Split (2)

train · ~239k rows (showing the first 191k)

id stringlengths 1 6	url stringlengths 16 1.82k	content stringlengths 37 9.64M
4500	https://math.answers.com/questions/Does_a_diagonal_of_a_parralelogram_divide_a_parallelogram_into_two_congruent_triangles	Does a diagonal of a parralelogram divide a parallelogram into two congruent triangles? - Answers Create 0 Log in Subjects>Science>Natural Sciences Does a diagonal of a parralelogram divide a parallelogram into two congruent triangles? Anonymous ∙ 14 y ago Updated: 6/10/2024 Yes, the diagonal splits the parallelogram into two equal triangle aka congruent the sides will stay the same, the two angles being divided are going to be split in half, one on each side, so its the same Wiki User ∙ 14 y ago Copy Show More Answers (2)Add Your Answer What else can I help you with? Search Continue Learning about Natural Sciences ### What divides a parallelogram into two congruent triangles? The term for the line that divides them is a diagonal. ### When a parallelogram is constructed in order to add forces what represents the resultant of the forces? In a parallelogram of forces, the resultant force is represented by the diagonal of the parallelogram drawn from the initial point of the forces to the opposite corner. The magnitude and direction of the resultant force are determined by the length and orientation of this diagonal in the parallelogram. ### What does jamaica's flag look like? Jamaica's flag has two diagonal yellow lines forming a cross on the flag, which section it off in four parts. The top and bottom sections are equal-sized green triangles, and the left and right sections are equal-sized black triangles. ### What is the diagonal of 12 inches and 5 inches? diagonal is 13 inch length of the rectangle of 12 and 5 inches sides ### The word used in muscle names that means diagonal to the midline is? The term you are looking for is "oblique." Muscles that are described as oblique run at an angle or diagonal to the midline of the body. Related Questions Trending Questions In PC 1 3 how many atoms bond to the central atom and how many lone pairs are there in it?What should kids do on snow days?Which activity is part of inflammatory response?What is the distance from Portland Maine to Los Angeles California in metric length?Is sunlight required for the synthesis of amino acids?What elements are in Vegemite?What does not have the affinity for water?What to wear at 40 degrees?What is stain used for electroscopic view of eye specimen?Is pulp a solid or a liquid?How is quantum theory opposed to Local Realism?How many milliseconds are there in a week?Write a sentence using silent mutation missense mutation frameshift mutation non sense mutation deletion dupication inversion transloction gene rearangement?What does the structure of the squids gills do?How many degrees does the world rotate in one week?How many centimeters are in a joule?Why can't human beings live without water?What are three plants found in the grassland?When was It Snows in Hell created?What causes the ash and water to mix and form a type of mud that sets like concreate when it stops floing when a lahar occurs? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
4501	https://www.youtube.com/watch?v=uHuhHKY_wO8	Digital SAT Math - Desmos Lesson #5 Inequalities Tutorllini Test Prep 28800 subscribers 2065 likes Description 100914 views Posted: 13 Nov 2022 Check out this calculator lesson - Digital SAT Math Desmos Lesson #5 Inequalities - created by an 800 SAT Math scorer. Learn how to use Desmos to solve questions more efficiently and even get answers to questions you know nothing about. You can learn more about Desmos by watching the Learn Desmos playlist here: You can download the Bluebook app to take Digital SAT Practice Tests here: You can check out Desmos Tools here: For more info on my tutoring services check out my website: digitalsat #digitalsatmath #desmos 103 comments Transcript: hi everyone welcome to my channel tutorlini test prep today we will be going over lesson number five on how to use Desmos for the digital SAT Math inequalities let's get started so what you can do for these types of points and inequalities or points and systems of inequalities questions is you can actually graph the inequality on Desmos and then you can graph the points and kind of guess and check to see what the answer might be so that's what we're going to do here we're first going to plot y is less than 6X Plus 2. so Y is less than six x plus two and you don't have to do this but sometimes I like to set the window so you see for these all the X values are kind of between pretty much 0 and 10 and then all the Y values are pretty much between like lowest possible maybe like 10 and 50. so I'm going to just manually set that on Desmos by clicking on this little gear and then setting the window so 0 to 10 and then 10 to 50. and that looks pretty good and you could type these points in manually I'm just for the sake of it I'm going to make a table I might not do this for the other ones but I'll do it for this one we can start to kind of guess and check so let me do 3 and 20. and I see that that is on the line so this one is a dashed inequality so that means points on the line are not included so a does not work so a is out so let's cross that out and let's try maybe three comma 16. okay 3 comma 16 is in the Shaded area so that's good so B and C are looking good right now uh let's check d three comma 24 okay three comma 24 is no good so D is out okay now I'll do root 3 comma 16 and then the second point in B is 5 comma 36 so let's try that 5 36. and that one is no good uh okay so right away B is out and you can right away pick C is the answer but um why don't we go ahead and type in the rest of them just to make sure that it's right so let's go back to here 5 28 and 740. so 528 and 740 and you can see all three points are in the solution set so this is our answer so we did a good job with that so let's just lock in really quick answer Choice C is our answer great so you can also do this you should have some idea of what to do now so you can also do this for systems of inequalities questions so two inequalities you just need to be a little bit more careful at which shaded region you're looking at so let's first type this in again as always this one's a little different but see if you can pause this video and solve this question using Desmos now that you have some idea of what to do so Y is greater than 14 and then the other one is 4X plus Y is less than 18. so 4X plus y is less than 18. and I don't really have a good sense of kind of what's going on here so why don't I zoom out a little bit uh so you see for a system there's kind of like four pizza slices and the solution is the pizza slice where both are shaded so it's like both red and blue so you see kind of like this area here so in particular I want to look at that area so I'll just kind of readjust a little bit and just make sure you're looking at this kind of pizza slice with your mind's eye okay all right so it says we want to do X comma 53. so so you could go through and manually type in like I don't know what's the first answer just negative nine so you could go through and manually type in like negative 9 comma 53. um just to make my life a little bit easier what I'm going to do is I'm going to make it a comma 53. and then I'll just type in a just so that I don't have to mess with the point as much so let's go through and guess and check all these so negative 9 53 and that is I'm not sure it's a little close let me zoom in yeah that's in the solution set it's in that pizza slice right so um so that's actually our answer um wow we got really lucky there that a was the answer let's go through and just kind of confirm that the other ones are wrong just for the sake of learning in the video so what does B say negative five okay so let's try negative five you see that is outside the red and blue pizza slice so that's no good so we can roll we can rule that out so let me switch to red here and cross that out and let's try five so let's try five and um as you can see that one is outside of the red and blue pizza slice so so that one's out and let's try nine comma 53. ah that one's outside of the red and blue pizza slice so so that one's wrong too so the answer for that question is a awesome all right let's do one more so this is another systems question and again we can just guess and check by by typing in the points so let's do it so why is uh again I always forget pause this video see if you can solve it now that you have some idea of what to do using Desmos so we're going to do y is less than or equal to Y is less than or equal to X plus seven and then Y is greater than or equal to by the way guys I'm typing in greater than and then equal on the keyboard and it'll make it an equal to automatically so very helpful if you didn't know that greater than or equal to X plus seven and then greater than or equal to negative 2x minus one okay great so let me just get a nice picture I'm going to zoom out a little bit and you can kind of see that the blue and red area where it's simultaneously shaded because again this is a system of inequalities question is this kind of e-store right most pizza slice kind of where I'm pointing with my mouse okay so the point needs to be in there to satisfy the system of inequality so let's again guess and check we'll just type in all of them so negative 14 comma zero okay that is not in the correct shaded area the rightmost pizza slice so that one's out and do I have red selected I don't know if I do okay and now let's try zero comma negative 14. so zero comma negative 14. and you see that is not in the rightmost pizza slice either so that is out let's try zero comma 14. that is also not in the rightmost pizza slice so that is out and now let's try 14 comma zero I bet that's our answer okay and as you see it is 14 comma zero is in the solution set another way to do this instead of typing it in manually is you could type in all the answer choices in order and then label them um and you could kind of look at all of them at once I don't know I think it's just easier for this one to type them in one at a time so I don't get distracted or mix them up so 14 comma zero is the only one in the correct shaded region where they're both red and blue are shaded so that's our answer so the answer to this question let me just check is d okay that completes the lesson please like And subscribe for more digital SAT Math content if you're interested in my tutoring Services the link to my website will be in the description I tutor all sections of the SAT and all math subjects from about 7th grade to AP slash early college level thanks for stopping by and good luck studying
4502	https://www.emc.school/math-tips/compensation-strategy-for-subtraction-made-easy	Compensation Strategy for Subtraction Made Easy 2025-2026 Academic Year has started! New students are always welcome!Register Today! Home About About Our SchoolOur FacultyCalendarFAQTestimonialsEMC Foundation Programs Programs All ProgramsApplied Math WorkshopsAcademic Year ProgramMath CompetitionsSummer ProgramsIn Person ProgramsCustom ProgramsMath Circles on Campus ScheduleEvents Blog School NewsMath Tips Sign In Sign In Current Students - CanvasParent Admin Portal Schedule Free EvaluationSchedule Free Evaluation Compensation Strategy for Subtraction Made Easy Subtraction can become a lot simpler with the Compensation Strategy! The trick is to adjust the number you subtract to make it an even ten. Then, after solving, you just "return" or add back what you adjusted. For example, if you need to subtract 38 from 92, round 38 up to 40 by adding 2. Now subtract 92 - 40 = 52, and finally add back the 2 you adjusted, so 52 + 2 = 54. Easy, right? This strategy is perfect for solving problems quickly and helps you build confidence with subtraction. It also works great for mental math! Want more practice and fun examples? Download our free guide to become a subtraction expert using Compensation! EMC Guide - Subtraction Strategy (Compensation) Download file here Event gallery: No items found. Subscribe to our Newsletter Sign your child up today and see their love of math blossom! Schedule Free Evaluation info@engagingmathcircles.com(408) 599-2663 About EMCOur FacultyCareersBlogFAQPrivacy Policy School ProgramsMath CompetitionsEventsSchool CalendarTerms of ServiceContact FAQTerms of ServicePrivacy PolicyContact Subscribe to EMC Newsletter Thank you! Your submission has been received! Oops! Something went wrong while submitting the form. Search our site © 2018-2025 Engaging Math Circles. All rights reserved. 📢 School Year Has Started!Students accepted year round. ‍ New Students: Book your free evaluation to find the perfect group fit. ‍ Existing Students: Register for classes via the Schedule page or Parent Portal. Get a sneak peek of both programs in a fun trial class! ‍Reserve your spot for June 7th-8th! Learn more
4503	https://math.stackexchange.com/questions/3349594/are-the-definite-and-indefinite-integrals-actually-two-different-things-where-i	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 Are the definite and indefinite integrals actually two different things? Where is the flaw in my understanding? Ask Question Asked Modified 5 years, 11 months ago Viewed 6k times 33 $\begingroup$ Some context: I'm an engineer, and I tend to have a rather unusual way of understanding and thinking about things, most likely related to my being autistic. I found this question on the HNQ and upon reading it I felt rather confused. I have always understood that the indefinite and definite integrals are two sides of the same underlying concept, that there is no meaningful difference between them other than that one evaluates to a function and the other to a quantity. Essentially, the definite integral is what you get from running the numbers on the result of an indefinite integral, and saying it's unrelated to the definite integral is like saying that $\sin(x)$ is fundamentally different from $\sin(a)$, assuming $x$ is a variable and $a$ a constant, for no reason other than that $x$ is a variable and $a$ a constant. To me, whether the definite and indefinite integrals are two sides of the same thing, or two closely related but different things, seems more like a question of philosophy than mathematics. Where is the flaw in my understanding? Is there one? calculus integration terminology Share edited Sep 11, 2019 at 15:51 HearthHearth asked Sep 9, 2019 at 14:42 HearthHearth 67355 silver badges1111 bronze badges $\endgroup$ 1 $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Aloizio Macedo – Aloizio Macedo ♦ 2019-09-12 18:52:28 +00:00 Commented Sep 12, 2019 at 18:52 Add a comment \| 5 Answers 5 Reset to default 34 $\begingroup$ The point is, there are three slightly different concepts here: 1) If $f$ and $F$ are functions with the same domain, and $F'(x)=f(x)$ on that domain, then we say that $F$ is an antiderivative of $f$. 2) If $f$ is a function on the interval $a \le x \le b$, then the definite integral of $f$, $\int_a^b f(x) \, dx$, is (loosely speaking) the area under the graph of $y=f(x)$ for $a \le x \le b$, or (more precisely speaking) a limit of Riemann sums. This definition does not involve any derivatives. 3) If $f$ is a function on the interval $a < x < b$ and $c$ is a point in that interval, then the function $G(x)=\int_c^x f(t) \, dt$ could be called an indefinite integral of $f$. This definition of the function $G$ comes directly from the definition in 2) of the definite integral, so it also does not involve any derivatives. You are saying that you don't see any difference between 2) and 3). You are correct! If your mental definition of "indefinite integral" is 3), then an indefinite integral is just a definite integral with some unknown limits. The question you're linking is saying that many calculus books use 1) as the definition of an "indefinite integral": that is, they say that if $F$ is an antiderivative of $f$, then $\int f(x) \, dx = F(x)+C$ is the indefinite integral of $f$. They can get away with this because of the fundamental theorem of calculus, which says (in part) that $$ \frac{d}{dx}\int_c^x f(t) \, dt=f(x) \, . $$ That is, the function $G(x)$ defined in 3) is an antiderivative of the function $f$, so these two notions of "indefinite integral" are closely related. But closely related does not mean identical! K B Dave's answer is giving you some examples of situations where the two definitions 1) and 3) diverge slightly. Specifically: We gave the definition 3) on an interval $a < x < b$, but it doesn't work very well on a more complicated domain. This means that definition 1) is more robust when $f$ has singularities. The fundamental theorem of calculus says that every "indefinite integral" in the sense of 3) is an antiderivative. It does not say that every antiderivative is an indefinite integral, which in fact is false; you can have functions $f$ and $F$ where $F'(x)=f(x)$, but there is no way to write $F(x)=\int_c^x f(t) \, dt$. Share edited Sep 9, 2019 at 16:22 answered Sep 9, 2019 at 16:07 MicahMicah 38.8k1515 gold badges8989 silver badges138138 bronze badges $\endgroup$ 6 1 $\begingroup$ Okay, it looks like what I fundamentally wasn't getting about the question that prompted this whole thing in the first place was that it was conflating "antiderivative" and "indefinite integral" to mean the same thing. My mental model is that antiderivative means 1), definite integral means 2), and indefinite integral means 3), and what I was thinking was that the question I linked (and people in the comments and other answers to this question) were saying that 2) and 3) were different things when they were saying that 1) and 2) were different things, which I understand better. $\endgroup$ Hearth – Hearth 2019-09-09 16:13:18 +00:00 Commented Sep 9, 2019 at 16:13 2 $\begingroup$ Part of the problem might also have been that, in my field, 1) and 3) are often considered the same thing too, though there are minor differences. Engineers working with the real world don't often have to consider singularities, and when we do, we can fudge them by abusing delta functions and everything works out because the real world is full of low-pass filters. $\endgroup$ Hearth – Hearth 2019-09-09 16:14:58 +00:00 Commented Sep 9, 2019 at 16:14 2 $\begingroup$ Yeah, on some level the point is that conflating 1) and 3) is computationally extremely handy, but conceptually confusing. As an engineer, you're more likely to care about the first half of that than the second; as a math educator, the person asking the other question is in the opposite situation. $\endgroup$ Micah – Micah 2019-09-09 16:17:24 +00:00 Commented Sep 9, 2019 at 16:17 5 $\begingroup$ Not all indefinite integrals are expressible as $$\int_c^x f(t)\, dt$$ The general form is actually $$\int_c^x f(t)\, dt + C$$ for some constant $C$. To understand the difference, note that the first expression must be $0$ for some $x$. Depending on the nature of $f$, the second expression need not be $0$ anywhere. Both expressions are antiderivatives of regular $f$, but only the second expression can represent all antiderivatives. $\endgroup$ Paul Sinclair – Paul Sinclair 2019-09-09 23:53:28 +00:00 Commented Sep 9, 2019 at 23:53 $\begingroup$ Another point worth stressing: if the function you started with is not continuous, then FTC does not apply. Granted, provided that it is at least locally integrable, it still holds almost everywhere. $\endgroup$ tomasz – tomasz 2019-09-10 10:54:20 +00:00 Commented Sep 10, 2019 at 10:54 \| Show 1 more comment 12 $\begingroup$ $\DeclareMathOperator{\dom}{dom}$ "Reasonable" functions $f$ can have antiderivatives $D^{-1}f$ that are not of the form $x\mapsto \int_{a}^xf(t)\mathrm{d}t$ for any $a$ in the domain of $f$. Let $U=(-\infty,0)\cup(0,\infty)$, and consider $$\begin{align} f&: U\to\mathbb{R} & x&\mapsto \frac{1}{x}\text{.} \end{align}$$ Then $$D^{-1}f(x)=\begin{cases} \ln(-x)+C_- & x < 0\ \ln(x) + C_+ & x > 0 \end{cases}$$ where $C_-$ and $C_+$ can differbecause the domain is not path-connected, there is no way to integrate from positive $x$ to negative $x$. Consider $$\begin{align} f&:\mathbb{R}\to \mathbb{R} & x&\mapsto \frac{1}{1+x^2}\text{.} \end{align}$$ Then $$\int_a^xf(t)\mathrm{d}t=\arctan x - \arctan a$$ but $$D^{-1}f(x)=\arctan x + C\text{;}$$ since the magnitude of $\arctan a$ is bounded by $\pi/2$, if $\lvert C \rvert \geq \pi/2$ then $D^{-1}f(x)$ is not of the form $\int_a^xf(t)\mathrm{d}t$. Share answered Sep 9, 2019 at 15:18 K B DaveK B Dave 9,49811 gold badge2020 silver badges3232 bronze badges $\endgroup$ 3 $\begingroup$ I think this might answer the question, but I also think I'm going to have to brush up on my mathematical terminology to understand it. $\endgroup$ Hearth – Hearth 2019-09-09 15:25:03 +00:00 Commented Sep 9, 2019 at 15:25 $\begingroup$ Yeah, my head hurts now. I think I'm going to have to come back to this later. $\endgroup$ Hearth – Hearth 2019-09-09 15:50:05 +00:00 Commented Sep 9, 2019 at 15:50 1 $\begingroup$ @Hearth: I noticed your comments in the chat. In fact, you are right, and Michael Bacthold is totally wrong. He arrogantly insinuates that mathematicians cannot help you. While some mathematical notations are troublesome to formalize in a first-order theory, there is no trouble formalizing them in a more flexible system. Any proper logician can assure you of this, and you can read a sketch of one possible way to do it here. Feel free to clarify any details in chat. $\endgroup$ user21820 – user21820 2020-04-16 10:59:06 +00:00 Commented Apr 16, 2020 at 10:59 Add a comment \| 3 $\begingroup$ As defined, the definite integral and the indefinite integral are totally different things. The definite integral $\int_a^b f(x)\ dx$ is defined by looking at approximations to $f$ which are piecewise linear functions, looking at the areas of the resulting trapezoids, and then examining what happens in the limit as these approximations get more and more accurate. The indefinite integral $\int f(x)\ dx$ is defined by searching for a function $g$ such that $g' = f$. As you can see, on the face of it, these two concepts appear to have nothing to do with each other whatsoever. The definite integral has nothing to do with derivatives. And the indefinite integral has nothing to do with piecewise linear approximations or areas of trapezoids. It just so happens that the definite integral and the indefinite integral are actually closely related: given some reasonable assumptions, if $F(x) = \int f(x)\ dx$, then $\int_a^b f(x)\ dx = F(b) - F(a)$. So the definite integral can, in fact, be found by finding the indefinite integral and then plugging some numbers into it. Likewise, the indefinite integral can be found by finding definite integrals where one endpoint is a variable. But the definitions are totally different. Ultimately, maybe it doesn't matter. The Dedekind real numbers and the Cauchy real numbers are defined in completely different ways, but they end up defining exactly the same thing (at least in ZFC). Definite integrals and indefinite integrals aren't exactly the same, but unless you're doing real analysis on "ill-behaved" functions, the distinction doesn't really have any consequences. Share answered Sep 9, 2019 at 16:33 Sophie SwettSophie Swett 10.9k3636 silver badges5656 bronze badges $\endgroup$ 1 1 $\begingroup$ It's not common to use trapezoids in defining the definite integral. Usually the definition is in terms of rectangles, as in the Riemann integral. But trapezoids can be useful in improving the accuracy of a numerical approximation of a definite integral. $\endgroup$ aschepler – aschepler 2019-09-10 07:02:23 +00:00 Commented Sep 10, 2019 at 7:02 Add a comment \| 2 $\begingroup$ Essentially, the definite integral is what you get from running the numbers on the result of an indefinite integral, and saying it's unrelated [...No-one would claim they are "unrelated"; merely that the are "different". "different" most certainly does not mean "unrelated"...] to the definite integral is like saying that $\sin(x)$ is fundamentally different from $\sin(a)$, assuming $x$ is a variable and $a$ a constant, for no reason other than that $x$ is a variable and a $a$ constant. I think this is your flaw. That $x$ is a variable and $a$ is a constant is a HUGE and very fundamental difference. The number $0.86602540378443864676372317075294...$ can be thought of in many ways but ultimately it is "what it is". One of the things that it is is one half of the positive number that when squared is $3$. Another thing it is is $\sin \frac \pi 3$. The number doesn't change what it is just because we describe it in different ways for different purpose. If $x$ is a variable then $\sin(x)$ is not a specific number (which $\sin a$ is). $\sin(x)$ is a theoretical possible unknown value of a number that would be sine of whatever number $x$ would be if it were pinned down. But as it isn't pinned down, it is an indefinite value. In a comment you claimed we don't call $\sin a$ the "definite sine" and $\sin x$ the "indefinite sine". We don't use those exact words but we most certainly do use those concepts and distinguish between them. To add to the confusion the function $\sin$, itself, is different than $\sin(x)$. $\sin = {(x,y)$ ordered pairs so that $y = \sin x}$ is the concept of the function itself; a collection of all ordered pairs, whereas $\sin(x)$ is an indefinite output value of the function for an indefinite input value, and whereas $\sin (a)$ is a definite output value of the function for a definite input value. These are three different applications of one concept or "three sides of the same coin". No-one is saying they are unrelated. But they are different things. ...... By the way, you've heard the joke about the a coworker who thinks the capital of Norway is in Sweden [a fundamentally impossible and inconsistent error; a capital of a country can't be in a different country so such an error is inconceivable for someone to make]? That's because he thinks Oslo is in Sweden [a stupid, but conceivable error] and Oslo is the capital of Norway [a fact; albeit one he is unaware of] and therefore he thinks that the capital of Norway [Oslo] is in Sweden. This is kind of the same thing. The definite actuality of Oslo, and the indefinite concept of the capitol of a country are two different things even though they may evaluate to the same thing (if the country you choose is specifically Norway-- but not Sweden). Share answered Sep 9, 2019 at 16:51 fleabloodfleablood 132k55 gold badges5252 silver badges142142 bronze badges $\endgroup$ 1 3 $\begingroup$ This is all good, except for the small problem that modern mathematics makes no distinction between a variable and a constant. Or could you give the definition of a variable element of $\mathbb{R}$ vs. a constant element of $\mathbb{R}$? See also the answer of Mike Shulman here $\endgroup$ Michael Bächtold – Michael Bächtold 2019-09-10 13:50:00 +00:00 Commented Sep 10, 2019 at 13:50 Add a comment \| -1 $\begingroup$ (there are several ways to define integrals, and they are not 100% equivalent, so I'll go the elementary way to try to be clear; for the same reason, I will only discuss continuous functions) We define the integral of the continuous function $f$ on the interval $[a,b]$ as $$\tag1 \int_a^b f(t)\,dt=\lim_{n\to\infty}\sum_{j=1}^n f(x_j)\,\Delta_j, $$ where $x_j=\frac{j(b-a)}n$ and $\Delta_j=1/n$ (more generally, one uses arbitrary partitions, but it is the same for continuous funcions). If you think about it, you will note that there is no trace of a derivative nor anti-derivative in $(1)$. An integral is an integral, it has a million applications, and if $f$ is nice enough it can be approximated very well by taking $n$ big enough on the right-hand-side of $(1)$ (and maybe tweaking the $x_j$, like using the mid-point rule). The relation with derivatives comes from Newton's wonderful Fundamental Theorem of Calculus: if $f$ is a continuous function, then the function $$\tag2 F(x)=\int_0^x f(t)\,dt $$ satisfies $F'(x)=f(x)$. Because of $(2)$, the notation $\int f(t)\,dt$ for an antiderivative of $f$ is so widespread. The fact that any two antiderivatives of $f$ differ by a constant leads to Barrow's Rule: $$\tag3 \int_a^b f(t)\,dt=F(b)-F(a). $$ So $(3)$ is the reason why we use antiderivatives to calculate integrals (when possible). To emphasize the fact that integral means $(1)$, and not $(2)$ nor $(3)$, consider the Error Function $$ \operatorname{erf}(x)=\frac1{\sqrt\pi}\int_{-x}^x e^{-t^2}\,dt, $$ which is one of many important functions defined by an integral. The function $\operatorname{erf}(x)$ is a multiple of an anti-derivative of $f(t)=e^{-t^2}$, and no explicit expression for it exists. Share edited Oct 3, 2019 at 10:34 answered Sep 9, 2019 at 15:59 Martin ArgeramiMartin Argerami 219k1717 gold badges161161 silver badges299299 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 calculus integration terminology 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 1 Confusion regarding the difference between Definite and Indefinite Integration Difference between $\int$ and $\int_{-\infty}^{\infty}$ Direct proof that integral of $1/x$ is $\ln(x)$ What area does the antiderivative represent? Related definite and indefinite sums and integrals 4 Is there any relation between summation and indefinite integration? 2 What is the difference between Definite Integral & Indefinite Integral on the basis of their connection with derivatives? 1 Exact differential equation proof - with indefinite integrals instead of definite integrals? 1 Evaluating indefinite integrals from tables of definite integrals 1 "Any two indefinite integrals of $f$ differ from each other by a definite integral of $f$" Is my explanation of this property correct? 1 Relationship between indefinite and definite integrals. Hot Network Questions Why is the fiber product in the definition of a Segal spaces a homotopy fiber product? Can Monks use their Dex modifier to determine jump distance? Help understanding moment of inertia What meal can come next? How to home-make rubber feet stoppers for table legs? Passengers on a flight vote on the destination, "It's democracy!" Lingering odor presumably from bad chicken Exchange a file in a zip file quickly Bypassing C64's PETSCII to screen code mapping How to understand the reasoning behind modern Fatalism? Can a state ever, under any circumstance, execute an ICC arrest warrant in international waters? The altitudes of the Regular Pentagon Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator I have a lot of PTO to take, which will make the deadline impossible How to use cursed items without upsetting the player? Two calendar months on the same page Why do universities push for high impact journal publications? Is it safe to route top layer traces under header pins, SMD IC? ICC in Hague not prosecuting an individual brought before them in a questionable manner? Is it ok to place components "inside" the PCB Singular support in Bezrukavnikov's equivalence Does clipping distortion affect the information contained within a frequency-modulated signal? Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Can a GeoTIFF have 2 separate NoData values? more hot questions Question feed
4504	https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/03%3A_Measurements/3.17%3A_Significant_Figures_in_Addition_and_Subtraction	Skip to main content 3.17: Significant Figures in Addition and Subtraction Last updated : Mar 20, 2025 Save as PDF 3.16: Significant Figures 3.18: Significant Figures in Multiplication and Division Page ID : 52725 ( \newcommand{\kernel}{\mathrm{null}\,}) How old do you think this calculator is? Calculators are great devices. Their invention has allowed for quick computation at work, school, and other places where manipulation of numbers needs to be done rapidly and accurately. But they are only as good as the numbers put into them. The calculator cannot determine how accurate each set of numbers is, and the answer given on the screen must be assessed by the user for reliability. Uncertainty in Addition and Subtraction Consider two separate mass measurements: 16.7g and 5.24g. The first mass measurement, (16.7g), is known only to the tenths place, or to one digit after the decimal point. There is no information about its hundredth place and so that digit cannot be assumed to be zero. The second measurement, (5.24g), is known to the hundredths place, or to two digits after the decimal point. When these masses are added together, the result on a calculator is 16.7+5.24=21.94g. Reporting the answer as 21.94g suggests that the sum is known all the way to the hundredths place. However, that cannot be true because the hundredths place of the first mass was completely unknown. The calculated answer needs to be rounded in such a way as to reflect the certainty of each of the measured values that contribute to it. For addition and subtraction problems, the answer should be rounded to the same number of decimal places as the measurement with the least number of decimal places. The sum of the above masses would be properly rounded to a result of 21.9g. Example 3.17.1 Determine the combined molecular mass of a glucose molecule and a maltose molecule. Glucose molecule = 180.156gmol Maltose molecule = 342.3gmol Solution 180.156+342.4=522.456 When adding and subtracting we know to look at the least number of decimals in our starting values; in this case 342.3 has only 1 digit after the decimal, so we need to round our answer to the same place. (522.456\to 522.5\frag{g}{mol} When working with whole numbers, pay attention to the last significant digit that is to the left of the decimal point, and round your answer to that same point. For example, consider the subtraction: 78,500m−362m. The calculated result is 78,138m. However, the first measurement is known only to the hundreds place, as the 5 is the last significant digit. Rounding the result to that same point means that the correct result is 78,100m. Example 3.17.2 What is 4200+540 = ? Solution 4200+540=4740 To determine where to round our answer, we look at our starting numbers to see which has the fewest number of decimal places. They both have 0 so we round to the nearest whole number, 4740. Summary For addition and subtraction problems, the answer should be rounded to the same number of decimal places as the measurement with the least number of decimal places. Review What is the basic principle to use in working with addition and subtraction? What do you pay attention to when working with whole numbers? 3.16: Significant Figures 3.18: Significant Figures in Multiplication and Division
4505	http://www.cs.emory.edu/~cheung/Courses/255/Syllabus/5-repr/fixed.html	Fixed point numbers The fixed point decimal number representation The decimal point: Decimal point = a point places in a decimal number representation to indicate the location of the digit whose weight = 1 Weights of the digits in a fixed point _decimal_ number: The digit that immediately preceeds the decimal point has weight = 10 0 = 1 The weight of digits moving towards left_increases_ by a factor of 10 The weight of digits moving towards right_decreases_ by a factor of 10 Example: Decimal number: 123.45 ^^^ ^^ \|\|\| \|\| \|\|\| \|+--- weight = 1/100 \|\|\| +---- weight = 1/10 \|\|+------- weight = 1 \|+-------- weight = 10 +--------- weight = 100 The fixed point binary number representation The "binary decimal" point: Binary decimal point = a point places in a binary number representation to indicate the location of the digit whose weight = 1 Weights of the digits in a fixed point _decimal_ number: The digit that immediately preceeds the decimal point has weight = 2 0 = 1 The weight of digits moving towards left_increases_ by a factor of 2 The weight of digits moving towards right_decreases_ by a factor of 2 Example: Binary number: 101.01 ^^^ ^^ \|\|\| \|\| \|\|\| \|+--- weight = 1/4 \|\|\| +---- weight = 1/2 \|\|+------- weight = 1 \|+-------- weight = 2 +--------- weight = 4 The value represented by 101.01 is: 1(4) + 0(2) + 1(1) + 0(1/2) + 1(1/4) = 5 1/4 = 5.25 (decimal) Converting: value <==> fixed point binary representation In the next 2 sections, I will show you how to convert: fixed point binary representation ==>value that is represented Given a (fractional) value ==> find the fixed point binary representation Convert: fixed point binary representation ==> value represented Method: Compute the value represented by a fixed point binary represention by adding the weighted sum of the value of the digits in the representation Example: Given the following fixed point binary representation: 10111.1011 The value represented is computed as: 10111.1011 \|\|\|\|\| \|\|\|\| \|\|\|\|\| \|\|\|+--- 1(1/16) \|\|\|\|\| \|\|+---- 1(1/8) \|\|\|\|\| \|+----- 0(1/4) \|\|\|\|\| +------ 1(1/2) \|\|\|\|\| \|\|\|\|+-------- 1(1) \|\|\|+--------- 1(2) \|\|+---------- 1(4) \|+----------- 0(8) +------------ 1(16) value represented = 16 + 4 + 2 + 1 + 1/2 + 1/8 + 1/16 = 23 11/16 = 23.6875 Convert: value ==> fixed point binary representation Method Split the value into 2 parts: An integral part A fractional part Find the binary representation of the integral part by repeatedly divide the value by 2 (to obtain the powers of 2 n) Find the binary representation of the fractional part by repeatedly multiply the value by 2 (to obtain the powers of 2-n) Example: Given the value 23.6875 Split the value into 2 parts: integral part = 23 fractional part = 0.6875 Convert the integral part 23 by repeated division: 23 /2 -------- 1 <---- remainder (= digit for the weight 2 0) 11 /2 -------- 1 <---- remainder (= digit for the weight 2 1) 5 /2 -------- 1 <---- remainder (= digit for the weight 2 2) 2 /2 -------- 0 <---- remainder (= digit for the weight 2 3) 1 /2 -------- 1 <---- remainder (= digit for the weight 2 4) 0 Binary representation for the value 23 = 10111 Convert the fractional part 0.6875 by repeated multiplication: 0.6875 x2 ------------ 1 <--- Overflow digit (= digit for the weight 2-1) 1.375 remove the overflow digit before continuing: ~~1~~.375 ==> 0.375 0.375 x2 ------------ 0 <--- Overflow digit (= digit for the weight 2-2) 0.75 remove the overflow digit before continuing: ~~0~~.75 ==> 0.75 0.75 x2 ------------ 1 <--- Overflow digit (= digit for the weight 2-3) 1.5 remove the overflow digit before continuing: ~~1~~.5 ==> 0.5 0.5 x2 ------------ 1 <--- Overflow digit (= digit for the weight 2-4) 1.0 remove the overflow digit before continuing: ~~1~~.0 ==> 0.0 Done (remainder of the value is 0 !) Binary representation for the value 0.6875 = 0.1011 Summary: Binary representation for the value 23 = 10111 Binary representation for the value 0.6875 = 0.1011 Therefore: Binary representation for the value 23.6875 = 10111.1011 Comment Although we can design a way to store fixed point fractional numbers inside the computer, there is a better way to represent fractional values: Floating point numbers So we never use fixed point fractional numbers. This material is just used as an introduction to floating point numbers. We will study that next.
4506	https://fskerman.github.io/expectation38.pdf	Degree Sequences of Random Bipartite Graphs Fiona Skerman 2010 A thesis submitted as a partial requirement of a Degree of Bachelor of Philosophy with Honours. Department of Mathematics The Australian National University October 28, 2010 Abstract This concerns three random bipartite graph models. For each random graph model a binomially based model is explicitly constructed which has the property that for ‘most’ degree sequences (s, t), the probability of (s, t) in the graph model is asymptotically approximated by the probability of (s, t) in the binomial model. This allows us to prove Theorems 7.4 and 7.5 which are the bipartite analogue of Theorem 2.6 in the paper by McKay and Wormald [MW97]. This construction is new as are Theorems 7.4 and 7.5. Acknowledgements It is an honor for me to thank both my co-supervisors Prof. Brendan Mckay and Dr. Judy-anne Osborn. I would like to thank Brendan Mckay for his unending fountain of mathematical insight and interesting ideas. I would also like to show my gratitude to Judy-anne Osborn for her time and help in writing the thesis. Their patience and under-standing made this thesis possible. I am also indebted to Dr. Adam Rennie for his advice on the art of mathematical writing and thank him for his help. Also to my fellow honours students, especially Brendan Fong, Howard Chuang, Tom Phelan, Ross Atkins, Wilson Ong and David Shellard with whom I shared an oﬃce. Additionally, I would like acknowledge my Mother and Father who have helped and in-spired me in ways too numerous to mention. Thanks also goes to the university for their generous support of my studies via the National Undergraduate Scholarship. i Contents I Literature review and background theory 1 1 Preliminaries 2 1.1 Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.1.1 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.1.2 Random bipartite graphs models . . . . . . . . . . . . . . . . . . . 3 1.1.3 Big ‘O’ notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.2 Motivating example: malaria transmission . . . . . . . . . . . . . . . . . . 8 2 Literature review 12 2.1 General random graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.1.1 Early results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.1.2 Threshold functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.1.3 Vertex degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.1.4 Approximation by binomial models . . . . . . . . . . . . . . . . . . 16 2.2 Bipartite random graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2.1 Vertex Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2.2 Enumeration results . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3 Probabilistic bounds and techniques 25 3.1 Concentration inequalities for independent random variables . . . . . . . . 26 3.2 Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.2.2 Azuma-Hoe↵ding inequality . . . . . . . . . . . . . . . . . . . . . . 28 3.2.3 Generalised Azuma-Hoe↵ding inequality . . . . . . . . . . . . . . . 28 3.2.4 Doob’s martingale process . . . . . . . . . . . . . . . . . . . . . . . 29 3.3 Probability generating functions . . . . . . . . . . . . . . . . . . . . . . . . 32 II New results: Degree sequences in random bipartite graphs 33 4 Graph p-model, Gp. 35 ii Contents 4.1 (", a)-regular degree sequences in Gp . . . . . . . . . . . . . . . . . . . . . . 35 4.1.1 Variation in the degree of each vertex. . . . . . . . . . . . . . . . . 35 4.1.2 Edge density in Gp . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 4.1.3 Bounding result on (", a)-regular degree sequences . . . . . . . . . . 41 4.2 Pathological degree sequences in Gp. . . . . . . . . . . . . . . . . . . . . . 42 4.2.1 Restricted distribution on the degrees of vertices . . . . . . . . . . . 42 4.2.2 Bounding a function on the white degrees: P j(Sj −S)2. . . . . . . 44 4.2.3 Likelihood of pathological degree sequences . . . . . . . . . . . . . . 48 5 Graph edge-model, GM. 50 5.1 Relation to graph p-model, Gp. . . . . . . . . . . . . . . . . . . . . . . . . . 51 5.2 (", a)-regular and pathological degree sequences in GM. . . . . . . . . . . . 51 6 Graph half -model, Gt. 53 6.1 (", a)-regular degree sequences in Gt . . . . . . . . . . . . . . . . . . . . . . 53 6.2 Pathological degree sequences in Gt . . . . . . . . . . . . . . . . . . . . . . 54 6.2.1 Locally ordered bipartite graphs . . . . . . . . . . . . . . . . . . . . 54 6.2.2 Decision tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 6.2.3 Graph ordered-half -model, Ga t . . . . . . . . . . . . . . . . . . . . . 57 6.2.4 Edge-uncovering martingale . . . . . . . . . . . . . . . . . . . . . . 58 6.2.5 Bounding Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . 61 6.2.6 Expectation of P j(Sj −λn)2 . . . . . . . . . . . . . . . . . . . . . 67 III New results: Approximation by binomial models 71 7 Binomial probability spaces 73 7.1 Deﬁnition of binomial models . . . . . . . . . . . . . . . . . . . . . . . . . 73 7.1.1 Binomial independent-model, Ip. . . . . . . . . . . . . . . . . . . . 73 7.1.2 Binomial p-model, Bp. . . . . . . . . . . . . . . . . . . . . . . . . . 73 7.1.3 Binomial edge-model, BM. . . . . . . . . . . . . . . . . . . . . . . . 74 7.1.4 Binomial half -model, Bt. . . . . . . . . . . . . . . . . . . . . . . . 76 7.2 Relation to random graph models . . . . . . . . . . . . . . . . . . . . . . . 78 7.2.1 Binomial edge-model, BM ⇠graph edge-model, GM. . . . . . . . . . 78 7.2.2 Binomial half -model, Bt ⇠graph half -model, Gt. . . . . . . . . . . 79 7.2.3 Binomial p-model, Bp. . . . . . . . . . . . . . . . . . . . . . . . . . 81 7.2.4 Integrated binomial model, Vp ⇠graph p-model Gp. . . . . . . . . . 84 7.3 Implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 iii Contents IV Appendix 94 8 Calculations on pathological degree sequences. 95 8.1 Generating functions for graph p-model, Gp. . . . . . . . . . . . . . . . . . 95 8.1.1 Preliminary expectations . . . . . . . . . . . . . . . . . . . . . . . . 95 8.1.2 Di↵erentials of A1, (Gp). . . . . . . . . . . . . . . . . . . . . . . . . 96 8.1.3 Expectation of P j(Sj −S)2 . . . . . . . . . . . . . . . . . . . . . . 96 8.1.4 Expectation of P j(Sj −np)2 . . . . . . . . . . . . . . . . . . . . . 97 8.2 Generating functions for graph half -model, Gt. . . . . . . . . . . . . . . . . 97 8.2.1 Di↵erentials of B, (Gt) . . . . . . . . . . . . . . . . . . . . . . . . . 97 8.3 Di↵erential operators and symbolic D notation . . . . . . . . . . . . . . . . 98 9 Results needed for Chapter 7. 99 9.1 Proof of Lemmas 7.8 and 7.9. . . . . . . . . . . . . . . . . . . . . . . . . . 99 9.2 Proof of Lemma 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Glossary of Notations 109 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 iv Introduction This thesis concerns random bipartite graphs. We deﬁne the key parameters of bipartite graphs. This follows the notation of [GM09]. For m, n 2 N, deﬁne N = mn and Im,n = {0, 1, . . . , n}m ⇥{0, 1, . . . , m}n. We deﬁne an (m + n)-tuple (s, t) 2 Im,n and denote the components of the m-tuple s, by s1, s2, . . . , sm and the n-tuple t, by t1, . . . , tn. Deﬁne the averages s = 1 n Pm j=1 sj and t = 1 m Pn k=1 tk. We can now deﬁne a bipartite graph on (m, n) vertices. It has a left vertex set U = {u1, u2, . . . , um} (drawn white) and right vertex set V = {v1, v2, . . . , vn} (drawn black). We also deﬁne an edge set E ✓{ (uj, vk) : uj 2 U, vk 2 V }. We call the number of edges incident with a vertex the degree of that vertex. Denote the degree of uj by sj and the degree of vk by tk. The degree sequence of the graph is then (s, t) as deﬁned above. We also deﬁne the edge-density, λ, of a bipartite graph. Observe that the number of edges in a bipartite graph can be determined by counting up the degrees of either side, so #edges = P j sj =: λmn. We illustrate these concepts in Figure 1. There is a natural bijection between bipartite graphs on (m, n) vertices and m ⇥n binary matrices where an edge between white vertex uj and black vertex vk corresponds to a ‘1’ in the jth row and kth edge. Likewise the absence of an edge indicates a ‘0’ in the corresponding position of the binary matrix. Hence this work can be thought of in this context. For instance, the degree of vertex uj a graph would become the row sum of the jth row in the corresponding matrix. This thesis concerns probability spaces over bipartite graphs. Let m, n be integers and let p = p(m, n) be a function into [0, 1]. We deﬁne a random bipartite graph with m white vertices and n black vertices of the other. In this random graph, each of the mn possible edges is present independently with probability p(m, n). We call the resulting probability space the graph p-model and denote it by Gp. v Contents • • • • • • • • • • • • • • • u1 v1 u2 v2 u3 v3 vn um−1 um Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q m m m m m m m m m m m m m m m m m m m m m i i i i i i i i i i i i i i i i i i i i i $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ sj = deg(uj) tk = deg(vk) #edges = P j sj = P k tk Figure 1: This gives a representation of a possible bipartite graph on (m, n) vertices. In the graph there are four edges incident with black vertex v2 and so t2 = 4. The graph p-model, Gp deﬁned above is one way to deﬁne a random bipartite graph. We also deﬁne two other models for random graphs, the graph edge-model, denoted GM, and the graph half -model, denoted Gt. In the graph edge-model, let M be an integer between zero and mn. We then set each possible bipartite graph G(m, n) with M edges to have equal probability. The graph half -model gets its name because we specify the degrees of the n black ver-tices. All bipartite graphs whose black degrees match those prescribed are then chosen with equal probability. We thus have three di↵erent random graph models, Gp, GM and Gt i.e. three di↵erent ways of choosing a graph at random. Of interest is the probability distribution of the degree sequence of a random graph as the number of vertices becomes very large. We consider the case when the ratio of white to black vertices is close to one. This is studied for each random graph model. Hence our problem becomes to ﬁnd the probability of a graph having a given degree se-quence (s, t) in each of our graph models. This we could calculate if we knew the number of bipartite graphs that can be constructed with degree sequence (s, t). Our starting point is a theorem by Greenhill and McKay [GM09] which has done this enumeration for bipartite graphs that satisfy certain conditions on the degree sequence. We term such degree sequences (", a)-regular. vi Contents We display the enumeration formula in the following theorem. Even without understand-ing every term seeing the general form will help explain the direction of this thesis. We refer the reader ahead to Theorem 2.8 for a full statement of the result. (Below, \|B(s, t)\| indicates the number of bipartite graphs with degree sequence (s, t) and the constant b in the error term is in the interval (0, 1 2).) Theorem 0.1 (Greenhill and McKay). For (", a)-regular⇤degree sequences: \|B(s, t)\| = ✓mn λmn ◆ −1 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ ⇥exp −1 2 ⇣ 1 − P j(sj −s)2 λ(1 −λ)mn ⌘⇣ 1 − P k(tk −t)2 λ(1 −λ)mn ⌘ + O(n−b) ! (0.1) A natural question which arises given this counting result is how common these well-behaved (", a)-regular degree sequences are in our random graph models. We show that (", a)-regular degree sequences account for the bulk of the probability space in each of the three random graph models. Result 1†: (", a)-regular degree Sequences For each random graph model, asymptotically, almost all degree sequences are (", a)-regular. Note that if a degree sequence (s, t) is both (", a)-regular and ⇣ 1 − P j(sj −s)2 λ(1 −λ)mn ⌘⇣ 1 − P k(tk −t)2 λ(1 −λ)mn ⌘ = O(n−b), then this simpliﬁes the formula by McKay and Greenhill above. We call such sequences non-pathological. Result 2‡: Pathological degree sequences For each random graph model, asymptotically, the probability a degree sequence is pathological tends to zero. After proving the previous two results we are in a good position. In each model, except for rare (i.e. pathological) degree sequences we have an asymptotic count for the number of ⇤(", a)-regular degree sequences, see Deﬁnition 2.7 †For the precise statement of this result for graph models Gp, GM and Gt see Lemmas 4.8, 5.1 and 6.1 respectively. ‡We show this result for models Gp, GM and Gt in Theorems 4.9, 5.2 and 6.20 respectively. vii Contents bipartite graphs with that degree sequence. From these counts we derive the approximate probability of ﬁnding a graph with that given non-pathological degree sequence. To get to this point comprises the bulk of the thesis, up until the end of Chapter 6. For all but pathological degree sequence the formula in (0.1) simpliﬁes and these patho-logical degree sequences are rare in our random bipartite graph models. This means that for ‘most’ degree sequences we are able to ﬁnd a simple asymptotic count for the number of bipartite graphs with that given degree sequence. We explicitly construct binomial models BM, Bt and Vp based on binomially distributed random variables subject to certain constraints§. Let the binomial models BM, Bt and Vp correspond to the graph models GM, Gt and Gp respectively. Then these binomial models have the property that, for each degree sequence, the prob-ability of that degree sequence occurring in the random graph model and the probability of it occurring in the corresponding binomial model are very close. This is an original result and is the culmination of all calculations in the thesis. We summarise this result below. Each random graph model can be approximated by one of these newly deﬁned binomial models in the following fashion. Result 3¶ There exist probability spaces, BM, Bt and Vp such that for any non-pathological degree sequence (s, t), PBM(s, t) = PGM(s, t)(1 + O(n−b)) PBt(s, t) = PGt(s, t)(1 + O(n−b)) PVp(s, t) = PGp(s, t)(1 + O(n−b)). (An explicit construction is given for each probability space.) Each of these newly constructed binomial spaces is based on binomial random variables subject to constraints on their sum. This has interesting theoretical implications. Con-sider the degree of a ﬁxed vertex in the random bipartite graph p-model, Gp. This vertex alone has a binomial distribution. Moreover, vertices of the same colour have independent degrees. However, vertices of di↵erent colours have degrees which are dependent. Our results will enable us to begin to quantify the magnitude of this dependence. §For the constructions see Deﬁnition 7.3 of BM, Deﬁnition 7.5 of Bt and Deﬁnition 7.7 of Vp. ¶This result appears as three separate Theorems in this thesis. We prove BM ⇠GM in Theorem 7.1, Bt ⇠Gt in Theorem 7.2 and lastly Vp ⇠Gp in Theorem 7.10. viii Part I Literature review and background theory 1 Chapter 1 Preliminaries 1.1 Deﬁnitions 1.1.1 Probability To be able to deﬁne what is meant by a random graph and also to make a meaningful sur-vey of previous results we need some probability theory. We give the necessary probability theory background for the content of the preliminary section (Part I). Later in Chapter 3 we detail the remaining probability results that are needed for the main body of the thesis. Many objects and functions studied in probability are special cases of familiar concepts from measure theory. We set out the some of the deﬁnitions we will require. Deﬁnition 1.1 (Probability space). Suppose we have a measure space which is deﬁned by the triple (⌦, ⌃, µ). If µ(⌦) = 1 then the triple (⌦, ⌃, µ) is a probability space and µ is a probability measure. Observe that if the measure µ satisﬁes µ(⌦) = k < 1, then it naturally determines a probability measure µ0(X) := 1 kµ(X). This is referred to as normalisation. In a probability space we will often denote the measure by P. Deﬁnition 1.2 (Random variable). Let (⌦, ⌃, P) be a probability space. Suppose we have a function X : ⌦! R, then X is a random variable if for each r 2 R, {! : X(w) r} 2 ⌃ (1.1) In this thesis we will deal only probability spaces (⌦, ⌃, P) where \|⌃\| is ﬁnite and ⌃= P(⌦). In this special case every subset of ⌦is an element of ⌃and so the requirement (1.1) will hold for any function X : ⌦! R. 2 1.1. Deﬁnitions The capital letters X, Y will be used to denote random variables, whilst the lowercase x, y used to denote particular events in ⌃. We denote the probability of the random variable X taking value x as P(X = x). Deﬁnition 1.3 (Expected value). Let X, be a random variable deﬁned on the ﬁnite probability space (⌦, ⌃, P). Then the expected value of X is deﬁned by E(X) := X x2⌦ xP(X = x). In this thesis we will ﬁnd that some of the random variables deﬁned on our random graph models have probability distributions that are well-known. For example, let Sj be the random variable which returns the degree of the jth white vertex in Gp. For any 1 j m we note in (2.2) that the degree of white vertex uj has a binomial distribution with parameters (n, p), in the graph p-model, Gp. We deﬁne the probability distributions we will encounter. In the following deﬁnitions we let [n] = {0, 1, . . . , n}. Deﬁnition 1.4 (Binomial distribution). Let ([n], P([n]), P) be a probability space and X be a real-valued random variable on this space. Then we say X is binomially distributed with parameters (n, p), if for each 0 r n, P(X = r) = ✓n r ◆ pr(1 −p)n−r. Deﬁnition 1.5 (Poisson distribution). Let (⌦, ⌃, P) be a probability space and X be a real-valued random variable on this space. Then we say X has a Poisson distribution with parameter c, if for each r 2 N, P(X = r) = 1 r!e−ccr. 1.1.2 Random bipartite graphs models In this section we deﬁne our models and state some of their elementary properties. First we will need a little more notation than has already been deﬁned on p.v. We will then deﬁne the random bipartite graph models, i.e. probability spaces over bipartite graphs. Notation The total number of edges in a bipartite graph can be determined from its degree sequence by summing the degrees of either side. Hence a necessary condition for 3 Chapter 1. Preliminaries an (m + n)-tuple, (s, t), to correspond to a degree sequence of a bipartite graph is that P j sj = P k tk. With this motivation we deﬁne the following subsets of Im,n, Em,n := {(s, t) 2 Im,n \| X j sj = X k tk} (1.2) Em,n,M := {(s, t) 2 Em,n \| X j sj = M} (for 0 M mn) (1.3) Em,nt0 := {(s, t) 2 Em,n \| t = t0} (for n-tuples t0). (1.4) Random variables This thesis will consider the following random variables on graphs. For 1 j m, let Sj be the random variable which returns the degree of the white vertex uj and deﬁne S := 1 m P j Sj. For 1 k n, let Tk be the random variable which returns the degree of the black vertex vk and deﬁne T := 1 n P k Tk. Deﬁne S := (S1, . . . , Sm) and T := (T1, . . . , Tn). The models The notation we deﬁned for our random bipartite graph models is consis-tent with that used for the general random graph models considered in [MW97]. There are two main random models for general graphs1, ˆ Gp and ˆ GM. Refer to Section 2.1 for a deﬁnition of these models and some important results in general (non-bipartite) random graph theory. As direct analogues of these in the bipartite case we deﬁne bipartite random graph mod-els; the graph p-model, Gp, and the graph edge-model, GM. We also deﬁne a model unique to bipartite graphs, the graph half -model, Gt. These three models are all probability spaces with the same domain, the set of all bipartite graphs on (m, n) vertices. Denote this domain by Bm,n and write 2Bm,n for the power set of Bm,n. Then we can express the probability spaces formally as (Bm,n, 2Bm,n, φ), where only the probability measure, φ, di↵ers between the three models. Graph p-model, Gp This model of a random bipartite graph appears in [GLS99], [Pal88] and [Pal84]. 1Refer to the ﬁrst section, The Basic Models, in the chapter on Random Graphs in the book, Modern Graph Theory, by Bollob´ as [Bol98] 4 1.1. Deﬁnitions Deﬁnition 1.6 (Graph p-model, Gp). The graph p-model, Gp(m, n) = (Bm,n, 2Bm,n, PGp) has domain the set of all labelled bipartite graphs on (m, n) vertices . A graph is chosen at random by selecting each of the possible mn edges independently with probability p. Hence in this model, the probability of a particular graph H with (m, n) vertices and \|H\| edges is, PGp(H) = p\|H\|qmn−\|H\|. For any degree sequence (s, t) 2 Em,n we are interested in the probability that a random graph in the graph p-model, Gp, has this degree sequence. The probability of the degree sequence (s, t) in the graph p-model written, PGp s, t + := PGp H : the degree sequence of H is (s, t) + . If we represent a bipartite graph as a binary matrix then the Gp(m, n) model corresponds to an m ⇥n matrix where each entry is independently chosen to be a 1 with probability p (and a zero with probability q = 1 −p). Thus for an (m + n)-tuple (s, t), PGp(s, t) is the probability that a binary matrix chosen at random in this way will have row sums s and column sums t. Graph edge-model, GM This model of a random bipartite graph appears in [GLS99], [ER61] and [Pal84]. Deﬁnition 1.7 (Graph edge-model, GM). The graph edge-modelGM = GM(m, n, M) has support the set of all labelled bipartite graphs on (m, n) vertices with M edges. We deﬁne each of the mn M + di↵erent graphs in the support to be equiprobable. Hence in the graph edge-model, the probability of a particular graph H with M edges vertices is PGM(G) = mn M +−1. Also notice that choosing a graph at random in the graph edge-model, GM corresponds to placing M ones in an m ⇥n binary matrix where each of the mn M + arrangements are equally likely. For any degree sequence (s, t) 2 Em,n,M we are interested in the probability that our random graph in the graph edge-model, GM, has this degree sequence. For the probability of the degree sequence (s, t) in the graph edge-model we will write, PGM s, t + := PGM({H : the degree sequence of H is (s, t)}). A consequence of the deﬁnitions of Gp and GM is that for 0 M mn and 0 < p < 1 then given any event A ⇢2Bm,n we have PGM(A) = PGp A \| M edges + . 5 Chapter 1. Preliminaries We now give an example. In Figure 1.1 we show the three bipartite graphs with degree sequence ((1, 3, 1, 1), (2, 1, 3)). We subsequently calculate the probability of this degree sequence occurring in each of the two random graph models we have deﬁned. Bipartite graphs with vertices (m, n) = (4, 3) Figure 1.1: The three possible bipartite graphs with degree sequence ((1,3,1,1), (2,1,3)). The total number of bipartite graphs on (4, 3) is 212 = 4096, of these, 12 6 + = 924 of these have precisely 6 edges. Hence, PGp= 1 2 ((1, 3, 1, 1), (2, 1, 3)) = 3 4096 PGM=6((1, 3, 1, 1), (2, 1, 3)) = 3 924 = 1 308 Graph half -model, Gt Unique to bipartite graphs we can deﬁne a third random graph model. In the graph half -model we ﬁx the degrees of the vertices on one side. Our con-vention will be to ﬁx the degrees of the black vertices. Deﬁnition 1.8 (Graph half -model, Gt.). The graph edge-model Gt = Gt(m, n, t) has support the set of all labelled bipartite graphs on (m, n) vertices such that the degrees of the black vertices agree with those in a speciﬁed n-tuple t. We deﬁne each of the di↵erent graphs in the support to be equiprobable. The graph half -model, Gt, is equivalent to the probability space over all m ⇥n binary matrices with column sums t where each matrix is weighted equally. Observe that in the graph half -model, the probability of a particular graph H on (m, n) vertices with black degree sequence matching the n-tuple t = (t1, . . . , tn) is ⇣m t1 +m t2 + . . . m tn +⌘−1 . 6 1.1. Deﬁnitions Let M = P k tk. For any degree sequence (s, t) 2 Em,n,M we are interested in the probability that our random graph in the half -model, Gt, has this degree sequence. The probability of the degree sequence (s, t) in the graph half-model is written, PGt(s, t) := PGt({H : the degree sequence of H is (s, t)}). A consequence of this deﬁnition is that for any n-tuple t and 0 < p < 1 then given any A ⇢Im,n,t we have PGt(A) = PGp A \| T = t + . Example Suppose we want to ﬁnd the probability of degree sequence ((1, 3, 1, 1), (2, 1, 3)) given we know that the degree sequence of the black vertices is (2, 1, 3), i.e. we want to calculate PGt=(2,1,3)((1, 3, 1, 1), (2, 1, 3)). We ﬁrst calculate the total number of di↵erent bipartite graphs on (4, 3) vertices with the prescribed degrees (2, 1, 3) for the black vertices. Consider vertex v1, there are two edges incident with this vertex creating 4 2 + = 6 distinguishable positions for these two edges. Hence there are 4 2 +4 1 +4 3 + = 96 distinguishable bipartite graphs with the required black vertices degree sequence. As illustrated by ﬁgure 1.1 on p6 there are three graphs with white vertex degree sequence (1, 3, 1, 1) and black vertex degree sequence (2, 1, 3). Hence, PGt=(2,1,3)((1, 3, 1, 1), (2, 1, 3)) = 3 96 = 1 32. Other models An interesting and general model is suggested by Palka in [Pal87]. A random graph is chosen with reference to an initial graph in the following way. Each edge of the initial graph remains with probability p (i.e. is deleted with probability q = 1 −p) and no new edges are added to the graph. There are two important special cases. In the bipartite case, if we take the initial graph, to be the complete graph on n vertices then this is the ˆ Gp(n) model deﬁned earlier and if we take the initial graph to be the complete bipartite graph on (m, n) vertices then we have the bipartite graph p-model, Gp, which we introduced in Deﬁnition 1.6. 1.1.3 Big ‘O’ notation We will be dealing with the asymptotic properties of real functions of two variables m and n. 7 Chapter 1. Preliminaries Deﬁnition 1.9 ( f(m, n) = O(g(m, n)) as m, n ! 1 subject to C). Suppose that f, g : N ⇥N ! R and C is a predicate on N ⇥N. Then we say ‘f(m, n) = O(g(m, n)) as m, n ! 1 subject to C’ if if there are constants A, N > 0 such that \|f(m, n)\| A\|g(m, n)\| whenever m, n ≥N and (m, n) satisﬁes C. The corresponding term for little ‘o’, f(m, n) = o(g(m, n)) as m, n ! 1 subject to C is deﬁned analogously. In our case m and n will be the number of white and black vertices respectively. We will consider asymptotic results, as our graphs get large, where we also require that the two sets of vertices are not too di↵erent in size. The following section is not required for understanding. It gives possible motivation for why one might want to deﬁne random bipartite graphs in the form of a toy example. A detailed calculation in the Gt model is also included. 1.2 Motivating example: malaria transmission Many biological systems are naturally modelled by random bipartite graphs. See for ex-ample [ZS02]. We deﬁne a toy example below. Say we are modelling a system of n mosquitoes and m people. Label the mosquitoes u1, u2, . . . , um and the people v1, v2, . . . , vn and draw each as a vertex of a graph. Malaria can be spread when a mosquito (uk) bites a person (vj) which we denote by drawing an edge between uk and vj. There are no edges between any two vertices ui and uj because there is human to human2 transmission of malaria (and similarily no mosquito to mosquito transmission ensures no edges between the vj and vk). This ensures that the graph drawn to convey this infor-mation will be bipartite. The requirement that each mosquito can only bite each person once ensures that the graph has no multiple edges. 2forgetting blood transfusions 8 1.2. Motivating example: malaria transmission • • • • • • • • • • • • • • • u1 Ev1 u2 Ev2 u3 Ev3 Evn um−1 um R R R R R R R R R R R R R R R R R R R R R m m m m m m m m m m m m m m m m m m m m m i i i i i i i i i i i i i i i i i i i i i $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ Here, sj, the degree of uj, is the number of people bitten by the mosquito, uj, and tk denotes the number of bites person vk received. We have deﬁned our bipartite graph representation of the malarial model and now can detail the interpretation of our random graph models: Gp, GM, Gt. We also illustrate the malarial interpretation of Gt with a de-tailed calculation. In the following three examples a bipartite graph is chosen at random from the bipartite graph models Gp, GM and Gt respectively. Suppose we interpret the random graph gener-ated as the bite pattern observed after a night with m mosquitos and n people left in the same room. We list the assumptions about the spread of malaria implicit in each model. Malaria interpretation of the graph p-model Gp. Assumption: Between each mosquito and human there is an equal and independent probability, p, of a bite occurring. Malaria interpretation of the graph edge-model GM. Assumption: Overnight there are M bites altogether. Between each mosquito and hu-man there is an equal likelihood of a bite occurring. Malaria interpretation of the graph half -model Gt. Assumption: Overnight the number of bites each person received is recorded in an or-9 Chapter 1. Preliminaries dered3 n-tuple t = (t1, t2, . . . , tn). All combinations of particular mosquitoes causing bites to the humans that yield the recorded bites (i.e. the correct t) are equally likely. Suppose we are interested in the number of bites made by each of the mosquitoes. For example we may want to know the likelihood that one mosquito makes greatly many more bites than the other mosquitoes. In this case it would become very critical whether or not this highly active mosquito was infected with malaria. We record the number of bites each mosquito receives in a m-tuple s = (s1, s2, . . . , sm). Then the probability that all mosquitoes make the number of bites encoding in s given that the people were bitten as in t is precisely PGt(s). Toy Example: Consider a system with ﬁve people in a room overnight with two mosquitos. The people, in alphabetical order, were bitten 1,1,2,2 and 1 times. It is unknown which mosquitoes caused these bites. We assume that any scenario, under which the people were bitten the prescribed number of times, is equally likely. That is, we assume that any bipartite graph with a degree sequence of (1,1,2,2,1) for the black vertices is equally likely. We draw each of these bi-partite graphs below and list the degree sequence of the white vertices for each graph. We ﬁnd the probability that the ﬁrst mosquito, u1, makes 3 bites and the second mosquito, u2 makes 4 bites. i.e. PGt(S = (3, 4)). PGt S = (3, 4) + = PGM=7 S = (3, 4) \| T = (1, 1, 2, 2, 1) + = 3 8 Similarly the probability that the mosquitos make 2 and 5 bites respectively is, PGt S = (2, 5) + = PGM=7 S = (2, 5) \| T = (1, 1, 2, 2, 1) + = 1 8 3i.e. tk is the number of bites received by the k-th person, E vk. 10 1.2. Motivating example: malaria transmission • • • • • • • s = (4, 3) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 • • • • • • • s = (3, 4) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 • • • • • • • s = (4, 3) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E R R R R R R R R R R R R R R R R R R R R R 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 • • • • • • • s = (5, 2) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E R R R R R R R R R R R R R R R R R R R R R 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 • • • • • • • s = (4, 3) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E m m m m m m m m m m m m m m m m m m m m m R R R R R R R R R R R R R R R R R R R R R 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 • • • • • • • s = (3, 4) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E R R R R R R R R R R R R R R R R R R R R R m m m m m m m m m m m m m m m m m m m m m 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 • • • • • • • s = (2, 5) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E m m m m m m m m m m m m m m m m m m m m m 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 • • • • • • • s = (3, 4) E E E E E E E E E E E E E E E E E E E E E E E E E Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E E E E E E E E E E E E E E E E E E E E E E E E E 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 m m m m m m m m m m m m m m m m m m m m m Figure 1.2: All bipartite graphs on (2, 5) vertices with black degree sequence (1, 1, 2, 2, 1) 11 Chapter 2 Literature review This thesis concerns the asymptotic behaviour of random bipartite graphs. The general ﬁeld of random graphs originated with an inﬂuential paper of Erd˝ os and R´ enyi, [ER61] (see Section 2.1.1). One of the key properties introduced in [ER61] was that small changes in the initial parameters could lead to abrupt changes in the asymp-totic behaviour of random graphs. This behaviour has been likened to phase transitions in physics and hence began the interest in what are termed threshold functions in random graphs. Threshold functions are discussed in 2.1.2. The particular aspect of random graphs that we consider is the asymptotic behaviour of vertex degrees. Results in this area for both the general and bipartite cases are (brieﬂy) surveyed in Sections 2.1.3 and 2.2.1 respectively. We also compile some graph theory results counting the asymptotic number of graphs with a particular degree sequence. Enumeration results have been used to construct binomial models which approximate the general random graph models in [MW97]. Some results from this paper are described in Section 2.1.4. This thesis could could be considered a bipartite analogue of [MW97]. We proceed from a bipartite graph enumeration result to show that the asymptotic probabilities of degree sequences in random bipartite graphs can be approximated by binomial models. The bipartite enumeration result we use is from [GM09]. We give details of this result in Section 2.2.2. The material in this chapter comes under three general headings: analogous results, pre-vious work not used in this thesis and previous work on which this thesis builds. The only section which we will directly make use of is Section 2.2.2 in which we give results on the asymptotic enumeration of bipartite graphs. Section 2.1.4 covers a result in general graph theory which is analogous to our work. The remainder of the literature review will not be needed in the thesis but serves to give context for our results. 12 2.1. General random graphs 2.1 General random graphs 2.1.1 Early results The ﬁeld of random graphs began with two inﬂuential papers of Erd˝ os and R´ enyi, [ER61] and [ER59]. A random graph. To deﬁne a random graph it is necessary to construct a domain from which it is sampled and a way in which it is sampled from that domain. For the traditional non-bipartite case, two models, ˆ Gp(n) and ˆ GM(n), are the most extensively studied. (These models are analogous to Gp(m, n) and GM(m, n) which we will later de-ﬁne for bipartite graphs. The hat over the G in our notation is not usual but used here to avoid confusion between the general case and the bipartite case.) There is an interesting construction introduced by Janson in [Jan94] which allows us to deﬁne both models ( ˆ Gp(n) and ˆ GM(n)) at the same time. Begin with n labelled vertices. Let t1, . . . , t(n 2) be times distributed uniformly at random in (0, 1), then labelled in as-cending order. We now add an edge at random at each time ti. This yields a random graph in ˆ Gp(n) at time p 2 (0, 1) but also a random graph in ˆ GM(n) at the M th time, tM. When Erd˝ os and R´ enyi’s paper [ER59] was published, an already famous topic in tradi-tional (non-random) graph theory was the chromatic number of graphs. Chromatic number, χ. The chromatic number of a graph G, denoted χ(G), is deﬁned to be the number of colours required to colour the vertices of G in such a way that any two vertices connected by an edge in G are coloured di↵erently. Theorem 2.1 (Erd˝ os and R´ enyi [ER61]). Let G be a random graph in ˆ GM(n) with pa-rameter M = M(n). Fix positive constants c < 1 2 and k. Then as n ! 1 the following results hold almost surely, χ(G) = 2 if M(n) = o(n), χ(G) = 3 if M(n) = cn, and χ(G) > n k if M(n) = n 2 + −o(n2(1−1/k)). This would be the ﬁrst result on the chromatic number of a random graph. Much later, it was shown by [SS87] and [ Luc91] that the chromatic value of a random graph G in ˆ Gp is almost surely within one of the expected chromatic value over all graphs in ˆ Gp. This is a seminal result in random graph theory. The paper [SS87] is also of particular interest to us as it applies a powerful probabilistic technique, Doob’s martingale process. A similar application of this technique is central to our proof that pathological degree sequences are 13 Chapter 2. Literature review rare in the graph half -model, Gt. We give an indication of the proof in [SS87] in Section 3.2.4. Notice that in Theorem 2.1, varying the value of the parameter M(n) changed the likely chromatic number of a graph picked at random from ˆ Gp. This is an example of a threshold function on random graphs, a class of functions introduced in [ER61]. 2.1.2 Threshold functions Erd˝ os and R´ enyi’s paper [ER61] sparked a ﬂurry of papers deriving the threshold functions for di↵erent properties of random graphs. Examples include .... These many isolated results were set in a general framework by Bollob´ as and Thomason in [BT87]. They gave a non-constructive proof showing the existence of threshold functions for a wide class of properties on graphs. We give this result below in Theorem 2.2, but prior to this we must make some deﬁnitions. In this section we work in the probability space ˆ GM(n) where every graph on n vertices with M edges is equally likely. Consider this space for di↵erent values of the parameter M(n). Let G(n) be the set of all labelled graphs on n vertices. A property Q is monotone increasing on G(n) if graph A 2 G(n) satisfying Q implies that any graph B 2 G(n) for which A is a subgraph must also satisfy Q. A non-trivial property is one such that the set of graphs G 2 G(n) for which the property holds as well as the set of graphs G0 2 G(n) for which the property fails are both non-empty. For example, the property of having a triangle as a subgraph is a non-trivial monotone increasing property on the set G(n) for n ≥3. Notice that if a property is non-trivial and monotone increasing then Q must hold for the complete graph and fail for the empty graph. Hence we have P ˆ GM=0(Q) = 0 and P ˆ GM=(n 2) (Q) = 1. This is important as it guarantees the existence of the function M ⇤(n) in Theorem 2.2. (The expression, P ˆ GM(Q), is the probability that a graph chosen at ran-dom from those with M edges will have property Q. The hat over the G is to indicate that we are working in the space of general random graphs and not random bipartite graphs.) We are now ready to state the Bollob´ as and Thomason’s Theorem showing general ex-istence of threshold functions. This appeared as Theorem 4 in their 1985 paper [BT87]. Their original theorem is for more general sets, but we give the result only as it applies to random graphs in ˆ GM. Theorem 2.2 (Bollob´ as & Thomason 85). Let Q be a monotone increasing property on the set of all labelled graph on n vertices, G(n). Deﬁne the function M ⇤(n) = max{l : 14 2.1. General random graphs P ˆ GM=l(Q) 1 2} and let w(n) ! 1. Then for any M(n) M ⇤(n)/w(n), P ˆ GM(Q) 1 −2−1/w, and for any M(n) ≥(M ⇤(n) + 1)w(n), P ˆ GM(Q) ≥1 −2−w. In the theorem the function M ⇤(n) is a threshold function for Q. Hence to show the existence of a threshold function for any property by this theorem it is suﬃcient to show that the property is non-trivial and monotone increasing. In many cases this is not hard1. 2.1.3 Vertex degrees There are two excellent surveys of this area, chapter 3 of [Bol01] and section two on [Pal88]. In the subject of vertex degrees in general random graphs it is the behaviour of two random variables that are the most studied. We deﬁne these below, Let X(r) be the random variable that returns the rth largest degree in the graph and let ⇣r be the random variable that returns the number of vertices of degree r. Number of vertices of degree r. We will ﬁrst consider the random variable ⇣r. For a ﬁxed r, the likely value of ⇣r is dependent on the expectation of ⇣r. The following result appears in [Bol01] and is from Bollob´ as’ paper [Bol82]. By almost surely we mean with probability tending to one as n ! 1. Lemma 2.3 (Bollob´ as 82). Fix t 2 R+ and " > 0. Let ˆ Gp = ˆ Gp(n) where p(1−p) ≥"n−3/2. Then almost surely, ⇣r = 0 if E ˆ Gp(⇣r) ! 0 ⇣r ≥t if E ˆ Gp(⇣r) ! 1. Also, if E ˆ Gp(⇣r) = c, 80 k n, P ˆ Gp ⇣r = k) ! 1 k!e−cck. 1For example consider the property of having a particular non-empty subgraph H on k vertices for any k n. This property is non-trivial and monotone increasing on the set G(n). (Why? Observe H is not a subgraph of the empty graph and is a subgraph of the complete graph so the property is non-trivial. Also, if H is a subgraph of the graph A and A is a subgraph of B then B contains all the edges of A and so H must be a subgraph of B. Hence the property is also monotone increasing and so we are done.) 15 Chapter 2. Literature review Note that this last results is often written as ⇣r ⇠Po(c) where Po(c) denotes the Poisson distribution with parameter c. Results similar to Lemma 2.3 hold for the distributions of the number of vertices with degree at least r and also for the number of vertices with degree at most r (see [Bol01][p.63]). The rth largest degree. Let d(r) be the random variable which takes the value of the rth largest degree. Naturally two particular values of r are the most studied, namely d(1) the maximum degree and d(n) the minimum degree. For suitable p these minimum and maximum degrees are unique. This was shown by Erd˝ os and Wilson in [EW77]. The following theorem appears in [Bol01]. Theorem 2.4 (Erd˝ os and Wilson 1977). Let ˆ Gp = ˆ Gp(n) for some p < 1 2 such that pn ln n ! 1. Let A be the event that a graph has both a unique minimum and a unique b maximum degree. Then as n ! 1, P ˆ Gp(A) ! 1. The probability that the maximum degree is unique depends on the value of the parameter p. For example, if p = o( 1 n ln n) then almost surely neither the maximum nor minimum degree is unique [p.68][Bol01]. There are results which show that sometimes this almost sure uniqueness can be extended beyond the maximum and minimum degrees. Fix i. Let Bi be the event that the largest i degrees are all unique, i.e. d(n−i+1) < . . . < d(n). Palka, in Lemma 2.1.2 of [p.17][Pal88] provides conditions under which P ˆ Gp(Bi) ! 1 as n ! 1. 2.1.4 Approximation by binomial models The paper by McKay and Wormald [MW97] constructs random models based on binomial variables that approximate the distribution of degree sequences in random graphs as the number of vertices gets arbitrarily large. This paper is of particular interest to us as their results, which concern general random graphs, are analogous to the results we prove for bipartite random graphs. We describe the enumeration result, Theorem 2.5, which forms the starting point for their paper [MW97]. Enumeration results Havel and Hakimi independently found the conditions under which an n-tuple is the degree sequence of a graph on n vertices in [Hav55] and [Hak62] 16 2.1. General random graphs respectively. In lieu of a simple closed form solution, much work has been done ﬁnding asymptotic results for the number of graphs with given degree sequence. We are interested in the following enumeration result, also by McKay and Wormald, in [MW90]. (Let \|G(d)\| denote the number of graphs on n vertices v1, . . . , vn such that the degree of each vi is di. Also, write λ = n 2 +−1 P i di and d = 1 n P i di.) Theorem 2.5 (McKay & Wormald 90). Suppose d is an n-tuple such that for each 0  i n, \|di−d\| n1/2+" and λ, 1−λ ≥ c log n for some c > 4 3. Let γ = (n−1)−2 Pn i=1(di−d)2. Then, \|Gn(d)\| = p 2 n Y i=1 ✓n −1 di ◆ exp ⇣1 4 − γ2 4λ2(1 −λ)2 + o(1) ⌘ . Binomial models McKay and Wormald construct binomially based models which will later be shown to approximate the random graph models closely. Each model is based on independent binomially distributed random variables with parameters (n −1, p) subject to certain constraints. Observe that in ˆ Gp if one considers a particular vertex vi then there are n −1 other ver-tices and an independent probability p of joining to each one. Thus the degree of vi is binomially distributed with parameters (n −1, p). We note however that the degrees of two di↵erent vertices vi and vj are not independent; for example if the degree of vi is n−1 then there must be an edge between vi and vj and so the degree of vj is at least 1. Also the sum of all the degrees is twice the number of edges in the graph, so in particular the sum of the degrees must be even. This motivates the ﬁrst deﬁnition. Let ˆ Ep be the space of n binomially distributed random variables with parameters (n−1, p) subject to even sum. McKay and Wormald show in Lemma 2.2 [MW97] that, PEp(d) = ✓1 2 + 1 2(q −p)2N ◆−1 pmq2N−m n Y i=1 ✓n −1 di ◆2 . Consider ˆ GM, the probability space in which every graph with M edges has equal proba-bility. The aim is to construct ˆ EM to approximate ˆ GM. Let ˆ EM be the space of n binomially distributed random variables with parameters (n−1, p) subject to sum M. Also deﬁne an integrated model over n-tuples with even sum, 17 Chapter 2. Literature review Pˆ Ip(d) = 2 V (p) 1 −(q −p)2N+ n Y i=1 ✓n −1 di ◆Z 1 0 Kp(p0)(p0)m(1 −p0)2N−m dp0, where Kp(p0) = s N ⇡pq exp ⇣ −(p −p0)2N pq ⌘ and V (p) = Z 1 0 Kp(p0) dp0. Results These binomial models are shown to provide good approximations to the dis-tribution of the degree sequences in the general random graph models ˆ Gp and ˆ GM. McKay and Wormald original theorem in [MW97] applies to all random variables on any normed space but we state it here only for real random variables. (Let ˆ In be the set of n-tuples with entries between 0 and n −1 and having even sum. The theorem holds only for what McKay and Wormald term acceptable values of the parameters p, M, see [MW97] for the deﬁnition.) Theorem 2.6 (McKay and Wormald [MW97]). For n ≥1, let Xn : ˆ In ! R be a random variable. Let !(n) be any function such that !(n) ! 1 and "(n) be any function such that "(n) ! 0. Then as n ! 1 subject to p = p(n) is acceptable, \|E ˆ Gp(Xn) −Eˆ Ip(Xn)\| = o(1)Eˆ Ip(\|Xn\|) + O ⇣ n−!(n) + exp(−"(n)(pqN)1/3) ⌘ max d2ˆ In \|Xn(d)\|. Similarly, as n ! 1 subject to M = M(n) is acceptable, \|E ˆ GM(Xn) −EEM(Xn)\| = o(1)EEM(\|Xn\|) + n−!(n) max d2In,m \|Xn(d)\|. The proof of this result (in [MW97]) involves showing that, P ˆ Gp ✓1 4 − γ2 4λ2(1 −λ)2 = o(1) ◆ (2.1) is asymptotically very close to 1. The importance of this relates to the enumeration result in Theorem 2.5. If the value of (2.1) is close to 1 then this means that for ‘most’ degree sequences in ˆ Gp, Theorem 2.5 provides a simple asymptotic enumeration for the number of graphs with that degree sequence. We will show the bipartite analogue of Theorem 2.6 in Theorems 7.4 and 7.5. 18 2.2. Bipartite random graphs 2.2 Bipartite random graphs Random bipartite graphs relate to another class of random graphs. Let us refer to the graph on three vertices with every edge present as a triangle. A graph which does not contain this graph as a subgraph is called triangle free. Asymptotically almost all triangle free graphs are bipartite [EKR76]. Also, as bipartite graphs contain no odd cycles, all bipartite graphs are triangle free. Hence results concerning triangle free graphs can imply results on bipartite graphs and vice-versa. There are many results for random bipartite graphs concerning what are called ‘match-ings’. A matching of a graph is a subset, E0, of the edge set such that each vertex in the graph is incident with at least one edge in the set E0. These matching results on random bipartite graphs are surveyed in []. Janson et. al. use results on matchings in random bipartite graphs to derive similar results in general random graphs, see [J LR00, p.85]. However, as matchings do not directly concern vertex degrees we will not include these results in this section. The work we shall focus on is divided into two sections, the ﬁrst concerns the vertex degrees in random bipartite graphs and the second is on enumeration results which count bipartite graphs by degree sequence. We include Section 2.2.1 on vertex degrees because it is this ﬁeld of work in which our thesis ﬁts. Thus it is important to survey previous results in this area to a give context to our work. The enumeration results in Section 2.2.2 are included for a vastly di↵erent reason. In this thesis we will approximate the probability of ﬁnding a particular degree sequence (s, t), in each of our random graph models. The asymptotic enumeration result forms the basis for these calculations. 2.2.1 Vertex Degrees We consider the properties of the degrees of the vertices in random graphs. This is the body of work into which our results will ﬁt so we make a brief survey of previously known results in the area. Many results steam from results in the general random graphs that can also be applied to the bipartite case. Except for the preliminary results none of these will be needed in the thesis so this section is entirely a literature survey. There are very few results on the vertex degrees of random bipartite graphs. Many of the results that have been found stem from similar results on the vertex degrees in general 19 Chapter 2. Literature review random graphs. Thus the results in this section can seem somewhat disjointed as they mostly follow from their counterparts in the general case rather than having been derived from other results in bipartite random graph theory. For completeness, we include all re-sults known to us in the area of asymptotics of vertex degrees in random bipartite graphs. Preliminary Consider the bipartite graph p-model, Gp(m, n). Fix a white vertex uj and let Sj be the random variable that returns the degree of uj. By Deﬁnition 1.6 of Gp, there is an independent probability p that uj is joined to each of the n black vertices. Hence, as noted in [Pal84], 80 r n, PGp Sj = r + = ✓n r ◆ prqn−r. (2.2) We say that Sj is binomially distributed with parameters n and p. Let ⇠r be the random variable which returns the number of white vertices with degree r. By (2.2), each of the m white vertices has degree r with probability n r + prqn−r. Hence, ⇠r is binomially distributed with parameters m and n r + prqn−r. By this we mean, 80 k n, PGp ⇠r = k + = ✓m k ◆ ✓n r ◆ prqn−r !k 1 − ✓n r ◆ prqn−r !m−k . (2.3) Result (2.3) appears in [Pal84]. Palka also notes that the converse is true for ⌘r, the number of black vertices with degree r. This random variable, ⌘r, is binomially distributed with parameters n and m r + prqm−r. Number of vertices of degree r Godbole et. al. in [GLS99] study random bipartite graphs in which the numbers of black and white vertices are equal (in our notation, m = n). They consider an n⇥n chessboard in which a rook is placed on each square with independent probability p(n). There is a natural bijection between this random chessboard and a random graph in Gp(n, n), where a rook in row j column k corresponds to an edge between vertices uj and vk. The degree of uj (resp. vk) then corresponds to the number of rooks in row j (resp. column k). Note any rooks in the same column or same row can be considered mutually threatening. Godbole et. al. deﬁne the random variable Yr to be the number of sets of r mutually threatening rooks. We observe that this corresponds to the number of vertices with degree h ≥r, weighted by h r + . Let ⇣r be the random variable which returns the number of vertices (either black or white ) of degree r. Then Yr can be written, Yr = ⇣r + (r + 1)⇣r+1 + r+2 r + ⇣r+2 + . . . n r + ⇣n. (2.4) We translate Theorem 2.1 of [GLS99] into our own notation. 20 2.2. Bipartite random graphs Theorem 2.1 (Godbole et. al. [GLS99]). Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Suppose also that m = n. Then as n ! 1, PGp Yr = 0 + = 8 < : 1 for p = o ⇣ 1 n1+ 1 r ⌘ 0 for p−1 = o(n1+ 1 r ) (2.5) There is an alternate way to generate a random chessboard; place M rooks onto the board such that all n2 M + possible arrangements are equally likely. This corresponds to the bipartite model GM(m, n) for m = n. Godbole et. al. prove a result in this model (i.e. BM) similar to Theorem 2.2. The paper also gives the following result on the limit of the distribution of Yr. Theorem 2.2 (Godbole et. al. [GLS99]). Let Gp = Gp(m, n) be as in Deﬁnition 1.7 where p = o(n−1). Suppose also that m = n. Then as n ! 1, PGp Yr = k + = e−cck k! where c = 2n n r + pr (2.6) Number of vertices with ﬁxed degree r. So the distribution of ⇣r converges to the Poisson distribution when the expectation of ⇣r is ﬁnite and converges to the normal distribution when the expectation of ⇣converges to inﬁnity, as o(m"). This has some interesting corollaries. The parameters in the probability distributions of ⇣r can be calcu-lated explicitly for particular values of p. This is done by Palka [Pal84] in the following corollaries. (We write X ⇠N(0, 1) to denote that the distribution of the random variable X converges to the normal distribution with expectation 0 and variance 1, see [Pal84] for precise deﬁnition.) Corollary 2.3. Fix 0 < β < 1. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Suppose also that m = ↵n. 1. Let r ≥2. Suppose np = βm−1/r. Then as m ! 1, ⇣r ⇠P(λ). i.e. PGp(⇣r = c) = λce−λ c! where λ = 1 r!βr(1 −↵1−r) 2. Let r ≥0. Suppose np = ln m −β ln ln m + o(ln ln m). Then as m ! 1, (⇣r −a) pa ⇠N(0, 1) where a = 8 < : 2 r!(ln m)r+β if ↵= 1 1 r!(ln m)r+β if 0 < ↵< 1 21 Chapter 2. Literature review 3. Let r ≥1 and 0 < y < 1. Suppose np = ln m −(1 −γ)r ln ln m + o(ln ln m). Then as m ! 1, (⇣r −a) pa ⇠N(0, 1) where a = 8 < : 2 r!(ln m)rγ if ↵= 1 1 r!(ln m)rγ if 0 < ↵< 1 4. Let r ≥0 and −1 < γ < 1. Suppose np = ln m −r ln ln m + γ + o(1). Then as m ! 1, ⇣r ⇠P(λ) where λ = 8 < : 2e−γ r! if ↵= 1 e−γ r! if 0 < ↵< 1 5. Let r ≥0. Suppose np = ln m −r ln ln m + f(m), where f(m) = o(ln ln m) tends to 1. Then as m ! 1, PGp ⇣r = 0 + ! 1 Let ⇠r (resp. ⌘r) be the number of white (resp. black) vertices of degree r. The ﬁnal result in [Pal84] concerns ⇠r and ⌘r for the case m = n, i.e. when the number of white vertices is equal to the number of black vertices. This result, shown in Theorem 2.4, gives the asymptotic distribution of ⇠r and ⌘r individually (rather than for the random variable ⇣r = ⇠r + ⌘r). Theorem 2.4. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Suppose also that m = n. Then, lim m!1 PGp ⇠r = i, ⌘r = j + = µi+j i!j! e−2µ where µ = e−γ/r! The limiting distribution of ⇠r and ⌘r is the bi-variate Poisson distribution with parameter µ [Pal84]. In a bipartite graph on (m, n) vertices the relative sizes of m and n are important. Palka, in [Pal87] shows that in Gp(m, n) that when the number of white vertices is suﬃciently large in comparison to the number of black vertices then the vertex of maximum degree is almost surely a black vertex. Indeed for suitable p and suitable ratio of m greater then n, Palka shows for ﬁxed i that almost surely the smallest i degrees will all be on white vertices. Similar results are shown by Rucinski for k-partite2graphs in [Ruc81]. In a bipartite graph the number of edges incident with each colour class of vertices is the same so the colour with fewer vertices will have a higher average degree, and so it makes sense that it is also more likely to host the vertex of maximum degree. 2A k-partite graph is one in which the vertices are partitioned into k sets, V1, . . . , Vk and the edge set E is a subset of {(va, vb) : va 2 Vi, vb 2 Vb for i 6= j}. 22 2.2. Bipartite random graphs 2.2.2 Enumeration results We are interested in results that give counts for the number of bipartite graphs with a given degree sequence. Much of the early enumeration work was done in the context of binary matrices with given row and column sums. Ryser [Rys63] found the necessary and suﬃcient conditions on the row and column sums that guarantee the existence of a binary matrix with such sums. A closed form solution for the precise number of bipartite graphs with given degree sequence is unknown but asymptotic results are given by [Bar10] and others. We shall use the result in [GM09]. McKay and Greenhill derived a formula for ﬁnding the number of bipartite graphs for some given degree sequences Theorem 2.1 of [GM09, p.4 ]. This result forms the foundation of the thesis. Before stating the theorem we deﬁne (", a)-regular degree sequences. Deﬁnition 2.5 (acceptable for a, m and n). Let fm,n(x) := (1 −2x)2 4x(1 −x) 1 + 5m 6n + 5n 6m ! . If fm,n(x) < a ln n then x is acceptable for a, m and n. Conditions 2.6. These conditions apply to an m-tuple s = s(m) = (s1, ..., sm) and an n-tuple t = t(n) = (t1, ..., tn). The conditions depend on the parameters a and ". For m, n ! 1, • sj −s and tk −t are uniformly O(n1/2+") for 1 j m and 1 k n. • P sj = P tk and λ = 1 mn P j sj = 1 mn P k tk is acceptable for a, m and n. Deﬁnition 2.7 ((", a)-regular). If m, n and (s, t) satisfy Conditions 2.6 then we say that (s, t) is (", a)-regular. We now state the enumeration result by Greenhill and McKay (stated as Theorem 2.1 in [GM09, p.4]). To maintain consistency with the notation of [GM09] we denote the number of graphs with degree sequence (s, t) by \|B(s, t)\|. Theorem 2.8 (Greenhill and McKay). Let a, b 2 R+ such that a + b < 1 2. Then there exists " = "0(a, b) > 0 such that the following holds. 23 Chapter 2. Literature review If (s, t) is (", a)-regular then as m, n ! 1, \|B(s, t)\| = ✓mn λmn ◆ −1 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ ⇥exp −1 2 ⇣ 1 − P j(sj −s)2 λ(1 −λ)mn ⌘⇣ 1 − P k(tk −t)2 λ(1 −λ)mn ⌘ + O(n−b) ! . (2.7) Greenhill and McKay also note that the error term O(n−b) is uniform. This will be im-portant in Section 7.3. Note that if a degree sequence (s, t), satisﬁes ⇣ 1 − P j(sj−s)2 λ(1−λ)mn ⌘⇣ 1 − P k(tk−t)2 λ(1−λ)mn ⌘ = O(n−b), then the enumeration in (2.7) simpliﬁes. This motivates the following deﬁnition. Deﬁnition 2.9 ((a, b, m, n, ")-pathological). A degree sequence is called non-(a, b, m, n, ")-pathological if it is both (", a)-regular and ⇣ 1 − P j(sj−s)2 λ(1−λ)mn ⌘⇣ 1 − P k(tk−t)2 λ(1−λ)mn ⌘ = O(n−b). Corollary 2.10. Let a, b > 0 be constants such that a + b < 1 2. Then there exists " = "0(a, b) > 0 such that the following holds. If (s, t) is not (a, b, m, n, ")-pathological then as m, n ! 1, \|B(s, t)\| = ✓mn λmn ◆ −1 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ 1 + O(n−b) + . (2.8) Hence, for any non-pathological degree sequence we now have the simple enumeration formula in Corollary 2.10. We prove in this thesis that in each of our random graph models (under suitable parameters) almost all graphs have non-pathological degree sequences. In particular we show this for models Gp, GM and Gt in Theorems 4.9, 5.2 and 6.20 respectively. 24 Chapter 3 Probabilistic bounds and techniques In this thesis at many points it will be necessary to show that a random variable is very likely to take a value close to its mean. To enable us to achieve this we make use of some concentration inequalities from probability theory. They are so-called because under suit-able conditions they can show that a random variable is likely to be concentrated about its mean. We will often show that the sum of a large number of random variables is very likely to fall close to the mean of the sum of those random variables. We will deal with two cases. Most random variables we deal with will be independent. This allows us to use the concentration results by Hoe↵ding and McDiarmid (Theorems 3.1 and 3.2 respectively). In the graph half -model, Gt, however, we will need to be able to manipulate random vari-ables which are not independent. For this situation we construct a martingale (weaker then independence) which allows us to use a generalised Azuma theorem (Theorem 3.8) to gain the necessary concentration results. Our method to construct this martingale follows the famous technique known as Doob’s martingale process. The theory for this technique forms Section 3.2.4 where we also run through the iconic proof by Shamir and Spencer in [SS87]. This proof shows the chromatic number in a general random graph is highly concentrated. It is a good, clean example of the Doob process invaluable to understand our application of it to the graph half -model which has some added complications. 25 Chapter 3. Probabilistic bounds and techniques 3.1 Concentration inequalities for independent ran-dom variables We use some general theorems from probability theory, which give ways to compute the likelihood that variables lie very close to their expectations. These Theorems were listed in the survey by Chung and Lu [pp.84-86][CL06]. The ﬁrst of these concentration equalities is due to Cherno↵and appears in his paper [Che81]. The second is done by McDiarmid in his paper [McD98]. Theorem 3.1 (Cherno↵1981). Let X1, . . . , Xn be independent random variables such that P(Xi = 1) = pi and P(Xi = 0) = 1 −pi. Deﬁne X = Pn i=1 Xi. This implies the following bounds, P(X E(X) −k) e−k2/2E(X) P(X ≥E(X) + k) e −k2 2(E(X)+k/3). And hence for positive k, P(\|X −E(X)\| ≥k) e−k2/2E(X) + e −k2 2(E(X)+k/3) < 2e −k2 2(E(X)+k/3). In most applications, pi = p for all i, in which case E(X) = np. The following Theorem is used to prove Lemma 4.4. It is quite powerful in this context because in Lemma 4.4 we have strong bounds on the di↵erence between each random variable and it’s expectation. Theorem 3.2 (McDiarmid 1998). Let X1, . . . , Xn be independent random variables sat-isfying \|Xi −E(Xi)\| ci for 1 i n. Deﬁne X = Pn i=1 Xi. Then, P(\|X −E(X)\| ≥k) e −k2 2 Pn i=1 c2 i . 3.2 Martingales 3.2.1 Preliminaries Deﬁnition 3.1 (σ-algebra). A set F is a σ-algebra if it satisﬁes the following conditions, • F 6= ? • A 2 F ) Ac 2 F 26 3.2. Martingales • A1, A2, . . . 2 F ) [1 i=1Ai 2 F Suppose we have a set P = {Yi}i2I which partitions ⌦, then there is a natural σ-algebra on ⌦corresponding the partition P. The following construction is from [CMZ09]. Deﬁnition 3.2 (σ-ﬁeld induced by partition Pi). Let Fi be the collection of all sets which may be deﬁned as unions of blocks in the partition Pi. Then we say Fi the σ-ﬁeld induced by the partition Pi. Cooper et. al. [CMZ09] also note that for a set of partitions {Pi}0in where each Pi+1 is a reﬁnement Pi, the set of σ-algebras {Fi}0in induced by the partitions {Pi}0in form a ﬁlter. A martingale is deﬁned with respect to the ﬁlter as follows [CMZ09]. Deﬁnition 3.3 (martingale with respect to ﬁlter). Let (⌦, F, P) be a probability space with a ﬁlter {Fi}0in. Suppose that X0, . . . , Xn are random variables such that for each 0 i n, Xi is Fi measurable. The sequence X0, . . . , Xn is a martingale provided that E Xi+1 \| Fi + = Xi for each 0 i < n. Conditional expectation The expectation of a random variable X conditional on a σ-algebra can be quite generally. However, we deﬁne only the special case which we will require, when the σ-algebra is induced by a partition and ⌦is a ﬁnite set. Note that in the probability space (⌦, ⌃, P) where \|⌦\| < 1 that any random variable Y : ⌃! R can be deﬁned by giving its value at each ! 2 ⌦. Deﬁnition 3.4 (Conditional expectation, E X \| F + ). This is deﬁned only for the special case that the σ-algebra F is induced by some partition {Yi}i2I of ⌦. We also assume that ⌦is ﬁnite. Let ! 2 ⌦. Now, because {Yi}i2I is a partition, 9! j 2 I such that ! 2 Yj. Then, E X \| Fi + (!) := P !02Yj X(!0)P(!0) P(Yj) . 27 Chapter 3. Probabilistic bounds and techniques We observe the following two special cases. Let F0 = {?, ⌦}. Then F0 is induced by the trivial partition {⌦}, and, E X \| F0 + (!) = P !02⌦X(!0)P(!0) P(⌦) = E(X). (3.1) Now suppose, conversely, that we have the ﬁnest possible partition, P = {{!} : ! 2 ⌦}. Let F⇤be the σ-algebra induced by P, then, E X \| F⇤+ (!) = P !02{!} X(!0)P(!0) P(!) = X(!). (3.2) The following lemma, which we use to construct martingales, is a special case of the tower of expectation property. For statement and proof of the more general case refer to [Wil91, p.88]. Lemma 3.5. Let {Yi}i2I and {Zj}j2J both be partitions of ⌦with ﬁnite index sets I and J. Suppose further that for each i 2 I, 9!Ji ⇢J such that {Zj}j2Ji is a partition of Yi. Let Fk and Fk+1 be the σ-algebras induced by partitions {Yi}i2I and {Zj}j2Ji respectively, according to Deﬁnition 3.2. Then, E ⇣ E X \| Fk+1 \| Fk ⌘ = E X \| Fk + . 3.2.2 Azuma-Hoe↵ding inequality This theorem is often referred to as Azuma’s inequality as it appeared in Azuma’s 1967 paper [Azu67], but it also appeared earlier in Hoe↵ding’s 1963 paper [Hoe63]. Theorem 3.6 (Azuma-Hoe↵ding). Suppose X0, . . . , Xn form a martingale such that \|Xk− Xk−1\| ck for each 1 k n. Then, P \|Xn −X0\| ≥r + e −r2 2 Pn k=1 c2 k . There is also a strengthened version of Theorem 3.6 proven by McDiarmid in [McD89]. He showed that the expression in the exponent could improved by a factor of 4. (That is, the right hand side becomes exp −2r2/Pt k=1 c2 k + for the same initial conditions.) 3.2.3 Generalised Azuma-Hoe↵ding inequality We also deﬁne a more general version of the Azuma-Hoe↵ding inequality, it applies even when the bound \|Xk −Xk−1\| ck does not hold on the entire space. A similar result appears in [Vu02, p.9] as Lemma 3.1. It also appears as Proposition 3 in [HHV09] who give a proof that it follows from Theorem 3.6. 28 3.2. Martingales Deﬁnition 3.7 (near-c-Lipschitz with exceptional probability ⌘). A martingale is near-c-Lipschitz with exceptional probability ⌘if, P(\|Xi −Xi−1\| ≥c) ⌘. Theorem 3.8 (Azuma-Hoe↵ding). Suppose X0, . . . , Xn forms a martingale which is near-c-Lipschitz with exceptional probability ⌘. Then, P \|Xn −X0\| ≥r + e −r2 2nc2 + ⌘. We will use Theorem 3.8 in Lemma 6.18 as part of our proof that pathological degree sequences are rare in the graph half -model, Gt. 3.2.4 Doob’s martingale process The method we will describe is often referred to as Doob’s martingale process and is often used to show that the value of a random variable is concentrated about its mean. Suppose we want to bound the di↵erence between the expected value of a function f over the whole domain (C, say) and the value of the function on a speciﬁc point in the domain (c 2 C). We set X0 = E[f(c)] and for a ﬁxed n, Xn = E[f(x)\|x 2 C]. The method then proceeds to deﬁne a series of random variables Xi for 0 i n such that Xi = E[f(x)\|x 2 Ci], where c = C0 ⇢. . . ⇢Ci ⇢Ci+1 ⇢. . . ⇢Cn = C. We are in e↵ect, zooming in on the value of f at c as we take the expected value of f over smaller and smaller subsets of C. The Xi’s then form a martingale and concentration results such as Azuma’s inequality can be used to bound \|X0 −Xn\|. Steps in Doob’s martingale process on the probability space (⌦, ⌃, P), • create a ﬁlter F0 ✓F1 ✓. . . ✓Fn (often via partitions of ⌦.) • deﬁne Xi = E(X \| Fi) • prove the Xi’s form a martingale with respect to the ﬁlter (via tower property). • bound the magnitude of successive di↵erences \|Xi −Xi−1\| for each 1 i n. • apply Azuma-Hoe↵ding Theorem 3.6 Shamir and Spencer used Doob’s martingale process to show that the chromatic number of a random graph is very concentrated in [SS87]. This is a pertinent example for two reasons. Firstly, it was one of the ﬁrst major results which used probabilistic techniques in combinatorics and piqued interest in the area. Secondly it is an example of a more general (widely used) technique called Doob’s martingale process. Later in of this thesis 29 Chapter 3. Probabilistic bounds and techniques we will use this technique to show Lemma 6.17. This will form part of our proof that pathological degree sequences are rare in the bipartite graph half -model, Gt. We work in the probability space ˆ Gp and begin by deﬁning a ﬁlter {Fi}0in−1 on this space. This ﬁlter is deﬁned via the partition induced by the following equivalence classes. i-equivalence of graphs Two graphs are i-equivalent, denoted H ⌘i G if the following condition holds. For all 1 x i and all 1 y n the edge {x, y} 2 H , {x, y} 2 G. That is, we require all edges which emanate from one of the ﬁrst i vertices to be the same in both graphs. We illustrate this deﬁnition with an example in 3.1. 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 G = G ⌘0 G ⌘1 G ⌘2 G ⌘3 Figure 3.1: The sets of graphs i equivalent to G for 0 i 3. The dashed lines indicate that edges that can be either absent or present in the graphs in that equivalence class. For each 0 i n −1 the equivalence relation deﬁnes a partition Pi. Observe that the partition Pi+1 is a reﬁnement of Pi. For 0 i n −1 let Fi be the σ-algebra induced by Pi according to Deﬁnition 3.2. Vertex uncovering martingale Given a random graph G 2 ˆ Gp(n), we deﬁne a se-quence of random variables X0, . . . , Xn−1 such that X0(G) = E ˆ Gp[χ] and Xn−1(G) = χ(G). We deﬁne, Xi(G) := E ˆ Gp χ\|Fi + (G) = E ˆ Gp χ(H)\|H ⌘i G + . The ﬁrst equality is our deﬁnition of Xi and the second follows by Deﬁnition 1.3 of con-ditional probability. When i = 0, Xi the expectation of the chromatic number over all random graphs in ˆ Gp(n). But when i = n −1, the (n −1)-equivalence of two graphs requires them to be the same, i.e. G is the only graph (n −1)-equivalent to G. Hence Xn−1 = χ(G). 30 3.2. Martingales Claim The Xi’s form a martingale. Proof. This is equivalent to proving the equality E ˆ Gp ⇣ E Xi+1 \| Fi+1 + \| Fi ⌘ = E ˆ Gp Xi+1 \| Fi + . But this is precisely the Tower Property of Expectation (see Lemma 3.5) and thus the random variables Xi, form a martingale. Hence we can apply Azuma’s lemma but ﬁrst we need to bound \|Xi+1 −Xi\|. Claim \|Xi+1 −Xi\| 1. Proof. We compare Xi+1 and Xi. Xi is the expected number of colours needed knowing the edges of the ﬁrst i vertices. The random variable Xi+1 is the expected number of colours needed knowing both edges of the ﬁrst i vertices and also the edges which join to the (i + 1)th vertex. This only a↵ects which colours are possible at the (i + 1)th vertex. Hence the expected value of colours can change by at most one. So we have \|Xi+1−Xi\| 1 as claimed. Concentration of Chromatic Number Result By the Azuma-Hoe↵ding Theorem 3.6, and because \|Xi+1 −Xi\| 1 for each 0 i n −2, we have, P ˆ Gp \|Xn−1 −X0\| ≥λ + e −λ2 2(n−1). Hence there is high probability that the chromatic number of G is close to the average chromatic number over all random graphs in ˆ Gp(n). For example, ﬁx " > 0, then, P ˆ Gp \|Xn−1 −X0\| ≥n1/2+"+ e −n1−2" (n−1) e−n2". Thus the chromatic number of a random graph G in ˆ Gp is almost surely within n1/2+" of the expected chromatic number as n ! 1. Shamir and Spencer show more than this. In [SS87] they prove that for s = 3, a random graph G in ˆ Gp (and for suitable p) there exists a function u(n) such that, limn!1P ˆ Gp u χ(G) u + s) = 1. (3.3) This was further strengthened by Luczak in [ Luc91] who proved (3.3) for the case s = 1. 31 Chapter 3. Probabilistic bounds and techniques 3.3 Probability generating functions Probability generating functions will be a useful tool at a couple of points within this thesis. Proofs of the statements can be found in [Wil06]. Suppose we have random variables X, Y deﬁned on some probability space (⌦, ⌃, P), for ⌦✓N. Then we say that the power series P(x) is a probability generating function for the random variable X if it satisﬁes (3.4). P(x) = p0 + p1x + p2x2 . . . where pi = P(X = i) (3.4) The expectation of X, E(X) may now be calculated in terms of P(x). E[X] = 0.p0 + 1.p1 + 2.p2 + 3.p3 + . . . = (1.p1 + 2.p2x + 3.p3x2 + . . .)x=1 = d dxP(x) + x=1. This is sometimes useful when calculating the expectation of discrete random variables. Similarly, E[X2] = 0.p0 + 1.p1 + 22.p2 + 32.p3 + . . . = d dxx(1.p1 + 2.p2x + 3.p3x2 + . . .)x=1 = d dxx d dxP(x) + x=1. Let Q(x) be then probability generating function for Y . Then if X and Y are independent random variables, E[XY ] = ⇣ x d dx + P(x) ⌘⇣ x d dx + Q(x) ⌘ x=1. 32 Part II New results: Degree sequences in random bipartite graphs 33 The main goal of this thesis is to show that in each of the random graph models the prob-ability of a non-pathological degree sequence (s, t) can be asymptotically approximated by the probability of an (m + n)-tuple in a specially constructed binomial model. In this part we will ﬁnd an asymptotic probability for non-pathological degree sequences in each of the random graph models. The starting point for this is Greenhill and McKay’s Theorem, our Theorem 2.8, which gives a count for the number of graphs with degree sequence (s, t). From this counting result we will derive the probability of a given degree sequence in each of our random graph models. Theorem 2.8 does not hold for all degree sequences, only (", a)-regular degree sequences. Hence our ﬁrst result, for each random graph model, is to show that these (", a)-regular degree sequences account for the bulk of the probability space. Our second result concerns a further restriction on (", a)-regular degree sequence. Observe that if an (", a)-regular degree sequence satisﬁes 1 − P j(sj −s)2 λ(1 −λ)mn ! 1 − P k(tk −t)2 λ(1 −λ)mn ! = O(n−b) (3.5) then the function on the degrees in the exponential term of (2.7) disappears into the error term, O(n−b). This is what we desire. For these well-behaved degree sequences (s, t), which we have called non-pathological, we have the simpliﬁed enumeration result, Corollary 2.10. So, our second result for each of our random graph models is that the probability of a degree sequence being pathological is asymptotically very low. This part is organised as follows. We begin by analysing the graph p-model, Gp, in Chapter 4. In Section 4.1, we show that degree sequences are likely to be (", a)-regular in this probability space. Then in Section 4.2 we show that most degree sequences are non-pathological. Both these steps are then repeated for the graph edge-model, GM in Chapter 5. This is done by considering the graph edge-model as a conditional case of the graph p-model. Lastly we show the same two results hold in the graph half -model, Gt. In this model we shall see that the degrees of the white vertices are not independent as they were for the graph p-model, Gp. Hence this case requires some more e↵ort and we work with mar-tingale concentration inequalities rather than the stronger concentration inequalities for independent random variables. 34 Chapter 4 Graph p-model, Gp. We work with our model of random graphs Gp where each edge of the complete bipartite graph Km,n is chosen independently with probability p = p(m, n). See Deﬁnition 1.6. In this model we will refer often to the following conditions. Conditions 4.1. These conditions apply to an m-tuple s = s(m, n) = (s1, ..., sm) and an n-tuple t = t(m, n) = (t1, ..., tn). The conditions depend on the parameters a and ". For m, n ! 