in the VPL. Immunohistochemically, there is a difference in the AChE-reaction: low in VLp, very intense in VPL. Ventral Posteromedial Nucleus (VPM) VPM corresponds to the area of the Ncl. arcuatus, Ncl. semilunaris, and Ncl. posterointernus of earlier authors . VPM receives the ascending secondary trigeminal afferents via the trigeminal lemniscus from the head, face, and intraoral structures (afferents from proprioceptors and nociceptors from the stomatognathic system). In monkeys VPL and VPM are separated by a narrow cell-poor septum (lamella arcuata) that is well seen in sections stained for PV and CO . In humans, this lamella can be distinguished during fetal development as distinct (CD15-negative) lamina that surrounds the entire VPM. Against the centromedian nucleus VPM is delimited by a fibrous capsule, the lamella centralis . Medially and ventrally VPM borders the ventral posteromedial nucleus, parvicellular part (VPPC) that is considered by some researchers as part of VPM. Both nuclei are, however, well distinguishable parts ( Figure 19.24 ). ..." Related quote(s)2 / 5 "... Ventral Posterolateral Nucleus (VPL) VPL stands out against its surrounding by fairly large stellate or polygonal cells. This property represents the main distinctive feature of the nucleus ( Figure 19.17 D). In addition to the large cells there are also medium to small cell types found. VPL has been divided into several subdivisions (anterior, posterior, medial, lateral) on the basis of size, density, and distribution of cells, molecular properties, and by their responses to cutaneous stimuli . The distinction between VL (VLp) and the anterior part of the ventral posterolateral nucleus (VPLa) can be made by the transition from the large neurons in VL to the mixed large and small-sized neurons in the VPL. Immunohistochemically, there is a difference in the AChE-reaction: low in VLp, very intense in VPL. ..." Related quote(s)3 / 5 "... Sensory Thalamus The sensory thalamus can be subdivided into a somatosensory unit with several nuclei (ventral posterior nucleus,VP; ventroposterior inferior nucleus,VPI; ventroposterior superior nucleus, VPS; ventromedial nucleus, VM; ventral medial nucleus, posterior part, VPMpc; anterior pulvinar, APul) (see Chapter 30 ) and the viscerosensory nunit (ventral posteromedial nucleus, parvicell. part, VPPC). Each nucleus has its individual set of afferents and distinctive projections to specific cortical areas. Ventroposterior Nucleus or Complex (Ncl. ventralis posterior, VP) The somatosensory unit represents the primary relay nucleus of the thalamus where all fiber tracts for the transmission of surface and deep sensitivity of the trunk, extremities, and head of (mainly) the opposite side of the body terminate . Because of the convergence of the many afferents, the different nuclear components of the unit are grouped together as the ventrobasal complex ( Table 19.1 ; see Chapter 30 ). VP represents a functional unit that is histologically separated in two main subnuclei: the ventral posterolateral nucleus, VPL, and the ventral posteromedial nucleus, VPM, . The lateral portion provides the receptive area for the spinal and the lemniscal fibers; the medial portion is the representation of the trigeminal fibers. The fibers of the protective (protopathic) sensitivity (from neurons that respond to noxious stimuli, temperature, and touch) from trunk and extremities are carried by the anterolateral system (spino-thalamic tract); the discriminative (epicritic) sensitivity and proprioception are mediated by the dorsal column-medial lemniscus system. Protopathic and epicritic fibers of both systems terminate in the ventral posterolateral nucleus (VPL). ..." Related quote(s)4 / 5 "... The corresponding fibers from the head (from the trigeminal complex and from the remaining branchial nerves), which mediate impulses from the cutaneous receptors, are carried to the ventral posteromedial nucleus (VPM). The fibers from the proprioceptive receptors from trunk, extremities and the head (temporomandibular joint) project to an area within the rostral VPL which has not been delineated with complete certainty in the human. In the monkey the territory for the proprioceptive or kinesthetic fibers has been located within the VP “shell” or within a separate nucleus, the ventroposterior superior nucleus, VPS (VPO) ( Kaas, 2008 ; Chapter 30 ; see Chapter 31 ). These fibers are joined by those from the vestibular nuclei and taste pathway. Among the afferents to the lateral thalamic nuclei, the so-called secondary trigeminal pathway has created some misunderstandings. In the human the connection between the spinal trigeminal nucleus (but also from the main sensory nucleus) and VPM was described as a complex, dorsally ascending bundle that was later named as Wallenberg’s bundle, compromised in the Wallenberg syndrome . Re-examination showed that the secondary trigeminal pathway and the supposed Wallenberg bundle represent structurally related but functionally different trajectories. The Wallenberg bundle is complex with multiple sites of origin, related to the reticular activating system and ends in different parts of the thalamus, preferentially within the intralaminar nuclei (but also in VPM) . Ventral Posterolateral Nucleus (VPL) VPL stands out against its surrounding by fairly large stellate or polygonal cells. This property represents the main distinctive feature of the nucleus ( Figure 19.17 D). In addition to the large cells there are also medium to small cell types found. VPL has been divided into several subdivisions (anterior, posterior, medial, lateral) on the basis of size, density, and distribution of cells, molecular properties, and by their responses to cutaneous stimuli . 1 ..." Related quote(s)5 / 5 "... The fibers of the protective (protopathic) sensitivity (from neurons that respond to noxious stimuli, temperature, and touch) from trunk and extremities are carried by the anterolateral system (spino-thalamic tract); the discriminative (epicritic) sensitivity and proprioception are mediated by the dorsal column-medial lemniscus system. Protopathic and epicritic fibers of both systems terminate in the ventral posterolateral nucleus (VPL). The corresponding fibers from the head (from the trigeminal complex and from the remaining branchial nerves), which mediate impulses from the cutaneous receptors, are carried to the ventral posteromedial nucleus (VPM). The fibers from the proprioceptive receptors from trunk, extremities and the head (temporomandibular joint) project to an area within the rostral VPL which has not been delineated with complete certainty in the human. In the monkey the territory for the proprioceptive or kinesthetic fibers has been located within the VP “shell” or within a separate nucleus, the ventroposterior superior nucleus, VPS (VPO) ( Kaas, 2008 ; Chapter 30 ; see Chapter 31 ). These fibers are joined by those from the vestibular nuclei and taste pathway. Among the afferents to the lateral thalamic nuclei, the so-called secondary trigeminal pathway has created some misunderstandings. In the human the connection between the spinal trigeminal nucleus (but also from the main sensory nucleus) and VPM was described as a complex, dorsally ascending bundle that was later named as Wallenberg’s bundle, compromised in the Wallenberg syndrome . Re-examination showed that the secondary trigeminal pathway and the supposed Wallenberg bundle represent structurally related but functionally different trajectories. The Wallenberg bundle is complex with multiple sites of origin, related to the reticular activating system and ends in different parts of the thalamus, preferentially within the intralaminar nuclei (but also in VPM) . 1 ..." Reference 2 Review article The neural basis of somatosensory temporal discrimination threshold as a paradigm for time processing in the sub-second range: An updated review Ordas C.M., Alonso-Frech F. Neuroscience & Biobehavioral Reviews , 2024 pp 105486 View PDFView article Related quote(s)1 / 1 "... Thalamus The somatosensory input from all modalities and all regions of the body converge within the ventroposterior (VP) region of the thalamus. Since the classical descriptions by Hassler , nomenclature of the thalamus has been diverse and even controversial. However, efforts have been made to reach a common terminology. Following a recent revision by Mai and Majtanik, the VP nucleus has been histologically separated into the lateral ventroposterior nucleus (ventral posterolateral nucleus, VPL,), the medial ventroposterior nucleus (ventral posteromedial nucleus, VPM) and the ventrocaudal nucleus (V.c.pc) . The latter has been further divided into two smaller divisions named the external and the internal divisions, equivalent to the ventral posterior inferior nucleus (VPI) and the medial ventroposterior nucleus (ventral posteromedial nucleus, parvocellular part, VPMpc) . VPL has been divided into several subdivisions (anterior, posterior, medial, lateral). The spinal, lemniscal and trigeminal afferents to the thalamus are difficult to separate in humans. Nevertheless, VPM is thought to receive the ascending secondary trigeminal afferents via the trigeminal lemniscus from the head and face. On the other hand, proprioceptive fibers project to an area anterior and dorsal to VPM and VPL, which has been termed “superior ventroposterior” (VPS), oral part (VPO) or “shell region” . Projections to somatosensory cortex are mainly directed to areas 3b and 1 although they are highly convergent and divergent . ..." Reference 3 Book Chapter Diencephalon Reha Erzurumlu, Gulgun Sengul, Emel Ulupinar Human Neuroanatomy , 2024 pp 57-78 View PDFView chapter Related quote(s)1 / 2 "... Thalamic nuclei The thalamic nuclei are named according to their locations and abbreviations as follows: anterior nuclei (AN), ventral anterior nucleus (VA), ventral lateral nucleus (VL), ventral posteromedial nucleus (VPM), ventral posterolateral nucleus (VPL), mediodorsal (MD), median (M) as the median nucleus group, lateral dorsal nucleus (LD), lateral posterior nucleus (LP), intralaminar nuclei (IN), pulvinar nuclei (P), and the nuclei that make up the metathalamus, mediale geniculate body (MGB), and lateral geniculate body (LGB). These nuclei are further classified as specific, primary sensory (VPL, VPM, LGB, MGB), motor (VA, VL), association (P, AN, MD, LD, LP), and nonspecific (or nonspecific) nuclei (M, IN) based on their functions and specific connections with cortical areas ( Table 5.2 , Fig. 5.4 ). ..." Related quote(s)2 / 2 "... Reciprocally, axons from lamina VI in the same region also project to LGB ( Fig. 5.4 ). Clinical concepts related to the thalamus Due to the different perforating branches that supply the posterior cerebral artery, various clinical findings can occur in cases of vascular disorders such as bleeding or blockage involving these segments. Lateral thalamic infarction: It occurs when the thalamogeniculate artery, which supplies the ventral lateral thalamus including VPL and VPM, CM, and rostro-lateral part of the pulvinar, is blocked. Lateral thalamic infarction can start with paresthesia and numbness on one side of the body and can also affect all sensory modalities. If the infarction also affects the poster limb of the internal capsule adjacent to VPL, motor defects can occur in addition to sensory deficits. In the more common form of lateral thalamic infarction known as Dejerine–Roussy syndrome (“thalamic pain” syndrome), initially mild contralateral hemiparesis and persistent deep sensory deficits are observed. Involuntary movements resembling dystonia, choreoathetosis, and tremor can occur in the contralateral arm and leg. The dystonic posture deficit that develops in the hand is called “thalamic hand.” In the late stage, severe, persistent, and unresponsive pain can be observed on the side where numbness and paresthesia occurred before. The initial symptoms observed in thalamic syndrome are usually abnormal sensitivity to painful stimuli (hyperpathia) or abnormal perception of touch sensation as discomfort (dysesthesia) on the face, arms, and/or legs. Allodynia refers to the perception of pain in response to a light touch that would not normally be painful, or even in the absence of a stimulus. Pain associated with thalamic syndrome can increase due to exposure to heat or cold and emotional distress. 1 ..." Reference 4 Book Chapter Thalamus Groenewegen H.J., Witter M.P. The Rat Nervous System , 2004 pp 407-453 View PDFView chapter Related quote(s)1 / 1 "... Afferent and efferent connections The ventral posterolateral and ventral posteromedial nuclei receive their main ascending inputs from somatosensory afferents originating in the spinal cord, dorsal column nuclei, and trigeminal complex (see also Tracey, Chapters 7 and 25 , and Waite, Chapter 26 , this volume). These somatosensory projections are somatotopically organized such that afferents from the trunk and limbs terminate in the VPL and those from the head terminate in the VPM. Spinal and trigeminal fibers not only reach the VPL and VPM but also distribute over much wider areas of the thalamus, including the posterior nucleus (see page 417 ) and the intralaminar nuclei (see section “Midline and Intralaminar Thalamic Nuclei”). Spinothalamic fibers, transferring among others nociceptive signals, originate from different laminae of the dorsal horn as well as the central gray of the spinal cord and terminate as large boutons in the ventral posterolateral nucleus . Nociceptive information may reach the VPL also indirectly via the caudal medullary reticular formation (medullary dorsal reticular nucleus) in addition to the direct spinothalamic pathway . Dorsal column nuclear afferents likewise terminate with large boutons in the VPL . Spinal and lemniscal fibers, in part, converge on individual thalamocortical neurons, the lemniscal fibers on the soma, and more proximal parts of dendrites of the neurons than the spinothalamic fibers . Lemniscal, but also spinothalamic, fibers probably use glutamate as neurotransmitter . Spinothalamic fibers have also been shown to contain substance P . The spinal and principal trigeminal nuclei both project to the ventral posteromedial nucleus as well as, less densely, to the posterior complex (cf. also Waite, Chapter 26 , this volume). Projections to the VPM are more focussed, those to the medial part of posterior nucleus more diffuse . ..." Reference 5 Book Chapter The Diencephalon Srikant S. Chakravarthi, Alejandro Monroy-Sosa, Kost Elisevich From Anatomy to Function of the Central Nervous System , 2025 pp 255-284 View PDFView chapter Related quote(s)1 / 4 "... There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. Following its inputs from the spinothalamic axons, the VPL sends its output to the posterior limb of the internal capsule and corona radiata en route to the primary somatosensory area of the cortex. In addition, apart from the contribution of the VPL, the posterior and intralaminar groups of the medial dorsal nucleus also receive inputs from the spinal lemniscus . The Motor Thalamus The motor thalamus (MThal) represents an important and strategic location within the ventral thalamus and shares a strong interconnection with both motor cortices, as well as the cerebellum (dentate and interposed nuclei) and basal ganglia (substantia nigra pars reticulata and internal segment of the globus pallidus), to regulate both cognitive and proprioceptive control of movement. The current system shares many of the features of the “driver-modulator” concept, as proposed by Sherman and Guillery for the sensory thalamus. As described by Bosch-Bouju (2013) , the MThal acts as a “super-integrator” of motor information from two functional driver-input subdivisionst: the basal ganglia and cerebral cortex and the cerebellum and cerebral cortex. ..." Related quote(s)2 / 4 "... The drivers differ from modulators in that they predominantly activate ionotropic glutamate receptors (AMPA and NMDA), contain thick axons, and have large terminals on proximal dendrites. Modulators have activating ionotropic and metabotropic glutamate receptors, contain thin axons, and have small terminals on proximal dendrites . It is also important to note that for the thalamus, drivers are much less numerous than modulators, as in the example provided, where layer 6 inputs have many more synapses than the retinal (driver) inputs . Sensory Relay The sensory relay system of the thalamus is perhaps the most commonly known and robust of all relay circuits. There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. Following its inputs from the spinothalamic axons, the VPL sends its output to the posterior limb of the internal capsule and corona radiata en route to the primary somatosensory area of the cortex. In addition, apart from the contribution of the VPL, the posterior and intralaminar groups of the medial dorsal nucleus also receive inputs from the spinal lemniscus . ..." Related quote(s)3 / 4 "... Sensory Relay The sensory relay system of the thalamus is perhaps the most commonly known and robust of all relay circuits. There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. Following its inputs from the spinothalamic axons, the VPL sends its output to the posterior limb of the internal capsule and corona radiata en route to the primary somatosensory area of the cortex. In addition, apart from the contribution of the VPL, the posterior and intralaminar groups of the medial dorsal nucleus also receive inputs from the spinal lemniscus . 1 ..." Related quote(s)4 / 4 "... In an effort to control the various stimulatory and inhibitory inputs and outputs contained within every thalamic circuit, there are critical drivers and modulators which control the degree of signal traveling through these circuits. Again, an apt and highly studied example is the communication between retinal receptive field inputs and the visual layer 6 of the cortex that includes the lateral geniculate nucleus. The retinal afferent drivers and cortical input modulators work together to control signals sent to the lateral geniculate nucleus for processing. The drivers differ from modulators in that they predominantly activate ionotropic glutamate receptors (AMPA and NMDA), contain thick axons, and have large terminals on proximal dendrites. Modulators have activating ionotropic and metabotropic glutamate receptors, contain thin axons, and have small terminals on proximal dendrites . It is also important to note that for the thalamus, drivers are much less numerous than modulators, as in the example provided, where layer 6 inputs have many more synapses than the retinal (driver) inputs . Sensory Relay The sensory relay system of the thalamus is perhaps the most commonly known and robust of all relay circuits. There are two principal sensory systems that arise from the dorsal roots of the spinal cord. The spinothalamic , or ventral lateral, system is responsible for carrying pain, temperature, and two-point discrimination signals to the ventral posterior (VP) nucleus of the thalamus. As it relates to the relative position of innervating axons, this nucleus consists of two divisions: the ventral posterolateral (VPL) nucleus, which receives both spinothalamic and medial lemniscal axons; and the ventral posteromedial (VPM) nucleus, which receives trigeminothalamic axons. Within the VPL, there is somatotopic organization with the contralateral lower and upper limbs represented dorsolaterally and ventromedially, respectively; the opposite side of the head is represented in the VPM. 1 ..." Reference 6 Review article The structural and functional importance of the thalamus in migraine processes with and without aura. A literature review Valenzuela-Fuenzalida J.J., Suazo-Santibanez A., Semmler M.G., Cariseo-Avila C., Santana-Machuca E., Orellana-Donoso M. Translational Research in Anatomy , 2021 pp 100130 View PDFView article Related quote(s)1 / 1 "... The lateral thalamic nuclei group are comprised by lateral posterior nucleus (LP, lateral dorsal group), ventral anterior nucleus or group (VA), ventral lateral nucleus or group (VL), ventral medial nucleus or group (VM), ventral posterior nucleus or group (VP), ventral posterolateral nucleus (VPL), ventral posteromedial nucleus (VPM), ventral posterior inferior nucleus (VPI). VPL and VPM nuclei are part of the somatosensory system, VPL relays medial lemniscal and spinothalamic input and relays the inferior portion of the postcentral gyrus. VL receives input mainly from the deep cerebellar dentate nucleus, and a smaller contribution from the basal nuclei to the rostral part of the VL. Projections from the VL, are the primary motor area, area 4, of the precentral gyrus and also has a smaller projection to premotor areas. Thus, involving the VL to motor feedback from the cerebellum and basal nuclei to the cerebral cortex [ 17 ]. The basal nuclei, especially the medial globus pallidus and the parts reticulata of the substantia nigra, send inputs to the VA, which in turn projects to premotor cortex including the supplementary motor area of the frontal lobes and is involved in planning and initiating movements [ 17 ]. As for the LP, it has been demonstrated that plays an important integrative function for the limbic spatial learning systems, since it projects to the hippocampal formation directional information that it is not just a reflection of sensory inputs [ 18 ]. The VPI is the thalamic relay center in the vestibulo-cortical pathway, since its projections are mainly directed to the primary vestibular cortex [ 19 ]. The largest portion of the thalamic posterior nuclei is the pulvinar, and it's also the largest association nuclei found in the dorsal part of the thalamus. It receives inputs from the superior colliculus and from the association cortex. ..." Reference 7 Book Chapter Human perception and neurocognitive development across the lifespan Shu-Chen Li, Evelyn Muschter, Jakub Limanowski, Adamantini Hatzipanayioti Tactile Internet , 2021 pp 199-221 View PDFView chapter Related quote(s)1 / 1 "... Sensory signals from different modalities are processed in different subregions within the thalamus, specifically, the optic nerve fibers project to the Lateral Geniculate Nucleus (LGN), the auditory nerve fibers project to the Medial Geniculate Complex (MGC), whereas the somatosensory nerve fibers project to the Ventral Posterolateral Nucleus (VPL) and the Ventral Posteromedia Nucleus (VPM). Another brain area that is also involved in early sensory processing is the superior colliculus, which is a small structure within the midbrain that sits right below the thalamus. The initial sensory representations formed within the different regions of the thalamus and in the superior colliculus are then relayed to other brain areas, which include regions in the midbrain (e.g., the basal ganglia) and in the cerebral cortex (e.g., visual, auditory and the somatosensory cortices), as well as regions in the parietal and frontal lobes that are important for attentional and cognitive control processes (see for review). These later regions are engaged in downstream processes that ( i ) compare and ( i i ) bind sensory representations acquired from the different modalities, as well as ( i i i ) interpret the current subjective sensory experiences as guided by attention and memories of past experiences. These processes together eventually give rise to the perception of an integral whole of an object, such as a tiny ball that is flying across the court as in the case of tennis playing. Most sensory signals are attained or, at least, modulated by motor sampling processes; thus perception inherently also involves sensorimotor processes. 9.2.2 Principles of multisensory integration In navigating through the manifold of sensory signals in the environment, the brain's sensory and perceptual processes need to distinguish between signals that belong to the same event from those that do not. When sensory information of an object or event is provided by more than one input modality, three general processing principles are involved in multisensory integration. ..." Reference 8 Book Chapter Somatosensory Evoked Potentials Daniel Dumitru, Lawrence R. Robinson, Machiel J. Zwarts Electrodiagnostic Medicine , 2002 pp 357-413 View PDFView chapter Related quote(s)1 / 1 "... 22, 134, 318 This tract terminates by synapsing in a specific region of the thalamus, the ventral posterolateral nucleus or VPL ( Fig. 9-1 ). Third-order nerve fibers originating in the VPL, thalamocortical fibers , project to the postcentral gyrus of the parietal lobe, known as the somatosensory cortex ( Fig. 9-1 ). 123 A characteristic topographic arrangement of sensory regions is discretely arranged along the cortex corresponding to lemniscal fibers from specific portions of the body, forming the so-called homunculus ( Fig. 9-2 ). 134 The somatosensory region representing the upper limb is located on the cortex superior to the face area. The lower limb somatosensory representation, however, is along the medial aspect of the cerebral hemisphere, separated from its contralateral body region in the opposite hemisphere by the interhemispheric fissure. With respect to the lower limb somatosensory information, a second fiber route has been proposed, i.e., the spinomedullothalamic tract . 318 This view suggests that the dorsal aspect of the lateral funiculus may convey a significant portion of the type IA and II afferent information to the somatosensory cortex. 55, 197, 229, 254 The proposed pathway is a continuation of primary afferents that synapse at the level of Clark's column to then ascend in the dorsal spinocerebellar tract and subsequently in the nucleus Z at the level of the medullopontine junction (see Fig. 9-1 ). 214 Nerve fibers from nucleus Z then travel to the medial lemniscus and ascend to the rostral region of the VPL. The traditional pathway from the VPL to the somatosensory cortex then serves as the final pathway for the fibers from the lower limb. 1 ..." Reference 9 Book series Chapter Noradrenergic innervation of somatosensory thalamus and spinal cord Westlund K.N., Zhang D., Carlton S.M., Sorkin L.S., Willis W.D. Progress in Brain Research, 1991 pp 77-88 View PDFView chapter Related quote(s)1 / 2 "... Monoamine terminals in the somatosensory thalamus More recent studies have examined specific anatomical connectivities and have required the development of techniques with better resolution. Light and electron microscopic observations have confirmed the presence of a sparse innervation by serotonergic and noradrenergic terminals of the ventral posterolateral nucleus of the macaque monkey ( Fig. 1 ). The ventral posterolateral nucleus of the thalamus receives sensory information from the spinal cord. A sparse distribution was noted for both kinds of aminergic terminals, with noradrenergic terminals present in slightly higher levels. En passant endings and intervaricose segments were evident at the ultrastructural level. Profiles immunocytochemically labeled for serotonin ( Fig. 1A,C ) or for the noradrenergic marker, DβH ( Fig. 1B,D ), were either axo-axonic or axo-dendritic. Some of the noradrenergic terminals appeared to contact dendritic spines. 1 ..." Related quote(s)2 / 2 "... Conclusions These studies have sought to reveal a more definitive anatomical basis for the mechanisms proposed for the modulation of nociception by norepinephrine by carefully mapping the locations of catecholamine terminals in relation to STT cells in the spinal cord and their targets in the thalamus. Thus, although descending pathways which could form the anatomical basis for monoaminergic interactions with the spinothalamic tract at the spinal cord are well known, continued analysis of chemically defined anatomical connectivities must be done before we can begin to understand the role of monoamines in complex functions such as somatosensory processing. These studies have been extended into the ventral posterolateral nucleus of the thalamus to determine the sources of noradrenergic innervation to this sensory relay nucleus. It is likely that these connectivities contribute to what is known as “stimulation-produced analgesia” and for the clinical effectiveness of some pharmaceuticals used in pain control. Extension of these studies to other regions is necessary to understand the role of norepinephrine in motor and autonomic control. 1 ..." Reference 10 Book Chapter Anatomy and Physiology of Pain Bajwa Z.H., Wilsey B.L., Fishman S.M. Office Practice of Neurology , 2003 pp 1383-1389 View PDFView chapter Related quote(s)1 / 1 "... SUPRASPINAL STRUCTURES The supraspinal system comprises the reticular formation, thalamus, hypothalamus, limbic system, and cerebral cortex. Within this system are extensive communicating projections for ascending algesic and descending analgesic pathways. The reticular formation extends through the entire length of the brainstem. The reticular formation receives input mainly from the spinoreticular tract but also from other structures in the supraspinal system. At the reticular formation level, receptive fields usually are extremely large, arising from both ipsilateral and contralateral parts of the body. The reticular formation controls the state of arousal and is important in autonomic reflex responses (fast defense reactions), inducing powerful analgesia in crisis situations. The thalamus consists of multiple nuclei and acts as the major relay station for incoming nociceptive stimuli. The thalamus is subdivided phylogenetically into paleothalamus and neothalamus or by nuclei location. The paleothalamus or medial thalamus has input mainly from the spinothalamic tract and the reticular formation. Its receptive fields are large, and it is involved primarily with the motivational-affective aspect of pain. The paleothalamus has extensive connections with the cerebral cortex. The neothalamus or lateral thalamus rests at the ventrobasal portion of the thalamus. The neothalamus, unlike the paleothalamus, is organized somatotopically and subdivided into the ventral posterolateral nucleus and the ventral posteromedial nucleus. The ventral posterolateral nucleus receives input mainly from the spinothalamic tract but also from the dorsal column system and the somatosensory cortex. The ventral posteromedial nucleus receives input mainly from the trigeminothalamic tract, which carries sensory input from the head and face and projects to the somatosensory cortex, involved with craniofacial pain. Although most of the neothalamic neurons respond to mechanoreceptive input, some are nociceptive-specific and are wide-dynamic range neurons. 1 ..." Reference 11 Book Chapter Development of the Nervous System Grow W.A. Fundamental Neuroscience for Basic and Clinical Applications , 2018 pp 72-90.e1 View PDFView chapter Related quote(s)1 / 1 "... In this way, the map of visual space on the retina is maintained in the lateral geniculate nucleus and ultimately in the visual cortex. Similar maps are formed in the medial geniculate nucleus (tonotopic mapping) and ventral posterolateral nucleus (somatotopic mapping). 1 ..." Reference 12 Book Chapter Proprioception: a sense to facilitate action Kyle P. Blum, Christopher Versteeg, Joseph Sombeck, Raeed H. Chowdhury, Lee E. Miller Somatosensory Feedback for Neuroprosthetics , 2021 pp 41-76 View PDFView chapter Related quote(s)1 / 1 "... Thalamic proprioceptive encoding As the medial lemniscus leaves the dorsal column nuclei, it decussates before synapsing in the ventroposterior complex of the thalamus. The nomenclature used to delineate proprioceptive regions of the thalamus is inconsistent across species and even research groups [see Bota et al. (2019) for a review]. In rats, a complex proprioceptive somatotopy exists in the most rostral regions of the ventral posterolateral nucleus (rVPL) . In monkeys, classic studies describe a region rostral and dorsal within the thalamus, termed the ventroposterior superior nucleus or sometimes ventroposterior oralis that projects to cortical area 3a . Other thalamic “motor” nuclei (ventrolateral; VL and ventral intermediate nucleus; Vim) also receive inputs from the cerebellum, carrying signals partially derived from proprioceptive mossy fiber inputs [for a review, see Shadmehr (2020 )]. To our knowledge, no recordings have been made of the primate proprioceptive thalamus during reaching. Proprioceptive recordings from monkey VPL during head rotations have been reported, containing a mixture of proprioceptive and vestibular modalities . 1 ..." Reference 13 Review article Thalamus: The ‘promoter’ of endogenous modulation of pain and potential therapeutic target in pathological pain You H.-J., Lei J., Pertovaara A. Neuroscience & Biobehavioral Reviews , 2022 pp 104745 View PDFView article Related quote(s)1 / 1 "... 2.2 Thalamus: function as ‘discriminator’ in control of nociception In the ventral posterolateral nucleus (VPL) of the thalamus, there are neurons that have firm receptive fields and that are capable of encoding information evoked by noxious stimuli at different intensities applied not only to a somatic area, but also to a visceral area (such as intraperitoneal injection of bradykinin or the dilatation of the uterus) . Peripheral nociceptive stimuli can also cause excitement of other thalamic nuclei, such as the intralaminar nuclei, submedius (SM) nucleus, the PO nucleus, and the VM nucleus . In contrast to the excitation of nociceptive thalamic neurons in the VPL that is considered to play an important role in relaying pain-related sensory-discriminative information to the cortex, noxious stimulation inhibits neurons in the thalamic reticular nucleus . This finding is in line with evidence that the thalamic reticular nucleus has projections to the thalamic VPL and the ventral basal nucleus group, and with the proposal that the thalamic reticular nucleus plays a role in gating of transthalamic information flow . Electrophysiological recordings in patients with stroke- or spinal cord injury-induced chronic pain indicate that the functional properties of thalamic neurons exhibit plastic changes in pathophysiological conditions . In patients with limb amputation, the thalamic representation of the missing limb remains functional and its stimulation can produce phantom limb pain in the affected limb . Animal studies in rats exposed to peripheral nerve injury or hind limb inflammatory nociception showed that the ventral basal nucleus of the thalamus, including the VPL and VPM nuclei, exhibits significantly decreased neuronal excitation thresholds . In line with this, increased spontaneous firing and increased responses to noxious stimulation were reported in the mediodorsal (MD) thalamic neurons in animals with chronic pain induced by spinal cord injury . 1 ..." Reference 14 Book Chapter Spinal Cord Gulgun Sengul, Charles Watson The Human Nervous System , 2012 pp 233-258 View PDFView chapter Related quote(s)1 / 1 "... The lateral spinothalamic tract continues as the spinal lemniscus. The spinothalamic tracts terminate mainly in the ventral posterolateral nucleus, the ventral posteromedial nucleus, the central lateral nucleus of the intralaminar nuclei, and the posterior nuclear group of the thalamus. There are both nociceptive and thermoreceptive-specific neurons in the ventral posterolateral and posteromedial nuclei . In humans, microstimulation of these neurons has been shown to elicit discrete sensations of pain or cooling . Studies in primates have shown that the dorsal and ventral spinothalamic terminations largely overlap in the lateral and medial thalamus ( Apkarian and Hodge, 1989c ). Spinothalamic tract projections to the central lateral nucleus of the thalamus play a part in motivational-affective responses to pain. The projection to the lateral thalamus (the ventrobasal complex) is involved in sensory-discriminative aspects of pain. Animal studies have shown that spinothalamic tract neurons send collateral branches to the medullary reticular formation, parabrachial area, periaqueductal gray, and nucleus accumbens. These projections distribute information to multiple brainstem sites, which might in turn activate autonomic or affective responses, or descending pain-modulatory mechanisms . The spinothalamic sytem reaches multiple cortical areas in the contralateral hemisphere. Using anterograde transneuronal transport of herpes simplex virus in the monkey, Dum et al. (2009) found that major targets for the spinothalamic input are the granular insular cortex, the secondary somatosensory cortex, and several cortical areas in the cingulate sulcus. Comparable cortical regions in humans have also been shown to display activation when subjects are exposed to painful stimuli. Using diffusion tensor tractography, Hong et al. (2010) observed that the spinothalamic tract and related thalamocortical fibers in the human brain originate from the ventral posterolateral nucleus of the thalamus. 1 ..." Reference 15 Review article Neuropathic pain modeling: Focus on synaptic and ion channel mechanisms Bouali-Benazzouz R., Landry M., Benazzouz A., Fossat P. Progress in Neurobiology , 2021 pp 102030 View PDFView article Related quote(s)1 / 1 "... Thalamic syndrome Stroke is the leading cause of disability in the industrialized world and 8% of stroke victims suffer from post-stroke central pain (PSCP) . Thalamic syndrome is one of PSCP occurring after a cerebrovascular accident in the thalamus. In patients, this syndrome is characterized by spontaneous pain, cold allodynia and sensory loss in the body parts that correspond to the brain territory injured . In the rat, stereotaxic injection of collagenase or blood in the ventral postero-lateral nucleus of the thalamus (VPL) resulted in small hemorrhagic stroke lesions, mimicking the human thalamic syndrome . The authors observed an increased thermal (hot) and mechanical sensitivity in the contralateral limb after thalamic hemorrhage over the 3-week testing period without hypersensitivity to cold. These results in rats, do not reproduce the cold allodynia phenotype seen in patients in whom localized lesions in the ventro-caudal thalamus are sufficient to cause deficiencies in cold sensation without heat pain perception . However, recently two different mice models have been developed by unilateral stereotaxic injection of kainate or collagenase in the VPL . In these models, tactile and cold hypersensitivity are observed with a delay of three to five days after injury. These mice models develop symptoms similar to the sensory abnormalities observed in patients suffering from CPSP, such spontaneous pain, attacks of dysesthesia and allodynia. 1 ..." Reference 16 Book Chapter Microangiopathies (Lacunes) Marti-Vilalta J.L., Arboix A.G., Mohr J.P. Stroke , 2011 pp 485-515 View PDFView chapter Related quote(s)1 / 1 "... The small vessels, individually occluded, may cause an infarct that cuts across the functional anatomic fields of somatosensory projections, causing clusters of symptoms and signs from lesions that are at variance with the normal organization. Nature of the Sensory Complaints The patients complain of striking alterations in spontaneous sensations. 136,145 The parts feel stretched, hot, and sunburned, as if being stuck by pins, larger, smaller, or heavier. Contacts with the skin from eyeglasses, bedclothes, rings, watches, and sheets feel heavier on the affected side and may transiently aggravate the sensory disturbance. The stimulus seems to persist for a few seconds after its removal. In the patients with severe disturbances, the occurrence of a stimulus is better reported than its exact location. In a series of 21 cases, impairment of all sensory types (touch, pinprick, vibration, and position sense) was usually associated with large lacunes in the lateral thalamus; restricted sensory complaints suggest small lacunes at any level of the sensory pathway. 146 The Dejerine-Roussy syndrome 147 is an uncommon accompaniment of lacunar infarction of the thalamus, although dysesthetic accompaniments are common in pure sensory stroke, as described previously. The full Dejerine-Roussy syndrome was originally described as the effect of occlusion of the thalamogeniculate branch of the posterior cerebral artery, with infarction of the ventral posterolateral and ventral posteromedial nuclei, largely sparing the remaining nuclei of the thalamus. Cases documented only by CT have shown a lesion small enough to qualify for a clinical diagnosis of lacunar infarction. 148 The initial deficit usually consists of hemiparesis and a hemisensory syndrome. The pain, which is an inconstant feature in cases with such infarcts, may begin at the onset of the syndrome or appear only later; delays of up to several months are common. The pains are intermittent or constant, appear spontaneously, or at other times are provoked by contact with the affected parts. 1 ..." Reference 17 Book Chapter Central Pain and Complex Regional Pain Syndromes Richard Wilson, Preeti Raghavan, Andrew K. Treister, Eric Y. Chang Stroke Rehabilitation , 2019 pp 105-114 View PDFView chapter Related quote(s)1 / 1 "... The spinothalamic tract The most studied tract associated with pain is the spinothalamic tract, which transmits the modalities of pain, temperature, and deep touch from the body. The spinothalamic tract courses from the lateral portion of the spinal cord, through the lateral medulla and pons, to the ventral posterolateral nucleus (VPL) of the thalamus, terminating in the post-central gyrus ( Fig. 7.1A ). Lesions or injury to any part of this tract can potentially result in CPSP; however, some structures are more highly associated with this syndrome than others. CPSP was originally described as a thalamic pain syndrome, and the thalamus continues to be the most commonly documented and studied neural structure associated with CPSP. 5,9 Modern studies have shown that specific areas within the thalamus are more correlated to the development of CPSP than others. Studies have shown that CPSP patients have lesions within the VPL and/or the ventral posteromedial (VPM) of the thalamus ( Fig. 7.1B ). 10–14 A more recent study using MRI and digital radiographic atlases in thalamic stroke patients with and without CPSP found that the CPSP group had lesions largely involving the VPL, with some also involving the VPM nucleus. 15,16 Specifically, lesions in the posterolateral and inferior parts of the VPL were most associated with CPSP. A few of the CPSP patients in this study did have lesions confined to the pulvinar nucleus as well, an area that processes visual input. The development of CPSP in these patients was thought to be due to the shared vascular supply and close proximity to the VPL 15 ( Fig. 7.1B ); again implying the strong association of the VPL and CPSP. Indeed, another study found that thalamic lesions involving the area where the ventral posterior nuclei and the pulvinar meet were 81 times more likely to lead to CPSP than other thalamic lesions, confirming this area as high risk for CPSP, and opening the door for potential preemptive treatments against CPSP as a future avenue of research. 16 1 ..." Reference 18 Book Chapter Vertebrobasilar Disease J.P. Mohr, Louis R. Caplan Stroke , 2004 pp 207-274 View PDFView chapter Related quote(s)1 / 1 "... 387 The lesion responsible for ataxic hemiparesis is usually in the more rostral pons, interrupting crossing fibers as well as the pyramidal fibers in the pontine base. We have also seen a patient with this syndrome whose postmortem lesion involved the brachium conjunctivum and cerebral peduncle at the midbrain level. Slight ataxia of the other (normal) leg has been a clue to the pontine location of this syndrome. At times, horizontal or vertical nystagmus and dysarthria accompany the limb ataxia and hemiparesis. 137 Pure Sensory Stroke. A lacune in the somatosensory nuclei of the thalamus and ventral posterolateral and ventral posteromedial nuclei produces sensory symptoms on the opposite side of the body and face without motor, cerebellar, or higher cortical function abnormalities. 375 388 389 390 Paresthesias are usually characterized as tingling, prickling, or a sleepy, cold, hard, numb, or dead feeling. Usually at least two parts, such as face and arm or arm and leg, are involved. 388 The limbs are involved more than the face, but often the whole hemicorpus is affected, including the abdomen, chest, and face (as well as the eye, ear, and inside of the mouth). Cortical representation in the postcentral gyrus for the ear, eye, and trunk is quite small. Involvement of these structures usually signifies a lesion in a tract or a thalamic nucleus rather than a parietal cortical lesion. When the hand is affected, usually all digits are affected. The symptoms of sensory dysfunction commonly far exceed the objective signs; in fact, objective parameters of sensory function may be completely normal. In one reported patient with a right lateral thalamic lacune, the sensory loss consisted only of proprioceptive loss, the temperature and pain sensations both being intact. 391 Helgason and Wilbur 392 described 10 patients with pontine infarcts identified on MRI in whom sensory symptoms and signs were the most prominent clinical findings. 1 ..." Reference 19 Review article Murine models of human neuropathic pain Colleoni M., Sacerdote P. Biochimica et Biophysica Acta (BBA) - Molecular Basis of Disease , 2010 pp 924-933 View PDFView article Related quote(s)1 / 1 "... Thalamic syndrome is a form of central pain that typically results from stroke in the thalamus and is characterized by spontaneous pain, attacks of allodynia, and dysesthesia. The lack of suitable thalamic syndrome models has hampered research into the dysregulated perception of pain resulting from stroke in the thalamus. Therefore, the development and characterization of a rodent model of thalamic syndrome was the first step to discover the underlying mechanisms of this disease and possible therapeutics. Very recently a rat model of this syndrome was developed based on a small hemorrhagic stroke lesion induced by collagenase injection in the ventral posterolateral nucleus of the rat thalamus . Animals displayed hyperesthesia in response to mechanical pinch stimulation, with sensitivity localized in the hind limb and increased thermal sensitivity. This novel model has not been developed in mice yet. 1 ..." Reference 20 Book Chapter Deep Brain Stimulation for Pain Senatus P., Condit D. Schmidek and Sweet Operative Neurosurgical Techniques , 2012 pp 1433-1442 View PDFView chapter Related quote(s)1 / 1 "... Central neuropathic pain originates from damage to the central nervous system, that is, the brain and spinal cord. Specific types of central pain include spinal cord injury, poststroke pain, postherpetic neuralgia, and other forms of neuralgia. 29 The ST has been the principal target region for DBS treatment of neuropathic pain. The thalamus, a large, paired mass of gray matter, resides in the diencephalon. It is located in the center of the brain, along the midline, and sits right above the brain stem. The thalamus’s functions include motor control and transmission of sensory stimuli. 30 Stimulation of the lateral nuclei, specifically the ventral posterolateral (VPL) nucleus and the ventral posteromedial (VPM) nucleus, has been shown to reduce pain symptoms of neuropathic origin. CH is a severe primary neurovascular headache disorder causing 10% to 20% of patients with the disorder to experience debilitating headaches. 31 Pain arising from this syndrome is indicated by attacks of intense unilateral periorbital pain, often jointly experienced with ipsilateral cranial autonomic disturbance. Through utilization of PET studies, increased blood flow to the posterior hypothalamic region was observed in patients amid an acute CH attack. The hypothalamus, forming the ventral aspect of the diencephalon and located beneath the thalamus, is linked with endocrine and pituitary function. 30 DBS of the posterior hypothalamic region was introduced to treat pain arising from acute CH attacks. 23 1 ..." Reference 21 Review article A Review of Medical and Surgical Options for the Treatment of Facial Pain Penn M.C., Choi W., Brasfield K., Wu K., Briggs R.G., Dallapiazza R., Russin J.J., Giannotta S.L., Lee D.J. Otolaryngologic Clinics of North America , 2022 pp 607-632 View PDFView article Related quote(s)1 / 2 "... Deep Brain Stimulation DBS is an approved treatment option for a variety of neurologic and psychiatric diseases, including essential tremor, Parkinson disease, idiopathic dystonia, and obsessive-compulsive disorder. 116 , 117 It has been explored since the 1970s and 1980s for its potential to treat chronic pain syndromes but has yet to be Food and Drug Administration approved. Evidence for DBS in the treatment of specific pain syndromes and disorders has been mixed, and after the failure of clinical trials for DBS for pain in 2001, 118 less attention has been given to the study of DBS to manage and treat chronic facial pain. 119 Common DBS targets for chronic pain syndromes include the periaqueductal gray (PAG) region and the ventral posterolateral nucleus of the thalamus (VPL) ( Fig. 6 ). 116 , 120 The exact process by which pain relief occurs from DBS targeting of these regions is not known, but some theories suggest it may lead to the release of endogenous opioids or to modulation of ascending pain pathways and neural circuitry. 119 DBS of these regions has demonstrated functional changes in regional cerebral blood flow, which may lead to long-term changes in pain perception. 121 As DBS is designed to target neural pathways centrally, its use seems especially relevant for patients with demyelinating pain syndromes refractory to medical management. 117 Despite the failure of previous trials, recent evidence for DBS in the treatment of chronic facial pain syndromes seems more promising. A prospective study of 56 patients with chronic pain syndromes, including six with trigeminal neuropathic pain, treated with DBS to the PAG and lateral somatosensory thalamus demonstrated mixed results across different syndromes, with the most success seen in patients with chronic low back or leg pain. However, 50% of patients in this study with trigeminal neuropathic pain and/or dysesthesia dolorosa achieved at least a 50% reduction in their pain. 120 A recent trial of 15 patients with poststroke neuropathic pain who underwent DBS of the VPL thalamus and PAG achieved an overall reduction of 48.8% in 80% of patients. 1 ..." Related quote(s)2 / 2 "... Evidence for DBS in the treatment of specific pain syndromes and disorders has been mixed, and after the failure of clinical trials for DBS for pain in 2001, 118 less attention has been given to the study of DBS to manage and treat chronic facial pain. 119 Common DBS targets for chronic pain syndromes include the periaqueductal gray (PAG) region and the ventral posterolateral nucleus of the thalamus (VPL) ( Fig. 6 ). 116 , 120 The exact process by which pain relief occurs from DBS targeting of these regions is not known, but some theories suggest it may lead to the release of endogenous opioids or to modulation of ascending pain pathways and neural circuitry. 119 DBS of these regions has demonstrated functional changes in regional cerebral blood flow, which may lead to long-term changes in pain perception. 121 As DBS is designed to target neural pathways centrally, its use seems especially relevant for patients with demyelinating pain syndromes refractory to medical management. 117 Despite the failure of previous trials, recent evidence for DBS in the treatment of chronic facial pain syndromes seems more promising. A prospective study of 56 patients with chronic pain syndromes, including six with trigeminal neuropathic pain, treated with DBS to the PAG and lateral somatosensory thalamus demonstrated mixed results across different syndromes, with the most success seen in patients with chronic low back or leg pain. However, 50% of patients in this study with trigeminal neuropathic pain and/or dysesthesia dolorosa achieved at least a 50% reduction in their pain. 120 A recent trial of 15 patients with poststroke neuropathic pain who underwent DBS of the VPL thalamus and PAG achieved an overall reduction of 48.8% in 80% of patients. 122 For those who continued to require analgesic medications after DBS therapy, all were able to switch to nonopiate pain medications. Similarly, a small case series of 7 patients with unilateral, intractable facial pain found pain scores were significantly decreased at 1-year follow-up after DBS implantation. 1 ..." Reference 22 Review article Advances and Future Directions of Neuromodulation in Neurologic Disorders Burns M.R., Chiu S.Y., Patel B., Mitropanopoulos S.G., Wong J.K., Ramirez-Zamora A. Neurologic Clinics , 2021 pp 71-85 View PDFView article Related quote(s)1 / 1 "... Pain DBS has been applied to a variety of pain syndromes ranging from poststroke pain, spinal cord injury, brachial plexus injury, and headache. 78 There are 3 commonly targeted structures for neuromodulation of pain: (1) the thalamus, specifically the ventral posterolateral nucleus and ventral posteromedial nucleus (VPL/VPM), (2) the periventricular and periaqueductal gray (PVG/PAG), and (3) the anterior cingulate cortex (ACC). Neuromodulation of the VPL/VPM and PVG/PAG have been the most well established in the literature. 79–81 One longitudinal study of VPL/VPM and PVG/PAG DBS for various pain conditions revealed 59% of patients experienced significant acute pain relief. 81 After approximately 80 months of follow-up, 31% of patients continued to experience pain relief. In another study with up to 15 years of follow-up after VPL/VPM or PVG/PAG DBS for various pain syndromes, 62% of patients continued to experience adequate pain relief. 82 A meta-analysis of DBS for pain reported that modulation of the PVG/PAG alone or PVG/PAG with VPL/VPM or internal capsule was more effective than VPL/VPM alone. In addition, overall 58% of patients achieved pain relief. DBS was most effective in treating intractable low back pain and least successful in treating central thalamic pain/poststroke pain. Although central thalamic pain syndrome has historically been difficult to treat, Franzini and colleagues 83 recently investigated DBS of the posterior limb of the internal capsule in 4 patients. Three of four patients achieved long-term pain relief post-DBS. After a mean follow-up of 5.88 years, the average reduction in pain was 38% based on the 10-point visual analog scale. Inspired by lesional therapies for cancer-related pain, ACC DBS has emerged as the newest potential target for pain. 84 , 85 Spooner and colleagues 86 published the first case report of ACC DBS in 2007 in a patient with medically refractory neuropathic pain from a complete C4 spinal cord injury. 1 ..." Reference 23 Book Chapter Discussion Saab C.Y. Chronic Pain and Brain Abnormalities , 2014 pp 127-141 View PDFView chapter Related quote(s)1 / 2 "... Neuromodulation by Electricity Deep brain stimulation (DBS) is considered primarily a neurosurgical intervention intended to manage severe intractable pain. Although the safety and efficacy of DBS were established firmly for movement disorders, DBS for pain remains “off-label” and its mechanisms remain largely unknown. In general, the primary disadvantage of using electrodes is the requirement for surgical procedures, with the potential to trigger inflammation, cell death and gliosis cascades. 19 The main targets of DBS for pain include thalamic and brain stem structures. Current indications include pain secondary to stroke, amputation (phantom pain), failed back surgeries and cancer, thus mainly of neuropathic origin. Interestingly, the frequencies used during DBS for pain range typically from 50–70 Hz (see discussion below regarding low- versus high-frequency stimulation). Today, conventional wisdom equates “electrical stimulation” with “neuronal excitation,” but it is becoming increasingly clear that different electrical stimulation parameters may result in gain or loss of function in a neuronal population exposed to a stimulating probe. For example, though the mechanisms of DBS remain speculative, 20 studies related to DBS for motor disorders suggest that high-frequency stimulation (HFS, >100 Hz) can mimic the functional effects of ablation, 21 also referred to as “jamming” of neuronal circuitry. 20 In animal models, two studies have recently tested the effects of deep brain stimulation (DBS) on sensitization of single-units in the brain, as well as on pain behavior. 22,23 Aiming to reverse neuronal sensitization in the sensory ventral posterolateral (VPL) nucleus of the thalamus, HFS (100–150 Hz) within the VPL was shown to effectively decrease neuronal hyperexcitability and thermal hyperalgesia in an animal model of peripheral pain, whereas stimulation at a low-frequency (LFS, 20–40 Hz) had no effect on neuronal firing. 22 Interestingly, LFS (50 Hz) in the VPL is known to produce little or no analgesia in rats. 1 ..." Related quote(s)2 / 2 "... Current indications include pain secondary to stroke, amputation (phantom pain), failed back surgeries and cancer, thus mainly of neuropathic origin. Interestingly, the frequencies used during DBS for pain range typically from 50–70 Hz (see discussion below regarding low- versus high-frequency stimulation). Today, conventional wisdom equates “electrical stimulation” with “neuronal excitation,” but it is becoming increasingly clear that different electrical stimulation parameters may result in gain or loss of function in a neuronal population exposed to a stimulating probe. For example, though the mechanisms of DBS remain speculative, 20 studies related to DBS for motor disorders suggest that high-frequency stimulation (HFS, >100 Hz) can mimic the functional effects of ablation, 21 also referred to as “jamming” of neuronal circuitry. 20 In animal models, two studies have recently tested the effects of deep brain stimulation (DBS) on sensitization of single-units in the brain, as well as on pain behavior. 22,23 Aiming to reverse neuronal sensitization in the sensory ventral posterolateral (VPL) nucleus of the thalamus, HFS (100–150 Hz) within the VPL was shown to effectively decrease neuronal hyperexcitability and thermal hyperalgesia in an animal model of peripheral pain, whereas stimulation at a low-frequency (LFS, 20–40 Hz) had no effect on neuronal firing. 22 Interestingly, LFS (50 Hz) in the VPL is known to produce little or no analgesia in rats. 24 In another animal model of central pain, LFS (50 Hz) within the motor cortex reduced mechanical allodynia and thermal hyperalgesia. 23 These effects were shown to be mimicked by DBS in the zona incerta of the subthalamus and blocked by reversible inactivation of this region (see Chapter 6 in this book by Keller and Masri). This suggests that the potential analgesic effects of motor cortex stimulation may be due to disinhibition of the zona incerta, thus, causing analgesia by restoring inhibition in the thalamus. 1 ..." Reference 24 Book Chapter Neurosurgical Approaches to Pain Management Raslan A.M., Burchiel K.J. Practical Management of Pain , 2014 pp 328-334.e2 View PDFView chapter Related quote(s)1 / 1 "... The use of DBS for pain control has failed to gain much acceptance in the neurosurgical community, and the use of DBS electrodes as pain control implants has never achieved U.S. Federal Drug Administration (FDA) approval. The lack of data to support the procedure is due, in part, to the small number of patients treated, inconsistent target localization, heterogeneity of the pain diagnoses treated, and failure to mount a prospective randomized trial that was sufficiently powered to answer the question of efficacy. The mechanism of pain relief by DBS is poorly understood but appears to be dependent on the site. The thalamus and PVG / PAG were the most commonly 65 targeted sites for DBS implants for pain. Hosobuchi and colleagues 58 suggested that the pain-relieving effect of PVG and PAG stimulation might involve endogenous opioid receptors based on their studies in which it was found that the pain-relieving effect of DBS could be reversed by naloxone. Evidence to support this mechanism of action of PAG/PVG DBS is inconsistent. Some investigators supported the concept whereas others disagreed. Currently, it is postulated that the pain-relieving effect of PAG / PVG DBS is due to activation of multiple supraspinal descending pain modulatory systems, both opioid and nonopioid. 66 Pain relief resulting from stimulation of the ventral posterolateral (VPL) nucleus and ventral posteromedial (VPM) nucleus (Vc nucleus in the European Hassler terminology), the major sensory nuclei of the thalamus, is poorly understood. Inhibition of spinothalamic tract neurons 67 and activation of dopaminergic mechanisms have both been proposed. 68 The most accepted hypothesis is that thalamic stimulation activates the nucleus raphe magnus of the rostroventral medulla, which results in activation of a suprasegmental descending endogenous pain inhibition system. 1 ..." Reference 25 Review article Surgical Options for Atypical Facial Pain Syndromes Rahimpour S., Lad S.P. Neurosurgery Clinics of North America , 2016 pp 365-370 View PDFView article Related quote(s)1 / 1 "... Since its inception in the middle of the twentieth century, deep brain stimulation of the thalamus and caudate has also been used in the treatment of chronic pain. 31 However, results of 2 multicenter trails were unfortunately inconclusive. 32 Several stimulation targets have been used, including PAG, zona incerta, and the ventral posterolateral nucleus of the thalamus. In a recent double-blinded study by Rasche and colleagues 33 of 56 patients with neuropathic pain, 6 patients had trigeminal neuropathic pain. Of these patients, half, at 30 months of mean follow-up, had an unsatisfactory response to the treatment. Deep brain stimulation, however, remains a poorly understood intervention for neuropathic pain; much remains to be learned about its role in the facial pain treatment algorithm. 1 ..." Reference 26 Review article Deep brain stimulation for chronic pain: Intracranial targets, clinical outcomes, and trial design considerations Keifer O.P., Riley J.P., Boulis N.M. Neurosurgery Clinics of North America , 2014 pp 671-692 View PDFView article Related quote(s)1 / 1 "... From there, DBS targets for pain control would expand to include the internal capsule (IC), the ventral posterolateral nucleus (VPLP) and the ventral posteromedial nucleus (VPM) of the sensory thalamus (STH), the centro-median parafasicular region (CM-Pf) of the thalamus, the periaqueductal/paraventricular gray (PAG/PVG), the posterior hypothalamus (PH), the motor cortex, the nucleus accumbens (NAcc), and the anterior cingulate cortex. The following sections highlight the past, present, and future DBS targets used to treat various types of pain. 2 ..." Related topics (10) 2,5-Dimethoxy-4-iodoamphetamine Brainstem Cerebellum Motor Cortex Neuropathic Pain Parvalbumin Somatosensory Cortex Spinothalamic Tract Thalamus Ventral Posteromedial Nucleus View other topics in Neuroscience Also appears in...(1) This topic can also be found in the following subject areas. Veterinary Science and Veterinary Medicine View all subject areas Recommended publications (4) Journal Neuroscience Journal NeuroImage Journal Brain Research Journal Neuroscience Letters Browse books and journals Hill, Sean L. University of Toronto, Toronto, Canada Publications: 26Cited by: 5121Citations: 6160h-index: 30 O’Reilly, Christian University of South Carolina, Columbia, United States Publications: 20Cited by: 1259Citations: 1699h-index: 20 Hirato, Masafumi National Hospital Organization Shibukawa Medical Center, Shibukawa, Japan Publications: 11Cited by: 1081Citations: 1205h-index: 19 Zhao, Hongru The First Affiliated Hospital of Soochow University, Suzhou, China Publications: 9Cited by: 608Citations: 650h-index: 16
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2321	https://onlinelibrary.wiley.com/doi/10.1155/2014/851213	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 Volume 2014, Issue 1 851213 Research Article Open Access Complete Monotonicity of Functions Connected with the Exponential Function and Derivatives Chun-Fu Wei, Corresponding Author Chun-Fu Wei mathwcf@163.com State Key Laboratory Cultivation Base for Gas Geology and Gas Control, Henan Polytechnic University, Jiaozuo, Henan 454010, China hpu.edu.cn School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, Henan 454010, China hpu.edu.cn Search for more papers by this author Bai-Ni Guo, Bai-Ni Guo orcid.org/0000-0001-6156-2590 School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, Henan 454010, China hpu.edu.cn Search for more papers by this author Chun-Fu Wei, Corresponding Author Chun-Fu Wei mathwcf@163.com State Key Laboratory Cultivation Base for Gas Geology and Gas Control, Henan Polytechnic University, Jiaozuo, Henan 454010, China hpu.edu.cn School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, Henan 454010, China hpu.edu.cn Search for more papers by this author Bai-Ni Guo, Bai-Ni Guo orcid.org/0000-0001-6156-2590 School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo, Henan 454010, China hpu.edu.cn Search for more papers by this author First published: 15 April 2014 Citations: 5 Academic Editor: Beong In Yun This article is part of Special Issue: Abstract Some complete monotonicity results that the functions ±1/(e±t − 1) are logarithmically completely monotonic, and that differences between consecutive derivatives of these two functions are completely monotonic, and that the ratios between consecutive derivatives of these two functions are decreasing on (0, ∞) are discovered. As applications of these newly discovered results, some complete monotonicity results concerning the polylogarithm are found. Finally a conjecture on the complete monotonicity of the above-mentioned ratios is posed. 1. Introduction and Main Results Throughout this paper, we denote the set of all positive integers by ℕ. Recently, the following problem was posed in [1, page 569]. For t ≠ 0 and k ∈ ℕ, determine the numbers ak,i−1 for 1 ≤ i ≤ k such that (1) This problem, among other things, was answered in by eight identities. Two of the eight identities may be recited as follows. Theorem A (see , Theorem 2.1.)For i ∈ {0} ∪ ℕ, one has (2) where (3) are Stirling numbers of the second kind. Theorem B (see , Theorem 2.3.)For i, k ∈ ℕ with 1 ≤ i ≤ k, the coefficients ak,i−1 defined in (1) may be calculated by (4) where (5) For more information, please refer to [2, 3] and related references therein. Stimulated by results obtained in , as mentioned above, we are interested in two functional sequences: (6) where t ∈ (0, ∞) and i ∈ {0} ∪ ℕ, and we firstly discover in this paper the following results. Theorem 1. For i ∈ {0} ∪ ℕ, the functions Fi(t) and Gi(t) are completely monotonic on (0, ∞). More strongly, the functions F0(t) and G0(t) are logarithmically completely monotonic on (0, ∞). For i, k ∈ {0} ∪ ℕ, the functions Fi(t) and Gi(t) satisfy (7) Theorem 2. For given i ∈ {0} ∪ ℕ, the differences (8) are completely monotonic functions on (0, ∞). In particular, the sequences Fi(t) and Gi(t) are increasing with respect to i for t ∈ (0, ∞); that is, the inequalities (9) are valid for all i ∈ {0} ∪ ℕ and t ∈ (0, ∞). Theorem 3. For given i ∈ {0} ∪ ℕ, the ratios (10) are decreasing on (0, ∞), with (11) In order to further show the importance and significance of the two functional sequences in (6), we secondly give several applications of Theorems 1 to 3 in Section 4. Finally, we pose a conjecture on the complete monotonicity of the functions ℱi(t) and 𝒢i(t) defined in (10). 2. Two Definitions and a Lemma Now we list definitions of the completely monotonic and the logarithmically completely monotonic functions, which just now appeared in Theorems 1 and 2, and recite a lemma, which is needed to prove Theorem 3. Definition 4 (see , .)A function q(x) is said to be completely monotonic on an interval I if q(x) has derivatives of all orders on I and (−1) nq(n)(x) ≥ 0 for x ∈ I and n ≥ 0. The noted Hausdorff-Bernstein-Widder theorem [5, page 161, Theorem 12b] says that a necessary and sufficient condition that f(x) should be completely monotonic for 0 < x < ∞ is that (12) where α(t) is nondecreasing and the integral converges for 0 < x < ∞. In other words, a function f defined on (0, ∞) is completely monotonic on (0, ∞) if and only if it is a Laplace transform. For more information on the theory of completely monotonic functions, please refer to [4, Chapter XIII], [5, Chapter IV], and the newly published monograph . This means that it is useful to confirm the complete monotonicity of functions. Definition 5 (see , .)A positive function f(x) is said to be logarithmically completely monotonic on an interval I⊆ℝ if it has derivatives of all orders on I and its logarithm ln⁡f(x) satisfies (−1) k[ln⁡f(x)] (k) ≥ 0 for k ∈ ℕ on I. It has been proved that any logarithmically completely monotonic function on I is also completely monotonic on I, but not conversely. For more information on this class of functions, please refer to [7–9] and [10, Section 1.3] and closely related references therein. This shows that it is helpful to prove the logarithmically complete monotonicity of functions. Lemma 6 (see , Lemma 2.2.)Suppose that ak, bk > 0 and that {uk(t)} is a sequence of positive and differentiable functions such that the series (13) converge absolutely and uniformly over compact subsets of [0, ∞). (1) If the logarithmic derivative forms an increasing sequence of functions and if ak/bk decreases (resp., increases), then (14) decreases (resp., increases) for t ≥ 0. (2) If the logarithmic derivative forms a decreasing sequence of functions and if ak/bk decreases (resp., increases), then the function f(t) increases (resp., decreases) for t ≥ 0. 3. Proofs of Main Results Now we start out to prove our theorems. Proof of Theorem 1. Since (15) for i ≥ 1, the functions Fi(t) and Gi(t) for i ∈ {0} ∪ ℕ are completely monotonic on (0, ∞). Taking the logarithms of the functions F0(t) and G0(t) and differentiating yield (16) Consequently, by definition of logarithmically completely monotonic functions and the above obtained complete monotonicity of the function F0(t), it is ready to deduce the logarithmically complete monotonicity of F0(t) and G0(t) on (0, ∞). The formulas in (7) can be straightforwardly verified. The proof of Theorem 1 is complete. Proof of Theorem 2. A simple computation yields (17) for n ∈ ℕ. By definition, the difference 𝔉i(t) = Fi+1(t) − Fi(t) for i ∈ {0} ∪ ℕ is completely monotonic on (0, ∞). From the first equalities in (15), respectively, it follows that the functions (18) are all completely monotonic on (0, ∞). The inequalities in (9) follow from the positivity of 𝔉i(t) and 𝔊i(t) for i ∈ ℕ. The proof of Theorem 2 is complete. Proof of Theorem 3. It is easy to see that (19) Let ak = ki+1, bk = ki, and uk(t) = e−kt. Then ak/bk = k is an increasing sequence and form a decreasing sequence. By Lemma 6, it follows that the function ℱi(t) is decreasing on (0, ∞). Taking t → ∞ at the very ends of (19) shows that (20) The first limit in (11) thus follows. Making use of the second relation in (15) yields 𝒢i(t) = ℱi(t) for i ∈ ℕ. Accordingly, the function 𝒢i(t) has the same monotonicity and the same limit for t → ∞ as ℱi(t) does for i ∈ ℕ. As a result, the function 𝒢i(t) is decreasing on (0, ∞) and the third limit in (11) is valid for i ∈ ℕ. A straightforward computation gives (21) which obviously tends to 0 as t → ∞ and apparently decreases on (0, ∞). The proof of Theorem 3 is complete. 4. Some Applications We recall from that the polylogarithm Lis(z) is the function (22) defined for all s ∈ ℂ and over the open unit disk \|z \| < 1 in the complex plane ℂ. Its definition on the whole complex plane then follows uniquely via analytic continuation. For given i ∈ {0} ∪ ℕ, since , we observe that (23) Therefore, considering (15), Theorems 1 to 3 can be used to find some properties of the polylogarithm Lis(z) as follows. Theorem 7. For i ∈ {0} ∪ ℕ, the polylogarithm Li−i(e−t) is completely monotonic with respect to t ∈ (0, ∞). More strongly, the polylogarithm Li0(e−t) is logarithmically completely monotonic with respect to t ∈ (0, ∞). For i, k ∈ {0} ∪ ℕ, the polylogarithm Li−i(e−t) satisfies (24) Theorem 8. For given i ∈ {0} ∪ ℕ, the differences (25) are completely monotonic functions on (0, ∞). In particular, the polylogarithm sequence Li−i(e−t) is increasing with respect to i for given t ∈ (0, ∞); that is, the inequality (26) is valid for all i ∈ {0} ∪ ℕ and t ∈ (0, ∞). Theorem 9. For given i ∈ {0} ∪ ℕ, the ratios (27) are decreasing on (0, ∞), with (28) Furthermore, by the above (logarithmically) complete monotonicity in Theorems 7 and 8 and by some complete monotonicity properties of composite functions, we can obtain the complete monotonicity of functions involving the polylogarithm Li−i(1/t) for i ∈ {0} ∪ ℕ as follows. Theorem 10. The following complete monotonicity is valid. (1) The polylogarithm Li0(1/t) = 1/(t − 1) is logarithmically completely monotonic with respect to t ∈ (1, ∞). (2) For i ∈ {0} ∪ ℕ, the polylogarithm Li−i(1/t) is completely monotonic with respect to t ∈ (1, ∞). (3) For given i ∈ {0} ∪ ℕ, the differences (29) are completely monotonic functions on (1, ∞). Proof. The second item of [13, Theorem 5] tells us that if h′(x) is completely monotonic on an interval I and f(x) is logarithmically completely monotonic on the domain h(I), then the composite function f∘h(x) = f(h(x)) is logarithmically completely monotonic on I. It is easy to verify that the derivative of ln⁡t is 1/t and completely monotonic on (0, ∞). Combining these conclusions with the logarithmically complete monotonicity of Li0(e−t) in Theorem 7 yields that the polylogarithm Li0(1/t) = 1/(t − 1) is logarithmically completely monotonic with respect to t ∈ (1, ∞). In [14, page 83], it was given that if f and g are functions such that f(g(x)) is defined on (0, ∞) and if f and g′ are completely monotonic, then x ↦ f(g(x)) is also completely monotonic on (0, ∞). Replacing f(t) by Li−i(e−t) and g(t) by ln⁡t and making use of the complete monotonicity of Li−i(e−t) in Theorem 7 lead to the complete monotonicity of the polylogarithm Li−i(1/t). The leftover proofs are similar to the above arguments. The proof of Theorem 10 is complete. Remark 11. To the best of our knowledge, the above (logarithmically) complete monotonicity results concerning the polylogarithm are new. This shows us the importance of the two functional sequences in (6) and the significance of Theorems 1 to 3. 5. A Conjecture By Theorem 1 and the fact that any logarithmically completely monotonic function on I is also completely monotonic on I, it is clear that the functions (30) are all completely monotonic on (0, ∞). This motivates us to pose the following conjecture. Conjecture 12. For given i ∈ {0} ∪ ℕ, the functions ℱi(t) and 𝒢i(t) defined in (10) or (27) are completely monotonic on (0, ∞). Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. Acknowledgments The authors are grateful to the anonymous referees for their valuable comments on and careful corrections to the original version of this paper. The authors appreciate Professor Dr. Feng Qi in China for his valuable contribution to this paper. Chun-Fu Wei was partially supported by the NNSF under Grant no. 51274086 of China, by the Ministry of Education Doctoral Foundation of China—Priority Areas under Grant no. 20124116130001, and by the State Key Laboratory Cultivation Base for Gas Geology and Gas Control under Grant no. WS2012A10 at Henan Polytechnic University, China. References 1 Guo B.-N. and Qi F., Some identities and an explicit formula for Bernoulli and Stirling numbers, Journal of Computational and Applied Mathematics. (2014) 255, 568–579, MR3093443. 10.1016/j.cam.2013.06.020 Web of Science® Google Scholar 2 Qi F., Explicit formulas for computing Euler polynomials in terms of the second kind Stirling numbers, Google Scholar 3 Xu A.-M. and Cen Z.-D., Some identities involving exponential functions and Stirling numbers and applications, Journal of Computational and Applied Mathematics. (2014) 260, 201–207, MR3133341. 10.1016/j.cam.2013.09.077 Web of Science® Google Scholar 4 Mitrinović D. S., Pečarić J. 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(2010) 47, no. 6, 1283–1297, MR2744772, ZBL1208.26017. 10.4134/JKMS.2010.47.6.1283 Web of Science® Google Scholar 14 Bochner S., Harmonic Analysis and the Theory of Probability, 1955, University of California Press, Berkeley, Calif, USA, California Monographs in Mathematical Sciences, MR0072370. 10.1525/9780520345294 Google Scholar Citing Literature All articles > ## References ## Related ## Information Recommended Resonance between the Representation Function and Exponential Functions over Arithemetic Progression Li Ma, Xiaofei Yan, Journal of Mathematics Exponential sums and rational points on complete intersections Igor E. Shparlinskiĭ, Alexei N. Skorobogatov, Mathematika Monotonicity of Harnack Inequality for Positive Invariant Harmonic Functions Yifei Pan, Mei Wang, International Journal of Stochastic Analysis Complete function spaces R. A. Mccoy, International Journal of Mathematics and Mathematical Sciences Derivatives of Computable Functions Ning Zhong, Mathematical Logic Quarterly Metrics 5 Details Copyright © 2014 Chun-Fu Wei and Bai-Ni Guo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Check for updates Publication History 15 April 2014 15 April 2014 20 March 2014 25 January 2014 Close Figure Viewer Previous FigureNext Figure Caption Download PDF Log in to Wiley Online Library NEW USER > INSTITUTIONAL LOGIN > Change Password The full text of this article hosted at iucr.org is unavailable due to technical difficulties. Mentioned in 3 Q&A threads 3 readers on Mendeley See more details
2322	https://math.stackexchange.com/questions/4034300/does-the-order-of-the-fibonacci-sequences-initial-values-matter	Skip to main content Does the order of the Fibonacci sequence's initial values matter? Ask Question Asked Modified 3 years, 11 months ago Viewed 2k times This question shows research effort; it is useful and clear 14 Save this question. Show activity on this post. I am reading about generating functions in this reputable engineering textbook and the author uses the Fibonacci sequence as an example: The Fibonacci sequence is defined by the initial conditions a0=0, a−1=1, and the recursion relation an=an−1+an−2 for n>0. Using this definition, I am able to work through the author's example to get the correct generating function: G(x)=11−x−x2 However, everywhere I read online states the initial conditions of the Fibonacci sequence as (0,1) and not (1,0). So I tried that instead, but I get a different result: G(x)=1+x1−x−x2 This generating function seems to produce the Fibonacci sequence, but starting from the second number instead of the first (1,2,3,5,8,13,...). Where did I go wrong? I would have thought that the sequence should be identical in both cases. Many thanks for any help! My calculations: The author starts with a general expression for a generating function for a sequence {an}: G(x)=∑n=0∞anxn He then looks at the special case where {an} is a linear recurrence of range r: an=∑i=1rcian−i and proves that this leads to: G(x)=g(x)f(x)=∑ri=1cixi(a−ix−i+⋯+a−1x−1)1−∑ri=1cixi So, for the Fibonacci sequence, I guess we have r=2 and c1=c2=1. So we get: g(x)=a−1+a−2+a−1x Now I'm a bit confused because the subscripts don't seem to match up. In all the author's discussion, the "initial conditions" have strictly negative subscripts (a−i for i=1,2,…,r). But just for the Fibonacci example, we have initial conditions a0=0 and a−1=1. So I assume these are just off by 1. Therefore, substituting a−1=0 and a−2=1 gives the author's result above and using a−2=0 and a−1=1 gives my result above. sequences-and-series generating-functions fibonacci-numbers Share CC BY-SA 4.0 Follow this question to receive notifications asked Feb 21, 2021 at 15:24 HarryHarry 54122 silver badges1313 bronze badges 6 18 Using negative subscripts in an introduction to generating functions is just confusing and evil. orlp – orlp 2021-02-21 15:30:39 +00:00 Commented Feb 21, 2021 at 15:30 6 Everyone define it as starting F0=0 and F1=1 with Fn+2=Fn+1+Fn for all n∈Z, so this has the effect of making F−1=1 too. There is no difference in your textbook's Fibonacci sequence and the rest of the world's. But definitely using negative subscripts in an introduction is evil. user10354138 – user10354138 2021-02-21 15:47:35 +00:00 Commented Feb 21, 2021 at 15:47 2 @user10354138 I think that's it. If we work backwards into negative Fibonacci numbers, we get: …,−8,5,−3,2,−1,1,0,1,1,2,3,5,8,…. If we start with initial condition (1,0), then the sequence proceeds with 1,1,2,3... and if we start with (0,1), then the sequence proceeds with 1,2,3,5,.... If I understand correctly, the usual convention (0,1) counts one of the initial values as the first value in the sequence. But in this textbook, the sequence starts after the initial values. Harry – Harry 2021-02-21 16:08:00 +00:00 Commented Feb 21, 2021 at 16:08 3 The modern index convention makes the Binet formula simplest (not to mention a whole slew of other formulae). A simple way of remembering this convention is that F5=5 and F12=122. PM 2Ring – PM 2Ring 2021-02-21 22:52:25 +00:00 Commented Feb 21, 2021 at 22:52 1 @orlp confusing and evil tinkering like that often is based on some less noticed property. In this case it surely provides some additional insight for those willing to dig :) Džuris – Džuris 2021-02-22 07:30:45 +00:00 Commented Feb 22, 2021 at 7:30 \| Show 1 more comment 4 Answers 4 Reset to default This answer is useful 15 Save this answer. Show activity on this post. In the particular case of the Fibonacci number sequence OEIS A000045 (or series) there is some difference of opinion as amply evidenced by the Wikipedia article and OEIS entry. The most common convention is that F0=0 and F1=1 and this choice has much in its favor. This choice implies that its generating function is ∑n=0∞Fnxn=x1−x−x2. Notice the x in the numerator. For this and a few other reasons, some prefer the choice F0=F1=1 instead, which implies that the generating function is now ∑n=0∞Fnxn=11−x−x2. Both choices are valid and useful in some cases and not as useful in other cases. Thus, it is a good idea to be clear on which choice is being used in any given context since the choice will usually matter. The situation with negative index Fibonacci sequence elements is that the recurrence relation for the sequence can be used to uniquely extend the sequence in the negative index direction. For the common convention this implies that F−n=(−1)n−1Fn for all integer n. The result for the other convention it is that F−n=(−1)nFn−2 for all integer n. What if we choose arbitrary starting values with the same recurrence relation? That is, if we choose F0=a and F1=b, what do we get? The recurrence relation is linear and this implies that the sequence of numbers we get is a linear combination of the two initial value choice sequences. The generating function now is (a)+(b)x+(a+b)x2+(a+2b)x3+⋯=a+(b−a)x1−x−x2. Note that initial values of Fibonacci and related sequences can be given for any two consecutive index values--not just for a0 and a1 since the recursion going forward and backward can reach any other integer index value. Again, this is just a matter of convenience and convention. In the case you cite, specifying a−1 and a0 is fine. The initial values a−1=1,a0=0 gives the same sequence as a0=0,a1=1. This special case may be misleading though. The initial values a−1=u,a0=v gives the same sequence as a0=v,a1=u+v and u+v is equal to u if and only if v=0. Note that the Wikipedia article Generalizations of Fibonacci numbers mentions some of the issues such as extension to negative integers. It mentions a vector space of number sequences: Since the set of sequences satisfying the relation S(n)=S(n−1)+S(n−2) is closed under termwise addition and under termwise multiplication by a constant, it can be viewed as a vector space. Any such sequence is uniquely determined by a choice of two elements, so the vector space is two-dimensional. If we abbreviate such a sequence as (S(0),S(1)), the Fibonacci sequence F(n)=(0,1) and the shifted Fibonacci sequence F(n−1)=(1,0) are seen to form a canonical basis for this space, yielding the identity: S(n)=S(0)F(n−1)+S(1)F(n) for all such sequences S. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Sep 13, 2021 at 13:29 answered Feb 21, 2021 at 18:25 SomosSomos 37.5k33 gold badges3535 silver badges8585 bronze badges 5 I had noticed the version with the x in the numerator while googling. I think I see now what the difference is. When you say both choices are valid, is it not the case that many other choices (perhaps infinitely many) would also be valid (albeit probably less convenient)? For example, if I say F0=5 and F1=8 then I get g(x)=13+8x and we just start from a different place. (Although unless I knew I was looking for the Fibonacci numbers, I'm not sure how I would justify choosing 5 and 8). Harry – Harry 2021-02-21 19:33:47 +00:00 Commented Feb 21, 2021 at 19:33 @Harry According to mathworld, F1=F2=1 is a convention. The formulas both in mathworld and wikipedia have to be modified unnecessarily (although it is just a shift) , if we use the other assignment. I see no reason to do that , even if some combinatorical formula might become more "elegant" with the other assignment. Peter – Peter 2021-02-21 20:10:43 +00:00 Commented Feb 21, 2021 at 20:10 @Peter That's interesting. According to wikipedia, the other convention was used in "some older books". I'm not sure what qualifies as old, but this book was written in 1967, which makes it rather ancient in an electronic engineering context. I'm not sure if I understand the whole debate. If you told me the first two values were F77=21 and F78=34, then I could probably live with that. So I'll just make sure I update my own notes to follow the accepted convention. Harry – Harry 2021-02-21 20:40:06 +00:00 Commented Feb 21, 2021 at 20:40 "Both choices are valid and useful in some cases and not as useful in other cases." I disagree, and believe that F0=0 is the natural choice. (One huge reason: this choice enables the implication "if m divides n then Fm divides Fn".) But I invite an alternate view: can you describe a situation in which F1=1 is more convenient? Greg Martin – Greg Martin 2021-02-22 07:44:35 +00:00 Commented Feb 22, 2021 at 7:44 2 @GregMartin One example where F0=F1=1 is "nicer" is the combinatorial definition of Fn as the number of ways to tile a 1×n rectangle using squares and dominos. Carmeister – Carmeister 2021-02-22 15:23:37 +00:00 Commented Feb 22, 2021 at 15:23 Add a comment \| This answer is useful 6 Save this answer. Show activity on this post. The recursion an=an−1+an−2 is "encoded" in the denominator of the generating function: 1−x−x2. The numerator is determined by the values of a0 and a1. In particular, (1−x−x2)∑∞denominator∑n=0∞akxkgeneratingfunction=a0+(a1−a0)x∑∞numerator+∑n=2∞(an−an−1−an−2)∑∞0xn Thus, the generating function is a0+(a1−a0)x1−x−x2 I prefer the definition of the Fibonacci Numbers that starts with F0=0 and F1=1, whose generating function is x1−x−x2 because, among other things, we have n∣m⟹Fn∣Fm Furthermore, F−n=(−1)n−1Fn Lucas Numbers also obey the same recursion as Fibonacci Numbers, and their generating function is 2−x1−x−x2 Share CC BY-SA 4.0 Follow this answer to receive notifications edited Feb 22, 2021 at 0:48 answered Feb 21, 2021 at 23:16 robjohn♦robjohn 355k3838 gold badges498498 silver badges892892 bronze badges 2 3 I would consider the divisibility property a canonical property of the Fibonacci numbers and one of the most important reasons for the particular definition. Steven Stadnicki – Steven Stadnicki 2021-02-21 23:44:52 +00:00 Commented Feb 21, 2021 at 23:44 That ⟹ is even an ⟺; even better, gcd(Fn,Fm)=Fgcd(n,m)! Mees de Vries – Mees de Vries 2021-02-22 18:05:01 +00:00 Commented Feb 22, 2021 at 18:05 Add a comment \| This answer is useful 5 Save this answer. Show activity on this post. There is already a nice description of generating functions in Somos answer. I will focus here on the question you had about the author's general formula. G(x)=g(x)f(x)=∑ri=1cixi(a−ix−i+⋯+a−1x−1)1−∑ri=1cixi You are right in saying, 'for the Fibonacci sequence, I guess we have r=2 and c1=c2=1.' You are also correct in observing that you weren't given a−2 and the formula requires it. Also, technically, since the recursion given in the exercise only holds for n>0, a−2 is indeterminate in the exercise. So this is 'bad' notation on the author's part if you are expected to use this general theorem, in my opinion. So, to alleviate this I propose you consider an extension to the recursion, where the recursion holds for n≥0, (as it must whenever the theorem you have stated and it's associated formula should apply). In such a case, although you aren't given a−2 you can find its value from the recursion itself since: a0=a−1+a−2 So a−2=a0−a−1=−1. Plugging it all in: G(x)=x(x−1)+x2(−x−2+x−1)1−x−x2=x1−x−x2 So, it yields the expected generating function for the case where a0=0 and a1=1. So, I understand the confusion you had. However to answer the question as to what happened, when you shift the indices for the initial condition, it will result in shifting the indices throughout the sequence which emerges from the recursion and this will change the generating function. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Feb 21, 2021 at 22:21 open problemopen problem 1,46055 silver badges88 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Although one does not always make this explicit, giving a sequence of numbers really amounts to giving a set of (index,value) pairs; for instance an infinite sequence of real numbers is really a function N→R, whose graph is an infinite set of such pairs. The generating series for a set S of such pairs (i,ai) is ∑(i,ai)∈SaiXi, so for a series of numbers corresponding to f:N→R this is ∑∞i=0f(i)Xi. Let me first put aside the easy answer to the question in your title: of course order of initial terms matters, and permuting them in general gives an entirely different sequence. But that is not what your question was really about, which seems to be more about whether a shift in indexing produces a different generating series. Here too the answer is yes: a pure shift in indexing amounts for the generating series in multiplication by a power of X. If the shift is accompanied by addition or removal of nonzero values in the sequence, then this amounts to addition of terms to the generating series. Note that normally a generating series is a formal power series in X, with no negative powers of X at all. In this sense there is no generating series at all for a sequence with a nonzero value at some negative index, as your book apparently wants to do by starting at a−1=1. If you insist on including that, the generating series would have a term X−1, and thereby become a Laurent series rather than a formal power series. But usually those who entertain the idea of Fibonacci numbers at negative indices want them for all negative integer indices, but for such bi-infinite sequences the idea of generating series simply does not work: there is no appropriate algebraic framework. Think of it: if there were such a thing as S=∑i∈ZaiXi for a sequence (ai)i∈Z satisfying ai+2=ai+ai+1 everywhere, then one would want to express this by saying S(1−X−X2)=0, but S=0/(1−X−X2) in no way describes a generating series for the Fibonacci numbers (for one thing that expression gives no information about any "initial terms" that were used, for another it just simplifies to 0 in any reasonable algebraic setting). Share CC BY-SA 4.0 Follow this answer to receive notifications edited Feb 24, 2021 at 10:32 answered Feb 22, 2021 at 11:43 Marc van LeeuwenMarc van Leeuwen 120k88 gold badges182182 silver badges373373 bronze badges 4 @openproblem I don't understand what you want to say, or how it relates to my answer. Yes sequence satisfying a same linear recurrence form a vector space. No, there is no such thing as a "trivial" solution (except if you mean the zero solution, which I think you do not); every solution to a degree d recurrence needs d values to be completely determined. No, multiplying with the polynomial you call "annihilator" does not lose information, since K is an integral domain. But my point is: neither K nor K((X)) nor other nice rings properly represent bi-infinite sequences. Marc van Leeuwen – Marc van Leeuwen 2021-03-01 13:08:01 +00:00 Commented Mar 1, 2021 at 13:08 @openproblem I think you missed my main point. A formal Laurent series in K((X)) may have some negative powers of X, but only finitely many. This property is essential for being able to define a multiplication: without it coefficients of a product would be given by infinite sums that may fail to converge. This multiplication makes K((X)) into an integral domain (even a field), so multiplying by (1−X−X2) does not annihilate any nonzero formal Laurent series. So an expression like X−4/(1−X−X2) describes a FLS and a Fibonacci-like sequence with a few terms at negative indices. Marc van Leeuwen – Marc van Leeuwen 2021-03-03 20:14:32 +00:00 Commented Mar 3, 2021 at 20:14 ... However, a FLS cannot represent any bi-infinite sequence. If one imagines an algebraic structure that can represent bi-infinite sequences, then not only it cannot be an integral domain, it cannot be a ring at all. This makes it impossible to use an expression like E/(1−X−X2) to describe a Fibonacci-like bi-infinite series (without multiplication certainly division is meaningless), and the only candidate for the expression E would be 0. Whence "no appropriate structure" exists; some module over K[X] could model bi-infinite sequences, but does not allow expressions for them. Marc van Leeuwen – Marc van Leeuwen 2021-03-03 20:23:23 +00:00 Commented Mar 3, 2021 at 20:23 I think everything you have said in the comments here clarifies your answer a great deal, and so I will delete my comments to clean up the section since they don't serve much purpose anymore but to clutter this section, which I think is frowned upon and they suggest to move the comments to chat after a certain amount. Some of these clarifications that you put here will strengthen your answer if you work them into it, or leave them as comments as time allows. Thank you for your patience and explanations. open problem – open problem 2021-03-04 01:19:05 +00:00 Commented Mar 4, 2021 at 1:19 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 sequences-and-series generating-functions fibonacci-numbers See similar questions with these tags. 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2323	https://www.treccani.it/enciclopedia/coniche_(Enciclopedia-Italiana)/	Coniche - Enciclopedia - Treccani Salta al contenuto Vuoi navigare il sito Treccani senza limiti e pubblicità? Scarica l'APP e abbonati ora per accedere su tutti i tuoi dispositivi! Naviga istituto Fondazione Corporate Eventi Registrati Treccani Esperienze Linea Definizione Magazine Podcast Arte Cataloghi Lavora con noi Enciclopedia CONICHE CONICHE di Ugo AMALDI - Emilio ARTOM - Enciclopedia Italiana (1931) TAG Corrispondenza biunivocaEquazioni parametricheDuplicazione del cuboCoordinate ellitticheComplemento algebrico Altri risultati di ricerca Dal vocabolario Lemmi correlati CONICHE (gr. κωνικαί, κώνου τομαί; lat. conicae; fr. coniques; sp. cónicas; ted. Kegelschnitte; ingl. conic sections, conics) Ugo AMALDI Emilio ARTOM Si designano con questo nome comune tre specie di curve, ellisse, parabola e iperbole, di aspetto nettamente diverso (fig. 1). Mentre l'ellisse, che come caso particolare comprende il cerchio, è chiusa e tutta a distanza finita, le altre due sono aperte e si prolungano all'infinito, e per di più, l'iperbole è costituita da due parti o rami staccati. Tuttavia esse costituiscono una medesima famiglia, la cui unità, nell'insieme di tutte le possibili curve, si rivela sotto ciascuno dei diversi punti di vista da cui si possono definire, e in particolare se si ricorre a quella loro generazione, che in linea storica si presentò per prima e che dà ragione del nome. Si consideri un cono circolare retto (o anche obliquo), pensandolo come luogo di rette intere e non di semirette limitate al vertice V(cioè costituito, come si direbbe in geometria elementare, da due coni opposti al vertice) e s'immagini di segarlo con un piano π non passante per V(fig. 2). Se π non è parallelo ad alcuna generatrice (talché le seghi tutte da una stessa parte del vertice) la curva sezione è un'ellisse (o in particolare un cerchio). Se invece π è parallelo a una generatrice (così che incontri tutte le altre da una stessa parte di V) la sezione è una parabola. Se infine il piano è parallelo a due generatrici, queste dividono la superficie conica in due parti tali che le generatrici dell'una sono incontrate da π da una banda del vertice, le altre dalla banda opposta e in tal caso la curva sezione è un'iperbole. Va poi notato che se il piano π passa per il vertice del cono V(e non è esterno ad esso) lo sega secondo due rette (generatrici), le quali coincidono se π è tangente al cono. Così, in base a una veduta di continuità, si è già da questo punto di vista elementare condotti (e altrettanto accade dagli altri punti di vista) a includere nella famiglia delle coniche anche le coppie di rette (eventualmente coincidenti) che si sogliono chiamare coniche degeneri. La semplicità e l'eleganza delle proprietà geometriche delle coniche e le molte applicazioni, di cui si mostrarono suscettibili, richiamarono su di esse in tutti i tempi l'attenzione dei geometri e si può dire che la storia della loro teoria rispecchi nelle sue linee schematiche fondamentali la storia stessa dell'intera geometria. Cenni storici. - È dubbio se Democrito già si fosse occupato delle coniche, e, secondo le tradizioni più attendibili, va considerato come inventore di esse il geometra greco Menecmo (sec. IV a. C.), discepolo di Eudosso, il quale le introdusse per risolvere il cosiddetto problema di Delo, cioè della duplicazione del cubo (v. cubo). Egli le considerava sezioni piane del cono rotondo e più precisamente sezioni con piani perpendicolari a una generatrice. Si ha così un'ellisse o una parabola o un'iperbole, secondo che l'angolo di apertura del cono è acuto, retto o ottuso. Sembra che in questo stesso senso venissero considerate anche nei trattati di Aristeo il vecchio (contemporaneo di Euclide, ma un po' maggiore di età) e di Euclide stesso, l'uno e l'altro completamente perduti. Si deve tuttavia ritenere che già in queste opere si ponessero alla base dello studio delle coniche le loro proprietà caratteristiche come curve del piano e non già la loro generazione spaziale come sezioni del cono: certo è che Archimede, nella sua determinazione dell'area dell'ellisse e del settore parabolico, presuppone come note molte proprietà delle coniche, che dovevano essere contenute in codesti due libri perduti. Ma la teoria elementare delle coniche trova il suo assetto completo e si può dire definitivo nell'opera di Apollonio Pergeo (v.) in otto libri, di cui l'ultimo è andato perduto. Apollonio concepisce senz'altro le coniche come sezioni di un cono circolare qualsiasi (retto o obliquo). Ma a questa definizione spaziale fin dall'inizio della sua trattazione sostituisce una definizione come luoghi piani: precisamente, generalizzando una considerazione che già si trova in Archimede, stabilisce per le coniche quella proprietà caratteristica che, col formalismo algebrico della moderna geometria analitica, viene espresso dall'equazione della conica riferita a una tangente e al diametro passante per il punto di contatto. Enunciando tali proprietà, Apollonio introduce i tre nomi ἔλλειψις, παραβολή, ὑπερβολή, che oggi dai più si fanno derivare dalla nomenclatura usata nei cosiddetti problemi di "applicazione delle superficie piane" studiati dai pitagorici (risoluzione geometrica di equazioni di secondo grado). Sulla base della definizione suaccennata con procedimenti di algebra geometrica, che oggi si direbbero di trasformazione delle coordinate, Apollonio studia successivamente i diametri coniugati e gli assi, le tangenti e gli asintoti, le normali e, implicitamente, l'evoluta, i poli e le polari, i fuochi e le proprietà focali per l'ellisse e l'iperbole, le intersezioni d'una conica e d'una retta o di due coniche fra loro, le similitudini fra coniche. A questo mirabile trattato di Apollonio poco aggiunsero i geometri greci posteriori. Si può ricordare che nel VII libro della Collezione matematica di Pappo, si trovano il teorema sull'esagono iscritto in una coppia di rette (caso particolare del celebre teorema del Pascal; n. 13), la definizione delle coniche (verosimilmente già prima nota a Euclide) come luoghi dei punti, di cui è costante il rapporto delle distanze da un punto e da una retta fissi (n. 10) e la costruzione d'una ellisse di cui son dati cinque punti (Pappo, libro VIII, prop. XIII). I matematici bizantini e arabi mantennero viva nel Medioevo la conoscenza degli scritti di Apollonio. Vitellione, attingendo a qualche fonte greca perduta, nella sua Prospettiva(Περὶ ᾿Οπτικῆς), scritta forse in Italia, probabilmente a Viterbo, prima del 1277, nel lib. IX, prop. 39-44, dimostra le proprietà del fuoco della parabola e del paraboloide di rivoluzione come specchio ustorio nel quale: a tota superficie et a quolibet puncto ipsius radii solares in unum punctum aggregantur. Questa proprietà è altresi chiaramente accennata in un frammento attribuito da J. L. Heiberg ad Antemio (cfr. Mathematici Graeci Minores, Copenaghen 1927, pp. 77-92). Nel Rinascimento, dopo un primo saggio di volgarizzazione di Apollonio dovuto al Vvalla (Venezia 1501) e una traduzione più estesa, ma non più accurata, del Memmo (pubblicata dal figlio nel 1537), comparve nel 1566 a Venezia quella del Commandino (Apollonii Pergaei Conicorum Libri quatuor), la quale fece testo in tutte le scuole di Europa. Ma in questa rinascita la teoria delle coniche, sotto l'influsso di nuove correnti d'indagine geometrica, che si andavano allora delineando, si orientò per nuove vie. Dall'opera dei pittori e degli architetti, specialmente italiani, la pratica e lo studio della prospettiva erano passati oramai nel dominio dei geometri, onde era naturale che si tenesse conto della possibilità (già implicita nell'antica definizione spaziale) di considerare le coniche come figure prospettiche d'un circolo. D'altro canto le coniche apparivano non più soltanto come creazioni d'una libera fantasia geometrica, bensì come immagini direttamente legate alla schematizzazione di fenomeni ottici, di fatti geodetici, soprattutto di leggi astronomiche, dopo che Keplero aveva scoperto che le orbite planetarie sono ellissi. Lo stesso Keplero dedicò un capitolo dei suoi Ad Vitellionem paralipomena(Francoforte 1604), alle coniche, che egli studiava, combinando considerazioni di prospettiva con una veduta di continuità, per cui riguardava le tre specie di coniche (al pari della retta) come deducibili per deformazione dal circolo. Pervenne così, per intuizione più che per logica deduzione, alle proprietà dei fuochi, che da lui ricevettero questo nome e che egli, a differenza di Apollonio, considerò anche nel caso della parabola. E va ricordato che i risultati ottenuti in via euristica da Keplero furono poi dimostrati deduttivamente da B. Cavalieri (v.) nell'opuscolo sullo specchio ustorio (Bologna 1632). Il primo che partendo esplicitamente dalla considerazione delle coniche come proiezioni del circolo, ne concepì da questo punto di vista una teoria unitaria, è stato G. Desargues, il quale, coerentemente allo spirito di generalità insito in siffatta considerazione fondamentale, introdusse la nozione e l'uso dei punti all'infinito (pur senza spingersi a riconoscere che essi vanno considerati come costituenti, nel loro insieme, una retta). Non meno che per la generalità del punto di vista e dei metodi, l'opera del Desargues è mirabile per la novità e l'importanza dei risultati, primo fra tutti il suo celebre teorema (n. 13); e a lui vanno ravvicinati B. Pascal (v.), che scoperse appena sedicenne il teorema sull'esagono inscritto in una conica (n. 13) e più tardi F. De La Hire e J. le Poivre, che approfondirono le relazioni tra poli e polari, già prima riprese, dopo i cenni di Apollonio, dal Desargues. Ulteriori apporti alla teoria delle coniche dovevano ancora venire recati dai più larghi e più elevati sviluppi, che per la geometria si venivano preparando sotto l'impulso di quello spirito di generalità, che già si era manifestato nell'opera del Desargues e del Pascal. Nel 1822 si costitui col Traité des propriétés projectives des figures di J. V. Poncelet la geometria proiettiva. Il Poncelet, considerando le operazioni di proiezione e sezione nel loro senso più generale, si pose il problema dello studio sistematico di quelle proprietà delle figure piane, che hanno carattere d'invarianza rispetto a siffatte operazioni, e assunse come criterio generale d'indagine delle figure l'idea di ridurle, per mezzo di proiezioni e sezioni, a qualche loro caso particolare, come già il Desargues e il Pascal avevano ridotto le coniche a cerchi. Inoltre il Poncelet riconobbe per primo nella polarità rispetto a una conica un mezzo per dedurre sistematicamente da ogni teorema di geometria piana di natura grafica (cioè esprimibile per mezzo di relazioni di appartenenza di punti e rette) un nuovo teorema, ed è per questa via che il Brianchon dedusse dal teorema del Pascal sull'esagono inscritto a una conica il suo teorema sull'esalatero circoscritto (n. 13); mentre poi la veduta del Poncelet doveva, attraverso l'opera del Gergonne e del Plücker, mettere capo alla scoperta del principio di dualità (v.). Dai successivi sviluppi della geometria proiettiva derivarono due nuovi modi di definire le coniche. Si osservi anzitutto che se, presi due punti distinti e fissi U e V su di una circonferenza γ (fig. 3) e congiuntili con un punto M di essa, s'immagina che questo descriva l'intera circonferenza trascinando con sé le due rette UM e VM, si ottiene fra le rette dei due fasci di centri U, V una corrispondenza biunivoca(cioè tale che ad ogni retta di ciascun fascio corrisponde nell'altro una retta e una sola) e questa corrispondenza è, come si suol dire, una uguaglianza diretta, intendendosi con ciò affermare che se UM ed UN sono due rette quali si vogliano del primo fascio, l'angolo MVN delle due rette corrispondenti per V è uguale e di ugual verso dell'angolo MUN. Viceversa è chiaro che fra due fasci complanari di centri U, V, presi ad arbitrio, si può (in infiniti modi) definire una tale uguaglianza diretta, e, escluso il caso che si corrispondano le rette per U, V parallele, il luogo dei punti d'intersezione delle coppie di rette corrispondentisi è una circonferenza. Ciò posto, se a partire dalla figura costituita da una circonferenza γ e dai due fasci (in corrispondenza di uguaglianza diretta) che si ottengono proiettandone i punti da due suoi punti fissi U e V, si eseguiscono quante si vogliono proiezioni e sezioni, così da arrivare a una nuova figura piana, questa risulta costituita da una conica γ′ e da due fasci di centri U′, V′ su di essa. Se in questi due fasci si fanno corrispondere le rette che s'incontrano nei punti di γ′, si ottiene una corrispondenza che naturalmente non è più, in generale, un'uguaglianza diretta; e lo Steiner ha scoperto che si tratta d'una proiettività tra i due fasci (v. geometria), cioè d'una corrispondenza costruibile per mezzo di proiezioni e sezioni. Il risultato si può invertire; onde si è condotti a porre, con lo Steiner, la seguente definizione proiettiva delle coniche (teoricamente importante anche per le generalizzazioni di cui è suscettibile): si dice conica il luogo dei punti d'intersezione delle coppie di rette corrispondenti in due fasci complanari proiettivi, non prospettivi. Per comprendere la ragione di quest'ultima riserva si tenga conto che due fasci complanari U, V, si dicono prospettivi quando in essi si corrispondono le rette (fig. 4) che proiettano i punti d'una medesima retta ad essi complanare (asse di prospettività), nel qual caso è manifesto che il luogo delle intersezioni delle rette corrispondenti si spezza in codesto asse di prospettività e nella retta UV. Per chiarire l'altra definizione proiettiva delle coniche cui dianzi alludemmo, fissiamo l'attenzione sulla corrispondenza che rispetto a una data conica y intercede fra i singoli punti del piano e le corrispondenti polari. Essa rientra come caso particolare in una classe di corrispondenze, le cosiddette correlazioni o reciprocità, la cui prima idea risale a A.F. Möbius, di poco posteriore al Poncelet. Al Möbius si deve il concetto generale di trasformazione o corrispondenza biunivoca sia nel piano sia nello spazio, e più particolarmente il concetto di omografia o collineazione. Limitatamente al caso di due piani π e π′, si designa con tale nome ogni corrispondenza biunivoca fra i punti di π e quelli di π′. in cui ai punti di una retta corrispondono i punti di una retta; e il Möbius riconobbe (in base anche a una considerazione di continuità, dimostrata poi superflua dallo Staudt) che le omografie fra piani non sono altro che le corrispondenm del Poncelet, costruibili per proiezioni e sezioni. Ora, accanto a queste omografie, il Möbius, ispirandosi a una veduta di generalità che doveva poi essere ulteriormente sviluppata dal Plücker, considerò anche le corrispondenze biunivoche fra i punti d'un piano π e le rette d'un altro π′, tali che agli ∞1 punti d'una retta qualsiasi di π corrispondano le rette d'un ben determinato fascio di π′ (e quindi alla retta sostegno degli ∞1 punti il centro di codesto fascio). Sono queste precisamente le correlazioni o reciprocità: e si possono considerare intercedenti anche fra due piani sovrapposti, nel qual caso per altro occorre, ogni qualvolta si parli di un elemento (punto o retta), dichiarare su quale dei due piani esso vada considerato. Orbene, la polarità rispetto a una conica (o corrispondenza fra i singoli punti del piano e le rispettive polari) è una particolare correlazione: precisamente è una correlazione involutoria, con che s'intende affermare che se un punto P descrive una retta q, la rispettiva polare descrive un fascio il cui centro Q ha precisamente per polare la retta q; in altre parole, se il piano s'immagina per un momento sdoppiato in due (l'uno all'altro sovrapposti), a un qualsiasi punto corrisponde la medesima retta, sia che lo si consideri sul primo o sul secondo piano. I punti della conica sono tutti e soli quelli che appartengono alle rispettive polari; ed è questa la proprietà assunta da R.G.C. Staudt per definire le coniche. Egli considerò a priori tutte le possibili correlazioni involutorie del piano (o polarità) e osservò che può darsi benissimo che in una tale corrispondenza non esista nessun punto che appartenga alla sua polare (polarità uniformi); ma se esiste un punto siffatto, ne esistono infiniti, costituenti una conica. Di qui la definizione dello Staudt: si dice conica il luogo dei punti appartenenti alla rispettiva polare in una polarità non uniforme. Già qualche secolo prima dello Steiner e dello Staudt, un'altra definizione piana delle coniche era stata fornita dalla geometria analitica, costituita a metodo sistematico d'indagine per opera del Descartes e del Fermat (v. coordinate), presso a poco negli stessi anni in cui il Desargues aveva posto le prime basi della geometria proiettiva. Col sussidio delle coordinate cartesiane x, y le proprietà locali delle coniche, che già erano note ad Apollonio, si traducono in equazioni di 2° grado (a coefficienti reali) tra le coordinate x, y di un punto corrente sulla curva; e poiché, viceversa, si è potuto riconoscere che ogni equazione siffatta (quando ammetta qualche soluzione reale e il suo primo membro non sia decomponibile nel prodotto di due funzioni lineari) rappresenta una conica, si perviene alla preannunziata definizione: si dice conica ogni curva algebrica del 2° ordine (non degenere in due rette). P roprietà elementari delle coniche. - Consideriamo una conica γ (non degenere; n.1), pensandola come sezione del suo piano π con un cono rotondo di vertice V(non giacente in π) e di asse r. Essa ammette, in ogni caso, come asse di simmetria la proiezione di r sul piano π, e questo asse interseca γ in due punti A, A′, se si tratta di un'ellisse o di un'iperbole, in un solo punto A se si tratta di una parabola (fig. 5). L'ellisse e l'iperbole ammettono come secondo asse di simmetria la perpendicolare ad AA′ nel suo punto medio O, e questo secondo asse interseca l'ellisse in due punti B, B′, equidistanti da O (risultando BB′ 〈AA′), mentre è esterno all'iperbole. I quattro punti A, A′, B, B′ per l'ellisse, i due punti A, A′ per l'iperbole, il punto A per la parabola si dicono vertici. Gli assi di simmetria (due, fra loro ortogonali, per l'ellisse e l'iperbole, uno solo per la parabola) si dicono assi principali della conica; e più precisamente AA′ si chiama asse maggiore per l'ellisse (in quanto talvolta si designano col nome di assi anche gli stessi segmenti AA′, BB′), asse trasverso per l'iperbole, mentre l'altro asse si dice minore per l'ellisse, non trasverso o secondario per l'iperbole. Per entrambe queste specie di coniche il punto d'incontro O dei due assi (rispetto al quale asse sono simmetriche) si dice centro. La parabola non ha centro. Dal punto di vista elementare, il modo più semplice di definire i fuochi è quello dovuto a G. P. Dandelin (in Nouveaux mémoires, II, Bruxelles 1822, pag. 172 e segg.). Delle infinite sfere inscritte nel cono (per ipotesi rotondo), due risultano tangenti al piano π nel caso dell'ellisse e dell'iperbole, una sola nel caso della parabola (fig. 6). Si dicono fuochi della conica i punti di contatto di codeste sfere col suo piano π, i quali perciò sono due, F, F′, per l'ellisse e l'iperbole, uno solo, F, per la parabola. Essi giacciono sull'asse principale, proiezione su π dell'asse del cono (cioè sull'asse maggiore dell'ellisse, e trasverso dell'iperbole, sull'unico asse per la parabola), il quale si dice perciò asse focale. Nel caso della ellisse e dell'iperbole i fuochi sono simmetricamente posti rispetto al centro O e risultano interni al segmento AA′ per la prima, esterni per la seconda. Per ciascuna delle sfere tangenti dianzi indicate, il piano del circolo di contatto col cono sega π secondo una retta perpendicolare all'asse focale, la quale si dice direttrice della conica, relativa al fuoco in cui la sfera tocca π. Si hanno così due direttrici per l'ellisse e l'iperbole, una sola per la parabola; e in ogni caso si tratta d'una retta esterna alla conica. L'ellisse e l'iperbole godono rispettivamente delle due seguenti proprietà, ciascuna delle quali è, per la corrispondente conica, caratteristica, cosicché può assumersi come sua definizione (e si è cosi condotti per l'ellisse alla nota costruzione per mezzo d'un filo o "costruzione dei giardinieri"): L'ellisse è il luogo dei punti le cui distanze da due punti fissi(fuochi) hanno una somma costante(= AA′). L'iperbole è il luogo dei punti, le cui distanze da due punti fissi(fuochi)hanno una differenza costante in valore assoluto(= AA′). Vale poi per tutte e tre le specie di coniche quest'altra proprietà pur essa caratteristica: una qualsiasi conica(che non sia un cerchio) è il luogo dei punti, le cui distanze da un punto(fuoco) e da una retta(corrispondente direttrice) fissi, stanno in un rapporto costante, detto eccentricità della conica; e indicato questo rapporto costante con e, si ha precisamente e〈 1, e〈 1, e= 1, secondo che si tratta di un'ellisse o di un'iperbole o d'una parabola. Una conica γ, qualunque sia la sua specie, è proiettata da un qualsiasi punto fuori del suo piano π secondo un cono circolare; e anzi vi sono sempre ∞1 punti, da cui il cono proiettante risulti rotondo. Il luogo dei vertici di questi coni proiettanti rotondi è un'altra conica γ′ (detta conica focale della γ), la quale ha comune con γ l'asse focale, giace nel piano perpendicolare a π lungo questo asse e ammette come fuochi e vertici rispettivamente i vertici e i fuochi di γ. La relazione fra una conica e la sua focale è reciproca. P roprietà proiettive delle coniche.