1, • p is acceptable for a, m and n. • m = o(n1+") and n = o(m1+"). 4.1 (", a)-regular degree sequences in Gp 4.1.1 Variation in the degree of each vertex. In this section we show that a graph, G, chosen at random in Gp has degree sequence (s, t), likely to satisfy the following for appropriate " > 0. For each 1 j m and 1 k n, sj −s and tk −t are O(n1/2+"). Lemma 4.1. Fix 0 < a < 1 2 and " > 0. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. 1. Let S be the random variable that returns the average degree of the white vertices u1, . . . , um. Then as m, n ! 1 subject to Conditions 4.1, PGp \|S −np\| ≥n3"+ e−n2" 35 Chapter 4. Graph p-model, Gp. 2. For 1 j m, let Sj be the random variable that returns the degree of the white vertex uj. Then as m, n ! 1 subject to Conditions 4.1, PGp \|Sj −np\| ≥n1/2+"/3+ e−n3"/2 Proof. (of 1.) We ﬁrstly observe that \|mS −mnp\| n1+2" n3"m (4.1) implies that \|S −np\| n3". We prove (4.1) using Cherno↵’s inequality (Theorem 3.1). Construct the random variable Xj,k by setting Xj,k = 1 if there is an edge between uj and vk in our random graph G and otherwise setting Xj,k = 0. Then mS = P j,k Xj,k = X, and EGp mS + = EGp X + = mnp. By Lemma 4.5, p > 1 ln n and so mnp < mn1+". Set k = n1+2", then by Cherno↵’s inequality (Theorem 3.1), PGp \|mS −mnp\| ≥n1+2"+ 2e −n2+4" 2(n2+"+n1+2"/3) e−n2" hence, PGp \|S −np\| ≥n3"+ e−n2" Proof. (of 2.) We again use Cherno↵’s inequality (Theorem 3.1), this time to bound the probability that Sj is concentrated about its expected value, np. We use the random variables Xj,k deﬁned above. The number of edges incident with uj is Sj = Pn k=1 Xj,k and EGp(Sj) = np. We can now apply Cherno↵’s inequality (Theo-rem 3.1), letting k = n1/2+"/3. Then for large enough n PGp Sj np −n1/2+"/3+ e−n1+2"/2np = e−n2"/2p 1 2e−n3"/2 (4.2) and PGp Sj ≥np + n1/2+"/3+ e −n1+2" 2(np+n1/2+"/3) 1 2e−n3"/2, (4.3) where the last steps on lines (4.2) and (4.3) use p < 1. Hence, PGp \|Sj −np\| ≥n1/2+"/3+ e−n3"/2. 36 4.1. (", a)-regular degree sequences in Gp Corollary 4.2. Fix 0 < a < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6 and let λ be the random variable which returns the edge density. Then as m, n ! 1 subject to Conditions 4.1, PGp \|λ −p\| ≥n−1+3"+ e−n2" Proof. Simply note that n−1\|S−np\| = \|λ−p\| so this is actually is an equivalent statement to Lemma 4.1a. Lemma 4.3. Fix 0 < a < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. 1. Fix any white vertex uj. Let Sj be the random variable that returns the degree of uj. Then as m, n ! 1 subject to Conditions 4.1, PGp \|Sj −S\| ≥n1/2+2"/5+ e−n4"/3 2. Fix any black vertex vk. Let Tk be the random variable that returns the degree of vk. Then as m, n ! 1 subject to Conditions 4.1, PGp \|Tk −T\| ≥n1/2+10"/11+ e−n5"/4 Proof. (of 1.) Let us ﬁx 1 j m and consider Sj. We have the following bounds on \|Sj −np\| and \|s −np\| by parts a and b of Lemma 4.1 respectively. PGp \|S −np\| ≥n1/2+"/3+ e−n3"/2 PGp \|Sj −np\| ≥n1/2+"/3+ e−n3"/2 Hence by the triangle inequality we conclude that for large enough n, PGp \|Sj −S\| ≥n1/2+2"/5+ e−n4"/3, as required. Proof. (of 2.) The symmetry of our bipartite graph allows us to swap the black ver-tices of our random graph for the white vertices. Hence by symmetry, any true state-ment on the variables m, n, uj, Sj, S implies the corresponding statement on the variables m, n, vk, Tk, T. Therefore our bounds on \|Sj −S\| in Lemma 4.1 (1) imply by symmetry, PGp \|Tk −T\| ≥m1/2+10"/11+ e−m4"/3. (4.4) 37 Chapter 4. Graph p-model, Gp. As we know that m = o(n1+") and n = o(m1+") the bound above translates into a bound in terms of n. In the working below c, c0 are arbitrary constants in R. m1/2+2"/5 (cn1+")1/2+2"/5 = c0n1/2+2"/5+"/2+2"2/5 n1/2+10"/11 e−m4"/3 e−(cn 1 1+" )4"/3 = e−n4"(1−"+"2−"3...)/3c4"/3 e−n5"/4 Hence (4.4) implies: PGp \|Tk −T\| ≥n1/2+10"/11+ e−n5"/4. We have our result. The last lemma showed that the degree of a particular vertex on either side is unlikely to vary too greatly from the average of the degrees on that side. This is an important step in showing that a degree sequence (s, t) for a random graph in the graph p-model, Gp is highly likely to be (", a)-regular. Lemma 4.4. Fix 0 < a < 1 2 and " > 0. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Let S1, . . . , Sm and T1, . . . , Tn be the random variables that return the degrees of the white vertices u1, . . . , um and the black vertices v1, . . . , vn respectively. Then as m, n ! 1 subject to Conditions 4.1, PGp 8j, k, Tk −T, Sj −S uniformly o(n1/2+") + ≥1 −e−n6"/5. Proof. By Lemma 4.3, parts (1) and (2) we have the following results: PGp \|Sj −S\| ≥n1/2+2"/5+ e−n4"/3, PGp \|Tk −T\| ≥n1/2+10"/11+ e−n5"/4. Hence, PGp 8j, \|Sj −S\| < n1/2+2"/5 and 8k, \|Tk −T\| < n1/2+10"/11+ ≥1 −me−n4"/3 −ne−n5"/4 ≥1 −e−n6"/5. 4.1.2 Edge density in Gp Recall the term edge density, denoted λ and deﬁned by λ = 1 mn P j sj, i.e. the number of edges divided by total number of possible edges in the complete bipartite graph. There is a condition on λ in Deﬁnition 2.7 of an (", a)-regular degree sequence. We show that this condition is likely to hold in Gp for suitable values of p. In particular, we show that 38 4.1. (", a)-regular degree sequences in Gp when p is acceptable for a, m and n it is highly likely that λ is acceptable for a+", m and n. First we prove two technical lemmas. The results are a property of the relationship between p and λ where p is a parameter of Gp and the random variable, λ, returns the edge density of a random graph in Gp. We prove a lemma showing the bounds placed on r(1 −r) by the assumption that r is acceptable for a, m and n. Lemma 4.5. Let n and m be positive integers. Suppose there exists some 0 < a < 1 2, such that r is acceptable for a, m and n. Then for n > e16, 1 ln n < r(1 −r) Proof. By the deﬁnition of acceptable for a, m and n we know r satisﬁes: (1 −2r)2 4r(1 −r) ⇣ 1 + 5m 6n + 5n 6m ⌘ a ln n We observe, (1 −2r)2 4r(1 −r) = 1 4r(1 −r) −1 and 1 + 5m 6n + 5n 6m ≥8 3. Hence, r(1 −r) > 4 16 + 3 ln n > 1 ln n for n > e16 as required. Lemma 4.6. Fix " > 0. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Then as m, n ! 1 subject to Conditions 4.1, PGp ⇣p(1 −p) λ(1 −λ) = 1 + O(n−1+4") ⌘ ≥1 −e−n2". Proof. We begin by noting the following rearrangement, p(1 −p) λ(1 −λ) = 1 −(p −λ) ⇣λ −(1 −p) p(1 −p) ⌘−1 . (4.5) 39 Chapter 4. Graph p-model, Gp. By Lemma 4.5, because p is acceptable for a, m and n: p(1 −p) > 1 ln n. (4.6) Also note that \|λ + p −1\| < 1. Thus by (4.6), 1 1 1λ −(1 −p) p(1 −p) 1 1 1 < ln n. (4.7) The expected value in the graph p-model, Gp of the edge-density, λ, is p. So, when λ is equal to its expectation, (4.5) is precisely 1. An earlier result, corollary 4.2 provides the following bound: PGp \|λ −p\| ≥n−1+3"+ e−n2". Hence by (4.5), (4.7) and noting that ln n < n" for large enough n, we have the result. We are now in a position to prove our result on acceptable degree sequences. Lemma 4.7. Fix " > 0 and 0 < a < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Then as m, n ! 1 subject to Conditions 4.1, PGp ⇣ λ is acceptable for a + ", m and n ⌘ ≥1 −e−n3"/2. Proof. For convenience we let c = ⇣ 1 + 5m 6n + 5n 6m ⌘−1 . Note that c > n−". The condition that p is acceptable for a, m and n can now be written 1 4pq < 1 + ac ln n. (4.8) By Lemma 4.6, PGp ⇣p(1 −p) λ(1 −λ) = 1 + O(n−1+4") ⌘ ≥1 −e−n2". We work now in the graph p-model, Gp; with probability greater than 1 −e−n3"/2 the following holds. 40 4.1. (", a)-regular degree sequences in Gp 1 4λ(1 −λ) = 1 4pq 1 + O(n−1+4") + (4.9)  1 4pq + O(n−1+4" ln n) (4.10) < 1 + ac ln n + O(n−1+4" ln n) (4.11) < 1 + ⇣ a + O ⇣n−1+4" ln n c ln n ⌘⌘ c ln n (4.12) < 1 + ⇣ a + O n−1+5"+⌘ c ln n (4.13) < 1 + a + " + c ln n (4.14) We justify the steps line by line. Line (4.9) holds with probability at least 1 −e−n2" by Lemma 4.6. Note this is the one step that holds only probabilistically. By Lemma 4.5 because p is acceptable for a, m and n, we know that 1 pq > ln n. This implies (4.10). Line (4.11) now follows by (4.8). Then (4.12) is a rearrangement of (4.11). Note that c > n−" because of our conditions on m and n. Thus we have (4.13). We are considering n tending to inﬁnity so " is larger than O(n−1/2+3"/2). Hence (4.14), which is exactly as we require. 4.1.3 Bounding result on (", a)-regular degree sequences When our parameter p is acceptable for a, m and n there is a high probability that a graph chosen randomly in Gp will have an (", a)-regular degree sequence. We state and prove this formally in the following lemma. Lemma 4.8. Fix 0 < a < 1 2 and " > 0 such that a0 = a + " < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Then as m, n ! 1 subject to Conditions 4.1, PGp (S, T ) is (", a)-regular + ≥1 −e−n7"/6 Proof. By Lemmas 4.4 and 4.7. Note that e−n7"/6 < e−n3"/2 + e−n6"/5 for large enough n. The result now follows. Recall that this is important as it is only graphs with (", a)-regular degree sequences (s, t) which satisfy Greenhill and McKay’s enumeration result, Theorem 2.8. Hence Lemma 4.8 shows that for a degree sequence (s, t), generated at random in the p-model, Gp, there is a high probability that we can apply Theorem 2.8. In the next section we will show that in the graph p-model, Gp, with a high probability, a random graph will have a degree sequence (s, t) that is non-pathological. 41 Chapter 4. Graph p-model, Gp. 4.2 Pathological degree sequences in Gp. We show that in the graph p-model, Gp, the probability of selecting a graph whose degree sequence is pathological is very small. This is done in Theorem 4.9 at the end of this sec-tion. Recall that a degree sequence is pathological if it is either not (", a)-regular or is such that the expression (3.5) is too large. Earlier in Section 4.1, we found that the ﬁrst of these possibilities was unlikely, i.e. we showed that the probability of a degree sequence not being (", a)-regular was small. In this section we concentrate on the other property that leads to (a, b, m, n, ")-pathological degree sequences. Thus, we analyse the left hand side of (3.5). In particular, we want to bound the probability in the graph p-model, Gp, that the left hand side of (3.5) exceeds O(n−b). We brieﬂy outline the steps we take to achieve this. By symmetry we will see it is suﬃcient to show that 1 − P j(Sj−S)2 λ(1−λ)mn is O(n−b) with high probability in Gp. This we do in two steps, we show that both of the following hold with high probability in Gp. P j(Sj −S)2 −pqmn λ(1 −λ)mn = O(n−b) (4.15) pq λ(1 −λ) = pqmn λ(1 −λ)mn = 1 + O(n−b) (4.16) Proving (4.15) comprises the entirety of Sections 4.2.1 and 4.2.2. It is a concentration result, we show that the function on the white degrees P j(Sj −S)2 is likely to be close to pqmn. In Section 4.2.1 we deﬁne a new random variable, S⇤ j , e↵ectively allowing us to work only with degree sequences whose white degrees are close to their mean. Then, in Section 4.2.2, working with these new random variables we use concentration inequalities to prove (4.15) holds with high probability in Gp. The result (4.16) follows more readily; indeed we have already proven a stronger result, Lemma 4.6. It was proven using bounds on λ(1 −λ) and \|p −λ\|. Finally, in Section 4.2.3, the two results (4.15) and (4.16) are combined with the result of Lemma 4.8, which bounds the probability that a degree sequence will not be (", a)-regular in the graph p-model, Gp. The combination of these yields Theorem 4.9 which states that the probability of a degree sequence being pathological is asymptotically very small in the graph p-model, Gp. 4.2.1 Restricted distribution on the degrees of vertices We will introduce a truncated version of our random variables for the degrees of the white vertices. 42 4.2. Pathological degree sequences in Gp. Deﬁnition 4.1 (truncated degree, S⇤ j ). S⇤ j = max np −n1/2+", min(np + n1/2+", Sj) + We will write S⇤:= (S⇤ 1, . . . , S⇤ m). The random variable S⇤ j has a truncated distribution of Sj. Note PGp S⇤ j = x + := 8 > > < > > : PGp Sj = x + if \|x −np\| < n1/2+" PGp Sj ≥np + n1/2+"+ if x = np + n1/2+" PGp Sj np −n1/2+"+ if x = np −n1/2+" 0 otherwise -6 ? 0.05 0.10 45 50 55 60 65 70 75 -6 ? 0.05 0.10 45 50 55 60 65 70 75 Figure 4.1: Probability distributions of s1 and s⇤ 1 for n = 100, p = 0.6 and " = 0. The advantage of our new random variables is twofold: each S⇤ j is at most pn away from the expected mean of the degrees of the white vertices, np, but also agrees with our original random variable Sj on a large part of the domain. We make this claim of likely agreement precise in the following lemma. Lemma 4.2. Deﬁne Dp = (np −n1/2+", np + n1/2+")m ⇢Rm. Then, PGp S 2 Dp + ≥1 −e−n4"/3 (4.17) and for any m-tuple x 2 Dp, PGp S⇤= x + = PGp S = x + . (4.18) Proof. Fix j. Note that by the deﬁnition of our new random variable S⇤ j , we have that for y 2 (np −n1/2+", np + n1/2+"), PGp S⇤ j = y + = PGp Sj = y + . 43 Chapter 4. Graph p-model, Gp. The extension to the m-dimensional probability space is then clear so we have (4.18). To prove (4.17) we recall that degree of each white vertex is highly likely to concentrated about its mean, np. By Lemma 4.1, PGp \|Sj −np\| ≥n1/2+"/3+ e−n3"/2 hence, PGp 8j, \|Sj −np\| n1/2+"/3) ≥1 −ne−n3"/2 ≥1 −e−n4"/3. (4.19) The result (4.17) follows directly from equation (4.19) and so we are done. We now work with these new random variables S⇤ j , where the distribution of the degree of each vertex is restricted. 4.2.2 Bounding a function on the white degrees: P j(Sj −S)2. The aim of this subsection is to show that the absolute di↵erence between the expressions P j(Sj −S)2 and pqmn is asymptotically likely to be small in the graph p-model, Gp. This is done in Lemma 4.6 at the end of this subsection. We begin with some algebraic trickery and note: m X j=1 (Sj −S)2 = −m(S −np)2 + m X j=1 (Sj −np)2. (4.20) Line (4.20) and the triangle inequality now allow us to write: 1 1 X j (Sj −S)2 −pqmn 1 1 \|m(S −np)2\| + 1 1 X j (Sj −np)2 − X j (S⇤ j −np)21 1 + 1 1 X j (S⇤ j −np)2 −EGp X j (S⇤ j −np)2+1 1 + 1 1EGp X j (S⇤ j −np)2+ −pqmn 1 1. (4.21) We note that when our m-tuple , S, for the degrees of the white vertices, lies within the restriction of all m-tuples, S 2 Dp the following equality holds, X j (Sj −np)2 = X j (S⇤ j −np)2. In particular, whenever S 2 Dp, then by (4.21), we have 1 1 X j (Sj −S)2 −pqmn 1 1 \|m(S −np)2\| + 1 1 X j (S⇤ j −np)2 −EGp X j (S⇤ j −np)2+1 1 + 1 1EGp X j (S⇤ j −np)2+ −pqmn 1 1. (4.22) 44 4.2. Pathological degree sequences in Gp. As we shall see next, each of three terms on the right hand side of (4.22) is asymptotically very likely to be small in the graph p-model, Gp. We prove probabilistic bounds for these three terms in Lemmas 4.3, 4.4 and 4.5 respectively. The ﬁrst bound follows directly from a bound shown in the previous section. To achieve the middle bound we use a concentra-tion inequality while the last term can be bounded by properties of expectation and of our random variables Sj and S⇤ j . We begin by bounding the ﬁrst of these three terms on the right-hand side of (4.22) and show that the magnitude of m(S −np)2 is small with high probability in Gp. Lemma 4.3. Fix 0 < a < 1 2 and " > 0. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Let S be the random variable that returns the average degree of the white vertices u1, . . . , um. Then as m, n ! 1 subject to Conditions 4.1, PGp \|m(S −np)2\| ≥n1+8"+ e−n2" Proof. This follows by a previous result. By Lemma 4.1 (1), PGp \|S −np\| ≥n3"+ e−n2". An immediate corollary of this is PGp m(S −np)2 ≥m(n3")2+ e−n2". Then, since n1+8" ≥m(n3")2 we have our result. Lemma 4.4. Fix 0 < a < 1 2 and " > 0. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. For 1 j m, let S⇤ j be the random variable which returns the truncated degree of the white vertex uj. Then as m, n ! 1 subject to Conditions 4.1, PGp \| X j (S⇤ j −np)2 −EGp X j (S⇤ j −np)2)\| ≥n3/2+4"+ e−n2". Proof. Fix a white vertex uj and consider its truncated degree S⇤ j . By Deﬁnition 4.1 (trun-cated degree) : \|S⇤ j −np\| n1/2+". This implies (S⇤ j −np)2 n1+2". Because the last re-sult is true always, we can say the same for the expectation, i.e. that EGp (S⇤ j −np)2+ n1+2". Since both (S⇤ j −np)2 and its expectation, EGp (S⇤ j −np)2+ , are positive, we have \|(S⇤ j −np)2 −EGp (S⇤ j −np)2+ \| \|max{(S⇤ j −np)2, EGp (S⇤ j −np)2+ }\| n1+2". (4.23) These observations above hold for all 1 j m. Deﬁne Xj = (S⇤ j −np)2. Then because the degrees (and hence the truncated degrees) of the white vertices are independent, X1, . . . , Xn are independent random variables. By (4.23), these random variables satisfy \|Xj −E(Xj)\| n1+2" for 1 j m. Hence we can apply Theorem 3.2 to get PGp \| X j (S⇤ j −np)2 −EGp X j (S⇤ j −np)2)\| ≥k + e −k2 2m(n1+2")2 e −k2 n"n1+"n2+4" e −k2 n3+6" . 45 Chapter 4. Graph p-model, Gp. Substituting k = n3/2+4" gives the required result. Lemma 4.5. Fix " > 0. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. For 1 j m let S⇤ j be the random variable which returns the truncated degree of the white vertex uj. Then as m, n ! 1 subject to m = o(n1+") and n = o(n1+"), \|EGp X j (Sj −np)2+ −pqmn\|  e−n7"/6. Proof. By calculations done in appendix, line (8.4), we know that EGp P j(Sj −np)2+ = pqmn. Hence, \|EGp X j (S⇤ j −np)2+ −pqmn\| = \|EGp X j (S⇤ j −np)2+ −EGp X j (Sj −np)2+ \|. (4.24) So it is equivalent to bound the right-hand-side of (4.24). The proof now follows by properties of expectation. Firstly, by the deﬁnition of expectation EGp X j (Sj −np)2+ = X x h PGp (S1, . . . , Sm) = x + X j (xj −np)2i EGp X j (S⇤ j −np)2+ = X x h PGp (S⇤ 1, . . . , S⇤ m) = x + X j (xj −np)2i . For any x 2 Dp, i.e. most degree sequences, x, of the white vertices, PGp S⇤= x + = PGp S = x + . Hence for x 2 Dp the corresponding terms in the sum below will cancel. Thus, \|EGp X j (Sj −np)2+ −EGp X j (S⇤ j −np)2+ \|  PGp (s, t) : s / 2 Dp + max x X j (xj −np)2  e−n6"/5mn2  e−n7"/6 as required. These three results (Lemmas 4.3,4.4 and 4.5) allow us to prove (4.15); indeed, we show a slightly stronger result in the following lemma. 46 4.2. Pathological degree sequences in Gp. Lemma 4.6. Fix 0 < a < 1 2 and " > 0 such that a = a0 + " < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. For 1 j m, let Sj be random variable that returns the degree of the white vertex uj, let S return the average of these degrees and let λ return the edge density. Then as m, n ! 1 subject to Conditions 4.1, PGp 1 1 1 P j(Sj −S)2 −pqmn λ(1 −λ)mn 1 1 1 ≥n−1/2+8" ! e−n6"/5 Proof. In the Lemmas 4.4, 4.3 and 4.5 we compiled the following probability bounds: PGp ⇣ \|m(S −np)2\| ≥ n3/2+5"⌘ e−n1/2−3" (4.25) PGp ⇣ \| X j (Sj −np)2 −EGp X j (Sj −np)2+ \| ≥ n3/2+3"⌘ e−n3"/2 (4.26) 1 1 1EGp X j (S⇤ j −np)2+ −pqmn 1 1 1  e−n1/2. (4.27) By (4.22), for any s 2 Dp, our expression of interest, 1 1 P j(sj −s)2 −pqmn 1 1, is bounded above by the sum of the three terms in (4.25), (4.26) and (4.27). Hence, PGp ⇣ \| X j (Sj−S)2−pqmn\| n3/2+5"+n3/2+3"+e−n1/2⌘ ≥1−e−n1/2−3"−e−n3"/2−PGp s / 2 Dp + By Lemma 4.2, PGp s / 2 Dp + e−n6"/5. Also, 1 −e−n1/2−3" −e−n3"/2 −e−n6"/5 ≥1 −e−n7"/6. So we have PGp ⇣ \| X j (Sj −S)2 −pqmn\| n3/2+6"⌘ ≥1 −e−n7"/6. By Lemma 4.7, λ is acceptable for a, m and n with probability at least 1−e−n3"/2. Hence, by Lemma 4.5, λ(1 −λ) > 1 ln n > n−"/2, with this same probability. Hence, 47 Chapter 4. Graph p-model, Gp. PGp ⇣\| P j(Sj −S)2 −pqmn\| λ(1 −λ)mn n−1/2+9"⌘ ≥1 −e−n8"/7. This Lemma 4.6 together with Lemma 4.6 will enable us to show that for a random graph in Gp its degree sequence (s, t), is unlikely to be pathological. Corollary 4.7. Fix 0 < a < 1 2 and " > 0 such that a := a0 + " < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. For 1 j m, let Sj be random variable that returns the degree of the white vertex uj, let S return the average of these degrees and let λ return the edge density. Then as m, n ! 1 subject to Conditions 4.1, PGp 1 1 11 − P j(Sj −S)2 λ(1 −λ)mn 1 1 1 ≥n−1/2+9" ! e−n7"/6 Proof. This follows directly by Lemmas 4.6 and 4.6. In the next section we will use this result, Corollary 4.7 and some symmetry arguments to show that pathological degree sequences are rare in the graph p-model, Gp. 4.2.3 Likelihood of pathological degree sequences Now that we have shown Corollary 4.7 we show that a similar probabilistic bound hold for the degrees of the black vertices. Lemma 4.8. Fix 0 < a < 1 2 and let " > 0 such that a0 := a + " < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. For 1 k n, let Tk be random variable that returns the degree of the black vertex vk, let T return the average of these degrees and let λ return the edge density. Then as m, n ! 1 subject to Conditions 4.1, PGp 1 1 11 − P k(Tk −T)2 λ(1 −λ)mn 1 1 1 ≥n−1/2+9" ! e−n8"/7. Proof. This proof will be done utilising symmetry. Note that because n = o(m1+"), ln n < (1 + ") ln m. Hence, a ln n < (a0 −")(1 + ") ln m < a0 ln m (4.28) 48 4.2. Pathological degree sequences in Gp. Now, because p is acceptable for a, m and n and by (4.28) we can conclude that p is acceptable for a0, n and m. Then by Corollary 4.7 swapping each instance of ‘S’ for ‘T’ and each instance of ‘m’ for ‘n’. We can conclude that as m, n ! 1 the following holds. PGp 1 1 11 − P k(Tk −T)2 λ(1 −λ)mn 1 1 1 ≥m−1/2+8" ! e−m6"/5. (4.29) Now the inequalities, m1/2+8" (cn1+")1/2+8" = c0n1/2+8"+"/2+8"2 n1/2+9" e−m7"/6 e−(cn 1 1+" )7"/6 = e−n7"(1−"+"2−"3...)/6c7"/6 e−n8"/7 complete the proof. Theorem 4.9. Fix a, b 2 R+ and " > 0 such that a+b < 1 2, a0 = a+" < 1 2 and b+17" < 1. Let Gp = Gp(m, n) be as in Deﬁnition. Then as m, n ! 1 subject to Conditions 4.1, PGp (S, T ) is (a, b, m, n, ")-pathological + e−n10"/11. Proof. Together, Corollary 4.7 and Lemma 4.8 imply, PGp 1 1 1 ⇣ 1 − P j(Sj −S)2 λ(1 −λ)mn ⌘⇣ 1 − P k(Tk −T)2 λ(1 −λ)mn ⌘1 1 1 ≥n−1+17" ! e−n9"/10 (4.30) Note that n−1+17" = O(n−b) for b < 1 −17". We are almost done. To ﬁnish, recall by Lemma 4.8, PGp (S, T ) is not (", a)-regular + e−n7"/6. 49 Chapter 5 Graph edge-model, GM. This chapter concerns the graph edge-model, GM (see Deﬁnition 1.7 on page 5). In this model every bipartite graph on (m, n) vertices with M edges is equally likely. We will derive a simple formula for the probability of any non-(a, b, m, n, ")-pathological degree sequence (s, t) in the graph edge-model, GM. The starting point will be the enumeration result by Greenhill and McKay, Theorem 2.8 as in the last chapter on the graph p-model, Gp. Hence the ﬁrst step will be (again) to show that this formula is likely to be applicable to a random degree sequence in the probabil-ity space. That is, we show that a degree sequence in GM is very likely to be (", a)-regular. Then, to ensure the enumeration result simpliﬁes we will also show that pathological de-gree sequences are rare in GM. These two results on (", a)-regular and pathological degree sequences are proven in Section 5.2. To show these two results we will prove similar probabilistic bounds in GM to those al-ready shown in the Gp model. The graph edge-model is the restriction of the Gp model to graphs with M edges. Hence these two probability spaces are related which allows us to derive Lemma 5.1. This lemma provides an upper bound on any event A occurring in GM in terms of the probability of the same event A occurring in Gp. Using Lemma 5.1 the results on GM in Section 5.2 in this chapter follow from similar results on Gp in the previous chapter. We will need to refer to the following conditions. Conditions 5.1. These conditions apply to an m-tuple s = s(m, n) = (s1, ..., sm) and an n-tuple t = t(m, n) = (t1, ..., tn). The conditions depend on the parameters a and ". For m, n ! 1, 50 5.1. Relation to graph p-model, Gp. • M mn is acceptable for a, m and n. • m = o(n1+") and n = o(m1+"). 5.1 Relation to graph p-model, Gp. Lemma 5.1. Let X be a random variable deﬁned on all bipartite graphs on (m, n) ver-tices. Then for p = M mn, we have PGM(X) mnPGp X + . Proof. The graph edge-model, GM, is the restriction of the graph p-model, Gp, to M edges, so PGM(X) = PGp X\|#edges = M + . Now, let A, B be any random variables. We note by Bayes Theorem (given P(B = 1) > 0): P ⇣ A \| (B = 1) ⌘ = P ⇣ A & (B = 1) ⌘ P(B = 1)  P(A) P(B = 1). (5.1) We work in the graph p-model, Gp, and set the value of our parameter p = M mn.1In (5.1) we substitute, for B, the indicator function, IM for the event that the graph has M edges. Then substitute X for A and we have the following inequality. PGM(X)  PGp= M mn (X) PGp= M mn (IM = 1). In the graph p-model, Gp, the number of edges in the graph is between 0 and mn inclusive; indeed the number of edges in the graph is the binomial distribution (mn, p). Here the most likely values for the number of edges are those close to the expected value, pmn. In particular, because we chose p = M mn then the probability that the number of edges in the graph is M is at least 1 mn. That is PGM= M mn (IM = 1) ≥ 1 mn. This yields the desired result. 5.2 (", a)-regular and pathological degree sequences in GM. Here, we show that in the graph edge-model, GM, (", a)-regular degree sequences are highly likely. 1Note by setting p = M mn, p is acceptable for a, m and n if and only if M is acceptable for a, m and n. 51 Chapter 5. Graph edge-model, GM. Lemma 5.1. Fix a 2 R+ and " > 0 such that a < 1 2. Let GM = GM(m, n) be as in Deﬁnition 1.7. Then as m, n ! 1 subject to Conditions 5.1, PGM( (S, T ) is not (", a)-regular ) e−n8"/7 Proof. By Lemma 4.8, PGp (S, T ) is (", a)-regular + ≥1 −e−n7"/6 Hence we know with probability at least 1 −e−n6"/5, that a degree sequence is likely to be (", a)-regular in the graph p-model, Gp. So by Lemma 5.1 in the graph edge-model, GM, we know a degree sequence is (", a)-regular with probability at least 1 −mne−n7"/6, i.e. with probability at least 1 −e−n8"/7. We also bound the probability that a random degree sequence is pathological in the graph edge-model, GM. Theorem 5.2. Fix a, b 2 R+ and " > 0 such that a+b < 1 2, a0 = a+" < 1 2 and b+17" < 1. Let GM = GM(m, n) be as in Deﬁnition 1.7. Then as m, n ! 1subjecttoConditions5.1, PGM( (S, T ) is (a, b, m, n, ")-pathological ) e−n11"/12. Proof. We proceed directly from Theorem 4.9 in the same manner as the proof of Lemma 5.1. 52 Chapter 6 Graph half -model, Gt. In our discussions pertaining to the graph half -model, Gt, we will often refer to the following set of conditions. Conditions 6.1. These conditions apply to an m-tuple s = s(m, n) = (s1, ..., sm) and an n-tuple t = t(m, n) = (t1, ..., tn). The conditions depend on the parameters a and ". For m, n ! 1, • tk −t is uniformly O(n1/2+") for 1 k n. • λ = 1 mn P k tk is acceptable for a, m and n. • m = o(n1+") and n = o(m1+"). 6.1 (", a)-regular degree sequences in Gt For a graph, G, chosen at random in Gt we show that with high probability the degree sequence (s, t) of G is (", a)-regular. Lemma 6.1. Fix a 2 R+ and " > 0 such that a < 1 2. Let Gt = Gt(m, n) be as in Deﬁnition 1.8. Then as m, n ! 1 subject to Conditions 6.1, PGt( (S, T ) is not (", a)-regular ) e−n6"/5 Proof. By the assumptions on the model, Gt, we already have many of the requirements of (", a)-regular. We have the required condition on the edge density, λ, and also that tk −t is uniformly o(n1/2+") for each graph in the domain of Gt. Hence it is suﬃcient to show that Sj −S is uniformly o(n1/2+") with probability greater than e−n6"/5 in Gt. We 53 Chapter 6. Graph half -model, Gt. shall use Theorem 3.1, a concentration result due to Cherno↵. Fix a white vertex uj. For 1 k n we consider the random variable Xj,k such that Xj,k = 1 if there is an edge between uj and vk and zero otherwise. In our model Gt the degrees of the vk’s are prescribed and any bipartite graph satisfying those degrees is equally likely. Hence the tk edges emanating from vertex vk have an equal chance of joining to any set of tk di↵erent vertices from u1 to um. Therefore, PGt(Xj,k = 1) = tk/m. (6.1) By the deﬁnition of our random variable, Sj = P k Xj,k. Note also that the expected value of Sj is λn by the following calculation, E(Sj) = P k P(Xj,k = 1) = m−1 P k Tk = m−1λmn = λn. The Xj,k are independent random variables, because the edges incident with one black vertex do not a↵ect the probability that edges will be incident with any other black vertex. We also note that since 0 Sj n we have 0 E(Sj) n. (This is a very crude bound but it suﬃces for this lemma.) Hence by Theorem 3.1, (letting k = n1/2+3"/4), PGt(\|Sj −λn\| ≥n1/2+3"/2) < 2e −n1+3"/2 2(E(X)+n1/2+3"/2/3) < e−n5"/4. Hence, PGt(8j, \|Sj −λn\| < n1/2+3"/2) ≥1 −me−n5"/4 ≥1 −e−n6"/5, (6.2) and we have our result. 6.2 Pathological degree sequences in Gt 6.2.1 Locally ordered bipartite graphs We will deﬁne a new probability space, G? t , over specially labelled bipartite graphs on (m, n) vertices with black degree sequence t. We will show that the probability that a random graph has any particular degree sequence (s, t) in Gt and Ga t . Hence bound the probability of non-pathological degree sequences in Gt we instead prove the corresponding bounds in G? t . In order to deﬁne the probability space, G? t , we deﬁne an allowable labelling for the graphs in the domain of G? t . We term these allowable labellings locally ordered. The name reﬂects the relation to edge ordered labellings which are well studied in the literature, see, for example [GK73]. 54 6.2. Pathological degree sequences in Gt Deﬁnition 6.1 (edge-ordering). An edge ordering L of a graph G is a bijection from the edges of G to {1, 2, . . . , \|E(G)\|}. Deﬁnition 6.2 (locally ordered). A bipartite graph G on (m, n) vertices together with an edge-labelling L is said to be locally ordered if, for each 1 k m, the subgraph (and sub-labelling) G{vl : l 6= k} is edge ordered. We will often write the locally ordered graph Go = (G, L) to refer to the graph G together with the locally orderedlabelling L. 1 P P P P P P P P P P P P P P P 2 1 Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ 2 1 3 P P P P P P P P P P P P P P P 1 2 ~ ~ ~ ~ n n n n Go 1 Go 1 v f f f f P P P P P P P P 2 1 Go 2 v f f f f Q Q Q Q Q Q Q Q ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ ⇣ 2 1 3 Go 3 v f f f f P P P P P P P P 1 2 Go 4 v f f f f } } } } u u u u u u u u u u u u u u u u Figure 6.1: Here, Go is locally ordered because Go 1, Go 2, Go 3 and Go 4 are edge-ordered 6.2.2 Decision tree For each triple of parameters t, m, n, we will construct a rooted tree. In this tree, the root node corresponds to the set of all graphs on (m, n) vertices with black degree sequence t. Also, the leaves of the tree each correspond to single graphs on (m, n) vertices with black degree sequence t; one leaf for each such graph. We illustrate the decision tree with black degree sequence t = (1, 1, 1) and vertices (m, n) = (3, 3) in Figure 6.4. To deﬁne the structure of a decision tree in general re-quires the following quagmire of deﬁnitions. We deﬁne sets of tuples. Below, Stk is the group of permutations of the numbers from 1 to tk. We also write M = P k tk. 55 Chapter 6. Graph half -model, Gt. Deﬁnition 6.3 (A, Al). A := A(t) = 5 a : \|a\| = M & 8k, (aP h<k th+1, . . . , aP hk th) 2 Stk Al := Al(t) = 5 (a1, . . . , al) \| (a1, . . . , aM) 2 A(t) Deﬁnition 6.4 (reference function R). This is deﬁned for a n-tuple, t, and integer 1  l n. We deﬁne, R(l, t) := (h, i) where l = X k<h tk + i. Deﬁnition 6.5 (Y a l ). Let l-tuple a 2 Al and for each 1 i l, let (hi, ri) = R(i, t). Then deﬁne Y a l = Y a l (m, n, t) by, Y a l (m, n, t) := 5 G 2 Bo m,n,t : 8 1 i l, (vhi, uri, ai) 2 Eo(G) . Each set of graphs Y a l , corresponds to a node at level l in our decision tree. In Figure 6.2 we give some examples that relate to the decision tree in Figure 6.4 on p60. Y (2) 1 = n o , , , , , , , , Y (1,1) 2 = n o Y (1,3) 2 = n o , , , , Y (1,1,1) 3 = n o Y (1,1,2) 3 = n o Y (1,3,2) 3 = n o Figure 6.2: We illustrate the sets Y a l = Y a l (3, 3, (1, 1, 1)) for various l-tuples a. Each Y a l is a subset of the set of all locally ordered bipartite graphs on (3, 3) vertices with black degree sequence t. (All edges shown are labelled ‘1’.) Decision tree construction for m, n, t. • nodes For each 0 l M, the set of nodes at level l in the tree is {Y a l }a2Al. 56 6.2. Pathological degree sequences in Gt • edges Let a 2 Al for some 1 l M. Set a0 = (a1, . . . , al−1) and deﬁne the parent of {Y a l } to be {Y a0 l−1}. An edge is drawn between each node and its parent node. Note that if the l-tuple a and (l+1)-tuple b agree in their ﬁrst l elements, then Y b l+1 ⇢Y a l . Furthermore for the l-tuple a deﬁne Ca l := {b 2 Al+1 : (b1, . . . , bl) = a}. Then the sets {Y b l+1}b2Ca l partition Y a l . Hence the nodes deﬁned form a decision tree as in [CL06, p.106]. We adopt the usual language of decision trees. For each b 2 Ca l we call the node Y b l+1 a child of the node Y a l . Also, two (l + 1)-tuples b, b0 2 Ca l are termed sibling nodes. 6.2.3 Graph ordered-half -model, Ga t Deﬁnition 6.6 (Bo m,n,t). Bo m,n,t := {locally ordered bipartite graphs on (m, n) vertices with black degree sequence t}. Deﬁnition 6.7 (Graph ordered-half -model, Ga t ). The graph ordered-half -model Ga t = G(m, n, a, t) has domain Bo m,n,t. Its support is the subset of these graphs, Y a \|a\|(m, n, t). All graphs in the support have equal probability. When a = ? the only restriction on the edges in the support of Ga t is that the black degree sequence is t. This is very similar to the graph half -model, Gt, except in G? t all edges have labels. Every graph in the support of Gt corresponds to the same number or labelled graphs in G? t . Hence, for any bipartite graph G with black degree sequence t, PGt(G) = PG? t ({Go = (G, L) : for some locally ordered labelling L}). (6.3) We illustrate some example calculations in Ga t in Figure 6.3. 57 Chapter 6. Graph half -model, Gt. PG(3) (1,1,1) ⇣ ⌘ = 1 9 PG(2) (1,1,1) ⇣ ⌘ = 0 PG(1,2) (1,1,1) ⇣ ⌘ = PG(1,2) (1,1,1) ⇣ ⌘ = PG(1,2) (1,1,1) ⇣ ⌘ = 1 3 Figure 6.3: We work in the probability spaces Ga t for t = (1, 1, 1) and various tuples a. The domain in each case is the set of all locally ordered bipartite graphs on (3, 3) vertices with black degree sequence (1, 1, 1). In the diagrams each edge has label ‘1’. Deﬁnition 6.8 (near-regular). An locally ordered bipartite graph, G, is called near-regular, (abbreviated nreg) if it satisﬁes, 8j, \|sj(G) −λn\| < n1/2+3"/2. Corollary 6.9. Fix a 2 R+ and " > 0 such that a < 1 2. Let G? t = G? t (m, n) be as in Deﬁnition 6.7. Then as m, n ! 1 subject to Conditions 6.1, PG? t (G is nreg) ≥1 −e−n6"/5. Proof. We have earlier observed, see (6.3), that any event on the degrees of a graph has the same probability in Gt and G? t . Now, note that by Deﬁnition 6.8 of a near-regular graph G, that the statement we are required to prove is precisely (6.2). So we are done. 6.2.4 Edge-uncovering martingale Fix 0 l M, then the sets {Y a l }a2Al partition our domain, Bm,n,t. Hence, the power set of these partitions, Fl := {Y a l }a2Al (6.4) is a σ-algebra. We are now in a position to deﬁne our martingale. For each 0 i M, we deﬁne the random variable, Xi = EG? t X j (Sj −λn)2 \| Fi + . (6.5) In the next lemma we show that these random variables do indeed form a martingale. Lemma 6.10. The random variables X0, . . . , XM as deﬁned in (6.5) form a martingale. 58 6.2. Pathological degree sequences in Gt Proof. Notice that P j(Sj −λn)2+2 < (mn2)2 < 1. This implies that each E(Xl)2 < 1 for each 0 l M. Hence by the Deﬁnition 3.3 of martingales it remains now to show, EG? t Xl+1 \| Fl + = Xl. (6.6) We show (6.6) by the following, EG? t Xl+1 \| Fl + = EG? t EG? t ( X j (Sj −λn)2 \| Fl+1) \| Fl + = EG? t ( X j (Sj −λn)2 \| Fl + (6.7) = Xl. The ﬁrst and last lines follow by the deﬁnitions of Xl+1 and Xl respectively. Because Fl ⇢Fl+1, line (6.7) follows by the tower of expectation property (see Lemma 3.5). In the next section we will ﬁnd a bound for \|Xi −Xi+1\|. Actually, we ﬁnd two bounds, a weaker one that holds at all points in the domain and stronger one that holds for most of the domain. This will then allow us to use the generalised Azuma Theorem, (our Theorem 3.8), to bound the di↵erence, \|X0 −XM\|. 59 Chapter 6. Graph half -model, Gt. pppppppppppppppppppppppppp ++++++++++++++++++++++++++++++++ ppppppppppppppppppppppppp ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ____ ppppppppppppppppppppppppp ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / (not included in ﬁgure) ⇤ E(P j(Sj −S)2 \| F0) = n , , , , , , , , , , , , , , , , , , , , , , , , , , o 4.3 ⇤ E(P j(Sj −S)2 \| F1) = n , , , , , , , , o 4.3 ⌅ E(P j(Sj −S)2 \| F2) = n , , o 6.3 ⇤ n , , o 3.3 ⇤ n , , o 3.3 ⌅ E(P j(Sj −S)2 \| F3) = n 9 o ⇤ n 5 o ⇤ n 5 o ⇤ n 5 o ⇤ n 5 o ⌅ n 0 o ⇤ n 5 o ⌅ n 0 o ⇤ n 5 o ⇤ Figure 6.4: The decision tree for bipartite graphs on (3, 3) vertices with black degree sequence (1, 1, 1). Let Xi = E(P j(Sj −S)2\|Fi). For each node at level i in the tree we display the value of Xi(G) for graphs G at that node. We have shaded the nodes at level i for which \|Xi+1(G) −Xi(G)\| ≥2 for graphs G at that node (such nodes are referred to to as bad in the text). Note this is not the same cut-o↵we use in our calculations. Figure 6.4: 60 6.2. Pathological degree sequences in Gt 6.2.5 Bounding Arguments Deﬁnition 6.11 (toxic). Fix 0 l M. Then l-tuple a 2 Al is toxic if PG? t G not near-regular \| G 2 Y a l + > n−1/2. Lemma 6.12. Fix 0 l M, 0 < a < 1 2 and " > 0. Let Gt = Gt(m, n) be as in Deﬁnition 6.7. Then as m, n ! 1 subject to Conditions 6.1, PG? t {Y a l : a toxic} + < n1/2e−n3"/2. Proof. This proof proceeds by contradiction. Assume that PG? t {Y a l : a toxic} + ≥n1/2e−n3"/2. Then, PG? t G not nreg + = X a2Al PG? t G not nreg \| G 2 Y a l + PG? t Y a l + (6.8) ≥ X toxic a2Al PG? t G not nreg \| G 2 Y a l + PG? t Y a l + (6.9) ≥n−1/2 X toxic a2Al PG? t Y a l + (6.10) > e−n3"/2 (6.11) We justify the calculation above line by line. Line (6.8) follows by Bayes rule and because the sets {Y a l }a2Al partition all graphs, Bm,n,t, in the domain of G? t . Line (6.9) then follows because all probabilities are non-negative. Deﬁnition 6.11 then implies (6.10). The last line, (6.11) follows by our assumptive hypothesis. To complete the proof, observe that (6.11) contradicts Corollary (6.9). Deﬁnition 6.13 (bad). Fix 1 l M. Then l-tuple a 2 Al is bad if Y a l is toxic or has a sibling node which is toxic. Lemma 6.14. Fix 0 l M, 0 < a < 1 2 and " > 0. Let Gt = Gt(m, n) be as in Deﬁnition 6.7. Then as m, n ! 1 subject to Conditions 6.1, PG? t {Y a l : a bad} + < n3/2+"e−n3"/2. Proof. This follows because there at most m children to any node. Hence, PG? t {Y a l : a bad} + < mPG? t {Y a l : a toxic} + < n3/2+"e−n3"/2. 61 Chapter 6. Graph half -model, Gt. Bijection between sibling nodes, φG!H Consider the two sets of bipartite graphs: Y a l and Y b l , where \|a\| = \|b\| = l. When a and b di↵er only in the last element we expect the graphs in the two sets to be somewhat similar.1 We make this idea precise by deﬁning a bijection φ from each graph in Y a l to one in Y b l with symmetric di↵erence at most four labelled edges. The intuitive idea of the bijection φ is we want to match each H 2 Y a l with the closest resembling H0 2 Y b l . Deﬁnition 6.15 (graph match bijection φa!b l ). φa!b l = φa!b l (t, m, n) : Y a l ! Y b l . The map is deﬁned when Y a l , Y b l are both subsets of locally ordered bipartite graphs on (m, n) vertices with black degree sequence t and ah = bh for all 1 h < l. Let (h, i) = ft(l) and ﬁx G 2 Y a l . We specify the graph φ(G) by giving the symmetric di↵erence 4 between the labelled edge sets of G and φ(G). There are two possibilities for edge labellings of G which give the following two symmetric di↵erences: if (vh, vbi, r) 2 E(Go) (for some r > i), then Eo(Go) 4 Eo(φ(Go)) = {(uai, vk, i), (ubi, vk, r), (uai, vk, r), (ubi, vk, i)} else, Eo(Go) 4 Eo(φ(Go)) = {(uai, vk, i), (ubi, vk, i)}. We shall give an example. Consider locally ordered bipartite graphs on (4,4) vertices, with black degree sequence t =(1,2,3,2). We illustrate φ = φ(4, 4, (1, 2, 3, 2)) by displaying the two graphs H0 1 and H0 2 before and after the map φ is applied. See Figure 6.5. 1In the language of decision trees, this is equivalent to saying Y a l and Y b l are sibling nodes. We deﬁne a bijection sending graphs at a node into any sibling node. 62 6.2. Pathological degree sequences in Gt • • • • • • • • Ho 1 1 2 1 2 1 3 1 2 QQQQQQQQQQQQQQQQQQQQQ DDDDDDDDDDDDDDDDDDDDDDDDD mmmmmmmmmmmmmmmmmmmmm QQQQQQQQQQQQQQQQQQQQQ φ ✏ ✏ / / • • • • • • • • φ(Ho 1) 1 2 1 3 1 2 1 2 QQQQQQQQQQQQQQQQQQQQQ DDDDDDDDDDDDDDDDDDDDDDDDD mmmmmmmmmmmmmmmmmmmmm QQQQQQQQQQQQQQQQQQQQQ Eo(Ho 1) 4 Eo(φ(Ho 1)) = (u1, v3, 2) + (u4, v3, 3) +(u1, v3, 3) + (u4, v3, 2) • • • • • • • • Ho 2 1 2 1 2 3 1 1 2 QQQQQQQQQQQQQQQQQQQQQ DDDDDDDDDDDDDDDDDDDDDDDDD QQQQQQQQQQQQQQQQQQQQQ QQQQQQQQQQQQQQQQQQQQQ φ ✏ ✏ / / • • • • • • • • φ(Ho 2) 1 2 1 2 1 3 1 2 QQQQQQQQQQQQQQQQQQQQQ DDDDDDDDDDDDDDDDDDDDDDDDD mmmmmmmmmmmmmmmmmmmmm QQQQQQQQQQQQQQQQQQQQQ Eo(Ho 2) 4 Eo(φ(Ho 2)) = (u2, v3, 3) + (u4, v3, 3) Figure 6.5: In the illustration the bijection map φ is, φ = φ((1),(2,1),(3,1))!((1),(2,1),(3,4)) 5 : Y ((1)(2,1)(3,1)) 5 ! Y ((1)(2,1)(3,4)) 5 . Bounds between a node and its child Lemma 6.16. Let G be an locally ordered bipartite graph on (m, n) vertices with black degree sequence t. Fix a white vertex uj. 1. Then, \| X j (sj(G) −λn)2 − X j (sj(φ(G)) −λn)2\| < 2n + 2. 2. Additionally, suppose that G is near-regular, then, \| X j (sj(G) −λn)2 − X j (sj(φ(G)) −λn)2\| < 2n1/2+" + 2. Proof. By property of φ either both G and φ(G) have the same edge set (with two pairs of labels interchanged) or G can be obtained from φ(G) by moving one edge. In the ﬁrst 63 Chapter 6. Graph half -model, Gt. case our lemma is trivially true, so we assume the second. Hence for some 1 r, q m and some 1 k n, we have E(G) 4 E(φ(G)) = (ur, vk) + (uq, vk). Assume, w.l.o.g that (ur, vk) 2 E(G), then, X j (sj(G) −λn)2 − X j (sj(φ(G)) −λn)2 = (sr(G) −λn)2 + (sq(G) −λn)2 −(sr(φ(G)) −λn)2 −(sq(φ(G)) −λn)2 = (sr(G) −λn)2 + (sq(G) −λn)2 −(sr(G) −1 −λn)2 + (sq(G) + 1 −λn)2 = 2(sr(G) −λn) −2(sq(G) −λn) + 2 (6.12) Note that (6.12) can be simpliﬁed further. Taking absolute values, \|2(sr(G) −λn) −2(sq(G) −λn) + 2\| = \|2(sr(G) −sq(G)) + 2\| 2n + 2. (6.13) The bound (6.13) follows because the degrees of any two white vertices must lie between zero and n and hence their absolute di↵erence is less than or equal to n. This proves part a. When we make the additional assumptions in (2) we can see that the last line (6.12) must be less than 2n1/2+" + 2. This follows because we required, in G, that \|(Sj(G) −λn)\| < n1/2+" for each j. This completes the proof of (2). For the following proof it will be convenient to deﬁne P i+1(G) = PG? t (H)/ P J⌘iH PG? t (J). This notation is consistent with that used in [SS87], on which we have modelled the method of our proof. Lemma 6.17. Let X : ⌦! R be the random variable, X(G) := P j(Sj(G) −λn)2. For 0 i M let Xi be the random variable deﬁned in (6.5). 1. Then, \|Xi+1(G) −Xi(G)\| < 2n + 2. 2. Additionally, suppose that we have a good i-tuple a. Then for G 2 Y a i , \|Xi+1(G) −Xi(G)\| < n1/2+2" 64 6.2. Pathological degree sequences in Gt Proof. (of 2.) We prove the more restricted case ﬁrst. Fix G 2 Y a i . Then doing average of averages,2 Xi(G) = X H⌘iG Xi+1(H)P i(H) (6.14) ) Xi+1(G) −Xi(G) = X H⌘iG Xi+1(G) −Xi+1(H) + P i(H). (6.15) Fix H ⌘i G, then, similarly to (6.14): Xi+1(G) = X J⌘i+1G X(J)P i+1(J) (6.16) Xi+1(H) = X K⌘i+1H X(K)P i+1(K). (6.17) We want to bound the di↵erence between (6.16) and (6.17). Consider the bijection: φ = φG!H i+1 : {J : J ⌘i+1 G} ! {K : K ⌘i+1 H}. Also note by the deﬁnition of the nodes in the decision tree, P i+1(J) = P i+1(φ(J)). Hence can apply this bijection to obtain Xi+1(G) −Xi+1(H) = X J⌘i+1G Xi+1(J) −Xi+1(φ(J)) + P i+1(J). (6.18) By Lemma 6.16, for each near-regular J ⌘i+1 G we have, 1 1Xi+1(J) −Xi+1(φ(J)) 1 1 < 2n1/2+" + 2, (6.19) 2Quote from Shamir and Spencer, this theorem is somewhat analogous to Theorem 5 in [SS87] from which the quote originates. We follow their method quite closely. Note that Shamir and Spencer ﬁnd a single bound for all G whereas we ﬁnd bounds for two separate cases. In their case the random variable X(G) is deﬁned to be the chromatic number of G, for more information on the proof by Shamir and Spencer refer to Section 3.2.4 in this thesis. 65 Chapter 6. Graph half -model, Gt. and, for each not near-regular J ⌘i+1 G we have, 1 1Xi+1(J) −Xi+1(φ(J)) 1 1 < 2n + 2. (6.20) By (6.18): 1 1Xi+1(G) −Xi+1(H) 1 1 = X nreg J⌘i+1G Xi+1(J) −Xi+1(φ(J)) + P i+1(J) + X not nreg J⌘i+1G Xi+1(J) −Xi+1(φ(J)) + P i+1(J) = 2n1/2+" + 2 + X nreg J⌘i+1G P i+1(J) + 2n + 2 + X not nreg J⌘i+1G P i+1(J) (6.21) = 2n1/2+" + 2 + 1 −n−1/2+ + 2n + 2 + n−1/2 (6.22) < n1/2+2" (6.23) We justify the above line by line. The ﬁrst line follows directly from (6.18) and the deﬁni-tion of near-regular graphs. Noting the bounds in (6.19) and (6.20) for near-regular and general graphs respectively yields (6.21). Then (6.22) follows by Lemma 6.14. So then by (6.15), 1 1Xi(G) −Xi+1(G) 1 1 < n1/2+2" which yields the desired result. Proof. (of 1.) The proof of this case follows the above proof exactly excepting that do not assume any J ⌘i+1 G are near-regular. So we have a weaker bound. For a general graph G from Lemma 6.16(1) we have, for each J ⌘i+1 G, 1 1Xi+1(J) −Xi+1(φ(J)) 1 1 < 2n + 2. And hence we have 1 1Xi(G) −Xi+1(G) 1 1 < 2n + 2, as required. 66 6.2. Pathological degree sequences in Gt Lemma 6.18. Fix 0 l M, 0 < a < 1 2 and " > 0. Let Gt = Gt(m, n) be as in Deﬁnition 6.7. For 1 j m, let Sj be the random variable that returns the degree of the white vertex uj. Then as m, n ! 1 subject to Conditions 6.1, PGt ⇣1 1 1 P j(Sj −λn)2 −EGt P j(Sj −λn)2+ λ(1 −λ)mn 1 1 1 > n−1/2+7"⌘ < e−n3"/2. Proof. Observe ﬁrst that by Lemma 4.5, λ(1−λ) > 1 ln n > n−"/2, and so 1 λ(1−λ)mn > n−2+2". Thus it is suﬃcient to prove, PGt ⇣1 1 1 X j (Sj −λn)2 −EGt X j (Sj −λn)2+1 1 1 > n3/2+5"⌘ < e−n3"/2. (6.24) Let c = n1/2+2" and ⌘= n7"/2e−6"/5. To prove (6.24) we will show that the random vari-ables, X0, . . . , XM form a near-c-Lipschitz martingale with exceptional probability ⌘. By Lemma 6.10, we already have that X0, . . . , XM form a martingale. Let G 2 Y a l for some good node Y a l . Then \|Xi −Xi+1\|(G) < c. This is shown in Lemma 6.17. Hence for a ﬁxed l we can bound the set of graphs {G0} for which \|Xi − Xi+1\|(G0) ≥c, by taking the union of all bad nodes at level l of the tree. So we bound the sum of probability of all bad nodes over all levels, l, of the decision tree, M X l=1 PG? t {Y a l : a bad} + < Mn3"/2e−n6"/5 < n7"/2e−6"/5 (6.25) In (6.25) the ﬁrst inequality follows by Lemma 6.14 and the second inequality follows because M mn and m = o(n1+"). This shows our values for ⌘and c hold and so X0, . . . , Xn form a near-c-Lipschitz martingale with exceptional probability ⌘. We can now substitute these values of c and ⌘into the generalised Azuma inequality (Theorem 3.8), to yield (6.24). 6.2.6 Expectation of P j(Sj −λn)2 The aim of this Section is to show that (a, b, m, n, ")-pathological degree sequences are rare in Gt. We already have Lemma 6.1 which shows non-(", a)-regular sequences are very rare. Hence our current task is showing 1 − P j(Sj−λn)2 λ(1−λ)mn + 1 − P k(Tk−λm)2 λ(1−λ)mn + = O(n−b) with high probability. So far we have shown in Lemma 6.18 that with high probabil-ity P j(Sj−λn)2−E(P j(Sj−λn)2) λ(1−λ)mn = O(n−b). Hence now we want to show that E(P j(Sj−λn)2) λ(1−λ)mn = 67 Chapter 6. Graph half -model, Gt. 1 + O(n−b). This is done in Lemma 6.19 below. Before we can prove this, however, we need to calculate some preliminary expectations on our random variables for the white vertex degrees: s1, . . . , sm. Fix Sj, we derived earlier in line (6.1) the probability that there is an edge between uj and vk is precisely m−1tk. Hence we can write a probability generating function, B(x), for Sj in Gt. B(x) = n Y k=1 tk mx + m −tk m + We can now calculate the expectation of Sj and of S2 j . By the theory of generating func-tions explained in Section 3.3 this requires the evaluation of some di↵erential operators on the function B(x). These are slightly messy and so are relegated to the appendix, the following two results are from lines (8.5) and (8.6). Note these hold for all j. EGt(Sj) = x d dxB(x) + x=1 = λn EGt(S2 j ) = x d dxx d dxB(x) + x=1 = λn + λ2n2 −m−2 P k t2 k (6.26) We are now in a position to prove our result. Lemma 6.19. Let Gt = Gt(m, n) be as in Deﬁnition 1.8. For 1 j m, let Sj be the random variable that returns the degree of the white vertex uj. Then as m, n ! 1 subject to Conditions 6.1, EGt(P j(Sj −λn)2) λ(1 −λ)mn = 1 + o(n−1+5"). Proof. We ﬁrst calculate the expectation of P j(Sj −λn)2 using the values for expectation from (6.26). EGt( X j (Sj −λn)2) = EGt( X j (Sj −λn)2) = EGt( X j S2 j ) −λ2mn2 = λmn + λ2mn2 −m−1 X k t2 k −λ2mn2 = λmn −m−1 X k t2 k (6.27) 68 6.2. Pathological degree sequences in Gt Fix 0 k n, by assumption, tk −λm = o(n1/2+") and so P k(tk −λm)2 = o(n2+2"). By Lemma 4.5, because λ is acceptable for a, m and n, λ(1 −λ) > ln n. Hence we have, P k(tk −λm)2 λ(1 −λ)mn = o(n3") (6.28) The calculations below use (6.28) to derive a value for P k t2 k, P k(tk −mλ)2 λ(1 −λ)mn = o(n3") o(n3")λ(1 −λ)mn = X k (tk −mλ)2 = X k (t2 k) −2mλ X k (tk) −λ2m2n ) X k t2 k = λ2m2n + o(n3")λ(1 −λ)mn. This approximate value for P k t2 k can then be used to bound E(P j(Sj−λn)2) λ(1−λ)mn . By inequality (6.27), EGt(P j(Sj −λn)2) λ(1 −λ)mn = λmn −λ2mn + λ(1 −λ)n λ(1 −λ)mn + o(n−3"m−1) = 1 + m−1 + o(n−3"m−1) = 1 + o(n−1+5") This is exactly what we wanted. We can are now ready to show that pathological degree sequences are rare in the graph half -model, Gt. Theorem 6.20. Fix a, b 2 R+ and " > 0 such that 0 < a + b < 1 2 and b + 11" < 1/2. Let Gt = Gt(m, n) be as in Deﬁnition 1.8. For 1 j m, let Sj be the random variable that returns the degree of the white vertex uj. Then as m, n ! 1 subject to Conditions 6.1, PGt (S, T ) is (a, b, m, n, ")-pathological + e−n7"/6. 69 Chapter 6. Graph half -model, Gt. Proof. By Lemma 6.1, PGt( (S, T ) is not (", a)-regular ) e−n6"/5. Hence it is suﬃcient to show that, PGt ✓ 1 − P j(Sj −S)2 λ(1 −λ)mn ◆✓ 1 − P k(Tk −T)2 λ(1 −λ)mn ◆ = O(n−b) ! ≥1 −e−n3"/2 (6.29) The assumptions of this model imply that the average degree of the white vertices is determined by the m-tuple t, and so S = λn. By two of our previous results, Lemmas 6.19 and 6.18, PGt 1 1 1 1 P j(Sj −λn)2 λ(1 −λ)mn 1 1 1 1 > n−1/2+7" + 1 + o(n−1+5") ! < e−n3"/2. (6.30) Finally, recall by the assumptions of the graph half -model, Gt, all random graphs in Gt have black degree sequence t. Hence, T = λm and for each 1 k n, Tk = tk. By (6.28), P k(tk −λm)2 λ(1 −λ)mn = o(n3"). (6.31) Combining (6.30) and (6.31) then yields, PGt 1 1 1 1 ✓ 1 − P j(Sj −S)2 λ(1 −λ)mn ◆✓ 1 − P k(Tk −T)2 λ(1 −λ)mn ◆1 1 1 1 > n−1/2+11" ! e−n3"/2 (6.32) In particular, because b + 11" < 1/2, we get n−1/2+11" = O(n−b). Thus, (6.32) implies (6.29) and so we are done. 70 Part III New results: Approximation by binomial models 71 In this section we will deﬁne simple binomial probability models in which the probability of an (m + n)-tuple (s, t) is asymptotically very close to the probability of a degree se-quence (s, t) occurring in the random graph models. There are three random bipartite graph models for which we ﬁnd a corresponding bino-mial model. These three models are the graph p-model, Gp, the graph edge-model, GM, and the graph half -model, Gt. (See Deﬁnitions 1.6, 1.7 and 1.8 respectively.) To achieve this we construct binomial models which we call the binomial p-model, Bp, binomial edge-model, BM, and binomial half -model, Bt, respectively. The latter two bi-nomial models BM and Bt straight away match very closely with GM and Gt. The ﬁrst case needs more work and so we deﬁne an integrated version of Bp, the binomial inte-grated-model, denoted Vp, which then matches the graph p-model, Gp very closely. We construct the binomial probability spaces Bp, BM and Bt as restrictions of a common probability space, the binomial independent-model, denoted Ip. The independent-model is deﬁned in terms of random variables S1, . . . , Sm which are independent binomials with parameters (n,p) and random variables T1, . . . , Tn which are independent binomials with parameters (m,p). The three binomial models are then deﬁned by considering di↵erent subspaces (corresponding to di↵erent constraints) of this independent-model Ip. We proceed by ﬁrst deﬁning our binomial models in Section 7.1. After this we show the correspondences between these binomial models and our random graph models in Section 7.2. The ordering of Section 7.2 reﬂects the diﬃculty in obtaining these correspondences and so we defer our discussion the graph p-model, Gp to Section 7.2.3, after the graph edge-model, GM and the graph half -model, Gt in Subsections 7.2.1 and 7.2.2 respectively. 72 Chapter 7 Binomial probability spaces 7.1 Deﬁnition of binomial models 7.1.1 Binomial independent-model, Ip. We deﬁne our binomial probability models in terms of a common model (with some added constraints). This core model, we call the binomial independent-model, Ip. Deﬁnition 7.1 (Binomial independent-model, Ip.). The binomial independent-model Ip = Ip(m, n) is an (m+n)-dimensional probability space with domain {0, 1, 2, . . . , n}m⇥ {0, 1, 2, . . . , m}n. For j = 1, . . . , m and k = 1, . . . , n we deﬁne Sj and Tk to be independent binomially distributed random variables with parameters (n, p) and (m, p) respectively. That is, we have, PIp(Sj = a) = n a + pa(1 −p)n−a and PIp(Tk = a) = m a + pa(1 −p)m−a. Our (m+n)-dimensional space has the random variables (S = (S1, . . . , Sm), T = (T1, . . . , Tn)). The probability of the (m + n)-tuple, (s, t), in the binomial independent-model, Ip is given by, PIp(S = s, T = t) = m Y j=1 ✓n sj ◆ psjqn−sj n Y k=1 ✓m tk ◆ ptkqm−tk. We will proceed to deﬁne our other binomial models as probability spaces on a subset of the domain of the independent-model, Ip. 7.1.2 Binomial p-model, Bp. Our constraint for the ﬁrst binomial model is to require that the only (m + n)-tuples (s, t) with non-zero probability in the space are those for which the sum of the ﬁrst m 73 Chapter 7. Binomial probability spaces terms is the same as the sum of the last n terms, i.e. (s, t) 2 Im,n. Note this is a sensible constraint. We are hoping to use these binomial models to approximate the random bipartite graph models. The (m+n)-tuples in the domain of Bp will correspond to degree sequences in random bipartite graphs on (m, n) vertices. Recall that the number of edges in a bipartite graph can calculated by summing up the degrees of the vertices on either side. In particular, these two sums, the sums of the degrees of the white vertices and the sums of the degrees of the black vertices should be equal. This is our motivation for deﬁning the binomial p-model, Bp. Deﬁnition 7.2 (Binomial p-model, Bp.). The binomial p-model Bp = Bp(m, n) has the same domain as Ip with support Im,n. We formally deﬁne the probability space, binomial p-model, Bp as a restriction of the space of independent binomials, Ip, as follows PBp S = s, T = t + := PIp ⇣ S = s, T = t \| X j Sj = X k Tk ⌘ . As we alluded to earlier, ﬁnding a corresponding binomial model for the graph p-model requires more work than for the other random graph models. This binomial p-model will be used as a stepping stone in constructing a probability model that approximates the graph p-model. Indeed in Deﬁnition 7.7 we will introduce a binomial integrated-model, Vp, which is a convolution of Bp with a normal distribution. This binomial integrated-model, Vp is shown to correspond to the graph p-model in Section 7.2.4. We will calculate the probability of an (m + n)-tuple, (s, t) in the binomial p-model. However, this result is left until Section 7.2.3 as it uses results from the following section where we introduce the binomial edge-model. 7.1.3 Binomial edge-model, BM. The random variables in the binomial edge-model, BM, inherit the property (P j Sj = P k Tk) from the binomial p-model just deﬁned. The parameter M will correspond to the number of edges in the bipartite graph. So in this model we have the extra constraint on (S, T ) which corresponds to the number of edges in the bipartite graph. Hence we require the stricter property (P j Sj = P k Tk = M), i.e. (S, T ) 2 Im,n,M. Deﬁnition 7.3 (Binomial edge-model, BM.). The binomial edge-model BM = BM(m, n) has support Im,n,M. We formally deﬁne the probability space, binomial edge-model, BM 74 7.1. Deﬁnition of binomial models as a restriction of the space of independent binomials, Ip, as follows PBM(S = s, T = t) := PIp ⇣ S = s, T = t \| X j Sj = X k Tk = M ⌘ . We now show an example calculation in the binomial edge-model, BM = BM(4, 3) where our parameter, M is equal to six. The probability of the (4 + 3)-tuple, (1, 3, 1, 1; 2, 1, 3) is then, PBM=6((1, 3, 1, 1)(2, 1, 3)) = 3 1 + pq23 3 + p33 1 + pq23 1 + pq24 2 + p2q24 1 + pq34 3 + p3q 12 6 + p6q612 6 + p6q6 = 3 1 +3 3 +3 1 +3 1 +4 2 +4 1 +4 3 + 12 6 +12 6 + = 18 772. Observe that in the calculation above, all appearances of p and q cancelled. We show this is true in general, i.e. the probability of an (m + n)-tuple (s, t) in the binomial edge-model, BM, is independent of p. In the following lemma, we calculate the probability of (s, t) in the binomial edge-model, BM, where (s, t) is an (m+n)-tuple that has some of the necessary properties of a degree sequence of a bipartite graph with M edges on vertices (m, n). Hence, we require that (s, t) 2 {0, 1, . . . , n}m ⇥{0, 1, . . . , m}n satisﬁes P j sj = P k tk = M, i.e. we require (s, t) 2 Im,n,M. Lemma 7.4. Fix an (m + n)-tuple (s, t) 2 Im,n,M. Then we have PBM(S = s, T = t) = ✓mn M ◆−2 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ . Proof. This follows by the short calculation given below. Note the ﬁrst line is the deﬁnition of the binomial edge-model, BM. Then, (7.1) follows by Bayes law. In (7.2) we factor the denominator using the independence of all random variables in the independent-model, Ip and can simplify the numerator because we have already required that (s, t) 2 Im,n,M. PBM(S = s, T = t) = PIp ⇣ S = s, T = t \| X j Sj = X k Tk = M ⌘ = PIp ⇣ S = s, T = t & P j Sj = P k Tk = M ⌘ PIp ⇣P j Sj = P k Tk = M ⌘ (7.1) = PIp(S = s, T = t) PIp ⇣P j Sj = M ⌘ PIp ⇣P k Tk = M ⌘ (7.2) 75 Chapter 7. Binomial probability spaces In (7.2), PBM(S = s, T = t) has been reduced to known quantities. Line (7.3) now follows by the deﬁnition of the independent-model, Ip. Lastly, in (7.3) all instances of p and q cancel and we have our result. = m Y j=1 ✓n sj ◆ psjqn−sj n Y k=1 ✓m tk ◆ ptkqm−tk✓mn M ◆ pMqmn−M+−2 (7.3) = ✓mn M ◆−2 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ (7.4) Later, Theorem 7.1 shows that for non-pathological degree sequences, (s, t), PBM(s, t) is asymptotically very close to the probability of (s, t) occurring in the graph edge-model, GM under suitable constraints on M. 7.1.4 Binomial half -model, Bt. We deﬁne a binomial model which will corresponds to the graph half -model, Gt. We term this the binomial half -model and denote it by Bt. Recall that the graph half -model, Gt, is a probability space unique to bipartite graphs. In this graph model, we ﬁx the degrees of our black vertices and are interested in the likely distribution of the degrees of the white vertices, given this constraint. The binomial half -model, Bt, inherits both the property that (P j Sj = P k Tk) from the binomial p-model, Bp, and the stricter property that (P j Sj = P k Tk = M) from the binomial edge-model, BM. Also, in the binomial half -model we have the yet stricter con-straint that the values of the Tk (and not just their sum) are ﬁxed. That is, we ﬁx the values of the random variables T = (T1, T2, . . . , Tn). The intuition behind this constraint is that the random variables, (T1, T2, . . . , Tn), in the binomial half -model, Bt, will correspond to the random variables for the degrees of the black vertices, (T1, T2, . . . , Tn), in the graph half -model, Gt. In particular, one of the parameters of the binomial half -model, Bt, is the n-tuple t and we work in the subset of the domain of Ip where T = t. We deﬁne this concept formally below. Deﬁnition 7.5 (Binomial half -model, Bt.). The binomial half -model Bt = Bt(m, n, t) has support In,m,t. We formally deﬁne binomial half -model, Bt, as a restriction of the 76 7.1. Deﬁnition of binomial models space Ip. Note t is a parameter of the space. PBt(S = s, T = t) := PIp ⇣ S = s, T = t \| T = t, X j Sj = X k tk ⌘ In the binomial half -model, Bt, the probability of obtaining an (m+n)−tuple is indepen-dent of the parameter p. We showed this for the edge-model in Lemma 7.4. Recall that the binomial half -model is deﬁned as a subspace of the binomial independent-model, Ip, which is dependent on p. We prove this result in the following lemma. Lemma 7.6. Fix an (m + n)-tuple (s, t) 2 Im,n,t and set M = P k tk. Then we have PBt(S = s, T = t) = ✓mn M ◆−1 m Y j=1 ✓n sj ◆ . Proof. The proof is very similar to that for Lemma 7.4. Note the ﬁrst line, (7.5) is the deﬁnition of the binomial half -model, Bt. Line (7.6) follows by Bayes law. In (7.7) we factor the denominator using the independence of all random variables in the independent-model, Ip and we can simplify the numerator because we have already required that (s, t) 2 Im,n,t. PBM(S = s, T = t) = PIp ⇣ S = s, T = t \| X j Sj = X k Tk & T = t ⌘ (7.5) = PIp ⇣ S = s & T = t & P j Sj = P k Tk & T = t ⌘ PIp ⇣P j Sj = M & T = t ⌘ (7.6) = PIp S = s + PIp T = t + PIp ⇣P j Sj = M ⌘ PIp ⇣ T = t ⌘ (7.7) We are almost done. Line (7.8) follows by the deﬁnition of the independent-model, Ip. Lastly, all instances of p and q cancel in (7.9) and we have our result. PBM(S = s, T = t) = m Y j=1 ✓n sj ◆ psjqn−sj ✓mn M ◆ pMqmn−M !−1 (7.8) = ✓mn M ◆−1 m Y j=1 ✓n sj ◆ (7.9) 77 Chapter 7. Binomial probability spaces 7.2 Relation to random graph models In the previous section we deﬁned some binomially based probability spaces. In this section we will show that the probability of a given (non-pathological) degree sequence (s, t) in each of the random graph models is asymptotically very close to the probability of the same (m + n)-tuple (s, t) in one of our newly deﬁned binomial models. Of these correspondences between our three graph models and the binomial models, two of these are easily shown. We do these cases ﬁrst. In Section 7.2.1 we show that the random graph edge-model, GM, can be approximated by our binomial edge-model, BM. Then in a similar fashion we show that the graph half -model, Gt, can be approximated by the binomial half -model, Bt, in Section 7.2.2. The third random graph model, the p-model, Gp, requires more work. We will construct another probability space, the bino-mial integrated-model, Vp, based on the binomial p-model, Bp. After some calculations we show that this newly deﬁned binomial integrated-model, Vp, can be used to approximate the probability of non-pathological degree sequences (s, t) in the graph p-model, Gp. This ﬁnal case is done is Section 7.2.3. 7.2.1 Binomial edge-model, BM ⇠graph edge-model, GM. We show that for non-pathologicaldegree sequences (s, t) the probability of that degree sequence occurring in the graph edge-model, GM, is very close to the probability of the same (m + n)-tuple in the binomial edge-model, BM. Theorem 7.1. Fix a, b 2 R+ such that a+b < 1 2. Let GM = GM(m, n) be as in Deﬁnition 1.7 and BM = BM(m, n) as in Deﬁnition 7.3. Then there exists " > 0 such that as m, n ! 1 subject to Conditions 5.1, for non-(a, b, m, n, ")-pathological degree sequence (s, t) 2 Im,n,M, PGM(s, t) = PBM(s, t)(1 + O(n−b)). Proof. We ﬁrst calculate the probability of a ﬁxed non-pathological degree sequence (s, t) in the graph edge-model, GM. From Deﬁnition 1.6 we have PGM(s, t) := PGM(H : the degree sequence of H is (s, t)). Hence PGM(s, t) can be calculated by taking the number of bipartite graphs with degree sequence (s, t) and dividing by the total number of labelled bipartite graphs on (m, n) vertices with M edges. There are mn M + graphs on (m, n) vertices with M edges. Let us denote the number of bipartite graphs with degree sequence (s, t) by \|B(s, t)\|. Hence, 78 7.2. Relation to random graph models PGM(s, t) := ✓mn M ◆−1 \|B(s, t)\|. (7.10) By Corollary 2.10, for p which is acceptable for a, m and n and non-(a, b, m, n, ")-pathological degree sequences, (s, t), we have an approximate count of the number of graphs with that degree sequence, i.e. an approximate value for \|B(s, t)\|. We substitute this value for \|B(s, t)\| in (7.10) to yield, PGM(s, t) = ✓mn M ◆−2 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ exp(O(n−b)). (7.11) The form of the right-hand side is very similar to that for the probability of an (m + n)-tuple (s, t) in the binomial edge-model BM. In particular by Lemma 7.4, PBM(s, t) = ✓mn M ◆−2 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ . (7.12) We now substitute the expression for PBM(s, t) in (7.12) into line (7.11) to yield the required result. 7.2.2 Binomial half -model, Bt ⇠graph half -model, Gt. In this section we show that the probability of an (", a)-regular degree sequence (s, t) in the graph half -model, Gt, is asymptotically very close to the probability of (s, t) in the binomial half -model, Bt. This proceeds by a direct proof which compares results for the probability of (m + n)-tuples (s, t) in the binomial half -model, Bt, to results for the probability of (", a)-regular degree sequences (s, t) in the graph half -model, Gt. Theorem 7.2. Let Gt = Gt(m, n) be as in Deﬁnition 1.8 and Bt = Bt(m, n) be as in Deﬁnition 7.5. Fix a non-pathological degree sequence (s(m, n), t(m, n)) 2 Im,n,t. Then as m, n ! 1 subject to Conditions 6.1, PGt s, t + = PBt s, t + 1 + O(n−b) + . Proof. We ﬁrst calculate the probability of a ﬁxed non-pathological degree sequence (s, t) in the graph half -model, Gt. From Deﬁnition 1.8 of the graph half -model, Gt we have 79 Chapter 7. Binomial probability spaces PGt(s, t) := PGt(H : the degree sequence of H is (s, t)). Hence PGt(s, t) can be calculated by taking the number of bipartite graphs with degree sequence (s, t) and dividing by the total number of labelled bipartite graphs on (m, n) vertices with black degree sequence t. There are Q k m tk + graphs on (m, n) vertices with black degree sequence t1. So we have PGt(s, t) := n Y k=1 ✓m tk ◆!−1 \|B(s, t)\|. (7.13) By Corollary 2.10 from [GM09], for acceptable for a, m and n values of the n-tuple t, and non-(a, b, m, n, ")-pathological degree sequences, (s, t), we have an approximate count for the number of graphs with that degree sequence, i.e. an approximate value of \|B(s, t)\|. We substitute this value for \|B(s, t)\| in (7.13) to yield, PGt(s, t) = ✓mn M ◆ n Y k=1 ✓m tk ◆!−1 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ exp(O(n−b)) = ✓mn M ◆−1 m Y j=1 ✓n sj ◆ exp(O(n−b)). (7.14) The form of the right-hand side is very similar to that for the probability of an (m + n)-tuple (s, t) in the binomial half -model, Bt. In particular by Lemma 7.6, PBM(s, t) = mn M +−1 Qm j=1 n Sj + . In light of this we may substitute PBt(s, t) into the equation (7.14) above to give the required result. 1To see this note the following. Fix a black vertex vk. There are tk edges incident with this vertex. Hence of the m white vertices, there are edge connecting vk to any tk distinct white vertices. So, there are m tk + possible arrangements for the edges incident with vk. Now, note the edges incident with one black vertex are disjoint from the edges incident with any other black vertex. Hence the total number of labelled bipartite graphs with black degree sequence t is the product: Qn k=1 m tk + . 80 7.2. Relation to random graph models 7.2.3 Binomial p-model, Bp. Given an (m + n)-tuple (s, t) we will calculate the probability of (s, t) in the binomial p-model, Bp but ﬁrst we need the following somewhat technical lemma. Lemma 7.3. Let 0 < a < 1/2 and suppose that pq > 1 log n. Then, n X i=1 ✓n i ◆ piqn−i = 1 2p⇡npq ⇣ 1 + O n−1/2+"+ ⌘ + O(e−n") Corollary 7.4. Let 0 < a < 1/2 and suppose that p is acceptable for a, m and n. Then, PIp X j Sj = X k Tk + = 1 2p⇡pqmn 1 + O(n−1) + + O(e−n") Proof. Each Sj is binomially distributed in Ip with parameters (n, p). Thus Pm j=1 Sj is binomially distributed with parameters (mn, p). Similarly, Pn k=1 Tk is also binomially distributed in Ip with parameters (mn, p). Thus we have, PIp X j Sj = X k Tk + = mn X i=1 ✓mn i ◆ piqmn−i. Hence it is suﬃcient to prove that, mn X i=1 ✓mn i ◆ piqmn−i = 1 2p⇡pqmn 1 + O(n−1) + + O(e−n") and so the result follows by Lemma 7.3 Lemma 7.5. Fix (s, t) 2 Im,n and set M = P j sj. Then let Bp = Bp(m, n) be as in Deﬁnition 7.2 and BM = BM(m, n, M) be as in Deﬁnition 7.3. Then as m, n ! 1 subject to Conditions 4.1, PBp(s, t) = m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ p2Mq2mn−2M2p⇡pqmn 1 + O(n−1/2 ln2 n) + . Proof. By Bayes law, PBp(s, t) = PBp ⇣X j Sj = M ⌘ PBp ⇣ (s, t) \| X j Sj = X k Tk = M ⌘ . 81 Chapter 7. Binomial probability spaces Then, PBp(s, t) = PIp ⇣X j Sj = X k Tk = M \| X j Sj = X k Tk ⌘ PBM(s, t) = PIp P j Sj = P k Tk = M & P j Sj = P k Tk + PIp(P j Sj = P k Tk) PBM(s, t) where the ﬁrst line follows by the deﬁnitions of Bp and BM and second line follows again by Bayes Law. Observe now that the two statements, X j Sj = X k Tk = M & X j Sj = X k Tk and X j Sj = M & X k Tk = M describe the same event. Also recall that in the Ip model, the Sj’s and the Tk’s are independent. Hence, PBp(s, t) = 1 PIp(P j Sj = P k Tk)PIp X j Sj = M + PIp X k Tk = M + PBM(s, t). (7.15) The values of each of the terms on the right hand side of (7.15) are known. The quantities, PIp P j Sj = M + and PIp P k Tk = M + follow straight from Deﬁnition 7.1 of the Ip model. The calculations for PBM(s, t) appear in Lemma 7.4. Substitute these values into (7.15) to yield, PBp(s, t) = 1 PIp ⇣P j Sj = P k Tk ⌘✓mn M ◆ pMqmn−M+2 ✓mn M ◆−2 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ . Lemma 7.4 provides an approximation for PIp(P j Sj = P k Tk). Hence we now have, PBp(s, t) = 2p⇡pqmn pMqmn−M+2 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ 1 + O(n−1/2 ln2 n), (7.16) which completes the proof. By Theorem 4.9 we have a formula to enumerate all bipartite graphs with any given non-pathological degree sequence. We use this below to calculate the asymptotic probability of a graph having degree sequence (s, t). 82 7.2. Relation to random graph models Lemma 7.6. Fix a, b 2 R+ and " > 0 such that 0 < a + b < 1 2. Let Gp = Gp(m, n) be as in Deﬁnition 1.6. Fix non-(a, b, m, n, ")-pathological (s, t) 2 Im,n. Then as m, n ! 1 subject to Conditions 4.1, PGp(s, t) = pMqmn−M ✓mn M ◆−1 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ 1 + O(n−b) + . Proof. By Bayes law and by the deﬁnitions of Gp and GM we have, PGp(s, t) = PGp X j Sj = M + PGp (s, t) \| X j Sj = M + = PGp X j Sj = M ⌘ PGM s, t + . Note the probability that a random graph in Gp has M edges is mn M + pMqmn−M. Also, recall that by line (7.10) we have a formula for PGM s, t + in terms of the number of bipartite graphs on (m, n) vertices, \|B(s, t)\|. Thus, PGp(s, t) = ✓mn M ◆ pMqmn−M ✓mn M ◆−1 \|B(s, t)\| = pMqmn−M ✓mn M ◆−1 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ 1 + O(n−b) + , where the second line follows by the result of Greenhill and Mckay, Corollary 2.10. 83 Chapter 7. Binomial probability spaces 7.2.4 Integrated binomial model, Vp ⇠graph p-model Gp. The binomial integrated-model, Vp is deﬁned in terms of the binomial p-model, BM. We construct Vp as a convolution of the binomial p-model, Bp, with a normal distribution, Kp. Currently the binomial p-model, Bp, is deﬁned only for p 2 [0, 1], so we let PBp(s, t) := 0, 8p / 2 [0, 1]. In the deﬁnition below we normalise the probability space by dividing through by V (p). Deﬁnition 7.7. The binomial integrated-model Vp = Vp(m, n, p) has support In,m. Let Kp = mn ⇡pq +1/2 exp −mn pq (p −p0)2+ and V (p) = R 1 0 Kp(p0)dp0. Then deﬁne, PVp(s, t) := 1 V (p) Z 1 −1 Kp(p0)PBp0(s, t)dp0. In this section we will calculate the probability of a given degree sequence (s, t) in the model Vp and show that this agrees very closely with the probability of this degree sequence occurring in our random graph p-model, Gp. This result will rely on the following two technical lemmas. Lemma 7.8. Let y := M−pmn ppqmn and assume that \|y\| < n4". Then as m, n ! 1 subject to p acceptable for a, m and n, ✓ pMqmn−M ✓mn M ◆◆−1 = p 2⇡pqmn exp y2 2 + O (mn)−1/2+"++ Lemma 7.9. Fix " > 0. Let y = M−pmn ppqmn and δ = p −p0. Write q = 1 −p and q0 = 1 −p0. Assume \|y\| < n4". Then as m, n ! 1 subject to p acceptable for a, m and n, Z 1 −1 p0 p +2M+1/2q0 q +2mn−2M+1/2 exp ⇣ −mn pq δ2⌘ dδ = r ⇡pq 2mn exp ⇣y2 2 ⌘⇣ 1 + O(n−1/2+12") ⌘ We defer the proof of both Lemma 7.8 and 7.9 to Section 9.1 of the Appendix. Assuming Lemmas 7.8 and 7.9 we are now in a position to show that for non-pathological (s, t), PVp(s, t) is a good asymptotic approximation for PGp(s, t). Theorem 7.10. Fix a, b 2 R+ and " < "0(a, b) such that a + b < 1 2, a = a0 + " < 1 2 and b + 17" < 1. Let Gp = Gp(m, n) be as in Deﬁnition 1.6 and Vp = Vp(m, n) be as in Deﬁnition 7.7. Let (s, t) 2 In,m,M for some 0 M mn. Suppose also that (s, t) is non-(a, b, m, n, ")-pathological and satisﬁes M−pmn ppqmn < n5". Then as m, n ! 1 subject to Conditions 4.1, PGp(s, t) = PVp(s, t) 1 + O(n−b) + . 84 7.2. Relation to random graph models Proof. PVp(s, t) = Z 1 −1 mn ⇡pq +1/2 exp −mn pq δ2+ PBp0(s, t)dδ (7.17) = Z 1 −1 mn ⇡pq +1/2 exp −mn pq δ2+ m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ ⇥p02Mq02mn−2M2 p ⇡p0q0mn 1 + O(n−1/2 ln2 n) + dδ (7.18) =pMqmn−M ✓mn M ◆ −1 m Y j=1 ✓n sj ◆ n Y k=1 ✓m tk ◆ 2mnpMqmn−M ✓mn M ◆ 1 + O(n−1/2 ln2 n) + ⇥ Z 1 −1 exp −mn pq δ2+⇣p0 p ⌘2M+1/2⇣q0 q ⌘2mn−2M+1/2 dδ (7.19) =PGp(s, t)2mnpMqmn−M ✓mn M ◆ 1 + O(n−1/2 ln2 n) + 1 + O(n−b) + ⇥ Z 1 −1 exp −mn pq δ2+⇣p0 p ⌘2M+1/2⇣q0 q ⌘2mn−2M+1/2 dδ (7.20) The ﬁrst line, (7.17), follows by Deﬁnition 7.7 of the binomial integrated-model, Vp. Then (7.18) follows by Lemma 7.5. Line (7.19) follows by rearranging and cancelling in (7.18). By Lemma 7.6 and because (s, t) is non-pathological we then get (7.20). Let y = M−pmn ppqmn and then by assumption \|y\| < n4". Now (7.21) follows by Lemma 7.9 (also note that the O(n−b) error term is larger). Hence, 85 Chapter 7. Binomial probability spaces PVp(s, t) = PGp(s, t)2mnpMqmn−M ✓mn M ◆ 1 + O(n−1/2 ln2 n) + 1 + O(n−b) + ⇥ r ⇡pq 2mn exp ⇣y2 2 ⌘ (7.21) = PGp(s, t)2mn ⇣p 2⇡pqmn exp y2 2 + O (mn)−1/2+"++⌘−1 ⇥ 1 + O(n−1/2 ln2 n) + 1 + O(n−b) +r ⇡pq 2mn exp ⇣y2 2 ⌘ (7.22) = PGp(s, t) ⇣ 1 + O (mn)−1/2+"+⌘⇣ 1 + O(n−1/2 ln2 n) ⌘⇣ 1 + O(n−b) ⌘ (7.23) = PGp(s, t) ⇣ 1 + O(n−b) ⌘ (7.24) Line (7.22) follows by Lemma 7.8. Rearrange to get (7.23). Then because O(n−b) is the largest error term we have (7.24) and we are done. 86 7.3. Implications 7.3 Implications We will show that the expectation of a random variable X : Im,n ! R in any one of the random bipartite graph models can be asymptotically approximated by the expectation of the same event in the corresponding binomial model. This is done for the graph edge-model, GM in Theorem 7.4 and the graph p-model, Gp in Theorem 7.5. The results in Theorems 7.4 and 7.5 form the bipartite analogue of Theorem 2.6 in [MW97], our Theo-rem 2.6. First we need some lemmas which bound the probability of pathological (m + n)-tuples in each of the binomial models. We begin by considering the binomial p-model, Bp. Lemma 7.1. Fix " > 0 and 0 < a < 1 2 such that a + " < 1 2 and b + 17" < 1. Let Bp = Bp(m, n) be as in Deﬁnition 7.2. Then as m, n ! 1 subject to Conditions 4.1, PBp (S, T ) is (a, b, m, n, ")-pathological + e−n11"/10. Proof. Many of the bounds we will show in Bp will follow directly from results in Gp. This is because, considered alone, the degrees of the white vertices in Gp are binomi-ally distributed with parameters (n, p). Also the degrees of uj and uj0 are independent 80 j, j0 k (except j = j0). Thus the distribution of the degree of uj in Gp is the same as that for the random variables Sj in Bp. (The dependence between the degrees of the vertices uj and vk in Gp is di↵erent to the dependence between Sj and Tk is di↵erent in Bp.) Hence by Lemma 4.3, PBp \|Sj −S\| ≥n1/2+2"/5+ = PGp \|Sj −S\| ≥n1/2+2"/5+ e−n4"/3, PBp \|Tk −T\| ≥n1/2+10"/11+ = PGp \|Tk −T\| ≥n1/2+10"/11+ e−n5"/4. Thus, PBp 8j, k, Tk −T, Sj −S uniformly o(n1/2+") + ≥1 −PGp \|Sj −S\| ≥n1/2+2"/5+ −PGp \|Tk −T\| ≥n1/2+10"/11+ ≥1 −e−n6"/5. (7.25) Deﬁne λ = 1 mn P j Sj. Then the distribution of λ in Bp and the edge density λ in Gp are the same. Thus by Lemma 4.7, PBp λ is acceptable for a, m and n + = PGp λ is acceptable for a, m and n + ≥1 −e−n3"/2. (7.26) 87 Chapter 7. Binomial probability spaces We can apply the same comparisons with the Gp method to deduce (by Lemmas 4.6 and 4.8 respectively), PBp 1 1 11 − P j(Sj −S)2 λ(1 −λ)mn 1 1 1 ≥n−1/2+8" ! e−n6"/5 and PBp 1 1 11 − P k(Tk −T)2 λ(1 −λ)mn 1 1 1 ≥n−1/2+9" ! e−n8"/7. And thus, PBp 1 1 1 1 ⇣ 1 − P j(Sj −S)2 λ(1 −λ)mn ⌘⇣ 1 − P k(Tk −T)2 λ(1 −λ)mn ⌘1 1 1 1 ≥n−1+17" ! e−n9"/8. (7.27) Hence by (7.25),(7.26) and (7.27), PBp (S, T ) is (a, b, m, n, ")-pathological + e−n6"/5 + e−n3"/2 + e−n9"/8 e−n11"/10. Next we work in the binomial edge-model, BM, and bound the probability that a random (m + n)-tuple in BM will be pathological. Corollary 7.2. Fix " > 0 and 0 < a < 1 2 such that a + " < 1 2 and b + 17" < 1. Let BM = BM(m, n) be as in Deﬁnition 7.3. Then as m, n ! 1 subject to Conditions 5.1, PBM (S, T ) is (a, b, m, n, ")-pathological + e−n12"/11. Proof. Similar to Lemma 5.1 we can conclude that for any event A ⇢Im,n,M, PBM(A) mnPBp= M mn (A). (7.28) Observe that as we have assumed Conditions 5.1, M mn is acceptable for a, m and n. Thus our choice of p = M mn implies that p is acceptable for a, m and n and we have Conditions 4.1. Let P ⇢Im,n,M be the set of (a, b, m, n, ")-pathological (m + n)-tuples in Im,n,M. By Lemma 7.1, PBp= M mn P + e−n11"/10. Now let the event A in (7.28) be our set P and this yields the asserted result. Lastly we consider the binomial integrated-model, Vp and bound the probability of the set of pathological (m + n)-tuples in Vp. 88 7.3. Implications Corollary 7.3. Fix " > 0 and 0 < a < 1 2 such that a + " < 1 2 and b + 17" < 1. Let Vp = Vp(m, n) be as in Deﬁnition 7.7. Then as m, n ! 1 subject to Conditions 4.1, PBM (S, T ) is (a, b, m, n, ")-pathological + e−n". Proof. Let P ⇢Im,n be the subset of (a, b, m, n, ")-pathological (m+n)-tuples in Im,n. By the deﬁnition of Vp and the change of variables δ = p0 −p we get, V (p).PVp P + = r mn ⇡pq Z 1 0 exp ⇣−mn pq (p0 −p)2⌘ PBp0 P + dp0 = r mn ⇡pq Z [−n"−1, n"−1] exp ⇣−mn pq δ2⌘ PBp0 P + dδ + r mn ⇡pq Z [−p, q]−[−n"−1, n"−1] exp ⇣−mn pq δ2⌘ PBp0 P + dδ (7.29) We will bound both terms on the right hand side of (7.29) seperately. The integral over the interval [−" ln n, " ln n] is bounded ﬁrst. Let a0 = a + " and note that a0 < 1/2. Now observe that by Deﬁnition 2.5 of acceptable that if p is acceptable for a, m and n then p + n"−1 is acceptable for a0, m and n. Hence for p0 2 [p −n"−1, p + n"−1], by Lemma 7.1, PBp0 P + e−n11"/10. Thus, Z [−n"−1, n"−1] exp ⇣−mn pq δ2⌘ PBp0 P + dδ < e−n11"/10 Z [−n"−1, n"−1] exp ⇣−mn pq δ2⌘ dδ. We also note that Z [−n"−1, n"−1] exp ⇣−mn pq δ2⌘ dδ  Z 1 0 exp ⇣−mn pq δ2⌘ dδ. Hence, r mn ⇡pq Z [−n"−1, n"−1] exp ⇣−mn pq δ2⌘ PBp0 P + dδ e−n12"/11. (7.30) We now bound the value of integral on the intervals [−p, −n"−1] and [n"−1, q] (which we denote [−p, q] −[−n"−1, n"−1] ). Observe that because Bp is a probability space, 8p0 2 [0, 1], PBp0 P + 1 and by deﬁnition 8p0 / 2 [0, 1], PBp0 P + = 0. 89 Chapter 7. Binomial probability spaces Hence by Lemma 9.6 we can bound the ‘tails’ of the integral as follows, Z [−p, q]−[−n"−1, n"−1] exp ⇣−mn pq δ2⌘ PBp0 P + dδ = O ✓1 u exp ⇣−u2 2 ⌘◆Z [−p, q] exp ⇣−mn pq δ2⌘ PBp0 P + dδ where u = r pq 2mnn1−" > n2"/3. And thus we have, r mn ⇡pq Z [−p, p]−[−" ln n, " ln n] exp ⇣−mn pq δ2⌘ PBp0 P + dδ e−n4"/3. (7.31) The asserted result now follows by (7.30) and (7.31). We are now ready to prove the asymptotic expectation of any random variable in the graph edge-model, GM can be approximated by its expectation in the binomial edge-model, BM. Theorem 7.4. Fix " > 0 and 0 < a < 1 2 such that a + " < 1 2 and b + 17" < 1. Let GM = GM(m, n) be as in Deﬁnition 1.7 and let BM = BM(m, n) be as in Deﬁnition 7.2. Let X : Im,n,M ! R be any random variable. Then as m, n ! 1 subject to Conditions 5.1, EGM X + −EGM X + = + O(n−b)EGM \|X\| + + max x2Im,n \|X(x)\|.O e−n10"/11+ Proof. Let P ⇢Im,n,M be the set of (a, b, m, n, ")-pathological(m+n)-tuples in Im,n,M. By the deﬁnition of expectation, EGM X + = X x2P c PGM(x)X(x) + X x2P PGM(x)X(x). (7.32) 90 7.3. Implications For non-pathological (s, t), PGM(s, t) is approximately PBM(s, t). Hence by Theorem 7.1, EGM X + = X x2P c PBM(x) 1 + O(n−b) + X(x) + max x2Im,n \|X(x)\|.O(PGM(P)) (7.33) = X x2Im,n PBM(x) 1 + O(n−b) + X(x) + max x2Im,n \|X(x)\|.O ⇣ PBM(P) 1 + O(n−b)) + PGM(P) ⌘ . (7.34) Consider the ﬁrst term on the right hand side of (7.34). X x2Im,n PBM(x) 1 + O(n−b) + X(x) = X x2Im,n PBM(x)X(x) + X x2Im,n PBM(x)O(n−b)X(x) (7.35) =EBM X + + O(n−b) X x2Im,n PBM(x)\|X(x)\| (7.36) =EBM X + + O(n−b)EBM \|X\| + (7.37) Here, line (7.35) follows by simple rearrangement. Then by the deﬁnition of expectation and the uniformity of the error term O(n−b) we get (7.36). Then (7.37) follows again by the deﬁnition of expectation. By Theorem 5.2 and Corollary 7.2 we have PGM(P) e−n11"/12 and PBM(P) e−n12"/11 respectively. Hence by (7.34) and (7.37), EGM X + =EBM X + + O(n−b)EBM \|X\| + + max x2Im,n \|X(x)\|.O e−n10"/11+ and we have our result. A similar result holds which relates asymptotic expectations in Gp and Vp. Theorem 7.5. Fix " > 0 and 0 < a < 1 2 such that a + " < 1 2 and b + 17" < 1. Let Gp = Gp(m, n) be as in Deﬁnition 1.6 and let Vp = Vp(m, n) be as in Deﬁnition 7.7. Let X : Im,n ! R be any random variable. Then as m, n ! 1 subject to Conditions 4.1, EGp X + −EVp X + = O(n−b)EVp \|X\| + + max x2Im,n \|X(x)\|O(e−n9"/10). 91 Chapter 7. Binomial probability spaces Proof. We begin with the following observations which will be necessary for the proof. Let y = P j Sj−pmn ppqmn and consider the likely magnitude of \|y\| in Gp. Note that y = q mn pq 1 1λ −p 1 1. Hence by Lemma 4.2, PGp(\|y\| > n4") e−n2". (Note that as Lemma 4.2 only concerns the random variables Sj and p (and not Tk’s) we also have that PBp(\|y\| > n4") e−n2" which implies PVp(\|y\| > n4") e−n" in a similar fashion to Corollary 7.3.) The remainder of the proof proceeds in the same way as the proof of Theorem 7.4. Let P ⇢Im,n be the set of (m + n)-tuples in Im,n that are either (a, b, m, n, ")-pathological or for which \|y\| > 4". By the deﬁnition of expectation, EGp X + = X x2P c PGp(x)X(x) + X x2P PGp(x)X(x). (7.38) By Theorem 7.10, for non-(a, b, m, n, ")-pathological (s, t), the values PGp(s, t) and PVp(s, t) are asymptotically very close. Hence, EGp X + = X x2P c PBp(x) 1 + O(n−b) + X(x) + max x2Im,n \|X(x)\|.O(PGp(P)) (7.39) Similarly to the proof of Theorem 7.4, line (7.39) implies, EGp X + −EVp X + = O(n−b)EVp \|X\| + + max x2Im,n \|X(x)\|.O ⇣ 1 + O(n−b) + PBp P + + O(PGp(P)) ⌘ (7.40) By Theorem 4.9 and Corollary 7.3 we have PGp(P) e−n10"/11 + e−n2" and PVp(P)  e−n" + e−n2" respectively. Hence our result follows from (7.40). Consider, any event A ⇢Im,n and the random variable IA deﬁned to be the indicator function of A. Then E(IA) = P(A). Hence Theorems 7.4 and 7.5 yield the following corollaries. Corollary 7.6. Under the same conditions and assumptions as Theorem 7.4, let A ⇢ Im,n,M then, PGM A + −PBM A + = O(n−b)PBM \|A\| + + O(e−n10"/11). Corollary 7.7. Under the same conditions and assumptions as Theorem 7.5, let A ⇢Im,n then, PGp A + −PVp A + = O(n−b)PVp \|A\| + + O(e−n9"/10). 92 7.3. Implications We have shown that degree sequences in random bipartite graph models can be approxi-mated by independent binomials subject to certain constraints. The independence of the random variables on which our binomial models are based suggests that the degrees of the white vertices are perhaps in some sense independent of the degrees of the black vertices. We make this idea precise in the following theorem concerning the graph edge-model, GM. Theorem 7.8. Fix a, b 2 R+ such that 0 < a + b < 1 2. Then ﬁx " = "0(a, b) > 0. Let GM = GM(m, n) be as in Deﬁnition 1.7 and BM = BM(m, n) be as in Deﬁnition 7.3. For 1 j m, let Sj be the random variable that returns the degree of the white vertex uj and for 1 k n, let Tk be the random variable that returns the degree of the black vertex uj. Let ES be an event deﬁned in terms of the random variables S1, . . . , Sm and ET be an event deﬁned in terms of the random variables T1, . . . , Tn. Then as m, n ! 1 subject to Conditions 5.1, PGM ES and ET + = 1 + O(n−b) + PGM ES + PGM ET + + O(e−n10"/11). Proof. PGM ES and ET + = 1 + O(n−b) + PBM ES and ET + + O(e−n10"/11) (7.41) = 1 + O(n−b) + PBM(ES)PBM(ET) + O(e−n10"/11) (7.42) = 1 + O(n−b) + PGM(ES)PGM(ET) + O(e−n10"/11) (7.43) We justify the calculation line by line. Deﬁne the random variable IES&ET to be the indica-tor function for the event ES&ES. Then the expectation, EGM(IES&ET ) = PGM(ES&ET). Thus (7.41), follows by Theorem 7.4. Line (7.42) follows by the independence of S and T in the binomial edge-model, BM. Line (7.43) then follows by reapplying Theorem 7.4 to (7.42). And thus we have our result. We also observe that such a theorem could not hold in the graph p-model, Gp. Recall that S (resp. T) is the random variable that returns the average degree of the white (respectively black) vertices. Now ﬁx an integer M, where 0 M mn and set ES = {S \| P j Sj = M} and ET = {T \| P k Tk = M}. Then ES = 1 mM = n mET. Hence for the Gp model we have a counter-example which shows there is no result in Gp which is analogous result to Theorem 7.8 in GM. 93 Part IV Appendix 94 Chapter 8 Calculations on the likelihood of pathological degree sequences. 8.1 Generating functions for graph p-model, Gp. 8.1.1 Preliminary expectations The aim of this section is to provide some calculations needed to show that pathological degree sequences are rare in the graph p-model, Gp. We calculate some preliminary ex-pectations on our random variables for the white vertices’ degrees: s1, . . . , sm. Fix a white vertex uj and consider the degree of that vertex, Sj. By deﬁnition of the graph p-model, Gp, the probability that there is an edge between uj and any black vertex vk is precisely p. As there are n black vertices, the random variable Sj is the standard binomial in (n, p). Hence we can write probability generating function, A1(x), for Sj in Gp. A1(x) = (px + q)n This generating function will allow us to calculate the expectation of Sj and of S2 j . By the theory of generating functions explained in Section 3.3 this requires the evaluation of some di↵erential operators on the function A1(x). 95 Chapter 8. Calculations on pathological degree sequences. 8.1.2 Di↵erentials of A1, (Gp). (xD)A1 = xpn(px + q)n−1 (xD)A1 + x=1 = pn (8.1) (xD)2A1 = x2D2A1 + xDA1 = x2p2n(n −1)(px + q)n−2 + xpn(px + q)n−1 (xD)2A1 + x=1 = p2n(n −1) + pn = pn(1 −p) + p2n2 = pnq + p2n2 (8.2) We can now calculate the following expectations of the functions on the degrees of the vertices. Note these hold for all j = 1, . . . , m. The ﬁrst two lines, the results on A1, are from the lines (8.1) and (8.2). E[Sj] = d dxA1(x) + x=1 = np(px + q)n−1+ x=1 = np E[S2 j ] = d dxx d dxA1(x) + x=1 = d dxxnp(px + q)n−1+ x=1 = np + n(n −1)p2 Thus by linearity of expectation, E[ X j S2 j ] = X j E[S2 j ] = mnp + mn(n −1)p2 E[ X j Sj] = X j E[Sj] = mnp 8.1.3 Expectation of P j(Sj −S)2 E[ X j (Sj −S)2] = E[ X j S2 j ] −2E[S X j Sj] + mE[S2] = E[ X j S2 j ] −2 mE[( X j Sj)2] + m m2E[( X j Sj)2] = E[ X j S2 j ] −1 mE[( X j Sj)2] = mnp + mn(n −1)p2 −1 m(mn(mn −1)p2 + mnp) = mnpq −npq (8.3) 96 8.2. Generating functions for graph half -model, Gt. 8.1.4 Expectation of P j(Sj −np)2 E[ X j (Sj −np)2] = E[S2 j ] −2E[np X j Sj] + mn2p2 = mnp + mn(n −1)p2 −2np(mnp) + mn2p2 = mnpq (8.4) 8.2 Generating functions for graph half -model, Gt. 8.2.1 Di↵erentials of B, (Gt) We work in the random graph model Gt. In this model the degrees of the black vertices, i.e. t, are given, and all bipartite graphs with this black degree sequence are equally likely. This section provides some calculations needed to show that pathological degree sequences are rare in the graph half -model, Gt. Recall that B(x) is the probability generating function for Sj, the degree of vertex uj in random graph model Gt. Similar to the previous graph model Gp, we ﬁnd the expectation of P j(Sj −S)2 in Gt. For this purpose we calculate up to the second order derivative of B(x). B = n Y k=1 tk mx + m −tk m + DB = 1 m n X h=1 th Y k6=h tk mx + m −tk m ++ DB(x) + x=1 = m−1 X k tk = λn (8.5) D2B(x) = 1 m2 X h6=l thtl Y k/ 2{h,l} tk mx + m −tk m ++ D2B(x) + x=1 = m−2 X h6=l thtl = m−2 ( X k tk)2 − X k t2 k + = λ2n2 −m−2 X k t2 k (xD)2B + x=1 = x2D2B + xDB + x=1 = λn + λ2n2 −m−2 X k t2 k (8.6) 97 Chapter 8. Calculations on pathological degree sequences. 8.3 Di↵erential operators and symbolic D notation This section is not strictly necessary for the thesis but is included here for the interested reader. Let A(x) be the probability generating function for some random variable X. Then the calculation of (xD)nA(x) + x=1 for large n 2 N, i.e. E(Xn) can be streamlined by noting the following relation. DxiD = ixi−1D + xiD2 (xD)2 = xDxD = x(xD2 + D) = x2D2 + xD (xD)3 = xD(xD + x2D) = xD + 3x2D2 + x3D3 (xD)4 = xD(xD + 3x2D2 + x3D3) = xD + 7x2D2 + 6x3D3 + x4D4 (xD)5 = xD(xD + 7x2D2 + 6x3D3 + x4D4) = xD + 15x2D2 + 25x3D3 + 10x4D4 + x5D5 The coeﬃcients above form the following pattern. 1 1 1 1 3 1 1 7 6 1 1 15 25 10 1 1 31 90 65 15 1 1 63 301 350 140 21 1 These are the Stirling numbers of the second kind and are deﬁned recursively by, ✓a a ◆ = ✓a 0 ◆ = 1, ✓i j ◆ = ✓i −1 j −1 ◆ + (j + 1) ✓i −1 j −1 ◆ . 98 Chapter 9 Results needed for Chapter 7. Lemmas 7.8 and 7.9 form part of our proof that non-pathological degree sequences have similar probabilities under the Gp and Vp models. We prove these results ﬁrst. The other result we will need to prove is Lemma 7.3 which is used to calculate the asymp-totic probability of choosing a particular (m+n)-tuple (s, t) in the binomial p-model, Bp. We prove this result in Section 9.2. 9.1 Proof of Lemmas 7.8 and 7.9. Asymptotic combinatorial results Some preliminary results from asymptotic com-binatorics are needed to show Lemmas 7.8 and 7.9. We state and sometimes re-develop these results below for the interested reader. The ﬁrst results are stated without proof. Stirling’s approximation is often used to bound factorials. A reference can be found on line (4.2) of [Odl95, p.1076]. Lemma 9.1 (Stirling’s approximation). n! = p 2⇡n ⇣n e ⌘n✓ 1 + O ⇣1 n ⌘◆ . The next lemma concerns rising and falling factorial approximations. This is an often used result but we give a proof here for completeness. Lemma 9.2. Let k 2 R and assume \| k n\| < c for some c < 1. Then, Note that we can now write for any k 2 R: 99 Chapter 9. Results needed for Chapter 7. 1. (n + k)! n! nk = exp k2 2n + O k n + k3 n2 ++ 2. (n −k)! n! n−k = exp k2 2n + O k n + k3 n2 ++ Proof. (of 1.) We use the trick of writing x = eln x and then applying the series expansion for natural log. (n + k)! n! nk = k Y i=1 ⇣ 1 + i n ⌘ = exp ✓ k X i=1 ln ⇣ 1 + i n ⌘◆ = exp ✓ k X i=1 i n + O ⇣i2 n2 ⌘◆ = exp ✓k2 2n + O ⇣k n + k3 n2 ⌘◆ . Proof. (of 2.) This proceeds similarly to the proof of part (1). Proof of Lemma 7.8. We will prove Lemma 7.8 ﬁrst and proceed by a series of pre-liminary lemmas. Lemma 9.3. Fix 0 < a < 1 2. Suppose p is acceptable for a, m and n and write q = 1−p. Then, (ppnqqn)−1(pn)!(qn)! n! = p 2⇡pqn ✓ 1 + O ⇣ln n n ⌘◆ Proof. In the calculations below, (9.1) follows by Stirling’s approximation (Lemma 9.1). Many terms in (9.1) then cancel to give us (9.2) as required. (ppnqqn)−1(pn)!(qn)! n! = (ppnqqn)−1p 2⇡ rpnqn n exp n −pn −qn +(pn)pn(qn)qn nn ✓ 1 + O ⇣1 pn + 1 qn ⌘◆ (9.1) = p 2⇡pqn ⇣ 1 + O ✓ln n n ◆⌘ . (9.2) 100 9.1. Proof of Lemmas 7.8 and 7.9. Lemma 9.4. Deﬁne y := M−pmn ppqmn and assume that \|y\| < n". Fix 0 < a < 1 2. Write q = 1 −p. Then as m, n ! 1 subject to p acceptable for a, m and n, (pmn + yppqmn)!(qmn −yppqmn)! (pmn)!(qmn)!pyppqmnq−yppqmn = exp ✓y2 2 + O (mn)−1/2+4"+◆ Proof. Note it is suﬃcient to prove the following two approximations. (pmn + yppqmn)! (pmn)!(pmn)yppqmn = exp ✓qy2 2 + O (mn)−1/2+4"+◆ (9.3) (qmn −yppqmn)! (qmn)!(qmn)−yppqmn = exp ✓py2 2 + O (mn)−1/2+4"+◆ . (9.4) In the proofs of both of these results we use the approximations developed for rising and falling factorials in Lemma 9.2. The last line in each of the calculations below follows by our assumption that \|y\| < n". We begin with the proof of (9.3). (pmn + yppqmn)! (pmn)!(pmn)yppqmn = exp y2pqmn 2pmn + O yppqmn pmn + y3(pqmn)3/2 (pmn)2 ++ = exp y2q 2 + O ypq ppmn + y3q3/2 ppmn ++ = exp y2q 2 + O (mn)−1/2+4"++ The proof of (9.4) proceeds similarly. We are now in a position to prove Lemma 7.8. This lemma was originally stated on p84 of this thesis but we restate it below for convenience. Lemma 7.8. We deﬁne y := M−pmn ppqmn and assume that \|y\| < n". Then as m, n ! 1 subject to p acceptable for a, m and n, ✓ pMqmn−M ✓mn M ◆◆−1 = p 2⇡pqmn exp ✓y2 2 + O (mn)−1/2+"+◆ . 101 Chapter 9. Results needed for Chapter 7. Proof. Observe that by the deﬁnition of y, M = yppqmn + pmn and M −mn = yppqmn −qmn. Hence by expansion and some rearrangement, ✓ pMqmn−M ✓mn M ◆◆−1 = (pmn)!(qmn)! (mn)!ppmnqqmn ⇥(pmn + yppqmn)!(qmn −yppqmn)! (pmn)!(qmn)!pyppqmnq−yppqmn In the expression above we can approximate the two fractions on the right hand side by Lemmas 9.3 and 9.4 respectively. This yields, pMqmn−M ✓mn M ◆+−1 = p 2⇡pqmn 1 + O 1 mn ++ exp ✓y2 2 + O ⇣ (mn)−1/2+"⌘◆ as required. Proof of Lemma 7.9. In the ﬁrst step in the proof of this result, we ﬁnd an exponential approximation to p0 p +2M+ 1 2 q0 q +2mn−2M+ 1 2. Lemma 9.5. Fix " > 0 and set δ = p −p0. Also assume \|δ\| < n"p pq mn. Write q = 1 −p and q0 = 1 −p0. Then as m, n ! 1 subject to Conditions 4.1, ⇣p0 p ⌘2M+ 1 2⇣q0 q ⌘2mn−2M+ 1 2 = exp (2pmn + 2yppqmn + 1 2) 1 pδ − 1 2p2δ2++ ⇥exp (2qmn −2yppqmn + 1 2) −1 qδ − 1 2q2δ2++ 1 + O(n−1+10") + . Proof. Similar to the proof of 9.2 we will use the trick of writing x = eln x and then applying the series expansion for natural log. This method is ﬁrst applied to p0 p , 102 9.1. Proof of Lemmas 7.8 and 7.9. p0 p =1 + δ p = exp ⇣ ln 1 + δ p +⌘ = exp ⇣ δ p −δ2 2p2 + O δ3 p3 +⌘ Similarly, q0 q =1 −δ q = exp ⇣ ln 1 + δ q +⌘ = exp ⇣ −δ q −δ2 2q2 + O δ3 p3 +⌘ . Hence we have, ⇣p0 p ⌘2M+ 1 2⇣q0 q ⌘2mn−2M+ 1 2 = exp ⇣ (2pmn + 2yppqmn + 1 2) 1 pδ − 1 2p2δ2+⌘ ⇥exp ⇣ (2qmn −2yppqmn + 1 2) −1 qδ − 1 2q2δ2+⌘ ⇥exp ⇣ O (2qmn −2yppqmn + 1 2)δ3 p3 +⌘ . (9.5) To bound the error term ﬁrst note that by our assumptions on \|δ\| we have, δ3 p3 < n3" p3 ⇣r pq mn ◆3 = n3"(mn)−3/2 s q3 p3. (9.6) The restrictions on m, n imply that (mn)−3/2 < n−3+4". Also by Lemma 4.5, p > 1 ln n and so in particular, q q3 p3 < (ln n)3/2. Hence by (9.6), δ3 p3 < n3"n−3+4"(ln n)3/2 < n−3+8". (9.7) The result now follows by (9.5) and (9.7). This Lemma appears, for example, in the preliminary theory of Bollob´ as’ book [Bol01]. Lemma 9.6. 1 σ p 2⇡ Z uσ −uσ exp −x2/2σ2+ dx = 1 + O ⇣1 u exp −u2/2 +⌘ In the following Corollary we put Lemma 9.6 into a more convenient form for our calcu-lations. 103 Chapter 9. Results needed for Chapter 7. Corollary 9.7. Let u  1 p 2a ⇥min{k − b 2a, k + b 2a}. Then, Z k −k exp −aδ2 + bδ + dx = r⇡ a exp ⇣b2 4a ⌘⇣ 1 + O ⇣1 u exp −u2/2 +⌘⌘ We are now ready to prove Lemma 7.9. It is restated here for convenience. Lemma 7.9. Fix " > 0. Let y = M−pmn ppqmn and δ = p −p0. Write q = 1 −p and q0 = 1 −p0. Assume \|y\| < n4". Then as m, n ! 1 subject to p acceptable for a, m and n, Z 1 −1 p0 p +2M+1/2q0 q +2mn−2M+1/2 exp ⇣ −mn pq δ2⌘ dδ = r ⇡pq 2mn exp ⇣y2 2 ⌘⇣ 1 + O(n−1/2+12") ⌘ Proof. The calculations begin, we work to simplify the integral. Z n5" −n5" p0 p +2M+1/21 −p0 1 −p +2mn−2M+1/2 exp −mn pq (p −p0)2+ dδ (9.8) = Z δ=ln n δ=−ln n exp −mn pq δ2+ exp (2pmn + 2yppqmn + 1 2) 1 pδ − 1 2p2δ2++ exp (2qmn −2yppqmn + 1 2) −1 qδ − 1 2q2δ2++ 1 + O(−n−1+10") + (9.9) = (1 + O(−n−1+10")) Z δ=ln n δ=−ln n exp −aδ2 + bδ + where, (9.10) a = 1 (2pq)2(8mnpq + 4yppqmn(1 −2p) + 2p2 −2p + 1), b = 1 2pq(4yppqmn −(2p −1)). Line 9.9 follows by Lemma 9.5. This then rearranges to give line 9.10. We now ﬁnd an approximation for b2 4a. By (9.10), 104 9.2. Proof of Lemma 7.3 b2 4a = (4yppqmn−(2p−1))2 4(8mnpq+4yppqmn(1−2p)+2p2−2p+1) = 16y2pqmn−(2p−1)8yppqmn+(2p−1)2 4(8mnpq+4yppqmn(1−2p)+2p2−2p+1) = y2 2 + ((1−2p)8yppqmn(y−1)+(2p−1)2−y(2p2−2p+1) 4(8mnpq+4yppqmn(1−2p)+2p2−2p+1) thus, b2 4a = y2 2 + O((mn)−1/2+2"). (9.11) We note also that by (9.10), a = 2mn pq + O((mn)−1/2+2"). Then by Corollary 9.7 we have, Z n5" −n5" p0 p +2M+1/21 −p0 1 −p +2mn−2M+1/2 exp −mn pq (p −p0)2+ dδ = r ⇡pq 2mn + O((mn)−1/2+2") exp ⇣y2 2 + O((mn)−1/2+2") ⌘ ⇥ ⇣ 1 + O ⇣1 u exp −u2/2 +⌘⌘ = ✓r ⇡pq 2mn exp ⇣y2 2 ⌘ (1 + O(n−1+10")) + O(n−1/2+4") ◆ ⇥ ⇣ 1 + O ⇣1 u exp −u2/2 +⌘⌘ where, (9.12) u = r pq 4mn ⇥ ⇣ n5" −y + O(n−1/2+4") ⌘ > n4" The right hand side of (9.12) then simpliﬁes to yield, r ⇡pq 2mn exp ⇣y2 2 ⌘ + O n−1+12"+ and so we are done. 9.2 Proof of Lemma 7.3 We state the Lemma again for convenience and provide the proof. 105 Chapter 9. Results needed for Chapter 7. Lemma 7.3 Let 0 < a < 1/2 and suppose that pq > 1 ln n. Then, n X j=0 ✓✓n j ◆ pjqn−j ◆2 = 1 2p⇡npq ⇣ 1 + O n−1/2+"+ ⌘ + O(e−n") Proof. Consider the region where the greater proportion of the sum lies. The largest term in the sum lies around j = np. Make the substitution j = np + x. and split the sum into two components: \|j −np\| < n3/5 and \|j −np\| n3/5. n X j=0 ✓✓n j ◆ pjqn−j ◆2 = X \|np−j\|<n3/5 ✓✓n j ◆ pjqn−j ◆2 + X \|np−j\|≥n3/5 ✓✓n j ◆ pjqn−j ◆2 We ﬁrst write a general outline of how the proof will proceed, writing E1, . . . for any error terms we pick up. We will then bound each of these errors in turn. Thus, n X j=0 ✓✓n j ◆ pjqn−j ◆2 = X \|np−j\|<n3/5 ✓✓n j ◆ pjqn−j ◆2 + E1 where E1 denotes the ‘tails’ of the sum. We can now use the Euler-Maclaurin summation formula to approximate the central part of the sum with a truncated normal distribution. Make the substitution x = np −j and write, Z n j=0 ✓✓n j ◆ pjqn−j ◆2 = X \|x\| 1 ln n. Then use the trick of writing z = eln z and using the series expansion for natural log. This yields ✓✓ n np + x ◆ pnp+xqnq−x ◆2 = 1 2⇡npq exp ✓ −2(np + x) log(1 + x np) −2(nq −x) log(1 −x nq) ◆⇣ 1 + O ⇣x n ⌘⌘ which simpliﬁes to, ✓✓ n np + x ◆ pnp+xqnq−x ◆2 = 1 2⇡npq exp ✓−x2 npq ◆ 1 + O n−1/5++ . (9.13) (9.14) Where the bound above is uniform. Thus the term, X \|np−j\|<n3/5 ✓✓n j ◆ pjqn−j ◆2 is the sum of a Gaussian. We now approximate this sum with an integral. To do this we apply the Euler-Maclaurin summation (for the deﬁnition of the terms B2, R2 etc. see [Odl95, p.1090]). The Euler-Maclaurin result means the summation over a function g can be approximated by an integral as follows, b X k=a g(k) = Z b a g(x)dx + B2 2 (g0(b) −g0(a)) + 1 2(g(a) + g(b)) + R2. In our case the error term will be R2, so we calculate the magnitude of this, 107 Chapter 9. Results needed for Chapter 7. \|R2\| \|B4\| 4! Z (pqn)1/2+" −(pqn)1/2+" \|g0000(x)\|dx 1 6! Z 1 −1 \|g0000(x)\|dx = 1 6! p⇡npq ✓ 12 (pqn)2 + 2npq 48 (pqn)3 + 3.16 4 1 (pqn)4(npq)2 ◆ =O(n−1/2) Hence we now have, X \|x\|<n 3 5 ⇣ n np+x + pnp+xqnq−x⌘2 = 1 2⇡npq 1 + O n−1/2+"++ Z \|x\|<n 3 5 exp ✓−x2 npq ◆ dx + O(e−n") And so E2 = O(n−1/2+"). One of our intermediate results to bound E2 can be used to bound E1. The error E1 corresponds to the tails of a binomial sum. As we know that the binomal distribution always decreases away from its mean, then we can bound E1 by n times its maximum height. E1 = 1 1 1 1 1 1 1 X \|x\|≥n 3 5 ✓✓n j ◆ pj(1 −p)n−j ◆2 1 1 1 1 1 1 1 n 1 1 1 1 1 2⇡npq exp ✓−n3/52 npq ◆1 1 1 1 = 1 2⇡pq exp −n6/5+"+ The error E3 corresponds to the tails of normal distribution. Thus we can bound it by Lemma 9.6. The magnitude of its error is absorbed into the other error terms. And so we are done. 108 Glossary of Notations Graph parameters See page v of the introduction for complete deﬁnitions. m #white vertices. n #black vertices. uj jth white vertex. vk kth black vertex. sj degree of uj. s s1, . . . , sm. s 1 m P j sj. tk degree of vk. t t1, . . . , tk. t 1 n P k tk. λ edge density, deﬁned to be λ = 1 mn P j sj. Random variables on bipartite graphs In each of our random graph models we consider the properties of random variables deﬁned on that space. Generally uppercase letters are used for the random variables corresponding to the deterministic parameters of bipartite graphs which are denoted by their lowercase counterparts. Sj degree of uj. S⇤ j truncated degree, see Deﬁnition 4.1. S S1, . . . , Sm. S 1 m P j sj. Tk degree of vk. T T1, . . . , Tk. T 1 n P k tk. λ edge density, deﬁned to be λ = 1 mn P j Sj. 109 GLOSSARY Notation and terms deﬁned to bound the likelihood of pathological degree sequences Gt. locally ordered see Deﬁnition 6.2. A, Al see Deﬁnition 6.3. reference function R see Deﬁnition 6.4. Y a l see Deﬁnition 6.5. toxic (node) see Deﬁnition 6.11. bad (node) see Deﬁnition 6.13. good (node) if not bad. martingale Xi Xi := EG? t P j(Sj −λn)2 \| Fi + , see Deﬁnition 4.1. σ-algebra Fl is the σ-algebra induced by the partition {Y a l }a2Al, see Deﬁnition 6.4 Probability Spaces on Bipartite Graphs Graph p-model, Gp. Deﬁnition 1.6 on p5 Graph edge-model, GM. Deﬁnition 1.7 on p5 Graph half -model, Gt. Deﬁnition 1.8 on p6 Graph ordered-half -model, Ga t . Deﬁnition 6.7 on p57 Binomially based Probability Spaces Binomial integrated-model, Ip. Deﬁnition 7.1 on p73 Binomial p-model, Bp. Deﬁnition 7.2 on p74 Binomial edge-model, BM. Deﬁnition 7.3 on p74 Binomial half -model, Bt. Deﬁnition 7.5 on p76 Binomial integrated-model, Vp. Deﬁnition 7.7 on p84 Terminology acceptable for a, m and n Deﬁnition 2.5 on p23 (", a)-regular Deﬁnition 2.7 on p23 (a, b, m, n, ")-pathological Deﬁnition 2.9 on p24 110 Bibliography [Azu67] Kazuoki Azuma. Weighted sums of certain dependent random variables. Tˆ ohoku Math. J. (2), 19:357–367, 1967. [Bar10] Alexander Barvinok. On the number of matrices and a random matrix with prescribed row and column sums and 0–1 entries. Adv. Math., 224(1):316–339, 2010. [Bol82] B´ ela Bollob´ as. Vertices of given degree in a random graph. J. Graph Theory, 6(2):147–155, 1982. [Bol98] B´ ela Bollob´ as. Modern graph theory, volume 184 of Graduate Texts in Mathe-matics. Springer-Verlag, New York, 1998. [Bol01] B´ ela Bollob´ as. Random graphs, volume 73 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, second edition, 2001. [BT87] B. Bollob´ as and A. Thomason. Threshold functions. Combinatorica, 7(1):35–38, 1987. [Che81] Herman Cherno↵. A note on an inequality involving the normal distribution. Ann. Probab., 9(3):533–535, 1981. [CL06] Fan Chung and Linyuan Lu. Concentration inequalities and martingale inequal-ities: a survey. Internet Math., 3(1):79–127, 2006. [CMZ09] Colin Cooper, Andrew McGrae, and Michele Zito. Martingales on trees and the empire chromatic number of random trees. In Miroslaw Kutylowski, Witold Charatonik, and Maciej Gebala, editors, Fundamentals of Computation Theory, volume 5699 of Lecture Notes in Computer Science, pages 74–83. Springer Berlin / Heidelberg, 2009. [EKR76] P. Erd˝ os, D. J. Kleitman, and B. L. Rothschild. Asymptotic enumeration of Kn-free graphs. In Colloquio Internazionale sulle Teorie Combinatorie (Rome, 1973), Tomo II, pages 19–27. Atti dei Convegni Lincei, No. 17. Accad. Naz. Lincei, Rome, 1976. 111 Bibliography [ER59] P. Erd˝ os and A. R´ enyi. On random graphs. I. Publ. Math. Debrecen, 6:290–297, 1959. [ER61] P. Erd˝ os and A. R´ enyi. On the evolution of random graphs. Bull. Inst. Internat. Statist., 38:343–347, 1961. [EW77] P. Erd˝ os and Robin J. Wilson. On the chromatic index of almost all graphs. J. Combinatorial Theory Ser. B, 23(2–3):255–257, 1977. [GK73] R. L. Graham and D. J. Kleitman. Increasing paths in edge ordered graphs. Period. Math. Hungar., 3:141–148, 1973. Collection of articles dedicated to the memory of Alfr´ ed R´ enyi, II. [GLS99] Anant P. Godbole, Ben Lamorte, and Jessica K. Sklar. If rooks could kill: vertex degrees in random bipartite graphs. In Combinatorics, graph theory, and algorithms, Vol. I, II (Kalamazoo, MI, 1996), pages 445–450. New Issues Press, Kalamazoo, MI, 1999. [GM09] Catherine Greenhill and Brendan D. McKay. Random dense bipartite graphs and directed graphs with speciﬁed degrees. Random Structures Algorithms, 35(2):222–249, 2009. [Hak62] S. L. Hakimi. On realizability of a set of integers as degrees of the vertices of a linear graph. I. J. Soc. Indust. Appl. Math., 10:496–506, 1962. [Hav55] V. Havel. A remark on the existence of ﬁnite graphs [czech]. ˇ Casopis Pest. Mat., 80:477–480, 1955. [HHV09] Ararat Harutyunyan, Paul Horn, and Jacques Verstraete. Independent domi-nationg sets in graphs of girth ﬁve. Preprint, March 2009. [Hoe63] Wassily Hoe↵ding. Probability inequalities for sums of bounded random vari-ables. J. Amer. Statist. Assoc., 58:13–30, 1963. [Jan94] Svante Janson. Orthogonal decompositions and functional limit theorems for random graph statistics. Mem. Amer. Math. Soc., 111(534):vi+78, 1994. [J LR00] Svante Janson, Tomasz Luczak, and Andrzej Rucinski. Random graphs. Wiley-Interscience Series in Discrete Mathematics and Optimization. Wiley-Interscience, New York, 2000. [ Luc91] Tomasz Luczak. A note on the sharp concentration of the chromatic number of random graphs. Combinatorica, 11(3):295–297, 1991. 112 Bibliography [McD89] Colin McDiarmid. On the method of bounded di↵erences. In Surveys in combi-natorics, 1989 (Norwich, 1989), volume 141 of London Math. Soc. Lecture Note Ser., pages 148–188. Cambridge Univ. Press, Cambridge, 1989. [McD98] Colin McDiarmid. Concentration. In Probabilistic methods for algorithmic dis-crete mathematics, volume 16 of Algorithms Combin., pages 195–248. Springer, Berlin, 1998. [MW90] Brendan D. McKay and Nicholas C. Wormald. Asymptotic enumeration by degree sequence of graphs of high degree. European J. Combin., 11(6):565–580, 1990. [MW97] Brendan D. McKay and Nicholas C. Wormald. The degree sequence of a random graph. I. The models. Random Structures Algorithms, 11(2):97–117, 1997. [Odl95] A. M. Odlyzko. Asymptotic enumeration methods. In Handbook of combina-torics, Vol. 1, 2, pages 1063–1229. Elsevier, Amsterdam, 1995. [Pal84] Z. Palka. On the degrees of vertices in a bichromatic random graph. Period. Math. Hungar., 15(2):121–126, 1984. [Pal87] Zbigniew Palka. Extreme degrees in random graphs. J. Graph Theory, 11(2):121–134, 1987. [Pal88] Zbigniew Palka. Asymptotic properties of random graphs. Dissertationes Math. (Rozprawy Mat.), 275:112, 1988. [Ruc81] Andrzej Ruci´ nski. The r-connectedness of k-partite random graph. Bull. Acad. Polon. Sci. S´ er. Sci. Math., 29(7-8):321–330, 1981. [Rys63] Herbert John Ryser. Combinatorial mathematics. The Carus Mathematical Monographs, No. 14. Published by The Mathematical Association of America, 1963. [SS87] E. Shamir and J. Spencer. Sharp concentration of the chromatic number on random graphs Gn,p. Combinatorica, 7(1):121–129, 1987. [Vu02] V. H. Vu. Concentration of non-Lipschitz functions and applications. Random Structures Algorithms, 20(3):262–316, 2002. Probabilistic methods in combina-torial optimization. [Wil91] David Williams. Probability with martingales. Cambridge Mathematical Text-books. Cambridge University Press, Cambridge, 1991. [Wil06] Herbert S. Wilf. generatingfunctionology. A K Peters Ltd., Wellesley, MA, third edition, 2006. 113 Bibliography [ZS02] Arif Zaman and Daniel Simberlo↵. Random binary matrices in biogeographical ecology—instituting a good neighbor policy. Environ. Ecol. Stat., 9(4):405–421, 2002. 114
4507	https://users.math.msu.edu/users/gnagy/teaching/12-spring/mth235/L03-235.pdf	Modeling with ﬁrst order equations (Sect. 2.3). ▶Main example: Salt in a water tank. ▶The experimental device. ▶The main equations. ▶Analysis of the mathematical model. ▶Predictions for particular situations. Salt in a water tank. Problem: Describe the salt concentration in a tank with water if salty water comes in and goes out of the tank. Main ideas of the test: ▶Since the mass of salt and water is conserved, we construct a mathematical model for the salt concentration in water. ▶The amount of salt in the tank depends on the salt concentration coming in and going out of the tank. ▶The salt in the tank also depends on the water rates coming in and going out of the tank. ▶To construct a model means to ﬁnd the diﬀerential equation that takes into account the above properties of the system. ▶Finding the solution to the diﬀerential equation with a particular initial condition means we can predict the evolution of the salt in the tank if we know the tank initial condition. Modeling with ﬁrst order equations (Sect. 2.3). ▶Main example: Salt in a water tank. ▶The experimental device. ▶The main equations. ▶Analysis of the mathematical model. ▶Predictions for particular situations. The experimental device. o r ri o instantaneously mixed V (t) q (t) q (t) Q (t) pipe tank water salt i The experimental device. Deﬁnitions: ▶ri(t), ro(t): Rates in and out of water entering and leaving the tank at the time t. ▶qi(t), qo(t): Salt concentration of the water entering and leaving the tank at the time t. ▶V (t): Water volume in the tank at the time t. ▶Q(t): Salt mass in the tank at the time t. Units: ri(t) = ro(t) = Volume Time , qi(t) = qo(t) = Mass Volume. V (t) = Volume, Q(t) = Mass. Modeling with ﬁrst order equations (Sect. 2.3). ▶Main example: Salt in a water tank. ▶The experimental device. ▶The main equations. ▶Analysis of the mathematical model. ▶Predictions for particular situations. The main equations. Remark: The mass conservation provides the main equations of the mathematical description for salt in water. Main equations: d dt V (t) = ri(t) −ro(t), Volume conservation, (1) d dt Q(t) = ri(t) qi(t) −ro(t) qo(t), Mass conservation, (2) qo(t) = Q(t) V (t), Instantaneously mixed, (3) ri, ro : Constants. (4) The main equations. Remarks: hdV dt i = Volume Time = h ri −ro i , hdQ dt i = Mass Time = h riqi −roqo i , h riqi −roqo i = Volume Time Mass Volume = Mass Time. Modeling with ﬁrst order equations (Sect. 2.3). ▶Main example: Salt in a water tank. ▶The experimental device. ▶The main equations. ▶Analysis of the mathematical model. ▶Predictions for particular situations. Analysis of the mathematical model. Eqs. (4) and (1) imply V (t) = (ri −ro) t + V0, (5) where V (0) = V0 is the initial volume of water in the tank. Eqs. (3) and (2) imply d dt Q(t) = ri qi(t) −ro Q(t) V (t). (6) Eqs. (5) and (6) imply d dt Q(t) = ri qi(t) − ro (ri −ro) t + V0 Q(t). (7) Analysis of the mathematical model. Recall: d dt Q(t) = ri qi(t) − ro (ri −ro) t + V0 Q(t). Notation: a(t) = ro (ri −ro) t + V0 , and b(t) = ri qi(t). The main equation of the description is given by Q′(t) = −a(t) Q(t) + b(t). Linear ODE for Q. Solution: Integrating factor method. Q(t) = 1 µ(t) h Q0 + Z t 0 µ(s) b(s) ds i with Q(0) = Q0, where µ(t) = eA(t) and A(t) = Z t 0 a(s) ds. Modeling with ﬁrst order equations (Sect. 2.3). ▶Main example: Salt in a water tank. ▶The experimental device. ▶The main equations. ▶Analysis of the mathematical model. ▶Predictions for particular situations. Predictions for particular situations. Example Assume that ri = ro = r and qi are constants. If r, qi, Q0 and V0 are given, ﬁnd Q(t). Solution: Always holds Q′(t) = −a(t) Q(t) + b(t). In this case: a(t) = ro (ri −ro) t + V0 ⇒ a(t) = r V0 = a0, b(t) = ri qi(t) ⇒ b(t) = rqi = b0. We need to solve the IVP: Q′(t) = −a0 Q(t) + b0, Q(0) = Q0. Predictions for particular situations. Example Assume that ri = ro = r and qi are constants. If r, qi, Q0 and V0 are given, ﬁnd Q(t). Solution: Recall the IVP: Q′(t) = −a0 Q(t) + b0, Q(0) = Q0. Integrating factor method: A(t) = a0t, µ(t) = ea0t, Q(t) = 1 µ(t) h Q0 + Z t 0 µ(s) b0 ds i . Z t 0 µ(s) b0 ds = b0 a0 ea0t −1 ⇒Q(t) = e−a0t h Q0 + b0 a0 ea0t −1 i . So: Q(t) = Q0 −b0 a0 e−a0t + b0 a0 . But b0 a0 = rqi V0 r = qiV0. We conclude: Q(t) = Q0 −qiV0 e−rt/V0 + qiV0. Predictions for particular situations. Example Assume that ri = ro = r and qi are constants. If r, qi, Q0 and V0 are given, ﬁnd Q(t). Solution: Recall: Q(t) = Q0 −qiV0 e−rt/V0 + qiV0. Particular cases: ▶Q0 V0 > qi; ▶Q0 V0 = qi, so Q(t) = Q0; ▶Q0 V0 < qi. 0 t Q Q Q Q Q Q = q v 0 i 0 0 0 0 ◁ Predictions for particular situations. Example Assume that ri = ro = r and qi are constants. If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter, ﬁnd t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value. Solution: This problem is a particular case qi = 0 of the previous Example. Since Q(t) = Q0 −qiV0 e−rt/V0 + qiV0, we get Q(t) = Q0 e−rt/V0. Since V (t) = (ri −ro) t + V0 and ri = ro, we obtain V (t) = V0. So q(t) = Q(t)/V (t) is given by q(t) = Q0 V0 e−rt/V0. Therefore, 1 100 Q0 V0 = q(t1) = Q0 V0 e−rt1/V0 ⇒ e−rt1/V0 = 1 100. Predictions for particular situations. Example Assume that ri = ro = r and qi are constants. If r = 2 liters/min, qi = 0, V0 = 200 liters, Q0/V0 = 1 grams/liter, ﬁnd t1 such that q(t1) = Q(t1)/V (t1) is 1% the initial value. Solution: Recall: e−rt1/V0 = 1 100. Then, −r V0 t1 = ln 1 100 = −ln(100) ⇒ r V0 t1 = ln(100). We conclude that t1 = V0 r ln(100). In this case: t1 = 100 ln(100). ◁ Predictions for particular situations. Example Assume that ri = ro = r are constants. If r = 5x106 gal/year, qi(t) = 2 + sin(2t) grams/gal, V0 = 106 gal, Q0 = 0, ﬁnd Q(t). Solution: Recall: Q′(t) = −a(t) Q(t) + b(t). In this case: a(t) = ro (ri −ro) t + V0 ⇒ a(t) = r V0 = a0, b(t) = ri qi(t) ⇒ b(t) = r 2 + sin(2t) . We need to solve the IVP: Q′(t) = −a0 Q(t) + b(t), Q(0) = 0. Q(t) = 1 µ(t) Z t 0 µ(s) b(s) ds, µ(t) = ea0t, We conclude: Q(t) = re−rt/V0 Z t 0 ers/V0 2 + sin(2s) ds.
4508	https://pmc.ncbi.nlm.nih.gov/articles/PMC6671332/	Tumor Suppressor Schwannomin/Merlin Is Critical for the Organization of Schwann Cell Contacts in Peripheral Nerves - 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 J Neurosci . 2008 Oct 15;28(42):10472–10481. doi: 10.1523/JNEUROSCI.2537-08.2008 Search in PMC Search in PubMed View in NLM Catalog Add to search Tumor Suppressor Schwannomin/Merlin Is Critical for the Organization of Schwann Cell Contacts in Peripheral Nerves Natalia Denisenko Natalia Denisenko 1 Inserm, Unité Mixte de Recherche en Santé 839, Paris, 75005, France, 2 Université Pierre et Marie Curie–Paris 6, Paris, 75005, France, 3 Institut du Fer à Moulin, Paris, 75005, France, Find articles by Natalia Denisenko 1,2,3,, Carmen Cifuentes-Diaz Carmen Cifuentes-Diaz 1 Inserm, Unité Mixte de Recherche en Santé 839, Paris, 75005, France, 2 Université Pierre et Marie Curie–Paris 6, Paris, 75005, France, 3 Institut du Fer à Moulin, Paris, 75005, France, Find articles by Carmen Cifuentes-Diaz 1,2,3,, Theano Irinopoulou Theano Irinopoulou 1 Inserm, Unité Mixte de Recherche en Santé 839, Paris, 75005, France, 2 Université Pierre et Marie Curie–Paris 6, Paris, 75005, France, 3 Institut du Fer à Moulin, Paris, 75005, France, Find articles by Theano Irinopoulou 1,2,3, Michèle Carnaud Michèle Carnaud 1 Inserm, Unité Mixte de Recherche en Santé 839, Paris, 75005, France, 2 Université Pierre et Marie Curie–Paris 6, Paris, 75005, France, 3 Institut du Fer à Moulin, Paris, 75005, France, Find articles by Michèle Carnaud 1,2,3, Evelyne Benoit Evelyne Benoit 4 Centre National de la Recherche Scientifique, Institut de Neurobiologie Alfred Fessard, Laboratoire de Neurobiologie Cellulaire et Moléculaire, Unité Propre de Recherche 9040, Gif sur Yvette, 91198, France, Find articles by Evelyne Benoit 4, Michiko Niwa-Kawakita Michiko Niwa-Kawakita 5 Inserm, Unité 674, Paris, 75010, France, and 6 Université Paris 7–Denis Diderot, Institut Universitaire d'Hématologie, Paris, 75010, France Find articles by Michiko Niwa-Kawakita 5,6, Fabrice Chareyre Fabrice Chareyre 5 Inserm, Unité 674, Paris, 75010, France, and 6 Université Paris 7–Denis Diderot, Institut Universitaire d'Hématologie, Paris, 75010, France Find articles by Fabrice Chareyre 5,6, Marco Giovannini Marco Giovannini 5 Inserm, Unité 674, Paris, 75010, France, and 6 Université Paris 7–Denis Diderot, Institut Universitaire d'Hématologie, Paris, 75010, France Find articles by Marco Giovannini 5,6, Jean-Antoine Girault Jean-Antoine Girault 1 Inserm, Unité Mixte de Recherche en Santé 839, Paris, 75005, France, 2 Université Pierre et Marie Curie–Paris 6, Paris, 75005, France, 3 Institut du Fer à Moulin, Paris, 75005, France, Find articles by Jean-Antoine Girault 1,2,3,✉, Laurence Goutebroze Laurence Goutebroze 1 Inserm, Unité Mixte de Recherche en Santé 839, Paris, 75005, France, 2 Université Pierre et Marie Curie–Paris 6, Paris, 75005, France, 3 Institut du Fer à Moulin, Paris, 75005, France, Find articles by Laurence Goutebroze 1,2,3 Author information Article notes Copyright and License information 1 Inserm, Unité Mixte de Recherche en Santé 839, Paris, 75005, France, 2 Université Pierre et Marie Curie–Paris 6, Paris, 75005, France, 3 Institut du Fer à Moulin, Paris, 75005, France, 4 Centre National de la Recherche Scientifique, Institut de Neurobiologie Alfred Fessard, Laboratoire de Neurobiologie Cellulaire et Moléculaire, Unité Propre de Recherche 9040, Gif sur Yvette, 91198, France, 5 Inserm, Unité 674, Paris, 75010, France, and 6 Université Paris 7–Denis Diderot, Institut Universitaire d'Hématologie, Paris, 75010, France ✉ Correspondence should be addressed to Jean-Antoine Girault, Inserm, Unité Mixte de Recherche en Santé 839, Institut du Fer à Moulin, 17 rue du Fer à Moulin, Paris, 75005, France. Email: jean-antoine.girault@inserm.fr N.D. and C.C.-D. contributed equally to this work. M. Giovannini's present address: The Center for Neural Tumor Research, House Ear Institute, Keck School of Medicine, University of Southern California, Los Angeles, CA 90057. ✉ Corresponding author. Received 2008 Jun 2; Revised 2008 Aug 1; Accepted 2008 Aug 10. Copyright © 2008 Society for Neuroscience 0270-6474/08/2810472-10$15.00/0 PMC Copyright notice PMCID: PMC6671332 PMID: 18923024 Abstract Schwannomin/merlin is the product of a tumor suppressor gene mutated in neurofibromatosis type 2 (NF2). Although the consequences of NF2 mutations on Schwann cell proliferation are well established, the physiological role of schwannomin in differentiated cells is not known. To unravel this role, we studied peripheral nerves in mice overexpressing in Schwann cells schwannomin with a deletion occurring in NF2 patients (P0–SCH–Δ39–121) or a C-terminal deletion. The myelin sheath and nodes of Ranvier were essentially preserved in both lines. In contrast, the ultrastructural and molecular organization of contacts between Schwann cells and axons in paranodal and juxtaparanodal regions were altered, with irregular juxtaposition of normal and abnormal areas of contact. Similar but more severe alterations were observed in mice with conditional deletion of the Nf2 gene in Schwann cells. The number of Schmidt–Lanterman incisures, which are cytoplasmic channels interrupting the compact myelin and characterized by distinct autotypic contacts, was increased in the three mutant lines. P0–SCH–Δ39–121 and conditionally deleted mice displayed exuberant wrapping of nonmyelinated fibers and short internodes, an abnormality possibly related to altered control of Schwann cell proliferation. In support of this hypothesis, Schwann cell number was increased along fibers before myelination in P0–SCH–Δ39–121 mice but not in those with C-terminal deletion. Schwann cell numbers were also more numerous in mice with conditional deletion. Thus, schwannomin plays an important role in the control of Schwann cell number and is necessary for the correct organization and regulation of axoglial heterotypic and glio-glial autotypic contacts. Keywords: neurofibromatosis type 2, nodes of Ranvier, paranodes, juxtaparanodes, Schmidt–Lanterman incisures, internodes Introduction Neurofibromatosis type 2 (NF2) is a dominant autosomic disease characterized by the occurrence of multiple schwannomas, as well as ependymomas and meningiomas. The Nf2 tumor suppressor gene codes for a protein closely related to ezrin, radixin, and moesin (ERM) termed schwannomin (Rouleau et al., 1993) or merlin (Trofatter et al., 1993). Cell type-specific mutations of schwannomin or its absence lead to tumor formation in various mouse tissues (McClatchey et al., 1998; Giovannini et al., 1999, 2000; Kalamarides et al., 2002). Schwannomin is thought to play a critical role at the plasma membrane, controlling cell–cell interactions and signaling pathways triggered by cell contacts (McClatchey and Giovannini, 2005; Okada et al., 2007). It stabilizes adherens junctions (Lallemand et al., 2003) and interacts with the cytoskeleton, cytoskeleton-associated proteins, many transmembrane and adaptor-scaffold proteins, and a variety of signaling proteins (Okada et al., 2007). However, besides its importance as a tumor suppressor, the physiological role of schwannomin is still poorly understood, and its function in differentiated Schwann cells is not known. During development of peripheral nerves, immature Schwann cells give rise to myelinating nonmyelinating cells (Sherman and Brophy, 2005). Nonmyelinating Schwann cells ensheath small-diameter axons (<1 μm), whereas myelination takes place in Schwann cells wrapped around large-diameter axons. In myelinated fibers, the voltage-gated Na+ channels are concentrated at nodes of Ranvier between adjacent Schwann cells, allowing the rapid saltatory conduction of action potentials. Myelination requires a complex series of interactions between myelinating Schwann cells and axons, which results in highly differentiated domains along the axon (Arroyo and Scherer, 2000). These domains are centered by nodes of Ranvier, which are flanked on either sides by paranodal junctions and juxtaparanodal regions. Paranodal junctions separate Na+ channels at the node and shaker-type K+ channels at the juxtaparanode and function as barriers to restrict the lateral diffusion of axonal membrane proteins and extracellular molecules between the node and the internodal space (Poliak and Peles, 2003). Although a number of molecules involved in axoglial contacts have been identified during the past few years (Girault and Peles, 2002; Poliak and Peles, 2003; Salzer, 2003), their organization remains poorly understood. Because of its proposed role in cell contacts, schwannomin is an interesting candidate for playing a role in the organization of axoglial contacts. Schwannomin-like immunoreactivity has been reported in paranodal regions and Schmidt–Lanterman incisures (SLIs) (Scherer and Gutmann, 1996), which are cytoplasmic channels interrupting the compact myelin. In neurons, schwannomin can associate with the cytoplasmic tail of the paranodal protein paranodin/Caspr (Denisenko-Nehrbass et al., 2003), whereas in Schwann cells, it is localized to the plasma membrane through a paxillin-mediated interaction with β1-integrin (Obremski et al., 1998; Fernandez-Valle et al., 2002). To address the function of schwannomin in vivo, we used mouse lines with targeted overexpression of mutated schwannomin in Schwann cells or bearing a conditional deletion of the Nf2 gene in Schwann cells. Our results indicate an important role of schwannomin in the organization of Schwann cell contacts and provide new clues about its function. Materials and Methods Mice. Transgenic mice overexpressing mutated forms of human schwannomin under the control of the P0 promoter were obtained as described previously (Giovannini et al., 1999). In P0–SCH–Δ39–121 mice (line 27), exons 2 and 3, which code for amino acid residues 39–121 within the FERM domain of isoform 1, have been deleted (supplemental Fig. 1, available at www.jneurosci.org as supplemental material). This deletion corresponds to a known mutation in Nf2 patients. In P0–SCH–ΔCter mice (line 3), the C-terminal residues beyond amino acid 314 were deleted (supplemental Fig. 1, available at www.jneurosci.org as supplemental material). Conditional deletion of schwannomin in Schwann cells was obtained as described previously (Giovannini et al., 2000) by crossing two lines of transgenic mice: one in which exon 2 of Nf2 was flanked by LoxP sequences (supplemental Fig. 1, available at www.jneurosci.org as supplemental material), and the other expressing the Cre recombinase under the control of the P0 promoter (P0–Cre B). The animals used for experiments (P0CreB; Nf2 flox2/flox2) had two floxed Nf2 alleles and were hemizygous for P0–Cre B. These animals are referred as P0 Cre;Nf2 flox2/flox2 mice. Electron microscopy and morphometry. Mice were anesthetized with pentobarbital and perfused with 0.9% NaCl, followed by 4% paraformaldehyde and 3% glutaraldehyde in 0.1 m phosphate buffer (PB). The sciatic and phrenic nerves were removed and placed in fresh fixative overnight at 4°C, rinsed in PB, postfixed in 2% OsO4 in PB, dehydrated in an ascending series of ethanol, and embedded in epoxy resin. Semithin sections (0.5 μm) were stained with toluidine blue and viewed with a Leica DMRAX light microscope. Morphometric analyses were performed on 0.5 μm semithin transversal sections from phrenic nerves of three different mice of each phenotype. G-ratios were calculated as ratios of internal to external perimeters of the myelin sheath, measured automatically after segmentation, in >100 fibers per section (Michailov et al., 2004). Ultrastructural studies were performed on transversal sections of the phrenic nerves and on longitudinal sections of the sciatic nerves. Ultrathin sections (40 nm) were cut, stained with Reynold's lead citrate and uranyl acetate, and viewed with a Philips CM-120 TEM electron microscope. Ten Ranvier nodes, from at least two different 12-month-old mice of each phenotype, were observed, and three paranodal regions were selected from 3- and 6-month-old mice. The size of 10 paranodal loops from 10 heminodes was evaluated by measuring the length of membrane of each loop in contact with the axonal membrane. Antibodies. Rabbit antibodies against paranodin (L51), Caspr2, and syndecan 4 have been described previously (Menegoz et al., 1997; Denisenko-Nehrbass et al., 2003; Goutebroze et al., 2003). The anti-neurofascin antibody reacting with NF155 and NF186 was generated by immunizing rabbits with the common intracellular region of the protein (residues 1065–1175; GenBank accession number AY061639) fused to glutathione S-transferase. The other antibodies were from the following sources: NF2 rabbit polyclonal antibodies (immunoblotting, A-19 sc-331; Santa Cruz Biotechnology); voltage-gated Na+ channel α subunit mouse monoclonal antibody (PAN Nav, clone K58/35); polyclonal β-catenin antibody (Sigma-Aldrich); Kv1.1 α subunit monoclonal antibody (clone K20/78; Millipore); neuronal class III β-tubulin (TUJ1) and nonphosphorylated neurofilament H (SMI-32) monoclonal antibodies (Covance Research Products); FITC-conjugated sheep anti-rabbit antibodies (Eurobio); cyanine 3 (Cy3)-conjugated goat anti-mouse antibodies (Invitrogen); and IRDye800CW-conjugated donkey anti-mouse and anti-rabbit antibodies (Rockland Immunochemicals). Immunofluorescence and quantitative studies. Immunostaining of cryostat sections (10 μm) or teased fibers of sciatic nerves was performed as described previously (Goutebroze et al., 2003). Images were acquired using a Leica epifluorescence microscope equipped with a CCD camera (Micromax; Roper Scientific), or a Leica SP2 confocal laser scanning microscope. For quantification of internodal length, nodes were identified by voltage-gated Na+ channel α subunit and paranodin immunostaining on teased fibers. The diameter of individual fibers was measured after immunolabeling with the Cy3-conjugated goat anti-mouse antibodies that underlined fibers surface. To determine the distance between Schmidt–Lanterman incisures, teased fibers were labeled with an antibody against β-catenin. Distances between nodes or incisures and axon diameters were measured using MetaMorph software (Molecular Devices). Three-dimensional reconstructions from confocal optical sections (0.162 μm apart) and surface rendering techniques to visualize immunoreactivity as solid objects were done using Imaris 4.0 software (Bitplane). Lysates preparation and immunoblotting. Sciatic nerves were dissected out and homogenized in a Dounce vessel containing 200 μl of a lysis buffer containing 150 m m NaCl, 50 m m Tris, pH 8, 0.5% deoxycholate, 0.1% SDS, 1% NP-40, and Complete proteases inhibitors (Roche Diagnostics). Homogenates were centrifuged 15 min at 4°C at 20,000 × g, and protein concentration in the supernatants were determined by the bicincholinic acid method (Sigma-Aldrich). Equal amounts of protein (20 μg) were fractionated by SDS-PAGE and transferred to nitrocellulose. Membranes were incubated with primary antibodies followed by appropriate IRDye-conjugated secondary antibodies, and developed and quantified using Odyssey (LI-COR Biosciences). Results In mice overexpressing mutated schwannomin in Schwann cells, myelin sheaths are preserved but nonmyelinated fibers are altered To investigate the role of schwannomin in peripheral nerves, we studied two transgenic mouse lines that overexpress in Schwann cells mutated forms of human schwannomin deleted of either amino acid residues 39–121 within the FERM domain (P0–SCH–Δ39–121 mice) or the C-terminal residues beyond amino acid 314 (P0–SCH–ΔCter mice) (Giovannini et al., 1999) (supplemental Fig. 1, available at www.jneurosci.org as supplemental material). In semithin transverse sections of phrenic nerves at low magnification at various ages (3, 6, and 12 months), the endoneurial space appeared to be increased in P0–SCH–Δ39–121 mice in which fibers often had an irregular shape (supplemental Fig. 2, available at www.jneurosci.org as supplemental material). However, no dramatic alteration of the number of fibers was found (Table 1). Myelination appeared globally normal, although myelin thickness was marginally increased in P0–SCH–ΔCter mice, as indicated by a slight reduction of the G-ratio (Table 1). Table 1. Anatomical parameters of phrenic nerves of wild-type, P0–SCH–ΔCter, P0–SCH–Δ39–121, and P0 Cre;Nf2 flox2/flox2 mice \| Genotype \| Number of fibers(n = 3–5 mice) \| G-ratio(n = 337–574; 3 mice) \| Percentage of fibers with double concentric rings(n = 4–5 mice) \| :--- :--- \| \| Wild type \| 269 ± 6 \| 0.698 ± 0.005 \| 5.4 ± 0.7 \| \| P0–SCH–ΔCter \| 266 ± 8 \| 0.685 ± 0.02 \| 5.2 ± 0.6 \| \| P0–SCH–Δ39–121 \| 254 ± 3 \| 0.700 ± 0.003 \| 12.3 ± 1.4 \| \| P0 Cre;Nf2 flox2/flox2 \| 239 ± 2 \| 0.652 ± 0.004 \| 15.5 ± 2.2 \| Open in a new tab Statistical analysis was done with ANOVA followed by Tukey's test. Number of fibers, F(3,15) = 5.905, p = 0.001; G-ratio, F(3,1845) = 38, p< 10−3; percentage of fibers with double concentric rings, F(2,10) = 15.2, p< 10−3. Mutant versus wild-type, p< 0.01, p< 0.001. Values are means ± SEM. Electron microscopy study of transverse sections of phrenic nerve revealed several abnormalities (Fig. 1). In register with increased endoneurial space, collagen pockets between fibers were observed in P0–SCH–Δ39–121 mice but not in P0–SCH–ΔCter or wild-type mice (data not shown). In wild-type mice, groups of unmyelinated fibers were regularly surrounded by a single Schwann cell process (Remak fibers) (Fig. 1A,B), whereas in P0–SCH–Δ39–121 mice, Schwann cell processes often made several turns around small-caliber axons (Fig. 1C,D). These abnormalities were not found in P0–SCH–ΔCter mice, although some features of aberrant myelination were occasionally present, such as partially undefasciculated bundles of unmyelinated axons surrounded by the same Schwann cell process (Fig. 1E,F). The observation of abnormal wrapping of nonmyelinated fibers indicated that regulation of autotypic cell contacts might be altered in the presence of mutated schwannomin. However, such abnormalities did not affect significantly the formation of myelin sheath. Figure 1. Open in a new tab Ultrastructural abnormalities of nonmyelinated fibers in mice overexpressing mutated schwannomin SCH–Δ39–121 and SCH–ΔCter. Electron micrographs from sections of phrenic nerves from 12-month-old wild-type (A, B), P0–SCH–Δ39–121 (C, D), and P0–SCH–ΔCter (E, F) mice. In wild-type mice, unmyelinated fibers are each surrounded by a single Schwann cell process (B, arrow). In P0–SCH–Δ39–121 mice, some unmyelinated small fibers are surrounded by several wraps of Schwann cell process (D). In P0–SCH–ΔCter mice, the endoneurial space encloses partially undefasciculated bundles of unmyelinated axons surrounded by the same process (F, asterisks). Scale bar, 20 μm. Overexpression of mutated schwannomin in Schwann cells alters the organization of paranodal and juxtaparanodal proteins We examined the role of schwannomin in interactions between Schwann cells and axons, by studying the distribution of proteins enriched at nodes of Ranvier and paranodal junctions, and playing crucial roles in their integrity. In longitudinal sections of sciatic nerves from P0–SCH–Δ39–121 and P0–SCH–ΔCter mice, nodal axonal proteins, Na+ channels (Fig. 2Aa–Af, arrows), and ankyrin G (Kordeli et al., 1990 and data not shown) appeared normally clustered, as in wild-type mice. The distribution of proteins enriched in Schwann cells microvilli that surround the axon at nodes of Ranvier (Melendez-Vasquez et al., 2001; Goutebroze et al., 2003), including syndecan 4 (Fig. 2Aa–Ac), syndecan 3, and ERM proteins (data not shown), was also indistinguishable in mutant and wild-type mice. In contrast, the localization of the paranodal protein paranodin/Caspr was altered in both P0–SCH–Δ39–121 and P0–SCH–ΔCter mice, in sciatic nerve sections (data not shown), and in teased fibers (Fig. 2Ba–Bc). The labeling was irregular with areas of intense immunoreactivity alternating with regions of low or absent staining. The limits of paranodin distribution, which normally ends rather abruptly at the borders of the paranodal junctions (Fig. 2Ba, arrowheads), with the exception of the labeling along the mesaxon (Fig. 2Ba, arrows), were less regular than in wild-type mice and appeared to protrude into the juxtaparanodal regions (Fig. 2Bb,Bc). Quantification of the number of paranodes with altered paranodin immunoreactivity revealed that the proportion of abnormal labeling tended to be higher in large fibers (58% in P0–SCH–ΔCter and 76% in P0–SCH–Δ39–121) than in small fibers (36 and 61%, respectively). We also examined the distribution of neurofascin, a cell adhesion molecule localized in the axon at nodes (NF186 isoform) and in glial loops at paranodes (NF155 isoform) (Tait et al., 2000). Alterations in the localization of paranodal neurofascin in teased fibers of sciatic nerves from P0–SCH–Δ39–121 and P0–SCH–ΔCter mice were similar to those observed for paranodin (Fig. 2Bd–Bf). Three-dimensional reconstructions of paranodin immunolabeling clearly showed its irregular and fragmented organization in mutant mice (Fig. 2C). Figure 2. Open in a new tab Nodal, paranodal, and juxtaparanodal proteins in the sciatic nerves of mice overexpressing mutated schwannomin. A, Immunolocalization of voltage-gated Na+ channels (a–f, Nav, in red), syndecan 4 (a–c, Synd4, in green), and Caspr2 (d–f, in green) were examined on longitudinal sections of sciatic nerves from wild-type (WT), P0–SCH–ΔCter, and P0–SCH–Δ39–121. Nav and Synd4 are correctly localized in the nodal regions of the mutant mice. In contrast, Caspr2 immunoreactivity is irregular in mutant mice compared with wild-type mice. In some cases, Caspr2 labeling extends into the paranodal regions and comes in contact of the nodal marker Nav. B, Immunolocalization of paranodin (a–c, Pnd), neurofascin (d–f), Caspr2 (g–i), and Kv1.1 K+ channels (j–l) was examined on teased fibers of sciatic nerves from wild-type (WT), P0–SCH–ΔCter, and P0–SCH–Δ39–121 mice. The abnormalities in the distribution of the paranodal (Pnd, Neurofascin) and juxtaparanodal (Caspr2, Kv1.1) proteins are similar to those observed on sciatic nerve sections. C, Three-dimensional reconstructions of paranodin immunoreactivity (in green) in teased fibers of sciatic nerves from wild-type (WT), P0–SCH–ΔCter, and P0–SCH–Δ39–121 mice, combined with the differential interference contrast (in gray). Images reveal the irregular and fragmented organization of the paranodes in mutant mice. D, Left, Protein levels of endogenous schwannomin (Sch) and of mutated forms of schwannomin in sciatic nerves of wild-type, P0–SCH–Δ39–121, P0–SCH–ΔCter, and P0 Cre;Nf2 flox2/flox2 mice. The mutated forms of schwannomin are expressed at higher levels than the endogenous schwannomin in both mutants. Endogenous schwannomin levels were not diminished in mice with a conditional deletion because the deletion was limited to Schwann cells. Right, Paranodin (Pnd), Caspr2, and neuronal class III β-tubulin (β-tubulin) in sciatic nerves from wild-type (WT), P0–SCH–ΔCter, and P0–SCH–Δ39–121 mice. Immunoblots were performed on lysates of sciatic nerves obtained from three different mice of each genotype (lanes 1–3) with specific primary antibodies, developed and quantified using the LI-COR Biosciences technology (Odyssey). β-Tubulin immunoreactivity was used to normalize the amount of loaded protein. The levels of paranodin as well as those of Caspr2 are virtually unchanged in P0–SCH–Δ39–121 and increased (by 170% for paranodin and 56% for Caspr2) in P0–SCH–ΔCter mice. Scale bars: A–C, 5 μm. A, Images acquired using a Leica epifluorescence microscope. B, Single-scan confocal images. We then examined the distribution of Caspr2, a protein normally enriched in juxtaparanodes (Poliak et al., 2001), which are differentiated regions of internodes flanking the paranodes. Caspr2 distribution was altered, and its intensity appeared diminished in both P0–SCH–Δ39–121 and P0–SCH–ΔCter mice compared with wild-type mice (Fig. 2Ad–Af,Bg–Bi). Areas of strong labeling alternated with almost negative patches, and the labeled regions were usually smaller and less intensely stained than in wild-type mice. At many nodes, Caspr2 labeling extended into the paranodal regions and came in contact of nodal markers (Fig. 2Ae,Af). Similar abnormalities were observed for Kv1.1 shaker-type K+ channels, which are normally enriched in the juxtaparanodal axonal membrane (Wang et al., 1993) (Fig. 2Bj–Bl). The distribution of paranodal and juxtaparanodal proteins displayed the same abnormalities in 3- and 6-month-old mutant mice as in 1-year old animals (data not shown), indicating that these perturbations were not age dependent. We evaluated the levels of expression of the endogenous and mutated forms of schwannomin in sciatic nerves extracts from wild-type and transgenic mice. Endogenous schwannomin expression was not diminished in mutant mice, and the levels of expression of mutated schwannomin were higher than those of the endogenous protein in both P0–SCH–Δ39–121 and in P0–SCH–ΔCter mice (Fig. 2D). To determine whether the altered immunoreactivity of paranodal and juxtaparanodal proteins resulted from a reduction in their levels of expression, we then examined the levels of paranodin/Caspr and Caspr2 in sciatic nerve extracts, using β-tubulin as a control for loading (Fig. 2E). The levels of paranodin and Caspr2 were virtually unchanged in P0–SCH–Δ39–121 and were slightly increased in P0–SCH–ΔCter mice (Fig. 2E). Thus, the abnormalities in the distribution of the paranodal and juxtaparanodal proteins could be related directly to the overexpression of the mutated forms of schwannomin. Overexpression of mutated schwannomin in Schwann cells alters the ultrastructure of paranodal regions We then used electron microscopy to determine whether the abnormalities observed by immunofluorescence corresponded to alterations in the ultrastructure of axoglial contacts. Ten nodes of Ranvier were examined in longitudinal sections of sciatic nerves from two 12-month-old mice of each genotype. In wild-type mice, the paranodal myelin loops were tightly and regularly positioned at paranodes, with a symmetrical organization across the heminode (Fig. 3A), and transverse bands, the ultrastructural hallmark of paranodal junctions, were clearly visible (Fig. 3B, arrowheads). In contrast, abnormal paranodal loops and interactions between axons and Schwann cells were frequently observed in mutant mice. Paranodal myelin loops were often asymmetrical in P0–SCH–Δ39–121 and P0–SCH–ΔCter mice with variable width, appearing either enlarged or shrunk from place to place (Fig. 3C–F). Enlarged loops, generally located close to the juxtaparanode (Fig. 3C, arrow), were observed in 33 and 46% of the heminodes in P0–SCH–Δ39–121 and P0–SCH–ΔCter mice, respectively. Loop width was more variable in the two lines of mutant mice than in wild-type mice (Fig. 4). Electron-dense transverse bands were visible at the level of the normal loops (Fig. 3C, inset, arrowheads) but were absent from those that were massively enlarged (Fig. 3C, arrow). Atrophic loops, filled with electron-dense material and devoid of transverse bands, were observed in 77% of the heminodes examined in P0–SCH–Δ39–121 mice but not in wild-type or P0–SCH–ΔCter mice (Fig. 3D). Formations similar to axon Schwann cell networks (Gatzinsky et al., 1991), in which axonal digitations protruded within large glial formations, were observed in mutant (Fig. 3E,F, arrows) but not in wild-type mice. Other features of mutant paranodes included protrusions of Schwann cell cytoplasm into the axon (P0–SCH–Δ39–121, 22% of the heminodes; P0–SCH–ΔCter, 14% of the heminodes) or axonal protrusions, bulging out of the axon limits into the cells (P0–SCH–Δ39–121, 22% of the heminodes; P0–SCH–ΔCter, 36% of the heminodes). Overall, only 22% of the heminodes were devoid of the features described above in P0–SCH–Δ39–121 and 28% in P0–SCH–ΔCter mice, whereas they were not observed in heminodes from matched wild-type mice. Similar features, in approximately the same proportions, were found in sciatic nerves from 3- and 6-month old mice, indicating that they were not secondary to a late degeneration of fibers. Figure 3. Open in a new tab Ultrastructural abnormalities of nodes of Ranvier in mice overexpressing mutated schwannomin SCH–Δ39–121 and SCH–ΔCter. Electron micrographs of longitudinal sections of sciatic nerves, at the level of nodes of Ranvier, from 12-month-old wild-type (A, B), P0–SCH–Δ39–121 (C, D), and P0–SCH–ΔCter (E, F) mice. In wild-type mice, paranodal loops are symmetrical across the axon (A), and transverse bands are clearly visible (B, inset, arrowheads). In paranodal regions of P0–SCH–Δ39–121 mice, enlarged loop-like structures devoid of transverse bands (C, arrow) can be found next to loops of regular size showing transverse bands (C, inset, arrowheads). Another type of abnormality encountered in the paranodal regions of this mutant line are loops severely reduced in size, enclosing electron-dense material (D), and lacking transverse bands (D, inset). In P0–SCH–ΔCter mice, paranodal regions display enlarged glial processes enclosing cellular digitations resembling axon Schwann cell networks (Gatzinsky et al., 1991) (E, arrow) or containing numerous organelles, in contact with axonal protrusions (F, arrows). Scale bars, 1 μm. Figure 4. Open in a new tab Quantification of paranodal loop width in sciatic nerves from wild-type (WT), P0–SCH–Δ39–121, P0–SCH–ΔCter, and P0 Cre;Nf2 flox2/flox2 mice. The width of 10 adjacent paranodal loops was measured in 10 heminodes for each genotype. Loop width was more variable in the three lines of mutant mice than in wild-type mice. Loop-like structures with width >400 nm were not included. Minimum, 25% percentile, median, 75% percentile, and maximum are indicated. Conditional deletion of schwannomin in Schwann cells alters axoglial contacts The results obtained in transgenic mice demonstrated that the targeted overexpression of mutated schwannomin in Schwann cells resulted in a marked disorganization of axoglial contacts. Importantly, many alterations found in P0–SCH–Δ39–121 mice were also observed in P0–SCH–ΔCter transgenic mice, albeit to a slightly lesser degree. Such alterations could theoretically result from a gain of function of mutated proteins or from a partial or complete loss of function with a dominant-negative effect. The similar phenotype of the two different mutations argued in favor of the second hypothesis. To distinguish more precisely between these two hypotheses, we used mutant mice in which NF2 exon 2 was conditionally deleted in Schwann cells (P0 Cre;Nf2 flox2/flox2) (supplemental Fig. 1, available at www.jneurosci.org as supplemental material). The line we used (P0–Cre B) had no major postnatal lethality, in contrast with the initially reported P0–Cre A line (Giovannini et al., 2000), allowing the study of the peripheral nervous system in adult mice. In these mice, a large proportion of Schwann cells is devoid of schwannomin, because deletion of exon 2 destabilizes the molecule (Giovannini et al., 2000). However, the total amount of schwannomin in sciatic nerve homogenates was not decreased (Fig. 2D) because the deletion occurred in Schwann cells, whereas the protein is also expressed in other cell types, including axons (Denisenko-Nehrbass et al., 2003). In P0 Cre B;Nf2 flox2/flox2 mice, analysis of semithin sections of sciatic nerve (Fig. 5A) showed a slight decrease (−12%) in the number of myelinated fibers (Table 1) compared with wild-type mice. The myelin thickness was slightly increased as indicated by a minor decrease in the G-ratio (Table 1). Electron microscopy of phrenic nerve sections showed some indirect evidence of axonal loss in P0 Cre B;Nf2 flox2/flox2 mice, including collagen pockets in the endoneurial space, fibers, and collapsed Schwann cells (Fig. 5B, asterisk, arrow, and arrowhead, respectively). Abnormal Schwann cell wrapping was evidenced by small axons surrounded by multiple wraps of nonmyelinating Schwann cells (Fig. 5C, arrow) and by bundles of undefasciculated small axons wrapped by a myelinating Schwann cell (data not shown). Thus, these mice displayed signs of moderate axonal loss and aberrant myelination. Figure 5. Open in a new tab Structural abnormalities in peripheral nerves from 12-month-old P0 Cre;Nf2 flox2/flox2 mice. A, Transversal semithin section of a phrenic nerve showing an increase of the endoneurial space between myelinated fibers and the presence of concentric myelin rings corresponding to Schmidt–Lanterman incisures (arrowheads). Scale bar, 20 μm. B, C, Electron micrographs from phrenic nerve transverse sections showing a regenerative fiber (B, arrow), a collapsed Schwann cell (B, arrowhead), a collagen pocket (B, asterisk), and a small axon surrounded by multiple wraps of nonmyelinating Schwann cells (C, arrow). Scale bar, 1 μm. D–G, Electron micrographs from longitudinal sections of nodes of Ranvier of sciatic nerve showing paranodal regions with irregular paranodal loops enclosing electron-dense material and lacking transverse bands (E, F) and enlarged loop-like structures, containing vesicular material (G). Scale bars, 2 μm. Electron microscopic study of longitudinal sections of sciatic nerves, at the level of nodes of Ranvier, revealed abnormalities of paranodes. The size of adjacent loops was irregular (Fig. 4). Atrophic paranodal loops (22%) and axonal protrusions (11%), sometimes with electron-dense material, were also present (Fig. 5D–F). Transverse bands were not visible at the level of these altered paranodal loops (Fig. 5E,F). Enlarged paranodal loops and axon Schwann cell networks containing mitochondria and other vesicular material were frequently observed (44% of the heminodes), generally located close to the juxtaparanode (Fig. 5G) (data not shown). We examined molecular markers of nodal regions in sections and teased fibers of sciatic nerves from P0 Cre B;Nf2 flox2/flox2 mice. There was no consistent difference in the distribution of proteins enriched in the nodal axon, including sodium channels (Fig. 6Aa–Ad) and proteins of glial microvilli syndecan 3 (data not shown) and 4 (Fig. 6Ac,Ad). In contrast, the distribution of the paranodal and juxtaparanodal markers was severely altered (Fig. 6B). Paranodin/caspr immunoreactivity was distributed irregularly, with a combination of areas of normal and decreased labeling, often not delineating paranodal regions and invading the nodal or juxtaparanodal spaces (Fig. 6Ba,Bb). We noticed that paranodin immunoreactivity was more frequently altered in large fibers than in small ones. The three-dimensional reconstructions further indicated the overall alterations of paranodin distribution (Fig. 6C). The labeling of paranodal neurofascin was less regular in P0 Cre;Nf2 flox2/flox2 than in wild-type mice, and its overall intensity was dramatically reduced (Fig. 6B), as confirmed by quantification (wild type, 61 ± 5; mutant, 28 ± 3; mean ± SEM, arbitrary units; n = 5; t test, p< 0.00035). Caspr2 immunoreactivity was also decreased, irregular, and abnormally located, often in contact with the nodal proteins (Fig. 6B). Kv1.1 immunoreactivity was similarly altered (data not shown). However, immunoblotting experiments on sciatic nerve extracts showed that the levels of paranodin as well as those of Caspr2 were virtually unchanged in P0 Cre;Nf2 flox2/flox2 mice (Fig. 6D), indicating that the alterations of immunoreactivity of the paranodal and juxtaparanodal proteins did not correspond to changes in their levels of expression. Thus, mice bearing a targeted deletion of schwannomin in Schwann cells displayed severe alterations of the organization of axoglial contacts. Figure 6. Open in a new tab Immunolocalization and expression of nodal, paranodal, and juxtaparanodal proteins in the sciatic nerves of P0 Cre;Nf2 flox2/flox2 mice. A, Immunolocalization of voltage-gated Na+ channels (a–d, Nav, in red) and syndecan 4 (c, d, Synd4, in green) was examined on longitudinal sections of sciatic nerves from wild-type (WT) and P0 Cre;Nf2 flox2/flox2 mice. Nav and Synd4 are correctly segregated in the nodal regions. B, Immunolocalization of paranodin/Caspr (Pnd), neurofascin (a, b) and Caspr2 (c, d) in teased fibers of sciatic nerves from wild-type (WT) and P0 Cre;Nf2 flox2/flox2 mice. The paranodal Pnd and neurofascin labeling is less regular in P0 Cre;Nf2 flox2/flox2 than in wild-type mice, and its overall intensity is reduced. Caspr2 immunoreactivity is also decreased and irregular. C, Three-dimensional reconstructions of paranodin immunoreactivity (in green) in teased fibers of sciatic nerves from wild-type (WT) and P0 Cre;Nf2 flox2/flox2 mice, combined with the differential interference contrast (in gray). The image reveals the irregular distribution of paranodin invading the juxtaparanodal and nodal regions. D, Protein expression levels of paranodin (Pnd), Caspr2, and neuronal class III β-tubulin (β-tubulin) in sciatic nerves from wild-type (WT) and P0 Cre;Nf2 flox2/flox2 mice. Immunoblots were performed on lysates of sciatic nerves with specific primary antibodies, developed, and quantified using the LI-COR Biosciences technology (Odyssey). β-Tubulin immunoreactivity was used to normalize the amount of loaded protein. The levels of paranodin and Caspr2 are unchanged in P0 Cre;Nf2 flox2/flox2 mice. Scale bars: A–C, 5 μm. A, Images acquired using a Leica epifluorescence microscope. B, Single-scan confocal images. Mutation of schwannomin in Schwann cells alters the number of Schmidt–Lanterman incisures, the internodal length, and the number of Schwann cells The compact myelin in Schwann cells is regularly interrupted by SLIs, which are helicoidal cytoplasmic channels thought to facilitate the communication between the cytoplasmic compartments on adaxonal and abaxonal sides of the myelin sheath. These incisures are characterized by the presence of Schwann cell autotypic contacts, including adherens junctions, tight junctions, and gap junctions, containing cadherins, claudins, and connexins, respectively (Arroyo and Scherer, 2000). On transverse sections, SLIs can appear as an empty space between double concentric myelin rings (Robertson, 1958). While studying transverse sections of phrenic nerves of P0 Cre;Nf2 flox2/flox2 mice, we noticed that myelin sheaths appeared more frequently as double concentric myelin rings than in wild-type mice (Fig. 5A), suggesting an increased number of SLIs. We counted the number of fibers with double concentric rings on transverse phrenic nerve sections (Table 1). Their number was significantly higher in P0–SCH–Δ39–121 and P0 Cre;Nf2 flox2/flox2 mice compared with wild-type or P0–SCH–ΔCter mutant mice (Table 1). This increased number of double concentric rings was observed in 3-, 6-, and 12-month-old mutant mice (data not shown). To quantify precisely the number of SLIs per Schwann cells in mutant mice, we measured internodal lengths and distances between adjacent SLIs (visualized by β-catenin immunostaining) on teased sciatic fibers (Fig. 7A). A characteristic feature of two of the mutant mouse lines was an increased proportion of very short internodes (<350 μm), which was significantly higher in P0–SCH–Δ39–121 transgenic mice (22%) and P0 Cre;Nf2 flox2/flox2 mice (63%) than in P0–SCH–ΔCter (0%) or wild-type (4%; χ 2 = 35.7; p< 0.0001) mice. The average internodal length was not significantly different between wild-type and P0–SCH–ΔCter and P0–SCH–Δ39–121 transgenic mice but was decreased in P0 Cre;Nf2 flox2/flox2 mice (Fig. 7B). Because a correlation between the caliber of a myelinated axon and the internodal length is well established (Friede, 1983), we normalized internodal distances to axon caliber. Although the average fiber diameter was smaller in P0 Cre;Nf2 flox2/flox2 than in wild-type mice, the internodal distance normalized to axonal diameter was significantly decreased (Fig. 7C,D). The distance between SLIs was also decreased in mutant mice, the diminution being more pronounced in P0–SCH–Δ39–121 and P0 Cre;Nf2 flox2/flox2 than in P0–SCH–ΔCter mice (Fig. 7E), confirming the increased number of SLIs in these mice. Figure 7. Open in a new tab Alteration of the number of Schmidt–Lanterman incisures and abnormal internodal length in the myelinated fibers of P0–SCH–Δ39–121, P0–SCH–ΔCter, and P0 Cre;Nf2 flox2/flox2 mice. A, Visualization of Schmidt–Lanterman incisures and nodes in teased fibers from sciatic nerves of wild-type (WT), P0–SCH–ΔCter, P0–SCH–Δ39–121, and P0 Cre;Nf2 flox2/flox2 mice. Nodes are delimited by voltage-gated Na+ channels immunolabeling (in red, arrowheads), whereas Schmidt–Lanterman incisures are revealed by β-catenin labeling (in green, arrows). Scale bar, 25 μm. B–E, Quantifications of the internodal length (B), axonal diameter (C), the ratio of internodal length to axonal diameter (D), and the distance between adjacent Schmidt–Lanterman incisures (E), in teased fibers of sciatic nerves from wild-type, P0–SCH–ΔCter, P0–SCH–Δ39–121, and P0 Cre;Nf2 flox2/flox2 mice. Data are means + SEM. Statistical analysis by one-way ANOVA, followed by Dunnett's multiple comparison test: p< 0.05; p< 0.01; p< 0.001. F, G, At postnatal day 1, more Schwann cells were aligned along fibers of P0–SCH–Δ39–121 mice than of wild-type or P0–SCH–ΔCter mice. Schwann cell nuclei were detected with DAPI staining (red) and fibers with anti-neurofilament antibody (green). Scale bar, 10 μm. Quantification was performed by calculating the number of nuclei per unit of fiber length and divided by the number of axons in the bundle (G). ANOVA: F(2,123) = 15.8; p< 0.0001. Post hoc Tukey's test: p< 0.001 compared with wild type. H, I, A similar approach was used in 3-month-old P0 Cre;Nf2 flox2/flox2 mice and wild-type littermate controls. Pairs of nuclei (arrows), sometimes close to a node (arrowhead), were observed in mutant but not wild-type mice; fibers contours are shown with nonspecific antibody (green). Scale bar, 30 μm. Quantification and statistical analysis by Student's t test: t = 10.7; n = 153; p< 0.0001 (I). J, Expression of β-catenin, NF155, and neuronal class III β-tubulin (β-tubulin) in sciatic nerves from wild-type (WT), P0–SCH–ΔCter, P0–SCH–Δ39–121, and P0 Cre;Nf2 flox2/flox2 mice. Immunoblots were performed on lysates of sciatic nerves with specific primary antibodies, developed, and quantified using the LI-COR Biosciences technology (Odyssey). Left, Immunoblot on lysates from three different mice of each wild-type and transgenic genotype (lines 1–3). Right, Immunoblot on lysates from wild-type and P0 Cre;Nf2 flox2/flox2 mice. The levels of β-catenin and NF155 are increased in sciatic nerve extracts from transgenic mice and P0 Cre;Nf2 flox2/flox2 mice compared with wild-type mice. To determine whether an increased proportion of very short internodes and the decreased distance between SLIs observed in P0–SCH–Δ39–121 mice could be associated with an increased proliferation of Schwann cells at the early onset of myelination, we counted Schwann cell nuclei along teased sciatic nerves derived from postnatal day 1 wild-type, P0–SCH–Δ39–121, and P0–SCH–ΔCter mice (Fig. 7F,G). At the early onset of the myelination, the number of Schwann cells along the nerve fibers was higher in P0–SCH–Δ39–121 mice than in wild-type or P0–SCH–ΔCter mice (Fig. 7F,G). Importantly, the number of Schwann cells was not altered in P0–SCH–ΔCter mice. We also looked for alterations in the number of Schwann cells in P0 Cre;Nf2 flox2/flox2 using 3-month-old mice (Fig. 7H,I). At this age, the cell density was much lower than in neonates. In wild-type mice, the Schwann cell nuclei were isolated and never in close vicinity of the node. In contrast, in P0 Cre;Nf2 flox2/flox2 mice, we often observed pairs of nuclei, close to the nodal region (Fig. 7H, right panel, arrows). Cell counts showed that the number of nuclei per length of fiber was significantly increased (Fig. 7I). These observations clearly supported the hypothesis of an increased number of Schwann cells during myelination in P0–SCH–Δ39–12 and P0 Cre;Nf2 flox2/flox2 but not in P0–SCH–ΔCter mice. We also measured, by immunoblotting, the amounts of β-catenin and NF155, the 155 kDa glial isoform of neurofascin, two proteins enriched in SLIs (Fannon et al., 1995; Chang et al., 2000). The levels of these two proteins were increased in sciatic nerve extracts from the three lines of mutant mice compared with wild-type mice (Fig. 7J). Altogether, these results show that the number of SLIs is altered by mutations in schwannomin and that overexpression of P0–SCH–Δ39–121 or conditional knock-out of schwannomin decreases the internodal length, i.e., increases the number of myelinating Schwann cell per length unit of axon. Discussion Our observations in mice overexpressing specifically mutated schwannomin in Schwann cells, under the control of P0 promoter, demonstrate an important role of this protein in the organization of axoglial interactions and Schwann cell autotypic contacts at the level of SLIs. It was shown previously that mice overexpressing SCH–Δ39–121, but not those overexpressing SCH–ΔCter, display Schwann cell hyperplasia and develop tumors (Giovannini et al., 1999). These findings were confirmed in mice with a selective deletion of schwannomin in Schwann cells (Giovannini et al., 2000). These previous observations, as well as those from other mutant mice (McClatchey et al., 1998) and a number of studies in cells in culture (McClatchey and Giovannini, 2005), demonstrated the tumor suppressor activity of schwannomin, providing a basis for the phenotype of patients with NF2. The present study shows that schwannomin is also important for the organization of myelinated fibers. The phenotype of schwannomin mutant mice included some abnormalities of nonmyelinated fibers, which may be attributable to the expression of P0 promoter in nonmyelinating Schwann cells (Zhang et al., 1995). A partial deficit in defasciculation of axon bundles by Schwann cells was observed in schwannomin mutants. Interestingly, it has been shown recently that focal adhesion kinase is involved in defasciculation (Grove et al., 2007), suggesting a possible link with schwannomin that is associated with focal adhesion kinase (FAK) and β1-integrins (Taylor et al., 2003) and can alter FAK activity (Poulikakos et al., 2006). Several other features suggested that Schwann cell membrane interactions were altered in schwannomin mutant mice. Nonmyelinating Schwann cell extension wrapped several fold around small axons in P0–SCH–Δ39–121 and P0 Cre;Nf2 flox2/flox2 mice, instead of stopping after a single turn. Exuberant myelination of large-caliber fibers was reported in a different line of P0 Cre;Nf2 flox2/flox2 mice, expressing higher levels of Cre than the line used in this study (Giovannini et al., 1999), and we observed a relative increase in myelin thickness in P0–SCH–ΔCter and P0 Cre;Nf2 flox2/flox2 mice. These findings suggest that the inhibitory signals that normally stop the extension of Schwann cell lamellipodia are defective when schwannomin is mutated or deleted. This could be a consequence of the ability of schwannomin to inhibit Pak1-mediated recruitment of Rac to the plasma membrane and matrix adhesions (Shaw et al., 2001; Morrison et al., 2007; Okada et al., 2007). Schwannomin may also have a regulatory role through its interactions with ErbB2 (Fernandez-Valle et al., 2002), a receptor for the axonal factor neuregulin-1, critical for the control of myelin thickness (Garratt et al., 2000; Michailov et al., 2004). Additional signs of altered axon–Schwann cell interactions included the frequent protrusions originating from one cell type into the other, including the formation of axon–Schwann cell networks. These complex interactions between axons and Schwann cells have been described in paranodal regions of other species (Gatzinsky et al., 1991). It is striking that we did not observe them in sciatic nerves of wild-type mice but only in mutant animals. Thus, it is tempting to suggest that their frequent occurrence also resulted from a deficient regulation of Schwann cell membrane expansion. In contrast with the exuberant axoglial interactions discussed above, we also observed atrophic paranodal loops containing electron-dense material. Thus, a distinctive feature of schwannomin mutant mice was the coexistence of apparently normal loops and loops with an increased size and/or atrophic loops. In these latter two types, transverse bands that are characteristic for intact axoglial contacts were usually not present. The combination of regions of normal and abnormal axoglial junctions provided the basis for the irregular distribution of juxtaparanodal and paranodal proteins, and their extension toward nodes and internodes. This phenotype was different from that described previously in other mutant mice in which the transverse bands were completely absent (Dupree et al., 1999; Bhat et al., 2001; Boyle et al., 2001; Mathis et al., 2001). In these mice, paranodal markers were diffusely distributed along the axon, and juxtaparanodal markers spread through paranodes to the vicinity of nodal markers. These observations provide evidence for an important role of schwannomin in the organization and stability of Schwann cells membrane domains. This role may be related to the partitioning of schwannomin in lipid rafts (Stickney et al., 2004), which are important for the segregation of membrane proteins and organization of axoglial contacts (Schafer et al., 2004). An increased number of incisures was found in the three Nf2 mutant genotypes, although it was much more striking in P0 Cre;Nf2 flox2/flox2 and, to a lesser degree, P0–SCH–Δ39–121 mice than in P0–SCH–ΔCter mice. A similar increased number of incisures was observed in mutant mice with altered Schwann cell–axon interactions, including myelin basic protein-deficient shiverer mice (Gould et al., 1995), sulfatide-deficient mice (Hoshi et al., 2007), and desert hedgehog-null mice (Sharghi-Namini et al., 2006). In contrast to desert hedgehog-null mice, both shiverer (Sharghi-Namini et al., 2006) and schwannomin mutant mice displayed an increase in β-catenin and NF155 levels, indicating that these various mutations altered SLI numbers through different mechanisms. Little is known about the regulation of the formation of SLIs in which cadherin-mediated autotypic interactions are a key component (Tricaud et al., 2005; Perrin-Tricaud et al., 2007). Schwannomin immunoreactivity has been reported in SLIs (Scherer and Gutmann, 1996), suggesting that alteration of SLI numbers in schwannomin mutant mice may result from the role of schwannomin in stabilizing cadherin-based adherens junctions (Lallemand et al., 2003). This may involve the associated protein erbin (Rangwala et al., 2005) and Rac1 (Tricaud et al., 2005). In support of this hypothesis, β-catenin appears to stabilize SLIs through an inhibition of Rac1 (Tricaud et al., 2005) and schwannomin is capable of suppressing the recruitment of Rac1 to the plasma membrane (Okada et al., 2005). Another conspicuous feature observed in schwannomin mutant mice was the shortening of internodal distances. The shortening of internodal distance was dramatic in P0 Cre;Nf2 flox2/flox2 mice, and an increased number of very short internodes was also found in P0–SCH–Δ39–121 mice. We found that reduced axon caliber could not account for these short internodes. Conversely, cell proliferation and apoptosis are known to control Schwann cell number in the developing peripheral nervous system (Jessen and Mirsky, 2005). Our observation of an increased number of Schwann cells aligned along axons before the onset of myelination in P0–SCH–Δ39–121 but not P0–SCH–ΔCter mutant mice provides an explanation for the abnormally short internodes. Moreover, an increased number of Schwann cells was found in P0 Cre;Nf2 flox2/flox2 mice. These alterations are likely to result from the impaired control of cell cycle in the absence of schwannomin (McClatchey et al., 1998; Giovannini et al., 1999, 2000). A similar mechanism has been proposed to explain the existence of short internodes in young mice deficient for PMP22, which is also implicated in cell proliferation and apoptosis (Neuberg et al., 1999). In support of this hypothesis, short internodes were not observed in P0–SCH–ΔCter mice that displayed no tendency to increased Schwann cell proliferation (Giovannini et al., 1999). The combination of the three types of mutant mice used in the present study allows a better understanding of the function of schwannomin. Most of the findings in P0–SCH–Δ39–121 and P0–SCH–ΔCter mice that overexpressed a mutated protein were qualitatively similar to those in P0 Cre;Nf2 flox2/flox2 that had a dramatic decrease in its expression levels (McClatchey et al., 1998; Giovannini et al., 1999, 2000). This indicated that the phenotype was attributable to a loss of function of the normal Nf2 gene product, because of its absence or a dominant-negative effect of overexpressed mutated protein. Moreover, it is striking that the tumorigenic (SCH–Δ39–121) and nontumorigenic (SCH–ΔCter) mutations had similar phenotypes, clearly dissociating the dual role of schwannomin in cell growth control and in cell–cell interactions. In addition, our results provide novel clues about the pathophysiology of the human familial NF2 disease, raising the possibility that peripheral nerves might be altered besides the occurrence of tumors. Our results suggest that some schwannomin mutations, even at the hemizygous stage, before the occurrence of tumors, may have a dominant-negative effect. It is also possible that a gene dosage effect occurs, especially taking into account the long lifespan of humans compared with mice. Neuropathies have been reported in NF2 patients that may not be all accounted for by microcompressions (Sperfeld et al., 2002). Our results suggest that primary alterations in Schwann cell biology may also play a role in these neuropathies and provide an incentive for further exploring this issue, which may have diagnostic and therapeutic implications. Footnotes This work was supported in part by grants of the National Multiple Sclerosis Society, Association pour la Recherche sur la Sclerose en Plaque, Association Française contre les Myopathies, Action Concertée Incitative Développement et Physiologie, Association pour la Recherche sur le Cancer, Foundation Schlumberger pour l'enseignement et la Recherche, and Agence Nationale de la Recherche-Neuro-05-NEUR-A05158DS. We are indebted to P. Ezan, A. Boisquillon, C. Fayet, P. Bozin, and E. Valjent for their help with some experiments. References Arroyo EJ, Scherer SS. On the molecular architecture of myelinated fibers. Histochem Cell Biol. 2000;113:1–18. doi: 10.1007/s004180050001. [DOI] [PubMed] [Google Scholar] Bhat MA, Rios JC, Lu Y, Garcia-Fresco GP, Ching W, St Martin M, Li J, Einheber S, Chesler M, Rosenbluth J, Salzer JL, Bellen HJ. Axon-glia interactions and the domain organization of myelinated axons requires neurexin IV/Caspr/Paranodin. Neuron. 2001;30:369–383. doi: 10.1016/s0896-6273(01)00294-x. [DOI] [PubMed] [Google Scholar] Boyle ME, Berglund EO, Murai KK, Weber L, Peles E, Ranscht B. Contactin orchestrates assembly of the septate-like junctions at the paranode in myelinated peripheral nerve. Neuron. 2001;30:385–397. doi: 10.1016/s0896-6273(01)00296-3. [DOI] [PubMed] [Google Scholar] Chang BJ, Cho IJ, Brophy PJ. A study on the immunocytochemical localization of neurofascin in rat sciatic nerve. J Vet Sci. 2000;1:67–71. [PubMed] [Google Scholar] Denisenko-Nehrbass N, Goutebroze L, Galvez T, Bonnon C, Stankoff B, Ezan P, Giovannini M, Faivre-Sarrailh C, Girault JA. Association of Caspr/paranodin with tumour suppressor schwannomin/merlin and beta1 integrin in the central nervous system. J Neurochem. 2003;84:209–221. doi: 10.1046/j.1471-4159.2003.01503.x. [DOI] [PubMed] [Google Scholar] Dupree JL, Girault JA, Popko B. Axo-glial interactions regulate the localization of axonal paranodal proteins. J Cell Biol. 1999;147:1145–1152. doi: 10.1083/jcb.147.6.1145. [DOI] [PMC free article] [PubMed] [Google Scholar] Fannon AM, Sherman DL, Ilyina-Gragerova G, Brophy PJ, Friedrich VL, Jr, Colman DR. Novel E-cadherin-mediated adhesion in peripheral nerve: Schwann cell architecture is stabilized by autotypic adherens junctions. J Cell Biol. 1995;129:189–202. doi: 10.1083/jcb.129.1.189. [DOI] [PMC free article] [PubMed] [Google Scholar] Fernandez-Valle C, Tang Y, Ricard J, Rodenas-Ruano A, Taylor A, Hackler E, Biggerstaff J, Iacovelli J. Paxillin binds schwannomin and regulates its density-dependent localization and effect on cell morphology. Nat Genet. 2002;31:354–362. doi: 10.1038/ng930. [DOI] [PubMed] [Google Scholar] Friede RL. Variance in relative internode length (l/d) in the rat and its presumed significance for the safety factor and neuropathy. J Neurol Sci. 1983;60:89–104. doi: 10.1016/0022-510x(83)90129-6. [DOI] [PubMed] [Google Scholar] Garratt AN, Voiculescu O, Topilko P, Charnay P, Birchmeier C. A dual role of erbB2 in myelination and in expansion of the schwann cell precursor pool. J Cell Biol. 2000;148:1035–1046. doi: 10.1083/jcb.148.5.1035. [DOI] [PMC free article] [PubMed] [Google Scholar] Gatzinsky KP, Berthold CH, Rydmark M. Axon-Schwann cell networks are regular components of nodal regions in normal large nerve fibres of cat spinal roots. Neurosci Lett. 1991;124:264–268. doi: 10.1016/0304-3940(91)90109-7. [DOI] [PubMed] [Google Scholar] Giovannini M, Robanus-Maandag E, Niwa-Kawakita M, van der Valk M, Woodruff JM, Goutebroze L, Mérel P, Berns A, Thomas G. Schwann cell hyperplasia and tumors in transgenic mice expressing a naturally occurring mutant NF2 protein. Genes Dev. 1999;13:978–986. doi: 10.1101/gad.13.8.978. [DOI] [PMC free article] [PubMed] [Google Scholar] Giovannini M, Robanus-Maandag E, van der Valk M, Niwa-Kawakita M, Abramowski V, Goutebroze L, Woodruff JM, Berns A, Thomas G. Conditional biallelic Nf2 mutation in the mouse promotes manifestations of human neurofibromatosis type 2. Genes Dev. 2000;14:1617–1630. [PMC free article] [PubMed] [Google Scholar] Girault JA, Peles E. Development of nodes of Ranvier. Curr Opin Neurobiol. 2002;12:476–485. doi: 10.1016/s0959-4388(02)00370-7. [DOI] [PubMed] [Google Scholar] Gould RM, Byrd AL, Barbarese E. The number of Schmidt–Lanterman incisures is more than doubled in shiverer PNS myelin sheaths. J Neurocytol. 1995;24:85–98. doi: 10.1007/BF01181552. [DOI] [PubMed] [Google Scholar] Goutebroze L, Carnaud M, Denisenko N, Boutterin MC, Girault JA. Syndecan-3 and syndecan-4 are enriched in Schwann cell perinodal processes. BMC Neurosci. 2003;4:29. doi: 10.1186/1471-2202-4-29. [DOI] [PMC free article] [PubMed] [Google Scholar] Grove M, Komiyama NH, Nave KA, Grant SG, Sherman DL, Brophy PJ. FAK is required for axonal sorting by Schwann cells. J Cell Biol. 2007;176:277–282. doi: 10.1083/jcb.200609021. [DOI] [PMC free article] [PubMed] [Google Scholar] Hoshi T, Suzuki A, Hayashi S, Tohyama K, Hayashi A, Yamaguchi Y, Takeuchi K, Baba H. Nodal protrusions, increased Schmidt–Lanterman incisures, and paranodal disorganization are characteristic features of sulfatide-deficient peripheral nerves. Glia. 2007;55:584–594. doi: 10.1002/glia.20487. [DOI] [PubMed] [Google Scholar] Jessen KR, Mirsky R. The origin and development of glial cells in peripheral nerves. Nat Rev Neurosci. 2005;6:671–682. doi: 10.1038/nrn1746. [DOI] [PubMed] [Google Scholar] Kalamarides M, Niwa-Kawakita M, Leblois H, Abramowski V, Perricaudet M, Janin A, Thomas G, Gutmann DH, Giovannini M. Nf2 gene inactivation in arachnoidal cells is rate-limiting for meningioma development in the mouse. Genes Dev. 2002;16:1060–1065. doi: 10.1101/gad.226302. [DOI] [PMC free article] [PubMed] [Google Scholar] Kordeli E, Davis J, Trapp B, Bennett V. An isoform of ankyrin is localized at nodes of Ranvier in myelinated axons of central and peripheral nerves. J Cell Biol. 1990;110:1341–1352. doi: 10.1083/jcb.110.4.1341. [DOI] [PMC free article] [PubMed] [Google Scholar] Lallemand D, Curto M, Saotome I, Giovannini M, McClatchey AI. NF2 deficiency promotes tumorigenesis and metastasis by destabilizing adherens junctions. Genes Dev. 2003;17:1090–1100. doi: 10.1101/gad.1054603. [DOI] [PMC free article] [PubMed] [Google Scholar] Mathis C, Denisenko-Nehrbass N, Girault JA, Borrelli E. Essential role of oligodendrocytes in the formation and maintenance of central nervous system nodal regions. Development. 2001;128:4881–4890. doi: 10.1242/dev.128.23.4881. [DOI] [PubMed] [Google Scholar] McClatchey AI, Giovannini M. Membrane organization and tumorigenesis–the NF2 tumor suppressor, merlin. Genes Dev. 2005;19:2265–2277. doi: 10.1101/gad.1335605. [DOI] [PubMed] [Google Scholar] McClatchey AI, Saotome I, Mercer K, Crowley D, Gusella JF, Bronson RT, Jacks T. Mice heterozygous for a mutation at the Nf2 tumor suppressor locus develop a range of highly metastatic tumors. Genes Dev. 1998;12:1121–1133. doi: 10.1101/gad.12.8.1121. [DOI] [PMC free article] [PubMed] [Google Scholar] Melendez-Vasquez CV, Rios JC, Zanazzi G, Lambert S, Bretscher A, Salzer JL. Nodes of Ranvier form in association with ezrin-radixin-moesin (ERM)- positive Schwann cell processes. Proc Natl Acad Sci U S A. 2001;98:1235–1240. doi: 10.1073/pnas.98.3.1235. [DOI] [PMC free article] [PubMed] [Google Scholar] Menegoz M, Gaspar P, Le Bert M, Galvez T, Burgaya F, Palfrey C, Ezan P, Arnos F, Girault JA. Paranodin, a glycoprotein of neuronal paranodal membranes. Neuron. 1997;19:319–331. doi: 10.1016/s0896-6273(00)80942-3. [DOI] [PubMed] [Google Scholar] Michailov GV, Sereda MW, Brinkmann BG, Fischer TM, Haug B, Birchmeier C, Role L, Lai C, Schwab MH, Nave KA. Axonal neuregulin-1 regulates myelin sheath thickness. Science. 2004;304:700–703. doi: 10.1126/science.1095862. [DOI] [PubMed] [Google Scholar] Morrison H, Sperka T, Manent J, Giovannini M, Ponta H, Herrlich P. Merlin/neurofibromatosis type 2 suppresses growth by inhibiting the activation of Ras and Rac. Cancer Res. 2007;67:520–527. doi: 10.1158/0008-5472.CAN-06-1608. [DOI] [PubMed] [Google Scholar] Neuberg DH, Sancho S, Suter U. Altered molecular architecture of peripheral nerves in mice lacking the peripheral myelin protein 22 or connexin32. J Neurosci Res. 1999;58:612–623. doi: 10.1002/(sici)1097-4547(19991201)58:5<612::aid-jnr2>3.0.co;2-x. [DOI] [PubMed] [Google Scholar] Obremski VJ, Hall AM, Fernandez-Valle C. Merlin, the neurofibromatosis type 2 gene product, and beta1 integrin associate in isolated and differentiating Schwann cells. J Neurobiol. 1998;37:487–501. [PubMed] [Google Scholar] Okada T, Lopez-Lago M, Giancotti FG. Merlin/NF-2 mediates contact inhibition of growth by suppressing recruitment of Rac to the plasma membrane. J Cell Biol. 2005;171:361–371. doi: 10.1083/jcb.200503165. [DOI] [PMC free article] [PubMed] [Google Scholar] Okada T, You L, Giancotti FG. Shedding light on Merlin's wizardry. Trends Cell Biol. 2007;17:222–229. doi: 10.1016/j.tcb.2007.03.006. [DOI] [PubMed] [Google Scholar] Perrin-Tricaud C, Rutishauser U, Tricaud N. P120 catenin is required for thickening of Schwann cell myelin. Mol Cell Neurosci. 2007;35:120–129. doi: 10.1016/j.mcn.2007.02.010. [DOI] [PubMed] [Google Scholar] Poliak S, Peles E. The local differentiation of myelinated axons at nodes of Ranvier. Nat Rev Neurosci. 2003;4:968–980. doi: 10.1038/nrn1253. [DOI] [PubMed] [Google Scholar] Poliak S, Gollan L, Salomon D, Berglund EO, Ohara R, Ranscht B, Peles E. Localization of Caspr2 in myelinated nerves depends on axon-glia interactions and the generation of barriers along the axon. J Neurosci. 2001;21:7568–7575. doi: 10.1523/JNEUROSCI.21-19-07568.2001. [DOI] [PMC free article] [PubMed] [Google Scholar] Poulikakos PI, Xiao GH, Gallagher R, Jablonski S, Jhanwar SC, Testa JR. Re-expression of the tumor suppressor NF2/merlin inhibits invasiveness in mesothelioma cells and negatively regulates FAK. Oncogene. 2006;25:5960–5968. doi: 10.1038/sj.onc.1209587. [DOI] [PubMed] [Google Scholar] Rangwala R, Banine F, Borg JP, Sherman LS. Erbin regulates mitogen-activated protein (MAP) kinase activation and MAP kinase-dependent interactions between Merlin and adherens junction protein complexes in Schwann cells. J Biol Chem. 2005;280:11790–11797. doi: 10.1074/jbc.M414154200. [DOI] [PubMed] [Google Scholar] Robertson JD. The ultrastructure of Schmidt–Lanterman clefts and related shearing defects of the myelin sheath. J Biophys Biochem Cytol. 1958;4:39–46. doi: 10.1083/jcb.4.1.39. [DOI] [PMC free article] [PubMed] [Google Scholar] Rouleau GA, Merel P, Lutchman M, Sanson M, Zucman J, Marineau C, Hoang-Xuan K, Demczuk S, Desmaze C, Plougastel B, Pulst SM, Lenoir G, Bijlsma E, Fashold R, Dumanski J, de Jong P, Parry D, Eldridge R, Aurias A, Delattre O, Thomas G. Alteration in a new gene encoding a putative membrane-organizing protein causes neuro-fibromatosis type 2. Nature. 1993;363:515–521. doi: 10.1038/363515a0. [DOI] [PubMed] [Google Scholar] Salzer JL. Polarized domains of myelinated axons. Neuron. 2003;40:297–318. doi: 10.1016/s0896-6273(03)00628-7. [DOI] [PubMed] [Google Scholar] Schafer DP, Bansal R, Hedstrom KL, Pfeiffer SE, Rasband MN. Does paranode formation and maintenance require partitioning of neurofascin 155 into lipid rafts? J Neurosci. 2004;24:3176–3185. doi: 10.1523/JNEUROSCI.5427-03.2004. [DOI] [PMC free article] [PubMed] [Google Scholar] Scherer SS, Gutmann DH. Expression of the neurofibromatosis 2 tumor suppressor gene product, merlin, in Schwann cells. J Neurosci Res. 1996;46:595–605. doi: 10.1002/(SICI)1097-4547(19961201)46:5<595::AID-JNR8>3.0.CO;2-E. [DOI] [PubMed] [Google Scholar] Sharghi-Namini S, Turmaine M, Meier C, Sahni V, Umehara F, Jessen KR, Mirsky R. The structural and functional integrity of peripheral nerves depends on the glial-derived signal desert hedgehog. J Neurosci. 2006;26:6364–6376. doi: 10.1523/JNEUROSCI.0157-06.2006. [DOI] [PMC free article] [PubMed] [Google Scholar] Shaw RJ, Paez JG, Curto M, Yaktine A, Pruitt WM, Saotome I, O'Bryan JP, Gupta V, Ratner N, Der CJ, Jacks T, McClatchey AI. The Nf2 tumor suppressor, merlin, functions in Rac-dependent signaling. Dev Cell. 2001;1:63–72. doi: 10.1016/s1534-5807(01)00009-0. [DOI] [PubMed] [Google Scholar] Sherman DL, Brophy PJ. Mechanisms of axon ensheathment and myelin growth. Nat Rev Neurosci. 2005;6:683–690. doi: 10.1038/nrn1743. [DOI] [PubMed] [Google Scholar] Sperfeld AD, Hein C, Schröder JM, Ludolph AC, Hanemann CO. Occurrence and characterization of peripheral nerve involvement in neurofibromatosis type 2. Brain. 2002;125:996–1004. doi: 10.1093/brain/awf115. [DOI] [PubMed] [Google Scholar] Stickney JT, Bacon WC, Rojas M, Ratner N, Ip W. Activation of the tumor suppressor merlin modulates its interaction with lipid rafts. Cancer Res. 2004;64:2717–2724. doi: 10.1158/0008-5472.can-03-3798. [DOI] [PubMed] [Google Scholar] Tait S, Gunn-Moore F, Collinson JM, Huang J, Lubetzki C, Pedraza L, Sherman DL, Colman DR, Brophy PJ. An oligodendrocyte cell adhesion molecule at the site of assembly of the paranodal axo-glial junction. J Cell Biol. 2000;150:657–666. doi: 10.1083/jcb.150.3.657. [DOI] [PMC free article] [PubMed] [Google Scholar] Taylor AR, Geden SE, Fernandez-Valle C. Formation of a beta1 integrin signaling complex in Schwann cells is independent of rho. Glia. 2003;41:94–104. doi: 10.1002/glia.10170. [DOI] [PubMed] [Google Scholar] Tricaud N, Perrin-Tricaud C, Brusés JL, Rutishauser U. Adherens junctions in myelinating Schwann cells stabilize Schmidt–Lanterman incisures via recruitment of p120 catenin to E-cadherin. J Neurosci. 2005;25:3259–3269. doi: 10.1523/JNEUROSCI.5168-04.2005. [DOI] [PMC free article] [PubMed] [Google Scholar] Trofatter JA, MacCollin MM, Rutter JL, Murrell JR, Duyao MP, Parry DM, Eldridge R, Kley N, Menon AG, Pulaski K, Haase VH, Ambrose CM, Munroe D, Bove C, Haines JL, Martuza RL, MacDonald ME, Seizenger BR, Short MP, Buckler AJ, Gusella JF. A novel moesin-, ezrin-, radixin-like gene is a candidate for the neurofibromatosis 2 tumor suppressor. Cell. 1993;72:791–800. doi: 10.1016/0092-8674(93)90406-g. [DOI] [PubMed] [Google Scholar] Wang H, Kunkel DD, Martin TM, Schwartzkroin PA, Tempel BL. Heteromultimeric K+ channels in terminal and juxtaparanodal regions of neurons. Nature. 1993;365:75–79. doi: 10.1038/365075a0. [DOI] [PubMed] [Google Scholar] Zhang SM, Marsh R, Ratner N, Brackenbury R. Myelin glycoprotein P0 is expressed at early stages of chicken and rat embryogenesis. J Neurosci Res. 1995;40:241–250. doi: 10.1002/jnr.490400213. [DOI] [PubMed] [Google Scholar] Articles from The Journal of Neuroscience are provided here courtesy of Society for Neuroscience ACTIONS View on publisher site PDF (1.3 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Materials and Methods Results Discussion 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
4509	https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:working-units/x2f8bb11595b61c86:appropriate-units/v/interpreting-units-in-formulas-novel	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
4510	https://brainly.com/question/10769862	[FREE] Why is \cos(n\pi) = (-1)^n true when n could be any integer? - brainly.com Advertisement 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 +82,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +11,3k Ace exams faster, with practice that adapts to you Practice Worksheets +8k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Why is cos(nπ)=(−1)n true when n could be any integer? 2 See answers Explain with Learning Companion NEW Asked by iamavampire • 08/16/2018 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 36356142 people 36M 4.6 6 Upload your school material for a more relevant answer Given,cos nπ=(−1)n We state why it is true. Lets check for some values of n. for n = 0 ⇒ cos 0=(−1)0=1 for n = 1 ⇒ cos π=(−1)1=−1 for n = -1 ⇒ cos\,-\pi=(-1)^{-1)=-1 for n = 2 ⇒ cos\,2\pi=(-1)^{2)=1 for n = -2 ⇒ cos\,-2\pi=(-1)^{-2)=1 So, Its clear from above . value of cos nπ is either 1 or -1. value of cos nπ is 1 when n is even integer and -1 when n is odd integer. Answered by aquialaska •3.1K answers•36.4M people helped Thanks 6 4.6 (10 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 36356142 people 36M 4.6 6 Big Ideas in Cosmology - Kim Coble, Kevin McLin, Lynn Cominsky Mathematics for Biomedical Physics - Jogindra M Wadehra Classical Mechanics - Peter Dourmashkin Upload your school material for a more relevant answer The equation cos(nπ)=(−1)n is true because the cosine function yields 1 for even integers and -1 for odd integers. This relationship follows from the periodic nature of the cosine function. Therefore, when evaluated at multiples of π, the values correspond to the outcomes of (−1)n. Explanation To understand why cos(nπ)=(−1)n holds true for any integer n, we start by analyzing the cosine function. When we graph the cosine function: The cosine function is periodic with a period of 2 π. This means that it repeats every 2 π radians. Notably, at integer multiples of π, the cosine function yields specific values: cos(0)=1, cos(π)=−1, and cos(2 π)=1. Even and Odd Integers: If n is an even integer (like 0, 2, 4, ...), we can write it as n=2 k for some integer k. Thus, cos(2 kπ)=cos(0)=1=(−1)2 k=1. If n is an odd integer (like 1, 3, 5, ...), we can express it as n=2 k+1. Therefore, cos((2 k+1)π)=cos(π)=−1=(−1)2 k+1=−1. Overall Result: This gives us the conclusion that whenever n is even, cos(nπ)=1, and whenever n is odd, cos(nπ)=−1. Therefore, we can summarize this relationship as: cos(nπ)=(−1)n. In summary, this relationship arises from the properties of the cosine function with respect to even and odd integers, yielding a value of 1 for even n and -1 for odd n. Examples & Evidence For example, if we plug in n=0, cos(0 π)=cos(0)=1 and (−1)0=1. For n=1, we have cos(π)=−1 which matches (−1)1=−1. The relationship can be confirmed through the periodicity of the cosine function and the values it takes at multiples of π, verifying that the equations hold for any integer. Thanks 6 4.6 (10 votes) Advertisement Community Answer This answer helped 95482 people 95K 3.4 8 its true because -1 to even power is positive, -1 to odd power is negative Answered by Lazerpent •85 answers•95.5K people helped Thanks 8 3.4 (14 votes) 5 Advertisement ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them? Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216 Choose an equivalent expression for 1 0 6÷1 0 4. A. 1 0 2 B. 1 0 3 C. 1 0 10 D. 1 0 24 How would you write 1 2−3 using a positive exponent? A. 1 2 3 B. 1 2 0 C. 1 1 2 3 D. 1 2 3 1 Is this equation correct? 6 3⋅7 3=4 2 3 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
4511	https://www.chegg.com/homework-help/questions-and-answers/4-three-masses-connected-together-two-thin-strings-20-mathrm-~kg-mass-hangs-freely-edge-ta-q108270721	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: 4. Three masses are connected together by a two thin strings. A 20 kg mass hangs freely over the edge of a table. It is connected via a pulley to a 10 kg mass that rests on the table. On the other side of the 10 kg mass a string is threaded over another pulley to the third 35 kg mass that rests on an inclined plane. The plane and the table top both have a. The acceleration of the 35 kg mass can be calculated using Newton's second law, which states that... Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
4512	https://ocw.tau.edu.ng/high-school/mathematics/exam-prep/derivative-as-a-function/relationship-between-behavior-of-f-sign-of-f/	Relationship Between Behavior of f & Sign of f' \| MIT OpenCourseWare \| Free Online Course Materials Subscribe to the OCW Newsletter Help\|Contact Us Subjects Biology Chemistry Engineering Humanities & Social Sciences Mathematics Physics Exam Preparation Biology Calculus Chemistry Physics IIT Joint Entrance Exam More Intro for Students Intro for Teachers Other MIT Resources What is Highlights? MIT OCW Home>Highlights for High School>Mathematics>Calculus Exam Preparation>Derivative as a Function>Relationship Between Behavior of f & Sign of f' Relationship Between Behavior of f & Sign of f' Exam Prep: Biology Exam Prep: Calculus Exam Prep: Chemistry Exam Prep: Physics Select Topic Select Sub-topic First Derivative Test First Derivative Test Increasing, decreasing, non-increasing, and non-decreasing functions are defined. First Derivative Test is explained and an example is given. 18.01 Single Variable Calculus, Fall 2005 Prof. Jason Starr Course Material Related to This Topic: Read lecture notes, section 2 on pages 1–2 Back to Top Subjects Biology Chemistry Engineering Humanities & Social Sciences Mathematics Physics Exam Prep Biology Calculus Chemistry Physics Tools Help & FAQs Contact Us Accessibility Site Map Privacy & Terms of Use RSS Feeds More Intro for Students Intro for Teachers Other MIT Resources What is Highlights? Featured Sites MIT OpenCourseWare OCW Educator MIT Crosslinks and OCW MITx and Related OCW Courses MIT+K12 Videos Teaching Excellence at MIT Outreach@MIT Open Education Consortium Our Corporate Supporters About MIT OpenCourseWare MIT OpenCourseWare is an online publication of materials from over 2,500 MIT courses, freely sharing knowledge with learners and educators around the world. Learn more » © 2001–2025 Massachusetts Institute of Technology Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use. Loading...
4513	https://sites.math.rutgers.edu/~sg1108/olympiad.html	Olympiad Math Resources Art of Problem Solving (AoPS) Canada IMO training British Math Olympiads Yufei Zhao Po-Shen Loh Alex Remorov Evan Chen Books The IMO Compendium: Dušan Djukić, Vladimir Janković, Ivan Matić, Nikola Petrović Problems From the Book, Titu Andreescu and Gabriel Dospinescu Straight From the Book, Titu Andreescu and Gabriel Dospinescu 110-geometry-problems-from-the IMO, Titu Andreescu and Cosmin Pohoata Lemmas in Olympiad Geometry, Titu Andreescu, Sam Korsky, and Cosmin Pohoata The Geometry of Remarkable Elements: Points, Lines, and Circles, Titu Andreescu, Dorin Andrica, Paul Blaga, and Dan Bränzei The Mathematical Reflections, Titu Andreescu and Maxim Ignatiuc New, Newer, and Newest Inequalities, Titu Andreescu and Marius Stanean 118 Inequalities for Mathematics Competitions, Titu Andreescu and Marius Stanean 113 Geometric Inequalities, Adrian Andreescu, Titu Andreescu, and Oleg Mushkarov Topics in Functional Equations, Titu Andreescu, Iurie Boreico, Oleg Mushkarov, and Nikolai Nikolov Topics in Geometric Inequalities, Titu Andreescu and Oleg Mushkarov Number Theory Concepts and Problems, Titu Andreescu, Gabriel Dospinescu, and Oleg Mushkarov A Romanian Problem Book, Titu Andreescu and Marian Tetiva Graphs, Radu Bumbacea Balkan Mathematical Olympiads, Mircea Becheanu and Bogdan Enescu Cuban Mathematical Olympiads, Robert Bosch (translated by Enrique Treviño The Art of Problem Solving, Volume 1: the Basics, Lehoczky and Rusczyk The Art of Problem Solving, Volume 2: and Beyond, Lehoczky and Rusczyk Problem Solving Strategies,Engel Mathematical Olympiad Treasures,Andreescu and Enescu Mathematical Olympiad Challenges, Andreescu and Gelca The Art and Craft of Problem Solving, Zeitz suryatejagavva
4514	https://ysu.edu/sites/default/files/mathematics-achievement-center/Calculus%203_Cheat_Sheet_Reduced.pdf	Derivatives Dxex = ex Dx sin(x) = cos(x) Dx cos(x) = −sin(x) Dx tan(x) = sec2(x) Dx cot(x) = −csc2(x) Dx sec(x) = sec(x) tan(x) Dx csc(x) = −csc(x) cot(x) Dx sin−1 = 1 p 1−x2 , x 2 [−1, 1] Dx cos−1 = −1 p 1−x2 , x 2 [−1, 1] Dx tan−1 = 1 1+x2 , −⇡ 2 x ⇡ 2 Dx sec−1 = 1 \|x\|p x2−1 , \|x\| > 1 Dx sinh(x) = cosh(x) Dx cosh(x) = −sinh(x) Dx tanh(x) = sech2(x) Dx coth(x) = −csch2(x) Dxsech(x) = −sech(x) tanh(x) Dxcsch(x) = −csch(x) coth(x) Dx sinh−1 = 1 p x2+1 Dx cosh−1 = −1 p x2−1 , x > 1 Dx tanh−1 = 1 1−x2 −1 < x < 1 Dxsech−1 = 1 xp 1−x2 , 0 < x < 1 Dx ln(x) = 1 x Integrals R 1 x dx = ln \|x\| + c R exdx = ex + c R axdx = 1 ln a ax + c R eaxdx = 1 a eax + c R 1 p 1−x2 dx = sin−1(x) + c R 1 1+x2 dx = tan−1(x) + c R 1 xp x2−1 dx = sec−1(x) + c R sinh(x)dx = cosh(x) + c R cosh(x)dx = sinh(x) + c R tanh(x)dx = ln \| cosh(x)\| + c R tanh(x)sech(x)dx = −sech(x) + c R sech2(x)dx = tanh(x) + c R csch(x) coth(x)dx = −csch(x) + c R tan(x)dx = −ln \| cos(x)\| + c R cot(x)dx = ln \| sin(x)\| + c R cos(x)dx = sin(x) + c R sin(x)dx = −cos(x) + c R 1 p a2−u2 dx = sin−1( u a ) + c R 1 a2+u2 dx = 1 a tan−1 u a + c R ln(x)dx = (xln(x)) −x + c U-Substitution Let u = f(x) (can be more than one variable). Determine: du = f(x) dx dx and solve for dx. Then, if a deﬁnite integral, substitute the bounds for u = f(x) at each bounds Solve the integral using u. Integration by Parts R udv = uv − R vdu Fns and Identities sin(cos−1(x)) = p 1 −x2 cos(sin−1(x)) = p 1 −x2 sec(tan−1(x)) = p 1 + x2 tan(sec−1(x)) = ( p x2 −1 if x ≥1) = (− p x2 −1 if x −1) sinh−1(x) = ln x + p x2 + 1 sinh−1(x) = ln x + p x2 −1, x ≥−1 tanh−1(x) = 1 2 ln x + 1+x 1−x , 1 < x < −1 sech−1(x) = ln[ 1+p 1−x2 x ], 0 < x −1 sinh(x) = ex−e−x 2 cosh(x) = ex+e−x 2 Trig Identities sin2(x) + cos2(x) = 1 1 + tan2(x) = sec2(x) 1 + cot2(x) = csc2(x) sin(x ± y) = sin(x) cos(y) ± cos(x) sin(y) cos(x ± y) = cos(x) cos(y) ± sin(x) sin(y) tan(x ± y) = tan(x)±tan(y) 1⌥tan(x) tan(y) sin(2x) = 2 sin(x) cos(x) cos(2x) = cos2(x) −sin2(x) cosh(n2x) −sinh2 x = 1 1 + tan2(x) = sec2(x) 1 + cot2(x) = csc2(x) sin2(x) = 1−cos(2x) 2 cos2(x) = 1+cos(2x) 2 tan2(x) = 1−cos(2x) 1+cos(2x) sin(−x) = −sin(x) cos(−x) = cos(x) tan(−x) = −tan(x) Calculus 3 Concepts Cartesian coords in 3D given two points: (x1, y1, z1) and (x2, y2, z2), Distance between them: p (x1 −x2)2 + (y1 −y2)2 + (z1 −z2)2 Midpoint: ( x1+x2 2 , y1+y2 2 , z1+z2 2 ) Sphere with center (h,k,l) and radius r: (x −h)2 + (y −k)2 + (z −l)2 = r2 Vectors Vector: ~ u Unit Vector: ˆ u Magnitude: \|\|~ u\|\| = q u2 1 + u2 2 + u2 3 Unit Vector: ˆ u = ~ u \|\|~ u\|\| Dot Product ~ u · ~ v Produces a Scalar (Geometrically, the dot product is a vector projection) ~ u =< u1, u2, u3 > ~ v =< v1, v2, v3 > ~ u · ~ v = ~ 0 means the two vectors are Perpendicular ✓is the angle between them. ~ u · ~ v = \|\|~ u\|\| \|\|~ v\|\| cos(✓) ~ u · ~ v = u1v1 + u2v2 + u3v3 NOTE: ˆ u · ˆ v = cos(✓) \|\|~ u\|\|2 = ~ u · ~ u ~ u · ~ v = 0 when ? Angle Between ~ u and ~ v: ✓= cos−1( ~ u·~ v \|\|~ u\|\| \|\|~ v\|\| ) Projection of ~ u onto ~ v: pr~ v~ u = ( ~ u·~ v \|\|~ v\|\|2 )~ v Cross Product ~ u ⇥~ v Produces a Vector (Geometrically, the cross product is the area of a paralellogram with sides \|\|~ u\|\| and \|\|~ v\|\|) ~ u =< u1, u2, u3 > ~ v =< v1, v2, v3 > ~ u ⇥~ v = $ $ $ $ $ $ ˆ i ˆ j ˆ k u1 u2 u3 v1 v2 v3 $ $ $ $ $ $ ~ u ⇥~ v = ~ 0 means the vectors are paralell Lines and Planes Equation of a Plane (x0, y0, z0) is a point on the plane and < A, B, C > is a normal vector A(x −x0) + B(y −y0) + C(z −z0) = 0 < A, B, C > · < x−x0, y−y0, z−z0 >= 0 Ax + By + Cz = D where D = Ax0 + By0 + Cz0 Equation of a line A line requires a Direction Vector ~ u =< u1, u2, u3 > and a point (x1, y1, z1) then, a parameterization of a line could be: x = u1t + x1 y = u2t + y1 z = u3t + z1 Distance from a Point to a Plane The distance from a point (x0, y0, z0) to a plane Ax+By+Cz=D can be expressed by the formula: d = \|Ax0+By0+Cz0−D\| p A2+B2+C2 Coord Sys Conv Cylindrical to Rectangular x = r cos(✓) y = r sin(✓) z = z Rectangular to Cylindrical r = p x2 + y2 tan(✓) = y x z = z Spherical to Rectangular x = ⇢sin(φ) cos(✓) y = ⇢sin(φ) sin(✓) z = ⇢cos(φ) Rectangular to Spherical ⇢= p x2 + y2 + z2 tan(✓) = y x cos(φ) = z p x2+y2+z2 Spherical to Cylindrical r = ⇢sin(φ) ✓= ✓ z = ⇢cos(φ) Cylindrical to Spherical ⇢= p r2 + z2 ✓= ✓ cos(φ) = z p r2+z2 Surfaces Ellipsoid x2 a2 + y2 b2 + z2 c2 = 1 Hyperboloid of One Sheet x2 a2 + y2 b2 −z2 c2 = 1 (Major Axis: z because it follows - ) Hyperboloid of Two Sheets z2 c2 −x2 a2 −y2 b2 = 1 (Major Axis: Z because it is the one not subtracted) Elliptic Paraboloid z = x2 a2 + y2 b2 (Major Axis: z because it is the variable NOT squared) Hyperbolic Paraboloid (Major Axis: Z axis because it is not squared) z = y2 b2 −x2 a2 Elliptic Cone (Major Axis: Z axis because it’s the only one being subtracted) x2 a2 + y2 b2 −z2 c2 = 0 Cylinder 1 of the variables is missing OR (x −a)2 + (y −b2) = c (Major Axis is missing variable) Partial Derivatives Partial Derivatives are simply holding all other variables constant (and act like constants for the derivative) and only taking the derivative with respect to a given variable. Given z=f(x,y), the partial derivative of z with respect to x is: fx(x, y) = zx = @z @x = @f(x,y) @x likewise for partial with respect to y: fy(x, y) = zy = @z @y = @f(x,y) @y Notation For fxyy, work ”inside to outside” fx then fxy, then fxyy fxyy = @3f @x@2y , For @3f @x@2y , work right to left in the denominator Gradients The Gradient of a function in 2 variables is rf =< fx, fy > The Gradient of a function in 3 variables is rf =< fx, fy, fz > Chain Rule(s) Take the Partial derivative with respect to the ﬁrst-order variables of the function times the partial (or normal) derivative of the ﬁrst-order variable to the ultimate variable you are looking for summed with the same process for other ﬁrst-order variables this makes sense for. Example: let x = x(s,t), y = y(t) and z = z(x,y). z then has ﬁrst partial derivative: @z @x and @z @y x has the partial derivatives: @x @s and @x @t and y has the derivative: dy dt In this case (with z containing x and y as well as x and y both containing s and t), the chain rule for @z @s is @z @s = @z @x @x @s The chain rule for @z @t is @z @t = @z @x @x @t + @z @y dy dt Note: the use of ”d” instead of ”@” with the function of only one independent variable Limits and Continuity Limits in 2 or more variables Limits taken over a vectorized limit just evaluate separately for each component of the limit. Strategies to show limit exists 1. Plug in Numbers, Everything is Fine 2. Algebraic Manipulation -factoring/dividing out -use trig identites 3. Change to polar coords if(x, y) ! (0, 0) , r ! 0 Strategies to show limit DNE 1. Show limit is di↵erent if approached from di↵erent paths (x=y, x = y2, etc.) 2. Switch to Polar coords and show the limit DNE. Continunity A fn, z = f(x, y), is continuous at (a,b) if f(a, b) = lim(x,y)!(a,b) f(x, y) Which means: 1. The limit exists 2. The fn value is deﬁned 3. They are the same value Directional Derivatives Let z=f(x,y) be a fuction, (a,b) ap point in the domain (a valid input point) and ˆ u a unit vector (2D). The Directional Derivative is then the derivative at the point (a,b) in the direction of ˆ u or: D~ uf(a, b) = ˆ u · rf(a, b) This will return a scalar. 4-D version: D~ uf(a, b, c) = ˆ u · rf(a, b, c) Tangent Planes let F(x,y,z) = k be a surface and P = (x0, y0, z0) be a point on that surface. Equation of a Tangent Plane: rF (x0, y0, z0)· < x −x0, y −y0, z −z0 > Approximations let z = f(x, y) be a di↵erentiable function total di↵erential of f = dz dz = rf· < dx, dy > This is the approximate change in z The actual change in z is the di↵erence in z values: ∆z = z −z1 Maxima and Minima Internal Points 1. Take the Partial Derivatives with respect to X and Y (fx and fy) (Can use gradient) 2. Set derivatives equal to 0 and use to solve system of equations for x and y 3. Plug back into original equation for z. Use Second Derivative Test for whether points are local max, min, or saddle Second Partial Derivative Test 1. Find all (x,y) points such that rf(x, y) = ~ 0 2. Let D = fxx(x, y)fyy(x, y) −f 2 xy(x, y) IF (a) D > 0 AND fxx < 0, f(x,y) is local max value (b) D > 0 AND fxx(x, y) > 0 f(x,y) is local min value (c) D < 0, (x,y,f(x,y)) is a saddle point (d) D = 0, test is inconclusive 3. Determine if any boundary point gives min or max. Typically, we have to parametrize boundary and then reduce to a Calc 1 type of min/max problem to solve. The following only apply only if a boundary is given 1. check the corner points 2. Check each line (0 x 5 would give x=0 and x=5 ) On Bounded Equations, this is the global min and max...second derivative test is not needed. Lagrange Multipliers Given a function f(x,y) with a constraint g(x,y), solve the following system of equations to ﬁnd the max and min points on the constraint (NOTE: may need to also ﬁnd internal points.): rf = λrg g(x, y) = 0(orkifgiven) Double Integrals With Respect to the xy-axis, if taking an integral, R R dydx is cutting in vertical rectangles, R R dxdy is cutting in horizontal rectangles Polar Coordinates When using polar coordinates, dA = rdrd✓ Surface Area of a Curve let z = f(x,y) be continuous over S (a closed Region in 2D domain) Then the surface area of z = f(x,y) over S is: SA = R R S q f 2 x + f 2 y + 1dA Triple Integrals R R R s f(x, y, z)dv = R a2 a1 R φ2(x) φ1(x) R 2(x,y) 1(x,y) f(x, y, z)dzdydx Note: dv can be exchanged for dxdydz in any order, but you must then choose your limits of integration according to that order Jacobian Method R R G f(g(u, v), h(u, v))\|J(u, v)\|dudv = R R R f(x, y)dxdy J(u, v) = $ $ $ $ @x @u @x @v @y @u @y @v $ $ $ $ Common Jacobians: Rect. to Cylindrical: r Rect. to Spherical: ⇢2 sin(φ) Vector Fields let f(x, y, z) be a scalar ﬁeld and ~ F (x, y, z) = M(x, y, z)ˆ i + N(x, y, z)ˆ j + P (x, y, z)ˆ k be a vector ﬁeld, Grandient of f = rf =< @f @x , @f @y , @f @z > Divergence of ~ F : r · ~ F = @M @x + @N @y + @P @z Curl of ~ F : r ⇥~ F = $ $ $ $ $ $ ˆ i ˆ j ˆ k @ @x @ @y @ @z M N P $ $ $ $ $ $ Line Integrals C given by x = x(t), y = y(t), t 2 [a, b] R c f(x, y)ds = R b a f(x(t), y(t))ds where ds = q ( dx dt )2 + ( dy dt )2dt or q 1 + ( dy dx )2dx or q 1 + ( dx dy )2dy To evaluate a Line Integral, · get a paramaterized version of the line (usually in terms of t, though in exclusive terms of x or y is ok) · evaluate for the derivatives needed (usually dy, dx, and/or dt) · plug in to original equation to get in terms of the independant variable · solve integral Work Let ~ F = Mˆ i + ˆ j + ˆ k (force) M = M(x, y, z), N = N(x, y, z), P = P (x, y, z) (Literally)d~ r = dxˆ i + dyˆ j + dzˆ k Work w = R c ~ F · d~ r (Work done by moving a particle over curve C with force ~ F ) Independence of Path Fund Thm of Line Integrals C is curve given by ~ r(t), t 2 [a, b]; ~ r0(t) exists. If f(~ r) is continuously di↵erentiable on an open set containing C, then R c rf(~ r) · d~ r = f(~ b) −f(~ a) Equivalent Conditions ~ F (~ r) continuous on open connected set D. Then, (a) ~ F = rf for some fn f. (if ~ F is conservative) , (b) R c ~ F (~ r) · d~ risindep.ofpathinD , (c) R c ~ F (~ r) · d~ r = 0 for all closed paths in D. Conservation Theorem ~ F = Mˆ i + Nˆ j + P ˆ k continuously di↵erentiable on open, simply connected set D. ~ F conservative , r ⇥~ F = ~ 0 (in 2D r ⇥~ F = ~ 0 i↵My = Nx) Green’s Theorem (method of changing line integral for double integral - Use for Flux and Circulation across 2D curve and line integrals over a closed boundary) H Mdy −Ndx = R R R(Mx + Ny)dxdy H Mdx + Ndy = R R R(Nx −My)dxdy Let: ·R be a region in xy-plane ·C is simple, closed curve enclosing R (w/ paramerization ~ r(t)) · ~ F (x, y) = M(x, y)ˆ i + N(x, y)ˆ j be continuously di↵erentiable over R[C. Form 1: Flux Across Boundary ~ n = unit normal vector to C H c ~ F · ~ n = R R R r · ~ F dA , H Mdy −Ndx = R R R(Mx + Ny)dxdy Form 2: Circulation Along Boundary H c ~ F · d~ r = R R R r ⇥~ F · ˆ udA , H Mdx + Ndy = R R R(Nx −My)dxdy Area of R A = H ( −1 2 ydx + 1 2 xdy) Gauss’ Divergence Thm (3D Analog of Green’s Theorem - Use for Flux over a 3D surface) Let: · ~ F (x, y, z) be vector ﬁeld continuously di↵erentiable in solid S ·S is a 3D solid ·@S boundary of S (A Surface) ·ˆ nunit outer normal to @S Then, R R @S ~ F (x, y, z) · ˆ ndS = R R R S r · ~ F dV (dV = dxdydz) Surface Integrals Let ·R be closed, bounded region in xy-plane ·f be a fn with ﬁrst order partial derivatives on R ·G be a surface over R given by z = f(x, y) ·g(x, y, z) = g(x, y, f(x, y)) is cont. on R Then, R R G g(x, y, z)dS = R R R g(x, y, f(x, y))dS where dS = q f 2 x + f 2 y + 1dydx Flux of ~ F across G R R G ~ F · ndS = R R R[−Mfx −Nfy + P ]dxdy where: · ~ F (x, y, z) = M(x, y, z)ˆ i + N(x, y, z)ˆ j + P (x, y, z)ˆ k ·G is surface f(x,y)=z ·~ n is upward unit normal on G. ·f(x,y) has continuous 1st order partial derivatives Unit Circle (cos, sin) Other Information pa p b = p a b Where a Cone is deﬁned as z = p a(x2 + y2), In Spherical Coordinates, φ = cos−1( q a 1+a ) Right Circular Cylinder: V = ⇡r2h, SA = ⇡r2 + 2⇡rh limn!inf(1 + m n )pn = emp Law of Cosines: a2 = b2 + c2 −2bc(cos(✓)) Stokes Theorem Let: ·S be a 3D surface · ~ F (x, y, z) = M(x, y, z)ˆ i + N(x, y, z)ˆ j + P (x, y, z)ˆ l ·M,N,P have continuous 1st order partial derivatives ·C is piece-wise smooth, simple, closed, curve, positively oriented · ˆ T is unit tangent vector to C. Then, H ~ Fc · ˆ T dS = R R s(r ⇥~ F ) · ˆ ndS = R R R(r ⇥~ F ) · ~ ndxdy Remember: H ~ F · ~ T ds = R c(Mdx + Ndy + P dz) Originally Written By Daniel Kenner for MATH 2210 at the University of Utah. Source code available at CheatSheet Thanks to Kelly Macarthur for Teaching and Providing Notes.
4515	https://stats.stackexchange.com/questions/582309/geometric-vs-arithmetic-mean	Geometric vs. arithmetic mean - Cross Validated Join Cross Validated By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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 Cross Validated helpchat Cross Validated Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Geometric vs. arithmetic mean [duplicate] Ask Question Asked 3 years, 2 months ago Modified3 years, 2 months ago Viewed 528 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This question already has answers here: Which "mean" to use and when? (5 answers) Arithmetic vs Geometric Mean (1 answer) Closed 3 years ago. I'm not sure if this is the best place for this question, but when averaging data, how do you know if the geometric or arithmetic mean is more appropriate for taking the average? Do you need to know something a priori about the data set to decide? For my particular problem, I am sampling the data in a log space and wish to average several data sets that should follow the same trends. mean geometric-mean Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Jul 18, 2022 at 2:07 user142054 user142054 asked Jul 18, 2022 at 1:58 user142054 user142054 2 1 It's a good place and we even already have it answered. See the linked threads or many other tagged as geometric-meanTim –Tim 2022-07-18 06:03:33 +00:00 Commented Jul 18, 2022 at 6:03 Thanks, that and the answers below are very helpful.user142054 –user142054 2022-07-18 06:05:07 +00:00 Commented Jul 18, 2022 at 6:05 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. There are lots of possible aggregations besides the arithmetic and geometric mean, popular other choices are e.g. the harmonic mean or the generalized mean, or, as a robust alternative, the median. For you, it might be of relevance that the logarithm of the geometric mean is the arithmetic mean of the logarithms. You might want to always consider at least the three Pythagorean means, harmonic (H M H M), geometric (G M G M), and arithmetic (A M A M) mean. Note, that their values on any set S S of numbers always satisfy H M(S)≤G M(S)≤A M(S).H M(S)≤G M(S)≤A M(S). Intuitively, small values have more weight in the harmonic and geometric mean, so if a value goes to zero, it draws the mean to zero, too, which is not the case with the arithmetic mean. A nice property of G M G M is that the geometric mean of ratios is the ratio of the geometric means: x 1 y 1…x n y n−−−−−−−−√n=x 1⋯x n−−−−−−−√n y 1⋯y n−−−−−−−√n.x 1 y 1…x n y n n=x 1⋯x n n y 1⋯y n n. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 18, 2022 at 5:41 frankfrank 11.5k 3 3 gold badges 23 23 silver badges 30 30 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. As with many things, it depends on what you want to get out of it. The neat thing about the arithmetic average is that if you have one, and you know the total number of values, you can multiply the two and get the total sum. If that property is important, you use the arithmetic average. Something neat about the geometric average is that it respects compounding, making it appropriate for financial comparisons of total wealth under various courses of action. But I don't know of a strict rule, other than that I can list many examples where one is appropriate and the other is not. But sometimes it's not even that clear. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 18, 2022 at 4:58 kqrkqr 924 5 5 silver badges 11 11 bronze badges Add a comment\| 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 236Which "mean" to use and when? 5Arithmetic vs Geometric Mean Related 1Average rate versus average period 4Difference between geometric and arithmetic mean 5Arithmetic or geometric mean to calculate an annual average price index? 3How is that possible that simple arithmetic mean works well even for strongly skewed distribution? 0Percentage Change values for Prices in a given basked: Gemoetric mean vs arithmetic mean to get "mean" Prices 12When summarizing data from a lognormal distribution, when does it make sense to report the arithmetic mean vs. geometric mean 2Is the use of geometric mean for water chemical concentrations appropriate? Hot Network Questions What's the expectation around asking to be invited to invitation-only workshops? Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation ICC in Hague not prosecuting an individual brought before them in a questionable manner? Numbers Interpreted in Smallest Valid Base Bypassing C64's PETSCII to screen code mapping Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? Why are LDS temple garments secret? Does the curvature engine's wake really last forever? With line sustain pedal markings, do I release the pedal at the beginning or end of the last note? The geologic realities of a massive well out at Sea how do I remove a item from the applications menu How different is Roman Latin? Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? What’s the usual way to apply for a Saudi business visa from the UAE? How to home-make rubber feet stoppers for table legs? Why does LaTeX convert inline Python code (range(N-2)) into -NoValue-? Do sum of natural numbers and sum of their squares represent uniquely the summands? An odd question Is direct sum of finite spectra cancellative? Who is the target audience of Netanyahu's speech at the United Nations? How to start explorer with C: drive selected and shown in folder list? For every second-order formula, is there a first-order formula equivalent to it by reification? Why do universities push for high impact journal publications? Origin of Australian slang exclamation "struth" meaning greatly surprised more hot questions Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Cross Validated Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4516	https://www.ck12.org/flexi/chemistry/calorimetry/how-can-the-final-temperature-in-calorimetry-be-calculated/	Flexi answers - How can the final temperature in calorimetry be calculated? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Chemistry Calorimetry Question How can the final temperature in calorimetry be calculated? Flexi Says: In calorimetry, the final temperature can be calculated using the formula: q = mcΔT where: - q is the heat energy absorbed or released, - m is the mass of the substance, - c is the specific heat capacity of the substance, and - ΔT is the change in temperature (final temperature - initial temperature). Rearranging the formula to solve for the final temperature gives: final temperature = initial temperature + q/(mc) This formula assumes that no heat is lost to the surroundings, which is an idealization. In real experiments, some heat loss usually occurs. Practice this conceptAnalogy / Example Try Asking: Eighty grams of brass (specific heat 0.092 cal/g-K) at 292 degrees C is added to 200 g of water (specific heat 1.0 cal/g-K) at 14 degrees C in an insulated container of negligible heat capacity. What is the final temperature in degree C?Hydrochloric acid is combined with a new cleaning solution in a calorimeter with 5 grams of water at 35 °C. The final temperature of the water is 50 °C. What is the heat change of the reaction? (c of water = 4.184)A piece of hot metal was put into a cup containing cold water. Given the information below, calculate the final temperature (oC) of the system. Cold Water Value mass (g) 100 specific heat capacity (J/goC) 1 Initial temperature (oC) 20 Hot metal Value mass (g) 100 specific heat capacity (J/goC) 0.4 Initial temperature (oC) 90 a 55.0 b 38.5 c 48.2 d 40.0 How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
4517	https://www.wordsmyth.net/school/klockner/?ac=5690&rid=5716	cabal \| Dictionaries and vocabulary tools for English language learners \| Wordsmyth Klockner Elementary SchoolLog In\|Register home\|subscription\|feedback\|about us\|blog\|widget\|FAQ Dictionaries Comprehensive Children's WILD (Illustrated)Search Tools Browse Search Filter Reverse Search A-Z Word Parts Grammatical PatternsPuzzle Solvers Anagram Solver Crossword SolverTeacher Tools Classes Students Lessons Assignments ReportsVocabulary Center Activities Wordlist Maker Writing Tool (Beta) Legacy activitiesMy Wordsmyth Lookup History My Wordlists Legacy activities Comprehensive Dictionary Suite Beginner's dictionary Intermediate dictionary Advanced dictionary 11th Amendment 13th Amendment 19th Amendment 22nd Amendment 3-D 4WD 5th Amendment (fifth amendment) A 1 A 2 A 3 a 1 a 2 a 3 a.1 a.2 a-1 a-2 A-1 (a one) AA A.A. AAA aah (ah) A and M aardvark aardwolf aardwolves Aaron AB 1 AB 2 A.B. Ab (av) ab- ABA aba abaca abaci aback abacus abacuses Abaddon abaft abalone abandon abandoned abandoned abandoning abandonment abandons abandon ship abase ↓ Next ↓ Advanced Dictionary More results Show multi-word results Browse in wordlist See entries that contain "cabal"Display options Show syllables Show Lookup History Double-Click Lookup Show Grammatical Patterns Show Word Combinations Show Word Parts Show Spanish support Show Chinese support Pronunciation Wordsmyth Phonics IPA Entry formats Standard ClassicLookup History cabal ca·bal ====== cabal ===== pronunciation:kə balparts of speech:noun, intransitive verbfeatures:Word Combinations (noun) part of speech:noun definition 1:a small group of people engaged in a secret plot, usu. of a political nature.The lieutenant colonel was a member of the cabal who plotted against Hitler.synonyms:camarilla, conspiracysimilar words:accomplices, band, crew, junta, league, party, ring definition 2:the plot or scheme of such a group.synonyms:connivance, conspiracy, plot, schemesimilar words:design, intrigue, plan, program, racket related words:collusion, confederacy, contrivance, faction, union Word CombinationsAbout this feature adjective + (n.)cabal conservative, evil, federal, foreign, international, loose, medical, powerful, right-wing, secret, shadowy, sinister, small verb + (n.)cabal accuse, control, create, destroy, exploit, expose, form, join, organize, plot (n.)cabal + verb control, exploit, plot, spread noun + (n.)cabal agent, banker, clique, conspiracy, conspirator, decision, elite, exile, government, intrigue, magistrate, meeting, network, officer, order, play, radical, vision, zealot part of speech:intransitive verb inflections:cabals, caballing, caballed definition:to form a cabal.synonyms:connive, conspire, plot related words:connive, plot HomeSend FeedbackHaving a problem?Suggest a WordPrivacy Policy©2025 Wordsmyth
4518	https://www.math.wichita.edu/~richardson/inradius.html	The Inradius of a Right Triangle With Integral Sides Bill Richardson September 1999 Let a = x2 - y2, b = 2xy, c = x2 + y2 with 0 < y < x, (x,y) = 1 and x and y being of opposite parity. Then (a, b, c) is a primative Pythagorean triple. Let triangle ABC, in the figure below, be a right triangle with sides a, b and hypotenuse c. Let the circle with center I be the inscribed circle for this triangle. We will prove that the inradius, r, is an integer. Proof. Let r be the inradius. Since the tangents to a circle from a point outside the circle are equal, we have the sides of triangle ABC configured as in the above figure. Thus, c = (a - r) + (b - r) = a + b - 2r and r = (a + b - c)/2. But a = x2 - y2, b = 2xy, c = x2 + y2 which gives r = (x2 - y2 + 2xy - x2 - y2)/2 = (2xy - 2y2)/2 = y(x - y). Thus, r is an integer given by r = y(x - y). An alternate proof uses the fact that AreaTriangleABC = AreaTriangleAIB + AreaTriangleBIC + AreaTriangleAIC. So, (½)ab = (½)rc + (½)ra + (½)rb and r = ab/(a + b + c) = (x2 - y2)2xy/(x2 - y2 + 2xy + x2 + y2) = [2xy(x - y)(x + y)]/[2x(x + y)] = y(x - y). Thus, r is an integer given by r = y(x - y).