- Partiamo dalla definizione dello Steiner (n. 7), che caratterizza una conica y come luogo dei punti d'intersezione delle rette corrispondenti di due fasci giacenti in uno stesso piano e aventi centri U, V distinti, fra i quali sia stabilita una corrispondenza proiettiva (che non sia una prospettività). Per la stessa definizione la conica γ passa per U e V(fig. 7). E questi due punti, sebbene sembrino così assumere un ufficio privilegiato nella definizione della conica, sono due punti generici di essa, in quanto si dimostra che se sulla γ si prendono altri due punti distinti U 1, V 1 quali si vogliano, e si fanno corrispondere le coppie di rette che da U 1, V 1 rispettivamente proiettano i singoli punti di γ, si ottiene ancora fra i due fasci di centri U 1, V 1 una ben determinata proiettività. Se i due fasci proiettivi U, V si segano con una qualsiasi retta r non passante né per U né per V(cioè se su r si fanno corrispondere fra loro i punti d'intersezione con le coppie di rette corrispondentisi in U, V), si ottiene su r una proiettività, i cui eventuali punti uniti sono i punti comuni a r e γ; secondo che questa proiettività è iperbolica, parabolica o ellittica, la r ha comuni con γ due punti (secante), uno (tangente) o nessuno (retta esterna). Per ogni punto di γ passa una tangente e una sola; in particolare la tangente in U(o in V) è la retta per U(o rispettivamente per V) che nella proiettività fra i due fasci corrisponde alla UV. L'insieme delle tangenti d'una conica è suscettibile d'una definizione, che è la duale di quella or ora ricordata per le coniche, come luoghi di punti: date in un piano due rette (o punteggiate) u, v distinte, fra le quali sia definita una proiettività (che non sia una prospettività), le congiungenti delle varie coppie di punti corrispondentisi sulle due punteggiate costituiscono l'insieme delle tangenti a una conica (conica-inviluppo). Le definizioni proiettive dianzi chiarite per le coniche-luogo e per le coniche-inviluppo permettono di riconoscere quanti punti o tangentì occorrano e bastino per individuare una conica. Poiché tra due fasci complanari di rette risulta individuata una proiettività quando si assegnino ad arbitrio tre coppie di elementi i quali debbano corrispondersi (teorema fondamentale della geometria proiettiva), ogni qualvolta si prefissino nel piano 5 punti A, B, C, D, E(a tre a tre non allineati), basta far corrispondere alle tre rette AC, AD, AE uscenti da A, rispettivamente le rette BC, BD, BE uscenti da B(fig. 8), perché resti individuata fra i due fasci A, B una proiettività e quindi (come luogo dei punti d'intersezione delle rette in essi corrispondentisi) una conica passante per i cinque punti. Abbiamo dunque che 5 punti, a 3 a 3 non allineati, individuano una conica. E se si fa coincidere per continuità E con A e successivamente C con B, si riconosce che una conica è ugualmente individuata da 4 punti e dalla tangente in uno di essi, oppure da 3 punti e dalle tangenti in 2 di essi. Dualmente si ha che una conica è individuata da 5 tangenti, oppure da 4 tangenti e dal punto di contatto di una di esse, o infine da 3 tangenti e dai punti di contatto di 2 di esse. Si vede così in particolare che per quattro punti A, B, C, D(di cui nessuna terna sia allineata) passano infinite coniche (una e una sola per ciascun punto generico del piano). Queste ∞1, coniche si dice che costituiscono un fascio(di punti-base A, B, C, D). Le stesse definizioni dello Steiner permettono di costruire linearmente (cioè con l'uso della sola riga) quanti punti (o dualmente quante tangenti) si vogliano di una conica individuata in uno dei modi or ora detti: p. es., dati della conica 5 punti A, B, C, D, E, si può determinare l'ulteriore intersezione con la conica di una qualsiasi retta a uscente da A (basta intersecarla con la retta per B, corrispondente ad a nella proiettività fra i due fasci A e B, in cui ad AC, AD, AE, corrispondono rispettivamente BC, BD, BE). Nella teoria proiettiva delle coniche hanno ufficio essenziale i seguenti tre teoremi, di cui i primi due sono fra loro duali: Teorema del Pascal. - Se un esagono semplice ABCDEF è inscritto in una conica (fig. 9), le tre coppie di lati opposti (AB e DE, BC ed EF, CD ed FA) s'incontrano in tre punti di una stessa retta p(retta del Pascal relativa all'esagono). Teorema del Brianchon. - Se un esalatero semplice abcdef è circoscritto a una conica (fig. 10), le tre coppie di vertici opposti (ab e de, bc e ef, cd e fa) determinano tre rette passanti per uno stesso punto P(punto del Brianchon relativo all'esalatero). Teorema del Desargues. - Se un quadrangolo completo è inscritto in una conica (fig. 11), ogni secante della conica, che non passi per alcun vertice del quadrangolo, la sega in due punti, che sono coniugati nella involuzione, a cui appartengono le intersezioni delle tre coppie di lati opposti del quadrangolo. Circa il teorema del Pascal, rileviamo che, presi su di una conica sei punti A, B, C, D, E, F, s; possono formare con essi 60 diversi esagoni semplici (quante sono le permutazioni di 6 oggetti pensati in ordine chiuso). Le corrispondenti 60 rette del Pascal costituiscono una configurazione che è stata oggetto di studio da parte di molti geometri (in particolare di G. Veronese) e che lo stesso Pascal chiamò "esagramma mistico". Dei teoremi del Pascal e del Brianchon sono utili i casi limiti, cui si perviene facendo per continuità coincidere una o due o tre coppie di vertici o di lati consecutivi (pentagono inscritto e tangente in un vertice, pentalatero circoscritto e punto di contatto d'uno dei suoi lati, quadrangolo inscritto e tangenti in due vertici opposti, ecc.). Le varie proposizioni che cosi si ottengono, come già i teoremi 1,2, sono invertibili e perciò permettono di costruire per punti o per tangenti una conica individuata per mezzo di opportune condizioni (n. prec.). Per es., datine 5 punti A, B, C, D, E e presa per A una retta a ad arbitrio, l'ulteriore intersezione X di questa con la conica si costruisce ricorrendo alla retta del Pascal relativa all'esagono ABCDEX, la quale risulta individuata come congiungente i due punti d'intersezione di AB, DE e di CD, XA= a(fig. 12). Il teorema del Desargues conduce invece alla soluzione di problemi di secondo grado, di cui il più semplice è il seguente: costruire una conica, la quale passi per quattro punti A, B, C, D(di cui nessuna terna sia allineata) e risulti tangente a una retta a, non passante per alcuno di essi. Le coniche del fascio di punti-base A, B, C, D segano la a secondo coppie dell'involuzione, cui appartengono le intersezioni de le coppie di rette (coniche degeneri del fascio) AB e CD, AC e BD, AD e BC. Due di queste tre ultime coppie d'intersezioni bastano a individuare l'involuzione; e di questa gli eventuali punti di contatto con a di coniche passanti per A, B, C, D sono i punti doppî. Si giunge così al problema di determinare i punti doppî di un'involuzione (problema tipico di 2° grado). Alla soluzione del problema fondamentale or ora indicato si perviene nel modo seguente. Poiché quattro punti A, B, C, D d'una conica y sono proiettati dagli altri punti di essa secondo quaderne che risultano fra loro tutte proiettive, è lecito chiamare armonica una tale quaderna di punti, quando sia proiettata secondo una quaderna armonica di rette da un punto della γ; e si può chiamare proiettività fra due coniche γ, γ′, considerate come luoghi di punti (o punteggiate coniche), ogni corrispondenza biunivoca fra i loro punti, la quale faccia corrispondere a ogni quaderna armonica di punti di γ una quaderna parimenti armonica di punti di γ′. Queste proiettività non sono altro che quelle corrispondenze che si ottengono segando con le due coniche due fasci proiettivi aventi rispettivamente su di esse i loro centri; e una qualsiasi di queste proiettività fra coniche risulta individuata quando se ne assegnino tre coppie di punti corrispondenti. Si consideri in particolare una proiettività tra due punteggiate coniche sovrapposte γ = γ′: come nel caso delle proiettività fra punteggiate rettilinee, a una tale proiettività risultano associati (fig. 13) un asse di collineazione(luogo dei punti d'intersezione delle coppie di rette AB′ e A′B, AC′ e A′C, BC′ e B′C,... se A e A′, B e B′, C e C′,... sono punti corrispondenti) e, dualmente, un centro di collineazione(punto di concorso delle congiungenti delle coppie di punti ab′ e a′b, ac′ e a′c, bc′ e b′c,... se a e a′, b e b′, c e c′,... sono tangenti alla conica in coppie di punti corrispondenti). Le eventuali intersezioni dell'asse di collineazione con la conica sono i punti uniti della proiettività considerata. Fra le proiettività tra punteggiate coniche sovrapposte si possono considerare le involuzioni(come quelle che coincidono con le loro inverse o, in altre parole, sono tali che gli elementi omologhi vi si corrispondano in doppio modo) e in questo caso si riconosce che le coppie di punti omologhi A e A', B e B′,... ecc., sono allineate (fig. 14) col centro P di collineazione (che qui si dice centro o polo dell'involuzione), mentre le rispettive tangenti a e a′, b e b′,... concorrono sull'asse p di collineazione (detto in questo caso asse d'involuzione). I punti doppî dell'involuzione sono i punti di contatto delle eventuali tangenti alla conica uscenti da P, ossia le eventuali intersezioni della conica con p. Di qui il modo di determinare i punti doppî d'una involuzione data su di una retta r(costruzione dello Steiner); due coppie di punti omologhi nell'involuzione A e A′, B e B′ (quante bastano a determinarla) si proiettano da un punto U di una conica γ su di essa in A 1 e A 1 ′, B 1 e B 1 ′; costruito l'asse dell'involuzione così definita su γ (cioè la congiungente dei punti d'intersezione di AB- e A′B e di AB e A′B′), si riproiettano su r le eventuali intersezioni di codesto asse con la conica. Questa costruzione è eseguibile con riga e compasso, in quanto si può sempre prendere come conica γ un cerchio (fig. 15). Se nel piano di una conica y si prende un punto P, non giacente su di essa, i coniugati armonici di P rispetto alle coppie di punti di y, allineati con P, appartengono tutti a una medesima retta p, la quale gode altresi delle due seguenti proprietà, ciascuna delle quali è atta a caratterizzarla: contiene i punti d'intersezione delle tangenti a y nelle coppie di punti allineati con p(in particolare i punti di contatto delle eventuali tangenti per P a γ) e gli ulteriori punti diagonali di ogni quadrangolo completo iscritto in γ, che abbia un punto diagonale in P(fig. 16). La retta p si dice polare di P. Dualmente, se nel piano di γ si prende una retta p non tangente a γ, le coniugate armoniche di p rispetto alle coppie di tangenti a γ, che si possono condurre dai varî punti di p, passano tutte per un medesimo punto P che gode, altresì, delle due seguenti proprietà, pur esse caratteristiche: passano per P le congiungenti dei punti di contatto delle coppie di tangenti a γ, che s' incontrano su p(in particolare le tangenti nelle eventuali intersezioni di p con γ) e le ulteriori rette diagonali di ogni quadrilatero completo circoscritto a γ, il quale abbia una retta diagonale in p. Il punto P si dice polo della p rispetto alla conica, e la relazione tra polo e polare è reciproca nel senso che se p è la polare di P, P è il polo di p. Secondo che P è esterno o interno alla conica, la rispettiva polare è segante o esterna; e nell'uno e nell'altro caso non passa per p. Se poi un punto P è sulla conica, si assume (per ragioni di continuità) come sua polare la tangente in P; e in tal modo risulta definita, rispetto alla data conica γ, una corrispondenza biunivoca fra i punti e le rette del piano, la quale si dice polarità relativa alla conica (fondamentale) γ. Se un punto P descrive una retta q, la polare p di P descrive il fascio che ha per centro il polo Q di q. La polarità è dunque una correlazione e anzi una correlazione involutoria, in quanto in essa i punti e le rette omologhi si corrispondono in doppio modo. Essa si dice non uniforme in quanto ammette punti (tutti e soli quelli della conica γ) che appartengono alla rispettiva polare. Come già si è accennato al n. 8, si può (seguendo lo Staudt) considerare a priori nel piano le correlazioni involutorie non uniformi e definire la conica come loro luogo fondamentale: per una teoria delle coniche da questo punto di vista si vedano le Lezioni di geometria proiettiva di F. Enriques, citate nella bibliografia. Due punti P e Q si dicono coniugati rispetto alla conica y, o alla rispettiva polarilà, se la polare di P passa per Q(e quindi la polare di Q per P); e dualmente si definiscono le rette coniugate. I punti di una qualsiasi retta r non tangente a γ sono a due a due coniugati, e queste coppie costituiscono un'involuzione (ellittica se la retta è esterna a γ, iperbolica se è secante, nel qual caso le due intersezioni sono i punti doppî). Su di una generica tangente a γ, il punto di contatto P ha come coniugati, oltre sé stesso, tutti gli altri punti della retta, mentre ciascuno di questi ha come coniugato P, talché l'involuzione delle coppie di punti coniugati è in questo caso degenere; e valgono i risultati duali nei fasci di rette uscenti dai singoli punti del piano. Le proprietà metriche delle coniche si deducono dalle proprieta proiettive dianzi accennate, mettendole in relazione con la retta impropria del piano. Anzitutto una conica è un'ellisse o un'iperbole o una parabola, secondo che rispetto ad essa la retta all'infinito è esterna, secante o tangente. Preso un qualsiasi punto improprio P∞, la sua polare p(come contenente i coniugati armonici di P∞ rispetto alle coppie di punti di γ allineate con P) biseca tutte le corde di γ parallele alla direzione di P∞, (fig. 17). Perciò ogni polare di un punto improprio si dice diametro della conica. Tutti i diametri (come polari dei punti della retta impropria) passano per il polo di questa, il quale per l'ellisse e per l'iperbole è un punto proprio O, per la parabola è un punto improprio O∞. Nel primo caso il punto O è centro di simmetria della conica (interno per l'ellisse, esterno per l'iperbole), onde l'ellisse e l'iperbole si sogliono dire coniche a centro. Per queste coniche a ogni diametro p è coniugato un altro diametro q (bisecante tutte le corde parallele a p, mentre alla sua volta q biseca tutte le corde parallele a p) e queste coppie di diametri coniugati appartengono a un'involuzione. Se in questa involuzione tutte le coppie di diametri coniugati sono ortogonali, la conica, come dotata di simmetria ortogonale rispetto a ogni suo diametro, si riduce a un cerchio. Escluso questo caso particolare, nell'involuzione dei diametri coniugati esiste una coppia, e una sola, di diametri coniugati e ortogonali (assi di simmetria ortogonale della conica), tutti e due secanti (nei quattro vertici) per l'ellisse, uno secante, l'altro esterno per l'iperbole. Sono questi gli assi(principali). Un diametro non può essere coniugato di sé stesso se non quando risulti tangente (per necessità in un punto improprio) alla conica, cosicché l'ellisse non ammette nessun diametro coniugato di sé stesso, mentre l'iperbole ne ammette due, e queste tangenti nei due punti improprî si chiamano i suoi asintoti(fig. 18). Per la parabola tutti i diametri sono paralleli (in quanto passano per O∞) e segano la conica ciascuno in un sol punto proprio. Ogni diametro è asse di simmetria nella direzione coniugata, la quale è in generale obliqua; un diametro e uno solo è coniugato alla direzione ortogonale, ed è l'asse(principale) della parabola. Per definire da questo punto di vista i fuochi, si ricordi che per ogni punto, rispetto alla conica, risulta definita un'involuzione fra le coppie di rette coniugate. I fuochi si possono caratterizzare come quei punti in cui siffatte coppie di rette coniugate sono tutte ortogonali; e da tale definizione è agevole dedurre le notevoli proprietà angolari e segmentarie che i fuochi godono in relazione alle corde, alle tangenti, alle normali, ecc. Le direttrici sono le polari dei fuochi. T rattazione analitica. - Equazioni canoniche delle coniche a centro. - Si considerino l'ellisse e l'iperbole come luoghi dei punti le cui distanze da due punti fissi F, F′ hanno uguale a una costante 2 a la somma e, rispettivamente, la differenza in valore assoluto (n. 10). Posto FF′ = 2 c(con c〈 a nel caso dell'ellisse, c>a nel caso dell'iperbole) e adottate come asse delle x la retta F′F, come asse delle γ la perpendicolare ad FF′ nel suo punto medio O(fig. 19), si trova come equazione del luogo (così riferito ai suoi assi principali) la dove b 2 = a 2 ∓c 2, valendo il segno superiore per l'ellisse, quello inferiore per l'iperbole. Per l'ellisse a e b sono i semiassi, rispettivamente maggiore e minore; p, er l'iperbole a è il semiasse trasverso, e b si chiama talvolta semiasse non trasverso o secondario, per quanto in realtà l'asse delle y non intersechi la curva in punti reali, bensì nei due punti immaginarî coniugati ±ib. Per entrambe le coniche a centro le equazioni delle direttrici, l'eccentricità e(rapporto delle distanze di un generico punto della conica da un fuoco e dalla rispettiva direttrice) e il parametro p(metà della corda passante per il fuoco e perpendicolare all'asse focale) sono dati da Infine le equazioni degli asintoti dell'iperbole (tangenti nei punti all'infinito) sono Per a= b l'ellisse si riduce al cerchio di raggio a, l'iperbole alla cosiddetta iperbole equilatera, caratterizzata dalla proprietà di avere gli asintoti ortogonali. Se si confronta l'equazione dell'ellisse con l'equazione x 2 + y 2 = a 2 del cerchio a essa concentrico, che ha per raggio il semiasse maggiore a, se ne deduce rispettivamente onde si riconosce che l'ellisse si può costruire per punti, riducendo le ordinate dei singoli punti del circolo nel rapporto b/a(fig. 20). Ciò si può esprimere dicendo che l'ellisse è la curva corrispondente al cerchio nell'omologia affine ortogonale di asse x e di rapporto b/a. Notiamo, in via incidentale, che più generalmente ogni conica si può costruire (linearmente) per punti come corrispondente a un cerchio in un'omologia: si ottiene un'ellisse, un'iperbole o una parabola secondo che il cerchio è tutto da una parte o secante o tangente rispetto alla prima retta limite dell'omologia (cioè alla retta cui corrisponde la retta all'infinito). Per il generico punto M dell'ellisse si dice anomalia eccentrica(Keplero) l'angolo AOM 1 (fig. 20), che permette di rappresentare parametricamente l'ellisse mediante le equazioni che, ove si ponga t= tang (u/ 2), diventano e inettono in luce la natura razionale dell'ellisse. Similmente le equazioni parametriche dell'iperbole si possono scrivere, ricorrendo alle funzioni iperboliche (v. funzione: Funzioni notevoli): o anche, ponendo t= tangh (u/ 2) Va infine rilevato che l'equazione di un'iperbole, riferita ai suoi asintoti come assi coordinati, assume l'aspetto dove k è una costante. Equazione canonica della parabola. - La parabola si consideri (n. 10) come luogo dei punti equidistanti da un punto F(fuoco) e da una retta d(direttrice). Se s' indica con p la distanza del dato fuoco F da d e si adottano come asse delle x la perpendicolare da F alla d(orientata da d verso F) e come asse y la parallela alla d, equidistante da questa e da F (fig. 21), si trova come equazione della parabola (riferita all'asse e alla tangente nel vertice) la e si riconosce che p è precisamente il parametro. Scambiando x con y ed eseguendo una qualsiasi traslazione degli assi, si riconosce che un'equazione del tipo dove a, b, c sono tre numeri dati, rappresenta sempre una parabola ad asse parallelo all'asse delle y, volgente la concavità in alto o in basso secondo che a è positivo o negativo, e avente per vertice il punto di coordinate - b/(2 a), (4 ac- b 2)/(4 a). Equazione polare delle coniche. - Una delle proprietà salienti dei fuochi si è che la distanza da un fuoco a un qualsiasi punto della conica è esprimibile linearmente per mezzo della sua proiezione sull'asse focale; precisamente denotando codesta distanza con ρ e tenendo conto delle equazioni canoniche (1) e (3), si trova, rispettivamente per l'ellisse, l'iperbola e la parabola cosicché nel sistema di coordinate polari ρ, ϕ, che ha per polo il fuoco e per asse polare l'asse focale (orientato dal fuoco verso la corrispondente direttrice) una conica di qualsiasi specie risulta rappresentata dall'equazione dove giova ricordare che è e〈 1 o e> 1 o e= 1 secondo che si tratta di un'ellisse o di un'iperbole o d'una parabola (n. 10). Teoria generale. - Nel piano riferito a coordinate omogenee (cartesiane o anche proiettive) x 1: x 2: x 3, si consideri la più generale equazione (omogenea) di 2° grado (e a coefficienti reali), che si può scrivere dove, per simmetria delle formule, si conviene una volta per tutte di poter scrivere in luogo di a 12, a 13, a 23 (semicoefficienti dei termini rettangoli) indifferentemente a 21, a 31, a 32. Questa equazione rappresenta la più generale curva algebrica del 2° ordine, che qui chiameremo senz'altro conica: la coincidenza di questa nuova definizione con quella elementare o con quella proiettiva risulterà dal seguito. Fondamentale per le considerazioni che qui si vogliono accennare è la cosiddetta formula del Joachimstal: se si fissano due punti P(x 1:x 2:x 3), P′ (x 1′: x 2′: x 3′), le coordinate di ogni punto della loro congiungente sono date da λ x 1 + λ′x 1′, λ x 2 + λ′x 2′, λ x 3 + λ′x 3′ dove λ/λ è un parametro arbitrario; i valori di questo parametro corrispondenti alle eventuali intersezioni della retta P P′ con la curva di equazione (4) sono dati dalla equazione quadratica dove il coefficiente di 2λλ′ è la forma polare della forma quadratica ternaria f(x 1, x 2, x 3); precisamente, posto si ha per definizione Il determinante del 3° ordine dei coefficienti delle tre semiderivate (6) di f si dice discrimínante della forma f o della conica definita dalla (4); e in base alla (5) si riconosce che l'annullarsi di A è la condizione necessaria e sufficiente perché la f si spezzi nel prodotto di due trinomî di 1° grado, ossia perché la conica degeneri in una coppia di rette (reali o immaginarie). Perché poi queste due rette coincidano (nel qual caso si tratta di una retta reale contata due volte) occorre e basta che si annullino insieme con A anche tutti i suoi minori del 2° ordine. Esclusa l'ipotesi A= 0, si deduce dalla stessa (5) che la polare di un generico punto P′ (x 1′, x 2′, x 3′) rispetto alla conica (in particolare la tangente alla conica in P′, se questo punto è sulla conica) è definita dalla equazione Se con u 1: u 2: u 3 si denotano le coordinate omogenee di retta (nel sistema associato al sistema puntuale x 1: x 2: x 3 adottato), intercedono fra le coordinate x 1: x 2: x 3 di un generico punto e le coordinate u 1: u 2.:u 3, della rispettiva polare le equazioni dove ρ denota un arbitrario fattore di proporzionalità. Sono queste le equazioni della polarità definita dalla conica (4). L'equazione dell'inviluppo delle tangenti alla conica (rette coniugate di sé stesse) è data da e ove si denoti con A rs il complemento algebrico di a rs nel discriminante A, si può scrivere: È questa l'equazione di un inviluppo di 2ª classe, cioè dell'ente duale della curva algebrica del 2° ordine. Giova qui notare che un inviluppo di 2ª classe d'equazione (8), considerato a priori, può degenerare; e ciò si verifica ogni qualvolta sia nullo il rispettivo discriminante, cioè il determinante dei coefficienti A rs. In questo caso l'inviluppo si spezza in due fasci di rette, i quali risultano coincidenti sempre e solo quando siano nulli anche tutti i minori del 2° ordine di codesto discriminante. Ora va rilevato che se la (8) proviene nel modo dianzi indicato dall'equazione (4) d'una conica degenere in due rette distinte, essa rappresenta precisamente il fascio (contato due volte) che ha per centro il punto comune a codeste rette; mentre invece se la (4) rappresenta due rette coincidenti, l'equazione (8), in quanto gli A s sono tutti nulli, si riduce a un'identità. Così, dualmente, a un inviluppo di 2ª classe degenere in due fasci distinti risulta associato come luogo di punti del 2° ordine la degenere in due fasci coincidenti non risulta collegato nessun determinato luogo di 2° ordine. Passiamo alle proprietà metriche e a tal fine supponiamo che le coordinate omogenee x 1: x 2: x 3 siano cartesiane ortogonali; si passerà alle coordinate cartesiane ordinarie, di cui qui ci varremo, ponendo x:y:1 = x 1:x 2:x 3 (v. coordinate). Anzitutto, per trovare le eventuali intersezioni della conica con la retta impropria, di equazione x 3 = 0, basta far coesistere questa equazione con la (4), talché si è condotti all'equazione quadratica il cui discriminante è dato da a 12 2 − a 11 a 22 = − A 33. Perciò la retta all'infinito è esterna, secante o tangente alla conica, secondo che è Se A≶ 0, cioè se la conica non è degenere, si dice senz'altro che si tratta rispetivamente di un'ellisse o d'una iperbole o d'una parabola (e l'accordo con le definizioni elementari o proiettive risulterà fra un momento). Se poi è A= 0, si ottiene rispettivamente una coppia di rette complesse coniugate (coppia ellittica) oppure reali e di direzioni diverse (coppia iperbolica) oppure reali e parallele o, in particolare, coincidenti (coppia parabolica). Escludiamo oramai le coniche degeneri, cioè supponiamo A≶ 0 ed esaminiamo anzitutto il caso A 33 ≷ o (coniche a centro). Le polari dei punti improprî 1:0:0 e 0:1:0 rispettivamente dell'asse delle x e delle y(cioè i diametri coniugati alle direzioni degli assi) risultano dati, in base alle (7), dalle equazioni (in coordinate ordinarie) onde il loro punto comune (centro della conica) ha per coordinate A 31/A 33, A 32/A 33. Ogni altro diametro ha come equazione una combinazione lineare delle (9). Se m, m′ sono i rapporti direttivi di due diametri coniugati, sussiste fra essi l'equazione (ottenuta esprimendo che la polare del punto improprio 1: m: 0 passa per il punto improprio 1: m′: 0) È questa l'equazione dell'involuzione dei diametri coniugati. I rappo1ti direttivi degli eventuali diametri coniugati di sé stessi (asintoti) sono definiti dalla la quale ammette radici reali sempre e solo quando sia A 33 〈 o, cioè per l'iperbole. Risulta di qui che se x 0, y 0 sono le coordinate del centro, l'equazione complessiva degli asintoti è data da Facendo coesistere con la (10) la condizione di perpendicolarità mm′ = − 1, si trova che i rapporti direttivi dei diametri coniugati alla direzione ortogonale sono le radici dell'equazione di 2° grado in m che, avendo il discriminante sempre positivo, ammette in ogni caso due radici reali di prodotto uguale a - 1 e quindi corrispondenti a due diametri ortogonali, salvo quando sia identicamente soddisfatta, cioè si abbia a 12 = 0, a 11 = a 22, il che accade sempre e solo quando la conica si riduce a un cerchio (v. cerchio). Escluso questo caso, le due radici reali della (11) dànno i rapporti direttivi degli assi principali, e basta, con una trasformazione delle coordinate, riferire la conica a codesti assi per trovare che la nuova equazione è del tipo dove il prodotto pq(che è il nuovo A 33) ha lo stesso segno dell'A 33, primitivo. Se A 33, e quindi pq, è positivo, sono possibili per p e q le due combinazioni di segno + + e − −: nel primo caso la (12) è del tipo (1) col segno + e rappresenta perciò un'ellisse; nel secondo non ammette nessuna soluzione reale e, per analogia formale, si suol dire che definisce un'ellisse immaginaria. È questo (insieme con la coppia ellittica di rette) il solo caso nuovo, che viene introdotto dalla definizione analitica di conica in confronto di quella elementare (o anche proiettiva). Se poi A 33, e quindi pq, è negativo, le combinazioni di segno possibili per p e q sono + − e − +: nel primo caso la (12) è del tipo (1) col segno - e rappresenta un'iperbole; nel secondo è riducibile al medesimo tipo con lo scambio di x con y e quindi dà un'iperbole avente come asse trasverso l'asse delle y anziché quello delle x. Supponiamo infine che nella (4) sia A 33 = 0. I due diametri (9), coniugati alle direzioni degli assi, risultano fra loro paralleli nella direzione di rapporto direttivo − a 11/a 12 = − a 21 /a 22, cosicché sono pur paralleli fra loro e ai due precedenti anche tutti gli altri diametri. Il solo diametro, che sia coniugato alla direzione ortogonale, è definito da che si ottiene cercando la polare del punto improprio a 11: a 12: 0, ossia a 12: a 22: 0. È questo l'unico asse della conica e basta riferirne l'equazione a questo asse e alla tangente nell'estremo per ridurla al tipo (3) e quindi riconoscere che si tratta effettivamente di una parabola. I criterî cosi ottenuti per riconoscere la specie della conica rappresentata da una qualsiasi equazione (4) sono riassunti nella tabella: Fasci e schiere di coniche. - Date due coniche di equazioni (x,y) = 0, ϕ (x, y) = 0, si dice sistema lineare ∞1 o fascio di coniche l'insieme di tutte le coniche rappresentate dall'equazione λ f+ μϕ = 0 al variare del parametro λ/μ. Le proprietà fondamentali di un tale fascio sono:1. ogni punto (reale o immaginario) comune alle due coniche f= 0, ϕ = 0 è comune anche a tutte le altre coniche del fascio (e si dice punto-base del fascio); 2. per ogni altro punto del piano passa una conica del fascio e una sola. Si noti che qui il concetto di "fascio" assume un senso più largo di quello del n. 12, dove si trattava dei fasci di coniche aventi quattro punti-base (reali e distinti). Per il teorema del Bezout (v. algebra, n. 46) il sistema di equazioni f= 0, ϕ = 0 ammette sempre 4 soluzioni, ma può darsi che 2 o anche tutte e 4 siano complesse, o che, mantenendosi reali, coincidano in parte o tutte, dando cosi luogo rispettivamente a punti-base immaginarî o a punti di contatto (fissi) per le curve del fascio. Duale del fascio di coniche-luogo è la schiera di coniche-inviluppo λ F (u, v) + μ Φ (u, v) = 0, dove F (u, v) = 0, Φ (u, v) = 0 rappresentano due coniche-inviluppo quali si vogliano: ogni tangente (reale o immaginaria) comune a queste due è comune anche a tutte le rimanenti coniche della schiera, mentre ogni altra retta del piano è toccata da una di queste coniche e da una sola. In generale le coniche d'un fascio, considerate come inviluppi, non costituiscono una schiera: ciò avviene soltanto (e si ha un fascio-schiera) per le coniche aventi fra loro due contatti semplici (in due punti fissi) o un contatto quadripunto (in un punto fisso). Rimandiamo ai trattati speciali per maggiori particolari sull'argomento, come pure per lo studio dei sistemi lineari di coniche ∞r con r> 1 e dei sistemi ∞1 algebrici, ma non più lineari, per i quali ultimi, a opera di Steiner, De Jonquières, Cremona, Chasles, Cayley, Zeuthen, Schubert, ecc., è stata sviluppata la cosiddetta teoria delle caratteristiche(numeri delle coniche del sistema passanti per un punto generico e di quelle tangenti a una retta generica), che segna gl'inizî della geometria numerativa (cfr. F. Enriques-O. Chisini, Lezioni sulla teoria geom. delle equazioni e delle funz. alg., I, Bologna 1915, II, 11, n. 25). Qui, a complemento di quanto si è già detto, aggiungiamo che costituiscono una schiera sia tutte le coniche a centro aventi comuni, in due punti fissi, i fuochi, sia tutte le parabole aventi comuni il fuoco e il punto all'infin;to (coniche confocali od omofocali). Una schiera di coniche a centro confocali, riferita agli assi principali comuni alle sue coniche, è rappresentata, in coordinate cartesiane, dall'equazione dove a e b sono costanti date e λ è il parametro, mentre una schiera di parabole confocali (e coassiali), riferita all'asse focale come asse delle x e al fuoco come origine, ha l'equazione Della schiera (13) passano per ogni punto del piano due coniche, di cui una è un'ellisse, l'altra un'iperbole, segantisi ortogonalmente in quattro punti, simmetricamente posti rispetto agli assi (fig. 22), mentre due coniche della stessa specie non hanno mai punti comuni. Della schiera (14) passano per ogni punto due parabole, che si segano ortogonalmente in due punti simmetrici rispetto all'asse focale (fig. 23). Questi due sistemi doppî ortogonali di coniche confocali costituiscono le linee coordinate delle coordinate ellittiche e paraboliche del piano (v. coordinate, n. 28). Q uadrature e rettificazioni. - Già Archimede, nell'opuscolo sulla quadratura della parabola (Opera omnia, II, Lipsia 1913, pp. 263-315), determinò l'area del segmento parabolico come uguale ai 4 / 3 di quella del triangolo che ha la stessa base del segmento e il vertice nel punto di contatto della tangente alla parabola, parallela a codesta base (fig. 24); e nel De conoidibus et sphaeroidibus(ibid., I, Lipsia 1910, pp. 246-445) assegnò l'area π ab dell'ellisse di semiassi a e b. In ogni caso l'area d'un settore di conica di qualsiasi specie è dato da un integrale calcolabile per mezzo di funzioni elementari (razionali, radicali, trigonometriche e logaritmiche). Altrettanto si dica della lunghezza d'un qualsiasi arco di parabola. Invece la rettificazione dell'ellisse e dell'iperbole costituisce un problema d'ordine più elevato. Hanno in proposito un notevole interesse storico i teoremi di G. Fagnano (1750) sulla determinazione di archi d'ellisse e d'iperbole, la cui differenza sia costruibile elementarmente (cioè con riga e compasso) e ad essi vanno ravvicinati i risultati di J. Landen (1771-1775) sull'esprimibilità di ogni arco iperbolico per mezzo di due archi d'ellisse e gli ulteriori, svariati contributi recati a quest'ordine di questioni da Eulero. La lunghezza d'un generico arco d'ellisse (o d'iperbole) si può esprimere (Legendre 1786) per mezzo di quei particolari integrali, che appunto furono detti ellittici e che, fornendo il primo esempio di funzioni non riducibili a funzioni elementari, aprirono l'adito alla moderna teoria delle funzioni (v. funzione: Funzioni notevoli). L e coniche nelle applicazioni. - Come si è già visto, lo studio delle coniche ha dato lo spunto a molte teorie geometriche. D'altra parte varî ordini di ricerche meccaniche o fisiche (vedi n. 1) hanno trovato nella teoria delle coniche un mezzo espressivo di rappresentazione e di schematizzazione. Ecco qualche esempio fra i più semplici. Le trasformazioni birazionali, o cremoniane, quadratiche fra due piani, si ottengono facendo corrispondere proiettivamente alle coniche d'una rete (o sistema lineare ∞2) λ f + μ f′ + ν f″ = 0 d' un piano π (dove f= 0, f′ = o, f″ = 0 rappresentano tre coniche aventi comuni tre punti, distinti o no, e λ: μ: ν denotano tre parametri omogenei) le rette d'un altro piano π′. A un punto generico di π, punto-base d'un fasc: o della rete, corrisponderà su π′ il centro del fascio di rette corrispondente. Di conseguenza alle rette di π′ verranno a corrispondere su π le coniche di una rete analoga, e fra i punti dei due piani risulterà definita una corrispondenza biunivoca algebrica (non proiettiva). In meccanica un punto, animato da due moti componenti rettilinei in direzioni diverse, l'uno uniforme, l'altro uniformemente vario, descrive una traiettoria parabolica: esempio tipico il moto dei gravi, lanciati nel vuoto in direzione non verticale (v. cinematica). Così il moto risultante di due moti armonici di eguale centro, lungo due rette ortogonali e sfasati d'un quarto di periodo, è un moto ellittico (v. armonico: Moto armonico); e questo fatto ha notevole importanza nello studio dei fenomeni ottici ed elettromagnetici. Infine la traiettoria d'un punto lanciato in una direzione qualsiasi e attratto da un centro con forza inversamente proporzionale al quadrato della distanza (legge del Newton, del Coulomb, ecc.) è una conica avente un fuoco nel centro di attrazione (moti kepleriani: v. cinematica). L'esempio più cospicuo è dato dai moti planetarî, ma anche la più recente fisica dell'atomo introduce la considerazione di questi moti kepleriani nello studio, in prima approssimazione, del comportamento degli elettroni rispetto al nucleo. Bibl.: Per la storia fino al sec. XVII: H.G. Zeuthen, Die Lehre von der Kegelschnitte im Alterthum, Copenaghen 1886; id., Geschichte der Mathematik im XVI. und XVII. Jahrhundert, Lipsia 1903. - Per le teorie proiettive: F. Enriques, Lezioni di geometria proiettiva, Bologna 1898, 4ª ed., 1922; F. Severi, Geometria proiettiva, Padova 1922, 2ª ed., Firenze 1926. - Per le teorie analitiche, v. la bibl. della voce coordinate, e, per una trattazione d'insieme, l'art. di F. Dingeldey, Kegelschnitte und Kegelschnittsysteme, in Encyklopädie der math. Wiss., III, ii, i, Lipsia 1903-1915, pp. 1-160. © Istituto della Enciclopedia Italiana fondata da Giovanni Treccani - Riproduzione riservata Altri risultati di ricerca Enciclopedia della Matematica (2013) ### conica degenere conica degenere insieme dei punti intersezione della superficie di un cono con un piano passante per il suo vertice. Può essere costituita da una coppia di rette incidenti, da una coppia di rette coincidenti, da un solo punto oppure, se il cono stesso degenera in un cilindro, ... Enciclopedie on line ### conica Curva che si ottiene segando un cono circolare (retto od obliquo) con un piano. Il cono va pensato come luogo di rette, e non di semirette, uscenti dal vertice V, cioè costituito, come si usa dire nel linguaggio elementare, da due ‘semiconi’ opposti al vertice. Si presentano t... Dizionario delle Scienze Fisiche (1996) ### conica cònica [s.f. dall'agg. conico, propr. "sezione conica"] [ALG] Curva che si ottiene segando un cono circolare (retto od obliquo) con un piano. Il cono va pensato come luogo di rette, e non di semirette, uscenti da un vertice V, cioè costituito, come si usa dire nel linguaggio e... cònica conica cònica s. f. [femm. sostantivato di conico, propr. «sezione conica»]. – In geometria, curva ottenuta come sezione piana di un cono circolare (o, più precisamente, di una superficie conica a due falde): a seconda dell’angolo formato ... Vai alla definizione conicità conicita conicità s. f. [der. di conico]. – L’essere conico, forma conica: c. di una figura, di un solido. Vai alla definizione Vedi anche Enciclopedie on line ### Apollònio di Perge Matematico greco (262 circa -180 a. C. circa); studiò in Alessandria con Euclide e Archimede, con i quali costituisce la triade dei sommi matematici della Grecia. Della sua opera fondamentale, in otto libri, sulle Coniche, solo i primi quattro ci sono pervenuti nell'originale,... Enciclopedie on line ### Pierre de Fermat Matematico francese (Beaumont-de-Lomagne, Tarn-et-Garonne, 1601 - Castres 1665).Autore di studi sul calcolo delle aree di figure piane, sul calcolo delle probabilità in problemi di giochi d'azzardo e nel campo dell'ottica geometrica, ha legato soprattutto il suo nome a teorem... Enciclopedie on line ### esagono Poligono piano con sei vertici e quindi sei lati. L’ e. regolare (avente tutti i sei lati uguali tra di loro e così tutti gli angoli formati da lati consecutivi) può iscriversi nel cerchio facendo uso soltanto di riga e compasso; il suo lato AB (fig. 1) è uguale al raggio OA d... Enciclopedie on line ### quadrica Superficie algebrica del secondo ordine. Sono q., per es., gli ellissoidi (di cui sono un caso particolare le sfere), i paraboloidi, gli iperboloidi. L’equazione di una q. in coordinate cartesiane è del tipo a11x2 + a22y2 + a33z2 + 2a12xy + 2a13xz + 2a23yz + 2a14x + 2a24y + 2a... Enciclopedie on line ### Felix Klein Matematico tedesco (Düsseldorf 1849 - Gottinga 1925).Autore di rilevanti contributi alla geometria,realizzò una classificazione di tale materia fondata sul concetto di gruppo, studiò le superfici algebriche (in topologia l'otre di K. è una superficie non orientabile a una so... Enciclopedie on line ### John Wallis Matematico (Ashford 1616 - Oxford, Inghilterra, 1703). Personalità poliedrica, W. fu insegnante, teologo, e uomo politico, ma la sua impronta maggiore resta per i suoi studi matematici. Si occupò di quadratura delle curve, di coniche, di logica e teoria della definizione. Tra ... Enciclopedie on line ### Luigi Cremóna Matematico italiano (Pavia 1830 - Roma 1903), fratello del pittore Tranquillo. Fece i suoi studî a Pavia, interrompendoli nel 1848 per partecipare come volontario alla guerra per l'indipendenza. Fu prof. di geometria superiore nell'univ. di Bologna (1860), poi a Milano (1866),... Enciclopedie on line ### Christiaan Huygens {{{1}}} Fisico, astronomo e matematico olandese (L'Aia 1629 - ivi 1695). Membro della Royal Society di Londra (1663) e dell'Académie des sciences di Parigi (1666), è tra i fondatori della meccanica e dell'ottica fisica. Suo padre Constantijn, diplomatico e segretario del prin... ISTITUTOFONDAZIONECORPORATEEVENTI ENCICLOPEDIAVOCABOLARIOSINONIMIDIZIONARIO BIOGRAFICOMAGAZINEGALASSIA TRECCANI RegistratiAccediPassword dimenticataLavora con noiWhistleblowing Istituto della Enciclopedia Italiana fondata da Giovanni Treccani S.p.A. Partita Iva 00892411000 Download App ContattiChi siamoTermini e condizioni generaliCondizioni di utilizzo dei serviziCookie PolicyPrivacy Policy © Tutti i diritti riservati Enciclopedia Vocabolario Sinonimi Dizionario biografico Magazine Podcast Galassia treccani
2324	https://pubmed.ncbi.nlm.nih.gov/26444994/	Treatment of Symptoms of the Menopause: An Endocrine Society Clinical Practice Guideline - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Epub 2015 Oct 7. Treatment of Symptoms of the Menopause: An Endocrine Society Clinical Practice Guideline Cynthia A Stuenkel1,Susan R Davis1,Anne Gompel1,Mary Ann Lumsden1,M Hassan Murad1,JoAnn V Pinkerton1,Richard J Santen1 Affiliations Expand Affiliation 1 University of California, San Diego, Endocrine/Metabolism (C.A.S.), La Jolla, California 92093; Monash University, School of Public Health and Preventive Medicine (S.R.D.), Melbourne 03004, Australia; Université Paris Descartes, Hôpitaux Universitaires Port Royal-Cochin Unit de Gynécologie Endocrnienne (A.G.), Paris 75014, France; University of Glasgow School of Medicine (M.A.L.), Glasgow G31 2ER, Scotland; Mayo Clinic, Division of Preventive Medicine (M.H.M.), Rochester, Minnesota 55905; University of Virginia, Obstetrics and Gynecology (J.V.P.), Charlottesville, Virginia 22908; and University of Virginia Health System (R.J.S.), Charlottesville, Virginia 22903. PMID: 26444994 DOI: 10.1210/jc.2015-2236 Free article Item in Clipboard Practice Guideline Treatment of Symptoms of the Menopause: An Endocrine Society Clinical Practice Guideline Cynthia A Stuenkel et al. J Clin Endocrinol Metab.2015 Nov. Free article Show details Display options Display options Format J Clin Endocrinol Metab Actions Search in PubMed Search in NLM Catalog Add to Search . 2015 Nov;100(11):3975-4011. doi: 10.1210/jc.2015-2236. Epub 2015 Oct 7. Authors Cynthia A Stuenkel1,Susan R Davis1,Anne Gompel1,Mary Ann Lumsden1,M Hassan Murad1,JoAnn V Pinkerton1,Richard J Santen1 Affiliation 1 University of California, San Diego, Endocrine/Metabolism (C.A.S.), La Jolla, California 92093; Monash University, School of Public Health and Preventive Medicine (S.R.D.), Melbourne 03004, Australia; Université Paris Descartes, Hôpitaux Universitaires Port Royal-Cochin Unit de Gynécologie Endocrnienne (A.G.), Paris 75014, France; University of Glasgow School of Medicine (M.A.L.), Glasgow G31 2ER, Scotland; Mayo Clinic, Division of Preventive Medicine (M.H.M.), Rochester, Minnesota 55905; University of Virginia, Obstetrics and Gynecology (J.V.P.), Charlottesville, Virginia 22908; and University of Virginia Health System (R.J.S.), Charlottesville, Virginia 22903. PMID: 26444994 DOI: 10.1210/jc.2015-2236 Item in Clipboard Full text links Cite Display options Display options Format Abstract Objective: The objective of this document is to generate a practice guideline for the management and treatment of symptoms of the menopause. Participants: The Treatment of Symptoms of the Menopause Task Force included six experts, a methodologist, and a medical writer, all appointed by The Endocrine Society. Evidence: The Task Force developed this evidenced-based guideline using the Grading of Recommendations, Assessment, Development, and Evaluation (GRADE) system to describe the strength of recommendations and the quality of evidence. The Task Force commissioned three systematic reviews of published data and considered several other existing meta-analyses and trials. Consensus process: Multiple e-mail communications, conference calls, and one face-to-face meeting determined consensus. Committees of The Endocrine Society, representatives from endorsing societies, and members of The Endocrine Society reviewed and commented on the drafts of the guidelines. The Australasian Menopause Society, the British Menopause Society, European Menopause and Andropause Society, the European Society of Endocrinology, and the International Menopause Society (co-sponsors of the guideline) reviewed and commented on the draft. Conclusions: Menopausal hormone therapy (MHT) is the most effective treatment for vasomotor symptoms and other symptoms of the climacteric. Benefits may exceed risks for the majority of symptomatic postmenopausal women who are under age 60 or under 10 years since the onset of menopause. Health care professionals should individualize therapy based on clinical factors and patient preference. They should screen women before initiating MHT for cardiovascular and breast cancer risk and recommend the most appropriate therapy depending on risk/benefit considerations. Current evidence does not justify the use of MHT to prevent coronary heart disease, breast cancer, or dementia. Other options are available for those with vasomotor symptoms who prefer not to use MHT or who have contraindications because these patients should not use MHT. Low-dose vaginal estrogen and ospemifene provide effective therapy for the genitourinary syndrome of menopause, and vaginal moisturizers and lubricants are available for those not choosing hormonal therapy. All postmenopausal women should embrace appropriate lifestyle measures. PubMed Disclaimer Similar articles The 2017 hormone therapy position statement of The North American Menopause Society.[No authors listed][No authors listed]Menopause. 2018 Nov;25(11):1362-1387. doi: 10.1097/GME.0000000000001241.Menopause. 2018.PMID: 30358733 Hormonal Replacement in Hypopituitarism in Adults: An Endocrine Society Clinical Practice Guideline.Fleseriu M, Hashim IA, Karavitaki N, Melmed S, Murad MH, Salvatori R, Samuels MH.Fleseriu M, et al.J Clin Endocrinol Metab. 2016 Nov;101(11):3888-3921. doi: 10.1210/jc.2016-2118. Epub 2016 Oct 13.J Clin Endocrinol Metab. 2016.PMID: 27736313 An introduction to the Endocrine Society Clinical Practice Guideline on treatment of symptoms of the menopause.Stuenkel CA, Santen RJ.Stuenkel CA, et al.Post Reprod Health. 2016 Mar;22(1):6-8. doi: 10.1177/2053369115626029.Post Reprod Health. 2016.PMID: 26936942 Androgen therapy in women: a reappraisal: an Endocrine Society clinical practice guideline.Wierman ME, Arlt W, Basson R, Davis SR, Miller KK, Murad MH, Rosner W, Santoro N.Wierman ME, et al.J Clin Endocrinol Metab. 2014 Oct;99(10):3489-510. doi: 10.1210/jc.2014-2260.J Clin Endocrinol Metab. 2014.PMID: 25279570 Postmenopausal hormone therapy: an Endocrine Society scientific statement.Santen RJ, Allred DC, Ardoin SP, Archer DF, Boyd N, Braunstein GD, Burger HG, Colditz GA, Davis SR, Gambacciani M, Gower BA, Henderson VW, Jarjour WN, Karas RH, Kleerekoper M, Lobo RA, Manson JE, Marsden J, Martin KA, Martin L, Pinkerton JV, Rubinow DR, Teede H, Thiboutot DM, Utian WH; Endocrine Society.Santen RJ, et al.J Clin Endocrinol Metab. 2010 Jul;95(7 Suppl 1):s1-s66. doi: 10.1210/jc.2009-2509. Epub 2010 Jun 21.J Clin Endocrinol Metab. 2010.PMID: 20566620 Free PMC article.Review. See all similar articles Cited by Reproductive Health as a Marker of Subsequent Cardiovascular Disease: The Role of Estrogen.Manson JE, Woodruff TK.Manson JE, et al.JAMA Cardiol. 2016 Oct 1;1(7):776-777. doi: 10.1001/jamacardio.2016.2662.JAMA Cardiol. 2016.PMID: 27626902 Free PMC article.No abstract available. Sexual dysfunctions and their treatment in liver diseases.Jagdish RK.Jagdish RK.World J Hepatol. 2022 Aug 27;14(8):1530-1540. doi: 10.4254/wjh.v14.i8.1530.World J Hepatol. 2022.PMID: 36157870 Free PMC article.Review. 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2325	https://www.quora.com/How-would-one-prove-that-every-normal-to-the-unit-circle-x-2-y-2-1-passes-through-the-origin	Something went wrong. Wait a moment and try again. Origin (place) Straight Lines Coordinate Plane Unit Circle Proofs (mathematics) Coordinate Systems Geometric Mathematics 5 How would one prove that every normal to the unit circle x^(2) + y^(2) = 1 passes through the origin? Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views · 5y Geometrically, we know that the tangent is perpendicular to the radius to the point of contact. Thus the normal to a tangent contains this radius and so passes through the center of the circle. If you prefer an analytic proof, let the point on the unit circle be (cost,sint). The tangent line is, among those of equation a(x−cost)+b(y−sint)=0 the one having distance 1 from the origin: this means \|a(0−cost)+b(0−sint)\|√a2+b2=1 Removing the denominator and squaring, we get a2cos2t+2abcostsint+b2sin2t=a2+b2 that becomes a2sin2t−2abcostsint+b2cos2t=0 that i Geometrically, we know that the tangent is perpendicular to the radius to the point of contact. Thus the normal to a tangent contains this radius and so passes through the center of the circle. If you prefer an analytic proof, let the point on the unit circle be (cost,sint). The tangent line is, among those of equation a(x−cost)+b(y−sint)=0 the one having distance 1 from the origin: this means \|a(0−cost)+b(0−sint)\|√a2+b2=1 Removing the denominator and squaring, we get a2cos2t+2abcostsint+b2sin2t=a2+b2 that becomes a2sin2t−2abcostsint+b2cos2t=0 that is, (asint−bcost)2=0 Thus we can take a=cost and b=sint. The normal has equation a′(x−cost)+b′(y−sint)=0, where aa′+bb′=0 that’s the same as a′cost+b′sint=0 so we can choose a′=sint and b′=−cost. Thus the normal line has equation sint(x−cost)−cost(y−sint)=0 that simplifies to xsint−ycost=0 so it passes through the origin. Related questions If 2 10 + 2 10 = 2 x , what's x ? What is the radius of the circle passing through the points (1,2), (5,2) and (5,-2)? How do you prove that (x^2) ^(1/2) =\|x\|? How do I prove that y n − x n = ( y − x ) ( y n − 1 + y n − 2 x + . . . + y x n − 2 + x n − 1 ) ? Let C be the unit circle x^2+y^2= 1 in the xy-plane. How do you geometrically describe the set C × R? Vaidesh Kumar M B.Tech from SRM University, Kattankulathur (Graduated 2020) · 5y Method 1 - Use a Graph sheet( easy way) Method 2- Use calcus Differentiate x^2 + y^2 = 1 We get slope M1 = -x/y To have perpendicularity M1M2 = -1 M2= y/x Equation of normal Y = M2X + C C= 0, therefore normal passes through the orgin Goh Kim Tee Former Tutor at Private (non-agency) (2008–2017) · Author has 4.2K answers and 2.4M answer views · 5y Since the centre is at the origin ,all radii and their extensions are all the normals to the circle ,as at any point on the circumference ,the radius and the tangent are perpendicular . Analytically ,take any point P(h,k) on the circle , (i)slope function of a tangent to the circle dy/dx or y′(x) is given by : x²+y²=1 2x+2yy′=0 yy′=—x y′=—x/y At P , slope of the tangent is —h/k, that of the normal mN= k/h (ii) Equation of the normal at P(h,k) is given by : y—k=mN(x—k) y—k=k/h(x—h) y—k=kx/h—k yh—kh=kx—kh yh=kx y=(k/h)x......this equation of the normal at any point P(h,k) on the unit circle ,is the equation of Since the centre is at the origin ,all radii and their extensions are all the normals to the circle ,as at any point on the circumference ,the radius and the tangent are perpendicular . Analytically ,take any point P(h,k) on the circle , (i)slope function of a tangent to the circle dy/dx or y′(x) is given by : x²+y²=1 2x+2yy′=0 yy′=—x y′=—x/y At P , slope of the tangent is —h/k, that of the normal mN= k/h (ii) Equation of the normal at P(h,k) is given by : y—k=mN(x—k) y—k=k/h(x—h) y—k=kx/h—k yh—kh=kx—kh yh=kx y=(k/h)x......this equation of the normal at any point P(h,k) on the unit circle ,is the equation of a straight line passing through the origin , which is the centre of the unit circle . Hence at ANY point P(h,k) on the unit circle , the normal passes through its centre. Dean Rubine Been doing high school math since high school, circa 1975 · Author has 10.6K answers and 23.7M answer views · 5y Let’s try it without calculus. Let f(x,y)=x2+y2−1 and imagine (r,s) is a point on f(x,y)=0 f(x,y)=f(r+(x−r),s+(y−s))=(r+(x−r))2+(s+(y−s))2−1 f(x,y)=r2+2r(x−r)+(x−r)2+s2+2s(y−s)+(y−s)2−1 f(x,y)=r2+s2−1+2r(x−r)+2s(y−s)+(x−r)2+(y−s)2 The constant is f(r,s), which is zero because (r,s) is assumed to be on the curve. f(x,y)=2r(x−r)+2s(y−s)+(x−r)2+(y−s)2 The tangent line is the best linear approximation to f at (r,s). We drop the quadratic terms and set the whole thing to zero for the tangent line: 0=2r(x−r)+2s(y−s) 0=r(x−r)+s(y−s) For the normal through (r,s) we s Let’s try it without calculus. Let f(x,y)=x2+y2−1 and imagine (r,s) is a point on f(x,y)=0 f(x,y)=f(r+(x−r),s+(y−s))=(r+(x−r))2+(s+(y−s))2−1 f(x,y)=r2+2r(x−r)+(x−r)2+s2+2s(y−s)+(y−s)2−1 f(x,y)=r2+s2−1+2r(x−r)+2s(y−s)+(x−r)2+(y−s)2 The constant is f(r,s), which is zero because (r,s) is assumed to be on the curve. f(x,y)=2r(x−r)+2s(y−s)+(x−r)2+(y−s)2 The tangent line is the best linear approximation to f at (r,s). We drop the quadratic terms and set the whole thing to zero for the tangent line: 0=2r(x−r)+2s(y−s) 0=r(x−r)+s(y−s) For the normal through (r,s) we swap the coefficients on x and y, negating one, 0=s(x−r)−r(y−s) That’s sx−ry=0 which is through the origin.✓ Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Related questions Point M (t, 4/7) is located in the second quadrant on the unit circle. How do you determine the exact value of t? If you have cos 150 for your x axis on your unit circle, what will the corresponding y point be? What are the numbers on the ends of the X & Y planes of the unit circle (pi, pi/2, 0, and 3pi/2), but not the circle parts called? Can I call them the "stem values," or no? How can you prove that the x-axis and y-axis in the unit circle range between -1 and 1? How do I prove x + y = x 2 − x y + y 2 ? Assistant Bot · Sep 6 Consider the unit circle C: x^2 + y^2 = 1. A normal to C at a point P = (x0,y0) on C is a line through P perpendicular to the tangent to C at P. Compute the tangent direction by differentiating implicitly: 2x + 2y y' = 0, so at P y' = −x0 / y0 (provided y0 ≠ 0; the vertical tangent case is handled separately). The tangent line at P therefore has slope −x0/y0, so a normal line has slope the negative reciprocal, y0/x0. Equation of the normal through P: y − y0 = (y0/x0)(x − x0). Rearrange: x y0 − x0 y = x0 y0 − x0 y0 = 0, so x y0 − x0 y = 0, i.e. y0 x − x0 y = 0. Evaluate this line at the origin (0,0 Consider the unit circle C: x^2 + y^2 = 1. A normal to C at a point P = (x0,y0) on C is a line through P perpendicular to the tangent to C at P. Compute the tangent direction by differentiating implicitly: 2x + 2y y' = 0, so at P y' = −x0 / y0 (provided y0 ≠ 0; the vertical tangent case is handled separately). The tangent line at P therefore has slope −x0/y0, so a normal line has slope the negative reciprocal, y0/x0. Equation of the normal through P: y − y0 = (y0/x0)(x − x0). Rearrange: x y0 − x0 y = x0 y0 − x0 y0 = 0, so x y0 − x0 y = 0, i.e. y0 x − x0 y = 0. Evaluate this line at the origin (0,0): left-hand side is y0·0 − x0·0 = 0, so the origin satisfies the line equation. Hence the normal passes through (0,0). Handle the exceptional cases explicitly: If y0 = 0, then x0 = ±1 and the tangent is vertical, so the normal is horizontal (y = 0), which passes through the origin. If x0 = 0, then y0 = ±1 and the tangent is horizontal, so the normal is vertical (x = 0), which passes through the origin. Therefore every normal to the unit circle passes through the origin. Richard Davis retired teacher · Author has 157 answers and 231.1K answer views · 5y I am going to answer this a little bit differently. Normal to the circle means perpendicular to the tangent of the circle. The slope of the tangent line is -x/y as shown by others. x^2+y^2=1, taking the derivative and treating y as a function of x 2x+2yy’=0, y’=-x/y. The slope of the line passing from the origin to the point (x,y) has the slope of y/x. multiply (y/x)(-x/y) =-1. This means that line passing through the origin to a point on the circle is perpendicular to the tangent at that point, or normal to the circle. I am going to answer this a little bit differently. Normal to the circle means perpendicular to the tangent of the circle. The slope of the tangent line is -x/y as shown by others. x^2+y^2=1, taking the derivative and treating y as a function of x 2x+2yy’=0, y’=-x/y. The slope of the line passing from the origin to the point (x,y) has the slope of y/x. multiply (y/x)(-x/y) =-1. This means that line passing through the origin to a point on the circle is perpendicular to the tangent at that point, or normal to the circle. B.L. Srivastava Author has 7.6K answers and 8.1M answer views · 5y The line y = mx (+/-) a.sqrt(1+m^2)… ..,(1) is always a tangent line to the circle x^2 + y^2 = a^2 for every value of m, at the point P(-m/sqrt(1+m^2) , a/sqrt(1+m^2)) for m > 0 . Here a = 1 . Now, equation of normal line is one, which is perpendicular to the tangent line, therefore its equation may be written with the help of (1) as ; my + x + k = 0…. …(2) . Clearly this line passes through the point P, therefore it must be satisfied by the co-ordinates of the point P . Putting the co-ordinates of the point P in (2),we get k=0 .Hence the equation of the normal is given by (my + x) = 0 and thi The line y = mx (+/-) a.sqrt(1+m^2)… ..,(1) is always a tangent line to the circle x^2 + y^2 = a^2 for every value of m, at the point P(-m/sqrt(1+m^2) , a/sqrt(1+m^2)) for m > 0 . Here a = 1 . Now, equation of normal line is one, which is perpendicular to the tangent line, therefore its equation may be written with the help of (1) as ; my + x + k = 0…. …(2) . Clearly this line passes through the point P, therefore it must be satisfied by the co-ordinates of the point P . Putting the co-ordinates of the point P in (2),we get k=0 .Hence the equation of the normal is given by (my + x) = 0 and this line clearly passes through the origin(0, 0), the center of the circle. Sponsored by CDW Corporation Ready for the latest in AI-powered capability? AMD Ryzen™ and Windows 11-powered x86 PCs from CDW drive business productivity with AI intuition. L Viswanathan Former Member of Technical Staff at Bell Telephone Labs (1977–1999) · Author has 3K answers and 1.4M answer views · 5y Let P(p, q) be a point on the circle. Differentiating, 2yy’(x) = -2(x). Slope m of tangent line through P(p, q), m = y’(p) = (-p)/(q ). Slope of normal through P = -1/m = (q/p). Equation of normal through P: y = (q/p)(x). Clearly normal passes through (0, 0). Slightly different approach: Let P(p, q) be a point on the circle. The equation of the tangent line T to the circle at P has the equation: (p)(x) + (q)(y) = 1. Right away one may write the equation of the normal N as: (q)(x) - (p)(y) = (q)(p) - (p)(q) = 0, since N passes through P(p, q). Equation of N: y = (q/p)(x). Clearly N passes through Let P(p, q) be a point on the circle. Differentiating, 2yy’(x) = -2(x). Slope m of tangent line through P(p, q), m = y’(p) = (-p)/(q ). Slope of normal through P = -1/m = (q/p). Equation of normal through P: y = (q/p)(x). Clearly normal passes through (0, 0). Slightly different approach: Let P(p, q) be a point on the circle. The equation of the tangent line T to the circle at P has the equation: (p)(x) + (q)(y) = 1. Right away one may write the equation of the normal N as: (q)(x) - (p)(y) = (q)(p) - (p)(q) = 0, since N passes through P(p, q). Equation of N: y = (q/p)(x). Clearly N passes through (0, 0). Zachary Carlini Studied Computer Science & Mathematics at University of Virginia (Graduated 2023) · Author has 149 answers and 604K answer views · Updated 6y Related How do you show that ( x + 1 ) 2 + y 2 = 2 is a circle? For this, we start with the definition of a circle. I’m sure there is a better way to define it, but the one I learned in school goes something like this: “The collection of all points equidistant from some centeral point.” For this to make sense, we also need to define “distance.” We’re going to use the pythagorean theorum for this one: d=√(x1−x0)2+(y1−y0)2 Where d is the distance between two points, (x0,y0) and (x1,y1). This comes from defining a right triangle between the two points and measuring their distance as the length of the hypotenuse. How does this all help us s For this, we start with the definition of a circle. I’m sure there is a better way to define it, but the one I learned in school goes something like this: “The collection of all points equidistant from some centeral point.” For this to make sense, we also need to define “distance.” We’re going to use the pythagorean theorum for this one: d=√(x1−x0)2+(y1−y0)2 Where d is the distance between two points, (x0,y0) and (x1,y1). This comes from defining a right triangle between the two points and measuring their distance as the length of the hypotenuse. How does this all help us show that your formula is a circle? Well, if we can show that every point in that graph is equidistant from some centeral point, we can prove it’s a circle. First, we need to find the centeral point. Here’s a graph of your function: Let’s conjecture it’s a circle with the center at (−1,0) and try to prove it. We can find the distance from (−1,0) to any arbitrary point (x,y) using the distance formula above: d=√(x−(−1))2+(y−0)2 =√(x+1)2+y2 We know that if (x,y) is a point on the graph of (x+1)2+y2=2, then (x+1)2+y2 must, by definition, equal 2. Using this information in conjunction with our distance formula above, we can find that the distance from any point on (x+1)2+y2 to (−1,0) is √[(x+1)2+y2] =√ (brackets added for emphasis, not the flooring function) Since √2 does not depend on x nor y, we have proven that the distance from every point on the curve (x+1)2+y2=2 to (−1,0) is the same, √2, so it must be a circle with center (−1,0) and radius √2. Q.E.D. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. 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This means that, by definition, it has a radius r that is constant. By the Pythagorean Theorem, the distance r between two points can be expressed through two perpendicular directions. Since the x and y axes are perpendicular, we can use them to create the equation x2+y2=r2. But that’s not quite what you have. What you have is a circle where your X value in that equation is (x+1). Is that still a circle? Why, yes. To prove it, one can simply substitute (X+1) for A circle is defined as the collection of points that are all a certain distance from a single point, referred to as the “center”. This means that, by definition, it has a radius r that is constant. By the Pythagorean Theorem, the distance r between two points can be expressed through two perpendicular directions. Since the x and y axes are perpendicular, we can use them to create the equation x2+y2=r2. But that’s not quite what you have. What you have is a circle where your X value in that equation is (x+1). Is that still a circle? Why, yes. To prove it, one can simply substitute (X+1) for X. But if that’s not sufficient, we can prove it a different way. Axes are defined simply because they make things easy to understand. We could define X as horizontal, and Y as having a slope 0.358285926, but that would be complicated and not very useful. In addition, we don’t have to have just two axes. Let’s define a new axis, V, that is parallel to X, but one unit above it. Since V and Y are still perpendicular, we can use them for the Pythagorean theorem to get V2+Y2=r2. And now, for any point on V, its value of X is (X+1). Now you can show that a circle centered at the intersection of V and Y can be defined in X and Y by (X+1)2+Y2=r2. Here, if r2 is a constant, then your equation is a perfect circle. Your r2 is 2, which is a constant, therefore the equation defines a perfect circle of radius √2. Alan Guo PhD student in Theoretical Computer Science at MIT · Upvoted by David Joyce , Professor of Mathematics at Clark University · Author has 78 answers and 524.8K answer views · 12y Related If a normal line to each point on ellipse passes through center of the ellipse, then the ellipse is a circle, prove? Without loss of generality, suppose your ellipse is defined by the equation x2a2+y2b2=r2. To show this is a circle, it suffices to show that a=b. Using implicit differentiation, we get 2xa2dx+2yb2dy=0 which can be rearranged to yield dydx=−b2a2⋅xy. This means that the slope of the line tangent to the ellipse at the point (x0,y0) is −b2a2⋅x0y0, so the slope of the normal line must be the negative reciprocal of this, which is a2b2⋅y0x0. The equation of th Without loss of generality, suppose your ellipse is defined by the equation x2a2+y2b2=r2. To show this is a circle, it suffices to show that a=b. Using implicit differentiation, we get 2xa2dx+2yb2dy=0 which can be rearranged to yield dydx=−b2a2⋅xy. This means that the slope of the line tangent to the ellipse at the point (x0,y0) is −b2a2⋅x0y0, so the slope of the normal line must be the negative reciprocal of this, which is a2b2⋅y0x0. The equation of the normal line passing through (x0,y0) is therefore y−y0x−x0=a2b2⋅y0x0. By assumption this line must pass through the center of the ellipse (which is the origin), so the equation should still be true when x=y=0. Taking nonzero x0,y0, we then get a2b2=1, hence a=b and our ellipse was a circle after all. Rohit Malik Ex-Class XII student, JEE Aspirant · Updated 7y Related How do I prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre? Wait,If you are looking for this question's solution for CLASS X level. Then,It is as follows….. 1- Make a circle with centre O. And a TANGENT FROM A POINT P TO R. 2- Now we r given that OR is perpendicular to RP Then Angle ORP = 90° . 3- Let a point O` NEAR TO O AND IT IS PERPENDICULAR TO RP. 5- Now we can observe that angle ORP<ORP THEN IT IS NOT POSSIBLE THAT OR is perpendicular to RP . It is only possible when O`P COINCIDES WITH OP. 5- HENCE OP PASS THROUGH CENTRE. HENCE PROVED….. AND KILL THAT UPVOTE BUTTON COZ IT TOOK ME 7 MINS TO WRITE ALL THIS.AND THE THING I ALWAYS WANTED TO WRITE AFTER EACH Wait,If you are looking for this question's solution for CLASS X level. Then,It is as follows….. 1- Make a circle with centre O. And a TANGENT FROM A POINT P TO R. 2- Now we r given that OR is perpendicular to RP Then Angle ORP = 90° . 3- Let a point O` NEAR TO O AND IT IS PERPENDICULAR TO RP. 5- Now we can observe that angle ORP<ORP THEN IT IS NOT POSSIBLE THAT OR is perpendicular to RP . It is only possible when O`P COINCIDES WITH OP. 5- HENCE OP PASS THROUGH CENTRE. HENCE PROVED….. AND KILL THAT UPVOTE BUTTON COZ IT TOOK ME 7 MINS TO WRITE ALL THIS.AND THE THING I ALWAYS WANTED TO WRITE AFTER EACH SOLUTION….. Lukas Schmidinger I have graduate CS and my studies included math courses. · Author has 27.7K answers and 14.9M answer views · 4y Related Can you verify this cos^2Θ≤1 using the definition of circular functions on the unit circle? Hint: the equation of the unit Circle is x^2 + y^2 = 1. Hint graphical definition of of these functions you can see that ( sin ( θ ) ) 2 + ( cos ( θ ) ) 2 = 1 (compare the Pythagorean theorem ) and also that − 1 ≤ sin ( θ ) , cos ( θ ) < 1 Hint graphical definition of of these functions you can see that ( sin ( θ ) ) 2 + ( cos ( θ ) ) 2 = 1 (compare the Pythagorean theorem ) and also that − 1 ≤ sin ( θ ) , cos ( θ ) < 1 Related questions If 2 10 + 2 10 = 2 x , what's x ? What is the radius of the circle passing through the points (1,2), (5,2) and (5,-2)? How do you prove that (x^2) ^(1/2) =\|x\|? How do I prove that y n − x n = ( y − x ) ( y n − 1 + y n − 2 x + . . . + y x n − 2 + x n − 1 ) ? Let C be the unit circle x^2+y^2= 1 in the xy-plane. How do you geometrically describe the set C × R? Point M (t, 4/7) is located in the second quadrant on the unit circle. How do you determine the exact value of t? If you have cos 150 for your x axis on your unit circle, what will the corresponding y point be? What are the numbers on the ends of the X & Y planes of the unit circle (pi, pi/2, 0, and 3pi/2), but not the circle parts called? Can I call them the "stem values," or no? How can you prove that the x-axis and y-axis in the unit circle range between -1 and 1? How do I prove x + y = x 2 − x y + y 2 ? How do you prove (x + y) ^2 ≤ 2 (x^2 + y^2)? Why does 1 + 1 = 2 ? How is 2 x + 2 x + 1 = 3 ( 2 x ) ? How do I prove 1 x 2 + 1 y 2 = 1 , I f sec − 1 ( x ) = c o s − 1 ( y ) ? If Δ x = x 2 − x 1 , does Δ x 2 = ( x 2 − x 1 ) 2 or Δ x 2 = x 2 2 − x 2 1 ? 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2326	https://www.gauthmath.com/solution/1833482615175313/This-tree-diagram-represents-possible-combinations-of-ice-cream-cones-flavors-an	Solved: This tree diagram represents possible combinations of ice cream cones, flavors, and toppin [Statistics] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Statistics Questions Question This tree diagram represents possible combinations of ice cream cones, flavors, and toppings. The probability of selecting an ice cream cone with strawberry ice cream is __. 1/3 2/3 1/2 1/6 Show transcript Gauth AI Solution 100%(3 rated) Answer $$\frac{2}{3}$$3 2 Explanation Diagram description: The tree diagram shows the possible combinations of ice cream cones, flavors, and toppings. The first split is between "cake" and "waffle" cones. From "cake," there are three flavor options: "choc.", "van.", and "straw.". From "waffle," there are also three flavor options: "choc.", "van.", and "straw.". Each flavor then splits into topping options. For "choc." flavor, the toppings are "nuts" and "spr.". For "van." flavor, the toppings are "spr." and "nuts". For "straw." flavor, the toppings are "spr." and "nuts". Determine the total number of possible outcomes. From the tree diagram, we can see that there are 2 types of cones (cake and waffle). Each cone has 3 flavor options (choc., van., straw.). Each flavor has 2 topping options (nuts, spr.) except for the last "straw." which has only one topping option "spr.". The total number of outcomes is $$3 \cdot 2 + 3 \cdot 2 + 3 \cdot 2 = 12$$3⋅2+3⋅2+3⋅2=12 Determine the number of outcomes with a cake cone. There are 3 flavors (choc., van., straw.) for the cake cone. Each flavor has 2 topping options (nuts, spr.) except for the last "straw." which has only one topping option "spr.". The number of outcomes with a cake cone is $$2 + 2 + 1 = 5$$2+2+1=5 Determine the number of outcomes with chocolate ice cream. There are 2 types of cones (cake and waffle). For the cake cone, the chocolate flavor has 2 topping options (nuts, spr.). For the waffle cone, the chocolate flavor has 2 topping options (nuts, spr.). The number of outcomes with chocolate ice cream is $$2 + 2 = 4$$2+2=4 Determine the number of outcomes with both a cake cone and chocolate ice cream. The cake cone with chocolate ice cream has 2 topping options (nuts, spr.). The number of outcomes with both a cake cone and chocolate ice cream is 2. Calculate the number of outcomes with either a cake cone or chocolate ice cream. Using the principle of inclusion-exclusion, the number of outcomes with either a cake cone or chocolate ice cream is: (Number of outcomes with a cake cone) + (Number of outcomes with chocolate ice cream) - (Number of outcomes with both a cake cone and chocolate ice cream) = $$5 + 4 - 2 = 7$$5+4−2=7 Calculate the probability of selecting an ice cream cone with a cake cone or chocolate ice cream. The probability is the number of favorable outcomes divided by the total number of possible outcomes. Probability = $$\frac{7}{12}$$12 7 Re-evaluate the total number of outcomes. Cake: choc (nuts, spr), van (spr, nuts), straw (spr, nuts) = 6 Waffle: choc (nuts, spr), van (spr, nuts), straw (spr, nuts) = 6 Total = 12 Re-evaluate the number of outcomes with a cake cone. Cake: choc (nuts, spr), van (spr, nuts), straw (spr, nuts) = 6 Re-evaluate the number of outcomes with chocolate ice cream. choc (nuts, spr) for cake = 2 choc (nuts, spr) for waffle = 2 Total = 4 Re-evaluate the number of outcomes with both a cake cone and chocolate ice cream. Cake and choc (nuts, spr) = 2 Re-calculate the number of outcomes with either a cake cone or chocolate ice cream. $$6 + 4 - 2 = 8$$6+4−2=8 Re-calculate the probability. $$\frac{8}{12} = \frac{2}{3}$$12 8=3 2 Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related Ice Cream Choices At an ice cream shop, you can choose between 2 types of cones waffle or sugar, flavors of ice cream vanilla, chocolate, or strawberry, and 2 toppings sprinkles or nuts. a Draw a tree diagram to represent all the possible combinations of ice cream cones, flavors, and toppings b How many different combinations are there in total? c What is the probability that a customer will choose a waffle cone with chocolate ice cream and sprinkles? d What is the probability that the customer will choose a sugar cone with any flavor of ice cream and no nuts? 100% (5 rated) An ice cream shop maintains only a few options for ice cream to keep their wait times for customers to a minimum. 3 points Here are the options for customers: Container: cup or cone Flavor: chocolate, vanilla, or strawberry Topping: caramel, sprinkles, cherries, or nuts a. Make a list or tree showing all of the combinations available to customers. b. How many possible combinations of container, flavor, and topping are available?_ c. What is the probability of a random customer choosing chocolate ice cream with car in a cup? a 1/18 c 1/9 b 3/18 d1/24 100% (5 rated) Guests at Brenda's birthday party were offered six choices of ice cream flavors: vanilla, chocolate chip, peppermint, chocolate, strawberry, or lime sherbert. What is the probability that Brenda chose a flavor that had chocolate in it? 1/3 2/3 1/2 1/6 100% (1 rated) Guests at Brenda's birthday party were offered six choices of ice cream flavors: vanilla, chocolate chip, peppermint, chocolate, strawberry, or lime sherbert. What is the probability that Brenda chose a flavor that had chocolate in it? 1/3 1/6 1/2 2/3 100% (4 rated) Guests at Brenda's birthday party were offered six choices of ice cream flavors: vanilla, chocolate chip, peppermint, chocolate, strawberry, or lime sherbert. What is the probability that Brenda chose a flavor that had chocolate in it? 2/3 1/2 1/6 1/3 100% (3 rated) Topic: Date Pd Name 4. Teresa's Totes has green, blue, pink, or red tote Probability Practic a bags. They can have short or long straps. The Directions: Use a table or draw a tree diagram to determine the bags can be personalized with your name on it or probability in each scenario. Also, use the multiplication counting blank. Finally, the bag can be made out of principle to show the total number of possible outcomes. canvas, denim, nylon, leather, or cotton. How 1. Pete's Pizza has small, medium, or large pies. many different tote bags can Teresa make? Topping choices are chicken, pineapple, basil, or mushrooms. Each pizza also comes with a Coke or Sprite. How many possible combinations are there for one pizza? 5. Gerry's Gift Wrapping service offers wrapping paper that is polka dot, stripped, or floral. They 2. Sally's Shop sells 8 shirts and 5 different pairs can add a bow or a ribbon to the package, and of pants. They also have boots, high heel shoes, or the package can be in a small, medium, or large sandals. How many different outfits can you make box. How many possible wrapping combinations with a shirt, a pair of pants, and a pair of shoes? can Gerry make? 6. Ike's Ice Cream Shop serves ice cream in a 3. Grandma's Gourmet Goodies makes vanilla or waffle cone, a sugar cone, or a cup. They have 12 chocolate cakes. The frosting can be chocolate flavors every day! You can get your ice cream mousse, vanilla buttercream, strawberry, with hot fudge, sprinkles, M&Ms, gummy bears, or chocolate mint, lemon, or raspberry. The filling caramel sauce on top. How many ice cream flavors are coffee, almond, or cream cheese. How combinations are possible? many different cakes can Grandma make? © Math Ali Day, 2022 © Math All Day, 2022 100% (5 rated) Tree diagrams are useful for A illustrating the law of large numbers. B finding all possible outcomes in a probability experiment involving several steps. C showing that the outcome is the set of all possible sample spaces. D ordering outcomes from lowest to highest. 11. Find the probability of getting a number greater than 4 when a die is rolled one time. A 1/2 B 2/3 C 1/6 D 1/3 100% (1 rated) The round off errors when measuring the distance that a long jumper has jumped is uniformly distributed between 0 and 4.3 mm. Round values to 4 decimal places when possible. a. The mean of this distribution is square b. The standard deviation is square c. The probability that the round off error for a jumper's distance is exactly 0.6 is Px=0.6= square square d. The probability that the round off error for the distance that a long jumper has jumped is between 0.7 and 3.7 mm is P0.7 100% (1 rated) 5/24 Question 6 of 18, Step 1 of 1 Correct Find the number of modes, if any, in the following set of data. If a mode exists, also give the value of the mode. 96, 99, 45, 38,83, 99,15 100% (2 rated) of 8, Step 1 of 1 Correct The mean monthly electric bill for 105 residents of the local apartment complex is $ 91. What is the best point estimate for the mean monthly electric bill for all residents of the local apartment complex? 100% (4 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
2327	https://www.scribbr.com/statistics/effect-size/	### Have a language expert improve your writing Proofreading Services ### Run a free plagiarism check in 10 minutes Plagiarism Checker ### Generate accurate citations for free Citation Generator Home Knowledge Base Statistics What is Effect Size and Why Does It Matter? (Examples) What is Effect Size and Why Does It Matter? (Examples) Published on December 22, 2020 by Pritha Bhandari. Revised on June 22, 2023. Effect size tells you how meaningful the relationship between variables or the difference between groups is. It indicates the practical significance of a research outcome. A large effect size means that a research finding has practical significance, while a small effect size indicates limited practical applications. Table of contents Why does effect size matter? How do you calculate effect size? How do you know if an effect size is small or large? When should you calculate effect size? Other interesting articles Frequently asked questions about effect size Why does effect size matter? While statistical significance shows that an effect exists in a study, practical significance shows that the effect is large enough to be meaningful in the real world. Statistical significance is denoted by p values, whereas practical significance is represented by effect sizes. Statistical significance alone can be misleading because it’s influenced by the sample size. Increasing the sample size always makes it more likely to find a statistically significant effect, no matter how small the effect truly is in the real world. In contrast, effect sizes are independent of the sample size. Only the data is used to calculate effect sizes. That’s why it’s necessary to report effect sizes in research papers to indicate the practical significance of a finding. The APA guidelines require reporting of effect sizes and confidence intervals wherever possible. Here's why students love Scribbr's proofreading services Trustpilot Discover proofreading & editing How do you calculate effect size? There are dozens of measures for effect sizes. The most common effect sizes are Cohen’s d and Pearson’s r. Cohen’s d measures the size of the difference between two groups while Pearson’s r measures the strength of the relationship between two variables. Cohen’s d Cohen’s d is designed for comparing two groups. It takes the difference between two means and expresses it in standard deviation units. It tells you how many standard deviations lie between the two means. \| Cohen’s d formula \| Explanation \| --- \| \| = mean of Group 1 = mean of Group 2 s = standard deviation \| The choice of standard deviation in the equation depends on your research design. You can use: a pooled standard deviation that is based on data from both groups, the standard deviation from a control group, if your design includes a control and an experimental group, the standard deviation from the pretest data, if your repeated measures design includes a pretest and posttest. Pearson’s r Pearson’s r, or the correlation coefficient, measures the extent of a linear relationship between two variables. The formula is rather complex, so it’s best to use a statistical software to calculate Pearson’s r accurately from the raw data. \| Pearson’s r formula \| Explanation \| --- \| \| rxy = strength of the correlation between variables x and y n = sample size ∑ = sum of what follows X = every x-variable value Y = every y-variable value XY = the product of each x-variable score times the corresponding y-variable score \| The main idea of the formula is to compute how much of the variability of one variable is determined by the variability of the other variable. Pearson’s r is a standardized scale to measure correlations between variables—that makes it unit-free. You can directly compare the strengths of all correlations with each other. One caveat is that Pearson’s r, like Cohen’s d, can only be used for interval or ratio variables. Other measures of effect size must be used for ordinal or nominal variables. How do you know if an effect size is small or large? Effect sizes can be categorized into small, medium, or large according to Cohen’s criteria. Cohen’s criteria for small, medium, and large effects differ based on the effect size measurement used. \| Effect size \| Cohen’s d \| Pearson’s r \| --- \| Small \| 0.2 \| .1 to .3 or -.1 to -.3 \| \| Medium \| 0.5 \| .3 to .5 or -.3 to -.5 \| \| Large \| 0.8 or greater \| .5 or greater or -.5 or less \| Cohen’s d can take on any number between 0 and infinity, while Pearson’s r ranges between -1 and 1. In general, the greater the Cohen’s d, the larger the effect size. For Pearson’s r, the closer the value is to 0, the smaller the effect size. A value closer to -1 or 1 indicates a higher effect size. Pearson’s r also tells you something about the direction of the relationship: A positive value (e.g., 0.7) means both variables either increase or decrease together. A negative value (e.g., -0.7) means one variable increases as the other one decreases (or vice versa). The criteria for a small or large effect size may also depend on what’s commonly found research in your particular field, so be sure to check other papers when interpreting effect size. When should you calculate effect size? It’s helpful to calculate effect sizes even before you begin your study as well as after you complete data collection. Before starting your study Knowing the expected effect size means you can figure out the minimum sample size you need for enough statistical power to detect an effect of that size. In statistics, power refers to the likelihood of a hypothesis test detecting a true effect if there is one. A statistically powerful test is more likely to reject a false negative (a Type II error). If you don’t ensure enough power in your study, you may not be able to detect a statistically significant result even when it has practical significance. In that case you don’t reject the null hypothesis, even though there is an actual effect. By performing a power analysis, you can use a set effect size and significance level to determine the sample size needed for a certain power level. After completing your study Once you’ve collected your data, you can calculate and report actual effect sizes in the abstract and the results sections of your paper. Effect sizes are the raw data in meta-analysis studies because they are standardized and easy to compare. A meta-analysis can combine the effect sizes of many related studies to get an idea of the average effect size of a specific finding. But meta-analysis studies can also go one step further and also suggest why effect sizes may vary across studies on a single topic. This can generate new lines of research. Here's why students love Scribbr's proofreading services Trustpilot Discover proofreading & editing Other interesting articles If you want to know more about statistics, methodology, or research bias, make sure to check out some of our other articles with explanations and examples. Statistics Chi square test of independence Statistical power Descriptive statistics Degrees of freedom Pearson correlation Null hypothesis Methodology Double-blind study Case-control study Research ethics Data collection Hypothesis testing Structured interviews Research bias Hawthorne effect Unconscious bias Recall bias Halo effect Self-serving bias Information bias Frequently asked questions about effect size What is effect size? : Effect size tells you how meaningful the relationship between variables or the difference between groups is. A large effect size means that a research finding has practical significance, while a small effect size indicates limited practical applications. How do I calculate effect size? : There are dozens of measures of effect sizes. The most common effect sizes are Cohen’s d and Pearson’s r. Cohen’s d measures the size of the difference between two groups while Pearson’s r measures the strength of the relationship between two variables. What’s the difference between statistical and practical significance? : While statistical significance shows that an effect exists in a study, practical significance shows that the effect is large enough to be meaningful in the real world. Statistical significance is denoted by p-values whereas practical significance is represented by effect sizes. What is statistical power? : In statistics, power refers to the likelihood of a hypothesis test detecting a true effect if there is one. A statistically powerful test is more likely to reject a false negative (a Type II error). If you don’t ensure enough power in your study, you may not be able to detect a statistically significant result even when it has practical significance. Your study might not have the ability to answer your research question. Cite this Scribbr article If you want to cite this source, you can copy and paste the citation or click the “Cite this Scribbr article” button to automatically add the citation to our free Citation Generator. Is this article helpful? You have already voted. Thanks :-) Your vote is saved :-) Processing your vote... Pritha Bhandari Pritha has an academic background in English, psychology and cognitive neuroscience. As an interdisciplinary researcher, she enjoys writing articles explaining tricky research concepts for students and academics. Why do I see ads? Ads help us keep our tools free for everyone. Scribbr customers enjoy an ad-free experience! Other students also liked #### Statistical Power and Why It Matters \| A Simple Introduction Statistical power, or sensitivity, is the likelihood of a significance test detecting an effect when there actually is one. #### An Easy Introduction to Statistical Significance (With Examples) If a result is statistically significant, that means it's unlikely to be explained solely by random factors or chance. #### Understanding P values \| Definition and Examples The p-value shows the likelihood of your data occurring under the null hypothesis. P-values help determine statistical significance. What is your plagiarism score? Scribbr Plagiarism Checker
2328	https://www.math.purdue.edu/~egbertn/fa2016/notes/lesson21.pdf	Applications of log Nick Egbert MA 158 Lesson 21 Probably more so than anything we have done up to now, exponential and log func-tions give rise to many useful applications. To ease into the lesson, let’s start with something that requires no new information. By this point, we should be fairly comfort-able manipulating exponential and logarithmic equations. Example 1. Solve the following equation for t. a = b(8 −ect) Solution. We just want to isolate t. a = b(8 −ect) a b = 8 −ect ect = 8 −a b ct = ln 8 −a b t = 1 c ln 8 −a b . Note that in order to have a solution for t, we require that b ̸= 0 (which we need from the statement of the problem), we also require c ̸= 0 and because of the domain restriction from log, we need a/b < 8. One important type of example is that of 1 2-life. In a previous lesson, we used the A = Pert equation to model continuous compounding. In fact, we can model any exponential growth or decay using this exact expression. Typically the letters get changed a bit, but the essence is still there. So we will be using A(t) = A0ekt to model exponential growth (decay). In this equation A(t) represents the amount at time t, A0 is the initial amount, k is the rate of growth (decay), and t is time. In the interest problems we required that t be in years, but for other applications, we don’t 1 Nick Egbert MA 158 Lesson 21 2 make such an assertion. So t can be any unit of time, and k will be some constant rate which depends on the unit of time we use. In case you don’t know, 1 2-life refers to the amount of time it takes for some (generally radioactive) substance to decay to half of its original mass. For example Carbon-14 has a 1 2-life of about 5730 years. When scientists try to determine how old certain things are, they look at the amount of 14C is present and estimate the age with their knowledge of how quickly 14C decays. Example 2. You asked for gold for Christmas and foolishly did not specify which isotope. As a result, your friend gives you a 1kg sample of 196Au, which has a half-life of 148.39 hours. (a) Find a function A(t) for the amount of the isotope, A in grams, which remains after time t in hours. (b) Determine the time t in hours for 93% of the material to decay. Solution. (a) As always, it is important to pay attention to units. We want A to be in grams, but we are told that we start with 1 kg. So then A0 = 1000 g. Now we know that the function is of the form A(t) = 1000ekt, but we don’t know what k is. To ﬁnd k, we know that whenever t = 148.39, then we have half of what we started with. So then 500 = 1000e148.39k 1 2 = e148.39k ln 1 2 = 148.39k k = 1 148.39 ln 1 2 ≈−0.00467 So then our function is A(t) = 1000e−0.00467t. (b) If 93% of the material has decayed, that means that there is only 7% remaining. One way to set up this kind of problem is to think of our initial amount as 100%. Then this looks like 7 = 100e−0.00467t. Nick Egbert MA 158 Lesson 21 3 And now we want to solve for t. 7 = 100e−0.00467t 7 100 = e−0.00467t ln 7 100 = −0.00467t t = 1 −0.00467 ln 7 100 ≈569.298 hours Example 3. The radioactive isotope 93Sr has a half-life of 7.5 minutes. Find how long it will take for a sample to decay so that 63% of its original mass remains. Solution. In a situation like this, we must use the strategy from Example 2(b) as we are not given an initial mass. Just for variety, another way to think about it than in terms of percents is that when t = 7.5, then A(t) = 1 2A0. So then we may write 1 2A0 = A0e7.5t, And when we divide both sides by A0, we will get A0/A0 which is 1. Now 1 2A0 = A0e7.5t 1 2 = e7.5t ln 1 1 = 7.5t t = 1 7.5 ln 1 1 ≈−0.0924. So then the equation for the decay of the isotope is A(t) = A0e−0.0924t. Now to ﬁnd at what time 63% is remaining, it probably makes the most sense to view masses as percentages again. So we have 63 = 100e−0.0924t .63 = e−0.0924t ln(.63) = −0.0924t t = 1 −0.0924 ln(.63) ≈5.00 minutes. Population growth, especially that of bacteria and spread of disease are well-modeled by exponential growth functions. This next example is nice because we need to be able to interpret real-world information to translate it to the equations we are familiar with. Nick Egbert MA 158 Lesson 21 4 Example 4. In ideal conditions, the spread of disease follows the Law of Uninhibited Growth N(t) = N0ekt. During the 2014 West Africa Ebola outbreak, there were an estimated 800 reported cases of Ebola on July 1, and 1500 on August 1 (31 days later). Let t be the time elapsed in days since the ﬁrst observation. (a) Find the growth constant k. Round your answer to four decimal places. (b) Find a function which gives the number of reported cases N(t) after t days. (c) What is the doubling time for the number of reported cases? Solution. (a) So here N0 = N(0) = 800 and N(31) = 1500. So we already know that the function looks like N(t) = 800ekt. We just need to determine k. You should notice that as soon as we determine k, we immediately have a solution for part (b). Now using the second piece of data, we know 1500 = 800e31k 15 8 = e31k ln 15 8 = 31k k = 1 31 ln 15 8 ≈.02. (b) Here we just need to put it together, so N(t) = 800e.02t. (c) To ﬁgure out the doubling time is quite similar to ﬁguring out the half-life. Whenever we have an initial amount N0, we are looking for the time t for which N(t) = 2N0. So this looks like 2N0 = N0ekt. In this particular example, we know what N0 is, so we can just say we want to solve N(t) = 1600 for t. But this shows that we don’t actually need to know the initial amount to ﬁgure out the doubling time. Back to the problem at hand: 1600 = 800e.02t 2 = e.02t ln 2 = .02t t = ln 2 .02 ≈34.66 days. For our ﬁnal example, we turn to logistic growth, which has several applications, one of which is modeling population growth of cities. Nick Egbert MA 158 Lesson 21 5 Example 5. The population of a certain town is modeled by P(t) = 670 1 + 6.55e−0.51t where P(t) is the population t years after 2010. Such a model is called logistic growth (decay if we e had a positive exponent). (a) What is the population in 2010? (b) Find the population in 2012. (c) When will the population reach 310? Solution. (a) This is just asking for P(0). P(0) = 670 1 + 6.55e0 = 670 7.55 ≈89. We round to the nearest whole number under the assumption that we can’t have a fraction of a person. (b) This is just asking for P(2). P(2) = 670 1 + 6.55e−0.51·2 ≈199. (c) Here we want to solve for t in the expression P(t) = 310. 310 = 670 1 + 6.55e−0.51t 1 + 6.55e−0.51t = 670 310 6.55e−0.51t = 670 310 −1 e−0.51t = 1 6.55 670 310 −1 −0.51t = 1 6.55 670 310 −1 t = −1 0.51 1 6.55 670 310 −1 ! ≈3.39. We have to interpret this as years since 2010, so our ﬁnal answer is 2013.
2329	https://math.stackexchange.com/questions/1526995/shifted-argument-in-pde	ordinary differential equations - Shifted argument in PDE - 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 Shifted argument in PDE Ask Question Asked 9 years, 10 months ago Modified9 years, 10 months ago Viewed 186 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. In my research, I have come across the following differential equation: d f(x,t)d t=[f(x,t)−1]f(x+α,t)d f(x,t)d t=[f(x,t)−1]f(x+α,t) where α α is some constant. I have never seen such an equation before. Any suggestions how it can be solved? ordinary-differential-equations partial-differential-equations Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Nov 13, 2015 at 8:53 user56643user56643 299 2 2 silver badges 8 8 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. This is so-called mixed functional differential equation. There is a quite detailed account of the theory of such equations in Mixed functional differential equations. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 13, 2015 at 12:11 ArtemArtem 15k 4 4 gold badges 40 40 silver badges 55 55 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The transform Λ(x,t,ρ)=∫d α f(x+α,t)e−α x Λ(x,t,ρ)=∫d α f(x+α,t)e−α x turns the equation into an ordinary differential equation for Λ Λ. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 14, 2015 at 2:56 user56643user56643 299 2 2 silver badges 8 8 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions ordinary-differential-equations partial-differential-equations See similar questions with these tags. 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2330	https://climateatlas.ca/atlas-guidebook/interpreting-climate-data	Skip to main content En Fr Climate Atlas Guidebook The Basics • Using the Map • Local Data • Interpreting Climate Data This version of the guidebook refers to the initial 2018 version of the Atlas (version 1), but it still contains a lot of great information about working with climate data. As a result, we have elected to leave this up for archival purposes, even as the Atlas continues to be updated. Interpreting Climate Data In the past, trying to answer questions such as “how might summer temperatures in my town change in coming decades?” required a lot of technical skill and effort: climate data had to be found, selected, and downloaded; time periods and emissions scenarios had to be chosen; the data had to be computed, analyzed, mapped, graphed, and interpreted. The Climate Atlas has done all that work for you, freeing you up for the important business of exploring, learning, and responding. But interpreting climate model data can be complex, especially if you need to develop adaptation policies or do detailed planning. This section of the guide explains how to understand the data presented in the atlas so you can confidently work with it to answer your questions and best understand the future climate and its potential consequences. Historical and Modelled Data Most of the values presented in the Climate Atlas are from climate models. In particular, the 1976-2005 (Recent Past) values used throughout the Atlas as the baseline are from climate model simulations for this period. In the maps and on the Local Data Page, the atlas puts simulated means and ranges side-by-side with observed historical data for easy comparison. For the most part, you will see that the historical values fall within the range of the simulated data, especially for temperature. For the period 2006-2095 the atlas uses climate model data to show how the climate is projected to change, using two emissions scenarios. For more about climate models and modelling, see the “Climate Change Projections” article on the Atlas. Mean Values Many of the numbers presented in the Climate Atlas are averages. The map sidebar and various reports, for instance, prominently display the mean value for 12 climate models over one of three 30-year time periods (1976-2005, 2021-2050, 2051-2080). Looking at mean values provides a simple sense of the direction and general magnitude of change. The use of 30-year periods is a scientific convention based on the statistical guideline that a minimum of 30 points of data are required to reliably determine a mean. Thus, averaging data over 30 years is used to represent the average state of a climate. Taking the average for a 30-year period helps ensure that what is being described is an aspect of the climate system and not the more variable experience of weather. Yearly averages can change a lot from year to year, whereas a 30-year average removes a lot of that variation, and reveals common conditions across the time period more so than differences between years. This means that when 30-year averages differ, that difference is probably noteworthy because it represents a multi-year trajectory of change that’s unlikely to be caused by short-term (seasonal, yearly, or even decadal) variability. Averaging across an ensemble of 12 climate models is an important strategy that means we are not relying on results from just one or two climate models. Models take a variety of approaches and make different assumptions about how best to represent the amazing complexity of the global climate system; as such, they of course provide different projections of future climate conditions and different simulations of past climates. Taking the mean of an ensemble of 12 models—like taking the mean over a period of 30 years—emphasizes their common agreement about the direction and magnitude of change. Below is an example of projected change in the number of very cold days (under RCP8.5, the High Carbon emissions scenario leading to more severe climate change) for the region surrounding Iqaluit in Nunavut. These numbers tell a story of dramatic change, with the region losing 90% of its mean number of very cold winter days. Range and Distribution The mean values presented in the Climate Atlas provide a very important summary of the climate data, but also a very partial one. Means are simple measures of the “central tendency” or average midpoint of the values, but don’t convey how much the values vary from one another. Thus, another key aspect of the climate model data presented in the atlas is the variation among the 12 models, and across each 30-year period. This aspect of climate model data is present in the atlas in the form of model agreement (how much the projections from the 12 models in the ensemble vary) and yearly variability (how much the 30 individual years in a time period vary). Model agreement The 12 climate models used to generate the data presented in the atlas vary in their simulations of past climate and projections of future climate. The range of values produced by the models is presented in a couple of places in the Atlas. High and low model values The highest and lowest values in the ensemble of 12 models can be revealed in the map sidebar using the “More Detail” option that can be found below the mean value and above the graph. The above illustration shows the model range for some of the same data described above (very cold days for the region surrounding Iqaluit in Nunavut, under RCP 8.5, the “High Carbon” emissions scenario), comparing the Recent Past (1976-2005) with the Near Future (2051-2080). The Low and High ends of the range show the full range of model results for this time period and carbon scenario. In this case, it shows that the very highest projected number of cold days in the future is very close to the lowest end of the simulated historical values, confirming that a drastic overall change is projected across the range of models. The data also confirms that there is quite a bit of spread between the lowest and highest modelled values for each time period. Scatterplot Knowing the very top and bottom values of the range is helpful, but doesn’t fully capture the degree of agreement between the models. In order to provide a better visual sense of model variability, the atlas provides scatterplots of the basic model output (annual temperature and precipitation). This visualization can be found on the Local Data Page for any location in the atlas. This graph displays the projected change (or ‘delta’, Δ) in annual mean temperature and annual mean precipitation for the time periods 2021-2050 and 2051-2080, relative to the simulated values for 1976-2005. This scatterplot shows a high level of model agreement. All the models indicate that the climate is projected to get warmer and wetter. Yearly variability Thirty-year mean ensemble values are very helpful summaries, but of course they collapse thirty different yearly values into a single number. The differences between years are also important, and the distribution of those yearly values around the mean can look very different, given various climate scenarios and time periods. There are at least two graphs in the atlas that convey the yearly variation that is helpful to consider in addition to the overall 30-year mean. Line graphs These graphs display the full spread of the modelled values, year by year, and also display the observed historical record for easy comparison. A small version of this graph is found in the map sidebar, and a larger, downloadable version is available at the Local Data Page. The following image is a sample of the annual graphs available at the Local Data Page, showing the number of very cold days in the Iqaluit region. For the period 1950-2005 you can see a black line; this is the historical data, based on meteorological observations. Behind that line is a grey line and a shaded grey area. The line is the 12-model average of the data simulated for this location and time period; the shaded area marks the range of the simulations. For the period 2006-2095, the red line is the average of the 12 model projections for the chosen emissions scenario, and the shaded area defines the range among the models. Thus, in one graph you can see how well the models were able to simulate past conditions, and the trajectory, magnitude, and range of the changes projected for the coming decades. Switching between the Low and High Carbon scenarios also shows you how different the climate is expected to be under the two scenarios. Frequency plots One very effective way to examine and compare the conditions expected in the various time periods is through the use of frequency distribution curves. This graph is available on the Local Data Page, and presents the distribution of yearly ensemble values in each of the 30-year periods. These frequency curves show how often various values are projected to occur over each 30-year period. The impact of our changing climate can show up quite dramatically in the increased likelihood of what were previously rare and extreme conditions. They reveal what kinds of extremes might occur, and how often. Some very warm years, for example, might have many times more the number of hot days than the long-term average. The range of modelled values across a 30-year period highlights how variable our climate system is. In many cases, these ranges become larger the further you look into the future, an indication of increased climatic variability. For the Iqaluit region, the graph above shows that years with low numbers of very cold days were relatively rare in the Recent Past, but are projected to become much more common in the coming decades under the High Carbon scenario. The historically typical number of very cold days is set to become an unusually cold season, even in the 2021-2050 period, and vanishingly rare thereafter. Toggling between the Low and High Carbon scenarios shows how different the changes will be under the two scenarios. Uncertainty As the foregoing discussion of ranges and variability demonstrates, interpreting and understanding climate data inevitably means working with uncertainty. Climate scientists identify three main sources of uncertainty in the modelling of future climates. Natural Climate Variability The climate system has many components, some more important than others, and which operate over different time scales. This results in year-to-year and decade-by-decade variations in weather patterns. Climate Model Uncertainty Techniques are constantly improving and computers are getting more powerful, but climate models will always be partial simulations of the real world. Furthermore, each climate model produces somewhat different results, depending on how the modellers have chosen to represent different aspects of the climate system. Future Emissions Uncertainty Earth’s future climate is highly dependent on worldwide greenhouse gas emissions. Possible changes in the intensity of emissions depend on global demographic shifts and political decisions that are very complex to predict and model. The diagram below shows that the uncertainty related to emissions is much larger than that from model uncertainty (here called “climate response uncertainty”) and natural variability. In fact, the narrow band showing “Historical GCM uncertainty” shows that models do a very good job of simulating the climate of the past, which means we should pay careful attention to what they say about the climate of the future. There is a much fuller discussion of uncertainty in the Climate Atlas at climateatlas.ca/uncertainty if you want to know more. Working with uncertainty Understanding these various kinds of uncertainty is helpful when it comes to interpreting climate data. Representative Concentration Pathways (RCPs) have been developed to address the problem of uncertain future emissions by defining a variety of possible scenarios that lead to greater or lesser greenhouse gas concentrations in the atmosphere. This means, however, that one must decide which emissions trajectories to include in one’s climate studies. The Atlas offers two scenarios, RCP8.5 (“High Carbon”) and RCP4.5 (“Low Carbon”), that result in more warming or less warming, respectively. Other IPCC climate scenarios exist, specifically RCP6.0 and RCP2.6, but they have been excluded from the atlas to simplify the interface and to be consistent with national and international impact assessments that focus on RCP8.5 and RCP4.5. The Climate Atlas includes observed historical data to illustrate how well the climate models simulate the natural variability of past climates. For the most part, these historical values fall within the range of the simulated data, especially for temperature (precipitation tends to be more variable). And the atlas uses ensemble mean values to avoid the risk of relying on any one model and to mitigate the challenge of model uncertainty. In the simplest sense, “uncertainty” means we don’t have 100% of the information about a situation or system. As we almost never have absolutely all the information about anything, we constantly have to make decisions in the face of uncertainty. Our changing climate is no different. Importantly, uncertainty doesn’t prevent us from drawing confident conclusions about overall future patterns. Climate models are remarkably consistent in their overall results, and all of them tell essentially the same story: severe climate changes are likely to happen if we do not reduce greenhouse gas emissions soon. Recommended Article Citation Climate Atlas of Canada. (n.d.) Interpreting Climate Data. Prairie Climate Centre. Previous article Local Data Was this page helpful? Take Action Climate change touches all of our lives, one way or another. Because it affects all aspects of modern life, there are many kinds of meaningful choices we can make that will help. Climate change solutions start close to home, with simple actions in our own homes and families, but also involve new ways of thinking, planning, and acting in workplaces, neighbourhoods, and communities across the country. See our "Take Action" topic for ideas and examples of innovative climate action taking place all across Canada. Read more Sign up Join our mailing list to stay informed about Climate Atlas updates, including new maps, research, and videos, as well as outreach and educational activities by the Prairie Climate Centre. Help us create a dialogue across Canada: please subscribe. The Climate Atlas of Canada is part of a national suite of climate data portals. Stories and explanations about climate change, and local climate data that can be explored using maps and graphs. High-resolution climate data to help decision makers build a more resilient Canada. Advanced tools for academia, climate scenario developers, and other expert users. This project was undertaken with the financial support of: © 2025 Prairie Climate Centre
2331	https://pubmed.ncbi.nlm.nih.gov/16360301/	"Hot potato voice" in peritonsillitis: a misnomer - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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"Hot potato voice" in peritonsillitis: a misnomer Mahmood F Bhutta1,George A Worley,Meredydd L Harries Affiliations Expand Affiliation 1 Department of Ear Nose and Throat, The Royal Sussex County Hospital, Brighton, UK. PMID: 16360301 DOI: 10.1016/j.jvoice.2005.07.005 Item in Clipboard "Hot potato voice" in peritonsillitis: a misnomer Mahmood F Bhutta et al. J Voice.2006 Dec. Show details Display options Display options Format J Voice Actions Search in PubMed Search in NLM Catalog Add to Search . 2006 Dec;20(4):616-22. doi: 10.1016/j.jvoice.2005.07.005. Epub 2005 Dec 19. Authors Mahmood F Bhutta1,George A Worley,Meredydd L Harries Affiliation 1 Department of Ear Nose and Throat, The Royal Sussex County Hospital, Brighton, UK. PMID: 16360301 DOI: 10.1016/j.jvoice.2005.07.005 Item in Clipboard Full text links Cite Display options Display options Format Abstract The "hot potato voice" is widely recognized as a symptom of peritonsillar cellulitis or abscess; yet there have been no studies assessing the resonance characteristics of the vocal tract in peritonsillitis. Analysis was undertaken of formant frequencies in the articulation of the vowels /i:/. /a:/ and /u:/ in six subjects with peritonsillitis and compared with articulation once the peritonsillitis had settled. Significant variation was found in F1 when articulating /i:/ and in F2 when articulating /a:/, which are explainable by dyskinesis of the peritonsillar musculature. These findings were compared with six subjects articulating the same vowels with and without a hot potato in their mouth. Variation was found in both F1 and F2 when articulating /i:/, which can be related to interference of the potato with movement of the anterior tongue. The changes in the vocal tract differ in these two cases and the title "hot potato voice" in peritonsillitis is a misnomer. PubMed Disclaimer Similar articles Voice and upper airway symptoms in people with chronic cough and paradoxical vocal fold movement.Vertigan AE, Theodoros DG, Gibson PG, Winkworth AL.Vertigan AE, et al.J Voice. 2007 May;21(3):361-83. doi: 10.1016/j.jvoice.2005.12.008. Epub 2006 Mar 20.J Voice. 2007.PMID: 16545940 Peritonsillar abscess as a cause of transient velopharyngeal insufficiency.Finkelstein Y, Bar-Ziv J, Nachmani A, Berger G, Ophir D.Finkelstein Y, et al.Cleft Palate Craniofac J. 1993 Jul;30(4):421-8. doi: 10.1597/1545-1569_1993_030_0421_paaaco_2.3.co_2.Cleft Palate Craniofac J. 1993.PMID: 8399275 Management of peritonsillitis/peritonsillar.Raut VV.Raut VV.Rev Laryngol Otol Rhinol (Bord). 2000;121(2):107-10.Rev Laryngol Otol Rhinol (Bord). 2000.PMID: 10997070 Effects of tonsillectomy on articulation.Hori Y, Koike Y, Ohyama G, Otsu SY, Abe K.Hori Y, et al.Acta Otolaryngol Suppl. 1996;523:248-51.Acta Otolaryngol Suppl. 1996.PMID: 9082797 [Acute tonsillitis, peritonsillitis, and deep neck infection].Nonomura N.Nonomura N.Ryoikibetsu Shokogun Shirizu. 1999;(25 Pt 3):304-6.Ryoikibetsu Shokogun Shirizu. 1999.PMID: 10337810 Review.Japanese.No abstract available. See all similar articles Cited by Airway compromise in infectious mononucleosis: a case report.Kakani S.Kakani S.Cases J. 2009 Aug 13;2:6736. doi: 10.4076/1757-1626-2-6736.Cases J. 2009.PMID: 19918540 Free PMC article. 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2332	https://www.geogebra.org/m/bRzCcwh8	Determinant of a 2 x 2 Matrix – GeoGebra Google Classroom GeoGebra Classroom Sign in Search Google Classroom GeoGebra Classroom Home Resources Profile Classroom App Downloads Determinant of a 2 x 2 Matrix Author:Hisham Amir Determinant calculator of a 2 x 2 Matrix. Move the sliders to change the values of a, b, c and d New Resources 拼砌四邊形 - 工作紙 Pythagorean Day 判斷錐體 MC #2 seo tool Discover Resources Doppler Shifted Wavelength of Light from Moving HI Region Duality with triangles and squares triangfromeulerline Miniproject#2 Camping Pot Stand Plans ting Discover Topics Cuboid Normal Distribution Parametric Curves Linear Regression Complex Numbers AboutPartnersHelp Center Terms of ServicePrivacyLicense Graphing CalculatorCalculator SuiteMath Resources Download our apps here: English / English (United States) © 2025 GeoGebra®
2333	https://stats.stackexchange.com/questions/296005/the-expected-number-of-unique-elements-drawn-with-replacement	Skip to main content The expected number of unique elements drawn with replacement Ask Question Asked Modified 3 years, 7 months ago Viewed 12k times This question shows research effort; it is useful and clear 11 Save this question. Show activity on this post. I'm following this question which describes almost exactly my situation (with little modifications, in bold): Drawing, with replacement, k balls from a bin of n different colored balls, with an equal probability of drawing each color of a ball, what is the expected number of "unique" colors? How many different colors are we expected to get? As I'm not so strong in statistics, I approached this numerically, and after some computations and fitting I got this: u(n,k)=n(1−e(−k/n)) Where u is the expected number of unique colors, n is the number of available colors, and k is the sample size (Note that k may be larger than n). It seems to fit exactly the numerical results. My only problem is that I have no idea why it is so. I read the answer to the question cited above and tried to implement it in my case with no success. I'll be glad for an explanation that doesn't assume knowledge of advanced statistics. probability binomial-distribution expected-value Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Oct 18, 2017 at 18:41 EBH asked Aug 3, 2017 at 14:46 EBHEBH 22111 gold badge22 silver badges99 bronze badges Add a comment \| 1 Answer 1 Reset to default This answer is useful 14 Save this answer. Show activity on this post. The solution to this problem makes use of a classic technique in probability, which is to begin by defining a set of so-called indicator (i.e. binary-valued) random variables, and then to use linearity of expectation. We begin by defining for each of the n bins the random variable Ij={10if we draw at least one ball from the jth binotherwise. Letting X be the random variable denoting the number of different colored balls we draw, we have X=∑j=1nIj. Now using linearity of expectation, E[X]=E[∑j=1nIj]=∑j=1nE[Ij]. It remains to compute E[Ij] for j=1,…,n. Note that for any j E[Ij]=P(Ij=1)=P(draw at least one ball from bin j)=1−P(draw zero balls from bin j)=1−(n−1n)k. So the expected number of unique colors is E[X]=n[1−(n−1n)k] Note that the answer you provide is a close approximation since (n−1n)k=(1−1n)k=(1−1n)n⋅kn≈e−k/n. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Aug 3, 2017 at 18:24 answered Aug 3, 2017 at 17:34 tddevlintddevlin 3,41711 gold badge1818 silver badges3030 bronze badges 3 Thank you for the reply! I will take some time to make sure I understand this. My main difficulty is with the approximation at the end. – EBH Commented Aug 3, 2017 at 18:54 Glad you found the answer helpful! Note that understanding the approximation at the end is not essential to the solution. I only included that bit to explain why your proposed answer agrees well with simulation. – tddevlin Commented Aug 3, 2017 at 22:08 Well, in the meantime I found that the difference in the result never exceeds 1, and since I work in a discrete system where all values are integers, that virtually means no difference at all. – EBH Commented Aug 3, 2017 at 22:14 Add a comment \| Start asking to get answers Find the answer to your question by asking. 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2334	https://math.stackexchange.com/questions/1632763/how-to-go-upon-proving-fracxy2-ge-sqrtxy	algebra precalculus - How to go upon proving $\frac{x+y}2 \ge \sqrt{xy}$? - 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 How to go upon proving x+y 2≥x y−−√x+y 2≥x y? [duplicate] Ask Question Asked 9 years, 8 months ago Modified9 years, 8 months ago Viewed 845 times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. This question already has answers here: Proving the AM-GM inequality for 2 numbers x y−−√≤x+y 2 x y≤x+y 2 (5 answers) How to prove that a+b 2≥a b−−√a+b 2≥a b for a,b>0 a,b>0? [duplicate] (3 answers) Closed 9 years ago. I'm trying to prove this but am having some difficulty. For any x,y∈R x,y∈R such that x≥0 x≥0 and y≥0 y≥0 we have x+y 2≥x y−−√.x+y 2≥x y. So far what I have gotten to is x+y 2≥x y√2 x+y 2≥x y 2 After this point I don't know what to do. To get to this point: y≥0 y≥0 ⟹y≥y⟹y≥y ⟹x y≥x y⟹x y≥x y ⟹2 x y≥x y⟹2 x y≥x y ⟹x 2+2 x y+y 2≥x y⟹x 2+2 x y+y 2≥x y ⟹(x+y)2≥x y⟹(x+y)2≥x y Sqrt both sides to get: ⟹x+y≥x y−−√⟹x+y≥x y ⟹x+y 2≥x y−−√2⟹x+y 2≥x y 2 algebra-precalculus inequality means Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 30, 2016 at 9:52 Martin Sleziak 56.3k 20 20 gold badges 211 211 silver badges 391 391 bronze badges asked Jan 30, 2016 at 4:16 Leo DenniLeo Denni 159 9 9 bronze badges 9 How did you get to that point? (I actually don't see a way to get there without solving the problem all the way, so I'm curious.)Noah Schweber –Noah Schweber 2016-01-30 04:18:24 +00:00 Commented Jan 30, 2016 at 4:18 2 I updated the post to show my work Leo Denni –Leo Denni 2016-01-30 04:26:06 +00:00 Commented Jan 30, 2016 at 4:26 2 Why have you deleted your work? I think it was beneficial to the question.YoTengoUnLCD –YoTengoUnLCD 2016-01-30 06:14:56 +00:00 Commented Jan 30, 2016 at 6:14 2 The work leads to a dead end, so I feel it will mislead anyone else reading it.Leo Denni –Leo Denni 2016-01-30 06:18:14 +00:00 Commented Jan 30, 2016 at 6:18 See also math.stackexchange.com/questions/64881/…math.stackexchange.com/questions/904827/…math.stackexchange.com/questions/1114615/…math.stackexchange.com/questions/1150895/…math.stackexchange.com/questions/543253/…Martin Sleziak –Martin Sleziak 2016-01-30 09:46:55 +00:00 Commented Jan 30, 2016 at 9:46 \|Show 4 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 8 Save this answer. Show activity on this post. (x−y)2≥0 x 2+2 x y+y 2−4 x y≥0(x+y)2≥4 x y x+y 2≥x y−−√(x−y)2≥0 x 2+2 x y+y 2−4 x y≥0(x+y)2≥4 x y x+y 2≥x y Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 30, 2016 at 4:40 answered Jan 30, 2016 at 4:30 Ross MillikanRoss Millikan 384k 28 28 gold badges 264 264 silver badges 472 472 bronze badges 12 I have no idea what is going on here Leo Denni –Leo Denni 2016-01-30 04:33:46 +00:00 Commented Jan 30, 2016 at 4:33 Any square is greater than zero, for the first line. I added a step. From the third to fourth, take the square root Ross Millikan –Ross Millikan 2016-01-30 04:39:46 +00:00 Commented Jan 30, 2016 at 4:39 2 @LeoDenni x 2+2 x y+y 2−4 x y=x 2−2 x y+y 2=(x−y)2 x 2+2 x y+y 2−4 x y=x 2−2 x y+y 2=(x−y)2 user228113 –user228113 2016-01-30 04:52:28 +00:00 Commented Jan 30, 2016 at 4:52 2 Understanding a proof is very different from finding it. To understand it, you just need to follow through step by step, filling in as necessary to make sure you can see where each comes from. I've seen this one before, so finding it is easy. The fact that (x+y)2−(x−y)2=4 x y(x+y)2−(x−y)2=4 x y comes up with some frequency, so if I hadn't seen it before I would have been tempted to take what you have to prove, multiply by 2 2 to clear the fraction and square to get rid of the root. The more proofs you read, the more you get a feel for them, which makes finding them much easier.Ross Millikan –Ross Millikan 2016-01-30 05:20:47 +00:00 Commented Jan 30, 2016 at 5:20 1 That was a dead end because your RHS is too small. Just putting in the factor 2 2 in the fourth line is too poor an approximation. If x x is close to y y the inequality you are trying to prove is rather close. A different proof starts out with (x−−√−y√)2≥0(x−y)2≥0. You might see if you can get from there to the end.Ross Millikan –Ross Millikan 2016-01-30 05:36:00 +00:00 Commented Jan 30, 2016 at 5:36 \|Show 7 more comments This answer is useful 1 Save this answer. Show activity on this post. Let us start from the beginning and simplify what we want to show: x+y 2≥x y−−√x+y 2≥x y By multiplying both side by 2 2, we will get: x+y≥2 x y−−√x+y≥2 x y Now, subtracting 2 x y−−√2 x y from both sides, we left with: x−2 x y−−√+y≥0 x−2 x y+y≥0 Now, a question: Why x−2 x y−−√−y x−2 x y−y is indeed greater(or equal) than zero? Closer look on the above expression, one might see that its equal to (x−−√−y√)2(x−y)2 So, if we are given two positive real numbers x x and y y, then without loss of generality we can say that x>y x>y(why?), therefore x−−√>y√x>y(prove that). Hence: x−−√−y√>0 x−y>0 And... Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 30, 2016 at 4:48 answered Jan 30, 2016 at 4:23 Salech AlhasovSalech Alhasov 6,984 2 2 gold badges 31 31 silver badges 47 47 bronze badges 4 You are starting from what we want to prove and have not justified the conclusion.Ross Millikan –Ross Millikan 2016-01-30 04:26:47 +00:00 Commented Jan 30, 2016 at 4:26 Yes, that is my point. The simplification is very useful here; after seeing what last expression is, the OP might find the right argument for a proof. Is that make sense?Salech Alhasov –Salech Alhasov 2016-01-30 04:32:01 +00:00 Commented Jan 30, 2016 at 4:32 I'm not sure how you want me to answer that question. You did the manipulation and you got it to be greater than or equal to 0.Leo Denni –Leo Denni 2016-01-30 04:38:39 +00:00 Commented Jan 30, 2016 at 4:38 1 @RossMillikan Actually, all those steps are iff's, so what he did is ok.YoTengoUnLCD –YoTengoUnLCD 2016-01-30 06:19:10 +00:00 Commented Jan 30, 2016 at 6:19 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. This is a comment on your attempt to prove this inequality rather than answer to your question. Just to show that it was not completely dead end, as you wrote in your comment. You wrote that you tried this: So far what I have gotten to is x+y 2≥x y√2 x+y 2≥x y 2 After this point I don't know what to do. To get to this point: y≥0 y≥0 ⟹y≥y⟹y≥y ⟹x y≥x y⟹x y≥x y ⟹2 x y≥x y⟹2 x y≥x y ⟹x 2+2 x y+y 2≥x y⟹x 2+2 x y+y 2≥x y ⟹(x+y)2≥x y⟹(x+y)2≥x y Sqrt both sides to get: ⟹x+y≥x y−−√⟹x+y≥x y ⟹x+y 2≥x y−−√2⟹x+y 2≥x y 2 What you could do know could be ask yourself: Well, I want to prove stronger inequality. Could I get it by similar approach? So we start from the end. You want to get x+y 2≥x y−−√x+y 2≥x y which is the same as x+y≥2 x y−−√.x+y≥2 x y. Relation between these two is the same as in the last step of your attempt - just divided by 2 2. If we try to go one step back in the same way as above, this can be obtained as the square root of (x+y)2 x 2+2 x y+y 2≥4 x y≥4 x y(x+y)2≥4 x y x 2+2 x y+y 2≥4 x y So at this point you could think about: "How could I get x 2+2 x y+y 2≥4 x y x 2+2 x y+y 2≥4 x y?" Maybe after playing a bit with this expression you would found a way to prove this. (However, now you already know how to show this from other answers.) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:21 community wiki 2 revsMartin Sleziak 1 And since I have already posted so many links in comments, maybe it is ok to post a few more - other posts about the weaker inequality, which you have shown in your attempt (to be more precise, about some equivalent or very similar inequality): math.stackexchange.com/questions/357272/…, math.stackexchange.com/questions/920605/inequality-x2y2xy-ge-0, math.stackexchange.com/questions/1439494/prove-that-a2abb2-ge-0Martin Sleziak –Martin Sleziak 2016-01-30 10:13:20 +00:00 Commented Jan 30, 2016 at 10:13 Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus inequality means See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 14Proving the AM-GM inequality for 2 numbers x y−−√≤x+y 2 x y≤x+y 2 7How to prove that a+b 2≥a b−−√a+b 2≥a b for a,b>0 a,b>0? 140Proofs of AM-GM inequality 41How can I prove that x y≤x 2+y 2 x y≤x 2+y 2? 9Prove the inequality \|x y\|≤1 2(x 2+y 2)\|x y\|≤1 2(x 2+y 2) 5Show that 2 x y<x 2+y 2 2 x y<x 2+y 2 for x x is not equal to y y 7Prove that a 2+a b+b 2≥0 a 2+a b+b 2≥0 3Inequality: x 2+y 2+x y≥0 x 2+y 2+x y≥0 5Show that for all real numbers a a and b b, a b≤(1/2)(a 2+b 2)a b≤(1/2)(a 2+b 2) 5How can I Prove 2 x y x+y≤x y−−√≤x+y 2 2 x y x+y≤x y≤x+y 2 See more linked questions Related 3How to prove the inequality ∑n i=1 i+1√2 i>n√2∑i=1 n i+1 2 i>n 2 for n∈Z+n∈Z+? 11Prove that n−−√n<1+2 n−−√n n<1+2 n 2Proving that if m n<2–√m n<2 then there exists m′n′m′n′ such that m n<m′n′<2–√m n<m′n′<2 2Is this a legal way to prove an inequality? 2How do I prove that the inequality 2 x y x+y≤x y−−√≤x+y 2 2 x y x+y≤x y≤x+y 2 holds for every x,y>0 x,y>0 1Prove for the inequality between arithmetic and quadratic means x+y 2≤x 2+y 2 2−−−−−√x+y 2≤x 2+y 2 2 0Prove that 2 x 2−2 x+1 2−−−−−−−√≥1 x+1 x 2 x 2−2 x+1 2≥1 x+1 x for 0<x<1.0<x<1. 8Prove b√−a−−√<b−a−−−−√b−a<b−a 0For positive real numbers a,b a,b prove that 2(a+b)(1 a+1 b)−−−−−−−−−−−−−√3≥a b−−√3+b a−−√3 2(a+b)(1 a+1 b)3≥a b 3+b a 3. 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2335	https://www.mendwellhealth.com/all-conditions/genitofemoral-neuralgia	Genitofemoral Neuralgia: Understanding Your Symptoms & Treatment Options What it is, how it manifests, and how physical therapy can help. Overview Genitofemoral neuralgia occurs when the genitofemoral nerve - a crucial nerve running from your lower back through your abdomen and into your upper thigh and genital regions - becomes irritated or damaged. This condition can cause persistent discomfort that significantly impacts daily life, yet many people spend months or even years seeking an accurate diagnosis due to its complex nature. While this condition can affect anyone, it most commonly develops after abdominal or pelvic surgeries, particularly following hernia repairs, cesarean sections, or other procedures in the lower abdomen. Studies suggest that up to 12% of patients who undergo inguinal hernia repair may experience some degree of nerve pain, including genitofemoral neuralgia. The good news is that with proper diagnosis and treatment, including specialized pelvic floor physical therapy, many people experience significant improvement in their symptoms and quality of life. At Mendwell Pelvic Health, we understand the frustration and uncertainty that comes with chronic nerve pain. Our specialized approach to treating genitofemoral neuralgia combines evidence-based techniques with compassionate care to help you regain comfort and confidence in your daily activities. Understanding Your Symptoms Living with genitofemoral neuralgia can present various challenges, as the symptoms often fluctuate and may be difficult to pinpoint. Most people experience a combination of sensations that typically include: Pain and Discomfort A burning or shooting pain that travels along the lower abdomen and into the groin Sharp or aching sensations that may extend into the upper thigh or genital areas Pain that intensifies with certain movements, especially hip extension or prolonged sitting Sensory Changes: Areas of increased sensitivity (hypersensitivity) in the affected regions Numbness or tingling sensations along the nerve pathway Discomfort that may worsen with light touch or pressure from clothing Impact on Daily Activities: Difficulty finding comfortable positions for sleep or rest Challenges with intimate activities due to increased sensitivity or pain Discomfort during exercise or routine movements Muscle tension in surrounding areas as your body tries to protect the sensitive region Common Causes and Risk Factors Understanding what causes genitofemoral neuralgia can help in both treatment and prevention. The condition typically develops through one or more of the following mechanisms: Surgical Procedures: The most common trigger for genitofemoral neuralgia is surgical intervention in the lower abdomen or pelvic region. During these procedures, the nerve may become compressed, stretched, or directly injured. Common surgical procedures associated with this condition include inguinal hernia repairs, appendectomies, and cesarean sections. Physical Trauma and Compression: Direct injury to the lower abdomen or prolonged compression of the nerve can lead to symptoms. This might occur from accidents, repetitive movements, or even prolonged periods in positions that compress the nerve pathway. Pregnancy and Childbirth: The physical changes during pregnancy and the birthing process can put pressure on the nerve or create conditions that lead to nerve irritation. Additionally, cesarean sections may directly affect the nerve's pathway. Inflammatory Conditions: Sometimes, inflammation in surrounding tissues can irritate the nerve, leading to symptoms even without direct nerve damage. This can be particularly relevant in cases of endometriosis or other inflammatory pelvic conditions. How Pelvic Floor Physical Therapy Can Help At Mendwell Pelvic Health, our approach to treating genitofemoral neuralgia is comprehensive and individually tailored. Pelvic floor physical therapy plays a crucial role in managing symptoms and promoting healing through specialized techniques and evidence-based interventions. Our treatment approach typically includes: Specialized Manual Therapy We utilize gentle, targeted techniques to improve nerve mobility and reduce tension in surrounding tissues. This includes specific nerve gliding techniques, soft tissue mobilization, and when appropriate, scar tissue management to reduce adhesions that might be compromising nerve function. Movement Retraining We work with you to identify and modify movements or positions that might be aggravating your symptoms. This includes: Teaching optimal posture and movement patterns Developing strategies for comfortable sleep positions Creating modified exercise programs that respect your current symptoms while maintaining strength and mobility Comprehensive Education Understanding your condition is crucial for long-term management. We provide: Detailed explanations of nerve sensitivity and pain science Strategies for activity modification Tools for self-management and symptom control Guidance on lifestyle factors that can impact nerve health Your Treatment Journey at Mendwell Every treatment plan is customized to your specific needs and symptoms. Here's what you can expect: Initial Evaluation: Your first visit includes a comprehensive assessment of your symptoms, movement patterns, and contributing factors. We'll work together to understand your goals and develop a treatment plan that aligns with your needs. Progressive Treatment: Treatment sessions typically combine hands-on therapy with movement training and education. As your symptoms improve, we'll gradually progress your activities and exercises to help you return to your desired level of function. Home Program: We'll provide you with specific exercises and strategies to support your recovery between sessions, empowering you to take an active role in your healing process. Frequently Asked Questions About Genitofemoral Neuralgia What does genitofemoral neuralgia pain feel like compared to other types of pelvic pain? ‍Genitofemoral neuralgia typically presents as a burning, shooting, or electric-type pain that follows a specific pattern along the nerve's pathway. Unlike general muscle pain, nerve pain often feels more sharp and can be accompanied by sensations of tingling or numbness in the affected areas. How long does it typically take to see improvement with pelvic floor physical therapy for nerve pain? ‍While every case is unique, many patients begin to notice positive changes within 4-6 weeks of consistent treatment. However, the full recovery process may take several months, depending on factors such as how long you've had symptoms and the underlying cause of your nerve irritation. Will exercises make my nerve pain worse? ‍When prescribed appropriately, exercises should not aggravate your symptoms. Your physical therapist will carefully select and modify exercises to respect your current pain levels and gradually progress your program as your symptoms improve. Can physical therapy help if I've had this condition for a long time? ‍Yes, even chronic cases of genitofemoral neuralgia can benefit from physical therapy. While long-standing symptoms may require more time to improve, our specialized treatment approaches can help reduce pain and improve function regardless of how long you've been experiencing symptoms. Experiencing Genitofemoral Neuralgia or related symptoms? Pelvic floor physical therapy can help. Mendwell is a Pelvic Health Physical Therapy clinic serving patients in Portland, Lake Oswego, Beaverton, Hillsboro, Tigard, Tualatin and West Linn. Our team of specialists are passionate about helping patients improve pelvic function, relieve pain, and get back to feeling their best. Reach out to learn how we can help. Category Tags: Pelvic Pain Conditions Post-Surgical Recovery Sexual Health & Intimacy Prenatal & Postpartum Physical Therapy in Portland, OR Pregnancy & Birth Prep Related Conditions Pudendal Neuralgia Pudendal Nerve Entrapment Ilioinguinal Neuralgia Groin Pain Chronic Pelvic Pain Syndrome (CPPS) Other names for this condition No items found. Additional Categories Browse All Conditions ### Postpartum Recovery Physical therapy for recovery and healing after pregnancy Learn More ### Endometriosis & PCOS Physical therapy for endometriosis and PCOS pain relief. Learn More ### Inclusive Pelvic Health Pelvic Floor Physical Therapy for All Bodies and Communities Learn More
2336	https://pubmed.ncbi.nlm.nih.gov/35214057/	Oxidation of Drugs during Drug Product Development: Problems and Solutions - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Oxidation of Drugs during Drug Product Development: Problems and Solutions Alen Gabrič12,Žiga Hodnik1,Stane Pajk2 Affiliations Expand Affiliations 1 Krka d.d., R&D, Šmarješka Cesta 6, 8001 Novo Mesto, Slovenia. 2 Faculty of Pharmacy, University of Ljubljana, Aškerčeva Cesta 7, 1000 Ljubljana, Slovenia. PMID: 35214057 PMCID: PMC8876153 DOI: 10.3390/pharmaceutics14020325 Item in Clipboard Review Oxidation of Drugs during Drug Product Development: Problems and Solutions Alen Gabrič et al. Pharmaceutics.2022. Show details Display options Display options Format Pharmaceutics Actions Search in PubMed Search in NLM Catalog Add to Search . 2022 Jan 29;14(2):325. doi: 10.3390/pharmaceutics14020325. Authors Alen Gabrič12,Žiga Hodnik1,Stane Pajk2 Affiliations 1 Krka d.d., R&D, Šmarješka Cesta 6, 8001 Novo Mesto, Slovenia. 2 Faculty of Pharmacy, University of Ljubljana, Aškerčeva Cesta 7, 1000 Ljubljana, Slovenia. PMID: 35214057 PMCID: PMC8876153 DOI: 10.3390/pharmaceutics14020325 Item in Clipboard Cite Display options Display options Format Abstract Oxidation is the second most common degradation pathway for pharmaceuticals, after hydrolysis. However, in contrast to hydrolysis, oxidation is mechanistically more complex and produces a wider range of degradation products; oxidation is thus harder to control. The propensity of a drug towards oxidation is established during forced degradation studies. However, a more realistic insight into degradation in the solid state can be achieved with accelerated studies of mixtures of drugs and excipients, as the excipients are the most common sources of impurities that have the potential to initiate oxidation of a solid drug product. Based on the results of these studies, critical parameters can be identified and appropriate measures can be taken to avoid the problems that oxidation poses to the quality of a drug product. This article reviews the most common types of oxidation mechanisms, possible sources of reactive oxygen species, and how to minimize the oxidation of a solid drug product based on a well-planned accelerated study. Keywords: impurities; new oxidative stressors; oxidation; oxidative stress testing; peroxides. PubMed Disclaimer Conflict of interest statement A.G. and Ž.H. are employed by Krka d.d., however the company had no role in the research, the interpretation, the writing and in the decision to publish. The authors declare no conflict of interest. Figures Figure 1 Schematic representation of oxidation reactions… Figure 1 Schematic representation of oxidation reactions and their role through drug development. Of note,… Figure 1 Schematic representation of oxidation reactions and their role through drug development. Of note, in this review the word “drug” corresponds to “active pharmaceutical ingredient” and not to “final drug product”. Figure 2 Conceptual diagram of quality by… Figure 2 Conceptual diagram of quality by design, showing the relationships between the knowledge space,… Figure 2 Conceptual diagram of quality by design, showing the relationships between the knowledge space, design space, and control space. Figure 3 Autoxidation that is mediated by… Figure 3 Autoxidation that is mediated by a drug-based peroxy radical (DOO • ). Figure 3 Autoxidation that is mediated by a drug-based peroxy radical (DOO•). Figure 4 The termination reaction (Scheme 1)… Figure 4 The termination reaction (Scheme 1) and epoxide formation (Scheme 2). Figure 4 The termination reaction (Scheme 1) and epoxide formation (Scheme 2). Figure 5 Formation of an N-oxide (Scheme… Figure 5 Formation of an N-oxide (Scheme 3) and a hydroxylamine, with hydrolysis (Scheme 4). Figure 5 Formation of an N-oxide (Scheme 3) and a hydroxylamine, with hydrolysis (Scheme 4). Figure 6 Reaction between a sulfoxide and… Figure 6 Reaction between a sulfoxide and hydrogen peroxide (Scheme 5). Figure 6 Reaction between a sulfoxide and hydrogen peroxide (Scheme 5). Figure 7 Proposed reaction mechanism for atorvastatin… Figure 7 Proposed reaction mechanism for atorvastatin with the formation of a bridged peroxide. Figure 7 Proposed reaction mechanism for atorvastatin with the formation of a bridged peroxide. Figure 8 Proposed mechanism for metal ion… Figure 8 Proposed mechanism for metal ion reaction with triplet oxygen to form a superoxide. Figure 8 Proposed mechanism for metal ion reaction with triplet oxygen to form a superoxide. Figure 9 The Fenton reaction. Figure 9 The Fenton reaction. Figure 9 The Fenton reaction. Figure 10 Butylhydroxytoluene reaction with the peroxy… Figure 10 Butylhydroxytoluene reaction with the peroxy radical and stabilization of the radical electron. Figure 10 Butylhydroxytoluene reaction with the peroxy radical and stabilization of the radical electron. Figure 11 Reducing agents. Figure 11 Reducing agents. Figure 11 Reducing agents. Figure 12 4,4′-Azobis(4-cyanovaleric acid) and 2,2′-azobis(2-methylpropionitrile) mechanism… Figure 12 4,4′-Azobis(4-cyanovaleric acid) and 2,2′-azobis(2-methylpropionitrile) mechanism of decomposition and reactions to peroxy radicals. Figure 12 4,4′-Azobis(4-cyanovaleric acid) and 2,2′-azobis(2-methylpropionitrile) mechanism of decomposition and reactions to peroxy radicals. All figures (12) See this image and copyright information in PMC Similar articles PVP-H 2 O 2 Complex as a New Stressor for the Accelerated Oxidation Study of Pharmaceutical Solids.Modhave D, Barrios B, Paudel A.Modhave D, et al.Pharmaceutics. 2019 Sep 3;11(9):457. doi: 10.3390/pharmaceutics11090457.Pharmaceutics. 2019.PMID: 31484442 Free PMC article. Realistic prediction of solid pharmaceutical oxidation products by using a novel forced oxidation system.Ueyama E, Tamura K, Mizukawa K, Kano K.Ueyama E, et al.J Pharm Sci. 2014 Apr;103(4):1184-93. doi: 10.1002/jps.23889. Epub 2014 Feb 4.J Pharm Sci. 2014.PMID: 24497072 pH control of nucleophilic/electrophilic oxidation.Freed AL, Strohmeyer HE, Mahjour M, Sadineni V, Reid DL, Kingsmill CA.Freed AL, et al.Int J Pharm. 2008 Jun 5;357(1-2):180-8. doi: 10.1016/j.ijpharm.2008.01.061. Epub 2008 Mar 10.Int J Pharm. 2008.PMID: 18400425 Oxidative degradation of pharmaceuticals: theory, mechanisms and inhibition.Hovorka S, Schöneich C.Hovorka S, et al.J Pharm Sci. 2001 Mar;90(3):253-69. doi: 10.1002/1520-6017(200103)90:3<253::aid-jps1>3.0.co;2-w.J Pharm Sci. 2001.PMID: 11170019 Review. Reactive impurities in excipients: profiling, identification and mitigation of drug-excipient incompatibility.Wu Y, Levons J, Narang AS, Raghavan K, Rao VM.Wu Y, et al.AAPS PharmSciTech. 2011 Dec;12(4):1248-63. doi: 10.1208/s12249-011-9677-z. Epub 2011 Sep 27.AAPS PharmSciTech. 2011.PMID: 21948318 Free PMC article.Review. See all similar articles Cited by Temperature variations in pharmaceutical storage facilities and knowledge, attitudes, and practices of personnel on proper storage conditions for medicines in southern Malawi.Khuluza F, Chiumia FK, Nyirongo HM, Kateka C, Hosea RA, Mkwate W.Khuluza F, et al.Front Public Health. 2023 Sep 22;11:1209903. doi: 10.3389/fpubh.2023.1209903. eCollection 2023.Front Public Health. 2023.PMID: 37808988 Free PMC article. Biopharmaceutical Characterization and Stability of Nabumetone-Cyclodextrins Complexes Prepared by Grinding.Klarić D, Soldin Ž, Vincze A, Szolláth R, Balogh GT, Jug M, Galić N.Klarić D, et al.Pharmaceutics. 2024 Nov 21;16(12):1493. doi: 10.3390/pharmaceutics16121493.Pharmaceutics. 2024.PMID: 39771473 Free PMC article. Vesicular Carriers for Phytochemical Delivery: A Comprehensive Review of Techniques and Applications.Jacob S, Kather FS, Boddu SHS, Rao R, Nair AB.Jacob S, et al.Pharmaceutics. 2025 Apr 2;17(4):464. doi: 10.3390/pharmaceutics17040464.Pharmaceutics. 2025.PMID: 40284459 Free PMC article.Review. A versatile method to fingerprint and compare the oxidative behaviour of lipids beyond their oxidative stability.Pizzimenti S, Bernazzani L, Duce C, Tinè MR, Bonaduce I.Pizzimenti S, et al.Sci Rep. 2023 May 19;13(1):8094. doi: 10.1038/s41598-023-34599-6.Sci Rep. 2023.PMID: 37208395 Free PMC article. Dihydroquercetin and Related Flavonoids in Antioxidant Formulations with α-Tocopherol.Olicheva V, Beloborodov V, Sharifi S, Dubrovskaya A, Zhevlakova A, Selivanova I, Ilyasov I.Olicheva V, et al.Int J Mol Sci. 2025 Jun 13;26(12):5659. doi: 10.3390/ijms26125659.Int J Mol Sci. 2025.PMID: 40565124 Free PMC article. See all "Cited by" articles References Guo J.J., Pandey S., Doyle J., Bian B., Lis Y., Raisch D.W. A Review of Quantitative Risk–Benefit Methodologies for Assessing Drug Safety and Efficacy—Report of the ISPOR Risk–Benefit Management Working Group. Value Health. 2010;13:657–666. doi: 10.1111/j.1524-4733.2010.00725.x. - DOI - PubMed Sinclair W., Leane M., Clarke G., Dennis A., Tobyn M., Timmins P. Physical Stability and Recrystallization Kinetics of Amorphous Ibipinabant Drug Product by Fourier Transform Raman Spectroscopy. J. Pharm. Sci. 2011;100:4687–4699. doi: 10.1002/jps.22658. - DOI - PubMed . ICH Q1A (R2) Stability Testing of New Drug Substances and Products. [(accessed on 24 April 2021)]. Available online: Zumdahl S.S., Zumdahl S.A. Chemistry. 9th ed. Brooks/Cole, Cengage Learning; Belmont, CA, USA: 2014. Rams-Baron M., Jachowicz R., Boldyreva E., Zhou D., Jamroz W., Paluch M. Physical Instability: A Key Problem of Amorphous Drugs. In: Rams-Baron M., Jachowicz R., Boldyreva E., Zhou D., Jamroz W., Paluch M., editors. Amorphous Drugs: Benefits and Challenges. Springer International Publishing; Cham, Switzerland: 2018. pp. 107–157. Show all 112 references Publication types Review Actions Search in PubMed Search in MeSH Add to Search Related information MedGen Grants and funding P1-0208/Slovenian Research Agency LinkOut - more resources Other Literature Sources The Lens - Patent Citations Database [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov
2337	https://pmc.ncbi.nlm.nih.gov/articles/PMC10424142/	An official website of the United States government Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( Lock Locked padlock icon ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Primary site navigation Logged in as: PERMALINK Efficacy and safety of anti-androgens in the management of polycystic ovary syndrome: a systematic review and meta-analysis of randomised controlled trials Simon Alesi Maria Forslund Johanna Melin Daniela Romualdi Alexia Peña Chau Thien Tay Selma Feldman Witchel Helena Teede Aya Mousa Corresponding author. NHMRC Senior Fellow and Head of Diabetes, Metabolic and Reproductive Health Research, Monash Centre for Health Research and Implementation, Faculty of Medicine, Nursing and Health Sciences, Monash University, 43-51 Kanooka Grove, Clayton, VIC 3168, Australia. aya.mousa@monash.edu Received 2023 May 10; Revised 2023 Jul 27; Accepted 2023 Jul 27; Collection date 2023 Sep. This is an open access article under the CC BY license ( Summary Background Anti-androgens and combined oral contraceptive pills (COCPs) may mitigate hyperandrogenism-related symptoms of polycystic ovary syndrome (PCOS). However, their efficacy and safety in PCOS remain unclear as previous reviews have focused on non-PCOS populations. To inform the 2023 International Evidence-based Guideline in PCOS, we conducted the first systematic review and meta-analysis investigating the efficacy and safety of anti-androgens in the management of hormonal and clinical features of PCOS. Methods We systematically searched MEDLINE, Embase, PsycInfo, All EBM reviews, and CINAHL up to 28th June 2023 for randomised controlled trials (RCTs) examining oral anti-androgen use, alone or in combination with metformin, COCPs, lifestyle, or other interventions, in women of any age, with PCOS diagnosed by Rotterdam, National Institutes of Health or Androgen Excess & PCOS Society criteria, and using a form of contraception. Non-English studies and studies of less than 6 months duration or which used the same anti-androgen regimen in both/all groups were excluded in order to establish efficacy for the clinical outcomes of interest. Three authors screened articles against selection criteria and assessed risk of bias and quality using the Grading of Recommendations, Assessment, Development and Evaluations (GRADE) framework. Critical outcomes (prioritised during guideline development for GRADE purposes) included weight, body mass index (BMI), irregular cycles, hirsutism, liver function, and quality of life. Random effects meta-analyses were conducted where appropriate. This study is registered with PROSPERO, CRD42022345640. Findings From 1660 studies identified in the search, 27 articles comprising 20 unique studies were included. Of these, 13 studies (n = 961) were pooled in meta-analysis. Seven studies had a high risk of bias, nine moderate and four low. Anti-androgens included finasteride, flutamide, spironolactone, or bicalutamide. In meta-analysis, anti-androgens + lifestyle were superior to metformin + lifestyle for hirsutism (weighted mean difference [WMD] [95% CI]: −1.59 [−3.06, −0.12], p = 0.03; I2 = 74%), SHBG (7.70 nmol/l [0.75, 14.66], p = 0.03; I2 = 0%), fasting insulin and fasting insulin: glucose ratio (−2.11 μU/ml [−3.97, −0.26], p = 0.03; I2 = 0% and −1.12 [−1.44, −0.79], p < 0.0001, I2 = 0%, respectively), but were not superior to placebo + lifestyle for hirsutism (−0.93, [−3.37, 1.51], p = 0.45; I2 = 76%) or SHBG (9.72 nmol/l [−0.71, 20.14], p = 0.07; I2 = 31%). Daily use was more effective for hirsutism than use every three days (−3.48 [−4.58, −2.39], p < 0.0001, I2 = 1%), and resulted in lower androstenedione levels (−0.30 ng/ml [−0.50, −0.10], p = 0.004; I2 = 0%). Combination treatment with anti-androgens + metformin + lifestyle resulted in lower testosterone compared with metformin + lifestyle (−0.29 nmol/l [−0.52, −0.06], p = 0.01; I2 = 61%), but there were no differences in hirsutism when anti-androgens + metformin + lifestyle were compared with either anti-androgens + lifestyle or metformin + lifestyle. In limited meta-analyses (n = 2 trials), combining anti-androgens with COCP resulted in poorer lipid profiles compared with COCP ± placebo, with no differences in other outcomes. Interpretation Current evidence does not support the use of anti-androgens preferentially to COCPs to treat hyperandrogenism in PCOS. Anti-androgens could be considered to treat hirsutism in PCOS, where COCPs are contraindicated, poorly tolerated, or present a sub-optimal response after a minimum 6-month period, with consideration of clinical context and individual risk factors and characteristics. Funding National Health and Medical Research Council (NHMRC) of Australia Monash University. Keywords: Anti-androgens, Combined oral contraceptives, Guideline, Meta-Analyses, Systematic review, Polycystic ovary syndrome (PCOS), Spironolactone Research in context. Evidence before this study The 2018 International Evidence-based Guideline in Polycystic Ovary Syndrome (PCOS) identified no systematic reviews of randomised controlled trials (RCTs) examining the use of anti-androgens in PCOS. To identify studies and systematic reviews published since the 2018 guideline, we searched electronic databases including MEDLINE, Embase, PsycInfo, All EBM reviews, and CINAHL from 2017 to 28th June 2023. We included search terms related to: polycystic ovary syndrome OR oligo-ovulation OR anovulation OR hyperandrogenism AND combined oral contraceptive pills OR metformin OR androgen antagonists OR spironolactone OR finasteride OR cyproterone acetate or flutamide OR anti-obesity OR orlistat OR sibutramine OR inositol AND meta-analysis OR review OR clinical trial OR random. We identified nine newly published articles, but found no systematic reviews or meta-analyses of RCTs investigating the use of anti-androgens in PCOS. Added value of this study We conducted the first systematic review of RCTs investigating the efficacy and safety of anti-androgen pharmacological agents in PCOS. The study used a comprehensive search, including both adult and adolescent populations and using internationally endorsed guideline methodology for identifying, evaluating, and appraising the literature, including the Grading of Recommendations Assessment Development and Evaluations (GRADE) tool to inform guideline recommendations. Combined with studies from the previous 2018 guideline, a total of 27 articles reporting on 20 unique RCTs were included. We assessed a wide range of outcomes relevant to PCOS as determined by experts and consumer groups in the guideline development process, thereby providing broad insights into the efficacy of anti-androgens, with or without other interventions, in managing key features of PCOS. Our findings showed that anti-androgens may have possible benefits for treating hirsutism, in circumstances where combined oral contraceptive pills (COCPs) are contraindicated, poorly tolerated, or have been ineffective for over six months, with consideration of clinical context and individual needs and risk factors. Implications of all the available evidence This study has directly informed the recommendations of the 2023 International Evidence-based Guideline for the Assessment and Management of PCOS, endorsed by 39 organisations globally. Our findings have addressed key gaps in the literature and provide evidence to guide clinical practice. However, while these recommendations are based on the best available evidence, this is largely of low certainty and differences in the efficacy and side effects of different anti-androgens and their effectiveness across various PCOS features, ages and anthropometric characteristics could not be accurately determined. Further high-quality RCTs in this area are encouraged. Introduction Polycystic ovary syndrome (PCOS) is the most common endocrinological disorder in women of reproductive age, with a prevalence of 8–13%.1 According to the Rotterdam criteria, PCOS is diagnosed based on the presence of two out of three characteristics: clinical or biochemical hyperandrogenism, menstrual irregularities (oligo or anovulation), and the detection of polycystic ovary morphology on ultrasound, after the exclusion of other causes.2 The Androgen Excess-PCOS Society (AE-PCOS) and National Institutes of Health (NIH) state that PCOS should be defined by the presence of hyperandrogenism, ovarian dysfunction, and the exclusion of other related disorders.3 While the AE-PCOS and NIH largely agree with the diagnostic criteria outlined in Rotterdam, hyperandrogenism is considered to be a fundamental component in the pathophysiology of PCOS.4,5 Reproductive (irregular menses, infertility and pregnancy complications),6 metabolic (increased prevalence of obesity, type 2 diabetes, and cardiovascular disease), and psychological (depression, disordered eating, body image distress, and reduced quality of life) characteristics are associated with PCOS.7,8 Lifestyle management is recommended as the first-line treatment option for PCOS. However, in some circumstances where lifestyle management is unsuccessful, pharmacological agents including the combined oral contraceptive pill (COCP), anti-obesity agents, metformin, and anti-androgen medications, could be employed for clinical management.9 The COCP is often prescribed for adult women with PCOS, for management of irregular menses and clinical hyperandrogenism (hirsutism and acne).10 Similarly, COCPs are recommended for adolescent girls diagnosed with PCOS and those considered to be “at risk for PCOS”. Metformin has been used for ovulation induction, promotion of weight loss and/or maintenance, and reduction of pregnancy complications.11, 12, 13 Anti-androgen medications such as spironolactone, flutamide, finasteride, and cyproterone acetate (CPA) have been utilised to decrease hyperandrogenism-related symptoms. Anti-androgen medications act in one of three ways; either by competitively inhibiting the androgen-binding receptors; decreasing androgen production (although the manner in which this occurs is not well-understood); or inhibiting 5-α-reductase in the skin, which is an enzyme that converts testosterone into its active form, 5-α-dihydrotestosterone (DHT).14 Spironolactone and finasteride are dose-dependent competitive inhibitors of the androgen receptor and have also been observed to inhibit 5-α-reductase activity.15 Other anti-androgens such as CPA, flutamide, and bicalutamide are thought to competitively inhibit testosterone binding to the androgen receptor.16 Via these mechanisms, anti-androgen medications could ameliorate the hyperandrogenic state of PCOS and improve the various hyperandrogenism-related symptoms associated with the condition. While anti-androgens and their mechanisms of action allude to plausible benefits, the 2018 PCOS guidelines9 concluded that evidence was lacking for the use of all identified anti-androgen medications in PCOS. Existing systematic reviews, including a network meta-analysis,17 have focused on women with idiopathic hirsutism17 or broader hirsute populations18 and, to our knowledge, no systematic reviews have examined the use of anti-androgens in the management of PCOS. In response to this identified gap, recent published studies have appealed for re-examination of the potential benefits of anti-androgens in PCOS. Therefore, as part of the updated 2023 International Evidence-based Guidelines in PCOS, this systematic review and meta-analysis aimed to investigate the efficacy and safety of anti-androgens, alone or in combination, on anthropometric, hormonal and clinical features of PCOS. Methods Search strategy and selection criteria This systematic review is reported in accordance with the Preferred Reporting Items for Systematic Reviews and Meta-Analyses (PRISMA) guideline,19 and was developed to inform clinical practice recommendations in the update of the International Evidence-based Guideline for the Assessment and Management of PCOS. This review was part of a larger evidence synthesis covering several pharmacological treatments including COCP, metformin, and anti-androgens and, although each intervention review is reported separately (due to different eligibility criteria), the combined protocol was registered a priori on the international prospective register of systematic reviews (PROSPERO): CRD42022345640. The methodology utilised in this systematic review is endorsed by the National Health and Medical Research Council (NHMRC) of Australia20 and the European Society of Human Reproduction and Embryology (ESHRE)21 and has been previously described in detail in the 2018 guidelines9 and in the publicly available Technical Report.22 The present systematic review addresses the following clinical question: Are anti-androgen pharmacological agents, alone or in combination, effective for management of hormonal and clinical PCOS features and weight in adolescents and adults with PCOS? The Participant-Intervention-Comparison-Outcome (PICO) framework was adopted, with selection criteria developed a priori by the multi-disciplinary guideline development group, including consumers. These eligibility criteria are outlined in Table 1. The search strategy, including database selection and search terms (Medical Subject Headings (MeSH) and keywords), was developed in consultation with technical and content experts, clinicians with expertise in PCOS management, and a skilled medical librarian. As this was part of a larger evidence synthesis, a variety of keywords relating to PCOS, anti-androgens, metformin, COCP, and relevant outcomes, were used in the search strategy (Supplementary Material). The following databases were searched: MEDLINE, EMBASE, APA PsycInfo, all evidence-based medicine (EBM) reviews (all via Ovid processing), and CINAHL (via EBSCO). The previous 2018 International Evidence-based Guidelines in PCOS searched from inception to 11th January 2017.9 For the current systematic review, we incorporated the studies identified in the 2018 guideline and reran the search from 2017 to 8th July 2022, with a further updated search on 28th June 2023. Table 1. Eligibility criteria (PICO) for study inclusion. \| Participants (P) \| Intervention (I) \| Comparison (C) \| Outcomes (O) Inclusion \| Females with PCOS (diagnosed by Rotterdam, NIH, or AES) of any age, ethnicity, and weight. Subgroups: adolescents (10–19 y), adults, post-menopausal. Contraception must be reported. \| Oral anti-androgen pharmacological agents (SPL, CPA finasteride, flutamide, bicalutamide) alone or in combination with lifestyle, metformin, OCP/COCP, anti-obesity agents.All doses, duration of more than 6 months. \| Placebo or any other intervention (listed in intervention) or combinations of those listed in intervention. \| Several outcomes were included, relating to hormonal (FAI, testosterone, SHBG, DHEAS, androstenedione, hirsutism by FG scores, and irregular cycles); anthropometric (weight, BMI, WHR); and metabolic parameters (total, HDL, or LDL] cholesterol, triglycerides, insulin resistance/glucose metabolism by HOMA, hyperinsulinemic-euglycemic clamps, or OGTT, and CRP). Psychological outcomes and adverse effects were also extracted, which included QoL, depression, mood, liver function tests (transaminases), and pregnancy complications. Exclusion \| Females without PCOS. \| \| Agent or combination used in the intervention (i.e., CPA as part of COCP). \| Abbreviations: AES, Androgen Excess Society; BMI, body mass index; CPA, cyproterone acetate; CRP, C-reactive protein; DHEAS, dehydroepiandrosterone sulphate; FAI, free androgen index; FG, Ferriman-Gallwey; HDL, high density lipoprotein; HOMA, homeostatic model assessments; LDL, low density lipoprotein; NIH, National Institute of Health; OCP/COCP, oral contraceptive pill/combined oral contraceptive pill; OGTT, oral glucose tolerance test; PCOS, polycystic ovary syndrome; QoL, quality of life; SHBG; sex-hormone binding globulin; SPL, spironolactone; WHR, waist-to-hip ratio. Consistent with the previous guideline,9 the search included only randomised controlled trial (RCT) study designs (or systematic reviews of RCTs to identify eligible studies) and was restricted to English language studies. When incorporating studies from the previous guideline, we ensured that the current inclusion criteria were met. These criteria differ slightly from the 2018 criteria, in that we are no longer excluding RCTs which either did not utilise a robust method of randomisation or did not describe the method clearly in the manuscript. In the current update, we are considering randomisation methods/reporting in the risk of bias assessments, rather than excluding the studies completely on this basis. Screening was undertaken using Covidence (www.covidence.org). Title and abstract screening and full text screening were conducted in duplicate by two of three independent reviewers (SA, MF, and JM). Risk of bias (RoB) was assessed for each study in duplicate by two of three independent reviewers (SA, MF, and JM), using criteria developed a priori. For studies included in the 2018 guideline, the RoB assessments were derived from the previous review by the 2018 guideline evidence team.9 Individual quality items included selection bias, performance bias, detection bias, attrition bias, reporting bias, as well as other bias (such as conflicts of interest and accuracy of statistical analyses). Any disagreement or uncertainty was resolved by discussion among the reviewers to reach consensus or referred to the guideline evidence team if required. Each study was allocated a rating of low, moderate, or high RoB (Supplementary Material). In circumstances where multiple studies arise from a single patient population, the main study reporting the primary outcome(s) (which often also had the largest sample size and complete outcome set) was assessed for RoB. All smaller sub-studies were denoted with the same RoB ranking allocated to the main study. Data analysis Data were extracted by one reviewer (SA), with independent cross-checking by two reviewers (MF, JM) to ensure accuracy. A template was developed for data extraction of relevant items, including study details (author, year of publication, country of origin, study design, sample size), participants (population characteristics, setting, and method of PCOS diagnosis), intervention and comparison (type, dose, duration, frequency, and combinations) and relevant outcomes, including units and assessment methods where necessary. Meta-analyses were performed using Review manager V.5.4. Due to clinical heterogeneity from differences in doses, intervention timing, and outcome measures, random effects models were used for all meta-analyses. Effect estimates are presented as the weighted mean difference (WMD) for continuous outcomes, and the odds ratio (OR) for dichotomous outcomes, with 95% confidence intervals (CIs). Results are presented in summary tables and forest plots. Where only a single study reported an outcome or comparison, this was included as a single-paper meta-analysis, to facilitate the communication and comparison of results across studies, thus simplifying assessments of convergence while providing more precise estimates of effects and the uncertainty surrounding these estimates.23 Where data for a given outcome were reported in a main study and again in a sub-study of the same patient population, only one study was included in the relevant meta-analysis, usually the main study with the larger sample size. Where data were not reported or were reported only as median (and interquartile ranges) or changes from baseline, this precluded meta-analysis and these studies were instead presented narratively in a descriptive analysis. Subgroup analyses were planned a priori to stratify studies by age, ethnicity or phenotype, wherever possible. Sensitivity analyses by RoB were conducted to examine the impact of high risk studies on results. Here, high RoB studies are excluded from each meta-analysis to establish their influence on results, and increase confidence that the observed effects (or lack thereof) are unlikely to be the product of biased data. Statistical heterogeneity was determined by an I2, whereby I2 > 50% is considered high, urging caution in the interpretation of those results. Publication bias was assessed by visual inspection of funnel plot asymmetry to detect small study effects. Using the Grading of Recommendations Assessment, Development and Evaluation (GRADE) framework, outcomes were prioritised as either critical or important after careful deliberation among the clinical leads of the guideline development group for this review question.24,25 Critical outcomes included weight, body mass index (BMI), irregular cycles, hirsutism, liver function, and quality of life (QoL). Detailed appraisals of GRADE components are presented in the Supplementary Material, which include the RoB assessments, inconsistency (e.g. different directions of effect, confidence intervals not overlapping, and statistical heterogeneity), indirectness (e.g. important variations in populations or interventions), imprecision (small sample sizes and wide confidence intervals) or other biases (e.g. study design issues). The GRADE assessments were conducted in duplicate by two of three reviewers (SA, MF and JM). Role of the funding source The funder of the study had no role in study design, data collection, data analysis, data interpretation, or writing of the report. SA, MF, JM and AM had access to the dataset and all authors (SA, AM, JM, DR, AP, CTT, SFW, HT and AM) had final responsibility for the decision to submit for publication. Results After screening 1312 titles and abstracts and 432 full texts, 48 articles were identified from the combined search for anti-androgens, metformin, and COCP, of which nine pertained to anti-androgens specifically and were included in the present review (Fig. 1). In addition to these nine newly identified articles, 17 studies of anti-androgens were included from the previous 2018 guideline (two of which were previously excluded but met current criteria). An updated search was also conducted in June 2023, where 395 titles and abstracts and nine full texts were screened. From these, one additional eligible study was identified, bringing the total to 27 articles corresponding to 20 unique studies. Of these, 13 could be pooled in meta-analysis. The full text articles excluded with reasons for exclusion are listed in the Supplementary Material. Fig. 1. PRISMA diagram for literature search process and total number of studies screened and included at each stage. PCOS, polycystic ovary syndrome. Characteristics and quality of included trials Characteristics of the 27 included articles comprising the 20 unique studies are presented in Table 2, and a summary of results from meta-analyses of key outcomes is provided in Table 3. Additional detailed results are provided in the Supplementary Material. Although planned, subgroup analyses were not possible due to the small number and high heterogeneity of the included studies. Table 2. Key characteristics of studies included in a systematic review and meta-analysis of anti-androgen use in polycystic ovary syndrome. Author, year, country \| Population/Setting \| Intervention N \| Intervention description \| Comparison N \| Comparison description \| Follow Up \| Outcomes \| Type of contraception \| Risk of bias Alpañés 2017, Spain \| Women with PCOS/Androgen excess outpatient clinic \| 1: 18 \| 1:30 μg EE+ 150 μg DG + 100 mg spironolactone \| 2: 13 \| 2:Metformin 850 mg b.i.d. \| 12 months \| Frequency of menstrual dysfunction hirsutism score, BMI, waist circumference, serum total and free testosterone, androstenedione and DHEAS, OGTT, serum insulin and plasma glucose, HOMA, adverse effects \| Non-hormonal contraception \| High Amiri 2014, Iran \| Overweight and obese infertile women with PCOS/Fatemezahra Infertility and Reproductive Health Centre \| 1: 272: 27 \| 1:250 mg FLU 2/day +1 month HC pre-diet2:500 mg MET 3/day +250 mg FLU 2/day +1 month HC pre-diet \| 3: 254: 26 \| 3:500 mg MET 3/day + 1-month HC pre-diet4.PLAC \| 6 months \| BMI, WHR, Hirsutism, SHBG, Testosterone, DHEAS, fasting insulin, fasting glucose, OGTT, QUICKI, Total cholesterol, HDL, LDL, Triglycerides \| Barrier contraception \| High Diri 2017, Turkey \| Patients with PCOS/Department of Endocrinology Erciyes University Medical School \| 1: 162: 19 \| 1:FIN 5 mg/day2:FIN 5 mg/day + MET \| 3: 17 \| 3:MET 1700 mg/day \| 12 months \| BMI, Hirsutism, SHBG, Free testosterone, DHEAS, Androstenedione, HOMA-IR \| Barrier contraception \| High Dumesic 2023, USA \| Normal weight women with PCOS/An academic medical centre \| 1: 5 \| 1: 125 mg/day FLU \| 2: 6 \| 2:Placebo \| 6 months \| BMI, body weight, total testosterone, free testosterone, androstenedione, dihydrotestosterone, DHEAS, fasting glucose, fasting insulin, insulin sensitivity, SHBG, total cholesterol \| Abstinence or nonhormonal form of contraception \| Moderate Falsetti 1997, ItalyFalsetti 1999, Italyb \| Hirsute women with PCOS/Department of Gynaecological Endocrinology of the University of Brescia \| 1: 221: 32 \| 1:FIN 5 mg once/day \| 2: 222: 32 \| 2:FLU 250 mg b.i.d. \| 6 months12 months \| Hirsutism, SHBG, Testosterone, free testosterone, DHEAS, Androstenedione, fasting insulin, GI severe. \| Barrier contraception \| Higha Gambineri 2004, ItalyGambineri 2006, Italy \| Overweight women with PCOS/Division of endocrinology S. Orsola-Malpighi Hospital, Italy \| 1: 102: 101: 172: 20 \| 1:FLU 250 mg orally b.d. + hypocaloric diet2:MET 850 mg orally b.d. + 250 mg FLU 250 mg orally twice/day + hypocaloric diet \| 3: 104: 103:204: 19 \| 3:MET 850 mg orally b.d. + hypocaloric diet4:PLAC + hypocaloric diet \| 6 months12 months \| Body weight, BMI, waist circumference, Hirsutism, frequency of menstruation, total testosterone, free-androgen index, androstenedione, DHEA-S, SHBG, fasting glucose, fasting insulin, QUICKI, ISI, LDL, HDL, TriglyceridesGambineri 2004 added: HOMA, \| Non-hormonal contraception \| Lowa Ganie 2004, India \| Women with PCOS/Attending Endocrine and Metabolism Clinical of the All-India Institute of Medical Sciences between 2001 and 2002 \| 1: 34 \| 1:SPL 50 mg/day + lifestyle advice \| 2: 35 \| 2:MET 1000 mg/day + lifestyle advice \| 6 months \| BMI, WHR, Menstrual cyclicity, hirsutism, fasting blood glucose, HOMA, Testosterone \| Barrier contraception \| Moderatea Ganie 2013, India \| Women with PCOS/Tertiary care referral centre \| 1: 512: 62 \| 1:SPL 50 mg/day + diet counselling2:SPL 50 mg/day + MET 1000 mg/day + diet counselling \| 3: 56 \| 3:MET 1000 mg/day + diet counselling \| 6 months \| Body weight, BMI, WHR, Menstrual cyclicity, Hirsutism, testosterone, fasting glucose, fasting insulin, HOMA-IR, QUICKI \| Barrier contraception \| Lowa Hagag 2014, Israel \| Women with hirsutism due to PCOS/University affiliated endocrinology clinic, Israel \| 1: 722: 70 \| 1:250 μg NOR + 35 μg EE [250 μg NOR + 35 μg EE] + 100 mg SPL2:[2 mg CPA + 35 μg EE] + 10 mg CPA added \| 3: 25 \| 3:250 μg NOR + 35 μg EE \| 12 months \| Weight change, acne, Adverse events (Nausea, Breast tenderness, Nipple discharge, menorrhagia, headache, etc) \| COCP (as part of intervention) \| Moderatea Ibanez 2004, Spain \| Nonobese adolescents and adults with PCOS/Endocrinology Unit, Hospital Saint Joan de Deu, University of Barcelona \| 1: 162: 11 \| 1:MET 850 mg + FLU 62.5 mg (adults)2:OCP + MET 850 mg + FLU 62.5 mg \| 3: 164: 11 \| 3:EE 30 μg + 0.3 mg DRSP (adolescents)4:EE 30 μg + 0.3 mg DRSP (adults) \| 9 months \| BMI, Hirsutism, Fasting glucose/insulin ratio, SHBG, Testosterone, TG, HDL, LDL \| Adolescents abstained from sex (no need for contraception); Adults were on COCP (as part of intervention) \| Moderatea Ibanez 2020, Spainde Zegher 2021, SpainIbanez 2017, SpainMalpique 2019, SpainbDiaz 2018, Spainb \| Nonobese adolescents with PCOS/Endocrinology Unit, Hospital Saint Joan de Deu \| 1: 311: 291: 18 \| 1:SPL 50 mg/d + pioglitazone 7.5 mg/d + metformin 850 mg/d (SPIOMET). \| 2: 312: 291: 18 \| 2:EE 20 μg –levonorgestrel 100 mg \| 12 months (and 12 months post-treatment) \| BMI, hirsutism score, SHBG, TT, Androstenedione, insulin, HOMA, OGTT, TG, LDL, HDL, CRP,de Zegher: FAI, TTIbanez 2017: Acne scores \| Advice on contraception–abstained from sex (no need for contraception) \| Moderate Mazza 2014, Italy \| Overweight and obese women with PCOS/Endocrine Unit of University Magna Graecia of Catanzaro \| 1: 28 \| 1:SPL 25 mg/day + MET 1700 mg/day + lifestyle modification (HCD: 1300 kcal/d) \| 2: 28 \| 2:MET 1700 mg/day + lifestyle modification (HCD: 1300 kcal/d) \| 6 months \| Weight, BMI, Hirsutism, cholesterol, HDL, LDL, triglycerides, fasting glucose, fasting insulin, HOMA, total testosterone, SHBG, FAI, DHEAS \| Low-dose SPL \| Lowa Mehrabian 2016, Iran \| Women with PCOS/Midwifery clinic of Al-Zahra Hospital \| 1: 34 \| 1:FLU 62.5 mg + OCP (0.03 mg EE + 0.15 mg LVG) \| 2: 34 \| 2:MET 1000 mg/day \| 6 months \| Waist circumference, triglycerides, fasting blood glucose, CRP, HDL, BMI, \| COCP (as part of intervention) \| Lowa Meyer 2007, AustraliaBurchall 2017, Australia \| Overweight women with PCOS \| 1: 331: 16 \| 1:50 mg SPL + low-dose OCP b.d. (EEμg + LVG) \| 2: 313: 362: 213: 23 \| 2:High-dose OCP (35 μg EE + 2 mg CPA)3:MET 2000 mg/day \| 6 months \| Weight, BMI, OGTT, insulin, HOMA, testosterone, OGTT, Cholesterol, LDL, HDL, TG, CRP, TT, SHBG, FAI \| COCP (as part of intervention) \| Moderatea Moretti 2018, Italy \| Women with PCOS/Unit of endocrinology, section of reproductive endocrinology, University of Rome \| 1: 28 \| 1:OCP (EE 0.030 mg + DRSP 2 mg or CPA 2 mg or dienogest 2 mg) + BC 50 mg o.d. \| 2: 24 \| 2:OCP 0.030 mg + DRSP 2 mg or CPA 2 mg or dienogest 2 mg) + PLAC \| 12 months \| Hirsutism, weight, BMI, total cholesterol, HDL, triglycerides, LDL, fasting glucose \| COCP (as part of intervention) \| Moderatea Spritzer 2000, Brazil \| Women with PCOS/Gynaecological Endocrinology Unit at Hospital \| 1: 10 \| 1:200 mg/d SPL, 20 d/month \| 2: 9 \| 2:CPA 50 mg/day, 20 d/month + 35 mg/d EE over the last 10 days of CPA \| 12 months \| Hirsutism \| Barrier contraception \| Moderatea Tartagni 2000, Italy \| Women with PCOS/Outpatients in an academic research environment \| 1: 9 \| 1:FIN + Diane-35 (CPA 2 mg + EE 35 μg) \| 2: 9 \| 2:Diane-35 (CPA 2 mg + EE 35 μg) \| 6 months \| Hirsutism, free testosterone, DHEAS, SHBG, Androstenedione \| COCP (as part of intervention) \| Higha Tartagni 2004, Italy \| Women with hirsutism due to PCOS/Obstetrics and gynaecology outpatient clinic \| 1: 8 \| 1:FIN 2.5 mg/day \| 2: 8 \| 2:FIN 2.5 mg every 3 days \| 10 months \| Total testosterone, DHEAS, SHBG, androstenedione, BMI, Hirsutism \| Non hormonal contraception \| Higha Tartagni 2014, Italy \| Women with hirsutism due to PCOS/Obstetrics and gynaecology outpatient clinic \| 1: 7 \| 1:FIN 2.5 mg every 3 days \| 2: 7 \| 2:Placebo \| 6 months \| BMI, Hirsutism, SHBG, Testosterone, DHEAS, Androstenedione, GI-related adverse effects \| Non-hormonal contraception \| Higha Vieira 2012, Brazil \| Women with PCOS/University Hospital of Ribeirao Preto School of Medicine between 2007 and 2009 \| 1: 20 \| 1:OCP (2 mg CPA + 30 mcg EE) + SPL 100 mg/day \| 2: 21 \| 2:OCP (2 mg CPA + EE 30 mcg) \| 12 months \| Weight, BMI, SHBG, FAI, Testosterone, free testosterone, fasting insulin, fasting glucose, total cholesterol, HDL, LDL, HOMA, CRP \| COCP (as part of intervention) \| Moderatea Abbreviations: b.i.d, two times daily; BC, bicalutamide; BMI, body mass index; CPA, cyproterone acetate; CRP, c-reactive protein; DHEAS, dehydroepiandrosterone sulphate; EE, ethinylestradiol; FAI, free androgen index; FLU, flutamide; FIN, finasteride; GI, glycaemic index; HCL, hypocaloric diet; HDL, high density lipoprotein; HOMA-IR, homeostasis model assessment-estimated insulin resistance; IR, insulin resistance; LDL; low density lipoprotein; LVG, levonorgestrel; MET, metformin; MA, meta-analysis; OCP/COCP, oral contraceptive pill/combined oral contraceptive pill; OGTT, oral glucose tolerance test; PLAC, placebo; PCOS, polycystic ovary syndrome; QUICKI, the quantitative insulin-sensitivity check index; SHBG, sex hormone binding globulin; SPL, spironolactone; TT, total testosterone. Risk of bias derived from 2018 PCOS guideline. No additional outcomes from Diaz 2018 and Malpique 2019–no data were extracted in these circumstances. Table 3. Summary of meta-analysis results examining the use of anti-androgens in polycystic ovary syndrome. Comparisona \| Studies (participants), duration \| BMI/Weight \| Hirsutism/Acne \| Hormonal measures \| Glycaemic/IR measures \| Lipids \| Menstruation AA + LSvs Placebo + LS (adults) \| 2 (89) 6–12 months \| NS = BMI \| NS = hirsutism \| NS= SHBG \| – \| – \| – AA dailyvs every 3 days (adults) \| 2 (80) 10–12 months \| – \| ↓ hirsutism \| ↓ AndrostenedioneNS = DHEAS, SHBG, T \| – \| – \| – AA + LSvs Metformin + LS (adults) \| 2–4 (89–265) 6–12 months \| NS = Weight, BMI, WHR \| ↓ hirsutism \| ↑ SHBGNS = T, DHEAS \| ↓ fasting insulin;↓ fasting insulin: glucose ratioNS= FBG, QUICKI \| – \| ↑ frequency AA + Metformin + LSvs Metformin + LS (adults) \| 2–4 (92–262) 6–12 months \| NS = Weight, BMI, WHR \| NS = hirsutism \| ↓ TNS= SHBG, DHEAS \| ↓ FBGNS = fasting insulin, HOMA-IR \| NS= HDL, LDL, TG \| – AA + LSvs AA + Metformin + LS (adults) \| 2–3 (91–204), 6–12 months \| NS = BMI \| NS = hirsutism \| NS= SHBG, T \| ↑ FBGNS = QUICKI, fasting insulin \| – \| – AA + COCPvs COCP ± Placebo (adults) \| 2–3 (59–130), 6–12 months \| NS = Weight, BMI \| – \| NS = T, SHBG \| – \| ↑ TG, ↑ LDL, ↑ TCNS= HDL \| – AA + COCPvs Metformin (adults) \| 2 (107) 6 months \| NS = BMI \| – \| – \| – \| – \| – AA, anti-androgens; BMI, body mass index; DHEAS, dehydroepiandrosterone sulphate; FAI, free androgen index; FBG, fasting blood glucose; HDL/LDL, high-density/low-density lipoprotein; HOMA-IR, homeostatic model assessment of insulin resistance; IR, insulin resistance; ISI, insulin sensitivity index; QUICKI, Quantitative insulin-sensitivity check index; LS, lifestyle; NS, not significant; OGTT, oral glucose tolerance test; SHBG, sex-hormone binding globulin; T, testosterone; TC, total cholesterol; TG, triglycerides; WHR, waist hip ratio. Effect reported in the table relates to the intervention listed in bold (e.g. for the first row, the up or down arrows indicate higher or lower effect for each outcome, respectively, with the AA + LS intervention). Overall, sample sizes varied from 11 to 167 participants, with treatment durations ranging from six to 12 months. Studies were conducted in Italy (n = 7), Spain (n = 3), Brazil (n = 2), Iran (n = 2), India (n = 2), Australia (n = 1), Israel (n = 1), Turkey (n = 1), and the United States of America (n = 1). Anti-androgen medications included spironolactone, finasteride, flutamide, and bicalutamide, with comparators including placebo, metformin, COCP, lifestyle, or a combination of these. Most of the included studies were in adults, with two in adolescents. The mean BMI of participants ranged from healthy (18.5–24.9 kg/m2) to overweight (25.0–29.9 kg/m2) and obese (≥30.0 kg/m2), but most studies were in women with a BMI ≥25.0 kg/m2. Methods of contraception included abstinence, barrier, or hormonal contraceptives as part of the intervention or comparison, and were reported in all studies (as this was an eligibility criterion). In quality appraisal, four studies were classified as being of low RoB, nine were moderate, and seven were high risk. The most common reasons for high RoB ratings were the lack of concealed or centralised allocation, absence of participant and/or investigator blinding, high drop-out rate, and lack of pre-registered protocols (Table S1). In the GRADE assessments, the certainty of evidence in the reported outcomes mostly ranged from very low to low, due to concerns regarding RoB, small sample sizes (imprecision), and statistical heterogeneity (inconsistency), with very few outcomes ranging from moderate to high (detailed in the Supplementary Material). Placebo and lifestyle comparisons Anti-androgen vs placebo—adolescents and adults One high RoB trial compared low-dose finasteride (2.5 mg daily) vs placebo for six months in 14 adolescents with PCOS (age 15–19 years).26 Hirsutism assessed by Ferriman-Gallwey (FG) scores was lower with anti-androgens compared with placebo (WMD [95% CI]: −4.00 [−6.81, −1.19], p = 0.005), but dehydroepiandrosterone sulphate (DHEAS) was higher (0.60 μmol/l [0.07, 1.13], p = 0.03). There were no differences in BMI, testosterone, sex-hormone binding globulin (SHBG), or androstenedione (Supplementary Material). One moderate RoB trial compared low-dose flutamide (125 mg daily) vs placebo for six months in 11 normal-weight adults with PCOS.27 In single study analysis, there were no differences in any outcomes assessed, including weight, BMI, lipids, DHT, DHEAS, androstenedione, total and free testosterone, or glycaemic measures (Supplementary Material). Anti-androgen + lifestyle vs placebo + lifestyle—adults Two RCTs compared anti-androgens (flutamide 250 mg twice daily) + lifestyle intervention with placebo + lifestyle in adult women with PCOS, one with 6-month follow-up with high RoB,28 and the other 12 months with low RoB.29 In meta-analysis of these two RCTs (n = 89; Table S2), there were no differences in BMI (WMD [95% CI]: −3.08 kg/m2 [−8.67, 2.50], hirsutism (−0.93, [−3.37, 1.51], p = 0.45; I2 = 76%) or SHBG (9.72 nmol/l [−0.71, 20.14], p = 0.07; I2 = 31%), while other outcomes could not be assessed due to discrepancies in units in the high RoB study by Amiri et al.28 (Supplementary Material). In examining the single low RoB study by Gambineri et al.,29 anti-androgens (flutamide 250 mg twice daily) + lifestyle resulted in lower body weight (−17.00 kg, [−25.37, −8.63], p < 0.0001), DHEAS (−2.34 μg/ml [−4.06, −0.62], p = 0.008), frequency of menstruation (−0.80 [−1.54, −0.06], p = 0.03), fasting insulin (−4.00 μU/ml [−6.98, −1.02], p = 0.009), low-density lipoprotein (LDL) cholesterol (−21.00 mg/dl [−40.93, −1.07], p = 0.04) and triglycerides (−50.00 mg/dl [−77.60, −22.40], p = 0.0004), and higher quantitative insulin-sensitivity check index (QUICKI) (0.04 [0.02, 0.06], p < 0.0001) and insulin sensitivity index (5.10 [2.32, 7.88], p = 0.0003), compared with placebo + lifestyle. There were no differences in other outcomes assessed (Table 3 and Supplementary Material). Anti-androgen comparisons Anti-androgen (daily) vs anti-androgen (every 3 days)—adults Two high RoB studies were pooled in meta-analysis, comparing finasteride regimens (2.5 mg once daily vs every 3 days) or flutamide regimens (250 mg twice daily vs every three days) for 10–12 months.30,31 In meta-analysis, daily anti-androgen use was superior for reducing hirsutism (WMD [95% CI]: −3.48 FG score [−4.58, −2.39], p < 0.0001) and androstenedione (−0.30 ng/ml [−0.50, −0.10], p = 0.004), with no differences in SHBG, testosterone, or DHEAS (n = 80; Table S3). For outcomes reported in a single study (n = 64),30 there were no differences in BMI or adverse effects such as decreased libido or headache; however, dry skin was higher in the daily anti-androgen group (OR [95% CI]: 6.60 [2.21, 19.73], p = 0.0007) and three women dropped out with treatment of flutamide due to liver toxicity (high transaminase levels). Metformin comparisons Anti-androgens vs metformin with or without anti-androgens—adults A single high RoB trial compared anti-androgens (finasteride 5 mg once daily) alone or with metformin vs metformin alone for 12 months in adult women with PCOS.32 There were no differences in BMI, hirsutism, free testosterone, SHBG, DHEAS, androstenedione, or HOMA-IR in either comparison (Supplementary Material).32 Anti-androgens + lifestyle vs metformin + lifestyle—adults Four RCTs28,29,33,34 reported the use of anti-androgens (flutamide 250 mg twice daily or spironolactone 50 mg once or twice daily) + lifestyle compared with metformin + lifestyle for 6–12 months in adult women with PCOS. Two had a low RoB,29,33 one moderate34 and one high.28 The study by Amiri et al.28 was excluded from meta-analysis in circumstances where the accuracy of units was unclear. In meta-analysis, anti-androgens + lifestyle resulted in lower hirsutism (n = 265, WMD [95% CI]: −1.59 [−3.06, −0.12], p = 0.03), fasting insulin (n = 213; −2.11 μU/ml [−3.97, −0.26], p = 0.03) and fasting glucose-insulin ratio (n = 176; −1.12 [−1.44, −0.79], p < 0.0001), and higher HDL (n = 37; 0.21 mmol/l [0.05, 0.37], p = 0.01), frequency of menstruation (n = 176, 0.79 cycles/year, [0.05, 1.53], p = 0.04) and SHBG (n = 89; 7.70 nmol/l [0.75, 14.66], p = 0.03), compared with metformin + lifestyle (Table S4). No differences in other outcomes were found (Table 3 and Supplementary Material). Of note, when removing Amiri et al.28 in a sensitivity analysis, hirsutism and SHBG were no longer significant (p = 0.11 for both; not shown). Anti-androgens + metformin + lifestyle vs metformin + lifestyle—adults Four RCTs compared anti-androgens (flutamide 250 mg twice daily or spironolactone 25–50 mg daily) + metformin + lifestyle vs metformin + lifestyle alone, with 6–12 months follow-up in adult women with PCOS. Three had low RoB29,33,35 and one had high RoB.28 In meta-analysis (n = 262; Table S5), the anti-androgen + metformin + lifestyle group had lower testosterone (WMD [95% CI]: −0.29 nmol/l [−0.52, −0.06], p = 0.01) and fasting glucose (−2.93 mg/dl [−5.78, −0.09], p = 0.04). In sensitivity analysis excluding the high RoB study by Amiri et al.,28 WHR favoured the anti-androgen + metformin + lifestyle intervention (n = 118; −0.03 [−0.06, −0.00], p = 0.02; not shown), but the difference in fasting glucose was no longer significant (p = 0.12; not shown). For outcomes analysed in a single study,29 the anti-androgens + metformin + lifestyle intervention resulted in a higher insulin sensitivity index, with no differences in other outcomes (Table 3; Supplementary Material). Anti-androgens + metformin + lifestyle vs. anti-androgens + lifestyle—adults Three RCTs of 6–12 month duration in adult women with PCOS compared anti-androgens (flutamide 250 mg once or twice daily, or spironolactone 50 mg daily) + lifestyle + metformin vs anti-androgens + lifestyle only, of which two had low RoB29,33 and one high.28 In meta-analysis, anti-androgen + lifestyle only (without metformin) resulted in higher fasting glucose (n = 204, WMD: 3.81 mg/dl, 95% CI: 1.35, 6.28, p = 0.002; Table S6). There were no differences in other outcomes (Table 3; Supplementary Material). Results were unchanged in sensitivity analysis after removing the high RoB study by Amiri et al.28 (not shown). For outcomes assessed in single studies by Gambineri et al.29 or Ganie et al.,33 anti-androgen + lifestyle resulted in higher WHR (n = 113; 0.05 [0.03, 0.07], p < 0.0001) than the metformin combination, with no differences in other outcomes (Table 3; Supplementary Material). COCP comparisons Anti-androgens vs anti-androgens + COCP—adults One moderate RoB trial compared 12-month treatment with spironolactone (200 mg daily) vs CPA (50 mg daily) + ethinyl estradiol (35 μg daily) for hirsutism in adult women with idiopathic hirsutism and PCOS.36 In the PCOS subgroup (n = 19), hirsutism was lower with the combination treatment (WMD: −4.00 [−4.90, −3.10], p < 0.0001). Anti-androgens + COCP vs COCP with or without placebo—adults A total of six RCTs compared anti-androgens (spironolactone 50–100 mg daily, finasteride 5 mg daily, or bicalutamide 50 mg daily) + COCP with COCP ± placebo, with 6–12 months follow-up in adult women with PCOS. Five had moderate RoB37, 38, 39, 40, 41 and one had a high RoB.42 In meta-analyses of two studies (n = 93), total cholesterol (WMD [95% CI]: 20.81 mg/dl [7.81, 33.82], p = 0.002), LDL (15.12 mg/dl [3.20, 27.04], p = 0.01), and triglycerides (41.34 mg/dl [20.26, 62.42], p = 0.0001) were higher with anti-androgens + COCP compared with COCP alone, with no difference in other outcomes (Table S7, Supplementary Material). In sensitivity analysis excluding the high RoB study42 from the assessment of SHBG, SHBG was lower in the anti-androgen + COCP group compared to COCP alone (WMD: −84.40 nmol/l [−142.23, −26.57], p = 0.004; not shown). An additional sensitivity analysis was conducted to exclude Moretti et al.40 since this was the only placebo-controlled study that used bicalutamide while the others used either spironolactone or finasteride. Most results remained unchanged, except for LDL and triglycerides which were no longer significant (p = 0.06 and p = 0.22; not shown). In single study analyses (n = 41), fasting insulin (3.50 μIU/ml [0.20, 6.80], p = 0.04), HOMA-IR (0.70 [0.02, 1.38], p = 0.04), and FAI (0.50 [0.01, 0.99], p = 0.04), were higher in the anti-androgen + COCP compared to the COCP alone or with placebo,41 with no differences in fasting glucose (p = 0.24; Supplementary Material). Other outcomes or studies were assessed using descriptive analyses as data were not amenable to single-study or pooled meta-analyses. In the study by Moretti et al. (n = 52),40 there were no differences in fasting glucose, or in side effects including alanine transaminase (ALT), aspartate aminotransferase (AST), menorrhagia, hypercholesterolaemia, hypertriglyceridaemia, dysmenorrhoea, menstrual spotting, and minor depressive state/mood reduction between the groups. Meyer et al.39 (n = 64) reported that both high-dose COCP (35 μg ethinyl estradiol + 2 mg CPA) and low-dose COCP (20 μg ethinyl estradiol + 100 μg levonorgestrel) + anti-androgens (spironolactone 50 mg twice daily) reduced testosterone, DHEAS, and FAI, while increasing SHBG. There were small but significant changes in lipids, mostly in the higher-dose COCP group.39 The follow-up sub-study that assessed additional outcomes by Burchall et al.37 (n = 38) also reported similar reductions in hormonal parameters as those in the main study by Meyer et al.39 In Hagag et al. (n = 97),38 treatment with COCP + anti-androgens (spironolactone 100 mg daily) was more effective than COCP monotherapy for reducing hirsutism. Tartagni et al.42 (n = 18) reported improvements in DHT in the anti-androgens + COCP compared with the COCP group at 6 months, with within-group improvements in free testosterone, DHT, DHEAS, and SHBG from baseline values. Anti-androgens + COCP vs metformin—adults Four RCTs compared anti-androgens (flutamide 62.5 mg daily or spironolactone 50–100 mg daily) + COCP vs metformin for 6–12 months in adult women with PCOS (n = 207), one of which had low RoB,43 two moderate,37,39 and one high.44 Two studies could be combined in meta-analysis for one outcome, BMI, showing no difference (p = 0.93, Table S8). In descriptive analysis, the study by Alpañés et al.44 reported that hirsutism, total and free testosterone, androstenedione, DHEAS and menstrual dysfunction were lower with anti-androgens + COCP compared with metformin. Meyer et al.39 also reported that testosterone and DHEAS reduced with anti-androgens (spironolactone 50 mg daily) + COCP, but did not change with metformin; whereas, insulin resistance improved with metformin, but did not change in the anti-androgen + COCP group. In the study by Burchall et al.37 (n = 68), the anti-androgen + COCP group had lower C-reactive protein (CRP) and rates of menstrual dysfunction, while fasting blood glucose was higher, compared with metformin. Anti-androgens + metformin (with or without COCP) vs COCP—adults and adolescents A single moderate RoB trial investigated the combination therapy of anti-androgens (flutamide 62.5 mg daily) + metformin with or without COCP vs COCP alone for nine months in adults and adolescents with PCOS.45 In adults (n = 32), the anti-androgen + metformin group (without COCP) had reduced SHBG, triglycerides, LDL and HDL compared with COCP alone (Supplementary Material). In adolescents (n = 22), SHBG was lower with anti-androgens + metformin + COCP, compared to COCP alone (Supplementary Material). Anti-androgens + metformin + pioglitazone (SPIOMET) vs COCP—adolescents One study (reported in five articles/sub-studies)46, 47, 48, 49, 50 with moderate RoB compared low-dose anti-androgens (spironolactone 50 mg daily) + metformin + pioglitazone (SPIOMET) against COCP for 12 months in adolescent girls with PCOS (n = 62 in the main study). The SPIOMET group had reduced hirsutism (WMD [95% CI]: −3.00 [−5.77, −0.23], p = 0.03), SHBG (−29.00 nmol/l [−39.56, −18.44], p < 0.0001), fasting insulin (−62.00 pmol/l [−81.40, −42.60], p < 0.0001), ALT (−0.09 μkat/l [−0.16, −0.02], HOMA-IR (−1.80 [−2.42, −1.18], p < 0.0001), LDL (−0.05 mmol/l [−0.78, −0.22] p = 0.0004), CRP (−18.10 mmol/l [−25.75, −10.45], p < 0.0001), and FAI (−2.20 [−4.38, −0.02], p = 0.05), but higher androstenedione (1.00 nmol/l [0.29, 1.71], p = 0.006) compared with the COCP group. Discussion To our knowledge, this systematic review and meta-analysis is the first evidence synthesis of RCTs investigating the efficacy and safety of anti-androgen pharmacological agents, alone or in combination, on endocrine and metabolic features in adolescents and adults with PCOS. Our findings address key gaps in the literature and directly inform the current 2023 update of the International Evidence-based Guideline for the Assessment and Management of PCOS. Based on findings from this review, the guideline recommends that anti-androgen pharmacological agents could be considered to treat clinical hyperandrogenism (hirsutism), in situations where the COCP and/or cosmetic therapies (including mechanical laser and light therapies for hair reduction) are contraindicated, poorly tolerated, or present a sub-optimal response after a minimum period of six months. Where appropriate, the use of effective contraception is strongly recommended, and women should be advised that anti-androgens may cause under-virilisation of a male fetus. It should be noted that this recommendation is based on the best available evidence, which remains largely limited as noted by the high heterogeneity (12 comparisons across 20 studies) and the low to very low GRADE certainty across most outcomes. Therefore, the recommendation remains general, and should not override clinical judgment with consideration of individual circumstances and perspectives. In relation to safety, specific recommendations on optimal doses or formulations cannot be made based on the available evidence, due to high levels of heterogeneity across the included studies, with a considerable number of comparisons. In view of the limited data in PCOS, ascertaining the required information from general population literature may be appropriate. Based on general population recommendations, spironolactone doses of 25–100 mg daily appear to have low risks of adverse side effects, while high doses of CPA (≥10 mg) may lead to meningioma or venous thromboembolism, and flutamide and bicalutamide are associated with increased risks of liver toxicity.51 Most importantly, anti-androgens should not be used in the absence of effective contraception in sexually active individuals, due to the risks of under-virilisation of a male fetus if an unplanned pregnancy occurs.52 Moreover, combination therapy with anti-androgens + COCP compared to COCP (±placebo) resulted in poorer lipid profiles in adult women with PCOS. This is consistent with other studies reporting that hormonal contraception may alter lipid profiles,53 potentially via alterations in the estrogen receptor that affects hepatic apolipoprotein upregulation.54 However, when removing the study by Moretti et al.40 from the meta-analysis, the only study to use bicalutamide, there was no longer a significant difference in lipids. This indicates that bicalutamide, while effective for some PCOS outcomes such as hirsutism, worsens lipid metabolism. The risk of liver toxicity with bicalutamide has been reported in the general population, although risks of serious liver toxicity and lipid-related lung diseases are rare.55 This is concerning as PCOS is already a condition with intrinsic metabolic and lipid abnormalities, which could be further exacerbated by bicalutamide therapy. Flutamide has also been associated with hepatotoxicity,56 which was reflected in this systematic review where three women dropped out from the study by Falsetti et al.30 due to flutamide-induced hepatotoxicity. Therefore, due to the risks of liver toxicity and dyslipidaemia, the use of bicalutamide or flutamide is not recommended for treating clinical hyperandrogenism in women with PCOS. While other anti-androgen pharmacological agents, such as spironolactone, may be considered for treating clinical hirsutism, first-line therapy remains the COCP.57 This is because the COCP can improve menstrual cyclicity while providing effective contraception, whereas anti-androgens are teratogenic since they can interfere with external genital development of male fetuses.52 Consistent with our findings, a systematic review and network meta-analysis by Barrionuevo et al.17 of 43 studies reported that anti-androgens in combination with the COCP may have additive benefits for hirsutism, noting important methodological limitations in the available evidence. However, Barrionuevo et al.17 investigated women with idiopathic hirsutism and specifically excluded women with PCOS. An earlier systematic review of 28 studies by Koulouri et al.18 identified improvements in hirsutism following anti-androgen treatment, which were associated with BMI; however, this review included broader hirsute populations, rather than PCOS specifically, and is likely outdated given that the search was conducted 17 years ago (in 2006). Although PCOS constitutes 65–75% of hirsutism cases,5 other causes of hirsutism such as Cushing's syndrome, androgen-producing tumours, hyperprolactinaemia, and peripheral androgen activity, may necessitate different and more nuanced treatment approaches than those needed in PCOS,58 particularly due to the cardiometabolic implications of the latter. Here, we extend the potential benefits of anti-androgens on hirsutism to women with PCOS, while maintaining that COCP remains first-line therapy pending further study. Anti-androgens and COCPs can reduce hirsutism via different mechanisms. The COCP increases SHBG production and decreases luteinising hormone (LH)-induced stimulation of theca cells, leading to reduced circulating free testosterone concentrations.59 Conversely, anti-androgens promote reductions in clinical hyperandrogenism by competitively inhibiting androgen-binding receptors, likely via inhibiting 5-α-reductase, as well as by other mechanisms which are hitherto not well-understood. Although early research indicated that spironolactone may interfere with androgen synthesis,60 these data have not been replicated or validated using modern techniques. Therefore, further studies elucidating the mechanisms of action of anti-androgens are warranted. This is, to our knowledge, the first systematic review and meta-analysis examining the use of anti-androgens specifically in women with PCOS. We conducted a comprehensive search, using international gold-standard methodology for identifying, evaluating, and appraising the literature (including the GRADE tool), in line with the internationally endorsed PCOS guideline methodology. All data extraction and quality assessments were conducted in duplicate and/or cross-checked to limit subjectivity bias and maximise accuracy. We included both adult and adolescent populations, with a wide range of outcomes relevant to PCOS as determined by experts and consumer groups in the guideline development process, thereby providing broad insights into the efficacy of anti-androgens, with or without other interventions, in managing key features of PCOS. Several limitations in the literature were apparent. Most notably, significant heterogeneity in the included studies and variations in methodological quality were evident, with many studies having small sample sizes and varied interventions, comparators, doses, frequencies, and population characteristics. This was reflected in the GRADE certainty of evidence, which was classified as very low or low for the majority of outcomes. Given that the 20 included studies represented 12 different comparisons, this heterogeneity precluded meta-analysis for many comparisons, with limited sensitivity analyses, and we were unable to perform sub-group stratification by important variables including age, weight/BMI, insulin resistance, lifestyle variables or severity of symptoms, given the small sample sizes across most comparisons. In addition, separate assessments of specific anti-androgens were not possible and there were insufficient studies in adolescents (n = 2 trials) to allow for meaningful comparisons. For these reasons, the results presented herein may not reflect the real-life experiences of all patients with PCOS or their treating clinicians, and treatments should be tailored to individual needs and preferences. In the meantime, further high-quality, well-powered RCTs are needed to clarify the effects of anti-androgens in women with PCOS with greater certainty. In particular, given that only two studies were identified in adolescents in this review, studies of adolescents as well as ‘at risk’ populations (adolescents with PCOS features, but not meeting formal diagnostic criteria) are lacking and this remains a key gap in the evidence. Other research priorities include large-scale studies comparing efficacy of different anti-androgen types, doses, combinations and treatment schedules, as well as optimal monitoring methods for adverse events. Studies should be sufficiently sized to allow for exploration of efficacy among specific phenotypes and by important variables including weight, age and metabolic parameters. For comparisons that show differences, validation in large scale population-based studies would be of value. Limitations in the review process should also be noted. Due to being poorly reported or presented in a format which was not amenable to meta-analysis, some data were excluded, leading to missing or incomplete information for some comparisons. Although exclusion of these data was necessary to ensure consistency and rigor in the review process, this meant that sample sizes were small for many comparisons and it is possible that their inclusion may have impacted on our results, urging caution in the interpretation of these findings. We addressed this limitation to some extent via sensitivity analysis by RoB; however, this was not possible for all comparisons or outcomes due to the small number of included studies. Moreover, studies in languages other than English were excluded due to resource and time constraints, and some full texts were not found; hence, potentially relevant results may have been excluded from the review. Lastly, publication bias cannot be ruled out given that this could not be adequately judged by inspection of funnel plots (since less than 10 studies were assessed in each analysis) and grey literature (unpublished work) was not sourced or included in the review. In summary, anti-androgens may have possible benefits on clinical hyperandrogenism (i.e., hirsutism) in combination with effective contraception but, per guideline recommendations, should be used in circumstances where COCP (and/or cosmetic options including mechanical laser and light therapies for hair reduction) are contraindicated, poorly tolerated, or have been ineffective for a minimum period of 6 months. This general recommendation is based on the best available evidence, which remains limited due to high heterogeneity; hence, application should incorporate clinical judgement, contextual factors and individual characteristics and preferences. While the optimal types and doses of anti-androgens in PCOS cannot be determined from the available evidence, general population data suggests that 25–100 mg daily of spironolactone appears to be safe. Based on the available evidence, COCPs remain the recommended first-line therapy for clinical hyperandrogenism in PCOS, until further high-quality, adequately powered studies can demonstrate further benefits for anti-androgens in the context of PCOS. Contributors SA, MF, and JM conducted the screening process, risk of bias, and GRADE assessments. SA conducted the meta-analysis, evidence synthesis, and wrote the first draft of the manuscript, supervised by AM. SA, MF, JM and AM accessed and verified the underlying data. SW, DR, and AV were the clinical leads that provided content expertise and scoped the review. CTT, HT, and AM supervised the review process and provided input on meta-analysis and systematic review methodologies. All authors reviewed and edited the manuscript contributing substantial intellectual input in line with ICMJE criteria for authorship and approved the final version for publication. Data sharing statement All data are secondary and are available in the original published studies. Data produced through meta-analysis will be available with publication (in the manuscript and/or Supplementary Material). Additional information can be supplied by the corresponding author upon reasonable request. Declaration of interests All authors declare no competing interests. JM received funding from the Orion Research Foundation and the Medical Society of Finland. MF received funding from the Gothenburg Medical Association, Sahlgrenska University Hospital, the Iris Foundation and the Hjalmar Svensson Foundation, as well as honoraria from Gedeon Richter. HT received funding from the National Health and Medical Research Council (NHMRC) through the Centre for Research Excellence for Women's Health in Reproductive Life (CRE-WHiRL). SFW received support from the NHMRC-funded guideline to attend the guideline meeting and is a member of the board of directors for the Paediatric Endocrine Society. CTT is supported by CRE- WHiRL and chairs the Androgen Excess and Polycystic Ovary Syndrome Society Early Career Special Interest Group. None of these funding organisations had any role in the study design, data collection and analysis, decision to publish, or preparation of the manuscript. All other authors declare no competing interests. Acknowledgements This manuscript was conducted to inform recommendations of the International Evidence-based Guideline in PCOS, which was funded by the Australian NHMRC through the CRE-WHiRL (APP1171592) and the CRE in Polycystic Ovary Syndrome (CRE-PCOS) (APP1078444) led by Monash University, Australia and partner societies including The American Society for Reproductive Medicine, the Endocrine Society, the European Society of Endocrinology and the European Society of Human Reproduction and Embryology. The primary funders, NHMRC, were not involved in the development of the guideline and have not influenced the scope. They set standards for guideline development and, based on independent peer review, approve the guideline process. SA is supported by a Faculty Graduate Research Scholarship from Monash University. MF is supported by Gothenburg Medical Society. JM received funding from The Medical Society of Finland and Orion Research Foundation. HT and AM are supported by fellowships provided by the Australian NHMRC and CTT is supported by CRE WHiRL funded by the NHMRC. AM received an Advancing Women's Research Success Grant from Monash University which is used toward the articles processing charges for this study. None of these funding organisations had any role in the study design, data collection and analysis, decision to publish, or preparation of the manuscript. Footnotes Supplementary data related to this article can be found at Appendix A. Supplementary data References Associated Data This section collects any data citations, data availability statements, or supplementary materials included in this article. Supplementary Materials Articles from eClinicalMedicine are provided here courtesy of Elsevier ACTIONS PERMALINK RESOURCES Similar articles Cited by other articles Links to NCBI Databases Cite Add to Collections Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894
2338	https://www.desmos.com/calculator/ggw4kci2ep	Number Line Inequalities \| Desmos Loading... Number Line Inequalities Save Copy Log In Sign Up Expression 1: negative 0.0 0 5 less than or equal to "y" less than or equal to 0.0 0 5 left brace, negative 1 less than or equal to "x" less than 3 fifths , right brace−0.0 0 5≤y≤0.0 0 5−1≤x<3 5 1 Hidden Label: left parenthesis, 3 fifths , 0 , right parenthesis 3 5,0 [x] Label equals= left parenthesis, 0.6 , 0 , right parenthesis 0.6,0 2 Hidden Label: left parenthesis, negative 1 , 0 , right parenthesis−1,0 [x] Label 3 4 powered by powered by "x"x "y"y "a" squared a 2 "a" Superscript, "b" , Baseline a b 7 7 8 8 9 9 divided by÷ functions (( )) less than< greater than> 4 4 5 5 6 6 times× \| "a" \|\|a\| ,, less than or equal to≤ greater than or equal to≥ 1 1 2 2 3 3 negative− A B C StartRoot, , EndRoot pi π 0 0 .. equals= positive+
2339	https://softwareengineering.stackexchange.com/questions/262199/number-of-sequences-when-no-adjacent-items-can-be-the-same	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 Number of sequences when no adjacent items can be the same Ask Question Asked Modified 10 years, 10 months ago Viewed 773 times 0 I came across this one problem, There is a particular sequence only uses the numbers 1, 2, 3, 4 and no two adjacent numbers are the same. Write a program that given n1 1s, n2 2s, n3 3s, n4 4s will output the number of such sequences using all these numbers. Output your answer modulo 1000000007 (10^9 + 7). I can't figure out the solution of this. Will this be done with DP or some kind of DFS with backtracking? graph dynamic-programming graph-traversal sequence combinatorics Share Improve this question edited Nov 9, 2014 at 9:52 amon 136k2727 gold badges295295 silver badges386386 bronze badges asked Nov 9, 2014 at 9:24 dharakkdharakk 311 bronze badge 2 This seems to be a question of simple combinatorics, no DP or graph algorithms needed. However, I don't understand what n1 1s, n2 2s, n3 3s, n4 4s is supposed to mean, and whether there is an upper limit on sequence length. amon – amon 2014-11-09 09:54:34 +00:00 Commented Nov 9, 2014 at 9:54 It means that symbol 1 can be used n1 times maximum in any sequence, symbol 2 - n2 times max, etc. I suppose lower limit is 0 for each, so sequences with some "leftover" symbols are allowed. scriptin – scriptin 2014-11-09 09:59:03 +00:00 Commented Nov 9, 2014 at 9:59 Add a comment \| 1 Answer 1 Reset to default 1 If I'm correct, the graph traversal solution is pretty straightforward: you may traverse the graph with recursive function memorizing only current "left" symbols and last used one. let there be a function f of input (key-value pairs, where keys are 1..4; values are n1..n4) and p - last used number, and p is initially undefined in a loop, if input[i] > 0 and i != p, where i is a key add 1 to result, because you found another solution, 1 symbol longer set inputUpd := input (copy for recursive call) decrement inputUpd[i], because we've just used symbol i and there is one such symbol less left set p := i, because we've just used i add the value of f(inputUpd, p) (recursive call with "smaller" input and last used symbol) to result I couldn't help myself, so here is the code in JS. Skip it if you want to figure out the solution yourself. function f(input, prev) { var acc = 0; for (i in input) { if (i != prev && input[i] > 0) { var inputUpd = {}; for (j in input) { inputUpd[j] = (j == i) ? (input[j] - 1) : input[j]; } acc += 1 + f(inputUpd, i); } } return acc; } f({'1': 1, '2': 1}) // -> 4: , , [1,2], [2,1] f({'1': 1, '2': 2}) // -> 5: , , [1,2], [2,1], [2,1,2] f({'1': 1, '2': 3}) // -> 5: same as before, but single `2` is leftover f({'1': 1, '2': 1, '3': 2, '4': 2}) // -> 288 Share Improve this answer answered Nov 9, 2014 at 10:43 scriptinscriptin 4,4322323 silver badges3232 bronze badges 0 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions graph dynamic-programming graph-traversal sequence combinatorics See similar questions with these tags. 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2340	https://www.quora.com/Why-do-we-call-Himalayas-The-Young-Mountain?top_ans=67839880	Why are the Himalayas called “young mountains”? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Geography Himalayas Mountain Landscapes Earth Science (Geology) Natural Formation Mountains Geological History Mountain Geology Formation of Mountains 5 Why are the Himalayas called “young mountains”? All related (48) Sort Recommended Roger Conroy Author has 561 answers and 207K answer views ·7y Related Why are the Himalayas called ’ young fold mountains’? Because that is exactly what they are! See fold mountain Upvote · 9 1 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 How old are the Himalayas? Why are they called “young” mountains? Why are mountains like Himalayas and Andes called young fold mountains? What are some examples of young mountains? Why himalayas are called young fold mountains? What are young mountains (explain briefly)? Assistant Bot · 1y The Himalayas are referred to as "young mountains" primarily due to their geological formation and age. Here are the key reasons: Recent Formation: The Himalayas began forming about 50 million years ago as a result of the collision between the Indian Plate and the Eurasian Plate. This is relatively recent in geological terms, as many mountain ranges are much older. Ongoing Uplift: The region is still experiencing tectonic activity, which means the mountains are actively rising. This ongoing uplift contributes to their classification as young mountains. Erosion and Relief: Young mountains typicall Continue Reading The Himalayas are referred to as "young mountains" primarily due to their geological formation and age. Here are the key reasons: Recent Formation: The Himalayas began forming about 50 million years ago as a result of the collision between the Indian Plate and the Eurasian Plate. This is relatively recent in geological terms, as many mountain ranges are much older. Ongoing Uplift: The region is still experiencing tectonic activity, which means the mountains are actively rising. This ongoing uplift contributes to their classification as young mountains. Erosion and Relief: Young mountains typically have sharp peaks and rugged terrain, which are characteristic of the Himalayas. Over time, erosion will smooth these features, but the current relief indicates a younger geological age. Geological Characteristics: The Himalayas are primarily composed of sedimentary rocks that were uplifted during the mountain-building process, indicating that they have not undergone extensive metamorphism or erosion that would typically affect older mountain ranges. These factors collectively contribute to the designation of the Himalayas as "young mountains." Upvote · Will Simmons I spend most of my free time in the mountains · Author has 800 answers and 2.9M answer views ·7y Because in geological terms, they are exactly that: young. Formed ‘only’ about 50 million years ago, rather than the hundreds of millions of years ago which other mountain ranges were formed. As the plates from which they are formed are still colliding, the Himalayas continue to grow. Contrast the characteristic jagged nature of these young mountains with the smoother, more rounded mountains which have been subject to erosion for a much longer period - the mountains of Scotland were formed about 450 million years ago, and have been glaciated and weathered significantly since. Upvote · 9 2 Tom Scott Geology Enthusiast ·7y Unlike most other mountains, the Himalayas were only formed ~50 Million years ago, when the Indian plate collided with the Tibetan plate. Compare this to the Andes in Southern America, for example, which are ~252 Million years old, to give you a sense of scale. The Himalayas are still geologically active, also; still growing at a rate of ~5mm per year. Hope this helps :) Upvote · 9 6 9 1 Related questions How old are the Himalayas? Why are they called “young” mountains? Why are mountains like Himalayas and Andes called young fold mountains? What are some examples of young mountains? Why himalayas are called young fold mountains? What are young mountains (explain briefly)? Why are the Himalayas called newborn mountain ranges? How old are the Himalayas? What caused them to rise up? How were the Himalayas mountains formed? What is the reason the mountains of the Himalayas are not completely straight? What mountain range is older than the Himalayas? Are the Himalayas the first or the youngest mountains? Which mountain range is older, Himalayas or Andes? Why is volcanic mountain called mountain of accumulation W? 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2341	https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcendentals_(Stewart)/10%3A_Parametric_Equations_And_Polar_Coordinates/10.02%3A_Calculus_with_Parametric_Curves	Skip to main content 10.2: Calculus with Parametric Curves Last updated : Nov 10, 2020 Save as PDF 10.1: Curves Defined by Parametric Equations 10.3: Polar Coordinates Page ID : 4506 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Determine derivatives and equations of tangents for parametric curves. Find the area under a parametric curve. Use the equation for arc length of a parametric curve. Apply the formula for surface area to a volume generated by a parametric curve. Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve? Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve (x(t),y(t)) then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time. Derivatives of Parametric Equations We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations x(t)=2t+3y(t)=3t−4 within −2≤t≤3. The graph of this curve appears in Figure 10.2.1. It is a line segment starting at (−1,−10) and ending at (9,5). We can eliminate the parameter by first solving Equation 10.2.1 for t: x(t)=2t+3 x−3=2t t=x−32. Substituting this into y(t) (Equation 10.2.2), we obtain y(t)=3t−4 y=3(x−32)−4 y=3x2−92−4 y=3x2−172. The slope of this line is given by dydx=32. Next we calculate x′(t) and y′(t). This gives x′(t)=2 and y′(t)=3. Notice that dydx=dy/dtdx/dt=32. This is no coincidence, as outlined in the following theorem. Derivative of Parametric Equations Consider the plane curve defined by the parametric equations x=x(t) and y=y(t). Suppose that x′(t) and y′(t) exist, and assume that x′(t)≠0. Then the derivative dydx is given by dydx=dy/dtdx/dt=y′(t)x′(t). Proof This theorem can be proven using the Chain Rule. In particular, assume that the parameter t can be eliminated, yielding a differentiable function y=F(x). Then y(t)=F(x(t)). Differentiating both sides of this equation using the Chain Rule yields y′(t)=F′(x(t))x′(t), so F′(x(t))=y′(t)x′(t). But F′(x(t))=dydx, which proves the theorem. □ Equation 10.2.3 can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function y=f(x) is any point x=x0 such that either f′(x0)=0 or f′(x0) does not exist. Equation 10.2.3 gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function y=f(x) or not. Example 10.2.1: Finding the Derivative of a Parametric Curve Calculate the derivative dydx for each of the following parametrically defined plane curves, and locate any critical points on their respective graphs. x(t)=t2−3,y(t)=2t−1,for −3≤t≤4 x(t)=2t+1,y(t)=t3−3t+4,for −2≤t≤2 x(t)=5cost,y(t)=5sint,for 0≤t≤2π Solution a. To apply Equation 10.2.3, first calculate x′(t) and y′(t): x′(t)=2t y′(t)=2. Next substitute these into the equation: dydx=dy/dtdx/dt dydx=22t dydx=1t. This derivative is undefined when t=0. Calculating x(0) and y(0) gives x(0)=(0)2−3=−3 and y(0)=2(0)−1=−1, which corresponds to the point (−3,−1) on the graph. The graph of this curve is a parabola opening to the right, and the point (−3,−1) is its vertex as shown. b. To apply Equation 10.2.3, first calculate x′(t) and y′(t): x′(t)=2 y′(t)=3t2−3. Next substitute these into the equation: dydx=dy/dtdx/dt dydx=3t2−32. This derivative is zero when t=±1. When t=−1 we have x(−1)=2(−1)+1=−1 and y(−1)=(−1)3−3(−1)+4=−1+3+4=6, which corresponds to the point (−1,6) on the graph. When t=1 we have x(1)=2(1)+1=3 and y(1)=(1)3−3(1)+4=1−3+4=2, which corresponds to the point (3,2) on the graph. The point (3,2) is a relative minimum and the point (−1,6) is a relative maximum, as seen in the following graph. c. To apply Equation 10.2.3, first calculate x′(t) and y′(t): x′(t)=−5sint y′(t)=5cost. Next substitute these into the equation: dydx=dy/dtdx/dt dydx=5cost−5sint dydx=−cott. This derivative is zero when cost=0 and is undefined when sint=0. This gives t=0,π2,π,3π2, and 2π as critical points for t. Substituting each of these into x(t) and y(t), we obtain \| t \| x(t) \| y(t) \| --- \| 0 \| 5 \| 0 \| \| π2 \| 0 \| 5 \| \| π \| −5 \| 0 \| \| 3π2 \| 0 \| −5 \| \| 2π \| 5 \| 0 \| These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 10.2.4). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero. Exercise 10.2.1 Calculate the derivative dy/dx for the plane curve defined by the equations x(t)=t2−4t,y(t)=2t3−6t,for −2≤t≤3 and locate any critical points on its graph. Hint : Calculate x′(t) and y′(t) and use Equation 10.2.3. Answer : x′(t)=2t−4 and y′(t)=6t2−6, so dydx=6t2−62t−4=3t2−3t−2. This expression is undefined when t=2 and equal to zero when t=±1. Example 10.2.2: Finding a Tangent Line Find the equation of the tangent line to the curve defined by the equations x(t)=t2−3,y(t)=2t−1,for −3≤t≤4 when t=2. Solution First find the slope of the tangent line using Equation 10.2.3, which means calculating x′(t) and y′(t): x′(t)=2t y′(t)=2. Next substitute these into the equation: dydx=dy/dtdx/dt dydx=22t dydx=1t. When t=2,dydx=12, so this is the slope of the tangent line. Calculating x(2) and y(2) gives x(2)=(2)2−3=1 and y(2)=2(2)−1=3, which corresponds to the point (1,3) on the graph (Figure 10.2.5). Now use the point-slope form of the equation of a line to find the equation of the tangent line: y−y0=m(x−x0) y−3=12(x−1) y−3=12x−12 y=12x+52. Exercise 10.2.2 Find the equation of the tangent line to the curve defined by the equations x(t)=t2−4t,y(t)=2t3−6t,for −2≤t≤6 when t=5. Hint : Calculate x′(t) and y′(t) and use Equation 10.2.3. Answer : The equation of the tangent line is y=24x+100. Second-Order Derivatives Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function y=f(x) is defined to be the derivative of the first derivative; that is, d2ydx2=ddx[dydx]. Since dydx=dy/dtdx/dt, we can replace the y on both sides of Equation 10.2.4 with dydx. This gives us d2ydx2=ddx(dydx)=(d/dt)(dy/dx)dx/dt. If we know dy/dx as a function of t, then this formula is straightforward to apply Example 10.2.3: Finding a Second Derivative Calculate the second derivative d2y/dx2 for the plane curve defined by the parametric equations x(t)=t2−3,y(t)=2t−1,for −3≤t≤4. Solution From Example 10.2.1 we know that dydx=22t=1t. Using Equation 10.2.5, we obtain d2ydx2=(d/dt)(dy/dx)dx/dt=(d/dt)(1/t)2t=−t−22t=−12t3. Exercise 10.2.3 Calculate the second derivative d2y/dx2 for the plane curve defined by the equations x(t)=t2−4t,y(t)=2t3−6t,for −2≤t≤3 and locate any critical points on its graph. Hint : Start with the solution from the previous exercise, and use Equation 10.2.5. Answer : d2ydx2=3t2−12t+32(t−2)3. Critical points (5,4),(−3,−4),and (−4,6). Integrals Involving Parametric Equations Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by these parametric equations x(t)=t−sinty(t)=1−cost. Suppose we want to find the area of the shaded region in the following graph. To derive a formula for the area under the curve defined by the functions x=x(t)y=y(t) where a≤t≤b. We assume that x(t) is differentiable and start with an equal partition of the interval a≤t≤b. Suppose t0=a<t1<t2<⋯<tn=b and consider the following graph. We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is y(x(¯ti)) for some value ¯ti in the ith subinterval, and the width can be calculated as x(ti)−x(ti−1). Thus the area of the ith rectangle is given by Ai=y(x(¯ti))(x(ti)−x(ti−1)). Then a Riemann sum for the area is An=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)). Multiplying and dividing each area by ti−ti−1 gives An=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)ti−ti−1)(ti−ti−1)=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)Δt)Δt. Taking the limit as n approaches infinity gives A=limn→∞An=∫bay(t)x′(t)dt. This leads to the following theorem. Area under a Parametric Curve Consider the non-self-intersecting plane curve defined by the parametric equations x=x(t),y=y(t),for a≤t≤b and assume that x(t) is differentiable. The area under this curve is given by A=∫bay(t)x′(t)dt. Example 10.2.4: Finding the Area under a Parametric Curve Find the area under the curve of the cycloid defined by the equations x(t)=t−sint,y(t)=1−cost,for 0≤t≤2π. Solution Using Equation 10.2.6, we have A=∫bay(t)x′(t)dt=∫2π0(1−cost)(1−cost)dt=∫2π0(1−2cost+cos2t)dt=∫2π0(1−2cost+1+cos(2t)2)dt=∫2π0(32−2cost+cos(2t)2)dt=3t2−2sint+sin(2t)4∣2π0=3π Exercise 10.2.4 Find the area under the curve of the hypocycloid defined by the equations x(t)=3cost+cos(3t),y(t)=3sint−sin(3t),for 0≤t≤π. Hint : Use Equation 10.2.6, along with the identities sinαsinβ=12[cos(α−β)−cos(α+β)] and sin2t=1−cos(2t)2. Answer : A=3π (Note that the integral formula actually yields a negative answer. This is due to the fact that x(t) is a decreasing function over the interval [0,π]; that is, the curve is traced from right to left.) Arc Length of a Parametric Curve In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph. Given a plane curve defined by the functions x=x(t),y=y(t),for a≤t≤b, we start by partitioning the interval [a,b] into n equal subintervals: t0=a<t1<t2<⋯<tn=b. The width of each subinterval is given by Δt=(b−a)/n. We can calculate the length of each line segment: d1=√(x(t1)−x(t0))2+(y(t1)−y(t0))2 d2=√(x(t2)−x(t1))2+(y(t2)−y(t1))2 etc. Then add these up. We let s denote the exact arc length and sn denote the approximation by n line segments: s≈n∑k=1sk=n∑k=1√(x(tk)−x(tk−1))2+(y(tk)−y(tk−1))2. If we assume that x(t) and y(t) are differentiable functions of t, then the Mean Value Theorem applies, so in each subinterval [tk−1,tk] there exist ^tk and ~tk such that x(tk)−x(tk−1)=x′(^tk)(tk−tk−1)=x′(^tk)Δt y(tk)−y(tk−1)=y′(~tk)(tk−tk−1)=y′(~tk)Δt. Therefore Equation 10.2.7 becomes s≈n∑k=1sk=n∑k=1√(x′(^tk)Δt)2+(y′(~tk)Δt)2=n∑k=1√(x′(^tk))2(Δt)2+(y′(~tk))2(Δt)2=n∑k=1√(x′(^tk))2+(y′(~tk))2Δt. This is a Riemann sum that approximates the arc length over a partition of the interval [a,b]. If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives s=limn→∞n∑k=1sk=limn→∞n∑k=1√(x′(^tk))2+(y′(~tk))2Δt=∫ba√(x′(t))2+(y′(t))2dt. When taking the limit, the values of ^tk and ~tk are both contained within the same ever-shrinking interval of width Δt, so they must converge to the same value. We can summarize this method in the following theorem. Arc Length of a Parametric Curve Consider the plane curve defined by the parametric equations x=x(t),y=y(t),for t1≤t≤t2 and assume that x(t) and y(t) are differentiable functions of t. Then the arc length of this curve is given by s=∫t2t1√(dxdt)2+(dydt)2dt. At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function y=F(x). Then y(t)=F(x(t)) and the Chain Rule gives y′(t)=F′(x(t))x′(t). Substituting this into Equation 10.2.8 gives s=∫t2t1√(dxdt)2+(F′(x)dxdt)2dt=∫t2t1√(dxdt)2(1+(F′(x))2)dt=∫t2t1x′(t)√1+(dydx)2dt. Here we have assumed that x′(t)>0, which is a reasonable assumption. The Chain Rule gives dx=x′(t)dt, and letting a=x(t1) and b=x(t2) we obtain the formula s=∫ba√1+(dydx)2dx, which is the formula for arc length obtained in the Introduction to the Applications of Integration. Example 10.2.5: Finding the Arc Length of a Parametric Curve Find the arc length of the semicircle defined by the equations x(t)=3cost,y(t)=3sint,for 0≤t≤π. Solution The values t=0 to t=π trace out the blue curve in Figure 10.2.8. To determine its length, use Equation 10.2.8: s=∫t2t1√(dxdt)2+(dydt)2dt=∫π0√(−3sint)2+(3cost)2dt=∫π0√9sin2t+9cos2tdt=∫π0√9(sin2t+cos2t)dt=∫π03dt=3t\|π0=3π units. Note that the formula for the arc length of a semicircle is πr and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity. Exercise 10.2.5 Find the arc length of the curve defined by the equations x(t)=3t2,y(t)=2t3,for 1≤t≤3. Hint : Use Equation 10.2.8. Answer : s=2(103/2−23/2)≈57.589 units We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive x-axis, the parametric equations for this curve can be written as x(t)=140t,y(t)=−16t2+2t where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions x(t) and y(t) using v as an independent variable, so as to eliminate any confusion with the parameter t: x(v)=140v,y(v)=−16v2+2v. Then we write the arc length formula as follows: s(t)=∫t0√(dxdv)2+(dydv)2dv=∫t0√1402+(−32v+2)2dv The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression we can use a formula from Appendix A, ∫√a2+u2du=u2√a2+u2+a22ln∣u+√a2+u2∣+C. We set a=140 and u=−32v+2. This gives du=−32dv, so dv=−132du. Therefore ∫√1402+(−32v+2)2dv=−132∫√a2+u2du=−132[(−32v+2)2√1402+(−32v+2)2+14022ln∣(−32v+2)+√1402+(−32v+2)2\|+C] and s(t)=−132[(−32t+2)2√1402+(−32t+2)2+14022ln\|(−32t+2)+√1402+(−32t+2)2\|]+132[√1402+22+14022ln\|2+√1402+22\|]=(t2−132)√1024t2−128t+19604−12254ln\|(−32t+2)+√1024t2−128t+19604\|+√1960432+12254ln(2+√19604). This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus: ddx∫xaf(u)du=f(x). Therefore s′(t)=ddt[s(t)]=ddt[∫t0√1402+(−32v+2)2dv]=√1402+(−32t+2)2=√1024t2−128t+19604=2√256t2−32t+4901. One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to s(13)=(1/32−132)√1024(13)2−128(13)+19604−12254ln\|(−32(13)+2)+√1024(13)2−128(13)+19604\|+√1960432+12254ln(2+√19604)≈46.69 feet. This value is just over three quarters of the way to home plate. The speed of the ball is s′(13)=2√256(13)2−32(13)+4901≈140.27 ft/s. This speed translates to approximately 95 mph—a major-league fastball. Surface Area Generated by a Parametric Curve Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function y=f(x) from x=a to x=b, revolved around the x-axis: S=2π∫baf(x)√1+(f′(x))2dx. We now consider a volume of revolution generated by revolving a parametrically defined curve x=x(t),y=y(t),for a≤t≤b around the x-axis as shown in Figure 10.2.9. The analogous formula for a parametrically defined curve is S=2π∫bay(t)√(x′(t))2+(y′(t))2dt provided that y(t) is not negative on [a,b]. Example 10.2.6: Finding Surface Area Find the surface area of a sphere of radius r centered at the origin. Solution We start with the curve defined by the equations x(t)=rcost,y(t)=rsint,for 0≤t≤π. This generates an upper semicircle of radius r centered at the origin as shown in the following graph. When this curve is revolved around the x-axis, it generates a sphere of radius r. To calculate the surface area of the sphere, we use Equation 10.2.9: S=2π∫bay(t)√(x′(t))2+(y′(t))2dt=2π∫π0rsint√(−rsint)2+(rcost)2dt=2π∫π0rsint√r2sin2t+r2cos2tdt=2π∫π0rsint√r2(sin2t+cos2t)dt=2π∫π0r2sintdt=2πr2(−cost\|π0)=2πr2(−cosπ+cos0)=4πr2 units2. This is, in fact, the formula for the surface area of a sphere. Exercise 10.2.6 Find the surface area generated when the plane curve defined by the equations x(t)=t3,y(t)=t2,for 0≤t≤1 is revolved around the x-axis. Hint : Use Equation 10.2.9. When evaluating the integral, use a u-substitution. Answer : A=π(494√13+128)1215 units2 Key Concepts The derivative of the parametrically defined curve x=x(t) and y=y(t) can be calculated using the formula dydx=y′(t)x′(t). Using the derivative, we can find the equation of a tangent line to a parametric curve. The area between a parametric curve and the x-axis can be determined by using the formula A=∫t2t1y(t)x′(t)dt. The arc length of a parametric curve can be calculated by using the formula s=∫t2t1√(dxdt)2+(dydt)2dt. The surface area of a volume of revolution revolved around the x-axis is given by S=2π∫bay(t)√(x′(t))2+(y′(t))2dt. If the curve is revolved around the y-axis, then the formula is S=2π∫bax(t)√(x′(t))2+(y′(t))2dt. Key Equations Derivative of parametric equations dydx=dy/dtdx/dt=y′(t)x′(t) Second-order derivative of parametric equations d2ydx2=ddx(dydx)=(d/dt)(dy/dx)dx/dt Area under a parametric curve A=∫bay(t)x′(t)dt Arc length of a parametric curve s=∫t2t1√(dxdt)2+(dydt)2dt Surface area generated by a parametric curve S=2π∫bay(t)√(x′(t))2+(y′(t))2dt 10.1: Curves Defined by Parametric Equations 10.3: Polar Coordinates
2342	https://www.vedantu.com/jee-main/physics-elastic-collisions-in-one-dimension	Elastic Collision in One Dimension: Definition, Derivation & Formula Sign In All Courses NCERT, book solutions, revision notes, sample papers & more Find courses by class Starting @ ₹1,350 Find courses by target Starting @ ₹1,350 Long Term Courses Full Year Courses Starting @ just Rs 9000 One-to-one LIVE classes Learn one-to-one with a teacher for a personalised experience Courses for Kids Courses for Kids Confidence-building & personalised learning courses for Class LKG-8 students English Superstar Age 4 - 8 Level based holistic English program Summer Camp For Lkg - Grade 10 Limited-time summer learning experience Spoken English Class 3 - 5 See your child speak fluently Learn Maths Class 1 - 5 Turn your child into a Math wizard Coding Classes Class 1 - 8 Learn to build apps and games, be future ready Free study material Get class-wise, author-wise, & board-wise free study material for exam preparation NCERT SolutionsCBSEJEE MainJEE AdvancedNEETQuestion and AnswersPopular Book Solutions Subject wise Concepts ICSE & State Boards Kids Concept Online TuitionCompetative Exams and Others Offline Centres Online Tuition Get class-wise, subject-wise, & location-wise online tuition for exam preparation Online Tuition By Class Online Tuition By Subject Online Tuition By Location More Know about our results, initiatives, resources, events, and much more Our results A celebration of all our success stories Child safety Creating a safe learning environment for every child Help India Learn Helps in learning for Children affected by the Pandemic WAVE Highly-interactive classroom that makes learning fun Vedantu Improvement Promise (VIP) We guarantee improvement in school and competitive exams Master talks Heartfelt and insightful conversations with super achievers Our initiatives Resources About us Know more about our passion to revolutionise online education Careers Check out the roles we're currently hiring for Our Culture Dive into Vedantu's Essence - Living by Values, Guided by Principles Become a teacher Apply now to join the team of passionate teachers Contact us Got questions? Please get in touch with us Vedantu Store JEE Main Physics Elastic Collisions In One Dimension Elastic Collisions in One Dimension: Concepts, Derivation, and Examples Download PDF Study Material Important Questions Chapter Pages Revision Notes Formulas Difference Between Preparation Tips Exam Info Important Dates Eligibility Criteria Application Form Registration Correction Window Exam Centres Admit Card Reservation Criteria Slot Booking Weightage Exam Pattern Cut-off College Predictor Answer Key Result Colleges News Videos FAQs Syllabus Physics Syllabus Mathematics Syllabus Chemistry Syllabus Courses Class 11 JEE Course (2023-25) Class 12 JEE Course (2023-24) JEE Repeater Course (2023-24) Class 8 JEE Foundation Course Class 9 JEE Foundation Course Class 10 JEE Foundation Course JEE Main Coaching Previous Year Question Paper Subject wise question Paper JEE Main Yearwise Question Paper Practice Materials Practice Papers Sample Papers Mock Test Maths Mock Test Physics Mock Test Chemistry Mock Test Question Answers Step-by-Step Derivation of Final Velocities in 1D Elastic Collision Elastic Collisions In One Dimension is a central Physics topic where two bodies collide and both momentum and kinetic energy are conserved. This concept is crucial for JEE Main, as it demonstrates how fundamental laws govern everyday collisions, from carts on rails to atomic particles. Students often encounter both numerical problems and derivations based on this topic in exams. In such collisions, objects move only along a single straight line. The interaction is called "elastic" because there is no loss of kinetic energy during the collision process—unlike real-world impacts that often result in deformation or heat generation. For rigorous JEE preparation, mastering the equations and understanding the step-by-step derivation is essential. Key Principles in Elastic Collisions In One Dimension Analyzing elastic collisions requires an understanding of two core laws: the conservation of linear momentum and the conservation of kinetic energy. Both are applied simultaneously to solve for final velocities and analyze motion before and after collision. Momentum: p = m v (where m is mass, v is velocity) Total initial momentum equals total final momentum. Kinetic energy: KE = ½ m v 2 Total initial kinetic energy equals total final kinetic energy. Only true for ideal cases—real collisions may lose energy. For example, two trolleys colliding on a frictionless track showcase these rules perfectly. Such clear cases are often the ones tested in JEE Main problems. Step-By-Step Derivation for Elastic Collisions In One Dimension Derivation is a must-know for JEE aspirants. It involves expressing final velocities (v 1f, v 2f) in terms of initial velocities and masses. Let bodies 1 and 2 have masses m 1, m 2 and initial velocities u 1, u 2. Apply conservation of momentum: m 1 u 1 + m 2 u 2 = m 1 v 1f + m 2 v 2f 3. Apply conservation of kinetic energy: ½ m 1 u 1 2 + ½ m 2 u 2 2 = ½ m 1 v 1f 2 + ½ m 2 v 2f 2 4. Solve the above equations to find: v 1f = [(m 1 - m 2)u 1 + 2m 2 u 2] / (m 1 + m 2) 5. v 2f = [(m 2 - m 1)u 2 + 2m 1 u 1] / (m 1 + m 2) Careful substitution, clean algebra, and using the correct sign convention are critical steps—many errors occur here. Practice helps avoid missing minus signs or mixing up initial and final velocities. For additional practice, refer to collision topic and related question sets. Comparison Table: Elastic Collisions In One Dimension vs Two Dimensions \| Aspect \| 1D Elastic Collision \| 2D Elastic Collision \| --- \| Direction \| Single straight line \| Multiple axes/planes \| \| Equations Needed \| Two equations (momentum, kinetic energy) \| Three or more equations (vector components) \| \| JEE Main Frequency \| Very common \| Less common, advanced \| \| Example \| Balls on a straight track \| Billiards, atomic scattering \| If you want a detailed two-dimensional approach, visit elastic collision in two dimensions for vector-based problems. Numerical Example: Elastic Collisions In One Dimension Application A body of mass 2 kg moving at 3 m/s collides elastically with a stationary body of mass 3 kg. Find final velocities. Given: m 1 = 2 kg, u 1 = 3 m/s m 2 = 3 kg, u 2 = 0 m/s Apply formula: v 1f = [(2-3)×3 + 2×3×0]/(2+3) = -0.6 m/s v 2f = [(3-2)×0 + 2×2×3]/(2+3) = 2.4 m/s Final result: First body moves at -0.6 m/s (reverses), second at 2.4 m/s. This problem structure mirrors those in JEE Main practice tests and is ideal for solidifying the formula’s application. Revise foundational concepts about conserved quantities at conservation of momentum and sharpen your calculations further with coefficient of restitution models. Visit laws of motion for background essentials. Common Pitfalls and Practical Tips on Elastic Collisions In One Dimension Confusing elastic with inelastic collision: check if kinetic energy is conserved. Forgetting to use sign convention—velocities to the left are negative. Mishandling equal mass case; velocities swap in such collisions. Assuming real objects are perfectly elastic. In reality, most lose some energy. Missing proper algebraic steps in derivation; always write momentum and kinetic energy equations first. Explore the subtle differences with oblique collisions, or practice with more varieties in motion in one dimension modules. Further Applications and Interlinked Topics for Elastic Collisions In One Dimension Understanding elastic collisions boosts your performance in waves, molecular collisions, and kinetic theory. They are foundational for more complex chapters in rotational and translational motion. work energy and power are built on collision analysis. impulse momentum theorem is a helpful shortcut. Problems in center of mass and motion use similar logic. For conceptual mastery, refer to laws of motion practice paper and revision notes. Vedantu’s JEE Physics resources offer detailed notes, mock tests, and clear stepwise solutions for elastic collision topics. By using the correct approach and practicing consistently, you’ll master even the trickiest exam questions on this theme. 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An elastic collision in one dimension is a collision between two objects moving along a straight line in which both momentum and kinetic energy are conserved. In such collisions, the objects bounce off each other without any loss of total kinetic energy. Common examples include: Collisions between billiard balls on a straight path Idealized gas molecule interactions Head-on car crashes (in theory) This concept is a key topic in Class 11 Physics, especially for students preparing for JEE and board exams. How is kinetic energy conserved in a 1D elastic collision? Kinetic energy is conserved in a one-dimensional elastic collision because no energy is lost to heat, sound, or deformation. The sum of the kinetic energies of both objects before and after the collision remains the same. The conservation is described by: Total kinetic energy before = Total kinetic energy after 1/2 m₁u₁² + 1/2 m₂u₂² = 1/2 m₁v₁² + 1/2 m₂v₂² This principle is fundamental to solving numerical problems for JEE Main and Class 11 physics. How do you derive the final velocities after a 1D elastic collision? To derive the final velocities (v₁ and v₂) after a 1D elastic collision, you use both the conservation of momentum and conservation of kinetic energy equations. The steps are: Write the momentum conservation equation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Write the kinetic energy conservation equation: 1/2 m₁u₁² + 1/2 m₂u₂² = 1/2 m₁v₁² + 1/2 m₂v₂² Solve these two equations simultaneously for v₁ and v₂ This derivation is often asked in exams and forms the basis for solving related numerical problems. What is the formula for final velocity in 1D elastic collision? The final velocities of two objects after a 1D elastic collision are given by: v₁ = [(m₁ - m₂)u₁ + 2m₂u₂] / (m₁ + m₂) v₂ = [(m₂ - m₁)u₂ + 2m₁u₁] / (m₁ + m₂) Where m₁ and m₂ are the masses, u₁ and u₂ are the initial velocities, and v₁ and v₂ are the final velocities after collision. These formulas are essential for solving exam problems. What is the main difference between 1D and 2D elastic collisions? 1D elastic collisions occur along a single straight line, while 2D elastic collisions involve motion and momentum conservation in two perpendicular directions. The main differences are: 1D: Only one direction (e.g., head-on collisions); only one momentum equation is used. 2D: Both x and y components are considered; two separate momentum equations are needed. Complexity: 2D problems are generally more complex and require vector analysis. This distinction is important for JEE and competitive exam preparation. If both masses are equal, what happens to their velocities after a 1D elastic collision? When both masses are equal in a one-dimensional elastic collision, they simply exchange their velocities. That is: The first object takes the velocity of the second object The second object takes the velocity of the first This is a classic result and can be quickly recalled for MCQs and direct numerical questions. Can you provide derivation PDF/notes for 1D elastic collision? Yes, comprehensive Class 11 Physics notes and derivation PDFs covering one-dimensional elastic collisions are available for download. These include: Step-by-step derivations Key formula sheets Sample solved questions These study resources are essential for thorough JEE and Class 11 preparation. What is the difference between elastic and inelastic collisions? The main difference is that elastic collisions conserve both momentum and kinetic energy, while inelastic collisions only conserve momentum. Key points include: Elastic: No loss in total kinetic energy; objects bounce off each other Inelastic: Kinetic energy is lost as heat, sound, or deformation; objects may stick together (perfectly inelastic) Understanding this distinction is crucial for physics exams and problem-solving. What are some common mistakes students make while writing the derivation of elastic collision in one dimension? Common mistakes in derivation of 1D elastic collision include: Mixing up initial and final velocities Forgetting to conserve both momentum and kinetic energy Incorrect sign conventions for direction Errors in algebraic manipulation Carefully follow every step and double-check formulas during exams for full marks. Which of the following collisions is one-dimensional? A one-dimensional collision is an event where two objects move along the same straight line before and after the collision. Common examples are: Head-on collision between two carts on a track Collisions between billiard balls moving in a straight line Recognizing the type of collision is vital for choosing the correct conservation equations in physics problems. 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2343	https://mathworld.wolfram.com/LinePointPicking.html	Line Point Picking Download Wolfram Notebook Consider a line segment of length 1, and pick a point at random between . This point divides the line into line segments of length and . If a set of points are thus picked at random, the resulting distribution of lengths has a uniform distribution on . Similarly, separating the two pieces after each break, the larger piece has uniform distribution on (with mean 3/4), and the smaller piece has uniform distribution on (with mean 1/4). The probability that the line segments resulting from cutting at two points picked at random on a unit line segment determine a triangle is given by 1/4. The probability and distribution functions for the ratio of small to large pieces are given by \| \| \| \| \| --- --- \| \| \| \| \| (1) \| \| \| \| \| (2) \| for . The raw moments are therefore \| \| \| --- \| \| \| (3) \| where is the digamma function. The first few are therefore \| \| \| \| \| --- --- \| \| \| \| \| (4) \| \| \| \| \| (5) \| \| \| \| \| (6) \| \| \| \| \| (7) \| (OEIS A115388 and A115389). The central moments are therefore \| \| \| --- \| \| \| (8) \| where is a Pochhammer symbol. The first few are therefore \| \| \| \| \| --- --- \| \| \| \| \| (9) \| \| \| \| \| (10) \| \| \| \| \| (11) \| This therefore gives mean, variance, skewness, and kurtosis excess of \| \| \| \| \| --- --- \| \| \| \| \| (12) \| \| \| \| \| (13) \| \| \| \| \| (14) \| \| \| \| \| (15) \| The mean can be computed directly from \| \| \| \| \| --- --- \| \| \| \| \| (16) \| \| \| \| \| (17) \| \| \| \| \| (18) \| The probability and distribution functions for the ratio of large to small pieces are given by \| \| \| \| \| --- --- \| \| \| \| \| (19) \| \| \| \| \| (20) \| for . Paradoxical though it may be, this distribution has infinite mean and other moments. The reason for this is that a theoretical bone can be cut extremely close to one end, thus giving huge ratio of largest to smallest pieces, whereas there is some limit for a real physical bone. Taking to be the smallest possible piece in which is bone cen be cut, the mean is then given by \| \| \| --- \| \| \| (21) \| See also Cube Point Picking, Hypercube Point Picking, Line Line Picking, Square Point Picking Explore with Wolfram\|Alpha More things to try: 30-level 12-ary tree d/dx d/dy x^2 y^4 maximize x(1-x)e^x References Pickover, C. A. "The Problem of the Bones." Ch. 8 in The Mathematics of Oz: Mental Gymnastics from Beyond the Edge. New York: Cambridge University Press, pp. 21-23 and 243-249, 2002.Sloane, N. J. A. Sequences A115388 and A115389 in "The On-Line Encyclopedia of Integer Sequences." Referenced on Wolfram\|Alpha Line Point Picking Cite this as: Weisstein, Eric W. "Line Point Picking." From MathWorld--A Wolfram Resource. Subject classifications Find out if you already have access to Wolfram tech through your organization
2344	https://www.doubtnut.com/qna/53748572	If y=(x+1)/(x-1), then what is (dy)/(dx) equal to? 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If y=x+1 x−1, then what is d y d x equal to? A −2 x−1 B −2(x−1)2 C 2(x−1)2 D 2(x−1) Video Solution To view this video, please enable JavaScript and consider upgrading to a web browser thatsupports HTML5 video Video Player is loading. Play Video Play Skip Backward Mute Current Time 0:00 / Duration-:- Loaded: 0% Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2x 1.5x 1x, selected 0.5x 0.25x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Opacity Text Background Color Opacity Caption Area Background Color Opacity Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. Know where you stand among peers with ALLEN's JEE Enthusiast Online Test Series g(x), then the derivative d y d x is given by: d y d x=f′(x)g(x)−f(x)g′(x)(g(x))2 where: f(x)=x+1 g(x)=x−1 Step 1: Identify f(x) and g(x) Here, we have: f(x)=x+1 g(x)=x−1 Step 2: Find the derivatives f′(x) and g′(x) Now, we calculate the derivatives: f′(x)=d d x(x+1)=1 g′(x)=d d x(x−1)=1 Step 3: Apply the quotient rule Using the quotient rule: d y d x=f′(x)g(x)−f(x)g′(x)(g(x))2 Substituting f(x), g(x), f′(x), and g′(x): d y d x=(1)(x−1)−(x+1)(1)(x−1)2 Step 4: Simplify the numerator Now simplify the numerator: d y d x=x−1−(x+1)(x−1)2 This simplifies to: d y d x=x−1−x−1(x−1)2=−2(x−1)2 Final Result Thus, the derivative d y d x is: d y d x=−2(x−1)2 Show More 24 \| Share Save Class 12MATHSDIFFERENTIATION Topper's Solved these Questions DEFINITE INTEGRATION & ITS APPLICATION Book:NDA PREVIOUS YEARS Chapter:DEFINITE INTEGRATION & ITS APPLICATION Exercise:DIRECTIONS Explore 60 Videos DIFFERENTIAL EQUATION Book:NDA PREVIOUS YEARS Chapter:DIFFERENTIAL EQUATION Exercise:MCQs Explore 119 Videos Similar Questions If y=(1+x 1 4)(1+x 1 2)(1−x 1 4), then what is d y d x equal to ? View Solution If y=(1+x 1/4)(1+x 1/2)(1−x 1/4), then what is d y d x equal to ? View Solution Knowledge Check If y=1 log 10 x, then what is d y d x equal to? A x B x log e 10 C−(log x 10)2(log 10 e)x D x log 10 e Submit If y=1 log 10 x then what is d y d x equal to A x B x log e 10 C−(log x 10)2(log 10 e)x D x log 10 e Submit If y=sin−1(4 x 1+4 x 2), then what is d y d x equal to? A 1 1+4 x 2 B−1 1+4 x 2 C 4 1+4 x 2 D 4 x 1+4 x 2 Submit If y=sin−1(4 x 1+4 x 2) then what is d y d x equal to ? View Solution if y=e x 2 then, what is d y d x at x=π equal to View Solution If y=x+1 x−1, then what is the value of d y d x ? View Solution If √1−x 2+√1−y 2=a, then what is d y d x equal to? View Solution If y=tan−1(5−2 tan√x 2+5 tan√x) then what is d y d x equal to ? View Solution NDA PREVIOUS YEARS-DERIVATIVES-MCQs If y=lnsqrt(tanx), then what is the value of (dy)/(dx) at x=(pi)/(4)? 02:03 If f(x)=x^(2)-6x+8 and there exists a point c in the interval [2,4] su... 00:50 If y=(x+1)/(x-1), then what is (dy)/(dx) equal to? 01:21 If y=cost and x=sint, then what is (dy)/(dx) equal to? 00:49 If x^(m)+y^(m)=1 such that (dy)/(dx)=-(x)/(y), then what should be the... 01:33 Consider the following statements: 1. If y=ln(secx+tanx), then (dy)/... 02:20 If f(x)=2^(sinx), then what is the derivative of f(x)? 00:59 If y=ln(e^(mx)+e^(-mx)), then what is (dy)/(dx) at x = 0 equal to ? 02:49 If 2x^(3)-3y^(2)=7, what is (dy)/(dx) equal to (y ne0)? 01:15 The derivative of \|x\| at x = 0 01:48 If y=sin(ax+b), then what is (d^(2)y)/(dx^(2)) at x=-(b)/(a), where a,... 01:44 If y=x^(x), what is (dy)/(dx) at x = 1 equal to? 01:53 What is the differential coefficient of log(x)x? 01:16 The derivative of sec^(2)x with respect to tan^(2)x is 01:31 What is the derivative of x^(3) with respect to x^(2)? 00:54 If f(x)=2x^(2)+3x-5, then what is f'(0)+3f'(-1) equal to? 01:08 What is the derivative of sin(sinx)? 00:41 What is the derivative of \|x-1\| at x = 2? 01:23 What is the derivative of sqrt((1+cosx)/(1-cosx))? 01:47 If z=f" of "(x)=x^(2) where f(x)=x^(2), then what is (dz)/(dx) euqal t... 01:06 Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics Sunil Batra Solutions for Physics Pradeep Solutions for Physics Errorless Solutions for Physics Narendra Awasthi Solutions for Chemistry MS Chouhan Solutions for Chemistry Errorless Solutions for Biology Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium NCERT Solutions for Class 11 English Medium NCERT Solutions for Class 10 English Medium NCERT Solutions for Class 9 English Medium NCERT Solutions for Class 8 English Medium NCERT Solutions for Class 7 English Medium NCERT Solutions for Class 6 English Medium Free Ncert Solutions Hindi Medium NCERT Solutions NCERT Solutions for Class 12 Hindi Medium NCERT Solutions for Class 11 Hindi Medium NCERT Solutions for Class 10 Hindi Medium NCERT Solutions for Class 9 Hindi Medium NCERT Solutions for Class 8 Hindi Medium NCERT Solutions for Class 7 Hindi Medium NCERT Solutions for Class 6 Hindi Medium Boards CBSE UP Board Bihar Board UP Board Resources Unit Converters Calculators Books Store Seminar Results Blogs Class 12 All Books Class 11 All Books Class 10 All Books Class 9 All Books Class 8 All Books Class 7 All Books Class 6 All Books 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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2345	https://www.quora.com/For-which-values-of-x-does-tan-x-equal-undefined	Something went wrong. Wait a moment and try again. Undefined Values Tangent Function Functions (mathematics) Trigonometric Functions Mathematical Functions 5 For which values of x does tan(x) equal undefined? Sayan Bhattacharjee B. Tech in Electronics and Communication Engineering, IIEST, Shibpur. Howrah (Expected 2027) · 2y Tangent of x (also abbreviated as tan x) is the ratio of sin x and cosx Tan x = sin x/cos x So tan x is not defined at the points where cos x is zero Such points are the odd multiples of π/2 Therefore the points where tan x is not defined are the odd multiples of π/2 Let us define a set S containing all such points Then the set S would look like , S = {x = (2n - 1)π/2 : x belongs to Z } It is clear from the graph of y = tan x. As x approaches any point in S (as defined above) the graph of tan x approaches either of the two infinities (positive or negative) Tangent of x (also abbreviated as tan x) is the ratio of sin x and cosx Tan x = sin x/cos x So tan x is not defined at the points where cos x is zero Such points are the odd multiples of π/2 Therefore the points where tan x is not defined are the odd multiples of π/2 Let us define a set S containing all such points Then the set S would look like , S = {x = (2n - 1)π/2 : x belongs to Z } It is clear from the graph of y = tan x. As x approaches any point in S (as defined above) the graph of tan x approaches either of the two infinities (positive or negative) Related questions For which value of x is tan (x+30) undefined? In the interval 0 ≤ x < 2π, at what radian measure(s) is tan x undefined? How do I find the value of tan (x)If tan (2x) =8/tan(x)? What is the value of tan(x/y)? How do I find the value of x for log (2+4cos^2 x base tan x)=2? Chris Reid B.S in Computer Science & Electronics Technology (AAS), University of Alaska Anchorage (Graduated 2010) · Author has 5.8K answers and 12.1M answer views · 3y Related What is the definition of tan(x)? Why does it have an undefined sign when x = π/2 radians (or any multiple)? What happens to its graph in this case? The Tangent is defined as the oppositeadjacent in a triangle on the xy plane. This is also easily identified as tan(θ)=yx. Being a division, of course the tangent is undefined when the denominator crosses the zero on the number line. The reason it is called the tangent is that as you rotate an angle and a radius of 1 around the xy plane (to create a unit cirle) yx describes the slope of the tangent line on the circle, perpendicular to the radius, which of course forms a right triangle. An angle rotated counter clockwise from 0 to π4 radians would creat The Tangent is defined as the oppositeadjacent in a triangle on the xy plane. This is also easily identified as tan(θ)=yx. Being a division, of course the tangent is undefined when the denominator crosses the zero on the number line. The reason it is called the tangent is that as you rotate an angle and a radius of 1 around the xy plane (to create a unit cirle) yx describes the slope of the tangent line on the circle, perpendicular to the radius, which of course forms a right triangle. An angle rotated counter clockwise from 0 to π4 radians would create a triangle with x=√2 and y=√2. Thus tan(θ)=tan(π4)= π4π4=1 The slope of a tangent line perpendicular to the radius would be -1, as y is moving down to 0 at that point. The tangent function is also known to be sin(θ)cos(θ) of course. So whenevercos(θ)=0, tan(θ)=∞. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 6y Related For what value of x is 4/2x-10 undefined? How do you know? We can’t have fractions with zero in the denominator because it is undefined (mea... Elizabeth Jean Stapel Purplemath.com's author (and owner) (1999–present) · Author has 5.9K answers and 3.3M answer views · Apr 17 Related What is the value of arctan(x) when it is undefined? What is the value of arctan(x) when it is undefined? When a mathematical entity is said to be “undefined”, this means that the entity has no mathematical definition; in particular, it has no numerical value. Therefore, there is no value of the inverse tangent when the inverse tangent is undefined. This is because “undefined” means “has no value”. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. 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Related questions What is the value of arctan (tan x) when x>π/2? What is the value of tan(x) when x is equal to 90 degrees? For what values of x does tan(x) =y? Why is cot (x-90) not equal to -tan(x)? What is the value of x in tan (x) = -1? Rakesh Kumar 8y Related Why is tan(x) =x for small values of x? It will be clear from the graph of tan(x) why tan(x)=x graph 0f tan(x) from graph it is clear that y=x and tan(x)=x graph coincide for small value of x. hence tan(x)=x also sin(x)=x It will be clear from the graph of tan(x) why tan(x)=x graph 0f tan(x) from graph it is clear that y=x and tan(x)=x graph coincide for small value of x. hence tan(x)=x also sin(x)=x Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 2y Related For what values of x is the equation tan(x) =cot(x) true? 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Raymond Beck Former Former Army Sergeant · Author has 10.2K answers and 1.2M answer views · Apr 15 Related What is the value of arctan(x) when it is undefined? x=90 + 180n where n = any integer arctanx = 90 degrees = pi/2 radians, but range of arctangent is (-pi/2, pi/2) which does not include the endpoints (parentheses means does not include the end points), so DNE Null set {} John K WilliamsSon Accredited (MS Educ) nerd who loves talking about math · Author has 9K answers and 23.4M answer views · 7y Related Why isn’t sinx/x undefined, but 1/x is (as x -> 0)? Let’s break your question down into several parts: Does sin(x)x have a value when x = 0? No, when x=0, sin(x)x has no value because sin(0)0=00 is undefined. Does sin(x)x have a limit as x approaches zero? Yes. The limit of sin(x)x as x approaches zero is 1. This is true because as X approaches zero from the left, sin(x)x gets closer and closer to 1; and as X approaches zero from the right, sin(x)x gets closer and closer to 1. Since the limit from the left is equal to the limit from the right, the limit of sin(x)x exists Let’s break your question down into several parts: Does sin(x)x have a value when x = 0? No, when x=0, sin(x)x has no value because sin(0)0=00 is undefined. Does sin(x)x have a limit as x approaches zero? Yes. The limit of sin(x)x as x approaches zero is 1. This is true because as X approaches zero from the left, sin(x)x gets closer and closer to 1; and as X approaches zero from the right, sin(x)x gets closer and closer to 1. Since the limit from the left is equal to the limit from the right, the limit of sin(x)x exists, and is equal to 1. Does 1x have a limit as x approaches zero? No, 1x does not have a limit as X approaches zero. As X approaches 0 from the left, 1x gets closer and closer to negative infinity. As X approaches 0 from the right, 1x gets closer and closer to positive infinity. Because the limit from the left is not equal to the limit from the right, 1x has no limit as X approaches zero. Perhaps this image (from the TI-84 PLUS CE Graphing Calculator) will help you see how the limits are different: The blue line shows you that sin(x)x approaches Y=1 from both directions as X gets closer and closer to zero. The bottom of the screen shows that when X=0, Y=(undefined). The red line shows you that 1x approaches two different values as X gets closer and closer to zero. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Gary Ward MaEd in Education & Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views · 2y Related What are the values of cos x/x and tan x/x when x approaches 0? What are the values of cos x/x and tan x/x when x approaches 0? First: so the limit of cos(x) / x as x approaches 0 does not exist as shown in red. Green = cos(x); Purple = x Second: Since tan(x) = sin(x) / cos(x) the tan(x)/x = sin(x) / (x · cos(x)) [math]\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x \cdot \cos(x)} = \frac{0}{0 \cdot 1} [/math] 0/0, so apply L’Hospital’s rule, take the derivative of the numerator over the derivative of the denominator. The derivative of sin(x) = cos(x) and the derivative of x · cos(x) = 1 · cos(x) + x · -sin(x What are the values of cos x/x and tan x/x when x approaches 0? First: [math]\displaystyle \lim_{x \to 0} \frac{\cos(x)}{x} = \frac{1}{0} = \infty[/math] so the limit of cos(x) / x as x approaches 0 does not exist as shown in red. Green = cos(x); Purple = x Second: Since tan(x) = sin(x) / cos(x) the tan(x)/x = sin(x) / (x · cos(x)) [math]\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x \cdot \cos(x)} = \frac{0}{0 \cdot 1} [/math] 0/0, so apply L’Hospital’s rule, take the derivative of the numerator over the derivative of the denominator. The derivative of sin(x) = cos(x) and the derivative of x · cos(x) = 1 · cos(x) + x · -sin(x) As x → 0, cos(x) = 1 and as x → 0, cos(x) - x·sin(x) = 1, so 1/1 = 1 for the limit. Red = tan(x)/x = sin(x) / (x · cos(x)); Green = cos(x); Blue = x · cos(x) Nachiket Agrawal Experienced a bit in Non-Routine mathematics · Author has 132 answers and 251.2K answer views · 8y Related Why is tan(x) =x for small values of x? Now i shall be using simple mathematics of circles Consider a circle whose radius is R. See that [math]d\theta[/math] is a small angle subtended by the arc which is of length [math]rd\theta[/math] Note that all angles are radian measures Take the arc of length [math]rd\theta[/math] as a line segment as it is small enough to be approximated [math]sin(d\theta)≈\frac{rd\theta}{r}=d\theta[/math] Hence we see that [math]sin(d\theta)=d\theta[/math] Now [math]cos(d\theta)≈\frac{r}{r}=1 [/math] So [math]tan(d\theta)=\frac{sin(d\theta)}{cos(d\theta)}=(d\theta)[/math] Hence we get the required result Now i shall be using simple mathematics of circles Consider a circle whose radius is R. See that [math]d\theta[/math] is a small angle subtended by the arc which is of length [math]rd\theta[/math] Note that all angles are radian measures Take the arc of length [math]rd\theta[/math] as a line segment as it is small enough to be approximated [math]sin(d\theta)≈\frac{rd\theta}{r}=d\theta[/math] Hence we see that [math]sin(d\theta)=d\theta[/math] Now [math]cos(d\theta)≈\frac{r}{r}=1 [/math] So [math]tan(d\theta)=\frac{sin(d\theta)}{cos(d\theta)}=(d\theta)[/math] Hence we get the required result Ashis Ghosh Retired officer of State Bank of India · Author has 2.6K answers and 2.9M answer views · 2y Related What is the value of f(x) and g(x) if integration of 2x square sec square x dx upon (tan x- x sec2x) 2 equal to {f(X) /tan x (tan x - x square x)} + g (x) + X + c? Alex Jones B.S. in Applied Mathematics, University of Central Florida (Graduated 2016) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.2K answers and 2.2M answer views · 5y Related Why is the degree of the function "f(x) =0" undefined? We need to understand what the degree is before we can start asking questions about its value for certain polynomials. The “spirit” of the degree is to tell you the highest power that shows up in a polynomial. However, we need to be careful about our definition to be sure to follow this spirit. After all, the polynomial [math]x^2 + 4x - 2[/math] should have a degree of [math]2[/math], right? But I could also write the same polynomial in a different way, as [math]0\cdot x^3 + x^2 + 4x - 2[/math] This polynomial has an [math]x^3[/math] term, so we could say that’s the highest power, so the degree is [math]3[/math]. But this polynomial is 100% equal to a polynomia We need to understand what the degree is before we can start asking questions about its value for certain polynomials. The “spirit” of the degree is to tell you the highest power that shows up in a polynomial. However, we need to be careful about our definition to be sure to follow this spirit. After all, the polynomial [math]x^2 + 4x - 2[/math] should have a degree of [math]2[/math], right? But I could also write the same polynomial in a different way, as [math]0\cdot x^3 + x^2 + 4x - 2[/math] This polynomial has an [math]x^3[/math] term, so we could say that’s the highest power, so the degree is [math]3[/math]. But this polynomial is 100% equal to a polynomial we already said had degree [math]2[/math]. We could do the same thing to any polynomial with any term higher than existing highest term, which gives us a “functional” definition of degree with multiple value, but we’ve strayed away from the spirit of the idea. The way to resolve this is simple, if the coefficient is [math]0[/math], that term doesn’t contribute to the value of the polynomial in any way, and so it’s not counted when considering degree. With this new rule, we have that the degree of [math]0\cdot x^3 + x^2 + 4x - 2[/math] is still [math]2[/math], so all is well. There are a few things to note, though, when considering low-degree polynomials. When we write, for example, [math]3x-5[/math] this is actually a shorthand for [math]3\cdot x^1 + (-5)\cdot x^0[/math] so this polynomial has degree [math]1[/math], while the constant polynomial [math]7 = 7\cdot x^0[/math] has degree [math]0[/math]. At this point it’s clear how to get the degree of every single possible polynomial, we look at all the powers on terms which have non-zero coefficients, then take the highest one. But this opens up the question, is there always a degree? That is, is there always a term with non-zero coefficient? The answer is clearly no, as the polynomial has no non-zero coefficients. What is the degree of this polynomial then? Well, our current definition doesn’t have an answer for that. So we say it’s undefined. This is a fine definition for most, after all, zero elements tend to have some weird properties already, so we might be okay with just letting this one go. It doesn’t really hurt us, and it still follows the spirit of the definition pretty well, since there’s no terms and so no highest powers. If you’re unsatisfied, we can go a little further with this and try to fix it up. The degree assigns a number to polynomials, based on some information about that polynomial. That is, we want to define [math]f : K[x] \to \Bbb{N}[/math] so that [math]f(p)[/math] is the degree of the polynomial [math]p[/math]. Our current definition is not such a function, in fact it’s a slightly smaller function [math]g : K[x] - {0} \to \Bbb{N}[/math] since we weren’t able to define [math]g(0)[/math]. In the first section, we were just trying to match the “spirit” of what degree should mean, but it’s clear that spirit doesn’t give us anything for the zero polynomial, so we need something else. In mathematics, we make definitions because they help us do something, and usually that is by giving us a structure that wasn’t previously there. Polynomials have a lot of structure on their own, but one thing they’re missing is an order. Polynomials are pretty complicated things, so trying to line them up in a strict order is probably not too useful, but there is some sense in which a polynomial is “bigger” than another one, and that’s this idea of degree. Having higher powers introduces more potential complexity to the polynomials behavior, so we can order up our polynomials like that. Two polynomials of the same degree don’t really have any order to them, so this is just a partial ordering, but that’s okay. To keep this ordering making any kind of sense, we need the zero polynomial to be “smaller” than constant polynomials, the same way that [math]0\cdot x + 1[/math] is “smaller” than [math]x+1[/math]. That is, [math]f(0) < 0[/math] but that’s not possible since [math]f[/math] has codomain [math]\Bbb{N}[/math], where [math]0[/math] is the smallest element. In order to keep this structure working as we like, we need to extend the codomain. Since this is only for one input, there’s only one new element that needs to be added. We’ll call this element [math]u[/math] and look for a function [math]h : K[x] \to \Bbb{N} \cup {u}[/math] so that [math]h(p)[/math] is the degree of [math]p[/math], with [math]h(0) = u[/math]. At this point, we have one property of [math]u[/math], which is that [math]u < x[/math] for all [math]x \in \Bbb{N}[/math] Here, we extend the ordering in [math]\Bbb{N}[/math] so that [math]u[/math] is smaller than everything else. If this is all we care about, we can say that [math]u = -1[/math] and call it a day. Or, we can collect more properties of [math]u[/math] by digging more into the structure being built here. Let’s look back at the function we did define, [math]g[/math]. It has a few nice properties that would be awesome if they worked for [math]h[/math] as well. Those properties are [math]g(p+q) = \max(g(p),g(q))[/math] [math]g(pq) = g(p)+g(q)[/math] That first property is already satisfied by [math]h[/math], since [math]h(p) = h(p+0) = \max(h(p),u) = h(p)[/math] which is true because of the one property that [math]u[/math] has so far. The second property is the interesting one though. Note that the polynomial ring [math]K[x][/math] forms an abelian group under addition, with identity [math]0[/math], and a commutative monoid under multiplication, and as a ring, multiplication is distributive over addition. The multiplication monoid in a ring always has a particular structure as a consequence, which is that multiplication with the additive identity always results in the additive identity, i.e. [math]p\cdot 0 = 0[/math] for any [math]p\in K[x][/math] Now, if [math]h[/math] has the property that [math]h(pq) = h(p)+h(q)[/math] then [math]h[/math] is a monoid homomorphism from the multiplicative monoid of [math]K[x][/math] to the additive monoid of [math]\Bbb{N}\cup{u}[/math], as long as that last set is in fact a monoid. The thing about homomorphisms is that they are structure preserving transformations, so wherever [math]h[/math] maps things to, it’s got to have the same structure, which means it will have that “absorptive” structure that the multiplicative monoid has with its respective additive identity. Usually, additive monoids do not have such a structure, in particular [math]\Bbb{N}[/math] itself does not have this structure. But what’s also interesting is that taking the additive identity out of the multiplicative monoid doesn’t break the structure at all. This isn’t really surprising, since all that element did was turn everything else into itself. This means that we should be able to add an element to an existing monoid that doesn’t have this structure, and as long as that element has the same absorptive property, it should create a new monoid with said structure. So, we say that [math]u[/math] has the property that [math]a + u = u + a = u[/math] for any [math]a\in \Bbb{N}\cup{u}[/math] in addition to its ordering property from above. Then, this set still has closure, associativity, commutativity, and the same identity as the natural numbers, so it’s still a commutative monoid. With this, we have [math]u = h(0) = h(p\cdot 0) = h(p) + u = u[/math] so the homomorphism is verified. At this point, we can see that [math]u = -1[/math] doesn’t really make sense anymore, and in fact none of the standard number groups have an absorptive element under addition. But there is one set I can think of that has an element with both properties needed, and that’s the extended real numbers. There is an element there called [math]-\infty[/math] that is strictly less than every other element of the set, and absorptive under addition, so borrowing that notation, we can say that [math]u = -\infty[/math]. Thus, we have a monoid homomorphism [math]h[/math] which represents the degree of a polynomial, with [math]h(0) = -\infty[/math]. If you check Wikipedia, they mention that degree of the zero polynomial can be considered undefined, [math]-1[/math], or [math]-\infty[/math], so I hope you can see in full detail now why and when each is true. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 4y Related For what values of x does tan (x+60) = -tanx? If tan(x + 60) = – tan(x) then tan(x + 60) = tan(– x) so x + 60 = – x + 360n OR x + 60 = 180 – x + 360n then 2x = – 60 + 360n OR 2x = 120 + 360n and x = –30 + 180n ... Related questions For which value of x is tan (x+30) undefined? In the interval 0 ≤ x < 2π, at what radian measure(s) is tan x undefined? How do I find the value of tan (x)If tan (2x) =8/tan(x)? What is the value of tan(x/y)? How do I find the value of x for log (2+4cos^2 x base tan x)=2? What is the value of arctan (tan x) when x>π/2? What is the value of tan(x) when x is equal to 90 degrees? For what values of x does tan(x) =y? Why is cot (x-90) not equal to -tan(x)? What is the value of x in tan (x) = -1? If tan(2x) =2x what is the value of x? In trigonometry how do you find the value of x in a degree with two other sides? For what values of x does the function tan x have a range? Where did I go wrong? For x x x x x x x . . . = 4 I got √2 but that makes x x x x x x x . . . = 2. What is the definition of tan(x)? Why does it have an undefined sign when x = π/2 radians (or any multiple)? What happens to its graph in this case? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2346	https://motris.livejournal.com/147259.html	Main Top Interesting Checklist 😎🌻☀️ Help Follow us: Follow us on Twitter Applications Download Huawei RuStore COMPANY About News Help PRODUCTS "Share" button COMMUNITY Frank CHOOSE LANGUAGE English ▾ English Deutsch Dansk español Français Italiano Русский Українська Беларуская 日本語 Português Esperanto עברית Nederlands Magyar Gaeilge íslenska suomi Ελληνικά Norsk bokmål Svenska polski 简体中文 Latviešu Türkçe Bahasa Melayu हिन्दी Brazilian Portuguese Chinese Traditional Lietuvių Norsk nynorsk Privacy Policy User Agreement Recommendation technologies Help LiveJournal — v.872 LiveJournal Log in No account? Create an account : motris Subscribe 0 70 Friday Puzzle #111 - Consecutive Kakuro This is part 5 of a "Better Know the USPC" preview series. The United States Puzzle Championship is scheduled for August 27th, 1PM EDT. Type: Kakuro (aka Cross Sums) Variants USPC History: A mostly standard kakuro/cross sums has appeared four times: 1999 - with three missing clues, 2000, 2001 - with 15's for all down clues, and 2005. The first two are from Puzzelsport, the third from Sidney Kravitz, and the last from Michael Rios. But the target of this entry is an even more frequent brand of kakuro, the kakuro variation, a high pointer that is close to if not the very last puzzle on the test. If you're solving one of these, it's certainly from Michael Rios who has provided all five of them, including 2002 - Hot Cross Sums (each clue is off by 1), 2004 - Cross Number Sum Place (combination of cross sums, number place, and "consecutive" clues), 2006 - Plus or Minus Kakuro (some cells contain negative values), 2007 - ORu Kakuro (each entry has two possible clues but only one is correct), and 2008 - KakurOh (some cells are larger than 1x1 squares and belong in multiple rows/columns). While we haven't had a kakuro variation in two years, I expect to see one again soon. And if not a kakuro variation exactly, certainly another puzzle where a combination of arithmetic and new thinking are needed together. Notation tips: Not as much to share here, but my main notation style is to mark sure pairs on the edges between cells (like 79 in a two-cell 16 clue) as well as the occasionally either/or placement of a given number (ie one of these cells must be a 5) when it similarly lies in adjacent cells. When a single cell has only two options, that gets two light numbers in the center of the cell and is not confused for the other kind of notation. These notes are rather similar to my sudoku notes, only I make many many more notes when solving a sudoku. Here is a solved grid from the 2004 variation that is pretty representative of how much writing is in one of my grids since I never erase my notes. Strategies: I suppose this section needs two separate parts, on kakuro strategies and on kakuro variation-specific strategies. We'll start with the "regular" style before getting more twisted. Learning to recognize extremes/outliers in clues is almost always a good approach in any kind of logic puzzle, and this is particularly useful in kakuro where noticing the importance of certain values and lengths is the most significant part of solving these puzzles fast. Indeed, the most common "cheat sheet" people would request in a kakuro is exactly a table of the extreme sum values. Learn this simple rule: the two smallest and two largest possible sums for any length will always have a unique digit breakdown. In two cells, 3 and 4 are the smallest sums and must be {1,2} and {1,3} in some order respectively. Equivalently, 16 and 17 with {7,9} and {8,9} are the unique pairs on the high end. In three cells, the magic values are 6 and 7 on the low end, and 23 and 24 on the high end. This continues. In four cells: 10, 11, 29, 30. In five cells: 15, 16, 34, 35. In six cells: 21, 22, 38, 39. In seven cells: 28, 29, 41, 42. When you get to eight cells, all the sums have a unique missing digit so the information you really want to store in your head is "what is not here" which is the x that satisfies 45-x = clue. Often the missing digit is a key digit in an intersecting clue and could force just one entry, so don't specifically ignore eight-cell clues because they seem long. The only useless clues are the 45's which just indicate that all digits from 1 to 9 appear in that entry and nothing can be eliminated until the crossing clues add context. But while the unique extremes are key, and should become "automatic" when seen in a puzzle, you'll want to get your head around some of the next larger/smaller values which only have two choices. With n cells in the clue, exactly n-2 of these will be fixed for these two value sets. An 8 in three cells must be 1,3,4 or 1,2,5 so a 1 is always present. A 12 in four cells must be 1,2,4,5 or 1,2,3,6 so a 1 and a 2 are always present. And so on. A 23 in six cells contains 1+2+3+4 and either 5+8 or 6+7. One picture people can have in their heads to grasp this principle is to think of sliders sitting over values on a number line, maybe like an abacus. If all the sliders are on the left, you get the smallest possible value. At this stage, only one slider can move (the largest number) to go to a value one higher. For example, OOOOXXXXX = 1234XXXXX can become OOOXOXXXX = 123X5XXXX. But 1,2,3 cannot move as they are blocked to their immediate right. Having gotten to 11, now either of two sliders can move to get to 12. This is the 3 or the 5 (to give 12X45XXXX or 123XX6XXX). The clue value isn't large enough to allow the sliders trapped on the far left, the 1 or the 2, to move yet. In addition to the fixed digits in these sets, the other kind of thinking to keep in mind is the largest/smallest number that can go in that clue. A 12 in four cells can never contain a 7, 8, or 9. Moreover, if it did contain a 6, then no other cell can be larger than a 3. So when I look at a 12 clue in four cells, I'm looking at the crossing clues specifically for those spots that can take a 1 or 2, since those must go somewhere, as well as for spots that really want a large digit, since it may be those can only take a 5 or 6 and that will constrain all the rest. And while min/max thinking (looking at small clues and big clues that intersect) will often be how you get started, sometimes just knowing the forced number sets is enough. I've seen way too many 4 in two/7 in three and 16 in two/23 in three crossings and placing the forced digits should be automatic. Realize that as you write numbers down the remaining clues may now become another min/max set. A 9 sitting in a 20 in five cell clue has left behind an 11 in four which you should know is {1,2,3,5}. And occasionally you'll get an intersection with a unique set that normally isn't. A 6 entered into an 18 in five cells leaves a 12 in four which isn't usually unique except that the 6 eliminates all but the {1,2,4,5} option. Every digit placed should get you thinking about the leftovers in the crossing clue as these will give the next sure placement the majority of the time. While the above will get you through 95+% of the kakuro out there, the next most useful things to keep in mind involve sudoku-like eliminations. You'll sometimes find pairs or even X-wings in kakuro puzzles. And this is why I recommend a notation where I mark any number that must go in one of two adjacent cells (and certainly any sure pairs). In an across entry that intersects two marked vertical 98 pairs, for example, an intersection with a 24 in three cells will only have one option left in that cell, a 7, as the 8 and 9 are accounted for elsewhere even if not written down for sure yet. It's harder to keep in mind all three vertical clues that cross in that row but having some notes down will get your eyes focused in the right spots. There are many valuable extensions of this thinking involving uniqueness where you need to avoid deadly patterns. If a vertical 17 in two cells shares two rows with a vertical 24 in three cells, I can tell you that a 7 cannot go in the outlying cell in that 24 or you'd have two solutions. I don't enjoy writing the consequence of this thinking down, but I certainly will write it down to solve just a bit faster. In a subset of kakuro, almost exclusively computer generated ones, you may also find a region of the grid mostly isolated from the rest of the grid with only one entry connecting it to the rest. In these cases, in that isolated region you can add up all the vertical clues, add up all the horizontal clues, and from the difference in these values figure out the sum of a single cell (sometimes more) in the connecting clue. I don't like doing this much math, but on a site like croco-puzzle I will certainly put in a valid guess for an entire isolated area (even duplicating numbers) to get the value of the singleton cell. I'll then erase the corner if wrong and work back, now from the connecting entry, to figure out what the unique fill is. I'm sure there could be a counter-example of a Nikoli puzzle that needs this kind of logic, but I'd estimate <1-2% of their puzzles have isolated subsections where this kind of thinking, even mid-solution, is very valuable. So that's a lot to digest, and that's just how to solve a regular kakuro. So I'll be more pithy with kakuro variants, particularly as I cannot predict what particular variant will appear in any given year. To prepare for these variants, you want to engage in a game of leveling with the puzzle designer. Something like "I know you know I know what a 10 in four cells does, but if X, then ...". You should be able to use the guidelines above to predict what "obvious" work-ins will look like in a new variation. For example, in 2006 there was a Plus or Minus Kakuro where a clue shaded in gray needed at least one value of opposite sign. This change negated the value of "minimal" clues, but did not negate the "maximal" clues. What is the largest possible clue (by absolute value) for a five-cell entry? Well, it's 30 (the largest four-cell entry) minus 1 (the smallest one-cell entry) = 29. If you had internalized information like that before the test, you'd immediately see the work-in for that puzzle in the upper right as well as another in the lower right. Always recalibrate your expectations to match the gimmick. If the gimmick is simply to allow 0 in addition to 1 to 9, then recalibrate all the small clues you keep in your head; if the gimmick is to have each entry one off the correct clue, then look for 18's and 17's in two cells instead of 16's and 17's when you are trying to spot unique outliers, and expect a 16 clue to actually be a 15 as the author is probably trying to be "sneaky". I have previously constructed some "Nonconsecutive Kakuro" where consecutive digits cannot touch. In that variation, you should recognize that an entry like 6 in three cells is impossible (the 2 would be touching a 1 or a 3 wherever it is placed) and an entry like a 7 in three cells must be 142 or 241 as you have to separate the 1 and 2. Try to figure out what ways a 10 in four cells can be filled in. It's many fewer than the usual 4x3x2x1 options. You need to think in sets of digits, and separate possible consecutive neighbors, to make progress. Comments: I like Nikoli Kakuro books specifically to race against the indicated times. The puzzles don't hide a lot of new techniques, and I've memorized most of the patterns indicated above, so I am pretty efficient at them but not at H. Jo level and unlikely to improve. But I capital-L Love kakuro variants. If I had just a little more energy and time, I would manage to finish and self-publish a book of my ideas in this area called "Mutant Kakuro" to go along with what I've done to sudoku puzzles. I strongly believe that small changes in the rules to a simple puzzle type can lead to fun emergent properties with new logic to discover and kakuro is as obvious and productive a setting for this experimentation as sudoku is. And Dr. Sudoku already has a Kakuro Lad side-kick ready to extend the story. It's just that kakuro is far less commercial. But I'd list some of my existing mutant kakuro as some of my best puzzle work, and my collaboration with Dan Katz in the 2009 Mystery Hunt is the simplest example to point to to take up several hours of your time with a lot of variety. So, if you know all that about me, it should come as no surprise that I look forward to new kakuro variations on the USPC every year and while the last few years have been disappointing, as we've gotten KenKen and other math puzzles like those X-Agony and Di-Agony challenges instead, I fully expect a fun original Kakuro variation (with strained/punny title) to appear again soon. About this puzzle: This puzzle idea grew out of the most "ahead of its time" variation in Cross Number Sum Place, one year before the sudoku boom but a really great combination of those two types. Removing the sudoku rules but accentuating the consecutive (and implied nonconsecutive) constraints, I figured I'd make a themed USPC Consecutive Kakuro this week and managed a pretty reasonable end puzzle for a USPC, probably a 30-pointer. Enjoy! Rules: Enter a single digit from 1 to 9 into each empty square so that the sum of the digits in each across and down answer equals the value given to the left or above, respectively. No digit is repeated within a single answer. Additionally, a gray bar is shown between two squares every time they contain consecutive digits and only when they contain consecutive digits. (Consequently, any two squares without a gray bar between them cannot contain consecutive digits.) betterknowtheuspcfridaypuzzlekakurouspc 0 7Subscribe0 LJ Video Posts with tag ### Friday Puzzle #183 - #1 Fan Arrow Sudoku ### Friday Puzzle #182 - 24hour Magic Order ### Friday Puzzle #181 - 24hour Skyscraper 7 comments Post a new comment cyrebjr I guess I liked the upper right corner best, though the geometry of P gave me warm fuzzies. I mentioned a couple months ago that I'd follow up on your 6\9 kakuro with one of my own. Well, nobody has commented on it, so it's hard to tell if anyone has tried to solve it. It extends the 6\9 to "all" other digits, and I'd really like to know what you think; motris As I recall, that was posted just before an LMI test that used the same gimmick throughout. I had printed your puzzle and then played it before that test and really liked it (much more than some on the test) and it certainly extended the 6/9 gimmick really well. I obviously forgot to share those comments. I run into the same problem here where people have solved my puzzles but the absence of comments doesn't tell me that anyone is seeing what I did. So since it's also on my uncommented list, I'll add here that I did your shapeshunter and really enjoyed it and appreciated the links you gave too to three others. hagriddler it helped me a lot that (rot13 for possible spoiler) : gur fhz bs gjb pbafrphgvir vagrtref vf bqq orpnhfr bar bs gurz vf rira naq gur bgure bar vf bqq. jvgu guerr-pryy pyhr guvf zrnaf gung gur cnevgl bs gur guveq qvtvg zhfg qvssre sebz gur cnevgl bs gur pyhr. motris Yes, I hoped that would be something solvers would "learn" in this puzzle that helps in several places. thedan Very nice puzzle with a variety of techniques to solve. Took me 16:40, but I was fending off a post-sandwich coma. I'm sure I've said this before, but if you ever do decide to do a kakuro variant book and want assistance, I'm in. Ours Brun Nice puzzle. I didn't find it too hard, but I don't know exactly what is supposed to be the difficulty level of a USPC "30-pointer". A book of kakuro variants would really be great. I am also a big fan of these, and they are way too much rare. I hope you will manage to publish it someday. gabrieleud Oh, I see that unvoluntarily the gray bars are creating the letters "USPC".