4519	https://www.restore.ac.uk/PEAS/finitepop.php	Finite Populations Corrections Skip to main content Menu Close About Resources Guidance Contact About this Resource P\|E\|A\|SFAQ\| home\| contact us\| index\| links and further resources\| 1 Stratification 1 Stratification 2 Clustering 2 Clustering 3 Weighting 3 Weighting 4 Design effects 4 Design effects 5 Non-response 5 Non-response 6 Imputation 6 Imputation 7 Subgroups 7 Subgroups 8 Check Data 8 Check Data 9 Finite pops 9 Finite pops 1 Income Dist.(FRS) 1 Income Dist.(FRS) 2 Internet Use(SHS) 2 Internet Use(SHS) 3 Smoking(SHlthS) 3 Smoking(SHlthS) 4 Equal Ops(WERS) 4 Equal Ops(WERS) 5 Drug Use(AAHS) 5 Drug Use(AAHS) 6 Delinquency(ESYTC) 6 Delinquency(ESYTC) R R SAS SAS Stata Stata SPSS SPSS text/print version 9: Finite Populations Corrections 9.1 Introduction>>9.2 Simple random sampling 9.3 FPC in stratified sampling>>9.4 FPC in two-stage sampling top 9.1 Introduction When making inferences from survey data we need to distinguish between inferences that only want to make statements about the units in the population from which our respondents are a sample or where we want to broaden the conclusions to go to wider groups In the first case our conclusions would only apply to the population that existed at the time. For example, if we were interested to know what was the total capacity of Scotland to accommodate tourists in 2004 we might do a survey of guesthouses and try to be estimate that fixed number. However, as part of the same survey, we might be interested to know whether, in the last year, guesthouses in seaside towns had higher occupancy than those elsewhere with a view to looking at the planning of guest house accommodation in future. In this case our inferences would need to go beyond the current sample and to look to predict trends for the future. In this case we would not be interested in exactly what had happened in the last year but we would want to know about the underlying mechanism or model might be producing these figures. In the first of these examples we would be more interested in making inferences for the current finite population but in the second we can be thinking of our sample as generalising more widely to a typical guest house that might be running in Scotland this year or in future years. It is only for the first kind of fixed population inference that it is appropriate to make the finite sample corrections that we will be discussing in the next section. top 9.2 Simplerandom sampling When we are interested in making inferences for a finite population and our sample begins to be a substantial proportion of that population then obviously our estimates will be more accurate and will tend to be perfect as we include more and more of the sample. In the case of simple random sampling there is a simple expression for how the design effects changes by having a finite population. If the sampling fraction is given byf then the design effect is reduced by a factor of (1-f). The design factor is reduced by the square root of this. If we put some numbers into this expression we will see that we need to have quite a substantial f before the standard errors get much smaller. For example, for a sample 1:1000 the finite population correction (FPC) gives a reduction of only 0.005%. For 1:100 the reduction is 0.5% and one needs to go up to a sample of 1:10 before a 5% reduction in the standard error can be gained from the finite population correction. A sample of 1:2 will give us a FPC that reduces the standard error by about 30%. So it is clear from this that unless one's sampling fraction is 5% or more, it is quite unnecessary to include the FPC in the calculation of standard errors or design effects from surveys. These results apply only to simple random sampling in a single stage, not to two stage clustering sampling where the results are more complicated (see below). top 9.3 FPCs in stratified sampling When a survey is designed with stratification and FPC are appropriate they are applied within each stratum. When one or more strata have high sampling fractions this can result in a considerable reduction in the standard errors of the estimates. An example would be surveys of organisations where a there are a few very large organisations. The stratum for these organisations is often given a high sampling fraction, even 100% in some cases, because they would be such major contributors to population totals. See the section on disproportionate stratification for a discussion of this. If all of the units in a stratum are selected its results will be known exactly and so it will not contribute anything to the standard errors of means or totals. If the sampling fraction is high then their contribution to standard errors will be much reduced compared to simple random sampling. top 9.4 FPC intwo-stage or multi-stahe sampling In two-stage sampling there will be different sampling fractions taken at each stage of the sampling. In this case the effect on the design effects and standard errors for surveys will be a complicated expression that involves the sampling fractions at each stage. In many cases the sampling fractions within individual clusters of a clustered sample will differ. For example if the PSUs are selected with probabilities proportional to size and then a fixed sample size is taken within each PSU, then the second stage sample fractions will be much larger in smaller PSUs. The way in which the FPC affect standard errors in this case will be a complicated expression and will depend on the extent to which the variables being analysed vary more within the PSUs or between PSUs. There are however one very common case where we don't need to worry about this. When the sampling fraction in the first stage (i.e. the sample of PSUs) is small enough to be ignored. If this is the case then the variation between the PSU means or totals will automatically incorporate any finite population correction that applies to the sub-sampling within the PSUs. Therefore if the first stage sampling fraction is small we can essentially ignore sub-sampling within the any later stages i.e. within clusters (PSUs). This is another advantage of the 'ultimate cluster' method of analysing clustered designs, see section 2.1 The only case we need to be concerned with is a two-stage design where we have sampled a substantial proportion of all the PSUs at the first stage and a substantial fraction of units at the second stage. The effect will be greatest where the variables being measured are highly variable between PSUs. In our experience this is uncommon, but a hypothetical example would be the following. A survey of primary school (P7) classes about what types of school meals they select with interest in this finite population here and now The primary sampling unit is schools, but in some schools there is more than one P7 class and to reduce work only a sample of classes is selected. The area being surveyed is divided into geographic strata and some strata have only a few schools, sometimes only a single school, so sampling fractions will be large. In these circumstances the finite population correction could make quite a big difference to the answers. To incorporate it fully you would need to use software that makes use of the full sampling design. SPSS, Stata version 9 and R allow you to do this. Overall, however, FPCs will usually be of minor importance to survey research and we have illustrated this in exemplar 4 and exemplar 5. peas project 2004/2005/2006 comprehensive training in research methods © 2022 National Centre for Research Methods About Resources Guidance Accessibility Privacy Statement Disclaimer and legal notices Connect with us National Centre for Research Methods (NCRM) Social Sciences Murray Building (Bldg 58) University of Southampton Southampton SO17 1BJ United Kingdom Contact © 2022 National Centre for Research Methods
4520	https://www.youtube.com/watch?v=1ND0sgIPcGs	Decomposition of Mercury (II) oxide : How to Balance Wayne Breslyn (Dr. B.) 887000 subscribers 206 likes Description 38092 views Posted: 18 Mar 2022 In this video we'll balance the equation HgO = Hg + O2 and provide the correct coefficients for each compound. To balance HgO = Hg + O2 you'll need to be sure to count all of atoms on each side of the chemical equation. Once you know how many of each type of atom you can only change the coefficients (the numbers in front of atoms or compounds) to balance the equation for Decomposition of Mercury (II) oxide. Important tips for balancing chemical equations: Only change the numbers in front of compounds (the coefficients). Never change the numbers after atoms (the subscripts). The number of each atom on both sides of the equation must be the same for the equation to be balanced. For a complete tutorial on balancing all types of chemical equations, watch my video: • Balancing Equations in 5 Easy Steps: • More Practice Balancing: For help with moles to grams conversions and more: • More Moles to Grams Practice: • Molar Mass in Three Easy Steps: • Understanding the Mole: • Moles - Gram Conversions: • How to Balance Chemical Equations: • Mole Ratio: • Reaction Stoichiometry: Drawing/writing done in InkScape ( Screen capture done with Camtasia Studio 4.0. Created on a Dell Dimension laptop computer with a Wacom digital tablet (Bamboo). 14 comments Transcript: let's balance the equation for the decomposition of mercury ii oxide that's hgo into mercury and oxygen gas here we'll also look at the type of reaction so let's start with the type of reaction we have hgo we have one thing it's breaking apart into two separate substances that means that this is a decomposition reaction this is breaking apart decomposing we could also say that it's a redox reaction because both the mercury and the oxygen atom they've changed their oxidation states let's balance the equation we have one mercury atom one oxygen atom in the reactants in the products we have one mercury and two oxygen atoms it looks like all we really need to do to start out is put a 2 in front of the mercury 2 oxide the hgo this 2 goes to everything here so we have 1 mercury times 2 that equals 2 and then we have 1 oxygen times two and that equals two we're almost there this mercury's by itself so if we put a two as the coefficient one times two that doesn't change anything else balances the mercury atoms we're done this equation is balanced if you wanted to write the states that would look like this this hgo mercury ii oxide this is a solid we're gonna have to heat this up when we do we'll get this liquid mercury and then we'll get oxygen gas this is dr b with the balanced equation for the decomposition of hgo mercury ii oxide we also looked at the type of reaction its decomposition and we looked at the states thanks for watching
4521	https://rbej.biomedcentral.com/articles/10.1186/1477-7827-9-87	Advertisement The emerging use of aromatase inhibitors for endometriosis treatment Reproductive Biology and Endocrinology volume 9, Article number: 87 (2011) Cite this article 16k Accesses 44 Citations 5 Altmetric Metrics details Abstract Endometriosis is defined as the growth of endometrial tissue outside of the uterine cavity. The disease occurs primarily in women of reproductive age but recurrent endometriosis is also detected in post-menopausal women. Regardless of age, endometriosis is associated with pain and reduces the quality of life for millions of women world-wide. Conventional therapies focus on reducing systemic levels of estrogen which results in cessation of endometriotic implant growth and pain symptoms associated with the disease. However, these treatments are not effective in all women and are not without side effects. Based upon the discovery that endometriotic tissue over-expresses aromatase, an enzyme critical for estrogen production, emphasis has been placed upon the use of aromatase inhibitors for the treatment of endometriosis and its associated symptoms. This article will review the rationale behind the use of aromatase inhibitors in treating endometriosis and summarize those studies which have evaluated the use of aromatase inhibitors in the treatment of endometriosis and its associated symptoms. Review Aromatase and estrogen biosynthesis Estradiol 17β (or estrogen) is the major biochemical driving force for endometriotic implant growth. In women of reproductive age, estrogen is derived primarily from the ovaries and the notion that systemic estrogen drives implant growth has long been considered dogma. However, substantial evidence also points to the endometriotic implant as an intracrine source of estrogen. This locally produced estrogen results from over-expression of P450 aromatase (referred to hence forth as aromatase) by endometriotic tissue (Figure 1). As a result, considerable emphasis has been placed upon the use of aromatase inhibitors to curtail endometriotic implant production of estrogen and subsequent implant growth. The following review highlights the discovery of endometriotic aromatase expression and the use of aromatase inhibitors in the treatment of endometriosis. Steroidogenic pathway leading to the production of estradiol. Elevated aromatase (P450 arom) expression by endometriotic implant tissue is proposed to lead to the local production of estradiol and subsequent implant growth. P450scc = side chain cleavage enzyme; P450c17 = 17 α-hydroxylase; 3β-HSD = 3β-hydroxysteroid dehydrogenase type 2; 17β-HSD-1 = 17β-hydroxysteroid dehydrogenase type 1. Aromatase expression in endometriotic tissue The first report describing expression of aromatase in peritoneal endometriotic implants was published in 1996 by Noble and colleagues . Since this initial report, numerous independent investigators have described the expression and cellular localization of aromatase transcript and protein in endometriotic tissue [2–8] as well as eutopic endometrium from women with the disease [2, 3, 5, 8–13]. The majority of these studies demonstrate that aromatase mRNA can be detected in most but not all endometriotic biopsies or eutopic endometrial biopsies from women with endometriosis; however, none of the endometrial biopsies from women without endometriosis expressed aromatase transcript. Within endometriotic implants and eutopic endometrium from women with endometriosis, aromatase transcript expression has been shown to be significantly greater in epithelial cells compared to stromal cells. Aromatase protein expression has been localized to both epithelial and stromal cells of the endometriotic implant and eutopic endometrium; however, the pattern, and relative level, of expression within each cell type is inconsistent. Epithelial cells do appear to be the major source of endometriotic/endometrial tissue aromatase protein expression. While the majority of the literature supports the elevated expression of aromatase in endometriotic tissue, a recent report by Colette and colleagues refutes the expression of aromatase in this tissue. In this study, human peritoneal, ovarian and rectovaginal endometriotic implants as well as matched eutopic endometrial biopsies were evaluated for aromatase protein and mRNA expression. In contrast to previous data, the findings from this study suggested that aromatase protein is not expressed in endometriotic tissue or in eutopic endometrium from women with the disease and only low but discernible levels of aromatase transcript were detected in ovarian endometriomas. The authors also raise the possibilities that aromatase transcript expression in ovarian endometriomas may be due to "contaminating" ovarian tissue and that elevated aromatase induction of estrogen production may result from local pelvic cavity tissues such as the peritoneum or adipose. While this explanation seems plausible for the discrepancy between the study by Colette and colleagues compared to previous studies evaluating aromatase expression in endometriotic or endometrial tissue, a more recent in vitro study supports the notion that aromatase is indeed expressed in endometriotic and endometrial cells from women with endometriosis. Using isolated stromal cells from endometriotic chocolate cysts and endometrial biopsies, Izawa and colleagues demonstrated that endometriotic stromal cells secrete estrogen and that this secretion could be increased by addition of testosterone to the media. Further, increased expression of aromatase transcript was confirmed in the endometriotic cell cultures and that this expression may be associated with epigenetic modifications of the aromatase gene. Molecular alterations leading to aberrant aromatase production by endometriotic stromal cells were first reported by Zeitoun and colleagues . Using isolated stromal cells from endometriotic and eutopic endometrial tissue, these investigators demonstrated that the stimulatory transcription factor, SF-1, was over-expressed in endometriotic stromal cells compared to stromal cells from eutopic endometrium. Further, expression of a transcription inhibitory factor, COUP-TF was not expressed in endometriotic stromal cells but was expressed in stromal cells from eutopic endometrium. As both of these factors compete for the same cis-acting element, it was proposed that over-expression of SF-1 coupled with the absence of COUP-TF in endometriotic stromal cells leads to expression of aromatase and local estrogen production by endometriotic stromal cells. In summary, the general consensus is that aromatase transcript and protein are elevated in endometriotic implants with stromal cells being the major source of the aberrant expression. Further, there is a great deal of variability in the expression of aromatase protein in the eutopic endometrium from women with the disease. The finding that endometriotic implant aromatase expression is elevated supports the notion of autonomous estrogen production by this tissue. These findings in turn have lead to the use of aromatase inhibitors in the medical management of endometriosis. Aromatase inhibitors in the medical management of endometriosis Anastrozole Anastrozole is a non-steroidal, competitive aromatase inhibitor which mimics normal enzyme substrate and competes for binding sites on endogenous aromatase enzyme . Marketed under the trade name Arimidex (AstraZeneca), anastrozol has been primary used for treatment of breast cancer after surgery and for metastases in postmenopausal women. As aromatase has been proposed to play a role in the pathogenesis of endometriosis, its use as a means to treat the disease has emerged. The first report describing the use of an aromatase inhibitor in the treatment of endometriosis was by Takayama and colleagues in 1998. In that study, a 57-year-old woman who presented with recurrent severe endometriosis after hysterectomy and bilateral salpingo-oophorectomy was administered oral anastrozole for 9 months. Anastrozole administration resulted in a significant reduction in pelvic pain as well as lesion size in this patient suggesting that aromatase inhibitors may be successful candidate drugs in the treatment of endometriosis. Additional studies evaluating the efficacy of anastrozole in treatment of endometriosis associated pain have been reported; their findings are discussed below and are summarized in Table 1. In 2004, Shippen and West reported treatment of two pre-menopausal women whom had sought surgical intervention for severe endometriosis and pain. Both patients were treated with anastrozole combined with progesterone, calcitriol and rofecoxib for six repeated 28-day cycles. Treatment resulted in a rapid, progressive elimination of symptoms over 3 months with the maintenance of remission of symptoms for over a year after treatment in both cases. Treatment was well-tolerated with no reports of adverse effects. Moreover, pregnancy was achieved after a year in both cases while in one case, a follow-up laparoscopy at 15 months after treatment confirmed the absence of disease. A larger assessment of the effect of anastrozole in combination with oral contraceptives was conducted by Amsterdam and co-workers in which fifteen premenopausal patients with documented refractory endometriosis and chronic pelvic pain were treated. Women were administered anastrozole (1 mg) combined with ethinyl estradiol (20 μg) and levonorgestrel (0.1 mg) daily for 6 months. Pelvic pain was assessed after each month while side effects, blood counts and bone density were monitored. Significant reduction in pelvic pain scores were noted in 14 of 15 patients and occurred as early as one month after treatment initiation. Treatment side effects were mild and improved over time, but serum estrogen levels were suppressed during treatment. Although the authors did not provide follow-up assessment to determine if and when symptoms returned after completion of therapy, these results do suggest that anastrozole in combination with oral contraceptives may be an effective therapy for premenopausal women with endometriosis which is refractory towards conventional therapies and who do not wish to conceive. The efficacy of anastrozole in combination with GnRH analogues in treating endometriosis-associated pain has also been conducted . Using a prospective, randomized design, eighty patients were randomized to receive either anastrozole (1 mg/day) plus 3.6 mg of goserelin or placebo plus goserelin every 4 weeks for 24 weeks. Patients were evaluated for pain at the end of the 24 week treatment period and then again at 6, 12, 18 and 24 weeks after the end of medical treatment. Compared to goserelin alone, co-treatment with anastrozole significantly increased the pain-free interval (over 2.4 months compared to 1.7 months) and decreased symptom recurrence rates (7.5% recurrence vs. 35% recurrence). Anastrozole in combination with goserelin suppressed serum estrogen levels as well as significantly reduced greater lumbar spine bone mineral density at the end of the treatment period compare to the goserelin-only group. However, this was not associated with a reduction in the menopausal quality of life, and the bone mineral density lose, although statistically different, was not determined to be of concern. The use of anastrozole alone has been reported for the treatment of rectovaginal endometriosis . Ten premenopausal women were treated with vaginally administered anastrozole (0.25 mg in a 2 gram vaginal suppository) for 6 months. During active treatment, pain was assessed daily while quality of life was assessed monthly and these assessments continued for 1 month after cessation of treatment to provide post-treatment values. Vaginally administered anastrozole had no effect on pelvic pain, dyspareunia, number of sexual contacts, duration and intensity of menstruation during or after treatment, but dysmenorrhea did improve in all but one patient. Anastrozole administration did result in improvement of quality of life, particularly with respect to physical and social functioning. Also of interest was the finding that serum estrogen levels were not affected by anastrozole administration, which, as pointed out by the authors, may be one reason for the lack of therapeutic efficacy of anastrozole in this study. An additional, more recent study which evaluated the use of anastrozole alone was conducted by Verma and Konje . Three pre-menopausal patients with refractory endometriosis and chronic pelvic pain were treated with anastrozole for a 6-month period (while one was treated with letrozole) after which pelvic pain, serum estradiol, FSH and LH were assessed in addition to bone density changes. In all patients, anastrozole (as well as letrozole) treatment was associated with a significant reduction in pelvic pain and this occurred independent of changes in systemic estradiol, LH or FSH levels. Bone density scores did not differ after treatment. The major side effect was irregular bleeding. These findings suggest that anastrozole (and letrozole) is beneficial in premenopausal patients with chronic pelvic pain with minimal side effects and that these benefits were not associated with changes in circulating estrogen levels. Letrozole Like anastrozole, letrozole (trade name Femera [Novartis]) is also a non-steroidal, competitive aromatase inhibitor which has been used primarily for treatment of breast cancer . Letrozole has been used alone or in combination with steroid analogs to treat endometriosis (summarized in Table 2). The first description of letrozole treatment in patients with endometriosis was reported in 2004. In that study , 10 premenopausal patients with endometriosis were subjected to an open-label, non-randomized design and were administered letrozole (2.5 mg daily) in combination with the progestin and norethindrone acetate (2.5 mg) for 6 months. Pelvic pain, endometriosis stage, bone density and serum estrogen, LH and FSH were evaluated before and after treatment administration. Administration of letrozole resulted in a significant reduction in pelvic pain and stage of endometriosis and was not associated with changes in serum estrogen levels or bone density. Ferrero and colleagues conducted a prospective, open-label, non-randomized trial consisting of 82 women with pain symptoms caused by rectovaginal endometriosis. Subjects were administered either letrozole plus norethisterone actetate or norethisterone acetate alone for 6 months. Changes in pain symptoms as well as side effects were monitored during the course of treatment and then again at a 12 month follow-up. At 3 months of treatment, chronic pelvic pain and deep dyspareunia were significantly decreased in both treatment groups compared to baseline values, while at 6 months, patients treated with both letrozole and norethisterone reported significantly less chronic pelvic pain and deep dyspareunia. However, at 6 months post-treatment, pain symptoms recurred in both groups. Study subjects also reported lower patient satisfaction with treatment and a higher percentage of adverse effects (such as weight gain, joint pain, migranes and spotting) in the letrozole group. Patients administered letrozole in combination with norethisterone acetate experienced a greater reduction in pain during active treatment, however, the combination therapy was associated with a higher cost and incidence of unwanted side effects and lower patient satisfaction. Based upon these findings and considering the higher cost of the combination regimen, the authors concluded that aromatase inhibitors should only be administered to patients who previously failed to respond to conventional therapies (such as progestins and/or oral contraceptives) and elect not to have surgical removal of the disease. Accordingly, these benefits and drawbacks should be taken into account when discussing combination letrozole and norethisterone acetate therapy for women with endometriosis-associated pain. In 2010, 3 additional small scale studies were conducted evaluating the efficacy of letrozole in the treatment of bladder endometriosis , colo-rectal endometriosis and recurrent ovarian endometriomas . Two premenopausal patients with bladder endometriosis were treated with letrozole (2.5 mg/day) and norethisterone actetate (2.5 mg/day) for 6 months. Pain and urinary symptoms were markedly improved in both patients. After interruption of the 6 month treatment period, one patient developed myalgia and severe arthralgia and pain and urinary symptoms returned a few months later while the other patient reported no adverse effects. In a separate study , 6 women with colo-rectal endometriosis who reported pain and intestinal symptoms were treated with letrozole (2.5 mg/day) and norethisterone acetate (2.5 mg/day) for 6 months. General pain, non-menstrual pelvic pain, deep dyspareunia, dyschezia, intestinal cramping and bloating were improved in all patients, and 4 of the 6 patients indicated that the treatment improved their gastrointestinal symptoms. Letrozole (2.5 mg/day) in combination with desogestrel (0.15 mg) and ethinyl estradiol (0.03 mg) has been reported to significantly improve pelvic pain in women with ovarian endometriomas . Treatment of these patients for 6 months with the combination therapy also induced a complete disappearance of the endometriomas but no change in bone density. Case reports have also suggested that letrozole either alone or in combination with steroids [29–31] is effective in treatment of endometriosis-associated pelvic pain. Razzi and colleagues reported that a 31 year old woman with recurrent endometriosis after subtotal hysterectomy with bilateral oophorectomy who was administered letrozole alone (2.5 mg) showed significant reduction in pelvic pain and dyspareunia during the 6 months of treatment with no adverse impact on bone density. Similar positive outcomes using letrozole in combination with progestins were noted in a post-menopausal patient with recurrent endometrioma and in a middle-aged woman with endometriosis and severe pelvic pain after hysterectomy and bilateral salpingo-oophorectomy . More recently, Sasson and Taylor reported the case of a post-menopausal woman with a large, recurrent abdominal wall endometrioma who was successfully treated with letrozole (5 mg) and medroxyprogesterone acetate. While the majority of the literature suggests that letrozole may be effective in treating patients with endometriosis, not all clinical trials using letrozole were successful. An open-label prospective study in 2007 reported that letrozole (2.5 mg/day) combined with norethisterone acetate (2.5 mg/day) quickly reduced the intensity of symptoms related to the presence of rectovaginal endometriosis. However, pain recurred at 3-month follow-up. Five of the initial 12 women treated, underwent surgery during the follow-up, and histological examination of rectovaginal nodules revealed the presence of active endometriotic lesions. In another open-label prospective study , 12 women with stage IV refractory endometriosis were treated with letrozole (2.5 mg) and desogestrel (75 μg). Unfortunately, all 12 patients developed ovarian cysts and were unable to complete the scheduled six-month treatment regime. However, at the interruption of treatment, all women reported significant improvements in dysmenorrhea and dyspareunia, but pain symptoms quickly recurred at 3-month follow up. Aromatase inhibitors and fertility restoration in women with endometriosis The majority of studies and trials evaluating the use of aromatase inhibitors in women with endometriosis have focused on relief of pelvic pain and/or effect on endometriotic implant size/severity of disease. However, one recent pilot study conducted by Lossl and colleagues concurrently evaluated the effect of anastrazole (coupled with goserelin) on endometriomal volume, CA125 levels and standard IVF (in vitro fertilization) fertility parameters. Twenty women with endometriosis undergoing IVF/ICSI were treated with goserelin (3.6 mg s.c.) on treatment days 1, 28 and 56 and anastrazole (1 mg daily) from days 1 to 69 of the study, with ovarian stimulation occurring on day 70. Both endometriomal volume and CA125 levels (as a marker of endometrioma activity) decreased by 29% and 61%, respectively during the combined down-regulation. The average number of oocytes retrieved was 7.5 and the fertilization rate was 78%. Nine of the twenty patients conceived while five of these had clinical pregnancies with three of them delivering healthy offspring. This study demonstrated that combined anastrazole and goserelin down-regulation markedly reduces endometriomal volume and disease activity which is compatible with pregnancy and delivery. Concern was expressed with the rate of pregnancy loss (from 5 chemical pregnancies to 3 patients delivering) but it should be emphasized that larger, controlled studies must be conducted to further evaluate this initial observation. This contention is supported by a recent report in which 159 infertile women undergoing controlled ovarian stimulation and artificial insemination treated with the aromatase inhibitor, letrozole in combination with FSH, exhibited comparable pregnancy rates with less cancelled cycles and less FSH required for stimulation compared to FSH-treated patients alone. Use of aromatase inhibitors in animal models of endometriosis In addition to human trials, animal studies have also evaluated aromatase inhibitors in the treatment of endometriosis and support the notion that inhibition of endometriotic implant aromatase activity leads to reduction in endometriotic implant mass. Aromatase has been shown to be expressed in endometriotic implants of mice with experimentally-induced endometriosis . Mice with experimentally-induced endometriosis in which the expression of aromatase has been genetically disrupted exhibit significantly smaller endometriotic implants validating the use of such animal models. In a side by side comparison , both anastrozole and letrozole significantly decreased endometriotic implant size in mice with experimentally-induced endometriosis. Both aromatase inhibitors decreased cell proliferation, but the impact on cell apoptosis was time-dependent with respect to time of intervention after disease establishment. Lastly, both anastrozole and letrozole decreased peritoneal fluid vascular endothelial cell growth factor levels while only anastrozole decreased peritoneal fluid PGE levels. The effect of anastrozole on experimental endometriosis in rats has also been evaluated. Alintas and colleagues conducted a randomized, placebo-controlled, single-blind study comparing the effect of anastrozole and raloxifene on endometriotic implant volume (size) after 8 weeks of treatment. The authors found that, compared to placebo, both anastrozole and raloxifene were found to be equally effective in reducing endometriotic implant volume. No difference was detected in implant size between the two drugs. The efficacy of letrozole in endometriosis treatment has also been examined in rats with surgically-induced endometriosis . In this study by Oner and colleagues, rats with surgically-induced endometriosis were randomly assigned to receive either one of two doses of metformin (100 or 200 mg/kg/day), letrozole (0.1 mg/kg/day) or placebo for 4 weeks after which endometriotic implant size and pelvic adhesions were determined. Metformin (both doses) and letrozole caused a significant reduction in endometriotic implant size. While letrozole did not affect pelvic adhesion, metformin reduced the severity of pelvic adhesions irrespective of dose. Collectively, these studies demonstrate that letrozole and anastrozole are effective in reducing endometriotic implant size in rodents with experimentally-induced endometriosis. The use of these and other genetically modified mice (tissue-specific aromatase knockout or over-expression) coupled with the use of aromatase inhibitors should allow for a thorough dissection of the mechanisms by which these inhibitors work to induced endometriotic implant regression. Conclusions The majority of the evidence in the literature supports the notion that aromatase expression is elevated in human endometriotic tissue, and this concept is further supported in studies using animal models for the disease. Based upon this expression, it has been proposed that elevated aromatase activity in endometriotic tissue leads to local estrogen production and endometriotic lesion growth which is associated with disease symptoms such as pelvic pain. Not surprising, reduction of aromatase activity by aromatase inhibitors has been associated with reduced endometriotic lesion size in both animal models and in human studies. In human subjects, aromatase alone, or in combination with steroids, appears to be effective in reduction of endometriosis-associated symptoms such as pelvic pain. Aromatase inhibitors appear to be most beneficial in the treatment of endometriosis in women with recurrent endometriosis who have not had success with more conventional treatment regimes such as gonadtropin releasing agonists/antagonists or steroidal analogues. However, one must keep in mind that aromatase inhibitors exhibit suboptimal tolerability and greater costs compared to some of the more conventional therapies. Clearly, aromatase inhibitor therapy may have a place in endometriosis treatment of a subset of patients suffering from the disease and benefits and limitations of these compounds must be discussed with patients. Future effort should be directed towards performing larger, multi-center trials with aromatase inhibitors to provide a more robust assessment of the efficacy of these compounds in the treatment of endometriosis and its associated symptoms. References Noble LS, Simpson ER, Johns A, Bulun SE: Aromatase expression in endometriosis. J Clin Endocrinol Metab. 1996, 81: 174-179. 10.1210/jc.81.1.174. CAS PubMed Google Scholar Kitawaki J, Noguchi T, Amatsu T, Maeda K, Tsukamoto K, Yamamoto T, Fushiki S, Osawa Y, Honjo H: Expression of aromatase cytochrome P450 protein and messenger ribonucleic acid in human endometriotic and adenomyotic tissues but not in normal endometrium. Biol Reprod. 1997, 57: 514-519. 10.1095/biolreprod57.3.514. Article CAS PubMed Google Scholar Matsuzaki S, Canis M, Pouly JL, Déchelotte PJ, Mage G: Analysis of aromatase and 17 beta-hydroxysteroid dehydrogenase type 2 messenger ribonucleic acid expression in deep endometriosis and eutopic endometrium using laser capture microdissection. Fertil Steril. 2006, 85: 308-313. 10.1016/j.fertnstert.2005.08.017. Article CAS PubMed Google Scholar Heilier JF, Donnez O, Van Kerckhove V, Lison D, Donnez J: Expression of aromatase (P450 aromatase/CYP19) in peritoneal and ovarian endometriotic tissues and deep endometriotic (adenomyotic) nodules of the rectovaginal septum. Fertil Steril. 2006, 85: 1516-1518. 10.1016/j.fertnstert.2005.10.041. Article CAS PubMed Google Scholar Velasco I, Rueda J, Acién P: Aromatase expression in endometriotic tissue and cell cultures of patients with endometriosis. Human Reprod. 2006, 12: 377-381. 10.1093/molehr/gal041. Article CAS Google Scholar Smuc T, Pucelj MR, Sinkovec J, Husen B, Thole H, Lanisnik Rizner T: Expression analysis of the genes involved in estradiol and progesterone action in human ovarian endometriosis. Gynecol Endocrinol. 2007, 23: 105-111. 10.1080/09513590601152219. Article CAS PubMed Google Scholar Acién P, Velasco I, Gutiérrez M, Martin-Beltrán M: Aromatase expression in endometriotic tissues and its relationship to clinical and analytical findings. Fertil Steril. 2007, 88: 32-38. 10.1016/j.fertnstert.2006.11.188. Article PubMed Google Scholar Hudelist G, Czerwenka K, Keckstein J, Haas C, Fink-Retter A, Gschwantler-Kaulich D, Kubista E, Singer CF: Expression of aromatase and estrogen sulfotransferase in eutopic and ectopic endometrium: evidence for unbalanced estradiol production in endometriosis. Reprod Sci. 2007, 14: 798-805. 10.1177/1933719107309120. Article CAS PubMed Google Scholar Kitawaki J, Kusuki I, Koshiba H, Tsukamoto K, Honjo H: Expression of aromatase cytochrome P450 in eutopic endometrium and its application as a diagnostic test for endometriosis. Gynecol Obstet Invest. 1999, 48 (Suppl 1): 21-28. Article CAS PubMed Google Scholar Kitawaki J, Kusuki I, Koshiba H, Tsukamoto K, Fushiki S, Honjo H: Detection of aromatase cytochrome P-450 in endometrial biopsy specimens as a diagnostic test for endometriosis. Fertil Steril. 1999, 72: 1100-1106. 10.1016/S0015-0282(99)00424-0. Article CAS PubMed Google Scholar Tarkowski R, Skrzypczak M, Winiarczyk S, Kotarski J, Jakowicki JA, Jakimiuk AJ: Aromatase (P450AROM) mRNA expression in normal, hyperplastic and malignant endometrium and aromatase activity in endometrial cancer tissue culture. Ginekol Pol. 2000, 71: 130-135. CAS PubMed Google Scholar Dheenadayalu K, Mak I, Gordts S, Campo R, Higham J, Puttemans P, White J, Christian M, Fusi L, Brosens J: Aromatase P450 messenger RNA expression in eutopic endometrium is not a specific marker for endometriosis. Fertil Steril. 2002, 78: 825-829. 10.1016/S0015-0282(02)03324-1. Article PubMed Google Scholar Kyama CM, Overbergh L, Mihalyi A, Meuleman C, Mwenda JM, Mathieu C, D'Hooghe TM: Endometrial and peritoneal expression of aromatase, cytokines, and adhesion factors in women with endometriosis. Fertil Steril. 2008, 89: 301-310. 10.1016/j.fertnstert.2007.02.057. Article CAS PubMed Google Scholar Colette S, Lousse JC, Defrère S, Curaba M, Heilier JF, Van Langendonckt A, Mestdagt M, Foidart JM, Loumaye E, Donnez J: Absence of aromatase protein and mRNA expression in endometriosis. Hum Reprod. 2009, 24: 2133-2141. 10.1093/humrep/dep199. Article CAS PubMed Google Scholar Izawa M, Harada T, Taniguchi F, Ohama Y, Takenaka Y, Terakawa N: An epigenetic disorder may cause aberrant expression of aromatase gene in endometriotic stromal cells. Fertil Steril. 2008, 89 (Suppl 5): 1390-1396. Article CAS PubMed Google Scholar Zeitoun K, Takayama K, Michael MD, Bulun SE: Stimulation of aromatase P450 promoter (II) activity in endometriosis and its inhibition in endometrium are regulated by competitive binding of SF-1 and COUP-TF to the same cis-acting element. Mol Endo. 1999, 13: 239-253. 10.1210/me.13.2.239. Article CAS Google Scholar Buzdar AU: Pharmacology and pharmacokinetics of the newer generation aromatase inhibitors. Clin Cancer Res. 2003, 9: 468S-472S. CAS PubMed Google Scholar Takayama K, Zeitoun K, Gunby RT, Sasano H, Carr BR, Bulun SE: Treatment of severe postmenopausal endometriosis with an aromatase inhibitor. Fertil Steril. 1998, 69: 709-13. 10.1016/S0015-0282(98)00022-3. Article CAS PubMed Google Scholar Shippen ER, West WJ: Successful treatment of severe endometriosis in two premenopausal women with an aromatase inhibitor. Fertil Steril. 2004, 81: 1395-1398. 10.1016/j.fertnstert.2003.11.027. Article PubMed Google Scholar Amsterdam LL, Gentry W, Jobanputra S, Wolf M, Rubin SD, Bulun SE: Anastrozole and oral contraceptives: a novel treatment for endometriosis. Fertil Steril. 2005, 84: 300-304. 10.1016/j.fertnstert.2005.02.018. Article CAS PubMed Google Scholar Soysal S, Soysal ME, Ozer S, Gul N, Gezgin T: The effects of post-surgical administration of goserelin plus anastrozole compared to goserelin alone in patients with severe endometriosis: a prospective randomized trial. Hum Reprod. 2004, 19: 160-167. 10.1093/humrep/deh035. Article PubMed Google Scholar Hefler LA, Grimm C, van Trotsenburg M, Nagele F: Role of the vaginally administered aromatase inhibitor anastrozole in women with rectovaginal endometriosis: a pilot study. Fertil Steril. 2005, 84: 1033-1036. 10.1016/j.fertnstert.2005.04.059. Article CAS PubMed Google Scholar Verma A, Konje JC: Successful treatment of refractory endometriosis-related chronic pelvic pain with aromatase inhibitors in premenopausal patients. Eur J Obstet Gynecol Reprod Biol. 2009, 143: 112-115. 10.1016/j.ejogrb.2008.12.002. Article CAS PubMed Google Scholar Ailawadi RK, Jobanputra S, Kataria M, Gurates B, Bulun SE: Treatment of endometriosis and chronic pelvic pain with letrozole and norethindrone acetate: a pilot study. Fertil Steril. 2004, 81: 290-296. 10.1016/j.fertnstert.2003.09.029. Article CAS PubMed Google Scholar Ferrero S, Camerini G, Seracchioli R, Ragni N, Venturini PL, Remorgida V: Letrozole combined with norethisterone acetate compared with norethisterone acetate alone in the treatment of pain symptoms cause by endometriosis. Hum Reprod. 2009, 24: 3033-3041. 10.1093/humrep/dep302. Article CAS PubMed Google Scholar Ferrero S, Biscaldi E, Luigi Venturini P, Remorgida V: Aromatase inhibitors in the treatment of bladder endometriosis. Gynecol Endocrinol. Ferrero S, Camerini G, Ragni N, Venturini PL, Biscaldi E, Seracchioli R, Remorgida V: Letrozole and norethisterone acetate in colorectal endometriosis. Eur J Obstet Gynecol Reprod Biol. 2010, 150: 199-202. 10.1016/j.ejogrb.2010.02.023. Article CAS PubMed Google Scholar Lall SS, Kamilya G, Mukherji J, De A, Ghosh D, Majhi AK: Aromatase inhibitors in recurrent ovarian endometriomas: report of five cases with literature review. Fertil Steril. 2010, 95: 291e15-e18. Google Scholar Razzi S, Fava A, Sartini A, De Simone S, Cobellis L, Petraglia F: Treatment of severe recurrent endometriosis with an aromatase inhibitor in a young ovariectomised woman. BJOG. 2004, 111: 182-184. 10.1046/j.1471-0528.2003.00038.x. Article PubMed Google Scholar Fatemi HM, Al-Turki HA, Papanikolaou EG, Kosmas L, De Sutter P, Devroey P: Successful treatment of an aggressive recurrent post-menopausal endometriosis with an aromatase inhibitor. Reprod Biomed Online. 2005, 11: 455-457. 10.1016/S1472-6483(10)61140-6. Article CAS PubMed Google Scholar Mousa NA, Bedaiwy MA, Casper RF: Aromatase inhibitors in the treatment of severe endometriosis. Obstet Gynecol. 2007, 109: 1421-1423. 10.1097/01.AOG.0000265807.19397.6d. Article PubMed Google Scholar Sasson IE, Taylor HS: Aromatase inhibitor for treatment of a recurrent abdominal wall endometrioma in a postmenopausal woman. Fertil Steril. 2009, 92: 1170.e1-1170.e4. 10.1016/j.fertnstert.2009.05.071. Article Google Scholar Remorgida V, Abbamonte HL, Ragni N, Fulcheri E, Ferrero S: Letrozole and norethisterone acetate in rectovaginal endometriosis. Fertil Steril. 2007, 88: 724-726. 10.1016/j.fertnstert.2006.12.027. Article CAS PubMed Google Scholar Remorgida V, Abbamonte LH, Ragni N, Fulcheri E, Ferrero S: Letrozole and desogestrel-only contraceptive pill for the treatment of stage IV endometriosis. Aust N Z J Obstet Gynaecol. 2007, 47: 222-225. 10.1111/j.1479-828X.2007.00722.x. Article PubMed Google Scholar Lossl K, Loft A, Freiesleben NLC, Bangsbφll , Andersen CY, Pedersen AT, Hartwell D, Andersen AN: Combined down-regulation by aromatase inhibitor and GnRH-agonist in IVF patients with endometriosis - A pilot study. Eur J Obstet Gynecol Reprod Biol. 2009, 144: 48-53. 10.1016/j.ejogrb.2009.02.001. Article CAS PubMed Google Scholar Bedaiwy MA, Shokry M, Mousa N, Claessens A, Esfandiari N, Botleib L, Casper R: Letrozole co-treatment in infertile women 40 years old and older receiving controlled ovarian stimulation and intrauterine insemination. Fertil Steril. 2009, 91: 2501-2507. 10.1016/j.fertnstert.2008.03.020. Article CAS PubMed Google Scholar Bilotas M, Meresman G, Stella I, Sueldo C, Barañao RI: Effect of aromatase inhibitors on ectopic endometrial growth and peritoneal environment in a mouse model of endometriosis. Fertil Steril. 2010, 93: 2513-2518. 10.1016/j.fertnstert.2009.08.058. Article CAS PubMed Google Scholar Fang Z, Yang S, Gurates B, Tamura M, Simpson E, Evans D, Bulun SE: Genetic or enzymatic disruption of aromatase inhibits the growth of ectopic uterine tissue. J Clin Endocrinol Metab. 2002, 87: 3460-3466. 10.1210/jc.87.7.3460. Article CAS PubMed Google Scholar Altintas D, Kokcu A, Kandemir B, Tosun M, Cetinkaya MB: Comparison of the effects of raloxifene and anastrozole on experimental endometriosis. Eur J Obstet Gynecol Reprod Biol. 2010, 150: 84-87. 10.1016/j.ejogrb.2010.02.004. Article CAS PubMed Google Scholar Oner G, Ozcelik B, Ozgun MT, Serin IS, Ozturk F, Basbug M: The effects of metformin and letrozole on endometriosis and comparison of the two treatment agents in a rat model. Hum Reprod. 2010, 25: 932-937. 10.1093/humrep/deq016. Article CAS PubMed Google Scholar Download references Author information Authors and Affiliations Departments of Obstetrics and Gynecology and Molecular and Integrative Physiology, University of Kansas Medical Center, 3901 Rainbow Boulevard, 66160, Kansas City, KS, USA Warren B Nothnick Search author on:PubMed Google Scholar Corresponding author Correspondence to Warren B Nothnick. Additional information Competing interests The author declares that they have no competing interests. Authors’ original submitted files for images Below are the links to the authors’ original submitted files for images. Authors’ original file for figure 1 Rights and permissions This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Reprints and permissions About this article Cite this article Nothnick, W.B. The emerging use of aromatase inhibitors for endometriosis treatment. Reprod Biol Endocrinol 9, 87 (2011). Download citation Received: 17 November 2010 Accepted: 21 June 2011 Published: 21 June 2011 DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Keywords Associated content Collection Aromatase inhibitors in reproductive biology and endocrinology Advertisement Reproductive Biology and Endocrinology ISSN: 1477-7827 Contact us Read more on our blogs Receive BMC newsletters Manage article alerts Language editing for authors Scientific editing for authors Policies Accessibility Press center Support and Contact Leave feedback Careers Follow BMC BMC Twitter page BMC Facebook page BMC Weibo page By using this website, you agree to our Terms and Conditions, Your US state privacy rights, Privacy statement and Cookies policy. Your privacy choices/Manage cookies we use in the preference centre. Follow BMC By using this website, you agree to our Terms and Conditions, Your US state privacy rights, Privacy statement and Cookies policy. Your privacy choices/Manage cookies we use in the preference centre. © 2025 BioMed Central Ltd unless otherwise stated. Part of Springer Nature.
4522	https://www.convertunits.com/from/microfarads/to/farad	Convert microfarads to farad - Conversion of Measurement Units Convert microfarad to farad [SI standard] More information from the unit converter How many microfarads in 1 farad? The answer is 1000000. We assume you are converting between microfarad and farad [SI standard]. You can view more details on each measurement unit: microfarads or farad The SI derived unit for capacitance is the farad. 1 microfarads is equal to 1.0E-6 farad. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between microfarads and farads. Type in your own numbers in the form to convert the units! Want other units? You can do the reverse unit conversion from farad to microfarads, or enter any two units below: Common capacitance conversions microfarads to gigafarad microfarads to centifarad microfarads to decifarad microfarads to kilofarad microfarads to terafarad microfarads to millifarad microfarads to nanofarad microfarads to ampere second/volt microfarads to puff microfarads to electrostatic unit Definition: Microfarad The SI prefix "micro" represents a factor of 10-6, or in exponential notation, 1E-6. So 1 microfarad = 10-6 farads. The definition of a farad is as follows: The farad (symbol F) is the SI unit of capacitance (named after Michael Faraday). A capacitor has a value of one farad when one coulomb of charge causes a potential difference of one volt across it. Its equivalent expressions in other SI units are: Since the farad is a very large unit, values of capacitors are usually expressed in microfarads (?F), nanofarads (nF), or picofarads (pF). The picofarad is comically called a "puff" in laboratory usage. Definition: Farad The farad (symbol F) is the SI unit of capacitance (named after Michael Faraday). A capacitor has a value of one farad when one coulomb of charge causes a potential difference of one volt across it. Its equivalent expressions in other SI units are: Since the farad is a very large unit, values of capacitors are usually expressed in microfarads (?F), nanofarads (nF), or picofarads (pF). The picofarad is comically called a "puff" in laboratory usage. Metric conversions and more ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! Convert · Metric · Capacitance · Dates · Salary · Chemistry · Forum · Search · Privacy · Bibliography · Contact © 2025 ConvertUnits.com
4523	https://projecteuclid.org/ebook/Download?urlid=10.3792%2Feuclid%2F9781429799850-20&isFullBook=False	Chapter 16 INVERSIONS IN CIRCLES Q: How does a geometer capture a lion in the desert? A: Build a circular cage in the desert, enter it, and lock it. Now perform an inversion with respect to the cage. Then you are outside, and the lion is locked in the cage. —A mathematical joke from before 1938 We will now study inversions in a circle ( which is the analogue of reflection in a line) and its applications. Though inversion in a circle can be defined on spheres and hyperbolic planes, it seems to have no significant applications on these surfaces. Therefore, in this chapter we will only consi der the case of the Euclidean plane. To study Chapter 16, the only results needed in Chapters 10 –15 are PROBLEM 13.4 : The AAA similarity and SAS similarity criteria for triangles on the plane . PROBLEM 15.1b : On a plane, if two lines through a point P intersect a circle at points A , A (possibly coincident ) and B , B (possibly coincident ), then \|PA \|  \|PA \| = \| PB \|  \|PB \| . If you are willing to assume these criteria for similar triangles, then you ca n work through Chapter 16 without Chapters 10 –15. EARLY HISTORY OF INVERSIONS Apollonius of Perga (c. 250 –175 B.C.) was famous in his time for work on astronomy (Navigation/Stargazing Strand), before his now well -known work on conic sections. Chapter 16 Inversions in Circles 202 Unfortunately, Apollonius’ original work on astronomy and most of his mathematical work (except for Conics , [AT : Apollonius] ) has been lost and we only know about it from a commentary by Pappus of Alexandria (290 –350 A.D.). According to Pappus, Apollonius investigated one particular family of circles and straight lines. Apollonius defined the curve: ck (A, B) is the locus of points P such that PA = k  PB , where A and B are two points in the Euclidean plane, and k is a positive constant. This curve is a straight line if k = 1 and a circle otherwise and is usually called an Apollonian circle . Apollonius proved (see Problem 16.1b ) that a circle c (with center C and radius r) belongs to the family {ck(A,B)} if and only if BC  AC = r2 and A and B are on the same ray from C. In modern terms, we use “ BC  AC = r2 and A and B are on the same ray from C ” as the definition of A and B being inversions of each other with respect to the circle c. There is indirect evidence (see [ HI : Calinger], page 181) that Apollonius used inversion in a circle to solve astronomical problems concerning celestial orbits. The theory of inversions was appar ently not carried on in a systematic way until the 19 th century, when the theory was developed purely geometrically from Euclid's Book III, but this (as far as we know) was not done in ancient times. We suggest that this was because Euclid’s Elements and A pollonius’ circles were parts of different historical strands. In Problem 16.4 we will explore a problem of Apollonius that uses inversions for its solutions. PROBLEM 16.1 INVERSIONS IN CIRCLES DEFINITIONS. An inversion with respect to a circle  is a transformation from the extended plane (the plane with , the “point at infinity,” added) to itself that takes C, the center of the circle, to  and vice versa, and that takes a point at a distance s from the ce nter to the point on the same ray (from the center) that is at a distance of r2/s from the center, where r is the radius of the circle. See Figure 16.1. We call ( P, P' ) an inversive pair because (as the reader can check) P and P' are taken to each othe r by the inversion. The circle  is called the circle of inversion . Figure 16.1 Inversion with respect to a circle P' r2/s CPChapter 16 Inversions in Circles 203 Note that an inversion takes the inside of the circle to the outside and vice versa and that the inversion takes any line through the center to itself. Because of this, inversion can be thought of as a reflection in the circle. See also part c for the clos e connection between inversions in the plane and reflections on a sphere. We strongly suggest that the reader play with inversions by using some dynamic geometry software such as Geo Gebra, for example . You may construct the image, P, of P under the invers ion through the circle  as follows (see Figure 16.2): If P is inside , then draw through P the line perpendicular to the ray CP. Let S and R be the intersections of this line with . Then P  is the intersection of the lines tangent to  at S and R. (T o construct the tangents, note that lines tangent to a circle are perpendicular to the radius of the circle.) If P is outside , then draw the two tangent lines from P to . Let S and R be the points of tangency on . Then P  is the inter - section of the line SR with CP. (The points S and R are the intersections of  with the circle with diameter CP . Why ?) Figure 16.2 Constructing inversive images a. Prove that these constructions do construct inversive pairs. The purpose of this part is to explore and better understand inversion, but it will not be directly used in the other parts of this problem. When P is inside , you need to prove that C-P-P are collinear. When P is outside , you need to prove that SR is perpendicular to CP . Next, we prove b. Apollonius’ Theorem. Define the curve c k(A, B) to be the locus of points P such that PA=k PB , where A and B are two points in the plane and k is a positive constant. A circle c (with center C and radius r ) belongs to the family {ck(A, B)} if and onl y if A and B are an inversive pair with respect to c. C P (P' ) P' (P)Chapter 16 Inversions in Circles 204 The following result demonstrates the close connection between inversion through a circle in the plane and reflections through great circles on a sphere. If you have studied Problem 14.4 (or assume it), then you can use this part c in your analysis of inve rsions in Problem 16.2 . c. Let  be a sphere tangent at its south pole to the plane  and let f:  →  be a stereographic projection from the north pole. If  is the circle that is the image under f of the equator and if g is the intrinsic (or extrinsic) reflection of the sphere through its equator (or equatorial plane), then show that the transformation f ○ g ○ f −1 is the inversion of the plane with respect to the circle . See Figure 16.3 . Imagine a sphere tangent to a plane at its south pole, S. Now use stereographic projection to project the sphere from the north pole, N, onto the plane; see Problem 14.4 . Stereographic projection was known already to Hipparchus (Greek, second century B.C.). Show that the triangle ∆SNP is si milar to ∆RNS , which is congruent to ∆QSN , which is similar to ∆SP’N . Figure 16.3 Stereographic projection and inversion PROBLEM 16.2 INVERSIONS PRESERVE ANGLES AND PRESERVE CIRCLES (AND LINES) Part c provides another route to prove that inversions are conformal (preserves angles, see Problem 14.4 ). In Problem 14.4 you showed that f, stereographic projection, is conformal. In addition, g (being an isometry) is conformal. Thus, the inversion f ○ g ○ f−1 is conformal. a. Show that an inversion takes each circle orthogonal to the circle of inversion to itself. See Figure 16.4. Two circles are orthogonal if, at each point of intersection, the angle between the tangent lines is 90°. (Note that, at these points , the radius of one circle is tangent to the other circle.) Chapter 16 Inversions in Circles 205 Figure 16.4 A circle orthogonal to inverts to itself b. Show that an inversion takes a circle through the center of inversion to a line not through the center, and vice versa. What happens in the special cases when either the circle or the straight line intersects the circle of inversion ? Figure 16.5 Circles through the center invert to lines Look at Figure 16.5 (where CP is a diameter of the circle) and prove that ∆CPQ and ∆CQ P are similar triangles. Note that the line is parallel to the line tangent to the circle at C. c. An inversion takes circles not through the center of inversion to circles not through the center. Note: The (circumference of a) circle inverts to another circle but the centers of these circles are on the same ray from C though not an inversive pair. Loo k at Figure 16.6, where PQ is a diameter of the circle. If P, Q, X invert to P, Q, X, then show that  P X Q =  P X Q = right angle by looking for similar triangles. Thus, argue that as X varies around the circle with diameter PQ , then X varies around the circle with diameter QP. P 'Chapter 16 Inversions in Circles 206 Figure 16.6 Circles invert to circles d. Inversions are conformal. Look at two lines that intersect and form an angle at P. Look at the images of these lines. Inversions were used in the 19 th century to solve a long -standing engineering problem (from the Motion/Machines Strand) that is the subject of Problem 16.3 . Other applications (that started with Apollonius ) are discussed in Problem 16.4 . More history and further expansions of the notion o f inversion are contained in the last section. PROBLEM 16.3 USING INVERSIONS TO DRAW STRAIGHT LINES At the beginning of Chapter 1, there is a brief history of attempts to find linkages that would draw a straight line. In this problem we explore the mathem atics behind this linkage. a. Show that for the linkage in Figure 16.7 the points P and Q are the inversions of each other through the circle of inversion with center at C and radius 𝑟 = √𝑠 2 − 𝑑 2. Draw the circle with center R and radius d and note that C, P, Q are collinear. Figure 16.7 A linkage for constructing an inversion Chapter 16 Inversions in Circles 207 b. Show that the point Q in the linkage in Figure 16.8 always traces a straight line. Figure 16.8 Linkage for drawing a straight line If we modify the Peaucellier -Lipkin linkage (see Figure 1.5) by changing the distance between the anchor points, then c. The point Q in the linkage in Figure 16.9 always traces the arc of a circle. Why? Show that the radius of the circle is expressed by r 2f / ( g2 − f 2), where r is as in part a. Figure 16.9 Peaucellier -Lipkin linkage modified to draw the arc of a circle PROBLEM 16.4 APOLLONIUS’ PROBLEM In Book IV of the Elements , Euclid shows how to construct the circle that passes through three given (non -collinear) points and also how to construct a circle tangent to three given straight lines (not passing the same point). Apollonius of Perga (c. 250 –175 B.C.) generalized this to Apollonius’ Problem: Given three objects, each of which may be a point, a line, or a circle, construct a circle that passes through each of the given points and is tangent to the given lines and circles. Solutions to this problem are discussed in Apollonius' On Tangencies (De Tactionibus ). Unfortunately, Apollonius’ work has not survived, but it has been “reconstructed” from both Arabic and Greek commentaries especially through the description of its contents from Pappus of Alexandria (ca. A.D. 300 ). In Book 7 of the Mathematical Collection ,Chapter 16 Inversions in Circles 208 Pappus described the contents of various works by Apollonius. Pappus presents a list of problems in Apollonius’ lost work, and on the basis of this information the work has been reconstructed at least four times . Francois Viete (1540 –1603) restored Apollonius De tactionibus and published it under the title Apollonius Gallus in 1600. Frans van Schooten , in a 1657 reconstruction, showed that Apollonius' problem can be solved by the algebraic methods of Descartes' Geometrie (1637). Joachim Jungius and Woldeck Weland (1622 –1641), in a reconstruction titled Apollonius Saxonicus , used a purely geometrical m ethod that they called “metagoge,” that is, the reduction of the general case of a problem to a special case of the same problem or a simpler problem (such as in Problem 16.4c below). Another reconstruction was completed by Robert Simson (1749). The author s of these reconstructions were doing mathematics, not the history of mathematics, as can be inferred from the fact that the “reconstructions” differ from each other and sometimes deal with generalizations of the problems that had actually been treated by Apollonius. In addition to the attempted reconstructions, there were many and varied solutions of the Apollonius problem produced by later mathematicians, including Isaac Newton (1643 –1727), A. van Roomen (1561 –1615), J. Casey (1820 –1881), R. Descartes, P. Fermat, Princess Elizabeth (1596 –1662), L. Euler (1707 –1783), N. Fuss (1755 –1826), L. N. M. Carnot (1753 –1823), J. D. Gergonne (1771 –1859), C. F. Gauss (1777 –1855), J. V. Poncelet (1788 –1867), A. L. Cauchy (1789 –1857), and Eduard Study (1862 –1930). You sh ould recognize some of these names. Poncelet and Cauchy solved Apollonius’ problem while first year students at the École Polytechnique. In 1679 P. Fermat formulated and solved an extension of Apollonius problem to 3 - space: Construct a sphere tangent to fo ur given spheres. This is called Fermat’s problem by some later authors. A large number of mathematicians were discussing this problem in 19 th century. Some further generalizations of Apollonius’ problem are discussed in H. S. M. Coxeter, "The Problem of A pollonius," American Mathematical Monthly 75 (1968), pp. 5 –15. In 2003, R. H. Lewis and S. Bridgett, in a paper entitled “Conic tangency equations and Apollonius problems in biochemistry and pharmacology” (Mathematics and Computers in Simulation , vol. 61; Jan. 2003, pp. 101 –114), discuss current applications of Apollonius’ problems. The applications involve bonding interactions in human bodies between protein molecules and hormone, drug, and other molecules. Apollonius’ problem can be discussed in 10 possi ble cases (letting P= point, L = line, C = circle): PPP, PPL, PPC, PLL, PLC, PCC, LLL, LLC, LCC, CCC. a. Solve the case PPP . Show that the solution implies that the perpendicular bisectors of the sides of a triangle all pass through the same point . We call this point the circumcenter of the triangle. See Figure 16.10. Chapter 16 Inversions in Circles 209 Figure 16.10 PPP Hint: If the three points are on a line, then that line (as a circle with infinite radius) is the solution. Otherwise the three points determine a triangle. b. Solve the case LLL . Show that the solution implies that the angle bisectors of a triangle pass all through the same point . This point is called the incenter of the triangle. Hint: If the three lines intersect in the same point, then there is no solution. If the three lines form a triangle, then there are four solutions. See Figure 16.11. What happens in other cases? Figure 16.11 LLL c. Solve the cases PPL and PPC . Outline of solution: If both points are on the given line or circle, the only solution is the given line or circle. If point A is not on the given line (or circle), then we can take A as the center of an inversion. Denote the other point B and the given li ne or circle . Under the inversion,  goes to another circle  (never a line, Why? ) and B goes to another point B. Let l be a line through B that is tangent to . Inverting this tangent l back to the original picture, we will have a solution. ( Why? ) I f B is on , then there exists one solution; if neither A or B are on , then there are two solutions. Chapter 16 Inversions in Circles 210 d. Solve the cases PLL , PLC , and PCC . Hint: Choose the given point as the center of an inversion. After the inversion the solution will be a line (in order to contain the inversive image of the point). See Figure 16.12. Depending on the original location of the given point, we get the following subcases: The point is on neither of the lines (line and circle, circles). Then those circles (line and circle , lines) in inversion will go to two circles and the problem reduces to constructing a tangent to two given circles. In this subcase there are either 0, 1, 2, 3, or 4 solutions possible. (Why?) The point is the point of tangency of circles (or circle and l ine). The point is the point of intersection of circles (or circle and line). The point is on one circle (or line) but not on the other. Figure 16.12 Case PCC : invert through and then find tangent to C’ and D’ e. Application of PCC. There is a story that in World War I troops used PCC to pinpoint the location of large enemy guns . The three separate observation points synchronize their clocks to the second and then note the exact second that they hear the gun's sound. How could these t imings be used to pinpoint the location of the gun? Hint: The speed of sound is approximately 340 meters/second (it varies 5% in somewhat predictable ways with temperature and atmospheric conditions). Draw a picture of the instant in time when the sound reaches the first observation point. f. Solve cases LLC , LCC , and CCC . Outline of solution: Each of these three cases can be reduced to either CCC or PCC by using an appropriate inversion. ( Do you see how? ) There are many subcases depending on how th e circles and lines relate to each other: inside, outside, intersecting. However, the overall strategy is to reduce these cases to PLC , PLL , or PCC in part d. We illustrate this with the subcase of LCC , where all circles and lines are disjoint, and the two circles are on the same side of the line and exterior to each other. See Figure 16.13. Chapter 16 Inversions in Circles 211 Figure 16.13 Reducing LCC to PLC Let r be the radius of the smaller circle, C. Shrink the other circle, D, to the circle, D, with the same center but with radius reduced by r. Construct another line parallel to L, on the side opposite to C and D, and at a distance of r. Now apply PLC to the point P, the circle D, and the line L, to get a circle E that is tangent to P, D, a nd L. The required solution to the original LCC is the circle E with the same center as E but with radius decreased by r. See Figure 16.13. EXPANSIONS OF THE NOTION OF INVERSIONS There was a rebirth of interest in inversions in the 19 th century. Jakob Steiner (1796 –1863) was among the first to start extensively using the technique of inversions in circles to solve geometric problems. Steiner had no early schooling and did not learn to read or write until he was age 14. Against the wishes of his pa rents, at age 18 he went to the Pestalozzi School at Yverdon, Switzerland, where his extraordinary geometric intuition was discovered. By age 28 he was making many geometric discoveries using inversions. At age 38 he occupied the chair of geometry establis hed for him at the University of Berlin, a post he held until his death. Steiner defined an inversive transformation to be any transform ation that is the composition of inversions and initiated inversive geometry , which is the study of properties of the ex tended plane that are preserved by inversive transformations. It follows from Problem 16.2 that inversive transformations preserve angles and take circles and lines to circles and lines. Jean Victor Poncelet (1788 –1867) showed that an inversion is a birati onal transformation , that is, a one -to -one transformation of the extended plane such that both the transformation and its inverse are of the form x´ = f (x, y), y´ = g (x, y), where f and g are rational functions. There were many other European mathematicians in the 19 th century who studied inversions and inversive geometry. Applications of inversions in physics were used by Chapter 16 Inversions in Circles 212 Lord Kelvin (Sir William Thomson) (1824 –1907) in 1845, and also by Joseph Liouville (1809 –1882) in 1847, who calle d inversions the transformations by reciprocal radii . In 1854 Luigi Cremona (1830 –1903) made a systematic study of birational transformations that carry the entire extended plane onto itself. These transformations are now often called Cremona transformatio ns . Since inversions take the entire extended plane to itself, they are Cremona transformations. These were subsequently studied by Max Noether (1844 –1921), who proved that a plane Cremona transformation (and thus inversions) could be constructed by a sequ ence of quadratic and linear transformations . In 1855, August Ferdinand Möbius (1790 –1868) undertook a systematic study of circular transformations (conformal transformations that map points on a circle to points on a circle) by purely geometrical means. H e defined what are now called Möbius transformations , which are often studied today in courses on complex analysis: A Möbius transformation is any transformation of the extended complex plane onto itself of the form M(z)=𝑎𝑧 +𝑏 𝑐𝑧 +𝑑 , where a, b, c, d are complex numbers and ad – bc  The following properties of Möbius transformations are proved in Sections 5.3 and 5.4 of [TX : Brannan]: • Every Möbius transformation is an inversive transformation (but no inversion can be a Möbius t ransformation because Möbius transformations preserve orientation and inversions do not). • Möbius transformations form a subgroup of the group of inversive transformation . • Every inversion F can be written in the form F(z)= M(z),where M is a Möbius transform ation . • Given any two sets of three points, z 1, z 2, z 3, and w 1, w 2, w 3, there is a unique Möbius transformation that maps z 1 to w 1, z 2 to w 2, and z 3 to w3. Möbius geometry is also connected to Laguerre geometry initiated by Edmond Laguerre (1834 –1868) and Minkowskian geometry initiated by Hermann Minkowski (1864 –1909), which is the geometry that Einstein used in the Theory of Relativity and Space/Time. In 190 0, Edward Kasner (1878 –1955) was apparently the first to study inversive geometry in accordance with Klein's Erlanger Program (see Chapter 11, page 153). Research on transformations that preserve circles continued into at least the middle of the 20 th centu ry. Continuing to today, inver - sions are used in the two Poincaré models of hyperbolic geometry (see Problems 17.2 through 17.5 ). For more discussion of inversive and related geometries, see [ TX : Brannan], Chapter 5, and [ HI : Kline], Section 39.3. Kline a lso makes connections to algebraic geometry. See also [ HM : Marchisotto] for related history in the 20 th century.
4524	https://thirdspacelearning.com/us/math-resources/topic-guides/number-and-quantity/inverse-operations/	High Impact Tutoring Built By Math Experts Personalized standards-aligned one-on-one math tutoring for schools and districts Request a demo In order to access this I need to be confident with: Addition and subtraction Multiplication and division Order of operations Exponents What are inverse operations? Solving equations with inverse operations Inverse operations and negative numbers Common Core State Standards How to find the unknown number using inverse operations Inverse operations examples ↓ Example 1: one-step equation Example 2: one-step equation Example 3: one-step equation Example 4: two-step equation Example 5: two-step equation Example 6: two-step equation Teaching tips for inverse operations Easy mistakes to make Related arithmetic lessons Practice inverse operations questions Inverse operations FAQs Next lessons Still stuck? Math resources Number and quantity Arithmetic Inverse operations Inverse operations Here you will learn about inverse operations, including what an inverse operation is and how to use inverse operations to solve one and two-step equations. Students will first learn about inverse operations as a part of operations and algebraic thinking in 3 rd grade and will expand on their knowledge with negative numbers in 6 th grade. What are inverse operations? Inverse operations are operations which reverse or “undo” another operation. They are also sometimes referred to as ‘opposite operations’. The four math operations are addition, subtraction, multiplication and division. Addition and subtraction are inverse operations. For example, \begin{aligned}& 4+5=9 \\ & 9-4=5 \end{aligned} Multiplication and division are inverse operations For example, \begin{aligned}& 5 \times 8=40 \\ & 40 \div 5=8 \end{aligned} [FREE] Arithmetic Worksheet (Grade 4 to 6) Use this quiz to check your grade 4 to 6 students’ understanding of arithmetic. 10+ questions with answers covering a range of 4th, 5th and 6th grade arithmetic topics to identify areas of strength and support! DOWNLOAD FREE x [FREE] Arithmetic Worksheet (Grade 4 to 6) Use this quiz to check your grade 4 to 6 students’ understanding of arithmetic. 10+ questions with answers covering a range of 4th, 5th and 6th grade arithmetic topics to identify areas of strength and support! DOWNLOAD FREE Solving equations with inverse operations You can use inverse operations to find the unknown number in one and two-step math equations. For example, You can find the unknown number to make the following equation, 56+x=174, true by subtracting. \begin{aligned}&\begin{aligned}& 56+x=174 \\ & 174-56=118 \end{aligned} \\ &x=118 \end{aligned} Note: You can solve inequalities and linear equations using inverse operations similarly to how you can solve equations. See also: Inequalities See also: Linear equations Inverse operations and negative numbers Using inverse operations with negative numbers works similarly to using inverse operations with positive numbers. However, you will need to pay special attention to the signs. Negative numbers are values that are less than zero. Adding a negative number is the same as subtracting its positive value, and vice versa. For example, 3+(- \, 5) is the same as 3-5, which equals 2. Subtracting a negative number is the same as adding its positive value. For example, 3-(- \, 3) is the same as 3+3, which equals 6. Multiplying a number by a negative number changes its sign. For example, 4 \times- \, 5=- \, 20 because multiplying a positive by a negative results in a negative product. What are inverse operations? Common Core State Standards How does this relate to 3 rd grade and 6 th grade math? Grade 3: Operations and Algebraic Thinking (3.OA.A.4)Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 \, \times \, ? = 48, 5 = \, ? \, \div 3, 6 \times 6 = \, ?. Grade 3: Operations and Algebraic Thinking (3.OA.C.7)Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 \times 5=40, one knows 40 \div 5=8 ) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers. Grade 6: The Number System (6.NS.C.5)Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation. Grade 6: Expressions and Equations (6.EE.B.7)Solve real-world and mathematical problems by writing and solving equations of the form x+p=q and p x=q for cases in which p, q and x are all nonnegative rational numbers. How to find the unknown number using inverse operations In order to find the unknown number using inverse operations: Identify the operation(s) being applied in the equation. Apply the inverse operation(s) and solve. Inverse operations examples Example 1: one-step equation Solve for the variable using inverse operations. x+5=12 Identify the operation(s) being applied in the equation. The variable has 5 added to it. 2Apply the inverse operation(s) and solve. The inverse operation of addition is subtraction, so subtract 5 from both sides of the equation. \begin{aligned} x+5&=12 \\ -5 & \; \; -5 \\ x&=7 \end{aligned} Example 2: one-step equation Solve for the variable using inverse operations. 4 t=36 Identify the operation(s) being applied in the equation. The variable is being multiplied by 4. Apply the inverse operation(s) and solve. The inverse operation of multiplication is division, so divide both sides of the equation by 4. \begin{aligned} 4 t &=36 \\ \div \, 4 & \; \; \div 4 \\ t &=9 \end{aligned} Example 3: one-step equation Solve for the variable using inverse operations. \cfrac{x}{3}=9 Identify the operation(s) being applied in the equation. The variable is being divided by 3. Apply the inverse operation(s) and solve. The inverse operation of division is multiplication, so multiply both sides of the equation by 3. \begin{aligned} \cfrac{x}{3} & =9 \\ \times \, 3 & \; \; \times 3 \\ x & =3 \end{aligned} Example 4: two-step equation Solve for y using inverse operations. 5 x+7=32 Identify the operation(s) being applied in the equation. The variable x is being multiplied by 5, then 7 is being added to the result. Apply the inverse operation(s) and solve. Because there are more than one operation present, apply the inverse operations in reverse order. The last operation performed was adding 7, so the first inverse operation will be to subtract 7 from both sides of the equation. \begin{aligned} 5 x+7 & =32 \\ -7 & \; \; -7 \\ 5 x & =25 \end{aligned} The first operation performed was multiplying by 5, so the next inverse operation will be to divide both sides of the equation by 5. \begin{aligned} 5 x & =25 \\ \div \, 5 & \;\; \div 5 \\ x & =5 \end{aligned} Example 5: two-step equation Solve for x using inverse operations. 3(y-2)=9 Identify the operation(s) being applied in the equation. The variable y is subtracted by 2, then the result is multiplied by three. Apply the inverse operation(s) and solve. Because there are more than one operation present, you will apply the inverse operations in reverse order. The last operation is multiplication by 3, so the first inverse operation will be to divide by 3. \begin{aligned} 3(y-2) & =9 \\ \div \, 3 & \;\; \div 3 \\ y-2 & =3 \end{aligned} The operation before multiplication was subtraction, so the next inverse operation will be adding 2. \begin{aligned} y-2 & =3 \\ +2 & \; \; +2 \\ y & =5 \end{aligned} Example 6: two-step equation Solve for x using inverse operations. 4 x-3=13 Identify the operation(s) being applied in the equation. The first operation present is multiplying 4 by x, then subtract 3 from the result. Apply the inverse operation(s) and solve. Because there are more than one operation present, we will apply the inverse operations in reverse order. The last operation performed is subtraction of 3, so the first inverse operation will be addition of 3 to both sides of the equation. \begin{aligned} 4 x-3 & =13 \\ +3 & \;\; +3 \\ 4 x & =16 \end{aligned} The first operation performed was multiplication of 4, so the next inverse operation will be division of 4 from both sides of the equation. \begin{aligned} 4 x & =16 \\ \div \, 4 & \;\; \div 4 \\ x & =4 \end{aligned} Teaching tips for inverse operations When creating lesson plans, be sure to include visual aids, such as number lines, to help students visualize the inverse operations. Provide examples of inverse operations when solving one- and two-step word problems, so that students have examples to refer back to when needed. Connect inverse operations to a topic students are already familiar with, like fact families. This will allow them to connect something they know with the new information they are learning. Provide students with practice, whether worksheets or other methods, with gradual complexity. This will allow students to become more comfortable with the concept, and feel confident moving to more difficult problems. Easy mistakes to make Confusing the inverse operationsStudents may be confused about what the inverse operation is when first learning. They may think because multiplication and addition are so closely related, that they are inverse operations. Provide students a ‘cheat sheet’ to refer to when needed. Forgetting to use the inverse operation on both sides of the equationWhen students begin using inverse operations to solve equations, it’s important to reinforce that what is done on one side of the equation must also be done on the other side of the equation. Incorrectly following the order of operationsStudents may believe that they are to follow the order of operations when applying the inverse operation to an equation. Students should remember that they are to reverse the order of operations as they are solving these equations for variables. Related arithmetic lessons Skip counting Number sense Two-step word problems Money word problems Calculator skills Practice inverse operations questions Solve for the variable using inverse operations. b+4=9 b=13 b=36 b=\cfrac{4}{9} b=5 Apply the inverse of addition to both sides of the equation. \begin{aligned} b+4 & =9 \\ -4 & \;\; -4 \\ b & =5 \end{aligned} Solve for the variable using inverse operations. \cfrac{y}{5}=9 y=\cfrac{9}{5} y=45 y=14 y=4 Apply the inverse of division to both sides of the equation. \begin{aligned} \cfrac{y}{5} & =9 \\ \times \, 5 & \;\; \times 5 \\ y & =45 \end{aligned} Solve for the variable using inverse operations. m-9=81 m=72 m=9 m=90 m=45 Apply the inverse of subtraction to both sides of the equation. \begin{aligned} m-9 & =81 \\ +9 & \;\; +9 \\ m &= 90 \end{aligned} Solve for the variable using inverse operations. 5 y+6=21 y=3 y=6 y=10 y=13 Apply the inverse operations in reverse order, with the inverse of addition (subtraction) first, and then the inverse of multiplication (division) second. \begin{aligned} 5 y+6 & =21 \\ -6 & \;\; -6 \\ 5 y &=15 \\ \div \, 5 & \;\; \div 5 \\ y & =3 \end{aligned} Solve for the variable using inverse operations. 7 z-9=12 z=\cfrac{3}{7} z=3 z=147 z=21 Apply the inverse operations in reverse order, with the inverse of subtraction (addition) first, and then the inverse of multiplication (division) second. \begin{aligned} 7 z-9 & =12 \\ +9 & \;\; +9 \\ 7 z & =21 \\ \div \, 7 & \;\; \div 7 \\ z & =3 \end{aligned} Solve for the variable using inverse operations. 2 a+8=20 a=6 a=2 a=14 a=18 Apply the inverse operations in reverse order, with the inverse of addition (subtraction) first, and then the inverse of multiplication (division) second. \begin{aligned} 2 a+8 & =20 \\ -8 & \;\; -8 \\ 2 a & =12 \\ \div \, 2 & \;\; \div 2 \\ a & =6 \end{aligned} Inverse operations FAQs What is the inverse function? The inverse function of a function f is a function that reverses the operation of f. What is the multiplicative inverse? The multiplicative inverse of a number is another number which, when multiplied with the original number, yields the product of 1. This is also commonly referred to as the reciprocal. What is the additive inverse? The additive inverse of a number is another number which, when added to the original number yields a sum of zero. The additive inverse is the negative of the original number. The next lessons are Properties of equality Addition and subtraction Multiplication and division Decimals Fractions Trigonometry Still stuck? At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts. Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. Find out how we can help your students achieve success with our math tutoring programs. Introduction What are inverse operations? ↓ Solving equations with inverse operations Inverse operations and negative numbers Common Core State Standards How to find the unknown number using inverse operations Inverse operations examples ↓ Example 1: one-step equation Example 2: one-step equation Example 3: one-step equation Example 4: two-step equation Example 5: two-step equation Example 6: two-step equation Teaching tips for inverse operations Easy mistakes to make Related arithmetic lessons Practice inverse operations questions Inverse operations FAQs Next lessons Still stuck? x [FREE] Common Core Practice Tests (3rd to 8th Grade) Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents. Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers! Download free We use essential and non-essential cookies to improve the experience on our website. Please read our Cookies Policy for information on how we use cookies and how to manage or change your cookie settings.Accept Privacy & Cookies Policy Privacy Overview This website uses cookies to improve your experience while you navigate through the website. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may affect your browsing experience. Necessary Always Enabled Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website. These cookies do not store any personal information. Non-necessary Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. It is mandatory to procure user consent prior to running these cookies on your website. SAVE & ACCEPT
4525	https://libguides.nwpolytech.ca/math/wordproblems	Skip to Main Content Math Hub Math Language Glossary Common Math Verbs Word Problems Math Anxiety Study Strategies Toggle Dropdown Learning Math Preparing for Tests The Guide Sections The Fundamentals Toggle Dropdown Order of Operations Calculating Percentages Fractions Dividing Fractions Transposition Inequalities Functions Laws of Exponents Absolute Value Significant Figures Rational Expressions and Equations Toggle Dropdown Rational Expressions and Non-Permissible Values Simplifying Rational Expressions Multiplying Rational Expressions Dividing Rational Expressions Adding and Subtracting Rational Expressions Rational Equations and Problem Solving Rational Expressions Quiz Logarithmic and Exponential Functions Toggle Dropdown Defining Logarithms Solving Exponential Equations Solving Problems Involving Exponential Functions Solving Problems Involving Logarithmic Functions Solving Quadratic Equations Toggle Dropdown Factoring Quadratic Equations Solving Quadratic Equations by Completing the Square The Quadratic Formula Sequences and Series Toggle Dropdown Arithmetic Sequences Arithmetic Series Geometric Sequences Geometric Series Set Theory Toggle Dropdown Understanding Sets and Set Notations Solving problems Involving Sets Probability Toggle Dropdown Odds and Probability Mutually Exclusive and Non-Mutually Exclusive Events Dependent and Independent Events Permutations and Combinations Toggle Dropdown The fundamental Counting Principle Permutations Combinations Trigonometry Toggle Dropdown Angles in Standard Position Trigonometric Ratios Transformations of Functions Composition of Functions Operations with Functions Inverse Functions Limits Toggle Dropdown Tangents Limit of a Function Properties of Limits Additional Evaluation Techniques Continuity Derivatives Toggle Dropdown Function of a Derivative Derivative Rules Trigonometric Functions Exponential and Logarithmic Functions Equation of Tangent Line Implicit Derivatives The Antiderivative Statistics Toggle Dropdown Normal Distribution Central Limit Theorem Confidence Intervals Sample Size Hypothesis Testing Hypothesis Testing Process Additional Resources Learning Portal Word Problems Have you ever looked at a math question and had no idea what it was asking you to do? Maybe you understand mathematical equations and algebraic expressions, but when a question is phrased as a sentence, you don’t know where to start. Maybe you just don’t understand the terms being used, or can’t remember what they mean. This module will help you understand the language of math so that you can apply your knowledge when solving math problems Top Tips ✓ Read carefully.Read the entire question before writing anything down or doing any math. You may need to read the question more than once before starting. ✓Determine what the question is asking.Write down what it is the question wants you to find. Assign a variable to this unknown and clearly define what the variable represents. ✓ Write down all the given information.Clearly write down all the information that is provided to you in the question. Make sure to include any units ✓ Look for keywords.Pick out the keywords that will help you translate the word problem into math. Highlight or underline these words. ✓ Sketch a picture or a diagram.A picture or a diagram can help you to visualize the problem. Clearly label your picture or diagram with all the given information and the unknown information you are trying to find. ✓ Choose the correct formula(s).Look for formulas that include the given and unknown information provided in your problem. You may need to use more than one equation or formula to get to your final answer. ✓ Check your answer.Look at your final answer, and think about if it makes sense. Ask yourself if the value is around what you would expect, and does the sign make sense? Understanding Math Language Steps to Solve Word Problems What are Math keywords and why are they important? Word problems show the real-world application of math concepts, so they are an essential part of learning math. However, many people find solving word problems difficult or intimidating. You may be comfortable with the math concepts, but how do you convert the words into a math problem? To help you translate words into math, there are specific keywords you can look for in a question. These keywords indicate the math operation(s) you should use to solve the problem. This is a list of commonly used keywords that are used to identify each math operation. Look for these keywords when solving word problems to help you translate the words into math. Addition add, sum, more, more than, increase, increased by, together, combined, total of, in total, plus, added to, also, in all, join, both, gain, and Example: Tan weighs 71 kilograms. Minh weighs 9 kilograms more than Tan. How much does Minh weigh? Solution: The words more than indicate this is an addition problem. To get Minh’s weight we must add 9 to Tan’s weight. 70 + 9 = 79 Therefore, Minh weighs 79 kilograms. Subtraction subtract, difference, difference between, less, less than, decrease, decreased by, minus, fewer, fewer than, reduce, deduct, left over, remaining, remove, take away, fell Example: Marcella has 6 fewer male cousins than female cousins. Let ƒ represent the number of female cousins. Write an expression for the number of male cousins. Solution: The word fewer indicates this is a subtraction problem. To get the number of male cousins, we need to subtract 6 from the number of female cousins. If we substitute ƒ for the number of female cousins, the expression for the number of male cousins is: ƒ - 6 Multiplication multiplied by, of, by, times, product, product of, factor of, double, triple, twice, rate Example: Kailey is putting in a flower garden. She wants to have eight times as many tulips as sunflowers. Let s represent the number of sunflowers. Write an expression for the number of tulips. Solution: The word times indicates this is a multiplication problem. To get the number of tulips, we need to multiply the number of sunflowers by 8. If we substitute s as the number of sunflowers, the expression for the number of tulips is: 8 Division divide, per, out of, ratio, rate, quotient of, percent, split, equal parts/groups, evenly, average, share, shared between, shared equally Example: Three friends went out to dinner and agreed to split the bill evenly. The bill was $79.35. How much should each person pay? Solution: The words split evenly indicate this is a division problem. To determine how much each friend paid we need to divide the total cost by 3. 79.35 ÷ 3 = 26.45 Therefore, each friend paid $26.45. Equals is, are, was, were, will be, gives, yields, answer, equates to, makes, produces, results, same as Example: Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens and ƒ represent the number of fives. Write an equation to represent the number of fives. Solution: The word is indicates the equals sign. We need to translate the rest of the words to determine the final equation. The words more than indicate addition, and the word times indicates multiplication. \| number of fives \| is \| 3 more than \| 6 times the number of tens \| --- --- \| \| ƒ \| = \| 3 + \| 6t \| Therefore, the final equation for the number of fives is: ƒ = 3 + 6t Examples are derivatives from: Prealgebra by Lynn Marecek & Mary Anne Anthony-Smith is licensed under CC BY 4.0 / A derivative from the original work Steps to solve word problems Do you have difficulty solving mathematical word problems? Do you need a strategy to tackle word problems? You’re not alone. Many students have difficulty with this area of math. This strategy can be utilized for all math word problems as well as math-related word problems in other courses such as chemistry or physics. After practicing the step-by-step method, you will find solving math problems less daunting. The GRASS Method Step 1 – GIVEN Read the question slowly and carefully. Reread the question. Write down or highlight/underline the GIVEN values. Step 2 – REQUIREDWrite down what the problem is asking you to find. This is the information that is REQUIRED. Step 3 – APPLYIdentify key words that will help you APPLY the correct math operation or formula. Step 4 – SOLVE(and check!) Perform the calculations by plugging in given values to SOLVE the problem. Think about your answer - does it make sense? Step 5 - STATEMENTWrite a STATEMENT that answers the question asked. Make sure to include units! Attribution Unless otherwise stated, the material in this guide is from the Learning Portalcreated by College Libraries Ontario. Content has been adapted for the NWP Learning Commons in June 2021. This work is licensed under a Creative Commons BY NC SA 4.0 International License. All icons on these pages are from The Noun Project. See individual icons for creator attribution. << Previous: Common Math Verbs Next: Math Anxiety >> Last Updated: Sep 11, 2025 10:11 AM URL: Print Page Login to LibApps Report a problem Subjects: Multidisciplinary Tags: math Grande Prairie Campus 10726 106 Avenue Grande Prairie, AB T8V 4C4 780-539-2911 1-888-539-4772 (Toll-Free) StudentInfo@NWPolytech.ca Fairview Campus 11235-98 Avenue PO Bag 3000 Fairview, AB T0H 1L0 780-835-6600 1-888-999-7882 (Toll-Free) StudentInfo@NWPolytech.ca National Bee Diagnostic Centre 1 Research Road PO Box 1118 Beaverlodge, AB T0H 0C0 780-357-7737 NBDC@NWPolytech.ca We acknowledge the Indigenous people and their ancestors, whose land we are on. © Northwestern Polytechnic - All rights reserved
4526	https://pubmed.ncbi.nlm.nih.gov/37615838/	Topical Treatment of Melanoma In Situ, Lentigo Maligna, and Lentigo Maligna Melanoma with Imiquimod Cream: A Systematic Review of the Literature - 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. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. To: Subject: Body: Format: [x] MeSH and other data Send email Cancel Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Add to My Bibliography My Bibliography Unable to load your delegates due to an error Please try again Add Cancel Your saved search Name of saved search: Search terms: Test search terms Would you like email updates of new search results? Saved Search Alert Radio Buttons Yes No Email: (change) Frequency: Which day? Which day? Report format: Send at most: [x] Send even when there aren't any new results Optional text in email: Save Cancel Create a file for external citation management software Create file Cancel Your RSS Feed Name of RSS Feed: Number of items displayed: Create RSS Cancel RSS Link Copy Full text links Europe PubMed Central Full text links Actions Cite Collections Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Permalink Permalink Copy Display options Display options Format Page navigation Title & authors Abstract Conflict of interest statement Figures Similar articles Cited by References Publication types Related information LinkOut - more resources Review Dermatol Ther (Heidelb) Actions Search in PubMed Search in NLM Catalog Add to Search . 2023 Oct;13(10):2187-2215. doi: 10.1007/s13555-023-00993-1. Epub 2023 Aug 24. Topical Treatment of Melanoma In Situ, Lentigo Maligna, and Lentigo Maligna Melanoma with Imiquimod Cream: A Systematic Review of the Literature Silvia Vaienti1,Paolo Calzari2,Gianluca Nazzaro3 Affiliations Expand Affiliations 1 Section of Dermatology and Venereology, Department of Medicine, University of Verona, Verona, Italy. 2 Department of Pathophysiology and Transplantation, University of Milan, Milan, Italy. 3 Dermatology Unit, Foundation IRCCS, Ca' Granda Ospedale Maggiore Policlinico, Via Pace 9, 20122, Milan, Italy. Gianluca.nazzaro@gmail.com. PMID: 37615838 PMCID: PMC10539275 DOI: 10.1007/s13555-023-00993-1 Item in Clipboard Review Topical Treatment of Melanoma In Situ, Lentigo Maligna, and Lentigo Maligna Melanoma with Imiquimod Cream: A Systematic Review of the Literature Silvia Vaienti et al. Dermatol Ther (Heidelb).2023 Oct. Show details Display options Display options Format Dermatol Ther (Heidelb) Actions Search in PubMed Search in NLM Catalog Add to Search . 2023 Oct;13(10):2187-2215. doi: 10.1007/s13555-023-00993-1. Epub 2023 Aug 24. Authors Silvia Vaienti1,Paolo Calzari2,Gianluca Nazzaro3 Affiliations 1 Section of Dermatology and Venereology, Department of Medicine, University of Verona, Verona, Italy. 2 Department of Pathophysiology and Transplantation, University of Milan, Milan, Italy. 3 Dermatology Unit, Foundation IRCCS, Ca' Granda Ospedale Maggiore Policlinico, Via Pace 9, 20122, Milan, Italy. Gianluca.nazzaro@gmail.com. PMID: 37615838 PMCID: PMC10539275 DOI: 10.1007/s13555-023-00993-1 Item in Clipboard Full text links Cite Display options Display options Format Abstract Introduction: The classical management of melanoma is surgery, but this can be challenging because of several factors, such as age, body area, lesion size, among others. Topical imiquimod may be a therapeutic option for the treatment of melanoma in situ and lentigo maligna melanoma due to its efficacy, tolerability, and non-invasiveness. The purpose of this systematic review is to assemble current evidence on the treatment of non-metastatic melanoma with topical imiquimod. Methods: The PubMed/MEDLINE and Cochrane Library databases were searched as the primary sources using the main search terms "imiquimod" combined with "lentigo maligna" and "melanoma" with the command "AND." Articles were identified, screened, and extracted for relevant data, following the PRISMA guidelines. Results: A total of 87 studies covering 1803 lesions treated with imiquimod cream were identified and included in this sytematic review. Forty-nine studies were case reports, 16 were retrospective analyses, 3 were open label trials, six were case series; one study was a controlled randomized trial, one was a randomized trial, and one was a single-arm phase III trial. Because of the high number of low-evidence studies, the overall risk of bias resulted high. In 55 studies, imiquimod 5% was used in monotherapy as the primary treatment; only in one study was imiquimod 3.75% introduced. In most cases, the topical treatment was applied once daily, with the exception of nine cases where an increased daily dosage was prescribed. The total duration of the treatment regimen was extremely variable and depended on body area and tolerability, with differences among patients of the same study. In six studies, imiquimod was used as neoadjuvant therapy before the surgical excision, and in 11 studies it was used after surgery as complementary or adjuvant therapy. In total, 1133 of the 1803 (62.8%) lesions were reported to be cleared after the treatment, taking into account that not all of the patients completed the treatment. Of these 1133 lesions, histological clearance was achieved in 645 (56.9%) lesions and clinical clearance only was achieved in 490 (43.2%) lesions; relapse occurred in 107 lesions. Conclusions: The heterogeneity of the studies included in this systematic review precludes the drawing of any relevant conclusions regarding the application of imiquimod. Its efficacy in melanoma in situ and lentigo maligna melanoma has been demonstrated, but further evidence from controlled studies concerning the modalities is missing. Keywords: Imiquimod; Lentigo maligna; Melanoma; Oncology; Skin cancer; Systematic review. © 2023. The Author(s). PubMed Disclaimer Conflict of interest statement The Authors do not have any conflict of interest to disclose. Figures Fig. 1 Flow chart of the selection… Fig. 1 Flow chart of the selection of the studies included in the review Fig. 1 Flow chart of the selection of the studies included in the review See this image and copyright information in PMC Similar articles Series of Fourteen Cases of Topical Imiquimod 5% in Lentigo Maligna: Treatment Modalities and Clues for Detecting Recurrences.Poveda-Montoyo I, Álvarez-Chinchilla P, Schneller-Pavelescu L, Hispán-Ocete P, Bañuls-Roca J.Poveda-Montoyo I, et al.Actas Dermosifiliogr. 2022 Apr;113(4):T407-T412. doi: 10.1016/j.ad.2021.07.012. Epub 2021 Dec 3.Actas Dermosifiliogr. 2022.PMID: 35623739 English, Spanish. Series of Fourteen Cases of Topical Imiquimod 5% in Lentigo Maligna: Treatment Modalities and Clues for Detecting Recurrences.Poveda-Montoyo I, Álvarez-Chinchilla P, Schneller-Pavelescu L, Hispán-Ocete P, Bañuls-Roca J.Poveda-Montoyo I, et al.Actas Dermosifiliogr. 2022 Apr;113(4):407-412. doi: 10.1016/j.ad.2021.07.018. Epub 2022 Mar 4.Actas Dermosifiliogr. 2022.PMID: 35431054 English, Spanish. Effectiveness of 5% Topical Imiquimod for Lentigo Maligna Treatment.Tio DCKS, van Montfrans C, Ruijter CGH, Hoekzema R, Bekkenk MW.Tio DCKS, et al.Acta Derm Venereol. 2019 Sep 1;99(10):884-888. doi: 10.2340/00015555-3241.Acta Derm Venereol. 2019.PMID: 31233181 Treatment of periocular lentigo maligna with topical 5% Imiquimod: a review.Neumann I, Patalay R, Kaushik M, Timlin H, Daniel C.Neumann I, et al.Eye (Lond). 2023 Feb;37(3):408-414. doi: 10.1038/s41433-022-02165-5. Epub 2022 Jul 14.Eye (Lond). 2023.PMID: 35835989 Free PMC article.Review. Evidence-Based Clinical Practice Guidelines for the Management of Patients with Lentigo Maligna.Robinson M, Primiero C, Guitera P, Hong A, Scolyer RA, Stretch JR, Strutton G, Thompson JF, Soyer HP.Robinson M, et al.Dermatology. 2020;236(2):111-116. doi: 10.1159/000502470. Epub 2019 Oct 22.Dermatology. 2020.PMID: 31639788 Review. See all similar articles Cited by Successful implementation of handheld reflectance confocal microscopy as the standard of care in the (surgical) management of lentigo maligna (melanoma).Elshot YS, Lasso Peña DJP, Zupan-Kajcovski B, Bekkenk MW, Balm AJM, Klop WMC, de Rie MA.Elshot YS, et al.J Eur Acad Dermatol Venereol. 2025 Mar;39(3):604-611. doi: 10.1111/jdv.20210. Epub 2024 Jun 26.J Eur Acad Dermatol Venereol. 2025.PMID: 38923079 Free PMC article. Imiquimod and Lentigo Maligna: Can Severe Inflammation Be the Endpoint in Short-Term Use?Bostancı S, Akay BN, Kuşçu DD, Öktem A.Bostancı S, et al.Balkan Med J. 2025 May 5;42(3):279-281. doi: 10.4274/balkanmedj.galenos.2024.2024-11-27. Epub 2025 Jan 17.Balkan Med J. 2025.PMID: 39818620 Free PMC article.No abstract available. Advancements in Melanoma Treatment: A Review of PD-1 Inhibitors, T-VEC, mRNA Vaccines, and Tumor-Infiltrating Lymphocyte Therapy in an Evolving Landscape of Immunotherapy.Mehta A, Motavaf M, Nebo I, Luyten S, Osei-Opare KD, Gru AA.Mehta A, et al.J Clin Med. 2025 Feb 12;14(4):1200. doi: 10.3390/jcm14041200.J Clin Med. 2025.PMID: 40004731 Free PMC article.Review. Lentigo Maligna Treatment-An Update.Ungureanu L, Vasilovici AF, Trufin II, Apostu AP, Halmágyi SR.Ungureanu L, et al.J Clin Med. 2024 Apr 25;13(9):2527. doi: 10.3390/jcm13092527.J Clin Med. 2024.PMID: 38731056 Free PMC article.Review. Hydroxypropyl Cellulose Hydrogel Containing Origanum vulgare ssp. hirtum Essential-Oil-Loaded Polymeric Micelles for Enhanced Treatment of Melanoma.Kamenova K, Iliev I, Prancheva A, Tuleshkov P, Rusanov K, Atanassov I, Petrov PD.Kamenova K, et al.Gels. 2024 Sep 29;10(10):627. doi: 10.3390/gels10100627.Gels. 2024.PMID: 39451280 Free PMC article. See all "Cited by" articles References Garbe C, Amaral T, Peris K, et al. European consensus-based interdisciplinary guideline for melanoma. Part 1: Diagnostics: update 2022. Eur J Cancer. 2022;170:236–55. 10.1016/j.ejca.2022.03.008. - PubMed Nazzaro G, Passoni E, Pozzessere F, Maronese CA, Marzano AV. Dermoscopy use leads to earlier cutaneous melanoma diagnosis in terms of invasiveness and size? A single-center, retrospective experience. J Clin Med. 2022;11(16):4912. doi: 10.3390/jcm11164912. - DOI - PMC - PubMed Pellegrini C, Botta F, Massi D, et al. MC1R variants in childhood and adolescent melanoma: a retrospective pooled analysis of a multicentre cohort. Lancet Child Adolesc Health. 2019;3:332–342. doi: 10.1016/S2352-4642(19)30005-7. - DOI - PMC - PubMed Moehrle M, Dietz K, Garbe C, Breuninger H. Conventional histology vs. three-dimensional histology in lentigo maligna melanoma. Br J Dermatol. 2006;154(3):453–459. doi: 10.1111/j.1365-2133.2005.07068.x. - DOI - PubMed Juhász MLW, Marmur ES. Reviewing challenges in the diagnosis and treatment of lentigo maligna and lentigo-maligna melanoma. Rare Cancers Ther. 2015;3(1):133–145. doi: 10.1007/s40487-015-0012-9. - DOI - PMC - PubMed Show all 124 references Publication types Review Actions Search in PubMed Search in MeSH Add to Search Related information Cited in Books MedGen LinkOut - more resources Full Text Sources Europe PubMed Central Full text links[x] Europe PubMed Central [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
4527	https://math.stackexchange.com/questions/3367086/the-structure-of-the-unit-circle-in-the-plane-f2-where-f-is-a-finite-field	cyclic groups - The structure of the unit circle in the plane $F^2$, where $F$ is a finite field with odd characteristic. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more The structure of the unit circle in the plane F 2 F 2, where F F is a finite field with odd characteristic. Ask Question Asked 6 years ago Modified5 years, 11 months ago Viewed 648 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. In what follows F F denotes the finite field G F(q)G F(q), where q=p k q=p k with p p an odd prime and k∈N∗k∈N∗. The unit circle (U C U C) on the vector space F 2 F 2 is defined as the set of couples (c,s)∈F 2(c,s)∈F 2 that satisfy c 2+s 2=1 c 2+s 2=1 (the names c c and s s have deliberately been chosen to their reference to the trigonometric functions sin sin and cos cos).Using the GAP computer system I looked for the number of elements of the UC and it resulted to be [q]4[q]4, where [n]4[n]4 denotes the nearest multiple of 4 4 to n∈N n∈N and n n odd (e.g. 4=84=8 and 4=84=8). Proving this didn't seem quite simple until I remembered the "`parametrisation"' of the U C U C by the equations c(t)=t 2−1 t 2+1 s(t)=2 t t 2+1 c(t)=t 2−1 t 2+1 s(t)=2 t t 2+1 If q mod 4=1 q mod 4=1 then for two values of t t the functions c(t)c(t) and s(t)s(t) are undefined, namely when t 2=−1 t 2=−1. This accounts for at most q−2 q−2 points. If q mod 4=−1 q mod 4=−1 then the functions c(t)c(t) and s(t)s(t) are defined ∀t∈F∀t∈F, so the parametrization accounts for q q points. The map ϕ:F→U C:t↦(c(t),s(t))ϕ:F→U C:t↦(c(t),s(t)) is injective on its image, its inverse given by: {1−x y 0 if y≠0 if y=0,x≠1{1−x y if y≠0 0 if y=0,x≠1 So for q mod 4=1 q mod 4=1 the image of ϕ ϕ contains exactly q−2 q−2 points and for q mod 4=−1 q mod 4=−1 exactly q q points. Since the only point of U C U C that is missing in the image of ϕ ϕ is (1,0)(1,0) (corresponding to the value t=±∞t=±∞ for the reals) the number of points of the U C U C is as originally stated. Endowing the U C U C with the composition law (c,s)∘(c′,s′)=(c c′−s s′,c s′+c′s)(c,s)∘(c′,s′)=(c c′−s s′,c s′+c′s) turns it into a finite abelian group, as can be verified by direct calculation. Futher tests in GAP shows that with this composition law the U C U C is a cyclic group. Finding a canonical generator of this group doesn't seem to be obvious, the only thing I found is that this group is isomorphic to the matrix group formed by the matrices of the form (c s−s c)(c−s s c) having determinant 1 1. Can one prove that this group is always a cyclic group? finite-fields cyclic-groups Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 23, 2019 at 18:06 Marc BogaertsMarc Bogaerts asked Sep 23, 2019 at 18:00 Marc BogaertsMarc Bogaerts 6,365 1 1 gold badge 19 19 silver badges 29 29 bronze badges 3 What's your question? What is a generator of this group as a function of q q?Randall –Randall 2019-09-23 18:03:24 +00:00 Commented Sep 23, 2019 at 18:03 This all seems like a very nice telling of the "circles in planes over finite fields share some characteristics of circles in the Euclidean plane" story, but is there a question?John Hughes –John Hughes 2019-09-23 18:03:52 +00:00 Commented Sep 23, 2019 at 18:03 @Randall I added an explicit question at the end: Does the group operation defined on the unit circle always result in a cyclic group?Marc Bogaerts –Marc Bogaerts 2019-09-23 18:08:29 +00:00 Commented Sep 23, 2019 at 18:08 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. This is the story of split/non-split torus. For p p odd prime (it will work the same way if you want to extend to finite fields with q=p k q=p k elements), since F∗p F p∗ is cyclic with p−1 p−1 elements If p≡3 mod 4 p≡3 mod 4 then −1−1 has no square root in F p F p, let i i be a square root (in the algebraic closure) then F p(i)=F p 2 F p(i)=F p 2 (the unique finite field with p 2 p 2 elements) and for x,y∈F p x,y∈F p x 2+y 2=(x+i y)(x−i y)=(x+i y)(x+i y)p x 2+y 2=(x+i y)(x−i y)=(x+i y)(x+i y)p thus x 2+y 2=1 x 2+y 2=1 iff (x+i y)p+1=1(x+i y)p+1=1, since F∗p 2 F p 2∗ is cyclic with p 2−1 p 2−1 elements, there are p+1 p+1 elements whose order divides p+1 p+1 and since F p 2 F p 2 is a 2 dimensional F p F p-vector space, 1,i 1,i is a basis and each of those p+1 p+1 elements is the form x+i y x+i y with x,y∈F p x,y∈F p thus a solution of x 2+y 2=1 x 2+y 2=1. In that case the group law of S O 2(F p)S O 2(F p) is that of {a∈F∗p 2,a p+1=1}{a∈F p 2∗,a p+1=1} If p≡1 mod 4 p≡1 mod 4 then −1=c 2∈F p−1=c 2∈F p and x 2+y 2=(x+c y)(x−c y)x 2+y 2=(x+c y)(x−c y) with the change of variable u=x+c y,v=x−c y,x=u+v 2,y=u−v 2 c u=x+c y,v=x−c y,x=u+v 2,y=u−v 2 c we have p−1 p−1 solutions for u v=1 u v=1 which gives p−1 p−1 solutions for x 2+y 2=1 x 2+y 2=1. In that case the group law of S O 2(F p)S O 2(F p) is that of F∗p F p∗ Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 1, 2019 at 22:01 answered Sep 23, 2019 at 19:47 reunsreuns 80.1k 3 3 gold badges 48 48 silver badges 138 138 bronze badges 2 This only answers the question for the special case where F F is a prime field and not the general case where q=p k q=p k where k>1 k>1.Marc Bogaerts –Marc Bogaerts 2019-09-24 12:20:25 +00:00 Commented Sep 24, 2019 at 12:20 @MarcBogaerts I didn't use that F p=Z/p Z F p=Z/p Z, in this answer there is everything you need to find the result for S O 2(F q)S O 2(F q) for q q odd depending on q≡mod 4 q≡mod 4 (the few facts I didn't show : F∗q,F q(i)∗F q∗,F q(i)∗ are cyclic and 1,i 1,i is a F q F q basis of F q(i)F q(i) for q≡3 mod 4 q≡3 mod 4)reuns –reuns 2019-09-24 12:22:24 +00:00 Commented Sep 24, 2019 at 12:22 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 finite-fields cyclic-groups 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 5Can you explain this relation between finite fields and circles? 4When do cos θ cos⁡θ and sin θ sin⁡θ exist over F p F p Related 3Product of Elements in a Finite Cyclic Group with Odd or Even Order 1The number of automorphisms of a finite field 0Can anyone explain how primitive roots work? 5Group structure of the unit circle in F q[−1−−−√]F q[−1] 3How do I find my composition law? 2Showing that for G=⟨x⟩G=⟨x⟩ if \|G\|\|G\| is odd, then ϕ ϕ is an injection. 0Representing the finite field as {i∗g+j}{i∗g+j} where g g is a generator Hot Network Questions Is it safe to route top layer traces under header pins, SMD IC? Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation Lingering odor presumably from bad chicken Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward" How do you create a no-attack area? If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church? What happens if you miss cruise ship deadline at private island? Does a Linux console change color when it crashes? How to start explorer with C: drive selected and shown in folder list? Drawing the structure of a matrix Are there any legal strategies to modify the nature and rules of the UN Security Council in the face of one, or more SC vetos? Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? The rule of necessitation seems utterly unreasonable Does "An Annotated Asimov Biography" exist? Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth? In the U.S., can patients receive treatment at a hospital without being logged? Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? Xubuntu 24.04 - Libreoffice How many color maps are there in PBR texturing besides Color Map, Roughness Map, Displacement Map, and Ambient Occlusion Map in Blender? Storing a session token in localstorage How to solve generalization of inequality problem using substitution? Exchange a file in a zip file quickly How to home-make rubber feet stoppers for table legs? How to fix my object in animation Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4528	https://proofwiki.org/wiki/Definition:Parity_of_Integer	Definition:Parity of Integer From ProofWiki Jump to navigation Jump to search Contents 1 Definition 1.1 Even Parity 1.2 Odd Parity 1.3 Same Parity 1.4 Opposite Parity 2 Also defined as 3 Also see 4 Sources Definition Let $z \in \Z$ be an integer. The parity of $z$ is whether it is even or odd. Even Parity An integer $z$ is of even parity if and only if: : $z = 2 n$ for some $n \in \Z$. Odd Parity An integer $z$ is of odd parity if and only if: : $z = 2 n + 1$ for some $n \in \Z$. Same Parity Two integers $z_1$ and $z_2$ have the same parity if and only if either: : $z_1$ and $z_2$ are both even or: : $z_1$ and $z_2$ are both odd. Opposite Parity Two integers $z_1$ and $z_2$ have opposite parity if and only if either: : $z_1$ is even and $z_2$ is odd or: : $z_2$ is even and $z_1$ is odd. Also defined as Some sources define parity as a property of a pair of integers $\set {z_1, z_2}$ thus: : If $z_1$ and $z_2$ are either both even or both odd, $z_1$ and $z_2$ have even parity : If $z_1$ is even and $z_2$ is odd, then $z_1$ and $z_2$ have odd parity. Also see Definition:Parity Group Definition:Parity Operator Odd Integer 2n + 1 Results about parity of integers can be found here. Sources 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: A Little Number Theory 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $2$ 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2$ 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): parity 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): parity 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): parity Retrieved from " Categories: Definitions/Parity of Integers Definitions/Integers Navigation menu Search
4529	https://www.sciencedirect.com/science/article/abs/pii/S0301211509007027	Guidelines for the management of ovarian cancer during pregnancy - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract Section snippets References (51) Cited by (68) European Journal of Obstetrics & Gynecology and Reproductive Biology Volume 149, Issue 1, March 2010, Pages 18-21 Review Guidelines for the management of ovarian cancer during pregnancy Author links open overlay panel Henri Marret a 1, Catherine Lhommé b 1, Fabrice Lecuru c 1, Michel Canis d 1, Jean Lévèque e 1, Francois Golfier f 1, Philippe Morice b 1 Show more Add to Mendeley Share Cite rights and content Abstract Adnexal masses may be detected during prenatal ultrasound, and ovarian cancer may be suspected during pregnancy. Even though such masses are rarely malignant (1/10,000 to 1/50,000 pregnancies), the possibility of borderline tumour or cancer must be considered. It is a common assumption by both patients and physicians that if an ovarian cancer is diagnosed during pregnancy, treatment necessitates sacrificing the well-being of the fetus. However, in most cases, it is possible to offer appropriate treatment to the mother without placing the fetus at serious risk. The care of a pregnant woman with cancer involves evaluation of sometimes competing maternal and fetal risks and benefits. These recommendations attempt to balance these risks and benefits; however, they should be considered advisory and should not replace specific interdisciplinary consultation with specialists in maternal–fetal medicine, gynecologic oncology and pediatrics, as well as imaging and pathology, as needed. Second level ultrasound including Doppler is needed. MRI is not often necessary, and CA 125 is of low contribution. We suggest surgery be performed after 15 weeks gestation for ovarian masses which (1) persist into the second trimester, (2) are greater than 5–10 cm in diameter, or (3) have solid or mixed solid and cystic ultrasound characteristics. During the antepartum period surgical staging and debulking, unilateral salpingo-oophorectomy on the side with the tumour, peritoneal cytology and exploration are necessary. Women found to have advanced stage epithelial ovarian cancer should consider having completion of the debulking of the reproductive organs at the conclusion of the pregnancy. If chemotherapy is indicated, we recommend delaying administration, if possible, until after the delivery or at least after 20 weeks in order to minimize the potential fetal toxicity. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Preliminary For optimal management of a suspected malignant adnexal tumour, a multidisciplinary approach involving specialists in oncology, and sometimes in paediatrics, is necessary whenever possible, as soon as it is diagnosed (Grade A). It should be kept in mind that managing cancer and pregnancy always requires a shared decision between the patient and at least a surgical oncologist, an obstetrician, a medical anatomopathologist and a paediatrician , , , , , . Diagnosis [7–29] Transcutaneous and transvaginal pelvic ultrasound imaging is an essential and reliable technique which assists the physician in the diagnosis of ovarian tumour during pregnancy. As about 10% of masses are complex, a second diagnostic test should be performed by a fully trained sonographer. The ultrasound examination should determine the origin of the mass, as well as its location, size and internal structure (existing vegetations or septa) and should classify it into one of the following five Management For asymptomatic adnexal masses, surgery should only be considered in pregnancy for suspicious masses or obvious malignant tumours (Grade B). In patients with suspicious ultrasound appearances of malignant ovarian tumour, the surgical staging procedure is threefold: -To establish a diagnosis: a definitive histological evidence or a reliable extemporaneous evidence should be obtained (epithelial malignant or not, or low malignant potential), without increasing the risk of dissemination (rupture) Conclusion As the incidence of invasive cancers and borderline ovarian tumours diagnosed during pregnancy is low, treatment strategies should ideally be discussed and structured during a “multidisciplinary meeting” involving specialists . The different cases should be recorded into the national database of the CNGOF (French College of Obstetricians and Gynaecologists (cancer.grossesse@tnn.aphp.fr)). Recommended articles References (51) G.B. Sherard et al. Adnexal masses and pregnancy: a 12-year experience Am J Obstet Gynecol (2003) H.F. Wong et al. Ovarian tumors of borderline malignancy: a review of 247 patients from 1991 to 2004 Int J Gynecol Cancer (2007) G. Leiserowitz et al. Adnexal masses in pregnancy: how often are they malignant? Gynecol Oncol (2006) L. Bernhard et al. Predictors of persistance of adnexal masses in pregnancy Obstet Gynecol (1999) J. Yazbek et al. The value of ultrasound visualization of the ovaries during the routine 11–14 weeks nuchal translucency scan Eur J Obstet Gynecol Reprod Biol (2007) L.M. Hill et al. The role of ultrasonography in the detection and management of adnexal masses during the second and third trimesters of pregnancy Am J Obstet Gynecol (1998) P. Glanc et al. The prevalence of incidental simple ovarian cysts > or =3 cm detected by transvaginal sonography in early pregnancy J Obstet Gynaecol Cancer (2007) H. Marret et al. Power Doppler index for preoperative ovarian tumors discrimination Gynecol Obstet Fertil (2007) B. Mol et al. Distinguishing the benign and the malignant adnexal mass: an external validation of prognostic models Gynecol Oncol (2001) J.L. Alcazar et al. Comparison of 2-dimensional and 3-dimensional power-Doppler imaging in complex adnexal masses for the prediction of ovarian cancer Am J Obstet Gynecol (2005) R. Sharony et al. Granulosa cell tumors of the ovary: do they have a unique ultrasonographic and color Doppler flow features? Int J Gynecol Cancer (2001) Y. Yinon et al. Clinical outcome of cystectomy compared with unilateral salpingo-oophorectomy as fertility-sparing treatment of borderline ovarian tumors Fertil Steril (2007) G. Zanetta et al. A prospective study of the role of ultrasound in the management of adnexal masses in pregnancy BJOG (2003) G. Dubernard et al. Accuracy of MR imaging combined with sonography for the diagnosis of persistent adnexal masses during pregnancy: about nine cases Gynecol Obstet Fertil (2005) F. Machado et al. Ovarian cancer during pregnancy: analysis of 15 cases Gynecol Oncol (2007) X.Y. Zhao et al. Ovarian cancer in pregnancy: a clinicopathologic analysis of 22 cases and review of the literature Int J Gynecol Cancer (2006) D.N. Platek et al. The management of a persistent adnexal mass in pregnancy Am J Obstet Gynecol (1995) G. Ferrandina et al. Management of an advanced ovarian cancer at 15 weeks of gestation: case report and literature review Gynecol Oncol (2005) M. Dede et al. Treatment of incidental adnexal masses at cesarean section: a retrospective study Int J Gynecol Cancer (2007) R.H. Young et al. Granulosa cell, Sertoli-Leydig cell, and unclassified sex cord-stromal tumor associated with pregnancy: a clinicopathological analysis of 36 cases Gynecol Oncol (1984) F. Amant et al. Gynecologic cancers in pregnancy: guidelines of an international consensus meeting Int J Gynecol Cancer (2009 May) G.S. Leiserowitz Managing ovarian masses during pregnancy Obstet Gynecol Surv (2006) K.M. Schmeler et al. Adnexal masses in pregnancy: surgery compared with observation Obstet Gynecol (2005) G. Condous et al. Should we be examining the ovaries in pregnancy? Prevalence and natural history of adnexal pathology detected at first-trimester sonography Ultrasound Obstet Gynecol (2004) G. Chiang et al. Imaging of adnexal masses in pregnancy J Ultrasound Med (2004) View more references Cited by (68) Ovarian cysts and cancer in pregnancy 2016, Best Practice and Research Clinical Obstetrics and Gynaecology Citation Excerpt : Malignant adnexal masses during pregnancy range between 0.8% and 13% [5–12]. The reported incidence of ovarian cancer (OC) in pregnancy varies from 1 in 15,000 to 1 in 32,000 and it is amongst the top five cancers diagnosed during pregnancy [13–17]. With the increasing maternal age, it is expected that more women will be diagnosed with OC in pregnancy in the future, therefore management guidelines should be formulated with a regular update and review of recent literature [15,16]. Show abstract Adnexal masses are diagnosed in 5% pregnancies and pose diagnostic and management challenges. Ultrasound and magnetic resonance imaging (MRI) are the mainstay as an evaluation procedure; surgery is warranted for persistent masses with a diameter of >5 cm and sonographic signs of possible malignancy. Optimal timing for a planned surgery is the second trimester and does not adversely affect neonatal outcome. Laparoscopy is safe in pregnancy. Management for ovarian cancer during pregnancy should be individualised and formulated by a multidisciplinary team in a specialised centre while also considering the patients' wishes to preserve pregnancy. The following options can be considered: (i) induced abortion followed by standard management of ovarian cancer, (ii) pregnancy-preserving surgery followed by chemotherapy, planned delivery and secondary surgical completion or (iii) neoadjuvant chemotherapy followed by surgery during the postpartum period. Standard chemotherapy administered in non-pregnant population can only be used during the first trimester of pregnancy. ### Gynaecological cancers in pregnancy 2012, Lancet Citation Excerpt : The present guideline for chemotherapy (if it has to be delivered) recommends the combination of bleomycin, etoposide, and cisplatin in non-pregnant patients.96 In pregnant patients, the indications for adjuvant chemotherapy are similar to those prevalent in non-pregnant patients.80 These patients do not need additional surgery and their fertility should be preserved whenever feasible. Show abstract Cervical and ovarian cancers are the most common gynaecological cancers diagnosed during pregnancy. In early-stage cervical cancer during the first and at the beginning of the second trimester, the two main considerations for management of the patient are the tumour size (and stage) and nodal staging. MRI and laparoscopic lymphadenectomy are useful for clinicians planning a potentially conservative approach. The management of patients with locally advanced cervical disease is controversial and should be discussed on a case-by-case basis according to the tumour size, radiological findings, the term of pregnancy, and the patient's wishes. Different histological types of malignant ovarian diseases arise during pregnancy and their management depends on the diagnosis (histological subtypes, tumour differentiation, and nodal status), the tumour stage, and the trimester of the pregnancy. In patients with peritoneal spread or high-risk early-stage disease, neoadjuvant chemotherapy with pregnancy preservation could be appropriate. ### Cancer in pregnancy: A challenging conflict of interest 2012, Lancet ### Multidisciplinary management of cancer during pregnancy 2020, JCO Oncology Practice ### Taxanes in combination with platinum derivatives for the treatment of ovarian cancer during pregnancy: A literature review 2017, International Journal of Clinical Pharmacology and Therapeutics ### Investigation and Management of Adnexal Masses in Pregnancy 2016, Scientifica View all citing articles on Scopus 1 On behalf of the French Working Group on Gynecological Cancers in Pregnancy SFOG (Société Française d’Oncologie Gynècologique), SFCP (Société Française de Chirurgie Pelvienne), CNGOF (Collège National des Gynécologues Obstétriciens Français). View full text Copyright © 2009 Elsevier Ireland Ltd. All rights reserved. Recommended articles Synergy between the solution based on the Transversal Method Of Lines and the Leveque solution in the platform of the Graetz/Nusselt problem International Communications in Heat and Mass Transfer, Volume 116, 2020, Article 104590 Antonio Campo, …, Yunesky Masip Macía ### Total rectosigmoidectomy versus partial rectal resection in primary debulking surgery for advanced ovarian cancer European Journal of Surgical Oncology (EJSO), Volume 42, Issue 3, 2016, pp. 383-390 F.Plotti, …, R.Angioli ### A case of incidental endometrial adenocarcinoma diagnosed in early pregnancy and managed conservatively Gynecologic Oncology Reports, Volume 28, 2019, pp. 101-103 I.Rizzuto, …, B.Rufford ### Peripartum hemorrhage and other obstetric and neonatal outcomes in pregnant women with epilepsy: A single-center study Epilepsy Research, Volume 171, 2021, Article 106566 Mehmet Murat Işıkalan, …, Ali Acar ### Cytoreductive surgery in advanced stage malignant ovarian germ cell tumors Journal of Gynecology Obstetrics and Human Reproduction, Volume 48, Issue 7, 2019, pp. 461-466 Alper Karalok, …, Taner Turan ### Management of intravenous leiomyomatosis: a comprehensive review of surgical and perioperative considerations International Journal of Gynecological Cancer, Volume 35, Issue 10, 2025, Article 102009 Emily Volfson, …, Rachel Soyoun Kim Show 3 more articles 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. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Settings Accept all cookies Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. You may also be able to exercise your privacy choices as described in our Privacy Policy Allow all Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. Cookie Details List‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Cookie Details List‎ Contextual Advertising Cookies [x] Contextual Advertising Cookies These cookies are used for properly showing banner advertisements on our site and associated functions such as limiting the number of times ads are shown to each user. Cookie Details List‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm my choices
4530	https://support.haltech.com/portal/en/kb/articles/knock-control-user-s-guide	Knock Control User's Guide Welcome to haltech Portal Skip to ContentSkip to MenuSkip to Footer Haltech Support Center Home My Area Knowledge Base Community Sign In Welcome to Haltech Support Center Knowledge Base Haltech Technical Library Functions and Tuning Knock Control User's Guide Introduction What is the Elite Knock Control function and how do you use it. Knock Detection and Knock Control are only available on the Nexus VCU and Elite 1000 and higher models. Hey, guess what? We've got a great article here to read and a video. Are you ready to learn something? Let's go! VIDEO - Knock Control. What it is and why you need it. What is Engine Knock ? A 2-stroke piston showing the peppering that is common with engine knock. Engine knock goes under a number of different names, pinging, detonation, etc.. and is quite literally uncontrolled explosions in your engine rather than a controlled and timed combustion event. Engine knock rapidly leads to broken engine components and typically manifests itself as peppered piston tops, followed by cracked pistons and heavily worn bearings. What you are basically hearing when an engine is knocking is the block ringing like a bell. Knock in most cases is started by a hot spot causing the air fuel mix to spontaneously ignite before the timed spark is introduced. Most of the time this is because of excessive ignition timing, but can be due to a mixture that is too lean, high component temperatures due to poor cooling, or poor engine design. The easiest way to cool the engine internals is to retard ignition timing, which diverts energy into the exhaust more than the internal components. What is Knock Detection? Knock Detection involves using a peizo-electric (mechanical to electric) sensor to monitor the frequency at which the engine block is vibrating, and determining whether or not the engine is actually knocking. For the Elite ECU we use a simple sensor that acts very much like a microphone, sending all frequencies through to the ECU to be processed. From within the ESP software the correct frequency to be used is then selected by the tuner, with all other frequencies being ignored. To do this the Knock Detection function must be enabled, and a sensor must be connected to the appropriate ECU input. There are two main types of knock sensors used in the industry:resonant and non-resonant. A typical resonant type knock sensor Resonant Knock Sensors This type of sensor is typically designed for the exact application and uses a small strip of metal that is designed to have the same resonant frequency as the engine block. These types simply output a signal as the metal strip vibrates. They can usually be identified as a screw-in style body with the connector sticking out away from the block and commonly a single wire connection. Some of these types of sensors can be used with the Elite ECU however the frequency selection will not work as they output a signal across off frequencies to the spectrogram. When running the spectrogram the tuner will notice vertical lines as knock is experienced so frequency selection is not important and can be set to just about anything. The resonant style sensor is also prone to false-tripping because we cannot tailor the exact frequency to suit the application. The Haltech HT-011100 Knock Sensor Non-Resonant Knock Sensors This type of sensor is more universal and by design will pick up a large range of frequencies making them more suitable to performance applications where we want to pick the frequency to be monitored. The Haltech Knock Sensor is a non-resonant type sensor. They can usually be identified by the through-bolt design and commonly have a 2-pin connection with the connector on the side of the sensor. There is one dedicated knock input (pin A21) on the Elite 1000 and 1500 models, and two dedicated knock inputs (pins A21 and A22) on the Elite 2000 and 2500 models. The knock sensor should connect to one of these pins, with the other pin of the knock sensor being wired to Signal Ground. Take note that NGK stress the fact that their sensors should be torqued to the correct specs.Quote:"The knock sensors have a specific torque. The torque change can affect the signal generated by the sensor. NTK recommends torque of 2.0 to 2.5 kgf.m and does not recommend the use of grease or washers on mounting the knock sensor." Knock Detection Settings and Tables Found in Main Setup / Functions / Knock Detection / Knock Detection section \| SETTING \| DESCRIPTION \| TYPICAL SETTINGS \| --- \| Detection - Start Spectrogram \| Turns On or Off the Knock Detection Spectrogram which is used to find the knock frequency. The spectrogram will run for 20 seconds and then turn off by itself. \| Turn on as required. \| \| Detection - Knock Frequency \| The knock frequency can be manually entered here. \| Found through testing. Changes with each engine. \| \| Detection - Start Angle (BTDC) \| The crank angle position at which the ECU will start to look for knock. \| Found through testing but start with -10 degrees. \| \| Detection - Duration \| The crank angle amount over which the ECU will look for knock. With a default duration of 40 degrees, and default Start Angle of -10, the ECU will look for knock between -10 and 30 degrees BTDC. \| Found through testing but start with a wide 60 degrees before narrowing it down afterwards. \| \| Momentary Knock - Ign Adv \| The amount of extra ignition timing to be added during the Momentary Knock process. \| Use enough timing to actually make the engine knock. Start with 10 degrees but more may be required. \| \| Momentary Knock - Duration \| The length of time over which the Momentary Knock process will run. \| use enough time to actually make the engine knock. Typically set to 10 seconds and if there is excessive knock the tuner should reduce engine load. \| \| Momentary Knock - Knock \| Starts the Momentary Knock process \| Use as required \| \| TABLE \| \| \| \| Knock Threshold \| The amount of knock signal level over which the engine is considered to be knocking. \| Typically set to around 3 to 5 db above the normal background noise of the engine when it is definitely not knocking. \| Knock Detection Process There are a number of different methods that tuners will use but the following process is one that I follow when setting upKnock Detection. There are two main parts to the setup, the knock frequency selection and the knock threshold table setup. What must be considered is that for the knock detection to work you must identify engine knock. In other words, you do have to make the engine knock so that you know what engine knock looks like for the ECU to be able to do something about it. There will be times when the knock detection process simply will not work, such as when fuel that has a high enough octane rating that it is not possible to make the engine knock. To complete the knock detection process the engine must already have a decent level of tune completed. Thefuelingshould be where the tuner wants, and a safe level of ignition timing should be entered that is reasonably ball-park. Knock Frequency Selection This part involves finding the best frequency for the ECU to monitor when detecting knock. This process only needs to be completed once, and yes you WILL need a dyno to do this process with any form of repeatability. Set the Ignition Base table to values that you feel will be relatively safe. Use the Ignition Overall Trim to take out another extra few degrees. Say -3.0 degrees top give us an extra bit of headroom. Warm the engine up to full operating temperatures. Your engine will not want to knock when everything is stone cold. Bring the engine up to around 1500rpm. This rpm is chosen so that there is enough oil pressure to prevent bearing damage during the testing procedure, the piston speed is not high enough to cause damage, and general engine/exhaust noise is low enough for knock to be audibly heard if required. Start the Spectrogram. Remember you only have around 20 seconds with each pass so move swiftly. Load the engine up, typically to half throttle or so. At the moment with the reduced ignition timing, you should see general background noise only at this point. Hit the Momentary Knock button and then look for any changes in the spectrogram pattern. Even a mildly trained ear should hear any knocking and correspondingly should see brighter spots appearing on the spectrogram. If no change is heard or seen, increase the Momentary Knock Ign Adv value by 5 and repeat until you do see a difference. Sometimes very large amounts of timing are required to make the engine knock. If knock is too excessive you should ease off the throttle to control how much knock is experienced. You do not need to destroy the engine to complete this task, you only need to make it knock enough to notice which frequency the knock is occurring at. Moving the mouse pointer over the spectrogram to the bright spots, typically red. This will highlight the frequency with a faint grey line. Enter the value it displays as the Knock Frequency. When completed turn Off the Spectrogram and hit OK. We are now done with the frequency selection. Knock Threshold Table Setup With the correct frequency chosen for the ECU to look at for knock, we now need to tell the system what amount of noise at that frequency is normal background and at what level that actual knock is at a level that we would like to do something about it. The Knock Threshold Table is where this is done. It is recommended to use the Knock Page Layout provided with ESP to do this next process. If your ESP software does not have this page, go to the top of the screen and select View / Load Page, select the screen resolution that best matches your laptop screen but if unsure go to the res1366x768 folder, then select the Knock.page file. Now view the Knock Threshold table which will look similar to this. To set this table we are simply going to give the engine a power run on the dyno. What we are trying to do is to set up the Knock Threshold table so that it is always around 3 to 5 db above the Knock Signal values at all times.This may take a number of power runs to complete. When this is done we will start to introduce more ignition timing. When we earlier used the Ignition Overall Trim to take out 3 degrees, now we will start to head back towards zero and even into positive values. Look for any instances where the Knock Signal exceeds the Knock Threshold values. Any time this does occur the Knock Count will increase, and you will see how far it exceeded the threshold by looking at the Knock Level values. At this point, the engine is considered to have knocked. If backing off the timing shows the knock signal to come back to normal then you have completed the setup. Ignition timing tuning from here can use the Knock Count and Knock Level channels to show then too much timing has been used. What is Knock Control ? Knock Control is pretty much the ECU doing something about knock when it is detected. This involves retarding the ignition timing when knock is detected. Pretty much all that needs to be adjusted is the amount of Short Term Retard, with a value high enough to stop the knocking being required. Usually the default 5 degrees is enough to do the job so very little input is required other than to turn it on and make some changes only if the default values are not sufficient. Knock Control Settings \| SETTING \| DESCRIPTION \| TYPICAL VALUES \| --- \| Mode \| Sets whether there is more than one bank to be controlled or not. \| Depend on the number of sensors. Unbanked when there is one sensor. \| \| Short Term Retard \| The amount of ignition timing retard to be applied when a knock event has been detected. \| Usually around 5 degrees is enough to stop the engine from knocking, but more can be used if an engine is more sensitive. \| \| Short Term Decay Rate \| The rate at which the Short Term Retard is reduced back to the normal amount of ignition timing. \| Usually start with 0.5 with a lower number holding the retard for longer, and a higher number for a shorter time. \| \| Hysterisis Time \| The amount of time after a knock event has triggered the short term retard before another event can trigger the short term retard again. \| Set to a relatively short time, around 0.1 to 0.3 seconds. \| Knock Control Long Term Trim Settings and Tables \| SETTING \| DESCRIPTION \| TYPICAL VALUES \| --- \| Enable Long Term Trim \| Turns On or Off the Long Term learning functionality \| Typically turned off by default. Currently the Elite ECU range do not reverse learn. i.e. it wont put back any learned values when the knocking has been eliminated. \| \| Long Term Retard \| The amount of retard to learn for the given cell for each given knock event. \| Use a small value, around 0.5 deg. using a smaller value means the system will learn too slowly. Set too high and it will learn too coarse. \| \| Max Long Term Retard \| The maximum amount of retard that any one cell can learn. \| Usually set to around 5 degrees, but if your engine is knock prone then allowing it to learn more can be advantageous. \| \| Reset \| Resets the learned values back to all zero. \| N/A. \| \| Apply To Base Table \| Applies the learned values to the Ignition Base table, and sets the table back to all zeroes when a single bank is used, or applies what it can when two banks are used and any differences are kept (not zero). \| Use as required. \| \| TABLES \| \| \| \| Bank 1 Long Term Trim \| Active when the Long Term Trim is enabled. The axis channels will be locked to use the same as the Ignition Base, however the values themselves can be adjusted to suit. \| N/A \| \| Bank 2 Long Term Trim \| Only available on the Elite 2000 and 2500 models, and if a second sensor has been connected and a second Bank has been configured. \| N/A \| Knock Detection and Control Channels Information about the channels available to be viewed in the ESP software. \| CHANNEL \| SHORT NAME \| DESCRIPTION \| --- \| Knock Threshold \| KnkThresh \| Taken from the Knock Threshold table. The amount of knock signal over which a knock event is considered to have occurred. \| \| Knock Sensor 1 Signal \| Knk1 Sig \| The signal from Knock Sensor 1. \| \| Knock Sensor 1 Level \| Knk1 Lvl \| The amount that Knock Signal 1 that has exceeded the threshold. \| \| Knock Sensor 1 Count \| Knk1 Count \| The amount of times the signal has exceeded the threshold on Knock Sensor 1. In other words how many times a knock event has been detected. \| \| Knock Sensor 2 Signal \| Knk2 Sig \| The signal from Knock Sensor 2. \| \| Knock Sensor 2 Level \| Knk2 Lvl \| The amount that Knock Signal 2 that has exceeded the threshold. \| \| Knock Sensor 2 Count \| Knk2 Count \| The amount of times the signal has exceeded the threshold on Knock Sensor 2. In other words how many times a knock event has been detected. \| \| Knock Detection Light Output State \| Knock LightOutputState \| Shows the state of the output to the Knock Light if one is enabled. \| \| Knock Control Bank 1 Ignition Correction \| Knock B1 IgnCorr \| The current amount of ignition correction being applied by Knock Sensor 1. The Short Term Trim \| \| Knock Control Bank 2 Ignition Correction \| Knock B2 IgnCor \| The current amount of ignition correction being applied by Knock Sensor 2. The Short Term Trim \| \| Knock Control Bank 1 Long Term Trim \| KnockB1 LTT \| The current amount of Long Term Trim being used for Bank 1. \| \| Knock Control Bank 2 Long Term Trim \| KnockB2 LTT \| The current amount of Long Term Trim being used for Bank 2. \| FAQs I can't see anything at all on my Knock Spectrogram. Check that the sensor is connected. The sensor itself may not be compatible. When in doubt the Haltech sensor is inexpensive and has shown to present a wide frequency spectrum to the ECU. I see the background noise on the spectrogram but I can't make my engine knock. Great, what are you complaining about then? This mostly happens with fuel with too high an octane to make the engine knock. A lower grade fuel may be temporarily required to obtain the correct knock frequency. The knock threshold table is difficult to set up. I don't change the timing but it will randomly detect knock. What should I do? Usually caused by excessive background noise (poor signal to noise ratio). Knock occurs at more than one frequency, usually in what is called harmonics. Selecting another frequency that has less background noise would be a good way to remove excess background noise. Updated:4 years ago Helpful? Follow Subscribe to receive notifications from this article. Functions and Tuning Knock Control What is knock and why you should control it Sequential Twin Turbo User's Guide Oil Pressure Sensor and Engine Protection Knock Control User's Guide Boost Control User's Guide Tuning with VE (Volumetric Efficiency) Direct Injection Engine Limiting Channel States Boost Control Explained View all Still can’t find an answer? Send us a ticket and we will get back to you. Submit a ticket Pages Home Shop Dealers Downloads Contact About us Employment Repairs Returns Warranty FAQ's Privacy Policy Account Login Register Manage Account Get in touch Haltech Australia 17 Durian Place, Wetherill Park NSW Australia 2164 +61 2 9729 0999 Sales enquiries: sales@haltech.com © 2020 Haltech Engine Management Systems. ABN 68 061 744 303. ISO9001 Quality Assurance compliant company SAFE & SECURE SHOPPING Haltech Australia 17 Durian Place, Wetherill Park NSW Australia 2164 +61 2 9729 0999 Sales enquiries: sales@haltech.com Pages Home Shop Dealers Downloads Contact About us Employment Repairs Returns Warranty FAQ's Privacy Policy Account Login Register Manage Account SAFE & SECURE SHOPPING © 2020 Haltech Engine Management Systems. ABN 68 061 744 303. ISO9001 Quality Assurance compliant company Accessibility Settings Accessibility Personas Choose a mode to personalise your experience ADHD Friendly More focus & fewer distractions Astigmatism Reduce glare and eye strain Blindness Optimize page for screen-readers & keyboard Color Blindness Easier to discern and distinguish colors Dyslexia Friendly Optimize page for dyslexic users Epilepsy Safe Reduces motion to prevent adverse reactions For The Elderly More clear and easy to work without distractions Low Vision Improve legibility by enhancing visual elements Motor Disabilities Efficient navigation using a keyboard & mouse Seizure Safe Clear flashes & reduces colors Keyboard Navigation (Motor) Navigate page with the keyboard Display Customize visual appearance Themes Choose the mode in which you want to view Default Night Mode Auto Aside Left Choose the position of the aside section on the page Aside Left Aside Right Full Width Choose the page width Default Full Width Content Scaling: 100% Adjust the size of the content to make it easier to read Font Adjust font size, spacing, and alignment Font Size: 100% Resize text for better readability Text Spacing: 100% Slide to adjust the space between each line Text Alignment Adjust the text alignment across the page Align Left Align Center Align Right Contrast Makes easier to read text and enhances color Monochrome Convert to black and white colors Low Saturation Decreases the intensity of colors High Saturation Increases the intensity of colors Low Contrast Softer colors, less difference Medium Contrast Balanced and clear High Contrast Bold and easy to see Highlights Choose your highlight style for reading the page Underline Links Increases the visibility of links with an underline Emphasize Focus Area Indicates the extent & clickability by adding a border Highlight Headings Increases the visibility of headings by adding a border Highlight Hovers Increases the visibility of content by adding a border Cursor Make your cursor bigger and easier to see Big Black Cursor Easy to see on light backgrounds Big White Cursor Easy to see on dark background Reading Tools Support tools for easier reading Select to Speak Tap to hear items read aloud Pause Animation Pause motion for a stable screen Text Magnifier Make text bigger for easier reading Reading Guide A bar moves with the cursor to help you read Reading Mask Masks the page to focus on the content Reset All Accessibility Settings
4531	https://www.malayajournal.org/articles/PAPER_20_2017.pdf	Malaya J. Mat. 5(2)(2017) 278–292 ai Type n−Variable Multi n−Dimensional Additive Functional Equation Matina J. Rassias,a ∗M. Arunkumar,b and E. Sathyac aDepartment of Statistical Science , University College London, 1-19 Torrington Place, #140, London, WC1E 7HB, UK. b,cDepartment of Mathematics, Government Arts College, Tiruvannamalai - 606 603, TamilNadu, India. Abstract In this paper, the authors investigated the general solution and generalized Ulam - Hyers stability of ai type n−variable multi n−dimensional additive functional equation 2h n ∑ i=1 ai x1i, n ∑ i=1 ai x2i, . . . . . . , n ∑ i=1 ai xni ! = n ∑ i=1 ai ! h n ∑ i=1 x1i, n ∑ i=1 x2i, . . . . . . , n ∑ i=1 xni ! + a1 − n ∑ i=2 ai ! h x11 − n ∑ i=2 x1i, x21 − n ∑ i=2 x2i, . . . . . . , xn1 − n ∑ i=2 xni ! where ai(i = 1, 2, . . . n) are different integers greater than 1, using two different technique. Keywords: Additive functional equations, Ulam - Hyers stability, Ulam - Hyers - Rassias stability, Ulam -Gavruta - Rassias stability, Ulam - JRassias stability. 2010 MSC: 39B52, 32B72, 32B82. c ⃝2016 MJM. All rights reserved. 1 Introduction During the last seven decades, the perturbation problems of several functional equations have been ex-tensively investigated by number of authors [1, 3, 20, 21, 30, 31, 34, 35]. The terminology generalized Ulam -Hyers stability originates from these historical backgrounds. These terminologies are also applied to the case of other functional equations. For more detailed deﬁnitions of such terminologies, one can refer to [8, 18, 22– 24]. One of the most famous functional equations is the additive functional equation f (x + y) = f (x) + f (y). (1.1) In 1821, it was ﬁrst solved by A.L. Cauchy in the class of continuous real-valued functions. It is often called an additive Cauchy functional equation in honor of Cauchy (see ). The additive function f (x) = cx is the solution of the additive functional equation (1.1). The solution and stability of various additive functional equations were discussed by D.O. Lee , K. Ravi, M. Arunkumar , M. Arunkumar [4–6, 8, 9]. W.G. Park, J.H. Bae [16, 27] investigate the general solution and the generalized Hyers-Ulam stability of the multi-additive functional equation and 2- variable ∗Corresponding author. E-mail addresses: matina@stats.ucl.ac.uk (Matina J. Rassias), annarun2002@yahoo.co.in (M. Arunkumar), sathya24mathematics@gmail.com ( E. Sathya). Matina J. Rassias et al. / ai Type n−Variable Multi n−... 279 quadratic functional equation of the forms f (x1 + x2, y1 + y2, z1 + z2) = ∑ 1≤i,j,k≤2 f (xi, yj, zk), (1.2) f (x + y, z + w) + f (x −y, z −w) = 2f (x, z) + 2f (y, w). (1.3) The stability of the functional equation (1.3) in fuzzy normed space was proved by M. Arunkumar et., al . Using the ideas in , the general solution and generalized Hyers-Ulam-Rassias stability of a 3- variable quadratic functional equation f (x + y, z + w, u + v) + f (x −y, z −w, u −v) = 2f (x, z, u) + 2f (y, w, v). (1.4) was discussed by K. Ravi and M. Arunkumar . Its solution is of the form f (x, y, z) = ax2 + by2 + cz2 + dxy + eyz + f zx. (1.5) Also, M. Arunkumar, S. Hema Latha established the general solution and generalized Ulam - Hyers stability of a 2 - variable Additive Quadratic functional equation f (x + y, u + v) + f (x −y, u −v) = 2f (x, u) + f (y, v) + f (−y, −v) (1.6) having solutions f (x, y) = ax + by (1.7) and f (x, y) = ax2 + bxy + cy2 (1.8) in Banach and Non Archimedean Fuzzy spaces respectively. Infact, M. Arunkumar et. al., introduced and discussed a 2 - variable AC - mixed type functional equation f (2x + y, 2z + w) −f (2x −y, 2z −w) = 4[ f (x + y, z + w) −f (x −y, z −w)] −6f (y, w) (1.9) having solutions f (x, y) = ax + by (1.10) and f (x, y) = ax3 + bx2y + cxy2 + dy3. (1.11) Recently, M.Arunkumar et.al., introduced and established the general solution and generalized Ulam -Hyers stability of a 2 - variable Associative functional equation g (x, u) + g (y + z, v + w) = g (x + y, u + v) + g (z, w) (1.12) having solutions g(x, y) = ax + by (1.13) using Banach and Intuitionistic Fuzzy Normed spaces, respectively. Inspired by the above results in this paper, the authors investigated the general solution generalized Ulam - Hyers stability of ai type n−variable multi n−dimensional additive functional equation 2h n ∑ i=1 ai x1i, n ∑ i=1 ai x2i, . . . . . . , n ∑ i=1 ai xni ! = n ∑ i=1 ai ! h n ∑ i=1 x1i, n ∑ i=1 x2i, . . . . . . , n ∑ i=1 xni ! + a1 − n ∑ i=2 ai ! h x11 − n ∑ i=2 x1i, x21 − n ∑ i=2 x2i, . . . . . . , xn1 − n ∑ i=2 xni ! (1.14) 280 Matina J. Rassias et al. / ai Type n−Variable Multi n−... having solution h(x1, x2, . . . , xn) = n ∑ i=1 cixi (1.15) where ai(i = 1, 2, . . . n) are different integers greater than 1, using Hyers direct and Alternative ﬁxed point methods. In particular, when n = 1, 2 in (1.14), we arrive 2h (a1 x11, a1 x21, . . . , a1 xn1) = a1h (x11, x21, . . . , xn1) + a1h (x11, x21 . . . , xn1) . (1.16) and 2h (a1 x11 + a2 x12, a1 x21 + a1 x22, . . . , a1 xn1 + a1 xn2) = (a1 + a2) h (x11 + x12, x21 + x22, . . . , xn1 + xn2) + (a1 −a2) h (x11 −x12, x21 −x22, . . . , xn1 −xn2) . (1.17) 2 General Solution In this section, the general solution of the functional equation (1.14) is given. Through out this section let as assume A and B be linear normed spaces. Lemma 2.1. If a mapping h : An →B satisﬁes the functional equation (1.14) then h is additive. Proof. Assume h : An →B be a mapping satisﬁes the functional equation (1.14). Replacing xmi = 0, i = 1, 2, . . . n, m = 1, 2, · · · n in (1.14), we get h(0, 0, . . . , 0) = 0. (2.1) Again replacing xmi = 0, i = 2, 3 . . . n, m = 1, 2, · · · n in (1.14), we obtain 2h(a1x11, a1x21, . . . , a1xn1) = (a1 + a2 + · · · + an)h(x11, x21, . . . , xn1) + (a1 −a2 −· · · −an)h(x11, x21, . . . , xn1) (2.2) for all x11, x21, . . . , xn1 ∈A. If we substitute (x11, x21, . . . , xn1) by (x, x . . . , x) in (2.2), we reach h(a1x, a1x, . . . , a1x) = a1 h(x, x, . . . , x) (2.3) for all x ∈A. Putting xmi = 0, i = 3, 4 . . . n, m = 1, 2, · · · n in (1.14), we obtain h(x12, 0, . . . , 0) = −h(−x12, 0, . . . , 0) (2.4) for all x12 ∈A. So one can show that h(ak 1x, ak 1x, . . . , ak 1x) = ak 1 h(x, x, . . . , x) (2.5) for all x ∈A and all k ∈N. Matina J. Rassias et al. / ai Type n−Variable Multi n−... 281 3 Stability Results: Banach Space: Hyers Method In this section, we investigate the generalized Ulam-Hyers stability of the functional equation (1.14). In this section, let we consider A be a normed space and B be a Banach space and deﬁne a mapping Dh : An →B by Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn) = 2h n ∑ i=1 ai x1i, n ∑ i=1 ai x2i, . . . . . . , n ∑ i=1 ai xni ! − n ∑ i=1 ai ! h n ∑ i=1 x1i, n ∑ i=1 x2i, . . . . . . , n ∑ i=1 xni ! − a1 − n ∑ i=2 ai ! h x11 − n ∑ i=2 x1i, x21 − n ∑ i=2 x2i, . . . . . . , xn1 − n ∑ i=2 xni ! for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈A. Theorem 3.1. Let ℓ= ±1 and ϑ, Θ : An →[0, ∞) be a function such that lim s→∞ 1 2sℓϑ asℓ 1 x11, . . . , asℓ 1 x1n, asℓ 1 x21, . . . , asℓ 1 x2n, asℓ 1 xn1, . . . , asℓ 1 xnn = 0 (3.1) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈A. Let h : An →B be a function satisfying the inequality ∥Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn)∥≤ n ∑ j=1 ϑj xj1, xj2, . . . , xjn (3.2) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈A. Then there exists a unique n−variable additive mapping A : An → B which satisﬁes (1.14) and ∥h(x, x, . . . , x) −A(x, x, . . . , x)∥≤1 a1 ∞ ∑ t=0 Θ(atℓ 1 x) atℓ 1 (3.3) where Θ atℓ 1 x and A(x, x, . . . , x) are deﬁned by Θ(atℓ 1 x) = 1 2 n ∑ j=1 ϑj   atℓ 1 x, 0, . . . , 0 \| {z } (n−1)−times    (3.4) and A(x, x, . . . , x) = lim s→∞ 1 asℓ 1 h(asℓ 1 x, asℓ 1 x, . . . , asℓ 1 x) (3.5) for all x ∈A, respectively. Proof. Given h : An →B be a function satisfying the inequality (3.2) for all x11, . . . , x1n, . . . , xn1, . . . , xnn ∈A. To establish this theorem, we have to show that (i) 1 as 1 h (as 1x, as 1x, . . . , as 1x) is a Cauchy sequence for every x ∈A; (ii) If A(x, x, . . . , x) = lim s→∞ 1 a1 h(as 1x, as 1x, . . . , as 1x) then A is additive on A; (iii) Further A satisﬁes (3.3), for all x ∈A; (iv) A is unique. 282 Matina J. Rassias et al. / ai Type n−Variable Multi n−... Replacing xmi = 0, i = 2, 3 . . . n, m = 1, 2, · · · n in (3.2), we get ∥2h(a1x11, a1x21, . . . , a1xn1) −(a1 + a2 + · · · + an)h(x11, x21, . . . , xn1) −(a1 −a2 −· · · −an)h(x11, x21, . . . , xn1)∥≤ n ∑ j=1 ϑj   xj1, 0, . . . , 0 \| {z } (n−1)−times    (3.6) for all x11, x21, . . . , xn1 ∈A. If we substitute xm1 = x, m = 1, 2, . . . n in (3.7), we arrive ∥2h(a1x, a1x, . . . , a1x) −2a1h(x, x, . . . , x)∥≤ n ∑ j=1 ϑj   x, 0, . . . , 0 \| {z } (n−1)−times    (3.7) for all x ∈A. Hence from (3.7), we reach 1 a1 h   a1x, a1x, . . . , a1x \| {z } n−times   −h  x, x, . . . , x \| {z } n−times   ≤ 1 2 × a1 n ∑ j=1 ϑj   x, 0, . . . , 0 \| {z } (n−1)−times    (3.8) for all x ∈A. It follows from (3.8) that 1 a1 h   a1x, a1x, . . . , a1x \| {z } n−times   −h  x, x, . . . , x \| {z } n−times   ≤1 a1 Θ(x) (3.9) where Θ(x) = 1 2 n ∑ j=1 ϑj   x, 0, . . . , 0 \| {z } (n−1)−times    for all x ∈A. Now replacing x by a1x and dividing by a1 in (3.9), we get 1 a2 1 h a2 1x, a2 1x, . . . , a2 1x −1 a1 h (a1x, a1x, . . . , a1x) ≤1 a2 1 Θ(a1x) (3.10) for all x ∈A. From (3.8) and (3.10), we obtain 1 a2 1 h a2 1x, a2 1x, . . . , a2 1x −h (x, x, . . . , x) ≤1 a1 Θ(x) + Θ(a1x) a1 (3.11) for all x ∈A. Proceeding further and using induction on a positive integer s, we get 1 as 1 h (as 1x, as 1x, . . . , as 1x) −h (x, x, . . . , x) ≤1 a1 s−1 ∑ t=0 Θ(at 1x) at 1 (3.12) for all x ∈A. In order to prove the convergence of the sequence 1 as 1 h (as 1x, as 1x, . . . , as 1x) , Matina J. Rassias et al. / ai Type n−Variable Multi n−... 283 replace x by ar 1x and dividing by ar 1 in (3.12), for any r, s > 0 , we deduce 1 ar+s 1 h ar+s 1 x, ar+s 1 x, . . . , ar+s 1 x −1 ar 1 h (ar 1x, ar 1x, . . . , ar 1x) = 1 ar 1 1 as 1 h (ar 1 · as 1x, ar 1 · as 1x, . . . , ar 1 · as 1x) −h (ar 1x, ar 1x, . . . , ar 1x) ≤1 a1 ∞ ∑ t=0 Θ(ar+s 1 x) ar+s 1 →0 as r →∞ for all x ∈A. Hence the sequence 1 as 1 h (as 1x, as 1x, . . . , as 1x) is a Cauchy sequence. Since B is complete, there exists a mapping A : An →B such that A(x, x, . . . x) = lim s→∞ 1 as 1 h (as 1x, as 1x, . . . , as 1x) , ∀x ∈A. Letting s →∞in (3.12), we see that (3.3) holds for all x ∈A. To prove that A satisﬁes (1.14), replacing xmi = as 1xmi, i = 1, 2, 3 . . . n, m = 1, 2, · · · n and dividing by as 1 in (3.2), we obtain 1 as 1 ∥Dh(as 1x11, . . . , as 1x1n, as 1x21, . . . , as 1x2n, as 1xn1, . . . , as 1xnn)∥ ≤1 as 1 n ∑ j=1 ϑj as 1xj1, as 1xj2, . . . , as 1xjn for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈A. Letting s →∞in the above inequality and using the deﬁni-tion of A(x, x, . . . , x), we see that DA(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn) = 0. Hence A satisﬁes (1.14) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈A. To prove that A(x, x, . . . x) is unique, let B(x, x, . . . x) be another n−variable additive mapping satisfying (1.14) and (3.3), then ∥A(x, x, . . . x) −B(x, x, . . . x)∥ = 1 as 1 ∥A(as 1x, as 1x, . . . as 1x) −B(as 1x, as 1x, . . . as 1x)∥ ≤1 2n {∥A(as 1x, as 1x, . . . as 1x) −h(as 1x, as 1x, . . . as 1x)∥ + ∥h(as 1x, as 1x, . . . as 1x) −B(as 1x, as 1x, . . . as 1x)∥} ≤2 a1 ∞ ∑ t=0 Θ(at+s 1 x) a(t+s) 1 →0 as s →∞ for all x ∈A. Thus A is unique. Hence for ℓ= 1 the Theorem holds. Now, replacing x by x a1 in (3.7), we reach 2h(x, x, . . . , x) −2a1h x a1 , x a1 , . . . , x a1 ≤ n ∑ j=1 ϑj   x a1 , 0, . . . , 0 \| {z } (n−1)−times    (3.13) for all x ∈A. Dividing the above inequality by 2, we obtain h(x, x, . . . , x) −a1h x a1 , x a1 , . . . , x a1 x ≤Θ x a1 (3.14) 284 Matina J. Rassias et al. / ai Type n−Variable Multi n−... where Θ x a1 = 1 2 n ∑ j=1 ϑj   x a1 , 0, . . . , 0 \| {z } (n−1)−times    for all x ∈A. The rest of the proof is similar to that of ℓ= 1. Hence for ℓ= −1 also the Theorem holds. This completes the proof of the theorem. The following Corollary is an immediate consequence of Theorem 3.1 concerning the Ulam-Hyers , Ulam-TRassias and Ulam-JMRassias stabilities of (1.14). Corollary 3.1. Let ρ and q be nonnegative real numbers. Let h : An →B be a function satisfying the inequality ∥Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn)∥≤            ρ, ρ n ∑ i=1 n ∑ m=1 \|\|xmi\|\|q, q ̸= 1; ρ n ∏ i=1 n ∏ m=1 \|\|xmi\|\|q + n ∑ i=1 n ∑ m=1 \|\|xmi\|\|nq, , nq ̸= 1; (3.15) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈A. Then there exists a unique n−variable additive function A : A → B such that ∥h(x, x . . . , x) −A(x, x, . . . x)∥≤                na1ρ 2\|a1 −1\|, na1ρ\|\|x\|\|q 2\|a1 −aq 1\| , na1ρ\|\|x\|\|nq 2\|a1 −anq 1 \| , (3.16) for all x ∈A. Now, we will provide an example to illustrate that the functional equation (1.14) is not stable for q = 1 in condition (ii) of Corollary 3.1. Example 3.1. Let ϑ : R →R be a function deﬁned by ϑ(x) = µx, if \|x\| <1 µ, otherwise where µ > 0 is a constant, and deﬁne a function h : Rn →R by h(x, x . . . , x) = ∞ ∑ n=0 ϑ(2nx) 2n f or all x ∈R. Then h satisﬁes the functional inequality \|Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn)\| ≤ 4 µ a1 (a1 −1)\|x\| (3.17) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈R. Then there do not exist a n−variable additive mapping A : Rn → R and a constant κ > 0 such that \|h(x, x . . . , x) −A(x, x, . . . x)\| ≤κ\|x\| f or all x ∈R. (3.18) Proof. Now \|h(x, x . . . , x)\| ≤ ∞ ∑ n=0 \|ϑ(an 1x)\| \|an 1\| = ∞ ∑ n=0 µ an 1 = a1µ a1 −1. Therefore, we see that h is bounded. We are going to prove that h satisﬁes (3.17). If xmi = 0, i = 1, 2, . . . , n, m = 1, 2, . . . , n then (3.17) is trivial. If \|xmi\| ≥1 a1 then the left hand side of (3.17) is less than 4 µ a1 a1 −1. Now suppose that 0 < \|xmi\| < 1 a1 . Then there exists a positive integer k such that 1 ak 1 ≤\|xmi\| < 1 ak−1 1 , (3.19) Matina J. Rassias et al. / ai Type n−Variable Multi n−... 285 so that ak−1 1 xmi < 1 a1 and consequently ak−1 1 (xmi), ak−1 1 (−xmi) ∈(−1, 1). Therefore for each p = 0, 1, . . . , k −1, we have ap 1(xmi), ap 1(−xmi) ∈(−1, 1) and 2ϑ ap 1 n ∑ i=1 ai x1i, ap 1 n ∑ i=1 ai x2i, . . . . . . , ap 1 n ∑ i=1 ai xni ! − n ∑ i=1 ai ! ϑ ap 1 n ∑ i=1 x1i, ap 1 n ∑ i=1 x2i, . . . . . . , ap 1 n ∑ i=1 xni ! − a1 − n ∑ i=2 ai ! ϑ ap 1x11 −ap 1 n ∑ i=2 x1i, ap 1x21 −ap 1 n ∑ i=2 x2i, . . . . . . , ap 1xn1 −ap 1 n ∑ i=2 xni ! = 0 for p = 0, 1, . . . , k −1. From the deﬁnition of h and (3.19), we obtain that 2h n ∑ i=1 ai x1i, n ∑ i=1 ai x2i, . . . . . . , n ∑ i=1 ai xni ! − n ∑ i=1 ai ! h n ∑ i=1 x1i, n ∑ i=1 x2i, . . . . . . , n ∑ i=1 xni ! − a1 − n ∑ i=2 ai ! h x11 − n ∑ i=2 x1i, x21 − n ∑ i=2 x2i, . . . . . . , xn1 − n ∑ i=2 xni ! ≤ ∞ ∑ p=0 1 an 1 2ϑ ap 1 n ∑ i=1 ai x1i, ap 1 n ∑ i=1 ai x2i, . . . . . . , ap 1 n ∑ i=1 ai xni ! − n ∑ i=1 ai ! ϑ ap 1 n ∑ i=1 x1i, ap 1 n ∑ i=1 x2i, . . . . . . , ap 1 n ∑ i=1 xni ! − a1 − n ∑ i=2 ai ! ϑ ap 1x11 −ap 1 n ∑ i=2 x1i, ap 1x21 −ap 1 n ∑ i=2 x2i, . . . . . . , ap 1xn1 −ap 1 n ∑ i=2 xni ! ≤ ∞ ∑ p=k 1 ap 1 2ϑ ap 1 n ∑ i=1 ai x1i, ap 1 n ∑ i=1 ai x2i, . . . . . . , ap 1 n ∑ i=1 ai xni ! − n ∑ i=1 ai ! ϑ ap 1 n ∑ i=1 x1i, ap 1 n ∑ i=1 x2i, . . . . . . , ap 1 n ∑ i=1 xni ! − a1 − n ∑ i=2 ai ! ϑ ap 1x11 −ap 1 n ∑ i=2 x1i, ap 1x21 −ap 1 n ∑ i=2 x2i, . . . . . . , ap 1xn1 −ap 1 n ∑ i=2 xni ! ≤ ∞ ∑ p=k 1 ap 1 4µ = 4 µ × a1 (a1 −1) ak 1 = 4 µ a1 (a1 −1)\|x\|. Thus h satisﬁes (3.17) for all xmi ∈R with 0 < \|xmi\| < 1 a1 . We claim that the additive functional equation (1.14) is not stable for q = 1 in condition (ii) Corollary 3.1. Indeed, assume the contrary that there exist a additive mapping A : Rn →R and a constant κ > 0 satisfying (3.18). Since h is bounded and continuous for all x ∈R, A is bounded on any open interval containing the origin and continuous at the origin. In view of Theorem 3.1, A must have the form A(x, x, . . . , x) = cx for any x in R. Thus, we obtain that \|h(x, x, . . . , x)\| ≤(κ + \|c\|) \|x\|. (3.20) 286 Matina J. Rassias et al. / ai Type n−Variable Multi n−... But, choose a positive integer i with iµ > κ + \|c\|. If x ∈ 0, 1 2i−1 , then 2px ∈(0, 1) for all p = 0, 1, . . . , i −1 . For this x, we get h(x, x, . . . , x) = ∞ ∑ p=0 ϑ(ap 1x) ap 1 ≥ i−1 ∑ p=0 µ(2px) 2p = iµx > (κ + \|c\|) x which contradicts (3.20). Therefore the additive functional equation (1.14) is not stable in sense of Ulam, Hyers and Rassias if q = 1, assumed in the inequality condition (ii) of (3.16). Now, we will provide an example to illustrate that the functional equation (1.14) is not stable for q = 1 n in condition (iii) of Corollary 3.1. Example 3.2. Let ϑ : R →R be a function deﬁned by ϑ(x) = µx, if \|x\| < 1 n µ n, otherwise where µ > 0 is a constant, and deﬁne a function h : Rn →R by h(x, x . . . , x) = ∞ ∑ n=0 ϑ(2nx) 2n f or all x ∈R. Then h satisﬁes the functional inequality \|Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn)\| ≤ 4 µ a1 n(a1 −1)\|x\| (3.21) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈R. Then there do not exist a n−variable additive mapping A : Rn → R and a constant κ > 0 such that \|h(x, x . . . , x) −A(x, x, . . . x)\| ≤κ\|x\| f or all x ∈R. (3.22) 4 Stability Results: Banach Space: Alternative Fixed Point Method In this section, we apply a ﬁxed point method for achieving stability of the functional equation (1.14) is present. Now, ﬁrst we will recall the fundamental results in ﬁxed point theory. Theorem 4.2. (Banach’s contraction principle) Let (X, d) be a complete metric space and consider a mapping T : X → X which is strictly contractive mapping, that is (A1) d(Tx, Ty) ≤Ld(x, y) for some (Lipschitz constant) L < 1. Then, (i) The mapping T has one and only ﬁxed point x∗= T(x∗); (ii)The ﬁxed point for each given element x∗is globally attractive, that is (A2) limn→∞Tnx = x∗, for any starting point x ∈X; (iii) One has the following estimation inequalities: (A3) d(Tnx, x∗) ≤ 1 1−L d(Tnx, Tn+1x), ∀n ≥0, ∀x ∈X; (A4) d(x, x∗) ≤ 1 1−L d(x, x∗), ∀x ∈X. Theorem 4.3. Suppose that for a complete generalized metric space (Ω, δ) and a strictly contractive mapping T : Ω→Ωwith Lipschitz constant L. Then, for each given x ∈Ω, either d(Tnx, Tn+1x) = ∞ ∀ n ≥0, or there exists a natural number n0 such that (FP1) d(Tnx, Tn+1x) < ∞for all n ≥n0 ; (FP2) The sequence (Tnx) is convergent to a ﬁxed to a ﬁxed point y∗of T (FP3) y∗is the unique ﬁxed point of T in the set ∆= {y ∈Ω: d(Tn0x, y) < ∞}; (FP4) d(y∗, y) ≤ 1 1−Ld(y, Ty) for all y ∈∆. Matina J. Rassias et al. / ai Type n−Variable Multi n−... 287 In this section, we take let us consider E and F to be a normed space and a Banach space, respectively and deﬁne a mapping Dh : E n →F by Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn) = 2h n ∑ i=1 ai x1i, n ∑ i=1 ai x2i, . . . . . . , n ∑ i=1 ai xni ! − n ∑ i=1 ai ! h n ∑ i=1 x1i, n ∑ i=1 x2i, . . . . . . , n ∑ i=1 xni ! − a1 − n ∑ i=2 ai ! h x11 − n ∑ i=2 x1i, x21 − n ∑ i=2 x2i, . . . . . . , xn1 − n ∑ i=2 xni ! for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈E. Theorem 4.4. Let h : E n →F be a mapping for which there exists a function ζ : E n →[0, ∞) with the condition lim k→∞ 1 τk i ζ(τk i x) = 0 (4.1) where τi = ( a1 i f i = 0; 1 a1 i f i = 1, (4.2) such that the functional inequality ∥Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn)∥≤ n ∑ j=1 ϑj xj1, xj2, . . . , xjn (4.3) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈E. If there exists L = L(i) < 1 such that the function x →Θ(x) = 1 2 n ∑ j=1 ϑj   x a1 , 0, . . . , 0 \| {z } (n−1)−times   , has the property 1 τi Θ(τix) = L Θ (x) . (4.4) for all x ∈E. Then there exists a unique additive mapping A : E →F satisfying the functional equation (1.14) and ∥h(x, x, . . . , x) −A(x, x, . . . , x)∥≤L1−i 1 −LΘ(x) (4.5) for all x ∈E. Proof. Consider the set Γ = { f / f : E n →F, f (0) = 0} and introduce the generalized metric on Γ, d( f, g) = inf{K ∈(0, ∞) :∥f (x, x, . . . , x) −g(x, x, . . . , x) ∥≤KΘ(x), x ∈E}. It is easy to see that (Γ, d) is complete. Deﬁne Υ : Γ →Γ by Υ f (x, x, . . . , x) = 1 τi f (τix, τix, . . . , τix), for all x ∈E. Now f, g ∈Γ, d( f, g) ≤K ⇒∥f (x, x, . . . , x) −g(x, x, . . . , x) ∥≤KΘ(x), x ∈E. ⇒ 1 τi f (τix, τix, . . . , τix) −1 τi g(τix, τix, . . . , τix) ≤1 τi KΘ(τix), x ∈E, ⇒ 1 τi f (τix, τix, . . . , τix) −1 τi g(τix, τix, . . . , τix) ≤LKΘ(x), x ∈E, ⇒∥Υ f (x, x, . . . , x) −Υg(x, x, . . . , x) ∥≤LKΘ(x), x ∈E, ⇒d(Υ f, Υg) ≤LK. 288 Matina J. Rassias et al. / ai Type n−Variable Multi n−... This implies d(Υ f, Υg) ≤Ld( f, g), for all f, g ∈Γ. i.e., T is a strictly contractive mapping on Γ with Lipschitz constant L. It follows from, (3.9) that ∥2h(a1x, a1x, . . . , a1x) −2a1h(x, x, . . . , x)∥≤ n ∑ j=1 ϑj   x, 0, . . . , 0 \| {z } (n−1)−times    (4.6) for all x ∈E. Now, from (4.6), we get 1 a1 h (a1x, a1x, . . . , a1x) −h (x, x, . . . , x) ≤ 1 2a1 Θ(x) (4.7) for all x ∈E. Using (4.4) for the case i = 0 it reduces to 1 a1 h (a1x, a1x, . . . , a1x) −h (x, x, . . . , x) ≤LΘ(x) for all x ∈E, i.e., d(Υh, h) ≤L ⇒d(Υh, h) ≤L = L1 < ∞. (4.8) Again replacing x = x ai in (4.6), we get h(x, x, . . . , x) −a1h x ai , x ai , . . . , x ai ≤1 2 n ∑ j=1 ϑj   x a1 , 0, . . . , 0 \| {z } (n−1)−times    (4.9) for all x ∈E. Using (4.4) for the case i = 1 it reduces to h(x, x, . . . , x) −a1h x ai , x ai , . . . , x ai ≤Θ(x) for all x ∈E, i.e., d(h, Υh) ≤1 ⇒d(h, Υh) ≤1 = L0 < ∞. (4.10) From (4.8) and (4.10), we arrive d(h, Υh) ≤L1−i. Therefore (FP1) holds. By (FP2), it follows that there exists a ﬁxed point A of Υ in Γ such that A(x, x, . . . , x) = lim k→∞ h(τk i x, τk i x . . . , τk i x) τk i , ∀x ∈E. (4.11) To order to prove A : E →F satisﬁes (1.14), replacing xmi = τk i xmi, i = 1, 2, 3 . . . n, m = 1, 2, · · · n in (4.3) and dividing by τk i , it follows from (4.1) that 1 τk i Dh(τk i x11, . . . , τk i x1n, τk i x21, . . . , τk i x2n, τk i xn1, . . . , τk i xnn) ≤1 τk i n ∑ j=1 ϑj τk i xj1, τk i xj2, . . . , τk i xjn for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈E. Letting k →∞in the above inequality and using the deﬁnition of A(x, x, . . . , x), we see that DA(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn) = 0. Hence A satisﬁes (1.14) for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈A. Matina J. Rassias et al. / ai Type n−Variable Multi n−... 289 By (FP3), A is the unique ﬁxed point of Υ in the set ∆= {A ∈Γ : d(h, A) < ∞}, such that ∥h(x, x, . . . , x) −A(x, x, . . . , x)∥≤KΘ(x) for all x ∈E and K > 0. Finally by (FP4), we obtain d(h, A) ≤ 1 1 −Ld(h, Υh) this implies d(h, A) ≤L1−i 1 −L which yields ∥h(x, x, . . . , x) −A(x, x, . . . , x)∥≤L1−i 1 −LΘ(x) this completes the proof of the theorem. The following corollary is an immediate consequence of Theorem 4.4 concerning the stability of (1.14). Corollary 4.2. Let h : E →F be a mapping and exists real numbers ρ and r such that ∥Dh(x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn)∥≤            ρ, ρ n ∑ i=1 n ∑ m=1 \|\|xmi\|\|q, q ̸= 1; ρ n ∏ i=1 n ∏ m=1 \|\|xmi\|\|q + n ∑ i=1 n ∑ m=1 \|\|xmi\|\|nq, , nq ̸= 1; (4.12) for all for all x11, . . . , x1n, x21, . . . , x2n, xn1, . . . , xnn ∈E. Then there exists a unique additive function A : E →F such that ∥h(x, x . . . , x) −A(x, x, . . . x)∥≤                nρ 2\|a1 −1\|, nρ\|\|x\|\|q 2\|a1 −aq 1\| , nρ\|\|x\|\|nq 2\|a1 −anq 1 \| , (4.13) for all x ∈E. Proof. Setting ϑ(x) =            ρ, ρ n ∑ i=1 n ∑ m=1 \|\|xmi\|\|q, ρ n ∏ i=1 n ∏ m=1 \|\|xmi\|\|q + n ∑ i=1 n ∑ m=1 \|\|xmi\|\|nq, , for all x ∈E. Now, 1 τk i ϑ(τk i x) =                  ρ τk i , ρ τk i n ∑ i=1 n ∑ m=1 \|\|τk i xi\|\|q, ρ τk i ( n ∏ i=1 n ∏ m=1 \|\|τk i xmi\|\|q + n ∑ i=1 n ∑ m=1 \|\|τk i xmi\|\|nq, ) , =              →0 as k →∞, →0 as k →∞, →0 as k →∞. Thus, (4.1) is holds. We, already have Θ(x) = 1 2 n ∑ j=1 ϑj   x a1 , 0, . . . , 0 \| {z } (n−1)−times   , 290 Matina J. Rassias et al. / ai Type n−Variable Multi n−... with the property 1 τi Θ(τix) = L Θ (x) for all x ∈E. Hence Θ(x) = 1 2 n ∑ j=1 ϑj   x a1 , 0, . . . , 0 \| {z } (n−1)−times   =            nρ 2 nρ 2 · aq 1 \|\|x\|\|q nρ 2 · anq 1 \|\|x\|\|nq. Also, 1 τi Θ(τix) =            nρ 2τi nρ 2τi \|\|τix\|\|q nρ 2τi \|\|τix\|\|nq. =            τ−1 i nρ 2 , τq−1 i nnρ\|\|x\|\|q 2 τnq−1 i nnρ\|\|x\|\|nq 2 =              τ−1 i Θ(x), τq−1 i Θ(x) τnq−1 i Θ(x). Hence the inequality (4.4) holds either, L = a−1 1 if i = 0 and L = 1 a−1 1 if i = 1. Now from (4.5), we prove the following cases for condition (i). Case:1 L = a−1 1 if i = 0 ∥h(x) −A(x)∥≤ a−1 1 1−0 1 −a−1 1 Θ(x) = nρ 2(a1 −1). Case:2 L = 1 a−1 1 or if i = 1 ∥h(x) −A(x)∥≤ 1 a−1 1 1−1 1 − 1 a−1 1 Θ(x) = nρ 2(1 −a1). Also the inequality (4.4) holds either, L = aq−1 1 for q < 1 if i = 0 and L = 1 aq−1 1 for q > 1 if i = 1. Now from (4.5), we prove the following cases for condition (ii). Case:3 L = aq−1 1 for q < 1 if i = 0 ∥h(x) −A(x)∥≤ a(q−1) 1 1−0 1 −a(q−1) 1 Θ(x) = nρ\|\|x\|\|q 2(a1 −aq 1) Case:4 L = 1 aq−1 1 for q > 1 if i = 1 ∥h(x) −A(x)∥≤ 1 a(q−1) 1 1−1 1 − 1 a(q−1) 1 Θ(x) = nρ\|\|x\|\|q 2(aq 1 −a1) . The proof of condition (iii) is similar to that of condition (ii). Hence the proof is complete. References J. Aczel and J. Dhombres, Functional Equations in Several Variables, Cambridge Univ, Press, 1989. M. Acikg¨ oz, A review on 2-normed structures, Int. J. Math. Anal., 1, No. 4 (2007), 187–191. Matina J. Rassias et al. / ai Type n−Variable Multi n−... 291 T. Aoki, On the stability of the linear transformation in Banach spaces, J. Math. Soc. Japan, 2 (1950), 64-66. M. Arunkumar, Solution and stability of Arun-additive functional equations, International Journal Mathe-matical Sciences and Engineering Applications, Vol 4, No. 3, August 2010, 33-46. M. Arunkumar, G. Ganapathy, S. Murthy, S. Karthikeyan, Stability of the generalized Arun-additive func-tional equation in Instutionistic fuzzy normed spaces, International Journal Mathematical Sciences and En-gineering Applications Vol.4, No. V, December 2010, 135-146. M. Arunkumar, C. Leela Sabari, Solution and stability of a functional equation originating from a chem-ical equation, International Journal Mathematical Sciences and Engineering Applications Vol. 5 No. II (March, 2011), 1-8. M. Arunkumar, V. Arasu, N. Balaji, Fuzzy Stability of a 2- Variable Quadratic Functional Equation, Interna-tional Journal Mathematical Sciences and Engineering Applications, Vol. 5 No. IV (July, 2011), 331-341. M. Arunkumar, S. Hema latha, C. Devi Shaymala Mary, Functional equation originating from arithmetic Mean of consecutive terms of an arithmetic Progression are stable in banach space: Direct and ﬁxed point method, JP Journal of Mathematical Sciences, Volume 3, Issue 1, 2012, Pages 27-43. M. Arunkumar, G. Vijayanandhraj, S. Karthikeyan, Solution and Stability of a Functional Equation Origi-nating From n Consecutive Terms of an Arithmetic Progression, Universal Journal of Mathematics and Math-ematical Sciences, Volume 2, No. 2, (2012), 161-171. M. Arunkumar, S. Hemalatha, 2-Variable AQ-Functional Equation, International Journal of Physical, Chemical and Mathematical Sciences (IJPCMS) - ISSN 2278 - 683X , Vol. 1; No. 2, (2012), 99 - 126. M. Arunkumar, Matina J. Rassias, Yanhui Zhang, Ulam - Hyers stability of a 2- variable AC - mixed type functional equation: direct and ﬁxed point methods, Journal of Modern Mathematics Frontier (JMMF), 1 (3), 2012, 10-26. M. Arunkumar, N. Maheshkumar, Ulam - Hyers, Ulam - TRassias, Ulam - JRassias stabilities of an additive functional equation in Generalized 2- Normed Spaces: Direct and Fixed Point Approach, Global Journal of Mathematical Sciences: Theory and Practical, Volume 5, Number 2 (2013), pp. 131-144. M. Arunkumar, Functional equation originating from sum of First natural numbers is stable in Cone Banach Spaces: Direct and Fixed Point Methods, International Journal of Information Science and Intelligent Sys-tem (IJISIS ISSN:2307-9142) , Vol. 2 No. 4 (2013), 89 - 104. M. Arunkumar, Perturbation of n Dimensional AQ - mixed type Functional Equation via Banach Spaces and Banach Algebra: Hyers Direct and Alternative Fixed Point Methods, International Journal of Advanced Math-ematical Sciences (IJAMS), Vol. 2 (1), 2014, 34-56. M. Arunkumar, C. Vijayalakshmi, G. Vijayanandhraj, 2 - Variable Associative functional equation, Pro-ceedings of the International Conference on Mathematical Methods and Computation, February 2014. J.H. Bae and W.G. Park , A functional equation orginating from quadratic forms, J. Math. Anal. Appl. 326 (2007), 1142-1148. D.G. Bourgin, Classes of transformations and bordering transformations, Bull. Amer. Math. Soc., 57, (1951), 223- 237. S. Czerwik, Functional Equations and Inequalities in Several Variables, World Scientiﬁc, River Edge, NJ, 2002. D.O. Lee, Hyers-Ulam stability of an addtiive type functional equation, J. Appl. Math. and Computing, 13 (2003) no.1-2, 471-477. P. Gavruta, A generalization of the Hyers-Ulam-Rassias stability of approximately additive mappings , J. Math. Anal. Appl., 184 (1994), 431-436. 292 Matina J. Rassias et al. / ai Type n−Variable Multi n−... D.H. Hyers, On the stability of the linear functional equation, Proc.Nat. Acad.Sci.,U.S.A.,27 (1941) 222-224. D.H. Hyers, G. Isac, Th.M. Rassias, Stability of functional equations in several variables, Birkhauser, Basel, 1998. S.M. Jung, Hyers-Ulam-Rassias Stability of Functional Equations in Mathematical Analysis, Hadronic Press, Palm Harbor, 2001. Pl. Kannappan, Functional Equations and Inequalities with Applications, Springer Monographs in Mathe-matics, 2009. S. Murthy, M. Arunkumar, G. Ganapathy, Fuzzy Stability Of A 3-D Additive Functional Equation: Hyers Direct And Fixed Point Method, Proceedings of National conference on Recent Trends in Mathematics and Computing (NCRTMC-2013), 69-79, ISBN 978-93-82338-68-0. B.Margoils and J.B.Diaz, A ﬁxed point theorem of the alternative for contractions on a generalized complete metric space, Bull. Amer. Math. Soc. 126 74 (1968), 305-309. W.G.Park, J.H.Bae, On the solution of a multi-additive functional equation and its stability, J. Appl. Math. and Computing, Vol. 22 No. 1 - 2, (2006), 517 - 522 V.Radu , The ﬁxed point alternative and the stability of functional equations, in: Seminar on Fixed Point Theory Cluj-Napoca, Vol. IV, 2003, in press. Matina J. Rassias, M. Arunkumar, S. Ramamoorthi, Stability of the Leibniz additive-quadratic functional equation in Quasi-Beta normed space: Direct and ﬁxed point methods, Journal Of Concrete And Applicable Mathematics (JCAAM), Vol. 14 No. 1-2, (2014), 22 - 46. J.M. Rassias, On approximately of approximately linear mappings by linear mappings, J. Funct. Anal. USA, 46, (1982) 126-130. Th.M. Rassias, On the stability of the linear mapping in Banach spaces, Proc.Amer.Math. Soc., 72 (1978), 297-300. K. Ravi, M. Arunkumar, On a n−dimensional additive Functional Equation with ﬁxed point Alternative, Proceedings of ICMS 2007,314-330. K.Ravi and M.Arunkumar, Stability of a 3- variable Quadratic Functional Equation, Journal of Quality Measurement and Analysis, July 4 (1), 2008, 97-107. K. Ravi, M. Arunkumar and J.M. Rassias, On the Ulam stability for the orthogonally general Euler-Lagrange type functional equation, International Journal of Mathematical Sciences, Autumn 2008 Vol.3, No. 08, 36-47. S.M. Ulam, Problems in Modern Mathematics, Science Editions, Wiley, NewYork, 1964. Received: April 13, 2016; Accepted: November 12, 2016 UNIVERSITY PRESS Website:
4532	https://web.stanford.edu/class/archive/cs/cs103/cs103.1142/lectures/00/Small00.pdf	Welcome to CS103! ●Lectures are recorded – sorry for being in such a packed room! ●Two Handouts ●Also available online if you'd like! ●Today: ●Course Overview ●Introduction to Set Theory ●The Limits of Computation Goals for this Course Goals for this Course ●How do we prove something with absolute certainty? ●Discrete Mathematics ●What problems can we solve with computers? ●Computability Theory ●Why are some problems harder to solve than others? ●Complexity Theory Course Staff Keith Schwarz (htiek@cs.stanford.edu) Kyle Brogle (broglek@stanford.edu) Berkeley Churchill (berkc@stanford.edu) Yifei Huang (yifei@stanford.edu) Jamie Irvine (jirvine@stanford.edu) Nicholas Isaacs (nisaacs@stanford.edu) Jeffrey Jacobs (jjacobs3@stanford.edu) Michael Kim (mpkim@stanford.edu) Stephen Macke (smacke@stanford.edu) Sathish Nagappan (srn@stanford.edu) Neha Nayak (nayakne@stanford.edu) Dilli Paudel (drpaudel@stanford.edu) Narek Tovmasyan (ntarmen1@stanford.edu) Course Staff Mailing List: cs103-aut1314-staff@lists.stanford.edu The Course Website “Prerequisite” CS106A Recommended Reading Online Course Notes Grading Policies 60% Assignments 15% Midterm 25% Final Let's Get Started! Introduction to Set Theory “The chemical elements” “Cute animals” “Cool people” “US coins.” “All the computers on the Stanford network.” “CS103 students” A set is an unordered collection of distinct objects, which may be anything (including other sets). , , , Set notation: Curly braces with commas separating out the elements Set notation: Curly braces with commas separating out the elements A set is an unordered collection of distinct objects, which may be anything (including other sets). , , , , , , These are the same set! These are the same set! A set is an unordered collection of distinct objects, which may be anything (including other sets). , , , , , , These are the same set! These are the same set! A set is an unordered collection of distinct objects, which may be anything (including other sets). Ø We use this symbol to denote the empty set. The empty set contains no elements. = Are these equal to one another? Ø Ø ≠ This set contains nothing at all. This set contains nothing at all. This set has one element, which happens to be the empty set. This set has one element, which happens to be the empty set. Are these equal to one another? 1 1 ≠ This is a number. This is a number. This is a set. It contains a number. This is a set. It contains a number. Membership , , , Is in this set? ∈ Membership , , , Is in this set? ∉ Set Membership ●Given a set S and an object x, we write x ∈ S if x is contained in S, and x ∉ S otherwise. ●If x ∈ S, we say that x is an element of S. ●Given any object and any set, either that object is an element of the set or it isn't. Infinite Sets ●Some sets contain infinitely many elements! ●The natural numbers, ℕ: { 0, 1, 2, 3, …} ●Some mathematicians don't include zero; in this class, assume that 0 is a natural number. ●The integers, ℤ: { …, -2, -1, 0, 1, 2, … } ●Z is from German “Zahlen.” ●The real numbers, ℝ, including rational and irrational numbers. ●e ∈ ℝ, π ∈ ℝ, 4 ∈ ℝ, etc. Describing Complex Sets ●Here are some English descriptions of infinite sets: “All even numbers.” “All real numbers less than 137.” “All negative integers.” ●We can't list the (infinitely many!) elements of these sets! ●How would we rigorously describe them? { n \| n ∈ ℕ and n is even } The set of all n n is a natural number and n is even Even Natural Numbers where { 0, 2, 4, 6, 8, 10, 12, 14, 16, … } Set Builder Notation ●A set may be specified in set-builder notation: { x \| some property x satisfies } ●For example: { r \| r ∈ ℝ and r < 137 } { n \| n is a power of two } { S \| S is a set of US currency } { a \| a is cute animal } Combining Sets Venn Diagrams A B 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Venn Diagrams A B A 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Venn Diagrams A B B 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Venn Diagrams A B A ∪ B 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Union { 1, 2, 3, 4, 5 } Venn Diagrams A B A ∩ B 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Intersection { 3 } Venn Diagrams A B A – B 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Difference { 1, 2 } Venn Diagrams A B A \ B 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Difference { 1, 2 } Venn Diagrams A B A Δ B 1 2 4 5 3 A = { 1, 2, 3 } B = { 3, 4, 5 } Symmetric Difference { 1, 2, 4, 5 } Venn Diagrams A B A Δ B Venn Diagrams Venn Diagrams for Three Sets Venn Diagrams for Four Sets A B C D Question to ponder: why can't we just draw four circles? Question to ponder: why can't we just draw four circles? Venn Diagrams for Five Sets Venn Diagrams for Seven Sets Subsets and Power Sets Subsets ●A set S is a subset of a set T (denoted S ⊆ T) if all elements of S are also elements of T. ●Examples: ●{ 1, 2, 3 } ⊆ { 1, 2, 3, 4 } ●ℕ ⊆ ℤ (every natural number is an integer) ●ℤ ⊆ ℝ (every integer is a real number) What About the Empty Set? ●A set S is a subset of a set T (denoted S ⊆ T) if all elements of S are also elements of T. ●Is Ø ⊆ S for any set S? ●Yes: This statement true for all sets S. ●Vacuous truth: A statement that is true because it does not apply to anything. ●“All unicorns are blue.” ●“All unicorns are pink.” Proper Subsets ●A set S is a subset of a set T (denoted S ⊆ T) if all elements of S are also elements of T. ●By definition, any set is a subset of itself. ●A proper subset of a set S is a set T such that T ⊆ S and T ≠ S. ●There are multiple notations for this: we either write T ⊊S or T ⊂ S. , , , , , S = ℘(S) = ℘(S) is the power set of S (the set of all subsets of S) ℘(S) is the power set of S (the set of all subsets of S) Ø What is (Ø)? ℘ Answer: {Ø} Cardinalities Cardinality ●The cardinality of a set is the number of elements it contains. ●We denote it \|S\|. ●Examples: ●\| { a, b, c, d, e} \| = 5 ●\| { {a, b}, {c, d, e, f, g}, {h} } \| = 3 ●\| { 1, 2, 3, 3, 3, 3, 3 } \| = 3 ●\| { n \| n ∈ ℕ and n < 137 } \| = 137 The Cardinality of ℕ ●What is \|ℕ\|? ●There are infinitely many natural numbers. ●\|ℕ\| can't be a natural number, since it's infinitely large. ●We need to introduce a new term. ●Definition: \|ℕ\| = ₀ ℵ. ●Pronounced “Aleph-Zero,” “Aleph-Nought,” or “Aleph-Null.” Consider the set S = { x \| x ∈ ℕ and x is even } What is \|S\|? How Big Are These Sets? , , , , , , Comparing Cardinalities ●Two sets have the same cardinality if their elements can be put into a one-to-one correspondence with one another. ●The intuition: , , , , , , , Comparing Cardinalities ●Two sets have the same cardinality if their elements can be put into a one-to-one correspondence with one another. ●The intuition: , , , , , We've run out of elements in the second set! Infinite Cardinalities 0 1 2 3 4 5 6 7 8 ... 0 2 4 6 8 10 12 14 16 ... n ↔ 2n S = { n \| n ∈ ℕ and n is even } \|S\| = \|ℕ\| = ℵ0 ℕ S Infinite Cardinalities 0 1 2 3 4 5 6 7 8 ... -3 -2 -1 ℕ ℤ 0 1 2 3 4 ... -4 n ↔ n / 2 (if n is even) n ↔ -(n + 1) / 2 (if n is odd) \|ℕ\| = \|ℤ\| = ℵ0 Infinite Cardinalities 0 1 2 3 4 5 6 7 ... 1 2 4 8 16 32 64 128 ... n ↔ 2n P = { n \| n ∈ ℕ and n is a power of two} \|P\| = \|ℕ\| = ℵ0 ℕ P Important Question Do all infinite sets have the same cardinality? Prepare for one of the most beautiful (and surprising!) results in mathematics... , , , , , S = ℘(S) = Ø \|S\| < \| (S)\| ℘ , S = ℘(S) = , , , Ø , , , , , , , , , , \|S\| < \| (S)\| ℘ S = {a, b, c, d} ℘(S) = { Ø, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {b, e} {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d} } \|S\| < \| ( ℘S)\| If S is infinite, what is the relation between \|S\| and \| ( ℘S)\|? Does \|S\| = \| ( ℘S)\|? If \|S\| = \| ( ℘S)\|, there has to be a one-to-one correspondence between elements of S and subsets of S. What might this correspondence look like? x0 x2 x3 x4 x5 { 0, 2, 4, ... } { 0, 2, 4, ... } { 0, 2, 4, ... } { 0, 2, 4, ... } { x0, x2, x4, ... } { x0, x3, x4, ... } { x4, ... } { x1, x4, ... } { 0, 2, 4, ... } { 0, 2, 4, ... } { “b”, “ab”, ... } { x0, x1, x2, x3, x4, x5, ... } x1 ... { x0, x5, ... } 0 1 2 3 4 5 ... x0 x1 x2 x3 x4 x5 ... x0 x2 x3 x4 x5 { 0, 2, 4, ... } ... Y Y Y N N N … Y Y Y N N N … Y N N N N N … Y Y N N N N … Y Y N N N N … Y Y Y Y Y Y … … … … … … … … x1 { 0, 2, 4, ... } { “b”, “ab”, ... } Y N Y N N … N Y N x0 x1 x2 x3 x4 x5 ... { 0, 2, 4, ... } Y Y Y N N N … Y Y Y N N N … Y N N N N N … Y Y N N N N … Y Y N N N N … Y Y Y Y Y Y … … … … … … … … Y N N N N Y … ... { 0, 2, 4, ... } { “b”, “ab”, ... } Y N Y N Y … N Y N x0 x2 x3 x3 x4 x1 x0 x1 x2 x3 x4 x5 ... Which row in the table is paired with this set? Which row in the table is paired with this set? x4 x5 { 0, 2, 4, ... } Y Y Y N N N … Y Y Y N N N … Y N N N N N … Y Y N N N N … Y Y N N N N … Y Y Y Y Y Y … … … … … … … … Y N N N N Y … ... { 0, 2, 4, ... } { “b”, “ab”, ... } Y N Y N Y … N Y N x0 x2 x3 x3 x4 x1 x0 x1 x2 x3 x4 x5 ... x4 x5 Flip all Y's to N's and vice-versa to get a new set Flip all Y's to N's and vice-versa to get a new set N Y Y Y N ... Y The Diagonalization Proof ●The complemented diagonal cannot appear anywhere in the table. ●In row n, the nth element must be wrong. ●No matter how we try to assign subsets of S to elements of S, there will always be at least one subset left over. ●Cantor's Theorem: Every set is strictly smaller than its power set: For any set S, \|S\| < \| ( ℘S)\| Infinite Cardinalities ●Recall: \|ℕ\| = ₀. ℵ ●By Cantor's Theorem: \|ℕ\| < \| (ℕ)\| ℘ \| (ℕ)\| < \| ( (ℕ))\| ℘ ℘℘ \| ( (ℕ))\| < \| ( ( (ℕ)))\| ℘℘ ℘℘℘ \| ( ( (ℕ)))\| < \| ( ( ( (ℕ))))\| ℘℘℘ ℘℘℘℘ … ●Not all infinite sets have the same size! ●There are infinitely many infinities! What does this have to do with computation? “The set of all computer programs” “The set of all problems to solve” Strings and Problems ●Consider the set of all strings: { “”, “a”, “b”, “c”, ..., “aa”, “ab”, “ac,” … } ●For any set of strings S, we can solve the following problem about S: Write a program that accepts as input a string, then prints out whether or not that string belongs to set S. ●Therefore, there are at least as many problems to solve as there are sets of strings. Every computer program is a string. So, there can't be any more programs than there are strings. From Cantor's Theorem, we know that there are more sets of strings than strings. There are at least as many problems as there are sets of strings. \|Programs\| \|Strings\| \|Sets of Strings\| \|Problems\| ≤ ≤ < \|Programs\| < \|Problems\| Every computer program is a string. So, there can't be any more programs than there are strings. From Cantor's Theorem, we know that there are more sets of strings than strings. There are at least as many problems as there are sets of strings. There are more problems to solve than there are programs to solve them. It Gets Worse ●Because there are more problems than strings, we can't even describe some of the problems that we can't solve. ●The set of all English phrases is no larger than the set of all strings, which is smaller than the set of all problems. ●Using more advanced set theory, we can show that there are infinitely more problems than solutions. ●In fact, if you pick a totally random problem, the probability that you can solve it is zero. But then it gets better... Where We're Going ●Given this hard theoretical limit, what can we compute? ●What are the hardest problems we can solve? ●How powerful of a computer do we need to solve these problems? ●Of what we can compute, what can we compute efficiently? ●What tools do we need to reason about this? ●How do we build mathematical models of computation? ●How can we reason about these models? Next Time ●Mathematical Proof ●What is a mathematical proof? ●How can we prove things with certainty?
4533	https://www.mcgillairflow.com/pdf/cas/designAdvisory01.pdf	Webster's Dictionary deﬁnes "funda-mentalism" as "a movement or point of view marked by a rigid adherence to fundamental or basic principles". At McGill AirFlow we take pride in having participated since 1951 in the develop-ment and establishment of duct sys-tem design fundamentals. This has included active participation in techni-cal and standards committees for pro-fessional organizations that serve the HVAC industry, such as ASHRAE, ASTM, AMCA, and SMACNA. This is the ﬁrst Design Advisory for McGill AirFlow's new Duct System Design Guide, being issued in bimonthly installments to CAS sub-scribers. These releases address real-world applications and design topics. Submit comments and questions about these releases to our Ask An HVAC Expert service! Chapter One of the Duct System Design Guide presents the fundamen-tals of duct system design — establish-ing a strong technical foundation that will aid in understanding and persevering over future topics. Design Advisory #1: CAS-DA1-2003 The Fundamentals of Duct System Design ©2003 McGill AirFlow Corporation 190 East Broadway Westerville, OH 43081 614/882-5455, Fax: 614/797-2175 E-mail: mafengineering@mcgillairﬂow.com To request the new Duct System Design Guide on CD along with McGill AirFlow's Products & Services Catalog please email us at: marketing@mcgillairﬂow.com Subject: DSDG CAS subscriber. Be sure to include your complete mailing address. Visit www.ductexpress.com! Request a quotation through our secure site. A bill of material is forwarded to our sales staff who will respond to your order within one business day. Over 300 products available for order online! Duct System Design Page 1.1 CHAPTER 1: Airflow Fundamentals for Supply Duct Systems 1.1 Overview This section presents basic airflow principles and equations for supply systems. Students or novice designers should read and study this material thoroughly before proceeding with the design sections. Experienced designers may find a review of these principles helpful. Those who are comfortable with their knowledge of airflow fundamentals may proceed to Chapter 2. Whatever the level of experience, the reader should find the material about derivations in Appendix A.3 interesting and informative. The two fundamental concepts, which govern the flow of air in ducts, are the laws of conservation of mass and conservation of energy. From these principles are derived the basic continuity and pressure equations, which are the basis for duct system designs. 1.2 Conservation of Mass The law of conservation of mass for a steady flow states that the mass flowing into a control volume must equal the mass flowing out of the control volume. For a one-dimensional flow of constant density, this mass flow is proportional to the product of the local average velocity and the cross-sectional area of the duct. Appendix A.3.1 shows how these relationships are combined to derive the continuity equation. 1.2.1 Continuity Equation The volume flow rate of air is the product of the cross-sectional area of the duct through which it flows and its average velocity. As an equation, this is written: Q = A x V Equation 1.1 where: Q = Volume flow rate (cubic feet per minute or cfm) A = Duct cross-sectional area (ft2) (=π D2/576 where D is diameter in inches) V = Velocity (feet per minute or fpm) The volume flow rate, velocity and area are related as shown in Equation 1.1. Knowing any two of these properties, the equation can be solved to yield the value of the third. The following sample problems illustrate the usefulness of the continuity equation. Duct System Design Page 1.2 Sample Problem 1-1 If the average velocity in a 20-inch diameter duct section is measured and found to be 1,700 feet per minute, what is the volume flow rate at that point? Answer: A = π x (202) / 576 = 2.18 ft2 V = 1,700 fpm Q = A x V = 2.18 ft2 x 1,700 fpm = 3,706 cfm Sample Problem 1-2 If the volume flow rate in a section of 24-inch duct is 5,500 cfm, what will be the average velocity of the air at that point? What would be the velocity if the same volume of air were flowing through a 20-inch duct? Answer: A24 = π x (242) / 576 = 3.14 ft2 Q = 5,500 cfm Q = A x V, therefore: V = Q/A V = (5,500 cfm) / 3.14 ft2 = 1,752 fpm (for 24-inch duct) A20 = π x (202) / 576 = 2.18 ft2 V = (5,500 cfm) / 2.18 ft2 = 2,523 fpm (for 20-inch duct) Sample Problem 1-3 If the volume flow rate in a section of duct is required to be 5,500 cfm, and it is desired to maintain a velocity of 2,000 fpm, what size duct will be required? Answer: V = 2,000 fpm Q = 5,500 cfm A = (5,500 cfm) / (2,000 fpm) = 2.75 ft2 D = (576 x A/π )0.5 = (576 x 2.75 ft2/π )0.5 = 22.45 inches Use: D = 22 inches then: Vactual = (5,500 cfm)/( π x (222) / 576) = 2083 fpm Duct System Design Page 1.3 1.2.2 Diverging Flows According to the law of conservation of mass, the volume flow rate before a flow divergence is equal to the sum of the volume flows after the divergence. Figure 1.1 and Equation 1.2 illustrate this point. Figure 1.1 Diverging Flow Qc = Qb + Qs Equation 1.2 where: Qc = Common (upstream) volume flow rate (cfm) Qb = Branch volume flow rate (cfm) Qs = Straight-through (downstream) volume flow rate (cfm) Duct System Design Page 1.4 Sample Problem 1-4 Figure 1.2 Multiple Diverging Flow The system segment shown in Figure 1.2 has four outlets, each delivering 200 cfm. What is the volume flow rate at points A, B and C? Answer: A = 800 cfm B = 600 cfm C = 400 cfm The total volume flow rate at any point is simply the sum of all the downstream volume flow rates. The volume flow rate of all branches and/or trunks of any system can be determined in this way and combined to obtain the total volume flow rate of the system. 1.3 Conservation of Energy The total energy per unit volume of air flowing in a duct system is equal to the sum of the static energy, kinetic energy and potential energy. When applied to airflow in ducts, the flow work or static energy is represented by the static pressure of the air, and the velocity pressure of the air represents the kinetic energy. Potential Duct System Design Page 1.5 energy is due to elevation above a reference datum and is often negligible in HVAC duct design systems. Consequently, the total pressure (or total energy) of air flowing in a duct system is generally equal to the sum of the static pressure and the velocity pressure. As an equation, this is written: TP = SP + VP Equation 1.3 where: TP = Total pressure SP = Static pressure VP = Velocity pressure Furthermore, when elevation changes are negligible, from the law of conservation of energy, written for a steady, non-compressible flow for a fixed control volume, the change in total pressure between any two points of a system is equal to the sum of the change in static pressure between the points and the change in velocity pressure between the points. This relationship is represented in the following equation: ∆ ∆TP = ∆ ∆SP + ∆ ∆VP Equation 1.4 Appendix A.3.2 shows the derivation of Equation 1.4 from the general equation of the first law of thermodynamics. Pressure (or pressure loss) is important to all duct designs and sizing methods. Many times, systems are sized to operate at a certain pressure or not in excess of a certain pressure. Higher pressure at the same volume flow rate means that more energy is required from the fan, and this will raise the operating cost. The English unit most commonly used to describe pressure in a duct system is the inch of water gauge ( inch wg). One pound per square inch ( psi), the standard measure of atmospheric pressure, equals approximately 27.7 inches wg. 1.3.1 Static Pressure Static pressure is a measure of the static energy of the air flowing in a duct system. It is static in that it can exist without a movement of the air stream. The air which fills a balloon is a good example of static pressure; it is exerted equally in all directions, and the magnitude of the pressure is reflected by the size of the balloon. The atmospheric pressure of air is a static pressure. At sea level, this pressure is equal to approximately 14.7 pounds per square inch. For air to flow in a duct system, a pressure differential must exist. That is, energy must be imparted to the system (by a fan or air handling device) to raise the pressure above or below atmospheric pressure. Air always flows from an area of higher pressure to an area of lower pressure. Because the static pressure is above atmospheric at a fan outlet, air will flow from the fan through any connecting ductwork until it reaches atmospheric pressure at the discharge. Because the static pressure is below atmospheric at a fan inlet, air will flow from the higher atmospheric pressure Duct System Design Page 1.6                         ρ ρ 4,005 V = 1,097 V = VP 2 2 through an intake and any connecting ductwork until it reaches the area of lowest static pressure at the fan inlet. The first type of system is referred to as a positive pressure or supply air system, and the second type as a negative pressure, exhaust, or return air system. Static Pressure Losses The initial static pressure differential (from atmospheric) is produced by adding energy at the fan. This pressure differential is completely dissipated by losses as the air flows from the fan to the system discharge. Static pressure losses are caused by increases in velocity pressure as well as friction and dynamic losses. Sign Convention When a static pressure measurement is expressed as a positive number, it means the pressure is greater than the local atmospheric pressure. Negative static pressure measurements indicate a pressure less than local atmospheric pressure. By convention, positive changes in static pressure represent losses, and negative changes represent regains or increases. For example, if the static pressure change as air flows from point A to point B in a system i s a positive number, then there is a static pressure loss between points A and B, and the static pressure at A must be greater than the static pressure at B. Conversely, if the static pressure change as air flows between these points is negative, the static pressure at B must be greater than the static pressure at A. 1.3.2 Velocity Pressure Velocity pressure is a measure of the kinetic energy of the air flowing in a duct system. It is directly proportional to the velocity of the air. For air at standard density (0.075 pounds per cubic foot), the relationship is: Equation 1.5 where: VP = Velocity pressure (inches wg) V = Air velocity (fpm) ρ ρ = Density (lbm/ft3) Appendix A.3.3 provides a derivation of this relationship from the kinetic energy term. This derivation also provides an equation for determining velocity pressures at nonstandard densities. Appendix A.1.6 provides a table of velocities and corresponding velocity pressures at standard conditions. Velocity pressure is always a positive number, and the sign convention for changes in velocity pressure is the same as that described for static pressure. From Equation 1.1, it can be seen that velocity must increase if the duct diameter (area) is reduced without a corresponding reduction in air volume. Similarly, the velocity must decrease if the air volume is reduced without a corresponding reduction in duct diameter. Thus, the Duct System Design Page 1.7 velocity and the velocity pressure in a duct system are constantly changing. 1.3.3 Total Pressure Total pressure represents the energy of the air flowing in a duct system. Because energy cannot be created or increased except by adding work or heat, there is no way to increase the total pressure once the air leaves the fan. The total pressure is at its maximum value at the fan outlet and must continually decrease as the air moves through the duct system toward the outlets. Total pressure losses represent the irreversible conversion of static and kinetic energy to internal energy in the form of heat. These losses are classified as either friction losses or dynamic losses. Friction losses are produced whenever moving air flows in contact with a fixed boundary. These are discussed in Section 1.4. Dynamic losses are the result of turbulence or changes in size, shape, direction, or volume flow rate in a duct system. These losses are discussed in Section 1.5. Referring to Equation 1.4, note that if the decrease in velocity pressure between two points in a system is greater than the total pressure loss, t he static pressure must increase to maintain the equality. Alternatively, an increase in velocity pressure will result in a reduction in static pressure, equal to the sum of the velocity pressure increase and the total pressure loss. When there is both a decrease in velocity and a reduction in static pressure, the total pressure will be reduced by the sum of these losses. These three concepts are illustrated in Figure 1.3. Figure 1.3 Conservation of Energy Relationship 1.4 Pressure Loss In Duct (Friction Loss) When air flows through a duct, friction is generated between the flowing air and the stationary duct wall. Energy must be provided to overcome this friction, and any energy converted irreversibly to heat is known as a friction loss. The fan initially provides this energy in the form of pressure. The amount of pressure necessary to overcome the friction in any section of duct depends on (1) the length of the duct, (2) the diameter of the duct, (3) the velocity (or volume) of the air flowing in the duct, and (4) the friction factor of the duct. Duct System Design Page 1.8                         1000 V D 1 2.56 = 100ft. P 1.8 1.18 ∆ 100ft. P ∆ The friction factor is a function of duct diameter, velocity, fluid viscosity, air density and surface roughness. For nonstandard conditions, see Section 1.5. The surface roughness can have a substantial impact on pressure loss, and this is discussed in Appendix A.3.5. These factors are combined in the Darcy equation to yield the pressure loss, or the energy requirement for a particular section of duct. Appendix A.3.4 discusses the use and application of the Darcy equation. 1.4.1 Round Duct One of the most important and useful tools available to the designer of duct systems is a friction loss chart (see Appendix A.4.1.1). This chart is based on the Darcy equation, and combines duct diameter, velocity, volume flow rate and pressure loss. The chart is arranged in such a manner that, knowing any two of these properties (at standard conditions), it is possible to determine the other two. The chart is arranged with pressure loss (per 100 feet of duct length) on the horizontal axis, volume flow rate on the vertical axis, duct diameter on diagonals sloping upward from left to right, and velocity on diagonals sloping downward from left to right. Examination of this chart (or the Darcy equation) reveals several interesting air flow properties: (1) at a constant volume flow rate, reducing the duct diameter will increase the pressure loss; (2) to maintain a constant pressure loss in ducts of different size, larger volume flow rates require larger duct diameters; and (3) for a given duct diameter, larger volume flow rates will increase the pressure loss. The following sample problems will give the reader a feel for these important relationships. Although there are many nomographs or duct calculators available to speed the calculation of duct friction loss problems, novice designers should use the friction loss charts to better visualize the relationships. The friction loss chart in the Appendix is approximated by Equation 1.6. Equation 1.6. where: = the friction loss per 100 ft of duct (inches wg) D = the duct diameter (inches) V = the velocity of the air flow in the duct (fpm) Sample Problem 1-5 What is the friction loss of a 150-foot long section of 18-inch diameter duct, carrying 2,500 cfm? What is the air velocity in this duct? Answer: From the friction loss chart, find the horizontal line that represents 2,500 cfm. Move across this line to the point where it intersects the diagonal line which represents an 18-inch diameter duct. From this point, drop down to the horizontal Duct System Design Page 1.9 wg inches 100ft. P 0.16 = 1000 1412 18 1 2.56 = 1.8 1.18                         ∆ (pressure) axis and read the friction loss. This value is approximately 0.16 inches wg. This represents the pressure loss of a 100-foot section of 18-inch diameter duct carrying 2,500 cfm. To determine the pressure loss for a 150-foot duct section, it is necessary to multiply the 100-foot loss by a factor of 1.5. Therefore, the pressure loss is 0.24 inches wg. At the intersection of the 2,500 cfm line and the 18-inch diameter line, locate the nearest velocity diagonal. The velocity is approximately 1,400 fpm. Equation 1.6 could have also been used to solve this problem. To use Equation 1.6 we must first calculate the velocity using Equation 1.1. To calculate the velocity, we have to determine the cross-sectional area of the 18-inch diameter duct. A = π x 182 / 576 = 1.77 ft2. The calculated velocity is Q/A = 2500/1.77 = 1412 fpm. The pressure loss per 100 ft is calculated as: For a 150 ft section the pressure is 1.5 x 0.16 = 0.24 inches wg. Sample Problem 1-6 Part of a system you have designed includes a 20-inch diameter, 500-foot duct run, carrying 3,000 cfm. You now discover there is only 16 inches of space in which to install this section. What will be the increase in pressure loss of the duct section if 16-inch duct is used in place of 20-inch? Answer: From the friction loss chart or Equation 1.6, find the pressure loss per 100 feet for 3,000 cfm flowing through a 20-inch diameter duct. This value is 0.14 inches wg, or 0.70 inches wg for 500 feet. Similarly, the friction loss for 500 feet of 16-inch diameter duct carrying 3,000 cfm is found to be 1.95 inches wg. Reducing this duct diameter by 4 inches results in an increase of pressure loss of 1.25 inches wg. Sample Problem 1-7 An installed system includes a 20-inch diameter, 500-foot duct run carrying 3,000 cfm. Due to unanticipated conditions downstream, it has become necessary to increase the volume flow rate in this duct section to 3,600 cfm. What will be the impact on the pressure loss of the section? Answer: From Sample Problem 1-6, the pressure drop of this section, as installed, is 0.70 inches wg. To find the new pressure drop, move up the 20-inch diameter line until it intersects the 3,600 cfm volume flow rate line. At this point, read down to the friction loss axis and find that the new friction loss is 0.19 inches wg per 100 Duct System Design Page 1.10 feet, or 0.95 inches wg for 500 feet. Therefore, this 20 percent increase in volume will result in a 36 percent increase in the pressure loss for the duct section. Sample Problem 1-8 A system is being designed so that the pressure loss in all duct sections is equal to 0.2 inches wg per 100 feet. What size duct will be required to carry (a) 500 cfm; (b) 1,500 cfm; (c) 5,000 cfm; (d) 15,000 cfm? Answer: From the friction loss chart, find the vertical line that represents 0.20 inches wg per 100 feet friction loss. Move up this line to the point where it intersects the horizontal line, which represents a volume flow rate of 500 cfm. At this point, locate the nearest diagonal line, representing duct diameter. It will either be 9-inch duct or 9.5-inch duct (if half-inch duct sizes are available). Similarly, for the other volumes, find the line representing duct size, which is nearest to the intersection of the 0.20 inches wg per 100 feet friction loss line and the appropriate volume. At 1,500 cfm, 14-inch duct is required; at 5,000 cfm, 22-inch duct is required; and at 15,000 cfm, 33-inch duct is required. In each case, the duct will carry the specified volume with a pressure loss of approximately 0.20 inches wg per 100 feet. 1.4.2 Flat Oval Duct Flat oval duct has the advantage of allowing a greater duct cross-sectional area to be accommodated in areas with reduced vertical clearances. Figure 1.4 shows a typical cross section of flat oval duct. References in Appendices A.1, A.2, and A.9 provide additional information about flat oval duct. Figure 1.4 Flat Oval Dimensions The Darcy equation is not applicable to flat oval duct, and there are no friction loss charts available for non-round duct shapes. To calculate the friction loss for flat oval duct, it is necessary to determine the equivalent round diameter of the flat oval size, and then determine the friction loss for the equivalent round duct. Duct System Design Page 1.11 ) (P ) (A 1.55 = D 0.25 0.625 eq 144 4 ) min x ( + ) min x (FS = A 2 π π 12 x FS) 2 ( + ) min x ( = P π π The equivalent round diameter of flat oval duct is the diameter of round duct that has the same pressure loss per unit length, at the same volume flow rate, as the flat oval duct. Equation 1.7 can be used to calculate the equivalent round diameter for flat oval duct with cross-sectional area A, and perimeter P. The equivalent round diameters for many standard sizes of flat oval duct are given in Appendix A.1.3.2. Equation 1.7 where: Deq = Equivalent round diameter (ft) A = Flat oval cross-sectional area (ft2) P = Flat oval perimeter (ft) The flat oval cross-sectional area is calculated using Equation 1.8. Equation 1.8 where: A = Cross-sectional area (ft2) FS = Flat Span (inches) = maj - min min = Minor axis (inches) maj = Major axis (inches) The perimeter of flat oval is calculated using Equation 1.9. Equation 1.9 where: P = Flat oval perimeter (ft) When calculating the air velocity in flat oval duct, it is necessary to use the actual cross-sectional area of the flat oval shape, not the area of the equivalent round duct. To use Equation 1.6 however, the air velocity is calculated using the equivalent round diameter cross sectional area. Duct System Design Page 1.12 Sample Problem 1-9 A 12-inch x 45-inch flat oval duct is designed to carry 10,000 cfm. What is the pressure loss per 100 feet of this section? What is the velocity? Answer: Equations 1.8 and 1.9 are used to calculate the area and perimeter of the flat oval duct. For 12 x 45 duct, A = 3.54 ft2 from: {(45 - 12) x 12 + (π x 122)/4}/144 and P = 8.64 ft from: {(π x 12) + (2 x (45-12))}/12. Substituting into Equation 1.7, Deq = 1.99 ft . 24 inches. From the friction loss chart, the pressure loss of 10,000 cfm air flowing in a 24-inch round duct is 0.50 inch wg per 100 feet. The velocity can be calculated from Equation 1.1: V = Q/A = 10,000 / 3.54 = 2,825 fpm Note that the velocity of the same air volume flowing in the 24-inch diameter round duct is 10,000 / 3.14, or 3,185 fpm. Alternatively, to use Equation 1.6, we must first calculate the velocity assuming the cross-sectional area is determined from the equivalent round ft A 2 Deq 3.14 = 576 ) 24 x ( 2 π = fpm V Deq 3185 = 3.14 10000 = wg inches 100ft. P 0.48 = 1000 3185 24 1 2.56 = 1.8 1.18                         ∆ Which is close to the 0.50 incheswg determined from the friction loss chart. Sample Problem 1-10 In Sample Problem 1-6, what size flat oval duct would be required in order to maintain the original (0.70 inches wg) pressure drop, and still fit within the 16-inch space allowance? Answer: In this situation, the available space will dictate the minor axis dimension of the Duct System Design Page 1.13 flat oval duct. It is always advisable to allow at least 2 to 4 inches for the reinforcement, which may be required on any flat oval or rectangular duct product. Therefore, we will assume that the largest minor axis that can be accommodated is 12 inches. Since we want to select a flat oval size which will have the same pressure loss as a 20-inch round duct, the major axis dimension can be determined by solving Equation 1.7 with Deq = 1.67 feet (20 inches), and min = 12 inches. Using Equations 1.8 and 1.9 for determining A and P, as functions of the major axis dimension (maj) and the minor axis dimension (min). Unfortunately, this requires an iterative solution. A simpler solution is to refer to the tables in Appendix A.1.3.2. These tables list the various available flat oval sizes and their respective equivalent round diameters. Since we already know that the minor axis must be 12 inches, we look for a flat oval size with a 12-inch minor axis and an equivalent round diameter of 20 inches. The required flat oval size is 12 inches x 31 inches. If it is determined that there is room for a 14-inch minor axis duct, the required size would be 14 inches x 27 inches (Deq = 20 inches). 1.4.3 Rectangular Duct Rectangular duct is fabricated by breaking two individual sheets of sheet metal (called L -sections) that have the appropriate duct dimensions (side and side adjacent) and joining them together by one of several techniques. Rectangular duct is also used when height restrictions are employed in a duct design. Equation 1.10 can be used to calculate the equivalent round diameter Deq of rectangular duct. The equivalent round diameter Deq for many standard sizes of rectangular duct are given in Appendix 1.4. Equation 1.10 where: Deq = Equivalent round diameter (inches) a = Duct side length (inches) b = Other duct side length (inches) When calculating air velocity in rectangular duct, it is necessary to use the actual cross-sectional area of the rectangular shape not the area of the equivalent round diameter. ) b + (a ) (ab 1.30 = D 0.250 0.625 eq Duct System Design Page 1.14 Sample Problem 1-11 A 12-inch x 45-inch rectangular duct is designed to carry 10,000 cfm. What is the pressure loss per 100 feet of this section? What is the velocity? Answer: Using Equation 1.10, Deq = 24 inches. From the friction loss chart, the loss of 10,000 cfm air flowing in a 24-inch round duct is 0.50 inch wg per 100 feet. The velocity can be calculated from Equation 1.1: fpm A Q = V 2,667 = 45)/144 x (12 10000 = Note that the velocity in the rectangular duct is less than in the flat oval with the same major and minor dimensions. (See Sample Problem 1-9) 1.4.4 Acoustically Lined and Double-wall Duct Applying an inner liner or a perforated inner metal shell sandwiching insulation between an inner and outer wall, increases the surface roughness that air sees and thus increases the friction losses of duct. Acoustically lined round and rectangular duct consist of a single-wall duct with an internal insulation liner but no inner metal shell. Double-wall duct that is acoustically insulated consists of a solid outer shell, a thermal/acoustical insulation, and a metal inner liner (either solid or perforated). The inner dimensions of lined duct or the metal inner liner dimensions of double-wall duct, are the nominal duct size dimensions that are used to determine the cross-sectional area for airflow calculations. The single-wall dimensions of lined duct or outer shell dimensions of double-wall duct depend on the insulation thickness. For a 1-inch thick insulation, the dimensions are 2 inches larger than the inner dimensions of lined duct or metal inner liner dimensions. Acoustically Lined Duct Correction factors to the friction loss determined from the friction loss chart or for Equation 1.6 have not been developed for internally insulated duct. Therefore the designer must use the Darcey equation as given in Appendix A.3.4. Assume an absolute roughness of ε = 0.015. Double-wall Duct Corrections factors to the friction loss determined from the friction loss chart or Equation 1.6 have been developed for when a perforated metal inner liner is used. Figure 1.5 is a chart, which gives the correction factors. This information is repeated in Appendix A.4.1.2. Note that these corrections are a function of duct diameter and velocity. If the duct shape is flat oval or rectangular, use the equivalent round diameter based on the perforated metal inner liner dimensions. If the inner shell of the double-wall duct does not use perforated metal, use the same friction loss as a single-wall duct of the same dimensions as the metal inner shell. Duct System Design Page 1.15 1.3 1.2 1.1 1.0 4” 8” 12” 18” 36” 72” 2,000 3,000 4,000 5,000 6,000 Correction factor to be applied to the friction loss of single-wall duct to calculate the friction loss of double-wall duct with a perforated metal inner liner Velocity (fpm) Figure 1.5 Correction Factors for Double-Wall Duct with Perforated Metal Inne r Liner When only thermal insulation is required, the metal inner liner may be specified as solid rather than perforated metal. In this case, the friction losses are identical to those for single-wall duct with a diameter equal to the metal inner liner diameter. For acoustically insulated flat oval or rectangular duct with a perforated metal inner liner, use the correction factors of the equivalent round diameter and the actual velocity (based on the metal inner liner of the flat oval or rectangular cross section). The reference in Appendix A.9.2 addresses friction losses for lined rectangular duct. Sample Problem 1-12 What is the friction loss of a 100-foot section of 22-inch diameter double-wall duct (with perforated metal inner liner), carrying 8,000 cfm? Answer: The pressure loss for 100 feet of 22-inch, single-wall duct, carrying 8,000 cfm is found from the friction loss chart to be 0.50 inches wg. The velocity is 3,000 fpm. From Figure 1.5, the correction factor for 22-inch duct (interpolated) at 3,000 fpm is approximately1.16.Therefore; the pressure loss of this section of double-wall duct is 0.50 x 1.16, or 0.58 inches wg. 1.4.5 Nonstandard Conditions All loss calculations thus far have been made assuming a standard air density of 0.075 pounds per cubic foot. When the actual design conditions vary appreciably from standard (i.e., temperature is " 30°F from 70°F, elevation above 1,500 feet, or moisture greater than 0.02 pounds water per pound dry air), the air density and viscosity will change. If the Darcy equation is used to calculate friction losses and the friction factor and velocity pressure are calculated using Duct System Design Page 1.16 actual conditions, no additional corrections are necessary. If a nomograph or friction chart is used to calculate friction losses at standard conditions, correction factors should be applied. The corrections for nonstandard conditions discussed above apply to duct friction losses only. Other corrections are applicable to the dynamic losses of fittings, as will be explained in the following section. For a more in-depth presentation of these and other correction factors, see Reference in Appendix A.9.2. Tables for determining correction factors are included in Appendix A.1.5. A temperature correction factor, Kt , can be calculated as follows:                 460) + T ( 530 = K a 0.825 t Equation 1.11 where: Kt = Nonstandard temperature correction factor Ta = Actual temperature of air in the duct (°F) An elevation correction factor, Ke , can be calculated as follows: ] (Z) ) 10 x (6.8754 -1 [ = K 4.73 -6 e Equation 1.12a Equation 1.12a can also be written as follows:             β β 29.921 = K 0.9 e Equation 1.12b where: Ke = Nonstandard elevation correction factor Z = Elevation above sea level (feet) β β = Actual barometric pressure (inches Hg) When both a nonstandard temperature and a nonstandard elevation are present, the correction factors are multiplicative. As an equation: K x K = K e t f Equation 1.13 where: Kf = Total friction loss correction factor The calculated duct friction pressure loss should be multiplied by the appropriate correction factor, Kt , Ke , or Kf , to obtain the actual pressure loss at the nonstandard conditions. Duct System Design Page 1.17 Sample Problem 1-13 The friction loss for a certain segment of a duct system is calculated to be 2.5 inches wg at standard conditions. What is the corrected friction loss if (a) the design temperature is 30°F; (b) the design temperature is 110°F; (c) the design elevation is 5,000 feet above sea level; (d) both (b) and (c). Answer: 1. Substituting into Equation 1.11: Kt = [530/(30 + 460)]0.825 = 1.07; Corrected friction loss = 2.5 x 1.07 = 2.68 inches wg. 2. Kt = [530/(110 + 460)]0.825 = 0.94; Corrected friction loss = 2.35 inches wg. 3. Substituting into Equation 1.12a: Ke = [1 - (6.8754 x 10-6)(5,000)] 4.73 = 0.85; Corrected friction loss = 2.13 inches wg. 4. Substituting into Equation 1.13: Kf = 0.94 x 0.85 = 0.80; Corrected friction loss = 2.00 inches wg. If moisture in the airstream is a concern, a humidity correction factor, Kh can be calculated as follows: ( ( ) ) 9 . 0 ws h P 378 . 0 1 K                 β β − − = = Equation 1.14 where: Pws = Saturation pressure of water vapor at the dew point temperature, (inches Hg) β β = Actual barometric pressure, (inches Hg) The total friction loss correction factor, Kf ,is expressed as: h e t f K x K x K = K Equation 1.15 1.5 Pressure Loss in Supply Fittings As mentioned in Section 1.3, pressure losses can be the result of either friction losses or dynamic losses. Section 1.4 discussed friction losses produced by air flowing over a fixed boundary. This section will address dynamic losses. Friction losses are primarily associated with duct sections, while dynamic losses are exclusively attributable to fittings or obstructions. Dynamic losses will result whenever the direction or volume of air flowing in a duct is altered or when the size or shape of the duct carrying the air is altered. Fittings of any type will produce dynamic losses. The dynamic loss of a fitting is generally proportional to the severity of the airflow disturbance. A smooth, large radius elbow, for example, will have a much lower dynamic Duct System Design Page 1.18 loss than a mitered (two-piece) sharp-bend elbow. Similarly, a 45° branch fitting will usually have lower dynamic losses than a straight 90° tee branch. 1.5.1 Loss Coefficients In order to quantify fitting losses, a dimensionless parameter known as a loss coefficient has been developed. Every fitting has associated loss coefficients, which can be det ermined experimentally by measuring the total pressure loss through the fitting for varying flow conditions. Equation 1.16a is the general equation for the loss coefficient of a fitting. VP TP = C ∆ Equation 1.16a where: C = Fitting loss coefficient ∆ ∆TP = Change in total pressure of air flowing through the fitting (inches wg) VP = Velocity pressure of air flowing through the fitting (inches wg) Once the loss coefficient for a particular fitting or class of fittings has been experimentally determined, the total pressure loss for any flow condition can be determined. Rewriting Equation 1.16a, we obtain: VP x C = TP ∆ Equation 1.16b From this equation, it can be seen that the total pressure loss is directly proportional to both the loss coefficient and the velocity pressure. Higher loss coefficient values or increases in velocity will result in higher total pressure losses for a fitting. A less efficient fitting will have a higher loss coefficient (i.e., for a given velocity, the total pressure loss is greater). 