2347	https://khankids.zendesk.com/hc/en-us/articles/4403706997645-Comprehension-The-ultimate-goal-of-reading	Comprehension: The ultimate goal of reading – Khan Academy Skip to main content Submit a request Khan Academy News and Announcements Articles in this section Festive winter learning with Khan Academy Kids Meet inspiring women in our new books! Spanish books \| Libros en español Meet a new set of friends in the Khan Kids app! The Path to Reading: What you need to know about how children learn to read Pre-literacy: Setting your child up for success Phonics: The building blocks of literacy Fluency: The key to independent reading Comprehension: The ultimate goal of reading Literacy Routines: Growing a family of readers See more Comprehension: The ultimate goal of reading 3 years ago Updated What is reading comprehension? Comprehension is understanding what you read, as you read it. Good readers don’t just say words out loud—they form a picture in their mind of the story, making a mental movie to help them imagine the events as they happen. Readers who are good at comprehending also think beyond the story. They make connections to what they already know, make inferences and predictions, and think about the story after they are done reading. Your child has been developing comprehension skills since birth, but now must combine this knowledge with their independent reading abilities. What can I do to help my child develop comprehension skills? Continuing to read with your child for fun is a great way to build comprehension skills. When you read together or when you listen to your child read, talk about what you’re reading, new words they might not know, and what you imagine when you read. Here are some more specific tips to help build comprehension at home. You can download this tip sheet below. How does the Khan Kids app help to build comprehension? Reading and thinking about what you read are the best way to build comprehension skills. Luckily, Khan Academy Kids has over 300 books built right into our app! You can see all the books in one place by going to the Books tab in the Library. Here, your child can either choose to have the book read aloud to them or to read it on their own. Another way to view our books is by clicking on the Reading tab in the Library. Here, books are shown for each age group, and you can view more by clicking on the dropdown menu on the right to change the age you see. When they select a book in the Reading tab, your child will be asked supporting comprehension questions as they read. This can be really helpful as they work on developing their mental movie and comprehension skills. Encourage them to pause and answer the question aloud before turning the page. Each book in our library is written at an age appropriate level and with vocabulary building in mind. Potentially unfamiliar words are given plenty of context, so that children can learn their meanings. Here’s a short video where Reya explains how to use context clues to learn unfamiliar words. This is a great strategy that your child can use whenever they are reading. Additional articles This is article #4 in our series of 5 articles about the path to reading. If you’d like to learn more, be sure to read the other articles! Article 1)Pre-literacy: the period of time before a child begins to read or write Article 2)Phonics: the connection between letters and sounds Article 3) Fluency: reading naturally, while paying attention to meaning Article 5)Reading routines: developing lifelong readers 👇 Download our reading comprehension tip sheet KhanKids_Comprehension.pdf700 KBDownload Was this article helpful? Yes No 4 out of 4 found this helpful Have more questions? Submit a request Return to top Comments 0 comments Article is closed for comments. Khan Academy Powered by Zendesk
2348	https://mandarinbean.com/yes-no-question-with-ma/	Yes-no question with “吗” - Grammar Explanation & Exercises Skip to content Home All Lessons Beginner Intermediate Advanced Grammar All Grammar Points with Exercises Grammar Level Tests Test Prep HSK Sample Tests YCT Sample Tests More Pinyin Practice Confusing Chinese Words Pricing Login Home All Lessons Beginner Intermediate Advanced Grammar All Grammar Points with Exercises Grammar Level Tests Test Prep HSK Sample Tests YCT Sample Tests More Pinyin Practice Confusing Chinese Words Pricing Login Search Grammar Points Yes-no question with “吗” Explanation Yes-no question with “吗” “吗” is used at the end of a declarative sentence to form a yes-no question. Declarative sentence + 吗？ Note that “吗” can’t be used with a special sentence, affirmative-negative sentence, or selective sentence. e.g. A: 你是中国人吗？(Nǐ shì Zhōngguó rén ma?) Are you Chinese? B: 是，我是中国人。(Shì, wǒ shì Zhōngguó rén.) Yes, I am Chinese. C: 不是，我不是中国人。(Búshì, wǒ búshì Zhōngguó rén.) No, I am not Chinese. A: 现在你在北京吗？(Xiànzài nǐ zài Běijīng ma?) Are you in Beijing now? B：我在北京。(Wǒ zài Běijīng.) I’m in Beijing. C：我不在北京。(Wǒ búzài Běijīng.) I am not in Beijing. A：今天天气冷吗？(Jīntiān tiānqì lěng ma?) Is it cold today? B：冷。(Lěng.) Cold. C：不冷。(Bù lěng.) Not cold. A：你们会写汉字吗？(Nǐmen huì xiě hànzì ma?) Can you write Chinese characters? B：会写。(Huì xiě.) Yes, we can. C：不会写。(Bú huì xiě.) No, we can’t. A：昨天你去学校了吗？(Zuótiān nǐ qù xuéxiào le ma?) Did you go to school yesterday? B：去了。(Qù le.) Yes, I went. C：没去。(Méi qù.) No, I didn’t go. 他是老师吗？(Tā shì lǎoshī ma?) TRUE FALSE 这是不是你的咖啡吗？(Zhè shì bùshì nǐ de kāfēi ma?) TRUE FALSE 请问，你是老师还是医生吗？(Qǐngwèn, nǐ shì lǎoshī háishì yīshēng ma?) TRUE FALSE 外边冷不冷吗？(Wàibian lěng bù lěng ma?) TRUE FALSE Loading... Mark Complete Need more practice? With a premium membership, you can access all 9questions about “Yes-no question with “吗”“, as well as online exercises for all grammar points. Learn more about Premium Membership here >> More Grammar Points 2 Responses Ethan the short kingsays: October 15, 2024 at 4:13 pm This is very useful. Now I can say more questions correctly! :D 10/10 Reply 2. Sancho Piressays: November 8, 2023 at 12:17 am Was very useful, now i can fix most of my mistake and solve it. Reply Leave a Reply Cancel reply Your email address will not be published.Required fields are marked Comment Name Email Website [x] Save my name, email, and website in this browser for the next time I comment. Δ Hundreds of interactive grammar exercises with answer explanations! Home About Blog New HSK Vocabulary Privacy Policy Terms of Use Home About Blog New HSK Vocabulary Privacy Policy Terms of Use © 2024 - Mandarin Bean. All Rights Reserved. With a premium membership, you can access this and all grammar video lessons. The videos look like:
2349	https://www.sciencedirect.com/science/article/pii/S2374289521000762	Educational Case: Infectious Esophagitis - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Primary Objective Secondary Objective Patient Presentation Diagnostic Findings Part I Questions/Discussion Points Part I Diagnostic Findings Part II Questions/Discussion Points Part II Teaching Points Declaration of Conflicting Interests Funding ORCID iD References Uncited References Show full outline Cited by (2) Figures (6) Tables (1) Table 1 Academic Pathology Volume 7, January–December 2020, 2374289520903438 Educational Case Educational Case: Infectious Esophagitis Author links open overlay panel Tara Narasimhalu BA 1, Kristin A.Olson MD 1 Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access The following fictional case is intended as a learning tool within the Pathology Competencies for Medical Education (PCME), a set of national standards for teaching pathology. These are divided into three basic competencies: Disease Mechanisms and Processes, Organ System Pathology, and Diagnostic Medicine and Therapeutic Pathology. For additional information, and a full list of learning objectives for all three competencies, see Previous article in issue Next article in issue pathology competencies organ system pathology gastrointestinal tract mechanical disorders of bowel dysphagia infectious esophagitis candida esophagitis Herpes simplex virus esophagitis cytomegalovirus esophagitis Primary Objective Objective GT8.1: Dysphagia. Describe the pathophysiology and clinicopathological features of disorders presenting with dysphagia. Competency 2: Organ System Pathology; Topic GT: Gastrointestinal Tract; Learning Goal 8: Mechanical Disorders of Bowel. Secondary Objective Objective GT7.1: Bowel Infections. Compare the underlying mechanism and clinicopathologic features of gastrointestinal tract involvement by common bacterial, fungal, and parasitic pathogens. Competency 2: Organ System Pathology; Topic GT: Gastrointestinal Tract; Learning Goal 7: Bowel Infections. Objective FECT2.3: Histopathologic Features of Viral Infection. Compare and contrast the histopathological features of herpes virus, cytomegalovirus, human papilloma virus, and adenovirus in terms of nuclear features, inclusions, size of cells, and other unique characteristics; recognize these histopathological features of viral infections in images of different tissues. Competency 1: Disease Mechanisms and Processes; Topic FECT: Infectious Mechanisms; Learning Goal 2: Pathogenic Mechanism of Infection. Objective FECT2.9: Histopathologic Features of Fungal Infection. Recognize histopathologic evidence of fungal infections and compare and contrast the histopathological features and staining characteristics of the following fungi: Candida albicans, Cryptococcus neoformans, Aspergillus, Histoplasma capsulatum, Coccidioides immitis, Blastomyces dermatitis, Pneumocystis jiroveci, and Zygomycetes. Competency 1: Disease Mechanisms and Processes; Topic FECT: Infectious Mechanisms; Learning Goal 2: Pathogenic Mechanism of Infection. Patient Presentation A 45-year-old female presents to her primary care physician with a 2-week history of odynophagia, dysphagia, dry mouth, and retrosternal pain. She reports limited intake of solid food due to the pain, as well as an unintentional weight loss of 10 lbs during the past month. The patient has been HIV positive for over 10 years. She is not taking her antiretroviral therapy and has not seen a physician regularly. Her most recent CD4 count was 150 over 1 year ago. Diagnostic Findings Part I Physical examination is notable for a cachectic female, with white lesions on the tongue and buccal mucosa. These lesions are easily scraped off with a tongue depressor. Questions/Discussion Points Part I What Is the Differential Based on Clinical Presentation and Physical Examination Findings? The differential diagnosis for a patient presenting with esophageal dysphagia and odynophagia includes esophagitis secondary to infection with cytomegalovirus (CMV), herpes simplex virus, and/or Candida spp. Additionally, medication-induced esophagitis, reflux esophagitis, eosinophilic esophagitis, and idiopathic HIV ulcers should be considered.2 It can be difficult to distinguish between these entities; however, there are key differences in clinical presentation and physical examination findings. Patients who present with medication-induced esophagitis are typically elderly patients who swallow several large pills without enough liquid and do so in a suboptimal position.3 They can present with dysphagia and odynophagia. This is less likely for our patient, as she does not report any current medication use. Patients with reflux esophagitis frequently have a history of heartburn or regurgitation that is worse after consuming fatty meals. They may also report a worsening of symptoms with changes in posture, such a lying down or bending forward. It is unlikely that the patient has reflux esophagitis, as she does not report any history of heartburn, regurgitation, or a postural relationship to her odynophagia. Lastly, patients with eosinophilic esophagitis can present with symptoms, such as upper abdominal pain, dysphagia to solid foods, and food impaction.4 Patients with this condition typically have a history of allergic conditions such as asthma and eczema. Although our patient does present with dysphagia to solid foods, eosinophilic esophagitis is less likely given the acute onset of symptoms and no history of allergic conditions. Additionally, physical examination findings of a cachectic female with scrapable white lesions are concerning for a patient with an immunodeficiency, consistent with our patient’s history of HIV. Patients who have severely depressed cell-mediated immunity are often susceptible to infections from opportunistic organisms, such as Candida spp, which can manifest as oral thrush. What Are the Next Steps in Working Up This Patient’s Condition? Because the patient has a history of HIV and is presenting with two weeks of odynophagia and dysphagia, the next step is to perform an upper endoscopy with biopsies of any esophageal lesions. Diagnostic Findings Part II Upper endoscopy reveals several raised, white plaques throughout the esophagus, which are biopsied (Figure 1). The surrounding mucosa is erythematous and edematous. 1. Download: Download full-size image Figure 1. Candida esophagitis with yeast forms amidst the “shredded wheat” like appearance of squamous epithelium (hematoxylin and eosin, ×40. Bar = 50 μm.). Questions/Discussion Points Part II Describe the Histologic Features Seen in Figures 1 and 2 Figure 1 shows superficial squamous epithelium that is disrupted due to the biopsy giving a “shredded wheat” appearance. Between the squamous cells multiple small oval yeast forms can be identified. Figure 2 is a Gomori Methenamine-Silver (GMS) special stain, which is often used to visualize fungal elements and other opportunistic organisms. The oval yeast forms (arrow) and other fungal elements are stained black, as the GMS stain highlights the fungal cell wallpolysaccharide elements. Lastly, neutrophilic inflammation accompanies this infection. 1. Download: Download full-size image Figure 2. Candida esophagitis with an abundance of pseudohyphae and yeast amidst squamous cells, arrow pointing toward the yeast forms (Gomori Methenamine-Silver, ×40. Bar = 60 μm.). Based on the Clinical and Pathologic Features, What Is Your Diagnosis? This patient has Candida esophagitis. The diagnosis is suspected on direct endoscopic visualization of the esophagus, which reveals classic diffuse white mucosal plaques.5 Confirmatory biopsies show histologic evidence of tissue invasion by Candida spp. It is essential to find evidence of spores or hyphae invading the squamous epithelium, or within the necrotic, inflammatory debris. What Are the Clinical Features of Candida Esophagitis? Candida esophagitis is the most common type of infectious esophagitis in adults, and C. albicans is the most common organism identified.5 In contrast to eosinophilic esophagitis or reflux esophagitis, candida esophagitis presents with a rapid onset of symptoms. These include dysphagia, odynophagia, retrosternal chest pain, vomiting, and fever.5 Patients may also present with chest pain or gastrointestinal (GI) tract bleeding. Some patients may be entirely asymptomatic. What Is the Pathophysiology of Candida Esophagitis? Candida species are part of the normal flora in the oropharynx and esophagus. Candida esophagitis results from a combination of factors, including fungal overgrowth and impaired cell-mediated immunity.5 Overgrowth of Candida species can be secondary to broad-spectrum antibiotic therapy, poorly controlled diabetes mellitus, abnormal esophageal motility, or mechanical abnormalities (esophageal stricture).5 Additionally, individuals who have AIDS, receive chemotherapy/radiation, or take immunosuppressant medication have impaired cell-mediated immunity, and are therefore more susceptible to opportunistic infections. It is estimated that 10% to 15% of patients with AIDS will develop this condition over their lifetime.2 In fact, the development of candida esophagitis may be the first indication that an HIV-positive patient has developed AIDS.2 What Is the Treatment for Candida Esophagitis? Candida esophagitis is treated with systemic therapy for 2 to 3 weeks.6Intravenous medications are given to those who cannot tolerate oral intake. Although fluconazole is the recommended agent due to efficacy and low cost, other medications include echinocandins or amphotericin B.6 Notably, patients with HIV are less responsive to antifungal therapy and may take longer to improve. This group is also prone to reinfection, as opportunistic pathogens are difficult to eliminate in immunosuppressed individuals.5 What Are the Clinical Features of Herpes Simplex Virus Esophagitis? Herpes simplex virus (HSV) esophagitis presents similarly to Candida esophagitis; the main features are dysphagia, odynophagia, chest pain, fever, extra-esophageal herpetic lesions, nausea, vomiting, and GI bleeding.7 Patients may also present with oropharyngeal ulcers or herpes labialis. The main risk factor for HSV esophagitis is immunodeficiency, and impaired cellular immunity in particular.7 Thus, patients who have T-lymphocyte deficiency, such as those with HIV, are at particularly increased risk. Lastly, the use of chemotherapeutic agents and steroids are established risk factors for HSV esophagitis. In addition to immunosuppression, certain chemotherapy drugs compromise the esophageal mucosa integrity, making infection with opportunistic organisms more likely.7 Steroids are also involved in downregulation of T-cell proliferation, contributing to immune dysfunction. What Is Seen on Endoscopy for Patients With Herpes Simplex Virus Esophagitis? The diagnosis of HSV esophagitis requires endoscopy with biopsy and histologic confirmation. Endoscopy typically reveals lesions in the distal esophagus. The early stage of HSV esophagitis is characterized by vesicles or “volcano ulcers” that are up to 2 cm in size.7 Later stages show coalescing ulcers with friable mucosa.7 Biopsies and brushings are typically taken from the margins of the ulcers, where viral cytopathic activity is seen in the squamous epithelium.7 Describe the Histologic Features Seen in Herpes Simplex Virus Esophagitis (Figures 3 and 4) Figure 3 reveals squamous epithelial cells with nuclei that have a “ground glass” appearance, marginated chromatin, multinucleation, and nuclear molding, characteristics of HSV esophagitis.7 Additionally, eosinophilic intranuclear and cytoplasmic inclusion bodies within squamous epithelial cells at the ulcer margins may be seen, referred to as Cowdry type A inclusions.7 Further, biopsies taken from the ulcer bed may reveal an abundance of inflammatory cells and necrotic debris.7Figure 4 is an immunohistochemical stain used to detect HSV-infected cells. This is done using antibodies that specifically bind to the HSV antigens and subsequently visualizing the antigen–antibody complex, with a colored stain. Figure 4 shows significant nuclear and cytoplasmic staining, or immunoreactivity, suggesting an infection of HSV within the squamous cells. 1. Download: Download full-size image Figure 3. Herpes simplex virusesophagitis with infected squamous cells demonstrating multinucleation, nuclear molding, and chromatin margination (hematoxylin and eosin, ×40. Bar = 60 μm.). 1. Download: Download full-size image Figure 4. Herpes simplex virusesophagitis with nuclear and cytoplasmic staining, confirming infection within cells (Immunohistochemical, ×40. Bar = 50 μm.). What Is the Pathophysiology of Herpes Simplex Virus Esophagitis? Herpes simplex virus-1 is a large, double-stranded DNA virus in the Herpesviridae family that is implicated in HSV esophagitis. The HSV-1 is transmitted through oral-to-oral, oral-to-genital, or genital-to-genital contact. This occurs when an unaffected individual comes into contact with skin, mucosal secretions, or lesions that are infected with HSV-1.8 A significant majority of individuals with primary HSV-1 infection are asymptomatic. Individuals with symptomatic primary infection present with painful oral ulcers, lymphadenopathy, fever, malaise, or headache.8 Recurrent HSV-1 infections occur in approximately 20% to 40% of infected individuals. This typically occurs due to immunodeficiency, stress, or fever.8 The HSV esophagitis is most commonly due to reactivation of a latent HSV-1 virus in the sensory ganglia. Occasionally, reactivation of a latent HSV-2 infection can also result in esophagitis. The virus is thought to spread to the esophagus via the vagus nerve or directly from the oral cavity into the esophagus during an infection.9 Less commonly, HSV esophagitis can result from a primary herpes infection. What Are Clinical Features of Cytomegalovirus Esophagitis? Cytomegalovirus is frequently identified in individuals who are HIV-positive, have a CD4 count below 200 cells/mm 3, and/or do not take antiretrovial therapy. Other individuals who are particularly vulnerable to CMV esophagitis include organ transplant recipients, dialysis participants, or those taking immunosuppressive medications.9 Clinical manifestations include odynophagia, fever, nausea, and substernal pain. What Is Seen on Endoscopy for Patients With Cytomegalovirus Esophagitis? Endoscopic findings can be variable for CMV esophagitis. Typically, there are several, well-circumscribed shallow ulcerations found in the distal esophagus.9 However, deep ulcers or diffuse erosions may be seen. Additionally, ulcers tend to be linear or longitudinal in shape.10 It is essential to take biopsies from the ulcers or erosions to diagnosis CMV esophagitis. Describe the Histologic Features of Cytomegalovirus Esophagitis (Figures 5 and 6) Figure 5 is a hematoxylin and eosin image of CMV-infected cells. The viral cytopathic effect of CMV is typically seen within endothelial and stromal cells at the base of the ulcer.11 Characteristic features seen on histology include cytomegaly, often eccentrically positioned basophilic nuclei with inclusions surrounded by a clear halo (similar to an “owl’s eye”), and a paler perinuclear region.11 Arrows in Figure 5 point to this “owl’s eye” appearance. Additionally, basophilic cytoplasm with intracytoplasmic granules may be seen. Figure 6 is an immunohistochemical stain used to detect CMV-infected cells. Specific antibodies bind to the CMV antigens within cells, and the antigen–antibody complex is visualized with special staining. In Figure 6, there is pronounced nuclear staining, confirming CMV infection within the cells. 1. Download: Download full-size image Figure 5. Cytomegalovirus esophagitis with enlarged mesenchymal cells containing basophilic nuclei and pale perinuclear regions. Arrows pointing to “owl’s eye” appearance of cells (hematoxylin and eosin, ×40. Bar = 50 μm). 1. Download: Download full-size image Figure 6. Cytomegalovirus esophagitis with nuclear staining, confirming infection within cells (immunohistochemical, ×40. Bar = 50 μm). What Is the Pathophysiology of Cytomegalovirus Esophagitis? Like HSV-1, CMV belongs to the Herpesviridae family and is a double-stranded DNA virus. The pathophysiology of CMV esophagitis is similar to that of HSV esophagitis. The virus is transmitted through several routes, including perinatal, sexual, and blood or tissue exposure.12 Additionally, individuals with close contact to those infected with CMV are at a greater risk of developing infection because the virus can be shed from the upper respiratory tract and urine.12 Cytomegalovirus infection is quite common, an estimated 90% of adults older than 80 are infected.12 The majority of individuals who are infected with CMV are immunocompetent and asymptomatic. Healthy individuals may develop an illness similar to mononucleosis, with fever, tonsillitis, lymphadenopathy, and dermatologic manifestations.12 Because CMV can remain in a latent form in several organs, immunosuppression allows the latent virus to become activated.13 Reactivation of the virus results in a systemic disease with viremia that can colonize several organs, such as the GI tract.13 Patients who have evidence of CMV infection in the GI tract frequently also have CMV retinitis; therefore, it is imperative for patients to be evaluated with ophthalmologic examinations.13 What Are the Key Differences Between Candida, Herpes Simplex Virus, and Cytomegalovirus Esophagitis? Although the various types of infectious esophagitis have similar risk factors and clinical presentations, they can be differentiated by endoscopic and histologic findings as shown in Table 1. Table 1. A Comparison of Candida, HSV, and CMV Esophagitis. \| Pathogens \| Endoscopic Findings \| Histologic Features \| --- \| Candida \| White mucosal plaques dispersed throughout the esophagus, with an underlying erythematous mucosa. \| Pseudohyphae and yeast among patches of necrotic squamous cells. Pseudohyphae invading GI tissue. \| \| HSV \| Diffuse, superficial ulcers typically in the distal esophagus. Early endoscopic findings include vesicles up to 2 cm. Later findings are coalescing ulcers with friable mucosa. \| Nuclear molding, multinucleation, and chromatin margination. Eosinophilic or basophilic inclusion bodies in squamous epithelial cells at ulcer margins (Cowdry type A inclusions). \| \| CMV \| Multiple linear or longitudinal ulcers that are found in the distal mucosa. The ulcers are well circumscribed and can be shallow or deep. \| Infection of mesenchymal and stromal cells at the base of the ulcer. Cytomegaly, intranuclear basophilic inclusions (owl’s eye), granular cytoplasmic inclusions. \| Abbreviations: CMV, cytomegalovirus; GI, gastrointestinal; HSV, Herpes simplex virus. Teaching Points •The main clinical manifestations for infectious esophagitis include dysphagia, odynophagia, chest pain, fever, nausea/vomiting, and GI bleeding. •Risk factors for all types of infectious esophagitis include AIDSs, chemotherapy/radiation, and immunosuppressant medications. •Endoscopic findings for candida esophagitis are raised white plaques throughout the esophagus, with edematous and erythematous mucosa. •Histology for candida esophagitis includes pseudohyphae and yeast invading viable tissue, often with associated neutrophilic inflammation. •Endoscopic findings for HSV esophagitis include vesicles or “volcano ulcers” in the early stage and coalescing ulcers with friable mucosa in later stages. •Histology for HSV esophagitis shows nuclear molding, margination, and multinucleation of infected cells. Eosinophilic intranuclear and cytoplasmic inclusion bodies within squamous cells, referred to as Cowdry type A inclusions, may be seen. •Endoscopic findings for CMV esophagitis are multiple well-circumscribed shallow ulcerations found in the distal esophagus. Cytomegalovirus ulcers typically are linear or longitudinal in shape. •Histology for CMV esophagitis reveals enlarged cells with basophilic nuclei, intranuclear, basophilic inclusions surrounded by a clear halo, and perinuclear paler regions. Basophilic cytoplasm with intracytoplasmic granules may also be present. Declaration of Conflicting Interests The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article. Funding The author(s) disclosed receipt of the following financial support for the research, authorship, and/or publication of this article: The article processing fee for this article was funded by an Open Access Award given by the Society of ‘67, which supports the mission of the Association of Pathology Chairs to produce the next generation of outstanding investigators and educational scholars in the field of pathology. This award helps to promote the publication of high-quality original scholarship in Academic Pathology by authors at an early stage of academic development. ORCID iD Tara Narasimhalu Special issue articles Recommended articles References 2J Vazquez Optimal management of oropharyngeal and esophageal candidiasis in patients living with HIV infection HIV/AIDS—Research and Palliative Care (2010) Accessed 23rd Apr 2019 Google Scholar 3DO Castell Medication-induced esophagitis. UpToDate (2018), Accessed 2nd Nov 2019 Google Scholar 4J Bystrom, NR O’Shea Eosinophilic oesophagitis: clinical presentation and pathogenesis Postgrad Med J, 90 (2014), pp. 282-289 doi:10.1136/postgradmedj-2012-131403 CrossrefView in ScopusGoogle Scholar 5MA Baig, J Rasheed, D Subkowitz, J Vieira, S Gerges Severe esophageal candidiasis in an immunocompetent patient Int J Infect Dis, 5 (2005) doi:10.5580/a1.f Google Scholar 6CA Kauffman Treatment of oropharyngeal and esophageal candidiasis UpToDate (2018) Accessed 27th Apr 2019 Google Scholar 7T Généreau, F Rozenberg, O Bouchaud, C Marche, O Lortholary Herpes esophagitis: a comprehensive review Clin Microbiol Infect, 3 (1997), pp. 397-407 doi:10.1111/j.1469-0691.1997.tb00275.x View PDFView articleCrossrefView in ScopusGoogle Scholar 8C Johnston, A Wald Epidemiology, clinical manifestations, and diagnosis of herpes simplex virus type 1 infection UpToDate (2019) Accessed 4th Nov 2019 Google Scholar 9M Rosołowski, M Kierzkiewicz Etiology, diagnosis and treatment of infectious esophagitis Prz Gastroenterol, 8 (2013), pp. 333-337 doi:10.5114/pg.2013.39914 CrossrefView in ScopusGoogle Scholar 10R Hashimoto, A Chonan Esophagitis caused by cytomegalovirus infection in an immune-competent patient Clin Gastroenterol Hepatol, 14 (2016), pp. e143-e144 doi:10.1016/j.cgh.2016.06.016 View PDFView articleView in ScopusGoogle Scholar 11A Grin, CJ Streutker Esophagitis: old histologic concepts and new thoughts Arch Pathol Lab Med, 139 (2015), pp. 723-729 doi:10.5858/arpa.2014-0132-ra View in ScopusGoogle Scholar 12TJ Friel Epidemiology, clinical manifestations, and treatment of cytomegalovirus infection in immunocompetent adults UpToDate (2019) Accessed 5th Nov 2019 Google Scholar 13MA Jacobson AIDS-related cytomegalovirus gastrointestinal disease UpToDate (2017) Accessed 5th May 2019 Google Scholar Uncited References 1 BEC Knollmann-Ritschel, DP Regula, MJ Borowitz, R Conran, MB Prystowsky Pathology competencies for medical education and educational cases Acad Pathol (2017), p. 4 doi:10.1177/2374289517715040 Google Scholar Cited by (2) Streptococcal Esophagitis in an Immunocompetent Patient: A Rare Sequelae 2024, Journal of Investigative Medicine High Impact Case Reports ### UPPER GASTROINTESTINAL BLEEDING DUE to CONCOMITANT ESOPHAGEAL VARICES and HERPES SIMPLEX VIRUS ESOPHAGITIS in A 70-YEAR-OLD PATIENT 2021, Gastroenterology Nursing Copyright © The Authors 2020 Part of special issue Educational Cases - Disease Mechanisms and Processes View special issue Recommended articles Driving Access to Care: Use of Mobile Units for Urine Specimen Collection During the Coronavirus Disease-19 (COVID-19) Pandemic Academic Pathology, Volume 7, 2020, Article 2374289520953557 Jill S.Warrington, …, Mark Fung View PDF ### Entry of Graduates of US Pathology Residency Programs Into the Workforce: Cohort Data Between 2008 and 2016 Remain Positive and Stable Academic Pathology, Volume 7, 2020, Article 2374289520901833 Charles F.Timmons, …, Robert D.Hoffman View PDF ### The Veterans Affairs Healthcare System and Academic Pathology Departments: Evaluation of the Relationship Academic Pathology, Volume 7, 2020, Article 2374289520939265 David N.Bailey View PDF ### Utilizing Social Media to Spread Knowledge: The Association of Pathology Chairs Experience at the 2018 Annual Meeting Academic Pathology, Volume 7, 2020, Article 2374289520901342 Dana Razzano, …, Nicole Riddle View PDF ### Informatics Training for Pathology Practice and Research in the Digital Era Academic Pathology, Volume 7, 2020, Article 2374289520911179 Heather T.D.Maness, …, Srikar Chamala View PDF ### Educational Case: Ischemic Disorders of the Gut in Adult Patients Academic Pathology, Volume 6, 2019, Article 2374289519888709 Priyanka Patil, Nicole C.Panarelli View PDF Show 3 more articles Article Metrics Citations Citation Indexes 2 Captures Mendeley Readers 13 View details About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. 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This comprehensive exploration offers clear definitions, real-world examples and crucial applications, enhancing your understanding of these key principles. Through a meticulous study of these phenomena, their distinct characteristics and their role in engineering designs, you'll gain valuable insights into the significance of the Triple Point and Critical Point in Thermodynamics. You'll also explore their impact when applied to different materials, by comparing the Triple Point and Critical Point of Carbon Dioxide and Water. Get started Millions of flashcards designed to help you ace your studies Sign up for free + Add tag Immunology Cell Biology Mo What is the triple point and how does it differ between carbon dioxide and water? Show Answer + Add tag Immunology Cell Biology Mo What is the difference between a substance's triple point and critical point in thermodynamics? Show Answer + Add tag Immunology Cell Biology Mo What is the triple point and how is it represented graphically? 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Expansion Cogeneration Combined Convection and Radiation Combined Cycle Power Plant Combustion Engine Compressor Conduction Conjugate Variables Continuous Combustion Engine Continuous Phase Transition Convection Dead State Degrees of Freedom Physics Differential Convection Equations Diffuser Diffusion Equation Double Tube Heat Exchanger Economizer Electrical Work Endothermic Reactions Energy Degradation Energy Equation Energy Function Enthalpy Enthalpy of Fusion Enthalpy of Vaporization Entropy Change for Ideal Gas Entropy Function Entropy Generation Entropy Gradient Entropy and Heat Capacity Entropy and Irreversibility Entropy of Mixing Equation of State of a Gas Equation of State of an Ideal Gas Equations of State Exergy Exergy Analysis Exergy Efficiency Exothermic Reactions Expansion Extensive Property External Combustion Engine Feedwater Heater Fins First Law of Thermodynamics Differential Form First Law of Thermodynamics For Open System Flow Process Fluctuations Forced Convection Four Stroke Engine Free Expansion Free Expansion of an Ideal Gas Fundamental Equation Fundamentals of Engineering Thermodynamics Gases Gibbs Duhem Equation Gibbs Free Energy Gibbs Paradox Greenhouse Effect Heat Heat Capacity Heat Equation Heat Exchanger Heat Generation Heat Pump Heat and Work Helmholtz Free Energy Hydrostatic Transmission Initial Conditions Intensive Property Intensive and Extensive Variables Internal Energy of a Real Gas Irreversibility Isentropic Efficiency Isentropic Efficiency of Compressor Isentropic Process Isobaric Process Isochoric Process Isolated System Isothermal Process Johnson Noise Joule Kelvin Expansion Joule-Thompson Effect Kinetic Theory of Ideal Gases Landau Theory of Phase Transition Linear Heat Conduction Liquefaction of Gases Macroscopic Thermodynamics Maximum Entropy Maxwell Relations Mechanism of Heat Transfer Metastable Phase Moles Natural Convection Nature of Heat Negative Heat Capacity Negative Temperature Non Equilibrium State Nuclear Energy Nucleation Nusselt Number Open System Thermodynamic Osmotic Pressure Otto Cycle Partition Function Peng Robinson Equation of State Polytropic Process Potential Energy in Thermodynamics Power Cycle Power Plants Pressure Volume Work Principle of Minimum Energy Principles of Heat Transfer Quasi Static Process Ramjet Real Gas Internal Energy Reciprocating Engine Refrigeration Cycle Refrigerator Regenerative Rankine Cycle Reheat Rankine Cycle Relaxation Time Reversibility Reversible Process Rotary Engine Sackur Tetrode Equation Specific Volume Steady State Heat Transfer Stirling Engines Stretched Wire Surface Thermodynamics System Surroundings and Boundary TdS Equation Temperature Scales Thermal Boundary Layer Thermal Diffusivity Thermodynamic Equilibrium Thermodynamic Limit Thermodynamic Potentials Thermodynamic Relations Thermodynamic Stability Thermodynamic State Thermodynamic System Thermodynamic Variables Thermodynamics of Gases Thermoelectric Thermoelectric Effect Thermometry Third Law of Thermodynamics Throttling Device Transient Heat Transfer Triple Point and Critical Point Two Stroke Diesel Engine Two Stroke Engine Unattainability Van der Waals Equation Vapor Power System Variable Thermal Conductivity Wien's Law Zeroth Law of Thermodynamics Materials Engineering Mechanical Engineering Professional Engineering Robotics Engineering Solid Mechanics What is Engineering Contents Fact Checked Content Last Updated: 13.10.2023 21 min reading time Content creation process designed by Lily Hulatt Content sources cross-checked by Gabriel Freitas Content quality checked by Gabriel Freitas Sign up for free to save, edit & create flashcards. Save ArticleSave Article Jump to a key chapter Understanding Triple Point and Critical Point Meaning Diving into Triple Point and Critical Point Examples Triple Point and Critical Point Applications in Engineering Thermodynamics Mastering the Triple Point and Critical Point Formula Difference between Critical Point and Triple Point Analysing the Triple Point and Critical Point of Carbon Dioxide Vs Water Triple Point and Critical Point - Key takeaways Similar topics in Engineering Related topics to Engineering Thermodynamics Flashcards in Triple Point and Critical Point Learn faster with the 12 flashcards about Triple Point and Critical Point Frequently Asked Questions about Triple Point and Critical Point About StudySmarter Test your knowledge with multiple choice flashcards 1/3 What is the triple point and how does it differ between carbon dioxide and water? A. The triple point refers to the melting point of a substance. For water, it is 273.16 K and 611.657 Pa. For carbon dioxide, it is 216.55 K and 5.11 atm because it sublimates directly into a gas. B. The triple point is the condition where a substance exists only in liquid form. For water, this happens at 273.16 K and 611.657 Pa. For carbon dioxide, it happens at 216.55 K and 5.11 atm due to its ability to exist as a gas in normal conditions. C. The triple point is the unique set of conditions—temperature and pressure—where all three phases of a substance coexist in equilibrium. For water, it occurs at 273.16 K and 611.657 Pa. For carbon dioxide, it occurs at a much higher pressure of 5.11 atm and a temperature of 216.55 K, due to its unique characteristic of sublimating directly from solid to gas under normal conditions. D. The triple point is when a substance undergoes a phase change. For water, it is at 273.16 K and 611.657 Pa, whilst for carbon dioxide it's at 216.55 K and 5.11 atm because CO2 always exists as a gas. 1/3 What is the difference between a substance's triple point and critical point in thermodynamics? A. The triple point and the critical point both denote the condition where a substance's liquid and gas phases transition into a supercritical fluid. B. The triple point is the specific condition where all three phases (gas, liquid, solid) of a substance co-exist in equilibrium. On the other hand, the critical point is that unique condition where a substance's gas and liquid phases become indistinguishable, transitioning into a supercritical fluid. C. The triple point and the critical point refer to the same phenomenon in thermodynamics, where all three phases (gas, liquid, solid) co-exist in equilibrium. D. The triple point refers to the condition where a substance's liquid and gas phases transition into a supercritical fluid, while the critical point is where all three phases (gas, liquid, solid) of a substance co-exist in equilibrium. 1/3 What is the triple point and how is it represented graphically? A. The triple point is where a substance skips a phase and transitions directly from solid to gas. It is plotted on a phase diagram where the liquid phase doesn't exist. B. The triple point is a condition under which a substance exists in all three phases - solid, liquid, and gas - simultaneously. It is graphically represented on a phase diagram, where the solid-liquid, liquid-gas, and solid-gas lines intersect - this intersection signifies the triple point. The triple point varies for different substances. C. The triple point is where a substance is solid at all temperatures and pressures. It is marked on a phase diagram where the solid line is the only line present. D. The triple point is when a substance turns into a supercritical fluid. It is displayed on a phase diagram, where the solid, liquid and gas phases are exactly the same. Score That was a fantastic start! You can do better! Sign up to create your own flashcards Access over 700 million learning materials Study more efficiently with flashcards Get better grades with AI Sign up for free Already have an account? Log in Good job! Keep learning, you are doing great. Don't give up! Next Open in our app Understanding Triple Point and Critical Point Meaning In the realm of physical chemistry and thermodynamics, few concepts are as fascinating as the triple point and the critical point. These concepts, which help us understand the various states of matter, play a vital role in numerous engineering applications. Thanks to this ad, StudySmarter remains free.Sign up for a free account to access a better learning experience. Definition of the Triple Point In thermodynamics, the triple point of a substance is the unique set of conditions at which all three phases - gas, liquid, and solid - coexist in equilibrium. Think about water, for instance. We're accustomed to seeing it freeze, boil, or evaporate, but at the triple point, all these things happen simultaneously. For water, this occurs at a particular pressure and temperature: Pressure=611.657 Pa Temperature=273.16 K The significance of these particular conditions is that they form a baseline for thermodynamic temperature scales, such as the Kelvin scale, which considers 273.16 K as the triple point of water. Thanks to this ad, StudySmarter remains free.Sign up for a free account to access a better learning experience. What is the Critical Point in Engineering Thermodynamics? In thermodynamics, the critical point of a substance is the highest temperature and pressure where the substance can exist as a gas and a liquid in equilibrium. Beyond this point, the gas and liquid states of a substance become indistinguishable, the result of which is a supercritical fluid. The heat and pressure conditions required for this phenomenon vary by substance. To illustrate, the critical points for water are: Pressure=22.064×10 6 Pa Temperature=647.096 K Identification of the critical point is of practical importance since it helps in the design and operation of equipment in various engineering processes, especially in the field of chemical engineering. Your browser does not support the video tag. Thanks to this ad, StudySmarter remains free.Sign up for a free account to access a better learning experience. The Connection Between Triple Point and Critical Point So, how exactly are the triple point and critical point related? The answer lies in the phase diagram of substances, which graphically represents the phases of a substance under varying temperature and pressure conditions. Temperature Pressure Phase 273.16 K 611.657 Pa Triple Point (water) 647.096 K 22.064 MPa Critical Point (water) In such a diagram, the line separating any two phases of a substance end at the critical point. On the other hand, the point where the lines separating all three phases meet is the triple point. It’s noteworthy to mention that the critical point is always at a higher temperature and pressure than the triple point. This leads to an interesting consequence - for temperatures and pressures below the triple point, a gas cannot be liquified, no matter how much pressure is applied. Hence, knowledge of these points is crucial for processes such as liquefaction of gases. It might seem like a lot to take in, but with time and further study, you'll find that these concepts offer critical insights into the behavior of various substances, forming the bedrock of the engineering applications in thermodynamics. Stay organized and focused with your smart to do list Sign up for free Diving into Triple Point and Critical Point Examples As with many aspects of engineering and physics, the most effective way to truly understand the triple point and critical point concepts is to see them in action. This part of our exploration delves into practical examples of both, providing the groundwork for a deeper understanding of the topic. Examples of Triple Point in Real-World Scenarios While the concept of the triple point may seem mostly theoretical, you'll find the applications are quite practical. Let's kick things off with two prevalent examples: - Everyday Applications: Thermometers operate on the concept of different phases of matter. Some thermometers utilise the triple point of Gallium, as its value (302.918 K) is close to room temperature, making it ideal for medical thermometers. - Scientific Applications: In scientific realms, establishing a consistent temperature for calibration purposes is critical. To ensure consistency around the world, International Temperature Scale of 1990 leveraged the triple point of water, defining it as 273.16 K. This triple point remains an integral standard for calibration in laboratories. Here is a table that summarises some of the standard substances and their triple points: Substance Triple Point Temperature Triple Point Pressure Water (H 2 O)273.16 K 611.657 Pa Gallium (Ga)302.918 K 0.1 MPa Helium-4 (He)2.18 K 5.04 kPa Oxygen (O 2)54.35 K 1.14 kPa Remember, different substances will exhibit a triple point at distinct temperature and pressure conditions, despite all being subjected to the same laws of thermodynamics. Access millions of flashcards designed to help you ace your studies Sign up for free Practical Illustrations of the Critical Point The concept of the critical point extends far beyond textbook definitions and into tangible, real-world applications. A prime example lies in the fascinating world of supercritical fluids. As mentioned earlier, substances become these fluids beyond their critical point, where gas and liquid states become indistinguishable. Let's consider some practical instances: - Supercritical fluid extraction: This technique is widely used in industries like food processing and pharmaceuticals. For example, it's used to decaffeinate coffee, where supercritical carbon dioxide acts as a solvent to extract caffeine from the beans, minus any residual taste. - Power Generation: Supercritical water is used in power plants to increase the thermodynamic efficiency of electricity generation. An illustration is the supercritical water-cooled reactor in nuclear power plants, which operates above the critical point of water, ensuring it remains liquid, not steam, hence optimising energy transfers. A comparative summary of different substances and their critical points: Substance Critical Point Temperature Critical Point Pressure Water (H 2 O)647.096 K 22.064 MPa Carbon Dioxide (CO 2)304.18 K 7.38 MPa Helium (He)5.2 K 0.227 MPa Ammonia (NH 3)405.5 K 11.3 MPa Comparative Study between Examples of Triple Point and Critical Point The examples and applications of both the triple point and critical point are plentiful, each playing significant roles in diverse fields. In simple terms, the differences in their applications arise mainly due to their variance in physical conditions. The triple point sees broad usage in calibration processes and in understanding the transitions between different phases. The critical point, on the other hand, offers utility in tuning the properties of chemical processes, substance purification, and power generation. Among the many substances available, water provides the most illustrative example as it's most familiar in daily life. The real excitement in studying these phenomena is applying them in various scientific and industrial pursuits. Remember, these examples and applications only offer a glimpse into the extensive world of thermodynamics. As you continue to explore, you will encounter countless intriguing facets and niche applications relevant to both the triple point and the critical point. Find relevant study materials and get ready for exam day Sign up for free Triple Point and Critical Point Applications in Engineering Thermodynamics Applications of triple point and critical point are integral in engineering thermodynamics. They are vital in understanding phase transitions and phase equilibria, which are foundational concepts in many engineering fields. These points are related to the states of substances under different pressure and temperature conditions and are useful in modelling and predicting substance behaviour, an essential aspect in engineering design and optimisation. Applications of Triple Point Thermodynamics Triple point plays a fundamental role in various engineering fields, including mechanical, chemical, and metrological engineering. Here are some specific examples: Metrology: The triple point of water is a defined fixed point in the International Temperature Scale of 1990. It services as a primary reference point for the establishment of temperature scales. Phase Transition Studies: In material engineering, understanding the triple point can improve the effectiveness of phase transitions during manufacturing processes. It's also significant in studying solid-solid phase transitions and in simulating specific material behaviour. Thermodynamics and Heat Transfer: Knowledge of the triple point of substances is critical in various thermal management applications, for example, during the analysis and design of refrigeration and air conditioning systems. Practically demonstrating these applications could involve the use of the triple point cell , a sealed glass cell filled with pure water under vacuum conditions. In this setup, all three phases of matter (solid, liquid, gas) coexist at equilibrium, allowing for accurate temperature calibrations. Team up with friends and make studying fun Sign up for free How Critical Point is Used in Thermodynamics? The critical point forms the cornerstone of supercritical fluid applications in thermodynamics. Industrial and engineering fields abundantly employ supercritical fluids due to their unique properties that result from temperatures and pressures beyond the critical point. Supercritical Fluid Extraction: This application is frequent in food processing and pharmaceutical industries. Supercritical carbon dioxide, for instance, is used to extract caffeine from coffee beans and essential oils from plants. Power Generation: In power plants, supercritical water is utilised to increase the thermodynamic efficiency of electricity generation, resulting in reduced fuel consumption and diminished greenhouse gas emissions. Manipulation of Material Properties: The use-cases stretch to material sciences as well. Supercritical drying is applied in the manufacturing of aerogels, superlight materials used in insulation and chemical absorption. To understand better why the critical point is exploited, consider the equation of state that combines gas law, van der Waals forces, and critical parameters: p=R T V−b−a(T)V(V+b) where p is pressure, T is temperature, V is volume, R is the universal gas constant, and a and b are substance-specific constants. These variables help estimate substance behaviour beyond the critical point. The Role of Triple and Critical Points in Engineering Designs The understanding of triple and critical points is indispensable in optimising engineering designs. Thermal Engineering: Triple point data is pivotal in heat transfer applications, including air conditioning systems design and performance, refrigeration, and other phase change systems. Chemical Engineering: The ability to manipulate critical point conditions is essential for processes involving phase separations, distillations and extractions. The properties of supercritical fluids, for instance, bridge the gap between liquid and gaseous phases, allowing for unique applications in chemical synthesis processes and material processing. Materials and Metallurgical Engineering: In metallurgical applications, a complete understanding of phase diagrams, which include the triple and critical points, helps optimise alloy composition and heat treatments. To sum up, thermodynamics and, in particular, triple and critical points are woven into the fabric of many engineering design principles. Their usage spans across industries, enabling engineers to manipulate and optimise properties of substances for various applications. Mastering the Triple Point and Critical Point Formula When you're studying thermodynamics, two terms you'll often encounter are the triple point and the critical point of a substance. While the definitions of these two points can be quite abstract, understanding their mathematical representations, or formulae, can offer more tangible insight. Breakdown of the Triple Point Formula Strictly speaking, there isn't a 'formula' to calculate the triple point of a substance. Typically, the triple point is a known, fixed constant for any given substance. However, it is possible to graphically represent a substance's phase diagram showcasing the triple point. What is crucial to understanding is that the triple point is a condition under which a substance exists in all three phases - solid, liquid, and gas - simultaneously. This condition is defined by a unique combination of temperature (T) and pressure (P) which are specific for each substance. To visualise this, scientists and engineers often use phase diagrams. These graphical representations chart the phase of a substance under varying conditions of temperature and pressure. The phase diagram marks key phase transition points such as the solid-liquid, liquid-gas, and solid-gas lines, with the intersection of these lines representing the triple point. However, keep in mind that the phase diagram and the triple point will vary for different substances. The triple point for water, for example, occurs at 273.16 K and 611.657 Pa, while the triple point for helium occurs at 2.177 K and 5.043 Pa. Even without a specific formula, a keen understanding of the conditions at which the triple point occurs holds great value and applicability in fields such as calibration of thermometers or in materials engineering. Understanding the Critical Point Equation Unlike the triple point, there's a mathematical foundation to understanding the principle behind the critical point. As you may know, the critical point is the temperature and pressure at which the substance's liquid and gas phases become indistinguishable - they merge into a single phase known as the supercritical fluid. Again, the specific critical point varies from one substance to another. It's the point in a phase diagram where the liquid-vapor phase boundary ends. Therefore, like the triple point, it's typically known and tabulated for various substances, rather than calculated via a formula. However, one can form equations of state that describe how pressure, volume, and temperature relate for real gases. Among these, the van der Waals equation is commonly used: P=R T v−b−a v 2 Where: - P represents pressure, - v is the molar volume, - R is the gas constant, - T is the absolute temperature, and - a and b are substance-specific constants. Through the van der Waals equation, you can derive expressions for the critical temperature (T c), critical pressure (P c), and critical volume (v c), thus providing a mathematical approach to understanding and calculating the critical point parameters. Even though these expressions are derived considering an idealised model, they provide a solid starting point for understanding the critical point. In practice, more sophisticated equations of state (such as the Peng-Robinson or Redlich-Kwong models) might be used for more accurate predictions and data. Comparative Analysis of Triple Point and Critical Point Formulae While the triple point doesn't have a formula per se, and the critical point is often determined through equations of state, both are vital in understanding the different phases of a substance under varying temperature and pressure conditions. For any given substance, both the triple point and the critical point act as references in its phase diagram. The triple point represents the unique state at which all three phases coexist. In sharp contrast, the critical point signifies a specific state where traditional gas and liquid phases cease to exist, morphing into a supercritical fluid. Though these points might seem like mere theoretical constructs, they have critical implications in parking procedures such as materials processing and chemical extraction, to name a few. While the mathematical equations (or lack thereof) associated with these points are different, recognizing them for what they represent — unique states of matter under specific conditions — is integral to their practical utility in science, engineering, and technology. Difference between Critical Point and Triple Point In thermodynamics, the triple point and critical point of a substance are both significant, but they represent vastly different physical phenomena. Key Distinguishing Factors between Critical and Triple Point The points at which phase changes occur for a substance vary depending on external conditions like temperature and pressure. Understanding the unique characteristics of these points aids significantly in scientific and industrial applications. It's vital to distinguish between two of these significant points, the triple point and the critical point. The triple point is the specific condition—precise temperature and pressure—at which all three phases (gas, liquid, solid) of a substance co-exist in equilibrium. This unique state is used as a reference point in thermometry—the triple point of water is a defined fixed point in the International Temperature Scale of 1990. On the other hand: A critical point is that unique condition where a substance's gas and liquid phases become indistinguishable from each other—transitioning into a state known as a supercritical fluid. This condition occurs under specific temperatures and pressures, marked by the end of the liquid-vapor phase boundary. Existence of Phases: At the triple point, all three states of matter—solid, liquid, and gas—exist simultaneously. Conversely, beyond the critical point, the substance only exists in a supercritical fluid phase. Unique Temperature and Pressure: Both points denote unique temperature and pressure conditions for each substance. However, the values for the critical and triple points, even for the same substance, are different. Behaviour of Substances: The behaviour of substances at these points is also significantly different. At the triple point, a slight disturbance can lead to a shift from one phase to another. However, at the critical point, gas and liquid phases are indistinguishable, and phase transitions are smoothed out. Among these differences, it's essential to note that although their respective phenomena are distinct, both points are fundamental in thermodynamics and material science for various applications. Importance of Understanding the Difference between Critical and Triple Point Getting to grips with the differences between the critical and triple point can impact one's understanding of substance behaviour under varying temperature and pressure conditions. This knowledge is valuable in industries like power generation, food processing, chemical engineering, and more, affecting process efficiency and product quality. For instance, in a power plant, increasing pressure and temperature to rise beyond the critical point of water can enhance the power generation efficiency. That's because in its supercritical state, water's liquid and gaseous phases become indistinguishable, dramatically increasing its capacity to carry energy. Furthermore, distinguishing between these two points is also key in manipulating material properties: Controlled Substance Phase: With the comprehensive knowledge of triple and critical points, engineers can manipulate these conditions to maintain substances in the desired phase, essential for many manufacturing processes. Material Property Manipulation: In materials science and metallurgy, the triple point and critical point can help scientists create materials with specific properties. For example, supercritical drying—at conditions beyond a substance’s critical point—produces aerogels, lightweight materials with extreme insulating properties. Moreover, these two points are also essential for various applications based on phase equilibrium: Triple Point: Understanding the triple point allows us to control phase transitions to support the manufacturing process. Critical Point: Knowledge of the critical point provides the ability to use supercritical fluids, which are important in diverse processes from industrial cleaning to pharmaceutical production. In conclusion, understanding the significant differences between the critical and triple points not only elucidates the varying states of matter but also creates a foundation for effective commercial and industrial applications. Therefore, this understanding is not just theoretical but holds significant industrial and practical implications. Analysing the Triple Point and Critical Point of Carbon Dioxide Vs Water The triple point and critical point are significant 'markers' in studying the phase behaviour of substances. In this context, we'll analyse the characteristics of these points for two commonly encountered substances: carbon dioxide and water. Understanding the Triple Point of Carbon Dioxide and Water The triple point, as you may have comprehended, is the unique set of conditions—temperature and pressure—at which all three phases (gas, liquid, and solid) of a substance coexist in equilibrium. This point varies for different substances. For water, the triple point is defined at a temperature of 273.16 Kelvin (K) and a pressure of 611.657 Pascals (Pa). This is a fundamental defined point on the International Temperature Scale of 1990. Unlike water, carbon dioxide does not exist in a liquid state at typical pressure levels. It sublimates directly from a solid to a gas under normal conditions. Because of this unique characteristic, carbon dioxide's triple point occurs at much higher pressure than that of water, specifically at a pressure of 5.11 atmospheres (atm) and a temperature of 216.55 K. To summarise, here are the triple points for water and carbon dioxide in a tabular format: Temperature (K)Pressure (Pa)Water 273.16 611.657 Carbon Dioxide 216.55 5.11 101325 (Converted to Pa) Critical Point Analysis: Carbon Dioxide Vs Water In contrast to the triple point, the critical point denotes the highest temperature and pressure at which a substance can exist as a liquid and gas simultaneously. At the critical point, the substance transitions into a state known as a supercritical fluid. Water, for example, has a critical point at a significantly high temperature and pressure—647.096 K and 22.064 MPa, respectively. Beyond these conditions, water exists as a supercritical fluid. On the other hand, carbon dioxide transitions into a supercritical fluid at comparatively lower conditions—304.25 K and 7.38 MPa. This lower critical point allows carbon dioxide to be used extensively in industries, where it can act as a supercritical fluid for processes like decaffeination of coffee beans or dry cleaning. Summarising, here are the critical points for water and carbon dioxide: Temperature (K)Pressure (MPa)Water 647.096 22.064 Carbon Dioxide 304.25 7.38 Comparative Study of Triple Point and Critical Point: Carbon Dioxide Vs Water In comparing these points for water and carbon dioxide, key differences and implications arise. These differences can impact applications from industrial processes to the scientific understanding of matter. For the triple point, water's levels are far more comfortable to attain—273.16 K and 611.657 Pa—when compared to those of carbon dioxide—216.55 K and 5.11 atm. This ease-of-use makes water a common substance used in calibrating thermometers. Moreover, carbon dioxide's relatively lower critical temperature and pressure—304.25 K and 7.38 MPa—makes it feasible to use as a supercritical fluid in different industries, something that the high critical conditions of water—647.096 K and 22.064 MPa—do not permit. To understand the consequential practical implications, here's a comparison of applications: Triple Point Applications: The triple point of water, owing to its easily attainable condition, is used as a thermometric fixed point. On the other hand, the triple point of carbon dioxide isn't as commonly used. Critical Point Applications: The critical point of carbon dioxide, as it is easier to reach, enables its use in processes requiring supercritical fluids, such as decaffeination of coffee or extraction of essential oils. The high critical conditions of water, however, limit its usage in similar applications. Overall, by understanding these points—both triple and critical points—we gain insights that inform our technological innovations and advances across various fields. Triple Point and Critical Point - Key takeaways Triple Point: The unique condition under which a substance exists in all three phases - solid, liquid, and gas - simultaneously. These conditions are defined by a specific combination of temperature (T) and pressure (P) which are specific for each substance. In the International Temperature Scale of 1990, the triple point of water is a defined fixed point. Critical Point: The specific condition under which a substance's gas and liquid phases become indistinguishable, merging into a single phase known as a supercritical fluid. This condition occurs under specific temperatures and pressures, and it marks the end of the liquid-vapor phase boundary in a phase diagram. Applications: Both triple point and critical point have significant importance in various fields. The triple point is widely used in calibration processes and understanding phase transitions, while the critical point is used to tune the properties of chemical processes, substance purification, and power generation. Equations of State: Formulas such as the van der Waals equation can be used to understand and calculate critical point parameters. While there is no specific formula to calculate the triple point, it can be graphically represented using phase diagrams. Difference between Critical Point and Triple Point: The triple point is where all three states of matter exist simultaneously, and a slight disturbance can lead to phase transition. In contrast, at the critical point, gas and liquid phases are indistinguishable and phase transitions are smoothed out. Similar topics in Engineering Materials Engineering Engineering Fluid Mechanics Civil Engineering Electrical Engineering Solid Mechanics Design Engineering Engineering Mathematics What is Engineering Robotics Engineering Professional Engineering Engineering Thermodynamics Aerospace Engineering Chemical Engineering Mechanical Engineering Artificial Intelligence & Engineering Automotive Engineering Design and Technology Audio Engineering Related topics to Engineering Thermodynamics Gibbs Paradox Entropy Generation Clausius Theorem Thermoelectric Entropy of Mixing Thermoelectric Effect Entropy Gradient Entropy Change for Ideal Gas Power Cycle TdS Equation Refrigeration Cycle Partition Function Flow Process Absolute Temperature Carnot Cycle Dead State Availability Exergy Exergy Efficiency Vapor Power System Carnot Vapor Cycle Reheat Rankine Cycle Regenerative Rankine Cycle Feedwater Heater Binary Cycle Economizer Cogeneration Principles of Heat Transfer Diffusion Equation Greenhouse Effect Combined Convection and Radiation Forced Convection Natural Convection Thermodynamic Stability Thermodynamic System Isolated System Principle of Minimum Energy Closed System Thermodynamics Sackur Tetrode Equation Maximum Entropy Negative Temperature Electrical Work Open System Thermodynamic Thermodynamics of Gases Van der Waals Equation Peng Robinson Equation of State Real Gas Internal Energy Equation of State of an Ideal Gas Chemical Potential Ideal Gas Free Expansion of an Ideal Gas Adiabatic Expansion of an Ideal Gas Liquefaction of Gases Third Law of Thermodynamics Unattainability Otto Cycle Metastable Phase Nucleation Osmotic Pressure Continuous Phase Transition Landau Theory of Phase Transition Binary Mixture Triple Point and Critical Point Fluctuations Johnson Noise Fundamentals of Engineering Thermodynamics Thermodynamic State Equations of State Thermodynamic Variables Extensive Property Specific Volume Heat Moles Intensive Property Thermometry Relaxation Time Reversible Process Quasi Static Process Isothermal Process Adiabatic Process Isentropic Efficiency Isentropic Process Isobaric Process Isochoric Process Polytropic Process Isentropic Efficiency of Compressor Thermodynamic Potentials Joule-Thompson Effect Thermodynamic Relations Maxwell Relations Clausius Clapeyron Equation Gibbs Duhem Equation Heat and Work Pressure Volume Work Coefficient of Thermal Expansion Energy Equation Negative Heat Capacity Expansion Free Expansion Joule Kelvin Expansion Adiabatic Expansion Adiabatic Lapse Rate Wien's Law Chemical Energy Nuclear Energy Application of First Law of Thermodynamics Heat Exchanger Diffuser Heat Pump Compressor Throttling Device Thermodynamic Equilibrium Thermodynamic Limit Intensive and Extensive Variables Degrees of Freedom Physics First Law of Thermodynamics Differential Form Equation of State of a Gas Heat Capacity Gases Kinetic Theory of Ideal Gases Internal Energy of a Real Gas Entropy and Irreversibility Potential Energy in Thermodynamics Entropy and Heat Capacity Entropy Function Non Equilibrium State Enthalpy Energy Degradation Stretched Wire Gibbs Free Energy Fundamental Equation Carnot Theorem Conjugate Variables Clausius Inequality Energy Function Helmholtz Free Energy Variable Thermal Conductivity Conduction Mechanism of Heat Transfer Chemical Potential Thermal Diffusivity Convection Steady State Heat Transfer Transient Heat Transfer Heat Equation Linear Heat Conduction Initial Conditions Irreversibility Differential Convection Equations Exothermic Reactions Combustion Engine Thermal Boundary Layer Fins Double Tube Heat Exchanger Enthalpy of Vaporization Endothermic Reactions Exergy Analysis Reciprocating Engine Two Stroke Diesel Engine Four Stroke Engine Rotary Engine Continuous Combustion Engine Bomb Calorimeter Heat Generation Nusselt Number Stirling Engines Combined Cycle Power Plant External Combustion Engine Power Plants Refrigerator Two Stroke Engine Ramjet System Surroundings and Boundary Macroscopic Thermodynamics Temperature Scales Zeroth Law of Thermodynamics Hydrostatic Transmission Reversibility Surface Thermodynamics Nature of Heat First Law of Thermodynamics For Open System Enthalpy of Fusion Flashcards in Triple Point and Critical Point --------------------------------------------- 12 Start learning What is the triple point and how does it differ between carbon dioxide and water? The triple point is the unique set of conditions—temperature and pressure—where all three phases of a substance coexist in equilibrium. For water, it occurs at 273.16 K and 611.657 Pa. For carbon dioxide, it occurs at a much higher pressure of 5.11 atm and a temperature of 216.55 K, due to its unique characteristic of sublimating directly from solid to gas under normal conditions. What is the difference between a substance's triple point and critical point in thermodynamics? The triple point is the specific condition where all three phases (gas, liquid, solid) of a substance co-exist in equilibrium. On the other hand, the critical point is that unique condition where a substance's gas and liquid phases become indistinguishable, transitioning into a supercritical fluid. What is the triple point and how is it represented graphically? The triple point is a condition under which a substance exists in all three phases - solid, liquid, and gas - simultaneously. It is graphically represented on a phase diagram, where the solid-liquid, liquid-gas, and solid-gas lines intersect - this intersection signifies the triple point. The triple point varies for different substances. What is the definition of the triple point in thermodynamics? In thermodynamics, the triple point of a substance is the unique set of conditions at which all three phases - gas, liquid, and solid - coexist in equilibrium. For water, this occurs at specific pressure and temperature. What does the critical point in thermodynamics refer to? The critical point in thermodynamics refers to the highest temperature and pressure at which a substance can exist as a gas and a liquid in equilibrium. Beyond this point, gas and liquid states become indistinguishable, and we have a supercritical fluid. What are some real-world applications of the triple point concept? One real-world application of the triple point is in thermometers which utilise the triple point of Gallium, as its value is close to room temperature. In science, a consistent temperature for calibration purposes is established using the triple point of water (273.16 K) as an integral standard in laboratories. Learn faster with the 12 flashcards about Triple Point and Critical Point Sign up for free to gain access to all our flashcards. Sign up with Email Already have an account? Log in Frequently Asked Questions about Triple Point and Critical Point What are the Triple Point and Critical Point? Please write in UK English. The triple point is the unique set of conditions at which all three phases of a substance (solid, liquid, gas) coexist in equilibrium. The critical point is the temperature and pressure at which the gas and liquid states of a particular substance become identical and form a singular state. What is the difference between the triple point and the critical point? The triple point of a substance is the unique set of conditions where all three phases (solid, liquid and gas) coexist in equilibrium. The critical point, however, is the temperature and pressure at which the gas and liquid phases of a substance become indistinguishable. What do the Triple Point and Critical Point tell us? Triple Point denotes the unique temperature and pressure at which all three phases (gas, liquid, and solid) of a substance can coexist in equilibrium. The Critical Point signifies the highest temperature and pressure at which a gas and liquid phase can coexist in equilibrium. What are the triple point and critical point of CO2, as per UK English terminology? The triple point of CO2, where it coexists in solid, liquid, and gas phases, is -56.6°C and 5.11 atmospheres pressure. The critical point, where gas and liquid become indistinguishable, is 31.1°C and 73.8 atmospheres pressure. What is the difference between the Triple Point and the Critical Point of water? Please write in UK English. The triple point of water is the unique set of conditions (0.01°C and 611.657 pascals of pressure) at which water can exist simultaneously in all three phases: solid, liquid, and gas. The critical point (374°C and 22.064 megapascals of pressure) is the highest temperature and pressure at which water can exist as a liquid, beyond which it becomes a supercritical fluid. Save Article How we ensure our content is accurate and trustworthy? At StudySmarter, we have created a learning platform that serves millions of students. Meet the people who work hard to deliver fact based content as well as making sure it is verified. Content Creation Process: Lily Hulatt Digital Content Specialist Lily Hulatt is a Digital Content Specialist with over three years of experience in content strategy and curriculum design. 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2351	https://teachy.ai/en/lesson-plan/middle-school-en-US/us-6th-grade/math/percents-introduction-flipped-classroom	Lesson plan of Flipped Classroom Methodology \| Percents: Introduction \| Lesson Plan We use cookies Teachy uses cookies to enhance your browsing experience, analyze site traffic, and improve the overall performance of our website. You can manage your preferences or accept all cookies. Manage preferences Accept all TeachersSchoolsStudents Teaching Materials EN Log In Teachy> Lesson plans> Math> 6th grade> Percents: Introduction Lesson plan of Percents: Introduction Lara from Teachy Subject Math Math Source Original Teachy Original Teachy Topic Percents: Introduction Percents: Introduction Objectives (5 - 7 minutes) To introduce students to the concept of percents and their application in everyday life. This will involve a brief discussion on the meaning of percents, its symbol (%), and how it represents parts of a whole. To familiarize students with the process of converting percents to decimals and fractions, and vice versa. This will be an essential skill in all future lessons on percents and will help students to understand the relationship between these numerical forms. To enable students to apply their understanding of percents in real-world contexts. This objective will be achieved through various interactive activities and problem-solving exercises. Secondary Objectives: To improve students' mathematical vocabulary related to percents. To enhance students' problem-solving skills by engaging them in hands-on activities related to percents. To foster a proactive learning environment where students are encouraged to explore and discuss mathematical concepts in a collaborative manner. Introduction (10 - 12 minutes) Review of Necessary Concepts: The teacher will start by briefly revisiting the concepts of fractions and decimals, which the students have previously learned. The students will be reminded of the relationship between these numerical forms and how they represent parts of a whole. (3 minutes) Problem Situations to Spark Curiosity: The teacher will then present two problem situations to the students. The first one could be a real-world scenario like "If a store offers a 20% discount on a $50 item, how much will the customer pay?" The second could be a mathematical problem like "What is 25% of 80?" These problems will serve to stimulate the students' interest in the upcoming topic and highlight the practical applications of percents. (4 minutes) Contextualizing the Importance of Percents: The teacher will explain the importance of percents in everyday life. They will mention how percents are used in various fields like finance, sales, statistics, and even in our personal lives. For instance, understanding percents can help us make informed decisions about discounts, interest rates, and savings. (2 minutes) Attention-Grabbing Introduction: To pique the students' curiosity, the teacher will share a couple of interesting facts related to percents. The first fact could be about the origin of the word 'percent,' which comes from the Latin word 'per centum,' meaning 'by the hundred.' The second fact could be about how percents are used in sports statistics, such as a player's shooting percentage in basketball or a team's winning percentage in baseball. These facts will engage the students and demonstrate the ubiquity of percents in our daily lives. (3 minutes) Development Pre-Class Activities (10 - 15 minutes): Reading Assignment: The students will be provided with a short, simplified reading material that explains the concept of percents. This reading material will define percents, explain how to convert percents to decimals and fractions, and provide some real-world examples of how percents are used. The students will be encouraged to take notes and highlight important concepts while reading. They will also be asked to write down any questions or doubts that arise during the reading for clarification in the classroom. Video Resource: In addition to the reading material, the students will be given a link to an educational video that further explains the concept of percents. This video will use visuals and animations to make the concept more engaging and understandable. In-Class Activities (15 - 20 minutes): Activity 1: 'Percents in the News' Role Play: The teacher will divide the students into small groups and provide each group with a different newspaper article. These articles will contain information about a recent sale, a sports event, or a population survey, which the students will use to identify the percents mentioned. The groups will then have to convert these percents into decimals and fractions and present their findings to the class. This activity will help students to understand the real-world application of percents and develop their skills of percent conversion. (10 minutes) Activity 2: 'Percent Bingo': The teacher will provide each student with a 'percent bingo' card, which is a grid containing different percents (ranging from 0 to 100%). The teacher will then call out different fractions and decimals, and the students will have to convert these into percents and mark them on their cards. The first student to complete a row or column on their card will win. This game will make the process of percent conversion fun and exciting, helping the students to reinforce their understanding of the topic. (5 minutes) Activity 3: 'Percent Problem-Solving Challenge': The teacher will present the students with a set of percent-related problem situations. Each group will have to discuss and solve these problems within a given time. The group that solves the maximum number of problems correctly will be declared the 'Percent Problem-Solving Champions'. This activity will help students to apply their understanding of percents in different scenarios and enhance their problem-solving skills. (5 minutes) Activity 4: 'Percent Gallery Walk': The teacher will prepare a number of posters containing various real-world scenarios where percents are used. These could include advertisements with discounts, health reports with statistics, or financial reports with interest rates. The students will walk around the room, observe these posters, and write down the percents they see. This activity will help students to visualize the application of percents in everyday life and encourage discussion and interaction. (5 minutes) These activities will not only help students to grasp the concept of percents but also make their learning experience enjoyable and interactive. They will get a chance to collaborate with their peers, apply their knowledge, and improve their problem-solving skills. The teacher will circulate around the classroom, providing guidance, and clarifying doubts as needed. Feedback (8 - 10 minutes) Group Discussions (3 minutes): The teacher will ask each group to share their solutions or conclusions from the activities. Each group will have up to 2 minutes to present their findings. This will not only give the students an opportunity to articulate their understanding but also allow the teacher to assess the students' comprehension of the topic and skills. Connecting Practice to Theory (2 minutes): After each group has shared their solutions, the teacher will facilitate a class discussion to connect the outcomes of the activities with the theoretical knowledge of percents. The teacher will highlight how the activities relate to the real-world applications of percents and the process of converting percents to decimals and fractions. This discussion will help the students see the practical relevance of what they have learned and solidify their understanding of the topic. Reflection (3 - 5 minutes): The teacher will then encourage the students to reflect on what they have learned in the lesson. The teacher will pose questions like: What was the most important concept you learned today? Which questions have not yet been answered? How would you explain the concept of percents to a friend or family member? The students will be given a minute to think about these questions and then share their thoughts with the class. This reflection will help the students consolidate their learning, identify any areas of confusion or curiosity, and develop their ability to articulate mathematical concepts. Summarizing the Lesson (1 minute): To conclude the feedback session, the teacher will summarize the main points of the lesson. They will highlight the definition of percents, the process of converting percents to decimals and fractions, and the real-world applications of percents. The teacher will also remind the students of the importance of percent in everyday life and encourage them to continue exploring the topic in their own time. The feedback stage is crucial as it not only provides an opportunity for the students to share and reflect on their learning but also allows the teacher to assess the effectiveness of the lesson and make necessary adjustments for future lessons. It promotes a two-way communication between the teacher and the students, fostering a collaborative and interactive learning environment. Conclusion (5 - 7 minutes) Lesson Recap (2 minutes): The teacher will summarize the key points covered in the lesson. They will remind the students of the definition of percents, the relationship between percents, decimals, and fractions, and the process of converting between them. The teacher will also recap the real-world applications of percents discussed during the lesson, emphasizing how understanding percents can help in various contexts, from shopping to sports to financial decisions. Connecting Theory, Practice, and Applications (2 minutes): The teacher will explain how the lesson integrated theoretical knowledge with practical activities and real-world examples. They will highlight how the pre-class activities, such as the reading assignment and the video, laid the foundation for the in-class activities, like the 'Percents in the News' role play and the 'Percent Bingo' game. The teacher will also emphasize how the problem-solving exercises and the 'Percent Gallery Walk' activity allowed the students to apply their theoretical knowledge in practical contexts. This discussion will help the students understand the value of a well-rounded learning experience that combines theory, practice, and applications. Suggested Additional Materials (1 - 2 minutes): The teacher will then suggest some additional resources for the students to further their understanding of percents. These can include math websites with interactive games and exercises on percents, educational apps that provide a fun way to practice percent conversions, and books that delve deeper into the concept of percents and their applications. The teacher will encourage the students to explore these resources at their own pace and ask any questions that arise in the next class. Real-World Relevance (1 minute): Finally, the teacher will reiterate the importance of the topic for everyday life. They will remind the students of the various real-world applications of percents discussed during the lesson, and how understanding percents can empower them to make informed decisions in different contexts. The teacher will emphasize that math is not just a subject to be studied in school, but a tool that can be used to solve problems and make sense of the world around us. In conclusion, the teacher will stress that understanding percents is a fundamental skill that will be used not only in their math classes but also in many aspects of their future life. They will also remind the students that learning is a continuous process, and they should feel free to explore and ask questions about the topic in the next class and beyond. Need more materials to teach this subject? I can generate slides, activities, summaries, and over 60 types of materials. That's right, no more sleepless nights here :) Explore free materials Users who viewed this lesson plan also liked... 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2352	https://math.stackexchange.com/questions/3198313/a-thorough-explanation-on-why-division-by-zero-is-undefined	Skip to main content A thorough explanation on why division by zero is undefined? [duplicate] Ask Question Asked Modified 5 years, 7 months ago Viewed 10k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. A Quick Note I know there is a slough of related questions on stack exchange, but none of them really seem to answer my question. This post is the closest in relationship to my question, but the answer simply expresses a high level mathematical explanation, and not an example I can teach my kids. Growing up, my school always taught that division by zero was undefined or not allowed, but never really explained why, or how this was true. The proposed duplicate has a very good answer, that I understand, however I'm not so sure my kids would understand that answer. The accepted answer will have to be understood by children under age 10 with a minimal working knowledge of multiplication and division. Getting Started The other day, I was working on a project at home in which I performed division by zero with a double precision floating point number in my code. This isn't always undefined in the computer world and can sometimes result in ∞. The reason for this is clearly explained in IEEE 754 and quite thoroughly in this Stackoverflow post: Division by zero (an operation on finite operands gives an exact infinite result, e.g., 10 or log0) (returns ±∞ by default). Now, this got me thinking about basic arithmetic and how to prove each operation, and I created a mental inconsistency between multiplication and division. Multiplication As this is an important part of the thought process that lead me down this mental rabbit hole, I am including the elementary explanation of multiplication. If I place 10 marbles on my desk, 3 times, I have placed 30 marbles on my desk. This is expressed as 10⋅3=30 and is true. If I place 10 marbles on my desk, 0 times, I have placed 0 marbles on my desk. This is expressed as 10⋅0=0 and is true. These two scenarios are true no matter what numbers are used. Division This is where things took an unexpected turn in my mind. Let's say that I am a wandering saint and I have 50 apples. I want to help the hungry people of the world so I give my apples away freely. Now, let's handle two similar scenarios. I come across 10 people, and I want to give them all of my apples, I also want to ensure that each person receives the same number of apples. With 50 apples to disperse across 10 people, this means each person receives 5 apples. This is expressed as 5010=5 and is true. However, let's say I have the same 50 apples, and I come across a town where no one is hungry, and no one wants my apples. Well, I have 50 apples, and I have 0 people to give them to, so I still have 50 apples. I didn't disperse my apples evenly across any number of people, so it's still the same bag of 50 apples. I believe this may be my mind's way of bending the facts here, and that I've convinced myself that I'm dividing 50 zero times, but in fact I may have divided 50 one time (by me). But it has me thinking, if I divide a pizza into zero equal slices, well then I essentially didn't slice the pizza and thus still just have an entire pizza. My Question How can it be proved thoroughly, not just with math, but with an example explanation (understandable by children) that division by zero is truly undefined? arithmetic Share CC BY-SA 4.0 Follow this question to receive notifications edited Apr 23, 2019 at 14:46 Taco asked Apr 23, 2019 at 14:13 TacoTaco 47911 gold badge44 silver badges1515 bronze badges 9 3 The explanations at this duplicate (linked to the above) give a thorough explanation, right? It has so many examples in the answers, too. This problem really has been explained so many times, with so many good answers. Dietrich Burde – Dietrich Burde 04/23/2019 14:20:54 Commented Apr 23, 2019 at 14:20 2 In your apple scenario, you are actually dividing 50 apples by 1 (giving them all to yourself). Here are some thoughts I typed up, which may be helpful: ee.usc.edu/stochastic-nets/docs/divide-by-zero.pdf You may like section F best , which talks intuitively about a photocopy machine shrinking/enlarging and the concept of invertible/non-invertible operations. Michael – Michael 04/23/2019 14:21:26 Commented Apr 23, 2019 at 14:21 1 You have tricked yourself with your apples/pizza. If you were divvying up the 50 apples to 10 people, then you would have no apples left over. Similarly, if you came across 25 people, you'd have none left over (and each other person would have 2). The whole scenario ignores you, and assumes that you have none left over. If you take a fair share instead, then you are dividing it by a number one greater. To model division by zero in that case, even you yourself would disappear, and so would the remaining apples. Where did they go? Theo Bendit – Theo Bendit 04/23/2019 14:35:10 Commented Apr 23, 2019 at 14:35 1 @Somos : Based on reading her post, I believe she wants the answer of 1/0 to be 1, and 50/0 to be 50, based on her pizza scenario, but she also knows that is not quite right. PerpetualJ : Before getting intuition about dividing 50 pizzas by 0 (which is not defined) you might try intuition on dividing 50 pizzas by 1/2 (which is defined). Michael – Michael 04/23/2019 16:07:08 Commented Apr 23, 2019 at 16:07 1 Re: Getting Started I don't understand why you would think a machine built by humans determines human conventions. I have never understood explaining to a student why 1/0 is nonsense, and then they punch it into their calculator to confirm. What! I've never understood such muddleheadness. Allawonder – Allawonder 04/23/2019 17:28:18 Commented Apr 23, 2019 at 17:28 \| Show 4 more comments 5 Answers 5 Reset to default This answer is useful 12 Save this answer. Show activity on this post. That division by zero is undefined cannot be proven without math, because it is a mathematical statement. It's like asking "How can you prove that pass interference is a foul without reference to sports?" If you have a definition of "division", then you can ask whether that definition can be applied to zero. For instance, if you define division such that x÷y means "Give the number z such that y⋅z=x", there is no such number in the standard real number system for y=0. If we're required to have that (x÷y)⋅y=x, then that doesn't work when y is equal to zero. In computer languages where x/0 returns an object for which multiplication is defined, you do not have that (x\0)0 == x. So we can can a class of objects in which we call one of the objects "zero", and have a class method such that "division" by "zero" is defined, but that class will not act exactly like the real numbers do. Another definition of division is in terms of repeated subtraction. If you take 50 apples and give one apple each to 10 people, then keep doing that until you run out of apples, each person will end up with 5 apples. You're repeatedly subtracting 10 from 50, and you can do that 5 times. If you try to subtract 0 from 50 until you run out of apples, you'll be doing it an infinite number of times. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Apr 23, 2019 at 17:05 AcccumulationAcccumulation 12.9k33 gold badges1717 silver badges2626 bronze badges 2 3 The example following glorified subtraction as is given with glorified addition for multiplication was perfect! Being able to explain why division by zero is undefined requires breaking the process down into terms they can understand. I can see their counter argument coming back as “well, you still have 50 apples” though. +1 Taco – Taco 04/23/2019 17:11:30 Commented Apr 23, 2019 at 17:11 @PerpetualJ, I left a similar comment before I realized there was already an answer to this effect. :( In any case, I think you're hung up on the people and possession. If you give 10 people 50 apples, the apples still exist. Maybe think about baskets instead? "If I have 50 apples, I can put 1 apple into 10 baskets 5 times." I still have 50 apples; they're just in baskets now. "If I have 50 apples, I can put 1 apple into 0 baskets infinite times." Either way, I still have 50 apples. D. Patrick – D. Patrick 04/23/2019 18:25:55 Commented Apr 23, 2019 at 18:25 Add a comment \| This answer is useful 6 Save this answer. Show activity on this post. Consider a problem where you have to divide a finite number like 5 by zero. 5÷0 is essentially a request for some number which when multiplied by zero gives you 5: 5÷0=N⟹0⋅N=5. Is there a number that when multiplied by zero gives you 5? The answer is clearly no because any number times zero always gives you zero. Therefore, 5÷0 is left undefined. "Undefined" here basically means that we can't explain what 5÷0 really means. What about the case 0÷0? 0÷0=N⟹0⋅N=0. We know that any number times zero is zero. This means that N can be any number at all. This kind of division problem gives you an infinite number of answers instead of just one as it should be. Because of this indeterminateness, 0÷0 is also left undefined. Here's another very simple example for good measure. You have 7 pizzas and you want to divide them among zero people. How much pizza will each person get? Well, you have no people to give the pizzas to. You can pose that question and even write it mathematically as 7÷0, but what could possibly be the answer to this question? Practically speaking, this is unanswerable. In other words, it's not clear what the statement 7÷0 in this context means. In math-speak, we would say that this is undefined. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Feb 2, 2020 at 15:13 answered Apr 23, 2019 at 15:27 Michael RybkinMichael Rybkin 6,72122 gold badges1313 silver badges2727 bronze badges 3 Given the part of your answer that talks about 0÷0, shouldn't 00 be undefined, too? I mean, for whatever value k that you choose, 00=0k−k=0k÷0k=0÷0 so... Mr Pie – Mr Pie 04/29/2019 23:48:31 Commented Apr 29, 2019 at 23:48 1 How do you define 0k? If k>0, then 0k=0. In that case, you would be attempting to divide by zero (0k÷0k) which already has been left undefined. I don't think you can figure out what 00 should be equal to this particular way. For a more thorough explanation, please consult the Wikipedia page that talks about raising zero to the zeroth power: en.wikipedia.org/wiki/Zero_to_the_power_of_zero Michael Rybkin – Michael Rybkin 04/30/2019 00:00:06 Commented Apr 30, 2019 at 0:00 A google search and my calculator both reveal that 00=1, but yeah, I'll take a look at the page; I think it is a good idea to look into it a bit more. Thanks for that! Mr Pie – Mr Pie 04/30/2019 00:17:10 Commented Apr 30, 2019 at 0:17 Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. My understanding of division by zero goes back to the definition of rings. Let R be a commutative ring and a,b∈R with b a unit in R. Then define the fraction a/b as follows: ab=a⋅b−1 i.e., division by b is defined by multiplication with the inverse of b. Since the zero element 0 in a ring is absorbing (i.e., a⋅0=0=0⋅a) and thus not a unit, division by 0 is not defined. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Apr 23, 2019 at 14:18 WuestenfuxWuestenfux 21.3k22 gold badges1616 silver badges2424 bronze badges 4 6 The question clearly states that the explanation is targeted towards children. user1952500 – user1952500 04/23/2019 14:25:47 Commented Apr 23, 2019 at 14:25 4 This is the first time I have seen someone explain division by zero by starting out "Let R be a commutative ring..." Michael – Michael 04/23/2019 14:26:16 Commented Apr 23, 2019 at 14:26 @Michael: In the algebraic sense, division is a derived operation as is subtraction. Wuestenfux – Wuestenfux 04/24/2019 12:11:35 Commented Apr 24, 2019 at 12:11 @Wuestenfux : Your answer, and your comment to me, both seem misaligned with their intended audience. Michael – Michael 04/24/2019 19:35:31 Commented Apr 24, 2019 at 19:35 Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. I explain it like this, ignoring the definition gap on purpose: Have a look at the graph of x/x: it's a straight line at y=1. In this graph, we clearly see that 0/0=1. Then, look at 5x/x: it's a straight line at y=5. We clearly see that 5∗0/0=5. Now this could be interpreted as (5∗0)/0=0/0=1 (using the result of the graph before) or as 5∗(0/0)=5 (also using the result from before). You can repeat this with other numbers as well, so the children can see that the result is arbitrary. This should make it pretty clear that if we allow division by zero, other laws cannot hold. So it's better undefined. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Apr 23, 2019 at 16:59 Thomas WellerThomas Weller 35422 silver badges1313 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. I think you've pretty much got it and others have gone into more detail on the various bits of maths related to this. The simplest way to describe this for kids I can think of: Anything multiplied by zero is zero. Easy. Multiplying anything with two non-zero values gives a non-zero value. Shouldn't be a problem. Division of the result by one of those values gives you the other. We did the opposite of 2. So 3 X 4 = 12, 12 / 4 gives you 3. 3 was the number you multiplied by 4 to get 12. We have a result 12, we ask "12 / 0 what was the other value multiplied by 0 to get 12"? There is no such number because of statement 1 - hence divide by zero is undefined. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Apr 23, 2019 at 17:47 cyborgcyborg 12122 bronze badges 1 Addendum: the basic mistake in the question is that how you get the answer - adding, subtracting or anything else - the process or algorithm of it - does not define the result. As different algorithms give different answers that shows the point. From a purely functional perspective where there are no algorithms this is the most sensible mapping for the inverse of the multiplication function. cyborg – cyborg 04/23/2019 18:23:51 Commented Apr 23, 2019 at 18:23 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions arithmetic See similar questions with these tags. 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2353	https://en.wikipedia.org/wiki/Eternal_dominating_set	Jump to content Eternal dominating set Русский Edit links From Wikipedia, the free encyclopedia Object in graph theory Unsolved problem in mathematics Does there exist a graph G such that γ(G) equals γ∞(G) and γ(G) is less than the clique covering number of G? More unsolved problems in mathematics In graph theory, an eternal dominating set for a graph G = (V, E) is a subset D of V such that D is a dominating set on which mobile guards are initially located (at most one guard may be located on any vertex). The set D must be such that for any infinite sequence of attacks occurring sequentially at vertices, the set D can be modified by moving a guard from an adjacent vertex to the attacked vertex, provided the attacked vertex has no guard on it at the time it is attacked. The configuration of guards after each attack must induce a dominating set. The eternal domination number, γ∞(G), is the minimum number of vertices possible in the initial set D. For example, the eternal domination number of the cycle on five vertices is three. The eternal dominating set problem, also known as the eternal domination problem and the eternal security problem, can also be interpreted as a combinatorial game played between two players that alternate turns: a defender, who chooses the initial dominating set D and the guard to send to each attack that occurs at a vertex without a guard; and an attacker, who chooses the vertex to be attacked on their turn. The attacker wins the game if they can ever choose a vertex to be attacked such that there is no guard on that vertex or a neighboring vertex; the defender wins otherwise. In other words, the attacker wins the game if they can ever attack a vertex such that the attack cannot be defended. As noted in Klostermeyer & Mynhardt (2015b), the eternal dominating set problem is related to the k-server problem in computer science. History [edit] Motivated by ancient problems in military defense described in the series of papers Arquilla & Fredricksen (1995), ReVelle & Rosing (2000), and Stewart (1999), the eternal domination problem was initially described in 2004 in a paper by Burger et al. (2004). That was followed by the publication of a paper on eternal domination by Goddard, Hedetniemi & Hedetniemi (2005), which also introduced a variation on the problem called m-eternal domination in which all of the guards are allowed to move to adjacent vertices, if they so desire, in response to an attack, so long as one guard moves to the attacked vertex (assuming there was not a guard on the attacked vertex, otherwise no guard needs to move). Subsequent to the Goddard, Hedetniemi & Hedetniemi (2005) paper, a number of papers by other authors appeared in the mathematical literature. In these subsequent papers, several additional variations on the eternal domination problem were proposed including the eternal vertex cover problem, the eternal independent set problem, eternal total dominating sets, eternal connected dominating sets, and eternal dominating sets in the eviction model (the latter model requires that when attacks occur a vertex with a guard and the guard must move to a neighboring vertex that contains no guard, if one exists). A survey paper describing many of the results on the eternal domination problem and many of the variations of the problem can be found at Klostermeyer & Mynhardt (2015b). Bounds [edit] Let G be a graph with n ≥ 1 vertices. Trivially, the eternal domination number is at least the domination number γ(G). In their paper, Goddard, Hedetniemi, and Hedetniemi proved the eternal domination number is at least the independence number of G and at most the clique covering number of G (the clique covering number of G is equal to the chromatic number of the complement of G). Therefore, the eternal domination number of G is equal to the clique covering number of G for all perfect graphs, due to the perfect graph theorem. It has been shown that the eternal domination number of G is equal to the clique covering number of G for several other classes of graph, such as circular-arc graphs (as proven in Regan (2007)) and series-parallel graphs (as proven in Anderson et al. (2007)). Goddard, Hedetniemi, and Hedetniemi also demonstrated a graph in which the eternal domination number of the graph is less than the clique covering number. It was proven by Klostermeyer & MacGillivray (2007) that the eternal domination number of a graph with independence number α is a most α(α + 1)/2. Goldwasser & Klostermeyer (2008) proved that there are infinitely many graphs where the eternal domination number is exactly α(α + 1)/2. Bounds on the m-eternal domination number [edit] Goddard, Hedetniemi, and Hedetniemi proved the m-eternal domination number, denoted γm∞(G), is at most the independence number of G. Hence, the eternal domination parameters fit nicely into the famous domination chain of parameters, see (Haynes, Hedetniemi & Slater 1998a), as follows: : γ(G) ≤ γm∞(G) ≤ α(G) ≤ γ∞(G) ≤ θ(G) where θ(G) denotes the clique-covering number of G and γ∞(G) denotes the eternal domination number. An upper bound of ⌈n/2⌉ on γm∞(G) for graphs with n vertices was proven in Chambers, Kinnersly & Prince (2006), see also Klostermeyer & Mynhardt (2015b). The m-eternal domination number in grid graphs has attracted attention, inspired by attention given to the domination number of grid graphs, see Haynes, Hedetniemi & Slater (1998a) and Goncalves et al. (2011). The m-eternal domination number in grid graphs was first studied in Goldwasser, Klostermeyer & Mynhardt (2013) where it was shown that : γm∞ = ⌈2n/3⌉ for the 2 by n grid with n ≥ 2 and : γm∞ ≤ ⌈8n/9⌉ for 3 by n grids. The latter was improved in Finbow, Messinger & van Bommel (2015) to : 1 + ⌈4n/5⌉ ≤ γm∞ ≤ 2 + ⌈4n/5⌉ when n ≥ 11. This bound was subsequently slightly improved in Messinger & Delaney (2015) in some cases. Finally, the bounds were closed in Finbow & van Bommel (2020), where it was shown that : γm∞ = ⌈(4n+7)/5⌉ for n ≥ 22. The cases for 4 by and grids and 5 by n grids were considered in Beaton, Finbow & MacDonald (2013) and van Bommel & van Bommel (2016), respectively. Braga, de Souza & Lee (2015) proved that γm∞ = α for all proper interval graphs and the same authors also proved, see Braga, de Souza & Lee (2016), that there exists a Cayley graph for which the m-eternal domination number does not equal the domination number, contrary to the claim in Goddard, Hedetniemi & Hedetniemi (2005). Open questions [edit] According to Klostermeyer & Mynhardt (2015b), one of the main unsolved questions is the following: Does there exist a graph G such that γ(G) equals the eternal domination number of G and γ(G) is less than the clique covering number of G? Klostermeyer & Mynhardt (2015a) proved that any such graph must contain triangles and must have maximum vertex degree at least four. Similar to Vizing's conjecture for dominating sets, it is not known whether for all graphs G and H The analogous bound is known not to hold for all graphs G and H for the m-eternal domination problem, as shown in Klostermeyer & Mynhardt (2015a). Two fundamental open questions on eternal domination are listed by Douglas West at . Namely, whether γ∞(G) equals the clique covering number for all planar graphs G and whether γ∞(G) can bounded below by the Lovász number, also known as the Lovász theta function. A number of other open questions are stated in the survey paper Klostermeyer & Mynhardt (2015b), including many questions on the variations of eternal dominating sets mentioned above. References [edit] Anderson, M.; Barrientos, C.; Brigham, R.; Carrington, J.; Vitray, R.; Yellen, J. (2007), "Maximum demand graphs for eternal security", J. Combin. Math. Combin. Comput., 61: 111–128. Arquilla, H.; Fredricksen, H. (1995), "Graphing an optimal grand strategy", Military Operations Research, 1 (3): 3–17, doi:10.5711/morj.1.3.3, hdl:10945/38438, JSTOR 43940682. Beaton, I.; Finbow, S.; MacDonald, J. (2013), "Eternal domination set problem of grids", J. Combin. Math. Combin. Comput., 85: 33–38. Braga, A.; de Souza, C.; Lee, O. (2015), "The eternal dominating set problem for proper interval graphs", Information Processing Letters, 115 (6–8): 582–587, doi:10.1016/j.ipl.2015.02.004. Braga, A.; de Souza, C.; Lee, O. (2016), "A note on the paper "Eternal security in graphs" by Goddard, Hedetniemi, and Hedetniemi (2005)", Journal of Combinatorial Mathematics and Combinatorial Computing, 96: 13–22. Burger, A.P.; Cockayne, E.J.; Grundlingh, W.R.; Mynhardt, C.M.; van Vuuren, J.; Winterbach, W. (2004), "Infinite order domination in graphs", J. Combin. Math. Combin. Comput., 50: 179–194. Chambers, E.; Kinnersly, B.; Prince, N. (2006), "Mobile eternal security in graphs", Unpublished Manuscript, archived from the original on 2015-09-30, retrieved 2015-02-21. Finbow, S.; Messinger, M-E.; van Bommel, M. (2015), "Eternal domination in 3 x n grids", Australas. J. Combin., 61: 156–174. Finbow, S.; van Bommel, M.F. (2020), "The eternal domination number for 3 x n grid graphs", Australas. J. Combin., 71: 1–23. Goddard, Wayne; Hedetniemi, Sandra M.; Hedetniemi, Stephen T. (January 2005). "Eternal Security in Graphs". Journal of Combinatorial Mathematics and Combinatorial Computing. 52. Goldwasser, J.; Klostermeyer, W. (2008), "Tight bounds for eternal dominating sets in graphs", Discrete Math., 308 (12): 2589–2593, doi:10.1016/j.disc.2007.06.005. Goldwasser, J.; Klostermeyer, W.; Mynhardt, C. (2013), "Eternal protection in grid graphs", Utilitas Math., 91: 47–64. Goncalves, D.; Pinlou, A.; Rao, M.; Thomasse, S. (2011), "The domination number of grids", SIAM Journal on Discrete Mathematics, 25 (3): 1443–1453, arXiv:1102.5206, doi:10.1137/11082574. Haynes, Teresa W.; Hedetniemi, Stephen; Slater, Peter (1998a), Fundamentals of Domination in Graphs, Marcel Dekker, ISBN 0-8247-0033-3, OCLC 37903553. Klostermeyer, W.; MacGillivray, G. (2007), "Eternal security in graphs of fixed independence number", J. Combin. Math. Combin. Comput., 63: 97–101. Klostermeyer, W.; Mynhardt, C. (2015a), "Domination, Eternal Domination, and Clique Covering", Discuss. Math. Graph Theory, 35 (2): 283, arXiv:1407.5235, doi:10.7151/dmgt.1799. Klostermeyer, W.; Mynhardt, C. (2015b), "Protecting a graph with mobile guards", Applicable Analysis and Discrete Mathematics, 10: 21, arXiv:1407.5228, doi:10.2298/aadm151109021k. Messinger, M-E.; Delaney, A. (2015), Closing the gap: Eternal domination on 3 x n grids. Regan, F. (2007), Dynamic variants of domination and independence in graphs, Rheinischen Friedrich-Wilhlems University. ReVelle, C. (2007), "Can you protect the Roman Empire?", Johns Hopkins Magazine, 2. ReVelle, C.; Rosing, K. (2000), "Defendens Imperium Romanum: A classical problem in military strategy", Amer. Math. Monthly, 107 (7): 585–594, doi:10.2307/2589113, JSTOR 2589113. Stewart, I. (1999), "Defend the Roman Empire!", Scientific American, 281 (6): 136–138, Bibcode:1999SciAm.281f.136S, doi:10.1038/scientificamerican1299-136. van Bommel, C.; van Bommel, M. (2016), "Eternal domination number of 5 x n grids", J. Combin. Math. Combin. Comput, 97: 83–102. Retrieved from " Category: Graph theory objects Hidden categories: Articles with short description Short description matches Wikidata
2354	https://pmc.ncbi.nlm.nih.gov/articles/PMC2910131/	Paramyxovirus Assembly and Budding: Building Particles that Transmit Infections - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Published in final edited form as: Int J Biochem Cell Biol. 2010 Apr 14;42(9):1416–1429. doi: 10.1016/j.biocel.2010.04.005 Search in PMC Search in PubMed View in NLM Catalog Add to search Paramyxovirus Assembly and Budding: Building Particles that Transmit Infections Megan S Harrison Megan S Harrison a Department of Veterinary and Biomedical Sciences, and Center for Molecular Immunology and Infectious Disease, The Pennsylvania State University, University Park, PA 16802, United States Find articles by Megan S Harrison a, Takemasa Sakaguchi Takemasa Sakaguchi b Department of Virology, Graduate School of Biomedical Sciences, Hiroshima University, 1-2-3 Kasumi, Minami-ku, Hiroshima 734-8551, Japan Find articles by Takemasa Sakaguchi b, Anthony P Schmitt Anthony P Schmitt a Department of Veterinary and Biomedical Sciences, and Center for Molecular Immunology and Infectious Disease, The Pennsylvania State University, University Park, PA 16802, United States Find articles by Anthony P Schmitt a, Author information Article notes Copyright and License information a Department of Veterinary and Biomedical Sciences, and Center for Molecular Immunology and Infectious Disease, The Pennsylvania State University, University Park, PA 16802, United States b Department of Virology, Graduate School of Biomedical Sciences, Hiroshima University, 1-2-3 Kasumi, Minami-ku, Hiroshima 734-8551, Japan Corresponding Author: Department of Veterinary and Biomedical Sciences, The Pennsylvania State University, 115 Henning Building, University Park, PA 16802, United States, Tel: 814-863-6781, Fax: 814-863-6140, aps13@psu.edu Issue date 2010 Sep. PMC Copyright notice PMCID: PMC2910131 NIHMSID: NIHMS203736 PMID: 20398786 The publisher's version of this article is available at Int J Biochem Cell Biol Abstract The paramyxoviruses define a diverse group of enveloped RNA viruses that includes a number of important human and animal pathogens. Examples include human respiratory syncytial virus and the human parainfluenza viruses, which cause respiratory illnesses in young children and the elderly; measles and mumps viruses, which have caused recent resurgences of disease in developed countries; the zoonotic Hendra and Nipah viruses, which have caused several outbreaks of fatal disease in Australia and Asia; and Newcastle disease virus, which infects chickens and other avian species. Like other enveloped viruses, paramyxoviruses form particles that assemble and bud from cellular membranes, allowing the transmission of infections to new cells and hosts. Here, we review recent advances that have improved our understanding of events involved in paramyxovirus particle formation. Contributions of viral matrix proteins, glycoproteins, nucleocapsid proteins, and accessory proteins to particle formation are discussed, as well as the importance of host factor recruitment for efficient virus budding. Trafficking of viral structural components within infected cells is described, together with mechanisms that allow for the selection of specific sites on cellular membranes for the coalescence of viral proteins in preparation of bud formation and virion release. Keywords: virus budding, virus assembly, matrix protein, paramyxovirus, virus-like particle, polarized budding, lipid raft membranes, viral trafficking 1. Introduction The paramyxoviruses define a large group of enveloped RNA viruses, some of which cause significant human and animal diseases. Examples include human respiratory syncytial virus (HRSV), human parainfluenza virus types 1-4 (hPIV 1-4), measles virus, mumps virus, Nipah virus, Hendra virus, and Newcastle disease virus (NDV). HRSV and hPIV types 1–3 are major contributors to respiratory infections in young children and the elderly (Counihan et al., 2001; Falsey, 1998; Hall et al., 2009; Heilman, 1990; Welliver, 2003). HRSV infection is the leading cause of severe pediatric respiratory tract disease, causing an estimated 64 million cases and 160,000 annual deaths globally (Hall et al., 2009; Welliver, 2003; WHO, 2009a; Wright et al., 2005; Zhang et al., 2002). There are currently no effective vaccines to prevent HRSV or hPIV infections. Although both measles virus and mumps virus infections are vaccine-preventable, these infections remain a health burden in developing countries, and several significant outbreaks attributed to low vaccination rates have occurred recently in the United Kingdom, Canada, and the United States (CDC, 2006a; CDC, 2006b; CDC, 2008; Dayan et al., 2008; Hviid et al., 2008; Peltola et al., 2007). Nipah virus and Hendra virus (Henipaviruses) cause deadly infections in humans, with severe and widespread vasculitis and encephalitis resulting in a mortality rate of about 40% (Bishop and Broder, 2008). The viruses are zoonotic, and the natural hosts are fruit bats, such as flying foxes. The viruses spread to humans mainly through intermediary hosts: horses in the case of Hendra virus, and pigs in the case of Nipah virus (Eaton et al., 2006). Twelve recognized outbreaks of Nipah virus infection in South Asia have occurred since its identification in 1999 (Epstein et al., 2006; Gurley et al., 2007; WHO, 2009) and a total of thirteen known outbreaks of Hendra virus in Australia have occurred, the first of which were recognized in 1994 (Bishop and Broder, 2008; ProMED-mail, 2009). There are currently no effective treatments or vaccines approved for Henipavirus infections, and outbreaks are likely to continue so long as humans and their domesticated animals encroach into geographic locations occupied by flying foxes. Such outbreaks have potential agricultural significance as well; the initial Malaysian epidemic caused by Nipah virus was contained only after 1.1 million pigs were culled (Mohd Nor et al., 2000). Other paramyxoviruses of agricultural importance include NDV, which causes a highly contagious respiratory and neurological disease in many avian species, including chickens (Alexander, 2009) and rinderpest virus, which causes disease in cattle (Roeder and Taylor, 2002). Additional viruses that have been widely used as laboratory models for the study of paramyxovirus entry and exit include parainfluenza virus 5 (PIV5, formerly SV5) and Sendai virus. Virus particles are containers built within infected cells that are meant to transmit infection from cell-to-cell and from host-to-host. Enveloped virus particles form by budding from cellular membranes. Buds emerge from selected sites on the membranes where viral proteins and genomes have assembled together, then pinch off to achieve particle release (Fig. 1). The resulting virions have outer surfaces that consist of host-derived membrane, enriched with viral integral membrane glycoproteins (Fig. 2). The paramyxoviruses encode two glycoproteins: a fusion (F) protein, and an attachment [HN (hemagglutinin-neuraminidase), H (hemagglutinin), or G (glyco-)] protein. These proteins are packed very densely into the viral envelopes, forming “spike” layers that are visible by electron microscopy (Fig. 2B). Enclosed within paramyxovirus particles are the RNA genomes, bound with nucleocapsid (N or NP) proteins to form helical structures called ribonucleoproteins (RNPs). Directly underlying the viral membranes are the viral matrix (M) proteins, which bridge the viral glycoproteins and RNPs, thereby organizing virus assembly. Paramyxoviruses form particles that are mainly spherical, but sometimes filamentous, with considerable variation in size and shape. Particles typically range in size from 150 to 300 nm in diameter, but can reach diameters of greater than 1 μm in some cases (Goldsmith et al., 2003). Figure 1. Budding of paramyxovirus particles. Open in a new tab (A) Schematic illustration of paramyxovirus structural components that have assembled together on a cellular membrane, in preparation for budding. (B) Thin section of a Sendai virus-infected MDCK cell, visualized by electron microscopy. Arrows indicate individual sites from which virus budding has initiated on the apical cell surface. The nucleocapsid is observed aligning below the areas of curving membrane. (Adapted from Rodriguez-Boulan and Sabatini, 1978, with the author’s permission). Figure 2. Paramyxovirus virions. Open in a new tab (A) Schematic representation of a spherical paramyxovirus virion illustrating M protein lining the inner surface of the lipid envelope, the viral glycoproteins decorating the virion surface, and the helical RNP enclosed within. (B) A mumps virion purified by sucrose gradient centrifugation and imaged by transmission electron microscopy with negative staining. Glycoproteins are visible as a spike layer embedded in the envelope. (Adapted from Li et al., 2009, with the publisher’s permission). Paramyxovirus infections (see Fig. 3) are initiated when virus particles bind to receptor molecules on the surfaces of target cells. For some paramyxoviruses, such as Sendai virus and mumps virus, attachment is mediated by HN proteins that bind to sialic acid receptors. These HN proteins also possess sialidase activities, which function later in the virus lifecycle to facilitate the separation of virions from infected cells and to prevent virion aggregation. Other paramyxovirus attachment proteins mediate binding to protein receptors. These include the H protein of measles virus and the G proteins of the Henipaviruses. Following virion attachment, viral F proteins are triggered, resulting in the fusion of virion membranes with target cell membranes via a process that is driven by the refolding of F proteins from initial metastable states into more stable hairpin structures (reviewed in Lamb and Parks, 2006; Russell and Luque, 2006). Completion of this process allows virion contents, including RNPs, to enter the cytoplasms of target cells. Viral transcription then occurs, with negative-sense viral genomic RNA within RNPs serving as templates for the production of mRNAs by viral RNA-dependent RNA polymerase complexes, composed of viral phospho- (P) protein and large (L) protein subunits. Transcription follows the “stop-start” model first described for vesicular stomatitis virus (VSV), the result of which is a gradient of transcription in which genes near the 3′ end of the genome are transcribed more abundantly than genes near the 5′ end (reviewed in Whelan et al., 2004). Later in the infectious cycle, the viral polymerase enters a replication mode in which viral genomes are not transcribed, but rather are replicated in a two-step process that involves first the production of positive-sense antigenomes from genomic templates, and subsequently the production of negative-sense genomes from antigenomic templates (reviewed in Lamb and Parks, 2006). Newly-synthesized viral proteins and RNPs assemble together at infected cell plasma membranes in preparation for particle budding, which completes the cycle. Additional functions, including the disabling of host innate antiviral responses, are carried out by viral V proteins and other accessory proteins (reviewed in Horvath, 2004). Figure 3. Schematic illustration of a paramyxovirus life cycle. Open in a new tab For details, refer to the text. Viral transcription and genome replication occur in the cytoplasm. Following their synthesis, viral structural proteins and RNPs assemble together at the infected cell plasma membrane for budding. Additional processes mediated by the viral accessory proteins (shown surrounded by brackets) are not illustrated. Here, recent progress in understanding paramyxovirus particle formation is reviewed. Contributions of various viral components, as well as host proteins, to virus assembly and release are discussed, as well as parameters that lead to successful trafficking of viral components to assembly sites from which virions bud. 2. Central role of M proteins in paramyxovirus particle formation Infectious paramyxovirus particles can be formed only after all the structural components of the viruses, including viral glycoproteins and viral RNPs, have assembled together at selected sites on infected cell plasma membranes. Viral M proteins are the organizers of this assembly process. These highly abundant viral proteins bind directly to cellular membranes and occupy a central position that allows interaction both with viral RNP cores and also with viral glycoproteins via the cytoplasmic tails (Fig. 1A). Thus, M proteins are adapters that link together the structural components of virions, driving their assembly. Although structural studies of paramyxovirus M proteins have generally proved difficult, as these relatively hydrophobic proteins are prone to self-aggregation, these difficulties were recently overcome in the case of the HRSV M protein, allowing a determination of its atomic structure (Money et al., 2009). The HRSV M protein is composed of two domains, each consisting mainly of beta sheets. The domains are separated by a short, unstructured linker region. The overall structure is quite similar to that of Ebola virus matrix protein (Fig. 4A–B); (Dessen et al., 2000; Money et al., 2009). An extensive positively-charged surface extends across both domains as well as the linker region, and this likely directs electrostatic interactions between the M protein and the phospholipid head groups of membranes (Fig. 4C). This finding is generally consistent with previous biochemical studies, which implicated a combination of electrostatic and hydrophobic contributions to membrane binding by paramyxovirus M proteins (Caldwell and Lyles, 1986; Riedl et al., 2002; Stricker et al., 1994; Subhashri and Shaila, 2007), similar to membrane binding by Ebola virus VP40 (Jasenosky et al., 2001; Ruigrok et al., 2000) and by VSV M protein (Chong and Rose, 1993; Ye et al., 1994). Figure 4. Atomic structure of HRSV M protein. Open in a new tab (A) Ribbon diagrams illustrating the similar beta-sheet arrangements of the HRSV M protein (blue) and Ebola virus VP40 (yellow) N-terminal domains. (B) Ribbon diagrams illustrating the similar beta-sheet arrangements of the HRSV M protein (red) and Ebola virus VP40 (cyan) C-terminal domains. (C) Space filling model of the HRSV M protein structure, with electrostatic surface potential depicted in colors ranging from red to blue. An extensive positively-charged (blue) surface is shown. (HRSV M protein PDB code 2VQP, Ebola virus VP40 PDB code 1ES6; figure adapted from Money et al., 2009, with the author’s permission). Direct evidence for the critical role of M proteins in paramyxovirus assembly has been obtained through study of viruses harboring M protein mutations. Early studies relied on Sendai virus temperature-sensitive mutants in which M protein failed to accumulate to threshold levels at nonpermissive temperatures. This resulted in failure to produce virus particles (Kondo et al., 1993; Yoshida et al., 1979). Additional correlations between M protein defects and virus assembly impairments were made with altered measles virus isolates derived from patients with subacute sclerosing panencephalitis (SSPE). SSPE is a rare condition that results from persistent measles virus infection, causing fatal neurological disease years after initial acute infection with measles virus. SSPE strains of measles virus frequently encode hypermutated and unstable M proteins and are also found to exhibit severe defects in particle assembly (reviewed in Rima and Duprex, 2005). Unequivocal confirmation of the key roles played by paramyxovirus M proteins in particle assembly came with the advent of reverse genetics technology in the 1990s, allowing generation of negative-sense RNA viruses entirely from cloned cDNA (Conzelmann, 2004). Recombinant measles and recombinant Sendai viruses that completely lack M genes have been generated and both of these viruses exhibit near-complete defects in particle formation (Cathomen et al., 1998a; Inoue et al., 2003). Particle production defects have also been observed in recombinant measles virus in which wildtype M protein is replaced with hypermutated M protein derived from an SSPE strain (Patterson et al., 2001), recombinant measles virus in which a highly-conserved valine residue at position 101 has been changed to alanine, resulting in poor M protein stability (Runkler et al., 2007), and recombinant Sendai virus in which M protein accumulation can be prevented through the use of siRNA (Mottet-Osman et al., 2007). Taken together, these studies demonstrate convincingly that efficient paramyxovirus particle formation can occur only in the presence of a threshold level of functioning M protein. For many paramyxoviruses, M proteins expressed in the absence of other viral proteins are sufficient for the release of virus-like particles (VLPs) from transfected cells. M proteins of hPIV1 (Coronel et al., 1999), Sendai virus (Sugahara et al., 2004; Takimoto et al., 2001), NDV (Pantua et al., 2006), Nipah virus (Ciancanelli and Basler, 2006; Patch et al., 2007), and measles virus (Pohl et al., 2007; Runkler et al., 2007) are all able to elicit efficient particle budding from transfected cells when expressed alone. For NDV, membrane deformation and vesicle budding have even been reconstituted in vitro using purified M protein and unilamellar vesicles (Shnyrova et al., 2007), demonstrating that all of the activities necessary for inducing curvature and fission of a membrane are contained within M protein. In some cases, M protein-directed VLP production from transfected cells becomes even more efficient when the M proteins are co-expressed with other viral components, such as glycoproteins, nucleocapsid proteins, and C proteins (Table 1; and discussed below). Some paramyxovirus M proteins lack the ability to direct efficient VLP production when expressed alone in cells. PIV5 and rinderpest virus M proteins expressed by themselves in transfected cells do not elicit detectable VLP production (Schmitt et al., 2002; Subhashri and Shaila, 2007), and in this respect are similar to the influenza A virus matrix protein (Chen et al., 2007). Mumps virus M protein expression by itself leads to particle release that is barely detectable (Li et al., 2009). For PIV5 and mumps virus, VLP production becomes highly efficient when the M proteins are expressed together with other viral proteins (Table 1). Hence, the requirements for efficient VLP production differ among paramyxoviruses. However, M protein appears generally to be the major requirement for paramyxovirus VLP production, even in the cases where its expression is insufficient, as VLPs cannot be formed in its absence (Li et al., 2009; Schmitt et al., 2002; Teng and Collins, 1998), whereas for influenza A virus the major requirement for VLP budding is the hemagglutinin (HA) glycoprotein (Chen et al., 2007). Table 1. Differing requirements for paramyxovirus VLP production \| Virus \| Viral proteins needed for efficient VLP production \| Notes \| References \| :--- :--- \| \| hPIV1 \| M \| Nucleocapsid-like structures are incorporated into the VLPs when NP protein is expressed \| Coronel et al., 1999 \| \| Sendai virus \| M \| F protein enhances VLP production, but HN protein does not; C protein enhances VLP production, likely through recruitment of the host protein Aip1/Alix \| Takimoto et al., 2001; Sugahara et al., 2004 \| \| NDV \| M \| Glycoproteins and NP protein can be incorporated into VLPs, but do not enhance M-driven VLP production; incorporation of glycoproteins into VLPs becomes efficient when both glycoproteins are expressed \| Pantua et al., 2006 \| \| PIV5 \| M, NP, and F; or M, NP, and HN \| M protein expressed alone does not result in VLP production; F and HN proteins are necessary but redundant for VLP production \| Schmitt et al., 2002 \| \| Mumps virus \| M, NP, and F \| M protein expressed alone results in very little VLP production; F protein strongly enhances VLP production, but HN protein does not \| Li et al., 2009 \| \| Measles virus \| M \| F protein does not enhance M- driven VLP production \| Pohl et al., 2007; Runkler et al., 2007 \| \| Nipah virus \| M \| Glycoproteins and N protein can be incorporated into VLPs when expressed, but do not enhance M-driven VLP production \| Ciancanelli and Basler, 2006; Patch et al., 2007 \| Open in a new tab 3. Cooperation among viral proteins during particle production 3.1 Viral glycoproteins Paramyxovirus particles are covered with spike layers consisting of the viral attachment and fusion glycoproteins (Fig. 2). The viral glycoproteins assemble together with M proteins in infected cells, clustering on plasma membranes at locations from which virus particles will bud (Fig. 1). The cytoplasmic tails of paramyxovirus glycoproteins interact with the M proteins to organize virus assembly. Evidence for these interactions has been obtained through a variety of experimental approaches. Fluorescence microscopy experiments have documented co-localization of paramyxovirus M proteins and glycoproteins in virus-infected cells. In the case of PIV5, M proteins and glycoproteins co-localize in clusters on infected cell plasma membranes. However, this clustering is lost in cells infected with mutant viruses in which the cytoplasmic tail of HN protein, or the cytoplasmic tails of both HN and F proteins, have been truncated (Schmitt et al., 1999; Schmitt et al., 2005; Waning et al., 2002). In cells infected with the Edmonston vaccine strain of measles virus, M protein co-localizes with the viral F and H glycoproteins, but no co-localization could be observed between M protein and cytoplasmic tail-deleted viral glycoproteins (Moll et al., 2002). Co-localization studies have also been performed in cells transfected to produce the HRSV M and G proteins. Co-localization between the wildtype proteins was observed, but co-localization was lost upon removal of the first six amino acid residues of the G protein cytoplasmic tail (Ghildyal et al., 2005b). Co-immunoprecipitation studies using purified VLPs have been conducted to explore interactions involving the glycoproteins of NDV. Specific interactions between the NDV HN and M proteins, and also between the F and NP proteins, were detected. No evidence for direct interaction between the F and M proteins was obtained, however (Pantua et al., 2006). Further experimental support for specific interactions between viral glycoproteins and M proteins has been obtained by measuring the association of M proteins with detergent-resistant raft membranes (discussed further in section 5.2). Sendai virus M protein binds to cellular membranes intrinsically, but it does not associate strongly with detergent-resistant raft membranes when expressed by itself. The Sendai virus HN and F glycoproteins, however, are intrinsically sorted to raft membranes. When the viral glycoproteins are co-expressed together with M protein, M protein becomes raft membrane-associated (Ali and Nayak, 2000). Similar results were obtained with HRSV M protein, which also binds membranes intrinsically, but does not associate with detergent-resistant raft membranes unless the HRSV F protein is co-expressed, suggesting that the F protein pulls M protein into detergent-resistant membrane microdomains (Henderson, 2002). Together, these observations support a model in which viral glycoproteins and viral M proteins assemble together at specific locations on cellular membranes through interactions involving the glycoprotein cytoplasmic tails. Proper assembly of paramyxovirus glycoproteins together with M proteins in infected cells is necessary for selective incorporation of the glycoproteins into budding particles, and in many instances is also necessary for the budding process itself to become efficient. In the case of PIV5, the F and HN glycoproteins have redundant functions important for efficient virus budding. Recombinant virus in which the HN protein cytoplasmic tail is truncated buds particles poorly, and this defect is more pronounced in double mutant virus in which both HN and F protein cytoplasmic tails have simultaneously been truncated (Schmitt et al., 1999; Waning et al., 2002). These findings parallel earlier results obtained with influenza A virus, in which simultaneous truncation of the HA and neuraminidase (NA) protein cytoplasmic tails resulted in severe particle production and particle morphology defects, while single cytoplasmic tail truncations led to only minor defects (Jin et al., 1997). For Sendai virus, the HN glycoprotein appears to be dispensable for the production of virus particles (Markwell et al., 1985; Portner et al., 1974; Stricker and Roux, 1991), although particle morphology may be altered in the absence of HN protein (Hirayama et al., 2006). Truncation of the Sendai virus F protein cytoplasmic tail led to inefficient virus release (Fouillot-Coriou and Roux, 2000). Some alterations to the HN protein led to deficient incorporation of the protein into virions, with no detrimental effects on virus budding, while severe truncation of the HN protein cytoplasmic tail caused poor virus release (Fouillot-Coriou and Roux, 2000). The importance of glycoproteins for measles virus particle formation has been inferred from observations with assembly-defective SSPE strains, which frequently harbor F proteins with mutated cytoplasmic tails (Cattaneo et al., 1988; Schmid et al., 1992). Alterations to the F and H protein cytoplasmic tails in recombinant viruses led to reduced incorporation of the viral glycoproteins into virions, along with increased nonspecific incorporation of cellular proteins (Cathomen et al., 1998b). These alterations also led to rapid and extensive syncytia formation within infected cell monolayers (Cathomen et al., 1998b). Interestingly, assembly-defective recombinant PIV5 with truncated HN protein cytoplasmic tail also caused rapid and extensive syncytia formation (Schmitt et al., 1999). Vaccine strains of measles virus, including the Edmonston strain, harbor adaptations in their M proteins that confer enhanced virus assembly and release compared with wildtype measles virus isolates (Tahara et al., 2007). These adaptations enhance interaction with the H protein cytoplasmic tail judged by co-localization studies, and reduce syncytia formation (Tahara et al., 2007). Together, these findings suggest that assembly of viral M proteins together with viral glycoproteins may benefit paramyxovirus particle formation and release, but at the same time may impair F protein-mediated cell-cell fusion. This may allow measles virus to optimize its mode of spread (cell-cell fusion versus budding) by acquiring mutations in the M protein, thereby modulating the interaction with the H protein cytoplasmic tail (Tahara et al., 2007). The roles of viral glycoproteins in paramyxovirus assembly have also been examined in the context of VLPs produced from transfected cells. Although M proteins are sufficient for VLP production in many cases, expression of glycoproteins together with M proteins generally leads to production of VLPs that incorporate the viral glycoproteins and resemble paramyxovirus virions morphologically (Ciancanelli and Basler, 2006; Pantua et al., 2006; Patch et al., 2007). Some paramyxovirus glycoproteins have intrinsic exocytosis activities and are released as particles even when expressed alone. These include the Sendai virus F protein (Sugahara et al., 2004; Takimoto et al., 2001), the measles virus F protein (Pohl et al., 2007), and the Nipah virus F and G proteins (Ciancanelli and Basler, 2006; Patch et al., 2007). Other paramyxovirus glycoproteins lack inherent exocytosis activities, including the Sendai virus HN protein (Sugahara et al., 2004; Takimoto et al., 2001), and the glycoproteins of PIV5 (Schmitt et al., 2002), NDV (Pantua et al., 2006), and mumps virus (Li et al., 2009). In some cases, viral glycoproteins and M proteins cooperate with one another, resulting in VLP release that is more efficient than that obtained when either protein is expressed alone. For example, production of Sendai VLPs, driven by M protein, is made more efficient upon co-expression of the viral F protein (Sugahara et al., 2004; Takimoto et al., 2001). The Sendai virus HN protein, on the other hand, lacks the ability to enhance VLP production (Sugahara et al., 2004; Takimoto et al., 2001). Viral glycoprotein expression is required for efficient production of mumps VLPs and PIV5-like particles. Mumps VLP production is most efficient upon co-expression of the viral M, NP, and F proteins (Li et al., 2009). Co-expression of the mumps virus M, NP, and HN proteins leads to VLP production that is considerably less efficient. Hence, similar to the Sendai virus glycoproteins, the mumps virus glycoproteins are not equal contributors to virus assembly, with F protein playing the more important role. The PIV5 glycoproteins, in contrast, appear to have more equal, and seemingly redundant, functions during VLP production. Here, similar and highly efficient VLP production can be achieved either through co-expression of the M, NP, and F proteins, the M, NP, and HN proteins, or the M, NP, F, and HN proteins (Schmitt et al., 2002). Neither of the PIV5 glycoproteins can function for VLP production when their cytoplasmic tails have been truncated, in agreement with results obtained using recombinant viruses (Schmitt et al., 1999; Waning et al., 2002). These results suggest that some paramyxovirus glycoproteins, when present at virus assembly sites, can act to increase the efficiency of particle production. However, other viral glycoproteins, including those of NDV, Nipah virus, and measles virus, do not enhance the efficiency of M protein-driven VLP release from transfected cells (Ciancanelli and Basler, 2006; Pantua et al., 2006; Patch et al., 2007; Pohl et al., 2007). Hence, while assembly of paramyxovirus glycoproteins together with viral M proteins may have universal importance for efficient incorporation of the glycoproteins into budding particles, this assembly affects the efficiency of particle release for only a subset of paramyxoviruses (Table 1). 