1.5.2 Elbows Table 1.1 shows typical loss coefficients for 8-inch diameter elbows of various construction. Table 1.1 Loss Coefficient Comparisons for Abrupt-Turn Fittings 90E E Elbows, 8-inch Diameter Fitting Loss Coefficient Die-Stamped/Pressed, 1.5 Centerline Radius 0.11 Five-Piece, 1.5 Centerline Radius 0.22 Mitered with Turning Vanes 0.52 Mitered 1.24 From Equation 1.4, we can determine the pressure loss: ∆ ∆TP = ∆ ∆SP + ∆ ∆VP Duct System Design Page 1.19 Since the elbow diameter and volume flow rate are constant, the continuity equation (Equation 1.1) tells us that the velocity will be constant. From Equation 1.5, the velocity pressure is a direct function of velocity, and so ∆ ∆VP = 0. Therefore, ∆ ∆SP = ∆ ∆TP . Note that whenever there is no change in velocity, as is the case in duct and constant diameter elbows, the change in static pressure is equal to the change in total pressure. Sample Problem 1-14 What is the total pressure loss of an 8-inch diameter die-stamped elbow carrying 600 cfm? What is the static pressure loss? Answer: From Equation 1.1: Q = A x V or V = Q/A A = π D2/576 = π (8)2 / 576 = 0.35 ft2 V = 600 / 0.35 V = 1,714 fpm From Equation 1.5 or Appendix A.1.6: wg inches VP 0.18 = 4,005 1,714 = 2             From Table 1.1: C = 0.11 (die-stamped elbow) From Equation 1.16b: ∆ TP = C x VP = 0.11 x 0.18 = 0.02 inches wg Sample Problem 1-15 A designer is trying to determine which 8-inch elbow to select for a location, which will have a design velocity of 1,714 fpm. What will be the implications, in terms of pressure loss, if the designer chooses (1) a die-stamped elbow, (2) a five-gore elbow, (3) a mitered elbow with turning vanes, or (4) a mitered elbow without turning vanes? Answer: From Sample Problem 1-14, we calculated the total pressure loss of an 8-inch die-stamped elbow at 1,714 fpm to be 0.02 inches wg. For the other elbows, we can determine the pressure loss from loss coefficients given in Table 1.1: Duct System Design Page 1.20 C2 = 0.22 ; C3 = 0.52; C4 = 1.24 From Equation 1.16b: ∆ TP2 = ∆ SP2 = 0.22 x 0.18 = 0.04 inches wg (five-gore) ∆ TP3 = ∆ SP3 = 0.52 x 0.18 = 0.09 inches wg (mitered with turning vanes) ∆ TP4 = ∆ SP4 = 1.24 x 0.18 = 0.22 inches wg (mitered without turning vanes) Therefore, using a five-gore elbow will increase the total pressure loss by 100 percent, but it will be a very modest 0.04 inches wg. Using the mitered elbow with vanes would result in a 350 percent increase over the die-stamped elbow, or a 125 percent increase over the five-gore elbow. The mitered elbow without turning vanes would have a loss of 0.22 inches wg, which is a tenfold increase over the die-stamped elbow. The increased pressure losses associated with the use of les s efficient fittings may or may not be critical to the operation of the system, depending on the location of the fittings. Succeeding chapters will note when there could be locations in a system where it is desirable to increase the losses of certain fittings. In general, unless the system has been carefully analyzed to determine the location of the critical path(s) and the excess pressures present in other paths, it is wise to always select fittings with the lowest pressure drop. The loss coefficients of most elbows vary as a function of diameter. The ASHRAE Duct Fitting Database Program (Appendix A.8.2) presents loss coefficients as a function of diameter for various elbow constructions. The loss coefficient drops sharply as diameters increase through approximately 24 inches, then only slightly from 24 inches through 60 inches. Also, eliminating turning vanes in mitered elbows more than doubles the pressure loss. Flat Oval Elbows Although the use of equivalent duct lengths as a measure of dynamic fitting losses is usually strongly discouraged, it provides acceptable approximations in the case of flat oval elbows. Data indicates that flat oval 90° elbows (hard or easy bend), with 1.5 centerline radius bends, have a pressure loss approximately equal to the friction loss of a flat oval duct with an identical cross section and a length equal to nine times the elbow major axis dimension, calculated at the same air velocity that is flowing through the elbow. For example, a 12-inch x 31-inch flat oval elbow would have a pressure loss approximately equal to that of a 12-inch x 31-inch flat oval duct, 23 feet long (9 x 31 inches) at the same velocity. For flat oval elbows that do not have a 1.5 centerline radius bend, use the loss coefficient for a round elbow of similar construction, with the diameter equal to the flat oval minor axis. Rectangular Elbows (see ASHRAE= = s Duct Fitting Database Program) Duct System Design Page 1.21 VP TP = C b b -c b ∆ Acoustically Lined/Double-wall Elbows For acoustically lined elbows or double-wall elbows with either a solid or perforated metal inner liner, the losses are the same as for standard single-wall elbows with dimensions equal to the metal inner liner dimensions of the acoustically lined or double-wall elbow. Elbows With Bend Angles Less Than 90° For elbows constructed with bend angles less than 90°, multiply the calculated pressure loss for a 90° elbow by the correction factor given in Table 1.2. Table 1.2 Elbow Bend Angle Correction Factor Angle CFelb 22.5° 0.31 30° 0.45 45° 0.60 60° 0.78 75° 0.90 1.5.3 Diverging-Flow Fittings: Branches The pressure losses in diverging-flow fittings are somewhat more complicated than elbows, for two reasons: (1) there are multiple flow paths and (2) there will almost always be velocity changes. First, consider the case of air flowing from the common (upstream) section to the branch. Referring to Figure 1.1, this is from c to b. (Refer to Appendix A.1.1 for clarification of upstream and downstream.) As is the case for elbows, loss coefficients are determined experimentally for diverging-flow fittings. However, it is now necessary to specify which flow paths the equation parameters refer to. By definition: Equation 1.17a where: Cb = Branch loss coefficient ∆ TPc-b = Total pressure loss, common-to-branch (inches wg) VPb = Branch velocity pressure (inches wg) Rewriting in terms of total pressure loss: ∆ TPc-b = Cb x VPb Equation 1.17b Therefore, the total pressure loss of air flowing into the branch leg of a diverging-flow fitting is Duct System Design Page 1.22 directly proportional to the branch loss coefficient and the branch velocity pressure. For duct and elbows, the total pressure loss is always equal to the static pressure loss, because there is no change in velocity. However, diverging-flow fittings almost always have velocity changes associated with them. If ∆VP is not zero, then the total and static pressure losses cannot be equal (Equation 1.4). For diverging-flow fittings, the static pressure loss of air flowing into the branch leg can be determined from Equation 1.17c: ∆ SPc-b = VPb (Cb + 1) - VPc Equation 1.17c where: ∆ SPc-b = Static pressure loss, common-to-branch (inches wg) VPb = Branch velocity pressure (inches wg) VPc = Common velocity pressure (inches wg) Cb = Branch loss coefficient (dimensionless) Equation 1.17c is derived from Equations 1.17a and 1.17b, as shown in Appendix A.3.6. As is the case for elbows, a comparison of loss coefficients gives a good indication of relative fitting efficiencies. The following samples compare loss coefficients of various diverging-flow fittings. Duct System Design Page 1.23 Table 1.3 Loss Coefficient Comparisons for Diverging-Flow Fittings Fitting Loss Coefficient (Cb) Y-Branch plus 45° Elbows 0.22 Vee Fitting 0.30 Tee with Turning Vanes plus Branch Reducers (Bullhead Tee with Vanes) 0.45 Tee plus Branch Reducers 1.08 Capped Cross with Straight Branches 4.45 Capped Cross with Conical Branches 4.45 Capped Cross with 1-foot Cushion Head 5.4 Capped Cross with 2-foot Cushion Head 6.0 Capped Cross with 3-foot Cushion Head 6.4 The loss coefficient, Cb, is for a Vb/Vc ratio of approximately 1.0. Duct System Design Page 1.24 A A , Q Q , VP , V , VP , V c b c b b b c c Sample Problem 1-16 What is the total pressure loss for flow from c to b in the straight tee shown below? What is the static pressure loss? Answer: ∆ TPc-b = Cb x VPb From Equation 1.17b ∆ SPc-b = VPb (Cb + 1) - VPc From Equation 1.17c Reference: ASHRAE Duct Fitting Database Number SD5-9 Given: Qc = 5,000 cfm Dc = 24 inches Qb = 2,000 cfm Db = 18 inches Qs = 3,000 cfm Ds = 24 inches Calculate: fpm A Q V c c c 1,592 = 3.14 5,000 = = c s b Duct System Design Page 1.25 wg inches 4,005 V VP c 2 c 0.16 = 4,005 1,592 = = 2                         fpm A Q V b b b 1,130 = 1.77 2,000 = = wg inches 4,005 V VP b 2 b 0.08 = 4,005 1,130 = = 2                         0.40 = 5,000 2,000 = Q Q c b 0.56 = 3.14 1.77 = A A c b Determine: Cb - Interpolated from the ASHRAE table = 2.14 Answer: ∆ TPc-b = 2.14 x 0.08 = 0.17 inches wg ∆ SPc-b = 0.08 (2.14 + 1) - 0.16 = 0.09 inches wg Duct System Design Page 1.26 Sample Problem 1-17 What would be the static and total pressure losses in Sample Problem 1-16, if a conical tee were substituted for the straight tee? A LO-LOSSTM tee? Reference: Conical Tee: ASHRAE Fitting SD5-10 LO-LOSST M Tee: ASHRAE Fitting SD5-12 Determine: Cb (conical) = 1.35 Cb (LO-LOSST M) = 0.79 Answer: ∆ TPc-b (conical) = 0.11 inches wg ∆ SPc-b (conical) = 0.03 inches wg ∆ TPc-b (LO-LOSST M) = 0.06 inches wg ∆ SPc-b (LO-LOSST M) = -0.02 inches wg In the preceding problem, the static pressure loss for a LO-LOSST M tee at the given conditions resulted in a negative number. Recall from Section 1.3.1 that a pressure change expressed as a positive number is a loss, while a pressure change expressed as a negative number represents an increase in pressure. This pressure increase is a common phenomenon in air handling systems and is known as static regain. It occurs for the LO-LOSST M fitting because the decrease in velocity pressure is greater than the total pressure loss of the fitting. Total Pressure Losses versus Static Pressure Losses Just as total pressure represents the total energy present at any point in a system, the total pressure loss of a fitting represents the true energy loss of the fitting for a given flow situation. Static pressure losses are useful for certain design methods, as we shall see later; however, they do not give an accurate indication of fitting efficiency. Sample Problem 1-18 illustrates this concept. Duct System Design Page 1.27 Sample Problem 1-18 Change the branch size in Sample 1-17 from 18 inches diameter to 12 inches diameter and recalculate the total and static pressure losses of the LO-LOSS TM tee: Reference: ASHRAE Fitting SD5-12 Given: Qc = 5,000 cfm Dc = 24 inches Qb = 2,000 cfm Db = 12 inches Qs = 3,000 cfm Ds = 24 inches Calculate: A A , Q Q , VP , V , VP , V c b c b b b c c Vc = 1,592 fpm VPc = 0.16 inches wg Vb = 2,548 fpm VPb = 0.40 inches wg Qb / Qc = 0.40 Ab / Ac = 0.25 Determine: Cb, Interpolate from ASHRAE Fitting SD5-12 = 0.21 Answer: ∆ TPc-b = CbxVPb (from Equation 1.17b) = 0.21x0.40 = 0.08 inches wg ∆ SPc-b = VPb (Cb + 1) - VPc (from Equation 1.17c) = 0.40(0.21+1)-0.16 = 0.32 inches wg In this problem, the total pressure loss is 0.08 inches wg, but the static pressure loss is 0.32 inches wg. If one were to look only at the static pressure, this would seem to be a very inefficient fitting. However, notice that due to flow and pressure conditions in the system, the velocity increased as the air moved from the common section into the branch, resulting in a velocity pressure increase of 0.24 inches wg. c s b Duct System Design Page 1.28 This situation is shown in Figure 1.3, Case II. The apparently large decrease in static pressure was caused by a large increase in velocity pressure. As Equation 1.4, this becomes: 0.08 = 0.32 + -0.24 (∆ TP = ∆ SP + ∆ VP) Conversely, when certain flow conditions are present, it is possible for a fitting to have a small static pressure loss but a relatively large total pressure loss. It is always advisable to calculate the total pressure loss in order to determine the total energy consumption of a fitting. Manifold Fittings The single-branch fittings discussed thus far are assumed to be factory fabricated, and constructed as a separate unit from the duct to which t hey would be attached. Occasionally, it is desirable to construct a manifold fitting, with a tap attached directly to the duct. This construction will generally result in a less efficient fitting, especially if the manifold is constructed in the field. Flat Oval Diverging-Flow Fittings Diverging-flow fittings of similar construction generally exhibit the same pressure loss for the same volume flow rate ratios and area ratios. Flat oval fittings exhibit similar pressure losses as round fittings. Testing is under way to develop a database of loss coefficients for flat oval diverging-flow fittings. Until this data is available, use the same loss coefficients as for the same construction of round. Acoustically Lined and Double-wall Diverging-Flow Fittings Whether a fitting has been acoustically lined or has a perforated metal inner shell, the difference in surface roughness is accounted for in the friction loss determination, since all friction loss calculations are base on fitting-to-fitting centerline dimensions. Therefore there is no need to increase the dynamic loss of a diverging flow-fitting that is either acoustically lined or one that has an inner metal shell, even if the shell is perforated. Determine the loss coefficient of the fitting as if it were a single-wall fitting with the dimension of the inner liner or metal inner shell. Rectangular Diverging-Flow Fittings (see ASHRAE= = s Duct Fitting Database Program) 1.5.4 Diverging-Flow Fittings: Straight-Throughs, Reducers, and Transitions Straight-Throughs The straight-through (downstream) leg of a diverging-flow fitting is that path followed by air flowing from c to s, as represented in Figure 1.1. The straight-through may have a constant diameter, such that Dc = Ds, or there may be a reducer attached to the straight -through, such that Dc > Ds. In the case of a constant diameter straight -through, there will always be a velocity reduction caused by a reduced volume (after the branch) flowing through the same diameter duct. If a reducer is attached to the straight -through, it can be sized to reduce, maintain, or increase the downstream velocity relative to the common velocity. Dynamic losses associated with air flowing straight through a diverging-flow fitting and/or a Duct System Design Page 1.29 reducer is very slight. This is understandable, since there is little physical disturbance of the airflow. The total pressure loss in a straight -through leg or reducer is often only a few hundredths of an inch wg. Perhaps the most important phenomenon associated with the straight-through flow situation is the potential for static regain. This situation is illustrated in Figure 1.3, Case I. A large reduction in velocity pressure and a small reduction in total pressure must (by Equation 1.4) result in an increase in static pressure, or static regain. The regain will be equal in magnitude to the velocity pressure loss minus the total pressure loss. Of course, if the total pressure loss is greater than the reduction in velocity pressure, as shown in Figure 1.3, Case III, there can be no static regain. Referring again to Sample Problem 1-16, we see that the velocity in the constant diameter straight-through leg is reduced from 1,592 fpm (VPc = 0.16 inches wg) in the duct before the straight-through to 955 fpm (VPs = 0.06 inches wg) in the duct after the straight-through, due to the reduced volume flow. This is a velocity pressure reduction of ∆ VPc-s = 0.10 inches wg. If we assume a total pressure loss of ∆TPc-s = 0.01 inches wg, then from Equation 1.4 we get: 0.01 = ∆SPc-s + 0.10 or ∆SPc-s = -0.09 inches wg The negative result indicates a static regain, or that the static pressure at point s will be 0.09 inches wg higher than the static pressure at point c. The loss coefficient data for reducers and straight-throughs is found in the ASHRAE Duct Fitting Database. When using loss coefficients to determine straight -through losses, Equations 1.17b and 1.17c are rewritten as follows: ∆ ∆TPc-s = Cs x VPs Equation 1.18a ∆ ∆SPc-s = VPs (Cs + 1) - VPc Equation 1.18b where: ∆ ∆TPc-s = Total pressure loss, common-to-straight-through (inches wg) ∆ ∆SPc-s = Static pressure loss, common-to-straight-through (inches wg) VPc = Common velocity pressure (inches wg) VPs = Straight-through velocity pressure (inches wg) Cs = Straight-through loss coefficient Sample Problem 1-19 Calculate the total and static pressure losses for the straight-through portion of the straight tee in Sample Problem 1-16, Reference: ASHRAE Fitting SD5-9 Duct System Design Page 1.30 Calculate: A A , Q Q , VP , V , VP , V c s c s s s c c Vc = 1,592 fpm VPc = 0.16 inches wg Vs = 955 fpm VPs= 0.06 inches wg Qs / Qc = 0.60 As / Ac = 1.0 Determine: Cs, Interpolate from ASHRAE Fitting SD5-9 = 0.20 Answer: ∆ TPc-s = Cs x VPs (Equation 1.18a ) = 0.20 x 0.06 = 0.01 inches wg ∆ SPc-s = VPs (Cs + 1) - VPc (Equation 1.18b) = 0.06(0.20 + 1) - 0.16 = -0.09 inches wg Reducers A stand-alone reducer will cause the velocity to increase, since after the reducer, the same volume of air will be flowing through a smaller diameter duct (see Sample Problem 1-2). The use of a reducer on its own is not consistent with any design methods presented in this manual, and should be fairly rare in most duct systems. However, when this fitting is used, the losses are calculated using the same charts and in the same manner as the straight -throughs. Losses are a function of upstream and downstream velocity. Reducing fittings should be constructed as shown in Figure 1.7, such that the length of the taper portion ( L) is equal to the difference between the common diameter and the straight -through diameter (Dc - Ds). Verify this with manufacturer's dimension sheets. Since reducers are very efficient fittings, the use of a longer taper section will not necessarily provide a significant improvement in performance. Figure 1.7 Reducing Fitting Construction Transitions Transitions between round and flat oval duct also produce dynamic pressure losses. As with other fittings, these losses can be quantified in terms of a loss coefficient. The loss coefficient Duct System Design Page 1.31 for round-to-flat oval or flat oval-to-round transitions depends on the flat oval aspect ratio (major axis/minor axis), the direction of airflow, and the air velocity. When round duct transitions to flat oval, the flat oval minor axis dimension is usually less than the original round diameter, while the flat oval major axis dimension is greater than the round diameter. The reverse is true in transitions from flat oval to round. Therefore, round/flat oval transitions usually involve both a reducer effect (round to flat oval minor or flat oval major to round) and an enlarger effect (round to flat oval major or flat oval minor to round). The change in dimension involving the flat oval major axis is normally much greater than the change to/from the flat oval minor axis. Therefore, in round-to-flat oval transitions the enlarger effect predominates while in flat oval-to-round transitions the reducer effect predominates. Dynamic losses which result from expanding areas (decreasing velocities) are always more severe than losses from reducing areas (increasing velocities). Therefore the flat oval-to-round transition is more efficient than the round-to-flat oval fitting. Figure A.24 in Appendix A.4.2 is a plot of loss coefficient (Cs) versus round duct velocity, for both round-to-flat oval and flat oval-to-round transitions. The curves are valid for any size flat oval and will be conservative for transitions involving flat oval with a low aspect ratio. Use the appropriate loss coefficient value in the following equations to determine static and total pressure losses for transition fittings. ∆ ∆TPc-s = Cs x VPs Equation 1.19a ∆ ∆SPc-s = VPs (Cs + 1) - VPc Equation 1.19b where: ∆ ∆TPc-s = Total pressure loss, common-to-straight-through (inches wg) ∆ ∆SPc-s = Static pressure loss, common-to-straight-through (inches wg) VPc = Common velocity pressure (inches wg) VPs = Straight-through velocity pressure (inches wg) Cs = Straight-through loss coefficient Sample Problem 1-20 In Sample Problem 1-10, we determined that a 12-inch H 31-inch flat oval duct would have the same pressure loss per unit length as a 20-inch round duct. What would be the impact on total and static pressure losses in a 100-foot section of 20-inch round duct carrying 5,000 cfm if 40 feet of this duct were replaced by a 12-inch H 31-inch flat oval duct? Assume the first 30 feet of duct is round, next 40 feet is flat oval, and the last 30 feet is round. Answer: Since the duct sizes are equivalent, 40 feet of 12-inch H 31-inch flat oval would have the same pressure loss per 100 feet as the section of 20-inch round duct it replaced. The addition of two transitions, one round-to-flat oval at the start of the Duct System Design Page 1.32 40-foot section and the other flat oval-to-round at the transition back to round duct would cause the only change in pressure loss. The velocity in the 20-inch round duct carrying 5,000 cfm is 2,294 fpm (VP = 0.33 inches wg). The flat oval duct cross-sectional area from Appendix A.1.3 is 2.36 ft2; therefore the velocity in the flat oval duct is 2,119 fpm (VP = 0.28 inches wg). From Appendix A.4.3.4, the loss coefficients at 2,294 fpm are Cr-o = 0.17 and Co-r = 0.06. Substituting into Equations 1.19a and 1.19b ∆ TPc-s{r-o} = 0.17 x 0.28 = 0.05 inches wg ∆ TP c-s{o-r} = 0.06 x 0.33 = 0.02 inches wg ∆ SPc-s{r-o} = 0.28 (0.17 + 1) - 0.33 = -0.00 (-0.002) inches wg ∆ SP c-s{o-r} = 0.33 (0.06 + 1) - 0.28 = -0.07 inches wg The flat oval section will therefore increase the total pressure loss by an additional 0 .07 inches wg (0.05 + 0.02), due to the combined effects of both transitions. As expected, since there was no net change in velocity in the round duct, ∆ VP = 0 and (by Equation 1.4) the combined static pressure loss (0.07 inches wg) is equal to the combined total pressure loss. 1.5.5 Miscellaneous Fittings Heel-Tapped Elbows The tee-type diverging-flow fittings discussed in Section 1.5.3 are generally used where the designer desires to direct a relatively small quantity of air at some angle relative to the main trunk duct, while maintaining a straight -through flow for the majority of the air. Occasionally, situations arise where the main air stream must be diverted at some angle, while a smaller quantity of air is required in a straight-through direction. In these situations, the use of a heel-tapped elbow will generally result in lower pressure losses, in both common and branch directions. ASHRAE Fitting SD5-21 presents loss coefficients (Cb) for both the straight-through tap and the elbow section of heel-tapped elbows as a function of velocity ratio. Use Equations 1.17b and 1.17c for determining the total and static pressure losses. If 65 percent or more of the airflow is diverted, then it is advisable to use a heel-tapped elbow. Sample Problem 1-21 A diverging-flow fitting must be selected which will split 10,000 cfm volume flow rate such that 7,000 cfm will flow at a 90E angle relative to the upstream direction, and 3,000 cfm will continue in the same direction as the upstream flow. Compare the performance of a (a) straight tee, (b) conical tee, (c) LOLOSSTM tee, and (d) heel-tapped elbow, and select the most efficient fitting for this situation. Assume it is desired to Duct System Design Page 1.33 maintain approximately constant velocity: Given: Qc = 10,000 cfm Qb = 7,000 cfm Qs = 3,000 cfm Dc = 28 inches Db = 23 inches Ds = 15 inches Ac = 4.28 sq ft Ab = 2.89 sq ft As = 1.23 sq ft Vc = 2,336 fpm Vb = 2,422 fpm Vs = 2,439 fpm VP = 0.34 inches wg VPb = 0.37 inches wg VPs = 0.37 inches wg Qb/Qc = 0.70 Qs/Qc = 0.30 Ab/Ac = 0.68 As/Ac = 0.29 Reference: Straight Tee ASHRAE Fitting SD5-9 Conical Tee ASHRAE Fitting SD5-10 LO-LOSST MTee ASHRAE Fitting SD5-12 Heel-tapped Elbow ASHRAE Fitting SD5-21 Determine: Coefficients Cb Cs Straight Tee 1.15 0.13 Conical Tee 0.62 0.13 LO-LOSST M Tee 0.33 0.13 Heel-tapped Elbow 0.65 0.09 Answer: Pressure Losses (inches wg) ∆ TPc-b ∆ TPc-s ∆ SPc-b ∆ SPc-s Straight Tee 0.43 0.05 0.46 0.08 Conical Tee 0.23 0.05 0.26 0.08 LO-LOSST M Tee 0.12 0.05 0.15 0.08 Heel-tapped Elbow 0.24 0.03 0.27 0.06 Duct System Design Page 1.34 In this case, the heel-tapped elbow and the conical tee will have nearly the same total pressure loss in the 90E bend direction. The heel-tapped elbow provides a significant performance increase over the straight tee but has a higher loss than the LO-LOSST M tee branch. The straight-through leg of the heel-tapped elbow is slightly more efficient than the straight -throughs of the tees. The best fitting though, based strictly on efficiency, would appear to be the LO-LOSST M tee. However, bear in mind that all three tee fittings will require substantial strai ght-through reducers (28 inches to 15 inches) which will generally be at least 12 inches long (Figure 1.7) and will add substantially to the cost of the fitting. If a compromise between cost and performance is desired, the heel-tapped elbow may still be the best choice for these flow situations. Also, increased loss may help balance the leg in which this fitting resides. Note that in Sample Problem 1-21 the total pressure is lower than the static pressure by 0.03 inches wg in all cases, due to the increase in velocity pressure (0.03 inches wg) from common to both straight-through and branch. In the form of Equation 1.4: ∆ TPc-b = ∆ SPc-b - 0.03 ∆ TPc-s = ∆ SPc-s - 0.03 Crosses Cross fittings are those which have two taps located at the same cross section of a main or trunk duct. Usually these taps will be constructed so that they discharge air in diametrically opposed directions. The pressure loss of the taps on a cross fitting depends, to a large extent, on the cross-sectional area reduction of the straight-through duct. For straight-through area reductions of less that 20 percent, the branch losses through either tap of a cross fitting will be the same as those for a single-branch fitting with identical tap construction. For example, a conical cross (a cross with conical taps) that has a straight -through of 24 inches, reducing to 22 inches, would have an area reduction of 16 percent. Since this is less than 20 percent, the branch losses for either tap would be found from the direct-read charts in ASHRAE Fittings SD5-9 through SD5-17. For crosses where the straight -through area reduction exceeds 20 percent, use the loss coefficients presented in ASHRAE Fittings SD5-23 through SD5-26. The three curves shown are for varying percentage area reductions and apply for all tap constructions. Use interpolation to find loss coefficients for area reductions between these curves. Use Equations 1.17b and 1.17c for determining the total and static pressure losses of each branch. Split Fittings Split fittings are diverging-flow fittings where the air divides into two branches, each of which turns at 90E to the main. There is no straight-through leg. The most common types of split fittings are the Vee, Y-branch and the bullhead tee. The Y-branches are the most efficient split fittings; however, they are more expensive than Vee's and may be more expensive than bullhead tees. Bullhead tees should always be specified with turning vanes. ASHRAE Fittings SD5-18, 19 and 21 presents drawings and loss Duct System Design Page 1.35 coefficients data for Y-branches, bullhead tees with turning vanes, bullhead tees without turning vanes, and capped crosses. The capped cross is discussed in the following section. Use Equations 1.17b and 1.17c for determining the total and static pressure losses. Capped Fittings Capped fittings are those in which the main or straight -through is completely closed off. The most common use for a capped fitting is in locations where it is expected that future expansion will require additional ducting, at which time the cap can be removed and the duct run continued. In general, the use of capped fittings is strongly discouraged. For both single-branch fittings and crosses, the performance is severely degraded if the main is capped. Where these fittings are unavoidable SD5-2 includes a curve for capped crosses. Use Equations 1.17b and 1.17c for determining the total and static pressure losses. Close-Coupled Fittings After air flows through any type of fitting, a certain length of straight duct is required to re-establish the flow profile of the airstream. Simply stated, it takes a certain distance for air to recover from a disturbance produced by a fitting. If the airstream encounters a second fitting before it has had a chance to recover from a previous disturbance, the effect of the second fitting will be more pronounced than if it had been located in a long run of straight duct. Generally, two elbows in series will have the same loss as the sum of the individual elbows. The exception to this is when the second elbow has an additional change in direction such that the air is not flowing parallel to the f irst flow. For this case, as much as an additional 100 percent of the combined losses should be added, unless the elbows are at least 10 diameters apart. The loss of two tees in series is a function of the spacing between the tees, although the loss coefficient of the upstream tee is not significantly affected. The loss coefficient of the downstream tee actually decreases at half-diameter spacing. At two-diameter spacing, however, the downstream loss coefficient is significantly higher. The loss coeffi cient gradually decreases back to its original value at 10 diameters. To account for the increased pressure loss at two-diameter spacing, add 100 percent of the calculated loss. This can be decreased as the diameter spacing between tees becomes greater. Appendix A.9.9 has a more detailed discussion of the effect of spacing of tees. Couplings Slip couplings, which are inserted inside duct sections and are therefore exposed to the air stream, are generally used to join two adjacent duct sections. Fittings, which connect directly to duct sections, do not require couplings, and fittings which are connected directly to other fittings usually have an outside coupling. Losses associated with duct couplings are very low. When slip couplings are separated by 10 to 20 feet of duct, their effects are negligible. However, in the event it is necessary to calculate the loss due to duct couplings, Appendix A.4.2 presents a table of loss coefficients versus coupling diameter. Use Equations 1.17b and 1.17c for determining the total and static pressure losses. As can be seen from the loss coefficient values, it is normally quite acceptable to ignore these losses when calculating system pressure losses. Experience has shown that even poorly made and undersized couplings have negligible losses. The resulting loss coefficient may be two to three times that of a slip coupling, Duct System Design Page 1.36 but this is still a very low value. Offsets Offsets are required to change the location of a duct run horizontally, vertically or both. This is most often necessary to avoid interference with some obstruction along the duct run. Offsets are usually constructed with two or more elbows, joined by a suitable length of straight duct. Due to the almost limitless number of offsets that could be created, there are no tables or charts in this manual for the calculation of these losses. It is suggested that offset losses be obtained by adding the losses of the individual elbows and duct, which form the offset and, if necessary, adding a factor for any close-coupling effects that may exist. Bellmouths The bellmouth fitting is used as an intake or entrance to a duct, usually from a plenum or fan housing. There is a substantial advantage in having a smooth radiused entrance, as opposed to a square-edged entrance. Loss coefficients for bellmouths are presented in ASHRAE Fittings SD1-1, SD1-2, and SD1-3. To calculate the total and static pressure losses, use (transition loss) Equations 1.19a and 1.19b. Expanders Increasing the duct diameter, upstream-to-downstream, in a supply air system is not a recommended design practice. Abrupt expanders are very inefficient fittings in that their loss coefficients are always 1.0 or greater. This means that the entire upstream velocity head is lost and unrecoverable. This can be shown by substituting a unity loss coefficient into Equations 1.18a and 1.18b: ∆ TPc-s = 1.0 x VPc = VPc ∆ SPc-s = VPc (1.0 - 1.0) + VPs = VPs Therefore, although the static pressure loss may be small, the total pressure loss is equal to (at least) the entire upstream velocity pressure. Exits Exits are fittings that discharge air into the surrounding environment. Refer to ASHRAE Fittings SD1-1 and SD1-2 (plenums), SD2-1 to SD2-6 (atmosphere), and SD7-1 to SD7-5 (fans) for loss coefficients of round exits. Refer to SR1-1 (plenums), SR2-1 to SR2-6 (atmosphere), and SR7-1 to SR7-17 (fans) for loss coefficients of rectangular exits. Increasing the duct size at an exit is advantageous in minimizing pressure loss. Obstructions In-line losses common to supply systems also must be taken into account. Refer to ASHRAE=s Duct Fitting Database CD9-1 to CD9-3 for loss coefficients of round dampers, CR9-1 to CR9-7 for loss coefficients of rectangular dampers, CD8-1 to CD8-8 for loss coefficients of round silencers, CR8-1 to CR8-4 for loss coefficients of rectangular silencers, CR8-5 to CR8-8 for loss coefficients of coils, CR8-9 to CR8-11 for loss coefficients of VAV boxes, and CD6-1 to CD6-4 for loss coefficients of other round obstructions. Duct System Design Page 1.37 1.5.6 Nonstandard Conditions In Section 1.4.5 equations were given for correcting the calculated friction loss of a system for nonstandard conditions of temperature and/or elevation. Since velocity pressure is a function of air density (see Appendix A.3.3), and since all dynamic fitting losses are a function of velocity pressure, an additional correction must be made to the calculated fitting losses whenever there are substantial variations from standard conditions. If a density-corrected velocity pressure (Appendix A.3.3) is used to calculate all dynamic fitting losses, then no further corrections (except friction loss corrections) are required. If the pressure losses are calculated assuming standard conditions, the results can be corrected by multiplying by the ratio of actual density diverging by standard density. For most HVAC applications, this ratio can be calculated as shown in Equation 1.20:             β β             ρ ρ ρ ρ 29.921 x 460) + T ( 530 = a std act Equation 1.20 where: ρ ρact = Actual density (lbm/ft3) ρ ρstd = Standard density (lbm/ft3) Ta = Actual air temperature (EF) β β = Actual barometric pressure (inches Hg) The corrected total, static, and velocity pressure can be calculated as follows:                 ρ ρ ρ ρ std act std act x TP = TP ∆ ∆ Equation 1.21                 ρ ρ ρ ρ std act std act x SP = SP ∆ ∆ Equation 1.22                 ρ ρ ρ ρ std act std act x VP = VP ∆ ∆ Equation 1.23 where: ∆ ∆TPact = Total pressure loss at actual conditions (inches wg) ∆ ∆TPstd = Total pressure loss at standard conditions (inches wg) Duct System Design Page 1.38 wg inches x TP TP std act std b, -c act b, -c 0.15 = 0.86 x 0.17 = = ρ ρ ρ ρ ∆ ∆ wg inches x SP SP std act std b, -c act b, -c 0.08 = 0.86 x 0.09 = = ρ ρ ρ ρ ∆ ∆ ∆ ∆SPact = Static pressure loss at actual conditions (inches wg) ∆ ∆SPstd = Static pressure loss at standard conditions (inches wg) ∆ ∆VPact = Change in velocity pressure calculated at actual conditions ( inches wg) ∆ ∆VPstd = Change in velocity pressure calculated at standard conditions (inches wg) Density correction factors are tabulated in Appendix A.1.5. Often, for systems operating a t normal HVAC temperatures (70E " 30EF) and elevations less than 1,500 feet above sea level, these corrections can be neglected. When Equations 1.21 and/or 1.22 are applied to the aggregate pressure loss of an entire system instead of to the individual fitting components, the resulting pressures are not necessarily accurate. This is because the friction correction factors (Section 1.4.5) are calculated in a different manner from the dynamic loss correction factors discussed above. The accuracy depends on the ratio of duct length versus number of fittings and on the deviation from standard conditions. Sample Problem 1-22 Determine the effect on the total and static pressure losses of the straight tee of Sample Problem 1-16 if the air temperature is 55EF and the elevation is 5,000 feet. Answer: From Sample Problem 1-16 at standard conditions: wg inches SP wg inches TP std b, -c std b, -c 0.09 = 0.17 = ∆ ∆ From Appendix A.1.5: 0.86 = ρ ρ ρ ρ std act (interpolation required) From Equation 1.21: From Equation 1.22:
4534	https://afteracademy.com/blog/graph-traversal-breadth-first-search	AfterAcademy Graph Traversal: Breadth-First Search In the previous blog i.e. T he Introduction to Graph in Programming , we saw what a graph is and we also saw some of the properties and types of graph. We also saw how to represent a graph i.e. using the Adjacency Matrix and Adjacency List. In this blog, we will learn about the Breadth-First Search i.e. BFS that is used to search some node in a graph by traversing it. Graph traversal is a process of visiting all the nodes from a source node only once in some defined order. The order of traversal of nodes of a graph is very important while solving some graph problems. Also, you must track the nodes that are already visited because, in traversal, you need to traverse a node only once. So, a proper list of the traversed nodes of the graph must be maintained. There are two ways of Graph traversal: In this blog, we will cover the BFS part. So, let's get started. Breadth-First Search Breadth-First Search or BFS is a graph traversal algorithm that is used to traverse the graph level wise i.e. it is similar to the level-order traversal of a tree. Here, you will start traversing the graph from a source node and from that node you will first traverse the nodes that are the neighbours of the source node. After traversing all the neighbour nodes of the source node, you need to traverse the neighbours of the neighbour of the source node and so on. Based on the source node, the whole graph can be divided into various levels i.e. the nodes that are at distance 1 from the source node are said to be at level 1. Similarly, the nodes that are at distance 2 from the source node are said to be at level 2 and so on. Based on the layers of the graph, the BFS can be performed by the following steps: NOTE: There can be more than one path from node i to node j because a graph can contain a cycle. But you should visit a node once. So, you have to keep a record of all the visited nodes so that one node can be visited only once. The following is an example of Breadth-First Search: BFS implementation In order to implement BFS, we need to take care of the following things: So, to apply the above conditions, we make the use of Queue data structure that follow First In First Out(FIFO) order. We will insert the nodes in the queue and mark it as visited and after that, all the neighbour nodes of that node will also be inserted in the queue. Since it follows FIFO order, the node entered first will be visited first and their neighbours will be added in the queue first. During insertion of nodes in the queue, we will check if the nodes are visited or not. If it is visited then we will not add those nodes in the queue. Otherwise, we will add the node in the queue. The following is the pseudocode of Breadth-First Search: BFS (G, s): //here, G is the graph and s is the source or initial node Q.enqueue( s ) //here, Q is queue for storing nodes. So, we are inserting node "s" in the queue Q also mark "s" as visited node while ( Q is not empty) //removing that vertex from queue whose neighbour will be visited now v = Q.dequeue( ) //this will return the node inserted first in the queue //processing all the neighbours of v //if the node "w" is already visited, then ignore it //if the node "w" is not visited, then insert it into the queue for all neighbours w of v in Graph G if w is not visited Q.enqueue( w ) //Stores w in Q to further visit its neighbour mark w as visited. Let's understand the above pseudocode with the help of one example. The following process will be followed in different iteration: Application of BFS These are some of the applications of Breadth-First Search. Graph questions Happy Learning, Enjoy Algorithms! Team AfterAcademy! AfterAcademy Data Structure And Algorithms Online Course - Admissions Open Recommended for You Reverse First K Elements Of A Queue Given an integer K and queue Q of integers. Write a program to reverse the order of the first K elements of the queue. The other elements will be in the same relative order. Smallest Multiple With 0s and 1s You are given an integer n, write a program to find the smallest multiple of n which consists of 0 and 1. The resultant number could be quite large so return it in the form of a string. This problem is a good example of graph-based problems. Largest Element In An Array Given an array arr[] of size n , write a program to find the largest element in it. To find the largest element we can just traverse the array in one pass and find the largest element by maintaining a max variable. Merge Two BST Given two binary search trees with root nodes as tree1 and tree2 of size n and m, write a program to return an array of integers that contains all the elements of tree1 and tree2 in non-decreasing order. The expected time complexity is O(m+n). Combinations Given two integers n and k, Write a program to return all possible combinations of k numbers out of 1 2 3 ... n. Elements in a combination must be in a non-descending order. The combinations themselves must be sorted in ascending order, i.e., the combination with the smallest first element should be printed first. Search in a 2-D Matrix You are given a matrix arr of m x n size. Write a program to searches for a value k in arr. This arr has the following properties: - Integers in each row are sorted from left to right. - The first value of each row is greater than the last value of previous row. This is a basic optimization problem that will clear the concept of searching. Connect With Your Mentors Janishar Ali Amit Shekhar Copyright 2022, MindOrks Nextgen Private Limited
4535	https://pmc.ncbi.nlm.nih.gov/articles/PMC10669870/	The RNA-Binding Function of Ribosomal Proteins and Ribosome Biogenesis Factors in Human Health and Disease - 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 Advanced Search Journal List User Guide New Try this search in PMC Beta Search 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 Biomedicines . 2023 Nov 4;11(11):2969. doi: 10.3390/biomedicines11112969 Search in PMC Search in PubMed View in NLM Catalog Add to search The RNA-Binding Function of Ribosomal Proteins and Ribosome Biogenesis Factors in Human Health and Disease Caterina Catalanotto Caterina Catalanotto 1 Department of Molecular Medicine, Sapienza University of Rome, 00185 Rome, Italy; caterina.catalanotto@uniroma1.it (C.C.); carlo.cogoni@uniroma1.it (C.C.) Find articles by Caterina Catalanotto 1, Christian Barbato Christian Barbato 2 National Research Council (CNR), Department of Sense Organs DOS, Institute of Biochemistry and Cell Biology (IBBC), Sapienza University of Rome, 00185 Rome, Italy; christian.barbato@cnr.it Find articles by Christian Barbato 2, Carlo Cogoni Carlo Cogoni 1 Department of Molecular Medicine, Sapienza University of Rome, 00185 Rome, Italy; caterina.catalanotto@uniroma1.it (C.C.); carlo.cogoni@uniroma1.it (C.C.) Find articles by Carlo Cogoni 1, Dario Benelli Dario Benelli 1 Department of Molecular Medicine, Sapienza University of Rome, 00185 Rome, Italy; caterina.catalanotto@uniroma1.it (C.C.); carlo.cogoni@uniroma1.it (C.C.) Find articles by Dario Benelli 1, Editor: Bernard Lebleu Author information Article notes Copyright and License information 1 Department of Molecular Medicine, Sapienza University of Rome, 00185 Rome, Italy; caterina.catalanotto@uniroma1.it (C.C.); carlo.cogoni@uniroma1.it (C.C.) 2 National Research Council (CNR), Department of Sense Organs DOS, Institute of Biochemistry and Cell Biology (IBBC), Sapienza University of Rome, 00185 Rome, Italy; christian.barbato@cnr.it Correspondence: dario.benelli@uniroma1.it Roles Bernard Lebleu: Academic Editor Received 2023 Oct 10; Revised 2023 Oct 27; Accepted 2023 Oct 30; Collection date 2023 Nov. © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( PMC Copyright notice PMCID: PMC10669870 PMID: 38001969 Abstract The ribosome is a macromolecular complex composed of RNA and proteins that interact through an integrated and interconnected network to preserve its ancient core activities. In this review, we emphasize the pivotal role played by RNA-binding proteins as a driving force in the evolution of the current form of the ribosome, underscoring their importance in ensuring accurate protein synthesis. This category of proteins includes both ribosomal proteins and ribosome biogenesis factors. Impairment of their RNA-binding activity can also lead to ribosomopathies, which is a group of disorders characterized by defects in ribosome biogenesis that are detrimental to protein synthesis and cellular homeostasis. A comprehensive understanding of these intricate processes is essential for elucidating the mechanisms underlying the resulting diseases and advancing potential therapeutic interventions. Keywords: ribosomal origins and evolution, RNA-binding proteins, ribosomal RNA, ribosomopathies, DBA, SBDS 1. Introduction RNA-binding proteins (RBPs) are a broad group of proteins that specifically bind to RNA molecules, forming ribonucleoprotein particles (RNPs) through interactions with their RNA-binding domains (RBDs), including RNA recognition motifs (RRMs), K homology (KH) domains, double-stranded RNA-binding domains (dsRBDs), intrinsically disordered regions (IDRs) and others . RBPs play a crucial role in numerous biological processes of all living organisms, ranging from transcription to splicing, from modification to intracellular trafficking, and from translation to decay. To date, more than 1500 RBPs have been identified in humans [2,3], and their activity and regulation are essential for the proper function of many biological processes. Therefore, it is not surprising that the impaired function of RBPs can lead to the development of a wide range of diseases, ranging from genetic disorders to cancer [4,5,6,7]. Among the plethora of RNPs, ribosomes can be considered their archetype, having appeared in their archaic form about 4 billion years ago and remaining highly conserved throughout extant life (Figure 1) [8,9,10]. Current ribosomes are macromolecules composed of ribosomal RNA (rRNA) and ribosomal proteins (RPs), whose composition is believed to have been selected during evolution to contribute to the improved fitness of RNA molecules with peptidyl transferase activity (PTA) and to ensure accurate and efficient reading of the genetic code. Numerous ribosome biogenesis factors (RBFs), also RBPs, participate in the proper assembly and function of the mature ribosome. Nowadays, ribosomes control the expression of almost all proteins present in living organisms, and the impairment of their functions results in the emergence of a set of diseases, including ribosomopathies and cancer. The objective of this study is to focus attention on the modified RNA-binding activity of RB and RBF as a causative factor leading to impaired ribosomal function in the development of certain ribosomopathies. In this context, we aim to provide an explanation for how the altered function of these proteins affects the translation of specific mRNAs, rather than all translatable ones. Figure 1. Open in a new tab Schematic depiction of the evolutionary distribution of ribosomal proteins across superkingdoms. Universal ribosomal proteins are depicted in white within the central area of the three intersecting circles. Ribosomal proteins specific to bacteria are located within the purple section of the lower circle, while those specific to eukaryotes can be found in the light blue section on the left. Ribosomal proteins exclusive to archaea are situated within the region emerging from the overlapping light blue and green circles. Notably, all archaea-specific ribosomal proteins are also present in eukarya, indicating the absence of any ribosomal proteins unique to archaea. The nomenclature of ribosomal proteins is consistent with the literature criteria . 2. From the RNA World to the Appearance of RNA-Binding Proteins through Ribozymes with Peptidyl Transferase Activity While contemporary eukaryotic ribosomes consist of approximately equal parts protein and rRNA macromolecules, accumulating evidence over time suggested that rRNA plays a central role in controlling peptide bond catalysis and ensuring the fidelity of genetic code reading. Initially met with skepticism, this perspective has now gained widespread acceptance, leading to the recognition of ribosomes as ribozymes. Indeed, the atomic structures of the large ribosomal subunits in different living species consistently reveal the absence of ribosomal proteins (RPs) at the core of the peptidyl transferase center (PTC) [12,13,14,15]. Similar conclusions were reached in the past by studying the dispensability of different ribosomal proteins while maintaining the integrity of the ribosomal PTC [16,17] and through experiments involving ribosomal inactivation by cleaving a single phosphodiester bond in the rRNAs (Figure 2) [18,19,20]. Figure 2. Open in a new tab Functional ribosome cores are universally conserved in all living systems. (a) A three-dimensional representation of the pseudo-symmetry within the peptidyl transferase center (PTC) is depicted in 2D structure and ribbons. The structural representations were extracted from the cryo-EM data obtained for ribosomes from E. coli (PDB ID: 7K00) and H. sapiens (PDB ID: 4UG0), as depicted on the left and right sides of the figure, respectively. The visualization was generated using UCSF Chimera . Two-dimensional structures of PTC were generated using R2DT software (Version 1.4) . Bases corresponding to the P-site and A-site are denoted by green and light blue letters, respectively, while bases delineating the PTC pore are highlighted in red. This color scheme is consistently maintained throughout the figure. P and A represent two of three active sites of ribosome corresponding to peptidyl and aminoacyl site, respectively. (b) In evidence, PTC mapped into the large ribosomal subunit and superimposed. It has been proposed that the essential functions performed by the RNA component of ribosomes during protein synthesis can be traced back to its ancestral role in the prebiotic environment of primordial Earth. According to the “RNA world” hypothesis, this primordial environment favored the emergence of biological replicators composed of RNA molecules exhibiting behavior similar to genetic algorithms; namely, RNA molecules that could iteratively replicate, mutate, and be selected based on improved fitness resulting from acquired mutations [23,24,25,26]. Some of these mutations enabled certain replicators to fold into structures with enzymatic activity, such as ribozymes capable of peptidyl transferase activity (PTA), ultimately leading to the synthesis of stochastic and non-finalized peptides. Among these peptides, those with RNA-binding activity could have played a pivotal role in the selection of riboswitches with PTA for several reasons: (1) Stabilizing one of the numerous diverse structures that RNA molecules can fold into, thus slowing down the folding process towards the most functional structure [27,28]. (2) Promoting conformational changes in RNA upon binding to even relatively small peptides, expanding the range of possible RNA structures, as exemplified by “riboswitches” . (3) Preventing RNA misfolding through chaperone-like activity [30,31]. In addition to these reasons, the inherent versatility of proteins, in comparison to RNA molecules, enables them to adopt a wide variety of shapes and three-dimensional structures. This expands their range of functions to the extent that they play a significant role in shaping the skeletons and structures of cells, tissues and organisms. Collectively, the examples described above provide a rationale for the evolution of living systems that led to the emergence of proteins, of which some were capable of binding RNA. Support for this hypothesis is derived from phylogenetic studies demonstrating that ribosomal proteins with RNA-binding activity are among the most ancient and universally conserved proteins [32,33]. It is important to consider that in the prebiotic environment, proto-peptides were not exclusively generated by ribozymes with PTA. In fact, ever since Stanley Miller’s pioneering work in 1953, it has become clear that amino acids coexisted with various other small organic compounds and numerous small peptides in the primordial environment [34,35,36,37,38]. Furthermore, there is evidence of traces of amino acids and simple organic compounds in cosmic dust and meteorites, even in the absence of living organisms [39,40]. However, the remarkable advancement of protein synthesis machinery, as we know it today, goes beyond PTA alone. Instead, it is rooted in the ability to combine PTA with the accurate reading of the genetic code. This has two important implications: first, during evolution, the selection process shifted from nucleic acids to proteins, and second, stochastic mutations that increased an organism’s fitness through the expression of a new protein could be retained in subsequent generations. To better understand the significance of this last feature, let us consider the existence of ribozymes that translate non-coded peptides with zero probability of being identical to each other. In this scenario, the proteins produced by a ribozyme would exhibit subtle differences from one another . Some of these newly synthesized peptides could enhance the fitness of certain RNA replicators. However, for the information contained in their sequences to be preserved in subsequent generations, the system must ensure a certain level of reading fidelity. In other words, the evolutionary success in perpetuating variants with higher fitness is achieved through the selection of proteins that reduce errors during the translation process, as in the case of proteins which, by directly binding rRNA, regulate the fidelity of the genetic code reading. Indeed, both rRNA and RBPs are directly involved in the accuracy of translation by, for example, regulating tRNA binding to the P site [41,42] and binding to rRNA [43,44,45], respectively. In this context, the reduced size of RPs could be attributed to their ancestral origins when translational accuracy was limited. A shorter amino acid chain decreases the likelihood of encoding an incorrect protein, ensuring the success of perpetuating variants with higher fitness. The evolutionary journey that transformed a ribozyme with limited accuracy in reading the genetic code into the present translation machinery, with an error frequency of one error in 10 3–10 4 polymerized amino acids, occurred through an extensive series of intermediate stages. Each of these steps was based on the reciprocal interaction between rRNA and RPs. Indeed, the Accretion Model of ribosomal evolution depicts rRNA that recursively accreted and froze over time, increasing in mass [8,46,47]. This model is reinforced by the discovery of temporal correlations (co-evolution) between the acquisition of rRNA elements and RP segments with which they interact . A three-dimensional comparative analysis of ribosomes has revealed sequential acquisitions of capabilities during evolution, including RNA folding, non-coded amino acid condensation to form peptides, subunit association, correlated subunit evolution, decoding and energy transduction . It should be noted that this model is not strictly unidirectional as it is also possible to observe the removal of eukaryotic expansion segments and the loss of RPs following genome reduction, as seen in microsporidia . Expanding the Accretion Model to the evolution of RPs also reveals a hierarchical increase in the complexity of their three-dimensional structures. They progress from simple short random coil (RC) peptides bound to rRNA to proteins elongated and coalesced into secondary structures such as β-sheets and α-helices, [12,17,41,51]. Some are essential for the correct overall folding and assembly of ribosomes, whereas others are not. A few RPs are positioned at or near the subunit interface, where they can influence ribosomal function, while the majority are located on the solvent-exposed surface and are distant from any functional site. Many of these RPs do not exhibit extensive interaction with rRNA, and similarly, it appears that only a few single-stranded nucleotides in rRNA participate in the interaction. Consequently, it seems unlikely that a single, specific interaction could significantly stabilize the RNA tertiary structure. Instead, it is more plausible that strong protein–protein contacts enable the formation of RP complexes that crosslink two or more segments of ribosomal RNA. In summary, we must recognize that the interplay between r-proteins and rRNA is essential for the optimal functioning of the ribosomal machinery. The functional importance of these interactions lies in their cooperative nature. rRNA acts as a molecular scaffold, guiding the precise positioning of ribosomal proteins within the ribosome. These interactions stabilize the ribosomal structure and provide functional surfaces for substrate binding and catalysis during translation. Conversely, ribosomal proteins assist in maintaining the correct folding of rRNA segments, contributing to the overall stability and function of the ribosome. 3. Overview of Ribosome and Ribosome Biogenesis The present-day ribosomes consist of both small (30S and 40S) and large (50S and 60S) subunits in prokaryotes and eukaryotes, respectively, and they exhibit a highly conserved three-dimensional structure in all living cells, as shown in Figure 3 . Figure 3. Open in a new tab The structure of ribosomes. A side-by-side comparison of the structures of prokaryotic (E. coli, pdb ID 7K00) and eukaryotic (H. sapiens, pdb ID: 6QZP) ribosomes. Ribosomes consist of the large (blue) and the small (red) subunit which comprise a combination of ribosomal RNA (rRNA) molecules and proteins. The rRNA molecules provide the basic building block of the ribosome, establishing its basic structure and functional characteristics. The ribosomal proteins contribute to the overall stability and integrity of the ribosome by bridging structural gaps and promoting the efficient synthesis of proteins [53,54,55]. At the center of the figure, rRNA backbone ribbons extracted from superimposed large ribosomal subunits display noticeable variations in size compared to their conserved three-dimensional conformation. The bacterium (E. coli) is green, and the eukaryote (H. sapiens) is blue. Their reciprocal association forms the 70S complex in prokaryotes and the 80S complex in eukaryotes, creating the integral A (aminoacyl), P (peptidyl) and E (exit) sites. The A site accepts incoming aminoacyl-tRNA, the P site holds the tRNA with the growing peptide chain and the E site accommodates the deacylated tRNA before it departs from the ribosome. The precise order in which amino acids are incorporated into nascent polypeptide chains depends on the accuracy with which the genetic information in the mRNA is encoded by the ribosomes through the coordinate action of the mRNA, tRNA and various translation factors. Achieving this accuracy requires the ribosomes to undergo a complex and energy-intensive process of biogenesis. In eukaryotes, ribosome biogenesis initiates in the nucleolus, which is a specialized nuclear region for ribosome production that involves all three primary RNA polymerases. RNA polymerase I transcribes rDNA to produce the 47S polycistronic precursor pre-rRNA (35S in yeast), which undergoes further processing to yield mature rRNAs: 18S, 5.8S and 28S. RNA polymerase II generates a class of messenger RNAs (mRNA) known as 5′-Terminal-Oligo-Pyrimidine (TOP)-mRNAs, encoding ribosomal proteins (RPs) and ribosome biogenesis factors (RBFs). These motifs help coordinate the regulation of all ribosome biogenesis and translation components . RNA polymerase III synthesizes the 5S rRNA, which becomes part of the large ribosomal subunit . The sequential assembly of RPs and rRNAs relies on a series of transient factors referred to as ribosomal assembly factors (RAFs) or ribosome biogenesis factors (RBFs). These factors include small nucleolar ribonucleoproteins (snoRNPs), nucleases, ATPases, GTPases, RNA helicases and other proteins without predicted enzymatic activity. In eukaryotes, more than 200 of these factors have been identified, and their coordinated interaction is essential for functional ribosome formation [58,59]. Some of these factors are associated with the pre-rRNA 47S to form the 90S pre-ribosome, inducing specific exo- and endonucleolytic cleavages in premature rRNA [60,61]. Others participate in concurrent post-transcriptional modifications of approximately 200 rRNA nucleotides. These modifications include pseudouridylation and 2′-O-ribose methylation. They are catalyzed by two types of small nucleolar ribonucleoprotein complexes: H/ACA box snoRNPs and C/D box snoRNPs, respectively [62,63,64], that together cover around 95% of identified rRNA post-transcriptional modifications, with the remaining 5% involving acetylation or other types of changes [65,66,67,68]. It is worth noting that the nucleotides targeted by snoRNPs are in crucial regions of the ribosome, including the peptidyl transferase and decoding centers. These modifications contribute to both the correct folding of rRNA and, consequently, the proper functioning of ribosomes [69,70,71,72]. Additional ribosome biogenesis factors known as “placeholders” temporarily bind to specific sites on nascent ribosomes until these sites are structurally ready for other factors to take over and prevent premature recruitment of subsequent factors, early formation of structures and potential folding issues . The initial stages of ribosome biogenesis in the nucleolus yield the pre-40S and pre-60S subunits, which are then exported to the cytoplasm for final maturation. The list of factors involved in the correct assembly of functional ribosomes is extensive and the detailed description for all of them is beyond the scope of this review, although their high number gives an idea of the complexity of the process, making it one of the most energy-intensive for cell growth . As a result, rigorous control mechanisms have evolved to ensure the quality of ribosome biogenesis through various cell signaling pathways, including c-Myc, MAPK/ERK and mTORC1. These pathways allow ribosome biogenesis rates to adapt to changing environmental conditions [75,76,77,78]. When normal mammalian cells receive stimuli promoting cell proliferation, they respond by increasing ribosome biogenesis and protein synthesis. The gained ribosome production enables them to meet the increased biosynthetic demands associated with cell division, ensuring that daughter cells possess the necessary cellular machinery for survival and normal function [79,80]. Conversely, exposure to various stressors (such as doxorubicin, replication stress, hypoxia and growth factor deprivation), or the compromised functioning of ribosomes themselves, leads to an immediate arrest of rRNA transcription and subsequent disruption of various steps in ribosome biogenesis. This is accomplished through the activation of nucleolar stress by various routes, involving factors like p53, ARF, PTEN and pRB [81,82,83,84,85,86,87]. In the context of cancer cells, the dysregulation of tumor suppressor genes and proto-oncogenes results in the upregulation of ribosome biogenesis . This, in turn, accelerates cell growth by altering the rate of cell cycle progression. Therefore, changes in ribosome biogenesis rates can be considered a consequence of neoplastic transformation . However, even alterations in protein synthesis levels alone can induce neoplastic transformation. Increased expression of proteins involved in the control of translation initiation, such as eIF4E, leads to changes in mRNA translation, resulting in tumor formation [90,91]. Additionally, evidence from ribosomal disorders suggests that changes in both the quantity and quality of ribosomes can, on their own, shift the pool of translated mRNAs toward promoting neoplastic transformation . However, despite stringent quality controls of ribosome biogenesis, since their first identification, ribosomes appear to be different . The next progressive technical improvements made to their study confirmed the marked diversity of ribosomal particles between different types of cells of the same organism or during the different stages of organism development to such an extent that today we speak of heterogeneous ribosomes [94,95,96]. Sources of diversity arise both from RP content and post-translational modifications (PTMs) of RPs and from rRNA sequences and their post-transcriptional modifications [98,99,100], as well as the type of non-ribosomal proteins bound to them and the substitution of RP paralogs [102,103]. To date there is no univocal and definitive vision regarding the role of the heterogeneous ribosomal architecture and the physiological role for some of their modifications is not yet fully known. However, ribosomes can be perceived as a hub for the integration of a set of spatiotemporal intra- and extracellular signals that would lead to dynamic variations in their composition [104,105,106,107]. At any time, the different combination of PTMs and/or RBPs bound to the ribosomes could change their binding affinity for specific structures or sequence motifs of specific mRNA resulting in alterations of their translational activity [108,109]. Several studies on different types of living organisms have highlighted a functional relationship in this sense [97,103,110,111]. 4. Ribosomopathies May Be Caused by Impaired RNA-Binding Activity of RPs and RBFs Based on what is described above, the cooperative interaction between RPs and rRNAs must be considered a prerequisite for the optimal functioning of the RNP complex. Due to the structural complexity of both the molecular machinery and the mechanisms controlling its correct function, it is generally difficult to define a single path for the appearance of the pathological phenotype resulting from their impairment. Indeed, although all ribosomopathies share defects in the production of ribosomes, their appearance can originate from mutations of genes that control different molecular mechanisms ranging from the transcription and modification of pre-ribosomal RNA (pre-rRNA) to its processing until ribosome assembly. As a rule, homozygous mutations of genes coding for RPs and RBFs are lethal, while their single-copy mutations can generate a group of heterogeneous diseases, defined as ribosomopathies, including Diamond-Blackfan anemia (DBA), Shwachman–Diamond syndrome (SDS), X-linked subtype of dyskeratosis congenita (DKCX) and 5q- myelodysplastic syndrome (5q- MDS) [112,113,114,115]. Contrary to what one might expect from dysfunctions of macromolecular complexes present ubiquitously in all cells, the resulting class of diseases shares a series of common and paradoxical characteristics, including hematopoietic defects and skeletal anomalies that relegate the pathological phenotype to specific areas of the body . To date, there is no univocal explanation to justify this phenomenon, although numerous data show that p53 through cell cycle arrest and apoptosis may be the etiological agent of the tissue-specific phenotype of ribosomopathies. As a matter of fact, depletion of p53 has the potential to rescue the clinical manifestations associated with various ribosomopathies [117,118,119]. Other studies instead highlight how mutations in RP or RBF can influence per se the levels of ribosomes and, consequently, also the rate of global protein synthesis [120,121,122,123]. At first glance, it might appear straightforward to assume that the reduced number of ribosomes leads to a global reprogramming of translation, resulting in lower levels of translation for all expressed mRNAs across different cell types. However, several experimental evidence suggest that the concentration of ribosomes plays a crucial role in the translational control of a specific subset of mRNAs, depending on their structural characteristics. Specifically, the 5′ untranslated region (UTR) and its length, as well as the length of the open reading frame (ORF), may determine the loading of mRNAs onto ribosomes [124,125,126]. The strongest evidence is the significant and specific reduction in translational control for those mRNAs expressing proteins involved in erythroid lineage commitment. For some ribosomopathies, detailed studies have revealed the molecular mechanisms that govern the function of proteins derived from genes affected by mutations, highlighting how their defective rRNA-binding capacity is also causative of the reduced ribosomal activity associated with ribosomopathies. Below, we present two emblematic examples of mutated genes expressing RP and RBF associated with DBA and SDS, respectively. 4.1. Impairment of Ribosomal Function for Rps19 Protein in Diamond-Blackfan Anemia DBA corresponds to a hereditary, uncommon form of pure red blood cell aplasia characterized by the incapability of erythropoiesis. The disease condition typically appears within the initial year of one’s life, with 95% of DBA cases diagnosed before 2 years of age . Fifty-five percent of DBA patients have autosomal dominant mutations associated with 1 of 19 genes encoding RPs belonging to small or large ribosomal subunits that cause haploinsufficiency of the corresponding gene expression product . In addition to RPs, several non-RP genes also contribute to the pathogenesis of DBA like GATA1, HSP70 and TSR2 [129,130]. The frequency of mutations differs among various genes, reaching a maximum value of 25-30%, affecting the RPS19 gene, which encodes for RP eS19 [131,132]. This protein is a component of the small ribosomal subunit where it interacts with the 18S rRNA through both a series of hydrogen bonds and electrostatic interactions . Both computational and experimental analyses showed that several mutations of the RPS19 gene reported so far may break some of these hydrogen bonds while other mutations appear to alter the surface electrostatic properties of the protein perturbing interactions between negatively charged rRNA and the positively charged surface of Rps19 [134,135]. Consequently, the inability of mutated RPS19 to interact with 18S rRNA would impair its correct assembly into ribosomes. In addition, Rps19 also participates in the processing of pre-rRNA and in the maturation of the 40S ribosomal subunit, favoring the formation of the decoding site. Its defective function or its depletion causes an impairment of translational fidelity and a reduction in 47S rRNA synthesis . The net result is a reduction in the amount of functional 80S and protein synthesis that induces a competition between the cytosolic mRNAs for the available ribosomes to the detriment of those mRNA essentials for differentiation of erythroblasts. Among these, the GATA1 mRNA, for which the presumed folded structure is assumed by its 5′-UTR, makes it particularly sensitive even to slight variations in protein synthesis [126,137,138]. It has also been hypothesized that the tissue-specific character of ribosomopathies may originate from the existence of specialized subgroups of ribosomes with increased affinity for 5′-UTR of transcripts mostly involved in erythroid differentiation. However, stoichiometric analysis of RPs isolated from ribosomes belonging to four cell models of erythropoiesis showed a consistent protein balance throughout the process of cellular differentiation, thus ruling out in this case the possibility of ribosomal diversity as a source of DBA ribosomopathy . 4.2. Impairment of Ribosomal Function for SBDS Protein in Shwachman–-Diamond Syndrome Shwachman–Diamond syndrome (SDS) is a rare autosomal recessive disorder characterized by bone marrow failure, exocrine pancreatic dysfunction and a predisposition to leukemia. The age of onset is heterogeneous, being able to occur in the prenatal, neonatal and infantile periods . Approximately 90% of SDS patients carry mutations in the Shwachman–Bodian–Diamond Syndrome (SBDS) gene, which encodes the SBDS protein of 28,764 Da [141,142]. However, ~10% of patients do not exhibit mutations in the SBDS gene. In these cases, other mutated genes such as DnaJ heat shock protein family (Hsp40) member C21 (DNAJC21), elongation factor-like 1 (EFL1) and signal recognition particle 54 (SRP54) are associated with an SDS-like phenotype . The expression products of these genes participate in a common pathway involved in the maturation of the 60S ribosomal subunit. Specifically, SBDS is involved in the late stage of 60S ribosomal subunit maturation in the cytoplasm. It collaborates with the GTPase elongation factor-like 1 (EFL1) to facilitate the release of eIF6 from the large ribosomal subunit [144,145,146,147]. The eIF6 protein binds to the sarcin–ricin loop (SRL) of 28S rRNA, uL14 and eL24 on the intersubunit face of the 60S subunit, inhibiting its association with the 40S particle [148,149]. SBDS occupies multiple sites on the 60S subunit, spanning from the P site to the peptidyl transferase center (PTC), while EFL1 binds to the forming GTPase-associated center (GAC) of the large subunit. The concerted action of SBDS and Efl1 enhances the rotational dynamics of SBDS through its flexible region, which interacts with rRNA to promote the release of eIF6 . The interactions of SBDS and EFL1 with the pre-60S subunit serve as a final quality control system for the integrity of both the P-site and the GAC. The subsequent release of eIF6 from the 60S subunits signifies that the 60S particle is mature and ready for translation. Certain disease-associated variants of SBDS have demonstrated impaired binding to ribosomal RNA, hindering proper folding into dynamic conformational states necessary for initiating the release of eIF6 from the 60S ribosomal subunits [115,150]. The increased retention of eIF6 inhibits the joining between ribosomal subunits, resulting in reduced protein synthesis. In transgenic mice, overexpression of eIF6 mimics the phenotypes of ribosomopathies characterized by defective hematopoiesis, while a series of suppressor mutations targeting eIF6 rescue the disease-associated phenotype [144,151]. Subsequent X-ray crystallography analysis of the binary complex 60S/eIF6 revealed that suppressor mutations affecting amino acids utilized by eIF6 to interact with the surface of the 60S subunit included those amino acids favoring binding with the SRL of 28S rRNA [148,149]. Additionally, for SDS, it has been demonstrated that the reduction in available mature ribosomes impacts translation at the level of re-initiation of mRNAs whose products are involved in the regulation of granulocytic differentiation [151,152,153]. 5. Conclusions The ancestral origins of the ribosome and its evolutionarily conserved essential functions make it possible to consider it as an excellent reference model for understanding the evolutionary meaning of RNA–protein interactions that are useful for the emergence of living organisms. Indeed, the primitive heart of the ribosome made up of RNA with ribozyme functions has represented a source for the synthesis of new proteins from which those capable of binding the RNA itself have, in turn, represented a driving force to further improve its enzymatic activity and reading fidelity. Their functions, however, are not attributable to the interaction of a single and specific couple of RNA/proteins but are the result of an intricate interconnected network composed in eukaryotes by 79 ribosomal proteins (RPs) and four ribosomal RNAs (rRNAs). The complexity of a system also results from its ability to keep the different and distant sites that compose it in communication with each other. Remarkably, the system is further complicated in its understanding by the numerous factors participating both in the assembly process and in the surveillance mechanism for the preservation of ribosomal integrity. Understanding the specific contribution of each RNA–protein interaction to the proper functioning of the above represents one of the major challenges in the current ribosome research which can be prosecuted through interdisciplinary and integrated approaches ranging from structural, phylogenetic and mathematical methods. Looking ahead, utilizing emerging methods such as in vivo single-molecule tracking and cryo–electron tomography [155,156] will afford a vital comprehension of the assembly process of ribosomes within their intricate native cellular milieu. Furthermore, the formulation of predictive models elucidating pathogenicity of missense mutations through artificial intelligence coupled with the functioning of complex cellular processes within a living cell will enable better integration and understanding of those exome sequencing data of patients revealing further gene variants associated with ribosomopathies that are negative for mutation of known genes. Author Contributions The co-authors contributed to the review paper in the following ways: Conceptualization, D.B. and C.C. (Carlo Cogoni); writing—original draft preparation, D.B.; reviewing the work critically for important intellectual content, C.C. (Carlo Cogoni); review and editing of the draft, D.B. and C.C. (Caterina Catalanotto); funding acquisition, C.B.; supervision, C.C. (Caterina Catalanotto), C.B., C.C. (Carlo Cogoni) and D.B. All authors have read and agreed to the published version of the manuscript. Conflicts of Interest The authors declare no conflict of interest. Funding Statement This work was supported by grants to Prof Paola Londei from “Sapienza” University of Rome for the project “Role of the translation factor eIF6 in ribosome biogenesis and differential translation of specific mRNAs (protocol number RM11715C7CE47750). Footnotes Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. References 1.Neelamraju Y., Hashemikhabir S., Janga S.C. The human RBPome: From genes and proteins to human disease. J. Proteom. 2015;127:61–70. doi: 10.1016/j.jprot.2015.04.031. [DOI] [PubMed] [Google Scholar] 2.Gerstberger S., Hafner M., Tuschl T. A census of human RNA-binding proteins. Nat. Rev. Genet. 2014;15:829–845. doi: 10.1038/nrg3813. [DOI] [PMC free article] [PubMed] [Google Scholar] 3.Van Nostrand E.L., Freese P., Pratt G.A., Wang X., Wei X., Xiao R., Blue S.M., Chen J.-Y., Cody N.A.L., Dominguez D., et al. A large-scale binding and functional map of human RNA-binding proteins. Nature. 2020;583:711–719. doi: 10.1038/s41586-020-2077-3. [DOI] [PMC free article] [PubMed] [Google Scholar] 4.Lukong K.E., Chang K.-W., Khandjian E.W., Richard S. RNA-binding proteins in human genetic disease. Trends Genet. 2008;24:416–425. doi: 10.1016/j.tig.2008.05.004. [DOI] [PubMed] [Google Scholar] 5.Wendel H.-G., Silva R.L., Malina A., Mills J.R., Zhu H., Ueda T., Watanabe-Fukunaga R., Fukunaga R., Teruya-Feldstein J., Pelletier J., et al. Dissecting eIF4E action in tumorigenesis. Genes Dev. 2007;21:3232–3237. doi: 10.1101/gad.1604407. [DOI] [PMC free article] [PubMed] [Google Scholar] 6.Kolb S.J., Battle D.J., Dreyfuss G. Molecular Functions of the SMN Complex. J. Child Neurol. 2007;22:990–994. doi: 10.1177/0883073807305666. [DOI] [PubMed] [Google Scholar] 7.Darnell J.C., Fraser C.E., Mostovetsky O., Stefani G., Jones T.A., Eddy S.R., Darnell R.B. Kissing complex RNAs mediate interaction between the Fragile-X mental retardation protein KH2 domain and brain polyribosomes. Genes Dev. 2005;19:903–918. doi: 10.1101/gad.1276805. [DOI] [PMC free article] [PubMed] [Google Scholar] 8.Hsiao C., Mohan S., Kalahar B.K., Williams L.D. Peeling the Onion: Ribosomes Are Ancient Molecular Fossils. Mol. Biol. Evol. 2009;26:2415–2425. doi: 10.1093/molbev/msp163. [DOI] [PubMed] [Google Scholar] 9.Mears J.A., Cannone J.J., Stagg S.M., Gutell R.R., Agrawal R.K., Harvey S.C. Modeling a Minimal Ribosome Based on Comparative Sequence Analysis. J. Mol. Biol. 2002;321:215–234. doi: 10.1016/S0022-2836(02)00568-5. [DOI] [PubMed] [Google Scholar] 10.Fox G.E. Origin and Evolution of the Ribosome. Cold Spring Harb. Perspect. Biol. 2010;2:a003483. doi: 10.1101/cshperspect.a003483. [DOI] [PMC free article] [PubMed] [Google Scholar] 11.Ban N., Beckmann R., Cate J.H., Dinman J.D., Dragon F., Ellis S.R., Lafontaine D.L., Lindahl L., Liljas A., Lipton J.M., et al. A new system for naming ribosomal proteins. Curr. Opin. Struct. Biol. 2014;24:165–169. doi: 10.1016/j.sbi.2014.01.002. [DOI] [PMC free article] [PubMed] [Google Scholar] 12.Stuart S.N., Chanson J.S., Cox N.A., Young B.E., Rodrigues A.S.L., Fischman D.L., Waller R.W. The complete atomic structure of the large ribosomal subunit at 2.4 A resolution. Science. 2000;306:1783–1786. doi: 10.1126/science.1103538. [DOI] [Google Scholar] 13.Armache J.-P., Jarasch A., Anger A.M., Villa E., Becker T., Bhushan S., Jossinet F., Habeck M., Dindar G., Franckenberg S., et al. Cryo-EM structure and rRNA model of a translating eukaryotic 80S ribosome at 5.5-Å resolution. Proc. Natl. Acad. Sci. USA. 2010;107:19748–19753. doi: 10.1073/pnas.1009999107. [DOI] [PMC free article] [PubMed] [Google Scholar] 14.Ben-Shem A., de Loubresse N.G., Melnikov S., Jenner L., Yusupova G., Yusupov M. The Structure of the Eukaryotic Ribosome at 3.0 Å Resolution. Science. 2011;334:1524–1529. doi: 10.1126/science.1212642. [DOI] [PubMed] [Google Scholar] 15.Khatter H., Myasnikov A.G., Natchiar S.K., Klaholz B.P. Structure of the human 80S ribosome. Nature. 2015;520:640–645. doi: 10.1038/nature14427. [DOI] [PubMed] [Google Scholar] 16.Dabbs E.R. Mutant Studies on the Prokaryotic Ribosome. In: Hardesty B., Kramer G.A., editors. Structure, Function, and Genetics of Ribosomes. Springer; New York, NY, USA: 1986. pp. 733–748. (Springer Series in Molecular Biology). [DOI] [Google Scholar] 17.Nomura M., Mizushima S., Ozaki M., Traub P., Lowry C.V. Structure and Function of Ribosomes and Their Molecular Components. Cold Spring Harb. Symp. Quant. Biol. 1969;34:49–61. doi: 10.1101/SQB.1969.034.01.009. [DOI] [PubMed] [Google Scholar] 18.Bowman C.M., Dahlberg J.E., Ikemura T., Konisky J., Nomura M. Specific Inactivation of 16S Ribosomal RNA Induced by Colicin. E3 In Vivo. Proc. Natl. Acad. Sci. USA. 1971;68:964–968. doi: 10.1073/pnas.68.5.964. [DOI] [PMC free article] [PubMed] [Google Scholar] 19.Senior B.W., Holland I.B. Effect of Colicin E3 upon the 30S Ribosomal Subunit of Escherichia coli. Proc. Natl. Acad. Sci. USA. 1971;68:959–963. doi: 10.1073/pnas.68.5.959. [DOI] [PMC free article] [PubMed] [Google Scholar] 20.Endo Y., Wool I.G. The site of action of alpha-sarcin on eukaryotic ribosomes. The sequence at the alpha-sarcin cleavage site in 28 S ribosomal ribonucleic acid. J. Biol. Chem. 1982;257:9054–9060. doi: 10.1016/S0021-9258(18)34241-8. [DOI] [PubMed] [Google Scholar] 21.Pettersen E.F., Goddard T.D., Huang C.C., Couch G.S., Greenblatt D.M., Meng E.C., Ferrin T.E. UCSF Chimera?A visualization system for exploratory research and analysis. J. Comput. Chem. 2004;25:1605–1612. doi: 10.1002/jcc.20084. [DOI] [PubMed] [Google Scholar] 22.Sweeney B.A., Hoksza D., Nawrocki E.P., Ribas C.E., Madeira F., Cannone J.J., Gutell R., Maddala A., Meade C.D., Williams L.D., et al. R2DT is a framework for predicting and visualising RNA secondary structure using templates. Nat. Commun. 2021;12:3494. doi: 10.1038/s41467-021-23555-5. [DOI] [PMC free article] [PubMed] [Google Scholar] 23.Woese C.R. On the evolution of the genetic code. Proc. Natl. Acad. Sci. USA. 1965;54:1546–1552. doi: 10.1073/pnas.54.6.1546. [DOI] [PMC free article] [PubMed] [Google Scholar] 24.Crick F. The origin of the genetic code. J. Mol. Biol. 1968;38:367–379. doi: 10.1016/0022-2836(68)90392-6. [DOI] [PubMed] [Google Scholar] 25.Orgel L. Evolution of the genetic apparatus. J. Mol. Biol. 1968;38:381–393. doi: 10.1016/0022-2836(68)90393-8. [DOI] [PubMed] [Google Scholar] 26.Gilbert W. Origin of life: The RNA world. Nature. 1986;319:618. doi: 10.1038/319618a0. [DOI] [Google Scholar] 27.Zuker M. On Finding All Suboptimal Foldings of an RNA Molecule. Science. 1989;244:48–52. doi: 10.1126/science.2468181. [DOI] [PubMed] [Google Scholar] 28.Buchmueller K.L., Weeks K.M. Near Native Structure in an RNA Collapsed State. Biochemistry. 2003;42:13869–13878. doi: 10.1021/bi035476k. [DOI] [PubMed] [Google Scholar] 29.Breaker R.R. Riboswitches and Translation Control. Cold Spring Harb. Perspect. Biol. 2018;10:a032797. doi: 10.1101/cshperspect.a032797. [DOI] [PMC free article] [PubMed] [Google Scholar] 30.Semrad K., Green R., Schroeder R. RNA chaperone activity of large ribosomal subunit proteins from Escherichia coli. RNA. 2004;10:1855–1860. doi: 10.1261/rna.7121704. [DOI] [PMC free article] [PubMed] [Google Scholar] 31.de la Cruz J., Karbstein K., Woolford J.L. Functions of Ribosomal Proteins in Assembly of Eukaryotic Ribosomes In Vivo. Annu. Rev. Biochem. 2015;84:93–129. doi: 10.1146/annurev-biochem-060614-033917. [DOI] [PMC free article] [PubMed] [Google Scholar] 32.Draper D.E., Reynaldo L.P. RNA binding strategies of ribosomal proteins. Nucleic Acids Res. 1999;27:381–388. doi: 10.1093/nar/27.2.381. [DOI] [PMC free article] [PubMed] [Google Scholar] 33.Timsit Y., Sergeant-Perthuis G., Bennequin D. Evolution of ribosomal protein network architectures. Sci. Rep. 2021;11:625. doi: 10.1038/s41598-020-80194-4. [DOI] [PMC free article] [PubMed] [Google Scholar] 34.Parker E.T., Cleaves H.J., Dworkin J.P., Glavin D.P., Callahan M., Aubrey A., Lazcano A., Bada J.L. Primordial synthesis of amines and amino acids in a 1958 Miller H2S-rich spark discharge experiment. Proc. Natl. Acad. Sci. USA. 2011;108:5526–5531. doi: 10.1073/pnas.1019191108. [DOI] [PMC free article] [PubMed] [Google Scholar] 35.Zepik H., Shavit E., Tang M., Jensen T.R., Kjaer K., Bolbach G., Leiserowitz L., Weissbuch I., Lahav M. Chiral Amplification of Oligopeptides in Two-Dimensional Crystalline Self-Assemblies on Water. Science. 2002;295:1266–1269. doi: 10.1126/science.1065625. [DOI] [PubMed] [Google Scholar] 36.Suwannachot Y., Rode B.M. Mutual amino acid catalysis in salt-induced peptide formation supports this mechanism’s role in prebiotic peptide evolution. Space Life Sci. 1999;29:463–471. doi: 10.1023/a:1006583311808. [DOI] [PubMed] [Google Scholar] 37.Leman L.J., Huang Z.-Z., Ghadiri M.R. Peptide Bond Formation in Water Mediated by Carbon Disulfide. Astrobiology. 2015;15:709–716. doi: 10.1089/ast.2015.1314. [DOI] [PubMed] [Google Scholar] 38.Leman L., Orgel L., Ghadiri M.R. Carbonyl Sulfide-Mediated Prebiotic Formation of Peptides. Science. 2004;306:283–286. doi: 10.1126/science.1102722. [DOI] [PubMed] [Google Scholar] 39.Kitadai N., Maruyama S. Origins of building blocks of life: A review. Geosci. Front. 2018;9:1117–1153. doi: 10.1016/j.gsf.2017.07.007. [DOI] [Google Scholar] 40.Frenkel-Pinter M., Samanta M., Ashkenasy G., Leman L.J. Prebiotic Peptides: Molecular Hubs in the Origin of Life. Chem. Rev. 2020;120:4707–4765. doi: 10.1021/acs.chemrev.9b00664. [DOI] [PubMed] [Google Scholar] 41.Wimberly B.T., Brodersen D.E., Clemons W.M., Morgan-Warren R.J., Carter A.P., Vonrhein C., Hartsch T., Ramakrishnan V. Structure of the 30S ribosomal subunit. Nature. 2000;407:327–339. doi: 10.1038/35030006. [DOI] [PubMed] [Google Scholar] 42.Yusupov M.M., Yusupova G.Z., Baucom A., Lieberman K., Earnest T.N., Cate J.H.D., Noller H.F. Crystal Structure of the Ribosome at 5.5 Å Resolution. Science. 2001;292:883–896. doi: 10.1126/science.1060089. [DOI] [PubMed] [Google Scholar] 43.Stern S., Wilson R.C., Noller H.F. Localization of the binding site for protein S4 on 16 S ribosomal RNA by chemical and enzymatic probing and primer extension. J. Mol. Biol. 1986;192:101–110. doi: 10.1016/0022-2836(86)90467-5. [DOI] [PubMed] [Google Scholar] 44.Malygin A.A., Yanshina D.D., Karpova G.G. Interactions of human ribosomal proteins S16 and S5 with an 18S rRNA fragment containing their binding sites. Biochimie. 2009;91:1180–1186. doi: 10.1016/j.biochi.2009.06.013. [DOI] [PubMed] [Google Scholar] 45.Agarwal D., Kamath D., Gregory S.T., O’Connor M. Modulation of Decoding Fidelity by Ribosomal Proteins S4 and S5. J. Bacteriol. 2015;197:1017–1025. doi: 10.1128/JB.02485-14. [DOI] [PMC free article] [PubMed] [Google Scholar] 46.Bokov K., Steinberg S.V. A hierarchical model for evolution of 23S ribosomal RNA. Nature. 2009;457:977–980. doi: 10.1038/nature07749. [DOI] [PubMed] [Google Scholar] 47.Petrov A.S., Bernier C.R., Hsiao C., Norris A.M., Kovacs N.A., Waterbury C.C., Stepanov V.G., Harvey S.C., Fox G.E., Wartell R.M., et al. Evolution of the ribosome at atomic resolution. Proc. Natl. Acad. Sci. USA. 2014;111:10251–10256. doi: 10.1073/pnas.1407205111. [DOI] [PMC free article] [PubMed] [Google Scholar] 48.Kovacs N.A., Petrov A.S., Lanier K.A., Williams L.D. Frozen in Time: The History of Proteins. Mol. Biol. Evol. 2017;34:1252–1260. doi: 10.1093/molbev/msx086. [DOI] [PMC free article] [PubMed] [Google Scholar] 49.Petrov A.S., Gulen B., Norris A.M., Kovacs N.A., Bernier C.R., Lanier K.A., Fox G.E., Harvey S.C., Wartell R.M., Hud N.V., et al. History of the ribosome and the origin of translation. Proc. Natl. Acad. Sci. USA. 2015;112:15396–15401. doi: 10.1073/pnas.1509761112. [DOI] [PMC free article] [PubMed] [Google Scholar] 50.Barandun J., Hunziker M., Vossbrinck C.R., Klinge S. Evolutionary compaction and adaptation visualized by the structure of the dormant microsporidian ribosome. Nat. Microbiol. 2019;4:1798–1804. doi: 10.1038/s41564-019-0514-6. [DOI] [PMC free article] [PubMed] [Google Scholar] 51.Peng Z., Oldfield C.J., Xue B., Mizianty M.J., Dunker A.K., Kurgan L., Uversky V.N. A creature with a hundred waggly tails: Intrinsically disordered proteins in the ribosome. Cell. Mol. Life Sci. 2014;71:1477–1504. doi: 10.1007/s00018-013-1446-6. [DOI] [PMC free article] [PubMed] [Google Scholar] 52.Melnikov S., Ben-Shem A., de Loubresse N.G., Jenner L., Yusupova G., Yusupov M. One core, two shells: Bacterial and eukaryotic ribosomes. Nat. Struct. Mol. Biol. 2012;19:560–567. doi: 10.1038/nsmb.2313. [DOI] [PubMed] [Google Scholar] 53.Klein D.J., Moore P.B., Steitz T.A. The Roles of Ribosomal Proteins in the Structure Assembly, and Evolution of the Large Ribosomal Subunit. J. Mol. Biol. 2004;340:141–177. doi: 10.1016/j.jmb.2004.03.076. [DOI] [PubMed] [Google Scholar] 54.Wilson D.N., Nierhaus K.H. Ribosomal Proteins in the Spotlight. Crit. Rev. Biochem. Mol. Biol. 2005;40:243–267. doi: 10.1080/10409230500256523. [DOI] [PubMed] [Google Scholar] 55.Timsit Y., Acosta Z., Allemand F., Chiaruttini C., Springer M. The Role of Disordered Ribosomal Protein Extensions in the Early Steps of Eubacterial 50 S Ribosomal Subunit Assembly. Int. J. Mol. Sci. 2009;10:817–834. doi: 10.3390/ijms10030817. [DOI] [PMC free article] [PubMed] [Google Scholar] 56.Cockman E., Anderson P., Ivanov P. TOP mRNPs: Molecular Mechanisms and Principles of Regulation. Biomolecules. 2020;10:969. doi: 10.3390/biom10070969. [DOI] [PMC free article] [PubMed] [Google Scholar] 57.Peña C., Hurt E., Panse V.G. Eukaryotic ribosome assembly, transport and quality control. Nat. Struct. Mol. Biol. 2017;24:689–699. doi: 10.1038/nsmb.3454. [DOI] [PubMed] [Google Scholar] 58.Klinge S., Woolford J.L., Jr. Ribosome assembly coming into focus. Nat. Rev. Mol. Cell Biol. 2019;20:116–131. doi: 10.1038/s41580-018-0078-y. [DOI] [PMC free article] [PubMed] [Google Scholar] 59.Dörner K., Ruggeri C., Zemp I., Kutay U. Ribosome biogenesis factors—From names to functions. EMBO J. 2023;42:e112699. doi: 10.15252/embj.2022112699. [DOI] [PMC free article] [PubMed] [Google Scholar] 60.Sloan K.E., Mattijssen S., Lebaron S., Tollervey D., Pruijn G.J., Watkins N.J. Both endonucleolytic and exonucleolytic cleavage mediate ITS1 removal during human ribosomal RNA processing. J. Cell Biol. 2013;200:577–588. doi: 10.1083/jcb.201207131. [DOI] [PMC free article] [PubMed] [Google Scholar] 61.Montellese C., Montel-Lehry N., Henras A.K., Kutay U., Gleizes P.-E., O’donohue M.-F. Poly(A)-specific ribonuclease is a nuclear ribosome biogenesis factor involved in human 18S rRNA maturation. Nucleic Acids Res. 2017;45:6822–6836. doi: 10.1093/nar/gkx253. [DOI] [PMC free article] [PubMed] [Google Scholar] 62.Watkins N.J., Bohnsack M.T. The box C/D and H/ACA snoRNPs: Key players in the modification, processing and the dynamic folding of ribosomal RNA. Wiley Interdiscip. Rev. RNA. 2012;3:397–414. doi: 10.1002/wrna.117. [DOI] [PubMed] [Google Scholar] 63.Sloan K.E., Warda A.S., Sharma S., Entian K.-D., Lafontaine D.L.J., Bohnsack M.T. Tuning the ribosome: The influence of rRNA modification on eukaryotic ribosome biogenesis and function. RNA Biol. 2017;14:1138–1152. doi: 10.1080/15476286.2016.1259781. [DOI] [PMC free article] [PubMed] [Google Scholar] 64.Taoka M., Nobe Y., Yamaki Y., Sato K., Ishikawa H., Izumikawa K., Yamauchi Y., Hirota K., Nakayama H., Takahashi N., et al. Landscape of the complete RNA chemical modifications in the human 80S ribosome. Nucleic Acids Res. 2018;46:9289–9298. doi: 10.1093/nar/gky811. [DOI] [PMC free article] [PubMed] [Google Scholar] 65.Piekna-Przybylska D., Decatur W.A., Fournier M.J. The 3D rRNA modification maps database: With interactive tools for ribosome analysis. Nucleic Acids Res. 2008;36:D178–D183. doi: 10.1093/nar/gkm855. [DOI] [PMC free article] [PubMed] [Google Scholar] 66.Sharma S., Lafontaine D.L. ‘View From A Bridge’: A New Perspective on Eukaryotic rRNA Base Modification. Trends Biochem. Sci. 2015;40:560–575. doi: 10.1016/j.tibs.2015.07.008. [DOI] [PubMed] [Google Scholar] 67.Natchiar S.K., Myasnikov A.G., Kratzat H., Hazemann I., Klaholz B.P. Visualization of chemical modifications in the human 80S ribosome structure. Nature. 2017;551:472–477. doi: 10.1038/nature24482. [DOI] [PubMed] [Google Scholar] 68.Boccaletto P., Stefaniak F., Ray A., Cappannini A., Mukherjee S., Purta E., Kurkowska M., Shirvanizadeh N., Destefanis E., Groza P., et al. MODOMICS: A database of RNA modification pathways. 2021 update. Nucleic Acids Res. 2022;50:D231–D235. doi: 10.1093/nar/gkab1083. [DOI] [PMC free article] [PubMed] [Google Scholar] 69.Green R., Noller H.F. In vitro complementation analysis localizes 23S rRNA posttranscriptional modifications that are required for Escherichia coli 50S ribosomal subunit assembly and function. RNA. 1996;2:1011–1021. [PMC free article] [PubMed] [Google Scholar] 70.Baxter-Roshek J.L., Petrov A.N., Dinman J.D. Optimization of Ribosome Structure and Function by rRNA Base Modification. PLoS ONE. 2007;2:e174. doi: 10.1371/journal.pone.0000174. [DOI] [PMC free article] [PubMed] [Google Scholar] 71.Jack K., Bellodi C., Landry D.M., Niederer R.O., Meskauskas A., Musalgaonkar S., Kopmar N., Krasnykh O., Dean A.M., Thompson S.R., et al. rRNA Pseudouridylation Defects Affect Ribosomal Ligand Binding and Translational Fidelity from Yeast to Human Cells. Mol. Cell. 2011;44:660–666. doi: 10.1016/j.molcel.2011.09.017. [DOI] [PMC free article] [PubMed] [Google Scholar] 72.Khoshnevis S., Dreggors-Walker R.E., Marchand V., Motorin Y., Ghalei H. Ribosomal RNA 2′-O-methylations regulate translation by impacting ribosome dynamics. Proc. Natl. Acad. Sci. USA. 2022;119:1593–1609.39. doi: 10.1073/pnas.2117334119. [DOI] [PMC free article] [PubMed] [Google Scholar] 73.Espinar-Marchena F.J., Babiano R., de la Cruz J. Placeholder factors in ribosome biogenesis: Please, pave my way. Microb. Cell. 2017;4:144–168. doi: 10.15698/mic2017.05.572. [DOI] [PMC free article] [PubMed] [Google Scholar] 74.Warner J.R. The economics of ribosome biosynthesis in yeast. Trends Biochem. Sci. 1999;24:437–440. doi: 10.1016/S0968-0004(99)01460-7. [DOI] [PubMed] [Google Scholar] 75.Stefanovsky V.Y., Pelletier G., Hannan R., Gagnon-Kugler T., Rothblum L.I., Moss T. An Immediate Response of Ribosomal Transcription to Growth Factor Stimulation in Mammals Is Mediated by ERK Phosphorylation of UBF. Mol. Cell. 2001;8:1063–1073. doi: 10.1016/S1097-2765(01)00384-7. [DOI] [PubMed] [Google Scholar] 76.Iadevaia V., Liu R., Proud C.G. mTORC1 signaling controls multiple steps in ribosome biogenesis. Semin. Cell Dev. Biol. 2014;36:113–120. doi: 10.1016/j.semcdb.2014.08.004. [DOI] [PubMed] [Google Scholar] 77.Mayer C., Grummt I. Ribosome biogenesis and cell growth: mTOR coordinates transcription by all three classes of nuclear RNA polymerases. Oncogene. 2006;25:6384–6391. doi: 10.1038/sj.onc.1209883. [DOI] [PubMed] [Google Scholar] 78.van Riggelen J., Yetil A., Felsher D.W. MYC as a regulator of ribosome biogenesis and protein synthesis. Nat. Rev. Cancer. 2010;10:301–309. doi: 10.1038/nrc2819. [DOI] [PubMed] [Google Scholar] 79.Derenzini M., Montanaro L., Trerè D. Ribosome biogenesis and cancer. Acta Histochem. 2017;119:190–197. doi: 10.1016/j.acthis.2017.01.009. [DOI] [PubMed] [Google Scholar] 80.Thomas G. An encore for ribosome biogenesis in the control of cell proliferation. Nature. 2000;2:E71–E72. doi: 10.1038/35010581. [DOI] [PubMed] [Google Scholar] 81.Rubbi C.P., Milner J. Disruption of the nucleolus mediates stabilization of p53 in response to DNA damage and other stresses. EMBO J. 2003;22:6068–6077. doi: 10.1093/emboj/cdg579. [DOI] [PMC free article] [PubMed] [Google Scholar] 82.Lindström M.S., Bartek J., Maya-Mendoza A. p53 at the crossroad of DNA replication and ribosome biogenesis stress pathways. Cell Death Differ. 2022;29:972–982. doi: 10.1038/s41418-022-00999-w. [DOI] [PMC free article] [PubMed] [Google Scholar] 83.Fumagalli S., Di Cara A., Neb-Gulati A., Natt F., Schwemberger S., Hall J., Babcock G.F., Bernardi R., Pandolfi P.P., Thomas G. Absence of nucleolar disruption after impairment of 40S ribosome biogenesis reveals an rpL11-translation-dependent mechanism of p53 induction. Nature. 2009;11:501–508. doi: 10.1038/ncb1858. [DOI] [PMC free article] [PubMed] [Google Scholar] 84.Burger K., Mühl B., Harasim T., Rohrmoser M., Malamoussi A., Orban M., Kellner M., Gruber-Eber A., Kremmer E., Hölzel M., et al. Chemotherapeutic Drugs Inhibit Ribosome Biogenesis at Various Levels. J. Biol. Chem. 2010;285:12416–12425. doi: 10.1074/jbc.M109.074211. [DOI] [PMC free article] [PubMed] [Google Scholar] 85.Grummt I. The nucleolus—Guardian of cellular homeostasis and genome integrity. Chromosoma. 2013;122:487–497. doi: 10.1007/s00412-013-0430-0. [DOI] [PubMed] [Google Scholar] 86.Sherr C.J. Divorcing ARF and p53: An unsettled case. Nat. Rev. Cancer. 2006;6:663–673. doi: 10.1038/nrc1954. [DOI] [PubMed] [Google Scholar] 87.Zhang C., Comai L., Johnson D.L. PTEN Represses RNA Polymerase I Transcription by Disrupting the SL1 Complex. Mol. Cell. Biol. 2005;25:6899–6911. doi: 10.1128/MCB.25.16.6899-6911.2005. [DOI] [PMC free article] [PubMed] [Google Scholar] 88.Pelletier J., Thomas G., Volarević S. Ribosome biogenesis in cancer: New players and therapeutic avenues. Nat. Rev. Cancer. 2018;18:51–63. doi: 10.1038/nrc.2017.104. [DOI] [PubMed] [Google Scholar] 89.Bustelo X.R., Dosil M. Ribosome biogenesis and cancer: Basic and translational challenges. Curr. Opin. Genet. Dev. 2018;48:22–29. doi: 10.1016/j.gde.2017.10.003. [DOI] [PubMed] [Google Scholar] 90.Lazaris-Karatzas A., Montine K.S., Sonenberg N. Malignant transformation by a eukaryotic initiation factor subunit that binds to mRNA 5′ cap. Nature. 1990;345:544–547. doi: 10.1038/345544a0. [DOI] [PubMed] [Google Scholar] 91.Ruggero D., Montanaro L., Ma L., Xu W., Londei P., Cordon-Cardo C., Pandolfi P.P. The translation factor eIF-4E promotes tumor formation and cooperates with c-Myc in lymphomagenesis. Nat. Med. 2004;10:484–486. doi: 10.1038/nm1042. [DOI] [PubMed] [Google Scholar] 92.Sulima S.O., Kampen K.R., De Keersmaecker K. Cancer Biogenesis in Ribosomopathies. Cells. 2019;8:229. doi: 10.3390/cells8030229. [DOI] [PMC free article] [PubMed] [Google Scholar] 93.Palade G.E. A small particulate component of the cytoplasm. J. Biophys. Biochem. Cytol. 1955;1:59–68. doi: 10.1083/jcb.1.1.59. [DOI] [PMC free article] [PubMed] [Google Scholar] 94.Shi Z., Barna M. Translating the Genome in Time and Space: Specialized Ribosomes, RNA Regulons, and RNA-Binding Proteins. Annu. Rev. Cell Dev. Biol. 2015;31:31–54. doi: 10.1146/annurev-cellbio-100814-125346. [DOI] [PubMed] [Google Scholar] 95.Dinman J.D. Pathways to Specialized Ribosomes: The Brussels Lecture. J. Mol. Biol. 2016;428:2186–2194. doi: 10.1016/j.jmb.2015.12.021. [DOI] [PMC free article] [PubMed] [Google Scholar] 96.Genuth N.R., Barna M. The Discovery of Ribosome Heterogeneity and Its Implications for Gene Regulation and Organismal Life. Mol. Cell. 2018;71:364–374. doi: 10.1016/j.molcel.2018.07.018. [DOI] [PMC free article] [PubMed] [Google Scholar] 97.Kondrashov N., Pusic A., Stumpf C.R., Shimizu K., Hsieh A.C., Xue S., Ishijima J., Shiroishi T., Barna M. Ribosome-Mediated Specificity in Hox mRNA Translation and Vertebrate Tissue Patterning. Cell. 2011;145:383–397. doi: 10.1016/j.cell.2011.03.028. [DOI] [PMC free article] [PubMed] [Google Scholar] 98.Krogh N., Jansson M.D., Häfner S.J., Tehler D., Birkedal U., Christensen-Dalsgaard M., Lund A.H., Nielsen H. Profiling of 2′-O-Me in human rRNA reveals a subset of fractionally modified positions and provides evidence for ribosome heterogeneity. Nucleic Acids Res. 2016;44:7884–7895. doi: 10.1093/nar/gkw482. [DOI] [PMC free article] [PubMed] [Google Scholar] 99.Locati M.D., Pagano J.F., Girard G., Ensink W.A., van Olst M., van Leeuwen S., Nehrdich U., Spaink H.P., Rauwerda H., Jonker M.J., et al. Expression of distinct maternal and somatic 5.8S, 18S, and 28S rRNA types during zebrafish development. RNA. 2017;23:1188–1199. doi: 10.1261/rna.061515.117. [DOI] [PMC free article] [PubMed] [Google Scholar] 100.Parks M.M., Kurylo C.M., Dass R.A., Bojmar L., Lyden D., Vincent C.T., Blanchard S.C. Variant ribosomal RNA alleles are conserved and exhibit tissue-specific expression. Sci. Adv. 2018;4:eaao0665. doi: 10.1126/sciadv.aao0665. [DOI] [PMC free article] [PubMed] [Google Scholar] 101.Simsek D., Tiu G.C., Flynn R.A., Byeon G.W., Leppek K., Xu A.F., Chang H.Y., Barna M. The Mammalian Ribo-interactome Reveals Ribosome Functional Diversity and Heterogeneity. Cell. 2017;169:1051–1065.e18. doi: 10.1016/j.cell.2017.05.022. [DOI] [PMC free article] [PubMed] [Google Scholar] 102.Samir P., Browne C.M., Rahul R., Sun M., Shen B., Li W., Frank J., Link A.J. Identification of Changing Ribosome Protein Compositions using Mass Spectrometry. Proteomics. 2018;18:e1800217. doi: 10.1002/pmic.201800217. [DOI] [PMC free article] [PubMed] [Google Scholar] 103.Segev N., Gerst J.E. Specialized ribosomes and specific ribosomal protein paralogs control translation of mitochondrial proteins. J. Cell Biol. 2018;217:117–126. doi: 10.1083/jcb.201706059. [DOI] [PMC free article] [PubMed] [Google Scholar] 104.Mukhopadhyay R., Ray P.S., Arif A., Brady A.K., Kinter M., Fox P.L. DAPK-ZIPK-L13a Axis Constitutes a Negative-Feedback Module Regulating Inflammatory Gene Expression. Mol. Cell. 2008;32:371–382. doi: 10.1016/j.molcel.2008.09.019. [DOI] [PMC free article] [PubMed] [Google Scholar] 105.Knight Z.A., Tan K., Birsoy K., Schmidt S., Garrison J.L., Wysocki R.W., Emiliano A., Ekstrand M.I., Friedman J.M. Molecular Profiling of Activated Neurons by Phosphorylated Ribosome Capture. Cell. 2012;151:1126–1137. doi: 10.1016/j.cell.2012.10.039. [DOI] [PMC free article] [PubMed] [Google Scholar] 106.Meyuhas O. Ribosomal Protein S6 Phosphorylation: Four Decades of Research. Int. Rev. Cell Mol. Biol. 2015;320:41–73. doi: 10.1016/bs.ircmb.2015.07.006. [DOI] [PubMed] [Google Scholar] 107.Simsek D., Barna M. An emerging role for the ribosome as a nexus for post-translational modifications. Curr. Opin. Cell Biol. 2017;45:92–101. doi: 10.1016/j.ceb.2017.02.010. [DOI] [PMC free article] [PubMed] [Google Scholar] 108.Thompson M.K., Rojas-Duran M.F., Gangaramani P., Gilbert W.V., Massachusetts Institute of Technology, of the UnitedStates The ribosomal protein Asc1/RACK1 is required for efficient translation of short mRNAs. eLife. 2016;5:e11154. doi: 10.7554/eLife.11154. [DOI] [PMC free article] [PubMed] [Google Scholar] 109.Gallo S., Ricciardi S., Manfrini N., Pesce E., Oliveto S., Calamita P., Mancino M., Maffioli E., Moro M., Crosti M., et al. RACK1 Specifically Regulates Translation through Its Binding to Ribosomes. Mol. Cell. Biol. 2018;38:e00230-18. doi: 10.1128/MCB.00230-18. [DOI] [PMC free article] [PubMed] [Google Scholar] 110.Vesper O., Amitai S., Belitsky M., Byrgazov K., Kaberdina A.C., Engelberg-Kulka H., Moll I. Selective Translation of Leaderless mRNAs by Specialized Ribosomes Generated by MazF in Escherichia coli. Cell. 2011;147:147–157. doi: 10.1016/j.cell.2011.07.047. [DOI] [PMC free article] [PubMed] [Google Scholar] 111.Shi Z., Fujii K., Kovary K.M., Genuth N.R., Röst H.L., Teruel M.N., Barna M. Heterogeneous Ribosomes Preferentially Translate Distinct Subpools of mRNAs Genome-wide. Mol. Cell. 2017;67:71–83.e7. doi: 10.1016/j.molcel.2017.05.021. [DOI] [PMC free article] [PubMed] [Google Scholar] 112.Draptchinskaia N., Gustavsson P., Andersson B., Pettersson M., Willig T.-N., Dianzani I., Ball S., Tchernia G., Klar J., Matsson H., et al. The gene encoding ribosomal protein S19 is mutated in Diamond-Blackfan anaemia. Nat. Genet. 1999;21:169–175. doi: 10.1038/5951. [DOI] [PubMed] [Google Scholar] 113.Ebert B.L., Pretz J., Bosco J., Chang C.Y., Tamayo P., Galili N., Raza A., Root D.E., Attar E., Ellis S.R., et al. Identification of RPS14 as a 5q- syndrome gene by RNA interference screen. Nature. 2008;451:335–339. doi: 10.1038/nature06494. [DOI] [PMC free article] [PubMed] [Google Scholar] 114.Bellodi C., Krasnykh O., Haynes N., Theodoropoulou M., Peng G., Montanaro L., Ruggero D. Loss of function of the tumor suppressor DKC1 perturbs p27 translation control and contributes to pituitary tumorigenesis. Cancer Res. 2010;70:6026–6035. doi: 10.1158/0008-5472.CAN-09-4730. [DOI] [PMC free article] [PubMed] [Google Scholar] 115.Finch A.J., Hilcenko C., Basse N., Drynan L.F., Goyenechea B., Menne T.F., Fernández A.G., Simpson P., D’Santos C.S., Arends M.J., et al. Uncoupling of GTP hydrolysis from eIF6 release on the ribosome causes Shwachman-Diamond syndrome. Genes Dev. 2011;25:917–929. doi: 10.1101/gad.623011. [DOI] [PMC free article] [PubMed] [Google Scholar] 116.Armistead J., Triggs-Raine B. Diverse diseases from a ubiquitous process: The ribosomopathy paradox. FEBS Lett. 2014;588:1491–1500. doi: 10.1016/j.febslet.2014.03.024. [DOI] [PubMed] [Google Scholar] 117.Jones N.C., Lynn M.L., Gaudenz K., Sakai D., Aoto K., Rey J.-P., Glynn E.F., Ellington L., Du C., Dixon J., et al. Prevention of the Neurocristopathy Treacher Collins Syndrome through Inhibition of P53 Function. Nat. Med. 2008;14:125–133. doi: 10.1038/nm1725. [DOI] [PMC free article] [PubMed] [Google Scholar] 118.Jaako P., Debnath S., Olsson K., Zhang Y., Flygare J., Lindström M.S., Bryder D., Karlsson S. Disruption of the 5S RNP–Mdm2 interaction significantly improves the erythroid defect in a mouse model for Diamond-Blackfan anemia. Leukemia. 2015;29:2221–2229. doi: 10.1038/leu.2015.128. [DOI] [PMC free article] [PubMed] [Google Scholar] 119.Tiu G.C., Kerr C.H., Forester C.M., Krishnarao P.S., Rosenblatt H.D., Raj N., Lantz T.C., Zhulyn O., Bowen M.E., Shokat L., et al. A p53-dependent translational program directs tissue-selective phenotypes in a model of ribosomopathies. Dev. Cell. 2021;56:2089–2102.e11. doi: 10.1016/j.devcel.2021.06.013. [DOI] [PMC free article] [PubMed] [Google Scholar] 120.Oliver E.R., Saunders T.L., Tarlé S.A., Glaser T. Ribosomal protein L24 defect in Belly spot and tail (Bst), a mouse Minute. Development. 2004;131:3907–3920. doi: 10.1242/dev.01268. [DOI] [PMC free article] [PubMed] [Google Scholar] 121.Cmejlova J., Dolezalova L., Pospisilova D., Petrtylova K., Petrak J., Cmejla R. Translational efficiency in patients with Diamond-Blackfan anemia. Haematologica. 2006;91:1456–1464. [PubMed] [Google Scholar] 122.Signer R.A., Magee J.A., Salic A., Morrison S.J. Haematopoietic stem cells require a highly regulated protein synthesis rate. Nature. 2014;509:49–54. doi: 10.1038/nature13035. [DOI] [PMC free article] [PubMed] [Google Scholar] 123.Ma H., Wang X., Cai J., Dai Q., Natchiar S.K., Lv R., Chen K., Lu Z., Chen H., Shi Y.G., et al. N6-Methyladenosine methyltransferase ZCCHC4 mediates ribosomal RNA methylation. Nat. Chem. Biol. 2019;15:88–94. doi: 10.1038/s41589-018-0184-3. [DOI] [PMC free article] [PubMed] [Google Scholar] 124.Lodish H.F. Model for the regulation of mRNA translation applied to haemoglobin synthesis. Nature. 1974;251:385–388. doi: 10.1038/251385a0. [DOI] [PubMed] [Google Scholar] 125.Mills E.W., Green R. Ribosomopathies: There’s strength in numbers. Science. 2017;358:eaan2755. doi: 10.1126/science.aan2755. [DOI] [PubMed] [Google Scholar] 126.Khajuria R.K., Munschauer M., Ulirsch J.C., Fiorini C., Ludwig L.S., McFarland S.K., Abdulhay N.J., Specht H., Keshishian H., Mani D., et al. Ribosome Levels Selectively Regulate Translation and Lineage Commitment in Human Hematopoiesis. Cell. 2018;173:90–103.e19. doi: 10.1016/j.cell.2018.02.036. [DOI] [PMC free article] [PubMed] [Google Scholar] 127.Da Costa L.M., Leblanc T.M., Mohandas N. Diamond-Blackfan anemia. Blood. 2020;136:1262–1273. doi: 10.1182/blood.2019000947. [DOI] [PMC free article] [PubMed] [Google Scholar] 128.Engidaye G., Melku M., Enawgaw B. Diamond Blackfan Anemia: Genetics, Pathogenesis, Diagnosis and Treatment. EJIFCC. 2019;30:67–81. [PMC free article] [PubMed] [Google Scholar] 129.Gripp K.W., Curry C., Olney A.H., Sandoval C., Fisher J., Chong J.X.-L., Pilchman L., Sahraoui R., Stabley D.L., Sol-Church K., et al. Diamond-Blackfan anemia with mandibulofacial dystostosis is heterogeneous, including the novel DBA genes TSR2 and RPS28. Am. J. Med. Genet. Part A. 2014;164:2240–2249. doi: 10.1002/ajmg.a.36633. [DOI] [PMC free article] [PubMed] [Google Scholar] 130.Mirabello L., Khincha P.P., Ellis S.R., Giri N., Brodie S., Chandrasekharappa S.C., Donovan F.X., Zhou W., Hicks B.D., Boland J.F., et al. Novel and known ribosomal causes of Diamond-Blackfan anaemia identified through comprehensive genomic characterisation. J. Med. Genet. 2017;54:417–425. doi: 10.1136/jmedgenet-2016-104346. [DOI] [PMC free article] [PubMed] [Google Scholar] 131.Flygare J., Aspesi A., Bailey J.C., Miyake K., Caffrey J.M., Karlsson S., Ellis S.R. Human RPS19, the gene mutated in Diamond-Blackfan anemia, encodes a ribosomal protein required for the maturation of 40S ribosomal subunits. Blood. 2006;109:980–986. doi: 10.1182/blood-2006-07-038232. [DOI] [PMC free article] [PubMed] [Google Scholar] 132.Ulirsch J.C., Verboon J.M., Kazerounian S., Guo M.H., Yuan D., Ludwig L.S., Handsaker R.E., Abdulhay N.J., Fiorini C., Genovese G., et al. The Genetic Landscape of Diamond-Blackfan Anemia. Am. J. Hum. Genet. 2019;104:356. doi: 10.1016/j.ajhg.2018.12.011. [DOI] [PMC free article] [PubMed] [Google Scholar] 133.Graifer D., Karpova G. Eukaryotic protein uS19: A component of the decoding site of ribosomes and a player in human diseases. Biochem. J. 2021;478:997–1008. doi: 10.1042/BCJ20200950. [DOI] [PubMed] [Google Scholar] 134.An K., Zhou J.-B., Xiong Y., Han W., Wang T., Ye Z.-Q., Wu Y.-D. Computational Studies of the Structural Basis of Human RPS19 Mutations Associated with Diamond-Blackfan Anemia. Front. Genet. 2021;12:650897. doi: 10.3389/fgene.2021.650897. [DOI] [PMC free article] [PubMed] [Google Scholar] 135.Gregory L.A., Aguissa-Touré A.-H., Pinaud N., Legrand P., Gleizes P.-E., Fribourg S. Molecular basis of Diamond Blackfan anemia: Structure and function analysis of RPS19. Nucleic Acids Res. 2007;35:5913–5921. doi: 10.1093/nar/gkm626. [DOI] [PMC free article] [PubMed] [Google Scholar] 136.Juli G., Gismondi A., Monteleone V., Caldarola S., Iadevaia V., Aspesi A., Dianzani I., Proud C.G., Loreni F. Depletion of ribosomal protein S19 causes a reduction of rRNA synthesis. Sci. Rep. 2016;6:35026. doi: 10.1038/srep35026. [DOI] [PMC free article] [PubMed] [Google Scholar] 137.Horos R., Ijspeert H., Pospisilova D., Sendtner R., Andrieu-Soler C., Taskesen E., Nieradka A., Cmejla R., Sendtner M., Touw I.P., et al. Ribosomal deficiencies in Diamond-Blackfan anemia impair translation of transcripts essential for differentiation of murine and human erythroblasts. Blood. 2012;119:262–272. doi: 10.1182/blood-2011-06-358200. [DOI] [PubMed] [Google Scholar] 138.Ludwig L.S., Gazda H.T., Eng J.C., Eichhorn S.W., Thiru P., Ghazvinian R., George T.I., Gotlib J.R., Beggs A.H., Sieff C.A., et al. Altered translation of GATA1 in Diamond-Blackfan anemia. Nat. Med. 2014;20:748–753. doi: 10.1038/nm.3557. [DOI] [PMC free article] [PubMed] [Google Scholar] 139.Papagiannopoulos C.I., Kyritsis K.A., Psatha K., Mavridou D., Chatzopoulou F., Orfanoudaki G., Aivaliotis M., Vizirianakis I.S. Invariable Ribosome Stoichiometry during Murine Erythroid Differentiation: Implications for Understanding Ribosomopathies. Front. Mol. Biosci. 2022;9:805541. doi: 10.3389/fmolb.2022.805541. [DOI] [PMC free article] [PubMed] [Google Scholar] 140.Han X., Lu S., Gu C., Bian Z., Xie X., Qiao X. Clinical features, epidemiology, and treatment of Shwachman-Diamond syndrome: A systematic review. BMC Pediatr. 2023;23:503. doi: 10.1186/s12887-023-04324-3. [DOI] [PMC free article] [PubMed] [Google Scholar] 141.Boocock G.R., Morrison J.A., Popovic M., Richards N., Ellis L., Durie P.R., Rommens J.M. Mutations in SBDS are associated with Shwachman–Diamond syndrome. Nat. Genet. 2003;33:97–101. doi: 10.1038/ng1062. [DOI] [PubMed] [Google Scholar] 142.Cipolli M. Shwachman-Diamond Syndrome: Clinical Phenotypes. Pancreatol. Off. J. Int. Assoc. Pancreatol. 2001;1:543–548. doi: 10.1159/000055858. [DOI] [PubMed] [Google Scholar] 143.Bezzerri V., Cipolli M. Shwachman-Diamond Syndrome: Molecular Mechanisms and Current Perspectives. Mol. Diagn. Ther. 2019;23:281–290. doi: 10.1007/s40291-018-0368-2. [DOI] [PubMed] [Google Scholar] 144.Menne T.F., Goyenechea B., Sánchez-Puig N., Wong C.C., Tonkin L.M., Ancliff P.J., Brost R.L., Costanzo M., Boone C., Warren A.J. The Shwachman-Bodian-Diamond syndrome protein mediates translational activation of ribosomes in yeast. Nat. Genet. 2007;39:486–495. doi: 10.1038/ng1994. [DOI] [PubMed] [Google Scholar] 145.Lo K.-Y., Li Z., Bussiere C., Bresson S., Marcotte E.M., Johnson A.W. Defining the Pathway of Cytoplasmic Maturation of the 60S Ribosomal Subunit. Mol. Cell. 2010;39:196–208. doi: 10.1016/j.molcel.2010.06.018. [DOI] [PMC free article] [PubMed] [Google Scholar] 146.Moore J.B., Farrar J.E., Arceci R.J., Liu J.M., Ellis S.R. Distinct ribosome maturation defects in yeast models of Diamond-Blackfan anemia and Shwachman-Diamond syndrome. Haematologica. 2010;95:57–64. doi: 10.3324/haematol.2009.012450. [DOI] [PMC free article] [PubMed] [Google Scholar] 147.Burwick N., Coats S.A., Nakamura T., Shimamura A. Impaired ribosomal subunit association in Shwachman-Diamond syndrome. Blood. 2012;120:5143–5152. doi: 10.1182/blood-2012-04-420166. [DOI] [PMC free article] [PubMed] [Google Scholar] 148.Gartmann M., Blau M., Armache J.-P., Mielke T., Topf M., Beckmann R. Mechanism of eIF6-mediated Inhibition of Ribosomal Subunit Joining. J. Biol. Chem. 2010;285:14848–14851. doi: 10.1074/jbc.C109.096057. [DOI] [PMC free article] [PubMed] [Google Scholar] 149.Klinge S., Voigts-Hoffmann F., Leibundgut M., Arpagaus S., Ban N. Crystal Structure of the Eukaryotic 60 S Ribosomal Subunit in Complex with Initiation Factor 6. Science. 2011;334:941–948. doi: 10.1126/science.1211204. [DOI] [PubMed] [Google Scholar] 150.Weis F., Giudice E., Churcher M., Jin L., Hilcenko C., Wong C.C., Traynor D., Kay R.R., Warren A.J. Mechanism of eIF6 release from the nascent 60S ribosomal subunit. Nat. Struct. Mol. Biol. 2015;22:914–919. doi: 10.1038/nsmb.3112. [DOI] [PMC free article] [PubMed] [Google Scholar] 151.Jaako P., Wong C.C., Adams D., Warren A.J. Attenuated Protein Synthesis Drives the Hematopoietic Defects in Shwachman-Diamond Syndrome. Blood. 2017;130:876. doi: 10.1182/blood.V130.Suppl_1.876.876. [DOI] [Google Scholar] 152.In K., Zaini M.A., Müller C., Warren A.J., von Lindern M., Calkhoven C.F. Shwachman–Bodian–Diamond syndrome (SBDS) protein deficiency impairs translation re-initiation from C/EBPα and C/EBPβmRNAs. Nucleic Acids Res. 2016;44:4134–4146. doi: 10.1093/nar/gkw005. [DOI] [PMC free article] [PubMed] [Google Scholar] 153.Jaako P., Faille A., Tan S., Wong C.C., Escudero-Urquijo N., Castro-Hartmann P., Wright P., Hilcenko C., Adams D.J., Warren A.J. eIF6 rebinding dynamically couples ribosome maturation and translation. Nat. Commun. 2022;13:1562. doi: 10.1038/s41467-022-29214-7. [DOI] [PMC free article] [PubMed] [Google Scholar] 154.Ruland J.A., Krüger A.M., Dörner K., Bhatia R., Wirths S., Poetes D., Kutay U., Siebrasse J.P., Kubitscheck U. Nuclear export of the pre-60S ribosomal subunit through single nuclear pores observed in real time. Nat. Commun. 2021;12:6211. doi: 10.1038/s41467-021-26323-7. [DOI] [PMC free article] [PubMed] [Google Scholar] 155.Xue L., Lenz S., Zimmermann-Kogadeeva M., Tegunov D., Cramer P., Bork P., Rappsilber J., Mahamid J. Visualizing translation dynamics at atomic detail inside a bacterial cell. Nature. 2022;610:205–211. doi: 10.1038/s41586-022-05255-2. [DOI] [PMC free article] [PubMed] [Google Scholar] 156.Erdmann P.S., Hou Z., Klumpe S., Khavnekar S., Beck F., Wilfling F., Plitzko J.M., Baumeister W. In situ cryo-electron tomography reveals gradient organization of ribosome biogenesis in intact nucleoli. Nat. Commun. 2021;12:5364. doi: 10.1038/s41467-021-25413-w. [DOI] [PMC free article] [PubMed] [Google Scholar] 157.Cheng J., Novati G., Pan J., Bycroft C., Žemgulytė A., Applebaum T., Pritzel A., Wong L.H., Zielinski M., Sargeant T., et al. Accurate proteome-wide missense variant effect prediction with AlphaMissense. Science. 2023;381:eadg7492. doi: 10.1126/science.adg7492. [DOI] [PubMed] [Google Scholar] Articles from Biomedicines are provided here courtesy of Multidisciplinary Digital Publishing Institute (MDPI) ACTIONS View on publisher site PDF (1.5 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases 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
4536	https://math.stackexchange.com/questions/3421845/find-all-angles-between-0-and-180-degrees-such-that-tan-theta-16	algebra precalculus - Find all angles between $0$ and $180$ degree's such that $tan(\theta)=-16$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find all angles between 0 and 180 degree's such that tan(\theta)=-16 Ask Question Asked 5 years, 11 months ago Modified5 years, 11 months ago Viewed 2k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. \begingroup Find all angles between 0 and 180 degrees such that \tan(\theta)=-16. I've been working on this problem for days and can't figure it out!! Need a fresh perspective, thanks in advance! algebra-precalculus Share Cite Follow Follow this question to receive notifications edited Nov 5, 2019 at 5:10 Allawonder 13.6k 1 1 gold badge 22 22 silver badges 28 28 bronze badges asked Nov 4, 2019 at 17:54 user637978 user637978 \endgroup 0 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. \begingroup Such an angle must fall in the second quadrant -- in other words, it must be obtuse. Then just take the arctangent to get 180°+\arctan(-16)=180°-\arctan(16) Share Cite Follow Follow this answer to receive notifications edited Nov 5, 2019 at 5:10 answered Nov 4, 2019 at 17:56 AllawonderAllawonder 13.6k 1 1 gold badge 22 22 silver badges 28 28 bronze badges \endgroup 3 \begingroup I think this is the most direct answer. The function \arctan(x) is defined to give results in the first quadrant for positive x and the fourth quadrant for negative x. But the tangent of an angle is x when the angle is \arctan(x) plus any integer multiple of 180 degrees. Moreover, in degrees the fourth-quadrant result is in the range from -90 to 0. so to get it into the range 0 to 180, it is necessary and sufficient to add 180.\endgroup David K –David K 2019-11-04 18:13:36 +00:00 Commented Nov 4, 2019 at 18:13 \begingroup are there typo's in this?\endgroup user637978 –user637978 2019-11-04 19:18:01 +00:00 Commented Nov 4, 2019 at 19:18 \begingroup@MathematicalMushroom No, I've checked again. What looks like a typo?\endgroup Allawonder –Allawonder 2019-11-05 05:09:51 +00:00 Commented Nov 5, 2019 at 5:09 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. \begingroup The reference angle is \arctan 16 \approx 86.42^{\circ}. The tangent is positive in the first quadrant but negative in the second, so the only solution is 180^{\circ} - 86.42^{\circ}=93.58^{\circ}. Share Cite Follow Follow this answer to receive notifications answered Nov 4, 2019 at 18:01 DeepakDeepak 27.5k 1 1 gold badge 30 30 silver badges 54 54 bronze badges \endgroup Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. \begingroup For a fresh perspective think of the tangent as the gradient of the half line rotating about the origin. I think there should be an angle between 90 and 180. This is because the gradient of the half line is negative there. Likewise in quadrant 4 too. Note your calc will not give this directly only the principle value. Share Cite Follow Follow this answer to receive notifications edited Nov 4, 2019 at 18:04 answered Nov 4, 2019 at 17:58 KarlKarl 4,783 4 4 gold badges 20 20 silver badges 24 24 bronze badges \endgroup Add a comment\| You must log in to answer this question. 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 Related 8Find \sin \theta and \cos \theta given \tan 2\theta 0tan \theta = -12/5 and 90^\circ < \theta < 180 ^\circ, then find the exact values of \cos \theta/2 and \tan \theta/2 6Finding all the values of \theta for which \tan(\theta)=\sqrt3; problem with understanding. 3Find the exact value of \sin (\theta) and \cos (\theta) when \tan (\theta)=\frac{12}{5} 0How to find \tan{\theta} when \theta=\arctan⁡{(8/3)} 3Find \cos{\theta} and \sin{\theta} such that the rotation through the acute angle \theta results in a new equation with no x'y' term 2Mercator projection - Use existing equation to solve for degrees 1How do I find all angles \theta between 0° and 180° such that \cos(\theta) = \frac {6} {7}? 4Find all angles \theta such that x_1\cos(0)+x_2\cos(\theta)+x_3\cos(2\theta)+\dots + x_{n+1}\cos (n\theta)=0 Hot Network Questions A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man Alternatives to Test-Driven Grading in an LLM world Do we need the author's permission for reference Storing a session token in localstorage ICC in Hague not prosecuting an individual brought before them in a questionable manner? Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation What happens if you miss cruise ship deadline at private island? Passengers on a flight vote on the destination, "It's democracy!" How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? In the U.S., can patients receive treatment at a hospital without being logged? For every second-order formula, is there a first-order formula equivalent to it by reification? Vampires defend Earth from Aliens In Dwarf Fortress, why can't I farm any crops? How to convert this extremely large group in GAP into a permutation group. Why do universities push for high impact journal publications? How to start explorer with C: drive selected and shown in folder list? Can a cleric gain the intended benefit from the Extra Spell feat? On being a Maître de conférence (France): Importance of Postdoc Program that allocates time to tasks based on priority How can the problem of a warlock with two spell slots be solved? Lingering odor presumably from bad chicken "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Weird utility function more hot questions Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
4537	https://math.stackexchange.com/questions/288546/combinatorial-interpretation-of-the-identity-n-choose-k-n-choose-n-k	combinatorics - Combinatorial interpretation of the identity: ${n \choose k} = {n \choose n-k}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Combinatorial interpretation of the identity: (n k)=(n n−k)(n k)=(n n−k) Ask Question Asked 12 years, 8 months ago Modified6 years, 2 months ago Viewed 7k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. What is the combinatorial interpretation of the identity: (n k)=(n n−k)(n k)=(n n−k)? Proving this algebraically is trivial, but what exactly is the "symmetry" here. Could someone give me some sort of example to help my understanding? EDIT: Can someone present a combinatorial proof? combinatorics binomial-coefficients symmetry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 28, 2013 at 2:50 CodeKingPlusPlusCodeKingPlusPlus asked Jan 28, 2013 at 2:17 CodeKingPlusPlusCodeKingPlusPlus 7,558 17 17 gold badges 72 72 silver badges 106 106 bronze badges 2 Excuse lack of my background, but I am curious about your second question. What do you mean by 'combinatorial proof'? Are you considering counting two different sets by a bijection? What do you have in mind?user123454321 –user123454321 2013-01-28 02:53:53 +00:00 Commented Jan 28, 2013 at 2:53 That seems sufficient and like the correct approach. I especially just do not want to see the algebraic proof.CodeKingPlusPlus –CodeKingPlusPlus 2013-01-28 02:57:23 +00:00 Commented Jan 28, 2013 at 2:57 Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 9 Save this answer. Show activity on this post. Choosing k k objects among n n objects is same as leaving n−k n−k objects among n n objects. (Notice that there is no essential difference between the words "choose" and "leave".) Digression: I also consider this as one of the reasons why 0!=1 0!=1 is a good definition. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 28, 2013 at 2:18 user123454321user123454321 2,548 2 2 gold badges 21 21 silver badges 32 32 bronze badges 1 I like the explanation with the words "choose" and "leave".CodeKingPlusPlus –CodeKingPlusPlus 2013-01-28 02:50:11 +00:00 Commented Jan 28, 2013 at 2:50 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. (n k)(n k) denotes the number of ways of picking k k objects out of n n objects, and specifying the k k objects that are picked is equivalent to specifying the n−k n−k objects that are not picked. To put it differently, suppose you have n n objects, and you want to partition them into two sets: a set A A of size k k, and a set B B of size n−k n−k. If you pick which objects go into set A A, the number of ways of doing so is denoted (n k)(n k), and if you (equivalently!) pick which objects go into set B B, the number of ways is denoted (n n−k)(n n−k). The point here is that the binomial coefficient (n k)(n k) denotes the number of ways partitioning n n objects into two sets one of size k k and one of size n−k n−k, and is thus a special case of the multinomial coefficient (n k 1,k 2,…k m)where k 1+k 2+…k m=n(n k 1,k 2,…k m)where k 1+k 2+…k m=n which denotes the number of ways of partitioning n n objects into m m sets, one of size k 1 k 1, one of size k 2 k 2, etc. Thus (n k)(n k) can also be written as (n k,n−k)(n k,n−k), and when written in this style, the symmetry is apparent in the notation itself: (n k)=(n k,n−k)=(n n−k,k)=(n n−k)(n k)=(n k,n−k)=(n n−k,k)=(n n−k) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 28, 2013 at 4:50 answered Jan 28, 2013 at 2:42 ShreevatsaRShreevatsaR 42.4k 8 8 gold badges 101 101 silver badges 138 138 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let's say you have 3 apples and you want to choose one of them. Then you are left with an apple, which is the same, in a sense, as choosing that apple. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 28, 2013 at 2:19 JebruhoJebruho 1,708 9 9 silver badges 23 23 bronze badges 2 3 I cannot make any sense of this sentence.CodeKingPlusPlus –CodeKingPlusPlus 2013-01-28 02:49:13 +00:00 Commented Jan 28, 2013 at 2:49 I think he meant "...choose two of them."Pedro –Pedro♦ 2013-01-28 06:13:32 +00:00 Commented Jan 28, 2013 at 6:13 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let S S be a set of n n elements. Choosing m m members of S S leaves a remainder of n−m n−m members in S S. Thus, each m m-subset of S S corresponds to exactly one (n−m)(n−m)-subset of S S, and conversely, to any such (n−m)(n−m)-subset, there corresponds exactly one m m-subset of S S. Therefore, (n m)=(n n−m)(n m)=(n n−m). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 3, 2019 at 15:14 answered Dec 26, 2018 at 16:02 Little RookieLittle Rookie 1,236 1 1 gold badge 18 18 silver badges 40 40 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 combinatorics binomial-coefficients symmetry 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 0Why is (n k)=(n n−k)(n k)=(n n−k)? Related 8Combinatorial argument for the identity k(n k)=n(n−1 k−1)k(n k)=n(n−1 k−1) 6Combinatorial reasoning for linear binomial identity 3Combinatorial proof of n!=(n k)k!(n−k)!n!=(n k)k!(n−k)! 4A combinatorial identity: ∑i+j=k(−1)i(n i)(n+j−1 n−1)=0∑i+j=k(−1)i(n i)(n+j−1 n−1)=0 15Interpretation of a combinatorial identity involving iterated binomial coefficients 2Proving identities using combinatorial interpretation of binomial coefficients 4Prove the identity ∑n k=0(−1)k(n k)=0∑k=0 n(−1)k(n k)=0 using a combinatorial interpretation of the positive and negative terms 2Combinatorial Intuition Behind Binomial Identity Hot Network Questions Proof of every Highly Abundant Number greater than 3 is Even Exchange a file in a zip file quickly Another way to draw RegionDifference of a cylinder and Cuboid Who is the target audience of Netanyahu's speech at the United Nations? How different is Roman Latin? Is existence always locational? Overfilled my oil Are there any legal strategies to modify the nature and rules of the UN Security Council in the face of one, or more SC vetos? Should I let a player go because of their inability to handle setbacks? Does the Mishna or Gemara ever explicitly mention the second day of Shavuot? With line sustain pedal markings, do I release the pedal at the beginning or end of the last note? Childhood book with a girl obsessed with homonyms who adopts a stray dog but gives it back to its owners How do you create a no-attack area? What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left? In the U.S., can patients receive treatment at a hospital without being logged? ICC in Hague not prosecuting an individual brought before them in a questionable manner? On being a Maître de conférence (France): Importance of Postdoc Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Repetition is the mother of learning How to start explorer with C: drive selected and shown in folder list? Xubuntu 24.04 - Libreoffice I have a lot of PTO to take, which will make the deadline impossible Interpret G-code Does the curvature engine's wake really last forever? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4538	https://www.laserskintreatment.co.uk/assets/files/005.Laser%20hair%20removal.L%20Marza%20JAN%20Nov%202014.pdf	CLINICAL ▼ 386 Journal of AESTHETIC NURSING ► November 2014 ► Volume 3 Issue 9 © 2014 MA Healthcare Ltd E xcessive facial hair (hirsutism) is a common prob lem for many women and often has a negative im pact on their quality of life (Housman et al, 2004). It is defined as an androgen-dependent, male pat tern of hair distribution in women and is typically meas ured using a scoring system such as the Ferriman-Galwey (Newson, 2010). Excessive facial hair can be treated using a variety of methods, however the most efficient, long-term treatment for this indication is laser hair removal. Lasers have been used since the mid-1990s for the pur pose of hair depilation, where the treatment was initially only available in a few clinics and procedures were very expensive. The first lasers used for hair removal were the ruby (694 nm) and alexandrite (755 nm) lasers, but with the introduction in the later years of light-based ma chines using broad-spectrum intense pulsed light (IPL) (590–1200 nm), along with changes in laser regulations, such treatments expanded into aesthetic salons and the price of treatments became affordable. Nowadays, the majority of skin types can be treated safely and light-based hair removal is one of the fastest growing, non-surgical treatments in the US and Europe (Haedersdal et al, 2011). However, despite these hair removal treatments being safe and effective, the inappropriate use of lasers have been carried out by both physicians and non-physicians (Moseley, 2014) and recommendations have been made that the key essentials should be training and competence and that laser practitioners should belong to a registered profession (Department of Health, 2013). Proper patient selection and preoperative preparation, an understanding of the principles of laser safety, experience with treating facial hair with a variety of lasers/light devices, and a thor ough understanding of laser-tissue interactions, are also vital to ensure safe and efficient results while minimising risks (Ibrahimi et al, 2011). Health professionals having acquired the above skills are therefore in a good position to carry out treatments for hair removal. This paper will focus on facial laser hair removal procedures carried out by laser nurse practitioners. Causes of excessive facial hair Excessive facial hair could be a result of hormone imbal ance, may be idiopathic (no explained cause), or can be a family trait (particularly in families with Mediterra nean, south Asian or middle Eastern ancestry) (Azziz et al, 2004). The majority of endocrine causes for hirsutism (75%) are linked to polycystic ovary syndrome (PCOS), which affects 4–8% of the female population and is the most common endocrine abnormality in women of re productive age (Azziz et al, 2004). However around one half of hirsute women have idiopathic hirsutism, which develops without the presence of excess androgen or any underlying medical condition(s) (British Association of Dermatologists (BAD), 2006). The incidence of hirsutism is estimated at 5–15% of the reproductive female popula tion (Guidelines Working Party, 2005). Medical treatments In its 2008 guidelines, The Endocrine Society recom mends that, following tests, premenstrual women suf fering from moderate-to-severe excessive facial hair should take oral contraceptives (Martin et al, 2008). If the response to this medication is suboptimal, this should then be followed by antiandrogen drugs after 6 months (Martin et al, 2008). For menstrual dysfunc tion and insulin resistance, oral contraceptive pills or Laser facial hair removal protocol and key consultation considerations LILIANA MARZA Clinical Laser Nurse Practitioner, The London Clinic, London. e: l.marza@thelondonclinic.co.uk Abstract Excessive facial hair in women can cause significant distress and may contribute to a reduced quality of life for those who are affected. Although there are a multitude of mechanical options used for hair depilation, over the years lasers have proved to be the most effective long-term option. However, the British Association of Dermatologists (BAD) recommends that facial hair removal should be carried out in specialised clinics by trained practitioners, as favourable treatment outcomes and the reduction of side effects is dependent on understanding how lasers work, choosing the appropriate parameters and using a suitable machine for each individual's needs. This article presents a facial hair removal treatment protocol from a clinical laser nurse perspective. Key words ► Laser ► Hair removal ► Hirsutism ► Polycystic ovary syndrome ► Quality of life Volume 3 Issue 9 ► November 2014 ► Journal of AESTHETIC NURSING 387 ▼ CLINICAL © 2014 MA Healthcare Ltd Figure 1: A 38-year-old hirsute woman with white and black hairs on her chin presenting with scarring and irritation caused by electrolysis and plucking insulin-sensitising drugs (e.g. metformin) can also be prescribed (Johnson, 2014). Lifestyle changes Hirsutism varies with both ethnicity and degree of obes ity (Azziz et al, 2004). Obesity has been found to increase androgenism and hirsutism independently and by ex acerbating PCOS (Teede et al, 2010). Lifestyle change is therefore the first line of treatment in the management of PCOS in women who are overweight. As little as 5–10% weight loss and daily exercise has been shown to have significant clinical benefits, improving psychologi cal outcomes and associated symptoms (Galletly et al, 1996). For this reason, prevention of weight gain should be emphasised to all patients with PCOS, regardless of their body mass (Galletly et al, 1996). Topical treatments Applied twice daily, eflornithine cream (Vaniqa, Almirall Limited) is licensed to treat hirsutism, specifically pre senting on the face (Guidelines Working Party, 2005). It works by inhibiting the rate-controlling step in the pro duction of putrescine in active hair follicle cells, slowing the rate of hair growth and making hair less visible (Lapi doth et al, 2010). Its long-term use has been reported to be safe (Hickman et al, 2001); the most frequent adverse reaction is a mild acne-like eruption, but there is no worsening of pre-existing acne. Mechanical and cosmetic treatments There are a multitude of mechanical treatments, all of which vary in their effectiveness, discomfort and cost, in cluding shaving, tweezing, electrolysis, waxing, epilation and laser hair removal. Although shaving hair on the face may not be acceptable to some women (as it is perceived as a male activity), waxing may cause ingrown hairs, and epilation cream may cause irritation, the advantage of these methods is that they are cheap and easily accessible. In contrast, electrolysis offers a more permanent hair re moval option by inserting a fine needle into the hair folli cle and using electricity to destroy it (Casey and Goldeberg, 2008). Electrolysis is suitable for small, localised regions of hair growth and side effects are oedema, erythema and scarring. Electrolysis is a more permanent method of hair removal but courses can be costly and lengthy (Burns et al, 2010). Evidence supports the use of electrolysis for perma nent hair removal in localised areas, and lasers (particularly alexandrite and diode) for permanent hair reduction (Som ani and Turvy, 2014). Figure 1 shows scarring on a 38-year-old woman's chin as a result of electrolysis and plucking. Laser treatments Although laser treatments are more expensive, they cover a larger surface area in one treatment and have quickly become the treatment of choice for the removal of un wanted hair (Casey and Goldeberg, 2008); however the recommended treatment parameters remain vague. Laser hair removal aims to destroy the hair root permanently by focusing on hairs that are in the anagen stage, targeting the melanin in the hair shaft and the bulb of the hair (Patil and Dhami, 2008). The procedure is based on the ability of melanin in the cells of the hair follicle to convert the energy emitted by the laser into heat, thereby giving rise to the local destruction of hair follicle (Styczynski et al, 2008). Laser hair removal allows for hair-free periods and thins the hairs, however it does not provide permanent hair removal (particularly for facial hairs) for women with an underlying medical condition. Laser hair removal is also rarely available on the NHS and may be expensive as several treatments are needed every couple of months. This procedure is not effective on blonde or white hairs. The BAD (2006) recommends that facial hair removal using a laser should be carried out at a specialist clinic by an operator who is properly qualified and has access to medical clinicians. The British Medical Laser Association (2014) also advises that such clinics should have access to a laser protection advisor. Patients for whom excessive facial hair contributes to a compromised quality of life may ben efit from core nursing skills such as empathy and reassur ance. Alternative treatments should always be discussed if laser treatment is not appropriate for the patient in ques tion. This may be the case for those with finer, lighter hairs and for patients who cannot afford the procedure. Consultation considerations Before starting treatment, practitioners must carry out a thorough consultation with patients to assess their in dividual needs and ensure an appropriate laser is used. Taking medical and treatment history At consultation, the patient’s individual needs are as sessed, treatment options are discussed and any ques Liliana Marza 388 Journal of AESTHETIC NURSING ► November 2014 ► Volume 3 Issue 9 CLINICAL ▼ © 2014 MA Healthcare Ltd tions the patient has are answered. Sunscreen and/or sun avoidance is universally recommended before the procedure because melanin is the principle chromo phore for laser hair removal (Casey and Goldeberg, 2008). Moreover, no plucking, waxing or electrolysis should be performed before treatment. Laser goggles designed for the specific wavelength emitted by the light source must be worn at all times during treatment to avoid injury to the eyes. Table 1 lists some of the as pects to address at consultation. Assessment Assessment at the time of consultation should cover: ► ►Relevant medical history, including allergies and medications being taken ► ►Existing and previous methods of dealing with exces sive hair, along with their frequency and any side ef fects. The responses to this will create a basis against which the patient's progress can be monitored ► ►Baseline photos taken with the patient’s consent ► ►Number of treatments, outcomes and potential risks ► ►Managing patient expectations ► ►Obtaining a signed informed consent form. When discussing treatment outcomes, practitioners should take into account the patient's skin phototype and hair characteristics, as well as the areas to be treated and any underlying medical condition(s). Contact infor mation should also be given to the patient should he or she have more questions or concerns. The price of the treatment should be clearly communicated at this stage. Expected treatment outcomes A review of 11 randomised controlled trials involving 444 people concluded that certain lasers (alexandrite and diode) offered 50% improvement up to 6 months post laser treatment (Haedersdal and Gotzsche, 2006). However, there were limited results for IPL and other lasers. Haedersdal and Gotzsche (2006) recommended further research to determine long-term results. McGill et al (2007) studied 60 PCOS patients who were suffering from facial hirsutism, who had six laser treat ments. Following treatment, the number of hairs de creased by 30–40%, which is less than in patients who do not have PCOS; however, 95% reported being ‘very satis fied with the treatment’ and the effect was sufficient to change patients’ lifestyle and increase their self-esteem. Patients with darker skin can also benefit from in creased numbers of treatments owing to the need to use reduced fluence, a longer wavelength and increased skin cooling to protect the skin from potential side ef fects (Alai et al, 2009). It is therefore important to as sess each individual case and give patients realistic ex pectations; this will ensure a high satisfaction rate is associated with the treatment. However, carrying out a skilful prediction of treatment outcomes will reflect both the practitioner’s clinical experience and his or her understanding of how lasers work. Adverse events Erythema (redness) and perifollicular oedema (inflam mation around hair follicle) are expected reactions to la ser hair removal (Sanchez et al, 2002). They are caused by the heat absorbed by the melanin in the hair and in the skin and can be addressed by applying cold packs dur ing and immediately after treatment (Willey et al, 2007). Changes in skin pigmentation (e.g. hyperpigmentation and hypopigmentation), scabbing and blistering are un desirable outcomes which may develop into scars (Bre adon and Barnes, 2007). To reduce further stimulation of the melanocytes it is reasonable to conclude that sun avoidance would reduce the risk of post-inflammatory pigmentary changes. Some authors have recommended sun avoidance for up to 3 months post laser treatment (Drosener and Adatto, 2004) to minimise additional in flammation of the treated area; however, there are no formal recommendations regarding sun exposure before and after laser treatment (Casey and Goldeberg, 2008). One risk when treating hairs on the face is paradoxi cal hair growth (hypertrichosis). This can present in a number of patients and research has demonstrated an association with Mediterranean skin type. The incidence of hypertrichosis has been reported to be from 0.6% (Ala jlan et al, 2005) to 10.5% (Willey et al, 2007). Subthera peutic thermal injury may cause induction of the hair cycle, particularly on face and neck (Willey et al 2007). Suboptimal energy seems to stimulate the cells responsi ble for normal follicular cycle, and induce longer, thicker hairs as a consequence of accelerating the transition from vellus to terminal follicular hair (Willey et al, 2007). A more aggressive treatment approach is recommended using higher energy in combination with cold packs to reduce the risk of adverse reactions (Willey et al, 2007). Table 1: Elements of consultation: communication, records, referrals Element Considerations Communication Verbal and written information Advice Reassurance Price Records Medical history, allergies, medication history Skin phototype assessment Previous treatments and methods employed to remove the hair Quality of life/limitations to social life Consent form explained and signed Baseline photos Referral To specialists (e.g. endocrinologist, dermatologist, dietitian) Volume 3 Issue 9 ► November 2014 ► Journal of AESTHETIC NURSING 389 ▼ CLINICAL © 2014 MA Healthcare Ltd Lasers and skin phototype Lasers should be used according to the patient's skin phototype, which can be assessed using the Fitzpatrick (1975) scale. With the advent of longer wavelength, long er pulse durations and efficient cooling devices, lasers can be used safely for the majority of skin phototypes. All devices may be used in patients with light skin (pho totypes I–III), however the alexandrite laser (755 nm) is the most appropriate for lighter skin and finer hairs (Sanchez et al, 2002). The two wavelengths for laser hair reduction on darker skin types are the diode (810 nm) and neodymium: yttrium-aluminium-garnet Nd:YAG (1064 nm) (Sanchez et al, 2002). Ruby lasers are rarely used these days due to higher risk of side effects. Each laser has a cooling system which can be by con tact, cold spray or cool air and must be tested before treatment as per the manufacturer’s advice. The cooling system protects the epidermis and allows the heat to be absorbed by the target chromophore, while protecting the surrounding areas. Using active cooling during laser hair removal limits heat accumulation in the epidermis, thus minimising collateral damage and reducing dis comfort to the patient. Moreover, the use of a cooling system permits delivery of higher fluences needed to an effective epilation. Cooling methods commonly used in clude gel cooling, cryogen cooling, contact cooling and forced air cooling (Motta, 2014). Laser treatment protocol Lasers and non-coherent light sources have been intro duced to cause damage to hair follicles on the basis of the principles of selective photothermolysis (Anderson and Parrish, 1983). For selective thermal damage of a pigment ed target structure to be achieved, sufficient fluence at a wavelength preferentially absorbed by the target must be delivered for a time that is equal to or less than the ther mal relaxation time of the target (Anderson and Parrish, 1983). Treatment outcomes are dependent on the patient’s characteristics, the laser used and the practitioner’s under standing the principles of selective photothermolysis. Although there is some discordance regarding the ap propriate number of treatments, most authors agree that, regardless of the laser used, multiple treatments are usually required to produce satisfactory results (Lepselter, 2004). In the guidelines set forth by the European Society for La ser Dermatology, Drosner and Adatto (2005) recommend ed 3–8 treatments. The ideal treatment interval remains a matter of debate; Bouzari et al (2005) suggest that a shorter treatment interval is more effective. Drosner and Adatto (2005) also recommend treatments every 4–8 weeks. Tak ing into account the above, Casey and Goldesberg (2008) recommended interval of 6–8 weeks, which may vary de pending on the region of the body that is being treated. Each treatment takes about 15 minutes and is well toler ated without any local anaesthesia. The area treated can be red in appearance for up to an hour, but make-up can be applied straight after the treatment. Cold compresses must be used to reduce the risk of side effects. Patch test It is good practice to do a patch test on the treatment area before starting the treatment and indemnity insurance will require this test to be carried out at least 24 hours before proceeding with treatment. The parameters of each laser are chosen to obtain efficient treatment with minimal side effects, taking into account skin type (using the Fitzpatrick scale) and hair colour, thickness and density, by manipulat ing three variables: wavelength, pulse duration and fluence (Patil and Dhami, 2008). Areas with thicker, denser and darker hairs will need to be treated with lower fluencies and longer wavelength, while finer, lighter and sparse hairs will need an increased fluence and shorter wavelength. Skin preparation Any make up and impurities should be removed before treatment to avoid absorption of heat by any particles that are not hair. Hairs will also need to be shaved or cut short before treatment (Casey and Goldeberg, 2008). Setting the laser parameters Spot size refers to the diameter of the laser beam on the skin (Patil and Dhami, 2008). For facial hairs, a diameter of 10–15 mm is most appropriate and the larger spot size of 18 mm should be used on larger areas such as the back and legs. Using a larger spot size reduces dermal scattering, leading to greater depth of penetration of the beam and a lower threshold fluence (Halachmi and Lapidoth, 2012). For sensitive areas such as upper lip, the spot size should be smaller than on the rest of the face to reduce discom fort, as larger spots cause more pain than a smaller spot size at an identical fluence (Eremia and Newman, 2000). The duration of each pulse is measured in milliseconds and is adjusted according to the thickness of hair. The root of an extra coarse hair will never be destroyed, however high the energy level is set to, if a shorter pulse is used (Pa til and Dhami, 2008). Therefore, the pulse duration should be primarily guided by the size of the chromophore and only secondarily by the competing chormophore. Whereas fine hairs (e.g. on the upper lip) will need a shorter pulse duration, coarser hairs (e.g. on the chin) require a longer pulse duration. By choosing an appropriate pulse duration, a smaller or a larger target having the same chromophore can be selected (Patil and Dhami, 2008). As a general rule, the larger the chromophore, the longer the thermal relaxa tion time, as large objects take a longer time to cool. Fluence is the amount of energy applied per treatment area and is measured in joules/cm2 (Patil and Dhami, 2008). The fluence of a laser should be tested at lower settings and increased until the desired outcome is ob served as advised by the manufacturer. For areas of high density and with coarse hairs (e.g. chin), the fluence should be lower then for areas where hair is sparse. 390 Journal of AESTHETIC NURSING ► November 2014 ► Volume 3 Issue 9 CLINICAL ▼ © 2014 MA Healthcare Ltd Repetition rate is the number of pulses per second and measures the speed of the treatment in hertz (Hz) (Patil and Dhami, 2008). The repetition rate will be set in line with the patient's skin type, density of hair and the skill of the practitioner. Combining laser and topical treatments For patients who are resistant to laser treatment or have reached the stage where hairs are fine or some hairs are white, combining laser hair removal with a topical treat ment will be most beneficial (Burns et al, 2010). Figure 2 shows a hirsute woman that would benefit from a com bination of treatments. Eflornithine cream applied twice-daily is licensed for use of hirsutism, specifically on the face (Guidelines Working Party, 2005); however, this will need to be prescribed by a medical practitioner. Although eflornithine cream can be recommended in combination with laser therapy, it should be stopped 3 days before and 3 days after laser treatment (Hamzavi and Owen, 2006). One of the key benefits of using topical eflornithine is that it can be used on vellus hairs, lighter and thinner hairs, and salt and pepper hairs, for which a laser is less effective (Lapidoth et al, 2010). This can increase the time between laser treatments, however it will take a couple of months of daily use for patients to see results and its effects only last while the cream is being applied. Conclusion Excessive facial hair is a common problem for many women and can have a significant impact on their quality of life. Laser hair removal is the most efficient method of long-term hair removal available, however outcomes will be dependent on the skill of the practitioner in choos ing the correct parameters and the appropriate machine. Multiple treatments will be needed and combining laser hair removal with topical treatment can render best re sults in some patients. It is important to note that los ing weight can also reduce excessive facial hair in obese women with PCOS. Giving patients realistic expecta tions based on individual characteristics will increase satisfaction with the treatment and its outcomes. References Alai NN, Saemi AM, Ang JM (2009) Laser-assisted hair removal. com/nj757l8 (Accessed 27 October 2014) Alajlan A, Shapiro J, Rivers JK, MacDonald N, Wiggin J, Lui H (2005) Paradoxical hypertrichosis after laser epilation. J Am Acad Dermatol 53(1): 85–8 Anderson RR, Parrish JA (1983) Selective photothermolysis: precise microsurgery by selective absorption of pulsed radiation. Science 220(4596): 524–7 Azziz R, Woods KS, Reyna R, Key TJ, Knochenhauer ES, Yildiz BO (2004) The prevalence and features of the polycystic ovary syndrome in an unselected population. J Clin Endocrinol Metab 89(6): 2745–9 Bouzari N, Nouri K, Tabatabai H, Abbasi Z, Firooz A, Dowlati Y (2005) The role of number of treatments in laser-assisted hair removal using a 755 nm alexandrite laser. J Drugs Dermatol 4(5): 573–8 Breadon JY, Barnes CA (2007) Comparison of adverse events of laser and light-assisted hair removal systems in skin types IV-VI. J Drugs Dermatol 6(1): 40–6 British Association of Dermatologists (2006) Patient Information leaflet. Hirsutism. (Accessed 16 October 2014) British Medical Laser Association (2014) Laser/IPL safety awareness. http:// tinyurl.com/n94dmzy (Accessed 16 October 2014) Bouzari N, Tabatabai H, Abbasi Z, Firooz A, Dowlati Y (2005) Hair removal using an 800-nm diode laser: Comparison at different treatment intervals of 45, 60, and 90 days. Int J Dermatol 44(1): 50–3 Burns T, Breathnach S, Cox N, eds. (2010) Rook’s Textbook of Dermatology. 8th edn. Wiley-Blackwell, Oxford Casey A, Goldeberg D (2008) Guidelines for laser hair removal. J Cosmet Laser Ther 10(1): 24–33. doi: 10.1080/14764170701817049 Department of Health (2013) Review of Regulation of Cosmetic Interventions. DH, London Drosner M, Adatto M (2005) Photo-epilation: guidelines for care from the European Society for Laser Dermatology (ESLD). J Cosmet Laser Ther 7(1): 33–8 Fitzpatrick TB (1975) "Soleil et peau" [Sun and skin]. Journal de Médecine Esthétique [published in French] 2: 33–4 Galletly C, Clark A, Tomlinson T, Blaney F (1996) A group program for obese, infertile women: weight loss and improved psychological health. J Psychosom Obstet Gynaecol 17(2): 125–8 Guidelines Working Party (2005) Medical Management of Facial Hirsutism. The Outcomes of a Guidelines Working Party. Primary Care Dermatology Society, Hatfield Haedersdal M, Beerwerth F, Nash JF (2011) Laser and intense pulsed light hair removal technologies: from professional to home use. Br J Dermatol 165(Suppl 3): 31–6. doi: 10.1111/j.1365-2133.2011.10736.x Haedersdal M, Gotzsche P (2006) Laser and photoepilation for unwanted hair growth. Cochrane Database Syst Rev 4: CD004684 Halachmi S, Lapidoth M (2012) Low-fluence vs. standard fluence hair removal: a contralateral control non-inferiority study. J Cosmet Laser Ther 14(1): 2–6. doi: 10.3109/14764172.2011.634421 Hamzavi I, Owen MR (2006) Eflornithine treatment of excess facial hair: a review. Dermatol Pract 13: 19–20 Hickman JG, Huber F, Palmisano M (2001) Human dermal safety studies with eflornithine HCl 13.9% cream (Vaniqa), a novel treatment for excessive facial hair. Curr Med Res Opin 16(4): 235–44 Housman TS, Derrow AE, Snively BM et al (2004) Women with excessive facial hair: a statistical evaluation and review of impact on quality of life. Cosm Dermatol 17(3): 157–65 Hunter MH, Carek PJ (2003) Evaluation and treatment of women with hirsutism. Am Fam Physician 67(12): 2565–72 Ibrahimi OA, Avram MM, Hanke CW, Kilmer SL, Anderson RR (2011) Laser hair removal. Dermatol Ther 24(1):94-107. doi: 10.1111/j.1529-8019.2010.01382.x Johnson NP (2014) Metformin use in women with polycystic ovary syndrome. Ann Transl Med 2(6): 56. doi: 10.3978/j.issn.2305-5839.2014.04.15 Lapidoth M, Dierickx C, Lanigan S et al (2010) Best practice options for hair removal in patients with unwanted facial hair using combination therapy with laser: guidelines drawn up by an expert working group. Dermatology 221(1): 34–42. doi: 10.1159/000315499 Lepselter J, Elman M (2004) Biological and clinical aspects in laser hair removal. J Dermatolog Treat 15(2): 72–83 Martin KA, Chang RJ, Ehrmann DA et al (2008) Evaluation and treatment of hirsutism in premenopausal women: an endocrine society clinical practice guideline. J Clin Endocrinol Metab 93(4): 1105–20. doi: 10.1210/jc.2007-2437 McGill DJ, Hutchison C, McKenzie E, McSherry E, Mackay IR (2007) Laser hair removal in women with polycystic ovary syndrome. J Plast Reconstr Aesthet Surg 60(4): 426–31 Moseley H (2014) How to improve safe delivery of intense pulsed light treatment. Photodermatol Photoimmunol Photomed 30(1): 52. doi: 10.1111/ phpp.12067 Newson L (2010) The basics. Hirsutism. (Accessed 16 October 2014) Patil UA, Dhami LD (2008) Overview of lasers. Indian Journal of Plastic Surgery 41(3): 101–13 Sanchez LA, Perez M, Azziz R (2002) Laser hair reduction in the hirsute patient: a critical assessment. Hum Reprod Update 8(2): 169–81 Styczynski P, Oblong J, Ahluwalia GS (2008) Removal of unwanted facial hair. In: Ahluwalia GS, ed. Cosmetic Applications of Laser and Light-Based Systems. William Andrew, New Y ork T eede H, Deeks A, Moran L (2010) Polycystic ovary syndrome: a complex condition with psychological, reproductive and metabolic manifestations that impact on health across the lifespan. BMC Med 8: 41. doi: 10.1186/1741-7015-8-41 Willey A, Torrontegui J, Azpiazu J, Landa N (2007) Hair stimulation following laser and intense pulsed light photo-epilation: review of 543 cases and ways to manage it. Lasers Surg Med 39(4): 297–301 Figure 2: A 43-year-old woman with excessive hair on her chin Liliana Marza
4539	https://math.stackexchange.com/questions/1869999/odds-of-10-thrown-dice-landing-all-the-same	probability - Odds of 10 thrown dice landing all the same - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Odds of 10 thrown dice landing all the same Ask Question Asked 9 years, 2 months ago Modified7 years, 2 months ago Viewed 12k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. What are the odds of throwing 10 six sided dice and landing all the same number. Also, how many throws would I need to do to achieve a 100 percent success of this happening. Is this even possible in terms of mathematical probability? probability Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jul 24, 2016 at 22:28 David_Shmij 416 3 3 silver badges 18 18 bronze badges asked Jul 24, 2016 at 22:17 Sean WinterSean Winter 21 1 1 gold badge 1 1 silver badge 1 1 bronze badge 5 5 However many times one throws, there cannot be certainty of success.André Nicolas –André Nicolas 2016-07-24 22:23:56 +00:00 Commented Jul 24, 2016 at 22:23 It seems straightforward to compute the chance of this happening in one throw. Did you work through the details?hardmath –hardmath 2016-07-24 22:27:57 +00:00 Commented Jul 24, 2016 at 22:27 The answers posted so far are bare assertions that don't explain how to find the solution. Michael Hardy –Michael Hardy 2016-07-25 00:26:15 +00:00 Commented Jul 25, 2016 at 0:26 It depends what you mean by throw. Will you throw all 10 at once or one at a time? Also I disagree with Andre Nicolas. If you get all the same the that is a certain success. Just keep going until you observe it. The number of throws cannot be determined beforehand to get to this state but if you count the states as you go then observe your desired goal condition, that is your answer. Power of observation. Using computer simulation , you can verify that it is possible to achieve this special condition of all 10 the same.David –David 2016-07-25 04:21:26 +00:00 Commented Jul 25, 2016 at 4:21 I know the level of certainty never hits 100% but it will be so close that so for all intents are purposes, it is basically 100%. Simulate 1 billion 10 dice rolls and what is the probability of never getting all 10 dice the same? Practically 0%.David –David 2016-07-25 04:28:26 +00:00 Commented Jul 25, 2016 at 4:28 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. To answer your first question: The probability that ten six-sided dice land with the same number is equal to the probability that nine of the six-sided dice have the same number as the tenth i.e. 1/6 9. Your second question does not make sense, unless you are asking how many rolls you need to make for the probability of ten six sided dice landing on the same number to happen almost surely. Then the answer is an infinite number of rolls. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 24, 2016 at 22:26 smccsmcc 5,828 12 12 silver badges 19 19 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The odds are 1 to 6 9−1. The probability is 1 6 9. This is because there are 6 choices for the face and 6 10 values for 10 dice. After n throws, the chances of getting all 10 the same at least once is 1−(6 9−1 6 9)n It never is exactly 1. However, to get a 50% chance of seeing all faces the same, you would need to throw the dice 6,985,327 times, and to get a 99% chance, 46,409,503 times. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jul 25, 2016 at 13:23 answered Jul 24, 2016 at 22:49 robjohn♦robjohn 355k 38 38 gold badges 499 499 silver badges 892 892 bronze badges 0 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. For your first question: The probability of throwing 10 six sided dice and having them all land on the same arbitrary number is = (1 6)10×(6 1) =1 10077696 assuming fair six-sided dice. Notice that if you are simply looking for answers to these kinds of questions, you can use WolframAlpha, for example try this. This related post can help with some methodology if required. 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 12:20 CommunityBot 1 answered Jul 24, 2016 at 22:32 David_ShmijDavid_Shmij 416 3 3 silver badges 18 18 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 probability 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 5Chance of 7 of a kind with 10 dice Related 5How do I calculate the odds of a given set of dice results occurring before another given set? 1What are the odds of rolling seven dice and all coming up equal? 1dice probability and correct statement to express the unsuccess 0Throw dice twice picking the best result. 0What is more probable? Six 6-sided dice landing in combinations 05 dice are roll what is the probability that three dice show the same number 1Odds of rolling 3 or more consecutive numbers on n number of dice Hot Network Questions Checking model assumptions at cluster level vs global level? An odd question A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man Is existence always locational? What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left? Is there a way to defend from Spot kick? Another way to draw RegionDifference of a cylinder and Cuboid Xubuntu 24.04 - Libreoffice Storing a session token in localstorage Is it safe to route top layer traces under header pins, SMD IC? Making sense of perturbation theory in many-body physics With with auto-generated local variables I have a lot of PTO to take, which will make the deadline impossible Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator Why are LDS temple garments secret? My dissertation is wrong, but I already defended. How to remedy? In Dwarf Fortress, why can't I farm any crops? Alternatives to Test-Driven Grading in an LLM world Proof of every Highly Abundant Number greater than 3 is Even Why do universities push for high impact journal publications? Analog story - nuclear bombs used to neutralize global warming Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Riffle a list of binary functions into list of arguments to produce a result Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4540	https://www.sohu.com/a/416397004_99940503	微专题07 分段函数的性质与应用 2020-09-04 07:41 来源: 每日一题学好高中数学分段函数是函数中比较复杂的一种函数，其要点在于自变量取不同范围的值时所使用的解析式不同，所以在解决分段函数的问题时要时刻盯着自变量的范围是否在发生变化。即“分段函数——分段看” 1 、分段函数的定义域与值域——各段的并集 2 、分段函数单调性的判断：先判断每段的单调性，如果单调性相同，则需判断函数是连续的还是断开的，如果函数连续，则单调区间可以合在一起，如果函数不连续，则要根据函数在两段分界点出的函数值（和临界值）的大小确定能否将单调区间并在一起。 3 、分段函数对称性的判断：如果能够将每段的图像作出，则优先采用图像法，通过观察图像 5 、遇到分段函数要时刻盯住变量的范围，并根据变量的范围选择合适的解析式代入，若变量的范围并不完全在某一段中，要注意进行分类讨论 6 、如果分段函数每一段的解析式便于作图，则在解题时建议将分段函数的图像作出，以便必要时进行数形结合。阅读 (1655) 搜狐“我来说两句” 用户公约点击登录搜狐小编发布推荐阅读据说，欧洲的小偷居然达成了一种共识，那就是可以去偷中国人的钱财和物品，但是绝对不... 吃一颗苹果 · 昨天 10:41 7 江西一女子考上大学，却交不起学费，向大舅借了4万，毕业后女子去还钱，谁料，大舅说钱... 骊歌声声慢 · 昨天 18:10 6 我年薪300万，回老家三婶问我收入，我谎称月薪9000，三婶：那你堂弟结婚，28万彩礼你来... 烟斗来了 · 昨天 04:29 11 主办方拒绝挂中国国旗，中方果断退出，其他国家：没有中国不开赛！朔方瞭望 · 今天 03:20 3 全世界都被普京骗了 , 打击乌克兰只是一个幌子 , 真正目标已布局4年！史谭a · 昨天 07:28 0 我嫌老公买的维C太苦，便拿着药瓶去医院，医生看完神色凝重地说：这里面不是维C，是米... 瓜汁橘长Dr · 09-26 02:20 2 此次爆炸仅次于911，美国联邦大楼被炸，9层大楼瞬间坍塌，168人死亡680人受伤！白衣说客 · 昨天 07:35 1 大尺度电影推荐，部部都是经典，值得你熬夜观看！梦新看世界 · 09-26 00:52 0 9月28日特讯！洪森发表涉华言论，罕见言辞引爆国际舆论小笛点评 · 今天 03:09 2 官方通报：作为“一把手”，他安排单位买500斤散装白酒存于办公楼，用于日常接待政事儿 · 昨天 09:23 86 “成人网站”上的女生，到底是怎样被偷拍的？女生一定要小心了！萌鑫看世界 · 今天 01:14 0 成都可乐教育怎么样？告别冗余与混乱：程序员必知的 3 个基础优化技巧绘梦者 · 09-17 05:08 0 太突然！清华赴美博士李昊然去世，年仅28岁，大学邮件曝光去世原因，我国大使馆紧急发声影像温度 · 昨天 23:56 13 港媒已经收到风声，中国在为国家紧急事态，提前做充分的准备澜风说 · 昨天 23:38 0 国庆黄金周还没到，就已经出现了3大反常现象，还有一个让人担忧平说财经张平 · 昨天 11:31 18 "驱逐"中国外交官，准备与中方战斗，叫嚣：中国必须收下货品，否则就干架史谭a · 今天 03:59 1 韩国女主持嘲讽：中国制造就是廉价、仿品，教授反问了一句，让她瞬间脸红相框框一切 · 09-26 11:57 8 快报快报！泽连斯基愿承认乌俄新边界峰火人生 · 昨天 12:01 30 今年国庆不一般！有1个坏消息，2个别大意，3个要谨慎，早做准备小娟教做菜 · 09-26 22:32 0 新中国成立后，只有3个人可以随时见毛主席，他们分别都是谁？南风一史 · 昨天 03:20 0 比房子更荒唐！个人收入开始严查？2025年后，账户收入超过这个数，就要多注意，别不当回事万公子 · 09-25 17:49 2 美媒：空警-700已不仅是一架预警机，为将来的中美冲突量身定制历农爱生活 · 今天 01:37 0 男驴友爬雪山坠亡后续：现场视频流出，死状极惨，目击者曝真相历史特别冷 · 09-26 06:50 9 陈某直播泄露军事秘密，将受严惩！光明网 · 今天 01:40 1 一夜之间突然倒戈！3名师长带领5万士兵赴美，还顺走600辆坦克朔方瞭望 · 今天 02:24 1 82万奖金变3万，我没闹，再没修进口机床，15天后老板出100万请来专家，见到是我，愣了瓜汁橘长Dr · 09-26 02:21 3 泽连斯基在联合国很没有面子，泽连斯基在联合国大会上发表演讲，直播显示，泽连斯基未... 小陈言社会 · 09-25 15:38 4 谢孟伟儿子已8岁，都36岁了还这么傻饭樊 · 今天 12:59 0 “禁毒女警”陈光明，将7000名毒贩送进牢房，下场极其悲惨微历史之中山古国 · 昨天 03:13 5 “一家几口都是公务员”，人口小县财政供养入不敷出！副部级退休官员建议按比例削减公... 有料新语 · 09-26 06:17 440 成都可乐教育：解锁并发任务的高效实现方式历历如绘 · 09-23 08:09 0 19000吨钢筋桥梁“瞬间崩塌”，大桥竣工前突然垮掉，75人当场身亡朔方瞭望 · 09-26 02:38 1 男子直播开飞机坠亡，曝遗体烧的只剩一条腿，细节曝光，早有预兆青橘罐头 · 昨天 23:44 1 网红“唐飞机”直播时坠机，飞机空中解体后坠毁！画面曝光，妻子：他不听劝奇趣阁 · 昨天 22:44 2 就在刚刚！中国雷霆宣言：台湾海峡全面军事管辖，外舰“自由航行”时代终结朔方瞭望 · 今天 03:57 2 歼-50再次低空试飞，曝光照片中一细节，透露出恐怖的研发实力！全民小侦探 · 09-26 07:42 8 三星堆新发现：祭祀坑年代厘清与青铜铸造技术创新中关村在线 · 昨天 05:20 0 县长被开除党籍、撤职、降为正科11年后，涉嫌严重违法，再次被查观察者网 · 今天 04:16 91 张召忠：错过统一的最好时机，该出手时没出手，看完恍然大悟小舟说史 · 昨天 02:58 127 七彩虹 python中对象是什么，以及一切皆对象的理解？梦幻绘境 · 09-25 07:21 0 金水区教学双杯赛郑州八中教师展风采河南日报 · 09-24 03:48 0 美媒无语了:美部长召集全球美军将领回国，原来就为了这点事？！多看看啊这个世界 · 昨天 23:26 19 清华大学本科毕业生李昊然去世，年仅27岁非常优秀，同学曝是自杀八卦娱V · 今天 00:17 0 警方通报“华南理工大学发生车祸致1死1伤” 环球时报新媒体 · 今天 11:57 0 连续吃5年西贝的毛毛“找到了”，满屏段子让官方尴尬，文章被删东山的枫 · 09-26 03:07 13 离谱！76岁男子交通违法279条，被记819分，本人：不知道要考驾照长春晚报 · 昨天 06:03 0 三大影业联手开火！AI公司MiniMax被告侵权油麦菜 · 今天 12:43 0 美国防长:我们可能要开战了！达文西看世界 · 09-26 06:56 4 以色列警告埃及：撤下中国导弹，停止用雷达照射以色列战机小陈言社会 · 今天 11:14 0 粉丝冲卡推人，追星乱象已超越理智底线陆弃 · 昨天 20:43 0 台湾要员曾放话：满足这3条件，台湾自动回归，实行“一国一制” 柠重 · 09-26 01:06 0 广州警方通报！新浪财经 · 今天 11:01 0 从图形化到Python的最佳学习路线图规划美食小琴 · 09-21 12:18 0 广东与广西的辖区调整，广西的1个县，为何划归了广东省？阿鑫历史说 · 09-26 06:00 0 中共中央批准，开除段成刚、龙翔党籍长安街知事 · 今天 06:12 0 许家印案目前进展情况：量刑预判与羁押状况，死刑的概率有多大四季日常诠释社会 · 今天 12:39 0 两个草包，把西贝害惨了简十一 · 昨天 02:34 0 周总理逝世，没人敢读悼词，叶剑英：论资格论威望，只有一人可以壹周历史 · 昨天 04:25 0 毛主席给我们攒下的三大红利已经消耗殆尽，以后得靠我们自己了！游游历史 · 昨天 03:14 0 原正部级唐仁健，被判死缓长安街知事 · 今天 05:06 0 “卖女孩的红烧肉”店名争议，官方回应深圳新闻网 · 今天 10:43 0 万达集团及老板王健林，再次被限制高消费！末流作家 · 今天 05:36 0 喊话4天，泽连斯基决定离职，乌恳求中国做一件事，流放名单公开北极星哨 · 09-26 06:53 0 违规买卖股票，展翔被罚没1.59亿元国际旅游岛商报 · 09-26 21:39 0 美国总统特朗普在联合国大会上，第二次提到了中国，他声称：“我很佩服中国，我给予中... 吃一颗苹果 · 09-26 00:28 0 阅兵迎外宾的车，50辆来自河南车主，不是国家缺车，是有三大意义鲸探所 · 昨天 03:19 0 南京市人大常委会原主任龙翔被“双开”，通报称其搞迷信活动红星新闻 · 今天 07:08 0 后续！官方通报8亿建厂遭强行接管，网友评论区炸锅！粤广辉哥 · 09-26 12:00 0 县城的无奈，如今只剩下“公务员经济”，也即将坚持不下去了！看看咋回事娱人有访 · 09-25 21:54 0 老公的弟弟结婚，仪式结束我正要去吃饭时，新娘的妈突然冲到我面前嚣张质问：伴娘的红... 瓜汁橘长Dr · 昨天 03:04 0 还好当年没答应爱因斯坦，要不然，云南可能就成了现在的巴勒斯坦凡人侃史 · 今天 11:43 0 全球首款“摄像”磁共振亮相，华山医院神经临床研究新项目启动文汇报 · 今天 11:55 0 医院各层干部形成权色钱交易网，招聘等十几种腐败触目惊心四季日常诠释社会 · 09-26 15:11 0 交管 12123 显示查封？查封车还能年审吗？怎么查车辆查封原因？越山向海 · 今天 13:07 0 以牺牲环境为代价换取经济增长、在污染治理中弄虚作假，实行“一票否决” 政知新媒体 · 今天 12:40 0 48女教师连续带出6届状元班，评优时却没有她，决绝之下辞职加入隔壁中学，校长后悔哭了烟斗来了 · 09-26 08:06 0 春波老师简化数学：高三差生别慌，这样学稳提分、能上本科教育研究观察者 · 09-21 01:12 0 普京心意已决：除了“消灭”乌克兰，还有一国将在2025不复存在？昭烈忠臣 · 09-26 00:36 0 微信群聊“八卦”被行拘，女教师坚称是闲聊中国新闻周刊 · 今天 09:46 0 40岁上海男子“被患精神分裂”11年，政审时才发现？“600号”复核：精神分裂→急性应激障碍鲁网 · 今天 02:43 0 已经到底了每日一题学好高中数学文章 0 总阅读 9.6万 24小时热文 1 乌媒：俄对乌发动大规模袭击，泽连斯基发声 57万阅读 2 从教室到食堂，狐友校园百科一站式全导航！ 160万阅读 3 普林斯顿中国博士后李昊然去世，毕业于清华，刚完成论... 324万阅读受贿2.68亿余元，唐仁健一审被判死缓 125万阅读搜狐号北青网 “北京建筑之旅”打卡手册在正阳门箭楼发布雷科技闹掰了！宗馥莉自立门户，“娃小宗”将取代“娃哈哈”？山西经济日报越来越挤的云栖大会，越来越 AI 的阿里巴巴 “中国版英伟达” IPO过会这支“概念股”已经连续涨停补链强基延链拓新——济南历城绘就产业发展新图景请输入正确的登录账号或密码安全提示暂不发送发送语音验证码
4541	https://mathworld.wolfram.com/TrilinearCoordinates.html	Trilinear Coordinates Given a reference triangle , the trilinear coordinates of a point with respect to are an ordered triple of numbers, each of which is proportional to the directed distance from to one of the side lines. Trilinear coordinates are denoted or and also are known as homogeneous coordinates or "trilinears." Trilinear coordinates were introduced by Plücker in 1835. Since it is only the ratio of distances that is significant, the triplet of trilinear coordinates obtained by multiplying a given triplet by any nonzero constant describes the same point, so For simplicity, the three polygon vertices , , and of a triangle are commonly written as , , and , respectively. Trilinear coordinates can be normalized so that they give the actual directed distances from to each of the sides. To perform the normalization, let the point in the above diagram have trilinear coordinates and lie at distances , , and from the sides , , and , respectively. Then the distances , , and can be found by writing for the area of , and similarly for and . We then have \| \| \| \| \| --- --- \| \| \| \| \| (2) \| \| \| \| \| (3) \| \| \| \| \| (4) \| \| \| \| \| (5) \| so \| \| \| --- \| \| \| (6) \| where is the area of and , , and are the lengths of its sides (Kimberling 1998, pp. 26-27). To obtain trilinear coordinates giving the actual distances, take , so we have the coordinates \| \| \| --- \| \| \| (7) \| These normalized trilinear coordinates are known as exact trilinear coordinates. The trilinear coordinates of the line \| \| \| --- \| \| \| (8) \| are \| \| \| --- \| \| \| (9) \| where is the point-line distance from polygon vertex to the line. The homogeneous barycentric coordinates corresponding to trilinear coordinates are , and the trilinear coordinates corresponding to homogeneous barycentric coordinates are . Important points of a triangle are called triangle centers, and the vector functions describing the location of the points in terms of side length, angles, or both, are called triangle center functions . Since by symmetry, triangle center functions are of the form \| \| \| --- \| \| \| (10) \| it is common to call the scalar function "the" triangle center function. Note also that side lengths and angles are interconvertible through the law of cosines, so a triangle center function may be given in terms of side lengths, angles, or both. Trilinear coordinates for some common triangle centers are summarized in the following table, where , , and are the angles at the corresponding vertices and , , and are the opposite side lengths. Here, the normalizations have been chosen to give a simple form. \| \| \| --- \| \| triangle center \| triangle center function \| \| circumcenter \| \| \| de Longchamps point \| \| \| equal detour point \| \| \| Feuerbach point \| \| \| incenter \| 1 \| \| isoperimetric point \| \| \| symmedian point \| \| \| nine-point center \| \| \| orthocenter \| \| \| triangle centroid \| , \| In trilinear coordinates, the coordinates of the vertices are 1:0:0 (), 0:1:0 (), and 0:0:1 (). Extensions along the sidelines by a distance have trilinears as illustrated above. Trilinear coordinates of points fractional distances , , and along the sidelines are given in the above figure, where . A point located a fraction of the distance along the sideline from to has trilinear coordinates \| \| \| --- \| \| \| (11) \| To determine the conversion of trilinear to Cartesian coordinates, orient the triangle with the axis parallel to the -axis and with its incenter at the origin, as illustrated above. Then \| \| \| \| \| --- --- \| \| \| \| \| (12) \| \| \| \| \| (13) \| where \| \| \| --- \| \| \| (14) \| is the inradius, is the triangle area, and \| \| \| --- \| \| \| (15) \| (Kimberling 1998, pp. 31-33). More generally, to convert trilinear coordinates to a vector position for a given triangle specified by the - and -coordinates of its axes, pick two unit vectors along the sides. For instance, pick \| \| \| \| \| --- --- \| \| \| \| \| (16) \| \| \| \| \| (17) \| where these are the unit vectors and . Assume the triangle has been labeled such that is the upper rightmost polygon vertex and . Then the vectors obtained by traveling and along the sides and then inward perpendicular to them must meet \| \| \| --- \| \| \| (18) \| Solving the two equations \| \| \| \| \| --- --- \| \| \| \| \| (19) \| \| \| \| \| (20) \| gives \| \| \| \| \| --- --- \| \| \| \| \| (21) \| \| \| \| \| (22) \| But and are unit vectors, so \| \| \| \| \| --- --- \| \| \| \| \| (23) \| \| \| \| \| \| And the vector coordinates of the point are then See also Areal Coordinates, Barycentric Coordinates, Exact Trilinear Coordinates, Major Triangle Center, Orthocentric Coordinates, Power Curve, Quadriplanar Coordinates, Reference Triangle, Regular Triangle Center, Triangle, Triangle Center, Triangle Center Function, Trilinear Line, Trilinear Polar, Trilinear Vertex Matrix, Tripolar Coordinates Explore with Wolfram\|Alpha More things to try: triangle properties 5!! distinct permutations of {1, 2, 2, 3, 3, 3} References Boyer, C. B. History of Analytic Geometry. New York: Yeshiva University, 1956.Casey, J. "The General Equation--Trilinear Co-Ordinates." Ch. 10 in A Treatise on the Analytical Geometry of the Point, Line, Circle, and Conic Sections, Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd ed., rev. enl. Dublin: Hodges, Figgis, & Co., pp. 333-348, 1893.Coolidge, J. L. A Treatise on Algebraic Plane Curves. New York: Dover, pp. 67-71, 1959.Coxeter, H. S. M. Introduction to Geometry, 2nd ed. New York: Wiley, 1969.Coxeter, H. S. M. "Some Applications of Trilinear Coordinates." Linear Algebra Appl. 226-228, 375-388, 1995.Kimberling, C. "Central Points and Central Lines in the Plane of a Triangle." Math. Mag. 67, 163-187, 1994.Kimberling, C. "Triangle Centers and Central Triangles." Congr. Numer. 129, 1-295, 1998.Wong, M. K. F. Int. J. Math. Educ. Sci. Tech. 27, 293-296, 1996.Wong, M. K. F. Int. J. Math. Educ. Sci. Tech. 29, 143-145, 1998. Referenced on Wolfram\|Alpha Trilinear Coordinates Cite this as: Weisstein, Eric W. "Trilinear Coordinates." From MathWorld--A Wolfram Resource. Subject classifications Find out if you already have access to Wolfram tech through your organization
4542	https://golem.ph.utexas.edu/category/2019/11/random_permutations_part_3.html	The n-Category Café A group blog on math, physics and philosophy Skip to the Main Content Enough, already! Skip to the content. Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser. « Topics in Category Theory: A Spring School \| Main \| Random Permutations (Part 4) » November 24, 2019 Random Permutations (Part 3) Posted by John Baez I’m trying to understand what a random permutation of a large set is typically like. While I’ve been solving some puzzles that shed a bit of light on this question, I now realize there are some other puzzles that would help more! When I started, I had a wrong mental image of a random permutation of a large finite set. I foolishly imagined it would be made of many small orbits. But that’s far from true! It’s easy to see why if you think about it. Given a random permutation of a large -element set, it’s very unlikely for to be mapped to itself by , or even by for some small . So we should instead imagine our random permutation as consisting mainly of quite large orbits! Here are the facts I’ve proved or at least stated so far: As , the probability that a random permutation of an -element set has fixed points approaches So there’s about a 37% chance of having no fixed points, a 37% chance of having one, an 18% chance of having two, a 6% chance of having three, and it drops off rapidly from there. I showed this in Part 1. The expected number of fixed points of a random permutation of a nonempty finite set is exactly 1. In the limit this follows from the previous remark. But in fact there’s an easy way to show it’s true for any . Do you see how? In the limit as , the probability that the shortest cycle in a random permutation of an -element set has length is So, there’s a 37% chance that the shortest cycle has length , a 22% chance that it has length , a 16% chance that it has length , a 12% chance that it has length , and it drops off slowly from there. I proved this in Part 2. This is potentially misleading to the unwary. It says a random permutation of a huge set is quite likely to have at least one short cycle. But you shouldn’t assume it has lots of short cycles. Indeed, remark 2) says that it has just one cycle of length 1, on average. In the limit as , the average length of the longest cycle in a random permutation is , where I discussed in this in my post on the Golomb–Dickman constant, which is the name for this number . Dickman proved more: he computed the probability that the longest cycle in a random permutation of an -element set has length , in the limit. He showed it drops off very rapidly as decreases. For example, the probability that the longest permutation has length is less than , for large . This is makes it very interesting to probe the nature of large cycles. But the details of these, and many other aspects of random permutations, remain mysterious to me! So, let me resume my puzzle series. These are puzzles for myself, or anyone who wants to help: Puzzle 5. What is the expected number of cycles in a random permutation of an -element set? What about the mode, or median? Apparently the expected number of cycles is As this is asymptotically . That’s very small compared to the size of our set! This goes along with the fact that typically there are some rather large cycles. A proof is here, but it’s not easy to follow unless you go back and figure out the notation. I should write up a nice one. I don’t know the mode or median. Puzzle 6. What is the expected length of a cycle in a random permutation of an -element set, asymptotically as ? What about the mode, or median? I haven’t worked this out, but I want the expected value to be asymptotically . Puzzle 7. What is the expected number of cycles of length in a random permutation of an -element set? The answer is apparently when . A proof is here, but I should explain it. Note that when we’re back to my claim that the expected number of fixed points is . Here are some more puzzles designed to shed light on the large cycles in a random permutation: Puzzle 8. If we choose an element of an -element set, what is the probability that it lies in a cycle of length ? The answer is very simple: , for . In other words, it’s equally likely to be on a cycle of any allowed size! A proof is here, and again I want to explain it. Puzzle 9. If we choose one element of an -element set, what is the expected length of the cycle it lies on? Thanks to Puzzle 8, the answer is : the average of the natural numbers from to . Note that there can be at most one cycle of length . I’ll call such a cycle a giant cycle, because it reminds me of the giant connected component of a random graph. Puzzle 10. How probable is the existence of a giant cycle? When is even, the probability is This is connected to the hundred prisoners problem. Something similar is true when is odd. As the probability approaches , or about 69.3%. Okay, now I need to process all this information, write up proofs for some of these things, but most importantly synthesize all this information to get a better mental picture of a random permutation. The way I see it now, when you have a random permutation of a large set, there’s a very good chance a substantial fraction of the elements lie on the largest cycle. If this happens, what about the rest? Is the permutation restricted to the rest of the set random in the same way—all elements of the permutation group equally likely? I think so. If so, we can recursively ‘chip away’ at a random permutation, repeatedly removing the largest cycle, getting a kind of fractal structure where each time we do this, the random permutation of the remaining elements is governed by all the same rules I’ve listed here! Posted at November 24, 2019 11:12 PM UTC TrackBack URL for this Entry: Some Related Entries Random Permutations (Part 2) — Nov 24, 2019 Random Permutations (Part 1) — Nov 24, 2019 Torsion: Graph Magnitude Homology Meets Combinatorial Topology — Apr 11, 2018 Lattice Paths and Continued Fractions II — Sep 22, 2017 Lattice Paths and Continued Fractions I — Sep 18, 2017 Schröder Paths and Reverse Bessel Polynomials — Aug 28, 2017 Mathematics and Magic: the de Bruijn Card Trick — Jan 05, 2015 Graph Colouring and Cartesian Closed Categories — Dec 05, 2014 5 Comments & 0 Trackbacks Re: Random Permutations (Part 3) But in fact there’s an easy way to show it’s true for any . There’s also a “hard” way of finding the expected number of fixed points: decompose the natural linear representation of (as permutation matrices) and use that to compute the average of its character. Posted by: unekdoud on November 25, 2019 9:21 AM \| Permalink \| Reply to this Re: Random Permutations (Part 3) Nice! For an action of a finite group on a finite set , we get a representation of on . For each , the trace is the number of fixed points of the action of on . The sum thus gives us the average number of fixed points for a randomly chosen . This sum equals where we’re taking the inner product using the normalized translation-invariant measure on . But this is where is the 1-dimensional trivial representation of . And this inner product is the multiplicity with which the representation appears as a summand of . When acts in a doubly transitive way on things simplify, since then is the direct sum of and some other irreducible representation. Thus contains with multiplicity 1, so the inner product is 1, so the average number of fixed points is 1. This happens for acting on the -element set. Posted by: John Baez on November 25, 2019 9:51 PM \| Permalink \| Reply to this Re: Random Permutations (Part 3) Hello, hello, I’ve been a while away; Akshay Venkatesh gave a fun public lecture about some of these questions, but lifted from random permutations to random braid links— about a month ago, viewable on youtube. Posted by: Jesse C. McKeown on November 25, 2019 2:36 PM \| Permalink \| Reply to this Re: Random Permutations (Part 3) Regarding this talk, Akshay Venkatesh suggested this paper by Andrew Granville: Posted by: Abel Wolman on November 26, 2019 12:33 AM \| Permalink \| Reply to this Re: Random Permutations (Part 3) Wow, that paper is great, Abel! Thanks! I’m sort of glad I didn’t know about this paper earlier, because I might have never gotten up the nerve to think about random partitions… this paper contains so much information about them, so nicely explained. But I love how it sets up and studies the analogy between cycle decompositions of permutations and prime factorizations of numbers. Posted by: John Baez on November 26, 2019 2:03 AM \| Permalink \| Reply to this Post a New Comment Access Keys: 0 : Accessibility Statement 1 : Main Page 2 : Skip to Content 3 : List of Posts 4 : Search p : Previous (individual/monthly archive page) n : Next (individual/monthly archive page)
4543	https://about.illinoisstate.edu/mshesso-test2/reporting-statistics-in-apa-style/	Professor and Coordinator Reporting Statistics in APA Style Reporting Statistics in APA Style A Short Guide to Handling Numbers and Statistics in APA Format The material in this guide is based on the sixth edition of the publication manual of the American Psychological Association: American Psychological Association. (2010). Publication manual of the American Psychological Association (6th ed.). Washington, DC: Author. Report The Results of All Hypothesis Tests Statistics are reported for all hypothesis tests, including tests that are not significant. The principle is that the reader should be able to join the author in deciding that an effect is not statistically significant based on the descriptive and inferential statistics. (See page 32 of the Publication Manual). Report Exact P-Values The preferred method of reporting P-values is to use an exact number, with two or three significant decimal places rather than as a range or category (e.g., NS, p > .05, or p < .05). The principle is that not everyone takes a strictly Pearsonian view of probabilities as absolutely, categorically significant or absolutely, categorically non-significant. Sir R. A. Fisher advocated the position that probabilities can be interpreted with varying degrees of signficance. By providing an exact p-value rather than a range, readers may adopt either approach to evaluating probabilities. Additionally, when evaluating the strength of an effect, in the absence of other measures of effect size, the p-value can convey the strength of the finding. For instance, a p-value of .054 is more encouraging as a line for further study (say, with a larger sample size) than a p-value of .67. Additionally, a p-value of .012 indicates a stronger effect than a p-value of .049. (See page 34 of the Publication Manual). Use Rounding Appropriately Round numbers to one or two decimal places, keeping in mind that fewer decimal places are easier to comprehend. Consider rescaling measurements that require more than two decimal places to report meaningful differences (e.g., convert meters to millimeters). (See page 113 of the Publication Manual). Scientific convention stipulates that, when rounding numbers, numbers should be rounded up as often as they are rounded down. To round a number to a given precision, examine the first digit to be truncated. If this digit is 1, 2, 3, or 4, round the number down. If this digit is 6, 7, 8, or 9, round the number up. If this digit is a 5, then you should look to the remaining digits beyond the 5 to see if they are all zeroes. If they are not all zeroes, then the number does not end in an exact 5 and should be rounded up. If all remaining digits to the right are zero (or there are no additional digits available to the right of the 5), then the number (in its current precision) is an exact 5. In this case, the number should be rounded up as often as it is rounded down; therefore, round so as to produce an even digit for the last digit. Consider these examples: \| \| \| --- \| \| Number \| Rounded to 2 decimal places \| \| 1.2349999 \| 1.23 \| \| 1.6762124 \| 1.68 \| \| 1.4256398 \| 1.43 \| \| 1.4250001 \| 1.43 \| \| 1.4250000 \| 1.42 \| \| 1.6750000 \| 1.68 \| \| 1.6850000 \| 1.68 \| \| \| \| --- \| \| Number \| Rounded to 2 decimal places \| \| 1.2349999 \| 1.23 \| \| 1.6762124 \| 1.68 \| \| 1.4256398 \| 1.43 \| \| 1.4250001 \| 1.43 \| \| 1.4250000 \| 1.42 \| \| 1.6750000 \| 1.68 \| \| 1.6850000 \| 1.68 \| Spell Out Numbers Or Use Numerals Appropriately Numbers less than 10 are typically spelled out (e.g., five, seven), but numbers greater than 10 are typically represented with digits (e.g., 452). Among other exceptions is the rule that numbers starting a sentence are always spelled out, regardless of size, such as Forty-seven participants refused consent.. (See Sections 4.31 and 4.32, pp. 111-112 in the Publication Manual). Statistical Abbreviations Statistical abbreviations (e.g., M, SD) are only to be used within parentheses or at the end of sentences (i.e., when the abbreviation is not being used as a part of speech within the sentence). When the statistic in question is functioning as a part of speech in the sentence (e.g., as the subject of the sentence or the object of a prepositional phrase), then the statistic name must be spelled out as a word and not abbreviated, such as mean or standard deviation. (See page 117 of the Publication Manual). Provide a Set of Minimally Sufficient Statistics The guiding principle for determining which information to include with a statistical test is that the reader should have enough information to verify the computations. The following list summarizes which information is necessary and provides an example of how it is typically presented. Descriptive Statistics Descriptive statistics are the building blocks used to augment other findings. The most frequently reported descriptive statistics are the sample size, mean, and standard deviation because they are usually the basis for computing inferential statistics. When means are reported, standard devations should always be reported as well, “A mean without a standard deviation is like a day without sunshine!” In addition, it is important to include the sample size on which the mean has been computed. Examples: The average reaction time for the 12 participants was 820 ms (SD = 192) in the treatment group, but the mean reaction time was only 642 ms (SD = 183) for the 11 participants in the control group. The 16 teenagers who volunteered for the pilot study were younger than expected, M = 14.2 years, SD = 1.3. The 16 teenagers who volunteered for the pilot study were younger than expected, M = 14.2 years, SD = 1.3. Note that abbreviations are only used for statistics when the statistics are reported within parentheses or at the end of a sentence. Note that there are no periods used in these abbreviations. Also note that when one or more statistics interrupt the sentence to provide supporting information, these statistics are placed within parentheses to separate them from the rest of the sentence. When the statistical information is included at the end of the sentence, then this material is separated by a comma, and the parentheses are not typically used. Correlations When correlations are reported, we need to know the sample size used to compute the correlation (which is not the same as the general sample size when there is missing data). When there are more than a few correlations, they are often displayed in a correlation matrix, which is a structured table, rather than being (laboriously) listed within the text. When correlations are listed in text, it is typical to include the degrees of freedom (n-2) and the significance level, expressed as an exact probability (or p-value). When correlations are listed in tables, one or more asterisks are often used to flag correlations significant at noted signficance levels (e.g., for p < .05, for p < .01). It is typical to present means and standard deviations with just about every statistical analysis, so if these descriptive statistics have not already been reported in the results section, it is typical to include them. Examples The correlation of peer reports (M = 4.2, SD = 2.1, N = 367) and self reports (M = 5.8, SD = 2.3) of victimization was highly significant, r(365) = .32, p = .008. \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| Table 1: Intercorrelations between measures of victimization \| \| \| \| \| \| \| \| \| \| \| \| Measure \| 1 \| 2 \| 3 \| 4 \| 5 \| \| 1. Peer (Schwartz et al, 1997) \| — \| .80 \| .21 \| .26 \| .34 \| \| 2. Peer (Perry et al., 1988) \| \| — \| .32 \| .21 \| .22 \| \| 3. Self report \| \| \| — \| .34 \| .07 \| \| 4. Diary \| \| \| \| — \| .08 \| \| 5. Observer \| \| \| \| \| — \| \| \| \| \| \| \| p < .05, p < .01 Adapted from Table 2 of Pellegrini, A. D. & Bartini, M. (2000). An empirical comparison of methods of sampling aggression and victimization in school settings. Journal of Educational Psychology, 92, ???-???. \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| Table 1: Intercorrelations between measures of victimization \| \| \| \| \| \| \| \| \| \| \| \| Measure \| 1 \| 2 \| 3 \| 4 \| 5 \| \| 1. Peer (Schwartz et al, 1997) \| — \| .80 \| .21 \| .26 \| .34 \| \| 2. Peer (Perry et al., 1988) \| \| — \| .32 \| .21 \| .22 \| \| 3. Self report \| \| \| — \| .34 \| .07 \| \| 4. Diary \| \| \| \| — \| .08 \| \| 5. Observer \| \| \| \| \| — \| \| \| \| \| \| \| p < .05, p < .01 Adapted from Table 2 of Pellegrini, A. D. & Bartini, M. (2000). An empirical comparison of methods of sampling aggression and victimization in school settings. Journal of Educational Psychology, 92, ???-???. \| \| \| \| \| \| \| \| \| \| \| Regression Regression is often reported to characterize the degree of linear relationship between one or more predictor variables and a criterion variable; thus, the standardized regression weights (betas) and their associated probabilities (p-values) are of primary importance because the beta-weights allow one to compare the strength of each predictor. In other contexts, though, the primary emphasis is on making predictions for individuals not represented in the data, in which case unstandardized regressions weights are to be preferred because they can be used with unstandardized variables. The multiple correlation coefficient (R2), which describes the overal proportion of variance in the criterion that can be explained by the linear regression equation, is reported to assess the regression equation overall in a more global sense than the individual beta-weights. It is important to note, however, that there is no clear concensus in the literature about the exact specifics on presenting regression. Examples A linear regression analysis revealed that social skills was a highly significant predictor of aggression scores (β = .40, p = .008), accounting for 16% of the variance in aggressive behavior. Achievement test scores were regressed on class size and number of writing assignments. These two predictors accounted for just under half of the variance in test scores (R2 = .49), which was highly significant, F(2,289) = 12.5, p=.005. Both the writing assignment (β = .46, p=.001) and the class size (β=.28, p = .014) demonstrated significant effects on the achievement scores. t Tests There are several different research designs that utilize a t-test for the statistical inference testing. The differences between one-sample t-tests, related measures t-tests, and independent samples t-tests are so clear to the knowledgeable reader that most journal editors eliminate the elaboration of which type of t-test has been used. Additionally, the descriptive statistics provided will identify further which variation was employed. It is important to note that we assume that all p-values represent two-tailed tests unless otherwise noted and that independent samples t-tests use the pooled variance approach (based on an equal variances assumption) unless otherwise noted. Examples The 36 study participants had a mean age of 27.4 (SD = 12.6) and were significantly older than the university norm of 21.2 years, t(35) = 2.95, p = 0.01. The 25 participants had an average difference from pre-test to post-test anxiety scores of -4.8 (SD = 5.5), indicating the anxiety treatment resulted in a highly significant decrease in anxiety levels, t(24) = -4.36, p = .005 (one-tailed). The 36 participants in the treatment group (M = 14.8, SD = 2.0) and the 25 participants in the control group (M = 16.6, SD = 2.5), demonstrated a significant difference in performance (t = -3.12, p = .01); as expected, the visual priming treatment inhibited performance on the phoneme recognition task. ANOVA tests The results of both one-way (one factor) ANOVAs and multi-way (more than one-factor) ANOVAs are reported with the same format and same descriptive statistics. The only difference is that for one-way ANOVA models, we only have the effects of one factor to report, but for multi-way ANOVA models, we need to report the effect of each main effect and all interaction effects included in the modeled analyses. Despite the practice of many journal editors and authors of excluding the non-significant effects, the sixth edition requires these effects to be reported and substantiated regardless of the significance status. We need to report the observed F-ratio, the numerator and denominator degrees of freedom, and the exact p-value. Additionally, we need means, standard deviations, and sample sizes for each cell (i.e., condition) in the study as the supporting descriptive statistics. From this information, we can confirm the ANOVA computations. Examples The 12 participants in the high dosage group had an average reaction time of 12.3 seconds (SD = 4.1); the 9 participants in the moderate dosage group had an average reaction time of 7.4 seconds (SD = 2.3), and the 8 participants in the control group had a mean of 6.6 (SD = 3.1). The effect of dosage, therefore, was highly significant, F(2,26) = 8.76, p=.012. The cell sizes, means, and standard deviations for the 3×4 factorial design are presented in Table 1. The main effect of Dosage was marginally significant (F[2,17] = 3.23, p = .067), as was the main effect of diagnosis category, F(3,17) = 2.87, p = .097. The interaction of dosage and diagnosis, however, has highly significant, F(6,17) = 14.2, p ≤ .0005. Chi-Square tests The results of all chi-square tests are reported in a similar way. The degrees-of-freedom are identified, with the sample size, within parentheses, and the p-value should be reported precisely as noted above. The descriptive statistics necessary to support the chi-square test vary according to which specific test was performed, but the frequencies of each category or combination of categories are typically sufficient. For instance, for the chi-square test of fixed proportions, we need to know the frequencies of each category. For the chi-square test of independence (of two categorical variables), we need to know the frequencies in the cross tabulation. Examples The sample included 30 respondents who had never married, 54 who were married, 26 who reported being separated or divorced, and 16 who were widowed. These frequencies were significantly different, χ2 (3, N = 126) = 10.1, p = .017. As can be seen by the frequencies cross tabulated in Table 1, there is a highly significant relationship between marital status and depression, χ2 (3, N = 126) = 24.7, p ≤ .0005. © 2025 Illinois State University, Normal, IL USA Privacy Statement \| Appropriate Use Policy \| Accessibility
4544	http://www.rod.beavon.org.uk/mixedind.htm	mixed ind Determination of sodium hydroxide and sodium carbonate in the presence of one another. If sodium carbonate is titrated with hydrochloric acid using methyl orange as the indicator, the titre is twice what it is if phenolphthalein is used as the indicator. The reason is that the two indicators do not change colour at the same pH. Methyl orange changes at about pH 3 – 4; this corresponds to the complete neutralisation of sodium carbonate with two moles of hydrochloric acid: Na 2 CO 3 + 2HCl à 2NaCl + H 2 O + CO 2. On the other hand phenolphthalein changes at pH 8.5 – 10; this colour change corresponds to one mole of hydrochloric acid reacting with sodium carbonate: Na 2 CO 3 + HCl à NaHCO 3 + NaCl. The titration curve for sodium carbonate has two vertical portions, corresponding to the two reactions given above. Sodium hydroxide has only one, and gives the same titre with both indicators. The result is that, if a mixture of sodium carbonate and sodium hydroxide is titrated with hydrochloric acid using phenolphthalein, the volume corresponds to the sodium hydroxide plus half the sodium carbonate. Addition of methyl orange to this mixture and further titration gives the volume of acid corresponding to half the sodium carbonate. Method: The mixed alkali solution used for analysis should be about 0.5 mol dm-3 in sodium hydroxide and 0.25 mol dm-3 in sodium carbonate. 1 Pipette 25.0 cm 3 of the mixed alkali solution into a 250 cm 3 graduated flask, make to the mark with pure water, and mix thoroughly. 2Pipette 25.0 cm 3 of this diluted solution into a 250 cm 3 conical flask, add 3 – 4 drops of phenolphthalein indicator, and titrate with standard 0.100 mol dm-3 hydrochloric acid until the pink colour is just discharged. Note the volume required. 2 Add to the same mixture 3 – 4 drops of methyl orange indicator (or screened methyl orange if preferred) and continue titration until the methyl orange changes from yellow to red. 3 Repeat to give three consistent sets of titres. Results: Final volume (methyl orange)/cm 3: Initial volume/cm 3: Titre (methyl orange)/cm 3: Final vol (phenolphthalein)/cm 3: Initial volume/cm 3: Titre (phenolphthalein)/cm 3: Mean titre (phenolphthalein)/cm 3: Mean titre (methyl orange)/cm 3: Calculation: If the first titre (phenolphthalein), x cm 3, represents the hydroxide plus half the carbonate, and the difference between the phenolphthalein and methyl orange titres, y cm 3, represents half the carbonate, then the hydroxide is equivalent to (x – y) cm 3 of acid, and the carbonate to 2 y cm 3 of acid. Find the concentration of each of the compounds in mol dm-3. Practicals IndexHome Page
4545	https://blog.powerscore.com/lsat/negating-compound-and-conditional-statements/	LSAT and Law School Admissions Blog Negating Compound and Conditional Statements The ability to logically negate a statement—whether conditional, causal, etc.—is critical to your success on the LSAT. It comes up most commonly in the Logical Reasoning section of the test, although any question stem using the word “EXCEPT” (always capitalized) will require you to logically negate that stem. The list does not stop here. Every time you apply the contrapositive of a conditional statement, you will need to reverse and negate the two conditions that constitute that statement. This is relevant to Must Be True, Justify, and Parallel Reasoning questions mostly, but can also be critical in other sections of the test. Negating statements is also useful in Assumption questions because proving the correct answer choice requires the application of the Assumption Negation Technique: the correct answer choice, when logically negated, must weaken the conclusion of the argument. And, of course, the ability to understand the logical opposite of a conditional statement will be directly relevant to many Cannot Be True questions, where the correct answer choice is the one that can be disproven using the information contained in the stimulus. Negating a Statement As described in The PowerScore Logical Reasoning Bible (and in Lesson 2 of our Full Length LSAT course), negating a statement consists of creating the logical opposite of that statement, i.e. a statement that denies the truth of the original. The logical opposite of Jon is wise is Jon is not wise. The affirmative and negative counterparts of the oppositional construct are mutually exhaustive as well as mutually exclusive. The same is true of quantified expressions. For instance, the logical opposite of I gained more than 20 points on my latest practice test is I did not gain more than 20 points on my latest practice test. How many points did you actually gain? The only possibility precluded is the possibility of gaining more than 20 points. You could have gained exactly 20 points, or maybe you did not gain any. In quantity constructs, all/not all, some/none, most/not most are logically proper oppositional constructs. By contrast, a construct such as All pleasure is good/No pleasure is good does not contain a logical opposition, because these opposites do not mutually exhaust the domain. Understand the Implications Formulating logical opposites may seem mechanistic, and in many ways it is. Nevertheless, it is important to understand the implications of your oppositional construct. Take, for instance, the logical opposite of I believe that stealing is moral. The correct form of the logical opposite (I do not believe that stealing is moral) amounts to a negation of the embedded clause (I believe that stealing is not moral). Similarly, when logically negating a statement such as It is rational not to eat food that is poisonous, consider the implications of the double negative construct (It is not rational not to eat food that is poisonous) and immediately simplify it: It is rational to eat food that is poisonous. Negative statements are generally less informative than affirmatives; they are more complex and harder to process. Simplification is key! Conditional Statements What happens when you need to negate a conditional statement? In that case, you need to show that the sufficient condition can occur even if the necessary condition does not occur, i.e. that the necessary condition is not, in fact, necessary. So, the logical opposite of “All that glitters is gold” is simply “All that glitters is not gold.” To put it another way, some things may glitter even if they are not gold, because not everything that glitters is gold. Quite often, the logical opposite of a conditional statement is formed using the phrase “even if.” The logical opposite of the statement Unless you practice, you will not succeed is You can succeed even if you do not practice. This is because the original statement posits practicing as a necessary condition for success (Succeed –> Practice). To contradict this statement, you need to show that practice is not a necessary condition for success. Note that even if is not, by itself, an indicator of a sufficient condition: it merely states that the lack of practice does not prevent one from succeeding, not that the lack of practice somehow ensures success. Compound Statements Now, consider what happens when negating compound sentences, such as conjunctions of the form X and Y and disjunctions of the form X or Y (or both). To negate a conjunction such as I love you and you love me requires showing that the two clauses cannot be both true: it is simply not the case that we both love each other. Using De Morgan’s laws for interrelating conjunctions and disjunctions, this would mean one of three things: either I don’t love you, or else you don’t love me, or else neither of us loves the other. The either/or construction does not preclude the possibility of both propositions being true, unless specifically told otherwise (e.g. either X or Y, but not both). To negate a disjunction such as Jon is either a doctor or a lawyer requires showing that Jon is neither a doctor nor a lawyer (i.e. that Jon is not a doctor and that Jon is not a lawyer). At the most basic level, negating a conjunction (and) requires the use of or, whereas negating a disjunction (or)requires the use of and. If you wish to practice sentential negation of propositions that are both conditional and compound, try these out: Only those who are both romantic and cautious can truly fall in love. You are not a realist unless you either believe in miracles or in yourself. You are neither a realist nor a humanist if you believe in miracles. It is rational not to believe in miracles, unless you are either romantic or idealistic. If you’d like to run your answers by us, feel free to use the Submit Comment button below. Good luck! arindom Could you please check my answers? 1) Even if you fall in love, you are not a romantic or you are not cautious. 2) Even if you are a realist, you do not believe in miracles and you do not believe in yourself. 3) Even if you believe in miracles, you are either a realist or a humanist. 4) Even if it is rational to believe in miracles, you are not a realist and not an idealist. Thanks. – Arindom Nicolay Siclunov Hi Arindom, Thanks for taking a stab at my quiz! However, your answers need to be tweaked a little. Take a look at the answer key: Only those who are both romantic and cautious can truly fall in love. (Fall in Love –> Romantic AND Cautious) Negates to:To fall in love, you don’t need to be both romantic and cautious. In other words, it is possible to fall in love even if you are not romantic or cautious. It may even be possible to fall in love if you are neither romantic nor cautious. The logical opposite of the statement would suggest that romance and caution are not necessary conditions for falling in love. You are not a realist unless you either believe in miracles or in yourself. (Realist –> Believe in miracles OR Believe in yourself) Negates to: You can be a realist even if you don’t believe in miracles AND you don’t believe in yourself. In other words, the sufficient condition can occur (being a realist) even if neither of the two necessary conditions were to occur. You are neither a realist nor a humanist if you believe in miracles. (Believe in miracles –> NOT Realist AND NOT Humanist) Negates to: It is possible to believe in miracles even if you are a realist or a humanist. It is also possible to believe in miracles if you are both. It is rational not to believe in miracles, unless you are either romantic or idealistic. (Rational to believe in miracles –> Romantic OR Idealistic) Negates to: It’s rational to believe in miracles even if you are neither romantic nor idealistic (i.e. NOT romantic AND NOT idealistic). In other words, you don’t have to be either romantic or idealistic in order to believe in miracles. Remember – when negating conditional statements, you need to show that the sufficient condition can occur even in the absence of the necessary condition. Leave a Reply Cancel reply
4546	https://pubmed.ncbi.nlm.nih.gov/33569613/	A systematic approach to imaging the pelvis in amenorrhea - 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. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. To: Subject: Body: Format: [x] MeSH and other data Send email Cancel Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Add to My Bibliography My Bibliography Unable to load your delegates due to an error Please try again Add Cancel Your saved search Name of saved search: Search terms: Test search terms Would you like email updates of new search results? Saved Search Alert Radio Buttons Yes No Email: (change) Frequency: Which day? Which day? Report format: Send at most: [x] Send even when there aren't any new results Optional text in email: Save Cancel Create a file for external citation management software Create file Cancel Your RSS Feed Name of RSS Feed: Number of items displayed: Create RSS Cancel RSS Link Copy Full text links Springer Full text links Actions Cite Collections Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Permalink Permalink Copy Display options Display options Format Page navigation Title & authors Abstract Similar articles References Publication types MeSH terms Related information LinkOut - more resources Review Abdom Radiol (NY) Actions Search in PubMed Search in NLM Catalog Add to Search . 2021 Jul;46(7):3326-3341. doi: 10.1007/s00261-021-02961-9. Epub 2021 Feb 10. A systematic approach to imaging the pelvis in amenorrhea Sze Yiun Teo1,Chiou Li Ong2 Affiliations Expand Affiliations 1 Department of Diagnostic and Interventional Imaging, KK Women's and Children's Hospital, Singapore, 229899, Republic of Singapore. teo.sze.yiun@singhealth.com.sg. 2 Department of Diagnostic and Interventional Imaging, KK Women's and Children's Hospital, Singapore, 229899, Republic of Singapore. PMID: 33569613 DOI: 10.1007/s00261-021-02961-9 Item in Clipboard Review A systematic approach to imaging the pelvis in amenorrhea Sze Yiun Teo et al. Abdom Radiol (NY).2021 Jul. Show details Display options Display options Format Abdom Radiol (NY) Actions Search in PubMed Search in NLM Catalog Add to Search . 2021 Jul;46(7):3326-3341. doi: 10.1007/s00261-021-02961-9. Epub 2021 Feb 10. Authors Sze Yiun Teo1,Chiou Li Ong2 Affiliations 1 Department of Diagnostic and Interventional Imaging, KK Women's and Children's Hospital, Singapore, 229899, Republic of Singapore. teo.sze.yiun@singhealth.com.sg. 2 Department of Diagnostic and Interventional Imaging, KK Women's and Children's Hospital, Singapore, 229899, Republic of Singapore. PMID: 33569613 DOI: 10.1007/s00261-021-02961-9 Item in Clipboard Full text links Cite Display options Display options Format Abstract This is a pictorial review on the radiological approach to patients with amenorrhea using a level-based framework. The prevalence of amenorrhea is 3 to 4% with wide-ranging causes involving multiple clinical disciplines. Normal menstruation depends on complex coordinated hormonal functions of the hypothalamic-pituitary-ovarian axis exerting its effect on an intact uterine end-organ and outflow tract. A disruption of any of these factors may result in amenorrhea. Categorizing the causes of primary and secondary amenorrhea into uterine, ovarian/gonadal, and intracranial levels provides a logical framework for its evaluation. A systematic level-based approach by targeted ultrasound of the pelvic structures is suggested, with different aims in primary versus secondary amenorrhea. Pelvic sonographic findings of various conditions within the uterine and ovarian/gonadal levels are illustrated. Conditions due to an intracranial cause result in downstream effects on the uterus and ovaries and can often be suspected based on a combination of clinical assessment, ultrasound findings, and laboratory investigations. By correlating pelvic ultrasound findings with underlying pathology, the clinical radiologist is able to provide useful diagnostic information in the management of these patients. Keywords: Amenorrhea; Primary amenorrhea; Secondary amenorrhea; Ultrasound. PubMed Disclaimer Similar articles Sonographic imaging of the paediatric female pelvis.Ziereisen F, Guissard G, Damry N, Avni EF.Ziereisen F, et al.Eur Radiol. 2005 Jul;15(7):1296-309. doi: 10.1007/s00330-005-2648-6. Epub 2005 Mar 9.Eur Radiol. 2005.PMID: 15756554 Uterine size and ovarian size in adolescents with functional hypothalamic amenorrhoea.Bumbuliene Z, Klimasenko J, Sragyte D, Zakareviciene J, Drasutiene G.Bumbuliene Z, et al.Arch Dis Child. 2015 Oct;100(10):948-51. doi: 10.1136/archdischild-2014-307504. Epub 2015 Jul 15.Arch Dis Child. 2015.PMID: 26177656 Use of pelvic ultrasound to monitor ovarian and uterine maturity in childhood onset anorexia nervosa.Lai KY, de Bruyn R, Lask B, Bryant-Waugh R, Hankins M.Lai KY, et al.Arch Dis Child. 1994 Sep;71(3):228-31. doi: 10.1136/adc.71.3.228.Arch Dis Child. 1994.PMID: 7979496 Free PMC article. Sonography of the pelvis in patients with primary amenorrhea.Rosenberg HK.Rosenberg HK.Endocrinol Metab Clin North Am. 2009 Dec;38(4):739-60. doi: 10.1016/j.ecl.2009.09.002.Endocrinol Metab Clin North Am. 2009.PMID: 19944290 Review. US of the pediatric female pelvis: a clinical perspective.Garel L, Dubois J, Grignon A, Filiatrault D, Van Vliet G.Garel L, et al.Radiographics. 2001 Nov-Dec;21(6):1393-407. doi: 10.1148/radiographics.21.6.g01nv041393.Radiographics. 2001.PMID: 11706212 Review. See all similar articles References The practice committee of the American Society for Reproductive Medicine (2006) Current evaluation of amenorrhea. Fertil Steril 86 (5 Suppl 1):S148-55. - DOI Speroff L, Fritz MA (2005) Amenorrhea. In: Speroff L, Fritz MA (ed) Clinical gynecologic endocrinology and infertility, 7 th edn. Philadelphia, USA: Lippincott Williams & Wilkins, pp 401-464 Garel L, Dubois J, Grignon A, Filiatrault D, Van Vliet G (2001) US of the pediatric female pelvis: a clinical perspective. Radiographics 21(6):1393-407. - DOI - PubMed Siegel MJ (1991) Pediatric gynecologic sonography. Radiology 179(3):593-600. - DOI - PubMed Langer JE, Oliver ER, Lev-Toaff AS, Coleman BG (2012) Imaging of the female pelvis through the life cycle. Radiographics 32(6):1575-97. - DOI - PubMed Show all 39 references Publication types Review Actions Search in PubMed Search in MeSH Add to Search MeSH terms Amenorrhea / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Ovary / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Pelvis / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Ultrasonography Actions Search in PubMed Search in MeSH Add to Search Uterus / diagnostic imaging Actions Search in PubMed Search in MeSH Add to Search Related information MedGen LinkOut - more resources Full Text Sources Springer Full text links[x] Springer [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
4547	http://www.cntronics.com/kbupload/kb_1374149961_0.pdf	1 摄氏、华氏温度对照表 Contrast Contrast Contrast Contrast between between between between Celsius Celsius Celsius Celsius Temperatures Temperatures Temperatures Temperatures and and and and Fahrenheit Fahrenheit Fahrenheit Fahrenheit Temperatures Temperatures Temperatures Temperatures 摄氏温度与华氏温度的最典型的换算式是： 5 5 5 5（x x x xº º º ºF-F-F-F- 50 50 50 50）=9( =9( =9( =9(y y y yº º º ºC-10) C-10) C-10) C-10) 其中x 代表华氏温度的值，y 代表摄氏温度的值。摄氏度 ºC 华氏度 ºF 摄氏度 ºC 华氏度 ºF 华氏度 ºF 摄氏度 ºC 华氏度 ºF 摄氏度 ºC 华氏度 ºF 摄氏度 ºC 华氏度 ºF 摄氏度 ºC -10 14.00 15 59.00 11 -11.67 36 2.22 61 16.11 86 30.00 -9 15.80 16 60.80 12 -11.11 37 2.78 62 16.67 87 30.56 -8 17.60 17 62.60 13 -10.56 38 3.33 63 17.22 88 31.11 -7 19.40 18 64.40 14 -10.00 39 3.89 64 17.78 89 31.67 -6 21.20 19 66.20 15 -9.44 40 4.44 65 18.33 90 32.22 -5 23.00 20 68.00 16 -8.89 41 5.00 66 18.89 91 32.78 -4 24.80 21 69.80 17 -8.33 42 5.56 67 19.44 92 33.33 -3 26.60 22 71.60 18 -7.78 43 6.11 68 20.00 93 33.89 -2 28.40 23 73.40 19 -7.22 44 6.67 69 20.56 94 34.44 -1 30.20 24 75.20 20 -6.67 45 7.22 70 21.11 95 35.00 0 32.00 25 77.00 21 -6.11 46 7.78 71 21.67 96 35.56 1 33.80 26 78.80 22 -5.56 47 8.33 72 22.22 97 36.11 2 35.60 27 80.60 23 -5.00 48 8.89 73 22.78 98 36.67 3 37.40 28 82.40 24 -4.44 49 9.44 74 23.33 99 37.22 4 39.20 29 84.20 25 -3.89 50 10.00 75 23.89 100 37.78 5 41.00 30 86.00 26 -3.33 51 10.56 76 24.44 101 38.33 6 42.80 31 87.80 27 -2.78 52 11.11 77 25.00 102 38.89 7 44.60 32 89.60 28 -2.22 53 11.67 78 25.56 103 39.44 8 46.40 33 91.40 29 -1.67 54 12.22 79 26.11 104 40.00 9 48.20 34 93.20 30 -1.11 55 12.78 80 26.67 105 40.56 10 50.00 35 95.00 31 -0.56 56 13.33 81 27.22 106 41.11 11 51.80 36 96.80 32 0.00 57 13.89 82 27.78 107 41.67 12 53.60 37 98.60 33 0.56 58 14.44 83 28.33 108 42.22 13 55.40 38 100.4 0 34 1.11 59 15.00 84 28.89 109 42.78 14 57.20 39 102.2 0 35 1.67 60 15.56 85 29.44 110 43.33 2 目前使用的温标主要有摄氏温度(Celsius (Celsius (Celsius (Celsius temperature temperature temperature temperature，记号t t t t，单位℃) ) ) )、热力学温度 (Thermodynamic (Thermodynamic (Thermodynamic (Thermodynamic temperature temperature temperature temperature，记号T T T T，单位K) K) K) K)、华氏温度(Fahrenheit (Fahrenheit (Fahrenheit (Fahrenheit temperature temperature temperature temperature，记号tF tF tF tF，单位℉) ) ) )、兰氏温度(Rankine (Rankine (Rankine (Rankine temperature temperature temperature temperature，记号TR TR TR TR，单位° ° ° °R) R) R) R)和雷氏温度(R (R (R (Ré é é éa a a aumur umur umur umur temperature temperature temperature temperature，记号tR tR tR tRé é é é，单位° ° ° °R R R Ré é é é) ) ) )，它们的相互换算关系如下： a a a a ℃=(4/5) =(4/5) =(4/5) =(4/5) a a a a ° ° ° °R R R Ré é é é = = = = [(9/5) [(9/5) [(9/5) [(9/5) a a a a +32] +32] +32] +32] ℉ b b b b ° ° ° °R R R Ré é é é = = = = (5/4) (5/4) (5/4) (5/4) b b b b ℃= = = = [(9/4) [(9/4) [(9/4) [(9/4) b b b b + + + + 32] 32] 32] 32] ℉ c c c c ℉= = = = (5/9)(c (5/9)(c (5/9)(c (5/9)(c ---- 32) 32) 32) 32)℃= = = = (4/9) (4/9) (4/9) (4/9) (c (c (c (c ---- 32) 32) 32) 32)° ° ° °R R R Ré é é é d d d d ℃= = = = (d (d (d (d + + + + 273.15) 273.15) 273.15) 273.15) K K K K e e e e K K K K = = = = (e (e (e (e ---- 273.15) 273.15) 273.15) 273.15)℃= = = = [1.80 [1.80 [1.80 [1.80× × × ×(e (e (e (e ---- 273.15) 273.15) 273.15) 273.15) + + + + 32] 32] 32] 32]℉= = = = (9/5) (9/5) (9/5) (9/5) e e e e ° ° ° °R R R R Formulas: C = (F - 32) 5 / 9, F = (C 9 / 5) + 32 Fahrenheit Celsius -47.2 -44 -45.4 -43 -43.6 -42 -41.8 -41 -40 -40 -38.2 -39 -36.4 -38 -34.6 -37 -32.8 -36 -31 -35 -29.2 -34 -27.4 -33 -25.6 -32 -23.8 -31 -22 -30 -20.2 -29 -18.4 -29 -16.6 -27 -14.8 -26 -13 -25 -11.2 -24 -9.4 -23 Fahrenheit Celsius -7.6 -22 -5.8 -21 -4 -20 -2.2 -19 -0.4 -18 1.4 -17 3.2 -16 5 -15 6.8 -14 8.6 -13 10.4 -12 12.2 -11 14 -10 15.8 -9 17.6 -8 19.4 -7 21.2 -6 23 -5 24.8 -4 26.6 -3 28.4 -2 30.2 -1 Fahrenheit Celsius 32 0 33.8 1 35.6 2 37.4 3 39.2 4 41 5 42.8 6 44.6 7 46.4 8 48.2 9 50 10 51.8 11 53.6 12 55.4 13 57.2 14 59 15 60.8 16 62.6 17 64.4 18 66.2 19 68 20 69.8 21 Fahrenheit Celsius 71.6 22 73.4 23 75.2 24 77 25 78.8 26 80.6 27 82.4 28 84.2 29 86 30 87.8 31 89.6 32 91.4 33 93.2 34 95 35 96.8 36 98.6 37 100.4 38 102.2 39 104 40 105.8 41 107.6 42 109.4 43
4548	https://thirdspacelearning.com/us/math-resources/topic-guides/geometry/surface-area-of-a-cube/	High Impact Tutoring Built By Math Experts Personalized standards-aligned one-on-one math tutoring for schools and districts Request a demo In order to access this I need to be confident with: Area of a square 2D shapes 3D shapes What is the surface area of a cube? Common Core State Standards How to calculate the surface area of a cube Surface area of a cube examples Example 1: surface area of a cube (integer side lengths) Example 2: surface area of a cube (one known edge of the cube) Example 3: surface area of a cube (worded problem) Example 4: surface area of a cube (area of a face given) Example 5: find the length of a cube given the surface area Example 6: find the length of a cube given the surface area (decimal solution) Teaching tips for surface area of a cube Easy mistakes to make Related surface area lessons Practice surface area of a cube questions Surface area of a cube FAQs Next lessons Still stuck? Math resources Geometry Surface area Surface area of a cube Surface area of a cube Here you will learn about the surface area of a cube, including how to calculate the surface area of a cube and how to find missing values of a cube given its surface area. Students will first learn about the surface area of a cube as a part of geometry in 8 th grade. What is the surface area of a cube? The surface area of a cube (cuboid) is the sum of the areas of all the faces of a cube. This is also referred to as the total surface area of a cube (tsa) or the lateral surface area of a cube. A cube is a three-dimensional figure that has six congruent square faces. This means that all the side faces are the same size. To find the area of each face, multiply the side lengths together. Then multiply the area of each of the square faces by six. The formula to calculate the surface area, S, of a cube is S=6x^{2} where x represents the side length (edge length) of the cube. The surface area of a cube formula can be used to find the surface area of any cube. Surface area is measured in square units, for example mm^{2}, \, cm^{2} or \, m^{2}. What is the surface area of a cube? Common Core State Standards How does this relate to 8 th grade math? Grade 8: Geometry (8.G.C.9)Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. [FREE] Surface Area Worksheet (Grade 6) Use this quiz to check your grade 6 students’ understanding of surface area. 10+ questions with answers covering a range of 6th grade surface area topics to identify areas of strength and support! DOWNLOAD FREE x [FREE] Surface Area Worksheet (Grade 6) Use this quiz to check your grade 6 students’ understanding of surface area. 10+ questions with answers covering a range of 6th grade surface area topics to identify areas of strength and support! DOWNLOAD FREE How to calculate the surface area of a cube In order to calculate the surface area of a cube: Write the formula for the surface area of the cube. Substitute any known value(s) into the formula. Complete the calculation. Write the solution, including the units. Surface area of a cube examples Example 1: surface area of a cube (integer side lengths) Find the surface area of the cube below. Write the formula for the surface area of the cube. S=6x^{2} 2Substitute any known value(s) into the formula. Here, x=5 and so S=6\times{5}^{2}. 3Complete the calculation. S=6\times{5}^{2}=6\times{25}=150 4Write the solution, including the units. As the unit of length is centimeters (cm), the unit of area is square centimeters (cm^{2}). S=150 \, cm^{2} Example 2: surface area of a cube (one known edge of the cube) Find the surface area of the cube. Write the formula for the surface area of the cube. S=6x^{2} Substitute any known value(s) into the formula. Substituting x=6 into the formula, we have S=6\times{6}^{2}. Complete the calculation. S=6\times{6}^{2}=6\times{36}=216 Write the solution, including the units. As the unit of length is centimeters (cm), the unit of area is square centimeters (cm^{2}). S=216 \, cm^{2} Example 3: surface area of a cube (worded problem) A cube structure has a side length of 7 \, m. Calculate the total surface area of the structure. Write the formula for the surface area of the cube. S=6x^{2} Substitute any known value(s) into the formula. Substituting x=7 into the formula, S=6\times{7}^{2}. Complete the calculation. S=6\times{7}^{2}=6\times{49}=294 Write the solution, including the units. As the unit of length is meters (m), the unit of area is square meters (m^{2}). S=294 \, m^{2} Example 4: surface area of a cube (area of a face given) The area of the face of a cube is 30 \, cm^{2}. Find the surface area of the cube. Write the formula for the surface area of the cube. S=6x^{2} Substitute any known value(s) into the formula. Knowing the area of one face of the cube, you can express this as x^2=30 as x is the side length of the cube, and you know the area, x^2. Substituting x^2=30 into the formula, you have S=6\times{30}. Complete the calculation. S=6\times{30}=180 Write the solution, including the units. As the unit of area is square centimeters (cm^{2}), you can use this in the solution. S=180 \, cm^{2} Example 5: find the length of a cube given the surface area The surface area of a cube is 24 \, ft^{2}. Find the length of the cube. Write the formula for the surface area of the cube. S=6x^{2} Substitute any known value(s) into the formula. Here you know that S=24 and so substituting this into the formula, 24=6\times{x}^{2}. Complete the calculation. To complete the calculation, divide both sides by 6 first, and then square root both sides to find x. \begin{aligned}24\div{6}&=x^{2} \\ 4&=x^{2} \\ x&=2 \end{aligned} Write the solution, including the units. As the unit of area is square feet (ft^2), the unit length will be in feet. x=2 \, ft Example 6: find the length of a cube given the surface area (decimal solution) The surface area of a cube is 483 \, mm^{2}. Find the length of the side x correct to 2 decimal places. Write the formula for the surface area of the cube. S=6x^{2} Substitute any known value(s) into the formula. As you know the surface area, you can substitute S=483 into the formula 483=6\times{x}^{2}. Complete the calculation. To complete the calculation, divide both sides by 6 first, and then square root both sides to find x. \begin{aligned}483\div{6}&=x^{2} \\ 80.5&=x^{2} \\ \sqrt{80.5}&=x \\ x&=8.972179222463… \end{aligned} Write the solution, including the units. x=8.97 \mathrm{~cm} \, (1 d p). Teaching tips for surface area of a cube When introducing the surface area of a cube, use a math net to break down the cube into individual cubes, showing that each side of a cube is a square. Allow students time to work through a variety of problems, on worksheets or using a technology tool. Students should have time to work independently, as well as with other students, to really reinforce learning and understanding. Use visuals and real life examples of cubes, such as a rubik’s cube, to allow students to physically hold and explore cubes. Easy mistakes to make Missing/incorrect units You should always include units in your answer. Surface area is measured in square units (for example, \mathrm{mm}^2, \mathrm{~cm}^2, \mathrm{~m}^2 etc.). Calculating with different units You need to make sure all measurements are in the same units before calculating the surface area. For example, you can’t have some in cm and some in m. Calculating volume instead of surface area Volume and surface area are different quantities. The volume of the cube is the three-dimensional space in a shape and is measured in cubic units. Surface area is the amount of space covering the outside of a 3D shape. To find surface area, find the area of one face and multiply it by six. Related surface area lessons Math nets Surface area of a prism Surface area of a rectangular prism Surface area of a triangular prism Surface area of a pyramid Surface area of a cone Surface area of a cylinder Surface area of a sphere Surface area of a hemisphere Practice surface area of a cube questions Find the surface area of the cube. 27 \, cm^3 12 \, cm^2 36 \, cm^2 54 \, cm^2 S=6x^{2} where x=3 \, cm. S=6\times{3}^{2}=6\times{9}=54 \, cm^{2}. Calculate the surface area of the cube below. Write your answer in square centimeters. 0.125 \, cm^3 15,000 \, cm^2 6 \, cm^2 1.5 \, cm^2 S=6x^{2} where x=0.5 \mathrm{~m}=50 \mathrm{~cm}. S=6\times{50}^{2}=6\times{2500}=15,000 \, cm^{2}. Find the surface area of the cube. Give your answer in cm^2. 96 \, cm^2 9,600 \, cm^2 960 \, cm^2 9.6 \, cm^2 S=6x^{2} where x=40 \mathrm{~cm}=0.4 \mathrm{~m}. S=6\times{40}^{2}=6\times{1600}=9,600 \, cm^{2}. The surface area of a cube is 150 \, cm^2. Find the length of the side of the cube. 135,000 \, cm 2.04 \, cm 5 \, cm 12.5 \, cm S=6x^{2} where S=150 \, cm^{2}. \begin{aligned}150&=6\times{x}^{2} \\ 25&=x^{2} \\ x&=\sqrt{25} \\ x&=5cm \end{aligned} The surface area of the cube is 6 \, m^2. Find the length of each side x. 2 \, m 1 \, m 0.3 \, m 0.41 \, m S=6x^{2} where S=6 \, m^{2}. \begin{aligned}6&=6\times{x}^{2} \\ 6\div{6}&=x^{2} \\ x^{2}&=1 \\ x&=\sqrt{1} \\ x&=1m \end{aligned} The surface area of a cube is 186 \, m^2. Find the length of each side. Write your answer to the nearest centimeter. 5.57 \, m 2.27 \, m 15.50 \, m 207,576 \, m S=6x^{2} where S=186 \, m^{2}. \begin{aligned}186&=6\times{x}^{2} \\ 31&=x^{2} \\ x&=\sqrt{31}=5.567764362830…=5.57 \, (2 d p). \end{aligned} Surface area of a cube FAQs What is a cube? A cube is a three-dimensional figure with six equal square faces and 8 vertices. What is the formula to find the volume of a cube? To find the volume of a cube, you will use the formula V=a^3 or V=\text { side } \times \text { side } \times \text { side }. The next lessons are Pythagorean theorem Congruence and similarity Transformations Still stuck? At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts. Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. Find out how we can help your students achieve success with our math tutoring programs. Introduction What is the surface area of a cube? Common Core State Standards How to calculate the surface area of a cube Surface area of a cube examples ↓ Example 1: surface area of a cube (integer side lengths) Example 2: surface area of a cube (one known edge of the cube) Example 3: surface area of a cube (worded problem) Example 4: surface area of a cube (area of a face given) Example 5: find the length of a cube given the surface area Example 6: find the length of a cube given the surface area (decimal solution) Teaching tips for surface area of a cube Easy mistakes to make Related surface area lessons Practice surface area of a cube questions Surface area of a cube FAQs Next lessons Still stuck? x [FREE] Common Core Practice Tests (3rd to 8th Grade) Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents. Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers! Download free
4549	https://m.fx361.com/news/2020/0511/6645860.html	Cloudflare Tunnel error \| m.fx361.com \| Cloudflare Please enable cookies. Error 1033 Ray ID: 9868aa350b15ae2f •2025-09-29 04:16:23 UTC Cloudflare Tunnel error What happened? You've requested a page on a website (m.fx361.com) that is on the Cloudflare network. The host (m.fx361.com) is configured as an Cloudflare Tunnel, and Cloudflare is currently unable to resolve it. What can I do? If you are a visitor of this website: Please try again in a few minutes. If you are the owner of this website: Ensure that cloudflared is running and can reach the network. You may wish to enable load balancing for your tunnel. Was this page helpful? Yes No Thank you for your feedback! Cloudflare Ray ID: 9868aa350b15ae2f• Your IP: Click to reveal 2600:1900:0:2106::401•Performance & security byCloudflare
4550	https://www.mdpi.com/1422-0067/24/23/16668	Previous Article in Journal Zn2+ Differentially Influences the Neutralisation of Heparins by HRG, Fibrinogen, and Fibronectin Previous Article in Special Issue Size-Exclusion Chromatography Combined with Ultrafiltration Efficiently Isolates Extracellular Vesicles from Human Blood Samples in Health and Disease All articles published by MDPI are made immediately available worldwide under an open access license. No special permission is required to reuse all or part of the article published by MDPI, including figures and tables. For articles published under an open access Creative Common CC BY license, any part of the article may be reused without permission provided that the original article is clearly cited. For more information, please refer to Feature papers represent the most advanced research with significant potential for high impact in the field. A Feature Paper should be a substantial original Article that involves several techniques or approaches, provides an outlook for future research directions and describes possible research applications. Feature papers are submitted upon individual invitation or recommendation by the scientific editors and must receive positive feedback from the reviewers. Editor’s Choice articles are based on recommendations by the scientific editors of MDPI journals from around the world. Editors select a small number of articles recently published in the journal that they believe will be particularly interesting to readers, or important in the respective research area. The aim is to provide a snapshot of some of the most exciting work published in the various research areas of the journal. Original Submission Date Received: . Journals IJMS Volume 24 Issue 23 10.3390/ijms242316668 Submit to this Journal Review for this Journal Propose a Special Issue ► ▼ Article Menu Article Menu Academic Editor Christopher Jackson Subscribe SciFeed Recommended Articles Related Info Links PubMed/Medline Google Scholar More by Authors Links on DOAJ Wolff, L. Horisberger, A. Moi, L. Karampetsou, M. P. Comte, D. on Google Scholar Wolff, L. Horisberger, A. Moi, L. Karampetsou, M. P. Comte, D. on PubMed Wolff, L. Horisberger, A. Moi, L. Karampetsou, M. P. Comte, D. /ajax/scifeed/subscribe Article Views 13128 Citations 25 Table of Contents Abstract Introduction Epidemiology Physiopathology Clinical Manifestations Treatments Conclusions Author Contributions Funding Data Availability Statement Conflicts of Interest References Altmetric share Share announcement Help format_quote Cite question_answer Discuss in SciProfiles Need Help? Support Find support for a specific problem in the support section of our website. Get Support Feedback Please let us know what you think of our products and services. Give Feedback Information Visit our dedicated information section to learn more about MDPI. Get Information clear JSmol Viewer clear first_page Download PDF settings Order Article Reprints Font Type: Arial Georgia Verdana Font Size: Aa Aa Aa Line Spacing:    Column Width:    Background: Open AccessReview Polyarteritis Nodosa: Old Disease, New Etiologies by Louis Wolff Louis Wolff SciProfilesScilitPreprints.orgGoogle Scholar 1, Alice Horisberger Alice Horisberger SciProfilesScilitPreprints.orgGoogle Scholar 2,3, Laura Moi Laura Moi SciProfilesScilitPreprints.orgGoogle Scholar 4, Maria P. Karampetsou Maria P. Karampetsou SciProfilesScilitPreprints.orgGoogle Scholar 5 and Denis Comte Denis Comte SciProfilesScilitPreprints.orgGoogle Scholar 6, 1 Department of Internal Medicine, Hôpital Universitaire de Bruxelles (H.U.B.), Université Libre de Bruxelles (ULB), 1050 Brussels, Belgium 2 Department of Medicine, Division of Rheumatology, Inflammation, and Immunity, Brigham and Women’s Hospital, Harvard Medical School, Boston, MA 02115, USA 3 Department of Medicine, Division of Immunology and Allergy, Lausanne University Hospital, University of Lausanne, 1011 Lausanne, Switzerland 4 Immunology and Allergology, Institut Central des Hôpitaux, Valais Hospital, 1951 Sion, Switzerland 5 Rheumatology Private Practice, 11635 Athens, Greece 6 Department of Medicine, Division of Internal Medicine, Lausanne University Hospital, University of Lausanne, 1005 Lausanne, Switzerland Author to whom correspondence should be addressed. Int. J. Mol. Sci. 2023, 24(23), 16668; Submission received: 13 October 2023 / Revised: 11 November 2023 / Accepted: 17 November 2023 / Published: 23 November 2023 (This article belongs to the Special Issue Key Players in Connective Tissue Diseases: The Immune System and Beyond) Download keyboard_arrow_down Download PDF Download PDF with Cover Download XML Download Epub Browse Figure Review Reports Versions Notes Abstract Polyarteritis nodosa (PAN), also known as panarteritis nodosa, represents a form of necrotizing vasculitis that predominantly affects medium-sized vessels, although it is not restricted to them and can also involve smaller vessels. The clinical presentation is heterogeneous and characterized by a significant number of patients exhibiting general symptoms, including asthenia, fever, and unintended weight loss. Although PAN can involve virtually any organ, it preferentially affects the skin, nervous system, and the gastrointestinal tract. Orchitis is a rare but specific manifestation of PAN. The absence of granulomas, glomerulonephritis, and anti-neutrophil cytoplasmic antibodies serves to distinguish PAN from other types of vasculitis. Major complications consist of hemorrhagic and thrombotic events occurring in mesenteric, cardiac, cerebral, and renal systems. Historically, PAN was frequently linked to hepatitis B virus (HBV) infection, but this association has dramatically changed in recent years due to declining HBV prevalence. Current epidemiological research often identifies a connection between PAN and genetic syndromes as well as neoplasia. This article provides a comprehensive review of PAN, specifically focusing on the progression of its clinical manifestations over time. Keywords: PAN; polyarteritis nodosa; panarteritis nodosa; monogenic; VEXAS; DADA2 1. Introduction PAN was first described in 1866 by Kussmaul and Maier. They reported an “intermittent nodular appearance affecting arteries throughout the body, sparing large vessels (the aorta and its branches), small vessels (arterioles, capillaries, and venules) and pulmonary vessels” . The distinct “pearl necklace” pattern led to the naming of this condition as polyarteritis nodosa. In 1952, pathologist Pearl Zeek laid out the first classification of PAN, distinguishing between hypersensitivity vasculitis, allergic vasculitis, PAN, and temporal arteritis, now, respectively, known as urticarial vasculitis, eosinophilic granulomatosis with polyangiitis (Churg–Strauss syndrome), PAN, and giant cell arteritis (Horton’s disease) . A pivotal development came in 1982 when autoantibodies directed against neutrophil cytoplasm antigens (ANCA) were identified in eight patients with clinical characteristics of vasculitis, introducing ANCA-associated vasculitis (AAV) as a distinct category of vasculitis separate from PAN . The most recent definition of PAN came from the 2012 Chapel Hill Consensus Conference (CHCC) where the disease was described as “necrotizing arteritis of the medium or small arteries without glomerulonephritis or vasculitis of arterioles, capillaries or venules and without ANCA” [4,5,6]. Over the past two decades, the medical community’s understanding of PAN has significantly evolved. While initially described as either primary vasculitis or Hepatitis B virus (HBV)-related, the current understanding recognizes PAN as a disease often secondary to genetic syndromes and malignant hematologic disorders . In this review, we examine the evolution of our understanding of PAN over the years by providing key aspects of epidemiology, pathophysiology, and clinical presentations and discussing how these translate into the current therapeutic approach. 2. Epidemiology The prevalence of PAN varies greatly across different countries, ranging from 2 to 31 per million inhabitants in Europe [8,9]. There are notable north–south and seasonal gradients in the occurrence of the disease. Historically, PAN was frequently linked to HBV infection. With the advent of its vaccine and the enhancement of public health measures in response to the AIDS crisis, the incidence of PAN has seen a significant decline, turning it from one of the most common vasculitis in the 1990s to one of the least common today [8,10]. In addition, the improvement in laboratory techniques for the detection of ANCA in the 1980s led to the reclassification of certain vasculitis that were initially diagnosed as PAN . The challenge in estimating the overall prevalence of PAN arises from a combination of factors, including the absence of serum markers, heterogeneous classification criteria, and a variety of predisposing genetic and environmental factors. In the cohort from Pagnoux et al., which included 348 patients enrolled between 1963 and 2005, the mean age was 51 ± 17 years old with a male to female ratio of 1.7, and no ethnic predominance was observed . A more recent cohort from Rohmer et al., comprising 197 patients identified with PAN from 2005 to 2019, had a mean age of 53.6 ± 18 years and a male predominance of 1.5 . In addition, over 300 cases of a PAN-like disease in children and young adults mainly presenting with fever, skin ulcers, early-onset stroke, peripheral neuropathy, hypogammaglobulinemia, cytopenias, and elevated acute phase reactants who were found positive for adenosine deaminase 2 mutations (DADA2 disease) have been described in the literature since 2014. In a study of 118 patients with idiopathic adult PAN, 4.3% were found to have biallelic pathogenic variants of ADA2, and several others had less clearly significant ADA2 variants . Finally, PAN may be one of the manifestations, or even the initial manifestation, of a rare autoinflammatory disease known as VEXAS syndrome. VEXAS syndrome was initially described in 2020 and is caused by somatic mutations in methionine-41 of UBA1, the major E1 enzyme that initiates ubiquitylation. Large exome sequencing efforts have identified the prevalence of VEXAS syndrome-associated pathogenic variants at approximately 1:4000 in men and 1:26,000 in women over the age of 50 . In the initial report describing VEXAS, 3 out of 12 patients met the criteria for PAN. Since then, another nine cases of medium-vessel vasculitis have been attributed to VEXAS syndrome [16,17]. 3. Physiopathology PAN is typically characterized by a segmental, necrotizing, and transmural inflammation, predominantly involving small- to medium-sized arteries, although any arterial size could theoretically be susceptible. The disease most commonly impacts the visceral and muscular arteries, including their branches. In patient biopsies, it is common to observe co-existing lesions of diverse stages of inflammation and scarring within a single sample . Because of the arterial inflammatory process, fibrinoid necrosis may develop, leading to the formation of microaneurysms. Over time, these complications can progress to chronic stages, characterized by fibrous scarring and vascular aneurysms, which can rupture and lead to severe bleeding [19,20]. During the acute phase, the cellular infiltrate, composed of macrophages, T lymphocytes, neutrophils, and eosinophils, is generally observed in the tunica media but can also invade the tunica interna and tunica externa . One distinctive feature of PAN, compared to other vasculitis, is the absence of granulomas. This disease is also characterized by the coexistence of different stages of vascular inflammations at the same time. The pathophysiology of PAN, not yet fully understood, may vary depending on the disease’s specific etiology. Serum cytokine profile analysis in PAN patients has revealed an elevation in interferon-alpha (IFN-⍺), interleukine-2 (IL-2), tumor necrosis factor-α (TNF⍺), and IL-1-ß compared to healthy individuals and those with granulomatosis with polyangiitis (GPA) . Immunohistochemical studies of muscle and nerve biopsies from patients showed the presence of macrophages (41%) and mostly CD4+ T lymphocytes (41%) [19,23]. Most of these studies, however, primarily focus on PAN associated with HBV. Viral infections remain a common trigger of PAN and should be excluded in all cases. In PAN associated with HBV, the HBs antigen is responsible for the formation of immune complexes [24,25], as suggested by animal models of hepatitis B antigen-associated PAN, which show an accumulation of immune complexes in blood vessels [26,27]. Hepatitis C virus (HCV) has also been linked to PAN, with HCV-associated PAN tending to present more severe and acute symptoms [28,29]. However, this only concerns 5% of patients with PAN, and the distinction with cryoglobulinemic vasculitis can sometimes be challenging . HIV infection has been associated with PAN, though HIV-associated PAN is generally less aggressive than HBV-associated PAN. The classical manifestation is mononeuritis multiplex and can occur at any stage of HIV infection . Although parvovirus B19 has been associated with PAN, a study using PCR tests found no higher prevalence of this infection in people with PAN compared to those without [32,33,34,35]. More recently, vasculitis has been associated with COVID-19 infection, but to date no cases of PAN have been reported [36,37]. COVID-19 vaccines have been associated with PAN manifestations [38,39,40]. Like other vasculitides, PAN can be induced by the use of certain drugs, such as minocycline . The association between PAN and neoplasia is well established, especially for hematological malignancies, such as hairy T cell leukemia, or, more recently, myelodysplastic syndrome (MDS) [42,43,44,45,46]. In a study by Roupie et al., out of 70 patients with MDS and vasculitis, 9% presented with PAN. MDS is associated with a pro-inflammatory state in 10–30%, which also tends to present with autoimmune and inflammatory disorders. MDS and certain chronic inflammatory diseases share common genetic markers (such as HLA-B27) and polymorphisms (such as IL-1) [47,48]. More recently, genetic forms of PAN have been described. In the early 2000s, cases of PAN-like vasculitis were described in patients with Familial Mediterranean Fever (FMF) [49,50]. In a nationwide study in Turkey, PAN prevalence in patients with FMF was 0.9% . FMF is caused by mutations in the MEFV gene that encodes for pyrin/marenostrin, which result in unregulated production of IL-1, leading to recurring inflammation, fever, and, sometimes, autoimmune manifestations . Patients with PAN associated with FMF present a higher incidence of perirenal hemorrhages and elevated levels of inflammation [49,50,51,53]. Another condition related to genetic forms of PAN is STING-associated vasculopathy, with onset in infancy (SAVI), which is a type I interferonopathy. This condition is caused by mutations in the TMEM173 gene that induce the inflammation of endothelial cells in children. It often presents PAN-like symptoms in affected children . A monogenic syndrome resulting from a deficiency in Adenosine Deaminase 2 (DAD2) has been described in familial cases of necrotizing vasculitis that resemble PAN . Since 2014, over 60 bi-allelic loss-of-function mutations in the ADA2 gene have been documented [56,57]. Vascular inflammation in DAD2 patients is believed to be caused by an imbalance in macrophages, favoring the M1 type over the M2 type. To date, more than 200 cases of this condition have been recorded [58,59]. In 2020, Beck et al. published a cohort study of 25 men exhibiting a somatic mutation affecting methionine-41 (p.Met41) in the UBA1 gene. Located on the X chromosome, this gene encodes for a critical enzyme involved in the initiation of ubiquitination. The syndrome associated with this mutation is known as VEXAS, an acronym for Vacuoles, E1 enzyme, X-linked, Autoinflammatory, and Somatic. Although PAN-like features were initially reported in 12% of these patients, more recent studies suggest a lower incidence [16,60] (Figure 1). Figure 1. Comparison of various etiologies of PAN across different cohorts over distinct time periods. Adapted from Refs. [7,10,12,61]. 4. Clinical Manifestations Signs and symptoms of PAN result from damage to the vascular walls, potentially affecting all organs. This section provides an overview of organ systems that can be impacted in PAN patients. Unless otherwise stated, the percentages and specifics of the manifestations come from the cohorts shown in Table 1. Table 1. Characteristics of PAN patients reported in different cohorts (PNP: peripheral neuropathy, CNS: central nervous system, FFS: Five Factor Score, 1996). Results are expressed as percentages. 4.1. General Symptoms (85–93%) General, non-specific symptoms such as asthenia, fever, weight loss, myalgia, and arthralgia are frequently the initial symptoms of PAN. They are present in over 9 out of 10 patients. 4.2. Neurologic (59–79%) Neurologic manifestations occur in more than two-thirds of patients, most commonly as motor and sensory mononeuritis multiplex of the peripheral nerves [12,64]. Peripheral neuropathy is typically distal, asymmetric, and can be rapid-onset, often associated with localized skin edema. Of note, deep sensation is rarely affected. Foot drop, an important and disabling complication, may be the initial presentation . Cranial nerves are affected in less than 1% of patients. Central manifestations such as strokes occur in 2 to 10% of patients, typically in the later stages of the disease [64,66,67]. Compared with classic PAN, DADA2 patients more frequently present with central neurologic manifestations, particularly stroke, usually at a young age. 4.3. Cutaneous (50–59%) Skin lesions, including nodules, purpura, necrotic ulcers, and livedo reticularis, are present in half of the patients . In cases of cutaneous manifestations suggestive of vasculitis, a skin biopsy is recommended. The biopsy should be deep enough to include the dermal layer where medium-sized arteries are located. A subset of PAN, known as cutaneous PAN (CPAN), is confined to the skin and requires different management [69,70]. 4.4. Renal (15–75%) Kidney involvement is typically characterized by stenosis and aneurysm, primarily affecting renal and interlobar arteries and less frequently affecting the smaller arcuate and interlobular arteries [71,72]. This can manifest as hypertension (up to 35% of patients), micro or macro hematuria, mild proteinuria, and renal infarct [12,72,73]. Despite up to 75% of patients experiencing renal involvement, renal insufficiency occurs in only 15% of cases [12,74]. Glomerular involvement is infrequent . Severe complications such as ruptured aneurysm and spontaneous perirenal hemorrhage, which require embolization or nephrectomy, are infrequent but can be life-threatening [76,77,78,79]. 4.5. Gastrointestinal (22 to 38%) Gastrointestinal manifestations of PAN are common, affecting up to 50% of patients. Abdominal pain is reported by one out of three patients [80,81]. Vascular inflammation in mesenteric arteries can be severe, leading to intestinal ischemia, perforation, and hemorrhage [82,83]. Reports have also shown gallbladder involvement, malabsorption with loss of weight, and pancreatitis in patients with PAN [84,85]. In rare but severe instances, hepatic aneurysm can occur, potentially triggering acute liver failure and resulting in high mortality [86,87,88,89,90]. Gastrointestinal manifestations are associated with a poor prognosis; the mortality rate is around 25% for patients with such involvement [12,82]. This association with high mortality is supported by a retrospective study from 1988, which found that digestive complications contributed to the death of 16% of PAN patients [91,92]. Diagnosing mesenteric arterial involvement in PAN can be challenging, and conventional angiography may be valuable, especially in patients with minimal clinical evidence of extraintestinal manifestations. 4.6. Genital (15 to 17%) Manifestations, such as testicular pain, with or without orchitis, have been described as potential symptoms specific to PAN, although similar symptoms have been noted in Behcet’s patients. Testicular biopsy can be useful for diagnosis . Interestingly, there are reports of ovarian artery dissection associated with PAN in women . 4.7. Cardiovascular (7 to 78%) Cardiac involvement predominantly affects the myocardium due to coronary artery vasculitis. The left anterior descending, circumflex branches, and right coronary arteries are most affected. Pericarditis is relatively rare, often resulting from pre-existing myocardial involvement [95,96,97]. Myocardial infarction due to coronary infarction is also unusual . Celiac artery involvement and new-onset hypertension are potential risk factors for coronary involvement . Heart failure often presents during the initial stages of the disease. Hypertrophic cardiomyopathy, which may be a result of uncontrolled hypertension, could trigger serious conditions like ventricular tachycardia and syncope. Mild diffuse interstitial myocarditis can be caused by focal necrosis . Among pediatric patients with PAN, cases of hemopericardium have been described [101,102,103]. In terms of vascular manifestations, large vessels can be affected due to necrosis of the vasa vasorum . Symptoms of arterial claudication can be indicative of stenosis or ischemia in the lower extremities [105,106,107]. 4.8. Other Manifestations Ophthalmic complications, including retinal vasculitis, are observed in PAN [108,109,110]. Unlike other forms of vasculitis, such as granulomatosis with polyangiitis (GPA), eosinophilic granulomatosis with polyangiitis (EGPA), or microscopic polyangiitis (MPA), pulmonary lesions are notably absent in PAN. However, an autopsy study revealed bronchial artery damage in 7 out of 10 patients, despite the absence of symptoms . Muscular manifestations in PAN can vary from nonspecific myalgia to paresis, and muscle biopsy shows inflammation related to PAN in up to 50% of cases . VEXAS syndrome presents general symptoms like fever or weight loss in 96% of patients, skin manifestations such as neutrophilic dermatosis or tender plaques (84%), pulmonary infiltrates (49%), chondritis (36%), and deep vein thrombosis (35%) . Hematologic manifestations include macrocytic anemia (96%) and vacuoles in bone marrow myeloid and erythroid cells . The vascular manifestations of VEXAS mimic small to large vessel vasculitis [16,63,113]. In the inaugural study of 25 VEXAS patients, 12% were diagnosed with PAN . A recent literature review showed that among nine cases of medium vessel vasculitis found, all were men with macrocytic anemia and skin lesions, six of whom had passed away prior to the article’s publication . Patients with DADA2 exhibit vasculitis, immunodeficiency, and hematological manifestations. Vasculitis appears as mucocutaneous manifestations in 75% of cases, including livedo reticularis (50%), PAN-like skin lesions with non-granulomatous necrotizing inflammation of medium-sized arteries (34%), digital necrosis (22%), nodules (14%), Raynaud’s phenomenon (8%), and aphthous ulcers (7%). Neurological manifestations occur in 51% of cases and may include ischemic strokes (27%), cranial nerve palsy (27%), hemorrhagic strokes (12%), and polyneuropathy (9%). General symptoms such as fever-elevated erythrocyte sedimentation rate or CRP are present in half of the patients. Immunodeficiency manifests as hypogammaglobulinemia (22%), low IgM (18%), low IgA (12%), and infections (20%). Viral infections (11%) are more frequent than bacterial ones (7%). Hematological diseases manifest as anemia (13%), neutropenia (7%), and thrombocytopenia (6%) due to bone marrow failure or autoimmune cytopenia. Lymphoproliferative symptoms (32%), including splenomegaly and lymphadenopathy, are common in patients with DADA2. Most symptoms and signs (85%) occur before the age of 12 [63,114] (see Table 1). 5. Treatments The treatment recommendations for PAN are primarily based on weak empirical evidence and are often drawn from recommendations for other forms of vasculitis, with modifications according to the disease severity. Mild PAN, characterized by non-life- or organ-threatening manifestations like constitutional symptoms, arthritis, or skin lesions, is differentiated from moderate to severe PAN, which involves more severe complications, such as arterial stenosis—particularly those involving the renal arteries and aorta—and ischemic complications that affect the heart, peripheral nervous system, and gastrointestinal system. To aid in risk stratification, the 1996 version of the Five Factor Score (FFS) can be used. This score assigns +1 point for each of the following: proteinuria greater > 1 g/day, serum creatinine > 140 µmol/L, cardiomyopathy, severe gastrointestinal involvement, and CNS involvement [115,116]. Treatment for mild PAN (FFS of 0) may include glucocorticoids (GC) only. The clinical benefit of supplementing glucocorticoids with an immunosuppressive agent is not definitively established, but it could potentially offset the high 40% relapse rate and function as a steroid-sparing strategy. Guidelines, however, show divergence in recommendations. The French protocol typically prescribes glucocorticoids as a standalone treatment, introducing immunosuppressants such as methotrexate or azathioprine only in instances of resistance or intolerance. In contrast, the ACR’s 2021 guidelines advocate for a combined approach from the beginning, recommending the incorporation of azathioprine (administered orally at 2–3 mg/kg/day) or methotrexate (preferably given subcutaneously at 0.3 mg/kg/week) with glucocorticoids. Moderate to severe PAN (FFS > 0) is treated with intravenous (IV) GC in conjunction with an immunosuppressive agent, preferably cyclophosphamide. The start of treatment marks the induction phase, lasting 3 to 6 months, aimed at achieving disease remission, defined by the American College of Rheumatology (ACR) as a complete absence of clinical manifestations, with or without immunosuppressive treatment. Initial treatment strategies recommend starting with at least 1 mg/kg/day of prednisone equivalent, capped at 60 mg/day. In patients with severe manifestations requiring rapid intervention, IV boluses of methylprednisolone are recommended. If remission is incomplete, the duration of cyclophosphamide therapy may be extended, although it is recommended not to exceed a period of 6 months given its potential toxicity [115,116]. Alternative therapies, including rituximab, mycophenolate mofetil, tocilizumab, anti-TNF alpha, JAK inhibitors, IV immunoglobulins, or plasma exchange, have not been well studied and their application is only reserved for certain refractory or relapsed patients [34,117,118,119,120,121,122,123,124,125]. A recent European retrospective study analyzed 42 patients treated for relapsed and/or refractory PAN. Tocilizumab, anti-TNF alpha, and rituximab achieved complete remission in 50%, 40%, and 33% of cases, respectively, with comparable safety profiles. These biotherapies may become first-line treatments in the future, but more data are needed . The induction phase is followed by the maintenance phase, with the objective of preventing relapse. Patients treated with cyclophosphamide with complete remission may be switched to azathioprine or methotrexate for 12 to 18 months [115,116]. In secondary forms of PAN, the therapeutic approach focuses on the underlying etiology. In the context of HBV-associated PAN, antiviral therapy is used as the primary intervention. In severe cases, management with GC and plasma exchange may be considered . When PAN is concomitant with MDS, interventions targeting the MDS are often effective in attenuating the vasculitic manifestations . From this perspective, Mekinian et al. showed that azacytidine successfully treated autoimmune manifestations in 9 out of 11 patients with MDS . In the case of DADA2-associated PAN, numerous treatments have been explored (azathioprine, cyclosporine, tacrolimus, cyclophosphamide, and methotrexate) with mitigated results. The ACR 2021 guidelines have now approved the use of steroids and anti-TNF alpha-agents (etanercept, infliximab, or adalimumab) following demonstration of their efficacy in PAN associated with DADA2 . Hematopoietic stem cell transplantation (HSCT) has been reported to treat cytopenia . Regarding VEXAS syndrome, several drugs have been tested with mixed results. GC in combination with azacytidine (possibly supplemented by HSCT) seems most effective for patients with MDS features. For those without myelodysplasia, JAK inhibitors or tocilizumab may be suitable [128,129,130]. Finally, vasculitis associated with primary immunodeficiency can be managed with biotherapies, HSCT, or IV immunoglobulin therapy [55,131]. 