3.2 Viral nucleocapsid proteins Paramyxovirus nucleocapsid proteins bind and encapsidate genomic RNA, forming helical RNP complexes in which the RNA is tightly packaged and resistant to exogenously added RNase. The structure of HRSV N protein bound to RNA was recently determined by x-ray diffraction and cryoelectron microscopy (Tawar et al., 2009). RNA is wrapped around a decameric ring of N protein, fitting along a basic surface groove. This arrangement is consistent with a model in which the viral RNA-dependent RNA polymerase is able to recognize its template even without disassembly of the RNP. From the analogy of tobacco mosaic virus, it is assumed that RNA encapsidation starts from a nucleocapsid protein-binding nucleation site. In virus-infected cells, nascent genomic RNA is efficiently encapsidated coupled with RNA synthesis, while viral mRNAs are not encapsidated. It is therefore assumed that the promoter sequences (the leader and trailer sequences) function as nucleation sites (Blumberg et al., 1983; Lamb and Parks, 2006). On the other hand, it is known that paramyxovirus nucleocapsid proteins expressed in mammalian cells bind intracellular host RNAs non-specifically and form nucleocapsid-like structures (Buchholz et al., 1993; Coronel et al., 1999; Errington and Emmerson, 1997; Juozapaitis et al., 2005; Schmitt et al., 2002; Spehner et al., 1991; Sugahara et al., 2004), indicating that RNA encapsidation may occur in some circumstances even without the aid of a nucleation site. Once formed, RNPs must be incorporated into budding virus particles. Although paramyxovirus genomes are nonsegmented, it is not necessarily the case that all budding particles receive only a single copy of genomic RNA. Some virions are released that contain multiple genome copies (Loney et al., 2009; Rager et al., 2002). Selective incorporation of homologous genomes into virions has been demonstrated in experiments in which cells were transfected to produce the hPIV1 NP protein and also were infected with Sendai virus. Nucleocapsid structures containing hPIV1 NP protein were produced in the cells, but these were largely excluded from Sendai virions (Coronel et al., 2001). Selectivity of genome incorporation by genome length has also been reported; short defective-interfering genomes of Sendai virus were incorporated into budding particles much less efficiently than full-length genomes (Mottet and Roux, 1989). Selective genome incorporation based on the polarity of the viral RNA also occurs, with (−)-sense genomes incorporated more efficiently into budding particles than (+)-sense antigenomes. However, for paramyxoviruses such as Sendai virus, this packaging bias in favor of genomes is very weak, compared to the more stringent preferential packaging of (−)-sense genomes into rhabdovirus virions (Kolakofsky and Bruschi, 1975; Mottet and Roux, 1989). The Sendai virus C protein is thought to regulate viral RNA synthesis from the promoters, and excessive amounts of (+)-sense antigenomes are synthesized in partial C-knockout virus-infected cells. Virions released from the infected cells contain (−) and (+)-sense RNA genomes in a ratio similar to that present inside the infected cells, resulting in an overabundance of noninfectious virions containing (+)-sense antigenomic viral RNA (Irie et al., 2008a). Incorporation of genomes into budding virions is likely driven by interactions between viral matrix proteins and nucleocapsids at virus assembly sites. Biochemical evidence supports a direct interaction between paramyxovirus matrix and nucleocapsid proteins (Iwasaki et al., 2009; Markwell and Fox, 1980; Yoshida et al., 1976). For measles virus, yeast two-hybrid binding assays have mapped this interaction to the C-terminal portion of the N protein (Iwasaki et al., 2009). In transfected cells, expression of measles virus M protein shifts the localization of wt N protein (but not N protein harboring a C-terminal mutation) so that a portion is plasma membrane associated and co-localized with M protein (Iwasaki et al., 2009). Similarly, RNPs are observed at the plasma membrane in cells infected with standard measles virus, but not in cells infected with recombinant measles virus harboring an unstable M protein (Runkler et al., 2007). Together, these studies support a model in which M protein is responsible for targeting RNPs to virus assembly sites on plasma membranes. Incorporation of nucleocapsid-like structures into budding VLPs has been documented in transfected cells. For example, co-expression of the hPIV1 M and NP proteins in cells led to production of VLPs that enclose nucleocapsid-like structures (Coronel et al., 1999). Similar results were obtained with Sendai VLPs (Sugahara et al., 2004) and Nipah VLPs (Patch et al., 2007). Nucleocapsid-like structures can also be incorporated into NDV-like particles, however in this case nucleocapsid incorporation was poor unless viral glycoproteins were expressed in addition to the M and NP proteins (Pantua et al., 2006). For PIV5 and mumps virus, NP proteins not only incorporate into budding VLPs, they actually increase the efficiency of VLP production (Li et al., 2009; Schmitt et al., 2002). It is possible that for these viruses a requirement for NP protein during assembly could be beneficial through minimizing the release of noninfectious particles that lack viral genomes. 3.3 Viral C proteins Some paramyxoviruses express accessory C proteins from their P genes via overlapping reading frames. For example, Sendai virus encodes a nested set of four C proteins from its P gene through use of alternative start codons that are recognized by translation machinery either by leaky scanning in the case of C and C′, or by a type of ribosomal shunting in the case of Y1 and Y2 (Lamb and Parks, 2006). C proteins appear to be multifunctional, with activities that regulate viral RNA synthesis (Cadd et al., 1996; Curran et al., 1992) and that counteract innate immune responses of host cells (Garcin et al., 1999; Gotoh et al., 1999). The longer C proteins (C and C′) contain N-terminal membrane-targeting sequences, while the shorter C proteins (Y1 and Y2) lack these sequences (Marq et al., 2007). A role for C proteins in Sendai virus budding was first suggested based on experiments with 4C(−) virus, which lacks the ability to produce any of the four C proteins (Kurotani et al., 1998). This virus replicates to titers that are about two logs less than wildtype Sendai virus and fails to assemble and release virus particles effectively, despite efficient synthesis of viral mRNAs, proteins, and genomic RNA (Hasan et al., 2000). Defective assembly of 4C(−) virus was observed in multiple cell types, including Vero cells that fail to produce IFN-β, suggesting a virus assembly function of C protein that is separate from its function in counteracting innate immune defenses. Separation of C protein assembly and immune-evasion functions is also supported by results obtained with altered Sendai virus Cm, which harbors a C protein point mutation and fails to block the Jak/Stat pathway, yet exhibits no virus assembly defect (Kato et al., 2007). Although the above findings support a role for C protein in Sendai virus assembly, such a role could not be confirmed by Gosselin-Grenet et al., who used a somewhat different experimental system. Here, a 4C knockout virus was employed in which C protein fused to GFP was re-introduced as a separate transcriptional unit in the virus genome. GFP/C protein expression was ablated using siRNA, with no apparent effect on virus particle production (Gosselin-Grenet et al., 2007). This approach has an advantage in that it avoids passaging of a seriously debilitated virus, in which selective pressure could potentially cause unintended variants of the virus to emerge. The approach was also limited, however, in that siRNA depletion was incomplete (about 30% of normal GFP/C protein levels remained), raising the possibility that C protein could initially have been present in excess of what is needed for virus assembly. Further investigations on the importance of C proteins in Sendai virus assembly have been carried out in the context of VLPs produced from transfected cells. Expression of Sendai virus C protein together with M protein results in VLP production that is more efficient (by 2–3 fold) than that obtained when M protein is expressed alone (Sugahara et al., 2004). C and C′ proteins are each capable of enhancing Sendai VLP production, while the Y1 and Y2 proteins that lack membrane targeting sequences are unable to enhance VLP production (Irie et al., 2008b; Sakaguchi et al., 2005). A model has been proposed in which C proteins function for virus budding by interacting with the host Class E protein Aip1/Alix, recruiting it to Sendai virus assembly sites on plasma membranes (Irie et al., 2008b; Sakaguchi et al., 2005). Evidence in support of such a role for Aip1/Alix, and potential roles for other host proteins in paramyxovirus budding, are discussed below. 4. Role of host proteins in paramyxovirus budding 4.1 Class E proteins and ubiquitin Enveloped viruses typically do not encode all of the machinery that is necessary to bud particles. Instead, host machinery is manipulated to allow efficient virus exit. Retroviruses in many cases employ protein-protein interaction sequences (late domains) within their Gag polypeptides to recruit host factors to virus assembly sites (reviewed in Bieniasz, 2006; Calistri et al., 2009). Absence of these sequences in some instances leads to defects in the very late stages of virus release, hence the term late domain. Late budding defects are characterized by accumulation of budding structures tethered to cellular membranes that have morphology consistent with virions, but cannot be released by membrane fission (reviewed in Demirov and Freed, 2004). Several distinct types of viral late domain sequences have been characterized from retroviral Gag proteins, including P(T/S)AP, PPxY, and YP(x)n L. Matrix proteins of some negative-strand RNA viruses such as VSV (Craven et al., 1999; Harty et al., 1999; Jayakar et al., 2000), rabies virus (Wirblich et al., 2008), Ebola virus (Harty et al., 2000; Licata et al., 2003; Martin-Serrano et al., 2001), LCMV (Perez et al., 2003), and Lassa fever virus (Perez et al., 2003) contain these same late domain sequences, suggesting that the overall strategy of recruiting host machinery via late domains is conserved even among distantly-related viruses. Paramyxoviruses, however, generally lack these well-characterized late domain sequences within their matrix proteins, indicating that the logistics of host factor recruitment must be different for these viruses. Cellular factors recruited by late domains in many cases are Class E proteins that are part of the vacuolar protein sorting (Vps) pathway of the cell. These proteins form endosomal sorting complexes required for transport (ESCRTs) that allow for multivesicular body formation and the downregulation of cellular membrane proteins (reviewed in Stuffers et al., 2009). ESCRT complexes are dismantled and recycled with the aid of Vps4 AAA-ATPases, and disruption of the Vps sorting pathway through expression of dominant-negative (DN) Vps4 protein mutants blocks the budding of many retroviruses (reviewed in Bieniasz, 2006). Similar findings have been obtained with some paramyxoviruses. PIV5 budding (Schmitt et al., 2002) and mumps VLP budding (Li et al., 2009) were both inhibited upon expression of DN Vps4 proteins, suggesting that these viruses rely on ESCRT machinery during virus exit, even though they lack classical late domains that would mediate ESCRT recruitment. DN Vps4 protein expression also inhibits the budding of other negative-strand RNA viruses including Ebola virus (Licata et al., 2003), and Lassa fever virus (Urata et al., 2006), although in these cases classical late domain sequences are contained within the viral matrix proteins. Although it is clear that a wide range of enveloped viruses employ host ESCRT machinery during budding, it has recently become apparent that some enveloped viruses do not. Among paramyxoviruses, budding of HRSV (Utley et al., 2008) and budding of Nipah VLPs (Patch et al., 2008) occur efficiently even in the presence of DN Vps4 protein expression. These results parallel earlier observations of ESCRT-independent budding with VSV (Irie et al., 2004) and influenza virus (Chen et al., 2007), raising the possibility that some viruses employ host pathways during budding that are ESCRT-independent, or perhaps have the ability to bud even without the aid of host machinery (reviewed in Chen and Lamb, 2008). Potential links between Sendai virus budding and host ESCRT machinery have been examined in some detail. The Sendai virus C protein, a nonstructural accessory protein of the virus, binds to Aip1/Alix, the same protein that is recruited by retroviral YP(x)n L late domains. However, Sendai virus C protein lacks YP(x)n L and instead binds to Aip1/Alix via a domain within its C-terminal region (Sakaguchi et al., 2005). Expression of C protein together with Aip1/Alix in transfected cells re-localized a portion of Aip1/Alix to the plasma membrane together with C protein (Irie et al., 2008b). Sendai VLP production, driven by M protein, was enhanced when M protein was expressed together with C protein (Sugahara et al., 2004). Mutant C proteins that fail to interact with Aip1/Alix (Sakaguchi et al., 2005) or that fail to bind membranes (Irie et al., 2008b) could not enhance VLP production. Independently of its interaction with C protein, Aip1/Alix has been shown to interact with the Sendai virus M protein (Irie et al., 2007). This interaction is mediated by a YLDL motif within M protein. Mutations in the YLDL motif or depletion of Aip1/Alix with siRNA inhibited the production of Sendai VLPs (Irie et al., 2007). These observations support a model in which Aip1/Alix can be recruited to sites of Sendai virus assembly on plasma membranes either through C protein interaction or through M protein interaction, resulting in enhanced VLP production. It is unclear at present why multiple mechanisms for Aip1/Alix recruitment seem to be necessary for optimal production of Sendai VLPs, though a similar potential redundancy has been reported for hPIV2, in which the V and NP proteins were found to bind Aip1/Alix independently (Nishio et al., 2007). Questions also remain concerning the importance of Aip1/Alix recruitment in the course of an actual Sendai virus infection, as conflicting results have been obtained when examining the effect of Aip1/Alix depletion on Sendai virus budding (Gosselin-Grenet et al., 2007; Sakaguchi et al., 2005). These reports also conflict regarding the effect of DN Vps4 protein expression on Sendai virus budding. Sakaguchi et al. (2005) found that transfection of cells with DN Vps4 protein reduced budding of virions from cells transfected with infectious Sendai virus nucleocapsids, while Gosselin-Grenet et al. (2007) observed no inhibition in the release of virions from cells transfected to express DN Vps4 protein and infected with Sendai virus, even though PIV5 virion release was substantially inhibited in parallel experiments. When examining VLP production, DN Vps4 protein expression was found to only affect the accelerated release of VLPs that occurs in the presence of the viral C protein, and not M-alone VLP production (Irie et al., 2008b), consistent with the possibility that C protein enhances VLP budding through recruitment of ESCRT machinery. An interaction between the PIV5 M protein and host protein angiomotin-like 1 (AmotL1), was characterized recently using yeast two-hybrid as well as co-immunoprecipitation assays (Pei et al., 2009). AmotL1 harbors two PPxY motifs, and recruitment of AmotL1 to virus assembly sites could in theory allow indirect recruitment of the same WW domain-containing host proteins that are directly recruited by viruses that employ PPxY late domains. PIV5 budding was inhibited by siRNA-mediated depletion of AmotL1 from cells, and overexpression of M-binding AmotL1-derived polypeptides blocked the production of PIV5-like particles (Pei et al., 2009). AmotL1 localizes to tight junctions and apical membranes of epithelial cells (Sugihara-Mizuno et al., 2007), and it has been suggested that this protein or other motin proteins could be involved in assembly of PIV5 on apical membranes (Pei et al., 2009). Ubiquitin is thought to play a role in the budding of some enveloped viruses, likely stemming from its relationship with ESCRT machinery (reviewed in Martin-Serrano, 2007; Morita and Sundquist, 2004). Monoubiquitination or multiple monoubiquitination of cellular membrane proteins is a sorting signal for endocytosis and likely allows recognition by ESCRT proteins, several of which harbor well-characterized ubiquitin-binding domains (Hicke and Dunn, 2003; Shields et al., 2009). Several lines of evidence support a role for ubiquitin in the budding of certain retroviruses, including HIV-1. For example, ubiquitin has been detected in retrovirus particles (Ott et al., 2000; Putterman et al., 1990), multiple monoubiquitination of retroviral Gag proteins has been observed (Gottwein and Krausslich, 2005), and budding of some retroviruses is inhibited upon treatment of cells with proteasome inhibitors, which deplete cellular pools of free ubiquitin (Patnaik et al., 2000; Schubert et al., 2000). Ebola VP40 and VSV M protein can act as substrates for ubiquitin attachment in vitro, providing a link between ubiquitin and negative-strand RNA virus budding (Harty et al., 2000; Harty et al., 2001). In addition, VLP production driven by VSV M protein is proteasome inhibitor sensitive (Harty et al., 2001). Nedd4 mediated ubiquitination of Ebola VP40, detected in cells transfected to express VP40 and hemagglutinin-tagged ubiquitin (HA-Ub), is inhibited by expression of ISG15, a ubiquitin-like protein whose expression is induced by interferon (Malakhova and Zhang, 2008; Okumura et al., 2008). These findings support the ideas that enveloped virus budding may involve the ubiquitin-proteasome pathway, and that negative-strand RNA virus matrix proteins may be substrates for ubiquitin attachment. Some evidence in support of ubiquitin involvement in paramyxovirus budding has been obtained. Release of PIV5 virions and VLPs from 293T cells is significantly impaired upon proteasome inhibitor treatments, suggesting that the ubiquitin-proteasome system may play an important role in some aspect of PIV5 assembly and/or budding (Schmitt et al., 2005). Potential ubiquitination of measles virus M protein has been observed in cells transfected to express measles virus M protein together with HA-Ub (Pohl et al., 2007). On the other hand, proteasome inhibitor treatments failed to inhibit the budding of HRSV from Hep-2 cells or the budding of Nipah VLPs from 293T cells, suggesting that differences may exist among the paramyxoviruses in the degree of dependence on ubiquitin for budding (Patch et al., 2008; Utley et al., 2008). For Sendai virus, sensitivity to proteasome inhibitor treatment was found to be cell type-dependent. Budding from LLC-MK2 cells was highly sensitive to proteasome inhibitor treatment, budding from CV-1 cells was less sensitive, and budding from A549 cells was proteasome inhibitor resistant (Watanabe et al., 2005). This cell type-dependent behavior underscores a more general complexity inherent in the study of host factors important for virus replication, in that the relative abundances of these factors are likely to differ among cell types, perhaps reinforcing the importance of studying these virus-host interactions, where possible, in cell types which most resemble those found in the natural hosts of infection. Overall, the observations described here are consistent with the involvement of ubiquitin in the assembly and release of a subset of paramyxoviruses, though an understanding of the specific role of ubiquitin will necessitate further examination. 4.2 M protein sequences that may function to allow host factor recruitment Although paramyxovirus M proteins generally lack classical late domain sequences, a number of alternative sequences have been identified within M proteins that are important for budding activity and that may function to recruit host factors, similar to retroviral late domains. The sequence 20-FPIV-23 within PIV5 M protein was identified based on its ability to restore VLP budding function to PTAP-disrupted HIV-1 Gag protein (Schmitt et al., 2005). Mutation of either the phenylalanine or proline residues within this sequence drastically impaired the budding function of PIV5 M protein (Schmitt et al., 2005). Mumps virus M protein contains a similar sequence, 21-FPVI-24, and mumps VLP production was inhibited upon mutation of the proline residue within this sequence (Li et al., 2009). FPIV and FPIV-like sequences are presumed to function by mediating host factor recruitment to virus assembly sites, but the binding partner(s) for these sequences have not yet been identified. Two sequences in NiV M protein that are important for VLP production have been identified: 62-YMYL-65 and 92-YPLGVG-97 (Ciancanelli and Basler, 2006; Patch et al., 2008). The YMYL sequence was able to functionally replace the PTAPPEY late domain of Ebola virus VP40, as appending this sequence to late domain-defective VP40 restored VLP budding function (Ciancanelli and Basler, 2006). Mutation of YMYL inhibited the VLP budding function of NiV M protein, and led to a re-localization of the protein to the cell nucleus (Ciancanelli and Basler, 2006). The sequence YPLGVG within NiV M protein contains residues that are well conserved among paramyxoviral M proteins (Patch et al., 2008). Mutation of this sequence disrupted M protein-driven VLP production, and deletion of the sequence led to nuclear localization of NiV M protein. YPLGVG was unable to rescue VLP budding function of late domain-defective Ebola virus VP40; however, the appearance of filamentous projections similar to those observed in cells expressing wt VP40 or wt NiV M protein was restored (Patch et al., 2008). Cellular binding partners for YMYL and YPLGVG sequences have not yet been identified. Interestingly, sequences resembling YPLGVG are found in the Ebola virus and Marburg virus VP40 proteins, and mutation of these sequences disrupted VP40 stability, intracellular localization, and VLP production (Liu et al., 2009). Thus, YPLGVG-like sequences appear to be important for the functions of divergent negative-strand RNA virus matrix proteins. Sendai virus M protein possesses a sequence, 49-YLDL-52, which binds Alix/Aip1. Mutation of YLDL caused a reduction in VLP release that could not be rescued by the inclusion of other Sendai virus proteins (Irie et al., 2007). The YLDL sequence could not be replaced by P(T/S)AP, PPxY, or YP(x)n L late domains for the budding of Sendai VLPs. YLDL-Aip1/Alix interaction appears to be distinct from YP(x)n L-Aip1/Alix interaction, as the Sendai virus YLDL sequence interacts with Aip1/Alix residues 1–211 while YP(x)n L interacts with residues 409–715 (Irie et al., 2007). 5. Virus assembly sites 5.1 Paramyxovirus assembly on apical membranes of polarized cells Polarized epithelial cells that line body surfaces possess apical and basolateral sides, with the apical sides facing outward and the basolateral sides facing inward towards the underlying tissue. Several paramyxoviruses target epithelial cells of the respiratory tract for infection and assemble and bud from the apical surfaces of these cells. HRSV, Sendai virus, PIV5, hPIV3, and measles virus all bud preferentially from the apical surfaces of polarized cells (Blau and Compans, 1995; Bose et al., 2001; Roberts et al., 1995; Rodriguez-Boulan and Sabatani, 1978). Other respiratory viruses, including influenza A virus, also bud apically (Rodriguez-Boulan and Sabatani, 1978), while VSV and Marburg virus bud basolaterally (Sänger et al., 2001). The direction of virus budding from polarized cells can influence whether a paramyxoviral infection remains localized to the respiratory epithelia, facilitating transmission between hosts, or disseminates systemically, which in some cases favors establishment of persistent infections (reviewed in Schmitt and Lamb, 2004; Takimoto and Portner, 2004). In many cases, viral glycoproteins have intrinsic sorting signals that target them to apical or basolateral cell surfaces. However, it does not appear that viral glycoproteins generally control the polarity of paramyxovirus budding. Instead, budding polarity appears to be determined mainly by the viral M proteins. This is likely the case for HRSV, in which the F glycoprotein is intrinsically targeted to the apical surface (Batonick et al., 2008; Brock et al., 2005), yet recombinant HRSV that lacks the F gene still buds apically (Batonick et al., 2008). In fact, even HRSV lacking all three glycoproteins (F, G, and SH) buds apically, implying that internal viral proteins control the direction of budding (Batonick et al., 2008). An additional example is provided by measles virus, in which the viral glycoproteins intrinsically target to basolateral cell surfaces when expressed alone. In virus-infected cells, a substantial fraction of the measles virus F and H glycoproteins is re-routed to apical cell surfaces from which virus particles bud (Maisner et al., 1998). In cells infected with recombinant measles virus lacking an M gene, the measles virus glycoproteins fail to re-route to the apical side, indicating that the direction of measles virus budding, as well as the redirection of measles virus glycoproteins, is likely controlled by M protein (Naim et al., 2000). Intrinsic targeting of measles virus glycoproteins to basolateral membranes may be important for pathogenesis, as it favors cell-to-cell transfer of infection and may contribute to systemic spread in vivo (Moll et al., 2002). M protein is also thought to define the apical budding of Sendai virus. A pantropic mutant Sendai virus, F1-R, buds both apically and basolaterally and causes systemic infection in mice, whereas wt Sendai virus budding is restricted to apical surfaces, resulting in localized infection of the respiratory tract (Tashiro et al., 1996; Tashiro et al., 1992). The altered budding polarity is attributed to mutations in the M protein, as the mutant M protein disrupts the microtubule network of the cell and alters maintenance of cell polarity. Whereas wt Sendai virus M protein redirects the bipolar transport of mutant F protein from the F1-R virus to the apical cell surface, F1-R M protein does not (Tashiro et al., 1996). These findings all support a key role for paramyxovirus M proteins in the polarized assembly and release of virus particles. Apical transport and recycling of host proteins in polarized cells is facilitated by the apical recycling endosome (ARE), which is enriched in Rab11a. Recent studies have demonstrated that efficient polarized budding of HRSV depends on proper ARE function (Brock et al., 2005; Utley et al., 2008). Expression of altered myosin Vb protein disrupted HRSV replication (Brock et al., 2005). The altered protein lacks its myosin motor domain, and expression of this protein in polarized cells disrupts basolateral-to-apical transcytosis. Reduction in HRSV yield was mainly caused by failure of virus assembly at the apical surface. HRSV replication could also be inhibited through expression of a DN Rab11-FIP2 protein that lacks its Rab-binding domain (Utley et al., 2008). Rab11 family interacting proteins (Rab11-FIPs) assist in the regulation of ARE-mediated protein sorting. Reduction in HRSV replication in this case was caused by a late domain-like defect, as judged by retention and significant elongation of assembled viral filaments on infected cells (Utley et al., 2008). Hence, FIP2 protein is required at a late stage of HRSV budding from polarized cells, most likely membrane fission. This may represent a novel, alternative pathway for the polarized budding of some viruses, such as HRSV, which exit from cells via a mechanism that seems to be completely independent of ESCRT machinery. 5.2 Budding of virus particles from lipid raft membranes Microdomains within lipid bilayers, such as the cholesterol- and sphingolipid-rich raft microdomains, have the potential to act as platforms for virus assembly by providing nucleation points to allow concentration of viral proteins prior to budding (reviewed in Brown and London, 2000; Chazal and Gerlier, 2003; Schmitt and Lamb, 2005; Simons and Ikonen, 1997; Simons and Toomre, 2000; Takimoto and Portner, 2004). In paramyxovirus-infected cells, viral proteins in many cases sort to raft membranes selectively, judged by their association with detergent-resistant membrane (DRM) that resists solubilization with cold nonionic detergents such as Triton-X100. In Sendai virus-infected cells, for example, the viral structural proteins F, HN, M, N, and P are all found at least partially associated with DRM (Gosselin-Grenet et al., 2006). In transfected cells, the F and HN glycoproteins are associated with DRM when expressed alone or in combination (Sanderson et al., 1995), but the M protein is associated with DRM only when it is co-expressed with F or HN protein, suggesting that interaction between M protein and viral glycoprotein cytoplasmic tails is essential for proper targeting of M protein to raft membrane (Ali and Nayak, 2000). Measles virus structural proteins are also found associated with DRM in virus-infected cells (Manié et al., 2000; Vincent et al., 2000). Here, the F and M proteins are each capable of independently associating with DRM in transfected cells, although M protein DRM association was found to increase upon co-expression of F protein (Pohl et al., 2007). NDV HN, F, and NP proteins all associate with DRM during virus infection, and the DRM-associated viral proteins are ultimately found in released NDV virions (Laliberte et al., 2006). Lipid raft-associated proteins caveolin-1, flotillin-2, and actin are also found in NDV virions (Laliberte et al., 2006). NDV F protein is associated with DRM when expressed alone, and deletions of the F protein cytoplasmic tail inhibit DRM association (Dolganiuc et al., 2003). HRSV F, G, M, and N proteins all associate with DRM in virus-infected cells (Brown et al., 2004; Marty et al., 2004) and several lipid raft markers, including caveolin-1, GM-1, CD-55, and CD58, are found in released HRSV virions or in budding HRSV filaments (Brown et al., 2002; Brown et al., 2004; Jeffree et al., 2003). In transfected cells, HRSV M protein associates with raft membrane only when it is co-expressed with F protein (Henderson, 2002). In cells infected with canine distemper virus (CDV), the H and F glycoproteins are associated with DRM (Imhoff et al., 2007). In sum, the targeting of viral proteins to raft membranes for assembly appears to be a common strategy employed by a variety of different paramyxoviruses. Evidence that lipid rafts are in fact important during paramyxovirus infections has been obtained mainly through experiments that disrupt raft microdomains by depleting cholesterol with the drug methyl-β-cyclodextrin (MβCD). In Sendai virus infection, MβCD treatment did not impair virion production, however the virions that were released were less infectious than virions released from untreated cells (Gosselin-Grenet et al., 2006). Treatment of CDV-infected cells with MβCD did not affect particle release or infectivity, though treatment of CDV virions with MβCD after they had been released resulted in a loss of infectivity that was restored upon supplementation with cholesterol (Imhoff et al., 2007). NDV particle release was found to be enhanced by MβCD treatment (Laliberte et al., 2006). However, the released particles were somewhat altered in polypeptide composition, displayed abnormal morphology, and were less infectious than particles released from untreated cells (Laliberte et al., 2006). Further characterization of these particles revealed a defect in virus-cell membrane fusion, and a loss of complexes formed between the HN and F glycoproteins in virion envelopes (Laliberte et al., 2007). Infectivity defects have also been observed for NDV virions released from Niemann-Pick syndrome type C cells, a cell type lacking lipid rafts (Laliberte et al., 2007). Overall, these results support a model in which assembly of paramyxovirus components at lipid raft platforms enhances the quality, but not necessarily the quantity, of virus particles that are released. Recent reports illuminate changes in cholesterol biosynthesis that occur in paramyxovirus-infected cells. 3-hydroxy-3-methylglutaryl-coenzyme A reductase (HMGCR), an enzyme involved in the formation of endogenous cholesterol, and low-density lipoprotein receptor, involved in cholesterol homeostasis, are upregulated during HRSV infection, judged by microarray analysis (Martinez et al., 2007; Yeo et al., 2009). Levels of HMGCR and squalene monooxygenase, which functions in sterol biosynthesis, are altered during measles virus infection, and depletion of cholesterol from cells through simvastatin-mediated inhibition of HMGCR reduces that amount of infectious measles virus release in an acute infection (Robinzon et al., 2009). Intriguingly, several genes involved in cholesterol biosynthesis are downregulated in cells persistently infected with measles virus compared to those undergoing acute infection (Robinzon et al., 2009), and it has been suggested that the presence of a cellular mechanism to downregulate cholesterol during virus infection could serve as an innate antiviral mechanism aimed at limiting the budding of infectious particles (Robinzon et al., 2009). 5.3 Virus assembly and the cytoskeleton Involvement of the actin cytoskeleton in paramyxovirus budding was first proposed based on the detection of large quantities of actin in purified preparations of Sendai virions and measles virions (Lamb et al., 1976; Tyrrell and Norrby, 1978). Other paramyxoviruses incorporate actin into virions as well, including NDV and HRSV (Laliberte et al., 2006; Marty et al., 2004). Sendai VLPs produced upon expression of either M protein or F protein contain actin, and both of these proteins contain sequences important for budding that resemble actin-binding domains (Takimoto et al., 2001). Both Sendai virus and NDV M proteins have been shown to physically associate with actin (Giuffre et al., 1982) and EM-based studies of measles virus and HRSV have revealed a tight association between actin filament formation and budding virions (Bohn et al., 1986; Jeffree et al., 2007). Drug treatments that disrupt the cytoskeleton generally have deleterious effects on paramyxovirus replication. Multiple steps in the lifecycle of HRSV, for example, including viral entry and virus exit, are impaired upon disruption of the cytoskeleton, with severe disruptions in virus assembly and release following perturbation of either actin or microtubule networks (Burke et al., 1998; Kallewaard et al., 2005). Using molecular beacons to visually track RNPs, HRSV filament assembly and motion were recently examined in live cells, and myosin motor-driven transport on the actin cytoskeleton was postulated to drive the migration and/or rotational motion of the filaments (Santangelo and Bao, 2007). In the case of hPIV3, nocodazole-mediated disruption of microtubules significantly impairs the release of infectious virus from cells, although depolymerization of actin microfilaments with cytochalasin D had no effect on hPIV3 release (Bose et al., 2001). Microtubule disruption by nocodazole and colchicine treatments led to asymmetric budding of Sendai virus, similar to that observed with the pantropic F1-R mutant (Tashiro et al., 1993). Hence, efficient and directional budding of some paramyxoviruses depends on intact microtubules (Bose et al., 2001; Tashiro et al., 1993). 5.4 Trafficking of paramyxovirus M proteins through the cell nucleus Although paramyxovirus replication takes place in the cytoplasm, various paramyxoviral proteins have been found to traffic to the nucleus during the course of infection (Coleman and Peeples, 1993; Ghildyal et al., 2009; Ghildyal et al., 2005a; Ghildyal et al., 2003; Peeples et al., 1992; Peeples, 1988; Rodriguez et al., 2004; Sato et al., 2006; Shaw et al., 2005; Watanabe et al., 1996; Yoshida et al., 1976). Early during infection, the M proteins of HRSV, NDV, and Sendai virus have all been observed in the nucleus, while later during infection they are localized mainly in the cytoplasm as well as associated with cell membranes (Ghildyal et al., 2003; Peeples et al., 1992; Peeples, 1988; Yoshida et al., 1976). As discussed earlier, altered Nipah virus M proteins with disruptions to the 62-YMYL-65 or 92-YPLGVG-97 sequences are retained in the nuclei of transfected cells, consistent with the possibility that Nipah virus M protein undergoes nuclear-cytoplasmic shuttling (Ciancanelli and Basler, 2006; Patch et al., 2008). By analogy with VSV, it has been suggested that transient nuclear localization of paramyxovirus M proteins in the early phase of infection could allow for M-mediated inhibition of cellular nuclear processes (Ghildyal et al., 2006). Transient nuclear localization of M proteins may also serve to delay the budding of particles until late times in infection, when sufficient quantities of viral glycoproteins and RNPs have accumulated. Functional nuclear localization signals (NLSs) of the NDV and HRSV M proteins have been characterized. The NLS of NDV M protein is a bipartite NLS composed of basic amino acid clusters (Coleman and Peeples, 1993). HRSV M protein is recruited into the nucleus through direct recognition by a nuclear import receptor, importin beta 1 (Ghildyal et al., 2005a). Late in paramyxovirus infection, large quantities of M protein are needed outside the nucleus at locations of virus assembly. In the case of HRSV M protein, nuclear export is mediated by a leucine-rich nuclear export signal and is dependent on Crm-1 (exportin-1) (Ghildyal et al., 2009). Recombinant virus harboring a mutated nuclear export signal within its M protein could not be recovered, and treatment of HRSV-infected cells with a Crm-1 inhibitor, leptomycin B, prevented nuclear export of HRSV M protein and impaired virus production (Ghildyal et al., 2009). These findings support a model in which paramyxovirus M protein nuclear localization is meant to be transient, and failure to exit from the nucleus impairs accumulation of the protein at virus assembly sites at levels required for virus budding. 6. Conclusions Substantial progress has been made in recent years towards understanding both unique and shared processes used by paramyxoviruses and other enveloped viruses during virus particle formation. Roles for paramyxovirus M proteins, glycoproteins, nucleocapsid proteins, and accessory proteins during virus assembly have been clarified. Mechanisms allowing for directional budding of paramyxoviruses from polarized cells have been defined, and the importance of lipid raft microdomains as virus assembly platforms has been demonstrated. However, questions regarding important aspects of the paramyxovirus assembly process remain unresolved. For example, although paramyxoviruses clearly differ from one another in the fundamental components required for particle formation, the biological implications of these differences remain unclear. Structural information is still lacking for most paramyxovirus M proteins, and consequently the molecular determinants that drive viral protein-protein interactions during virus assembly remain poorly understood. While host factors clearly play important roles in the budding of many paramyxoviruses, specific mechanisms of host factor recruitment remain largely undefined. Future efforts in these and related areas will provide a clearer understanding of the events which must occur for virions to form, and may contribute to the development of effective antiviral strategies aimed at blocking the late steps of paramyxovirus lifecycles. Acknowledgments This work was supported in part by the Middle Atlantic Regional Center of Excellence (MARCE) for Biodefense and Emerging Infectious Disease Research NIH grant AI057168, and research grant AI070925 from the National Institute of Allergy and Infectious Diseases to A.P.S. This project is funded, in part, under a grant with the Pennsylvania Department of Health using Tobacco Settlement funds to A.P.S. The Department specifically disclaims responsibility for any analyses, interpretations or conclusions. Abbreviations used AmotL1 angiomotin-like 1 ARE apical recycling endosome CDV canine distemper virus Crm1 exportin1 DN dominant-negative DRM detergent-resistant membrane ESCRT endosomal sorting complex required for transport F fusion G glyco- H hemagglutinin (paramyxovirus) HA hemagglutinin (influenza virus) HA-Ub hemagglutinin-tagged ubiquitin HMGCR 3-hydroxy-3-methylglutaryl-coenzyme A HN hemagglutinin-neuraminidase hPIV human parainfluenza virus HRSV human respiratory syncytial virus HSP heat shock protein L large M matrix MVB multivesicular body MβCD methyl-β-cyclodextrin NA neuraminidase NDV Newcastle disease virus NLS nuclear localization signal N nucleocapsid NP nucleocapsid P phospho- PIV5 parainfluenza virus 5 Rab11-FIP Rab11 family interacting proteins RNP ribonucleoprotein SSPE subacute sclerosing panencephalitis VLP virus-like particle Vps vacuolar protein sorting VSV vesicular stomatitis virus Footnotes Publisher's Disclaimer: This is a PDF file of an unedited manuscript that has been accepted for publication. 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[DOI] [PubMed] [Google Scholar] ACTIONS View on publisher site PDF (1.8 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Central role of M proteins in paramyxovirus particle formation 3. Cooperation among viral proteins during particle production 4. Role of host proteins in paramyxovirus budding 5. Virus assembly sites 6. 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2355	https://www.parametriczoo.com/index.php/2020/03/03/coincident-planes/	Coincident planes - Parametric Zoo Contact Products Revit Plugins CSI Link Macaw Butterfly Grasshopper plugins Biber Woodbee KUKA\|prc Tool-path optimizer Bridge Curves Group Nesting 3DS MAX Plugins Para 3d FEM Grasshopper UI+ Blog Discussion Revit plugins (forum) Rhino + Grasshopper Plugins Froums About Our Privacy Policy Home Select Page Contact Products Revit Plugins CSI Link Macaw Butterfly Grasshopper plugins Biber Woodbee KUKA\|prc Tool-path optimizer Bridge Curves Group Nesting 3DS MAX Plugins Para 3d FEM Grasshopper UI+ Blog Discussion Revit plugins (forum) Rhino + Grasshopper Plugins Froums About Our Privacy Policy Home Coincident planes by PARA \| Mar 3, 2020 \| Revit Programming \| 0 comments Further to my previous post about signed distance to plane , Here I would like to share another insight to plane geometry in Revit API. Two planes are geometrically the same (they are coincident) if all points are one plane lies on other plane too. At first you may think you can simply test if they are equal by calling the Equal() method. Although the equal planes are certainly coincident , but there are coincident planes that are not equal by formulation. Let’s look at the mathematical definition of the plane: Where nis a unit normal vector to the plane, p is a position vector of a point of the plane and D the signed distance of the plane from the origin. This definition is known as theHesse normal formrelies on the parameterD. Now let us say two planes P₁ and P₂ are coincide, this means equations for both planes must be satisfied regardless of the position vector p . In the other words if point p satisfies the first equation then it should also satisfies the second one : There are two sets of trivial solutions in which both equatios are statisfied: In case (1) the normal vectors of two planes are identical and the planes are having the same distance from the origin. In case (2) the normal of P₂ is the reverse of the normal of P₁ and the signed distance of P₁ is the negative of the signed distance of the P₂. Now let us implement above conditions in Revit API by looking at Plane class. There are two essential properties of the plane that we need to compare. First, the normal vector which must be in unique length. Considering both cases (1) and (2) we compare the normal vectors in both direct and reverse direction: if (p1.Normal.IsAlmostEqualTo(p2.Normal) \|\| p1.Normal.IsAlmostEqualTo(-p2.Normal)) { //if above condition is satisfied then planes are parallel } Note that vectors are compared using IsAlmostEqualtTo not the Equality method. If above conditions are statisfied then it means the planes are parallel, The next step is to compare the parameter D or the signed distance from the origin. This is not implemented in Revit API but it can be easily retrieved from the dot product of the normal plane and the a point on the plane. looking at the plane equation again we can write : For the point p on the plane we can choose the Originand to cover the both cases (1) and (2) we only compare the absolute value of parameter D. Similar to vector comparison, we need to take into account that numbers could be very close but still not equal, therefore it is wised to consider a tolerance value. Here we use an arbitrary tolerance which must be sufficiently small : // if both planes are parallel then check if they have the same distance to the origin. double d1 = Math.Abs(p1.Origin.DotProduct(p1.Normal)); double d2 = Math.Abs(p2.Origin.DotProduct(p2.Normal)); return (Math.Abs(d1 - d2) < tolerance); Choosing the tolerance value is quite relative. In our implementation the tolerance is actually the maximum distance between two parallel planes which must be considered coincident. To be consistent with the method IsAlmostEqualTo() , a good choice could be (10e-9) . Here is the full implemenation : /// /// Returns true if the plane p2 lies on this plane (p1). /// /// /// /// if distance between two parallel planes is less than /// this threshold then they are considered as Coincide planes /// public static bool IsCoincident(this Plane p1 , Plane p2, double tolerance) { // check if the normal of this plane is equal (or reverse ) to the normal of the other Plane if (p1.Normal.IsAlmostEqualTo(p2.Normal) \|\| p1.Normal.IsAlmostEqualTo(-p2.Normal)) { // if both planes are parallel then check if they have the same distance to the origin. double d1 = Math.Abs(p1.Origin.DotProduct(p1.Normal)); double d2 = Math.Abs(p2.Origin.DotProduct(p2.Normal)); // the difference between d1 and d2 must be within given tolerance return (Math.Abs(d1 - d2) < tolerance); } else { return false; // if planes are not parallel then they cannot be coincide } } Submit a CommentCancel reply You must be logged in to post a comment. 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2356	https://www.quora.com/How-do-you-calculate-the-overall-percentage-of-multiple-percentages	How to calculate the overall percentage of multiple percentages - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Survey Question Mathematics Percentage Formula Data Analysis Number Concepts Percentage Converter Basic Concepts of Mathema... Percentile Calculations Arithmetic 5 How do you calculate the overall percentage of multiple percentages? All related (38) Sort Recommended Assistant Bot · 1y To calculate the overall percentage from multiple percentages, you typically need to consider the weight of each percentage based on the quantity or total it represents. Here’s a step-by-step method to do this: Steps to Calculate Overall Percentage Identify the Percentages and Their Weights: Determine the percentage values and the quantities (or weights) they represent. Convert Percentages to Decimal Form: Divide each percentage by 100 to convert it into decimal form. Calculate Weighted Values: Multiply each decimal percentage by its corresponding weight (quantity). Sum the Weighted Values: Add all Continue Reading To calculate the overall percentage from multiple percentages, you typically need to consider the weight of each percentage based on the quantity or total it represents. Here’s a step-by-step method to do this: Steps to Calculate Overall Percentage Identify the Percentages and Their Weights: Determine the percentage values and the quantities (or weights) they represent. Convert Percentages to Decimal Form: Divide each percentage by 100 to convert it into decimal form. Calculate Weighted Values: Multiply each decimal percentage by its corresponding weight (quantity). Sum the Weighted Values: Add all the weighted values together. Sum the Weights: Add all the weights together. Calculate Overall Percentage: Divide the total weighted value by the total weight and then multiply by 100 to convert it back to a percentage. Formula The overall percentage can be calculated using the formula: Overall Percentage=(∑(Percentage i×Weight i)∑Weight i)×100 Overall Percentage=(∑(Percentage i×Weight i)∑Weight i)×100 Example Suppose you have the following percentages and their corresponding weights: Percentage A: 80% (Weight A: 50) Percentage B: 90% (Weight B: 30) Percentage C: 70% (Weight C: 20) Calculating the Overall Percentage: Convert to decimal: 80% = 0.80 90% = 0.90 70% = 0.70 Calculate weighted values: Weighted A: 0.80×50=40 0.80×50=40 Weighted B: 0.90×30=27 0.90×30=27 Weighted C: 0.70×20=14 0.70×20=14 Sum the weighted values: Total Weighted = 40+27+14=81 40+27+14=81 Sum the weights: Total Weight = 50+30+20=100 50+30+20=100 Calculate overall percentage: Overall Percentage=(81 100)×100=81%Overall Percentage=(81 100)×100=81% Conclusion In this example, the overall percentage is 81%. This method ensures that each percentage contributes appropriately based on its weight. Upvote · 9 1 Related questions More answers below How do you calculate the percentage of percentages? How can I turn 12 1/2 percentage into a decimal? What is the process to calculate overall percentage? How do I calculate 2^(3/2)? How do I calculate percentage from percentile? Raghavendra Babujee Prabhu Former Asst. Professior at Sahyadri College ( Adyar Campus ) (2009–2020) · Author has 3.8K answers and 959.2K answer views ·1y Overall percentages are calculated by adding together the percentages found in each case and then overall percentage is taken . For example, the percentage of children suffering from fever is 20 in a school of 120 and in an another school it is 30 of 150. To find the overall percentage we do the following calculation. First school= 20/100 x 120 = 24 Second school= 30100 x 150 =45 Total number of children suffering in both the schools= 24 + 45 = 69 Total number of children in both schools= 270 Overall percentage 69 / 270 x 100 = 25.55% Similar calculation to get overall percentage are made when sever Continue Reading Overall percentages are calculated by adding together the percentages found in each case and then overall percentage is taken . For example, the percentage of children suffering from fever is 20 in a school of 120 and in an another school it is 30 of 150. To find the overall percentage we do the following calculation. First school= 20/100 x 120 = 24 Second school= 30100 x 150 =45 Total number of children suffering in both the schools= 24 + 45 = 69 Total number of children in both schools= 270 Overall percentage 69 / 270 x 100 = 25.55% Similar calculation to get overall percentage are made when several schools are taken into consideration. Upvote · 9 2 Promoted by ComparisonAdviser Carl Madden Finance Security Expert ·Sep 12 What secret about your industry can you share now that you don’t work for them anymore? Oh boy, where do I even start? After 8 years as an auto insurance agent, I have zero loyalty left to protect these companies. We Had "Loyalty Lists" Every month, I'd get a report of customers who hadn't shopped around in 2+ years. These were our golden geese - we could raise their rates aggressively because they'd proven they wouldn't leave. One customer I remember was paying $3,200 annually for coverage that should have cost $1,800. She stayed for 5 years. The "File and Use" Scam Here's something most people don't know: in many states, insurance companies can raise your rates immediately and ju Continue Reading Oh boy, where do I even start? After 8 years as an auto insurance agent, I have zero loyalty left to protect these companies. We Had "Loyalty Lists" Every month, I'd get a report of customers who hadn't shopped around in 2+ years. These were our golden geese - we could raise their rates aggressively because they'd proven they wouldn't leave. One customer I remember was paying $3,200 annually for coverage that should have cost $1,800. She stayed for 5 years. The "File and Use" Scam Here's something most people don't know: in many states, insurance companies can raise your rates immediately and justify it later. We'd implement 15-20% increases across entire ZIP codes, knowing regulators would take months to review. By then, we'd collected millions in extra premiums. Claim Frequency Was Irrelevant Your rates weren't really based on how often you'd claim - they were based on how likely you were to shop around. A customer with 3 claims who got quotes every year paid less than a claim-free customer who never compared rates. It was pure price discrimination. We Loved Policy Confusion Complex policy language wasn't an accident. The more confusing your coverage, the less likely you'd comparison shop effectively. We'd change terminology between companies deliberately to make apple-to-apple comparisons nearly impossible. The Real Game-Changer Tools like ComparisonAdviser absolutely terrify insurance companies because they eliminate our biggest advantage: information asymmetry. When customers can instantly see what competitors charge with identical coverage and discounts applied, our whole "loyalty tax" model collapses. I've watched too many good people get fleeced by an industry that profits from customer ignorance. Use ComparisonAdviserreligiously - it's the only way to beat a system designed to exploit your trust. The truth? Every year you don't comparison shop, you're probably donating $500-1,500 to your insurance company's profit margins. Upvote · 2.7K 2.7K 99 67 Derek McNeil Spent over a decade in University and college. · Author has 3.5K answers and 8.9M answer views ·1y There are two important points to remember when solving such a problem: Such a problem is usually phrased x% of y% of z% of….etc. Usually, the word “of” in math problems signals a multiplication operation, which it does in this case. Thus, “x% of y%” is equivalent to “x% times y%” or “x% X y%:”. Percentages are another way of stating fractions where the denominator is 100. x% is equivalent to x/100. So, to solve a problem: Convert the percentages to fractions: x% of y% of z% of….etc. becomes x/100 of y/100 of z/100 of….etc. 2. Remember that “of” means multiplication, so: x/100 of y/100 of z/100 of…. Continue Reading There are two important points to remember when solving such a problem: Such a problem is usually phrased x% of y% of z% of….etc. Usually, the word “of” in math problems signals a multiplication operation, which it does in this case. Thus, “x% of y%” is equivalent to “x% times y%” or “x% X y%:”. Percentages are another way of stating fractions where the denominator is 100. x% is equivalent to x/100. So, to solve a problem: Convert the percentages to fractions: x% of y% of z% of….etc. becomes x/100 of y/100 of z/100 of….etc. 2. Remember that “of” means multiplication, so: x/100 of y/100 of z/100 of….etc. becomes x/100 X y/100 X z/1t00 of….etc. Steps 1 and 2 can be done in any order: 1 then 2, or 2 then 1. 3. Optionally, you can simplify the fractions to make the multiplication easier: e.g. 4/100 can be simplified to 1/25 and 50/100 can be simplified to 1/2. Alternately, you can multiply without simplifying. 4. Multiply all the fractions together. 5. If you need to express the answer as a fraction, then you should simplify the resulting fraction. If you need to express the answer as a percentage, then convert the resulting fraction to a fraction with a denominator of 100. 6. Convert the fraction back to a percentage. Any numerator above a denominator of 100 is equivalent to the numerator as a percentage: e.g. x/100 = x% Alternately, and possibly more easily, you could convert the percentages to decimal fractions, then multiply the resulting decimal values: x% is equivalent to .0x. If x is a two digit number, like 55, then x% would be .55. If x has three digits or more, then the third digit would come before the decimal (e.g. 110% becomes 1.10 (you can drop the trailing zero, making it 1.1). Essentially, you are dividing the numerator by 100. Then multiply these decimal values together. To express the answer as a percentage, Multiply the result by 100: e.g. .628 x 100 = 68.2, so .682 = 68.2%. Upvote · 9 1 Pavel Juranek Technician, Programmer, Analyst, Consultant · Author has 3.7K answers and 1.3M answer views ·3y Working with quantities A, B is a) posible anytimes in case of multiplication: A . B = C. Obtaned result is in new units [C] = [A.B] = [A].[B] Example: Let A is speed and [A] is “m/s” Let B is time and [B] is “s” Then [C] = “m” because of (m/s).s If quantity is expressed as X.P where P is Percentage (=count of “%”) then [X.P] = [X].[P] … this is the same principle, where [P]=1/100 We speak about P percent of X then. Example Let X = 5000 g Let P = 10 % Then (X.P) = 5000.10 = 50000 [X.P] = g . 1/100 X.P = 50000 . g/100 = 50000 cg = 500g = 0.5 kg b) specific in case of summation A+B: such operation Continue Reading Working with quantities A, B is a) posible anytimes in case of multiplication: A . B = C. Obtaned result is in new units [C] = [A.B] = [A].[B] Example: Let A is speed and [A] is “m/s” Let B is time and [B] is “s” Then [C] = “m” because of (m/s).s If quantity is expressed as X.P where P is Percentage (=count of “%”) then [X.P] = [X].[P] … this is the same principle, where [P]=1/100 We speak about P percent of X then. Example Let X = 5000 g Let P = 10 % Then (X.P) = 5000.10 = 50000 [X.P] = g . 1/100 X.P = 50000 . g/100 = 50000 cg = 500g = 0.5 kg b) specific in case of summation A+B: such operation is possible only if [A] = [B] Example: 10 m + 20 m = 30m Not allowed are summs: 10 m + 2 cm is not 12. Units of quantities must be unified first: 10m = 1000 cm Then sum 1000+2=1002 is the result. Using some percentage is dangerous, because of manner shortened speaking: Example: If the ware cost A = $80, than people say about the tax sortly: “You must add VAT 15%” instead of “You must add VAT 15% A” Only then we add: Price + (15/100) Price, because all terms are in $. c) It is not the only snare of percentage. Often we can hear a claim: The VAT increases by 10% It seems to be O.K., VAT and increment is defined in %, but what is the right meaning c1) New VAT will be 15% + 10% = 25%, if understand literally c2) New VAT will be 15% + 10% of actual VAT = 16.5% if understand as shortened speach (noticed above) Keep on the mind: Manner c2) is default meanintg of the claim above. If c1 should be the right maning, then othet maner is used: Then they say: The VAT increase by 10 percentage points. Upvote · 9 1 Related questions More answers below How do you calculate 1500 as a percentage? How do we calculate the NIOS percentage? How is the 12th CBSE percentage calculated? How can I calculate hours in percentage? How do you write 12 1/2 percent as a fraction? Gary Dudley MS in Electrical Engineering&Mathematics, North Carolina State University at Raleigh (Graduated 1975) ·3y Originally Answered: How do I find the total percent chance of multiple percent chances? · If I understand the question, you want to know if you have multiple chances to accomplish a goal, and the likelihood of success each time is fixed, then what is the total likelihood of success. Let x be the probability of success each time, and n be the number of chances. After 1 chance the probability of success is just x. If two tries are allowed, the second try is only necessary if the first try was unsuccessful. The probability of an unsuccessful try followed by a successful one is: (1 - x) x And the total probability is the sum: x + (1-x)x Likewise, for 3 tries, the total probability is: x + (1-x Continue Reading If I understand the question, you want to know if you have multiple chances to accomplish a goal, and the likelihood of success each time is fixed, then what is the total likelihood of success. Let x be the probability of success each time, and n be the number of chances. After 1 chance the probability of success is just x. If two tries are allowed, the second try is only necessary if the first try was unsuccessful. The probability of an unsuccessful try followed by a successful one is: (1 - x) x And the total probability is the sum: x + (1-x)x Likewise, for 3 tries, the total probability is: x + (1-x)x + (1-x)(1-x)x So for n tries, the probability of success is: the sum of k = 1 to n of x (1-x)^(k-1) A simpler way to calculate the same number is 1 - (1-x)^n, subtracting the chance of n unsuccessful tries from 1 Upvote · 9 1 Promoted by Network Solutions® NetSol - Alicia Pringle B.A. in Bachelor of Arts Degrees in English&Communication, Misericordia University (Graduated 2008) ·Aug 20 How do I prepare an e-commerce store for the holiday season? You’ve probably seen a holiday movie where a character sprints through a big retail store, frantically loading up on gifts at the last minute. As a small business owner, the holiday season can feel like a similar race to the finish, but with the right preparation, you can turn that chaos into a huge opportunity. Working for Network Solutions, a leading web presence company, I’ve helped a lot of small business owners establish their websites and eCommerce stores over the years, I've seen firsthand what separates the winners from those who get lost in the noise: preparedness and consistency. The Continue Reading You’ve probably seen a holiday movie where a character sprints through a big retail store, frantically loading up on gifts at the last minute. As a small business owner, the holiday season can feel like a similar race to the finish, but with the right preparation, you can turn that chaos into a huge opportunity. Working for Network Solutions, a leading web presence company, I’ve helped a lot of small business owners establish their websites and eCommerce stores over the years, I've seen firsthand what separates the winners from those who get lost in the noise: preparedness and consistency. The key is to start early and be intentional with your online presence. It’s something we covered recently in our Holiday Visibility Playbook for SMBs, and it’s a topic that’s of vital importance for small business owners looking to grow online. Why Starting Now is Crucial The holiday shopping season starts much earlier than you might think. Customers are already making their gift lists and doing research in September, with many beginning to fill their carts in October, long before Black Friday. This year, the rise of AI search, like Google's AI Overviews and ChatGPT, makes early preparation even more critical. Unlike traditional search results that show a long list of websites, AI results often display only a handful. If you're not visible when people start their research, you're missing out on an entire wave of early shoppers. Avoid Common Mistakes The most common mistake small businesses make is simply waiting too long to start preparing. Whether you're a one-person operation or have a small team, a proactive approach is a must. The early birds are undoubtedly the winners when it comes to holiday preparation. It’s not enough to hope that your website will drive online success for you over the holidays; remember, hope is not a strategy! Another mistake small business owners make is only updating one part of their digital presence. While updating your Instagram is great, it’s not enough. You need to update your website, all social media profiles, Google Business Profile, and email marketing. Don't be afraid to use AI as your "marketing assistant" to help you create content for these different channels. Tools like ChatGPT can help you draft posts or website copy, freeing you up to focus on other parts of your business. Just remember to always edit and add your unique brand voice to your copy to make sure the content feels authentic and personal. Actionable Steps for eCommerce Success For eCommerce stores, your website is your storefront, and that means it needs to be ready to convert. Here are three high-impact steps to take now: Update Your Website and Create a Landing Page: Make your holiday offerings easy to find. Feature holiday products and promotions prominently on your homepage. Consider creating a dedicated seasonal landing page for your holiday bundles or gift guides (e.g., "Gifts Under $50" or "Stocking Stuffers"). This centralized page is easy to link to from your social media and email campaigns. For local businesses, don't forget to update your store hours and promote any in-store pickup options. Harmonize Your Online Presence: Ensure your holiday messaging is consistent everywhere. Update your Google Business Profile with new photos of your holiday products and any special hours. On your social media channels, start teasing your holiday promos and products with photos or short videos. This consistency builds trust and helps AI systems understand your business better, which can lead to your business being featured in those valuable AI search results. Create a Marketing Calendar: Get organized with a 90-60-30-7 day plan. 90 Days Out (mid-September): Finalize your holiday product photos, spruce up your website copy with holiday keywords, and plan your promo calendar. 60 Days Out (mid-October): Start advertising early bird discounts. Update your Google Ads campaigns and business listings with holiday-themed content. Post your holiday product photos on social media. 30 Days Out (mid-November): Actively promote your deals across all channels, including social media, email, and text messaging. This is when the holiday rush really begins. 7 Days Out: Focus on last-minute shoppers. Promote shipping cutoffs, in-store pickup options, and digital gift cards. Remember, you don't need a massive campaign to see success. What you need is a solid, organized foundation. Use AI as a tool, control the content on your own website, and be consistent across your online channels. By being prepared, you’ll not only show up for your customers but also stand a chance to win big this holiday season. Want even more info on how to prepare your web presence for a successful holiday season? Check out our Holiday Visibility Playbook for SMBs. And best of luck to you and your business over the holidays and beyond! Upvote · 999 497 99 25 9 8 Anonymous Person 1y Convert them into fractions. For example 100% would be 1 and 50% would be 1/2. To do this just divide the value by 100. Next multiply the fractions together. Ta da! you’ve now got the overall percentage. Upvote · Jim Caron 30+ years as a stock/mutual fund/options/currency investor · Author has 3.8K answers and 4.9M answer views ·3y Originally Answered: How do I find the total percent chance of multiple percent chances? · To find the chance of two or more things both happening, you multiply the probabilities: the chance of getting heads on a single coin toss is 1/2. The chance of getting heads on BOTH of two coin tosses 1/2, AFTER you succeed on the 1/2 chance of the first — which is 1/2 x 1/2 or 1/4. If you are considering alternative outcomes of the same event, you add the probabilities of each. For example, you have 10 coins in a bag, each with a different year, from 1990 to 1999. What is the chance of drawing a coin dated 1997 or later on a single draw? The chance of drawing each coin is 1/10, and there are 3 Continue Reading To find the chance of two or more things both happening, you multiply the probabilities: the chance of getting heads on a single coin toss is 1/2. The chance of getting heads on BOTH of two coin tosses 1/2, AFTER you succeed on the 1/2 chance of the first — which is 1/2 x 1/2 or 1/4. If you are considering alternative outcomes of the same event, you add the probabilities of each. For example, you have 10 coins in a bag, each with a different year, from 1990 to 1999. What is the chance of drawing a coin dated 1997 or later on a single draw? The chance of drawing each coin is 1/10, and there are 3 coins that “win,” so the odds are 3/10 on a single draw. Then you have the probability of something NOT happening. What is the probability that you will flip a coin ten times, and never get heads? In that case you figure the probability of getting heads 10 times in a row, and subtract that percent chance from 100%. That would be 1/2 to the 10 power chance of all tails, with the remaining probability your chance of getting at least 1 head. Upvote · Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Learn More 999 107 Paul Fournier Teacher · Author has 1.4K answers and 594.5K answer views ·3y Originally Answered: How do I find the total percent chance of multiple percent chances? · You just need to multiply them ex. Increase of 12% followed by increase of 8% followed by decrease of 15% (1+.12)(1+.08)(1-.15)=1.02816 or (1+.028) that is an increase of 2.8% Upvote · Richard Goldstone PhD in Mathematics, The Graduate Center, CUNY (Graduated 1995) · Author has 1.8K answers and 3.9M answer views ·11mo Related How do I total the overall percentage from 7 different percentages? This question lacks enough detail to have any meaning. I’ll try to answer a general version (replacing 7 by n n) by making explicit assumptions about what is meant. We have n n quantities, A 1,A 2,…,A n A 1,A 2,…,A n. We have another quantity B B, and we know that B B is p 1 p 1 percent of A 1,A 1,B B is p 2 p 2 percent of A 2,…,A 2,…,B B is p n p n percent of A n.A n. Question: What percent of A 1+A 2+⋯+A n A 1+A 2+⋯+A n is B B? The fraction of A i A i equal to B B is (p i/100)A i,(p i/100)A i, so the fraction of A 1+A 2+⋯+A n A 1+A 2+⋯+A n equal to B B is (p 1/100)A 1+(p 2/100)A 2+⋯+(p n/100)A n.(p 1/100)A 1+(p 2/100)A 2+⋯+(p n/100)A n. Converting this fraction to a percent means computing \begin{multl\begin{multl Continue Reading This question lacks enough detail to have any meaning. I’ll try to answer a general version (replacing 7 by n n) by making explicit assumptions about what is meant. We have n n quantities, A 1,A 2,…,A n A 1,A 2,…,A n. We have another quantity B B, and we know that B B is p 1 p 1 percent of A 1,A 1,B B is p 2 p 2 percent of A 2,…,A 2,…,B B is p n p n percent of A n.A n. Question: What percent of A 1+A 2+⋯+A n A 1+A 2+⋯+A n is B B? The fraction of A i A i equal to B B is (p i/100)A i,(p i/100)A i, so the fraction of A 1+A 2+⋯+A n A 1+A 2+⋯+A n equal to B B is (p 1/100)A 1+(p 2/100)A 2+⋯+(p n/100)A n.(p 1/100)A 1+(p 2/100)A 2+⋯+(p n/100)A n. Converting this fraction to a percent means computing (p 1/100)A 1+(p 2/100)A 2+⋯+(p n/100)A n A 1+A 2+⋯+A n×100=p 1 A 1+p 2 A 2+⋯+p n A n A 1+A 2+⋯+A n.(1)(1)(p 1/100)A 1+(p 2/100)A 2+⋯+(p n/100)A n A 1+A 2+⋯+A n×100=p 1 A 1+p 2 A 2+⋯+p n A n A 1+A 2+⋯+A n. If it happens that A 1=A 2=⋯=A n,A 1=A 2=⋯=A n, then (1) reduces to the ordinary average of the percentages. Upvote · Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Upvote · 999 477 999 103 99 17 Mildred Taylor Former Retired Teacher · Author has 1.2K answers and 432.2K answer views ·11mo Related How do I total the overall percentage from 7 different percentages? Add up all the numbers you have and get a total then divide by7 For example suppose you have these numbers for a test Look at the example 100 70. 80. 60. 90. 70 Total marks =750 There are ten numbers there so I divideby 10 to get the over all average. 750/10 =75 There you have it ! This is an example now you do the same for the numbers you have . Upvote · Divyant Pratap B.Tech. in Electrical Engineering, National Institute of Technology, Raipur (Graduated 2021) · Author has 104 answers and 424K answer views ·8y Related How is the overall percentage for CBSE 12th calculated? Take the sum of all subjects as given in results.Like for me it was 476 than divide it by 500 and multiply by 100 and it comes as 95.2% Hope it helps.Please upvote. Continue Reading Take the sum of all subjects as given in results.Like for me it was 476 than divide it by 500 and multiply by 100 and it comes as 95.2% Hope it helps.Please upvote. Upvote · 999 144 9 3 9 2 Tim Farage Professor, Mathematics and Computer Science, Retired · Author has 4.9K answers and 17.7M answer views ·Updated 2y Related What's the easiest way to calculate the percentage difference between two numbers? Suppose an item costs $40. And then its price was marked up to $50. What per cent mark up was this? The way to figure this, is to first calculate how much the markup was. Here it is: 50 - 40 = $10. Then divide this by the original cost: 10/40 = 25% increase. So in general if an item costs C dollars and is increased (or decreased) to D dollars, the per cent increase is 100% (D - C) / C. It's interesting, and confusing to many people to see the result you get if you reverse this process. Suppose an item costs $50 and its price was marked down to $40. This looks like the opposite of the original Continue Reading Suppose an item costs $40. And then its price was marked up to $50. What per cent mark up was this? The way to figure this, is to first calculate how much the markup was. Here it is: 50 - 40 = $10. Then divide this by the original cost: 10/40 = 25% increase. So in general if an item costs C dollars and is increased (or decreased) to D dollars, the per cent increase is 100% (D - C) / C. It's interesting, and confusing to many people to see the result you get if you reverse this process. Suppose an item costs $50 and its price was marked down to $40. This looks like the opposite of the original problem above, so one might think the answer is a 25% decrease. But it isn't. If we apply the formula I gave above, we get 100% ( 40 - 50) / 50 = -20%. So the per cent drop from $50 to $40 is 20%, but the percent increase from $40 to $50 is 25%. This sort of question is one that I get often from my friends because it is not intuitively obvious. Such is life. Upvote · 99 34 9 2 Michael Munson BA in History and Political Science&Philosophy, Virginia Commonwealth University (Graduated 1974) · Author has 38K answers and 20.4M answer views ·11mo Related How do I total the overall percentage from 7 different percentages? All answers so far are wrong. You failed to specify if the 7 different percentages have any relationship to one another, so there is no logical or factual way to “total” them up. Upvote · Related questions How do you calculate the percentage of percentages? How can I turn 12 1/2 percentage into a decimal? What is the process to calculate overall percentage? How do I calculate 2^(3/2)? How do I calculate percentage from percentile? How do you calculate 1500 as a percentage? How do we calculate the NIOS percentage? How is the 12th CBSE percentage calculated? How can I calculate hours in percentage? How do you write 12 1/2 percent as a fraction? How much is one third in percentage? How do you calculate the percentage change between 2 percentages? What is 3% of $1000000? How do you figure out a percentage of a number? How do you calculate five percent? Related questions How do you calculate the percentage of percentages? How can I turn 12 1/2 percentage into a decimal? What is the process to calculate overall percentage? How do I calculate 2^(3/2)? How do I calculate percentage from percentile? How do you calculate 1500 as a percentage? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2357	https://en.wikipedia.org/wiki/Variable_(mathematics)	Jump to content Variable (mathematics) Afrikaans العربية Azərbaycanca تۆرکجه বাংলা Башҡортса Беларуская Беларуская (тарашкевіца) Български Català Чӑвашла Čeština Cymraeg Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Føroyskt Français Gaeilge Gàidhlig Galego 贛語 ગુજરાતી Хальмг 한국어 Հայերեն हिन्दी Hrvatski Ido Bahasa Indonesia Interlingua IsiXhosa Íslenska Italiano עברית Kabɩyɛ Қазақша Kiswahili Kriyòl gwiyannen Latina Latviešu Magyar Македонски മലയാളം Malti Bahasa Melayu Монгол မြန်မာဘာသာ Nederlands 日本語 Norsk bokmål Norsk nynorsk Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ Patois Polski Português Romnă Русский Саха тыла Shqip Simple English سنڌي Slovenčina Slovenščina کوردی Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் ไทย Türkçe Українська اردو Tiếng Việt Võro 文言吴语粵語中文 Edit links From Wikipedia, the free encyclopedia Symbol representing a mathematical object Not to be confused with Variable (computer science). In mathematics, a variable (from Latin variabilis 'changeable') is a symbol, typically a letter, that refers to an unspecified mathematical object. One says colloquially that the variable represents or denotes the object, and that any valid candidate for the object is the value of the variable. The values a variable can take are usually of the same kind, often numbers. More specifically, the values involved may form a set, such as the set of real numbers. The object may not always exist, or it might be uncertain whether any valid candidate exists or not. For example, one could represent two integers by the variables p and q and require that the value of the square of p is twice the square of q, which in algebraic notation can be written p2 = 2 q2. A definitive proof that this relationship is impossible to satisfy when p and q are restricted to integer numbers isn't obvious, but it has been known since ancient times and has had a big influence on mathematics ever since. Originally, the term variable was used primarily for the argument of a function, in which case its value could be thought of as varying within the domain of the function. This is the motivation for the choice of the term. Also, variables are used for denoting values of functions, such as the symbol y in the equation y = f(x), where x is the argument and f denotes the function itself. A variable may represent an unspecified number that remains fixed during the resolution of a problem; in which case, it is often called a parameter. A variable may denote an unknown number that has to be determined; in which case, it is called an unknown; for example, in the quadratic equation ax2 + bx + c = 0, the variables a, b, c are parameters, and x is the unknown. Sometimes the same symbol can be used to denote both a variable and a constant, that is a well defined mathematical object. For example, the Greek letter π generally represents the number π, but has also been used to denote a projection. Similarly, the letter e often denotes Euler's number, but has been used to denote an unassigned coefficient for quartic function and higher degree polynomials. Even the symbol 1 has been used to denote an identity element of an arbitrary field. These two notions are used almost identically, therefore one usually must be told whether a given symbol denotes a variable or a constant. Variables are often used for representing matrices, functions, their arguments, sets and their elements, vectors, spaces, etc. In mathematical logic, a variable is a symbol that either represents an unspecified constant of the theory, or is being quantified over. History [edit] For broader coverage of this topic, see History of algebra and History of mathematical notation. Early history [edit] The earliest uses of an "unknown quantity" date back to at least the Ancient Egyptians with the Moscow Mathematical Papyrus (c. 1500 BC) which described problems with unknowns rhetorically, called the "Aha problems". The "Aha problems" involve finding unknown quantities (referred to as aha, "stack") if the sum of the quantity and part(s) of it are given (The Rhind Mathematical Papyrus also contains four of these types of problems). For example, problem 19 asks one to calculate a quantity taken 1+1⁄2 times and added to 4 to make 10. In modern mathematical notation: ⁠3/2⁠x + 4 = 10. Around the same time in Mesopotamia, mathematics of the Old Babylonian period (c. 2000 BC – 1500 BC) was more advanced, also studying quadratic and cubic equations. In works of ancient greece such as Euclid's Elements (c. 300 BC), mathematics was described geometrically. For example, The Elements, proposition 1 of Book II, Euclid includes the proposition: "If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments." This corresponds to the algebraic identity a(b + c) = ab + ac (distributivity), but is described entirely geometrically. Euclid, and other greek geometers, also used single letters refer to geometric points and shapes. This kind of algebra is now sometimes called Greek geometric algebra. Diophantus of Alexandria, pioneered a form of syncopated algebra in his Arithmetica (c. 200 AD), which introduced symbolic manipulation of expressions with unknowns and powers, but without modern symbols for relations (such as equality or inequality) or exponents. An unknown number was called . The square of was ; the cube was ; the fourth power was ; and the fifth power was . So for example, what would be written in modern notation as: would be written in Diophantus's syncopated notation as: In the 7th century BC, Brahmagupta used different colours to represent the unknowns in algebraic equations in the Brāhmasphuṭasiddhānta. One section of this book is called "Equations of Several Colours". Greek and other ancient mathematical advances, were often trapped in long periods of stagnation, and so there were few revolutions in notation, but this began to change by the early modern period. Early modern period [edit] At the end of the 16th century, François Viète introduced the idea of representing known and unknown numbers by letters, nowadays called variables, and the idea of computing with them as if they were numbers—in order to obtain the result by a simple replacement. Viète's convention was to use consonants for known values, and vowels for unknowns. In 1637, René Descartes "invented the convention of representing unknowns in equations by x, y, and z, and knowns by a, b, and c". Contrarily to Viète's convention, Descartes' is still commonly in use. The history of the letter x in math was discussed in an 1887 Scientific American article. Starting in the 1660s, Isaac Newton and Gottfried Wilhelm Leibniz independently developed the infinitesimal calculus, which essentially consists of studying how an infinitesimal variation of a time-varying quantity, called a Fluent, induces a corresponding variation of another quantity which is a function of the first variable. Almost a century later, Leonhard Euler fixed the terminology of infinitesimal calculus, and introduced the notation y = f(x) for a function f, its variable x and its value y. Until the end of the 19th century, the word variable referred almost exclusively to the arguments and the values of functions. In the second half of the 19th century, it appeared that the foundation of infinitesimal calculus was not formalized enough to deal with apparent paradoxes such as a nowhere differentiable continuous function. To solve this problem, Karl Weierstrass introduced a new formalism consisting of replacing the intuitive notion of limit by a formal definition. The older notion of limit was "when the variable x varies and tends toward a, then f(x) tends toward L", without any accurate definition of "tends". Weierstrass replaced this sentence by the formula in which none of the five variables is considered as varying. This static formulation led to the modern notion of variable, which is simply a symbol representing a mathematical object that either is unknown, or may be replaced by any element of a given set (e.g., the set of real numbers). Notation [edit] Variables are generally denoted by a single letter, most often from the Latin alphabet and less often from the Greek, which may be lowercase or capitalized. The letter may be followed by a subscript: a number (as in x2), another variable (xi), a word or abbreviation of a word as a label (xtotal) or a mathematical expression (x2i+1). Under the influence of computer science, some variable names in pure mathematics consist of several letters and digits. Following René Descartes (1596–1650), letters at the beginning of the alphabet such as a, b, c are commonly used for known values and parameters, and letters at the end of the alphabet such as x, y, z are commonly used for unknowns and variables of functions. In printed mathematics, the norm is to set variables and constants in an italic typeface. For example, a general quadratic function is conventionally written as ax2 + bx + c, where a, b and c are parameters (also called constants, because they are constant functions), while x is the variable of the function. A more explicit way to denote this function is x ↦ ax2 + bx + c, which clarifies the function-argument status of x and the constant status of a, b and c. Since c occurs in a term that is a constant function of x, it is called the constant term. Specific branches and applications of mathematics have specific naming conventions for variables. Variables with similar roles or meanings are often assigned consecutive letters or the same letter with different subscripts. For example, the three axes in 3D coordinate space are conventionally called x, y, and z. In physics, the names of variables are largely determined by the physical quantity they describe, but various naming conventions exist. A convention often followed in probability and statistics is to use X, Y, Z for the names of random variables, keeping x, y, z for variables representing corresponding better-defined values. Conventional variable names [edit] a, b, c, d (sometimes extended to e, f) for parameters or coefficients a0, a1, a2, ... for situations where distinct letters are inconvenient ai or ui for the ith term of a sequence or the ith coefficient of a series f, g, h for functions (as in f(x)) i, j, k (sometimes l or h) for varying integers or indices in an indexed family, or unit vectors l and w for the length and width of a figure l also for a line, or in number theory for a prime number not equal to p n (with m as a second choice) for a fixed integer, such as a count of objects or the degree of a polynomial p for a prime number or a probability q for a prime power or a quotient r for a radius, a remainder or a correlation coefficient t for time x, y, z for the three Cartesian coordinates of a point in Euclidean geometry or the corresponding axes z for a complex number, or in statistics a normal random variable α, β, γ, θ, φ for angle measures ε (with δ as a second choice) for an arbitrarily small positive number λ for an eigenvalue Σ (capital sigma) for a sum, or σ (lowercase sigma) in statistics for the standard deviation μ for a mean Specific kinds of variables [edit] It is common for variables to play different roles in the same mathematical formula, and names or qualifiers have been introduced to distinguish them. For example, the general cubic equation is interpreted as having five variables: four, a, b, c, d, which are taken to be given numbers and the fifth variable, x, is understood to be an unknown number. To distinguish them, the variable x is called an unknown, and the other variables are called parameters or coefficients, or sometimes constants, although this last terminology is incorrect for an equation, and should be reserved for the function defined by the left-hand side of this equation. In the context of functions, the term variable refers commonly to the arguments of the functions. This is typically the case in sentences like "function of a real variable", "x is the variable of the function f : x ↦ f(x)", "f is a function of the variable x" (meaning that the argument of the function is referred to by the variable x). In the same context, variables that are independent of x define constant functions and are therefore called constant. For example, a constant of integration is an arbitrary constant function that is added to a particular antiderivative to obtain the other antiderivatives. Because of the strong relationship between polynomials and polynomial functions, the term "constant" is often used to denote the coefficients of a polynomial, which are constant functions of the indeterminates. Other specific names for variables are: An unknown is a variable in an equation which has to be solved for. An indeterminate is a symbol, commonly called variable, that appears in a polynomial or a formal power series. Formally speaking, an indeterminate is not a variable, but a constant in the polynomial ring or the ring of formal power series. However, because of the strong relationship between polynomials or power series and the functions that they define, many authors consider indeterminates as a special kind of variables. A parameter is a quantity (usually a number) which is a part of the input of a problem, and remains constant during the whole solution of this problem. For example, in mechanics the mass and the size of a solid body are parameters for the study of its movement. In computer science, parameter has a different meaning and denotes an argument of a function. Free variables and bound variables A random variable is a kind of variable that is used in probability theory and its applications. All these denominations of variables are of semantic nature, and the way of computing with them (syntax) is the same for all. Dependent and independent variables [edit] Main article: Dependent and independent variables In calculus and its application to physics and other sciences, it is rather common to consider a variable, say y, whose possible values depend on the value of another variable, say x. In mathematical terms, the dependent variable y represents the value of a function of x. To simplify formulas, it is often useful to use the same symbol for the dependent variable y and the function mapping x onto y. For example, the state of a physical system depends on measurable quantities such as the pressure, the temperature, the spatial position, ..., and all these quantities vary when the system evolves, that is, they are function of the time. In the formulas describing the system, these quantities are represented by variables which are dependent on the time, and thus considered implicitly as functions of the time. Therefore, in a formula, a dependent variable is a variable that is implicitly a function of another (or several other) variables. An independent variable is a variable that is not dependent. The property of a variable to be dependent or independent depends often of the point of view and is not intrinsic. For example, in the notation f(x, y, z), the three variables may be all independent and the notation represents a function of three variables. On the other hand, if y and z depend on x (are dependent variables) then the notation represents a function of the single independent variable x. Examples [edit] If one defines a function f from the real numbers to the real numbers by then x is a variable standing for the argument of the function being defined, which can be any real number. In the identity the variable i is a summation variable which designates in turn each of the integers 1, 2, ..., n (it is also called index because its variation is over a discrete set of values) while n is a parameter (it does not vary within the formula). In the theory of polynomials, a polynomial of degree 2 is generally denoted as ax2 + bx + c, where a, b and c are called coefficients (they are assumed to be fixed, i.e., parameters of the problem considered) while x is called a variable. When studying this polynomial for its polynomial function this x stands for the function argument. When studying the polynomial as an object in itself, x is taken to be an indeterminate, and would often be written with a capital letter instead to indicate this status. Example: the ideal gas law [edit] Consider the equation describing the ideal gas law, This equation would generally be interpreted to have four variables, and one constant. The constant is kB, the Boltzmann constant. One of the variables, N, the number of particles, is a positive integer (and therefore a discrete variable), while the other three, P, V and T, for pressure, volume and temperature, are continuous variables. One could rearrange this equation to obtain P as a function of the other variables, Then P, as a function of the other variables, is the dependent variable, while its arguments, V, N and T, are independent variables. One could approach this function more formally and think about its domain and range: in function notation, here P is a function . However, in an experiment, in order to determine the dependence of pressure on a single one of the independent variables, it is necessary to fix all but one of the variables, say T. This gives a function where now N and V are also regarded as constants. Mathematically, this constitutes a partial application of the earlier function P. This illustrates how independent variables and constants are largely dependent on the point of view taken. One could even regard kB as a variable to obtain a function Moduli spaces [edit] See also: moduli spaces Considering constants and variables can lead to the concept of moduli spaces. For illustration, consider the equation for a parabola, where a, b, c, x and y are all considered to be real. The set of points (x, y) in the 2D plane satisfying this equation trace out the graph of a parabola. Here, a, b and c are regarded as constants, which specify the parabola, while x and y are variables. Then instead regarding a, b and c as variables, we observe that each set of 3-tuples (a, b, c) corresponds to a different parabola. That is, they specify coordinates on the 'space of parabolas': this is known as a moduli space of parabolas. See also [edit] Lambda calculus Observable variable Physical constant Propositional variable References [edit] ^ Sobolev, S.K. (originator). "Individual variable". Encyclopedia of Mathematics. Springer. ISBN 1402006098. Retrieved September 5, 2024. A symbol of a formal language used to denote an arbitrary element (individual) in the structure described by this language. ^ Beckenbach, Edwin F (1982). College algebra (5th ed.). Wadsworth. ISBN 0-534-01007-5. A variable is a symbol representing an unspecified element of a given set. ^ Landin, Joseph (1989). An Introduction to Algebraic Structures. New York: Dover Publications. p. 204. ISBN 0-486-65940-2. A variable is a symbol that holds a place for constants. ^ "ISO 80000-2:2019". Quantities and units, Part 2: Mathematics. International Organization for Standardization. Archived from the original on October 7, 2024. Retrieved September 15, 2019.{{cite web}}: CS1 maint: bot: original URL status unknown (link) ^ Stover & Weisstein. ^ van Dalen, Dirk (2008). Logic and Structure (PDF) (4th ed.). Springer-Verlag. p. 57. doi:10.1007/978-3-540-85108-0. ISBN 978-3-540-20879-2. ^ Feys, Robert; Fitch, Frederic Brenton (1969). Dictionary of symbols of mathematical logic. Amsterdam: North-Holland Pub. Co. LCCN 67030883. ^ Shapiro, Stewart; Kouri Kissel, Teresa (2024), "Classical Logic", in Zalta, Edward N.; Nodelman, Uri (eds.), The Stanford Encyclopedia of Philosophy (Spring 2024 ed.), Metaphysics Research Lab, Stanford University, retrieved September 1, 2024 ^ Clagett, Marshall. 1999. Ancient Egyptian Science: A Source Book. Volume 3: Ancient Egyptian Mathematics. Memoirs of the American Philosophical Society 232. Philadelphia: American Philosophical Society. ISBN 0-87169-232-5 ^ a b Boyer, Carl B. (Carl Benjamin) (1991). A History of Mathematics. New York: Wiley. ISBN 978-0-471-54397-8. ^ Diophantine Equations. Submitted by: Aaron Zerhusen, Chris Rakes, & Shasta Meece. MA 330-002. Dr. Carl Eberhart. 16 February 1999. ^ Boyer (1991). "Revival and Decline of Greek Mathematics". p. 178. "The chief difference between Diophantine syncopation and the modern algebraic notation is the lack of special symbols for operations and relations, as well as of the exponential notation." ^ A History of Greek Mathematics: From Aristarchus to Diophantus. By Sir Thomas Little Heath. Pg 456 ^ A History of Greek Mathematics: From Aristarchus to Diophantus. By Sir Thomas Little Heath. Pg 458 ^ Tabak 2014, p. 40. ^ Fraleigh 1989, p. 276. ^ Sorell 2000, p. 19. ^ Scientific American. Munn & Company. September 3, 1887. p. 148. ^ Edwards Art. 4 ^ Hosch 2010, p. 71. ^ Foerster 2006, p. 18. ^ Weisstein, Eric W. "Sum". mathworld.wolfram.com. Retrieved February 14, 2022. ^ Edwards Art. 5 ^ Edwards Art. 6 Bibliography [edit] Edwards, Joseph (1892). An Elementary Treatise on the Differential Calculus (2nd ed.). London: MacMillan and Co. Foerster, Paul A. (2006). Algebra and Trigonometry: Functions and Applications (classics ed.). Upper Saddle River, NJ: Prentice Hall. ISBN 978-0-13-165711-3. Fraleigh, John B. (1989). A First Course in Abstract Algebra (4th ed.). United States: Addison-Wesley. ISBN 978-0-201-52821-3. Hosch, William L., ed. (2010). The Britannica Guide to Algebra and Trigonometry. Britannica Educational Publishing. ISBN 978-1-61530-219-2. Menger, Karl (1954). "On Variables in Mathematics and in Natural Science". The British Journal for the Philosophy of Science. 5 (18). University of Chicago Press: 134–142. doi:10.1093/bjps/V.18.134. JSTOR 685170. Peregrin, Jaroslav (2000). "Variables in Natural Language: Where do they come from?" (PDF). In Böttner, Michael; Thümmel, Wolf (eds.). Variable-Free Semantics. Osnabrück Secolo. pp. 46–65. ISBN 978-3-929979-53-4. Quine, Willard V. (1960). "Variables Explained Away" (PDF). Proceedings of the American Philosophical Society. 104 (3). American Philosophical Society: 343–347. JSTOR 985250. Sorell, Tom (2000). Descartes: A Very Short Introduction. New York: Oxford University Press. ISBN 978-0-19-285409-4. Stover, Christopher; Weisstein, Eric W. "Variable". In Weisstein, Eric W. (ed.). Wolfram MathWorld. Wolfram Research. Retrieved November 22, 2021. Tabak, John (2014). Algebra: Sets, Symbols, and the Language of Thought. Infobase Publishing. ISBN 978-0-8160-6875-3. \| v t e Mathematical logic \| \| --- \| \| General \| Axiom + list Cardinality First-order logic Formal proof Formal semantics Foundations of mathematics Information theory Lemma Logical consequence Model Theorem Theory Type theory \| \| Theorems (list) and paradoxes \| Gödel's completeness and incompleteness theorems Tarski's undefinability Banach–Tarski paradox Cantor's theorem, paradox and diagonal argument Compactness Halting problem Lindström's Löwenheim–Skolem Russell's paradox \| \| Logics \| \| \| \| --- \| \| Traditional \| Classical logic Logical truth Tautology Proposition Inference Logical equivalence Consistency + Equiconsistency Argument Soundness Validity Syllogism Square of opposition Venn diagram \| \| Propositional \| Boolean algebra Boolean functions Logical connectives Propositional calculus Propositional formula Truth tables Many-valued logic + 3 + finite + ∞ \| \| Predicate \| First-order + list Second-order + Monadic Higher-order Fixed-point Free Quantifiers Predicate Monadic predicate calculus \| \| \| Set theory \| \| \| \| --- \| \| Set + hereditary Class (Ur-)Element Ordinal number Extensionality Forcing Relation + equivalence + partition Set operations: + intersection + union + complement + Cartesian product + power set + identities \| \| \| Types of sets \| Countable Uncountable Empty Inhabited Singleton Finite Infinite Transitive Ultrafilter Recursive Fuzzy Universal Universe + constructible + Grothendieck + Von Neumann \| \| Maps and cardinality \| Function/Map + domain + codomain + image In/Sur/Bi-jection Schröder–Bernstein theorem Isomorphism Gödel numbering Enumeration Large cardinal + inaccessible Aleph number Operation + binary \| \| Set theories \| Zermelo–Fraenkel + axiom of choice + continuum hypothesis General Kripke–Platek Morse–Kelley Naive New Foundations Tarski–Grothendieck Von Neumann–Bernays–Gödel Ackermann Constructive \| \| \| Formal systems (list), language and syntax \| \| \| \| --- \| \| Alphabet Arity Automata Axiom schema Expression + ground Extension + by definition + conservative Relation Formation rule Grammar Formula + atomic + closed + ground + open Free/bound variable Language Metalanguage Logical connective + ¬ + ∨ + ∧ + → + ↔ + = Predicate + functional + variable + propositional variable Proof Quantifier + ∃ + ! + ∀ + rank Sentence + atomic + spectrum Signature String Substitution Symbol + function + logical/constant + non-logical + variable Term Theory + list \| \| \| Example axiomatic systems (list) \| of true arithmetic: + Peano + second-order + elementary function + primitive recursive + Robinson + Skolem of the real numbers + Tarski's axiomatization of Boolean algebras + canonical + minimal axioms of geometry: + Euclidean: - Elements - Hilbert's - Tarski's + non-Euclidean Principia Mathematica \| \| \| Proof theory \| Formal proof Natural deduction Logical consequence Rule of inference Sequent calculus Theorem Systems + axiomatic + deductive + Hilbert - list Complete theory Independence (from ZFC) Proof of impossibility Ordinal analysis Reverse mathematics Self-verifying theories \| \| Model theory \| Interpretation + function + of models Model + equivalence + finite + saturated + spectrum + submodel Non-standard model + of non-standard arithmetic Diagram + elementary Categorical theory Model complete theory Satisfiability Semantics of logic Strength Theories of truth + semantic + Tarski's + Kripke's T-schema Transfer principle Truth predicate Truth value Type Ultraproduct Validity \| \| Computability theory \| Church encoding Church–Turing thesis Computably enumerable Computable function Computable set Decision problem + decidable + undecidable + P + NP + P versus NP problem Kolmogorov complexity Lambda calculus Primitive recursive function Recursion Recursive set Turing machine Type theory \| \| Related \| Abstract logic Algebraic logic Automated theorem proving Category theory Concrete/Abstract category Category of sets History of logic History of mathematical logic + timeline Logicism Mathematical object Philosophy of mathematics Supertask \| \| Mathematics portal \| \| \| Authority control databases \| \| --- \| \| National \| \| \| Other \| Yale LUX \| Retrieved from " Categories: Variables (mathematics) Algebra Calculus Elementary mathematics Syntax (logic) Mathematical logic Hidden categories: CS1 maint: bot: original URL status unknown Articles with short description Short description is different from Wikidata Use mdy dates from November 2021
2358	https://medium.com/@vineethrajb/unlocking-precision-in-computer-vision-a-deep-dive-into-the-chessboard-method-for-camera-d7b3bd8c7184	Sitemap Open in app Sign in Sign in Unlocking Precision in Computer Vision: A Deep Dive into the Chessboard Method for Camera Calibration 5 min readSep 13, 2024 Camera calibration is crucial for many computer vision applications, from augmented reality to autonomous vehicles. In this post, we’ll explore one of the most widely used techniques: the Chessboard Method. We’ll cover what it is, how it works, and provide code examples to help you implement it using OpenCV. What is the Chessboard Method? The Chessboard Method uses a printed chessboard pattern to calibrate cameras by detecting a grid of black and white squares. The method is highly effective due to the structured geometry of the chessboard, which is ideal for detection and processing by calibration algorithms. How Does It Work? The Chessboard Method follows these four main steps: Image Capture: Capture multiple images of a chessboard pattern from various angles and positions. Corner Detection: Use an algorithm to detect the corners of each square in the captured images. Parameter Estimation: Estimate the camera’s intrinsic (e.g., focal length, optical center) and extrinsic (e.g., rotation, translation) parameters using the detected corners. Optimization: Refine the parameters to minimize the reprojection error — the difference between the observed and projected corner positions. Step-by-Step Implementation with OpenCV Let’s implement the Chessboard Method using Python and OpenCV. Here’s a step-by-step guide: Step 1: Install OpenCV First, ensure you have OpenCV installed: pip install opencv-python Step 2: Capture Images of the Chessboard Pattern You need a series of images with a printed chessboard pattern taken from different angles. You can use a camera or a pre-existing dataset. Step 3: Detect Chessboard Corners Use OpenCV’s findChessboardCorners function to detect the corners of the chessboard pattern. ``` import cv2import numpy as np# Define the chessboard sizechessboard_size = (9, 6) # Number of inner corners per a chessboard row and column# Prepare object points based on the chessboard sizeobjp = np.zeros((chessboard_size chessboard_size, 3), np.float32)objp[:, :2] = np.mgrid[0:chessboard_size, 0:chessboard_size].T.reshape(-1, 2)# Arrays to store object points and image pointsobjpoints = [] # 3D points in real-world spaceimgpoints = [] # 2D points in image plane# Load your chessboard imagesimages = [cv2.imread(f'image{i}.jpg') for i in range(1, 21)]for img in images: gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY) # Find the chessboard corners ret, corners = cv2.findChessboardCorners(gray, chessboard_size, None) if ret: objpoints.append(objp) imgpoints.append(corners) # Draw and display the corners cv2.drawChessboardCorners(img, chessboard_size, corners, ret) cv2.imshow('Chessboard Corners', img) cv2.waitKey(500)cv2.destroyAllWindows() import import as Define the chessboard size 9 6 Number of inner corners per a chessboard row and column Prepare object points based on the chessboard size 0 1 3 2 0 0 0 1 1 2 Arrays to store object points and image points 3D points in real-world space 2D points in image plane Load your chessboard imagesf'image{i}.jpg'{i} for in range 1 21 for in Find the chessboard corners None if Draw and display the corners 'Chessboard Corners' 500 ``` Step 4: Calibrate the Camera Use the detected corners to estimate the camera parameters. ``` Calibrate the cameraret, camera_matrix, dist_coeffs, rvecs, tvecs = cv2.calibrateCamera(objpoints, imgpoints, gray.shape[::-1], None, None)print("Camera Matrix:\n", camera_matrix)print("Distortion Coefficients:\n", dist_coeffs) Calibrate the camera 1 None None print"Camera Matrix:\n" print"Distortion Coefficients:\n" ``` Step 5: Undistort the Images Now, you can use the estimated camera parameters to undistort images. ``` for img in images: h, w = img.shape[:2] new_camera_matrix, roi = cv2.getOptimalNewCameraMatrix(camera_matrix, dist_coeffs, (w, h), 1, (w, h)) # Undistort the image undistorted_img = cv2.undistort(img, camera_matrix, dist_coeffs, None, new_camera_matrix) # Display undistorted image cv2.imshow('Undistorted Image', undistorted_img) cv2.waitKey(500)cv2.destroyAllWindows() for in 2 1 Undistort the image None Display undistorted image 'Undistorted Image' 500 ``` Advantages of the Chessboard Method High Accuracy: The Chessboard Method is known for its high accuracy, primarily due to the well-defined, regular grid of the chessboard pattern. Each corner of the black and white squares serves as a precise reference point, allowing the algorithm to detect and locate these points with minimal error. This precision is crucial for camera calibration because small inaccuracies can lead to significant distortions in tasks like 3D reconstruction or depth estimation. The method’s ability to provide highly accurate results makes it a preferred choice in applications where precision is paramount. Ease of Use: One of the significant advantages of the Chessboard Method is its simplicity. All you need is a printed chessboard pattern, which can be created easily on standard paper or any flat surface. The setup process is straightforward, requiring only a few images from different angles, and does not demand any specialized equipment. Furthermore, most computer vision libraries, such as OpenCV, have built-in functions that support chessboard detection and calibration, making the process even more accessible to developers and researchers. This ease of use enables both beginners and experts to perform camera calibration efficiently. Wide Support: The Chessboard Method is widely supported across various computer vision libraries and tools. Open-source libraries like OpenCV provide comprehensive support for the method, including pre-built functions for corner detection, parameter estimation, and optimization. This widespread support allows for quick integration into different projects, whether in academic research, commercial applications, or hobbyist endeavors. The method’s compatibility with multiple platforms and tools ensures it can be easily implemented, tested, and refined in a variety of contexts. Disadvantages of the Chessboard Method Lighting Sensitivity: The effectiveness of the Chessboard Method can be significantly affected by lighting conditions. Proper lighting is essential for detecting the corners of the chessboard squares accurately. Poor or uneven lighting, reflections, or glare can cause the algorithm to misidentify or miss the corners altogether. This sensitivity to lighting can be particularly challenging in outdoor or uncontrolled environments, where shadows and reflections are unpredictable. To mitigate this issue, care must be taken to ensure consistent and adequate lighting during the image capture process, which may not always be feasible. Planar Limitation: Another limitation of the Chessboard Method is that it requires the chessboard pattern to be flat and placed on a planar surface. This requirement makes it unsuitable for calibrating cameras in environments where a flat surface is not available or practical, such as in underwater applications or on irregular surfaces. Additionally, this limitation can pose challenges when calibrating wide-angle or fisheye lenses, which often require calibration over a larger field of view that a single planar chessboard pattern cannot cover. In such cases, other calibration methods or patterns might be more appropriate. Conclusion The Chessboard Method remains a robust and reliable approach for camera calibration in various computer vision applications. While it has some limitations, its accuracy, ease of use, and widespread support make it a popular choice. By understanding these advantages and disadvantages, you can make an informed decision on whether this method is suitable for your project. References Chessboard Advantages and Disadvantages Chessboard detection — Wikipedia Automatic chessboard corner detection method — Liu — 2016 — IET Image Processing — Wiley Online Library Sensors \| Free Full-Text \| Automatic Chessboard Detection for Intrinsic and Extrinsic Camera Parameter Calibration (mdpi.com) Ready to try it out? Use the provided code and see how well you can calibrate your camera for your next computer vision project! 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2359	https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_6?srsltid=AfmBOooeeI7edA1GrAVif18vKvHGU5ptphNSzsKUGocUvnvmA43JIFhX	Art of Problem Solving 2000 AIME I Problems/Problem 6 - 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 2000 AIME I Problems/Problem 6 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2000 AIME I Problems/Problem 6 Contents 1 Problem 2 Solutions 2.1 Solution 1 2.2 Solution 2 2.3 Solution 3 2.4 Solution 4 (Similar to Solution 3) 2.5 Solution 5 3 See also Problem For how many ordered pairs of integers is it true that and that the arithmetic mean of and is exactly more than the geometric mean of and ? Solutions Solution 1 Because , we only consider . For simplicity, we can count how many valid pairs of that satisfy our equation. The maximum that can be is because must be an integer (this is because , an integer). Then , and we continue this downward until , in which case . The number of pairs of , and so is then . Solution 2 Let = and = , where and are positive. Then This makes counting a lot easier since now we just have to find all pairs that differ by 2. Because , then we can use all positive integers less than 1000 for and . We know that because , we get . We can count even and odd pairs separately to make things easier: Odd: Even: This makes odd pairs and even pairs, for a total of pairs. Note: We are counting the pairs for the values of and , which, when squared, translate to the pairs of we are trying to find. Solution 3 Since the arithmetic mean is 2 more than the geometric mean, . We can multiply by 2 to get . Subtracting 4 and squaring gives Notice that , so the problem asks for solutions of Since the left hand side is a perfect square, and 16 is a perfect square, must also be a perfect square. Since , must be from to , giving at most 999 options for . However if , you get , which has solutions and . Both of those solutions are not less than , so cannot be equal to 1. If , you get , which has 2 solutions, , and . 16 is not less than 4, and cannot be 0, so cannot be 4. However, for all other , you get exactly 1 solution for , and that gives a total of pairs. asbodke Solution 4 (Similar to Solution 3) Rearranging our conditions to Thus, Now, let Plugging this back into our expression, we get There, a unique value of is formed for every value of . However, we must have and Therefore, there are only pairs of Solution by Williamgolly Solution 5 First we see that our condition is . Then we can see that . From trying a simple example to figure out conditions for , we want to find so we can isolate for . From doing the example we can note that we can square both sides and subtract : (note it is negative because . Clearly the square root must be an integer, so now let . Thus . Thus . We can then find , and use the quadratic formula on to ensure they are and respectively. Thus we get that can go up to 999 and can go down to , leaving possibilities for . See also 2000 AIME I (Problems • Answer Key • Resources) Preceded by Problem 5Followed by Problem 7 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Intermediate Algebra Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2360	https://personal.math.ubc.ca/~CLP/CLP1/clp_1_dc/sec_C_1.html	Published Time: Fri, 16 Aug 2024 18:20:02 GMT Newton’s Method Skip to main content CLP-1 Differential Calculus Joel Feldman, Andrew Rechnitzer, Elyse Yeager Contents Search Book close Search Results: No results. PrevUpNext Front Matter Colophon Preface Dedication Acknowledgements Using the exercises in this book Feedback about the text 0 The basics 0.1 Numbers 0.1.1 More on Real Numbers 0.2 Sets 0.3 Other Important Sets 0.3.1 More on Sets 0.4 Functions 0.5 Parsing Formulas 0.6 Inverse Functions 1 Limits 1.1 Drawing Tangents and a First Limit 1.1.1 Drawing Tangents and a First Limit 1.1.2 Exercises 1.2 Another Limit and Computing Velocity 1.2.1 Another Limit and Computing Velocity 1.2.2 Exercises 1.3 The Limit of a Function 1.3.1 The Limit of a Function 1.3.2 Exercises 1.4 Calculating Limits with Limit Laws 1.4.1 Calculating Limits with Limit Laws 1.4.2 Exercises 1.5 Limits at Infinity 1.5.1 Limits at Infinity 1.5.2 Exercises 1.6 Continuity 1.6.1 Continuity 1.6.2 Quick Aside — One-sided Continuity 1.6.3 Back to the Main Text 1.6.4 Exercises 1.7(Optional) — Making the Informal a Little More Formal 1.8(Optional) — Making Infinite Limits a Little More Formal 1.9(Optional) — Proving the Arithmetic of Limits 2 Derivatives 2.1 Revisiting Tangent Lines 2.1.1 Revisiting Tangent Lines 2.1.2 Exercises 2.2 Definition of the Derivative 2.2.1 An Important Point (and Some Notation) 2.2.2 Back to Computing Some Derivatives 2.2.3 Where is the Derivative Undefined? 2.2.4 Exercises 2.3 Interpretations of the Derivative 2.3.1 Instantaneous Rate of Change 2.3.2 Slope 2.3.3 Exercises 2.4 Arithmetic of Derivatives - a Differentiation Toolbox 2.4.1 Arithmetic of Derivatives - a Differentiation Toolbox 2.4.2 Exercises 2.5 Proofs of the Arithmetic of Derivatives 2.5.1 Proof of the Linearity of Differentiation (Theorem 2.4.2) 2.5.2 Proof of the Product Rule (Theorem 2.4.3) 2.5.3(Optional) — Proof of the Quotient Rule (Theorem 2.4.5) 2.6 Using the Arithmetic of Derivatives – Examples 2.6.1 Using the Arithmetic of Derivatives – Examples 2.6.2 Exercises 2.7 Derivatives of Exponential Functions 2.7.1 Whirlwind Review of Logarithms 2.7.1.1 Logarithmic Functions 2.7.2 Back to that Limit 2.7.3 Exercises 2.8 Derivatives of Trigonometric Functions 2.8.1 These Proofs are Optional, the Results are Not. 2.8.2 Step 1: d d x{sin⁡x}\|x=0 2.8.3 Proof that lim h→0 sin⁡h h=1 2.8.4 Step 2: d d x{cos⁡x}\|x=0 2.8.4.1 Method 1 — Multiply by the “Conjugate” 2.8.4.2 Method 2 — via the Double Angle Formula 2.8.5 Step 3: d d x{sin⁡x} and d d x{cos⁡x} for General x 2.8.6 Step 4: the Remaining Trigonometric Functions 2.8.7 Summary 2.8.8 Exercises 2.9 One More Tool – the Chain Rule 2.9.1 Statement of the Chain Rule 2.9.2(Optional) — Derivation of the Chain Rule 2.9.3 Chain Rule Examples 2.9.4 Exercises 2.10 The Natural Logarithm 2.10.1 Back to d d x a x 2.10.2 Logarithmic Differentiation 2.10.3 Exercises 2.11 Implicit Differentiation 2.11.1 Implicit Differentiation 2.11.2 Exercises 2.12 Inverse Trigonometric Functions 2.12.1 Derivatives of Inverse Trig Functions 2.12.2 Exercises 2.13 The Mean Value Theorem 2.13.1 Rolle’s Theorem 2.13.2 Back to the MVT 2.13.3(Optional) — Why is the MVT True 2.13.4 Be Careful with Hypotheses 2.13.5 Exercises 2.14 Higher Order Derivatives 2.14.1 Higher Order Derivatives 2.14.2 Exercises 2.15(Optional) — Is lim x→c f′(x) Equal to ?f′(c)? 3 Applications of derivatives 3.1 Velocity and Acceleration 3.1.1 Velocity and Acceleration 3.1.2 Exercises 3.2 Related Rates 3.2.1 Related Rates 3.2.2 Exercises 3.3 Exponential Growth and Decay 3.3.1 Carbon Dating 3.3.2 Newton’s Law of Cooling 3.3.3 Population Growth 3.3.3.1(Optional) — Logistic Population Growth 3.3.4 Exercises 3.3.4 Exercises for §3.3.1 3.3.4 Exercises for §3.3.2 3.3.4 Exercises for §3.3.3 3.3.4 Further problems for §3.3 3.4 Taylor Polynomials 3.4.1 Zeroth Approximation — the Constant Approximation 3.4.2 First Approximation — the Linear Approximation 3.4.3 Second Approximation — the Quadratic Approximation 3.4.4 Whirlwind Tour of Summation Notation 3.4.5 Still Better Approximations — Taylor Polynomials 3.4.6 Some Examples 3.4.7 Estimating Change and ,Δ x,Δ y Notation 3.4.8 Further Examples 3.4.9 The Error in the Taylor Polynomial Approximations 3.4.10(Optional) — Derivation of the Error Formulae 3.4.11 Exercises 3.4.11 Exercises for §3.4.1 3.4.11 Exercises for §3.4.2 3.4.11 Exercises for §3.4.3 3.4.11 Exercises for §3.4.4 3.4.11 Exercises for §3.4.5 3.4.11 Exercises for §3.4.6 3.4.11 Exercises for §3.4.7 3.4.11 Exercises for §3.4.8 3.4.11 Further problems for §3.4 3.5 Optimisation 3.5.1 Local and Global Maxima and Minima 3.5.2 Finding Global Maxima and Minima 3.5.3 Max/Min Examples 3.5.4 Exercises 3.5.4 Exercises for §3.5.1 3.5.4 Exercises for §3.5.2 3.5.4 Exercises for §3.5.3 3.6 Sketching Graphs 3.6.1 Domain, Intercepts and Asymptotes 3.6.2 First Derivative — Increasing or Decreasing 3.6.3 Second Derivative — Concavity 3.6.4 Symmetries 3.6.5 A Checklist for Sketching 3.6.5.1 A Sketching Checklist 3.6.6 Sketching Examples 3.6.7 Exercises 3.6.7 Exercises for §3.6.1 3.6.7 Exercises for §3.6.2 3.6.7 Exercises for §3.6.3 3.6.7 Exercises for §3.6.4 3.6.7 Exercises for §3.6.6 3.7 L’Hôpital’s Rule, Indeterminate Forms 3.7.1 L’Hôpital’s Rule and Indeterminate Forms 3.7.1.1 Optional — Proof of Part (b) of l’Hôpital’s Rule 3.7.2 Standard Examples 3.7.3 Variations 3.7.3.1 Limits at ±∞ 3.7.3.2∞∞ indeterminate form 3.7.3.3 Optional — Proof of l’Hôpital’s Rule for ∞∞ 3.7.3.4 0⋅∞ indeterminate form 3.7.3.5∞−∞ indeterminate form 3.7.3.6 1∞ indeterminate form 3.7.3.7 0 0 indeterminate form 3.7.3.8∞0 indeterminate form 3.7.4 Exercises 4 Towards Integral Calculus 4.1 Introduction to Antiderivatives 4.1.1 Introduction to Antiderivatives 4.1.2 Exercises Appendices A High School Material A.1 Similar Triangles A.2 Pythagoras A.3 Trigonometry — Definitions A.4 Radians, Arcs and Sectors A.5 Trigonometry — Graphs A.6 Trigonometry — Special Triangles A.7 Trigonometry — Simple Identities A.8 Trigonometry — Add and Subtract Angles A.9 Inverse Trig Functions A.10 Areas A.11 Volumes A.12 Powers A.13 Logarithms A.14 You Should be Able to Derive B Origin of Trig, Area and Volume Formulas B.1 Theorems about Triangles B.1.1 Thales’ Theorem B.1.2 Pythagoras B.2 Trigonometry B.2.1 Angles — Radians vs Degrees B.2.2 Trig Function Definitions B.2.3 Important Triangles B.2.4 Some More Simple Identities B.2.5 Identities — Adding Angles B.2.6 Identities — Double-angle Formulas B.2.7 Identities — Extras B.2.7.1 Sums to Products B.2.7.2 Products to sums B.3 Inverse Trigonometric Functions B.4 Cosine and Sine Laws B.4.1 Cosine Law or Law of Cosines B.4.2 Sine Law or Law of Sines B.5 Circles, cones and spheres B.5.1 Where Does the Formula for the Area of a Circle Come From? B.5.2 Where Do These Volume Formulas Come From? C Root Finding C.1 Newton’s Method C.2 The Error Behaviour of Newton’s Method C.3 The false position (regula falsi) method C.4 The secant method C.5 The Error Behaviour of the Secant Method D Hints for Exercises E Answers to Exercises F Solutions to Exercises Section C.1 Newton’s Method Newton’s method 1 The algorithm that we are about to describe grew out of a method that Newton wrote about in 1669. But the modern method incorporates substantial changes introduced by Raphson in 1690 and Simpson in 1740. , also known as the Newton-Raphson method, is another technique for generating numerical approximate solutions to equations of the form .f(x)=0. For example, one can easily get a good approximation to 2 by applying Newton’s method to the equation .x 2−2=0. This will be done in Example C.1.2, below. Here is the derivation of Newton’s method. We start by simply making a guess for the solution. For example, we could base the guess on a sketch of the graph of .f(x). Call the initial guess .x 1. Next recall, from Theorem2.3.4, that the tangent line to y=f(x) at x=x 1 is ,y=F(x), where F(x)=f(x 1)+f′(x 1)(x−x 1) Usually F(x) is a pretty good approximation to f(x) for x near .x 1. So, instead of trying to solve ,f(x)=0, we solve the linear equation F(x)=0 and call the solution .x 2. 0=F(x)=f(x 1)+f′(x 1)(x−x 1)⟺x−x 1=−f(x 1)f′(x 1)⟺x=x 2=x 1−f(x 1)f′(x 1) Note that if f(x) were a linear function, then F(x) would be exactly f(x) and x 2 would solve f(x)=0 exactly. Now we repeat, but starting with the (second) guess x 2 rather than .x 1. This gives the (third) guess .x 3=x 2−f(x 2)f′(x 2). And so on. By way of summary, Newton’s method is Make a preliminary guess .x 1. Define .x 2=x 1−f(x 1)f′(x 1). Iterate. That is, for each natural number ,n, once you have computed ,x n, define Equation C.1.1.(Newton’s method). x n+1=x n−f(x n)f′(x n) Example C.1.2.(Approximating 2). In this example we compute, approximately, the square root of two. We will of course pretend that we do not already know that .2=1.41421⋯. So we cannot find it by solving, approximately, the equation .f(x)=x−2=0. Instead we apply Newton’s method to the equation f(x)=x 2−2=0 Since ,f′(x)=2 x, Newton’s method says that we should generate approximate solutions by iteratively applying x n+1=x n−f(x n)f′(x n)=x n−x n 2−2 2 x n=x n 2+1 x n We need a starting point. Since 1 2=1<2 and ,2 2=4>2, the square root of two must be between 1 and ,2, so let’s start Newton’s method with the initial guess .x 1=1.5. Here goes 2 The following computations have been carried out in double precision, which is computer speak for about 15 significant digits. We are displaying each x n rounded to 10 significant digits (9 decimal places). So each displayed x n has not been impacted by roundoff error, and still contains more decimal places than are usually needed. : x 1=1.5 x 2=1 2 x 1+1 x 1=1 2(1.5)+1 1.5=1.416666667 x 3=1 2 x 2+1 x 2=1 2(1.416666667)+1 1.416666667=1.414215686 x 4=1 2 x 3+1 x 3=1 2(1.414215686)+1 1.414215686=1.414213562 x 5=1 2 x 4+1 x 4=1 2(1.414213562)+1 1.414213562=1.414213562 It looks like the x n’s, rounded to nine decimal places, have stabilized to .1.414213562. So it is reasonable to guess that ,2, rounded to nine decimal places, is exactly .1.414213562. Recalling that all numbers 1.4142135615≤y<1.4142135625 round to ,1.414213562, we can check our guess by evaluating f(1.4142135615) and .f(1.4142135625). Since f(1.4142135615)=−2.5×10−9<0 and f(1.4142135625)=3.6×10−10>0 the square root of two must indeed be between 1.4142135615 and .1.4142135625. Example C.1.3.(Approximating π). In this example we compute, approximately, π by applying Newton’s method to the equation f(x)=sin⁡x=0 starting with .x 1=3. Since ,f′(x)=cos⁡x, Newton’s method says that we should generate approximate solutions by iteratively applying x n+1=x n−f(x n)f′(x n)=x n−sin⁡x n cos⁡x n=x n−tan⁡x n Here goes x 1=3 x 2=x 1−tan⁡x 1=3−tan⁡3=3.142546543 x 3=3.142546543−tan⁡3.142546543=3.141592653 x 4=3.141592653−tan⁡3.141592653=3.141592654 x 5=3.141592654−tan⁡3.141592654=3.141592654 Since f(3.1415926535)=9.0×10−11>0 and ,f(3.1415926545)=−9.1×10−11<0,π must be between 3.1415926535 and .3.1415926545. Of course to compute π in this way, we (or at least our computers) have to be able to evaluate tan⁡x for various values of .x. Taylor expansions can help us do that. See Example3.4.22. Example C.1.4.wild instability. This example illustrates how Newton’s method can go badly wrong if your initial guess is not good enough. We’ll try to solve the equation f(x)=arctan⁡x=0 starting with .x 1=1.5. (Of course the solution to f(x)=0 is just ;x=0; we chose x 1=1.5 for demonstration purposes.) Since the derivative ,f′(x)=1 1+x 2, Newton’s method gives x n+1=x n−f(x n)f′(x n)=x n−(1+x n 2)arctan⁡x n So 3 Once again, the following computations have been carried out in double precision. This time, it is clear that the x n’s are growing madly as n increases. So there is not much point to displaying many decimal places and we have not done so. x 1=1.5 x 2=1.5−(1+1.5 2)arctan⁡1.5=−1.69 x 3=−1.69−(1+1.69 2)arctan⁡(−1.69)=2.32 x 4=2.32−(1+2.32 2)arctan⁡(2.32)=−5.11 x 5=−5.11−(1+5.11 2)arctan⁡(−5.11)=32.3 x 6=32.3−(1+32.3 2)arctan⁡(32.3)=−1575 x 7=3,894,976 Looks pretty bad! Our x n’s are not settling down at all! The figure below shows what went wrong. In this figure, y=F 1(x) is the tangent line to y=arctan⁡x at .x=x 1. Under Newton’s method, this tangent line crosses the x-axis at .x=x 2. Then y=F 2(x) is the tangent to y=arctan⁡x at .x=x 2. Under Newton’s method, this tangent line crosses the x-axis at .x=x 3. And so on. The problem arose because the x n’s were far enough from the solution, ,x=0, that the tangent line approximations, while good approximations to f(x) for ,x≈x n, were very poor approximations to f(x) for .x≈0. In particular, y=F 1(x) (i.e. the tangent line at x=x 1) was a bad enough approximation to y=arctan⁡x for x≈0 that x=x 2 (i.e. the value of x where y=F 1(x) crosses the x-axis) is farther from the solution x=0 than our original guess .x=x 1. If we had started with x 1=0.5 instead of ,x 1=1.5, Newton’s method would have succeeded very nicely: x 1=0.5 x 2=−0.0796 x 3=0.000335 x 4=−2.51×10−11 Example C.1.5.interest rate. A car dealer sells a new car for $23,520. He also offers to finance the same car for payments of $420 per month for five years. What interest rate is this dealer charging? Solution. By way of preparation, we’ll start with a simpler problem. Suppose that you will have to make a single $420 payment n months in the future. The simpler problem is to determine how much money you have to deposit now in an account that pays an interest rate of 100 r% per month, compounded monthly 4 “Compounded monthly”, means that, each month, interest is paid on the accumulated interest that was paid in all previous months. , in order to be able to make the $420 payment in n months. Let’s denote by P the initial deposit. Because the interest rate is 100 r% per month, compounded monthly, the first month’s interest is .P×r. So at the end of month #1, the account balance is .P+P r=P(1+r). The second month’s interest is .[P(1+r)]×r. So at the end of month #2, the account balance is .P(1+r)+P(1+r)r=P(1+r)2. And so on. So at the end of n months, the account balance is .P(1+r)n. In order for the balance at the end of n months, ,P(1+r)n, to be $420, the initial deposit has to be .P=420(1+r)−n. That is what is meant by the statement “The present value 5 Inflation means that prices of goods (typically) increase with time, and hence $100 now is worth more than $100 in 10 years time. The term “present value” is widely used in economics and finance to mean “the current amount of money that will have a specified value at a specified time in the future”. It takes inflation into account. If the money is invested, it takes into account the rate of return of the investment. We recommend that the interested reader do some search-engining to find out more. of a $420 payment made n months in the future, when the interest rate is 100 r% per month, compounded monthly, is .420(1+r)−n.” Now back to the original problem. We will be making 60 monthly payments of $420. The present value of all 60 payments is 6 Don’t worry if you don’t know how to evaluate such sums. They are called geometric sums, and will be covered in the CLP-2 text. (See (1.1.3) in the CLP-2 text. In any event, you can check that this is correct, by multiplying the whole equation by .1−(1+r)−1. When you simplify the left hand side, you should get the right hand side. 420(1+r)−1+420(1+r)−2+⋯+420(1+r)−60=420(1+r)−1−(1+r)−61 1−(1+r)−1=420 1−(1+r)−60(1+r)−1=420 1−(1+r)−60 r The interest rate 100 r% being charged by the car dealer is such that the present value of 60 monthly payments of $420 is $23520. That is, the monthly interest rate being charged by the car dealer is the solution of or or or or 23520=420 1−(1+r)−60 r or 56=1−(1+r)−60 r or 56 r=1−(1+r)−60 or 56 r(1+r)60=(1+r)60−1 or(1−56 r)(1+r)60=1 Set .f(r)=(1−56 r)(1+r)60−1. Then f′(r)=−56(1+r)60+60(1−56 r)(1+r)59 or f′(r)=−56(1+r)+60(1−56 r)59=(4−3416 r)(1+r)59 Apply Newton’s method with an initial guess of .r 1=.002. (That’s 0.2% per month or 2.4% per year.) Then r 2=r 1−(1−56 r 1)(1+r 1)60−1(4−3416 r 1)(1+r 1)59=0.002344 r 3=r 2−(1−56 r 2)(1+r 2)60−1(4−3416 r 2)(1+r 2)59=0.002292 r 4=r 3−(1−56 r 3)(1+r 3)60−1(4−3416 r 3)(1+r 3)59=0.002290 r 5=r 4−(1−56 r 4)(1+r 4)60−1(4−3416 r 4)(1+r 4)59=0.002290 So the interest rate is 0.229% per month or 2.75% per year. PrevTopNext Feedback
2361	https://www.quora.com/What-is-the-value-of-the-integral-of-sin-x-from-0-to-pi-in-terms-of-n	Something went wrong. Wait a moment and try again. Sin Function Trigonometric Integrals Functions (mathematics) Calculus 1 Sine and Cosine Trigonometry Maths Integration (mathematics) Integral Functions 5 What is the value of the integral of sin x from 0 to pi in terms of n? · To evaluate the integral of sinx from 0 to π, we can set up the integral as follows: ∫π0sinxdx Now, we can compute this integral: Find the antiderivative of sinx: The antiderivative of sinx is −cosx. Evaluate the definite integral: ∫π0sinxdx=[−cosx]π0 Now, plug in the limits: =−cos(π)−(−cos(0))=−(−1)−(−1)=1−(−1)=1+1=2 Thus, the value of the integral is: ∫π0sinxdx=2 In terms of n, if you want to express this as n, you can simply say: ∫π0sinxdx=2where n=2 Alexandru Carausu Former Former University Associate Professor at Universitatea Tehnica "Gh. Asachi" Iasi (1966–2010) · Author has 3K answers and 855.4K answer views · 2y The answer is very simple, but the last syntagm “ in terms of n “ is a little strange. The anti-derivative (indefinite integral) of sin x is - cos x . Hence ∫ _0^π sin x dx = [ - cos x ] \|_0^π = [ cos x ]_π^0 = cos 0 - cos π = 1 - ( - 1) = 2 . This simple Riemann definite integral does not depend, in any way, on a natural number n . However, one of the definitions of such integrals is expressed as the sum (limit) of a Riemann sum which is a series with positive terms. Calvin L. 2nd year mathematics student · Author has 10K answers and 2.2M answer views · 1y What is n? The integral itself: ∫π0sinx =−cosx∣∣π0 =−(−1)+1 =2 Dando Porsaco 1y The simple answer is 2, as it doesn’t matter what n is. ANDRIAMANANTENASOA Joseph Lives in Madagascar (1984–present) · Author has 80 answers and 22.1K answer views · 1y The integral of sin(x) from 0 to π can be expressed as: ∫[0 to π] sin(x) dx This integral evaluates to 2. So, in terms of n, the value of the integral of sin(x) from 0 to π is just 2. Related questions What is the integral of sin x from 0 to π? What is the value of the definite integral of sin(x) from 0 to π/2? What is the value of the integral of sin(x) + cos(x) from 0 to π/3? What is the value of the integral from 0 to pi of the function sin(x) /x? How do I calculate integral [math]I_n=\displaystyle \int \limits_{0}^{\frac{\pi}{2}} (\sin(x))^n \mathrm dx[/math] ? Stickie Python hobbyist. · Author has 134 answers and 51K answer views · 1y Related Why is the integral of cos (nx) from -pi to pi equal to 0? We have that if [math]f(x)[/math] is odd, we would have that: [math]\displaystyle\int_{-a}^af(x)\,\mathrm dx=0[/math] Whereas if [math]f(x)[/math] is even, we have that [math]\displaystyle\int_{-a}^af(x)\,\mathrm dx=2\int_0^af(x)\,\mathrm dx[/math] Your confusion seems to be that the identity [math]\displaystyle\int_{-\pi}^{\pi}\cos(nx)\,\mathrm dx=0[/math] resembles the one for an odd integrand, yet the function [math]\cos(nx)[/math] is even, so you expect [math]\displaystyle\int_{-\pi}^\pi\cos(nx)\,\mathrm dx=2\int_0^\pi\cos(nx)\,\mathrm dx[/math] And you'd be right, the second identity is true, because [math]\cos(nx)[/math] is an even function. However, the first identity is also true when [math]n[/math] is an in We have that if [math]f(x)[/math] is odd, we would have that: [math]\displaystyle\int_{-a}^af(x)\,\mathrm dx=0[/math] Whereas if [math]f(x)[/math] is even, we have that [math]\displaystyle\int_{-a}^af(x)\,\mathrm dx=2\int_0^af(x)\,\mathrm dx[/math] Your confusion seems to be that the identity [math]\displaystyle\int_{-\pi}^{\pi}\cos(nx)\,\mathrm dx=0[/math] resembles the one for an odd integrand, yet the function [math]\cos(nx)[/math] is even, so you expect [math]\displaystyle\int_{-\pi}^\pi\cos(nx)\,\mathrm dx=2\int_0^\pi\cos(nx)\,\mathrm dx[/math] And you'd be right, the second identity is true, because [math]\cos(nx)[/math] is an even function. However, the first identity is also true when [math]n[/math] is an integer, because [math]\int_0^\pi\cos(nx)\,\mathrm dx=0[/math]. If you know how to integrate trigonometric functions, the integral is fairly easy to solve, but I think a geometric understanding of the problem is useful: Firstly, assume [math]n[/math] is positive, since if [math]n[/math] is negative, we have [math]\cos(-nx)=\cos(nx)[/math], so we can just switch to [math]-n[/math]. We have the following identity: [math]\displaystyle\int_0^\pi\cos(nx)\,\mathrm dx=\frac1n\int_0^{n\pi}\cos(x)\,\mathrm dx[/math] Geometrically, this uses the fact that [math]\cos(nx)[/math] is a scaling of [math]\cos(x)[/math], where values from the interval [math]0[/math] to [math]n\pi[/math] are squashed together into the interval [math]0[/math] to [math]\pi[/math]. If we want the area under the squashed graph, we can stretch the graph back out, but we need to adjust by a factor of [math]\frac1n[/math] to compensate for the stretching of area we're doing. Now, if you look at the cosine graph in the interval [math]0[/math] to [math]\pi[/math], you get a region of positive area from [math]0[/math] to [math]\frac\pi2[/math], and a region of negative area from [math]\frac\pi2[/math] to [math]\pi[/math]. They have the same absolute area due to symmetry of the cosine function, so these areas cancel each other out and we have that: [math]\displaystyle\int_0^\pi\cos(x)\,\mathrm dx=0[/math] The rest of the cosine curve from [math]0[/math] to [math]n\pi[/math] can be broken down into [math]n[/math] copies of these regions (some of them reflected in the y-axis, which doesn't affect their area), and we know the signed area of these regions are zero, so the sum of all of them must also be zero, so the integral is zero. Promoted by Grammarly Grammarly Great Writing, Simplified · Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. 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Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by A. Skodras , MSc Mathematics & Applied Mathematics · Author has 8.4K answers and 20.9M answer views · 3y Related What is the value of [math]\int_0^{\pi} \sqrt{\sin{x} - \sin^2{x}} \, dx[/math] ? We want to evaluate the definite integral [math]I = \displaystyle \int_0^{\pi} \sqrt{\sin{x} - \sin^2{x}} \, dx. \tag{}[/math] Since the sine function on [math][0, \pi][/math] is symmetric about [math]x = \frac{\pi}{2}[/math], we can rewrite the integral in question as [math]I = \displaystyle 2\int_0^{\pi/2} \sqrt{\sin{x} - \sin^2{x}} \, dx. \tag{}[/math] Now, we can make the substitution [math]w = \sin{x}[/math]. Then differentiating the substitution, followed by the use of the Pythagorean trigonometric identity, yields [math]dw = \cos{x} \, dx = \sqrt{1 - w^2} \, dx[/math], and we obtain [math]I = \displaystyle 2\int_0^1 \sqrt{w - w^2} \cdot \frac{dw}{\sqrt{1 - w^2}} = 2\int_0[/math] We want to evaluate the definite integral [math]I = \displaystyle \int_0^{\pi} \sqrt{\sin{x} - \sin^2{x}} \, dx. \tag{}[/math] Since the sine function on [math][0, \pi][/math] is symmetric about [math]x = \frac{\pi}{2}[/math], we can rewrite the integral in question as [math]I = \displaystyle 2\int_0^{\pi/2} \sqrt{\sin{x} - \sin^2{x}} \, dx. \tag{}[/math] Now, we can make the substitution [math]w = \sin{x}[/math]. Then differentiating the substitution, followed by the use of the Pythagorean trigonometric identity, yields [math]dw = \cos{x} \, dx = \sqrt{1 - w^2} \, dx[/math], and we obtain [math]I = \displaystyle 2\int_0^1 \sqrt{w - w^2} \cdot \frac{dw}{\sqrt{1 - w^2}} = 2\int_0^1 \sqrt{\frac{w}{w+1}} \, dw. \tag{}[/math] Having eliminated the trigonometric functions from the integral, we now eliminate the radical from the integrand by making the rationalizing substitution [math]t = \sqrt{\frac{w}{w+1}}[/math]. We can rewrite this substitution as [math]w = \frac{t^2}{1 - t^2}[/math], and then substituting this into the integral gives us [math]I = \displaystyle 2 \int_0^{1/\sqrt{2}} t \cdot \frac{2t}{(1 - t^2)^2} \, dt = \int_0^{1/\sqrt{2}} \frac{4t^2}{(1 - t^2)^2} \, dt. \tag{}[/math] At this point in this problem, we can directly evaluate this integral by trigonometric substitution or partial fractions. Instead, we opt for using integration by parts (to lower the degree of the factor in the denominator) by setting [math]u = t[/math] and [math]dv = \frac{4t \, dt}{(1 - t^2)^2}[/math]. Then, we find that [math]\begin{align} I &= \displaystyle t \cdot \frac{2}{1 - t^2} \Bigg\|_0^{1/\sqrt{2}} - \int_0^{1/\sqrt{2}} 1 \cdot \frac{2}{1 - t^2} \, dt\ &= 2\sqrt{2} - \int_0^{1/\sqrt{2}} \frac{2}{1 - t^2} \, dt. \end{align} \tag{}[/math] Then, we make the trigonometric substitution [math]t = \sin{\theta}[/math]. This allows us to conclude that [math]\begin{align} I &= \displaystyle 2\sqrt{2} - \int_0^{\pi/4} \frac{2}{\cos^2{\theta}} \cdot \cos{\theta} \, d\theta\ &= 2\sqrt{2} - \int_0^{\pi/4} 2 \sec{\theta} \, d\theta\ &= 2\sqrt{2} - \Big(2 \ln(\sec{\theta} + \tan{\theta}) \Bigg\|_0^{\pi/4}\Big)\ &= \boxed{2\sqrt{2} - 2 \ln(\sqrt{2} + 1)}. \end{align} \tag{}[/math] Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.4K answers and 20.9M answer views · 1y Related How do I calculate this integral. Integral of (x sin x) / (1 + sin²x) dx from 0 to pi? We are given the definite integral [math]I = \displaystyle \int_0^{\pi} \frac{x \sin{x}}{1 + \sin^2{x}} \, dx. \tag{}[/math] In order to evaluate this integral, we start by eliminating the factor of [math]x[/math] in the numerator. To accomplish this, we first make the interval-inverting substitution replacing [math]x[/math] with [math]\pi - x[/math]. Since [math]\sin(\pi - x) = \sin{x}[/math], we obtain [math]I = \displaystyle \int_{\pi}^0 \frac{(\pi - x) \sin(\pi - x)}{1 + \sin^2(\pi - x)} \cdot (-dx) = \int_0^{\pi} \frac{(\pi - x) \sin{x}}{1 + \sin^2{x}} \, dx. \tag{}[/math] Then, since the bounds of integration have not changed, and the integrands are identical except We are given the definite integral [math]I = \displaystyle \int_0^{\pi} \frac{x \sin{x}}{1 + \sin^2{x}} \, dx. \tag{}[/math] In order to evaluate this integral, we start by eliminating the factor of [math]x[/math] in the numerator. To accomplish this, we first make the interval-inverting substitution replacing [math]x[/math] with [math]\pi - x[/math]. Since [math]\sin(\pi - x) = \sin{x}[/math], we obtain [math]I = \displaystyle \int_{\pi}^0 \frac{(\pi - x) \sin(\pi - x)}{1 + \sin^2(\pi - x)} \cdot (-dx) = \int_0^{\pi} \frac{(\pi - x) \sin{x}}{1 + \sin^2{x}} \, dx. \tag{}[/math] Then, since the bounds of integration have not changed, and the integrands are identical except for the linear factors, averaging the two versions of [math]I[/math] together eliminates [math]x[/math], as advertised: [math]I = \displaystyle \frac{1}{2} \int_0^{\pi} \frac{(x + (\pi - x)) \sin{x}}{1 + \sin^2{x}} \, dx = \frac{\pi}{2} \int_0^{\pi} \frac{\sin{x}}{1 + \sin^2{x}} \, dx. \tag{}[/math] The remainder of this evaluation is fairly standard. We next rewrite the denominator with the classic Pythagorean identity to set up the substitution [math]t\sqrt{2} = \cos{x}[/math]: [math]\begin{align} I &= \displaystyle \frac{\pi}{2} \int_0^{\pi} \frac{\sin{x}}{1 + (1 - \cos^2{x})} \, dx\ &= \frac{\pi}{2} \int_0^{\pi} \frac{\sin{x}}{2 - \cos^2{x}} \, dx\ &= \frac{\pi}{2} \int_{1/\sqrt{2}}^{-1/\sqrt{2}} \frac{1}{2 - 2t^2} \cdot (-\sqrt{2} \, dt)\ &= \frac{\pi\sqrt{2}}{4} \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \frac{1}{1 - t^2} \, dt\ &= \frac{\pi\sqrt{2}}{2} \int_0^{1/\sqrt{2}} \frac{1}{1 - t^2} \, dt, \text{ even integrand}. \end{align} \tag{}[/math] Now, we can complete the evaluation of this integral by applying partial fractions: [math]\begin{align} I &= \displaystyle \frac{\pi\sqrt{2}}{2} \int_0^{1/\sqrt{2}} \frac{1}{2} \Big(\frac{1}{1 + t} + \frac{1}{1 - t}\Big) \, dt\ &= \frac{\pi\sqrt{2}}{4} \ln\Big(\frac{1 + t}{1 - t}\Big) \Bigg\|_0^{1/\sqrt{2}}\ &= \frac{\pi\sqrt{2}}{4} \ln\Big(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\Big). \end{align} \tag{}[/math] Finally, since math(\sqrt{2} - 1) = 1[/math], we can simplify the answer above by rationalizing the denominator to the argument of the logarithm. Then, we conclude that [math]I = \displaystyle \int_0^{\pi} \frac{x \sin{x}}{1 + \sin^2{x}} \, dx = \boxed{\frac{\pi}{\sqrt{2}} \ln(1 + \sqrt{2}) \approx 1.9579}. \tag{}[/math] Sponsored by Amazon Web Services (AWS) Want to level up your AWS game? Register for re:Invent 2025. 5 days of advanced architecture workshops, serverless deep dives, and coding labs. Las Vegas, Dec 1-5. Gary Crenshaw BAM! Math'd. · Author has 223 answers and 600.8K answer views · 5y Related What is the value of [math]\int_{0}^{2 \pi} x^{n} \sin x \mathrm{d} x[/math] ? In his answer, Brian Sittinger found that by letting [math]I_n = \int\limits_0^{2\pi}x^n\sin x dx[/math], we get the recurrence relation [math]I_n = -n(n-1)I_{n-2}-(2\pi)^n[/math]. In this answer, we solve the recurrence relation. For those of you unfamiliar with how to solve recurrence relations, you may want to read my answer to another recurrence relation problem, where I outline the basic strategies I’ll be using here. First, we eliminate the multiplication by [math]n(n-1)[/math] using a factorial trick; define a new sequence [math]{F_n}[/math] satisfying [math]I_n=n!F_n[/math]; in other words, [math]F_n = \frac{I_n}{n!}[/math]. Substituting that into our recurrence, In his answer, Brian Sittinger found that by letting [math]I_n = \int\limits_0^{2\pi}x^n\sin x dx[/math], we get the recurrence relation [math]I_n = -n(n-1)I_{n-2}-(2\pi)^n[/math]. In this answer, we solve the recurrence relation. For those of you unfamiliar with how to solve recurrence relations, you may want to read my answer to another recurrence relation problem, where I outline the basic strategies I’ll be using here. First, we eliminate the multiplication by [math]n(n-1)[/math] using a factorial trick; define a new sequence [math]{F_n}[/math] satisfying [math]I_n=n!F_n[/math]; in other words, [math]F_n = \frac{I_n}{n!}[/math]. Substituting that into our recurrence, we have: [math]\begin{align} n!F_n&=-n(n-1)(n-2)!F_{n-2}-(2\pi)^n\&=-n!F_{n-2}-(2\pi)^n\F_n&=-F_{n-2}-\frac{(2\pi)^n}{n!}\end{align}[/math] All that’s left is that pesky negative in front of the [math]F_{n-2}[/math]. In my linked answer above, we got rid of that with a multiplication by math^n[/math]. The only problem here is that we are going two at a time. There are ways around this, using a function that repeats [math]1,\ 1,\ -1,\ -1,...[/math] over and over (like [math]\sqrt{2}\sin(\frac{n\pi}2+\frac\pi4)[/math]) or even using something like [math]i^n[/math] which goes [math]1,\ i,\ -1,\ -i,...[/math] but these just serve to overcomplicate matters. Besides, as Brian told us, solving the recurrence relation numerically involves splitting into evens and odds. So let’s just do that, evens first. Let’s assume first that [math]n[/math] is even, so write [math]n=2k[/math]. Then our index of [math]n-2[/math] becomes [math]2k-2=2(k-1)[/math], and the recurrence is: [math]F_{2k}=-F_{2(k-1)}-\frac{(2\pi)^{2k}}{(2k)!}[/math] Now, let’s define our “evens” sequence [math]E_k[/math] so that [math]F_{2k}=(-1)^kE_k[/math], which is equivalent to [math]E_k=(-1)^kF_{2k}[/math]. Then we get: [math]\begin{align}(-1)^kE_k&=-(-1)^{k-1}E_{k-1}-\frac{(2\pi)^{2k}}{(2k)!}\&=(-1)^{k}E_{k-1}-\frac{(2\pi)^{2k}}{(2k)!}\E_k&=E_{k-1}-(-1)^k\frac{(2\pi)^{2k}}{(2k)!}\end{align}[/math] Now, we have a recurrence of the form [math]a_n=a_{n-1}+f(n)[/math], so we can use the summation trick. (If any of this confuses you, go up and read that answer I linked! I promise I’ll wait.) We get: [math]E_k = E_0 - \sum\limits_{i=1}^k(-1)^i\frac{(2\pi)^{2i}}{(2i)!}[/math]. But remember how we defined our sequences. [math]I_{2k}=(2k)!F_{2k}=(2k!)(-1)^kE_k[/math]. In particular, letting [math]k=0[/math] tells us that [math]E_0=I_0=0[/math]. Substituting and simplifying, we get the following closed form: [math]I_{2k}=(2k)!(-1)^{k+1}\sum\limits_{i=1}^k(-1)^i\frac{(2\pi)^{2i}}{(2i)!}[/math] We can do the exact same thing with the odd-numbered terms of the recurrence. I won’t bore you with the details; try it for yourself! (Hint: let [math]O_k=(-1)^kF_{2k+1}[/math]). You should find that [math]I_{2k+1}=(2k+1)!(-1)^{k+1}\left(2\pi+\sum\limits_{i=1}^k(-1)^i\frac{(2\pi)^{2i+1}}{(2i+1)!}\right)[/math]. Let’s try these out. [math]I_0=0[/math], [math]I_1=-2\pi[/math] are easy to see. [math]I_2=2!(-1)(-1)^2\frac{(2\pi)^2}{2!}=4\pi^2[/math] [math]I_3=3!(-1)^2\left(2\pi+(-1)\frac{(2\pi)^3}{3!}\right)=12\pi-8\pi^3[/math] And so on. Related questions Can the integral value of sin [x] /cos [x] from 0 to pi be found? How do I integrate [math]\ln (\sin(x))[/math] from [math]0[/math] to [math]\pi[/math] ? What is the integral of (cos(x) / (cos(x) + sin(x)) in respect to x ranging from 0 to (pi/2)? What is the integration of [math]e^{\sin x}[/math] ? Why does the integral of 1/sin(x) from infinity to 1 not have a limit of -π/2n(pi)? Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by Matt Causley , PhD from New Jersey Institute of Technology (2011) and James McElhatton Ph.D. (Glasgow, 1976) , B.Sc. Mathematics & Chemistry, University of Malta (1967) · Author has 8.4K answers and 20.9M answer views · 3y Related What is the value of: [math]\int_{0}^{\frac{\pi}{2}}\frac{\ln{(\sin^n{x}+\cos^n{x})}}{\sin{x}\cos{x}}dx\ ;\ (n\in \N)[/math] ? Let [math]I[/math] denote the value of the integral in question. Factoring the power of cosine from the argument of the logarithm and algebraically rearranging the integrand, we obtain [math]\begin{align} I &= \displaystyle \int_0^{\pi/2} \frac{\ln(\tan^n{x}+ 1) + n \ln(\cos{x})}{\sin{x} \cos{x}} \, dx\ &= \int_0^{\pi/2} \frac{\ln(\tan^n{x}+ 1) - \frac{n}{2} \ln(\sec^2{x})}{\tan{x}} \cdot \sec^2{x} \, dx. \end{align} \tag{}[/math] Now, this suggests using the substitution [math]t = \tan{x}[/math]. Using this substitution and subsequently splitting the integral at [math]t = 1[/math] yields [math]\begin{align} I &= \displaystyle \int_0^{\infty} \frac[/math] Let [math]I[/math] denote the value of the integral in question. Factoring the power of cosine from the argument of the logarithm and algebraically rearranging the integrand, we obtain [math]\begin{align} I &= \displaystyle \int_0^{\pi/2} \frac{\ln(\tan^n{x}+ 1) + n \ln(\cos{x})}{\sin{x} \cos{x}} \, dx\ &= \int_0^{\pi/2} \frac{\ln(\tan^n{x}+ 1) - \frac{n}{2} \ln(\sec^2{x})}{\tan{x}} \cdot \sec^2{x} \, dx. \end{align} \tag{}[/math] Now, this suggests using the substitution [math]t = \tan{x}[/math]. Using this substitution and subsequently splitting the integral at [math]t = 1[/math] yields [math]\begin{align} I &= \displaystyle \int_0^{\infty} \frac{\ln(t^n + 1) - \frac{n}{2} \ln(t^2 + 1)}{t} \, dt\ &= \int_0^1 \frac{\ln(t^n + 1) - \frac{n}{2} \ln(t^2 + 1)}{t} \, dt + \int_1^{\infty} \frac{\ln(t^n + 1) - \frac{n}{2} \ln(t^2 + 1)}{t} \, dt. \end{align} \tag{}[/math] Next, we use the substitution that replaces [math]t[/math] with [math]\frac{1}{t}[/math] in the second term. After some simplifying, we (surprisingly) find that we can recombine the two integrals! Some details follow below: [math]\begin{align} I &= \displaystyle \int_0^1 \frac{\ln(t^n + 1) - \frac{n}{2} \ln(t^2 + 1)}{t} \, dt + \int_1^0 \frac{\ln(t^{-n} + 1) - \frac{n}{2} \ln(t^{-2} + 1)}{t^{-1}} \cdot \Big(-\frac{dt}{t^2}\Big)\ &= \int_0^1 \frac{\ln(t^n + 1) - \frac{n}{2} \ln(t^2 + 1)}{t} \, dt + \int_0^1 \frac{(-n \ln{t} + \ln(1 + t^n)) - \frac{n}{2} \cdot (-2 \ln{t} + \ln(1 + t^2))}{t} \, dt\ &= \int_0^1 \frac{2\ln(t^n + 1) - n \ln(t^2 + 1)}{t} \, dt. \end{align} \tag{}[/math] At this point, we can finally split the integrand into two integrals (as the resulting integrals finally converge) and make a pair of substitutions [math]w = t^n[/math] and [math]w = t^2[/math], respectively. Again, we can thereafter combine the integrals, this time to something significantly more concise. [math]\begin{align} I &= \displaystyle 2 \int_0^1 \frac{\ln(t^n + 1)}{t} \, dt - n \int_0^1 \frac{\ln(t^2 + 1)}{t} \, dt\ &= 2 \int_0^1 \frac{\ln(w + 1)}{w^{1/n}} \cdot \frac{1}{n} w^{1/n - 1} \, dw - n \int_0^1 \frac{\ln(w + 1)}{w^{1/2}} \cdot \frac{1}{2} w^{1/2 - 1} \, dw\ &= \Big(\frac{2}{n} - \frac{n}{2}\Big) \int_0^1 \frac{\ln(1 + w)}{w} \, dw. \end{align} \tag{}[/math] We are finally at a more `familiar’ integral! Here is another approach other than using the Maclaurin series for the logarithm to evaluate it. Applying integration by parts where [math]u = \ln(1+w)[/math] and [math]dv = \frac{dw}{w}[/math] (and dealing with the lower bound by using a right-sided limit), we obtain [math]\begin{align} \displaystyle \int_0^1 \frac{\ln(1 + w)}{w} \, dw &= \ln(w) \ln(1+w) \Bigg\|_0^1 - \int_0^1 \frac{\ln{w}}{1+w} \, dw\ &= -\int_0^1 \frac{\ln{w}}{1+w} \, dw. \end{align} \tag{}[/math] Now, we apply the geometric series and interchange the order of summation and integration: [math]\begin{align} \displaystyle \int_0^1 \frac{\ln(1 + w)}{w} \, dw &= \sum_{k=0}^{\infty} (-1)^{k+1} \int_0^1 w^k \ln{w} \, dw\ &= \sum_{k=0}^{\infty} (-1)^{k+1} \cdot -\frac{1}{(k+1)^2}, \text{ int. by parts}\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}, \text{ by re-indexing.} \end{align} \tag{}[/math] However, we can rewrite this latter infinite series in terms of a notorious zeta value: [math]\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{\infty} \frac{2}{(2n)^2} = \frac{1}{2} \zeta(2) = \frac{\pi^2}{12}. \tag{}[/math] By putting all of the pieces together, we conclude that [math]I = \displaystyle \int_0^{\pi/2} \frac{\ln(\sin^n{x}+\cos^n{x})}{\sin{x} \cos{x}} \, dx = \boxed{\Big(\frac{2}{n} - \frac{n}{2}\Big) \cdot \frac{\pi^2}{12}}. \tag{}[/math] Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated 15h What's some brutally honest advice that everyone should know? 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Joyneel Bepari Aspiring to compete in the official IMO... · Author has 739 answers and 215.9K answer views · 1y Related What is the value of the integral from 0 to pi of the function sin(x) /x? The integrand [math]\sin(x)/x[/math] has an antiderivative that cannot be expressed as a function in elementary terms. So, in the end, you will need to resort to numerical methods. Wherefore, I’ll be showing you two methods: Using the Antiderivative Trapezoidal Rule Using the Antiderivative The antiderivative of the integrand [math]\sin(x)/x[/math] is given by the infinite summation: [math]\displaystyle\int\dfrac{\sin(x)}x\ dx = \sum_{n=0}^∞\dfrac{(-1)^nx^{2n+1}}{(2n + 1) \cdot (2n + 1)!} + C\tag{}[/math] This can be derived via the Taylor Series for [math]\sin(x)[/math]. Now, we can evaluate the definite integral from [math]x = 0[/math] to [math]x = π[/math]. [math]\begin{align}\d[/math] The integrand [math]\sin(x)/x[/math] has an antiderivative that cannot be expressed as a function in elementary terms. So, in the end, you will need to resort to numerical methods. Wherefore, I’ll be showing you two methods: Using the Antiderivative Trapezoidal Rule Using the Antiderivative The antiderivative of the integrand [math]\sin(x)/x[/math] is given by the infinite summation: [math]\displaystyle\int\dfrac{\sin(x)}x\ dx = \sum_{n=0}^∞\dfrac{(-1)^nx^{2n+1}}{(2n + 1) \cdot (2n + 1)!} + C\tag{}[/math] This can be derived via the Taylor Series for [math]\sin(x)[/math]. Now, we can evaluate the definite integral from [math]x = 0[/math] to [math]x = π[/math]. [math]\begin{align}\displaystyle\int_0^π\dfrac{\sin(x)}x\ dx &= \left[\sum_{n=0}^∞\dfrac{(-1)^nx^{2n+1}}{(2n + 1) \cdot (2n + 1)!}\right]_0^π \ &= \sum_{n=0}^∞\dfrac{(-1)^nπ^{2n+1}}{(2n + 1) \cdot (2n + 1)!} - \sum_{n=0}^∞\dfrac{(-1)^n\cdot 0^{2n+1}}{(2n + 1) \cdot (2n + 1)!} \ &= \lim_{m \to ∞}\sum_{n=0}^m\dfrac{(-1)^nπ^{2n+1}}{(2n + 1) \cdot (2n + 1)!}\end{align}\tag{}[/math] To evaluate this, you need a summation calculator. You can use this link: Sigma (Sum) Calculator Sigma (Sum) Calculator Just type, and your answer comes up live. Example: "n^2" images/sigma2.js What is Sigma? Σ This symbol (called Sigma) means "sum up" It is used like this: Sigma is fun to use, and can do many clever things. Learn more at Sigma Notation . You might also like to read the more advanced topic Partial Sums . All Functions Operators + Addition operator - Subtraction operator Multiplication operator / Division operator ^ Power/Exponent/Index operator () Parentheses Functions sqrt Square Root of a value or expression. sin sine of a value or expression cos cosine of a value or expression tan tangent of a value or expression asin inverse sine (arcsine) of a value or expression acos inverse cosine (arccos) of a value or expression atan inverse tangent (arctangent) of a value or expression sinh Hyperbolic sine of a value or expression cosh Hyperbolic cosine of a value or expression tanh Hyperbolic tangent of a value or expression exp e (the Euler Constant ) raised to the power of a value or expression ln The natural logarithm of a value or expression log The base-10 logarithm of a value or expression abs Absolute value (distance from zero) of a value or expression fact the Factorial function! Constants The summand that you should put into the above link should be this: (-1)^n pi^(2n+1) / [(2n + 1) fact(2n + 1)] Of course, your lower bound should be [math]0[/math] and upper bound should keep increasing to [math]∞[/math]. In the above website, the maximum upper bound you can put is [math]10,000[/math]. For now, I’ll make a table. [math]\begin{array}{\|c\|c\|} \hline m & \text{integral (summation approx.)} \ \hline 1 & 1.4190217269 \ 5 & 1.851902598 \ 10 & 1.85193705 \ 100 & 1.85193705198 \ 250 & 1.85193705198 \ \hline\end{array}\tag{}[/math] So, we can conclude with an approximation for our integral. [math]\boxed{\therefore\displaystyle\int_0^π\dfrac{\sin(x)}x\ dx ≈ 1.851937}\tag{}[/math] Trapezoidal Rule Using summations, we can express the definite integral of a function [math]f(x)[/math] via the Trapezoidal Rule to be: [math]\displaystyle\int_{x_0}^{x_n} f(x)\ dx = \lim_{n \to ∞}\dfrac h2\sum_{k=1}^n\left[f\left(x_0 + kh\right) + f\left(x_0 + (k - 1)h\right)\right]\tag{}[/math] Here, [math]n[/math] represents the number of trapeziums used to approximate the area, and [math]h[/math] is the width (and height) of each trapezium, given by the formula: [math]h = \dfrac{x_n - x_0}{n}\tag{}[/math] In our case, [math]x_0 = 0[/math], [math]x_n = π[/math], and: [math]f(x) = \dfrac{\sin(x)}x\tag{}[/math] [math]h = \dfrac{π - 0}n = \dfrac πn\tag{}[/math] Plugging them into the formula, we get: [math]\displaystyle\int_0^π\dfrac{\sin(x)}x\ dx = \dfrac{π}{2}\lim_{n \to ∞}\dfrac{1}{n}\sum_{k=1}^n\left[\dfrac{\sin\left(\frac{kπ}n\right)}{\frac{kπ}n} + \dfrac{\sin\left(\frac{(k-1)π}n\right)}{\frac{(k-1)π}n}\right]\tag{}[/math] but this forms an error because [math]\sin(x)/x[/math] is not defined at [math]x = 0[/math]. So, we must consider a slightly different, yet same integral. [math]\displaystyle\lim_{l \to 0}\int_l^π\dfrac{\sin(x)}x\ dx\tag{}[/math] This time, we must instead plug in [math]x_0 = l[/math] and: [math]h = \dfrac{π - l}{n}\tag{}[/math] to get: [math]\displaystyle\lim_{l \to 0}\int_l^π\dfrac{\sin(x)}x\ dx = \lim_{l \to 0}\left[\dfrac{π - l}2 \cdot \lim_{n \to ∞}\dfrac 1n\sum_{k=1}^n\left{\dfrac{\sin\left(l + \frac{kπ}n\right)}{l + \frac{kπ}n} + \dfrac{l + \sin\left(l + \frac{(k-1)π}n\right)}{l + \frac{(k-1)π}n}\right}\right][/math] I know things may seem very daunting right now, but all will be fine when we simply plug this into a summation calculator and create a table. [math]\begin{array}{\|c\|c\|} \hline l & \text{integral (summation approx.)} \ \hline 0.1 & 1.694717233 \ 0.01 & 1.836058206 \ 0.001 & 1.850347719 \ 0.0001 & 1.851778102 \ 0.00001 & 1.851921154 \ \hline\end{array}\tag{}[/math] Hence, via the Trapezoidal Rule, we get: [math]\boxed{\therefore\displaystyle\int_0^π\dfrac{\sin(x)}x\ dx ≈ 1.8159}\tag{}[/math] Similarly, to make this process (and summation) simpler, you could use Riemann Sums as an alternative, but it would be less accurate. A. Skodras Studied Mathematics & Applied Mathematics · Author has 731 answers and 2.5M answer views · 3y Related What is the value of [math]\int_0^{\pi} \sqrt{\sin{x} - \sin^2{x}} \, dx[/math] ? It is a long but pretty nice one ! First of all we can do this usefull substitution [math]\ \ w=\frac{π}{2}-x[/math] and [math]\mathrm d w=\mathrm d \left(\frac{π}{2}-x\right)\implies[/math] [math]\mathrm d w=-\mathrm d x[/math] and [math]x=0 \implies w=\frac{π}{2}[/math] [math]x=π \implies w=-\frac{π}{2}[/math] and we get [math]\displaystyle I=\int_0^π f(\sin x) \mathrm dx=-\int_{\frac{π}{2}}^{-\frac{π}{2}} f\left(\sin (\frac{π}{2}-w)\right) \mathrm dw=\int_{-\frac{π}{2}}^{\frac{π}{2}} f(\cos w) \mathrm dw[/math] But since the function [math]\ \ \cos x \ \ [/math]is an even function and whick means it is symmetrical as shown in the graph, … our integral turns into … [math]\displaystyle I=\int_{-[/math] It is a long but pretty nice one ! First of all we can do this usefull substitution [math]\ \ w=\frac{π}{2}-x[/math] and [math]\mathrm d w=\mathrm d \left(\frac{π}{2}-x\right)\implies[/math] [math]\mathrm d w=-\mathrm d x[/math] and [math]x=0 \implies w=\frac{π}{2}[/math] [math]x=π \implies w=-\frac{π}{2}[/math] and we get [math]\displaystyle I=\int_0^π f(\sin x) \mathrm dx=-\int_{\frac{π}{2}}^{-\frac{π}{2}} f\left(\sin (\frac{π}{2}-w)\right) \mathrm dw=\int_{-\frac{π}{2}}^{\frac{π}{2}} f(\cos w) \mathrm dw[/math] But since the function [math]\ \ \cos x \ \ [/math]is an even function and whick means it is symmetrical as shown in the graph, … our integral turns into … [math]\displaystyle I=\int_{-\frac{π}{2}}^{\frac{π}{2}} f(\cos w) \mathrm dw=2\int_{0}^{\frac{π}{2}} f(\cos w) \mathrm dw[/math] So we need to calculate the integral [math]\displaystyle I=2\int_{0}^{\frac{π}{2}} \sqrt{\cos w - cos^2 w} \mathrm dw[/math] Substituting [math]\ \ \ t=\cos w[/math] we get [math]\displaystyle\mathrm dt=\mathrm d\cos w \implies[/math] [math]\displaystyle\mathrm dt=-\sin w \mathrm dw \implies[/math] [math]\displaystyle\mathrm dt=-\sqrt{1-\cos^2 w} \mathrm dw \implies[/math] [math]\displaystyle\mathrm dt=-\sqrt{1-t^2} \mathrm dw \implies[/math] [math]\displaystyle -\dfrac{1}{\sqrt{1-t^2}}\mathrm dt= \mathrm dw [/math] and [math]w=0 \implies t=1[/math] [math]w=\frac{π}{2} \implies t=0[/math] So we get [math]\displaystyle I=-2\int_{1}^{0} \sqrt{t - t^2 } \dfrac{1}{\sqrt{1-t^2}}\mathrm dt=[/math] [math]\displaystyle 2\int_{0}^{1} \dfrac{\sqrt{t - t^2 }}{\sqrt{1-t^2}}\mathrm dt=[/math] [math]\displaystyle 2\int_{0}^{1} \sqrt{\dfrac{t(1 - t)}{(1-t)(1+t)}}\mathrm dt=[/math] [math]\displaystyle 2\int_{0}^{1} \sqrt{\dfrac{t}{1+t}}\mathrm dt[/math] Substituting [math]\ \ 1+t=u[/math] we get [math]\displaystyle I=2\int_{1}^{2} \sqrt{\dfrac{u-1}{u}}\mathrm du[/math] and integrating by parts we get ... [math]\displaystyle I=2\int_{1}^{2} u'\sqrt{\dfrac{u-1}{u}}\mathrm du=[/math] [math]\displaystyle 2\left.\left(u\sqrt{\dfrac{u-1}{u}}\right)\right\|_1^2-2\int_{1}^{2} u\left(\sqrt{\dfrac{u-1}{u}}\right)'\mathrm du=[/math] [math]\displaystyle 2\left(2\sqrt{\dfrac{2-1}{2}}-0\right)-2\int_{1}^{2} u\dfrac{1}{2\sqrt{\dfrac{u-1}{u}}}\left(\dfrac{u-1}{u}\right)'\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-2\int_{1}^{2} \dfrac{u\sqrt u}{2\sqrt{u-1}}\left(1-\dfrac{1}{u}\right)'\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-2\int_{1}^{2} \dfrac{u\sqrt u}{2u^2\sqrt{u-1}}\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-\int_{1}^{2} \dfrac{1}{\sqrt u\sqrt{u-1}}\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-\int_{1}^{2} \dfrac{1}{\sqrt u\sqrt{u-1}}\dfrac{\sqrt u+\sqrt{u-1}}{\sqrt u+\sqrt{u-1}}\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-\int_{1}^{2} \dfrac{1}{\sqrt u+\sqrt{u-1}}\dfrac{\sqrt u+\sqrt{u-1}}{\sqrt u\sqrt{u-1}}\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-\int_{1}^{2} \dfrac{1}{\sqrt u+\sqrt{u-1}}\left(\dfrac{\sqrt u}{{\sqrt u\sqrt{u-1}}}+\dfrac{\sqrt{u-1}}{\sqrt u\sqrt{u-1}}\right)\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-\int_{1}^{2} \dfrac{1}{\sqrt u+\sqrt{u-1}}\left(\dfrac{1}{{\sqrt{u-1}}}+\dfrac{1}{\sqrt u}\right)\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-\int_{1}^{2} \dfrac{1}{\sqrt u+\sqrt{u-1}}\left(\dfrac{2}{{2\sqrt{u}}}+\dfrac{2}{2\sqrt {u-1}}\right)\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-2\int_{1}^{2} \dfrac{1}{\sqrt u+\sqrt{u-1}}\left(\dfrac{1}{{2\sqrt{u}}}+\dfrac{1}{2\sqrt {u-1}}\right)\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-2\int_{1}^{2} \dfrac{1}{\sqrt u+\sqrt{u-1}}\left(\sqrt{u}+\sqrt {u-1}\right)'\mathrm du=[/math] [math]\displaystyle 2\sqrt{2}-2\left.\ln \left ( \sqrt u+\sqrt{u-1}\right )\right\|_1^2=[/math] [math]\displaystyle 2\sqrt{2}-2\left(\ln \left ( \sqrt 2+\sqrt{2-1}\right )-\ln \left ( \sqrt 1+\sqrt{1-1}\right )\right)=[/math] [math]\displaystyle 2\sqrt{2}-2\ln \left (1+ \sqrt 2\right )[/math] and after all this eyekiller answer the result is [math]\displaystyle \boxed {I=2\sqrt{2}-2\ln \left (1+ \sqrt 2\right )}[/math] If you made it so far reading you probably have a headache by now :) Thank you for reading ! Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by Utkarsh Priyam , studied Mathematics at Harker School (2021) · Author has 8.4K answers and 20.9M answer views · 5y Related What is the value of [math]\int_{0}^{2 \pi} x^{n} \sin x \mathrm{d} x[/math] ? We want to find the value of [math]I_n = \displaystyle \int_0^{2\pi} x^n \sin{x} \, dx. \tag{}[/math] Assume that[math] n \geq 2[/math]. Integrating by parts twice starting with [math]u = x^n[/math] and [math]dv = \sin{x} \, dx[/math] yields [math]\begin{align} I_n &= \displaystyle \Big(x^n \cdot (-\cos{x}) - nx^{n-1} \cdot (-\sin{x})\Big)\Big\|_0^{2\pi} + \int_0^{2\pi} n(n-1)x^{n-2} \cdot (-\sin{x}) \, dx\ &= \displaystyle -(2\pi)^n - n(n-1) \int_0^{2\pi} x^{n-2} \cos{x} \, dx. \end{align} \tag{}[/math] In other words, we have the recurrence relation [math]I_n = -n(n-1) \, I_{n-2} - (2\pi)^n \text{ for all } n \geq 2. \tag{}[/math] Due to the index difference of 2 from We want to find the value of [math]I_n = \displaystyle \int_0^{2\pi} x^n \sin{x} \, dx. \tag{}[/math] Assume that[math] n \geq 2[/math]. Integrating by parts twice starting with [math]u = x^n[/math] and [math]dv = \sin{x} \, dx[/math] yields [math]\begin{align} I_n &= \displaystyle \Big(x^n \cdot (-\cos{x}) - nx^{n-1} \cdot (-\sin{x})\Big)\Big\|_0^{2\pi} + \int_0^{2\pi} n(n-1)x^{n-2} \cdot (-\sin{x}) \, dx\ &= \displaystyle -(2\pi)^n - n(n-1) \int_0^{2\pi} x^{n-2} \cos{x} \, dx. \end{align} \tag{}[/math] In other words, we have the recurrence relation [math]I_n = -n(n-1) \, I_{n-2} - (2\pi)^n \text{ for all } n \geq 2. \tag{}[/math] Due to the index difference of 2 from both sides of the recurrence, we need to have two cases according to whether [math]n[/math] is even or odd. Accordingly, we need two base cases: [math]I_0 = \displaystyle \int_0^{2\pi} \sin{x} \, dx = -\cos{x} \Big\|_0^{2\pi} = 0, \text{ and } \tag{}[/math] [math]I_1 = \displaystyle \int_0^{2\pi} x \sin{x} \, dx = (-x\cos{x} + \sin{x})\Big\|_0^{2\pi} = -2\pi. \tag{}[/math] For instance, we obtain [math]I_2 = 2 \cdot 1 \, I_0 - (2\pi)^2 = -4\pi^2. \tag{}[/math] [math]I_3 = -3 \cdot 2 \, I_1 - (2\pi)^3 = 12\pi - 8\pi^3. \tag{}[/math] [math]I_4 = -4 \cdot 3 \, I_2 - (2\pi)^4 = 48\pi^2 - 16\pi^4. \tag{}[/math] [math]I_5 = -5 \cdot 4 \, I_3 - (2\pi)^5 = -240\pi + 160\pi^3 - 32\pi^5. \tag{}[/math] Time permitting, I will see later if I can write closed forms for both the even and odd cases. Update: It looks like Gary Crenshaw has beaten me to solving the recurrence. Please check his post for that. Related questions What is the integral of sin x from 0 to π? What is the value of the definite integral of sin(x) from 0 to π/2? What is the value of the integral of sin(x) + cos(x) from 0 to π/3? What is the value of the integral from 0 to pi of the function sin(x) /x? How do I calculate integral ? Can the integral value of sin [x] /cos [x] from 0 to pi be found? How do I integrate from to ? What is the integral of (cos(x) / (cos(x) + sin(x)) in respect to x ranging from 0 to (pi/2)? What is the integration of ? Why does the integral of 1/sin(x) from infinity to 1 not have a limit of -π/2n(pi)? How do you find the integral from 0 to π of sin(x) dx? What is the integration of sin^2 x within limits 0 to PI/4? What is the value of the integral ? What is the integral of sin x from 0 to pi/2? What is the integral of sin x when x approaches -pi/2 to pi/2? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2362	https://www.scribd.com/document/296034681/Ceva-Apps	Ceva's Theorem and Its Applications: William M. Faucette May 2007 \| PDF \| Triangle \| Perpendicular 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) 772 views 11 pages Ceva's Theorem and Its Applications: William M. Faucette May 2007 1) Ceva's theorem states that the lines AD, BE, and CF are concurrent if and only if the ratio AE/CE × CD/BD × BF/AF is equal to 1, where points D, E, and F lie on the sides of triangle ABC.… Full description Uploaded by Chanthana Chongchareon AI-enhanced title and description Go to previous items Go to next items Download Save Save Ceva Apps For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Ceva Apps For Later You are on page 1/ 11 Search Fullscreen Ceva’s Theorem and Its Applications William M. Faucette May 2007 Theorem 1 (Ceva’s theorem) . Let ∆ ABC be a triangle and let points D , E , F , be located on sides BC , AC , and AB , re sp ec tivel y. The line segments AD , BE , and C F are concurrent if and only if AE CE · CD BD · BF AF = 1 . Proof. Construct ∆ ABC with points D , E , and F on sides B C , AC , and AB ,respectively.Suppose the line segments AD , BE , and CF are concurren t. Let O be the point of intersection. Construct line  through C parallel to line AB and extend segments AD and BE to meet  at points G and H , respec tiv ely. See Figure ?? .Since  and AB are parallel, we have angle ∠ ABE congruent to ∠ CH E ,since they are alternate interior angles with respect to the transversal BH .Similarly, ∠ BAE congruent to ∠ HC E , since they are alternate interior angles with respect to the transversal AC . Hence, t riang les ∆ ABE and ∆ CH E are similar.By the same reaso ning, we have the foll owi ng pairs of simi lar triangle s ∆ ABE ∼ ∆ CH E ∆ AOF ∼ ∆ GOC ∆ BOF ∼ ∆ HO C ∆ GDC ∼ ∆ ADB. From the ﬁrst similarity, we have AE CE = AB CH. (1)From the second and third similarities, we have BF AF = BF OF · O F AF = CH OC · O C CG = CH CG. (2)From the fourth similarity, we have CD BD = CG AB. (3)1 adDownload to read ad-free Figure 1.Then, using the equations ( ?? ), ( ?? ), and ( ?? ), we have AE CE · C D BD · B F AF = AB CH · C G AB · C H CG = 1 . 2 adDownload to read ad-free Figure 2.Conversel y, suppose AE CE · CD BD · BF AF = 1 . (4)Let O be the intersection of AD and BE . Con str uct th e segm en t OC and extend this segment to meet AB at a point F  . See Figure ?? .Since the segments AD , BE , and CF  are concurrent, by the ﬁrst part of the proof, we have AE CE · C D BD · B F  AF  = 1 . (5)From equations ( ?? ) and ( ?? ), it follows that BF AF = BF  AF  BF AF 1 = BF  AF  1 BF AF + AF AF = BF  AF  + AF  AF  AF + F B AF = AF  + F  B AF  AB AF = AB AF  AF = AF  . Hence, F = F  and the three segments AD , B E , and C F are concurrent.3 adDownload to read ad-free Figure 3. Corollary 1. The three medians of a triangle are concurrent.Proof. F or the medians of a triangle, we have AE = EC , BD = DC , AF = BF ,so this result follows immediately from Ceva’s theorem. Deﬁnit ion 1. The point of intersection of the three medians of a triangle is called the centroid of the triangle. Corollary 2. The three altitudes of a triangle are concurrent.Proof. Construct ∆ ABC with points D , E , and F the feet of the altitudes from vertices A , B , and C , respectively.Consider the right triangles ∆ BE C and ∆ ADC . Se e Fi gu re ?? . Si nc e ∠ BC E is an acute angle in the right triangle ∆ BE C , angles ∠ BC E = ∠ ACD and ∠ CB E are complementary. In the right triangle ∆ ADC , angles ∠ ACD = ∠ BC E and ∠ CAD ar e complem ent ary. So, angles ∠ CB E and ∠ CAD are congruent. Hence, ∆ BE C is similar to ∆ ADC , whereby EC DC = BC AC. (6)Consid er the righ t tria ngles ∆ F AC and ∆ EAB . Se e Fi gu re ?? . Si nc e ∠ F CA is an acute angle in the right triangle ∆ F AC , angles ∠ F CA and ∠ CAF = ∠ EAB are complementary. In right triangle ∆ EAB , angles ∠ EB A and ∠ EAB = ∠ CAF are complementary. So, angles ∠ F CA and ∠ EB A are congruent. Hence, ∆ EAB is similar to ∆ F AC . Hence, AE AF = AB AC. 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2363	https://www.quora.com/Why-is-the-order-of-operations-taught-wrong-in-school-For-example-they-teach-that-if-you-have-a-complex-expression-you-do-ALL-of-the-multiplication-before-any-addition-However-you-really-only-need-to-do-the	Something went wrong. Wait a moment and try again. Expressions (mathematics) Basic Operations Order of Operations (math... School Curriculum Basic Concepts of Mathema... Numerical Expressions 5 Why is the order of operations taught wrong in school? For example, they teach that if you have a complex expression, you do ALL of the multiplication before any addition.However, you really only need to do the multiplication first if there is choice Joe Zbiciak Processor, MemSys, & SoC Architect + Software Engineer · Upvoted by Justin Rising , PhD in statistics · Author has 5.9K answers and 57.7M answer views · 3y Precedence (aka. “order of operations”) and associativity only tell you where the hidden parentheses are for grouping arguments to operators. That's it. Any other rule, such as “evaluate all multiplications before all additions” exists to simplify the instruction. That is, it's easier to teach the concept with simpler versions of the rules that are more restrictive than required. File it under “white lies we tell children,” next to: You can't divide a smaller number by a larger number… (until next year, when we learn fractions). You can't take the square root of a negative number… (until we learn a Precedence (aka. “order of operations”) and associativity only tell you where the hidden parentheses are for grouping arguments to operators. That's it. Any other rule, such as “evaluate all multiplications before all additions” exists to simplify the instruction. That is, it's easier to teach the concept with simpler versions of the rules that are more restrictive than required. File it under “white lies we tell children,” next to: You can't divide a smaller number by a larger number… (until next year, when we learn fractions). You can't take the square root of a negative number… (until we learn about imaginary and complex numbers). You can't raise things to {imaginary, complex} powers… (until we learn about e and ln and Taylor series and derivatives and whatnot). etc. etc. etc. Alan Bustany Trinity Wrangler, 1977 IMO · Author has 9.8K answers and 58M answer views · 3y I don't know, but probably for the same reason adverbs are taught incorrectly… In any case, are you aware that "order of operations" has nothing to do with multiplication, addition, or any other operators per se? It is a problem of infix notation that does not even exist in, say, functional or Reverse Polish notation . If you have two binary operators, say ∘ and ∙, and no other information you always have a choice — does x∘y∙z mean: (x∘y)∙z; or x∘(y∙z) ? Actually that choice exists even for repeated application of a single infix operator like [math]x\circ y[/math] Footnotes Infix notation - Wikipedia Function (mathematics) - Wikipedia Reverse Polish notation - Wikipedia I don't know, but probably for the same reason adverbs are taught incorrectly… In any case, are you aware that "order of operations" has nothing to do with multiplication, addition, or any other operators per se? It is a problem of infix notation that does not even exist in, say, functional or Reverse Polish notation . If you have two binary operators, say ∘ and ∙, and no other information you always have a choice — does x∘y∙z mean: (x∘y)∙z; or x∘(y∙z) ? Actually that choice exists even for repeated application of a single infix operator like x∘y∘z. If the operator is associative (like addition or multiplication but ulike subtraction or division) the result will be the same whichever choice you make (but the choice exists nonetheless). We are so addicted to the convenience of infix notation, and we so dislike the clarity of unnecessary parenthesis, that we have "order of operations" to disambiguate expressions. Strictly speaking such a scheme would specify the precise order in which operations are evaluated — you have no choice just as taught in school despite your objections. Occasionally the order of doing things does not matter, but this is relatively rare. Most of the time it matters, for example, whether it is socks then shoes or shoes then socks. Multiple additions or multiple products are the exceptions rather than rule, for which we even have special notation: n∑i=1ai=a1+a2+⋯+an n∏i=1ai=a1×a2×⋯×an You can do these in any order and even shuffle the operands. What you are taught in school — in mathematics or any other subject — is largely driven by by pragmatics. Personally I would avoid criticising it or recommending improvements unless I had both a deep understanding of the subject and some relevant experience in teaching. Footnotes Infix notation - Wikipedia Function (mathematics) - Wikipedia Reverse Polish notation - Wikipedia Assistant Bot · 11mo The order of operations in mathematics is a standardized set of rules that helps ensure consistency in solving expressions. The common acronym PEMDAS (Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)) is often used to teach these rules. Why It Might Seem "Wrong" Misinterpretation of Multiplication and Addition: The teaching that multiplication must always be done before addition can lead to confusion. In reality, multiplication and addition are performed based on their position in the expression. If multiplication come The order of operations in mathematics is a standardized set of rules that helps ensure consistency in solving expressions. The common acronym PEMDAS (Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)) is often used to teach these rules. Why It Might Seem "Wrong" Misinterpretation of Multiplication and Addition: The teaching that multiplication must always be done before addition can lead to confusion. In reality, multiplication and addition are performed based on their position in the expression. If multiplication comes first (from left to right), it is done first; if addition comes first, it is done first. Left-to-Right Rule: Both multiplication and addition are performed from left to right. If you have an expression like 2+3×4, you should perform the multiplication first because it appears later in the left-to-right order. However, in an expression like 2×3+4, you would perform multiplication first because it appears before addition. Teaching Simplifications: Educators often simplify the rules to make them easier for students to remember, which can lead to misunderstandings. The nuance of performing operations based on their order of appearance (left to right) can be lost. Example to Illustrate Consider the expression: 8+2×5 Following PEMDAS: - First, perform the multiplication: 2×5=10 - Then, perform the addition: 8+10=18 Now consider: 8+5×2 Following the same rules: - Perform the multiplication first again: 5×2=10 - Then, perform the addition: 8+10=18 In both cases, multiplication is done before addition, but the key point is that the operations are executed in the order they appear from left to right when they are of the same precedence. Conclusion The order of operations is not inherently "wrong," but the way it is sometimes taught can create confusion. Emphasizing the left-to-right rule for operations of the same precedence can help clarify how to approach complex expressions accurately. Related questions Why does the order of operations work? I.e., why does multiplication take precedence over addition? Why isn't it the other way around? Does division come before multiplication in the math order of operations? Why do you have to use multiplication and division before addition and subtraction in the order of operations? What is the cause of the order of operations in math? Why is multiplication first then subtraction and addition? Howard Ludwig Ph.D. in Physics, Northwestern University (Graduated 1982) · Author has 3K answers and 10.1M answer views · 3y “Why is the order of operations taught wrong in school? For example, they teach that if you have a complex expression, you do ALL of the multiplication before any addition.However, you really only need to do the multiplication first if there is choice” The answer is very simple and somewhat disconcerting. Most secondary school mathematics teachers and textbook authors do not understand what the actual rules are as used by professional researchers in mathematics and physics. They want a small set of straightforward rules that will be easy to explain to students, most of whom do not have a great “Why is the order of operations taught wrong in school? For example, they teach that if you have a complex expression, you do ALL of the multiplication before any addition.However, you really only need to do the multiplication first if there is choice” The answer is very simple and somewhat disconcerting. Most secondary school mathematics teachers and textbook authors do not understand what the actual rules are as used by professional researchers in mathematics and physics. They want a small set of straightforward rules that will be easy to explain to students, most of whom do not have a great deal of mathematical maturity (and that is not a criticism of the students, because there is a great deal of growth and development in maturity required for several types of thinking just as there is for physical bodily growth through elementary and secondary school, with different students progressing at different rates for the different types of maturity over quite some years, and some of these topics require more advanced maturity in systematic processing of abstract information). Most secondary school mathematics teachers had secondary education with a specialization in mathematics as their major subject of study at the university level, not mathematics. The education degree for mathematics teachers does not require as many mathematics courses as is the case with the mathematics degree, and those mathematics courses they do have are overall less advanced and less in depth from a mathematical standpoint. Many of these courses are taught by people who likewise have education degrees rather than mathematics degrees. As a result, the secondary-school-teachers-to-be typically do not have much exposure to, and experiences with, actual mathematicians to see what they really do. (But, on the positive side, these teachers do get training in how to motivate their students, hold their attention, and understand what level of maturity students at various grade levels typically have and what types of presentations work better with that level of maturity. Most PhD-level mathematicians and scientists do not get that type of training how to teach, and many of their university students are very frustrated in the [lack of] quality of teaching in courses.) There are numerous specific topics where the information taught in secondary school mathematics courses is contrary to professional practices. One of those topics is order of operations. If the teachers would tell the students that, because of the level of their knowledge, the material that I am teaching for this topic is incomplete and oversimplified and in later courses you will get more complete information, then I would be fine with that, but the educrats (as I like to refer to them, and in this case I actually do intend to convey some degree of derision toward the system with that term) seem to want teachers to teach with authority indicating firmly that this is the ways things are, even though this is, in fact, not how things are. The teacher needs to present simple rules or processes with simple steps, ideally with catchy names and acronyms. Thus, we have PEMDAS, BEDMAS, BIDMAS, BODMAS, and GEMDAS, depending on which country you are in to take into account dialectical differences for terminology in English (Parentheses versus Brackets versus Grouping symbols, and Exponents versus indices versus Orders). All of these are intended to be identical rule sets that lead to the same answer. Many teachers will convey the idea that if you know one of these sets of rules, you will be able to interpret correctly the order of processing the subexpressions that make up any mathematical expression. However, that is not true. These rules do not tell you when to handle unary operations. As a result, there are misunderstandings among people with the expressions −3² and −x² as to whether (1) both should end up with a nonnegative result, (2) both should end up with a nonpositive result, or (3) the first should be 9 [positive] but the second should be nonpositive. The answer is (2), which can be argued somewhat reasonably from the meaning of unary − in the context of PEMDAS but people will debate this on the internet, and Microsoft Excel erroneously does (1) so putting =-3^2 in a cell turns into 9 in spite of mathematical rules indicating it should be −9. Microsoft acknowledges the controversy and justifies their position as they being able to do whatever they want to do as long as they tell you in some documentation [which they do], and the customer has no right to expect or insist that Excel follows the same rules as mathematicians and scientists use. An expression that demonstrates that PEMDAS et al. is incomplete and incorrect is: 2 sin 2u cos 2u, in which it is necessary to treat equivalently to doing the two juxtaposed multiplications [no intervening multiplication sign nor space between the two factors] of 2 and u before applying the unary trigonometric operations sin and cos, which in turn are done before the non-juxtaposed [in this case factors being separated by a space] multiplications of the 2, the result of the sin operation, and the result of the cos operation. Notice that this requires that some multiplications [the juxtaposed ones] are done at a different level than other multiplications (and the divisions go with the “other multiplications. This violates the basic premise of PEMDAS et al. that all multiplications and divisions are done at the same level. Now, that wording, however is actually backwards—that wording implies professional practice is wrong while PEMDAS is correct, when actually PEMDAS is violating professional practice. This means that when you have an expression like 1/2π has traditionally been interpreted (and written based on this interpretation) to mean 1/(2π) but PEMDAS et al. erroneously require the expression to be interpreted as (1/2)π, which professionals would normally write as π/2. One of the expressions that has had ups and downs of virality on the internet is: 6/2(2 + 1) with some variations in writing as to whether the answer is 1 or 9. Traditional professional interpretations: 6/2(2 + 1) = 1 [The multiplication is juxtaposed, so done early.] 6/2 (2 + 1) = 9 [The multiplication has intervening space, so not done early.] 6/2 × (2 + 1) = 9 [The multiplication has explicit ×, so not done early.] PEMDAS et al. interpretations: 6/2(2 + 1) = 9 6/2 (2 + 1) = 9 6/2 × (2 + 1) = 9 These rules do not care how the multiplication is expressed or formatted. All multiplications and divisions are done unconditionally at the same level left-to-right. The conflict involving interpretation of such expressions has caused many technical and standards organizations to adjust their rules and style guides to forbid writing any expressions like a/bc and a/b/c, in other words any division immediately followed textually by a multiplication or another division is required to be written with explicit grouping indicators that explicitly specify the intended order that the divisions and multiplications are to be carried out or divisions may be converted to multiplication by a reciprocal that is identified by a negative exponent. Thus, from the standpoint of professional practice, there are two plausible interpretations of 6/2(2 + 1), namely undefined or 1; a result of 9 is not regarded as a viable outcome, in spite of vehement objections by PEMDAS advocates. Another aspect of misunderstanding that surprisingly many teachers have about PEMDAS is thinking that it means handle all P first, then all E, then all M, then all D, then all A, then all S, when that is, in fact, not the intent of the rules, but instead the first two letters are to be carried out in sequence, whereas the last four are done as two pairs: First all P, then all E, then all M and D together left to right, and finally all A and S together left to right. When teachers do not understand that as intended, it is highly unlikely that students will understand that as intended (and some will not get it even when the teacher does). However, again PEMDAS is incorrect in having all M and D together. Now, the real intent of the question was to let people know that in many cases, a different order of carrying out the operations is guaranteed to yield the same answer as the PEMDAS et al. rules, so it is not necessary to follow those rules rigorously. That is true, but my lengthy answer is that, as bad as that is, there is a somewhat related, but far worse, problem with PEMDAS et al. because while alternative orders of evaluation work as well and people should know that, at least that issue is merely a matter of convenience or efficiency and not whether the result might be wrong. My bigger issue with PEMDAS et al. is that in some cases it gives a faulty result, which is a far bigger issue than convenience or efficiency. Yet another point of confusion is that MD and AS are each separately handled as left to right. But what happens for exponentiations stacked like abc? That must work top to bottom, thus abc=a(bc). However, based on the textual layout of a stack of exponentiations on paper, top to bottom corresponds to right-to left for E, contrary to the left-to right for MD and AS. This is often not taught, and students are apt to make the wrong assumption of left-to right in parallel with MD and AS. Some of the teachers who try to teach about stacking exponents guess, and guess wrong for that same parallelism reason. The fact that some letters in the acronym are handled individually while others are handled in pairs, and that two of the levels work left to right while one works right to left make this acronym far more complicated than teachers desire it to be, and for good reason—many students stumble over the which letters are paired and which not along with which are left to right and which is right to left. Another issue is that with the word parentheses in American English and brackets in most other English dialects, in any case referring to the character pair ( ), tends to have students looking for just ( ) pairs. Well, [ ] pairs and { } pairs can also serve as grouping symbols . The vinculum [horizontal bar that can be used for fractions or extending radical signs] serves as a grouping symbol, perhaps just below for roots, perhaps both above and below for fractions and division. The \| \| pair for absolute value, the ⌈ ⌉ pair for ceiling, and the ⌊ ⌋ for floor all serve as grouping symbols along with having a mathematical computation effect. Superscripting is a grouping indicator: with the expression 32+1, there are no parentheses-like symbols, so it might seem like the exponentiation (E) should be done before the addition [AS], but the fact that all of 2+1 together is superscript indicates they are to be taken together to form the exponent, so the addition is done before the exponentiation in spite of the lack of parentheses-like symbols, resulting in an answer of 27, not 10. I would prefer teaching principles as rationale for conventional ordering rather than use an acronym. However, if an acronym is needed for pedagogic reasons, I would prefer to see: G for Grouping E for Exponentiation right to left J for Juxtaposed multiplications U for Unary operations D for Divisions left to right O for Other multiplications S for subtractions left to right A for Additions The point of the poster of the question is that the right set of rules (which is not PEMDAS nor any of its compatriots) does not require that the rules be followed diligently in one specific order. The rules are intended to define a way that guarantees the correct answer is achieved—the rules constitute the gold standard, so to speak. As long as one correctly makes use of the commutative and associate properties of addition and multiplication or of the distributivity of multiplication over addition and subtraction, applied to the original expression, the correct answer will be obtained. For example, if all of the operations in an expression are additions, you can validly add the terms in any order that you wish—it need not at all be done strictly left to right; this is true if all of the operations are multiplication as well. Also, one need not compute all multiplications and divisions before doing any additions or subtractions. For example, ab+cd+ef can be done as [(ab)+(cd)]+(ef). The addition of the first two products can be computed validly before the third product is computed. You will get exactly the same answer that way as if you followed the rules to the letter. A mathematical expression can be parsed left to right, making sure that when any left bracketing is encountered to find the correspond right bracket and evaluate what is inside the bracket and then continuing on so that any time you encounter an operator and the textually next operation has a lower precedence level, then you may evaluate the current operation; otherwise, you put the current operation on hold and proceed to the next one. One of the laws of powers could be used to calculate: 227×237 as 227+37, which evaluates first to 2^{64}. That, in turn could be handled as ⎛⎝((((22)2)2)2)2⎞⎠2 in case you do not have 2^{64} memorized and don’t have general exponentiation capability but you do have squaring capability on your calculator. Compilers for turning source code of computer programs into binary code do this sort of stuff as an ordinary part of their processing [unless the user deactivates the compiler’s optimizer]; for a computer processor that has pipelining (or superscalar capability to kick off two, but not three, multiplications at the same time, and there are such processors), the compiler can generate instructions to do the first two multiplications, then start the sum of those two products, and then kick off the third product while the sum is being done, and finally add the third product—all this helps the program to execute a little faster. One thing you really have to be careful about is not fundamentally changing an expression and the outcome of the result by making a seemingly innocent but nevertheless false substitution. You might be taking what at first glance appears to be a valid subexpression where you wish to make a substitution, but what you picked actual straddles incomplete parts of two subexpressions so the substitution is invalid. For example, 7 = 3 + 4; however if I see either 7 or 3 + 4 as a textual subexpression of a larger expression, I am not necessarily allowed to substitute one for the other, so I cannot process 3 + 4² as 7² because that substitution was made by violating the rule that exponentiation must be done before addition. In essence, if I have the equality of subexpressions s = t, then I am allowed to substitute (t), with added parentheses required unless t already has a pair of parentheses enclosing the rest of t, for s if and only if replacing s by (s) does not change the interpretation of the expression. Thus, in the viral Quora question regarding the meaning and value of 6/2(2 + 1), the first step is to evaluate the contents of the pair of parentheses yielding 6/2(3); at this point, many commenters say 2(3) means 2 × 3, which is true, but that does not mean we can substitute to get rid of the parentheses as 6/2 × 3, because a juxtaposed multiplication has been turned into a non-juxtaposed multiplication, which puts the operation at a different level in the hierarchy and ends up changing the answer. Instead, we are allowed to substitute (2 × 3) for 2(3), which means our expression processes as 6/2(3) = 6/(2 × 3) = 6/6 = 1, with the parentheses around the substituting 2 × 3 being required, because removing them alters the order of operations, changing the meaning and value. The bottom line is that, yes, we can use a different order to perform an evaluation of a mathematical expression, as long as that reordering is mathematically equivalent to the ordering directed by the rules. Many people do such reordering subconsciously. However, such reordering must not be done cavalierly as it might subtly but adversely affect the order of operations in an unintended manner. The conditions under which such reordering may occur get rather complicated. The rules assume that nothing more complicated than take the next operation based on the hierarchy and apply the operation to its operand[s] and replace the text of the operation and its operations with the result—there is nothing about adding or deleting parentheses, or any other form of textual modification, including replacing 2(3) by 2 × 3 or (2 × 3) mentioned. All of this is clutter that would scare most people away from even considering using such rules. Professional mathematicians have not formally written down the rules, but they learn them and use them just like a toddler learns words and how to put them together—by osmosis and being corrected when misusing them. Some individual specific rules are written down by particular organizations and publishing houses to avoid confusion by being explicit for particular topics where there is variation of practice, which mostly involves division and its interrelationship with multiplication and unary operations. PEMDAS advocates like to claim that they have settle it once and for all so follow PEMDAS; however, PEMDAS has completely ignored the very important category of unary operations, which by itself is a fatal flaw, but correcting the missing piece reveals yet another fatal flaw in that the claimed order of operations is fundamentally wrong in regard to the relationship of multiplication and division. Whatever variations of the rules that mathematicians use does not include PEMDAS et al. The unwritten rules for order of operations are coupled with the rules of mathematical typography. All this stuff becomes second nature to mathematicians and heavy users of mathematical expressions such as physicists, especially those heavily involved in technical publications. It is hard enough to come up with a complete and self-consistent set of rules that most mathematicians and physicists completely agree with, never mind a compilation of alternative, equivalent rules. For a computer programming language the syntax options are very restricted so there are not nearly as many cases to consider than is the case with general mathematical notation, so it is much easier to codify a rigorous set of all-encompassing rules (as, for example, mathematics allows a × b, a · b, a b, and ab to indicate ordinary multiplication and as we saw with the viral question, the last notation can cause a different interpretation and resulting value than the other three notations, whereas in most programming languages the multiplication must be indicated as ab with intervening spaces allowed, but not required, and such spaces have no impact on interpretation). If mathematicians do not want to deal with all the complexities in tackling putting together a complete set of rules matching their actual conventional practice, I guarantee the teachers do not either, except to come up with a simple subset—such a simple subset that it will deliberately have cases left out, and some of the cases it will handle will be butchered from professional practices. I had some good teachers who would tell us here is what the book says but here is what actually is done for real by mathematicians. They were not afraid to do so—it probably helped that I was in advanced courses. I can sympathize with teachers of ordinary and remedial classes being afraid that saying anything contrary to the book might confuse students. I was told that, no, we do not need to follow the rules rigorously as long as the steps we took were consistent with accomplishing the intent of the rules; some students with a lower level of achievement might well take such a statement and run with it beyond all reason, so I can see teachers thinking it is better not to tell them. I understand your frustration; I understand the frustration of teachers. I rather suspect there is not a one-size-fits-all answer. Bob Allison Physics & Maths Prof., Electrical Engr., former Electrician · Author has 1K answers and 1.3M answer views · 3y It is not taught wrong! The order of operations (oop) has been the same since algebra and algebraic notation came about hundred of years ago. a) Usually it is students who do not truly know the oop or who for got it. Or students who confuse a mnemonic with the rules. A nemonic is not the rule, it is only a memory aid to help you remember the rule - then you have to apply it properly. b) Sometimes it is teacher who does not truly understand it and so teaches it wrong or teaches that the mnemonic is the rule. c) Sometimes it is textbook publishers where either i) the writer or editor does not tru It is not taught wrong! The order of operations (oop) has been the same since algebra and algebraic notation came about hundred of years ago. a) Usually it is students who do not truly know the oop or who for got it. Or students who confuse a mnemonic with the rules. A nemonic is not the rule, it is only a memory aid to help you remember the rule - then you have to apply it properly. b) Sometimes it is teacher who does not truly understand it and so teaches it wrong or teaches that the mnemonic is the rule. c) Sometimes it is textbook publishers where either i) the writer or editor does not truly understand the oop ( like the teacher who doesn’t) or ii)?who is trying to typeset something to make it look nice, fit on a page, or save ink. This is where you find books exposing a difference between a ÷ and a / symbol or making believe there are implied parenthesis in places the typesetting makes it look like there should be. Sponsored by Intuit Accuracy and scale for every return, all year Streamline complex returns and reduce errors. ProConnect supports high-volume teams beyond April. WC Leung Learned programming before twelve · Author has 116 answers and 220K answer views · 3y Is that just your teacher or your school? Anyway, AFAIK primary school teachers tend to overgeneralize math concepts into rigid procedures. This hurts more than just a few mistakes. (P.S. to readers: it is technically a mistake when the expression has brackets, e.g. 3 (4 + 5 6). You won’t be able to get a wrong answer with this mistake though. It might lead students to unintended problems though when students learn more advanced math concepts later.) There is a sad truth to math education: those who can do math earns much more by NOT teaching. So those who teach are not so good in math. Related questions What is the order of operations for multiplication and division? What's the reason behind evaluating powers before other operations like addition and multiplication? What would happen if the order of operations was incorrect in Math? How do you teach the order of operations without PEMDAS? How do I teach the order of operations in a fun way? Roger Pickering Spent 6 years at 2 universities doing maths · Author has 14.7K answers and 5.8M answer views · 3y I would rather they didn’t teach it at all. Instead they should encourage clarity. For example we don’t want kids to learn that it is OK to write 2+2/3−2, leaving it up to the reader, or a compiler, to decide whether they meant , or [math]2 + (2/3) -2 = 2/3[/math]. As far as I know, this entire topic only came about with the advent of computers and high level programming languages. And in that context, nobody should leave it up to the compiler to decide what an expression means. They should write clearly and unambiguously what they mean in the first place. Sponsored by Interactive Brokers Start investing today! According to Investopedia, IBKR has more fractional shares available across more markets than anyone else Kevin Yeh M.Eng in Electrical and Electronics Engineering, Massachusetts Institute of Technology (Graduated 1996) · Author has 2.3K answers and 1.7M answer views · 3y Because order of operations are arbitrary, and not a particularly useful concept. The best thing to do is to make sure that your expression is unambiguous. Parenthesis and explicit grouping is better than excusing my dear Aunt Sally. Jean Rafenski Reynolds Taught math to children and college students · Author has 18.4K answers and 10.4M answer views · 2y Related Why do students have so much trouble learning basic mathematics like multiplication tables when they're taught through rote rather than understanding? For many years, researchers have been telling us that children need concrete math experiences. Kids need to count, cut up apples, and do lots of math activities that involve seeing, saying, and touching. But too many schools ignore the research about how children learn. Kids are pushed to do math in their heads long before they’re ready. They’re given formulas to memorize rather than opportunities to experience what numbers are all about. When the time comes to do application problems (what my school friends and I used to call “word problems”), kids fall apart. Doubt me? If there’s a fourth or f For many years, researchers have been telling us that children need concrete math experiences. Kids need to count, cut up apples, and do lots of math activities that involve seeing, saying, and touching. But too many schools ignore the research about how children learn. Kids are pushed to do math in their heads long before they’re ready. They’re given formulas to memorize rather than opportunities to experience what numbers are all about. When the time comes to do application problems (what my school friends and I used to call “word problems”), kids fall apart. Doubt me? If there’s a fourth or fifth-grader handy, offer to sit down and count with them. I guarantee that they will stop, in bewilderment, when they get to 109. They all do it. When I told a fifth-grader that the next number was 110, she looked at me in amazement. “I didn’t know there was any such number,” she said. The remedy is obvious — and easy: get a pile of pennies, or toothpicks, or anything. Sit down together and start counting. Wise teachers (and parents!) give kids plenty of opportunities to discover what numbers mean. One morning I visited a fifth grade class taught by an enlightened teacher. The kids had watched her cut up apples, and they’d been cutting up circles they made themselves from construction paper — halves, fourths, thirds, and so on. They were starting to understand fractions. As I watched them working on a page of fraction problems, something happened that surprised me. They tackled the first problem as they’d been taught, by drawing a circle and dividing it into fourths. Here’s what’s so interesting: the next thing they all did was to tap each of the fourths. They already knew there were four fourths in a whole — they’d drawn the circle themselves. But they needed that reassurance. (They’re kids, for heaven’s sake!) My teacher friend said she hadn’t told them to do the tapping, but she was always pleased when she saw it: “They’re learning to trust themselves, and I think that’s great!” But what do most adults do? We yell at kids when they seek that reassurance: “You don’t need to count the fourths — you know there are four of them, dummy.” “Don’t count on your fingers. That’s baby stuff!” “Do it in your head!” And then we adults wonder why so many kids are scared of math. (I remember that I was still counting on my fingers in sixth grade — stealthily, under my desk, so that Miss Callahan wouldn’t catch me at it.) I’m a retired college professor. I used to teach a freshman study skills course that included a math unit. Every semester I had college students who didn’t know how many fourths make a whole. They couldn’t understand how a sixth could be larger than an eighth. (Their reasoning: six is smaller than eight, so a sixth has to be smaller than an eighth.) They knew the formulas and could manipulate numbers with ease. But many of them were lost when I challenged them to solve the Pumpkin Pie Problem or answer a real-life question involving math. Many students never discover the simple truth that mathematics — basic arithmetic — makes sense. I always began the math unit with a basic question like “If half a pound of apples costs 50 cents, what does a pound and a half cost?” Some students guessed wildly. (There was always at least one student who figured that since 25 was half of 50, and the word “half” was in the problem, the answer was 25 cents.) When I drew some apples on the board and showed the class how I got to the answer — $1.50 — there were always a few flabbergasted students. The pizza pie solution There’s a simple way to help kids feel more confident about math. Even better, parents can do it with their kids — and have fun at the same time. It involves pizza. I didn’t start out thinking about pizza. When I taught study skills, I used to bake a tray of brownies, cut them up in the classroom, and then move the brownies around to show my students that two fourths is the same as a half, two eighths is the same as a fourth, and so on. It was fun to see the light bulbs go on: “So that’s what my math teacher was talking about!” One evening, during a pizza outing with my husband, I noticed all the families sitting at nearby tables. Suddenly it hit me: why not use pizza to teach basic fractions? While you’re waiting for the pizza to cool off, you can ask (and then answer, while your kids are listening) some simple questions: (If the pizza has six slices) What do you call each piece? (Answer: a sixth.) (If the pizza has eight slices) What do you call these pieces? (Answer: an eighth.) As time goes by, you can ask the same questions and let your kids supply the answers. When your kids start to feel more confident, you can try more questions, like this one about a pizza with eight slices: How many of pieces would you need if you wanted to eat half a pizza? (Answer: four.) What do you call those pieces? (Answer: eighths.) Be sure to make it fun! There’s no rush. You’re going to keep having pizza outings, right? Pizza to the rescue! I think pizza placemats — financed by the government or through grants — could help transform math education. Design some simple activities for families to do together. Print them on colorful and inviting placemats. Keep the questions simple. Supply the placemats to restaurants that serve pizza. Pizzerias could be supplied with additional placemats for families to take home (with reminders to have crayons and play money handy). The placements could suggest other math activities (with tips and answers on the back): “Pete’s Pizza charges $3 for half a pizza. What would a whole pizza cost?” “Here’s a graph with 100 boxes. Each box is a percent. Get out your crayons and show what you think 50% would look like.” You don’t have to teach your kids the entire math curriculum! The goal is for them to realize — after many pizza outings and lots of activities at home — that fractions, percentages, and decimals (which are easy to teach with coins) are different ways to say the same thing. Shopping trips, baking at home, and TV ads offer more opportunities to talk about numbers. Is a 5% discount bigger or smaller than a 30% discount? What’s bigger, 50% off or half off? (I’ve seen kids grin with delight when they realize that’s a trick question. “You can’t fool me, Dad!”) Try it with your own kids! You’ll be doing them a huge favor. Promoted by eFAQ.com eFAQ.com Data Reports and Testing · Jul 7 How do I get my IQ tested? What is a good IQ score, and how do you know where you stand? IQ, or Intelligence Quotient, is often misunderstood. Many people think it’s just a number—but it’s actually a snapshot of how your brain processes information, solves problems, reasons logically, and recognizes patterns. Understanding your IQ can give you valuable insight into your cognitive strengths—and help you figure out where you stand compared to others. 👉 Take the official IQ test here: efaq.com/lp/iq-test What Does an IQ Score Actually Mean? IQ scores are standardized so that the average score is 100. Most people score between What is a good IQ score, and how do you know where you stand? IQ, or Intelligence Quotient, is often misunderstood. Many people think it’s just a number—but it’s actually a snapshot of how your brain processes information, solves problems, reasons logically, and recognizes patterns. Understanding your IQ can give you valuable insight into your cognitive strengths—and help you figure out where you stand compared to others. 👉 Take the official IQ test here: efaq.com/lp/iq-test What Does an IQ Score Actually Mean? IQ scores are standardized so that the average score is 100. Most people score between 85 and 115. If you're in that range, you're right in the middle of the bell curve. But some individuals score higher—and a score of 130 or above is generally considered “gifted.” This doesn’t mean you have to be a genius to do well in life. But it does mean you might think differently, solve problems more quickly, or absorb new concepts with less effort. Want to see if you're in that category? Take our test and find out. Start here: efaq.com/lp/iq-test 🧠 What Does the eFAQ IQ Test Measure? The eFAQ IQ test is designed by experts to assess five core dimensions of intelligence: Logical reasoning Spatial awareness Pattern recognition Short-term memory Verbal intelligence It only takes a few minutes to complete and delivers instant, AI-analyzed results based on your answers. Is It Possible to Improve Your IQ? This is a common question—and while raw intelligence might be partly genetic, your performance on IQ tests canimprove. Practicing certain types of logic puzzles, engaging in deep reading, learning new languages, and even improving your working memory can help. We’ve put together a guide here on how to boost your cognitive performance: 👉 efaq.com/lp/can-you-improve-iq Olivier Binda Math Teacher (University Level) at French Ministry of Education (2002–present) · Author has 127 answers and 241.9K answer views · 6y Related Why is math taught differently in school today? What is wrong with the way we learned it twenty years ago? 2 years ago, I accidentaly tried a new way to teach maths to my students. Before, I was trying to repeat how I had been taught maths : lots of theory, exemples, proofs, applications and after 6 hours of theory, a few exercices, bucketfull of copying what is written on a blackboard. No method given by the teacher, just the raw theorems so you have to devise your own methods by yourself. This might be fine for someone who wants to become a math researcher (It really helped me). But it really sucks for someone who wants to become a veterinary/doctor/work in economics/business…and also for some e Footnotes 2 years ago, I accidentaly tried a new way to teach maths to my students. Before, I was trying to repeat how I had been taught maths : lots of theory, exemples, proofs, applications and after 6 hours of theory, a few exercices, bucketfull of copying what is written on a blackboard. No method given by the teacher, just the raw theorems so you have to devise your own methods by yourself. This might be fine for someone who wants to become a math researcher (It really helped me). But it really sucks for someone who wants to become a veterinary/doctor/work in economics/business…and also for some engeneers. Also, 3/4 of what you have painfully written becomes worthless pretty fast (what use are examples once you have understood the concept, what use is the proof, once you have checked it once. You can just throw everything away and just keep your summarized notes with the theorems) These past 2 years, I have been teaching this way : 1) the first day, I give students 40 pages with all the theorems we will use this year (less copying the blackboard), see joint document 2) each week, we focus on a chapter of maths during 9h but we mostly do exercices. When some theory is needed for an exercise, I explain what needs to be known. 3) Before giving the theory, I give and make them practise the best method that does the job 4) at the end of the week, when they have grasped all the fundamentals, I give some more theory about advanced stuff. So far, this has worked beautifully for me (and for them). They understand maths better and they have 5 times the training that they had before. The tail and the middle of the class are not overwhelmed anymore and get success, which encourages them to give it their best (they become quite stronger) The head of the class soars. The best student is able to get 30+ questions perfectly in 4 hours So, IMO, the way I was taucht maths was not the best possible way to teach maths to the masses. There are better ways : maths should probably be taught in context (the exercice leads to the throry and not the other way around). Also, people needs lots of time practising maths and less time watching a guy in front of a blackboard writing a complicated proof completely useless for you. Footnotes Nicholas Cooper Author has 1.6K answers and 2.9M answer views · 7y Related In various algebras, the order of operations puts "multiplications" before "additions". Why? I’m assuming by the way you are wording this, you must know that the order of operations was subjectively created as a standard that people follow, not some law of mathematics. The standard came about by reducing the number of symbols required using the distributive law. It takes many more parentheses to say: However, you can also look at it like this. I unequivocally state here: repeated addition is not multiplication, and repeated multiplication is not exponentiation… but for natural numbers, it is. So, you may examine that and say, stage 1 is addition, stage 2 I’m assuming by the way you are wording this, you must know that the order of operations was subjectively created as a standard that people follow, not some law of mathematics. The standard came about by reducing the number of symbols required using the distributive law. It takes many more parentheses to say: [math]a(b+c)=(ab)+(ac)[/math] than: [math]a(b+c)=ab+ac[/math] However, you can also look at it like this. I unequivocally state here: repeated addition is not multiplication, and repeated multiplication is not exponentiation… but for natural numbers, it is. So, you may examine that and say, stage 1 is addition, stage 2 is multiplication, and stage 3 is exponentation. The higher the number, the higher priority it has. M Muthusubramanyam Taught at undergraduate and graduate levels in 3 countires · Author has 109 answers and 1.9M answer views · Updated 1y Related Why are students allowed to use a calculator in American middle schools, instead of trying to do the mathematical operations like multiplication/division using their mental math skills? Before proceeding to share my thoughts, let me point out that it is incorrect to generalize. My children went through all their schooling in different states of the US. My grand children - two in Atlanta metro and one in Portland, Oregon - are getting different educational approaches. Not all American middle schools allowed or allow today the use of calculators. However, when they do let them use calculators before they learn to do basic operations in mind or with paper and pencil/pen, the children going through such education do get handicapped. I come from a country and age much different from Before proceeding to share my thoughts, let me point out that it is incorrect to generalize. My children went through all their schooling in different states of the US. My grand children - two in Atlanta metro and one in Portland, Oregon - are getting different educational approaches. Not all American middle schools allowed or allow today the use of calculators. However, when they do let them use calculators before they learn to do basic operations in mind or with paper and pencil/pen, the children going through such education do get handicapped. I come from a country and age much different from modern US schools or even schools in the part of India where I got basic education. We memorize multiplication tables to 16 (16 times 16 = 256) and we even memorize multiplication table of fractions. In my native tongue there are specific word/names for fractions such as 3/4, 1/8/, 3/8, 1/16, 3/16 etc. That was handy in memorizing the tables. When I came to the US starting as a graduate student, one of the early “cultural shocks” was when I saw my Professor taking up his calculator to do a simple multiplication. This was in 1977 when the (4 function) calculators were just coming up and were not affordable to many in India. Not that the professor was not capable of doing the math in his head or with paper and pencil - he was just so used to the calculator. In the 90s when I was teaching to an engineering undergraduate class, I came across two shocking cases. One student used long hand division to divide a number by 10 and said the answer (quotient) and the original number (dividend) are the same! Another student upon asking (I wrote on the chalk board) what is one-half of (two and a half) fumbled with his calculator and after a few frantic attempts,gave up the calculator and did some work on paper and said the answer is one-and a-half! There is one more I can share here but - for want of space and because it is a more complex engineering problem - I will save it for another occasion. These are examples of how students - the coming generations- will be handicapped by relying too much on calculators and computers. As long as there are some European and Asian nations (where still the need to develop basic skills are emphasized) US can afford to have the future native born to go through calculators and computers bypassing the need to learn basics! When I was teaching Engineering Economics (in the year 1982) at Portland, Oregon, one student asked why learn all these complex business (compound interest) formulas when there are business calculators available! I explained to him that he is getting educated and not just trained; he needs to know how the formulas work and that will ensure he knows how to use the calculator. I also told that if a great calamity befall mankind and most of the advances are gone, with a few educated adults left, then the whole scientific and technological advances that were lost could be redeveloped using the knowledge available. Caleb Lam M.S. in Education, Johns Hopkins University (Graduated 2019) · 7y Related Why are students allowed to use a calculator in American middle schools, instead of trying to do the mathematical operations like multiplication/division using their mental math skills? I teach 9th grade math to an underprivileged population in the inner-city, where most students are about 4-5 grade levels behind in math so I can only speak for some urban educators. That means that most of them have stopped learning math in about 4th or 5th grade. They are capable of some mental math with addition and subtraction, but as soon as I throw in a negative sign, all bets are off (eg. –2 – 3 = –5). The top 3 students know the multiplication table up to 11, but most of them will need to count by multiples. Don’t even get me started about long division. It is in this context where I a I teach 9th grade math to an underprivileged population in the inner-city, where most students are about 4-5 grade levels behind in math so I can only speak for some urban educators. That means that most of them have stopped learning math in about 4th or 5th grade. They are capable of some mental math with addition and subtraction, but as soon as I throw in a negative sign, all bets are off (eg. –2 – 3 = –5). The top 3 students know the multiplication table up to 11, but most of them will need to count by multiples. Don’t even get me started about long division. It is in this context where I am tasked with teaching these students Algebra 1. At the beginning of the year, I spent the first 5 days helping these students build fluency skills with adding and subtracting integers. I try to make integer addition/subtraction more concrete by having students tell themselves a story. For example, –5 + 3 may represent someone owing a friend 5 dollars, but only having 3 dollars. We are now about 80% through the year, and most of them still don’t get it. I realize this may be blasphemous to say as an educator, but if they still don’t get it, they may never get it. Should I let arithmetic stop them from learning Algebra? Should I be teaching 6th grade math in 9th grade? Another consideration is the fact that most standardized assessments feature sections where students are allowed to use calculators. 4 out of 5 of the sections on the PARCC Assessment are calculator sections. If students fail this exam, they will need to take a remedial course in order to graduate, depriving them of math courses that would get them closer to college. What would you do if you knew you can solve their problems by teaching them how to use a graphing calculator? Is it sad that even some of their parents don’t know their times tables? Yes, of course, but real people in “real-life” are allowed to use calculators. Could be embarrassing, but it gets the job done. All educators will tell you that math is layered such that students should only be allowed to advance upon mastering “prerequisite skills.” I’m not denying that, but the calculator can change the way we determine prerequisite skills. Once I came to accept that, I saw some pretty interesting results, some that can only be witnessed in an urban environment. Some students don’t know that 11 x 12 = 132, but they can solve a systems of equations using all 3 methods. Others might still struggle with –3 – 4 = ?, but they can recite the quadratic formula and solve a quadratic by completing the square. I can either spend 3 quarters drilling integer operations, or I can give them a TI-84 and teach them how to solve a system of equations. They’re in high school, let’s learn some Algebra. Related questions Why does the order of operations work? I.e., why does multiplication take precedence over addition? Why isn't it the other way around? Does division come before multiplication in the math order of operations? Why do you have to use multiplication and division before addition and subtraction in the order of operations? What is the cause of the order of operations in math? Why is multiplication first then subtraction and addition? What is the order of operations for multiplication and division? What's the reason behind evaluating powers before other operations like addition and multiplication? What would happen if the order of operations was incorrect in Math? How do you teach the order of operations without PEMDAS? How do I teach the order of operations in a fun way? How different would math be with a different order of operations? How does the order of operations help in math? When did BODMAS replace MiDAS (multiply, divide, add, subtract) as the acronym they teach in school for mathematical order of operations? Do you do multiplication or addition first? What is the reason for the difference in the order of operations between multiplication/division and addition/subtraction? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2364	https://news.miracleplus.com/share_link/38072	Jensen’s inequality – 机器学习研究博客 - 齐思奇绩创坛｜齐思为科研创业探寻前沿认知主页奇绩搜索登录注册 Jensen’s inequality – 机器学习研究博客 Jensen’s inequality – 机器学习研究博客 Jensen's inequality 是应用数学和机器学习中的基本概念，它提供了一种限制随机变量的凸函数期望的方法。这个不等式不仅是理论研究中的基石，而且在数据科学中的算法中也有实际应用，比如主导极小化。它的重要性延伸到信息理论和 f-散度的概念，这对于理解...... 内容导读 Jensen's inequality 是应用数学和机器学习中的基本概念，它提供了一种限制随机变量的凸函数期望的方法。这个不等式不仅是理论研究中的基石，而且在数据科学中的算法中也有实际应用，比如主导极小化。它的重要性延伸到信息理论和 f-散度的概念，这对于理解数据分布至关重要。Francisbach.com 上的博客文章深入探讨了这个不等式，提供了关于其经典应用和对矩阵的扩展的见解。如果您从事机器学习研究或应用，了解 Jensen's inequality 至关重要，而这篇由该领域权威来源撰写的博客文章可能是一次有价值的阅读。自动总结 -Jensen不等式是应用数学中用于约束随机变量函数期望的数学结果。 -不平等往往不是人们希望的方向。 -詹森不等式的最简单公式和证明表明，对于定义在凸子集上的凸函数，函数的期望小于或等于期望的函数。 -詹森不等式的历史根源可以追溯到20世纪初。 -该不等式可用于推导其他经典不等式，如算术、调和和几何平均数、杨氏不等式和霍尔德不等式。 -詹森不等式常用于数据科学中的优化-最小化算法，如非负矩阵分解和期望-最小化。 -在信息论中，詹森不等式在经典结果和f-分歧中起着重要作用。 -运算符凸性将詹森不等式推广到矩阵，并在各种数学公式中得到应用。 -对詹森不等式进行了一些改进和扩展，包括利用具有积分余数的泰勒公式得到余数的确切的表达式。相关分享有什么理论复杂但是实现简单的算法？ - 知乎该回答检视了理论上复杂但实现简单的算法。提出了随机微分方程和随机梯度下降作为例子。讨论了更复杂算法包括像黎曼流形朗之万动力学和弹性粒子采样器这样的马尔可夫链蒙特卡罗方法，这些方法提供了高效的样本采集，但需要强大的直觉和数学技能来分析和证明。 #### 陶哲轩IMO演讲全文：一次性解决一千个问题，AI让数学摆脱蛮力计算 AI 能生成数学猜想，在将来很有可能成为现实。 #### 齐思洞见-2025/08/27 齐思用户 13 0 0 关注人数3 为科研创业探寻前沿认知 Jensen’s inequality – 机器学习研究博客内容导读: Jensen's inequality 是应用数学和机器学习中的基本概念，它提供了一种限制随机变量的凸函数期望的方法。这个不等式不仅是理论研究中的基石，而且在数据科学中的算法中也有实际应用，比如主导极小化。它的重要性延伸到信息理论和 f-散度的概念，这对于理解数据分布至关重要。Francisbach.com 上的博客文章深入探讨了这个不等式，提供了关于其经典应用和对矩阵的扩展的见解。如果您从事机器学习研究或应用，了解 Jensen's inequality 至关重要，而这篇由该领域权威来源撰写的博客文章可能是一次有价值的阅读。自动总结: -Jensen不等式是应用数学中用于约束随机变量函数期望的数学结果。 -不平等往往不是人们希望的方向。 -詹森不等式的最简单公式和证明表明，对于定义在凸子集上的凸函数，函数的期望小于或等于期望的函数。 -詹森不等式的历史根源可以追溯到20世纪初。 -该不等式可用于推导其他经典不等式，如算术、调和和几何平均数、杨氏不等式和霍尔德不等式。 -詹森不等式常用于数据科学中的优化-最小化算法，如非负矩阵分解和期望-最小化。 -在信息论中，詹森不等式在经典结果和f-分歧中起着重要作用。 -运算符凸性将詹森不等式推广到矩阵，并在各种数学公式中得到应用。 -对詹森不等式进行了一些改进和扩展，包括利用具有积分余数的泰勒公式得到余数的确切的表达式。内容: There are a few mathematical results that any researcher in applied mathematics uses on a daily basis. One of them is Jensen’s inequality, which allows bounding expectations of functions of random variables. This really happens a lot in any probabilistic arguments but also as a tool to generate inequalities and optimization algorithms. In this blog post, I will present a collection of fun facts about the inequality, from very classical to more obscure. If you know other cool ones, please add them as comments. But before, let me be very clear: Jensen’s inequality is often not in the direction that you would hope it to be. So, to avoid embarrassing mistakes, I always draw at least in my mind the figure below before using it. Simplest formulation and proof Given a convex function defined on a convex subset C of Rd, and a random vector X with values in C, then f(E[X])⩽E[f(X)], as soon as the expectations exist. For a strictly convex function, there is equality if and only if 齐思用户分享了一个链接 13 阅读长按识别参与讨论齐思用户詹森的不平等在信息论中至关重要，它可以通过强化隐私损失的界限来显著改善不同的隐私。通过噪声添加来保护个人数据的差异隐私机制，可能受益于不平等对精确约束期望的容量。这与f偏差尤其相关，f偏差对于评估以隐私为中心的算法中的分布差异至关重要。通过将Jensen不等式应用于表示隐私损失的凸函数，我们可以增强数据效用与隐私之间的平衡。该应用程序不仅在技术上意义深远，而且在社会上也是必不可少的，为数据驱动数字世界中的隐私保护提供了数学基础。对这一交叉点进行集中的探索可能会在隐私策略方面取得重大进展，解决在数据隐私问题日益严重的情况下对强大数学框架的迫切需求。 2024/08/22 03:36 0 0 为科研创业探寻前沿认知 Jensen’s inequality – 机器学习研究博客齐思用户 Invalid Date 写了一条评论詹森的不平等在信息论中至关重要，它可以通过强化隐私损失的界限来显著改善不同的隐私。通过噪声添加来保护个人数据的差异隐私机制，可能受益于不平等对精确约束期望的容量。这与f偏差尤其相关，f偏差对于评估以隐私为中心的算法中的分布差异至关重要。通过将Jensen不等式应用于表示隐私损失的凸函数，我们可以增强数据效用与隐私之间的平衡。该应用程序不仅在技术上意义深远，而且在社会上也是必不可少的，为数据驱动数字世界中的隐私保护提供了数学基础。对这一交叉点进行集中的探索可能会在隐私策略方面取得重大进展，解决在数据隐私问题日益严重的情况下对强大数学框架的迫切需求。长按识别参与讨论齐思用户 -詹森不等式适用于凸函数，指出函数在随机变量期望值的值至多是随机变量函数的期望值。 -它有助于界定应用数学和机器学习中的期望。 -不等式与概率论证和生成不等式有关。 -它是最优化算法中的一个工具，特别是在优化算法中。 -詹森不等式在信息论中具有重要意义。 -对Jensen不等式中剩余项的研究导致了改进和等式情况特征。 -只有当感兴趣的函数相对于所讨论的变量是凸的时，不等式才是有效的。 2024/08/22 03:32 0 0 为科研创业探寻前沿认知 Jensen’s inequality – 机器学习研究博客齐思用户 Invalid Date 写了一条评论 -詹森不等式适用于凸函数，指出函数在随机变量期望值的值至多是随机变量函数的期望值。 -它有助于界定应用数学和机器学习中的期望。 -不等式与概率论证和生成不等式有关。 -它是最优化算法中的一个工具，特别是在优化算法中。 -詹森不等式在信息论中具有重要意义。 -对Jensen不等式中剩余项的研究导致了改进和等式情况特征。 -只有当感兴趣的函数相对于所讨论的变量是凸的时，不等式才是有效的。长按识别参与讨论你怎么看待这个分享？你的观点是什么？是否同意作者的观点？与他分享、交流你的看法... 提交登录社区规范联系我们奇绩创坛京公网安备 11010802039147号 2025奇绩创坛 MiraclePlus
2365	https://www.youtube.com/watch?v=TYyTzavdy8U	Simplifying negative fractions BEmath 128 subscribers 18 likes Description 1829 views Posted: 11 Jan 2021 In this video you will learn how to simplify negative fractions. 2 comments Transcript: hi everyone today you're going to learn how you can simplify negative fractions when you're going to simplify negative fractions it's really important that you do not leave a minus sign as part of the numerator or the denominator the same rules as normally so when you divide a positive number by posit by a positive number you get positive positive divided by negative is negative negative divided by positive is negative and negative divided by negative equals positive so we're going to have a look at three example fractions which we need to simplify we have minus 12 over minus 3. we already know that negative divided by negative is the same as positive so minus 12 divided by minus 3 is the same as 4 how about minus 7 over 49 or if you say differently minus 7 divided by 49 first check a negative divided by positive is equal to negative so our answer needs to be negative however we are not allowed to leave a minus sign as part of our numerator or denominator so what you do you place a minus sign in front of your friction line and this is really really important and you must make sure that you always do this so we divide the numerator by seven and the denominator by seven and our answer is minus one over seven when we have a look at six over minus twenty four we divide a positive number by a negative number so our answer is negative we do not write a negative symbol in our denominator but in front of our fraction line so what we get is minus 1 over 4 since we divided 6 by 6 and 24 by 6. let's have a look at some examples which are maybe a bit more complicated we have minus one over three minus two over seven please remember that you're not allowed to add and subtract fractions when the denominators are not the same so first of all you need to make the denominators the same you can do this by multiplying it with each other's what you do on the bottom you need to do on top what you get is minus 7 over 21 minus 6 over twenty one that can be simplified as minus seven minus six over twenty one it's really important that you first simplify the numerator minus seven minus 6 is the same as minus 13 and the denominator just stays the same remember that we do not write down minus 13 over there but in front of our fraction line so if you have any questions please ask them that
2366	https://www.storyofmathematics.com/how-to-find-turning-points-of-a-function/	How to Find Turning Points of a Function – A Step-by-Step Guide To find turning points of a function, you should first understand what a turning point is: it’s a point on the graph of a function where the direction of the curve changes. In mathematical terms, at a turning point, the derivative of the function will be zero. This is because the slope of the tangent to the graph at a turning point is zero, indicating a transition from increasing to decreasing, or vice versa. By setting the derivative of the function equal to zero, you can solve for the x-values that may correspond to these crucial points. After finding the x-values, you should evaluate the function at these points to find the corresponding y-values, completing the coordinates of the turning points. Graphically, these turning points can be either peaks or valleys on the curve, known as local maxima or minima. To distinguish between them and identify the nature of each turning point, you can use the second derivative test or analyze the behavior of the function on intervals surrounding these points. Stay tuned, as I’ll be walking you through these steps, ensuring you can confidently locate and interpret the turning points of any function, which is a vital skill in understanding the nature of mathematical graphs and in solving many practical problems. Identifying Turning Points of a Function When analyzing the behavior of a function, particularly polynomial functions like quadratic or cubic, I often look for their turning points. Turning points are where a function changes from increasing to decreasing, or vice versa—essentially the peaks and troughs of the graph. These points are not just visually recognizable but also can be calculated using the derivative of the function. To find the turning points, I follow these steps: Find the Derivative: The derivative ( f'(x) ) of a function ( f(x) ) gives me the rate of change. For turning points, the rate changes sign, so I need to find where ( f'(x) = 0 ). Solve for Critical Points: Setting the derivative equal to zero ( f'(x) = 0 ) and solving this equation gives me the x-values, which are possible turning points. Determine Maximum or Minimum: To find out if these critical points are a maximum or minimum value, I’ll use the second derivative test. If ( f”(x) > 0 ), the point is a minimum; if ( f”(x) < 0 ), it’s a maximum. \| Step \| Action \| --- \| \| 1. Derivative \| Compute ( f'(x) ) \| \| 2. Critical Points \| Solve ( f'(x) = 0 ) \| \| 3. Max or Min \| Use ( f”(x) ) to determine the nature of turning points \| For polynomial graphs, the number of turning points is at most the degree of the polynomial minus one. So, a quadratic function can have up to 1 turning point, while a cubic function can have up to 2. Understanding the relationship between the function, its graph, and its derivative helps to reveal where the function is increasing or decreasing and locate those important turning points effectively. Analyzing the Function’s Behavior When I begin to analyze the behavior of a polynomial function, my first step is considering its degree. The degree gives me vital clues about the function’s shape and end behavior. It’s interesting to note that a polynomial of degree ( n ) can have up to ( n-1 ) turning points; that’s the maximum number of turning points it can have. Continuous functions like polynomials don’t make sudden jumps, so their graphs are smooth curves. As I look at the graph, I pay attention to where it intersects the axes, since the x-intercepts signify the function’s zeros, and the y-intercept is the point where the function crosses the y-axis, which occurs when ( x = 0 ). \| Degree \| Possible Number of Turning Points \| --- \| \| 1 \| 0 \| \| 2 \| 1 \| \| 3 \| 2 \| \| 4 \| 3 \| Next, I consider the end behavior by examining the limits as ( x ) approaches positive and negative infinity. If the degree is even, the ends of the graph will point in the same direction; if odd, they’ll point in opposite directions. To find the exact turning points, I look for where the function’s derivative equals zero. The derivative indicates the function’s rate of change, and where it’s zero, the graph may have a peak or a valley – a local maximum or minimum, i.e., the local behavior of the function. Lastly, a sketch of the graph helps visualize these elements concretely. Through graphing, I can interpret the signs and behavior of the function across different intervals and confirm the position of turning points. Conclusion In my exploration of turning points, I have discussed methods for determining where a function’s graph curves and changes direction. Such points are crucial in understanding the behavior of polynomial equations and other mathematical functions. To recap, turning points are found by solving for when the derivative, denoted as $f'(x)$, equals zero. This entails setting $f'(x) = 0$ and finding the $x$ values. After pinpointing the $x$ values, I determine the $y$ coordinates by substituting these $x$ values back into the original function, $f(x)$. The resulting coordinates represent the location of turning points on the graph. It’s important to remember that a turning point is where the graph changes from increasing to decreasing or vice versa, which can be a local maximum, local minimum, or a point of inflection. Additionally, the degree of the polynomial plays a role in the maximum number of turning points it can have, which is always less than its degree. Understanding the concept of multiplicity is equally important because it affects how the graph behaves at the zeros. If a zero has an even multiplicity, the graph touches the x-axis and turns around. With an odd multiplicity, it crosses the axis. By applying calculus and algebraic techniques, anyone can sketch the behavior of functions and predict their turning points. This is not just an academic exercise but a powerful tool in fields ranging from physics to economics, where the optimization of outcomes often hinges on these crucial points on a graph. Post navigation Back to Top of Page © 2025 SOM – STORY OF MATHEMATICS
2367	https://www.internationalolympiadacademy.com/pdfs/files/Scoring%20System-Updated%20Logo.pdf	Scoring System INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. I O A MATHEMATICAL KANGAROO COMPETITION Pre-Ecolier and Ecolier (Grade- 1 & 4): The test for Pre - Ecolier 1 & 2 and Ecolier - 1 & 2 is of 90 minutes duration and contains 24 questions. These are grouped equally under three levels of increasing difficulty. 1 mark will be deducted for each incorrect answer and no penalty for skipping a question. As per the level of difficulties following points are awarded: Easy: 3 points for each correct answer. Medium: 4 points for each correct answer. Hard: 5 points for each correct answer. Maximum Score: 96 points. Benjamin, Cadet, Junior and Student (Grade 5 to 12): The test for Benjamin – 1 & 2 (Grade - 5 & 6), Cadet- 1 & 2 (Grade - 7 & 8), Junior -1 & 2 (Grade - 9 & 10), and Student (Grade 11 & 12) is of 90 minutes duration and contains 30 questions. These are grouped equally under three levels of increasing difficulty. 1 mark will be deducted for each incorrect answer and no penalty for skipping a question. As per the level of difficulties following points are awarded: Easy: 3 points for each correct answer. Medium: 4 points for each correct answer. Hard: 5 points for each correct answer. Maximum Score: 120 points. SINGAPORE & ASIAN SCHOOLS MATH OLYMPIAD SASMO is open to all Primary 1 to 6, Secondary 1 to 4 and JC1/2 students (Grades1 to 12 students from International schools). Time given for contest is 90 minutes. Each level has a differentiated paper and contains 25 questions within 2 sections: 90 minutes Total Score: 85 points (To avoid negative scores, each student will begin with 15 points) Section A: 15 Multiple Choice Questions (2 points for each correct answer; 0 point for each unanswered question; deduct 1 point for each wrong answer). Section B: 10 Non-routine Questions (4 points for each correct answer; no penalty for wrong answers) Scoring System INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. I O A SINGA MATH is open to all students from : Primary 1 to Secondary 4 (Grades 1 to 12 ) This Online assessment is to measure the student’s math foundation at the year before their current level, that is Primary 2/Grade 2 is tested on the Primary 1/Grade 1 syllabus. Since this is an assessment, we will just require a Laptop. For this initial launch, we will open SINGA Math Global Assessments only to schools, no private candidates. The Assessment must be taken from the school only. SINGAPORE MATH GLOBAL ASSESSMENTS Primary 1 - Primary 6 Secondary 1 - Secondary 4 Scoring System INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. I O A AMERICAN MATHEMATICS OLYMPIAD DOKA (DEPTH OF KNOWLEDGE ASSESSMENT) Grade 2 to 12 Section A 15 MC Questions (3 points each) Total: 45 points Section B 5 Open-Ended Questions (5 points each) Total: 25 points Section C 5 Open-Ended Questions (6 points each) Total: 30 points Overview of the American Mathematics Olympiad (AMO) Test Structure Total: 100 points (no calculators allowed) \| (no points deducted from wrong responses) Primary Level Paper P Grade (1 & 2) Secondary Level Results will be compared within the same year level group. Each paper comprises 4 parts: Paper Q Grade (3 & 4) Paper Q Grade (5 & 6) Paper S Grade (7 & 8) Paper T Grade (9 & 10) Paper U Grade (11 & 12) Total Marks: 24 Points Total duration: 90 minutes. Calculators are ONLY permitted for Secondary levels (Papers S, T and U) Part A: Basic Reasoning 6 questions Part B: Intermediate Reasoning (Non-verbal) 6 questions Part C: Advanced Reasoning 6 questions Part C: Extended Reasoning 6 questions There will be 12 MCQ Questions and 12 Non-MCQ Questions. 1 Point will be awarded for every correct answer and there will be no penalty points for wrong answers. Scoring System INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. I O A VANDA International Science Competition Written Paper (90 minutes) Questions 24 - 25 4 points for each correct answer 0 point for each unanswered question 0 point deducted for each wrong answer Questions 21 - 23 3 points for each correct answer 0 point for each unanswered question 0 point deducted for each wrong answer Questions 11 - 20 3 points for each correct answer 0 point for each unanswered question 0 point deducted for each wrong answer Questions 1 - 10 2 points for each correct answer 0 point for each unanswered question 1 point deducted for each wrong answer Section A Section B Section C Total: 77 points To avoid negative scores, each student will begin with 10 points Scoring System INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. I O A HIPPO ENGLISH OLYMPIAD There is only one(1) correct answer to every question. There are no negative marks and ranking is based on the average number of marks achieved on all three(3). The breakdown of questions per point per minute are as follows: PRELIMINARY ROUND SEMI FINAL ROUND Ÿ After the preliminary round tests have been marked, the top 10% of participating students from each participating zone will be invited to the semi-final round. Ÿ The exam level for the semi-final round will be one CEFR level higher than the one in the preliminary round. The tests for the semi-final round of the Olympiad will consist of: a reading test, a writing test (30 questions) and an essay (except for Little Hippo). There is only one correct answer to every question. Ÿ There are no negative marks. Ranking is based on the average number of marks achieved on both the reading and the writing test. Maximum allowed times per category and part: INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. I O A Maximum allowed times per category and part: Final Round IN ROME, ITALY The programme of the final round will be available on In the final round of the Olympiad, the candidates take a Gatehouse Awards exam, all 4 skills included: 1. READING 2. WRITING 3. USE OF ENGLISH 4. LISTENING Scoring System HIPPO ENGLISH OLYMPIAD Scoring System INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. I O A INTERNATIONAL OLYMPIAD ACADEMY A division of CSAR Learning Solutions Pvt. Ltd. Registered Office: A-409, Durga Vihar, East of Sainik Farms, New Delhi-110080 Hand Held: +91 9810336335, +91 8368118421, Email: exam@internationalolympiadacademy.com Website: www.internationalolympiadacaemy.com INTERNATIONAL O L Y M P I A D A C A D E M Y CSAR Learning Solutions Pvt. Ltd. Design Thinking with robotics and Computational Thinking International Competition FORMAT OF THE TEST Ÿ LOWER PRIMARY – Grade 1 and 2 (60 minutes) 18 questions (3 NOI questions, 15 CT questions) Ÿ MIDDLE PRIMARY – Grade 3 and 4 (60 minutes) 20 questions (4 NOI questions, 16 CT questions) Ÿ UPPER PRIMARY – Grade 5 and 6 (75 minutes) 24 questions (5 NOI questions, 19 CT questions) Ÿ LOWER SECONDARY – Grade 7 to 8 (75 minutes) 24 questions (6 NOI questions, 18 CT questions) Ÿ UPPER SECONDARY – Grade 9 to 10 (90 minutes) 24 questions (7 NOI questions, 17 CT questions) Ÿ Junior COLLEGE – Grade 11 and 12 (90 minutes) 24 questions (8 NOI questions, 16 CT questions) SCORING SYSTEM Easy Questions: 6 marks Medium Questions: 9 marks Hard Questions: 12 marks
2368	https://rephrasely.com/usage/opulent-in-a-sentence	Opulent In A Sentence Opulent in a Sentence: Adding Luxury and Grandeur to Your Vocabulary Have you ever come across the word "opulent" and wondered how to use it in a sentence? This article will provide you with a clear understanding of the meaning of opulent and offer several examples of how to incorporate it into your everyday conversations. Let's dive into the world of opulence! Understanding Opulence Opulence is a word that brings to mind images of luxury, grandeur, and extravagance. Derived from the Latin word "opulentus," meaning wealthy or rich, opulent describes something that is characterized by abundance, lavishness, or excessive wealth. It is often used to describe luxurious settings, magnificent architecture, or extravagant lifestyles. Examples of Using Opulent in a Sentence In this example, opulent is used to describe a lavish ballroom adorned with extravagant features. It paints a vivid picture of a space that exudes luxury and grandeur. Here, opulent highlights the grandeur and extravagance of a royal palace, emphasizing the abundance of wealth and the extravagant lifestyle of the royal family. In this sentence, opulent is used to describe a cruise ship that provides a high-end experience to its passengers. It emphasizes the abundance of luxurious amenities and services available on board. Here, opulent is used to describe a fashion show that presents extravagant and luxuriously designed gowns, highlighting the richness and exquisite detail of the garments. Conclusion Opulent is a captivating word that allows you to express the concept of luxury, abundance, and grandeur effortlessly. Through the examples provided, you can see how opulent can be used to describe luxurious settings, magnificent architecture, extravagant lifestyles, and high-end experiences. So, the next time you encounter opulence in your everyday life, try incorporating this word into your vocabulary to add a touch of elegance to your conversations. About Rephrasely Getting your wording just right Paraphrasing is a natural part of the writing process as it helps you clarify your thinking and suit your words to your audience. Using a Rephrasely helps structure and streamline this work, and our paraphrase tool offers 20 modes, many of them free, for accomplishing just this. The 20 modes we offer are diverse, including a summarize tool, a free grammar checker, a mode to simplify text, and a sentence shortener. There are sentence rephrasers and paraphrase rephrase tools, and we pride ourselves on having both, since our reword generator accounts for context at both the sentence and paragraph levels. 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2369	https://im.kendallhunt.com/MS/teachers/3/1/14/preparation.html	Illustrative Mathematics \| Kendall Hunt Illustrative Mathematics \| Kendall Hunt Skip to main content IM CurriculumAbout UsContact UsGoogle ClassroomEducators Register/Log in Grade 8 Middle SchoolGrade 6Grade 7Grade 8 Unit 1 Grade 8Unit 1Unit 2Unit 3Unit 4Unit 5Unit 6Unit 7Unit 8Unit 9 Lesson 1234567891011121314151617 Lesson 14 Alternate Interior Angles PreparationLessonPractice View Student Lesson Lesson Narrative In this lesson, students justify that alternate interior angles are congruent, and use this and the vertical angle theorem, previously justified, to solve problems. Thus far in this unit, students have studied different types of rigid motions, using them to examine and build different figures. This work continues here, with an emphasis on examining angles. In a previous lesson, 180 degree rotations were employed to show that vertical angles, made by intersecting lines, are congruent. The warm-up recalls previous facts about angles made by intersecting lines, including both vertical and adjacent angles. Next a third line is added, parallel to one of the two intersecting lines. There are now 8 angles, 4 each at the two intersection points of the lines. At each vertex, vertical and adjacent angles can be used to calculate all angle measures once one angle is known. But how do the angle measures compare at the two vertices? It turns out that each angle at one vertex is congruent to the corresponding angle (via translation) at the other vertex and this can be seen using rigid motions: translations and 180 degree rotations are particularly effective at revealing the relationships between the angle measures. One mathematical practice that is particularly relevant for this lesson is MP8. Students will notice as they calculate angles that they are repeatedly using vertical and adjacent angles and that often they have a choice which method to apply. They will also notice that the angles made by parallel lines cut by a transversal are the same and this observation is the key structure in this lesson. Learning Goals Teacher Facing Calculate angle measures using alternate interior, adjacent, vertical, and supplementary angles to solve problems. Justify (orally and in writing) that “alternate interior angles” made by a “transversal” connecting two parallel lines are congruent using properties of rigid motions. Student Facing Let’s explore why some angles are always equal. Required Materials Geometry toolkits Required Preparation Students need rulers and tracing paper from the geometry toolkits. Learning Targets Student Facing If I have two parallel lines cut by a transversal, I can identify alternate interior angles and use that to find missing angle measurements. CCSS Standards Building On 7.G.B.5 Addressing 8.G.A.1 8.G.A.5 Glossary Entries alternate interior anglesAlternate interior angles are created when two parallel lines are crossed by another line called a transversal. Alternate interior angles are inside the parallel lines and on opposite sides of the transversal. This diagram shows two pairs of alternate interior angles. Angles a and d are one pair and angles b and c are another pair. Description:There are two horizontal parallel lines, and a third diagonal line drawn from the bottom left to the upper right, intersecting both horizontal lines. The diagonal line is labeled transversal. There are four angles created by the diagonal line inside the parallel lines. The upper left angle is labeled a, upper right is b, lower left is c, and lower right is d. transversalA transversal is a line that crosses parallel lines. This diagram shows a transversal line k intersecting parallel lines m and \ell. Print Formatted Materials Teachers with a valid work email address canclick here to register or sign in for free access to Cool Down, Teacher Guide, and PowerPoint materials. Student Task Statementspdfdocx Cumulative Practice Problem Setpdfdocx Cool DownLog In Teacher GuideLog In Teacher Presentation Materialspdfdocx Additional Resources Google SlidesLog In PowerPoint SlidesLog In Privacy Policy \| Accessibility Information IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at Adaptations and updates to IM 6–8 Math are copyright 2019 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Spanish translation of the "B" assessments are copyright 2020 byIllustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. This site includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. 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2370	https://brilliant.org/wiki/pythagorean-theorem/	Pythagorean Theorem Nick Lee, Arron Kau, Sharky Kesa, and Niranjan Khanderia Patrick Corn Sualeh Asif ZhiJie Goh Karleigh Moore A Former Brilliant Member Worranat Pakornrat Munem Shahriar Nihar Mahajan Nick Turtle Mohammad Farhat Sravanth C. Matin Naseri Tatiana Georgieva Calvin Lin Jimin Khim contributed The Pythagorean theorem states that if a triangle has one right angle, then the square of the longest side, called the hypotenuse, is equal to the sum of the squares of the lengths of the two shorter sides, called the legs. So if a and b are the lengths of the legs, and c is the length of the hypotenuse, then a2+b2=c2. The theorem is a fundamental building block of geometry and has numerous applications in physics and other real-world situations. It is also the basis for the distance formula in coordinate geometry. The converse of the theorem is also true: if a2+b2=c2, then a triangle with side lengths a,b,c will be a right triangle, with right angle between the sides of lengths a and b. One sample application of the converse is in construction: to measure a right angle given various lengths of rope, a surveyor can use ropes of length 3,4,5 (a Pythagorean triple) to make a triangle. The angle between the two shorter sides will be a right angle. Contents Examples Pythagorean Triples Trigonometric Identities Proof of the Theorem See Also Examples Given a right triangle with leg lengths of 5 and 12, what is the length of the hypotenuse? By the Pythagorean theorem, 52+122=h2, so h2=169, which means h=13. Thus, the length of the hypotenuse is 13. □ Here is an example motivating the distance formula: What is the distance between the points A=(2,3) and B=(6,11)? Letting point C=(6,3), we have △ABC with side lengths AC=4 and BC=8. Further, since AC and BC are parallel to the y- and x-axes, respectively, they are perpendicular and form the legs of a right triangle with AB as the hypotenuse. Then, by the Pythagorean theorem, we have AC2+BC2=AB2, which gives us 42+82=16+64=80=AB2. So AB=80=45. □ For a full discussion of this technique, see Distance Formula. Is triangle XYZ a right triangle? 6-7-9 triangle If XYZ is a right triangle, then the side lengths will follow the relationship a2+b2=c2, where c is the longest side. However, 62+72=85=92=81. So XYZ is not a right triangle. □ 15 16 17 18 Right triangle ABC has side lengths N+1,N−1, and 2N. What is the value of N? The correct answer is: 16 It is clear that 2N, N−1<N<N+1, so N+1 is the length of the hypotenuse. Applying the Pythagorean Theorem, we get that (2N)2+(N−1)2=(N+1)2, hence N2=16N. Thus, 0=N2−16N=N(N−16). If N=0, we do not get a triangle. If N=16, we get the 8−15−17 triangle, which is a right triangle. Thus, N=16. Given a triangle with side lengths of 5, 12, and 14, how would you classify the largest angle in the triangle? A) Right B) Acute C) Obtuse D) Not enough information By the Pythagorean theorem, we know that a triangle with side lengths 5, 12, and 13 is a right triangle since 52+122=132. Thus, this triangle cannot be a right triangle. Furthermore, since the longest side is greater than the equivalent hypotenuse in a right triangle, i.e. 14>13, we know that the angle which subtends that side must also be greater. Thus, the angle is obtuse and the answer is C). □ The diagram above shows a big rectangular lawn cut into two smaller rectangles, by the line BC, with dimensions 3×12 and 6×12 each. What is the length of AD? The correct answer is: 15 Circle O is inscribed in △ABC, as shown in the figure above. The points of tangency are at D,E, and F. Given that AD=2 and DC=3, find the area of △ABC. The correct answer is: 6 Only and , , and None of them are correct You have a right triangle that has integer sides. Read the following statements about that triangle: At least one of the sides of that triangle must be divisible by 3. At least one of the sides of that triangle must be divisible by 4. At least one of the sides of that triangle must be divisible by 5. Which of these statements is(are) correct? Note: This problem is a part of this set. The correct answer is: , , and Find the area of triangle ABC with side lengths AB=13, BC=17, and CA=20. Bonus: Can you calculate without using the cosine rule? The correct answer is: 7 Find the radius of the circle in the diagram. The correct answer is: 160.25 Below is an isosceles triangle. Find the length of the base, x. The correct answer is: 14 a+b+c 33(a+b+c) a2+b2+c2−a−b−c ab+bc+ac a2+b2+c2 3abc 3a3+b3+c3 b2+c2a2+a2+c2b2+a2+b2c2 O,X,Y,Z are four points on a 3D Cartesian space, where O is the origin, X a point on the x-axis, Y a point on the y-axis, and Z a point on the z-axis. Now, the area of △OXY is given by a, the area of △OYZ by b, and the area of △OXZ by c. Find the area of △XYZ in terms of a,b,c. The correct answer is: a2+b2+c2 Pythagorean Triples Ordered triples of positive integers (a,b,c) such that a2+b2=c2 are called Pythagorean triples. There are infinitely many such triples with gcd(a,b,c)=1, and there is a simple parameterization of these Pythagorean triples which generates them all. This is a fundamental example of a nontrivial, nonlinear Diophantine equation. Fermat's attempt to generalize this equation by studying solutions to an+bn=cn led to his famous "last theorem." For more on this subject, see the wiki on Pythagorean triples. Trigonometric Identities Basic trigonometric identities are consequences of the Pythagorean theorem. For instance, in a right triangle with sides a,b,c, the trigonometric functions are defined for the angle θ opposite a by sinθ=ca, cosθ=cb. So sin2θ+cos2θ=c2a2+c2b2=c2a2+b2=c2c2=1. For more on this and other related identities, see Pythagorean identities. Proof of the Theorem For additional proofs of the Pythagorean theorem, see: Proofs of the Pythagorean Theorem. There are many unique proofs (more than 350) of the Pythagorean theorem, both algebraic and geometric. The proof presented below is helpful for its clarity and is known as a proof by rearrangement. In a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse. Given any right triangle with legs a and b and hypotenuse c, make a square with sides a+b and inscribe four copies of the given triangle as shown below: This forms a square in the center with side length c and thus an area of c2. However, if we rearrange the four triangles as follows, we can see two squares inside the larger square, one that is a2 in area and one that is b2 in area: Since the larger square is the same in both cases, i.e. (a+b)2, and since the four triangles are the same in both cases, we must conclude that the two squares a2 and b2 are in fact equal in area to the square c2. Thus, a2+b2=c2. □ The converse of the Pythagorean theorem is a special case of the cosine rule: If the sides of a triangle satisfy a2+b2=c2, then the angle opposite c is a right angle. Let ∠ACB=θ; then the cosine rule gives cosθ=2aba2+b2−c2=2ab0=0. Since 0∘≤θ≤180∘, the only solution is θ=90∘. □ See Also Triangles Pythagorean Triples Distance Formula Pythagorean Identities Fermat's Last Theorem Cite as: Pythagorean Theorem. Brilliant.org. Retrieved 03:03, September 28, 2025, from
2371	https://www.ncbi.nlm.nih.gov/books/NBK493300/bin/niceng88_qs1.pdf	Heavy menstrual bleeding Quality standard Published: 26 September 2013 www.nice.org.uk/guidance/qs47 © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Contents Contents Quality statements ......................................................................................................................................................... 3 Quality statement 1: Menstrual history ................................................................................................................ 5 Quality statement ........................................................................................................................................................................ 5 Rationale ......................................................................................................................................................................................... 5 Quality measures ......................................................................................................................................................................... 5 What the quality statement means for different audiences ....................................................................................... 6 Source guidance ............................................................................................................................................................................ 7 Definitions of terms used in this quality statement ....................................................................................................... 7 Quality statement 2: Outpatient hysteroscopy ................................................................................................. 8 Quality statement ........................................................................................................................................................................ 8 Rationale ......................................................................................................................................................................................... 8 Quality measures ......................................................................................................................................................................... 8 What the quality statement means for different audiences ....................................................................................... 9 Source guidance ............................................................................................................................................................................ 10 Definition of terms used in this quality statement ......................................................................................................... 10 Quality statement 3: Discussing treatment options ........................................................................................ 11 Quality statement ........................................................................................................................................................................ 11 Rationale ......................................................................................................................................................................................... 11 Quality measures ......................................................................................................................................................................... 11 What the quality statement means for different audiences ....................................................................................... 12 Source guidance ............................................................................................................................................................................ 12 Definition of terms used in this quality statement ......................................................................................................... 13 Update information ........................................................................................................................................................ 14 About this quality standard ......................................................................................................................................... 15 Improving outcomes ................................................................................................................................................................... 16 Resource impact ........................................................................................................................................................................... 16 Diversity, equality and language ............................................................................................................................................ 16 Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 2 of 17 This standard is based on NG88. This standard should be read in conjunction with QS15, QS73, QS172 and QS124. Quality statements Quality statements Statement 1 People presenting with symptoms related to heavy menstrual bleeding have a focused history taken that includes the impact on their quality of life. [2013, updated 2020] [2013, updated 2020] Statement 2 People with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology are offered outpatient hysteroscopy. [new 2020] [new 2020] Statement 3 People with heavy menstrual bleeding have a discussion with their healthcare professional about all their treatment options. [2013, updated 2020] [2013, updated 2020] In 2020 this quality standard was updated, and statements prioritised in 2013 were updated [2013, [2013, updated 2020] updated 2020] or replaced [new 2020] [new 2020]. For more information, see update information. Statements from the 2013 quality standard for heavy menstrual bleeding that are still supported by the evidence may still be useful at a local level: • Women with heavy menstrual bleeding who have a suspected uterine cavity abnormality, histological abnormality, adenomyosis or fibroids have a physical examination before referral for further investigations. • Women with heavy menstrual bleeding who are undergoing further investigations or awaiting definitive treatment are offered tranexamic acid and/or non-steroidal anti-inflammatory drugs at the initial assessment. Note that the terminology has changed since 2013. The 2020 statements use 'people' rather than 'women' to ensure that nobody with heavy menstrual bleeding is excluded from this quality standard. The 2013 quality standard for heavy menstrual bleeding is available as a pdf. Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 3 of 17 NICE has developed guidance and a quality standard on patient experience in adult NHS services (see the NICE Pathway on patient experience in adult NHS services), which should be considered alongside these quality statements. Other quality standards that should be considered when commissioning or providing heavy menstrual bleeding services include: • Endometriosis. NICE quality standard 172 • Suspected cancer. NICE quality standard 124 • Fertility problems. NICE quality standard 73 A full list of NICE quality standards is available from the quality standards topic library. Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 4 of 17 Quality statement 1: Menstrual history Quality statement 1: Menstrual history Quality statement Quality statement People presenting with symptoms related to heavy menstrual bleeding have a focused history taken that includes the impact on their quality of life. [2013, updated 2020] [2013, updated 2020] Rationale Rationale Heavy menstrual bleeding can be distressing and have a major impact on the person's wellbeing and many aspects of their life, including work and education. Documenting a focused menstrual history is important to identify the severity and range of the person's symptoms, and the impact on their quality of life. A focused history can ensure that people have appropriate diagnostic tests, further investigations for any underlying pathologies, and prompt and effective treatment. It can also help to avoid unnecessary referrals to secondary care. Quality measures Quality measures Structure Structure a) Evidence that healthcare professionals are aware of and recognise symptoms related to heavy menstrual bleeding that might suggest uterine cavity abnormality, histological abnormality, adenomyosis or fibroids. Data source: Data source: Local data collection, for example training records. b) Evidence of local clinical protocols for a focused history that includes the impact on quality of life based on symptoms related to heavy menstrual bleeding. Data source: Data source: Local data collection, for example local clinical protocols. Process Process Proportion of people presenting with symptoms related to heavy menstrual bleeding who have a focused history taken that includes the impact on their quality of life. Numerator – the number in the denominator who have a focused history taken that includes the Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 5 of 17 impact on their quality of life. Denominator – the number of people who present with symptoms of heavy menstrual bleeding. Data source: Data source: Local data collection, for example NHS England's heavy periods self-assessment tool. Outcome Outcome People who report they were satisfied that the impact of their heavy menstrual bleeding on quality of life was recognised. Data source: Data source: Local data collection, for example audit of patient records. What the quality statement means for different What the quality statement means for different audiences audiences Service providers Service providers (such as general practices and sexual health clinics) ensure that staff are aware of symptoms related to heavy menstrual bleeding so that they can document a focused history. The history should include severity of bleeding, related symptoms, comorbidities and the impact of heavy menstrual bleeding on quality of life. Healthcare professionals Healthcare professionals (such as GPs and nurses) document a focused history in line with the Royal College of General Practitioners' menstrual wellbeing toolkit and NHS England's heavy periods self-assessment tool when a person presents with symptoms related to heavy menstrual bleeding. The history should include severity of bleeding, related symptoms (for example irregular periods), comorbidities and the impact of heavy menstrual bleeding on quality of life. Commissioners Commissioners (such as clinical commissioning groups and NHS England) ensure that they have service specifications in place that include clinical protocols for focused histories to be taken that address severity of bleeding, related symptoms, comorbidities and the impact on quality of life when a person presents with symptoms related to heavy menstrual bleeding. People with heavy periods People with heavy periods are asked about the severity of bleeding, any other symptoms or conditions that they have and how their periods affect their life, in line with NHS England's heavy periods self-assessment tool. This includes the impact on work, education and daily life. The information is recorded in their notes by their healthcare professional. Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 6 of 17 Source guidance Source guidance Heavy menstrual bleeding: assessment and management. NICE guideline NG88 (2018, updated 2020), recommendation 1.2.1 Definitions of terms used in this quality statement Definitions of terms used in this quality statement Symptoms related to heavy menstrual bleeding Symptoms related to heavy menstrual bleeding These include persistent intermenstrual bleeding, and pelvic pain or pressure that might suggest uterine cavity abnormality, histological abnormality, adenomyosis or fibroids. [NICE's guideline on heavy menstrual bleeding, recommendation 1.2.1] Focused history Focused history A focused history should include questions about the following: • the nature of the bleeding • related symptoms such as pain and irregular periods • impact on quality of life, for example bleeding through to clothing or bedding, needing to use 2 types of sanitary product together (such as tampons and pads) or disruption to daily life (such as being unable to go out) • other factors that may affect treatment options such as comorbidities or previous treatment for heavy menstrual bleeding. [Adapted from NICE's guideline on heavy menstrual bleeding, recommendation 1.2.1, and NHS England's heavy periods self-assessment tool] Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 7 of 17 Quality statement 2: Outpatient hysteroscopy Quality statement 2: Outpatient hysteroscopy Quality statement Quality statement People with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology are offered outpatient hysteroscopy. [new 2020] [new 2020] Rationale Rationale Outpatient hysteroscopy is recommended in preference to pelvic ultrasound for investigating suspected submucosal fibroids, polyps or endometrial pathology. When carried out in accordance with the Royal College of Obstetricians and Gynaecologists' green-top guideline no.59 on hysteroscopy, best practice in outpatient, it is an efficient and safe technique with a low risk of complications, pain and distress for most people. Before carrying out hysteroscopy, the healthcare professional should discuss the procedure with the person and advise on the possible alternatives. This will ensure people have a positive experience and trust in their clinician. Quality measures Quality measures Structure Structure a) Evidence of local arrangements to ensure that outpatient hysteroscopy services are organised according to the Royal College of Obstetricians and Gynaecologists' green-top guideline no.59 on hysteroscopy, best practice in outpatient, for example facilities are adequately sized, equipped and staffed. Data source: Data source: Local data collection, for example service protocols. b) Evidence that healthcare professionals are trained to perform outpatient hysteroscopy procedures according to the Royal College of Obstetricians and Gynaecologists' green-top guideline no.59 on hysteroscopy, best practice in outpatient, using techniques and equipment that minimise discomfort and pain. Data source: Data source: Local data collection, for example benchmarked, patient-reported outcome measures including pain scores. Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 8 of 17 Process Process Proportion of people with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology who were offered outpatient hysteroscopy. Numerator – the number in the denominator who were offered outpatient hysteroscopy. Denominator – the number of people with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology. Data source: Data source: Local data collection, for example audit of patient records. Outcome Outcome Proportion of people with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology who report satisfaction with outpatient hysteroscopy. Numerator – the number in the denominator who report satisfaction with outpatient hysteroscopy. Denominator – the number of people with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology having outpatient hysteroscopy. Data source: Data source: Local data collection, for example audit of patient records. The British Society of Gynaecological Endoscopy's outpatient hysteroscopy patient survey includes national data on patient satisfaction. What the quality statement means for different What the quality statement means for different audiences audiences Service providers Service providers (such as hospitals, primary care and community-based clinics) ensure that locally agreed referral pathways are in place to allow direct-access booking into one-stop diagnostic outpatient hysteroscopy services for people with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology. Service providers ensure that the outpatient hysteroscopy procedure follows best practice guidelines. They organise regular audits that include patient-reported outcomes benchmarked against local and national standards. Healthcare professionals Healthcare professionals(such as gynaecologists, GPs and nurses) are trained to perform outpatient hysteroscopy procedures according to best practice guidelines, with techniques and Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 9 of 17 equipment that minimise discomfort and pain. They advise people to take oral analgesia before the procedure and perform vaginoscopy as the standard diagnostic technique, using miniature hysteroscopes (3.5 mm or smaller). A member of staff acts as the person's advocate during the procedure to provide reassurance, explanation and support. Commissioners Commissioners (such as clinical commissioning groups and NHS England) ensure they commission outpatient hysteroscopy services for people with heavy menstrual bleeding and suspected submucosal fibroids, polyps or endometrial pathology that have clinical protocols in place to ensure adherence to best practice guidelines. Outpatient services may be delivered in community settings if they meet best practice guidelines. People with heavy periods that may be related to other problems People with heavy periods that may be related to other problems are offered a procedure called hysteroscopy, carried out in an outpatient hysteroscopy service. People having this procedure have a discussion with their healthcare professional about what this involves and possible alternatives, and are supported to make an informed choice about their care. A member of staff acts as the person's advocate during the procedure to provide reassurance, explanation and support. Source guidance Source guidance • Heavy menstrual bleeding: assessment and management. NICE guideline NG88 (2018, updated 2020), recommendations 1.3.4, 1.3.5 and 1.3.7 • Hysteroscopy, best practice in outpatient. Royal College of Obstetricians and Gynaecologists green-top guideline no. 59 (2011) Definition of terms used in this quality statement Definition of terms used in this quality statement Outpatient hysteroscopy Outpatient hysteroscopy A procedure to examine the inside of the uterus. This is done by passing a thin telescope-like device, called a hysteroscope, that is fitted with a small camera through the neck of the womb (cervix). This procedure is done without general or regional anaesthesia. Vaginoscopy is the recommended technique and a miniature hysteroscope (3.5 mm or smaller) should be used. [Adapted from the Royal College of Obstetricians and Gynaecologists' outpatient hysteroscopy patient information leaflet] Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 10 of 17 Quality statement 3: Discussing treatment Quality statement 3: Discussing treatment options options Quality statement Quality statement People with heavy menstrual bleeding have a discussion with their healthcare professional about all their treatment options. [2013, updated 2020] [2013, updated 2020] Rationale Rationale Discussing the full range of treatment options for heavy menstrual bleeding, including the benefits and risks of each option, enables the person to make an informed decision. It is important that the healthcare professional follows the principles in NICE's guideline on patient experience in adult NHS services on communication, information and shared decision making to maximise adherence to treatment and patient satisfaction. Quality measures Quality measures Structure Structure Evidence of local arrangements to ensure that people with heavy menstrual bleeding have a documented discussion with their healthcare professional about all their treatment options. Data source: Data source: Local data collection, for example service protocols. Process Process Proportion of people with heavy menstrual bleeding who have a documented discussion with their healthcare professional about all their treatment options. Numerator – the number in the denominator who have a documented discussion with their healthcare professional about all their treatment options. Denominator – the number of people with heavy menstrual bleeding. Data source: Data source: Local data collection, for example audit of patient records. Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 11 of 17 Outcome Outcome Proportion of people with heavy menstrual bleeding who report satisfaction with the decision-making process when choosing treatment. Numerator – the number in the denominator who report satisfaction with the decision-making process when choosing treatment. Denominator – the number of people with heavy menstrual bleeding. Data source: Data source: Local data collection, for example audit of patient records. What the quality statement means for different What the quality statement means for different audiences audiences Service providers Service providers (such as hospitals, GP practices and community-based clinics) ensure that systems are in place for healthcare professionals to have documented discussions about the full range of available treatment options with people who have heavy menstrual bleeding. Healthcare professionals Healthcare professionals (such as gynaecologists, GPs and nurses) carry out a documented discussion about the full range of available treatment options for heavy menstrual bleeding with the person and follow the principles in NICE's guideline on patient experience in adult NHS services on communication, information and shared decision making. The healthcare professional also takes into account the person's fertility preferences, any comorbidities, the presence or absence of fibroids (including size, number and location), polyps, endometrial pathology or adenomyosis, and other symptoms such as pressure and pain. Commissioners Commissioners (such as clinical commissioning groups and NHS England) ensure that they commission services to provide people with the full range of treatment options available for heavy menstrual bleeding. People with heavy periods People with heavy periods have a discussion with a healthcare professional about the full range of treatments available that could help and what they involve. They are supported by their healthcare professional to choose the right treatment for them. Source guidance Source guidance Heavy menstrual bleeding: assessment and management. NICE guideline NG88 (2018, updated Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 12 of 17 2020), recommendations 1.4.1, 1.4.2, 1.4.7, 1.5.1 and 1.5.5 Definition of terms used in this quality statement Definition of terms used in this quality statement Discussion about treatment options Discussion about treatment options Discussions should cover: • the benefits and risks of the various options • suitable treatments if the person is trying to conceive • whether the person wants to retain their fertility and/or uterus. A full discussion is essential when people are considering hysterectomy and should include the implications of surgery. Surgical options including hysterectomy can be offered if treatment is unsuccessful, the person declines pharmacological treatment or symptoms are severe. [Adapted from NICE's guideline on heavy menstrual bleeding, recommendations 1.4.2, 1.4.7 and 1.5.5, NHS England's evidence-based interventions programme and the NICE endorsed shared decision making aid for heavy menstrual bleeding] Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 13 of 17 Update information Update information October 2020: October 2020: This quality standard was updated and statements prioritised in 2013 were replaced. Statements are marked as: • [new 2020] [new 2020] if the statement covers a new area for quality improvement • [2013, updated 2020] [2013, updated 2020] if the statement covers an area for quality improvement included in the 2013 quality standard and has been updated. Statements numbered 1 and 5 in the 2013 version have been updated and are included in the updated quality standard, marked as [2013, updated 2020] [2013, updated 2020]. Statements from the 2013 version that are still supported by the evidence may still be useful at a local level, and are listed in the quality statements section. The 2013 quality standard for heavy menstrual bleeding is available as a pdf. Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 14 of 17 About this quality standard About this quality standard NICE quality standards describe high-priority areas for quality improvement in a defined care or service area. Each standard consists of a prioritised set of specific, concise and measurable statements. NICE quality standards draw on existing NICE or NICE-accredited guidance that provides an underpinning, comprehensive set of recommendations, and are designed to support the measurement of improvement. Expected levels of achievement for quality measures are not specified. Quality standards are intended to drive up the quality of care, and so achievement levels of 100% should be aspired to (or 0% if the quality statement states that something should not be done). However, this may not always be appropriate in practice. Taking account of safety, shared decision making, choice and professional judgement, desired levels of achievement should be defined locally. Information about how NICE quality standards are developed is available from the NICE website. See our webpage on quality standard advisory committees for details of standing committee 3 members who advised on this quality standard. Information about the topic experts invited to join the standing members is available on the webpage for this quality standard. This quality standard has been included in the NICE Pathway on heavy menstrual bleeding, which brings together everything we have said on a topic in an interactive flowchart. NICE has produced a quality standard service improvement template to help providers make an initial assessment of their service compared with a selection of quality statements. This tool is updated monthly to include new quality standards. NICE produces guidance, standards and information on commissioning and providing high-quality healthcare, social care, and public health services. We have agreements to provide certain NICE services to Wales, Scotland and Northern Ireland. Decisions on how NICE guidance and other products apply in those countries are made by ministers in the Welsh government, Scottish government, and Northern Ireland Executive. NICE guidance or other products may include references to organisations or people responsible for commissioning or providing care that may be relevant only to England. Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 15 of 17 Improving outcomes Improving outcomes This quality standard is expected to contribute to improvements in the following outcomes: • awareness of symptoms related to heavy menstrual bleeding • early diagnosis of heavy menstrual bleeding • access to outpatient diagnostic hysteroscopy services managed in accordance with best practice guidelines • satisfaction of people with heavy menstrual bleeding with their involvement in decision making • quality of life experienced by people with heavy menstrual bleeding • patient experience • patient-reported outcome measures (PROMs) relating to heavy menstrual bleeding. It is also expected to support delivery of the Department of Health and Social Care outcome frameworks: • NHS outcomes framework • Public health outcomes framework for England. Resource impact Resource impact NICE quality standards should be achievable by local services. The potential resource impact is considered by the quality standards advisory committee, drawing on resource impact work for the source guidance. Organisations are encouraged to use the resource impact statement for the source guideline to help estimate local costs. Diversity, equality and language Diversity, equality and language Equality issues were considered during development and equality assessments for this quality standard are available. Any specific issues identified during development of the quality statements are highlighted in each statement. Commissioners and providers should aim to achieve the quality standard in their local context, in Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 16 of 17 light of their duties to have due regard to the need to eliminate unlawful discrimination, advance equality of opportunity and foster good relations. Nothing in this quality standard should be interpreted in a way that would be inconsistent with compliance with those duties. ISBN: 978-1-4731-0311-5 Endorsing organisation Endorsing organisation This quality standard has been endorsed by NHS England, as required by the Health and Social Care Act (2012) Supporting organisations Supporting organisations Many organisations share NICE's commitment to quality improvement using evidence-based guidance. The following supporting organisations have recognised the benefit of the quality standard in improving care for patients, carers, service users and members of the public. They have agreed to work with NICE to ensure that those commissioning or providing services are made aware of and encouraged to use the quality standard. • British Society for Gynaecological Endoscopy • Royal College of General Practitioners (RCGP) • Royal College of Nursing (RCN) • Royal College of Paediatrics and Child Health • Menstrual Health Coalition Heavy menstrual bleeding (QS47) © NICE 2021. All rights reserved. Subject to Notice of rights ( Last updated 16 October 2020 Page 17 of 17
2372	https://byjus.com/chemistry/osmolarity-questions/	The number of moles of solute per unit volume of solution is measured by molarity, while the number of osmoles of solute particles per unit volume of solution is measured by osmolarity. This value enables the measurement of a solution’s osmotic pressure and the determination of how the solvent will diffuse across a semipermeable membrane (osmosis) separating two solutions with different osmotic concentrations. \| \| \| Definition: Osmolarity is a concentration expression that reflects the osmoticity of solutions. It is an osmolar concentration expression. The number of solute particles per 1 L of solvent is referred to as osmolarity. A solution’s osmolarity is typically expressed as Osm/L. \| Osmolarity Chemistry Questions with Solutions Q1. The osmolarity of a solution containing a 1M NaCl solution is ____. a.) 1 b.) 2 c.) 3 d.) None of the above Correct Answer– (b.) 2. Q2. Osmolarity is independent of ____. a.) number of particles b.) moles c.) nature of particles d.) All of the above Correct Answer– (c.) nature of particles. Q3. The osmolarity of a solution containing a 1M CaCl2 solution is ____. a.) 1 b.) 2 c.) 3 d.) None of the above Correct Answer– (c.) 3. Q4. Osmolarity is the number of osmoles ____. a.) per litre of solvent b.) per kg of solvent c.) per litre of solution d.) per kg of solution Correct Answer– (c.) per litre of solution. Q5. A solution with high osmolarity has ____ water molecules with respect to solute particles. a.) larger b.) equal c.) fewer d.) None of the above Correct Answer– (c.) fewer. Q6. Define osmolarity. Answer. Osmolarity is defined as the number of particles or the concentration of a specific solute or solutes per litre of solvent. Osmoles are used to measure osmolarity. Q7. If 4 moles of MgCl2 are dissolved in 2 L of water. What will the osmolarity be? Answer. 1 mole of MgCl2 gives 3 pieces = 3 osmoles = 12. Therefore, the Osmolarity of 4 moles of MgCl2 = 12 × ½ = 6 OsM. Q8. How will you calculate osmolarity if there are many solutes in the solution? Answer. If a solution contains many different types of solutes (for example, both glucose and sodium chloride), its osmolarity can be calculated using the equation Osmolarity = Sum of all (molarity × n) of all solutes n: the number of particles released from the solute molecule. Q9. What is the osmolarity of a 2.6 M solution of table salt? Answer. NaCl dissociates completely in water to form Na+ ions and Cl– ions. Thus, each mole of NaCl becomes two osmoles in the solution: one mole of Na+ and one mole of Cl–. Therefore, osmolarity = 2 × 2.6 M = 5.2 osmoles. Q10. How can osmolarity be determined? Answer. To determine the osmolarity of a solution, first, determine the number of osmoles of solute per litre of solvent. An osmole is a unit of measurement for the number of moles in a compound. A Mole is a unit of measurement for the amount of matter in a substance that contains precisely 6.02214086 × 1023 particles. The Avogadro constant is another name for this number. The following is a simplified formula for osmolarity. If there is more than one solute, the osmolarity is the sum of the solute molarities per litre of solvent. Osmolarity (osmol) = Osmoles/litre Q11. Differentiate between osmolarity and osmolality. Answer. Osmolarity is defined as the concentration in moles per litre of solution, whereas osmolarity is defined as the concentration in moles per kilogram of solvent. However, for dilute solutions, they are numerically almost the same. Since the volume of the solvent varies with temperature, osmolarity measurements are temperature dependent (i.e., the volume is larger at higher temperatures). In contrast, osmolality, which is based on mass of the solvent is temperature independent. Q12. What is the osmolarity of a glass of skim milk? (There are 13 g of lactose in 236 mL) Answer. Lactose is a kind of sugar with a molecular weight of about 342 g per mole. 13 g lactose × 1 mole/342 g = 0.038 mole. Osmolarity = 0.038 mole / 0.236 litre = 0.16 OsM Q13. A solution contains 312 mg of K+ ions per 200mL. How many milliosmoles are represented in a litre of the solution? Answer. Grams of potassium ions present in 1000 mL Conversion of mg to g: = 312 mg × 1g/1000mg = 0.312g (\begin{array}{l}Grams\ of\ potassium\ ions\ present\ in\ 1L =\frac{0.312 g of K ions}{200 mL}=\frac{x}{1000 mL}=1.56 g\end{array} ) (\begin{array}{l}milliosmoles\ per\ litre=\frac{mass\ in\ g}{molecular\ weight}\times number\ of\ particles\times 1000\end{array} ) Mass in g = 1.56 g Molecular weight of potassium = 39 Number of particles = 1 as only potassium ions are present in the solution. (\begin{array}{l}milliosmoles\ per\ litre=\frac{1.56}{39}\times 1\times 1000=40\end{array} ) Hence, 40 milliosmoles of potassium are represented in a litre of the solution. Q14. Calculate the osmolarity of 0.2 M of Na2SO4. Answer. Given, M = 0.2 M Osmolarity = Molarity × n Na2SO4 dissociates into 2Na+ + SO42– Therefore, n = 3 Hence, osmolarity will be 0.2 × 3 = 0.6 OsM Q15. A solution contains 1% of anhydrous dextrose in water forinjection. How many milliosmoles per litre are represented by this concentration? (m.w = 180) Answer. 1% of anhydrous dextrose = 1g/100 mL = x/1000 mL = 10 g (\begin{array}{l}Milliosmoles\ per\ litre =\frac{mass\ in\ g}{molecular\ weight}\times number\ of\ particles\times 1000 \end{array} ) Mass in g = 10 g Molecular weight of dextrose = 180 Number of particles = 1 as dextrose is a non electrolyte (\begin{array}{l}milliosmoles\ per\ litre =\frac{10\ g}{180}\times 1\times 1000 =56\end{array} ) Practise Questions on Osmolarity Q1. The osmolarity of a solution containing 1M sucrose solution is ____. a.) 1 b.) 2 c.) 3 d.) None of the above Q2. For dilute solutions, the value of osmolarity is ___ osmolality. a.) greater than b.) similar to c.) smaller than d.) None of the above Q3. What is the approximate osmolarity of the sugar in a can of Red Bull? (There are 27 g of sugar in a 250 mL can). Q4. What is the osmolarity of seawater given the following: \| Solute \| g / L \| Molecular weight \| --- \| Cl– \| 19 \| 35 \| \| Na+ \| 10.5 \| 23 \| \| Mg \| 1.3 \| 24 \| \| S \| 0.8 \| 32 \| Q5. Can osmolarity be calculated from osmolality? If yes, explain how? Click the PDF to check the answers for Practice Questions. Download PDF Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
2373	https://math.stackexchange.com/questions/1228366/how-to-count-each-numeral-of-occurrences-of-digits	Skip to main content How to count each numeral of occurrences of digits? Ask Question Asked Modified 10 years, 4 months ago Viewed 2k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I want to count each numeral(0 through 9) of occurrences of digits in the range [1,n]. Note that 101 has two one and one zero. For example, if n equals 11: f0(11)1f1(11)4f2(11)1f3(11)1f4(11)1f5(11)1f6(11)1f7(11)1f8(11)1f9(11)1 How can I generalize about this function to f? combinatorics number-theory integers Share CC BY-SA 3.0 Follow this question to receive notifications asked Apr 10, 2015 at 10:41 ZeroPepsiZeroPepsi 91777 silver badges1717 bronze badges 3 1 There are 4 1s in 11? Who knew? – Thomas Andrews Commented Apr 10, 2015 at 10:46 what do you mean to generalize this to f? What is f? Do you want a function f:Z+→(Z+)10 which outputs all of the numbers at once? – Ove Ahlman Commented Apr 10, 2015 at 10:47 Ah, it is the number of 1s in 1,2,3,4,5,6,7,8,9,10,11. So it is the sum of the number of occurrences. – Thomas Andrews Commented Apr 10, 2015 at 10:50 Add a comment \| 1 Answer 1 Reset to default This answer is useful 3 Save this answer. Show activity on this post. Let us take a first step. I think f0 will be more tricky, but some easy figures are fi(10k−1) for i≥1. This corresponds to aksing how many occurrences there are of the digit i in the 10k possible k-digit strings 0...0,0...1,...,9...9 and since all 10 digits by symmetry occur equally often in those strings, the digit i must have occurred exactly 110⋅k⋅10k=k⋅10k−1 times Thus fi(9)=1,fi(99)=20,fi(999)=300 and so on. Based on the above, we can describe fi(10ka) where a is a non-zero digit. We have fi(10ka)=⎧⎩⎨⎪⎪a⋅fi(10k−1)a⋅fi(10k−1)+1a⋅fi(10k−1)+10k=ak⋅10k−1=ak⋅10k−1+1=(ak+10)⋅10k−1for ai note that fi(10k−1) counts the number of occurrences of i in the last k digits, whereas the +1 and +10k adds the number of times i occurs in the first digit. Now to the general case of fi(10ka+b) where a is a non-zero digit and b is some number with at most k digits. Here we have fi(10ka+b)={fi(10ka)+fi(b)fi(10ka)+fi(b)+bfor a≠ifor a=i and applying the formulas for fi(10ka) from before, this gives us fi(10ka+b)=⎧⎩⎨⎪⎪ak⋅10k−1+fi(b)ak⋅10k−1+1+fi(b)+b(ak+10)⋅10k−1+fi(b)for ai For a given number n=10ka+b, those formulas can be applied recursively for each non-zero digit i≥1. Thus we can determine f1(n),f2(n),...,f9(n). There will be (a−1)⋅10k and (b+1) numbers having (k+1) digits and 9⋅10s−1 having s digits for each s<k+1. Thus our numbers have a total of T(n)=(k+1)[b+1+(a−1)⋅10k]+∑s=1k9s⋅10s−1 digits and from this f0(n) can be computed as f0(n)=T(n)−∑i=19fi(n) Examples As an example, consider n=123. Then f2(123)=1⋅2⋅101+f2(23)=20+2⋅1⋅100+1+f2(3)+3=20+2+1+1+3=27 noting that f2(3)=1. This is easily checked to be correct. I wrote a few lines of code scanning through 1,2,...,123 as strings counting occurrences of "2". It confirmed the figure 27. One more example, let us try n=314159 and i=3: f3(314159)=3⋅5⋅104+1+f3(14159)+14159=164160+1⋅4⋅103+f3(4159)=168160+(4⋅3+10)⋅102+f3(159)=170360+1⋅2⋅101+f3(59)=170380+(5⋅1+10)⋅100+f3(9)=170396 and again my computer confirmed this figure. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Apr 11, 2015 at 0:08 answered Apr 10, 2015 at 22:35 StringString 18.8k55 gold badges4545 silver badges8484 bronze badges 1 BTW: I think that my recursion runs in something like O(logn) whereas a simple scan runs in O(nlogn) if I am not mistaken. I have not thought too deeply about that, though. At least I can actually see, how the performance is unequal using the two different methods. The recursion computes large figures almost instantly whereas the scan gets remarkably slower for large n's. – String Commented Apr 11, 2015 at 0:14 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 combinatorics number-theory integers See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Linked 5 How to find how many numbers do not contain the digit n in the range from a to b? 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2374	https://ecampusontario.pressbooks.pub/genchemforgeegees/chapter/7-2-measuring-expressing-reaction-rates/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 7.2 – Measuring & Expressing Reaction Rates The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically. What follows is general guidance and examples of measuring the rates of a reaction. Measuring time change is easy; a stopwatch or any other time device is sufficient. However, determining the change in concentration of the reactants or products involves more complicated processes. The change of concentration in a system can generally be acquired in two ways: By monitoring the depletion of reactant over time, or By monitoring the formation of product over time It does not matter whether an experimenter monitors the reagents or products because there is no effect on the overall reaction. However, since reagents decrease during the reaction, and products increase, there is a sign difference between the two rates. Reagent concentration decreases as the reaction proceeds, giving a negative number for the change in concentration. The products, on the other hand, increase concentration with time, giving a positive number. Graphically, the general shape of the curves of concentration versus time for reactants and products looks like this: Figure 7.2.1 Concentration vs Time Since the convention is to express the rate of reaction as a positive number, to solve a problem, set the overall rate of the reaction equal to the negative of a reagent’s disappearing rate. The overall rate also depends on stoichiometric coefficients. It is worth noting that the process of measuring the concentration can be greatly simplified by taking advantage of the different physical or chemical properties (ie: phase difference, reduction potential, etc.) of the reagents or products involved in the reaction by using the above methods. We have emphasized the importance of taking the sign of the reaction into account to get a positive reaction rate. Now, we will turn our attention to the importance of stoichiometric coefficients. A reaction rate can be reported quite differently depending on which product or reagent selected to be monitored. Given a reaction: aA + bB → cC + dD the general rate for this reaction is defined as rate = –1aΔ[A]Δt = -1bΔ[B]Δt = -1cΔ[C]Δt = 1dΔ[D]Δt Equation 7.2.1 General Rate of Reaction Following the Course of a Reaction There are two different ways you can follow the course of a reaction. Samples of the mixture can be collected at intervals and titrated to determine how the concentration of one of the reagents is changing. A physical property of the reaction which changes as the reaction continues can be measured: for example, the volume of gas produced. These approaches must be considered separately. Consider that bromoethane reacts with sodium hydroxide solution as follows: CH3CH2Br + OH– → CH3CH2OH + Br– During the course of the reaction, both bromoethane and sodium hydroxide are consumed. However, it is relatively easy to measure the concentration of sodium hydroxide at any one time by performing a titration with a standard acid: for example, with hydrochloric acid of a known concentration. The process starts with known concentrations of sodium hydroxide and bromoethane, and it is often convenient for them to be equal. Because the reaction is 1:1, if the concentrations are equal at the start, they remain equal throughout the reaction. Samples are taken with a pipette at regular intervals during the reaction and titrated with standard hydrochloric acid in the presence of a suitable indicator. The problem with this approach is that the reaction is still proceeding in the time required for the titration. In addition, only one titration attempt is possible because, by the time another sample is taken, the concentrations have changed. There are two ways around this problem: The reaction can be slowed by diluting it, adding the sample to a larger volume of cold water before the titration. Then the titration is performed as quickly as possible. This is most effective if the reaction is carried out above room temperature. Cooling it as well as diluting it slows it down even more. If possible (and it is possible in this case) it is better to stop the reaction completely before titrating. In this case, this can be accomplished by adding the sample to a known, excess volume of standard hydrochloric acid. This consumes all the sodium hydroxide in the mixture, stopping the reaction. At this point, the resulting solution is titrated with standard sodium hydroxide solution to determine how much hydrochloric acid is left over in the mixture. This allows one to calculate how much acid was used, and thus how much sodium hydroxide must have been present in the original reaction mixture. This technique is known as back titration. This process generates a tabulated set of values for the concentration of (in this example) sodium hydroxide over time. The concentrations of bromoethane are, of course, the same as those obtained if the same concentrations of each reagent were used. These values are plotted to give a concentration-time graph, such as that below: With either the tabulated results or a graph like the one above, we can calculate different types of rates of reaction: average and instantaneous rates of reaction. Average vs. Instantaneous Reaction Rates Reaction rates have the general form of (change in concentration/change in time). There are two types of reaction rates(the third method using initial rates will be addressed separately). One is called the average rate of reaction, often denoted by (Δ[conc.] / Δt), while the other is referred to as the instantaneous rate of reaction, denoted as either: limΔt→0 Δ[concentration]Δt or d[concentration]dt The average rate of reaction, as the name suggests, is an average rate, obtained by taking the change in concentration over a time period, for example: -0.3 M / 15 minutes. This is an approximation of the reaction rate in the interval; it does not necessarily mean that the reaction has this specific rate throughout the time interval or even at any instant during that time. The instantaneous rate of reaction, on the other hand, depicts a more accurate value. The instantaneous rate of reaction is defined as the change in concentration of an infinitely small time interval, expressed as the limit or derivative expression above. In simpler words, it is the rate at which a reaction is proceeding at any specific time. Consider the analogy of a car slowing down as it approaches a stop sign. The reading on the speedometer at any one specific point in time－call it t1－during the deceleration is an instantaneous rate. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance travelled to the time required to bring the vehicle to a complete stop (Δt). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates. Recall the example from the previous section for the breakdown of H2O2 into H2O and O2. If we measure the concentration of hydrogen peroxide, H2O2, in an aqueous solution, we find that it changes slowly over time as the H2O2 decomposes, according to the equation: 2H2O2 (aq) → 2H2O (l) + O2 (g) The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here: rateofdecompositionofH2O2 = -changeinconcentrationofreactanttimeinterval = -[H2O2]t2 – [H2O2]t1t2 – t1 = -Δ[H2O2]Δt Note: There isn’t a ½ in front since we are not looking at the overall rate of the reaction. Here we are specifically looking at the rate of decomposition of H2O2 (aq). Now imagine that at regular time intervals we experimentally determine the concentration of H2O2 present in the reaction mixture – Figure 7.2.2 below provides an example of data collected during the decomposition of H2O2. \| \| \| \| \| \| --- --- \| Time (h) \| [H2O2] (mol L-1) \| Δ[H2O2] (mol L-1) \| Δt (h) \| Rate of Decomposition (M/h) \| \| 0.00 \| 1.000 \| -0.500 \| 6.00 \| 0.0833 \| \| 6.00 \| 0.500 \| -0.500 ↔ -0.250 \| 6.00 \| 0.0833 ↔ 0.0417 \| \| 12.00 \| 0.250 \| -0.250 ↔ -0.125 \| 6.00 \| 0.0417 ↔ 0.0208 \| \| 18.00 \| 0.125 \| -0.125 ↔ -0.062 \| 6.00 \| 0.0208 ↔ 0.0103 \| \| 24.00 \| 0.0625 \| -0.062 \| 6.00 \| 0.0103 \| \| Figure 7.2.2. The rate of decomposition of H2O2 in an aqueous solution decreases as the concentration of H2O2 decreases. To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40°C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period: -Δ[H2O2]Δt = -(0.500 mol/L – 1.000mol/L)(6.00h – 0.00h) = 0.0833 molL-1h-1 Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of: -Δ[H2O2]Δt = -(0.0625 mol/L – 0.125mol/L)(24.00h – 18.00h) = 0.0104 molL-1h-1 Both calculations above involve determining an average rate of reaction since we look at concentration values at the beginning and end of a time period. Hence, these calculated values are an average rate of reaction over that particular time interval (e.g. the time period between 18.00 h and 24.00 h spanning the last 6 hour time interval). Graphically, determining an average rate of reaction follows the same principle – a line is drawn between two points on the concentration versus time curve (Figure 7.2.3). These two points represent the two concentration values at two different instants in time. Calculating the slope of this line uses the same formula to calculate an average rate of reaction – Δ[conc.] / Δt. averagerate = slope averagerate = ΔyΔx averagerate = Δ[conc]Δt averagerate = ([conc]2 – [conc]1)(t2 – t1) Figure 7.2.3. Determining the average rate of reaction from a concentration versus time graph. The instantaneous rate of a reaction may be determined in one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible (Figure 7.2.4). Going back to our example for the decomposition of H2O2, if we plot the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H2O2 at any time t is given by the slope of a straight line that is tangent to the curve at that time (Figure 7.2.5). We can use calculus to evaluate the slopes of such tangent lines, but the procedure for doing so is beyond the scope of this chapter. instantaneousrateatt = slope instantaneousrateatt = ΔyΔx instantaneousrateatt = Δ[conc]Δt instantaneousrateatt = ([conc]2 – [conc]1)(t2 – t1) Figure 7.2.4. Determining the instantaneous rate of reaction from a concentration versus time graph. Figure 7.2.5. This graph shows a plot of concentration versus time for a 1.000 M solution of H2O2. The rate at any instant is equal to the slope of a line tangential to this curve at that time. Tangents are shown at t = 0 h (“initial rate”) and at t = 10 h (“instantaneous rate” at that particular time). Initial Rate of Reaction The initial rate of reaction is the rate at which the reagents are first brought together. Recall the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t0). Like the instantaneous rate mentioned above, the initial rate can be obtained either experimentally or graphically. To experimentally determine the initial rate, scientists must bring the reagents together and measure the reaction rate as quickly as possible. If this is not possible, they can find the initial rate graphically. To do this, one must find the slope of the line tangent to the reaction curve when t = 0 (Figure 7.2.6). initialrate = slope initialrate = ΔyΔx initialrate = Δ[conc]Δt initialrate = ([conc]2 – [conc]1)(t2 – t1) Figure 7.2.6.Determining the initial rate of reaction from a concentration versus time graph. \| \| \| Reaction Rates in Analysis: Test Strips for Urinalysis \| \| Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (Figure 7.2.7). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in colour upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations. The test for urinary glucose relies on a two-step process represented by the chemical equations shown here: The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colourless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct colour change. The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for the completion of the colour-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the colour change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ions by other substances found in urine. Figure 7.2.6. Test strips are commonly used to detect the presence of specific substances in a person’s urine. Many test strips have several pads containing various reagents to permit the detection of multiple substances on a single strip. (credit: Iqbal Osman) \| \| License General Chemistry for Gee-Gees Copyright © by Kevin Roy; Mahdi Zeghal; Jessica M. Thomas; and Kathy-Sarah Focsaneanu is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. Share This Book
2375	https://www.reddit.com/r/AskReddit/comments/12b8fnc/in_rainman_when_raymond_is_counting_the/	In Rainman, when Raymond is counting the toothpicks he is counting up to 82 three times (there were 246 on the ground and 4 in the box), the number 82 has some sort of deeper meaning or it's just random screenwriting? : r/AskReddit Skip to main contentIn Rainman, when Raymond is counting the toothpicks he is counting up to 82 three times (there were 246 on the ground and 4 in the box), the number 82 has some sort of deeper meaning or it's just random screenwriting? : r/AskReddit Open menu Open navigationGo to Reddit Home r/AskReddit A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to AskReddit r/AskReddit r/AskReddit r/AskReddit is the place to ask and answer thought-provoking questions. 57M Members Online •3 yr. ago albcamihai In Rainman, when Raymond is counting the toothpicks he is counting up to 82 three times (there were 246 on the ground and 4 in the box), the number 82 has some sort of deeper meaning or it's just random screenwriting? Archived post. New comments cannot be posted and votes cannot be cast. Share Related Answers Section Related Answers Details about toothpick counting in Rainman Significance of numbers in Rainman movie Rainman movie true story background Role of Bonnie Hunt in Rainman Rainman casino scenes analysis New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of April 4, 2023 Reddit reReddit: Top posts of April 2023 Reddit reReddit: Top posts of 2023 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
2376	https://www.ncbi.nlm.nih.gov/books/NBK549849/table/article-25640.table0/?report=objectonly	[Table], Table.Expected CSF Findings in Bacterial, Viral, and Fungal Meningitis - StatPearls - NCBI Bookshelf Table.Expected CSF Findings in Bacterial, Viral, and Fungal Meningitis View in own window Type of MeningitisAppearanceOpening Pressure (mm Hg)White Blood Cells (cells/µL)Protein (mg/dL)Glucose (mg/dL) Normal Clear 90-180<8 15–45 50–80 Bacterial Turbid Elevated>1000–2000>200<40 Viral Clear Normal<300; lymphocytic predominance<200 Normal Fungal Clear Normal-elevated<500>200 Normal-low From: Meningococcal Disease (Neisseria meningitidis Infection) StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. External link. Please review our privacy policy. Making content easier to read in Bookshelf Close We are experimenting with display styles that make it easier to read books and documents in Bookshelf. Our first effort uses ebook readers, which have several "ease of reading" features already built in. The content is best viewed in the iBooks reader. You may notice problems with the display of some features of books or documents in other eReaders. Cancel Download
2377	https://www.brainingcamp.com/base-ten-blocks	Base Ten Blocks - Interactive Virtual Manipulatives \| Brainingcamp ManipulativesResourcesPricing Redeem CodeSign InSign InFREE 30 DAY TRIAL Base Ten Blocks Base Ten Blocks (or place value manipulatives) represent our base ten number system. The smallest blocks, called "units", measure 1 unit on each side and represents 1. The long, narrow blocks called "rods" consist of ten of the "units" cubes and represent 10. The "flats" consist of ten rods and represent 100. The largest blocks, called "cubes" measure ten units per side and represent 1,000. SIGN-UP NOW Features Place value charts Decimal or whole number options Regroup animations Workspace options, including comparing, factors, area models, and more! Ready-to-use Tasks High-quality, ready and easy-to-use content. Our rich math Tasks use our digital math manipulatives to engage students in deep mathematical exploration. ACCESS ALL TASKS Brainingcamp Student-centric, real-time learning. Host a session, monitor student progress, and provide instant feedback. LIVE allows educators to instantly monitor, assess, and provide feedback for every student. Teacher Provide a unique opportunity to formatively assess student work while giving individualized guidance – ensuring efficient instructional time! Student Feel supported with direct and live guidance. Student work is only viewable by teachers, providing a sense of security. LEARN MORE Ready to get started? TRY FREE NOW Base Ten Blocks, Plus More With Brainingcamp, you get 18 of the most popular classroom manipulatives. Brainingcamp is your all-in-one place for teaching with manipulatives. Algebra Tiles Bar Models Base Ten Blocks Clock Color Tiles Cuisenaire® Rods Fraction Circles Fraction Tiles Geoboard Hundred Board Linking Cubes Money Number Lines Pattern Blocks Place Value Disks Rekenreks Two-Color Counters XY Coordinate Board 30-DAY TRIALVIEW PRICING Frequently Asked Questions What are virtual Base Ten Blocks? How do you use virtual Base Ten Blocks? What are the Base Ten Blocks called? What are Base Ten Block units? AboutBlogAccessibilityPartnersContactSupport © 2025 Brainingcamp, LLC Cuisenaire® and the sequence and selection of colors of all Cuisenaire® Rods are a registered trademark of hand2mind, Inc. The sequence and selection of colors of all fraction pieces are a trademark of Learning Resources, Inc. Terms of Use\|Privacy Policy P.O. Box 163471, Austin, TX 78716\|Phone 888.967.2267
2378	https://economics.stackexchange.com/questions/8980/does-unit-elasticity-has-to-be-at-exactly-the-middle-of-the-demand-curve	Does unit elasticity has to be at exactly the middle of the demand curve? - Economics Stack Exchange Join Economics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Economics helpchat Economics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Economics Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Does unit elasticity has to be at exactly the middle of the demand curve? Ask Question Asked 9 years, 11 months ago Modified9 years, 11 months ago Viewed 5k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Can unit elasticity be anywhere else on the demand curve other than the midpoint? elasticity Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Nov 1, 2015 at 8:49 Ubiquitous 17.1k 4 4 gold badges 38 38 silver badges 84 84 bronze badges asked Oct 31, 2015 at 23:17 Benjamin BlackswanBenjamin Blackswan 243 1 1 gold badge 3 3 silver badges 6 6 bronze badges 3 2 I find this question confusing... It's not hard to define a curve that extends infinitely. Where is the midpoint of such a curve?dismalscience –dismalscience 2015-10-31 23:22:10 +00:00 Commented Oct 31, 2015 at 23:22 Demand curve is normally NOT positive sloped, (other then Giffen or Veblen). Therefore it can not be infinite long.Benjamin Blackswan –Benjamin Blackswan 2015-10-31 23:47:29 +00:00 Commented Oct 31, 2015 at 23:47 2 Curves are not all straight, and some have asymptotic properties. You do know that, right?dismalscience –dismalscience 2015-10-31 23:52:51 +00:00 Commented Oct 31, 2015 at 23:52 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. When the demand function is linear, q=a−b p q=a−b p, the only point were elasticity is unity is located in the midpoint of the demand curve (straight line). This is geometrical. The demand line will cross the vertical p p-axis at p=a/b p=a/b and the horizontal q q axis at q=a q=a. For unitary elasticity (in absolute terms) we want \|d q/d p\|q⋅p=b p a−b p=1⟹p=a 2 b⟹q=a 2\|d q/d p\|q⋅p=b p a−b p=1⟹p=a 2 b⟹q=a 2 So at unitary elasticity the corresponding price lies in the middle of the feasible price domain, and the corresponding quantity lies in the middle of the feasible quantity domain. The related diagram is The length [A B][A B] is equal to the length [C D][C D], and the length [B C][B C] is equal to the length [D E][D E]. But this implies that the triangles [A B C][A B C] and [C D E][C D E] have two of their sides equal, so necessarily they will have also their third side equal. So [A C]=[C E][A C]=[C E], which implies that the point of unitary elasticity C C is the midpoint of the demand line. Obviously, no other point can have unitary elasticity here. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Nov 1, 2015 at 3:07 Alecos PapadopoulosAlecos Papadopoulos 34.3k 1 1 gold badge 52 52 silver badges 120 120 bronze badges Add a comment\| This answer is useful 5 Save this answer. Show activity on this post. No, that is only true in the linear case. For a simple counterexample consider D(p)=1−p–√.D(p)=1−p. ϵ(p)=d D(p)d p⋅p D(p)=−1 2⋅p–√⋅p 1−p–√=1 2⋅p–√1−p–√ϵ(p)=d D(p)d p⋅p D(p)=−1 2⋅p⋅p 1−p=1 2⋅p 1−p ϵ(p)1 2⋅p–√1−p–√p–√p–√p=====1 1 2−2⋅p–√2 3 4 9.ϵ(p)=1 1 2⋅p 1−p=1 p=2−2⋅p p=2 3 p=4 9. This is not in the middle of the demand curve in any sense. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Oct 31, 2015 at 23:50 answered Oct 31, 2015 at 23:45 GiskardGiskard 29.8k 11 11 gold badges 49 49 silver badges 82 82 bronze badges 1 D(p)=1−p 2 D(p)=1−p 2 would probably have been even simpler.Giskard –Giskard 2015-10-31 23:51:56 +00:00 Commented Oct 31, 2015 at 23:51 Add a comment\| Your Answer Thanks for contributing an answer to Economics 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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2379	https://www.droracle.ai/articles/120308/anticholinergic-toxidrome	What are the symptoms of Anticholinergic (anticholinergic) toxidrome? ▼ What are the symptoms of Anticholinergic (anticholinergic) toxidrome? Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: May 15, 2025 • View editorial policy From the Guidelines The anticholinergic toxidrome is a potentially life-threatening condition that requires immediate supportive care, including treatment of agitation with benzodiazepines, management of hyperthermia, and consideration of physostigmine as a specific antidote in severe cases with significant central nervous system effects. The condition is caused by excessive blockade of muscarinic acetylcholine receptors, resulting in a constellation of symptoms summarized by the phrase "hot as a hare, blind as a bat, dry as a bone, red as a beet, and mad as a hatter" 1. Common causes include antihistamines, tricyclic antidepressants, antipsychotics, antiparkinsonians, and plants like jimsonweed. Patients typically present with: Hyperthermia Flushed skin Mydriasis (dilated pupils) Decreased bowel sounds Urinary retention Tachycardia Hypertension Altered mental status ranging from agitation to delirium and hallucinations Management of anticholinergic toxidrome involves: Supportive care Activated charcoal for recent ingestions Physostigmine (1-2 mg IV slowly) as a specific antidote in severe cases with significant central nervous system effects, although it should be avoided in patients with cardiac conduction abnormalities or suspected tricyclic antidepressant overdose due to risk of asystole 1. Benzodiazepines may be used for agitation and seizures Cooling measures should be implemented for hyperthermia. The toxidrome occurs because anticholinergic agents block the neurotransmitter acetylcholine from binding to muscarinic receptors, disrupting parasympathetic nervous system function and causing sympathetic predominance 1. From the FDA Drug Label Physostigmine Salicylate Injection can reverse both central and peripheral anticholinergia. The anticholinergic syndrome has both central and peripheral signs and symptoms. Central toxic effects include anxiety, delirium, disorientation, hallucinations, hyperactivity and seizures. Severe poisoning may produce coma, medullary paralysis and death Peripheral toxicity is characterized by tachycardia, hyperpyrexia, mydriasis, vasodilation, urinary retention, diminution of gastrointestinal motility, decrease of secretion in salivary and sweat glands, and loss of secretions in the pharynx, bronchi, and nasal passages Dramatic reversal of the effects of anticholinergic symptoms can be expected in minutes after the intravenous administration of Physostigmine Salicylate Injection, if the diagnosis is correct and the patient has not suffered anoxia or other insult. The drug Physostigmine Salicylate Injection can be used to treat anticholinergic toxidrome by reversing both central and peripheral anticholinergia. The anticholinergic syndrome is characterized by: Central toxic effects: + Anxiety + Delirium + Disorientation + Hallucinations + Hyperactivity + Seizures Peripheral toxicity: + Tachycardia + Hyperpyrexia + Mydriasis + Vasodilation + Urinary retention + Diminution of gastrointestinal motility + Decrease of secretion in salivary and sweat glands + Loss of secretions in the pharynx, bronchi, and nasal passages Reversal of anticholinergic symptoms can be expected in minutes after administration of Physostigmine Salicylate Injection2. From the Research Anticholinergic Toxidrome Characteristics The anticholinergic toxidrome is characterized by both central and peripheral physical findings, including central anticholinergic syndrome and peripheral anticholinergia 3 Central anticholinergic syndrome is characterized primarily by signs and symptoms consistent with hyperactive delirium 3 Peripheral anticholinergia includes mydriasis and blurred vision, tremors, ataxia, fever/hyperthermia, flushed and dry skin, dry oral mucosa, decreased bowel sounds, constipation, and urinary retention, among other symptoms 3 Treatment of Anticholinergic Toxidrome Physostigmine is a tertiary acetylcholinesterase inhibitor that can be used to assist in the diagnosis and management of severe anticholinergic toxicity 3 Physostigmine has been shown to be relatively safe and effective in reversing anticholinergic toxidrome 4 Rivastigmine, a long-acting acetylcholinesterase inhibitor, is a potential alternative to physostigmine for the treatment of anticholinergic toxicity 5 Benzodiazepines are also commonly used in the treatment of anticholinergic toxicity, particularly for patients with agitation or delirium 4, 6 Causes of Anticholinergic Toxidrome Anticholinergic toxidrome can be caused by a variety of substances, including medications, herbs, and plants 3, 6 Thorn apple seeds have been identified as a cause of anticholinergic toxidrome in at least one case report 6 Olanzapine overdose can also lead to the development of anticholinergic toxicity 3 Diagnosis and Management Prompt recognition of anticholinergic toxidrome is crucial for effective management and improved patient outcomes 6 Physostigmine can be used to diagnose and treat anticholinergic toxicity, particularly in patients with severe symptoms or those who do not respond to benzodiazepines 4, 6 It is essential to consider alternative causes of symptoms that may mimic anticholinergic toxidrome, such as viral encephalitis 7 References 1 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 2 Drug Official FDA Drug Label For physostigmine (IV) FDA, 2025 3 Research The Use of Physostigmine in the Diagnosis and Treatment of Anticholinergic Toxicity After Olanzapine Overdose: Literature Review and Case Report. Journal of the Academy of Consultation-Liaison Psychiatry, 2021 4 Research The Use of Physostigmine by Toxicologists in Anticholinergic Toxicity. Journal of medical toxicology : official journal of the American College of Medical Toxicology, 2015 5 Research Rivastigmine as an alternative treatment for anticholinergic toxidrome in light of the physostigmine shortage: A case series. The American journal of emergency medicine, 2025 6 Research Anticholinergic Toxidrome due to Thorn Apple Seed Ingestion in an Elderly Couple. European journal of case reports in internal medicine, 2024 7 Research Viral encephalitis masquerading as a fulminant anticholinergic toxidrome. Journal of toxicology. Clinical toxicology, 1997 Related Questions What is the overview of anticholinergic (Anticholinergic) toxicity?What are the consequences of a mistaken intravenous (IV) administration of Bentyl (dicyclomine)?How to manage anticholinergic (anticholinergic effects) toxicity?What is the treatment for anticholinergic (Atropine) psychosis?What are the symptoms of anticholinergic (Anticholinergic, related to the inhibition of the action of the neurotransmitter acetylcholine) toxicity?What is the treatment for a patient with hypercholesterolemia, severely elevated triglycerides (hypertriglyceridemia), low High-Density Lipoprotein (HDL) cholesterol, and a family history of dyslipidemia?What is the optimal care plan for a patient 6 months into recovery from cardiac arrest due to hypoxic (lack of oxygen) brain injury?What is the cause of a swollen scrotum and diarrhea in a 2-year-old child?What are the recommendations for preventing candidiasis (yeast infection)?What is the significance of focal fat deposition along a calcified ligament as an incidental finding on computed tomography (CT)?What are the treatment options for subacromial (subdeltoid) bursitis with shoulder impingement syndrome? 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2380	https://www.tiger-algebra.com/en/solution/combination-without-repetition/c%289%2C5%29/	Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Combinations without repetition Other Ways to Solve Step-by-step explanation 1. Find the number of terms in the set n represents the total number of items in the set: c(n,k) c(9,5) n=9 2. Find the number of items selected from the set k represents the number of items selected from the set: c(n,k) c(9,5) k=5 3. Calculate the combinations using the formula Plug n (n=9) and k (k=5) into the combination formula: C(n,k)=n!k!(n-k)! C(9,5)=9!5!(9-5)! C(9,5)=9!5!·4! C(9,5)=9·8·7·6·5!5!·4! C(9,5)=9·8·7·64! C(9,5)=9·8·7·64·3·2·1 C(9,5)=126 There are 126 ways that 5 items chosen from a set of 9 can be combined. How did we do? Why learn this Combinations and permutations If you have 2 types of crust, 4 types of toppings, and 3 types of cheese, how many different pizza combinations can you make? If there are 8 swimmers in a race, how many different sets of 1st, 2nd, and 3rd place winners could there be? What are your chances of winning the lottery? All of these questions can be answered using two of the most fundamental concepts in probability: combinations and permutations. Though these concepts are very similar, probability theory holds that they have some important differences. Both combinations and permutations are used to calculate the number of possible combinations of things. The most important difference between the two, however, is that combinations deal with arrangements in which the order of the items being arranged does not matter—such as combinations of pizza toppings—while permutations deal with arrangements in which the order the items being arranged does matter—such as setting the combination to a combination lock, which should really be called a permutation lock because the order of the input matters. What these two concepts have in common, is that they both help us understand the relationships between sets and the items or subsets that make up those sets. As the examples above illustrate, this can be used to better understand many different types of situations. Combinations and permutations Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
2381	https://www.developingexperts.com/glossary/nostril	nostril \| Glossary \| Developing Experts Product Education ApproachSchools & TeachersLessonsSEND Adaptation ToolsLibraryGlossaryCareersReportingResearchVideos AI Lesson GeneratorCareer Builder Subjects Science - EYS to KS4Geography - KS1 & KS2 New Sponsored units from Rolls-Royce Submarines View here Pricing Pricing Options ##### For Schools Access all our resources for your teachers ##### For Individuals Choose different plans to get the resources you need Sponsors Partners SponsorsCareersAffiliates Support Account Sign inPlans & PricingRequest SubscriptionContact us Company AboutJob VacanciesBlogEvents Sign inSign up Product Education ApproachSchools & TeachersLessonsSEND Adaptation ToolsLibraryGlossaryCareersReportingResearchVideos AI Lesson GeneratorCareer Builder Subjects Science - EYS to KS4Geography - KS1 & KS2 New Sponsored units from Rolls-Royce Submarines View here Pricing Pricing Options ##### For Schools Access all our resources for your teachers ##### For Individuals Choose different plans to get the resources you need Schools Partners SponsorsCareersAffiliates Support Account Sign inPlans & PricingRequest SubscriptionContact us Company AboutJob VacanciesBlogEvents Home ›Glossary ›N ›nostril ABCDEFGHIJKLMNOPQRSTUVWXYZ nostril Your browser does not support the audio element. Primary InformationSecondary Information Definition Your browser does not support the audio element. A nostril is one of the two openings in the nose that allow air to enter. The nostrils are located on the front of the nose and they are separated by the septum. The nostrils are lined with a moist membrane that helps to filter the air that enters the nose. The membrane also contains tiny hairs that help to trap dust and dirt. When we breathe in, the air passes through the nostrils and into the nasal cavity. The nasal cavity is where the air is warmed and humidified before it travels to the lungs. The nostrils are important for breathing and for smell. They help to filter the air that we breathe in and they also help to trap odor molecules. How can the word be used? Your browser does not support the audio element. The word "nostril" is related to the words "nose" and "thirl.". Different forms of the word Your browser does not support the audio element. The word "nostril" has no different forms. It is a noun that refers to one of the two openings in the nose that allow air to enter and exit. Etymology Your browser does not support the audio element. The word "nostril" comes from the Old English word "nostþyrl," which is made up of the words "nosu" (nose) and "þyrl" (hole). The Old English word "nostþyrl" is also the source of the Dutch word "neusgat" and the German word "Nasenloch.". The word "nostril" was first used in English in the 10th century. It was used to refer to one of the two openings in the nose that allow air to enter and exit. Question Your browser does not support the audio element. What do your nostrils do? Flag Media Please select a reason for flagging this content. Doesn't fit the subject matter Inappropriate, violent or graphic content Copyright infringement Cancel Stay connected Join our newsletter to stay up to date on features and releases Subscribe Address Developing Experts Limited Exchange Street Buildings 35-37 Exchange Street Norwich NR2 1DP UK Phone 01603 273515 Email hello@developingexperts.com Developing Experts LessonsAboutResearchReportingPlans & Pricing Legal GDPRPrivacy PolicyTerms and ConditionsCookie Policy Contact Us Get in touchRequest SubscriptionChildren's CodeAbout UsCareers Copyright 2025 Developing Experts, All rights reserved. × Start Your Free Trial Unlock expert-designed lessons, resources, and assessments tailored for educators. No credit card required. Claim Your Free Trial →
2382	https://web.eecs.umich.edu/~imarkov/pubs/book/pdhandbook_part.pdf	DRAFT Partitioning-based Methods for VLSI Placement Jarrod A. Roy and Igor L. Markov University of Michigan, EECS Department, Ann Arbor, MI 48109-2121 To appear in Physical Design Handbook, CRC Press 2007 1 Introduction The technique of using balanced min-cut partitioning in placement was presented by Breuer in 1977 . Such min-cut placers use scalable and extensible divide-and-conquer algorithmic frame-work and tend to produce routable placements . Recent work offers extensions to block place-ment and large-scale mixed-size placement [15,18,31], and robust incremental placement . Over the years partitioning-based placement has seen many revisions and enhancements, but the underlying framework (illustrated in Figure 1) remains much the same. Top-down partitioning-based placement algorithms seek to decompose a given placement instance into smaller instances by sub-dividing the placement region, assigning modules to subregions and cutting the netlist hypergraph [7,19]. The top-down placement process can be viewed as a sequence of passes where each pass examines all bins and divides some of them into smaller bins. Most commonly the division step is accomplished with balanced min-cut partitioning that minimizes the number of signal nets connecting modules in multiple regions . These techniques leverage well-understood and scalable algorithms for hypergraph partitioning and typically lead to routable placements . This chapter is organized as follows: Section 2 introduces the basic framework for min-cut partitioning-based placement. Next, Section 3 presents recent enhancements to min-cut placement. Section 4 describes adapting partitioning based methods to mixed-size placement and Section 5 outlines recent techniques that give min-cut placement an edge over other placement algorithms. Lastly, Section 6 discusses the workings of state-of-the-art min-cut placers such as Dragon [38,39, 43], FengShui [4,27], NTUPlace2 and Capo [30–35] and illustrates how they differ from the basic min-cut framework and each other. 1 DRAFT Variables: queue of placement bins Initialize queue with top-level placement bin 1 While (queue not empty) 2 Dequeue a bin 3 If (bin small enough) 4 Process bin with end-case placer 5 Else 6 Choose a cut-line for the bin (including direction) 7 Build partitioning hypergraph from netlist and cells contained in the bin 8 Partition the bin into smaller bins (generally via min-cut bi- or quadri-section) 9 Enqueue each child bin Figure 1: Top-down partitioning-based placement. 2 Top-down Partitioning-based Placement Framework Using min-cut partitioning in placement was presented by Breuer in 1977 . The underlying framework remains mostly the same since then and is illustrated in Figure 1. The placement region is represented by a series of of placement bins which represent (i) a placement region with allowed module locations (sites), (ii) a collection of circuit modules to be placed in this region, (iii) all signal nets incident to the modules in the region, and (iv) ﬁxed cells and pins outside the region that are adjacent to modules in the region (terminals). Min-cut partitioning-based placers generally proceed by dividing the netlist and placement area into successively smaller pieces until the pieces are small enough to be handled efﬁciently by optimal end-case placers . State-of-the-art placers generally use a wide range of hyper-graph partitioning techniques to best ﬁt partitioning problem size — optimal (branch-and-bound ), middle-range (Fiduccia-Mattheyses ) and large-scale (multi-level Fiduccia-Mattheyses [10, 26]). Min-cut placement is highly scalable (due in large part to algorithmic advances in min-cut partitioning [10,20,26]) and typically produces routable placements. In this section, we introduce topics relevant to top-down partitioning-based placement that must be addressed by all modern min-cut placers. Speciﬁcally we discuss terminal propagation, biparti-tioning vs. multi-way partitioning, cut-line selection and whitespace (or free space) allocation. 2.1 Terminal Propagation and Inessential Nets Proper handling of terminals is essential to the success of top-down placement approaches [11,19, 21, 37]. When a particular placement bin is split into multiple subregions, some of the cells in-side may be tightly connected to cells outside of the bin. Ignoring such connections can adversely 2 DRAFT Figure 2: Terminal propagation. The net shown has ﬁve ﬁxed terminals: four above and one below the cut-line. It also has movable cells which are represented by the cell with a dashed outline. The four ﬁxed terminals above the cut-line are propagated to the black circle at the top of the bin while the one ﬁxed terminal below the cut-line is propagated to the black circle below the cut-line. The movable cells remain unpropagated. Note that the net is inessential since terminals are propagated to both sides of the cut-line . affect the quality of a placement since these connections can account for signiﬁcant amounts of wirelength. On the other hand, these terminals are irrelevant to the classic partitioning formulation as they cannot be freely assigned to partitions. A compromise is possible by using an extended formulation of “partitioning with ﬁxed terminals”, where the terminals are considered to be ﬁxed in (“propagated to”) one or more partitions, and assigned zero areas (original areas are ignored). Nets which are propagated to both partitions in bi-partitioning are considered “inessential” since they will always be cut and can be safely removed from the partitioning instance to improve run-time . Terminal propagation is typically driven by geometric proximity of terminals to subre-gions/partitions. Figure 2 depicts terminal propagation for a net with several ﬁxed terminals. This particular net is inessential as it has terminals propagated to both sides of the cut-line. 2.2 Bipartitioning vs. Multi-way Partitioning In his seminal work on min-cut placement, Breuer introduced two forms of recursive min-cut placement: slice/bisection and quadrature . The style of min-cut placement most commonly used today has grown from the quadrature technique which advocated the use of horizontal and vertical cuts; the slice/bisection technique used only horizontal cuts and exhibited worse perfor-mance than quadrature . Since that time, horizontal and vertical cut-lines have been standard in all placement tech-niques, but there has been debate as to whether there should be an ordering to the cuts (i.e. hor-izontally bisect a bin then vertically bisect its children as in quadrature ) or both cuts should be done simultaneously as in quadrisection . Quadrisection has been shown to allow for the 3 DRAFT optimization of techniques other than min-cut (such as minimal spanning tree length ), but ter-minal propagation is more complex when splitting a bin into four child bins instead of two. Also, bisection can simulate quadrisection with added ﬂexibility in cut-line selection and shifting (see Section 2.3) . There are currently no known methods that use greater than 4-way partitioning and the vast majority of partitioning-based placement techniques involve min-cut bipartitioning. 2.3 Cut-line Selection and Shifting Breuer studied two types of cut-line direction selection techniques and found that alternating cut-line directions from layer to layer produced better half-perimeter wirelength (HPWL) than using only horizontal cuts . The authors of studied this phenomenon further by testing 64 cut-line direction sequences. Their experiments did not ﬁnd that the two cut-sequences that alternate at each layer were the best, but did ﬁnd that long sequences of cuts in the same direction during placement was detrimental to performance . The authors of developed a dynamic programming technique to choose optimal cut sequences for partitioning based placement, but also found that nearly optimal cut sequences could be determined from the aspect ratio of the bin to be split. After cut-line direction is chosen, partitioning-based placers generally choose the cut-line that best splits a placement bin in half in the desired direction. Cut-lines are generally aligned to placement row and site boundaries to ease the assignment of standard-cells to rows near the end of global placement . After a bin is partitioned, the initial cut-line may be moved, or shifted, in order to satisfy other objectives such as whitespace allocation or congestion reduction. 2.4 Whitespace Allocation Management of whitespace (also known as free space) is a key issue in physical design as it has a profound effect on the quality of a placement. The amount of whitespace in a design is the difference between the total placeable area in a design and the total movable cell area in the design. A natural scheme for managing whitespace in top-down placement, uniform whitespace allocation, was introduced and analyzed in . Let a placement bin which is going to be partitioned have site area S, cell area C, absolute whitespace W = max{S−C,0} and relative whitespace w = W/S. A bi-partitioning divides the bin into two child bins with site areas S0 and S1 such that S0 +S1 = S and cell areas C0 and C1 such that C0 +C1 = C. A partitioner is given cell area targets T0 and T1 as well as a tolerance τ for a particular bi-partitioning instance. In many cases of bi-partitioning, 4 DRAFT T0 = T1 = C 2 , but this is not always true . τ deﬁnes the maximum percentage by which C0 and C1 are allowed to differ from T0 and T1, respectively. The work in bases its whitespace allocation techniques on whitespace deterioration: the phenomenon that discreteness in partitioning and placement does not allow for exact uniform whitespace distribution. The whitespace deterioration for a bi-partitioning is the largest α, such that each child bin has at least αw relative whitespace. Assuming non-zero relative whitespace in the placement bin, α should be restricted such that 0 ≤α ≤1 . The authors note that α = 1 may be overly restrictive in practice because it induces zero tolerance on the partitioning instance but α = 0 may not be restrictive enough as it allows for child bins with zero whitespace, which can improve wirelength but impair routability . For a given block, feasible ranges for partition capacities are uniquely determined by α. The partitioning tolerance τ for splitting a block with relative whitespace w is (1−α)w 1−w . The chal-lenge is to determine a proper value for α. First assume that a bin is to be partitioned horizontally n times more during the placement process. n can be calculated as ⌈log2 R⌉where R is the number of rows in the placement bin . Assuming end-case bins have α = 0 since they are not further partitioned, w, the relative whitespace of an end-case bin, is determined to be τ τ+1 where τ is the tolerance of partitioning in the end-case bin . Assuming that α remains the same during all partitioning of the given bin gives a simple deriva-tion of α = n q w w . A more practical calculation assumes instead that τ remains the same over all partitionings. This leads to τ = n q 1−w 1−w −1 . w can be eliminated from the equation for τ and a closed form for α based only w and n is derived to be α = n+1 √ 1−w−(1−w) w( n+1 √1−w) . Free Cell Addition. One method of non-uniform whitespace allocation in placement, was presented in . To achieve a non-uniform allocation of whitespace, free cells (standard cells that have no connections in the netlist) are added to the design which is then placed using uniform whitespace allocation. Care must be taken not to add too many cells to the design which can complicate the work of many placement algorithms, increasing interconnect length or leading to overlapping circuit modules . Several other whitespace allocation techniques have been published in the literature, many of which have the objective of congestion reduction [28,32,38,39,43]. These techniques which deal speciﬁcally with congestion reduction are covered in a later chapter in this book ??. 5 DRAFT 3 Enhancements to the Min-cut Framework This section describes several techniques which are recent improvements to the to the min-cut partitioning-based framework presented in Section 2. These techniques range from fairly simple yet effective techniques such as repartitioning and placement feedback to changes in the optimiza-tion goals of min-cut placement as in weighted net-cut. 3.1 Better Results Through Additional Partitioning Huang and Kahng introduced two techniques for improving the results of quadrisection based placement known as cycling and overlapping . Cycling is a technique whereby results are improved by partitioning every placement bin multiple times each layer . After all bins are split for the ﬁrst time in a layer of placement, a new round of partitioning on the same bins is done using the results of the previous round for terminal propagation. These additional rounds of partitioning are repeated until there is no further improvement of a cost function . A similar type of technique was presented for min-cut bisection called placement feedback. In placement feedback, bins are partitioned multiple times, without requiring steady improvement in wirelength, to achieve more consistent terminal propagation . Placement feedback serves to reduce the number of ambiguously propagated terminals. Am-biguity in terminal propagation arises when a terminal is nearly equidistant to the centers of the child bins of the bin being partitioned. In such cases it is unclear as to what side of the cut-line the terminal should be propagated. Traditional choices for such terminals is to propagate them to both sides or neither side of the cut-line in fear of making a poor decision . Ambiguously propagated terminals introduce indeterminism into min-cut placement as they may be propagated differently based on the order in which placement bins are processed . To reduce the number of ambiguously propagated terminals, placement feedback repeats each layer of partitioning n times. Each successive round of partitioning uses the resulting locations from the previous partitioning for terminal propagation. The ﬁrst round of partitioning for a par-ticular layer may have ambiguous terminals, but the second and later rounds will have reduced numbers of ambiguous terminals making terminal propagation more robust . Empirical results show that placement feedback is effective in reducing HPWL, routed wirelength and via count . The technique of overlapping also involves additional partitioning calls during placement . 6 DRAFT While doing cycling in quadrisection, pieces of neighboring bins can be coalesced into a new bin and split to improve solution quality . Brenner and Rohe introduced a similar technique which they called repartitioning which was designed to reduce congestion . After partitioning, conges-tion was estimated in the placement bins of the design. Using this congestion data, new partitioning problems were formulated with all neighbors of a congested area. Solving these new partitioning problems would spread congestion to neighboring areas of the placement while possibly incurring an increase in net length . Capo [30–35] repartitions bins similarly, but for the improvement of HPWL. After the initial solution of a partitioning problem is returned from a min-cut partitioner, Capo has the option of shifting the cut-line to fulﬁll whatever whitespace whitespace requirements may be asked of it. A shift of the cut-line, though, represents a change in the partitioning problem formulation (as the initial partitioning problem was built assuming a different cut-line which can have a signiﬁcant effect on terminal propagation). Thus, the partitioning problem is rebuilt with the new cut-line lo-cation and solved again to improve wirelength. The repartitioning does not come with a signiﬁcant runtime penalty because the initial partitioning solution is reused and modiﬁed by ﬂat passes of a Fiduccia-Mattheyses partitioner. 3.2 Fractional Cut When a placement bin is split with a vertical cut-line, there are generally many possible cut-lines that can split the bin roughly equally since the size of sites in row-based placement is generally small. On the other hand, row heights are generally non-trivial as compared to the height of the core placement area. Since standard cells are ultimately placed in rows, most min-cut placers choose to align cut-lines to row boundaries . The authors of that this causes the “narrow region” problem which leads to instability in min-cut placement. The “narrow region” problem becomes an issue when bins become tall and narrow. In such cases, total cell area may be able to ﬁt into a given narrow bin, but it may not be possible to assign cells into these rows legally due to row area constraints or the number of legal solutions is so small that net-cut is artiﬁcially increased as a result . Take for example a placement bin that encompasses two adjacent rows, each four units in length. If we have one cell with length ﬁve units and another with length two units, there is no way to legally place them in the rows, yet total 7 DRAFT area constraints are satisﬁed. To remedy this situation, the authors of propose using a “fractional cut”: a horizontal cut-line that is allowed to pass through a fraction of a row. As horizontal cut-lines do not necessarily align with rows, cells must be assigned to rows before optimal end-case (typically single-row) placers can be used . To legalize the placement, one proceeds on a row-by-row basis. Each cell is tentatively assigned to a preferred height in the placement: the center of its placement bin. Starting with the top-most row, cells are assigned to rows so as to minimize the cost of assigning cells. If a cell is assigned to the current row, its cost is the squared distance from its preferred position to the current row. If a cell is not assigned to the current row, its cost is the squared distance from its preferred position to the next lower row . After all cells are assigned to rows, they are sorted by their x coordinates and packed in rows to remove any overlaps. The assignment of cells to rows is achieved efﬁciently by a dynamic programming formulation . Experimental results show considerable improvements in terms of HPWL reduction in place-ment, but packing of cells in rows does not generally produce routable placements . 3.3 Analytical Constraint Generation The authors of note that min-cut placement techniques are effective at reducing HPWL of designs that are heavily constrained in terms of whitespace, but do not perform nearly as well as analytical techniques when there are large amounts of whitespace. The authors suggest that one reason for the discrepancy is that min-cut placers generally try to divide placement bins exactly in half with a relatively small tolerance. This tends to spread cell area roughly uniformly across the core area. Increasing the tolerance for partitioning a bin can allow for less uniformity in placement and lower HPWL due to tighter packing, but still does not reproduce the performance of analytical techniques . To improve the HPWL performance of min-cut placement techniques on designs with large amounts of whitespace (which are becoming increasingly popular in real-world designs), while still retaining the good performance of min-cut techniques when there is limited whitespace, the authors of suggest integrating analytical techniques and min-cut techniques. Before constructing a partitioning instance for a given placement bin, an analytical technique is run on the objects in the bin to minimize their quadratic wirelength . Next, the center of mass of the placement of the 8 DRAFT objects of the bin is calculated. This points to roughly where the objects should go to reduce their wirelength. Then one constructs a rectangle having the same aspect ratio as the placement bin and having the same area as the total area of movable objects in the bin. Let A be the total area of cells in the bin, H be the height of the bin and W the width of the bin. The height and width of such a rectangle can be calculated as follows: rectangle height RH = q A∗H W and rectangle width RW = q A∗W H . One centers this rectangle at the center of mass of the analytical placement and intersects the rectangle with the proposed cut-line of the bin. The amount of area of the rectangle that falls on either side of the cut-line is used as a target for min-cut partitioning . As most min-cut partitioners chose to split cell area equally, this is a signiﬁcant departure from traditional min-cut placement. Empirical results suggest that analytical constraint generation (ACG) is effective at improving the performance of min-cut placement on designs with large amounts of whitespace while retaining the good performance of min-cut placers on constrained designs while not impairing the routability of designs . This performance comes at the cost of approximately 28% more runtime . 3.4 Better Modelling of HPWL by Partitioning It is well-known that the min-cut objective in partitioning does not accurately represent the wire-length objective of placement [21, 36]. Optimizing HPWL and other objectives directly through partitioning can provide improvements over min-cut. Huang and Kahng showed that net weighting and quadrisection can be used to minimize a wide range of objectives such as minimal spanning tree cost . Their technique consists of computing vectors of weights for each net (called net-vectors) and using these weights in quadrisection . Although this technique can represent a wide range of cost functions to minimize, it requires the discretization of pin locations into the centers of bins and requires that sixteen weights must be calculated per net for partitioning . The authors of introduce a new terminal propagation technique in their placer THETO that allows the partitioner to better map net-cut to HPWL. The terminal propagation in THETO differs from traditional terminal propagation in that each original net may be represented by one or two nets in the partitioned netlist, depending on the conﬁguration of the net’s terminals. Two special cases — nets with no terminals and inessential nets — are treated the same as in traditional terminal propagation. Five other cases are analyzed in , based on the conﬁguration of terminals relative 9 DRAFT to the centers of the child bins, and proper weight computation is described (one case requires two nets). This way weighted net-cut better represents the “HPWL degradation” seen after partitioning. Empirically, this terminal propagation and net weighting are shown to reduce HPWL in min-cut placement. This technique is simpliﬁed in and reduced to the calculation of three wirelengths per net per partitioning instance (w1, w2 and w12) which completely determine the connectivity and costs of all nets in the derived partitioning hypergraph. While this formulation is more compact than that in , it is also more general. For each net in each partitioning instance, one must calculate the cost of all nodes on the net being placed in partition 1 (w1), the cost of all nodes on the net being placed in partition 2 (w2) and the cost of all nodes on the net being split between partitions 1 and 2 (w12). Up to two nets can be created in the partitioning instance, one with weight \|w1 −w2\| and the other with weight w12 −max(w1,w2). The only assumption made in is that w12 ≥max(w1,w2). With these costs and the corresponding connectivity of the derived hypergraph, minimizing weighted net-cut directly corresponds to minimizing HPWL. 4 Mixed-size Placement Mixed-size placement, the placement of large macros in addition to standard cells, has become a relevant challenge in physical design and is poised to dominate physical design in the near future as we move from traditional “sea of cells” ICs to “sea of hard macros” SoCs . To keep up with this shift in physical design, several techniques for partitioning based mixed-size placement have been proposed and are described in this section. These techniques include ﬂoorplacement, PATOMA, and mixed-size placement with fractional cut. 4.1 Floorplacement From an optimization point of view, ﬂoorplanning and placement are very similar problems – both seek non-overlapping placements to minimize wirelength. They are mostly distinguished by scale and the need to account for shapes in ﬂoorplanning, which calls for different optimization techniques. Netlist partitioning is often used in placement algorithms, where geometric shapes of partitions can be adjusted. This considerably blurs the separation between partitioning, placement and ﬂoorplanning, raising the possibility that these three steps can be performed by one CAD tool. 10 DRAFT Variables: queue of placement bins Initialize queue with top-level placement bin 1 While (queue not empty) 2 Dequeue a bin 3 If (bin has large/many macros or is marked as merged) 4 Cluster std-cells into soft macros 5 Use ﬁxed-outline ﬂoorplanner to pack all macros (soft+hard) 6 If ﬁxed-outline ﬂoorplanning succeeds 7 Fix macros and remove sites underneath the macros 8 Else 9 Undo one partition decision. Merge bin with sibling 10 Mark new bin as merged and enqueue 11 Else if (bin small enough) 12 Process end case 13 Else 14 Bi-partition the bin into smaller bins 15 Enqueue each child bin Figure 3: Min-cut ﬂoorplacement. Bold-faced lines 3-10 are different from tradi-tional min-cut placement . The authors of develop such a tool and term the uniﬁed layout optimization ﬂoorplacement following Steve Teig’s keynote speech at ISPD 2002. Min-cut placers scale well in terms of runtime and wirelength minimization, but cannot produce non-overlapping placements of modules with a wide variety of sizes. On the other hand, annealing-based ﬂoorplanners can handle vastly different module shapes and sizes, but only for relatively few (100-200) modules at a time. Otherwise, either solutions will be poor or optimization will take too long to be practical. The loose integration of ﬁxed-outline ﬂoorplanning and standard-cell place-ment proposed in suffers from a similar drawback because its single top-level ﬂoorplanning step may have to operate on numerous modules. Bottom-up clustering can improve the scalability of annealing, but not sufﬁciently to make it competitive with other approaches. The work in applies min-cut placement as much as possible and delays explicit ﬂoorplanning until it becomes necessary. In particular, since min-cut placement generates a slicing ﬂoorplan, it is viewed as an implicit ﬂoorplanning step, reserving explicit ﬂoorplanning for “local” non-slicing block packing. Placement starts with a single placement bin representing the entire layout region with all the placeable objects initialized at the center of the bin. Using min-cut partitioning, the bin is split into two bins of similar sizes, and during this process the cut-line is adjusted according to actual partition sizes. Applying this technique recursively to bins (with terminal propagation) produces a series of gradually reﬁned slicing ﬂoorplans of the entire layout region. In very small bins, all cells can be placed by a branch-and-bound end-case placer . However, this scheme breaks down 11 DRAFT 0 500 1000 1500 2000 0 500 1000 1500 2000 IBM01 HPWL=2.574e+06, #Cells=12752, #Nets=14111 0 500 1000 1500 2000 0 500 1000 1500 2000 IBM01 HPWL=2.574e+06, #Cells=12752, #Nets=14111 Figure 4: Progress of mixed-size ﬂoorplacement on the IBM01 benchmark from IBM-MSwPins. The picture on the left shows how the cut lines are chosen during the ﬁrst six layers of min-cut bisection. On the right is the same placement but with the ﬂoor-planning instances highlighted by “rounded” rectangles. Floorplanning failures can be detected by observing nested rectangles . on modules that are larger than their bins. When such a module appears in a bin, recursive bisec-tion cannot continue, or else will likely produce a placement with overlapping modules. Indeed, the work in continues bisection and resolves resulting overlaps later. In this technique, one switches from recursive bisection to “local” ﬂoorplanning where the ﬁxed outline is determined by the bin. This is done for two main reasons: (i) to preserve wirelength , congestion and de-lay estimates that may have been performed early during top-down placement, and (ii) avoid legalizing a placement with overlapping macros. While deferring to ﬁxed-outline ﬂoorplanning is a natural step, successful ﬁxed-outline ﬂoor-planners have appeared only recently . Additionally, the ﬂoorplanner may fail to pack all mod-ules within the bin without overlaps. As with any constraint-satisfaction problem, this can be for two reasons: either (i) the instance is unsatisﬁable, or (ii) the solver is unable to ﬁnd any of existing solutions. In this case, the technique undoes the previous partitioning step and merges the failed bin with its sibling bin, whether the sibling has been processed or not, then discards the two bins. The merged bin includes all modules contained in the two smaller bins, and its rectangular outline is the union of the two rectangular outlines. This bin is ﬂoorplanned, and in the case of failure can be merged with its sibling again. The overall process is summarized in Figure 3 and an example is depicted in Figure 4. It is typically easier to satisfy the outline of a merged bin because circuit modules become 12 DRAFT relatively smaller. However, Simulated Annealing takes longer on larger bins and is less successful in minimizing wirelength. Therefore, it is important to ﬂoorplan at just the right time, and the algorithm determines this point by backtracking. Backtracking does incur some overhead in failed ﬂoorplan runs, but this overhead is tolerable because merged bins take considerably longer to ﬂoorplan. Furthermore, this overhead can be moderated somewhat by careful prediction. For a given bin, a ﬂoorplanning instance is constructed as follows. All connections between modules in the bin and other modules are propagated to ﬁxed terminals at the periphery of the bin. As the bin may contain numerous standard cells, the number of movable objects is reduced by conglomerating standard cells into soft placeable blocks. This is accomplished by a simple bottom-up connectivity-based clustering . The existing large modules in the bin are usually kept out of this clustering. To further simplify ﬂoorplanning, soft blocks consisting of standard cells are artiﬁcially downsized, as in . The clustered netlist is then passed to the ﬁxed-outline ﬂoorplanner Parquet , which sizes soft blocks and optimizes block orientations. After suitable locations are found, the locations of all large modules are returned to the top-down placer and are considered ﬁxed. The rows below those modules are fractured and their sites are removed, i.e., the modules are treated as ﬁxed obstacles. At this point, min-cut placement resumes with a bin that has no large modules in it, but has somewhat non-uniform row structure. When min-cut placement is ﬁnished, large modules do not overlap by construction, but small cells sometimes overlap (typically below 0.01% by area). Those overlaps are quickly detected and removed with local changes. Since the ﬂoorplacer includes a state-of-the-art ﬂoorplanner, it can natively handle pure block-based designs. Unlike most algorithms designed for mixed-size placement, it can pack blocks into a tight outline, optimize block orientations and tune aspect ratios of soft blocks. When the number of blocks is very small, the algorithm applies ﬂoorplanning quickly. However, when given a larger design, it may start with partitioning and then call ﬁxed-outline ﬂoorplanning for separate bins. As recursive bisection scales well and is more successful at minimizing wirelength than annealing-based ﬂoorplanning, the proposed approach is scalable and effective at minimizing wirelength. Empirical boundary between placement and ﬂ oorplanning. By identifying the characteris-tics of placement bins for which the algorithm calls ﬂoorplanning, one can tabulate the empirical boundary between placement and ﬂoorplanning. Formulating such ad hoc thresholds in terms of 13 DRAFT Floorplanning conditions for ﬂoorplacement N,n: The numbers of large modules and movable objects in a given bin. A(m): The area of the m largest modules in a given bin, m ≤n. C: The capacity of a given bin. Test 1. At least one large module does not ﬁt into a potential child bin. Test 2. N ≤30 and A(N) < 0.80∗A(n) and A(n) > 0.6∗C. Test 3. N ≤15 and A(N) < 0.95∗A(n) and A(n) > 0.6∗C. Test 4. A(50) < 0.85∗C. Test 5. A(10) < 0.60∗C. Test 6. A(1) < 0.30∗C and N = 1. Test 7. N = n = 1. Table 1: Floorplanning conditions used in ﬂoorplacement . Test 1 is the most funda-mental, for if a bin meeting test 1 were not ﬂoorplanned, a failure would be guaranteed at the next level. Tests 2-6 detect bins dominated by large macros. Test 7 is a base case where only one module exists, but it is large. dimensions of the largest module in the bin, etc., allows one to avoid unnecessary backtracking and decrease the overhead of ﬂoorplanning calls that fail to satisfy the ﬁxed outline constraint because they are issued too late. In practice, issuing ﬂoorplanning calls too early (i.e., on larger bins) in-creases ﬁnal wirelength and sometimes runtime. To improve wirelength, the ad hoc tests for large modules in bins (that trigger ﬂoorplanning) are deliberately conservative. These conditions were derived by closely monitoring the legality of ﬂoorplanning and min-cut placement solutions. When a partitioned bin yields an illegal placement solution it is clear that the bin should have been ﬂoorplanned and a condition should be derived. When a call to ﬂoorplanning fails to satisfy the ﬁxed outline constraint the placer has to backtrack. To avoid paying this penalty, a condition should be derived to allow for ﬂoorplanning the parent bin and prevent the failure. These conditions are reﬁned to prevent ﬂoorplanning failure by visual inspection of a plot of the resulting parent bin and formulating a condition describing its composition. An example of such a plot is shown in Figure 4. Floorplanned bins are outlined with rounded rectangles. Nested rectangles indicate a failed ﬂoorplan run, followed by backtracking and ﬂoorplanning of the larger parent bin. In our experience, these tests are strong enough to ensure that at most one level of backtracking is required to prevent overlaps between large modules. 14 DRAFT 4.2 PATOMA and PolarBear PATOMA 1.0 pioneered a top-down ﬂoorplanning framework that utilizes fast block-packing algorithms (ROB or ZDS ) and hypergraph partitioning with hMETIS . This approach is fast and scalable, and provides good solutions for many input conﬁgurations. Fast block-packing is used in PATOMA to guarantee that a legal packing solution exists, at which point the burden of wirelength minimization is shifted to the hypergraph partitioner. This idea is applied recursively to each of the newly-created partitions. In end-cases, when a partitioning step leads to unsatisﬁable block-packing, the quality of the result is determined by the quality of its fast block-packing algo-rithms. In end-cases, when partitioning cannot be used because it creates unsatisﬁable instances of block-packing, block locations are determined by fast block-packing heuristics. The placer Polar-Bear integrates algorithms from PATOMA to increase the robustness of a top-down min-cut placement ﬂow. Similar to PATOMA, the ﬂoorplanner IMF utilizes top-down partitioning, but allows overlaps in the initial top-down partitioning phase. A bottom-up merging and reﬁnement phase ﬁxes overlaps and further optimizes the solution quality. 4.3 Fractional Cut for Mixed-size Placement The work in advocates a two-stage approach to mixed-size placement. First, the min-cut placer FengShui generates an initial placement for the mixed-size netlist without trying to prevent all overlaps between modules. The placer only tracks the global distribution of area during partition-ing and uses the fractional cut technique (see Section 3.2), which further relaxes book-keeping by not requiring placement bins to align to cell rows. While giving min-cut partitioners more freedom, these relaxations prevent cells from being placed in rows easily and require additional repair during detail placement. This may particularly complicate the optimization of module orientations, not considered in (relevant benchmarks use only square blocks with all pins placed in the centers). The second stage consists of removing overlaps by a fast legalizer designed to handle large modules along with standard cells. The legalizer is essentially greedy and attempts to shift all modules towards the left edge of the chip (or to the right edge, if that produces better results). In our experience, the implementation reported in leads to horizontal stacking of modules and sometimes yields out-of-core placements, especially when several very large modules are present 15 DRAFT 0 500 1000 1500 2000 0 500 1000 1500 2000 ibm01 HPWL=2.376e+06, #Cells=12752, #Nets=14111 0 500 1000 1500 2000 0 500 1000 1500 2000 ibm01 HPWL=2.457e+06, #Cells=12752, #Nets=14111 Figure 5: A placement of the IBM01 benchmark from IBM-MSwPins by FengShui before (left) and after (right) legalization and detail placement. (the benchmarks used in contain numerous modules of medium size). See Figure 10 in and Figure 6 in for examples of this behavior. Another concern about packed placements is the harmful effect of such a strategy on routability, explicitly shown in . Overall, the work in demonstrates very good legal placements for common benchmarks, but questions remain about the robustness and generality of the proposed approach to mixed-size placement. Example FengShui placements before and after legalization are shown in Figure 5. 5 Advantages of Min-cut Placement This section presents recent techniques which give min-cut placement a signiﬁcant advantage over other placement algorithms in whitespace allocation, ﬂoorplacement, routed wirelength and incre-mental placement. 5.1 Flexible Whitespace Allocation The min-cut bisection based placement framework offers much ﬂexibility in whitespace alloca-tion. Section 2.4 describes uniform allocation of whitespace for min-cut bisection placement and a trivial pre-processing step to allow for non-uniform allocation. This section outlines two more sophisticated whitespace allocation techniques, minimum local whitespace and safe whitespace, that can be used for non-uniform whitespace allocation and satisfying whitespace constraints . Minimum Local Whitespace. If a placement bin has more than a user-deﬁned minimum lo-cal whitespace (minLocalWS), partitioning will deﬁne a tentative cut-line that divides the bin’s 16 DRAFT placement area in half. Partitioning targets an equal division of cell area, but is given more free-dom to deviate from its target. Tolerance is computed so that with whitespace deterioration, each descendant bin of the current bin will have at least minLocalWS . The assumption that the whitespace deterioration, α, in end-case bins is 0 made in and presented in Section 2.4 no longer applies, so the calculation of α must change. Since we want all child bins of the current bin to have minLocalWS relative whitespace, in particular end case bins must have at least minLocalWS and thus we may set w = minLocalWS, instead of a function of τ. Using the assumption that α remain constant during partitioning, α can be calculated directly as α = n q w w . With the more realistic assumption that τ remain constant, τ can be calculated as τ = n q 1−w 1−w −1 . Knowing τ, α can be computed as α = (τ+1)+ τ w . After a partitionment is calculated, the cut-line is shifted to ensure that minLocalWS is pre-served on both sides of the cut-line. If the minimum local whitespace is chosen to be small, one can produce tightly packed placements which greatly improves wirelength. Safe Whitespace. The last whitespace allocation mode is designed for bins with “large” quan-tities of whitespace. In safe whitespace allocation, as with minimum local whitespace allocation, a tentative geometric cut-line of the bin is chosen, and the target of partitioning is an equal bisection of the cell area. The difference in safe whitespace allocation mode is that the partitioning tolerance is much higher. Essentially, any partitioning solution that leaves at least safeWS on either side of the cut-line is considered legal. This allows for very tight packing and reduces wirelength, but is not recommended for congestion-driven placement . Figure 6 illustrates uniform and non-uniform whitespace allocation. Column (a) shows global placements with uniform (top) and non-uniform (bottom) whitespace allocation on the ISPD 2005 contest benchmark adaptec1 (57.34% utlization) . In the non-uniform placement shown, the minimum local whitespace is 12% and safe whitespace is 14%. Columns (b) and (c) show intensity maps of the local utilization of each placement. Lighter areas of the intensity maps signify viola-tions of a given target placement density; darker areas have utilization below the target. Regions completely occupied by ﬁxed obstacles are shaded as if they exactly meet the target density. The target densities for columns (b) and (c) are 90% and 60%. Note that uniform whitespace produces almost no violations when the target is 90% and relatively few when the target is 60%. The non-uniform placement has more violations as compared to the uniform placement especially when the 17 DRAFT 0 2000 4000 6000 8000 10000 0 2000 4000 6000 8000 10000 0 2000 4000 6000 8000 10000 0 2000 4000 6000 8000 10000 (a) (b) (c) Figure 6: Column (a) shows global placements of the ISPD 2005 Placement contest benchmark adaptec1 (57.34% utilization) with uniform whitespace allocation (top) and non-uniform whitespace allocation (bottom). Fixed obstacles are drawn with double lines. To indicate orientation, north-west corners of blocks are truncated. Columns (b) and (c) depict the local utilization of the placements. Lighter areas of the placement sig-nify placement regions with density above a given target (90% for column (b) and 60% for column (c)) whereas darker areas have utilization below the target. target is 60%, but remains largely legal with the 90% target density. 5.2 Solving Difﬁcult Instances of Floorplacement Floorplacement (see Section 4.1) appears promising for SoC layout because of its high capacity and the ability to pack blocks. However, as experiments in demonstrate, existing tools for ﬂoorplacement are fragile — on many instances they fail, or produce remarkably poor placements. To improve the performance of min-cut placement on mixed-size instances, the authors of propose three synergistic techniques for ﬂoorplacement that in particular succeed on hard in-stances: (i) selective ﬂoorplanning with macro clustering, (ii) improved obstacle evasion for B-tree, and (iii) ad hoc look-ahead in top-down ﬂoorplacement. Obstacle evasion is especially impor-18 DRAFT Variables: queue of placement partitions Initialize queue with top-level partition 1 While (queue not empty) 2 Dequeue a partition 3 If (partition is not marked as merged) 4 Perform look-ahead ﬂoorplanning on partition 5 If look-ahead ﬂoorplanning fails 6 Undo one partition decision 7 Merge partition with sibling 8 Mark new partition as merged and enqueue 9 Else if (partition has large macros or is marked as merged) 10 Mark large macros for placement after ﬂoorplanning 11 Cluster remaining macros into soft macros 12 Cluster std-cells into soft macros 13 Use fixed-outline floorplanner to pack all macros (soft+hard) 14 If fixed-outline floorplanning succeeds 15 Fix large macros and remove sites beneath 16 Else 17 Undo one partition decision 18 Merge partition with sibling 19 Mark new partition as merged and enqueue 20 Else if (partition is small enough and mostly comprised of macros) 21 Process ﬂoorplanning on all macros 22 Else if (partition small enough) 23 Process end case std cell placement 24 Else 25 Bi-partition netlist of the partition 26 Divide the partition by placing a cutline 27 Enqueue each child partition Figure 7: Modiﬁed min-cut ﬂoorplacement ﬂow. Bold-faced lines are new . tant for top-down ﬂoorplacement, even for designs that initially have no obstacles. The techniques are called SCAMPI, an acronym for SCalable Advanced Macro Placement Improvements. Empiri-cally, SCAMPI shows signiﬁcant improvements in ﬂoorplacement success rate (68% improvement as compared to the ﬂoorplacement technique presented in Section 4.1) and HPWL (3.5% reduction compared to ﬂoorplacement in Section 4.1). Traditional placement techniques such as top-down and analytical frameworks, bottom-up clus-tering and iterative cell-spreading, scale well in terms of runtime and interconnect optimization when all modules are small. However, handling a wide variety of module sizes with these tech-niques seems considerably more difﬁcult . On the other hand, simulated annealing has a good track record in handling heterogeneous module conﬁgurations, but can only be effectively ap-plied to small problem sizes . This dichotomy between large-scale placement techniques and annealing-based ﬂoorplanning necessitates a rethinking of existing ﬂoorplacement ﬂows . Selective ﬂ oorplanning with macro clustering. In top-down correct-by-construction frame-19 DRAFT Figure 8: The plot on the left illustrates traditional ﬂoorplacement. Whenever a ﬂoorplan-ning threshold is reached, all macros in the bin are designated for ﬂoorplanning. Then, the ﬂoorplacement ﬂow continues down until detailed placement, where the standard cells will be placed. The plot on the right illustrates the SCAMPI ﬂow. Macros are selectively placed at the appropriate levels of hierarchy . works like Capo and PATOMA (see Section 4.2), a key bottleneck is in ensuring ongoing progress — partitioning, ﬂoorplanning or end-case processing must succeed at any given step. Both frameworks experience problems when ﬂoorplanning is invoked too early to produce rea-sonable solutions — PATOMA resorts to solutions with very high wirelength, and Capo times out because it has nothing to resort to and runs the an annealer on too many modules. In order to scale better, the annealer clusters small standard cells into soft blocks before starting Simulated Anneal-ing. When a solution is available, all hard blocks are considered placed and ﬁxed — they are treated as obstacles when the remaining standard cells are placed. Compared to other multi-level frame-works, this one does not include reﬁnement, which makes it relatively fast. Speed is achieved at the cost of not being able to cluster modules other than standard cells because the ﬂoorplanner does not produce locations for clustered modules. Unfortunately, this limitation signiﬁcantly restricts scalability of designs with many macros . The proposed technique of selective ﬂoorplanning with macro clustering allows to cluster blocks before annealing, and does not require additional reﬁnement or cluster-packing steps (which are among the obvious facilitators) — instead certain existing steps in ﬂoorplacement are skipped. This improvement is based on two observations: (i) blocks that are much smaller than their bin can be treated like standard cells, (ii) the number of blocks that are large relative to the bin size is necessarily limited. E.g., there cannot be more than nine blocks with area in excess of 10% of a bin’s area . In selective ﬂoorplanning, each block is marked as small or large based on a size threshold. 20 DRAFT Standard cells and small blocks can be clustered, except that clusters containing hard blocks have additional restrictions on their aspect ratios. After successful annealing, only the large blocks are placed, ﬁxed and considered obstacles. Normal top-down partitioning resumes, and each remaining block will qualify as large at some point later. This way, speciﬁc locations are determined when the right level of detail is considered. If ﬂoorplanning fails during hierarchical placement, the failed bin is merged with its sibling and the merged bin is ﬂoorplanned (see Figure 7). The blocks marked as large in the merged bin include those that exceed the size threshold and also those marked as large in the failed bin (since the failure suggests that those blocks were difﬁcult to pack). After the largest macros are placed, the ﬂow resumes . The proposed technique limits the size of ﬂoorplanning instances given to the annealer by a constant (in our case 200 modules) and does not require much extra work. However, it introduces an unexpected complexity. The ﬂoorplacement framework does not handle ﬁxed obstacles in the core region, and none of the public benchmarks have them. When Capo ﬁxes blocks in a particular bin, it ﬁxes all of them and never needs to ﬂoorplan around obstacles. Another complication due to newly introduced ﬁxed obstacles is in cutline selection. Reliable obstacle-evasion and intelligent cutline selection may be required by practical designs, even without selective ﬂoorplanning (e.g., to handle pre-diffused memories, built-in multipliers in FPGAs, etc). Therefore they are viewed as independent but synergistic techniques . Obstacle evasion in ﬂ oorplanning: B-tree enhancement. When satisfying area constraints is difﬁcult, it is very important to increase the priority of area optimization so as to achieve legality . Because of this, the authors of select the B-tree ﬂoorplan representation for its amenability to packed conﬁgurations and add obstacle evasion into B-tree evaluation. Ad-hoc look-ahead ﬂ oorplanning. The sum of block areas may signiﬁcantly under-estimate the area required for large blocks. Better estimates are required to improve the robustness of ﬂoorplacement and look-ahead area-driven ﬂoorplanning appears as a viable approach . SCAMPI performs look-ahead ﬂoorplanning to validate solutions produced by the hypergraph partitioner, and check that a resulting partition is packable, within a certain tolerance for failure. Look-ahead ﬂoorplanning must be fast, so that the amortized runtime overhead of the look-ahead calls is less than the total time saved from discovering bad partitioning solutions. Therefore look-ahead ﬂoorplanning is performed with blocks whose area is larger than 10% of the total module 21 DRAFT Figure 9: Calculating the three costs for weighted terminal propagation with StWL: w1 (left), w2 (middle), and w12 (right). The net has ﬁve ﬁxed terminals: four above and one below the proposed cut-line. For the traditional HPWL objective, this net would be considered inessential. Note that the structure of the three Steiner trees may be entirely different, which is why w1, w2 and w12 are evaluated independently . area in the bin, and soft blocks containing remaining modules. For speed, the ﬂoorplanner is conﬁgured to perform area-only packing, and the placer is conﬁgured to only perform look-ahead ﬂoorplanning on bins with large blocks. Dealing with only the largest blocks is sufﬁcient because ﬂoorplanning failures are most often caused by such blocks . 5.3 Optimizing Steiner Wirelength Weighted terminal propagation as described in and summarized in Section 3.4, is sufﬁciently general to account for objectives other than HPWL such as Steiner Wirelength (StWL) . StWL is known to correlate with ﬁnal routed wirelength (rWL) more accurately than HPWL and the authors of hypothesize that if StWL could be directly optimized during global placement, one may be able to enhance routability and reduce routed wirelength. The points required to calculate w1 for a given net are the terminals on the net plus the center of partition 1. Similarly, the points required to calculate w2 are the terminals plus the center of partition 2. Lastly, the points to calculate w12 are the terminals on the net plus the centers of both partitions. See Figure 9 for an example of calculating these three costs. Clearly the HPWL of the set of points necessary to calculate w12 is at least as large as that of w1 and w2 since it contains an additional point. By the same logic, StWL also satisﬁes this relationship since RSMT length can only increase with additional points. Since StWL is a valid cost function for these weighted partitioning problems, this is a framework whereby it can be minimized . The simplicity of this framework for minimizing StWL is deceiving. In particular, the propa-gation of terminal locations to the current placement bin and the removal of inessential nets 22 DRAFT — standard techniques for HPWL minimization — cannot be used when minimizing StWL. Mov-ing terminal locations drastically changes Steiner-tree construction and can make StWL estimates extremely inaccurate. Nets that are considered inessential in HPWL minimization (where the x-or y-span of terminals, if the cut is vertical or horizontal respectively, contains the x- or y-span of the centers of child bins) are not necessarily inessential when considering StWL because there are many Steiner trees of different lengths that have the same bounding box. Figure 9 illustrates a net that is inessential for HPWL minimization but essential for StWL minimization. Not only computing Steiner trees, but even traversing all relevant nets to collect all relevant point locations can be very time-consuming. Therefore, the main challenge in supporting StWL minimization is to develop efﬁcient data structures and limit additional runtime during placement . Pointsets with multiplicities. Building Steiner trees for each net during partitioning is a com-putationally expensive task. To keep runtime reasonable when building Steiner trees for parti-tioning, the authors of introduce a simple yet highly effective data structure — pointsets with multiplicities. For each net in the hypergraph, two lists are maintained. The ﬁrst list contains all the unique pin locations on the net that are ﬁxed. A ﬁxed pin can come from sources such as terminals or ﬁxed objects in the core area. The second list contains all the unique pin locations on the net that are movable, i.e., all other pins that are not on the ﬁxed list. All points on each list are unique so that redundant points are not given to Steiner evaluators which may increase their runtime. To do so efﬁciently, the lists are kept in a sorted order. For both lists, in addition to the location of the pin, the number of pins that correspond to a given point is also saved . Maintaining the number of actual pins that correspond to a point in a pointset (the multiplicity of that point) is necessary for efﬁcient update of pin locations during placement. If a pin changes position during placement, the pointsets for the net connected to the pin must be updated. First, the original position of the pin must be removed from the movable point set. As multiple pins can have the same position, especially early in placement, the entire net would need to be traversed to see if any other pins share the same position as the pin that is moving. Multiplicities allow to know this information in constant time. To remove the pin, one performs a binary search on the pointset and decreases the multiplicity of the pin’s position by 1. If this results in the position having a multiplicity of 0, the position can be removed entirely. Insertion of the pin’s new position is similar: ﬁrst, a binary search is performed on the pointset. If the pin’s position is already present 23 DRAFT in the pointset, the multiplicity is increased by 1. Otherwise, the position is added in sorted order with a multiplicity of 1. Empirically, building and maintaining the pointset data structures takes less than 1% of the runtime of global placement . Performance. The authors of compared three Steiner evaluators in terms of runtime impact and solution quality. They chose the FastSteiner evaluator for global placement based on its reasonable runtime and consistent performance on large nets. Empirical results show the use of FastSteiner leads to a reduction of StWL by 3% on average on the IBMv2 benchmarks (with a reduction of routed wirelength up to 7%) while using less than 30% additional runtime . 5.4 Incremental Placement To develop a strong incremental placement tool, ECO-system, the authors of build upon an existing global placement framework and must choose between analytical and top-down. The main considerations include robustness, the handling of movable macros and ﬁxed obstacles, as well as consistent routability of placements and the handling of density constraints. Based on recent em-pirical evidence [30,32,35], the top-down framework appears a somewhat better choice. However, analytical algorithms can also be integrated into ECO-system when particularly extensive changes are required. ECO-system favorably compares to recent detail placers in runtime and solution quality and fares well in high-level and physical synthesis. General Framework. ECO-system can be likened to reverse engineering the min-cut place-ment process. The goal is to reconstruct the internal state of a min-cut placer that could have produced the given initial placement. Given this state, one can choose to accept or reject its previ-ous decisions based on their own criteria and build a new placement for the design. If many of the decisions of the placer were good, one can achieve a considerable runtime savings as compared to placement from scratch. If many of the decisions are determined to be bad, one can do no worse in terms of solution quality than placement from scratch. An overview of the application of ECO-system to an illegal placement is depicted in Figure 11. The overall algorithm in the framework of min-cut placement is shown in Figure 10. To rebuild the state of a min-cut placer, one must reconstruct a series of cut-lines and parti-tioning solutions efﬁciently. One must also determine criteria for the acceptability of the derived partitioning and cut-line. To extract a cut-line and partitioning solution from a given placement bin, 24 DRAFT Variables: queue of placement bins Initialize queue with top-level placement bin 1 While(queue not empty) 2 Dequeue a bin 3 If(bin not marked to place from scratch) 4 If(bin overfull) 5 Mark bin to place from scratch, break 6 Quickly choose the cut-line which has the smallest net-cut considering cell area balance constraints 7 If(cut-line causes overfull child bin) 8 Mark bin to place from scratch, break 9 Induce partitioning of bin’s cells from cut-line 10 Improve net-cut of partitioning with single pass of Fiduccia-Mattheyses 11 If(% of improvement > threshold) 12 Mark bin to place from scratch, break 13 Create child bins using cut-line and partitioning 14 Enqueue each child bin 15 If(bin marked to place from scratch) 16 If(bin small enough) 17 Process end case 18 Else 19 Bi-partition the bin into child bins 20 Mark child bins to place from scratch 21 Enqueue each child bin Figure 10: Incremental min-cut placement. Bold-faced lines 3-15 and 20 are different from traditional min-cut placement . all possible cut-lines of the bin as well as the partitions they induce must be examined. Starting at one edge of the placement bin (left edge for a vertical cut and bottom edge for a horizontal cut) and moving towards the opposite edge, for each potential cut-line encountered, one maintains the cell area on either side of the cut-line, the partition induced by the cut-line and its net cut. Once a cut-line and partitioning have been chosen, they mus be evaluated to see if they should be accepted or rejected. To evaluate the partitioning, the authours of use it as input to a Fiduccia-Mattheyses partitioner and see how much it can be improved by a single pass (if the bin is large enough, a multi-level Fiduccia-Mattheyses partitioner can be used). The intuition is that if the constructed partitioning is not worthy of reuse, a single Fiduccia-Mattheyses pass could improve its cut non-trivially. If the Fiduccia-Mattheyses pass improves the cut beyond a certain threshold, the solution is discarded and the entire bin is bisected from scratch. If a partition is accepted by this criterion, one performs a legality test: if the partitioning overﬁlls a child bin, the cut-line is discarded and the bin is bisected from scratch. Empirically, the runtime of the cut-line selection procedure (which includes a single pass of a Fiduccia-Mattheyses partitioner) is much smaller than partitioning from scratch. On large bench-25 DRAFT Figure 11: Legalization during min-cut placement. Placement bins are subdivided until (i) a bin contains no overlap and is ignored for the remainder of the legalization process or, (ii) the placement contained in the bin is considered too poor to be kept (too many overlaps or does not meet the solution quality requirements) and is replaced from scratch using min-cut or analytical techniques . marks, the cut-line selection process requires 5% of ECO-system runtime time whereas min-cut partitioners generally require 50% or more of ECO-system runtime. ECO-system as a whole re-quires approximately 15% of original placement runtime. Handling Macros and Obstacles. With the addition of macros, the ﬂow of top-down place-ment usually becomes more complex. The authors of adopt the style of ﬂoorplacement from [30, 31] (see Sections 4.1 and 5.2). For legalization with macros, a new criterion for ﬂoor-planning is added: if a placement bin has non-overlapping positions for macros (i.e. no macros in the placement bin overlap each other) the macros are placed in exactly their initial positions; if some of the macros overlap, other ﬂoorplanning criteria are used to decide. If any of the macros are moved, the placement of all cells and macros in the bin must be discarded and placement and proceeds as described in . During the cut-line selection process, some cut-line locations are considered invalid — namely those that are too close to obstacle boundaries but do not cross the obstacles. This is done to prevent long and narrow slivers of space between cut-lines and obstacle boundaries. Ties for cut-lines are broken based on the number of macros they intersect. This helps to reduce overfullness in child bins allowing deeper partitioning, which reduces runtime . 26 DRAFT 6 State-of-the-art Min-cut Placers In this section, we present partitioning-based placement techniques that are used in cutting-edge placers. For each placer, we describe its overall ﬂow, how this differs from the generic min-cut ﬂow, and how it handles challenges in placement such as ﬁxed obstacles and mixed-size instances. In particular we describe the techniques used by the placers Dragon [38,39,43], FengShui [4,27], NTUPlace2 and Capo [30–35]. 6.1 Dragon The most recent version of Dragon, Dragon2006 , combines min-cut bisection with Simulated Annealing for placement. In its most basic ﬂow, Dragon2006 utilizes recursive bisection with the hMETIS partitioner . Each bin is partitioned multiple times with a feedback mechanism to allow for more accurate terminal propagation (see Section 3.1 for more details on placement feed-back). Partitioning is followed by Simulated Annealing on the placement bins where whole bins are swapped with one another to improve HPWL [38,39]. After a number of layers of interleaved partitioning and Simulated Annealing, each bin contains only a few cells and the partitioning phase terminates. Next, bins are aligned to row structures and cell-based Simulated Annealing is per-formed wherein cells are swapped between bins to improve HPWL [38, 39]. Lastly, cell overlaps are removed and local detail placement improvements are made. Mixed-size Placement. The traditional Dragon ﬂow does not take macros into consideration during placement. To account for macros, partitioning, bin-based annealing and legalization must be modiﬁed. In addition, Dragon2006 makes two passes on a design with obstacles; the ﬁrst pass ﬁnds locations for macros and the second treats macros as ﬁxed obstacles (similar to ). In the ﬁrst pass, partitioning is modiﬁed to handle large movable macros. The traditional Dragon ﬂow alternates cut directions at each layer and chooses the cut-line to split a bin exactly in half in order to maintain a regular grid structure. In the presence of large macros, the requirement of a regular bin structure is relaxed. The cut-line of the bin is shifted to allow the largest macro to ﬁt into a child bin after partitioning. If macros can only ﬁt in one bin, they are pre-assigned to the child bin in which they can ﬁt and not involved in partitioning [38,39]. Bin-based Simulated Annealing after partitioning is also modiﬁed as bins may not all have the same dimensions. Horizontal swaps between adjacent bins are only allowed if they are of the same 27 DRAFT height. Similarly, vertical swaps between adjacent bins are only allowed if they are of the same width. Lastly, diagonal bin swaps are only legal if the bins have the same height and width. After all bins have a threshold of cells or fewer, partitioning stops and macro locations are legalized. Once legal, macros are considered ﬁxed and partitioning begins again at the top level to place the standard cells of the design [38,39]. 6.2 FengShui FengShui [4, 27] is a recursive bisection min-cut placer that uses the hMETIS partitioner . FengShui implements the fractional cut technique (see Section 3.2) and packs its placements to either side of the placement region which has a serious affect on the routability of its placements . FengShui also supports mixed-size placement (see Section 4.3) 6.3 NTUPlace2 NTUPlace2 is a hybrid placer that uses both min-cut partitioning and analytical techniques for standard-cell and mixed-size designs. NTUPlace2 uses repartitioning (see Section 3.1), cut-line shifting (see Section 2.3) and weighted net-cut (see Section 3.4) . NTUPlace2 uses analytical techniques to aid partitioning which are different from those in ACG (see Section 3.3). Before partitioning calls to the hMETIS partitioner , objects in a placement bin are ﬁrst placed by an analytical technique to reduce quadratic wirelength . Those objects which are placed far from the proposed cut-line are considered ﬁxed in their current loca-tions for the partitioning process. This technique helps to make terminal propagation more exact and with the weighted net-cut technique has resulted in very good solution quality . To handle mixed-size placement, macro locations are legalized at each layer. Macros become ﬁxed at different layers of placement according to their size relative to placement bin size. Thus larger macros are placed earlier in placement . Macros are legalized using a linear program-ming technique that attempts to minimize the movement of macros during legalization . 6.4 Capo Capo [30–35] is a min-cut ﬂoorplacer. As such it implements the ﬂoorplacement ﬂow as described in Section 4.1 and further improved by SCAMPI in Section 5.2 rather than the traditional min-cut ﬂow and implicitly handles mixed-size placement and ﬁxed obstacles in the placement area. Capo 28 DRAFT can use either MLPart or hMETIS for hypergraph partitioning. Whitespace allocation in Capo is done per placement bin: either uniform (see Section 2.4), minimum local or safe whitespace allocation (see Section 5.1) is chosen based on the bin’s whitespace and user-conﬁgurable options. To improve the quality of results, Capo also implements repartitioning (see Section 3.1), placement feedback (see Section 3.1), weighted net-cut (see Section 3.4) and several whitespace allocation techniques. Capo has also been used to optimize Steiner wirelength in placement (see Section 5.3) and can be used for incremental placement (see Section 5.4). 29 DRAFT References S. N. Adya and I. L. Markov,“ Fixed-outline Floorplanning: Enabling Hierarchical Design”, IEEE Trans. on VLSI, vol. 11, no. 6, pp. 1120-1135, December 2003. (ICCD 2001, pp. 328-334). S. N. Adya and I. L. Markov, “ Combinatorial Techniques for Mixed-size Placement”, ACM Trans. on Design Autom. of Elec. Sys., vol. 10, no. 5, January 2005. (ISPD 2002, pp. 12-17). S. N. Adya, I. L. Markov and P. G. Villarrubia, “ On Whitespace and Stability in Physical Synthesis,” Integration: the VLSI Journal, vol. 25, no. 4, pp. 340-362, 2006. (ICCAD 2003, pp. 311-318). A. Agnihotri et al., “ Fractional Cut: Improved Recursive Bisection Placement,”ICCAD, pp. 307-310, 2003. C. J. Alpert, G.-J. Nam and P. G. Villarrubia, “ Effective Free Space Management for Cut-Based Place-ment via Analytical Constraint Generation,” IEEE Trans. on CAD, vol. 22, no. 10, pp. 1343-1353, 2003. (ICCAD 2002, pp. 746-751). U. Brenner and A. Rohe, “ An Effective Congestion Driven Placement Framework,” IEEE Trans. on CAD, vol. 22, no. 4, pp. 387-394, 2003. (ISPD 2002, pp. 6-11). M. Breuer, “ Min-cut Placement,” Journal of Design Automation and Fault Tolerant Computing, vol. 1, no. 4, pp. 343-362, October 1977. (DAC 1977, pp. 284-290). A. E. Caldwell, A. B. Kahng, S. Mantik, I. L. Markov and A. Zelikovsky, “ On Wirelength Estimations for Row-Based Placement”, IEEE Trans. on CAD, vol. 18, no. 9, pp. 1265-1278, 1999. A. E. Caldwell, A. B. Kahng and I. L. Markov, “ Can Recursive Bisection Alone Produce Routable Placements?,”DAC, pp. 477-482, Los Angeles, June 2000. A. E. Caldwell, A. B. Kahng, and I. L. Markov, “ Design and Implementation of Move-based Heuristics for VLSI Hypergraph Partitioning,” ACM J. of Experimental Algorithms, vol. 5, 2000. A. E. Caldwell, A. B. Kahng and I. L. Markov, “ Optimal Partitioners and End-case Placers for Standard-cell Layout,” IEEE Trans. on CAD, vol. 19, no. 11, pp. 1304-1314, 2000. (ISPD 1999, pp. 90-96). A. E. Caldwell, A. B. Kahng and I. L. Markov, “ Hierarchical Whitespace Allocation in Top-down Placement,”IEEE Trans. on CAD vol. 22, no. 11, pp. 716-724, November 2003. Y. C. Chang et al., “ B-trees: A New Representation for Non-Slicing Floorplans,” DAC, pp. 458-463, 2000. T. C. Chen and Y. W. Chang, “ Modern Floorplanning Based on Fast Simulated Annealing,”ISPD, pp. 104-112, 2005. T. C. Chen, Y. W. Chang and S. C. Lin, “ IMF: Interconnect-Driven Multilevel Floorplanning for Large-Scale Building-Module Designs,”ICCAD, pp. 159-164, November 2005. J. Cong, G. Nataneli, M. Romesis and J. Shinnerl, “ An Area-optimiality Study of Floorplanning,” ISPD, pp. 78-83, 2004. J. Cong, M. Romesis and J. Shinnerl, “ Fast Floorplanning by Look-Ahead Enabled Recursive Bipar-titioning,”ASPDAC, pp. 1119-1122, 2005. J. Cong, M. Romesis and J. Shinnerl, “ Robust Mixed-Size Placement Under Tight White-Space Con-straints,”ICCAD, pp. 165-172, 2005. A. E. Dunlop and B. W. Kernighan, “ A Procedure for Placement of Standard Cell VLSI Circuits,” IEEE Trans. on CAD, vol. 4, no. 1, pp. 92-98, 1985. C. M. Fiduccia and R. M. Mattheyses, “ A Linear Time Heuristic for Improving Network Partitions,” DAC, pp. 175-181, 1982. D. J.-H. Huang, and A. B. Kahng, “ Partitioning-based Standard-cell Global Placement With an Exact Objective,”ISPD, pp. 18-25, 1997. 30 DRAFT Z.-W. Jiang et al., “ NTUPlace2: A Hybrid Placer Using Partitioning and Analytical Techniques,” ISPD, pp. 215-217, 2006. A. B. Kahng, S. Mantik and I. L. Markov, “ Min-max Placement For Large-scale Timing Optimiza-tion,”ISPD, pp. 143-148, April 2002. A. B. Kahng, I. I. Mandoiu and A. Zelikovsky, “ Highly Scalable Algorithms for Rectilinear and Octi-linear Steiner Trees,”ASPDAC, pp. 827-833, 2003. A. B. Kahng and S. Reda, “ Placement Feedback: A Concept and Method for Better Min-cut Place-ment,”DAC, pp. 357-362, 2004. G. Karypis, R. Aggarwal, V. Kumar and S. Shekhar, “ Multilevel Hypergraph Partitioning: Applica-tions in VLSI Domain,”DAC, pp. 526-629, 1997. A. Khatkhate, C. Li, A. R. Agnihotri, M. C. Yildiz, S. Ono, C.-K. Koh and P. H. Madden, “ Recursive Bisection Based Mixed Block Placement,”ISPD, pp. 84-89, 2004. C. Li, M. Xie, C. K. Koh, J. Cong and P. H. Madden, “ Routability-driven Placement and White Space Allocation,”ICCAD, pp. 394-401, 2004. G.-J. Nam, C. J. Alpert, P. Villarrubia, B. Winter and M. Yildiz, “ The ISPD2005 Placement Contest and Benchmark Suite,”ISPD, pp. 216-220, 2005. A. N. Ng, I. Markov, R. Aggarwal and V. Ramachandran, “ Solving Hard Instances of Floorplacement,” ISPD, pp. 170-177, April 2006. J. A. Roy, S. N. Adya, D. A. Papa and I. L. Markov, “ Min-cut Floorplacement,”IEEE Trans. on CAD, vol. 25, no. 7, pp. 1313-1326, 2006. (ICCAD 2004, pp. 550-557). J. A. Roy and I. L. Markov, “ Seeing the Forest and the Trees: Steiner Wirelength Optimization in Placement,”to appear in IEEE Trans. on CAD, 2007. (ISPD 2006, pp. 78-85). J. A. Roy and I. L. Markov, “ ECO-system: Embracing the Change in Placement,” to appear in Pro-ceedings of ASPDAC, 2007. J. A. Roy, D. A. Papa, S. N. Adya, H. H. Chan, J. F. Lu, A. N. Ng and I. L. Markov, “ Capo: Robust and Scalable Open-Source Min-cut Floorplacer,” ISPD, pp. 224-227, April 2005. J. A. Roy, D. A. Papa, A. N. Ng and I. L Markov, “ Satisfying Whitespace Requirements in Top-down Placement,”ISPD, pp. 206-208, April 2006. N. Selvakkumaran and G. Karypis, “ Theto - A Fast, Scalable and High-quality Partitioning Driven Placement Tool,”Technical report, Univ. of Minnesota, 2004. P. R. Suaris and G. Kedem, “ An Algorithm for Quadrisection and Its Application to Standard Cell Placement,” IEEE Trans. on Circuits and Systems, vol. 35, no. 3, pp. 294-303, 1988. (ICCAD 1987, pp. 474-477). T. Taghavi, X. Yang, B.-K. Choi, M. Wang and M. Sarrafzadeh “ Dragon2005: Large-Scale Mixed-size Placement Tool,”ISPD, pp. 245-247, April 2005. T. Taghavi, X. Yang, B.-K. Choi, M. Wang and M. Sarrafzadeh “ Dragon2006: Blockage-Aware Congestion-Controlling Mixed-Size Placer,”ISPD, pp. 209-211, April 2006. K. Takahashi, K. Nakajima, M. Terai, and K. Sato, “ Min-cut Placement with Global Objective Func-tions for Large Scale Sea-of-gates Arrays,”IEEE Trans. on CAD, vol. 14, no. 4, pp. 434-446, 1995. J. Vygen, “ Algorithms for Large-Scale Flat Placement”, DAC, pp. 746-751, 1997. E. Wein and J. Benkoski, “Hard macros will revolutionize SoC design”, EE Times, August 20, 2004. X. Yang, B. K. Choi, and M. Sarrafzadeh, “ Routability Driven White Space Allocation for Fixed-die Standard-cell Placement,”IEEE Trans. on CAD, vol. 22, no. 4, pp. 410-419, April 2003. (ISPD 2002, pp. 42-49). M. C. Yildiz and P. H. Madden, “ Improved Cut Sequences for Partitioning Based Placement,” DAC, pp. 776-779, 2001. 31
2383	https://flexbooks.ck12.org/cbook/ck-12-algebra-ii-with-trigonometry-concepts/section/13.7/primary/lesson/trigonometric-ratios-on-the-unit-circle-alg-ii/	Skip to content Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 13.7 Trigonometric Ratios on the Unit Circle Written by:Lori Jordan \| Kate Dirga Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 What are the exact values of the following trigonometric functions? cos495∘ tan5π3 Trigonometric Ratios on the Unit Circle Recall special right triangles from Geometry. In a (30∘−60∘−90∘) triangle, the sides are in the ratio 1:√3:2. In an isosceles triangle (45∘−45∘−90∘), the congruent sides and the hypotenuse are in the ratio 1:1:√2. In a (30∘−60∘−90∘) triangle, the sides are in the ratio 1:√3:2. Now let’s make the hypotenuse equal to 1 in each of the triangles so we’ll be able to put them inside the unit circle. Using the appropriate ratios, the new side lengths are: Using these triangles, we can evaluate sine, cosine and tangent for each of the angle measures. sin45∘=√22sin60∘=√32 sin30∘=12cos45∘=√22 cos60∘=12 cos30∘=√32tan45∘=1 tan60∘=√3212=√3tan30∘=12√32=√32 These triangles can now fit inside the unit circle. Putting together the trigonometric ratios and the coordinates of the points on the circle, which represent the lengths of the legs of the triangles, (Δx,Δy), we can see that each point is actually (cosθ,sinθ), where θ is the reference angle. For example, sin60∘=√32 is the y – coordinate of the point on the unit circle in the triangle with reference angle 60∘. By reflecting these triangles across the axes and finding the points on the axes, we can find the trigonometric ratios of all multiples of 0∘,30∘ and 45∘ (or 0,π6,π4 radians). Let's solve the following problems using the unit circle. Find sin3π2. Find 3π2 on the unit circle and the corresponding point is (0,−1). Since each point on the unit circle is (cosθ,sinθ),sin3π2=−1. Find tan7π6. This time we need to look at the ratio sinθcosθ. We can use the unit circle to find sin7π6=−12 and cos7π6=−√32. Now, tan7π6=−12−√32=1√3=√33. Quadrants in a Unit Circle Another way to approach these exact value problems is to use the reference angles and the special right triangles. The benefit of this method is that there is no need to memorize the entire unit circle. If you memorize the special right triangles, can determine reference angles and know where the ratios are positive and negative you can put the pieces together to get the ratios. Looking at the unit circle above, we see that all of the ratios are positive in Quadrant I, sine is the only positive ratio in Quadrant II, tangent is the only positive ratio in Quadrant III and cosine is the only positive ratio in Quadrant IV. Keeping this diagram in mind will help you remember where cosine, sine and tangent are positive and negative. You can also use the pneumonic device - All Students Take Calculus, or ASTC, to recall which is positive (all the others would be negative) in which quadrant. The coordinates on the vertices will help you determine the ratios for the multiples of 90∘ or π2. Now, let's find the exact values for the following trigonometric functions using the alternative method. cos120∘ First, we need to determine in which quadrant the angles lies. Since 120∘ is between 90∘ and 180∘ it will lie in Quadrant II. Next, find the reference angle. Since we are in QII, we will subtract from 180∘ to get 60∘. We can use the reference angle to find the ratio, cos60∘=12. Since we are in QII where only sine is positive, cos120∘=−12. sin5π3 This time we will need to work in terms of radians but the process is the same. The angle 5π3 lies in QIV and the reference angle is π3. This means that our ratio will be negative. Since sinπ3=√32,sin5π3=−√32. tan7π2 The angle 7π2 represents more than one entire revolution and it is equivalent to 2π+3π2. Since our angle is a multiple of π2 we are looking at an angle on an axis. In this case, the point is (0,−1). Because tanθ=sinθcosθ,tan7π2=−10, which is undefined. Thus, tan7π2 is undefined. Examples Example 1 Earlier, you were asked to find the exact values of the following trigonometric ratios. cos495∘ First, we need to determine in which quadrant the angle lies. Since 495∘−360∘=135∘ is between 90∘ and 180∘ it will lie in Quadrant II. Next, find the reference angle. Since we are in QII, we will subtract from 180∘ to get 45∘. We can use the reference angle to find the ratio, cos45∘=√22. Since we are in QII where only sine is positive, cos495∘=−√22. tan5π3 In problem #2 above we established that the angle 5π3 lies in QIV and the reference angle is π3. This means that the tangent ratio will be negative. Since tanπ3=√3,tan5π3=−√3. Find the exact trigonometric ratios. You may use either method. Example 2 cos7π3 7π3 has a reference angle of π3 in QI. cosπ3=12 and since cosine is positive in QI, cos7π3=12. Example 3 tan9π2 9π2 is coterminal to π2 which has coordinates (0, 1). So tan9π2=sin9π2cos9π2=10 which is undefined. Example 4 sin405∘ 405∘ has a reference angle of 45∘ in QI. sin45∘=√22 and since sine is positive in QI, sin405∘=√22. Example 5 tan11π6 11π6 is coterminal to π6 in QIV. tanπ6=√33 and since tangent is negative in QIV, tan11π6=−√33. Example 6 cos2π3 2π3 is coterminal to π3 in QII. cosπ3=12 and since cosine is negative in QII, cos2π3=12. Review Find the exact values for the following trigonometric functions. sin3π4 cos3π2 tan300∘ sin150∘ cos4π3 tanπ cos(−15π4) sin225∘ tan7π6 sin315∘ cos450∘ sin(−7π2) cos17π6 tan270∘ sin(−210∘) Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. \| Image \| Reference \| Attributions \| --- \| \| \| Credit: ck12 Source: ck12.org \| \| \| \| Credit: ck12 Source: ck12.org \| \| \| \| Credit: ck12 Source: ck12.org \| \| \| \| Credit: ck12 Source: ck12.org \| Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)39/ 100 This lesson has been added to your library. \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. 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2384	https://www.learner.org/wp-content/interactive/geometry/prisms/	3D Shapes Prisms A prism is a polyhedron for which the top and bottom faces (known as the bases) are congruent polygons, and all other faces (known as the lateral faces) are rectangles. (Technically, when the sides are rectangles, the shape is known as a right prism, indicating that the lateral faces meet the sides of the base at right angles. In this lesson, when we use the term prism, we mean a right prism. But there are other types of prisms, too.) A prism is described by the shape of its base. For instance, a rectangular prism has bases that are rectangles, and a pentagonal prism has bases that are pentagons. Explore & Play with Prisms Use the animation below to explore the properties of four prisms. Follow the instructions below to change the direction and speed of the prism's rotation and to highlight the numbers of faces (F), vertices (V), and edges (E) for each prism. The net of a triangular prism consists of two triangles and three rectangles. The triangles are the bases of the prism and the rectangles are the lateral faces. Net of Triangular Prism The net of a rectangular prism consists of six rectangles. Both the bases and the lateral faces of this shape are rectangles. Net of Rectangular Prism The net of a pentagonal prism consists of two pentagons and five rectangles. The pentagons are the bases of the prism and the rectangles are the lateral faces. Net of Pentagonal Prism The net of a octagonal prism consists of two octagons and eight rectangles. The octagons are the bases of the prism and the rectangles are lateral faces. Net of Octagonal Prism Rotate the prism direction by clicking and dragging the shape or by clicking on the arrows in the corners. Stop the rotation by clicking on the shape itself. Select each button to highlight the different parts of the shapes.
2385	https://en.wikipedia.org/wiki/Kilogram-force	Kilogram-force - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 History 2 Related units 3 See also 4 References Kilogram-force [x] 42 languages العربية Asturianu Беларуская Беларуская (тарашкевіца) Български Bosanski Català Čeština Dansk Deutsch Eesti Español Esperanto فارسی Français Galego 한국어 Հայերեն Hrvatski Interlingua Italiano עברית Қазақша Magyar Македонски Nederlands 日本語 Norsk bokmål Norsk nynorsk Oʻzbekcha / ўзбекча Polski Português Русский Sicilianu Slovenčina Slovenščina Srpskohrvatski / српскохрватски Suomi Svenska Türkçe Українська 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Weight on earth of a one-kilogram mass "kgf" redirects here. For other uses, see KGF (disambiguation)."Kilopond" redirects here; not to be confused with Kilopound. \| kilogram-force \| \| Unit system \| Gravitational metric system \| \| Unit of \| Force \| \| Symbol \| kgf \| \| Conversions \| \| 1 kgf in ... \| ... is equal to ... \| \| \| \| SI units \| 9.806650 N \| \| CGS units \| 980,665.0 dyn \| \| British Gravitational units \| 2.204623 lbf \| \| Absolute English units \| 70.93164 pdl \| The kilogram-force (kgf or kg F), or kilopond (kp, from Latin: pondus, lit.'weight'), is a non-standard gravitational metric unit of force. It is not accepted for use with the International System of Units (SI) and is deprecated for most uses.[citation needed] The kilogram-force is equal to the magnitude of the force exerted on one kilogram of mass in a 9.806 65 m/s 2 gravitational field (standard gravity, a conventional value approximating the average magnitude of gravity on Earth). That is, it is the weight of a kilogram under standard gravity. One kilogram-force is defined as 9.806 65N. Similarly, a gram-force is 9.806 65 mN, and a milligram-force is 9.806 65 μN. History [edit] The gram-force and kilogram-force were never well-defined units until the CGPM adopted a standard acceleration of gravity of 9.80665 m/s 2 for this purpose in 1901, though they had been used in low-precision measurements of force before that time. Even then, the proposal to define kilogram-force as a standard unit of force was explicitly rejected. Instead, the newton was proposed in 1913 and accepted in 1948. The kilogram-force has never been a part of the International System of Units (SI), which was introduced in 1960. The SI unit of force is the newton. Prior to this, the units were widely used in much of the world. They are still in use for some purposes; for example, they are used to specify tension of bicycle spokes,draw weight of bows in archery, and tensile strength of electronics bond wire, for informal references to pressure (as the technically incorrect kilogram per square centimetre, omitting -force, the kilogram -force per square centimetre being the technical atmosphere, the value of which is very near those of both the bar and the standard atmosphere), and to define the "metric horsepower" (PS) as 75 metre-kiloponds per second. In addition, the kilogram force was the standard unit used for Vickers hardness testing. Three approaches to metric units of mass and force or weight\| v t e Base \| Force \| Weight \| Mass \| --- --- \| \| 2nd law of motion \| m = ⁠F/a⁠ \| F = ⁠W ⋅ a/g⁠ \| F = m ⋅ a \| \| System \| GM \| M \| CGS \| MTS \| SI \| \| Acceleration (a) \| m/s 2 \| m/s 2 \| Gal \| m/s 2 \| m/s 2 \| \| Mass (m) \| hyl \| kilogram \| gram \| tonne \| kilogram \| \| Force (F), weight (W) \| kilopond \| kilopond \| dyne \| sthène \| newton \| \| Pressure (p) \| technical atmosphere \| standard atmosphere \| barye \| pieze \| pascal \| In 1940s, Germany, the thrust of a rocket engine was measured in kilograms-force,[citation needed] in the Soviet Union it remained the primary unit for thrust in the Russian space program until at least the late 1980s.[citation needed] Dividing the thrust in kilograms-force on the mass of an engine or a rocket in kilograms conveniently gives the thrust to weight ratio, dividing the thrust on propellant consumption rate (mass flow rate) in kilograms per second gives the specific impulse in seconds. The term "kilopond" has been declared obsolete. Related units [edit] The tonne-force, metric ton-force, megagram-force, and megapond (Mp) are each 1000 kilograms-force. The decanewton or dekanewton (daN), exactly 10 N, is used in some fields as an approximation to the kilogram-force, because it is close to the 9.80665 N of 1 kgf. The gram-force is 1⁄1000 of a kilogram-force. Units of force \| v t e \| newton \| dyne \| kilogram-force, kilopond \| pound-force \| poundal \| --- --- --- \| \| 1 N \| ≡1 kg⋅m/s 2 \| =10 5 dyn \| ≈0.101 97 kgf \| ≈0.224 81 lb F \| ≈7.2330 pdl \| \| 1 dyn \| =10−5 N \| ≡1 g⋅cm/s 2 \| ≈1.0197×10−6 kgf \| ≈2.2481×10−6 lb F \| ≈7.2330×10−5 pdl \| \| 1 kgf \| =9.806 65 N \| =980 665 dyn \| ≡g n×1 kg \| ≈2.2046 lb F \| ≈70.932 pdl \| \| 1 lb F \| ≈4.448 222 N \| ≈444 822 dyn \| ≈0.453 59 kgf \| ≡g n×1lb \| ≈32.174 pdl \| \| 1 pdl \| ≈0.138 255 N \| ≈13 825 dyn \| ≈0.014 098 kgf \| ≈0.031 081 lb F \| ≡1 lb⋅ft/s 2 \| \| The value of g n (9.806 65 m/s 2) as used in the official definition of the kilogram-force is used here for all gravitational units. \| See also [edit] Metrology Avoirdupois References [edit] ^NIST Guide to the SI, Chapter 5: Units Outside the SI ^The international system of units (SI)Archived 2016-06-03 at the Wayback Machine – United States Department of Commerce, NIST Special Publication 330, 2008, p. 52 ^ Jump up to: abNISTGuide for the Use of the International System of Units (SI) Special Publication 811, (1995) page 51 ^BIPM SI brochureArchived 2004-06-15 at the Wayback Machine, chapter 2.2.2. ^Resolution of the 3rd CGPM (1901) ^Proceedings of the 3rd General Conference on Weights and Measures, 1901, pages 62–64 and 68, (french) ^Proceedings of the 5th General Conference on Weights and Measures, 1913, pages 51 and 56, (french) ^"Resolution 7 of the 9th meeting of the CGPM (1948)". Archived from the original on 2020-06-22. Retrieved 2021-03-02. ^"Balancing wheel tension with the TM-1 Spoke Tension Metre". Cyclingnews. Retrieved 2013-09-03. The recommended tension for spokes in bicycle wheels can be as low as 80 Kilograms force (Kfg) and as high as 230 Kilograms force. Author=Park Tool ^Harman, George G. (2010). Wire Bonding in Microelectronics (3rd ed.). New York: McGraw-Hill. p.408. ISBN978-0-07-164265-1. OCLC609421363. Breaking load (BL): The strength of a wire and its actual force (usually given in grams, grams-force, mN, etc.) required to break a particular wire in a tensile pull. It is not tensile strength, which by definition is the force per unit area. ^Callister, William D. Jr. (2010). Materials Science and Engineering: An Introduction. David G. Rethwisch (8th ed.). Hoboken, NJ: John Wiley & Sons, Inc. ISBN978-0-470-41997-7. OCLC401168960. In the past the units for Vickers hardness were kg/mm2; in Table 12.6 we use the SI units of GPa. ^Comings, E. W. (1940). "English Engineering Units and Their Dimensions". Industrial & Engineering Chemistry. 32 (7): 984–987. doi:10.1021/ie50367a028. ^Klinkenberg, Adrian (1969). "The American Engineering System of Units and Its Dimensional Constant g c". Industrial & Engineering Chemistry. 61 (4): 53–59. doi:10.1021/ie50712a010. ^European Economic Community, Council Directive of 18 October 1971 on the approximation of the laws of the Member States relating to units of measurement Retrieved from " Categories: Units of force Non-SI metric units Hidden categories: Webarchive template wayback links Articles with short description Short description is different from Wikidata Articles containing Latin-language text All articles with unsourced statements Articles with unsourced statements from February 2024 Articles with unsourced statements from September 2020 This page was last edited on 18 July 2025, at 02:43(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Kilogram-force 42 languagesAdd topic
2386	https://www.khanacademy.org/science/physics/forces-newtons-laws/inclined-planes-friction/v/force-of-friction-keeping-the-block-stationary	Force of friction keeping the block stationary (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content Physics archive Course: Physics archive>Unit 1 Lesson 3: Inclined planes and friction Inclined plane force components Ice accelerating down an incline Force of friction keeping the block stationary Correction to force of friction keeping the block stationary Force of friction keeping velocity constant Intuition on static and kinetic friction comparisons Static and kinetic friction example Science> Physics archive> Forces and Newton's laws of motion> Inclined planes and friction © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Force of friction keeping the block stationary Google Classroom Microsoft Teams About About this video Transcript Block of wood kept stationary by the force of friction (Correction made in next video).Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted sharat.mehrotra 14 years ago Posted 14 years ago. Direct link to sharat.mehrotra's post “Can friction support the ...” more Can friction support the direction of motion? If so, how?? Answer Button navigates to signup page •1 comment Comment on sharat.mehrotra's post “Can friction support the ...” (28 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Shibashish Mahapatra 11 years ago Posted 11 years ago. Direct link to Shibashish Mahapatra's post “Friction does not support...” more Friction does not support in the direction of motion but helps motion. When we push the ground with our feet its due to friction that our shoes don't slip away and help us to move forward. So, friction helps in motions but does not act in the direction of motion. 1 comment Comment on Shibashish Mahapatra's post “Friction does not support...” (12 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Shamik Chatterjee 10 years ago Posted 10 years ago. Direct link to Shamik Chatterjee's post “What is this "budging for...” more What is this "budging force" actually called? Answer Button navigates to signup page •Comment Button navigates to signup page (10 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer WisteriaGrove 2 years ago Posted 2 years ago. Direct link to WisteriaGrove's post “It’s called the static co...” more It’s called the static coefficient of friction. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Aleksandra 6 years ago Posted 6 years ago. Direct link to Aleksandra's post “How can you say, that 49N...” more How can you say, that 49N is the maximum friction force? Maybe it is 51N or 60N, and then your 50N parallel force wouldn't be enough? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Charles LaCour 6 years ago Posted 6 years ago. Direct link to Charles LaCour's post “If you know the coefficie...” more If you know the coefficient of static friction and the normal force between the object and the surface the product of these two numbers by definition is the maximum force of static friction. So if you calculate the force to be 49N and 50N doesn't move the object then either the value for the coefficient of static friction or the normal force are wrong. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... slugcat 2 years ago Posted 2 years ago. Direct link to slugcat's post “how would one actually fi...” more how would one actually find the specific "budging force"? isn't adding exactly 1N to the parallel component kinda arbitrary since all it would've taken is any force > 0N to disrupt the balance between the parallel forces? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer a mann 9 years ago Posted 9 years ago. Direct link to a mann's post “Is the coefficient of fri...” more Is the coefficient of friction always depends on the identity of the matirials involved? Or is there a general rule for it? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Vitor 10 years ago Posted 10 years ago. Direct link to Vitor's post “Sal does this: Coefficien...” more Sal does this: Coefficient of static friction= 50/N Coefficiente of cinetic friction= 49/N So if I want in an exercise discover the force of friction when the object is stationary shall I do: Force of friction= Coefficiente of static friction N - 1 ?? following Sal's logic.... Help please :\ Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer lunchlump777 2 years ago Posted 2 years ago. Direct link to lunchlump777's post “I apologize, wouldn't the...” more I apologize, wouldn't the budging force be any number greater than the force of friction? Assuming this object is in a state of exact(underline) "equilibrium"(balance would be more accurate but less fun to say) wouldn't practically negligible variables be worth considering? Of course, seeing as my opinion runs contrary to the entirety of modern scientific convention, I'm most likely just missing something or misinterpreting a statement. Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Charles LaCour 2 years ago Posted 2 years ago. Direct link to Charles LaCour's post “I never liked the term "b...” more I never liked the term "budging force" either because you are correct that it takes a little more than the budging force to cause it to have a net force and start to move. The way you use it in a problem is an accurate representation of what physically happens so it is a mathematically useful concept even if the description doesn't make perfect sense. Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Kailas Cassidy 10 years ago Posted 10 years ago. Direct link to Kailas Cassidy's post “how were you able to assu...” more how were you able to assume that 1 N was the extra parallel force needed to budge the block Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer invinciblepsn 8 years ago Posted 8 years ago. Direct link to invinciblepsn's post “would rubbing two stones ...” more would rubbing two stones produce sparks even in vacuum? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Andrew M 8 years ago Posted 8 years ago. Direct link to Andrew M's post “Hmm, interesting question...” more Hmm, interesting question, albeit unrelated to this video. I think that the sparks are very hot, tiny bits of stone, and if that's the case, you would still see them in space. I could be wrong. I'm thinking that an alternate possibility is that the sparks arise because the hot bits of stone heat the air around them, sort of like a very tiny flame. If that's the case, then no sparks in space. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Abhilasha Jha 11 years ago Posted 11 years ago. Direct link to Abhilasha Jha's post “why can't the car acceler...” more why can't the car accelerate without friction? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Andrew M 11 years ago Posted 11 years ago. Direct link to Andrew M's post “Where is the force going ...” more Where is the force going to come from to accelerate it? Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript In the last video, we had a ten kilogram mass sitting on top of an inclined plane at a 30 degree angle And in order to figure out what would happen to this block we broke down the force of gravity on this block into the components that are parallel to the surface of the plane and perpendicular to the surface of the plane and for a perpendicular component, we got 49 times the square root of 3 N downwards That's 98 times this quantity over here, downwards But we said look! We don't see this block of ice accelerating downwards into this wedge because the wedge is supporting it So there must be a counteracting force that the wedge is exerting on the block And that counteracting force is the normal force of the wedge on the block of ice And that is exactly opposite to the force of gravity in this direction The normal force of gravity is the normal force of the wedge And these completely balance each other out in that normal, perpendicular direction And that is why this block is not accelerating either in that direction or in this direction over here But the one component of the force of gravity that did not seem to have any offset (at least the way we set up the problem in the last video) is the force/component that is parallel to the surface of the plane And we figured that out to be 49N, it was essentially the weight of the block times the sin of this angle And we said look, if there is no other forces then it would be accelerated in this direction And to figure out the rate of acceleration you take the force in that direction divided by the mass of the block and you would get 4.9 m/s^2 Now let's say that wasn't happening. Let's say that you were to look at this system right over here and the block was just stationary And now, for the sake of argument, let's assume it is not ice on ice Let's assume that they are both made out of wood And now all of a sudden we have a situation where the block is stationary If it is stationary, what is necessarily the case? Well we already determined that if it is not accelerating in this normal direction there must be zero net forces on it But if it is stationary as a whole then there must be zero net forces in the parallel component too So there must be some force counteracting this 49 N that wants to take it down the slope So there must be some force counteracting the component of gravity that wants to accelerate it down the slope And the question is what is this force? We're dealing with a situation now where we're dealing with a stationary block, a block that is not accelerating So what is that force? I think you know from experience maybe what is the difference between a block of wood on top of wood and a block of ice on top of ice A block of ice on top of ice is much more slippery; there is no friction between ice and ice, but there is friction between wood and wood To make it a little bit more tangible, maybe we'd put some sandpaper on the surface over here And then it becomes a little bit clearer. The force that is keeping this block from sliding down in this situation is the force of friction and the force of friction will always act in a direction opposite to the motion if there was not any friction or the potential acceleration if there was not any action So what is the force of friction in this case? Well, this block is completely stationary. It's not accelerating down the ramp The force of friction over here is going to be 49 N, upwards, up the ramp Now I want think about, this is something that can be determined experimentally as long as you have some way of measuring force, you can do this experimentally But the interesting question here is how much do I have to push on this block until it starts to move down the ramp? How much do I have to push on it? Let's say you are able to experimentally determine that if you can apply another 1 N on this then all of a sudden, I can at least get the box to start accelerating down Not the rate which it would do naturally, but I can just start to nudge it down if I give it another push of one N in the parallel direction So what is the total force--so exactly one N So the total force at this point that's acting on it in order to just start to budge it I'll call this the budging force Remember here isn't a traditional class F sub B for the budging force The budging force is in the parallel direction If I'm applying 1 N in this direction and it already has 49 N due to the component of gravity in this direction, then my budging force is 50 N And so an interesting thing that you can determine based on the materials that are coming in contact with each other is just how much force you need to just start to overcome friction In this case, it's the budging force. That's a term that I made up And the interesting ratio which tends to hold for given materials pretty well is the ratio between the amount of force just to budge it and the amount of force between the two objects between how much force they are exerting on each other And in this case, the amount of force that is being exerted by this, by the wedge on the block is the normal force, is 49 sqrt of 3 Maybe I should say the magnitude of the budging force over the magnitude of the force that is putting these two things in contact In this case, it is 49 square root of 3 N Over the magnitude of the normal force, and that is 49 sqrt of 3 We call this the coefficient of static friction We're gonna use this a little bit more deeply in other problems but it tends to hold true for different materials so that in the future if you have a different mass, you have a different incline but you have the same materials given the normal force, you can figure out the budging force You can figure out exactly how much force you need to put if you know this, which you usually figure out experimentally So what would be the value in this case? You have 50 N over 49 square root of 3 N Let's get the calculator out So I have 50 divided by 40 times the square root of 3 Gives me .72--I'll just round to two significant digits--0.72 This is 0.72 And then you can use this information. This is the coefficient of static friction We call it the coefficient of static friction because this deals with the ratio of the force of friction relative to the normal force I guess the force just to overcome the force of friction, just kind of get right over that the most friction that can be applied by kind of the abrasiveness of the 2 things when the object is stationary I'll do a whole video on how this is different when an object is stationary to when it's moving A lot of times, they're very very very close, but for certain materials you have a very--at least, a noticeably different coefficient of friction when the object is stationery as opposed to when it is moving So I'll leave you there. In the next few videos, we'll use coefficient of friction or calculate coefficient of friction to do some more problems Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video 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 Accept All Cookies Strictly Necessary Only Cookies Settings 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. 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2387	https://golfguidebook.com/golf-ball-size-guide/	Published Time: 2023-02-22T09:48:58+00:00 Golf Ball Size Guide: Diameter, Width, Volume, Circumference, Weight + Design Specifications Explored Skip to content Menu Menu Golf 101 Golf Rules Golf Games Golf Fitness Random and Fun Golf Lingo and Terms Golf History + Facts Travel Guides Golf Instruction Beginners Advice Advanced Golf Advice Golf Swing Improvement Troubleshooting Shots Short Game Putting Golf Shot Types Golf Gear Golf Club Guides Golf Club Advice Golf Gear Guides Pro Golfer WITB Golf Gear Reviews Golf Data + Stats Golf Handicap Data Golf Club Distance Charts Golf Ball Compression Charts Golf Swing Speed Charts News About Meet The Team Contact Us Menu Golf 101 Golf Rules Golf Games Golf Fitness Random and Fun Golf Lingo and Terms Golf History + Facts Travel Guides Golf Instruction Beginners Advice Advanced Golf Advice Golf Swing Improvement Troubleshooting Shots Short Game Putting Golf Shot Types Golf Gear Golf Club Guides Golf Club Advice Golf Gear Guides Pro Golfer WITB Golf Gear Reviews Golf Data + Stats Golf Handicap Data Golf Club Distance Charts Golf Ball Compression Charts Golf Swing Speed Charts News About Meet The Team Contact Us Golf Ball Size Guide: Diameter, Width, Volume, Circumference, Weight + Design Specifications Explored Tommy Devoy Last Updated: July 18, 2023 Golf Charts and Reference Data Golf balls are a strange part of the game. So crucial to every round of golf, but how much do you know about them? For some golfers, golf ball size and dimensions are inconsequential; they just take a ball out of their bag and play. For others, it is crucial, as many golfers often have a preferred type of golf ball. The reality is there is a lot more to it than meets the eye regarding golf ball size. There are several elements to the size of a golf ball in cm and in g and in this article, we are going to explore them. We will look at: Circumference of a golf ball Radius of a golf ball Volume of a golf ball Width of a golf ball Golf ball weight Hopefully, we will answer how big golf balls are and any more questions you have as to the dimensions of a golf ball. Importantly we will also look at how all this can affect your game. Let’s get started! What is the circumference of a golf ball? The circumference is the length of the perimeter of a circle, or the golf ball. An easy way to think of this is if you wrapped a piece of string entirely around the middle of the ball, how long would that piece of string be? Well, the answer for the circumference of a golf ball is 134.05mm, or 13.4cm, or 5.277 inches. What is the radius of a golf ball? If you remember learning math at school, you may remember that circumference and radius of a sphere (ball) are closely linked. But what exactly is the radius? The radius refers to the distance from the edge of the ball to the center. The radius of a golf ball is equal to 21.35mm, or 2.135cm, or 0.84 inches. Now that we know those two values, can you remember the math that links radius and circumference? What is the width of a golf ball? The width of a ball is sometimes referred to as the diameter and is simple to work out. Often when looking at the dimensions of a golf ball, the radius will be listed but not the diameter. However, the diameter is simply 2x the radius! So the width or diameter of a golf ball is 42.67mm, or 4.267cm, or 1.680 inches. What is the volume of a golf ball? Another dimension of a golf ball that is closely linked to golf ball size is the volume of a golf ball, i.e. the amount of ‘space’ it takes up. The volume of a golf ball is 40.68 cm3 (cubic centimeters) or 2.48 cubic inches. To put that into context, the volume of a can of coke is 330cm3. What Is The Weight Of A Golf Ball? When thinking about the size of something, its weight is often an aspect that comes to mind as a key question. So when answering the question of golf ball size, we should also consider its weight as well. How much does a golf ball weigh then? Well, unlike the other aspects of golf ball size, golf ball weight can vary. The upper limit of golf ball weight is 45.9g, but they can weigh less than that. However, most golf balls weigh no less than 45g. How Do Golf Ball Dimensions Affect Your Golf Game? So we have answered the critical questions of golf ball size and specifications. But the question is, how much does it affect your game? The answer is not massively as golf ball size is regulated and standardized by both USGA (United States Golf Association) and the R&A, the two major regulatory bodies in golf. However, don’t be fooled into thinking all golf balls are the same! There are considerable variations in golf ball design, and other design specifications can impact how a golf ball performs. So keep reading for a guide on other golf ball design specifications and how to find the right golf ball for you. Golf ball compression An aspect of a golf ball’s design specifications that affects how a golf ball performs is how much it compresses at impact. Golf ball compression is how much the ball squeezes when impacted by the clubhead. The softer the ball the more its shape changes at impact. The harder the ball the less it changes shape at impact. The reality is that golf ball size and dimensions change during the shot. Manufacturers rate golf ball compression on a scale from 30 to 120. 30 represents the softest golf balls, while 120 represents the hardest. Whether to go for a softer or harder ball depends on what you want out of your game. If you are someone who is looking toachieve greater distance on your shots, then a softer golf ball is the one for you. The increased compression from a softer ball allows more energy to be transferred from the clubface to the ball. This means golfers who have less energy in their swing, i.e. have lower swing speeds, can attain greater distances with a softer ball. Softer Balls = More Spin Another advantage of softer golf balls is that they gain more spin. This makes them useful for approach shots, and your short game, as the increased spin can give you the extra control you need to get the ball closer to the hole. So what’s the downside of using a softer golf ball? Softer balls are on the expensive side. Some high-end soft balls could set you back $60 for 12. Harder balls may not offer as much distance or spin but they are cheaper! So if you’re someone who loses a few balls a round, it may not be worth spending the money on a softer ball. For the most part, a golf ball in the mid-range of compression is usually a good option. They offer some of the benefits of a soft ball, at a more affordable price. When you are next weighing up which golf ball to buy, and thinking about golf ball size, remember compression is a big part of the performance. Golf ball dimples Surely if golf ball size is standardized then the number of dimples on a ball is standardized too? No, the number of dimples on a golf ball is not regulated. On average, a golf ball will have around 330-340 dimples, but one ball was made with 1700 dimples! But how do dimples affect a ball’s flight? Interestingly a smooth golf ball with no dimples would only travel half the distance of a standard golf ball. This is becausethe dimples create smooth air around the ball as it flies. This effect decreases air resistance and helps the ball fly further. Do more expensive balls have more or fewer dimples then? Not particularly. Most golf balls have several dimples in the standard range. Some golf balls aimed at beginners will have a slightly higher number of dimples, as this helps to achieve a higher ball flight. It is the design and pattern of the dimples on a ball that has the greatest effect on how a ball flies. The size, depth, and pattern of dimples can affect the ball’s velocity, spin, and trajectory after impact. There is no particular pattern of dimples to look out for when buying golf balls, as the differences are often very subtle. The thing to bear in mind is that the most expensive balls will have had the most research and design work done on them, so they will generally have the most optimized performance. Generally, however, only the most experienced golfers will be able to tell the difference. When it comes to choosing golf balls, find the ones that work best for you and stick to them. So that’s our guide to golf ball size. We have looked at golf ball size in cm and g while assessing the width, volume circumference, and weight of a standard golf ball. We have also explored some other aspects of golf ball design specifications such as compression and dimples. The number and choice of available golf balls can seem excessive and several golf ball manufacturers claim to have the best ball! The reality is the best ball for you is the type that you feel most comfortable with and the one that produces the best results. Or you could be a golfer who couldn’t care less about the type of golf ball you use, and just pull out a ball and get on with it. If you are looking to find a ball that suits you, try a few different types that vary in design specifications. After a few rounds, you will have found a ball that you feel comfortable using and that produces the best results. Once you’ve found your ideal ball, then stock up! As we all know every golfer loses a few golf balls now and again! If you are looking for a good guide that tells you which golf ball to use, check out the golf guidebook article below for a comprehensive, experts guide. What Golf Ball Should I Use? An Expert Guide Categories Golf Charts and Reference Data How To Increase Club Head Speed: The 6 Secrets To More Power 60 Funny Golf Team Names For Your Next Team Game Tommy Devoy Leave a Comment Cancel reply Comment Name Email Website Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. Tommy Devoy Here's my full profile Article First Published: February 22, 2023 Article Last Updated: July 18, 2023 Meet The Whole Team! Golf Guidebook was founded in 2021 and is run by a group of golf fanatics, coaches, and caddies who believe in sharing know-how and their love of the game. © 2025 Golf Guidebook +44 (0)1346 517061 Golf Guidebook \| Uncle Paulies Ltd, Main Street, Fraserburgh, UK, AB43 9RT Links Meet The Team Our Newsletter Affiliate Disclaimer Terms of UseContact UsSitemap Income disclosure: We are reader-supported and earn affiliate commissions when you buy via links found on golfguidebook.com.
2388	https://www.youtube.com/watch?v=RolB4X9nt8E	Art of Problem Solving: An Inequality Word Problem Art of Problem Solving 103000 subscribers 60 likes Description 20949 views Posted: 24 Dec 2011 Art of Problem Solving's Richard Rusczyk solves a word problem with inequalities. This video is part of our AoPS Prealgebra and Algebra curriculums. Take your math skills to the next level with our advanced materials: 📚 AoPS Prealgebra Textbook: 🖥️ AoPS Prealgebra 1 Course (Textbook Chapters 1-7): 🖥️ AoPS Prealgebra 2 Course (Textbook Chapters 8-15): 📚 AoPS Introduction to Algebra Textbook: 🖥️ AoPS Introduction to Algebra A Course: 🔔 Subscribe to our channel for more engaging math videos and updates Transcript: I don't make a whole lot of money making free videos. So I have to have a second job to make ends meet. I make blickets. At the end of each week, they give me these options for a pay plan. I go there and I have a whole lot of blickets that I've made, and they give me these options, and it's a real pain because I have to sit there and I have to calculate each one of the three options to figure out which one's the highest. I want a nice little rule, a nice little rule so I can look at how many blickets I have and not have to compute it each time, and know which option I should choose. That's what we're going to figure out here. We're going to figure out what rules I should use to choose from these three options each week. And we'll start by comparing them two at a time. So I'm going to compare Option X to Option Y. I'm going to see, when is Option X better than Option Y. Now, for Option X, I make $375, plus $9 for each blicket. And I'm going to let b stand for the number of blickets I make. And I'm going to see when that is better, is greater, than $200 plus 10b. $200 plus 10b is what I make for Option Y. Now, we're going to isolate the variable here, we're going to subtract 9b from both sides, subtract 200 from both sides. Subtract 200 from both sides, we're left with 175 on the left. Subtract 9b from both sides, we're left with a b on the right. And this tells us that if I make less than 175 blickets, then Option X is better for me than Option Y. Now let's compare Option X to Option Z. I'm going to see when X is better than Z. We already know that we make - that I make -- $375 plus 9b for Option X. And I want to see when that's better than Option Z, which is $450 plus $6 per blicket. So that's 6b. Once again, we're going to isolate the variable. I'm going to subtract 6b from both sides. I'm going to subtract 375 from both sides. Subtracting 6b will leave 3b over here. Subtracting 375 will leave 75 over there. And I can isolate the b by dividing both sides by 3, and I get b is greater than 25. So if b is greater than 25, then I should choose Option X. So, I see that when b is between 25 and 175, then X is better than both Y and Z. If it's less than 175, it's better than Y. If it's greater than 25, it's also better than Z. So Option X is the winner from 25 through 175. And at 175, it doesn't matter which I choose, X or Y. At 25, doesn't matter which I choose, X or Z. So 25 through 175, I can just pick X, and that'll be best for me. But I still have to compare Option Y and Option Z. So let's go ahead and do that. I want to see, when is Option Y better than Option Z? Well, Option Y is 200 plus 10b. Option Z, of course, is 450 plus 6b. And again, we just isolate b. We're going to subtract 6b from both sides, that will leave me 4b over here. Subtract 200, that'll leave me 250 over there. And then we divide by 4. Divide both sides by 4 to isolate b, and we get b is greater than, let's see, 4 goes into here 62 and 2/4, which is 62 and a half. So if I make more than 62 and a half blickets - I don't even know what a blicket is, so how can I tell what a half a blicket is? - but, if I make more than 62 and a half blickets, then Option Y is better. Now, I know that Option X is the best all the way up to 175, so I'm not going to pick Option Y until I get all the way up to 175. But everything from 175 on, everything greater than 175, Option Y is the winner. Option Z is better than Option Y when b is less than 62 and a half. Now, I know I'm not going to pick Option Z all the way down to 25. Anything from 25 up to 62 and a half, anything from 25 on, Option X is better. So, I want Option Z only when b is less than 25. So, everything up to 25, I'm going to choose Option Z. And now that I know which option to choose based on how many blickets I make, I need to go make some blickets.
2389	https://mathworld.wolfram.com/Tetrahedron.html	Tetrahedron Download Wolfram Notebook In general, a tetrahedron is a polyhedron with four sides. If all faces are congruent, the tetrahedron is known as an isosceles tetrahedron. If all faces are congruent to an equilateral triangle, then the tetrahedron is known as a regular tetrahedron (although the term "tetrahedron" without further qualification is often used to mean "regular tetrahedron"). A tetrahedron having a trihedron all of the face angles of which are right angles is known as a trirectangular tetrahedron. A general (not necessarily regular) tetrahedron, defined as a convex polyhedron consisting of four (not necessarily identical) triangular faces can be specified by its polyhedron vertices as , where , ..., 4. Then the volume of the tetrahedron is given by \| \| \| (1) \| Specifying the tetrahedron by the three polyhedron edge vectors , , and from a given polyhedron vertex, the volume is \| \| \| (2) \| If the edge between vertices and is of length , then the volume is given by the Cayley-Menger determinant \| \| \| (3) \| Consider an arbitrary tetrahedron with triangles , , , and . Let the areas of these triangles be , , , and , respectively, and denote the dihedral angle with respect to and for by . Then the four face areas are connected by \| \| \| (4) \| involving the six dihedral angles (Dostor 1905, pp. 252-293; Lee 1997). This is a generalization of the law of cosines to the tetrahedron. Furthermore, for any , \| \| \| (5) \| where is the length of the common edge of and (Lee 1997). Given a right-angled tetrahedron with one apex where all the edges meet orthogonally and where the face opposite this apex is denoted , then \| \| \| (6) \| This is a generalisation of Pythagoras's theorem which also applies to higher dimensional simplices (F. M. Jackson, pers. comm., Feb. 20, 2006). Let be the set of edges of a tetrahedron and the power set of . Write for the complement in of an element . Let be the set of triples such that span a face of the tetrahedron, and let be the set of , so that and . In , there are therefore three elements which are the pairs of opposite edges. Now define , which associates to an edge of length the quantity , , which associates to an element the product of for all , and , which associates to the sum of for all . Then the volume of a tetrahedron is given by \| \| \| (7) \| (P. Kaeser, pers. comm.). The analog of Gauss's circle problem can be asked for tetrahedra: how many lattice points lie within a tetrahedron centered at the origin with a given inradius (Lehmer 1940, Granville 1991, Xu and Yau 1992, Guy 1994). There are a number of interesting and unexpected theorems on the properties of general (i.e., not necessarily regular) tetrahedron (Altshiller-Court 1979). If a plane divides two opposite edges of a tetrahedron in a given ratio, then it divides the volume of the tetrahedron in the same ratio (Altshiller-Court 1979, p. 89). It follows that any plane passing through a bimedian of a tetrahedron bisects the volume of the tetrahedron (Altshiller-Court 1979, p. 90). Let the vertices of a tetrahedron be denoted , , , and , and denote the side lengths , , , , , and . Then if denotes the area of the triangle with sides of lengths , , and , the volume and circumradius of the tetrahedron are related by the beautiful formula \| \| \| (8) \| (Crelle 1821, p. 117; von Staudt 1860; Rouché and Comberousse 1922, pp. 568-576 and 643-664; Altshiller-Court 1979, p. 249). Let be the area of the spherical triangle formed by the th face of a tetrahedron circumscribed in a sphere of radius and let be the angle subtended by edge . Then \| \| \| (9) \| as shown by J.-P. Gua de Malves around 1740 or 1783 (Hopf 1940). The above formula provides the means to calculate the solid angle subtended from a vertex by the opposite face of a regular tetrahedron by substituting (the dihedral angle) into the above formula. Consequently, \| \| \| (10) \| \| (11) \| or approximately 0.55129 steradians. See also Disphenoid, Isosceles Tetrahedron, Regular Tetrahedron Explore this topic in the MathWorld classroom Explore with Wolfram\|Alpha More things to try: tetrahedron 1/4 (4 - 1/2) Dynamic References Altshiller-Court, N. "The Tetrahedron." Ch. 4 in Modern Pure Solid Geometry. New York: Chelsea, pp. 48-110 and 250, 1979.Balliccioni, A. Coordonnées barycentriques et géométrie. Claude Hermant, 1964.Couderc, P. and Balliccioni, A. Premier livre du tétraèdre à l'usage des éléves de première, de mathématiques, des candidats aux grandes écoles et à l'agrégation. Paris: Gauthier-Villars, 1935.Crelle, A. L. "Einige Bemerkungen über die dreiseitige Pyramide." Sammlung mathematischer Aufsätze u. Bemerkungen 1, 105-132, 1821.Dostor, G. Eléments de la théorie des déterminants, avec application à l'algèbre, la trigonométrie et la géométrie analytique dans le plan et l'espace, 2ème ed. Paris: Gauthier-Villars, pp. 252-293, 1905.Gardner, M. "Tetrahedrons." Ch. 19 in The Sixth Book of Mathematical Games from Scientific American. Chicago, IL: University of Chicago Press, pp. 183-194, 1984.Geometry Technologies. "Tetrahedron." A. "The Lattice Points of an -Dimensional Tetrahedron." Aequationes Math. 41, 234-241, 1991.Guy, R. K. "Gauß's Lattice Point Problem." §F1 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 240-241, 1994.Hopf, H. "Selected Chapters of Geometry." ETH Zürich lecture, pp. 1-2, 1940. J. R. "The Law of Cosines in a Tetrahedron." J. Korea Soc. Math. Ed. Ser. B: Pure Appl. Math. 4, 1-6, 1997.Lehmer, D. H. "The Lattice Points of an -Dimensional Tetrahedron." Duke Math. J. 7, 341-353, 1940.Rouché, E. and de Comberousse, C. Traité de Géométrie, nouv. éd., vol. 1: Géométrie plane. Paris: Gauthier-Villars, 1922.Rouché, E. and de Comberousse, C. Traité de Géométrie, nouv. éd., vol. 2: Géométrie dans l'espace. Paris: Gauthier-Villars, 1922.von Staudt, K. G. C. "Ueber einige geometrische Sätze." J. reine angew. Math. 57, 88-89, 1860.Xu, Y. and Yau, S. "A Sharp Estimate of the Number of Integral Points in a Tetrahedron." J. reine angew. Math. 423, 199-219, 1992. Referenced on Wolfram\|Alpha Tetrahedron Cite this as: Weisstein, Eric W. "Tetrahedron." From MathWorld--A Wolfram Resource. Subject classifications
2390	https://www.cliffsnotes.com/cliffs-questions/1941377	[Solved] Gaseous iodine pentafluoride, IF5 , can be prepared by the reaction of solid iodine and gaseous fluorine:... \| CliffsNotes Lit NotesStudy GuidesDocumentsQ&AAsk AIChat PDFLog InSign Up Literature NotesStudy GuidesDocumentsHomework QuestionsChat PDFLog InSign Up Gaseous iodine pentafluoride, IF5 , can be prepared by the reaction of solid iodine and gaseous fluorine:... Answered step-by-step AI Answer Available See Answer Questions & AnswersChemistry Gaseous iodine pentafluoride, IF5 , can be prepared by the reaction of solid iodine and gaseous fluorine:... Question Answered Asked by BaronKomodoDragonPerson355 Gaseous iodine pentafluoride, IF5 , can be prepared by the reaction of solid iodine and gaseous fluorine: I2(s)+5F2(g)→2IF5(g) A 5.20 −L flask containing 10.4 g I2 is charged with 10.4 g F2 , and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 ∘C . Part A Part complete What is the partial pressure of IF5 in the flask? P = 0.516 atm Previous Answers Completed Part B Part complete What is the mole fraction of IF5 in the flask? X = 0.543 Previous Answers Completed Part C Part complete Draw the Lewis structure of IF5 . Draw the Lewis dot structure for IF5 . Lone pair was added to the atom. A iodine atom with no charge and 1 lone pair and no bonds. Use arrow keys to change atom position on the canvas. Press space to add lone pair to the atom. Press ctrl+ right square bracket or ctrl+ left square bracket to navigate bonds, electrons and charge applied to the atom. Press tab to go through atoms in the structure. Press esc to quit editing this element. Press del to delete the atom from the canvas. Previous Answers Completed Part D What is the total mass of reactants and products in the flask? Answer & Explanation Solved by AI Get the explanation ------------------- Unlock access to this and over 10,000 step-by-step explanations See Answer Related Q&A Q Please see attachments for details A (1) A (2) D (3) C (4) B (5) B (6) C (7) E (8) A Solved by verified expertQ Please see attachments for details A The correct answers are 14. B) 15. C) 16. A) 17. D) 18. D) 120 o Solved by verified expertQ Se solicita dibujar a mano (tinta azul) un total de 2 compuestos ismeros bajo las siguientes instrucciones: a) Todos cumplan con la frmula qumica C16H32Br2 y al menos cuatro carbonos de esos dieciseis A Los dos isómeros son: Isómero A a) 3,9-dibromo-4,4,7,7-tetrametildodecano (cadena principal dodecano con dos grupos gem-dimetilo en C-4 y C-7; Br en C-3 y C-9) b) CH 3 CH 2 CH(Br)C(CH 3 )2CH 2 C Solved by verified expertQ The isotope ^126 I can decay by ec, B-, and B+ transitions . Can you draw the decay scheme and what kinds of radiation can one expect from a ^126 I source A The Decay Scheme for 126 I would show 126 I decaying to 126 Te via both Electron Capture (EC) and positron emission (beta-plus) and to 126 Xe via beta-minus decay. From a 126 I source, one can Solved by verified expertQ A Answer: Third option Solved by verified expertQ draw the correct bond line structure for the following compound (CH3CH2)2CCH2 A Bond-line structure: Solved by verified expertAll Questions 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. Quick Links Literature NotesStudy GuidesDocumentsHomework Questions Company About CliffsNotesContact us Do Not Sell or Share My Personal Info Legal Service TermsPrivacy policyCopyright, Community Guidelines & other legal resourcesHonor CodeDisclaimer CliffsNotes, a Learneo, Inc. business © Learneo, Inc. 2025 AI homework help Explanations instantly
2391	https://brainly.com/question/36353631	[FREE] A solution of H₃PO₄ has pKₐ₁ = 2.15, pKₐ₂ = 7.20, and pKₐ₃ = 12.35, with a concentration of 0.950 M. - brainly.com 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 +65,9k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +48k Ace exams faster, with practice that adapts to you Practice Worksheets +7k Guided help for every grade, topic or textbook Complete See more / Chemistry Expert-Verified Expert-Verified A solution of H₃PO₄ has pKₐ₁ = 2.15, pKₐ₂ = 7.20, and pKₐ₃ = 12.35, with a concentration of 0.950 M. Determine the pH of this solution. 1 See answer Explain with Learning Companion NEW Asked by pbrowns232 • 08/25/2023 0:01 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 4190392 people 4M 0.0 0 Upload your school material for a more relevant answer The pH of a 0.950M solution of H₃PO₄ is calculated to be approximately 0.02, based on its first pKa value. The question is asking for the pH of a solution of H₃PO₄. For a strong acid like H₃PO₄, we can assume it completely dissociates in water. Given its pKₐ values: pKₐ₁=2.15,pKₐ₂=7.20 and pKₐ₃=12.35, we consider only the first pKa value because the first dissociation will have the most significant effect on the pH of the solution. With a concentration (C) of 0.950M, the pH can be calculated by the formula pH = -log[C(H₃O⁺)] = -log(0.950). When you calculate the -log(0.950), the answer should be approximately 0.02. Therefore, the pH of the H₃PO₄ solution is predicted to be 0.02. Learn more about pH Calculation here: brainly.com/question/35353518 SPJ11 Answered by onuka •24.1K answers•4.2M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 4190392 people 4M 0.0 0 Upload your school material for a more relevant answer The pH of a 0.950 M solution of H₃PO₄ is approximately 0.02, calculated using the formula pH = -log[C(H₃O⁺)]. The first dissociation has the strongest impact on the pH due to the high concentration of hydrogen ions produced. As such, the solution is characterized as highly acidic. Explanation To find the pH of a 0.950 M solution of phosphoric acid (H₃PO₄), we need to consider its dissociation and the given pKₐ values: pKₐ₁ = 2.15, pKₐ₂ = 7.20, and pKₐ₃ = 12.35. H₃PO₄ is a weak acid that dissociates in three steps: H₃PO₄ ⇌ H⁺ + H₂PO₄⁻ (First dissociation, represented by pKₐ₁) H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ (Second dissociation, represented by pKₐ₂) HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ (Third dissociation, represented by pKₐ₃) In this case, the first dissociation has the most significant impact on the pH, as H₃PO₄ is relatively strong in its first dissociation but weak overall. For the first dissociation, we can use the formula for pH: pH = -log[H⁺] To calculate the concentration of hydronium ions [C(H₃O⁺)] after the first dissociation: Assuming complete dissociation for an initial concentration of 0.950 M, we can approximate: [C(H₃O⁺)] ≈ 0.950 M (after the first dissociation). Now we can substitute this into the pH formula: pH = -log(0.950) ≈ 0.02 (As this is relatively close to the value calculated with the pKₐ) Therefore, the pH of the H₃PO₄ solution is approximately 0.02. This shows that the solution is quite acidic due to the presence of strong hydrogen ion concentration from the first dissociation of the acid. Examples & Evidence For similar acidic solutions, you might compare with other strong acids like hydrochloric acid (HCl), where the pH can be calculated similarly with direct concentration inputs. For instance, a 1 M HCl solution has a pH of 0 due to complete dissociation. This information is based on well-established principles of acid-base chemistry, where the pH is determined from the concentration of hydrogen ions produced during dissociation, taking into account the strength of the acid and its dissociation constants. Thanks 0 0.0 (0 votes) Advertisement pbrowns232 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers 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. Community Answer 4.8 268 How did Mendeleev come up with the first periodic table of the elements? (1 point) A He determined the mass of atoms of each element. B He estimated the number of electrons in atoms of each element. C He arranged the elements by different properties to find a pattern. D He organized the elements by their atomic number. New questions in Chemistry How many molecules are in 0.400 moles of N O 3 ? What type of chemical bond holds C a 2+ and O 2− together in CaO? A. metallic bond B. covalent bond C. ionic bond D. hydrogen bond Ionic compounds are compounds that A. consist only of metallic elements. B. have atoms that transfer electrons. C. exist as individual ions. D. have atoms that share electrons. What is the term for a combination of two or more substances in any proportions in which the substances do not combine chemically to form a new substance? A. mixture B. compound C. molecule D. element What happens to the acidity of solutions A and B after dry ice is added? \| Acidity Changes after Dry Ice Is Added \| \| Time (sec) \| \| pH of Solution A \| pH of Solution B \| \| 0 \| \| 11.0 \| 8.5 \| \| \| \| 10.8 \| 8.0 \| \| \| \| 10.2 \| 7.2 \| \| \| \| 9.6 \| 6.5 \| \| \| \| 9.0 \| 5.8 \| A. The pH of solution A decreases, and the pH of solution B increases. B. The pH of solution A increases, and the pH of solution B decreases. C. The pH of solution A increases, and the pH of solution B increases. D. The pH of solution A decreases, and the pH of solution B decreases. 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2392	https://www.expii.com/t/molecular-formulas-overview-examples-8637	Expii Molecular Formulas — Overview & Examples - Expii A molecular formula consists of the specific elements that make up a compound and the quantities they are in. For example, ethane has the formula C(_2)H(_6). Chemistry Chemical Formulas Molecular Formulas — Overview & Examples A molecular formula consists of the specific elements that make up a compound and the quantities they are in. For example, ethane has the formula C2H6. Molecular Formulas — Overview & Examples Go to Topic Explanations (2) Cassie Gates Video 1 (Video) Molecular Formula By The Science Classroom Molecular formulas tell us the elements and the number of elements in compounds. It is a type of chemical formula. Octane has the molecular formula C8H18, which means there are eight carbons and eighteen hydrogens. The empirical formula describes the lowest ratio of elements in a compound. For octane, the empirical formula is C4H9. Empirical formulas can be used to determine the molecular formula since it is all based on ratios. A simple ratio example can be described with a burger. A burger is made up of one bun to one patty. The ratio is 1:1. If a triple-sized burger were made it would consist of three buns to three patties. This has a ratio of 3:3. A way to write this as a molecular formula would be B3P3 where the empirical formula is B1P1. Common chemistry problems will give you an empirical formula and either the molecular weight (amu) or the molar mass (grams/mole). Given these two pieces of information, you can determine the molecular formula. The steps to solve a molecular formula include: Find the empirical formula. Determine the empirical formula's molar mass. Divide the compound's molar mass by the empirical formula's molar mass. Multiply the empirical formula subscripts by the nearest whole number from step 3. In the given example the empirical formula is CH2O and the compound molar mass is 180 g/mol. The empirical molar mass is 30 g/mol. Dividing the molar mass by the empirical molar mass gives a value of 6. If we multiply our subscripts by six we get C6H12O6. This is the molecular formula and corresponds to the compound of glucose, a sugar. Report Share 1 Like Related Lessons Determining Molecular Formulas — Calculation & Examples Structural Formulas — Overview & Examples Determining Empirical Formulas — Calculation & Examples Molecular Models — Ball-and-Stick Model & Space-Filling Model View All Related Lessons Adrien Video 1 Like the secret Krabby Patty formula in Spongebob, a molecular formula is a kind of recipe that tells you the ratios of different ingredients, or elements, in a molecule. There are a few types of molecular formulas. An empirical formula describes the simplified ratio of elements in a molecule, regardless of how many atoms are actually found in that molecule. A structural formula shows the bond connections, instances of multiple bonds, and lone pairs of electrons. They are also called Lewis dot structures. The most common type of molecular formula, and what people usually mean when they say "molecular formula," just lists the exact number of atoms of each element present in a molecule. For example, C6H12O6 is the molecular formula for glucose or sugar. It tells us what elements are in this substance and how many atoms per molecule. CH2O is the empirical formula for the same molecule. It still tells us the elements' ratio but doesn't tell us how big a molecule is. Sometimes, the molecular and empirical formulas are the same. A great example is water (H2O). Water has two hydrogens and one oxygen. So, the formula is already irreducible. The same is true for any molecule with a 1:2 ratio. So, another example is carbon dioxide (CO2). It is also common in organic chemistry. An example is methane (CH4) gas. It's the gas that is commonly used to heat homes. Sometimes it's called natural gas. There are also compounds that have 1:1 ratios. For example, soon we'll learn about acids and bases. Many strong acids have 1:1 ratios. For example, halogen-based acids, like hydrochloric acid (HCl), all have 1:1 ratios. So, they all have one proton to donate when they dissolve in solution. Report Share 1 Like You've reached the end TOP How can we improve? General Bug Feature Send Feedback
2393	https://artofproblemsolving.com/wiki/index.php/Mass_points?srsltid=AfmBOoqBmrp2Vi7OWsJbNw2iRQnEBnrrUUXKwBvPPqQs8zI8qz7osVRW	Art of Problem Solving Mass points - 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 Mass points Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Mass points Mass points is a technique in Euclidean geometry that can greatly simplify the proofs of many theorems concerning polygons, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local coordinate system to identify points by the ratios into which they divide line segments. Mass points are generalized by barycentric coordinates. Mass point geometry was invented by Franz Mobius in 1827 along with his theory of homogeneous coordinates. The technique did not catch on until the 1960s when New York high school students made it popular. The technique greatly simplifies certain problems. Contents [hide] 1 Uses 2 Examples 3 Example 1 3.1 Solution 4 Example 2 4.1 Solution 5 Problems 6 Videos Uses Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversely proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever). The way to systematically assign weights to the points involves first choosing a point for the entire figure to balance around. From there, WLOG a first weight can be assigned. From the first weight, others can be derived using a few simple rules. Given a line with point on it, and the mass put on a point P is denoted as , If two points balance, the product of the mass and distance from a line of balance of one point will equal the product of the mass and distance from the same line of balance of the other point. In other words, . If two points are balanced, the point on the balancing line used to balance them has a mass equal to the sum of the masses of the two points. That is, . Examples Example 1 Consider a triangle with its three medians drawn, with the intersection points being corresponding to and respectively. Let the centroid of triangle be . Prove is and the corresponding identities for medians from and . Solution Thus, if we label point with a weight of , must also have a weight of since and are equidistant from . By the same process, we find must also have a weight of 1. Now, since and both have a weight of , must have a weight of (as is true for and ). Thus, if we label the centroid , we can deduce that is - the inverse ratio of their weights. Example 2 has point on , point on , and point on . , , and intersect at point . The ratio is and the ratio is . Find the ratio of Solution Throughout this solution, let denote the weight at point . Since , let , which makes . Now, look at . Since (this is a general property commonly used in many mass points problems, in fact it is the same property we used above to determine ), we have . Then, (another property of mass points). Finally, we have . Problems 2019 AMC 8 Problems/Problem 24 2016 AMC 10A Problems/Problem 19 2013 AMC 10B Problems/Problem 16 2004 AMC 10B Problems/Problem 20 2016 AMC 12A Problems/Problem 12 2009 AIME I Problems/Problem 5 2009 AIME I Problems/Problem 4 2003 AIME I Problems/Problem 15 2001 AIME I Problems/Problem 7 2011 AIME II Problems/Problem 4 1992 AIME Problems/Problem 14 1988 AIME Problems/Problem 12 1989 AIME Problems/Problem 15 1985 AIME Problems/Problem 6 1971 AHSME Problems/Problem 26 Videos The Central NC Math Group's lecture on Mass Points and Barycentric Coordinates Video from Double Donut This article is a stub. Help us out by expanding it. 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2394	https://www.hsa-haiku.org/	of America About the HSA How to Join Regions Officers Contests Young Members Meetings & Events Facebook Twitter Member Memorials Bylaws Society Frogpond Journal Member Anthologies HSA Newsletter Books Education Resources HSA Definitions Mentorship Program Haiku Bibliographies Haiku haiku senryu haibun renku rengay student haiku haiku book awards Frogpond awards Home Welcome to the Haiku Society of America web site! The Haiku Society of America maintains a safe space free of discrimination and free of bullying or harassment of any kind. We work together honoring our differences and celebrating our strengths, individual qualities, and contributions respectfully. We pledge to treat everyone with equality and respect at all times. On the Edge Haiku Society of America Virtual Conference November 21-23, 2025 Proposals due by August 15, 2025 Call for Proposals Submission period: June 15th – August 15th 2025. The Haiku Society of America is accepting proposals for the 2025 HSA Conference, November 21st -23rd, 2025. See complete details: 2025 National Virtual Conference 2025 Conference Committee: Crystal Simone Smith (HSA President), Yvette Nicole Kolodi (Conference Coordinator - SCHSG Moderator), Marjorie Pezzoli - (Program Coordinator - SCHSG Promotions), and Michael Dudley (HSA International Coordinator). ~ ~ ~ Haiku Society of America 2025 Rengay Award in Honor of Garry Gay Judged by Jonathan Roman & Agnes Eva Savich Judges Commentary & Honorable Mention Awards 2025 First Place sweat bees drowned migrants look up head down they shave his hair her grandpa's corrido crying is part of the joy ICE van idling outside the taquería a baker’s dozen huddled masses she knows it by heart they cross the river with the moon on their backs Anton R–kelian, San Diego, CA 1, 3 & 5 Orense Nicod, Paris, France 2, 4 & 6 2025 Second Place Maiden Voyage early morning mist the shipwright's hands pungent with oil a whisper of white oak curls from the blade the dinghy made of ancient Huon pine rigged for a sail tacking planks to the steam-bent ribs scent of red cedar a last coat of varnish warming twilight starboard side . . . polishing the fog bell for its maiden voyage Ron C. Moss, Tasmania, Australia 1, 3 & 5 Paula Sears, Exter, New Hampshire 2, 4 & 6 2025 Third Place Between Worlds deep in the marsh one watches, one weaves nest-making geese a grey heron pauses where the river bends soft as fog hard as seed cattail fluff clatter of white storks a courting dance ruffles the reeds half dead half living lightning struck pine empty duck blind where a hunter’s shadow used to fall John Thompson, Sonoma, CA 1,3 & 5 Neena Singh, Lucknow, Uttar Pradesh, India 2,4 & 6 ~ ~ ~ Haiku Society of America Student Haiku Awards in Memorial of Nicholas A. Virgilio Student Haiku Awards for 2025 The judges for the 2025 competition were Eavonka Ettinger and John Pappas. See the web page of comments from the judges. Download a free PDF copy of the award winning student haiku & senryu up to 2022 to share with teachers and students (does not include award haiku since 2022). Here are the winning haiku for 2025 (in alphabetical order): autumn leaves her last words heard in a voicemail Phoebe Bain – Grade 12, Colorado Springs, CO ~ blood moon keys between my knuckles like my mother before me Brynn Jensen – Grade 12, Colorado Springs, CO ~ grandma's kitchen mooncake softens in my mouth Frances McIlvoy – Grade 8, Atlanta, GA ~ autumn wind the scent of the bonfire on a borrowed hoodie Evelyn Reynolds – Grade 8, Atlanta, GA ~ war victory the names of the dead misspelled Tenzin Tinley – Grade 12, Colorado Springs, CO ~ empty hallway one shoe sideways after the argument Connor Wong – Grade 12, Newport Coast, CA ~ ~ ~ HSA Haiku Awards 2024 Judged by P.H. Fischer & Annette Makino Judges Commentary ~ First Place ~ blue skies— finding the nowhere I'd rather be Matthew Markworth, OH, USA ~ Second Place ~ first fireflies the little boy in my voice Frank Hooven, PA, USA ~ Third Place ~ spring equinox a canoe flipped upside up Matthew Markworth, OH, USA ~ ~ ~ HSA Senryu Awards 2024 Judged by Bill Cooper & Helen Ogden Judges Commentary First Place garden clean-up one invasive species roots out another Brad Bennett, MA, USA Second Place water park grand opening a toddler finds the puddle Matthew Markworth, OH, USA Third Place “on your left” the Progressive Bike Club out for a ride David Green, IL, US ~ ~ ~ HSA Haibun Awards 2024 Judged by Peter Newton & Barbara Sabol Judges Commentary First Place: by Dylan Stover, OH, USA heartwood for David "Woody" Stover It began with a young beech tree on a windy day, mid-spring. Acting upon an impulse I cannot now explain, I pressed my ear to the smooth, gray bark and started listening. To my surprise, there was sound: a secret inner creaking, like a stifled moan, as the crown twisted in the breeze. It was voice—tree voice. Each limb, as it swayed high in the canopy, was sending reverberations down through the acoustic body of the tree and into my ear as I crouched patiently at its side. hand at my chest the doctor suspects a murmur That’s when I became a listener of trees: I quickly discovered that smooth bark was best; the thicker the cork layer, the fainter the sound. Lithe ones were more melodious, aged ones more laconic, terse. But all speaking. Then one day I noticed a pileated woodpecker hammering away in the upper branches of an ash tree. Sneaking up to the bole, I put my ear against the bark: ta-tum ta-tum ta-a-TUM ta-tum… The wood trembled at each jab. Even the scrapes of the bird’s claws were amplified: every movement echoed inside me, as if the bird, the tree, and I were unified in a single, ringing vibration. a simple procedure to remove the weevil —then silence ~ ~ ~ Second Place: by Dian Duchin Reed, CA, USA What I'm Doing on My Summer Vacation My yard is not very big, but it’s big enough to shelter a million ants, keep the bees in business, and lure hummingbirds down to flirt with red flowers. Who knew that aphids came in a rainbow of colors? When I sit still, I might see a gopher pushing dirt out of its hole, then taking a break while its head soaks in the warmth of the sun. I’m learning the towhee’s cat alarm and the crows’ hawk taunts. I sometimes hear coyotes singing along with a passing siren. Did I mention the opossum babies who ride on their mother’s back at dawn? The skunk’s evening saunter? School’s about to start, and I haven’t even scratched the surface. The millipedes and Jerusalem crickets will have to wait till next summer. the mockingbird plays its whole repertoire endless afternoon ~ ~ ~ Third Place: by J Hahn Doleman, CA, USA True Places Never Are Lighter than a tuft of seafoam, yet tasting heavy as iron, this palmful of mycelium dust dissolves on my tongue as we trudge across an ancient caldera high above the territory of mapped consciousness. thinning air an obsidian cairn warm to the touch Under the retreating sun, still shining like a new doubloon, our breath becomes visible as we follow the lost footsteps of Klamath and Paiute. Traversing the first sequence of switchbacks, a jagged mountain scrimshaw leading to the summit, our boots awaken prehistoric ghosts beneath the crunch and scuttle of igneous rock. On one flank of the volcano, bleached skeletons of mountain hemlock blur into a Danse Macabre, their trunks swaying like mizzen masts in a typhoon. A whale-sized andesite obelisk crests above us, spouting luminescent fractals from an invisible blowhole. the universe still expanding rhyolite fragments This white-haired alp appears to slumber as gusts of wind from an invisible, eternal Victrola play across its skin, spinning out records of our past. We test our sea legs on the glacier, a frozen brig drifting within its own concept of time, as meltwater runoff rushes its way to the ocean and a world we will never quite fathom. open crevasse a Pandora moth enshrined in ice ~ ~ ~ ~ ~ ~ Haiku Society of America Merit Books Awards 2025 for books published in 2024 Patricia J. Machmiller and Scott Mason, judges Judges' comments will be added to the web site after they are published in the Autumn issue of Frogpond. First Place Francine Banwarth. Bare Necessities: Selected Haiku of Francine Banwarth. Taylorville, IL; Brooks Books, 2024 Second Place Jennifer Hambrick. A Silence or Two. Winchester, VA; Red Moon Press, 2024 Third Place Debbie Strange. Random Blue Sparks. Ormskirk, Great Britain; Snapshot Press., 2024 Honorable Mentions (not ranked but in alphabetical order by author) Roberta Beary. Carousel. Ormskirk, Great Britain; Snapshot Press., 2024 Deborah P Kolodji. Vital Signs. Cuttlefish Books, 2024 paul m. Magnolia Diary. Champaign, IL; Modern Haiku Press, 2024 Peter Yovu. Shine Shadow. Winchester, VA; Red Moon Press, 2024 ~ ~ ~ HAIKU ANTHOLOGY AWARDS: Best Janice Doppler, editor. One Thread: Zoka in Contemporary Haiku. Massachusetts; self-published, 2024 Honorable Mention Susan Antolin, Garry Gay, and Carolyn Hall, editors. The San Francisco Haiku Anthology, Volume Two. Spare Poems Press 2024 ~ ~ ~ HAIBUN BOOK AWARDS: Best Bob Lucky. My Wife & Other Adventures. Winchester, VA; Red Moon Press, 2024 Honorable Mentions Joe McKeon. A Man on Horseback. Winchester, VA; Red Moon Press, 2024 ~ ~ ~ Nicholas A. Virgilio Memorial Haiku and Senryu Competition Anthology edited by Randy M. Brooks designed by Ignatius Fay © 2022 HAIKU Society of America To commemorate the 30th Anniversary of the Nicholas A. Virgilio Memorial Haiku and Senryu Competition, the executive committee of the Haiku Society of America published this anthology of award-winning haiku and senryu. The student observations, insights, experiences, emotions and insights evident in these haiku and senryu are a wonderful testament to the fresh voices and vivid imagery of young people. We believe the judges’ commentaries add a valuable layer of meaning as we see how leaders, editors, writers and members of the Haiku Society of America carefully consider the significance of each award-winning poem. Download your PDF copy for a teacher in your area. ~ ~ ~ It's time to renew your membership or join the Haiku Society of America for 2025. 2025 Membership Link The Haiku Society of America is a not-for-profit organization founded in 1968 to promote the writing and appreciation of haiku in English. The HSA has been meeting regularly since its inception and sponsors meetings, readings, publications and contests. The HSA has over 1000 members around the country and overseas. Membership is open to all readers, writers, and students of haiku. Join today. ~ ~ ~ HSA Newsletter Submission Guidelines We welcome our the newly elected editor for the Haiku Society of America Newsletter: Evan Vandermeer Please use the Google Form for submissions (more specific guidelines are on the form). Google Form for HSA News Submission ~ ~ ~ See the Latest Issue of Frogpond: Frogpond 48.2 • 2025 ~ ~ ~ HSA MEMBERS ANTHOLOGY 2024 The Haiku Society of America published the 2024 HSA Members' Anthology. Huddleston, Edward Cody, Editor. Hauling the Tide: Haiku Society of America Members’ Anthology 2024. New York; Haiku Society of America, 2024. Book design by Tanya McDonald. All members of HSA receive a copy of the annual members' anthology. ~ ~ ~ Events & Gatherings of Interest for HSA Members We want to feature events, conferences and gatherings of interest to members of the Haiku Society of America. Please follow this new feature on our meetings page. brooksbooks@gmail.com Specific details about these events change frequently, but these links should help you seek current information. Some of these links are to organizations with several events. We invite HSA members and haiku event planners to send us information about additional events and conferences. Please email URLs or updated information to the HSA webmaster at: brooksbooks@gmail.com Ancestral Haiku The Alluvian World Music Series Traverse City, MI 05/17/2025 Atlanta Haiku Festival Atlanta, GA 04/19/2025 Australian Haiku Society British Haiku Society The Confluence Haiku Writing Workshop Calgary, Canada 02/16/2025 Cradle of American Haiku Edinburgh Haiku Circle Gadigal Ginko Wareamah, Cockatoo Island Sydney Harbour, Australia 02/26/2025 Golden Triangle Haiku Workshop 01/15/2025 Haiku Canada Weekend Lennoxville, Canada 05/16 - 18, 2025 Haiku Down Under The Haiku Foundation's Events 04/10 - HaikuLife Film Festival 04/17 - International Haiku Poetry Day Haiku-la-Vier - Berlin, Germany 02/23/2025 Haiku North America 09/24 - 28, 2025 - San Francisco, CA Haiku Northwest 06/21/2025 Haiku Poets of Northern California 2024 events Haiku Poets of Northern California Two Autumns Reading 08/17/2025 Haiku Society of America meetings Haiku Society of America - Southeast Region Ekphrastic Haiku - A Workshop 04/26/2025 Haiku Society of America - Southwest Region Writing Arizona - an Immersive Workshop 09/11 - 09/14/2025 Haiku Walk for World Poetry Day Cladagh Glen, Northern Ireland 03/16/2025 Hailstones Haiku Circle Hyper Japan Festival London, UK 07/18-20, 2025 Japan Fair 2025, Bellevue, WA 07/12-13, 2025 Japan America Society of Washington DC Wild Apricot Haiku Group Kukai 01/18/2025 John P. Humers Japanese Stroll Garden Japanese Tales, Haiku and Refelction Locust Valley, NY 07/12/2025 Kaji Aso Studio Events Boston, MA Kukai Haiku Exchange Japan Culture and Information Center Brussels, Belgium 02/21/2025 New Zealand Poetry Society HaikuNewz Nick Virgilio Haiku Association events North Carolina Haiku Society Panorama International Literature Festival Karnataka, India 01/01 - 30, 2025 Peregrine Haiku Society Workshop Cincinnati, OH 08/14/2025 Renga: Japanese Collaborative Poetry Morikami Museum and Japanese Gardens Palm Beach, FL 04/05/2025 Seabeck Haiku Getaway 10/23-26, 2025 Seattle Cherry Blossom Festival 04/11-13, 2025 Tanka Society of America Triveni Haikai Upaya Zen Centre - Online Haiku Workshop Santa Fe, NM 10/21 - 23, 2025 Vancouver Cherry Blossom Festival Haiku Invitational Contest Vancouver, Canada 03/01 - 06/01, 2025 Wild Graces Haiku Gathering Deerfield, NH 08/23, 9am-5pm, 2025 Woodend Haiku Festival Woodend, Australia April 2025 World Haiku Association Writer's Journey: Haiku Walking in Japan Tokyo to Kyoto, 12/03 - 12/08 Yukei Teikei Haiku Society 2025 Events ~ ~ ~ Haiku Society of America Member Memorials HSA Member Memorials Haiku Society of America often features short memorials of members who have recently died. Usually these memorials are provided by HSA Regional Coordinators and featured in the Haiku Society of America Newsletter. We want to honor these HSA members who have died on this archive of HSA Member Memorials. This is a new feature on the HSA web site. We will include brief memorials of HSA members. If you would like to send information about an HSA member who has recently died please send complete information to your HSA Regional Coordinator. ~ ~ ~ Check out our Frogpond history: the Archive of Frogpond Journal Archive of Frogpond Journal (all issues from 1978-2020) Haiku Society of America is pleased to provide access to PDF copies of back issues of Frogpond. This includes all but the most recent issues published in the last two years. ~ ~ ~ ~ ~ ~ Support HSA with a Donation The Haiku Society of America is a not-for-profit organization that is dependent on membership dues and much appreciated donations. 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2395	https://www.thoughtco.com/converting-millimeters-to-centimeters-609313	Converting Millimeters to Centimeters Example Problem Skip to content Menu Home Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Search Close Search the site GO Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Contact Us Editorial Guidelines Privacy Policy Science, Tech, Math› Science› Chemistry› Converting Millimeters to Centimeters Worked unit conversion example problem Print Larry Washburn/Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. Learn about ourEditorial Process Updated on February 06, 2019 Close This example problem demonstrates how to convert millimeters to centimeters, which is actually much easier than you may have thought. The Problem Express 312 millimeters in centimeters. In order to calculate this problem, you'll need to know how many millimeters make up a single centimeter. The following is the equation for centimeters to millimeters: 1 centimeter = 10 millimeters From here, you can set up the conversion so the desired unit will be canceled out. In this case, we want cm to be the remaining unit. distance in cm = (distance in mm) x (1 cm/10 cm) distance in cm = (312/10) cm distance in cm = 3.12 cm The Answer 312 millimeters is 3.12 centimeters. Cite this Article Format mlaapachicago Your Citation Helmenstine, Anne Marie, Ph.D. "Converting Millimeters to Centimeters." ThoughtCo, Jun. 25, 2024, thoughtco.com/converting-millimeters-to-centimeters-609313.Helmenstine, Anne Marie, Ph.D. (2024, June 25). Converting Millimeters to Centimeters. Retrieved from Helmenstine, Anne Marie, Ph.D. "Converting Millimeters to Centimeters." ThoughtCo. (accessed September 29, 2025). copy citation Sponsored Stories After a storm, aviation crews help get the lights back on Duke Energy \| illumination Jeff Bezos Says the 1-Hour Rule Makes Him Smarter. I Gave It A Go.Blinkist Magazine North Charleston New Policy for Senior Drivers thequotegeneral.com Why everyone is quitting Gmail?Proton Converting Millimeters to Meters Example Problem Converting Centimeters to Meters (cm to m) Converting Cubic Centimeters to Liters Converting Cubic Feet to Liters Meter Definition and Unit Conversions Converting Cubic Inches to Cubic Centimeters A List of Common General Chemistry Problems Metric Unit Prefixes Sponsored Stories Discover Top Internet Providers for Seniors HighSpeedInternet.com Get the same tax-smart tech as the elite Betterment Senior Singles Want to Chat – Are You In?fiestadates.com What Really Happens If You Drink Bottled Water Daily Water Health Experts Converting Cubic Meters to Liters mbar to atm - Converting Millibars to Atmospheres How to Calculate the Rate of Metal Corrosion How to Convert Feet to Meters Converting Kilometers to Meters How to Convert Angstroms to Nanometers Converting Angstroms to Meters Converting Cubic Inches to Liters ThoughtCo Follow Us Science, Tech, Math Humanities Languages Resources About Us Advertise Careers Privacy Policy Editorial Guidelines Contact Terms of Service Your Privacy Choices ThoughtCo is part of the People Inc.publishing family. By clicking “Accept All Cookies”, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts. Cookies Settings Accept All Cookies
2396	https://www.smartick.com/blog/mathematics/fractions/singapore-bar-model-percentages/	Try it for free! Try it for free! Accelerate your child’s learning Smartick is a fun way to learn math! Start your 7-day free trial! Nov20 Singapore Bar Model and Percentages In today’s post, we’re going to look at percentagesthrough the Singapore bar model. To define a part of a whole, we can use a fraction, but we can also use a percentage. The whole divides into 100 equal parts, therefore each part is 1/100 of the whole or 1%. Let’s look at an example of percentages with the Singapore bar Model: There are 400 people at a football stadium, and 60% of them are women. How many women are there in total? In the Singapore bar model, the complete bar represents 100%, therefore the bar is 100 units. The percentage of 60% means that there are 60 units out of 100. Let’s take a look at how we can work out 60% of 400 We know that in this case, 100 units represents 400 -> 100 units = 400 Therefore, to find out 1 unit -> 1 unit = 400/100 = 4 Now we know that 1 unit is 4 y, and we want to know what 60 units represent -> 60 x 4 = 240 60% of 400 is 240. Therefore, the number of women at the football stadium in total is 240 Now, using the same example, we’re going to see how to find the whole when we have just one part: There are 240 women at a football stadium, which is 60% of the total number of people there. How many people are there in total? We know that 60 units are 240 -> 60 units = 240 Therefore 1 unit is -> 240/60 = 4 We know that 1 unit is 4 and we want to know what 100 units represent-> 100 x 4 = 400 The total number of people at the football stadium is 400 Let’s look at an example of a comparison problem: Sam has 70 euros and Tom has 30% more money than Sam. How much money does Tom have? In this example, the bar representing Sam’s money acts as a base, 100%, because Tom’s money is based on the relation to the money that Sam has. In this case, Tom has 30% more money than Sam. Therefore, the amount of money Tom has is 130% of the money Sam has. The percentage 130% is 130 units out of 100, and 100 units are equivalent to the €70 that Sam has. 100 units = €70 1 unit = 70/100 = €0.70 130 units = 130 x 0.70 = €91 Therefore, Tom has €91 Let’s take a look at another example of comparison: Julie has 50 books in her library, and Amy has 20% fewer books than Julie. How many books does Amy have in her library? We know how many books Julie has, so we can use that amount as a base. Therefore, 50 books are 100%. Amy has 20% fewer books than Julie, therefore, Amy has 80% of the books that Julie has. 100 units = 50 books 1 unit = 50/100 = 0.5 80 units = 80 x 0.5 = 40 Amy has 40 books in her library I hope you’ve enjoyed learning about the concept of percentages and how to solve them. If you want to keep practicing, log in to Smartick. Learn More: Using Singapore Bars to Help Solve Problems II Rule of Three for Calculating Percentages Using Singapore Bars to Solve Algebraic Equations Let’s Learn about Ratios with Singaporean Bar Models Singapore Bars Applied to Fractions Fun is our brain’s favorite way of learning Diane Ackerman Smartick is a fun way to learn math 15 fun minutes a day Adapts to your child’s level Millions of students since 2009 Start your 7-day free trial Author Recent Posts Smartick Content Creation Team.A multidisciplinary and multicultural team made up of mathematicians, teachers, professors and other education professionals! They strive to create the best math content possible. Latest posts by Smartick (see all) Attention Games: The Key to Fostering Sustained Concentration in Children - 01/31/2025 Mathematical Formulas: What Are They, How Are They Made and Types of Formulas? - 11/29/2024 The Language of Functions and Graphs - 07/01/2024 Add a new public comment to the blog: Cancel reply The comments that you write here are moderated and can be seen by other users.For private inquiries please write to [email protected]
2397	https://math.arizona.edu/~sethuram/588/lecture1.pdf	LECTURE 1: PRELIMINARIES AND HYDRODYNAMICS OF INDEPENDENT RANDOM WALKS Before discussing a basic example, illustrating possibilities in the study of ‘hy-drodynamics of stochastic particle systems’, we recall some basic notions in Markov chains. 1. Markov chains 1.1. Construction. A family of random variables {Xn : n ≥0} taking values on a countable state space E is called a ‘discrete time Markov chain’ if the ‘stationary Markov property’ is satisﬁed: P(Xn = xn\|X0 = x0, . . . , Xm = xm) = P(Xn = xn\|Xm = xm) = P(Xn−m = xn\|X0 = xm) for all x0, . . . , xm ∈E and n > m ≥0. When n = 1 and m = 0, the last quantity on the right-side represents a ‘transition probability’ of the Markov chain, denoted p(x, y) = P(X1 = y\|X0 = x). We now construct a ‘continuous time Markov chain’ on E with ‘skeleton’ {Xn : n ≥0} and transition probability vanishing on the diagonal, that is p(x, x) = 0 for all x ∈E. Let {λx : x ∈E} be a collection of positive numbers, and let {Wn : n ≥0} be a collection of independent identically distributed exponential random variables with rate 1, independent in particular the skeleton discrete time chain. Deﬁne now the process {Zt : t ≥0} as follows: Initially, Z0 = X0 ∈E. After time λ−1 X0W0, the process jumps to value X1, and after a subsequent time λ−1 X1W1, the process jumps to value X2, and so on. Let Tk = Pk i=0 λ−1 Xi Wi for k ≥0. Then, Zt =          x for 0 ≤t < T0 X1 for T0 ≤t < T1 . . . . . . Xn for Tn−1 ≤t < Tn for 0 ≤t < T∞= limn↑∞Tn. Suﬃcient conditions for T∞= ∞include the cases if the state space E is ﬁnite, or if supx∈E λx < ∞. We will assume from now on that the process is ‘regular’, that is T∞= ∞, so that the process Zt is deﬁned for all time t ≥0. One can show that {Zt : t ≥0} satisﬁes the stationary Markov property, which in this context is equivalent to P(Zt = y\|Zt0 = x0, . . . , Ztm = xm, Zs = x) = P(Zt = y\|Zs = x) = P(Zt−s = y\|Z0 = x) (1.1) for all x, y, x0, . . . , xm ∈E, 0 ≤t0 < · · · tm < s < t and m ≥0. Conversely, given a process {Zt : t ≥0} satisfying (1.1) and also the ‘jump property’ that there exists a sequence of strictly increasing stopping times {Tn : n ≥0} such that T0 > 0 = T−1 and Zt is constant on intervals [Tn, Tn+1) and ZT − n ̸= ZTn for n ≥0, one can determine a unique skeleton discrete time Markov 1 2 LECTURE 1 chain {Xn : n ≥0} and positive jump parameters {λx : x ∈E} such that Xn = ZTn, p(x, y) = P(ZTn+1 = y\|ZTn = x), and Tn −Tn−1 are independent exponentials with rates λXn for n ≥0. Chains with the same skeleton and jump parameters have the same joint distributions. The condition ZT − n ̸= ZTn ensures p vanishes on the diagonal. 1.2. Generators, Chapman-Komogorov equations. In the discrete time set-ting, we can deﬁne the transition operator P = (p(x, y) : x, y ∈E), which is a matrix when E is ﬁnite. Then, the nth powers give the nth step probabilities, P n(x, y) = P(Xn = y\|X0 = x). Computing the t-time probabilities for the continuous time Markov chain Zt is more complicated. Deﬁne the transition probability Pt(x, y) = P(Zt = y\|Z0 = x). Then, by the Markov property we have Pt+s(x, y) = X z∈E Pt(x, z)Ps(z, y) or in terms of operators Pt+s = PtPs. Given regularity of the process, the transition functions are diﬀerentiable in time, and satisfy the backward equation d dtPt(x, y) = X z∈E λxp(x, z)[Pt(z, y) −Pt(x, y)] P0(x, y) = δx,y. Similarly, one has the forward equation d dtPt(x, y) = X z∈E Pt(x, z)λzp(z, y) −Pt(x, y)λy. Integral versions of both the backward and forward equations can be rigorously derived from decomposing the transition probability on ﬁrst and last jump times respectively, or analytically. Exercise 1.1. Review this deriviation. Deﬁne the operator L(x, y) = λxp(x, y) for y ̸= x −λx for y = x. Then, compactly expressed, the backward and forward equations become d dtPt = LPt and d dtPt = PtL. Also, limt↓0 t−1[Pt −I] = L and Pt(x, y) = δx,y + tL(x, y) + o(t). When the space E is ﬁnite, L is a ‘generator’ matrix, that is L(x, y) ≥0 for x ̸= y and L(x, x) = −P y̸=x L(x, y), and by solving the ODE’s, one obtains Pt = etL which can be computed in some cases. More generally, the ‘Trotter-Kato’ formula holds Pt = limn↑∞(I + t nL)n. Let f : E →R be a bounded function on the state space. In the discrete time case, deﬁne Pf(x) = P y∈E p(x, y)f(y), which is the conditional expectation of LECTURE 1 3 f(X1) given X0 = x. Then, P nf(x) = P y∈E p(n)(x, y)f(y) is the conditional expec-tation of f(Xn) given X0 = x, where p(n)(x, y) is the n-fold convolution, or nth step transition probability. In the continuous case, deﬁne Ptf(x) = P y∈E Pt(x, y)f(y) which is the conditional distribution of f(Zt) given Z0 = x. In this framework, the generator L is often expressed in terms of its action on f: (Lf)(x) = X y∈E L(x, y)[f(y) −f(x)] = X y∈E λxp(x, y)[f(y) −f(x)]. 1.3. Invariant measures. Let µ be a probability measure on E. In the discrete time situation, deﬁne µP(x) = P y∈E µ(y)p(y, x). Hence, we see that µP n is the distribution at time n when the initial state is distributed according to µ. In the continuous model, deﬁne µPt(x) = X y∈E µ(y)Pt(y, x), and µL(x) = X y∈E µ(y)L(y, x). We say that µ is an ‘invariant measure’ for discrete time chains if µP = µ, and for continuous time chains if µPt = µ for all t ≥0. We also say that µ is a ‘reversible’ invariant measure in discrete time chains if µ(x)p(x, y) = µ(y)p(y, x) for all x, y ∈E. In continuous time chains, µ is ‘reversible’ when Pt is self-adjoint in L2(µ), that is P x∈E µ(x)f(x)Pt(x, y)g(y) = P x∈E µ(x)g(x)Pt(x, y)f(y), or in terms of the inner product on L2(µ), ⟨f, Ptg⟩µ = ⟨Ptf, g⟩µ for all f, g ∈L2(µ). One can verify that in the continuous time setting that µ is invariant is equivalent to µL = 0, and µ is reversible is equivalent to µ(x)L(x, y) = µ(y)L(y, x) for all x, y ∈E, or in terms of inner products ⟨f, Lg⟩µ = ⟨Lf, g⟩µ for all f, g ∈L2(µ). Exercise 1.2. Review these equivalences. Hint: The Trotter-Kano formula is useful in this exercise. In the ﬁnite state space case, invariant measures always exist, and if the skeleton chain can reach every state from any state in ﬁnite time, that is p is irreducible, the invariant measure is unique. However, in the countable state case there may be no invariant measures. Reversibility has the following interesting implication. Fix a time t > 0, and consider Rs = Zt−s for 0 ≤s ≤t. Suppose that initially Z0 is distributed according to an invariant measure µ. Then, it can be seen that Rs is a continuous time Markov chain with transition probability Qt(x, y) = (µ(y)/µ(x))Pt(y, x). In particular, when µ is reversible, Qt(x, y) = Pt(x, y) and in this case the ‘forward in time’ and ‘backward in time’ chains have the same distribution! We remark that it is sometimes easier to ﬁnd directly a reversible measure and therefore an invariant measure by checking the reversibility conditions. 1.4. Examples. We will content ourselves for the moment with two basic contin-uous time examples, the two-state Markov chain, and random walk. Example 1.3. Let E = {0, 1} correspond to states ‘on’ and ‘oﬀ’, or sometimes ‘empty’ and ‘occupied’. Here, state 0 can transition to state 1 and vice versa. Let 4 LECTURE 1 λ0 and λ1 be the corresponding jump rates. The skeleton chain probabilities are p(0, 1) = p(1, 0) = 1. The generator matrix is L = −λ0 λ0 λ1 −λ1 . and correspondingly, it is an exercise in diagonalization to compute Pt = etL and to ﬁnd the unique invariant measure. Exercise 1.4. Compute the invariant measure and Pt. Example 1.5. Let E = Td N the d-dimensional torus where TN = Z \ NZ, or integers modulo N. Let also λx ≡1, and p be a ﬁnite-range, translation-invariant transition probability: For all x, y ∈E, p(x, y) = 0 if \|x −y\| ≥R some R < ∞, and p(x, y) = p(0, y −x) =: p(y −x). We will also assume that p is irreducible. For instance, the nearest-neighbor, symmetric case is one possibility. Then, L(x, y) = p(x, y) for x ̸= y and L(x, x) = −P y̸=x L(x, y). In particular, the uniform distribution µ(x) ≡N −1 is the unique invariant measure. Moreover, µ is reversible exactly when p(x) = p(−x) for all x ∈E. Now let Rt be the number of jumps before time t. Since the jump rates are all 1, we see that Rt is a Poisson process with rate 1. In particular, the transition probability Pt(y −x) := Pt(x, y) = E p(Rt)(x, y) = X n≥0 e−1 n! p(n)(x, y). We now consider a sequence of chains {Z(N) t : t ≥0}N≥1 on a sequence of torii. Deﬁne m = P x∈E xp(x) be the mean displacement of the position. Then, it is not diﬃcult to extablish the following law of large numbers, lim N↑∞ Z(N) Nt N = m in probability. Here, m ∈Td whre T is the unit circle. Exercise 1.6. Show this LLN by say variance computations. Noting R(N) t is Poisson with variance t is useful. One can do it also by regeneration, and other methods. Also, if m = 0, let σ be the matrix of covariances, σi,j = P x xixjp(x) for 0 ≤i, j ≤d. We have the central limit theorem, Z(N) N 2t N ⇒N(0, σt). Exercise 1.7. There are a few ways to show this CLT. One way is to write Z(N) N 2tN = Z(N) T R(N) N2t q TR(N) N2t · q TR(N) N2t N and to use a random index CLT. For instance, in this case, R(N) t is independent of the displacements. LECTURE 1 5 Finally, when m = P x xp(x) ̸= 0, we say the walk is asymmetric, and when m = 0 and p(·) is not symmetric, we say the walk is mean-zero asymmetric, and when p(·) is symmetric, we say the walk is symmetric. 2. Hydrodynamics of independent random walks We would like to understand the space-time evolution of the mass in a system of particles with a conservation law. Perhaps the simplest model is that of non-interacting random walks on a d-dimensional torus with N locations. When N is large, and one looks at the system from afar, after long times, one can more discern the motion of the bulk of the mass rather than individual components. The goal is to make precise the evolution of the mass in this scale in terms of a continuum equation. We will be working with a Markov chain on E = NTd N , where N = {0, 1, 2, . . .}, which govern the motion of K independent random walks on Td N as in Example 1.5. Since we are interested in the ‘mass’ of particles, we will consider the occupation numbers at each location on the lattice Td N. That is, let Zi t be the position of the ith particle at time t. Deﬁne ηt(x) = K X i=1 1(Zi t = x). We now observe that the process ηt = {ηt(x) : x ∈Td N} is a Markov chain. Indeed, given independence and the Markov property of the individual particle movement, by splitting over all possibilities, the Markov property of ηt can be deduced. 2.1. Associated invariant measures. What are the invariant measures for the process? Since the process ηt, corresponding to K particles, is irreducible, there is a unique invariant measure. It is not so easy to characterize it immediately. We will come back to this question later. In fact, the following analysis will be easier if we relax the assumption there are K particles in the system. If we do not specify the initial number of particles, then ηt is no longer irreducible, since there is no birth or death present: For instance, a system with 10 particles cannot evolve into one with 20 random walks. However, we can more easily specify in nice form several invariant measures for this ‘relaxed’ system. Recall the Poisson distribution with parameter α, qα(k) = e−ααk/k! for k ≥0. Its moment generating function is given by X k≥0 eλke−α αk k! = eα(eλ−1). For a positive function ρ : Td →R+, deﬁne the product measure νN ρ(·) on NTd N by νN ρ(·)(η(x) = k) = qρ(x/N)(k). When ρ(·) ≡α is constant, we denote νN ρ(·) = νN α . The process {ηt : t ≥0} belongs to the space of right-continuous paths with left limits in E = NTd N , D([0, ∞); NTd N ). We will denote by Pµ and Eµ the probability measure and expectation with respect to the evolution of the process when initially 6 LECTURE 1 η0 is distributed according to µ. On the other hand, Eµ will refer to the expectation with respect to µ on E. Proposition 2.1. The measures νN α are invariant for the Markov chain ηt. Proof. We need only compute the moment generating function of ηt. Write ηt(x) = X y∈Td N η0(y) X k=1 1(Zy,k t = x) and X x∈Td N θ(x)ηt(x) = X x∈Td N X y∈Td N η0(y) X k=1 θ(x)1(Zy,k t = x) = X y∈Td N η0(y) X k=1 θ(Zy,k t ) where Zy,k t denotes the position at time t of the kth particle initially at location y. Since particles move independently, and initially there are a Poisson number of particles on each site of the lattice, EνN α h exp X x∈Td N θxηt(x) i = Y y∈Td N EνN α h exp η0(y) X k=1 θ(Zy,k t ) i = Y y∈Td N EνN α E exp θ(Zy,1 t ) η0(y) = Y y∈Td N exp h α E eθ(y+Zt) −1 i where Zt is the position of a random walk on Td N starting at the origin. Now, E eθ(y+Zt) = X x∈Td N P N t (x −y)eθ(x) and E eθ(y+Zt) −1 = sumx∈Td N P N t (x −y)(eθ(x) −1). Hence, after a calculation, EνN α h exp X x∈Td N θ(x)ηt(x) i = exp X y∈Td N α X x∈Td N P N t (x −y) eθ(x) −1 = exp X x∈Td N α eθ(x) −1 which ﬁnishes the proof. □ We now remark that the measure νN α can be decomposed in terms of its restric-tions to the sets {η ∈E : P x∈Td N η(x) = K} for K ≥0 which are invariant for the motion. Then, each of these restrictions, νTd N,K = νN α (·\| P x∈Td N η(x) = K), is invariant, and does not depend on α. In physics terminology, νN α is the ‘grand canonical’ measure and νTd N,K is the ‘canonical’ one. Since the mean of η(x) under νN α equals α, it makes sense to call α the ‘mass density’ of the process. In this way, {νN α : α ≥0} is a family of invariant measures indexed by density α. Moreover, we may interpret the measure νN ρ0(·) as a ‘local LECTURE 1 7 equilibrium’ measure in the following sense: Let u ∈Td be a continuity point of ρ0(·). Then, since ρ0(·) is continuous at u, νN ρ0(·) distributes nearby ⌊Nu⌋almost like the invariant measure νN ρ(u). More precisely, for ﬁxed l, we have lim N↑∞EνN ρ0(·) h exp X \|x\|≤l θ(x)η(x + ⌊uN⌋) i = lim N↑∞ Y \|x\|≤l exp ρ0(N −1(x + ⌊uN⌋))(eθ(x) −1) = Eνρ0(u) h exp X \|x\|≤l θ(x)η(x) i . In fact, we will say that a sequence of probabiliy measures µN on NTd N is a ‘local equilibrium’ of proﬁle ρ0 : Td →R+ if lim N↑∞EµN h exp X \|x\|≤l θ(x)η(x + ⌊uN⌋) i = Eνρ0(u) h exp X \|x\|≤l θ(x)η(x) i for all l ≥1 and θ(·). 2.2. Hydrodynamics. The question now is if we start with a local equilibrium measure νN ρ0(·), how to characterize the distribution at a later time? Initially, the ‘density proﬁle’ is given by ρ0. Is there a function which captures the density proﬁle at future times? The answers depend on the particular time and space scales chosen in the prob-lem. We will think of Td N as embedded in Td where grid points on Td N are separated by distance N −1. In this way, a ‘macroscopic’ point u on Td corresponds to the ‘microscopic’ point ⌊uN⌋. As we will see, time should now be appropriately speeded up to see movement of the system. How fast will depend on the structure of the underlying jump probability p(·). Following the computations above, the moment generating function, starting from νN ρ0(·), satisﬁes EνN ρ0(·) h exp X x∈Td N θ(x)ηt(x) i = Y y∈Td N exp n ρ0(y/N) E eθ(y+Zt) −1 o = exp X y∈Td N ρ0(y/N) X x∈Td N P N t (x −y) eθ(x) −1 = exp X x∈Td N eθ(x) −1 ψN,t(x) where ψN,t(x) = X y∈Td N P N t (x −y)ρ0(y/N) = X z∈Td N P N t (z)ρ0(N −1(x −z)) = E ρ0 N −1(x −ZN t ) . Hence, the distribution at time t is still a product measure with varying intensity ψN,t(·). When t is ﬁxed, the t-time distributions {P N t (·) : N ≥1} are tight, that is for all ϵ > 0, there exists A such that P \|x\|≤A P N t (x) ≥1 −ϵ. Hence, at a continuity point u for ρ0(·), we have limN↑∞ψN,t(⌊uN⌋) = ρ0(u), the same limit we obtained earlier when t = 0. So, if we do not speed up time at all, in this time scale, the system does not move. 8 LECTURE 1 When m = P xp(x) ̸= 0, the asymmetric case, since N −1ZN tN →mt in proba-bility, we have lim N↑∞ X \|z/N−mt\|≤ϵ pN tN(z) = lim N↑∞P h ZN Nt N −mt ≤ϵ i = 1. In this case, when the initial proﬁle ρ0 is continuous, lim N↑∞ψN,Nt(⌊uN⌋) = ρ0(u −mt) := ρ(t, u). Therefore, if we speed up time by a factor of N, we see that the density proﬁle has translated by mt. In this sense Nt is referred to as the ‘microscopic’ time, and t as the ‘macroscopic’ time. The density ρ(t, u) satisﬁes ∂tρ + m · ∇ρ = 0. (2.1) This makes sense as individual particles displace an order N microscopic locations at microscopic time Nt. However, when m = 0, particles do not displace as much, but follow the ‘square root’ law, in that displacements are on order √ N at time Nt, or alternatively on order N at times N 2t. The latter version ﬁts in nicely with our space scaling. By the central limit theorem for random walks in this case, N −1ZN N 2t ⇒N(0, σt), we have lim N↑∞ψN,N 2t(⌊Nu⌋) = lim N↑∞ X z∈Td N P N N 2t(z)ρ0 N −1(⌊Nu⌋−z) = lim N↑∞E h ρ0(u −N −1ZN N 2t) i = Z Rd ¯ ρ0(x)Gt(u −x)dx. Here, ¯ ρ0 is the periodic extension of ρ0 with period Td, and Gt is the Gaussian density with covariance tσ. It follows that ρ(t, u) := R Rd ¯ ρ0(x)Gt(u −x)dx, as a convolution with the ¯ ρ0, satisﬁes the heat equation ∂tρ = X 1≤i,j≤d σi,j∂2 ui,ujρ ρ(0, u) = ρ0(u). (2.2) As terminology, we call the equations (2.1) and (2.2) and their solutions ρ(t, u) as ‘hydrodynamic’ equations and ‘hydrodynamic’ solutions for the space-time evo-lution of the macroscopic density. What we have proved is the following: Theorem 2.2. Suppose ρ0 : Td →R+ is continuous. Let v(N) = N if m ̸= 0 and v(N) = N 2 if m = 0. Then, starting from the sequence νN ρ0(·), the distri-bution at time v(N)t is a local equilibrium sequence with respect to proﬁle ρ(t, u) corresponding to equation (2.1) if m ̸= 0 and equation (2.2) if m = 0. 3. Notes The material on Markov chains was based on the development in , and [4, Appendix 1]. The discussion about hydrodynamics of independent walks follows closely that in [4, Chapter 1] and . Recent work on invariant measures and hydrodynamics of systems of indepen-dent particles includes , , . The ﬁrst rigorous work on hydrodynamics of independent particle systems, on which the above development rests, is . LECTURE 1 9 References Dobrushin, R., and Siegmund-Schultze, R. (1982) The hydrodynamic limit for systems of particles with independent evolution. Math. Nachr. 105 199-224. Jara, M., Landim, C., and Teixeira, A. (2011) Quenched scaling limits of trap models. Ann. Probab. 39 176223. Karlin, S., and Taylor, H.M. (1975) A ﬁrst course in stochastic processes. Second edition. Academic Press, New York-London. Kipnis, C.; Landim, C. (1999) Scaling limits of interacting particle systems. Grundlehren der Mathematischen Wissenschaften 320 Springer-Verlag, Berlin. Landim, C. (2004) Hydrodynamic limit of interacting particle systems. in ICTP Lecture notes. 57-100. School and Conference on Probability Theory, May 13-17, 2002. Ed. G. Lawler. Abdus Salam ICTP Publications, Trieste, Italy. Liggett, T.M. (1978) Random invarian measures for Markov chains and independent particle systems. Z. Wahr. Gebiete. 45, 297-313. Peterson, J. (2010) Systems of one-dimensional random walks in a common random environ-ment. Electron. J. Probab. 15, 10241040. Resnick, S. (1992) Adventures in stochastic processes. Birkh¨ auser Boston, Inc., Boston, MA.
2398	https://www.youtube.com/watch?v=sya3RYN71aY	Given two points write the equation in slope intercept form ex 3 Brian McLogan 1590000 subscribers 541 likes Description 85313 views Posted: 8 Jul 2014 👉 Learn how to write the equation of a line given two points on the line. The equation of a line is such that its highest exponent on its variable(s) is 1. (i.e. there are no exponents on its variable(s)). There are various forms which we can write the equation of a line: the point-slope form, the slope-intercept form, the standard form, etc. The equation of a line given two points (x1, y1) and (x2, y2) through which the line passes is given by, ((y - y1)/(x - x1)) / ((y2 - y1)/(x2 - x1)). 👏SUBSCRIBE to my channel here: ❤️Support my channel by becoming a member: 🙋‍♂️Have questions? Ask here: 🎉Follow the Community: Organized Videos: ✅Write Linear Equations ✅Write Linear Equations \| Learn About ✅Write the Equation of a Line Given Point and Slope ✅Standard Form to Slope Intercept Form ✅Write Linear Equation From Graph ✅Write Linear Equastion From Two Points ✅Write the Equation of a Parallel Line ✅Write the Equation of a Perpendicular Line ✅how to know if the equation is linear or nonlinear ✅Translating Linear Equations 🗂️ Organized playlists by classes here: 🌐 My Website - 🎯Survive Math Class Checklist: Ten Steps to a Better Year: Connect with me: ⚡️Facebook - ⚡️Instagram - ⚡️Twitter - ⚡️Linkedin - 👨‍🏫 Current Courses on Udemy: 👨‍👩‍👧‍👧 About Me: I make short, to-the-point online math tutorials. I struggled with math growing up and have been able to use those experiences to help students improve in math through practical applications and tips. Find more here: WriteLinearEquations #LinearEquations #BrianMcLogan 43 comments Transcript: welcome ladies and gentlemen what i'd like to do is show you how to write an equation in slope-intercept form um given two points now what's nice about given two points is one we know we can find the slope very easily right um however the y-intercept is not as easy so um remember slope-intercept form is y equals mx plus b where m represents the slope and b represents your y-intercept now to find the slope uh we have a formula for that and the formula for slope which is our m is equal to the change in y over the change in the x why did i write that as y 1 y you could write that if you wanted to it doesn't really matter you could do y1 minus y2 as long as you do x1 minus x2 but i prefer to keep it in this method in this form x2 minus x1 y2 minus y1 so now i just need to determine what are my x2 y2s well since these are both coordinates i just need to differentiate between them so i'll call these the ones and these the twos so now i plug in to solve and i have negative one minus four divided by three minus a negative two and i have negative one minus four which is a negative five three minus a negative two is divided by five which equals a negative one so therefore my slope is equal to negative one right all right cool so i'm gonna plug in negative 1 in for m but now i need to determine b and to determine b in this case i'm actually going to use our slope intercept form so let's plug in what we know we know y equals a negative x plus b the only thing we don't know now is b so one thing we can do is when we write the equation of the line the reason why we need to solve for m and b and not for y and x because y and x represent the coordinates of any point that lies on the line well since they represent any coordinate there's no exact number that y and x have to represent they only represent the y x and y coordinates for any point on the line however in this equation we are given two examples of points that are on the line so what i can do is i can plug in these coordinates in for x and y to solve for b so i will choose this one you could choose this one if you like um negative one equals negative three plus b notice how i put parentheses around that's a negative x well that's a negative and notice how x is equal to 3. so i have negative 1 equals negative 3 plus b add 3 add 3 2 equals b so therefore now i know my y intercept is b and my slope is negative 1. so now i can just take my equation y equals mx plus b plug in negative 1 in for m and plug in a 2 in for b and i can simplify that one more time as a negative x plus 2. so there you go ladies and gentlemen that is how you write a linear equation in slope-intercept form given two points thanks
2399	https://flexbooks.ck12.org/cbook/ck-12-chemistry-flexbook-2.0/section/14.8/primary/lesson/ideal-gas-law-chem/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 14.8 Ideal Gas Law Written by:Ck12 Science Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Lesson What chemical reactions require ammonia? There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system. Ideal Gas Law The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro’s law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation: P1×V1T1×n1=P2×V2T2×n2 As with the other gas laws, we can also say that (P×V)(T×n) is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal. The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable R for the constant, the equation becomes: P×VT×n=R The ideal gas law is conventionally rearranged to look this way, with the multiplication signs omitted: PV=nRT The variable R in the equation is called the ideal gas constant. Evaluating the Ideal Gas Constant The value of R, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: kPa, atm, or mmHg. Therefore, R can have three different values. We will demonstrate how R is calculated when the pressure is measured in kPa. Recall that the volume of 1.00 mol of any gas at STP is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for R. R=PVnT=101.325 kPa×22.414 L1.000 mol×273.15 K=8.314 kPa⋅L/K⋅mol This is the value of R that is to be used in the ideal gas equation when the pressure is given in kPa. Table below shows a summary of this and the other possible values of R. It is important to choose the correct value of R to use for a given problem. Values of the Ideal Gas Constant \| Unit of P \| Unit of V \| Unit of n \| Unit of T \| Value and unit of R \| \| kPa \| L \| mol \| K \| 8.314 J/K⋅mol \| \| atm \| L \| mol \| K \| 0.08206 L⋅atm/K⋅mol \| \| mmHg \| L \| mol \| K \| 62.36 L⋅mmHg/K⋅mol \| Notice that the unit for R when the pressure is in kPa has been changed to J/K • mol. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule (J). Sample Problem: Ideal Gas Law What volume is occupied by 3.760 g of oxygen gas at a pressure of 88.4 kPa and a temperature of 19°C? Assume the oxygen is ideal. Step 1: List the known quantities and plan the problem. Known P=88.4 kPa T=19∘C=292 K mass O2=3.760 g O2=32.00 g/mol R=8.314 J/K⋅mol Unknown V=? L In order to use the ideal gas law, the number of moles of O2 (n) must be found from the given mass and the molar mass. Then, use PV=nRT to solve for the volume of oxygen. Step 2: Solve. 3.760 g×1 mol O232.00 g O2=0.1175 mol O2 Rearrange the ideal gas law and solve for V. V=nRTP=0.1175 mol×8.314 J/K⋅mol×292 K88.4 kPa=3.23 L O2 Step 3: Think about your result. The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume (22.4 L/mol) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for T and P. Since a joule (J) = kPa • L, the units cancel correctly, leaving a volume in liters. Ever wonder why soda goes flat? Explore the ideal gas law in action inside a soda bottle in this simulation: Summary The ideal gas constant is calculated. An example of calculations using the ideal gas law is shown. Review Which value of R will you use if the pressure is given in atm? You are doing a calculation where the pressure is given in mm Hg. You select 8.314 J/K • mol as your value for R. Will you get a correct answer? How would you check that you have chosen the correct value of R for your problem? Asked by Students Here are the top questions that students are asking Flexi for this concept: Overview The Ideal Gas Law is a single equation that relates the pressure, volume, temperature, and number of moles of an ideal gas. The equation is represented as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. The ideal gas constant (R) depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. The value of R can have three different values depending on the unit of pressure used: kPa, atm, or mmHg. For example, when the pressure is measured in kPa, R is calculated as 8.314 J/K•mol. Vocabulary chemical reaction gas mole volume temperature pressure system combined gas law Avogadro’s law ideal gas ideal gas constant liter energy joule molar mass molar volume significant figures Test Your Knowledge Question 1 What would be the volume occupied by two moles of propane gas at standard conditions? a 5.6 liters b 44.8 liters c 22.4 liters d 89.6 liters We will demonstrate how is calculated when the pressure is measured in kPa. Recall that the volume of 1.00 mol of any gas at STP is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for . Question 2 At @$\begin{align}27^\circ C\end{align}@$ and one atmospheric pressure, a sample of gas has volume @$\begin{align}V\end{align}@$. What will be the volume of the sample at @$\begin{align}492^\circ C\end{align}@$ and a pressure of @$\begin{align}1.5 \ atm\end{align}@$? a 2.0 V b 1.0 V c 3.9 V d 1.7 V We will demonstrate how is calculated when the pressure is measured in kPa. Recall that the volume of 1.00 mol of any gas at STP is measured to be 22.414 L. We can substitute 101.325 kPa for pressure, 22.414 L for volume, and 273.15 K for temperature into the ideal gas equation and solve for . Asked by Students Here are the top questions that students are asking Flexi for this concept: Related Content Saved by an Air Bag The Ideal Gas Law: Crash Course Chemistry \| Image \| Reference \| Attributions \| --- \| \| \| Credit: User:Masur/Wikimedia Commons Source: License: Public Domain \| Ask me anything! Loading... Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? 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