6. Conclusions Over the past two decades, the understanding of PAN has significantly evolved. Though PAN was previously considered primarily as either idiopathic or HBV-related, the landscape now encompasses a broader spectrum that includes PAN associated with infections, paraneoplastic syndromes, and newly recognized classifications such as DADA2, interferonopathies, and VEXAS syndrome. Identification of the underlying causes of PAN is critical for directing treatment strategies. While conventional immunosuppressants, such as cyclophosphamide, are often the standard of care for primary PAN, secondary forms may respond better to more specific agents: anti-TNF alpha for DADA2, JAK inhibitors for VEXAS, and targeted MDS treatment for PAN associated with MDS or allogeneic stem cell transplantation. Author Contributions Conceptualization, L.W. and D.C.; methodology, L.W., A.H., L.M., M.P.K. and D.C.; writing—original draft preparation, L.W. and D.C.; writing—review and editing, M.P.K., A.H. and L.M.; supervision, D.C. All authors have read and agreed to the published version of the manuscript. Funding This research received no external funding. Data Availability Statement No new data were created. Conflicts of Interest The authors declare no conflict of interest. References Stone, J.H. Vasculitis: A collection of pearls and myths. Rheum. Dis. Clin. N. Am. 2007, 33, 691–739. [Google Scholar] [CrossRef] Zeek, P.M. Periarteritis nodosa; a critical review. Am. J. Clin. Pathol. 1952, 22, 777–790. [Google Scholar] [CrossRef] [PubMed] Davies, D.J.; Moran, J.E.; Niall, J.F.; Ryan, G.B. Segmental necrotising glomerulonephritis with antineutrophil antibody: Possible arbovirus aetiology? BMJ 1982, 285, 606. [Google Scholar] [CrossRef] Karadag, O.; Jayne, D.J. Polyarteritis nodosa revisited: A review of historical approaches, subphenotypes and a research agenda. Clin. Exp. Rheumatol. 2018, 36 (Suppl. 111), 135–142. [Google Scholar] [PubMed] Jennette, J.C.; Falk, R.J.; Bacon, P.A.; Basu, N.; Cid, M.C.; Ferrario, F.; Flores-Suarez, L.F.; Gross, W.L.; Guillevin, L.; Hagen, E.C.; et al. 2012 Revised International Chapel Hill Consensus Conference Nomenclature of Vasculitides. Arthritis Rheum. 2012, 65, 1–11. [Google Scholar] [CrossRef] Guillevin, L.; Mahr, A.; Cohen, P. Les vascularites nécrosantes systémiques: Classification et stratégies actuelles de traitement. Rev. Méd. Interne 2003, 24, 172–182. [Google Scholar] [CrossRef] [PubMed] Rohmer, J.; Trefond, L.; Nguyen, Y.; Durel, C.A.; Lacout, C.; Maurier, F.; Rouzaud, D.; Cohen, P.; Lazaro, E.; Mekinian, A.; et al. Caractéristiques cliniques et évolution à long-terme des périartérite noueuses systémiques diagnostiquées depuis 2005. Rev. Méd. Interne 2020, 41, A33–A34. [Google Scholar] [CrossRef] Mahr, A.; Guillevin, L.; Poissonnet, M.; Aymé, S. Prevalences of polyarteritis nodosa, microscopic polyangiitis, Wegener’s granulomatosis, and Churg-Strauss syndrome in a French urban multiethnic population in 2000: A capture–recapture estimate. Arthritis Care Res. 2004, 51, 92–99. [Google Scholar] [CrossRef] Lane, S.E.; Watts, R.; Scott, D.G.I. Epidemiology of systemic vasculitis. Curr. Rheumatol. Rep. 2005, 7, 270–275. [Google Scholar] [CrossRef] Sönmez, H.E.; Armağan, B.; Ayan, G.; Barut, K.; Batu, E.D.; Erden, A.; Ugurlu, S.; Bilginer, Y.; Kasapcopur, O.; Karadag, O.; et al. Polyarteritis nodosa: Lessons from 25 years of experience. Clin. Exp. Rheumatol. 2019, 5, 52–56. [Google Scholar] Saadoun, D.; Vautier, M.; Cacoub, P. Medium- and Large-Vessel Vasculitis. Circulation 2021, 143, 267–282. [Google Scholar] [CrossRef] [PubMed] Pagnoux, C.; Seror, R.; Henegar, C.; Mahr, A.; Cohen, P.; Le Guern, V.; Bienvenu, B.; Mouthon, L.; Guillevin, L. Clinical features and outcomes in 348 patients with polyarteritis nodosa: A systematic retrospective study of patients diagnosed between 1963 and 2005 and entered into the French Vasculitis Study Group Database. Arthritis Rheum. 2010, 62, 616–626. [Google Scholar] [CrossRef] [PubMed] Rohmer, J.; Nguyen, Y.; Trefond, L.; Agard, C.; Allain, J.S.; Berezne, A.; Charles, P.; Cohen, P.; Gondran, G.; Groh, M.; et al. Clinical features and long-term outcomes of patients with systemic polyarteritis nodosa diagnosed since 2005: Data from 196 patients. J. Autoimmun. 2023, 139, 103093. [Google Scholar] [CrossRef] [PubMed] Schnappauf, O.; Sampaio Moura, N.; Aksentijevich, I.; Stoffels, M.; Ombrello, A.K.; Hoffmann, P.; Barron, K.; Remmers, E.F.; Hershfield, M.; Kelly, S.J.; et al. Sequence-Based Screening of Patients With Idiopathic Polyarteritis Nodosa, Granulomatosis With Polyangiitis, and Microscopic Polyangiitis for Deleterious Genetic Variants in ADA2. Arthritis Rheumatol. 2020, 73, 512–519. [Google Scholar] [CrossRef] Beck, D.B.; Bodian, D.L.; Shah, V.; Mirshahi, U.L.; Kim, J.; Ding, Y.; Magaziner, S.J.; Strande, N.T.; Cantor, A.; Haley, J.S.; et al. Estimated Prevalence and Clinical Manifestations of UBA1 Variants Associated With VEXAS Syndrome in a Clinical Population. JAMA 2023, 329, 318–324. [Google Scholar] [CrossRef] Beck, D.B.; Ferrada, M.A.; Sikora, K.A.; Ombrello, A.K.; Collins, J.C.; Pei, W.; Balanda, N.; Ross, D.L.; Cardona, D.O.; Wu, Z.; et al. Somatic Mutations in UBA1 and Severe Adult-Onset Autoinflammatory Disease. N. Engl. J. Med. 2020, 383, 2628–2638. [Google Scholar] [CrossRef] [PubMed] Watanabe, R.; Kiji, M.; Hashimoto, M. Vasculitis associated with VEXAS syndrome: A literature review. Front. Med. 2022, 9, 983939. [Google Scholar] [CrossRef] De Virgilio, A.; Greco, A.; Magliulo, G.; Gallo, A.; Ruoppolo, G.; Conte, M.; Martellucci, S.; de Vincentiis, M. Polyarteritis nodosa: A contemporary overview. Autoimmun. Rev. 2016, 15, 564–570. [Google Scholar] [CrossRef] Matsumoto, T.; Kobayashi, S.; Ogishima, D.; Aoki, Y.; Sonoue, H.; Abe, H.; Fukumura, Y.; Nobukawa, B.; Kumasaka, T.; Mori, S.; et al. Isolated necrotizing arteritis (localized polyarteritis nodosa): Examination of the histological process and disease entity based on the histological classification of stage and histological differences from polyarteritis nodosa. Cardiovasc. Pathol. 2007, 16, 92–97. [Google Scholar] [CrossRef] Lie, J.T. Systemic and isolated vasculitis. A rational approach to classification and pathologic diagnosis. Pathol. Annu. 1989, 24 Pt 1, 25–114. [Google Scholar] Kallenberg, C.G.; Brouwer, E.; Weening, J.J.; Tervaert, J.W.C. Anti-neutrophil cytoplasmic antibodies: Current diagnostic and pathophysiological potential. Kidney Int. 1994, 46, 1–15. [Google Scholar] [CrossRef] Grau, G.E.; Roux-Lombard, P.; Gysler, C.; Lambert, C.; Lambert, P.H.; Dayer, J.M.; Guillevin, L. Serum cytokine changes in systemic vasculitis. Immunology 1989, 68, 196–198. [Google Scholar] Cid, M.-C.; Grau, J.M.; Casademont, J.; Campo, E.; Coll-Vinent, B.; López-Soto, A.; Ingelmo, M.; Urbano-Márquez, A. Immunohistochemical characterization of inflammatory cells and immunologic activation markers in muscle and nerve biopsy specimens from patients with systemic polyarteritis nodosa. Arthritis Rheum. 1994, 37, 1055–1061. [Google Scholar] [CrossRef] [PubMed] Guillevin, L.; Ronco, P.; Verroust, P. Circulating immune complexes in systemic necrotizing vasculitis of the polyarteritis nodosa group. Comparison of HBV-related polyarteritis nodosa and Churg Strauss Angiitis. J. Autoimmun. 1990, 3, 789–792. [Google Scholar] [CrossRef] Prince, A.M.; Trepo, C. Role of immune complexes involving SH antigen in pathogenesis of chronic active hepatitis and polyarteritis nodosa. Lancet 1971, 297, 1309–1312. [Google Scholar] [CrossRef] Trepo, C.G.; Zuckerman, A.J.; Bird, R.C.; Prince, A.M. The role of circulating hepatitis B antigen/antibody immune complexes in the pathogenesis of vascular and hepatic manifestations in polyarteritis nodosa. J. Clin. Pathol. 1974, 27, 863–868. [Google Scholar] [CrossRef] Fye, K.H.; Becker, M.J.; Theofilopoulos, A.N.; Moutsopoulos, H.; Feldman, J.-L.; Talal, N. Immune complexes in hepatitis B antigen-associated periarteritis nodosa. Detection by antibody-dependent cell-mediated cytotoxicity and the Raji cell assay. Am. J. Med. 1977, 62, 783–791. [Google Scholar] [CrossRef] Carson, C.W.; Conn, D.L.; Czaja, A.J.; Wright, T.L.; Brecher, M.E. Frequency and significance of antibodies to hepatitis C virus in polyarteritis nodosa. J. Rheumatol. 1993, 20, 304–309. [Google Scholar] [PubMed] Saadoun, D.; Terrier, B.; Semoun, O.; Sene, D.; Maisonobe, T.; Musset, L.; Amoura, Z.; Rigon, M.R.; Cacoub, P. Hepatitis C virus-associated polyarteritis nodosa. Arthritis Care Res. 2010, 63, 427–435. [Google Scholar] [CrossRef] Cacoub, P.; Maisonobe, T.; Thibault, V.; Gatel, A.; Servan, J.; Musset, L.; Piette, J.C. Systemic vasculitis in patients with hepatitis C. J. Rheumatol. 2001, 28, 109–118. [Google Scholar] [PubMed] Patel, N.; Patel, N.; Khan, T.; Patel, N.; Espinoza, L.R. HIV infection and clinical spectrum of associated vasculitides. Curr. Rheumatol. Rep. 2011, 13, 506–512. [Google Scholar] [CrossRef] Leruez-Ville, M.; Laugé, A.; Morinet, F.; Guillevin, L.; Dény, P. Polyarteritis nodosa and parvovirus B19. Lancet 1994, 344, 263–264. [Google Scholar] [CrossRef] [PubMed] Gherardi, R.; Belec, L.; Mhiri, C.; Gray, F.; Lescs, M.; Sobel, A.; Guillevin, L.; Wechsler, J. The spectrum of vasculitis in human immunodeficiency virus-infected patients. A clinicopathologic evaluation. Arthritis Rheum. 1993, 36, 1164–1174. [Google Scholar] [CrossRef] [PubMed] Viguier, M.; Guillevin, L.; Laroche, L. Treatment of parvovirus B19-associated polyarteritis nodosa with intravenous immune globulin. N. Engl. J. Med. 2001, 344, 1481–1482. [Google Scholar] [CrossRef] [PubMed] Finkel, T.H.; Török, T.J.; Ferguson, P.J.; Durigon, E.L.; Zaki, S.R.; Leung, D.Y.; Harbeck, R.J.; Gelfand, E.W.; Saulsbury, F.T.; Hollister, J.R.; et al. Chronic parvovirus B19 infection and systemic necrotising vasculitis: Opportunistic infection or aetiological agent? Lancet 1994, 343, 1255–1258. [Google Scholar] [CrossRef] [PubMed] Ramadan, S.M.; Kasfiki, E.V.; Kelly, C.W.; Ali, I. An interesting case of small vessel pathology following coronavirus infection. BMJ Case Rep. 2020, 13, e237407. [Google Scholar] [CrossRef] [PubMed] Vacchi, C.; Meschiari, M.; Milic, J.; Marietta, M.; Tonelli, R.; Alfano, G.; Volpi, S.; Faltoni, M.; Franceschi, G.; Ciusa, G.; et al. COVID-19-associated vasculitis and thrombotic complications: From pathological findings to multidisciplinary discussion. Rheumatology 2020, 59, e147–e150. [Google Scholar] [CrossRef] Su, H.-A.; Hsu, H.-T.; Chen, Y.-C. Cutaneous polyarteritis nodosa following ChAdOx1 nCoV-19 vaccination. Int. J. Dermatol. 2022, 61, 630–631. [Google Scholar] [CrossRef] [PubMed] Ohkubo, Y.; Ohmura, S.-I.; Ishihara, R.; Miyamoto, T. Possible case of polyarteritis nodosa with epididymitis following COVID-19 vaccination: A case report and review of the literature. Mod. Rheumatol. Case Rep. 2022, 7, 172–176. [Google Scholar] [CrossRef] Ohmura, S.-I.; Ohkubo, Y.; Ishihara, R.; Otsuki, Y.; Miyamoto, T. Medium-vessel Vasculitis Presenting with Myalgia Following COVID-19 Moderna Vaccination. Intern. Med. 2022, 61, 3453–3457. [Google Scholar] [CrossRef] Kermani, T.A.; Ham, E.K.; Camilleri, M.J.; Warrington, K.J. Polyarteritis nodosa-like vasculitis in association with minocycline use: A single-center case series. Semin. Arthritis Rheum. 2012, 42, 213–221. [Google Scholar] [CrossRef] [PubMed] Carpenter, M.T.; West, S.G. Polyarteritis nodosa in hairy cell leukemia: Treatment with interferon-alpha. J. Rheumatol. 1994, 21, 1150–1152. [Google Scholar] [PubMed] Roupie, A.L.; Guedon, A.; Terrier, B.; Lahuna, C.; Jachiet, V.; Regent, A.; de Boysson, H.; Carrat, F.; Seguier, J.; Terriou, L.; et al. Vasculitis associated with myelodysplastic syndrome and chronic myelomonocytic leukemia: French multicenter case-control study. Semin. Arthritis Rheum. 2020, 50, 879–884. [Google Scholar] [CrossRef] Veitch, D.; Tsai, T.; Watson, S.; Joshua, F. Paraneoplastic polyarteritis nodosa with cerebral masses: Case report and literature review. Int. J. Rheum. Dis. 2014, 17, 805–809. [Google Scholar] [CrossRef] [PubMed] Hamidou, M.A.; Boumalassa, A.; Larroche, C.; El Kouri, D.; Blétry, O.; Grolleau, J.-Y. Systemic medium-sized vessel vasculitis associated with chronic myelomonocytic leukemia. Semin. Arthritis Rheum. 2001, 31, 119–126. [Google Scholar] [CrossRef] Fain, O.; Hamidou, M.; Cacoub, P.; Godeau, B.; Wechsler, B.; ParIès, J.; Stirnemann, J.; Morin, A.-S.; Gatfosse, M.; Hanslik, T.; et al. Vasculitides associated with malignancies: Analysis of sixty patients. Arthritis Rheum. 2007, 57, 1473–1480. [Google Scholar] [CrossRef] Komrokji, R.S.; Kulasekararaj, A.; Al Ali, N.H.; Kordasti, S.; Bart-Smith, E.; Craig, B.M.; Padron, E.; Zhang, L.; Lancet, J.E.; Pinilla-Ibarz, J.; et al. Autoimmune diseases and myelodysplastic syndromes. Am. J. Hematol. 2016, 91, E280–E283. [Google Scholar] [CrossRef] Saif, M.W.; Hopkins, J.L.; Gore, S.D. Autoimmune phenomena in patients with myelodysplastic syndromes and chronic myelomonocytic leukemia. Leuk. Lymphoma 2002, 43, 2083–2092. [Google Scholar] [CrossRef] Balbir-Gurman, A.; Nahir, A.M.; Braun-Moscovici, Y. Vasculitis in siblings with familial Mediterranean fever: A report of three cases and review of the literature. Clin. Rheumatol. 2006, 26, 1183–1185. [Google Scholar] [CrossRef] Ozen, S.; Ben-Chetrit, E.; Bakkaloglu, A.; Gur, H.; Tinaztepe, K.; Calguneri, M.; Turgan, C.; Turkmen, A.; Akpolat, I.; Danaci, M.; et al. Polyarteritis nodosa in patients with Familial Mediterranean Fever (FMF): A concomitant disease or a feature of FMF? Semin. Arthritis Rheum. 2001, 30, 281–287. [Google Scholar] [CrossRef] Familial Mediterranean fever (FMF) in Turkey: Results of a nationwide multicenter study. Medicine 2005, 84, 1–11. [CrossRef] [PubMed] Ozen, S. The changing face of polyarteritis nodosa and necrotizing vasculitis. Nat. Rev. Rheumatol. 2017, 13, 381–386. [Google Scholar] [CrossRef] Yalçınkaya, F.; Özçakar, Z.B.; Kasapçopur, O.; Oztürk, A.; Akar, N.; Bakkaloğlu, A.; Arısoy, N.; Ekim, M.; Özen, S. Prevalence of the MEFV gene mutations in childhood polyarteritis nodosa. J. Pediatr. 2007, 151, 675–678. [Google Scholar] [CrossRef] [PubMed] Liu, Y.; Jesus, A.A.; Marrero, B.; Yang, D.; Ramsey, S.E.; Montealegre Sanchez, G.A.; Tenbrock, K.; Wittkowski, H.; Jones, O.Y.; Kuehn, H.S.; et al. Activated STING in a vascular and pulmonary syndrome. N. Engl. J. Med. 2014, 371, 507–551. [Google Scholar] [CrossRef] [PubMed] Caratsch, L.; Schnider, C.; Moi, L.; Theodoropoulou, K.; Candotti, F.; Hofer, M. Adenosine deaminase 2 deficiency: A disease with multiple presentations. Rev. Med. Suisse 2022, 18, 669–673. [Google Scholar] [CrossRef] Navon Elkan, P.; Pierce, S.B.; Segel, R.; Walsh, T.; Barash, J.; Padeh, S.; Zlotogorski, A.; Berkun, Y.; Press, J.J.; Mukamel, M.; et al. Mutant Adenosine Deaminase 2 in a Polyarteritis Nodosa Vasculopathy. N. Engl. J. Med. 2014, 370, 921–931. [Google Scholar] [CrossRef] Zhou, Q.; Yang, D.; Ombrello, A.K.; Zavialov, A.V.; Toro, C.; Zavialov, A.V.; Stone, D.L.; Chae, J.J.; Rosenzweig, S.D.; Bishop, K.; et al. Early-Onset Stroke and Vasculopathy Associated with Mutations in ADA2. N. Engl. J. Med. 2014, 370, 911–920. [Google Scholar] [CrossRef] Kaljas, Y.; Liu, C.; Skaldin, M.; Wu, C.; Zhou, Q.; Lu, Y.; Aksentijevich, I.; Zavialov, A. Human adenosine deaminases ADA1 and ADA2 bind to different subsets of immune cells. Cell Mol. Life Sci. 2016, 74, 555–570. [Google Scholar] [CrossRef] Zavialov, A.V.; Gracia, E.; Glaichenhaus, N.; Franco, R.; Zavialov, A.V.; Lauvau, G. Human adenosine deaminase 2 induces differentiation of monocytes into macrophages and stimulates proliferation of T helper cells and macrophages. J. Leukoc. Biol. 2010, 88, 279–290. [Google Scholar] [CrossRef] Zakine, E.; Rodrigues, F.; Papageorgiou, L.; Georgin-Lavialle, S.; Mekinian, A.; Terrier, B.; Kosmider, O.; Hirsch, P.; Jachiet, M.; Bouaziz, J.; et al. Caractéristiques cliniques et histologiques des manifestations cutanées du syndrome VEXAS: Une étude rétrospective centralisée de 59 cas. Rev. Méd. Interne 2022, 43, A350–A351. [Google Scholar] [CrossRef] Frohnert, P.P.; Sheps, S.G. Long-term follow-up study of periarteritis nodosa. Am. J. Med. 1967, 43, 8–14. [Google Scholar] [CrossRef] Georgin-Lavialle, S.; Terrier, B.; Guedon, A.; Heiblig, M.; Comont, T.; Lazaro, E.; Lacombe, V.; Terriou, L.; Ardois, S.; Bouaziz, J.; et al. Further characterization of clinical and laboratory features in VEXAS syndrome: Large-scale analysis of a multicentre case series of 116 French patients. Br. J. Dermatol. 2021, 186, 564–574. [Google Scholar] [CrossRef] [PubMed] Meyts, I.; Aksentijevich, I. Deficiency of Adenosine Deaminase 2 (DADA2): Updates on the Phenotype, Genetics, Pathogenesis, and Treatment. J. Clin. Immunol. 2018, 38, 569–578. [Google Scholar] [CrossRef] Tervaert, J.W.C.; Kallenberg, C. Neurologic manifestations of systemic vasculitides. Rheum. Dis. Clin. N. Am. 1993, 19, 913–940. [Google Scholar] [CrossRef] [PubMed] Neves, F.S.; Lin, K. Bilateral Foot Drop in Polyarteritis Nodosa. N. Engl. J. Med. 2012, 367, e9. [Google Scholar] [CrossRef] [PubMed] de Boysson, H.; Guillevin, L. Polyarteritis Nodosa Neurologic Manifestations. Neurol. Clin. 2019, 37, 345–357. [Google Scholar] [CrossRef] [PubMed] Provenzale, J.M.; Allen, N.B. Neuroradiologic findings in polyarteritis nodosa. AJNR Am. J. Neuroradiol. 1996, 17, 1119–1126. [Google Scholar] Morgan, A.J.; Schwartz, R.A. Cutaneous polyarteritis nodosa: A comprehensive review. Int. J. Dermatol. 2010, 49, 750–756. [Google Scholar] [CrossRef] Stewart, M.; Lo, A.; Shojania, K.; Au, S.; Seidman, M.A.; Dutz, J.P.; Chan, J. Cutaneous polyarteritis nodosa diagnosis and treatment: A retrospective case series. J. Am. Acad. Dermatol. 2022, 87, 1370–1373. [Google Scholar] [CrossRef] Sunderkötter, C.H.; Zelger, B.; Chen, K.-R.; Requena, L.; Piette, W.; Carlson, J.A.; Dutz, J.; Lamprecht, P.; Mahr, A.; Aberer, E.; et al. Nomenclature of Cutaneous Vasculitis: Dermatologic Addendum to the 2012 Revised International Chapel Hill Consensus Conference Nomenclature of Vasculitides. Arthritis Rheumatol. 2018, 70, 171–184. [Google Scholar] [CrossRef] el-Reshaid, K.; Kapoor, M.M.; El-Reshaid, W.; Madda, J.P.; Varro, J. The spectrum of renal disease associated with microscopic polyangiitis and classic polyarteritis nodosa in Kuwait. Nephrol. Dial. Transplant. 1997, 12, 1874–1882. [Google Scholar] [CrossRef] [PubMed] Maritati, F.; Iannuzzella, F.; Pavia, M.P.; Pasquali, S.; Vaglio, A. Kidney involvement in medium- and large-vessel vasculitis. J. Nephrol. 2016, 29, 495–505. [Google Scholar] [CrossRef] [PubMed] Pourafshar, N.; Sobel, E.; Segal, M. A case of isolated renal involvement of polyarteritis nodosa successfully treated with steroid monotherapy. BMJ Case Rep. 2016, 2016, bcr2016215702. [Google Scholar] [CrossRef] [PubMed] Kaur, J.S.; Goldberg, J.P.; Schrier, R.W. Acute renal failure following arteriography in a patient with polyarteritis nodosa. JAMA 1982, 247, 833–834. [Google Scholar] [CrossRef] Fujimoto, S.; Yamamoto, Y.; Saita, M.; Hisanaga, S.; Morita, S.; Tanaka, K.; Sumiyoshi, A.; Koono, M. Renal failure associated with polyarteritis nodosa. Nihon Jinzo Gakkai Shi 1990, 32, 739–744. [Google Scholar] Launay, D.; Michon-Pasturel, U.; Boumbar, Y.; Dubrulle, F.; Bouroz-Joly, J.; Hachulla, E.; Lemaitre, L.; Devulder, B. Bilateral spontaneous perirenal hematoma: An unusual complication of polyarteritis nodosa. Rev. Med. Interne 1998, 19, 666–669. [Google Scholar] [CrossRef] Nandwani, G.M.; Musker, M.P.; Chaplin, B.J.; El Madhoun, I.; Akbani, H. Spontaneous perirenal haemorrhage in polyarteritis nodosa. J. Coll. Physicians Surg. Pak. 2013, 23, 445–447. [Google Scholar] Miyagawa, T.; Iwata, Y.; Oshima, M.; Ogura, H.; Sato, K.; Nakagawa, S.; Yamamura, Y.; Kitajima, S.; Toyama, T.; Hara, A.; et al. Polyarteritis nodosa with perirenal hematoma due to the rupture of a renal artery aneurysm. CEN Case Rep. 2021, 10, 244–249. [Google Scholar] [CrossRef] Allen, A.W.; Waybill, P.N.; Singh, H.; Brown, D.B. Polyarteritis nodosa presenting as spontaneous perirenal hemorrhage: Angiographic diagnosis and treatment with microcoil embolization. J. Vasc. Interv. Radiol. 1999, 10, 1361–1363. [Google Scholar] [CrossRef] Levine, S.M.; Hellmann, D.B.; Stone, J.H. Gastrointestinal involvement in polyarteritis nodosa (1986–2000): Presentation and outcomes in 24 patients. Am. J. Med. 2002, 112, 386–391. [Google Scholar] [CrossRef] Castelhano, R.; Win, K.M.; Carty, S. Celiac artery aneurysm causing an acute abdomen. BMJ Case Rep. 2021, 14, e240533. [Google Scholar] [CrossRef] [PubMed] Gendreau, S.; Porcher, R.; Thoreau, B.; Paule, R.; Maurier, F.; Goulenok, T.; Frumholtz, L.; Bernigaud, C.; Ingen-Housz-Oro, S.; Mekinian, A.; et al. Characteristics and risk factors for poor outcome in patients with systemic vasculitis involving the gastrointestinal tract. Semin. Arthritis Rheum. 2021, 51, 436–441. [Google Scholar] [CrossRef] [PubMed] Waisayarat, J.; Niyasom, C.; Vilaiyuk, S.; Molagool, S. Polyarteritis Nodosa with Cytomegalovirus Enteritis and Jejunoileal Perforation: Report of a Case with a Literature Review. Vasc. Health Risk Manag. 2022, 18, 595–601. [Google Scholar] [CrossRef] Dillard, B.M.; Black, W.C. Polyarteritis nodosa of the gallbladder and bile ducts. Am. Surg. 1970, 36, 423–427. [Google Scholar] Ito, M.; Sano, K.; Inaba, H.; Hotchi, M. Localized necrotizing arteritis. A report of two cases involving the gallbladder and pancreas. Arch. Pathol. Lab. Med. 1991, 115, 780–783. [Google Scholar] [PubMed] Empen, K.; Jung, M.-C.; Engelhardt, D.; Sackmann, M. Successful treatment of acute liver failure due to polyarteritis nodosa. Am. J. Med. 2002, 113, 349–351. [Google Scholar] [CrossRef] [PubMed] Herskowitz, M.M.; Flyer, M.A.; Sclafani, S.J.A. Percutaneous transhepatic coil embolization of a ruptured intrahepatic aneurysm in polyarteritis nodosa. Cardiovasc. Interv. Radiol. 1993, 16, 254–256. [Google Scholar] [CrossRef] Parent, B.A.; Cho, S.W.; Buck, D.G.; Nalesnik, M.A.; Gamblin, T.C. Spontaneous rupture of hepatic artery aneurysm associated with polyarteritis nodosa. Am. Surg. 2010, 76, 1416–1419. [Google Scholar] [CrossRef] Roberto, M.; Meytes, V.; Liu, S. Ruptured hepatic aneurysm as first presenting symptom of polyarteritis nodosa. Oxf. Med Case Rep. 2018, 2018, omx100. [Google Scholar] [CrossRef] Stambo, G.W.; Guiney, M.J.; Cannella, X.F.; Germain, B.F. Coil embolization of multiple hepatic artery aneurysms in a patient with undiagnosed polyarteritis nodosa. J. Vasc. Surg. 2004, 39, 1122–1124. [Google Scholar] [CrossRef] Guillevin, L.; Lhote, F.; Gallais, V.; Jarrousse, B.; Royer, I.; Gayraud, M.; Benichou, J. Gastrointestinal tract involvement in polyarteritis nodosa and Churg-Strauss syndrome. Ann. Med. Interne 1995, 146, 260–267. [Google Scholar] Cacoub, P.; Huong, L.T.; Guillevin, L.; Godeau, P. Causes of death in systemic vasculitis of polyarteritis nodosa. Analysis of a series of 165 patients. Ann. Med. Interne 1988, 139, 381–390. [Google Scholar] Teichman, J.M.; Mattrey, R.F.; Demby, A.M.; Schmidt, J.D. Polyarteritis nodosa presenting as acute orchitis: A case report and review of the literature. J. Urol. 1993, 149, 1139–1140. [Google Scholar] [CrossRef] [PubMed] Tao, Y.; Matsubara, S.; Yagi, K.; Kinoshita, K.; Fukunaga, T.; Yamamoto, A.; Uno, M. Intra-abdominal hemorrhage due to segmental arterial mediolysis of an ovarian artery pseudoaneurysm and concomitant aneurysmal subarachnoid hemorrhage: Illustrative case. J. Neurosurg. Case Lessons 2022, 4, CASE22233. [Google Scholar] [CrossRef] [PubMed] Kastner, D.; Gaffney, M.; Tak, T. Polyarteritis nodosa and myocardial infarction. Can. J. Cardiol. 2000, 16, 515–518. [Google Scholar] [PubMed] Bae, Y.D.; Choi, H.J.; Lee, J.C.; Park, J.J.; Lee, Y.J.; Lee, E.B.; Song, Y.W. Clinical features of polyarteritis nodosa in Korea. J. Korean Med. Sci. 2006, 21, 591–595. [Google Scholar] [CrossRef] [PubMed] Schrader, M.L.; Hochman, J.S.; Bulkley, B.H. The heart in polyarteritis nodosa: A clinicopathologic study. Am. Hear. J. 1985, 109, 1353–1359. [Google Scholar] [CrossRef] Blétry, O.; Godeau, P.; Charpentier, G.; Guillevin, L.; Herreman, G. Cardiac manifestations of periarteritis nodosa. Incidence of non-hypertensive cardiomyopathy. Arch. Mal. Coeur Vaiss. 1980, 73, 1027–1036. [Google Scholar] Lai, J.; Zhao, L.; Zhong, H.; Zhou, J.; Guo, X.; Xu, D.; Tian, X.; Zhang, S.; Zeng, X. Characteristics and Outcomes of Coronary Artery Involvement in Polyarteritis Nodosa. Can. J. Cardiol. 2020, 37, 895–903. [Google Scholar] [CrossRef] Papadopoulos, D.P.; Moyssakis, I.; Votteas, V.E. Polyarteritis nodosa and hypertrophic obstructive cardiomyopathy. A true association? Clin. Rheumatol. 2004, 23, 57–58. [Google Scholar] [CrossRef] Schafigh, M.; Bakhtiary, F.; Kolck, U.W.; Zimmer, S.; Greschus, S.; Silaschi, M. Coronary Artery Aneurysm Rupture in a Patient With Polyarteritis Nodosa. JACC Case Rep. 2022, 4, 1522–1528. [Google Scholar] [CrossRef] Reimold, E.W.; Weinberg, A.G.; Fink, C.W.; Battles, N.D. Polyarteritis in Children. Arch. Pediatr. Adolesc. Med. 1976, 130, 534–541. [Google Scholar] [CrossRef] Holt, S.; Jackson, P. Ruptured coronary aneurysm and valvulitis in an infant with polyarteritis nodosa. J. Pathol. 1975, 117, 83–87. [Google Scholar] [CrossRef] Iino, T.; Eguchi, K.; Sakai, M.; Nagataki, S.; Ishijima, M.; Toriyama, K. Polyarteritis nodosa with aortic dissection: Necrotizing vasculitis of the vasa vasorum. J. Rheumatol. 1992, 19, 1632–1636. [Google Scholar] [PubMed] Aggarwal, A.; Shukla, A. Polyarteritis nodosa presenting as peripheral vascular disease and acute limb ischemia. J. Postgrad. Med. 2017, 63, 47–49. [Google Scholar] [CrossRef] Holscher, C.M.; Stonko, D.P.; Weaver, M.L.; Reifsnyder, T. Successful bilateral popliteal-plantar bypasses for polyarteritis nodosa induced ischemia. J. Vasc. Surg. Cases Innov. Tech. 2020, 7, 152–156. [Google Scholar] [CrossRef] [PubMed] Zahoor, S.; Siddique, S.; Mahboob, H.M. An unusual presentation of polyarteritis nodosa: A case report. Reumatol. Clin. 2022, 18, 124–126. [Google Scholar] [CrossRef] [PubMed] Akova, Y.A.; Jabbur, N.S.; Foster, C.S. Ocular presentation of polyarteritis nodosa. Clinical course and management with steroid and cytotoxic therapy. Ophthalmology 1993, 100, 1775–1781. [Google Scholar] [CrossRef] [PubMed] Miloslavsky, E.; Unizony, S. The heart in vasculitis. Rheum. Dis. Clin. N. Am. 2014, 40, 11–26. [Google Scholar] [CrossRef] Hsu, C.T.; Kerrison, J.B.; Miller, N.R.; Goldberg, M.F. Choroidal infarction, anterior ischemic optic neuropathy, and central retinal artery occlusion from polyarteritis nodosa. Retina 2001, 21, 348–351. [Google Scholar] [CrossRef] Matsumoto, T.; Homma, S.; Okada, M.; Kuwabara, N.; Kira, S.; Hoshi, T.; Uekusa, T.; Saiki, S. The lung in polyarteritis nodosa: A pathologic study of 10 cases. Hum. Pathol. 1993, 24, 717–724. [Google Scholar] [CrossRef] [PubMed] Fort, J.G.; Griffin, R.; Tahmoush, A.; Abruzzo, J.L. Muscle involvement in polyarteritis nodosa: Report of a patient presenting clinically as polymyositis and review of the literature. J. Rheumatol. 1994, 21, 945–948. [Google Scholar] [PubMed] Grambow-Velilla, J.; Braun, T.; Pop, G.; Louzoun, A.; Soussan, M. Aortitis PET Imaging in VEXAS Syndrome: A Case Report. Clin. Nucl. Med. 2022, 48, e67–e68. [Google Scholar] [CrossRef] [PubMed] Fayand, A.; Sarrabay, G.; Belot, A.; Hentgen, V.; Kone-Paut, I.; Grateau, G.; Melki, I.; Georgin-Lavialle, S. Multiple facets of ADA2 deficiency: Vasculitis, auto-inflammatory disease and immunodeficiency: A literature review of 135 cases from literature. Rev. Med. Interne 2018, 39, 297–306. [Google Scholar] [CrossRef] [PubMed] Chung, S.A.; Gorelik, M.; Langford, C.A.; Maz, M.; Abril, A.; Guyatt, G.; Archer, A.M.; Conn, D.L.; Full, K.A.; Grayson, P.C.; et al. 2021 American College of Rheumatology/Vasculitis Foundation Guideline for the Management of Polyarteritis Nodosa. Arthritis Rheumatol. 2021, 73, 1384–1393. [Google Scholar] [CrossRef] Terrier, B.; Darbon, R.; Durel, C.-A.; Hachulla, E.; Karras, A.; Maillard, H.; Papo, T.; Puechal, X.; Pugnet, G.; Quemeneur, T.; et al. French recommendations for the management of systemic necrotizing vasculitides (polyarteritis nodosa and ANCA-associated vasculitides). Orphanet J. Rare Dis. 2020, 15, 351. [Google Scholar] [CrossRef] Ribi, C.; Cohen, P.; Pagnoux, C.; Mahr, A.; Arène, J.-P.; Puéchal, X.; Carli, P.; Kyndt, X.; Le Hello, C.; Letellier, P.; et al. Treatment of polyarteritis nodosa and microscopic polyangiitis without poor-prognosis factors: A prospective randomized study of one hundred twenty-four patients. Arthritis Rheum. 2010, 62, 1186–1197. [Google Scholar] [CrossRef] Samson, M.; Puéchal, X.; Mouthon, L.; Devilliers, H.; Cohen, P.; Bienvenu, B.; Ly, K.H.; Bruet, A.; Gilson, B.; Ruivard, M.; et al. Microscopic polyangiitis and non-HBV polyarteritis nodosa with poor-prognosis factors: 10-year results of the prospective CHUSPAN trial. Clin. Exp. Rheumatol. 2017, 35 (Suppl. 103), 176–184. [Google Scholar] Hadjadj, J.; Canzian, A.; Karadag, O.; Contis, A.; Maurier, F.; Sanges, S.; Sartorelli, S.; Denis, L.; de Moreuil, C.; Durel, C.-A.; et al. Use of biologics to treat relapsing and/or refractory polyarteritis nodosa: Data from a European collaborative study. Rheumatology 2022, 62, 341–346. [Google Scholar] [CrossRef] Guillevin, L.; Fain, O.; Lhote, F.; Jarrousse, B.; Huong, D.L.T.; Bussel, A.; Leon, A. Lack of superiority of steroids plus plasma exchange to steroids alone in the treatment of polyarteritis nodosa and Churg-Strauss syndrome. A prospective, randomized trial in 78 patients. Arthritis Rheum. 1992, 35, 208–215. [Google Scholar] [CrossRef] Machet, L.; Vincent, O.; Machet, M.C.; Barruet, K.; Vaillant, L.; Lorette, G. Cutaneous periarteritis nodosa resistant to combined corticosteroids and immunosuppressive agents. Efficacy of treatment with intravenous immunoglobulins. Ann. Dermatol. Venereol. 1995, 122, 769–772. [Google Scholar] [PubMed] Seri, Y.; Shoda, H.; Hanata, N.; Nagafuchi, Y.; Sumitomo, S.; Fujio, K.; Yamamoto, K. A case of refractory polyarteritis nodosa successfully treated with rituximab. Mod. Rheumatol. 2015, 27, 696–698. [Google Scholar] [CrossRef] [PubMed] Ribeiro, E.; Cressend, T.; Duffau, P.; Grenouillet-Delacre, M.; Rouanet-Larivière, M.; Vital, A.; Longy-Boursier, M.; Mercié, P. Rituximab Efficacy during a Refractory Polyarteritis Nodosa Flare. Case Rep. Med. 2009, 2009, 738293. [Google Scholar] [CrossRef] [PubMed] Al-Homood, I.A.; Aljahlan, M.A. Successful use of combined corticosteroids and rituximab in a patient with refractory cutaneous polyarteritis nodosa. J. Dermatol. Dermatol. Surg. 2017, 21, 24–26. [Google Scholar] [CrossRef] Boistault, M.; Corbeto, M.L.; Quartier, P.; Arcobé, L.B.; Durall, A.C.; Aeschlimann, F.A. A young girl with severe polyarteritis nodosa successfully treated with tocilizumab: A case report. Pediatr. Rheumatol. Online J. 2021, 19, 168. [Google Scholar] [CrossRef] [PubMed] Pașa, V.; Popa, E.; Poroch, M.; Cosmescu, A.; Bacusca, A.I.; Slanina, A.M.; Ceasovschih, A.; Stoica, A.; Petroaie, A.; Ungureanu, M.; et al. The “Viral” Form of Polyarteritis Nodosa (PAN)-A Distinct Entity: A Case Based Review. Medicina 2023, 59, 1162. [Google Scholar] [CrossRef] Mekinian, A.; Grignano, E.; Braun, T.; Decaux, O.; Liozon, E.; Costedoat-Chalumeau, N.; Kahn, J.-E.; Hamidou, M.; Park, S.; Puéchal, X.; et al. Systemic inflammatory and autoimmune manifestations associated with myelodysplastic syndromes and chronic myelomonocytic leukaemia: A French multicentre retrospective study. Rheumatology 2015, 55, 291–300. [Google Scholar] [CrossRef] Boyadzhieva, Z.; Ruffer, N.; Kötter, I.; Krusche, M. How to treat VEXAS syndrome: A systematic review on effectiveness and safety of current treatment strategies. Rheumatology 2023, 62, 3518–3525. [Google Scholar] [CrossRef] Diarra, A.; Duployez, N.; Fournier, E.; Preudhomme, C.; Coiteux, V.; Magro, L.; Quesnel, B.; Heiblig, M.; Sujobert, P.; Barraco, F.; et al. Successful allogeneic hematopoietic stem cell transplantation in patients with VEXAS syndrome: A 2-center experience. Blood Adv. 2022, 6, 998–1003. [Google Scholar] [CrossRef] Comont, T.; Heiblig, M.; Rivière, E.; Terriou, L.; Rossignol, J.; Bouscary, D.; Rieu, V.; Le Guenno, G.; Mathian, A.; Aouba, A.; et al. Azacitidine for patients with Vacuoles, E1 Enzyme, X-linked, Autoinflammatory, Somatic syndrome (VEXAS) and myelodysplastic syndrome: Data from the French VEXAS registry. Br. J. Haematol. 2021, 196, 969–974. [Google Scholar] [CrossRef] Sais, G.; Vidaller, A.; Servitje, O.; Jucgla, A.; Peyri, J. Leukocytoclastic vasculitis and common variable immunodeficiency: Successful treatment with intravenous immune globulin. J. Allergy Clin. Immunol. 1996, 98, 232–233. [Google Scholar] [CrossRef] [PubMed] Figure 1. Comparison of various etiologies of PAN across different cohorts over distinct time periods. Adapted from Refs. [7,10,12,61]. Table 1. Characteristics of PAN patients reported in different cohorts (PNP: peripheral neuropathy, CNS: central nervous system, FFS: Five Factor Score, 1996). Results are expressed as percentages. \| Characteristics \| Pagnoux et al. (1963 to 2005) \| Sönmez et al. (1990 to 2015) \| Rohmer et al. (2005 to 2019) \| Georgin-Lavialle et al. (VEXAS) \| Meyts et al. (ADA2) \| --- --- --- \| \| General symptoms \| 93.1 \| \| 85 \| 95.7 \| 50 \| \| Fever \| 63.8 \| 53.7 \| 54 \| 64.6 \| \| \| Loss of weight \| 69.5 \| 53.7 \| 50 \| 54.5 \| \| \| Myalgia \| 58.6 \| 46.2 \| 50 \| \| \| \| All cutaneous \| 49.7 \| 67.2 \| 59 \| 83.6 \| 75 \| \| Nodules \| 17.2 \| \| \| \| 14 \| \| Purpura \| 22.1 \| \| \| \| \| \| Livedo \| 16.7 \| 17.9 \| \| \| 50 \| \| Panniculitis \| \| \| 7.5 \| 12.9 \| \| \| Renal \| 50.6 \| 47.7 \| 20 \| 9.5 \| \| \| Hematuria \| 15.2 \| \| \| \| \| \| Proteinuria \| 21.6 \| \| \| \| \| \| Hypertension \| 34.8 \| 41.7 \| \| \| 21 \| \| Orchitis \| 17 \| 14.9 \| 16 \| \| 4 \| \| Neurologic \| 79.0 \| 43.2 \| 59 \| \| \| \| PNP \| 74.1 \| \| \| 5.2 \| 9 \| \| Mononeuritis \| 70.7 \| \| \| 2.6 \| \| \| CNS \| 4.6 \| \| \| \| 53 \| \| Digestive \| 37.9 \| 22.3 \| 28 \| 13.8 \| 33 \| \| Abdominal pain \| 35.6 \| 37.3 \| \| 8.6 \| 12 \| \| Bleeding \| 3.4 \| \| \| 0.9 \| \| \| Perforation \| 4.3 \| \| \| 0.9 \| 2 \| \| Cardiovascular \| 22.4 \| \| 39 \| \| \| \| Pericarditis \| 5.5 \| \| \| 4.3 \| \| \| Distal necrosis \| 6.3 \| 13.5 \| \| \| 22 \| \| Thrombo-embolism \| \| \| \| 35.3 \| \| \| Ophthalmic \| 8.6 \| \| \| 40.5 \| \| \| Retinal vasculitis \| 4.3 \| \| \| \| \| \| Pulmonary \| \| 2.9 \| 8 \| 49.1 \| \| \| Cough \| 5.7 \| \| \| \| \| \| Lung infiltrate \| 3.4 \| \| \| 40.5 \| \| \| Pleural effusion \| 3.4 \| \| \| 9.5 \| \| \| Chondritis \| \| \| \| 36.2 \| \| \| Arthralgia \| 48.9 \| 58.2 \| \| 28.4 \| \| \| Arthritis \| \| 17.9 \| \| \| \| \| \| \| --- \| \| \| Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. \| © 2023 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license ( Share and Cite MDPI and ACS Style Wolff, L.; Horisberger, A.; Moi, L.; Karampetsou, M.P.; Comte, D. Polyarteritis Nodosa: Old Disease, New Etiologies. Int. J. Mol. Sci. 2023, 24, 16668. AMA Style Wolff L, Horisberger A, Moi L, Karampetsou MP, Comte D. Polyarteritis Nodosa: Old Disease, New Etiologies. International Journal of Molecular Sciences. 2023; 24(23):16668. Chicago/Turabian Style Wolff, Louis, Alice Horisberger, Laura Moi, Maria P. Karampetsou, and Denis Comte. 2023. "Polyarteritis Nodosa: Old Disease, New Etiologies" International Journal of Molecular Sciences 24, no. 23: 16668. APA Style Wolff, L., Horisberger, A., Moi, L., Karampetsou, M. P., & Comte, D. (2023). Polyarteritis Nodosa: Old Disease, New Etiologies. International Journal of Molecular Sciences, 24(23), 16668. Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here. Article Metrics No Article Access Statistics For more information on the journal statistics, click here. Multiple requests from the same IP address are counted as one view. Zoom \| Orient \| As Lines \| As Sticks \| As Cartoon \| As Surface \| Previous Scene \| Next Scene Back to TopTop
4551	https://math.stackexchange.com/questions/3028013/lim-x-to-0-sin-x-tan-x	limits - $\lim_{x\to 0^+} (\sin x)^ {\tan x}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more lim x→0+(sin x)tan x lim x→0+(sin⁡x)tan⁡x Ask Question Asked 6 years, 9 months ago Modified6 years, 9 months ago Viewed 4k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I have stumbled on this question in one of my problem sets from Cal I and I'm not sure how to proceed after the last step. lim x→0+(sin x)tan x lim x→0+(sin⁡x)tan⁡x // Applying exponential rule x=e ln(x)x=e ln⁡(x) lim x→0+e x p[ln((sin x)tan x)]lim x→0+e x p[ln⁡((sin⁡x)tan⁡x)] // Using natural logarithm property to bring the exponent to the front lim x→0+e x p[(tan x)ln(sin x)]lim x→0+e x p[(tan⁡x)ln⁡(sin⁡x)] // Using an algebra trick where : x=1 1 x x=1 1 x lim x→0+e x p[ln(sin x)1 tan x]lim x→0+e x p[ln⁡(sin⁡x)1 tan⁡x] // After this step, we were taught to check for Hospital's rule, however in this case when you plug the value 0 0 in the numerator, you get ln(sin 0)ln⁡(sin⁡0) which is equal to ln(0)ln⁡(0). This is the part that is confusing me since ln(0)ln⁡(0) is undefined. // Could someone please explain to me why this whole limit is equal to 1 1? Does it have to do with the fact that x x is approaching 0 0 from the right side? limits Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Dec 6, 2018 at 4:01 Larry 5,170 4 4 gold badges 22 22 silver badges 49 49 bronze badges asked Dec 6, 2018 at 3:55 DBlykDBlyk 35 1 1 silver badge 5 5 bronze badges 1 1 So your limit is ∞∞∞∞ and with l'Hospital lim ln sin x cot x=−1 2 lim sin 2 x=0 lim ln⁡sin⁡x cot⁡x=−1 2 lim sin⁡2 x=0 Nosrati –Nosrati 2018-12-06 04:02:11 +00:00 Commented Dec 6, 2018 at 4:02 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Remember what L'Hospital's rule says: it's not a direct substitution of x=0 x=0, it's the limit: what happens as x x approaches 0 0, and specifically from the right (since we have 0+0+)? Generally, where the limits exist and all the other nuances associated with the definition, the rule states lim x→c f(x)g(x)=lim x→c f′(x)g′(x)lim x→c f(x)g(x)=lim x→c f′(x)g′(x) if the original limit is one of the indeterminate forms 0/0 0/0 or ±∞/±∞±∞/±∞. Now, in your problem, since the exponential function is continuous, we can say lim x→0+e x p[ln(sin x)1 tan x]=e x p[lim x→0+(ln(sin x)1 tan x)]lim x→0+e x p[ln⁡(sin⁡x)1 tan⁡x]=e x p[lim x→0+(ln⁡(sin⁡x)1 tan⁡x)] So notice, as x→0 x→0 from the right - not simply x=0 x=0 - sin(x)→0 sin⁡(x)→0. Since ln(x)→−∞ln⁡(x)→−∞ as x→0 x→0 from the right, and we have sin(x)→0 sin⁡(x)→0, it follows that ln(sin(x))→−∞ln⁡(sin⁡(x))→−∞ as x→0+x→0+. Further, as x→0+x→0+, tan(x)→0 tan⁡(x)→0 and thus 1/tan(x)→+∞1/tan⁡(x)→+∞. Thus, we can apply L'Hospital's rule since the limit as-is is an indeterminate form ±∞/±∞±∞/±∞. Differentiate the numerator and denominator (preferably after swapping cot(x)=1/tan(x)cot⁡(x)=1/tan⁡(x) for clarity's sake) and try to find the limit as x→0+x→0+ after. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Dec 6, 2018 at 4:05 PrincessEevPrincessEev 51.3k 5 5 gold badges 60 60 silver badges 142 142 bronze badges 1 2 So derivative of ln(sin x)=cot x ln⁡(sin⁡x)=cot⁡x and the derivative of cot x=−csc 2 x cot⁡x=−csc 2⁡x. So as x→0 x→0 the term becomes e 0=1 e 0=1 Hence the limit is one.Winodd Dhamnekar –Winodd Dhamnekar 2018-12-06 06:29:57 +00:00 Commented Dec 6, 2018 at 6:29 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 limits See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 2Limit of the function lim x→0 sin 2 x+a sin x x 3 lim x→0 sin⁡2 x+a sin⁡x x 3? 4Evaluate lim x→0 tan(tan x)−sin(sin x)tan x−sin x lim x→0 tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡x 3Evaluate limit lim x→0 ln(tan 2 x)ln(tan 3 x)lim x→0 ln⁡(tan⁡2 x)ln⁡(tan⁡3 x) 6Limit of the exponential functions: lim x→0 e x−e x cos x x+sin x lim x→0 e x−e x cos⁡x x+sin⁡x 2Find the lim n→0(cos(x)+tan 2(x))csc(x)lim n→0(cos⁡(x)+tan 2⁡(x))csc⁡(x) 1How to use L Hospital's rule for lim x→∞x−−√sin(1 x)lim x→∞x sin⁡(1 x) 0For what value(s) of a a does this lim x→0 a sin x−a tan x tan x−sin x lim x→0 a sin⁡x−a tan⁡x tan⁡x−sin⁡x equal 1?1? 5Evaluating lim x→π 6(1−2 sin(x))tan(π 6−x)lim x→π 6(1−2 sin⁡(x))tan⁡(π 6−x) 4Finding lim x→0 tan(x)1/x lim x→0 tan⁡(x)1/x 2Evaluating lim x→π/2 tan 11 x−tan 7 x tan 9 x+tan 13 x lim x→π/2 tan⁡11 x−tan⁡7 x tan⁡9 x+tan⁡13 x without L'Hopital Hot Network Questions On Chern roots and the Chern polynomial Clinical-tone story about Earth making people violent Numbers Interpreted in Smallest Valid Base Languages in the former Yugoslavia Lingering odor presumably from bad chicken Can a cleric gain the intended benefit from the Extra Spell feat? Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done? How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? How to start explorer with C: drive selected and shown in folder list? Determine which are P-cores/E-cores (Intel CPU) Is direct sum of finite spectra cancellative? Are there any world leaders who are/were good at chess? Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation ConTeXt: Unnecessary space in \setupheadertext Is existence always locational? What were "milk bars" in 1920s Japan? Cannot build the font table of Miama via nfssfont.tex Do sum of natural numbers and sum of their squares represent uniquely the summands? Do we need the author's permission for reference Does the curvature engine's wake really last forever? Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Riffle a list of binary functions into list of arguments to produce a result My dissertation is wrong, but I already defended. How to remedy? The geologic realities of a massive well out at Sea more hot questions Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4552	https://www.desmos.com/calculator/kxmpaisq9d	Expression 1: "f" left parenthesis, "x" , right parenthesis equals "x" cubed plus 1fx=x3+1 1 Expression 2: "g" left parenthesis, "y" , right parenthesis equals left parenthesis, "y" minus 1 , right parenthesis Superscript, 1 third , Baselinegy=y−113 2 Compute the integral from a to b: 3 Expression 4: Start integral from "a" to "b" , end integral, "g" left parenthesis, "t" , right parenthesis "d" "t"∫bagtdt equals= 2.2 8 1 2 6 5 2 2 9 2 22.28126522922 4 Expression 5: "a" equals 2.6a=2.6 negative 10−10 1010 5 Expression 6: "b" equals 4.3b=4.3 negative 10−10 1010 6 Visualize the area above the curve: 7 Expression 8: left brace, "g" left parenthesis, "y" , right parenthesis greater than 0 : 0 , "g" left parenthesis, "y" , right parenthesis less than 0 : "g" left parenthesis, "y" , right parenthesis , right brace less than "x" less than left brace, "g" left parenthesis, "y" , right parenthesis greater than 0 : "g" left parenthesis, "y" , right parenthesis , "g" left parenthesis, "y" , right parenthesis less than 0 : 0 , right brace left brace, "a" less than "y" less than "b" , "b" less than "y" less than "a" , right bracegy>0: 0, gy<0: gy<x< gy>0: gy, gy<0: 0a<y<b, b<y<a 8 9 powered by powered by "x"x "y"y "a" squareda2 "a" Superscript, "b" , Baselineab 77 88 99 divided by÷ functions (( )) less than< greater than> 44 55 66 times× \| "a" \|\|a\| ,, less than or equal to≤ greater than or equal to≥ 11 22 33 negative− A B C StartRoot, , EndRoot piπ 00 .. equals= positive+
4553	https://www.intmath.com/functions-and-graphs/the-hypotenuse-of-a-right-triangle.php	Interactive Mathematics Thank you for booking, we will follow up with available time slots and course plans. On this page Introduction to Geometry Functions and Graphs 1. Introduction to Functions 2. Functions from Verbal Statements 3. Rectangular Coordinates 4. The Graph of a Function 4a. Domain and Range of a Function 4b. Domain and Range interactive applet 4c. Comparison calculator BMI - BAI 5. Graphing Using a Computer Algebra System 5a. Online graphing calculator (1): Plot your own graph (JSXGraph) 5b. Online graphing calculator (2): Plot your own graph (SVG) 6. Graphs of Functions Defined by Tables of Data 7. Continuous and Discontinuous Functions 8. Split Functions 9. Even and Odd Functions Related Sections Math Tutoring Need help? Chat with a tutor anytime, 24/7. Chat now Online Math Solver Solve your math problem step by step! Online Math Solver IntMath Forum Get help with your math queries: See Forum The hypotenuse of a right triangle The hypotenuse of a right triangle In geometry, the hypotenuse is the side of a right triangle that is opposite to the right angle. The other two sides are called the legs of the triangle. The length of the hypotenuse of a right triangle can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. The term "hypotenuse" is derived from Greek, and it means "stretched under." In a right triangle, the hypotenuse is always the longest side. It is also sometimes referred to as the "major" or "longest" side. The two shorter sides are known as "legs," and they are usually labeled with letters corresponding to their respective angles: A for angle A, B for angle B, and so on. To find the length of the hypotenuse, you can use either the Pythagorean theorem or trigonometry. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This formula is represented by: c^2 = a^2 + b^2 where c is the length of the hypotenuse, and a and b are lengths of the other two sides. This theorem is also sometimes called "Pythagoras's theorem." You can also use trigonometry to find missing sides or angles in a right triangle. The most common trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). These ratios are defined as follows: sin(A) = opposite/hypotenuse cos(A) = adjacent/hypotenuse tan(A) = opposite/adjacent Where A is an angle in a right triangle, and opposite and adjacent refer to sides that are opposite and adjacent to angle A, respectively. Conclusion: In conclusion, in geometry, the hypotenuse is the side of a right triangle that is opposite to the rightangle .It maybefound usingthe Pythagoreantheorem . Trigonometrycan also be usedto find missing partsin a righttriangle .When tryingto findthe lengthofa line ,you must firstdetermineifthattriangleis arighttriangle .If so ,you c anusethe PythagoreanTheoremto solvefor c . However ,if you don'tknow ifthetriangleis arighttriangleor not ,you shoulddefaultto usingtrigonometryinstead. FAQ What is hypotenuse in simple words? In simple terms, the hypotenuse is the longest side of a right angled triangle. It is the side opposite of the right angle and is usually denoted by the letter 'h'. To find the length of the hypotenuse, you can use either the Pythagorean theorem or trigonometry. Where is the hypotenuse? The hypotenuse is always the side of a right triangle that is opposite to the right angle. In other words, it is the longest side of the triangle. It is also sometimes referred to as the "major" or "longest" side. How do you find the hypotenuse with Pythagorean theorem? The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This formula is represented by: c^2 = a^2 + b^2 where c is the length of the hypotenuse, and a and b are lengths of the other two sides. How do you find the hypotenuse example? A common example of finding the length of the hypotenuse is using the Pythagorean theorem. In the formula, c^2=a^2+b^2, c is the hypotenuse and a and b are the other two sides. So, to find c, you would solve for c by taking the square root of a^2+b^2. For example, if a=3 and b=4, then c=5 because 3^2+4^2=25 and the square root of 25 is 5. What is the hypotenuse of a 1-1-sqrt 2 triangle? The hypotenuse of a 1-1-sqrt 2 triangle is sqrt 2. This can be found by using the Pythagorean theorem which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In this case, the length of side a is 1, the length of side b is 1, and the length of side c is sqrt 2. Therefore, c^2=1^2+1^2=2 which means that the hypotenuse is sqrt 2. What is an example of a right triangle? A common example of a right triangle is a 3-4-5 triangle. In this triangle, the length of side a is 3, the length of side b is 4, and the length of side c is 5. The hypotenuse is the longest side and is equal to 5. This can be verified by using the Pythagorean theorem which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In this case, c^2=3^2+4^2=9+16=25 and the square root of 25 is 5. What are the different types of right triangles? There are two different types of right triangles: the 3-4-5 triangle and the 30-60-90 triangle. The 3-4-5 triangle is a right triangle where the length of side a is 3, the length of side b is 4, and the length of side c is 5. The hypotenuse is the longest side and is equal to 5. The 30-60-90 triangle is a right triangle where the length of side a is 30, the length of side b is 60, and the length of side c is 90. The hypotenuse is the longest side and is equal to 90. What are the different parts of a right triangle? The three parts of a right triangle are the hypotenuse, side a, and side b. The hypotenuse is the longest side and is opposite of the right angle. Side a and side b are the other two sides of the triangle that form the right angle. What is an obtuse triangle? An obtuse triangle is a triangle where one of the angles is greater than 90 degrees. This means that the length of the hypotenuse will be greater than the length of either side a or side b. What is an acute triangle? An acute triangle is a triangle where all of the angles are less than 90 degrees. This means that the length of the hypotenuse will be less than the length of either side a or side b. What is an equilateral triangle? An equilateral triangle is a triangle where all of the sides are equal in length. This means that the lengths of side a, side b, and the hypotenuse will all be equal. What is an isosceles triangle? An isosceles triangle is a triangle where two of the sides are equal in length. This means that either side a and side b will be equal, or side a and the hypotenuse will be equal, or side b and the hypotenuse will be equal. Functions and Graphs Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.
4554	https://www.cs.princeton.edu/courses/archive/spr05/cos598E/bib/covertree.pdf	COVER TREES FOR NEAREST NEIGHBOR ALINA BEYGELZIMER, SHAM KAKADE, AND JOHN LANGFORD ABSTRACT. We present a tree data structure for fast nearest neighbor operations in general -point metric spaces. The data structure requires space regardless of the metric’s structure. If the point set has an expansion constant in the sense of Karger and Ruhl [KR02], the data structure can be constructed in time. Nearest neighbor queries obeying the expansion bound require time. In addition, the nearest neighbor of points can be queried in time. We experimentally test the algorithm showing speedups over the brute force search varying between 1 and 2000 on natural machine learning datasets. 1. INTRODUCTION Nearest neighbor search is a fundamental problem with a number of applications in peer-to-peer networks, lossy data compression, vision, dimensionality reduction, computational biology, machine learning, and physical simulation. The standard setting is as follows: Given a set of points in some metric space , the problem is to preprocess so that given a query point , one can efﬁciently ﬁnd a point which minimizes . For general metrics, ﬁnding (or even approximating) the nearest neighbor of a point requires time. However, the metrics and datasets of practical interest typically have some structure which may be exploited to yield signiﬁcant computational speedups. Most research has focused on the special case when the metric is Euclidean. Although many strong theoretical guarantees have been made here, this case is quite limiting in a variety of settings when the data do not naturally lie in a Euclidean space or when the data is embedded in a high dimensional space, but has a low dimensional intrinsic structure [TSL00, RS00] (as is common in many machine learning problems). Several notions of metric structure and algorithms exploiting this structure have been proposed [Cla99, KR02, KL04a]. Karger and Ruhl [KR02] stated a notion of the intrinsic dimensionality of a dataset and an efﬁcient randomized algorithm for metric spaces in which this dimension is small. Krauthgamer and Lee [KL04a] suggested a more robust notion of intrinsic dimensionality (based on the theory of analysis in metric spaces [GKL03]), and presented a simpler deterministic data structure, called a navigating net, that is efﬁcient with respect to this notion. A navigating net is a leveled directed acyclic graph where each consequent level “covers” the dataset on a ﬁner scale; adjacent levels are connected by pointers allowing navigation between scales. Both of these algorithms have query time (assuming the Karger-Ruhl abstract dimension is constant). In machine learning applications, most of these theoretically appealing algorithms are not used in prac-tice (in both the Euclidean and the general case) for a variety of reasons. When the Euclidean dimension is small, one typical approach uses KD-trees (see [FBL77]). If the metric is non-Euclidean (or the Euclidean dimension is large), ball trees [Omo87] [Uhl91] provide compelling performance in many practical applica-tions [GM00]. These methods currently have only trivial query time guarantees of , although improved performance may be provable given some form of structure. We propose a simple data structure, a cover tree, for exact and approximate nearest neighbor operations. The data structure improves over the results in [KR02, KL04a] by making the space requirement linear in the dataset size. This space bound is independent of any dimensionality assumptions while a navigating net beygel@us.ibm.com, sham@gatsby.ucl.ac.uk, jl@tti-c.org. 1 COVER TREES FOR NEAREST NEIGHBOR 2 [KL04a] has linear space only when the abstract dimensionality is assumed to be constant with respect to the dataset size. As we observe experimentally (see Section 4), it is common for the Karger-Ruhl dimension to grow with the dataset size, so this latter assumption seems unrealistic in practice. Our algorithms are simple since the data structure being manipulated is a tree — in fact, our data structure (as a graph) is a subgraph of a navigating net [KL04a] (with slightly different parameters). We also provide experiments (and public code [Lan04]) suggesting this approach may be competitive with current practical approaches. 1.1. Intrinsic Dimensionality. We consider both notions of intrinsic dimensionality, proposed by Karger and Ruhl [KR02] and Krauthgamer and Lee [KL04a]. As usual, we denote the closed ball of radius around in by , and when clear from context, we just write . Karger and Ruhl [KR02] consider classes of metrics that satisfy a growth bound. The expansion constant of is deﬁned as the smallest value such that for every and . If is arranged uniformly on some surface of dimension , then , which suggests deﬁning the expansion dimension of as dimKR . However, as previously observed in [KR02, KL04a], some metrics that should intuitively be considered low-dimensional turn out to have large growth constants. For example, adding a single point in a Euclidean space may make the KR-dimension grow arbitrarily. The doubling constant (deﬁned in [KL04a] and motivated by the theory of analysis on metric spaces) provides a more robust alternative. The doubling constant is the minimum value such that every ball in can be covered by balls in of half the radius. The doubling dimension of is then deﬁned as dimKL . This notion of dimensionality is strictly more general than the KR dimension, as shown in [GKL03]. Furthermore, the doubling dimension of -dimensional Euclidean spaces is . A drawback (so far) of working with the doubling dimension is that only weaker results have been provable, and even those apply only to approximate nearest neighbors. The algorithm in [KL04a] depends on the aspect ratio (the ratio of the largest to the smallest interpoint distance). Although the scheme in [KL04b] eliminates this dependence and makes the space consumption , independent of the doubling dimension (while the algorithm in [KL04a] is exponential in the dimension), no linear-space algorithm is known even in the case when the dimension is constant. 1.2. New Algorithmic Results. We present the cover tree data structure and algorithms for using it. First, let us look at how our algorithm fares when no dimensionality assumption is made. For this case, the query time is . Cover Tree Ball Tree Nav. Net [KR02] Construction Space Construction Time In our analysis, we focus primarily on the expansion constant, because this permits results on exact nearest neighbor queries. If is the expansion constant of , then we can state the dependence on explicitly: Cover Tree Nav. Net [KR02] Construction Space Construction Time Insertion/Removal Query Batch Query It is important to note that the algorithms here (as in [KL04a] but not [KR02]) work for arbitrary metrics (with no assumptions on the structure); only the analysis is done with respect to the assumptions. The main advantage of our data structure is its linear space bound independent of any assumptions made about the dataset. The algorithms are also simple since the structure they manipulate is a tree. Comparison of time complexity in terms can be subtle (see the discussion at the end of Section 3.1). Also, such a COVER TREES FOR NEAREST NEIGHBOR 3 comparison is somewhat unfair since past work did not explicitly try to optimize the dependence. We make the dependence precise because further progress must optimize it. The algorithms easily extend to approximate nearest neighbor queries for sets with a bounded doubling dimension, as in [KL04a]. The query times of both algorithms are , where is the aspect ratio and is the approximation parameter. Note that, unlike the navigating net, our space is independent of the doubling dimension. Finally, we provide several algorithms of practical interest. These include a lazy construction (which amortizes the construction cost over queries), a batch construction (which is empirically superior to a se-quence of single point insertions), and a batch query (which amortizes the query time over multiple queries). 1.3. Experimental Results. We experimentally compared our algorithm to an optimized brute force search and to the algorithm [Cla02] on a number of benchmark machine learning datasets. This appears to be the ﬁrst such empirical study looking directly at the viability of algorithms based on intrinsic dimensionality. Figure 4.1(a) shows speedups ranging from a factor of 1 to 2000, depending on the dataset. For fairness, we (1) report every dataset tested, (2) include the time to construct the cover tree. See Section 4 for details. 2. CONSTRUCTING AND USING A COVER TREE 2.1. The Cover Tree Datastructure. A cover tree on a data set is a leveled tree where each level is a “cover” for the level beneath it. Each level is indexed by an integer scale which decreases as the tree is descended. Let denote the set of nodes at level . A cover tree on a dataset obeys the following invariants for all : (1) (nesting) . (2) (covering tree) For every , there exists a satisfying , and exactly one such is a parent of . (3) (separation) For all , . These invariants are essentially the same as used in navigating nets [KL04a], except for (2) where we require only one parent of a node rather than all possible parents. (For every node in , a navigating net keeps all nodes in that are within distance , where is some constant.) Despite potentially throwing out most of the links in a navigating net, all runtime properties can be maintained. We make use of an implicit and an explicit representation of the cover tree. The implicit representation consists of the (inﬁnitely many) levels and the pointers from every node to its children in the level beneath it. The level consists of a single node, called the root of the tree. It is simplest to describe the algorithms in terms of this implicit representation. However, we must use and analyze the explicit representation, which takes only space. First, note that if a point ﬁrst appears in level then it is in all levels below (and, as the following proof shows, is a child of itself in all of these levels). The explicit representation of the tree coalesces all nodes in which the only child is a self-child. This implies that every explicit node either has a parent other than the self-parent or a child other than the self-child, which immediately gives an space bound, independent of the growth constant . Theorem 2.1. (Space bound) A cover tree requires space at most . Proof. Every point has at most one parent other than itself in the explicit tree. To see this, assume and are two parents of . The scale at which and are parents must be different by the covering tree invariant. Nesting implies that is a sibling of the parent at some lower scale . If is the parent at the lower scale, then separation implies which implies that can not be a parent at scale . Every time a point is a parent of itself, it also has another point as a child. Consequently, there are at most links and points implying the space bound. 2.2. Single Point Operations. We now present the basic algorithms for cover trees and prove their correct-ness. The runtime analysis is given in Section 3. COVER TREES FOR NEAREST NEIGHBOR 4 Algorithm 1 Find-Nearest (cover tree , query point ) (1) set . (2) for from down to (a) consider the set of children of : { Children (b) form next cover set: (3) return . 2.2.1. Finding the nearest neighbor of a point. To ﬁnd the nearest neighbor of a point in a cover tree, we descend through the tree level by level, keeping track of a subset of nodes that may contain the nearest neighbor of as a descendant. The algorithm iteratively constructs by expanding to its children in then throwing away any child that cannot lead to the nearest neighbor of . For simplicity of exposition, it is easier to think of the tree as having an inﬁnite number of levels (with containing only the root of the tree, and with ). In what follows, let Children be the set of children of node and let be the distance to the nearest point in a set . Note that although the algorithm is stated using an inﬁnite loop over the implicit representation, it only needs to operate on the explicit representation. Theorem 2.2. If is a cover tree on , then Find-Nearest returns the nearest neighbor of in . Proof. For any in the distance between and any descendant is bounded by . Consequently, step 2(b) can never throw out a grandparent of the nearest neighbor of . Eventually, there are no descendants of not in , and the nearest neighbor must be in . This algorithm can be used for proximity searches in a more general sense. A function Bound deﬁned on ﬁnite sets is said to be monotonic if for any sets and , Bound Bound whenever . Example 2.3. (-nearest neighbors) Given a point and , ﬁnd the set of all points in . In this case, Bound for all . Example 2.4. (-nearest neighbors) Given a point and a number , return the closest points to in . In this case, Bound is the distance to the th closest point in . The algorithm can be modiﬁed as follows to work with any monotonic bound: (1) In Step 3, ﬁnd the subset of points of interest in by brute force. (2) In Step 2(b), substitute Bound for . 2.2.2. Approximating the nearest neighbor of a point. The Cover Tree can also be used to approximate nearest neighbors. Given a point and some , we want to ﬁnd a point satisfying . The main idea is to maintain a lower bound as well as an upper bound, stopping when the interval implied by the bounds is sufﬁciently small. When analyzed with respect to the doubling constant, the proof of the time bound is essentially the same as in [KL04a]. Moreover, the space bound is now linear (independent of the doubling constant), giving a strict improvement over the results in [KL04a]. Algorithm: The only change is in line 2, where instead of descending the tree until no point in is explicit, we stop as soon as . Proof of correctness: Suppose that the descent terminated in level . Then either or all points in are implicit (in which case we actually return the exact nearest neighbor). Let us consider the former case. Since is at distance at most from the exact nearest neighbor of (Theorem 2.2), and satisﬁes the triangle inequality, we have . Combining with COVER TREES FOR NEAREST NEIGHBOR 5 Algorithm 2 Insert(point , cover set , level ) (1) set Children (2) if then return “no parent found” (3) else (a) set . (b) if Insert(, , ) = “no parent found” and (i) pick satisfying (ii) insert into Children (iii) return “parent found” (c) else return “no parent found” , this gives , or . Hence, we have The time complexity follows from inspection of Lemma 2.6 in [KL04a]. In particular, an approximate query takes at most , where is the doubling constant and is the aspect ratio. 2.2.3. Single Point Insertion. The insertion algorithm (algorithm 2) is similar to the ﬁnd nearest neighbor algorithm although the algorithm is stated recursively. Here, is a subset of the points at level , which may contain as a descendant (after insertion). The algorithm starts with , the root node of . The proof of correctness implicitly shows that the datastructure always exists. Theorem 2.5. If is a cover tree on , then Insert returns a cover tree on . Proof. Let us prove that the algorithm is guaranteed to insert any not already contained in the cover tree. (If is in the tree, this can be determined with a single invocation of the search procedure.) The set starts non-empty. Since is not already in the tree, is nonzero, and the condition in line 2 must eventually hold. Since the root is has scale , there will be some minimal scale between and the scale where line 2 ﬁrst holds such that and so 3.(b) holds. We now prove that the insertion maintains all the cover tree invariants. If is inserted in level , we know that , and thus we can always ﬁnd a parent with , satisfying the covering tree invariant. Once is inserted in level , it is implicitly inserted in every level beneath it (as a child of itself in the previous level), maintaining the nesting invariant. Next we show that doing so does not violate the separation condition in lower levels. To prove the separation condition in level , consider . If , then . If , then at some iteration , some parent of , say , was eliminated (in Step 3a), which implies that . Using the covering tree invariant at level we have which proves the desired separation . Separation at levels below is proved similarly. 2.2.4. Single Point Removal. The removal (Algorithm 3) is similar to insertion, with extra complexity due to coping with children of removed nodes. Theorem 2.6. Given a cover tree on , Remove returns a cover tree on . Proof. As before, sets maintain points in level closest to , as we descend through the tree decrementing . The recursion stops when it reaches the level below which is always implicit. For each level explicitly containing , we remove from and from the list of children of its parent in . This does not disturb the nesting and the separation invariants. For each child of (by this time has already been removed from the list of its children), we go up the tree looking for a new parent. More precisely, if there exists a node such that we make a parent of ; otherwise, we COVER TREES FOR NEAREST NEIGHBOR 6 Algorithm 3 Remove(point , cover sets , level ) (1) set Children (2) set (3) Remove(, , , ) (4) if then (a) remove from (b) Remove from Children(Parent(p)) (c) for every Children (i) set . (ii) while (A) insert into (and ) (B) set (iii) pick satisfying (iv) make point to insert in level and repeat, propagating up the tree until a parent is found. Insertion does not violate the separation and the nesting constraints, since (otherwise we would not be inserting in ). This propagation process is guaranteed to terminate since is covered by the root (at the scale of the root). Hence the covering tree invariant is enforced for all children of . 2.3. Batch and Lazy Variants. In this section, we present batch and lazy variants of practical interest. 2.3.1. Batch Construction. It is common to start with some large dataset (instead of receiving points one by one in an online manner). It is thus natural to try to amortize the construction cost over points. We now present such a batch algorithm. The analysis is given in the next section. Theoretically, the algorithm has the same guarantees as a series of single point insertions; however, preliminary empirical studies show that it is considerably faster (roughly twice as fast, although a controlled comparison has yet to be done). At each step in the recursive construction, the algorithm has a point , a set of points NEAR which must be inserted beneath , and a set of points FAR which might be inserted beneath . The algorithm ﬁrst ﬁnds the near and far sets for itself (as a child), and then does the same for the remaining elements of NEAR. It then returns the created node, together with any unused elements of FAR. To describe the algorithm, we need a helper function, SPLIT which splits the ele-ments of into points satisfying and points satisfying . All such points are removed from . The batch construction algorithm (algorithm 4) starts with a call to Construct( . The proof is removed due to space limitations (see [BKL04]). Theorem 2.7. (Batch Correctness) Construct returns a valid cover tree for . Proof. Nesting holds because we explicitly construct the self-child of a point. Covering holds because all children of are always in the near set and thus at distance at most . To show separation at level , note that for any points satisfying , the procedure is ﬁrst called on either or . Assume without loss of generality that it is called on . Then is in NEAR (otherwise it would already be in the tree contradicting the assumption that came ﬁrst), and the algorithm makes either a child or a grandchild of at steps 3(b) or 3(d)(iii). 2.3.2. Finding nearest neighbors of many points. We can modify the previous query algorithm to simul-taneously ﬁnd the nearest neighbors of multiple points (similar to a method used in KD-trees [GM00]). The set of query points is given by a cover tree (which can be done by preprocessing the query set). This allows the descent of the search tree to be amortized over all queries improving the time complexity from to . In Algorithm 5, we abuse notation by letting denote the subtree of the query tree COVER TREES FOR NEAREST NEIGHBOR 7 Algorithm 4 Construct(point , point sets NEAR FAR , level ) (1) if NEAR (2) then return (3) else (a) SELF, NEAR = Construct SPLIT NEAR (b) add SELF to Children (c) while NEAR (i) pick in NEAR (ii) CHILD, UNUSED = Construct SPLIT NEAR,FAR (iii) add CHILD to Children (iv) let NEW-NEAR, NEW-FAR =SPLIT UNUSED (v) add NEW-FAR to FAR, and NEW-NEAR to NEAR. (d) return FAR . Algorithm 5 Find-All-Nearest (query cover tree , cover set ) (1) if then for each return as the nearest neighbor of . (2) else (a) if then (i) Set { Children . (ii) Set . (iii) Find-All-Nearest ( , ) (b) else for each Children Find-All-Nearest ( , ) rooted at node . Since a point can appear in different levels, a subscript disambiguates as necessary. denotes the set of leaves of the explicit part of the subtree. The procedure is recursive, starting with the root of the query cover tree and . The correctness argument is similar to Theorem 2.2. 2.3.3. Lazy Construction. When few queries are done, the overhead of the construction may be greater than the brute force query cost. This drawback disappears with lazy construction. The basic idea is to make the nodes of a cover tree be queries modiﬁed to hold a list of points. Now every point is in the list of node at some level must satisfy the following invariants: (1) (closeness) . (2) (minimal level) There does not exist with . Note, in particular, that we do not impose any separation constraint on nodes in the same layer. The cover tree over completed queries is constructed using a variant of a single point insertion (see Section 2.2.3) modiﬁed to satisfy the above invariants. We only need the following modiﬁcations: (1) For every in , at every level , compute (and record) the distance to every point attached to . (2) At the insertion level (say level ) attach every point within distance from level . Remove all newly attached points from higher level lists. Note that levels for which might have been implicit in the cover tree may not necessarily be implicit in the lazy cover tree. This is actually desirable because it organizes points relative to in a ﬁner manner. Theorem 2.8. (Correctness) The above modiﬁcations preserve the lazy cover tree invariants. COVER TREES FOR NEAREST NEIGHBOR 8 Proof. Closeness is preserved because we explicitly only attach points within distance . Minimal level is preserved because we only attach points from higher levels. Queries on a lazy cover tree are exactly the same as on a cover tree except that every point in the list of every node is treated as a leaf. Several observations are in order. (1) If every point is queried, the resulting cover tree is exactly the same as would be constructed using single point insertion. (2) The set of distance calculations required for insertion is a subset of the set required for a query. This holds because the distance constraint of step 3(a) in Insert is stronger than the distance constraint in step 2(b) of Find-Nearest. Consequently, the query can fully amortize the insertion. 3. THE RUNTIME ANALYSIS In this section, the distinction between implicit and explicit representation (see Section 2.1) is important. 3.1. Query analysis. We start with three lemmas about some structural properties of the data structure. Lemma 3.1. (Width bound) The number of children of any node is bounded by . Proof. Let be in level . The number of its children is at most , which is certainly bounded by . The idea of the proof is to bound the number of disjoint balls of radius that we can pack into . Each of these balls can cover at most one point in , thereby bounding the number of children. For any child of , since , we have implying The balls must be disjoint for all , since the points in are at least apart. We also know that each is contained within , since . Then the number of disjoint balls around the children that can be packed into is bounded by which gives a bound on the number of children of . The following lemma is useful in bounding the depth of the tree. It says that if there is a point in some annulus centered around , then the volume growth of a sufﬁciently large ball around containing the annulus is non-trivial. In other words, it gives a lower bound on the volume growth in terms of the growth constant , while the deﬁnition of gives an upper bound. Lemma 3.2. (Growth Bound) For all points and , if there exists a point such that , then Proof. Since , we have And since the balls and are disjoint and are subsets of , we have The result follows by combining these inequalities. Using this, we can prove a bound on the explicit depth of any point , deﬁned as the number of explicit grandparent nodes on the path from the root to in the lowest level in which is explicit. Lemma 3.3. (Depth Bound) The maximum depth of any point in the explicit representation is . COVER TREES FOR NEAREST NEIGHBOR 9 Proof. Deﬁne . First let us show that if point is a grandparent of , then . If for some , then any of its grandchildren is at most away implying . Nesting says that , since . Now let us consider the grandparents of in levels , , , . There are at most four of these, due to the tree property. In fact, there can be no other unique grandparents above level in . Recall that if , then . If is also in , the well-separateness constraint implies that there can be no other point in which is also in . Nesting implies that there are no other grandparents in , else these grandparents would also be in . Thus any annulus can only contain unique grandparents of up to level . Now we just need to bound the number of non-empty around containing all points in . To do this, apply the growth bound with where is the nearest neighbor of to discover . Then, ﬁnd the next nearest point satisfying , and apply the growth bound with to discover since each application of the growth bound is disjoint (note that this process may signiﬁcantly undercount points). This process can be repeated at most before the lower bound exceeds the upper bound of . Upon termination, every point can be associated with the maximal satisfying . The set of points associated with every step in the process lie in at most 4 annuli . Consequently, there are at most nonempty annuli around any . This is since . The number of explicit grandparents in is constant, completing the proof. We can now state and prove the main theorem. Theorem 3.4. (Query Time) If the dataset has expansion constant , the nearest neighbor of can be found in time . Proof. Let be the last explicit considered by the algorithm. Lemma 3.3 bounds the explicit depth of any point in the tree (and in particular any point in ) by . Consequently, the number of iterations is at most . In each iteration, at most time is required to determine which elements need explicit descent, implying a bound of on the query time. Also note that in Step 2(a), the number of children encountered is at most using Lemma 3.1. Step 2(b) never does more work than Step 2(a). Step 3 requires at most work. Consequently, the running time is bounded by ﬁnishing the proof, provided that we can show that . Consider any constructed during the -th iteration. Recall that { Children , and let . We have where the ﬁrst equality follows by deﬁnition of and the second from . First suppose that . Then we have Now since (as a consequence of ), and (by assumption), we also have . Hence , and . We are left with the case . Consider a point which is also in . As in the proof of Lemma 3.1, we will bound the number of disjoint balls of radius that can be packed into . Any such ball can contain at most one point in (due to the separation constraint), implying a bound on . We have COVER TREES FOR NEAREST NEIGHBOR 10 and thus . Comparing the time complexity of navigating nets and cover trees in terms of its dependence on the expansion constant is non-trivial. Our data structure does run-time computations which were done in the preprocessing stage of the navigating nets algorithm. Navigating nets can be run in a more greedy (depth ﬁrst search) mode, while cover trees use a from of a fused depth and breadth ﬁrst search. The tradeoff is even more subtle because the radius of the balls used to form the covers in the navigating nets is larger than the radius used in the cover tree, implying that a node may have to maintain more children. 3.2. Insertion and Removal. Theorem 3.5. Any insertion or removal takes time at most . Proof. First we show that all but one node in each cover set are either expanded to their children or removed in the next two cover sets. To see why, note that each is contained in a ball of radius around the point we are inserting (by deﬁnition). Fix and assume that some node appears (either explicitly or implicitly) in all of , , . Then no other node can appear in , since the separation constraint in level says that while the maximum distance between and any other node in can be at most . Thus is either removed or expanded to its children, in which case it has to consume one level of its explicit depth. Let be the maximum explicit depth of any point, given by Lemma 3.3. Then the total number of cover sets with explicit nodes is at most , where the ﬁrst term follows from the fact that any node that is not removed must be explicit at least once every three iterations, and the additional accounts for a single point that may stay implicit for more than three iterations. Thus the total amount of work in Steps 1 and 2 is proportional to . Step 3 requires work no greater than step 1. For every , is a valid set of children for a hypothetical node at level , and thus from Lemma 3.1. Multiplying these bounds together we get the result. To obtain the bound for the removal, we can use a similar argument to show that at most one point can be propagated up more than twice in the search for a parent. Thus Step 5 in Algorithm 3 takes at most steps total. Other steps require work no greater than for insertion, completing the proof. 3.3. Batch Operations. Theorem 3.6. (Construction Time) Construct requires time at most . Proof. (sketch) The maximum amount of work in level associated with point is proportional to the number of siblings within distance since all elements in level doing work related to must have a parent at level , for which is in the FAR set (and thus at distance ). The maximum number of such siblings is using an argument similar to the growth bound. The maximum number of explicit levels which nearby points can have is (similar to the insertion proof). Multiplying max sibling level max siblings/level number of points, we get O(c^{6}n\log n) work completing the proof. Theorem 3.7. (Batch Query Time) Find-All-Nearest (, ) requires time . Proof. (sketch) The proof is similar to the query proof, except that the cover set at level has a radius increased by due to the need to the need to cover the nearest neighbor of grandchildren of the query cover tree. This increases the size of the cover set by a factor of which increases the -complexity by . The dependence on rather than is due to amortizing the descent of the cover tree over the leaf nodes. In particular, Step 2(a)(i) is executed on each explicit node at most once. All other steps have the same or lower time complexity. 4. EXPERIMENTAL RESULTS We tested the algorithm on several datasets drawn from the UCI machine learning and KDD archives [UCI], the KDD 2004 championship [KDDCup], the Mnist handwritten digit recognition dataset [mnist], COVER TREES FOR NEAREST NEIGHBOR 11 and the Isomap “Images” dataset [isomap]. For each dataset, we queried for the {1,2,3,5,10}-nearest neigh-bors of each point using the Euclidean metric (results with an metric were qualitatively similar). The results, compared to an optimized brute force algorithm, are summarized in Figure 4.1(a). We used the following optimizations of the basic algorithms to obtain the results in Figure 4.1(a): (1) Similar to [Moo00], we cache summary information in each node; in particular, we cache the maxi-mum distance to any grandchild, the distance to the parent, and the scale. (2) Since the upper bound can be tightened as a cover set is expanded, the order of expansion is im-portant. Ideally, we might expand from the closest to furthest on the theory that closer nodes are more likely to have closer children. We used an lazy sorting which recurses only on the low direction of quicksort. (3) Our theoretical results use scales that are powers of . It turns out that a smaller base provides speedups in practice. (4) We relax the separation invariant in batch construction. A natural question is whether the expansion constant is a relevant quantity for analysis. Since it is deﬁned as the worst-case expansion over all points, it may not be the best measure of hardness of NNS. Figure 4.1(c) illustrates this point by showing two datasets on 5000 points each with the same worst-case expansion con-stant but different distributions of expansion across points, and not surprisingly, very different speedups. Figure 4.1(d) suggests that, for example, the 80th percentile (over datapoints) expansion constant seems to be a better predictor of performance. Figure 4.1(e) shows the speedups from Figure 4.1(a) (as stalactite im-pulses) against the respective normalized expansion constants and their 80% versions (stalagmite impulses). The 80th percentile (dashed stalagmites) are more predictive of the speedup. Finally, we did experiments comparing cover trees to Clarkson’s data structure [Cla02] developed for the same setting as ours (see also [Cla99]). For each dataset, we did exact nearest neighbor queries of every point using the “d” method in [Cla02] that was reported to be uniformly superior to all other methods available in the package. We included the construction time when evaluating both algorithms and used the same timing mechanisms and the same implementation of the distance functions. Our algorithm was signiﬁcantly faster on almost every dataset tested; the speedups (ranging from 0.8 to 30) are shown in Figure 4.1(e). The respective constuction times are shown in Figure 4.1(f). Notice considerable savings in the construction times of the cover tree on the onlu two datasets (bio_test and bio_train), on which is superior (for 1-nearest neighbor). It should be noted, however, that the -nearest neighbor implemen-tation in is via a reduction to ﬁxed-radius queries; a better scheme might be possible, but it is not straightforward. Figure 4.2 shows the speedup of the cover tree over for strings under the edit distance. For more details, the code, and the datasets see [Lan04, BKL04]. Acknowledgement. We would like to thank Piotr Indyk, David Karger, Robert Krauthgamer, James Lee, and Alexander Gray for helpful discussions and comments. REFERENCES [AMN+98] S. Arya, D. M. Mount, N. S. Netanyahu, R. Silverman and A. Wu. An optimal algorithm for approximate nearest neighbor searching, Journal of the ACM, 45(6) : 891–923, 1998. [AM] S. Arya and D. M. Mount. ANN: Library for Approximate Nearest Neighbor Searching, [BKL04] A. Beygelzimer, S. Kakade, J. Langford. Cover trees for nearest neighbor, preprint, 2004. Available at [Cla99] K. Clarkson: Nearest Neighbor Queries in Metric Spaces. Discrete & Computational Geometry, 22(1): 63-93 (1999) [Cla02] K. Clarkson: Nearest Neighbor Searching in Metric Space: Experimental Results for sb(S), 2002, [DIIM04] M. Datar, N. Immorlica, P. Indyk, and V. Mirrokni. Locality-sensitive hashing scheme based on p-stable distributions, Proceedings of the 20th Annual Symposium on Computational Geometry, 2004. [FBL77] J. H. Friedman, J. L. Bentley, and R. A. Finkel, “An algorithm for ﬁnding best matches in logarithmic expected time”, ACM Transactions on Mathematical Software, 3(3):209-226, September 1977. COVER TREES FOR NEAREST NEIGHBOR 12 (a) 1 10 100 1000 iris bupa wine glass pima ion pen_test opt_test pen_train opt_train letter corel images phy_train bio_test bio_train covtype phy_test mnist 10K 100K 1M 10M 100M speedup 1,2,3,5,10-Nearest Neighbor Speedup (b) 1 10 100 1000 100 1000 10000 100000 speedup number of points Dataset Size vs. Speedup, K=2 (c) 1000 2000 3000 4000 5000 0 1000 2000 3000 4000 5000 expansion constant c number of points with expansion at most c Expansion constant over 5000 datapoints mnist bio_test (d) 1 10 100 1000 0 1000 2000 3000 4000 5000 speedup expansion constant Expansion constant vs Speedup, n = 5000 worst-case 80th per. n / c (e) 0.1 1 10 100 iris bupa wine glass pima ion pen_test opt_test pen_train opt_train letter corel phy_train bio_test bio_train phy_test covtype mnist (f) 0.1 1 10 100 iris bupa wine glass pima ion pen_test opt_test pen_train opt_train letter corel phy_train bio_test bio_train phy_test covtype mnist FIGURE 4.1. a) The speedup of the cover tree algorithm over the brute force search when querying for the nearest (1,2,3,5,10) neighbors (left to right in the graph) of every point in the dataset (logscale); 1-nearest neighbor query corresponds to self-search. Datasets are sorted by byte size in ascending order; the size is shown using a dashed line. (b) The speedup versus the number of data points in each dataset. Larger datasets have larger potential speedups. (c) The cumulative distribution of expansion constants across points for two datasets with the same maximum expansion. We achieve very little speedup on the ’mnist’ dataset and about a factor of speedup on the bio_test dataset. (d) A graph showing speedup vs the worst case expansion constant and speedup vs the 80th percentile (over datapoints) expansion constant on various 5000 point datasets obtained as preﬁxes of datasets form [UCI, KDDCup, mnist, isomap]. (e) The speedup (logscale) over the data structure [Cla02] for the (1,2) neighbors (solid and dashed lines respectively). One datapoint (images) is missing due to parsing issues with the code. (f) Construction times. The speedup over the data structure [Cla02]. Notice considerable savings in the construction times on the two datasets (bio_test and bio_train), on which outper-forms the cover tree. COVER TREES FOR NEAREST NEIGHBOR 13 1 10 100 male counties verbs adj female verb.exc noun.exc index.adv places mcds index.adj words surnames countysub usenet index.noun FIGURE 4.2. The speedup of the cover tree algorithm over the data structure [Cla02] when querying for the nearest neighbor of very point in the dataset (i.e., self-search); points are strings under edit distance. Dashed spikes show the corresponding speedup in the con-struction times. Datasets are sorted by byte size in ascending order, from 8K to 1.4M. [GM00] A. Gray and A. Moore. N-Body Problems in Statistical Learning, Proceedings of NIPS, 2000. [GKL03] A. Gupta, R. Krauthgamer, and J. R. Lee. Bounded geometries, fractals, and low-distortion embeddings, Proceedings of the 44th Annual IEEE Symposium on Foundations of Computer Science, 534–543, 2003. [isomap] Isomap datasets, [KR02] D. Karger and M. Ruhl. Finding Nearest Neighbors in Growth Restricted Metrics, Proceedings of STOC, 2002. [KDDCup] The 2004 KDD-cup dataset, [KL04b] R. Krauthgamer and J. Lee. The black-box complexity of nearest neighbor search, Proceedings of the 31st International Colloquium on Automata, Languages and Programming (ICALP), 2004. [KL04a] R. Krauthgamer and J. Lee. Navigating nets: Simple algorithms for proximity search, Proceedings of the 15th Annual Symposium on Discrete Algorithms (SODA), 791–801, 2004. [Lan04] J. Langford, Cover Tree code written in C++, [Moo00] A. Moore, Using the Triangle Inequality to Survive High Dimensional Data, Proceedings of the Twelfth Conference on Uncertainty in Artiﬁcial Intelligence, 2000. [mnist] The MNIST database of handwritten digits, [Omo87] S. M. Omohundro, Efﬁcient Algorithms with Neural Network Behavior. J. of Complex Systems, 1(2):273-347, 1987. [RS00] Sam Roweis and Lawrence Saul, “Nonlinear dimensionality reduction by locally linear embedding”, Science v. 290 no. 5500, Dec. 22, 2000. pp. 2323-2326. [TSL00] Josh Tenenbaum, Vin de Silva and John Langford. A Global Geometric Framework for Nonlinear Dimensionality Reduction. Science 290, 2319-2323, 2000 [UCI] UCI machine learning repository ( and KDD Archive ( [Uhl91] J. K. Uhlmann, Satisfying general proximity/similarity queries with metric trees. Information Processing Letters, 40:175-179, 1991.
4555	https://math.stackexchange.com/questions/4775415/why-is-the-counting-methodology-i-used-incorrect	probability - Why is the counting methodology I used incorrect? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Why is the counting methodology I used incorrect? Ask Question Asked 2 years ago Modified2 years ago Viewed 185 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. The question: Consider an n×n grid. The grid squares from the middle row are shaded in. What is the probability that a randomly selected rectangle contains at least one shaded square? My (incorrect) method: Using complementary probability, P(contains one shaded square) = 1 - P(contains no shaded square) P(contains no shaded square) = # of rectangles with no shaded square / # of possible rectangles In general, we can generate a rectangle by choosing any x,y such that 0 < x,y <= n. The # of rectangles with no shaded square in the top half = n∗(n−1)/2 n∗(n−1)/2, we multiply by 2 by symmetry to get n(n−1)n(n−1) total rectangles with no shaded square in both halves. The # of total rectangles is just n 2 n 2, so 1−(n−1)/n=1/n 1−(n−1)/n=1/n. What is specifically wrong with this counting method? I'm overcounting something, but I'm not sure what. I realize that in both the numerator and denominator, I am counting (for example) a 5x6 rectangle and a 6x5 rectangle separately. Is this an issue? probability combinatorics Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Sep 25, 2023 at 18:16 user1114958 user1114958 9 What is the middle down when n n is even?Thomas Andrews –Thomas Andrews 2023-09-25 18:20:02 +00:00 Commented Sep 25, 2023 at 18:20 "The # of total rectangles is just n 2 n 2" looks too low: that is the number of small squares in the original grid. Is n n odd here?Henry –Henry 2023-09-25 18:23:30 +00:00 Commented Sep 25, 2023 at 18:23 1 It seems like the number of total rectangles is (n+1 2)2,(n+1 2)2, not n 2.n 2. There are n 2 n 2 1×1 1×1 rectangles.Thomas Andrews –Thomas Andrews 2023-09-25 18:24:25 +00:00 Commented Sep 25, 2023 at 18:24 1 Possible duplicate here.Peter Phipps –Peter Phipps 2023-09-25 18:36:48 +00:00 Commented Sep 25, 2023 at 18:36 1 You can in fact reduce this to starting with an n×1 n×1 rectangle with the middle square shaded (the other dimension cancels out), so with n n odd you get 2 n−1 2 n+1 2 2 n(n+1)2=n−1 2 n 2 n−1 2 n+1 2 2 n(n+1)2=n−1 2 n Henry –Henry 2023-09-26 00:31:17 +00:00 Commented Sep 26, 2023 at 0:31 \|Show 4 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. Requested from comments: You can in fact reduce this to starting with an n×1 n×1 rectangle with the middle square shaded (the other dimension cancels out), with n(n+1)2 n(n+1)2 total rectangles and (assuming n n is odd) n−1 2 n+1 2 2 n−1 2 n+1 2 2 rectangles on one side of the shaded square and the same number on the other side so you get a probability of 2 n−1 2 n+1 2 2 n(n+1)2=n−1 2 n.2 n−1 2 n+1 2 2 n(n+1)2=n−1 2 n. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 26, 2023 at 20:41 HenryHenry 171k 10 10 gold badges 139 139 silver badges 297 297 bronze badges Add a comment\| You must log in to answer this question. 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 10Analysis of how-many-squares and rectangles are are there on a chess board? 22005×2005 2005×2005 checkerboard with rows shaded, probability random rectangle includes a shaded square? Related 1Combinations and probabilities 0Method to solve probability of chips 1Infinite row of squares, some of them white with probability p p. What is the average width of a white rectangle? 3Explanation needed for why my solution is incorrect. 0A seemingly impossible combinatorics problem! 22005×2005 2005×2005 checkerboard with rows shaded, probability random rectangle includes a shaded square? 2What is the probability that randomly selected rectangles will not contain both of the two red squares? 15 distinct balls, I pick 2, Bob picks 3, probability that 1 of my ball is among the balls Bob picks Hot Network Questions What is a "non-reversible filter"? Calculating the node voltage How different is Roman Latin? Does the curvature engine's wake really last forever? Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done? Interpret G-code Determine which are P-cores/E-cores (Intel CPU) With line sustain pedal markings, do I release the pedal at the beginning or end of the last note? I have a lot of PTO to take, which will make the deadline impossible Gluteus medius inactivity while riding Clinical-tone story about Earth making people violent Lingering odor presumably from bad chicken Matthew 24:5 Many will come in my name! Childhood book with a girl obsessessed with homonyms who adopts a stray dog but gives it back to its owners Storing a session token in localstorage An odd question What meal can come next? Xubuntu 24.04 - Libreoffice Are there any world leaders who are/were good at chess? Direct train from Rotterdam to Lille Europe How to home-make rubber feet stoppers for table legs? Do sum of natural numbers and sum of their squares represent uniquely the summands? Making sense of perturbation theory in many-body physics The geologic realities of a massive well out at Sea more hot questions Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4556	https://blog.sina.com.cn/s/blog_7bf985a00102z2q2.html	小学四年级数学认识多位数知识点_殷璠_新浪博客推荐：公告：请您及时更换请请请您正在使用的模版将于2周后被下线，请您及时更换 × 加载中… 新浪首页登录注册首页博文目录关于我个人资料殷璠微博加好友发纸条写留言加关注博客等级：博客积分：158 博客访问：806 关注人气：0 获赠金笔：0 赠出金笔：0 荣誉徽章：正文字体大小：大中小小学四年级数学认识多位数知识点 (2022-04-01 09:55:50) 1.我国计数是从右起，每4个数位为一级;国际计数是每3个为一节。 (1)什么叫数位、计数单位、数级?整数数位的排列顺序是怎样的?从个位起依次说出各个数位。把计数单位按一定的顺序排列起来，它们所在的位置，叫作数位。计数单位有：个、十、百、千、万、十万、百万、千万、亿、十亿、百亿、千亿。从个位起，每四个数位是一级，一共分为个级、万级、亿级。 (2)每相邻两个计数单位之间有什么关系? 10个一万是十万;10个十万是一百万;10个一百万是一千万;10个一千万是一亿。每相邻的两个计数单位之间的进率都是10，这种计数方法叫十进制计数法。 2.复习多位数的读、写法。 (1)多位数的读法。从高位读起，一级一级地往下读。读亿级或万级的数，先按照个级的读法读，再在后面加上一个“亿”字或“万”字。每级中间有一个0或连续几个0，都只读一个零;每级末尾的零都不读。 (2)多位数的写法。先写亿级，再万级，最后写个级，哪个数位上一个单位也没有，就在那一位上写0。 3.复习数的改写及省略。改写。可以将万位、亿位后面的4个0、8个0省略，换成“万”或“亿”字，这样就将整万或整亿的数改写成用“万”或“亿”作单位的数。省略。省略时一般用“四舍五入”的方法。是“舍”还是“入”，要看省略部分的尾数最高位是小于5、等于5还是大于5。 4.比大小位数不同，位数多的数就大; 位数相同，左起第一位的数大的那个数就大; 如果左起第一位上的数相同，就比较左起第二位上的数。练习题由五个十万、四个万、六个百和三个一组成的数是( )。 30300300是( )位数，在这个数中，左边第一个数字3表示3个( )。 ( )个一万是十万，( )百万是一千万。 100个一千是( )，10个( )是一亿。由三个千、三个一组成的数是( )，读作( )。 gt;gt;gt;参考答案由五个十万、四个万、六个百和三个一组成的数是( 540603 )。 30300300是( 八 )位数，在这个数中，左边第一个数字3表示3个( 千万 )。 ( 10 )个一万是十万，( 10个 )百万是一千万。 100个一千是( 十万 )，10个( 一千万 )是一亿。由三个千、三个一组成的数是( 3003 )，读作( 三千零三 )。认识多位数知识点就先到这儿了，我会持续为大家更新最新的内容，希望大家学有所成。分享：打开微信“扫一扫” 打开网页后点击屏幕右上角“分享”按钮 0 喜欢 0 赠金笔阅读(138)┊收藏(0)┊喜欢▼┊打印┊举报/Report 加载中，请稍候...... 前一篇：锻炼学生的分析综合能力后一篇：小学数学学情分析报告新浪BLOG意见反馈留言板欢迎批评指正新浪简介 \| About Sina \| 广告服务 \| 联系我们 \| 招聘信息 \| 网站律师 \| SINA English \| 产品答疑 Copyright © 1996 - 2022 SINA Corporation, All Rights Reserved 新浪公司版权所有博客转载原文长微博分享到新浪微博含“”的博文含“”的博主含“”的音乐含“”的视频最近喜欢了的博主：加载中… \| \| 赠金笔关闭 \| \| --- \| \| \| \| \| \| 金笔兑换今日土豪榜今日提名榜兑换服务：金笔道具赠送数量：1支 10支 20支 50支 100支其他现有金笔：0支还需兑换：1支 x 1=1(积分) 确认点击了解更多规则 \| \| \| \| 提示关闭 \| \| --- \| \| \| \| \| \| 下载新浪博客APP 纸条升级为私信，我在APP上等你~ \| \|
4557	https://www.astro.umd.edu/~dphamil/ASTR498/expansions.pdf	Expansions for Small Quantities These truncated Taylor series expansions are valid for argument x << 1. General Functions (1 −x)n ≈1 −nx + n(n −1)x2/2 −... ex ≈1 + x + x2/2 + ... ln(1 + x) ≈x −x2/2 + ... Trigonometric Functions sin x ≈x −x3/6 + ... cos x ≈1 −x2/2 + ... tan x ≈x + x3/3... csc x ≈1/x + x/6 + ... sec x ≈1 + x2/2 + ... cot x ≈1/x −x/3 −... Inverse Trigonometric Functions sin−1 x ≈x + x3/6 + ... cos−1 x ≈π/2 −x −... tan−1 x ≈x −x3/3 + ... csc−1 x ≈1/x + 1/(6x3) + ... sec−1 x ≈π/2 −1/x −... cot−1 x ≈π/2 −x + ... Hyperbolic Functions sinh x ≈x + x3/6 + ... cosh x ≈1 + x2/2 + ... tanh x ≈x −x3/3 + ... sech x ≈1 −x2/2 + ... csch x ≈1/x −x/6 + ... coth x ≈1/x + x/3 + ... Inverse Hyperbolic Functions sinh−1 x ≈x −x3/6 + ... cosh−1 x ≈ln(2x) −1/(4x2) −... tanh−1 x ≈x + x3/3 + ... sech−1 x ≈ln(2/x) −x2/4 −... csch−1 x ≈1/x −1/(6x3) + ... coth−1 x ≈1/x + 1/(3x3) + ...
4558	https://cameroncounts.wordpress.com/wp-content/uploads/2013/12/ec.pdf	Enumerative Combinatorics The LTCC lectures Peter J. Cameron Autumn 2013 Abstract These are the notes of my lecture course on Enumerative Combina-torics at the London Taught Course Centre in Autumn 2013. Thanks to all who attended for their support. There are ten sections, as fol-lows: • Subsets, partitions, permutations • Formal power series • Catalan numbers • Unimodality • q-analogues • Symmetric polynomials • Group actions • Species • M¨ obius inversion • Cayley’s theorem Exercises are included at the end of the sections. 1 Subsets, Partitions, Permutations Enumerative combinatorics is concerned with counting discrete structures of various types. There is a great deal of variation both in what we mean by “counting” and in the types of structures we count. Typically each structure has a “size” measured by a non-negative integer n, and “counting” may mean 1 (a) ﬁnding an exact formula for the number f(n) of structures of size n; (b) ﬁnding an approximate or asymptotic formula for f(n); (c) ﬁnding an analytic expression for a generating function for f(n); (d) ﬁnding an eﬃcient algorithm for computing f(n) exactly or approxi-mately; (e) ﬁnding an eﬃcient algorithm for stepping from one of the counted ob-jects to the next (in some natural ordering). In this course I will mostly be concerned with the ﬁrst three goals; discussing algorithms would lead too far aﬁeld. The exception to this is one particularly important algorithm, a recurrence relation, in which the value of f(n) is computed from n and the earlier values f(0), . . . , f(n −1). An asymptotic formula for f(n) is an analytic function g(n) such that f(n)/g(n) →0 as n →∞. There are several types of generating functions; the most important for us are the ordinary generating function X n≥0 f(n)xn, and the exponential generating function X n≥0 f(n)xn n! . If you want to learn the state-of-the-art in combinatorial enumeration, I recommend the two volumes of Richard Stanley’s Enumerative Combina-torics, or the book Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick. The On-line Encyclopedia of Integer Sequences is another valu-able resource for combinatorial enumeration. 1.1 Subsets The three most important objects in elementary combinatorics are subsets, partitions and permutations; we brieﬂy discuss the counting functions for these. First, subsets. The total number of subsets of an n-element set is 2n. This can be used by noting that this number f(n) satisﬁes the recurrence relation f(n) = 2f(n −1); this is proved by observing that any subset of {1, . . . , n −1} can be extended to a subset of {1, . . . , n} in two diﬀerent ways, either including the element n or not. 2 The binomial coeﬃcient n k is the number of k-element subsets of {1, . . . , n}. The formula is n k = n(n −1) · · · (n −k + 1) k(k −1) · · · 1 = n! k! (n −k)!. Note that there are k factors in both numerator and denominator. We have n 0 = n n = 1. We can extend the deﬁnition to all non-negative inte-gers n and k by deﬁning n k = 0 for k > n: this ﬁts with the counting interpretation. The recurrence relation for binomial coeﬃcients is Pascal’s Triangle n k = n −1 k −1 + n −1 k for 0 < k < n. For the ﬁrst term on the right counts subsets containing n, while the second counts subsets not containing n. Counting subsets by cardinality gives n X k=0 n k = 2n. There is a huge literature on “binomial coeﬃcient identities”. A few examples are given as exercises. Anticipating our discussion of formal power series in the next chapter, we now discuss generating functions for binomial coeﬃcients. n X k=0 n k xk = (1 + x)n. This is the Binomial Theorem for non-negative integer exponents. If we write (1 + x)n = (1 + x) · · · (1 + x) and expand the product, then we obtain the term in xk by choosing x from k of the brackets and 1 from the remaining n −k, which can be done in n k ways; each contributes 1 to the coeﬃcient of xk, so the theorem holds. 3 If we multiply this equation by yn and sum, we obtain the bivariate generating function for the binomial coeﬃcients: X n≥0 n X k=0 n k xkyn = X n≥0 (1 + x)nyn = 1 1 −(1 + x)y = 1 1 −y · 1 1 −xy/(1 −y) = X k≥0 yk (1 −y)k+1 xk, so we obtain the other univariate generating function for binomial coeﬃcients: X n≥k n k yn = yk (1 −y)k+1. This formula is actually a rearrangement of the Binomial Theorem for neg-ative integer exponents. The basis of this connection is the following evalu-ation, for positive integers m and k: −m k = −m(−m −1) · · · (−m −k + 1 k! = (−1)k (m + k −1) · · · (m + 1)m k! = (−1)k m + k −1 k . 1.2 Partitions In this case and the next, we are unable to write down a simple formula for the counting numbers, and have to rely on recurrence relations or other techniques. The Bell number B(n) is the number of partitions of a set of cardinality n. We reﬁne this in the same way we did for subsets. The Stirling number of the second kind, S(n, k), is the number of partitions of an n-set into k parts. 4 Thus, S(0, 0) = 1 and S(0, k) = 0 for k > 0; and if n > 0, then S(n, 0) = 0, S(n, 1) = S(n, n) = 1, and S(n, k) = 0 for k > n. Clearly we have n X k=1 S(n, k) = B(n) for n > 0. The recurrence relation replacing Pascal’s is: S(n, k) = S(n −1, k −1) + kS(n −1, k) for 1 ≤k ≤n. It turns out that we can turn this into a statement about a generating function, but with a twist. Let (x)k = x(x −1) · · · (x −k + 1) (k factors). Then we have xn = n X k=1 S(n, k)(x)k for n > 0. It is possible to ﬁnd a traditional generating function for the index n: X n≥k S(n, k)yn = yk (1 −y)(1 −2y) · · · (1 −ky). Also, the exponential generating function for the index n is X n≥k S(n, k)xn n! = (exp(x) −1)k k! . Summing over k gives the e.g.f. for the Bell numbers: X n≥0 B(n)xn n! = exp(exp(x) −1). 1.3 Permutations The number of permutations of an n-set (bijective functions from the set to itself) is the factorial function n! = n(n −1) · · · 1 for n ≥0. The exponen-tial generating function for this sequence is 1/(1 −x), while the ordinary generating function has no analytic expression (it is divergent for all x ̸= 0). 5 Any permutation can be decomposed uniquely into disjoint cycles. So we reﬁne the count by letting u(n, k) be the number of permutations of an n-set which have exactly k cycles (including cycles of length 1). Thus, n X k=1 u(n, k) = n! for n > 0. The numbers u(n, k) are the unsigned Stirling numbers of the ﬁrst kind. The reason for the name is that it is common to use a diﬀerent count, where a permutation is counted with weight equal to its sign (as deﬁned in elementary algebra, for example the theory of determinants). Let s(n, k) be the sum of the signs of the permutations of an n-set which have k cycles. Since the sign of such a permutation is (−1)n−k, we have s(n, k) = (−1)n−ku(n, k). The numbers s(n, k) are the signed Stirling numbers of the ﬁrst kind. We have n X k=1 s(n, k) = 0 for n > 1. This is related to the algebraic fact that, for n > 1, the permutations with sign + form a subgroup of the symmetric group of index 2 (that is, containing half of all the permutations), called the alternating group. We will mainly consider signed Stirling numbers below, though it is some-times convenient to prove a result ﬁrst for the unsigned numbers. As usual we take s(n, 0) = 0 for n > 0 and s(n, k) = 0 for k > n. We have s(n, n) = 1, s(n, 1) = (−1)n−1(n−1)!, and the recurrence relation s(n, k) = s(n −1, k −1) −(n −1)s(n −1, k) for 1 ≤k ≤n. From this, we ﬁnd a generating function: n X k=1 s(n, k)xk = (x)n. Putting x = 1 in this equation shows that indeed the sum of the signed Stirling numbers is zero for n > 1. Note that this is the inverse of the relation we found for the Stirling numbers of the second kind. So the matrices formed by the Stirling numbers of the ﬁrst and second kind are inverses of each other. 6 Exercises 1. Let A be the matrix of binomial coeﬃcients (with rows and columns indexed by N, and (i, j) entry i j ), and B the matrix of “signed binomial coeﬃcients” (as before but with (i, j) entry (−1)i−j i j ). Prove that A and B are inverses of each other. What are the entries of the matrix A2? 2. Prove that n X k=0 n k 2 = 2n n . 3. (a) Prove that the following are equivalent for sequences (a0.a1, . . .) and (b0, b1, . . .), with exponential generating functions A(x) and B(x) respec-tively: (ii) b0 = a0 and bn = n X k=1 S(n, k)ak for n ≥1; (i) B(x) = A(exp(x) −1). (b) Prove that the following are equivalent for sequences (a0.a1, . . .) and (b0, b1, . . .), with exponential generating functions A(x) and B(x) respec-tively: (i) b0 = a0 and bn = n X k=1 s(n, k)ak for n ≥1; (ii) B(x) = A(log(1 + x)). 4. Construct a bijection between the set of all k-element subsets of {1, 2, . . . , n} containing no two consecutive elements, and the set of all k-element subsets of {1, 2, . . . , n −k + 1}. Hence show that the number of such subsets is n−k+1 k . In the UK National Lottery, six numbers are chosen randomly (with-out replacement, order unimportant) from the set {1, . . . , 49}. What is the probability that the selection contains no two consecutive numbers? 7 2 Formal power series Probably you recognised in the last chapter a few things from analysis, such as the exponential and geometric series; you may know from complex analysis that convergent power series have all the nice properties one could wish. But there are reasons for considering non-convergent power series, as the following example shows. Recall the generating function for the factorials: F(x) = X n≥0 n!xn, which converges nowhere. Now consider the following problem. A permu-tation of {1, . . . , n} is said to be connected if there is no number m with 1 ≤m ≤n −1 such that the permutation maps {1, . . . , m} to itself. Let Cn be the number of connected permutations of {1, . . . , n}. Any permutation is composed of a connected permutation on an initial interval and an arbitrary permutation of the remainder; so n! = n X m=1 Cm(n −m)!. Hence, if G(x) = 1 − X n≥1 Cnxn, we have F(x)G(x) = 1, and so G(x) = 1/F(x). Fortunately we can do everything that we require for combinatorics (ex-cept some asymptotic analysis) without assuming any convergence proper-ties. 2.1 Formal power series Let R be a commutative ring with identity. A formal power series over R is, formally, an inﬁnite sequence (r0, r1, r2, . . .) of elements of R; but we always represent it in the suggestive form r0 + r1x + r2x2 + · · · = X n≥0 rnxn. 8 We denote the set of all formal power series by R. The set R has a lot of structure: there are many things we can do with formal power series. All we require of any operations is that they only require adding or multiplying ﬁnitely many elements of R. No analytic properties such as convergence of inﬁnite sums or products are required to hold in R. (a) Addition: We add two formal power series term-by-term. (b) Multiplication: The rule for multiplication of formal power series, like that of matrices, looks unnatural but is really the obvious thing: we multiply powers of x by adding the exponents, and then just gather up the terms contributing to a ﬁxed power. Thus X anxn · X bnxn = X cnxn, where cn = n X k=0 akbn−k. Note that to produce a term of the product, only ﬁnitely many addi-tions and multiplications are required. (c) Inﬁnite sums and products: These are not always deﬁned. Suppose, for example, that A(i)(x) are formal power series for i = 0, 1, 2, . . .; assume that the ﬁrst non-zero coeﬃcient in A(i)(x) is the coeﬃcient of xni, where ni →∞as i →∞. Then, to work out the coeﬃcient of xn in the inﬁnite sum, we only need the ﬁnitely many series A(i)(x) for which ni ≤n. Similarly, the product of inﬁnitely many series B(i) is deﬁned provided that B(i)(x) = 1 + A(i)(x), where A(i) satisfy the condition just described. (d) Substitution: Let B(x) be a formal power series with constant term zero. Then, for any formal power series A(x), the series A(B(x)) ob-tained by substituting B(x) for x in A(x) is deﬁned. For, if A(x) = P anxn, then A(B(x)) = P anB(x)n, and B(x)n has no terms in xk for k < n. (e) Diﬀerentiation: of formal power series is always deﬁned; no limiting process is required. The derivative of P anxn is P nanxn−1, or alter-natively, P(n + 1)an+1xn. 9 (f) Negative powers: We can extend the notion of formal power series to formal Laurent series, which are allowed to have ﬁnitely many negative terms: X n≥−m anxn. Inﬁnitely many negative terms would not work since multiplication would then require inﬁntely many arithmetic operations in R. (g) Multivariate formal power series: We do not have to start again from scratch to deﬁne series in several variables. For R is a commutative ring with identity, and so R can be deﬁned as the set of formal power series in y over R. As hinted above, R is indeed a commutative ring with identity: veri-fying the axioms is straightforward but tedious, and I will just assume this. With the operation of diﬀerentiation it is a diﬀerential ring. Recall that a unit in a ring is an element with a multiplicative inverse. The units in R are easy to describe: Proposition 2.1 The formal power series P rnxn is a unit in R if and only if r0 is a unit in R. Proof If (P rnxn) (P snxn) = 1, then looking at the constant term we see that r0s0 = 1, so r0 is a unit. Conversely, suppose that r0s0 = 1. Considering the coeﬃcient of xn in the above equation with n > 0, we see that n X k=0 rksn−k = 0, so we can ﬁnd the coeﬃcients sn recursively: sn = −s0 n X k=1 rksn−k ! . This argument shows the very close connection between ﬁnding inverses in R and solving linear recurrence relations. 10 2.2 Example: partitions We are considering partitions of a number n, rather than of a set, here. A partition of n is an expression for n as a sum of positive integers arranged in non-increasing order; so the ﬁve partitions of 4 are 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1. Let p(n) be the number of partitions of n. Theorem 2.2 (Euler’s Pentagonal Numbers Theorem) p(n) = X k≥1 (−1)k−1 (p(n −k(3k −1)/2) + p(n −k(3k + 1)/2)) , where the sum contains all terms where the argument n −k(3k ± 1)/2 is non-negative. This is a very eﬃcient recurrence relation for p(n), allowing it to be computed with only about √n arithmetic operations if smaller values are known. For example, if we know p(0) = 1, p91) = 1, p(2) = 2, p(3) = 3, p(4) = 5, then we ﬁnd p(5) = p(4) + p(3) −p(0) = 7, p(6) = p(5) + p(4) −p(1) = 11, and so on. I will give a brief sketch of the proof. Step 1: The generating function. X n≥0 p(n)xn = Y k≥1 (1 −xk)−1. For on the right, we have the product of geometric series 1 + xk + x2k + · · ·, and the coeﬃcient of xn is the number of ways of writing n = P kak, which is just p(n). 11 Step 2: The inverse of the generating function. We need to ﬁnd Y k≥1 (1 −xk). The coeﬃcient of xn in this product is obtained from the expressions for n as a sum of distinct positive integers, where sums with an even number of terms contribute +1 and sums with an odd number contribute −1. For example, 9 = 8 + 1 = 7 + 2 = 6 + 3 = 5 + 4 = 6 + 2 + 1 = 5 + 3 + 1 = 4 + 3 + 2, so there are four sums with an even number of terms and four with an odd number of terms, and so the coeﬃcient is zero. Step 3: Pentagonal numbers appear. It turns out that the following is true: The numbers of expressions for n as the sum of an even or an odd number of distinct positive integers are equal for all n except those of the form k(3k ± 1)/2, for which the even expressions exceed the odd ones by one if k is even, and vice versa if k is odd. This requires some ingenuity, and I do not give the proof here. This shows that the expression in Step 2 is equal to 1 + X k≥1 (−1)k xk(3k+1)/2 + xk(3k−1)/2 , and we immediately obtain the required recurrence relation. Exercises 1. Suppose that R is a ﬁeld. Show that R has a unique maximal ideal, consisting of the formal power series with constant term zero. Describe all the ideals of R. 2. Suppose that A(x), B(x) and C(x) are the exponential generating func-tions of sequences (an), (bn) and (cn) respectively. Show that A(x)B(x) = C(x) if and only if cn = n X k=0 n k akbn−k. 12 3. (a) Let (an) be a sequence of integers, and (bn) the sequence of partial sums of (an) (in other words, bn = n X i=0 ai). Suppose that the generating function for (an) is A(x). Show that the generating function for (bn) is A(x)/(1 −x). (b) Let (an) be a sequence of integers, and let cn = nan for all n ≥ 0. Suppose that the generating function for (an) is A(x). Show that the generating function for (cn) is x(d/dx)A(x). What is the generating function for the sequence (n2an)? (c) Use the preceding parts of this exercise to ﬁnd the generating function for the sequence whose nth term is n X i=1 i2, and hence ﬁnd a formula for the sum of the ﬁrst n squares. 3 Catalan numbers In the last chapter, as in most of this course, we treated power series as formal objects: even diﬀerentiation involves no limiting processes. However, if the coeﬃcients are complex numbers, and the series converge in some neighbour-hood of the origin, then analytic methods can be used. These methods can be very powerful. We will see them at work in the derivation of a formula for the Catalan numbers, and then give a few examples of combinatorial objects counted by Catalan numbers. 3.1 Analysis A complex function which is analytic in some neighbourhood of the origin is represented by a convergent power series in a disc about the origin. If an analytic relation between functions holds in a suitable disc, then any connection between the coeﬃcients which can be derived will also be true in the world of formal power series. The most important formal power series to which this principle can be applied are (a) The binomial series (1 + x)a = X n≥0 a n xn, where a is any complex 13 number, and the binomial coeﬃcient is deﬁned as a n = a(a −1) · · · (a −n + 1) n! . (b) The exponential series exp(x) = X n≥0 xn n! .. (c) The logarithmic series log(1 + x) = X n≥1 (−1)n−1xn n . Here is a simple example. The identity (1 + x)a(1 + x)b = (1 + x)a+b, valid for \|x\| < 1, gives rise to the Vandermonde convolution n X k=0 a k b n −k = a + b n . 3.2 Example: Catalan numbers The Catalan numbers are one of the most important sequences of combi-natorial numbers, with a large range of occurrences in apparently diﬀerent counting problems. I will introduce them with one particular occurrence, and then give a number of diﬀerent places where they arise. The derivation of the formula for them is on the border between formal and analytic methods, and multivariate versions of this method are useful in areas such as lattice path problems. Problem Given an algebraic structure with a (non-associative) binary op-eration ◦, in how many diﬀerent ways can a product of n terms be evaluated by inserting brackets? For example, the product a ◦b ◦c ◦d has ﬁve evaluations: ((a ◦b) ◦c) ◦d, (a ◦(b ◦c)) ◦d, (a ◦b) ◦(c ◦d), a ◦((b ◦c) ◦d), a ◦(b ◦(c ◦d)). Let Cn be the number of evaluations of a product of n terms, for n ≥1, so that C1 = C2 = 1, C3 = 2, C4 = 5. Let c(x) = X n≥1 Cnxn be the generating function. 14 In a bracketing of n terms, the last application of ◦will combine some product of the ﬁrst m terms with some product of the last n −m terms, for some m with 1 ≤m ≤n −1. So we have the recurrence relation Cn = n−1 X m=1 CmCn−m for n > 1. Combined with the initial condition C1 = 1, this determines the sequence. Now consider the product c(x)2. The recurrence relation shows that the terms in xn in c(x)2 are the same as those in c(x) for n > 1; only the terms in x diﬀer, with c(x) containing 1x and c(x)2 containing 0x. So we have c(x) = x + c(x)2. We can rearrange this as a quadratic equation: c(x)2 −c(x) + x = 0. The solution of this equation is c(x) = 1 2 1 ± √1 −4x . The choice of sign in the square root is determined by the fact that c(0) = 0, so we must take the negative sign: c(x) = 1 2 1 −√1 −4x . From this expression it is possible to extract the coeﬃcient of xn. Ac-cording to the Binomial Theorem, (1 −4x)1/2 = X n≥0 1/2 n (−4x)n, and so Cn = −1 2(−4)n 1/2 n . Now 1/2 n = (1/2)(−1/2) · · · (−(2n −3)/2) n! 15 = 1 2n(−1)n−11 · 3 · (2n −3) n! = 1 2n(−1)n−1 1 n (2n −2)! 2n−1((n −1)!)2 = −2(−1 4)n 1 n 2n −2 n −1 , so ﬁnally we obtain Cn = 1 n 2n −2 n −1 . The result and its proof call for a few remarks. First, are these manipulations really valid? (a) We have used here the Binomial Theorem for exponent 1/2, which is proved analytically by observing that the function (1+x)1/2 is analytic in the interior of the unit disc (it has a branchpoint at x = −1), and then using the formula for the coeﬃcient of xn in the Taylor series (diﬀerentiate n times, put x = 0, divide by n!). (b) It is clear, from back substitution, that the function c(x) = 1 2(1 − √1 −4x) does indeed satisfy the equation c(x) = x + c(x)2; so its coeﬃcients satisfy the recurrence relation and initial condition for the Catalan numbers Cn. Since these data determine the numbers uniquely, our ﬁnal formula is indeed valid. Second, this is a case where, even once you know the formula for the Cata-lan numbers, it is diﬃcult to show directly that they satisfy the recurrence relation. (Spend a few moments trying; you will be convinced of this!) And third, it is not at all obvious that n divides the binomial coeﬃ-cient 2n −2 n −1 ; but since Cn counts something, it is an integer, and so this divisibility is indeed true. 3.3 Other Catalan objects Here are a small selection of the many objects counted by Catalan numbers. The obvious ways of verifying this for a class of objects are either (a) to verify the Catalan recurrence and initial condition; or 16 (b) to ﬁnd a bijection to a known class of Catalan objects. There are sometimes other less obvious ways, as we will see in the case of Dyck paths. Where possible I have given an illustration of the ﬁve Catalan objects counted by C4. Binary trees A binary tree has a root of degree 2; the other vertices have degree 1 or 3. So every non-root vertex is either a leaf or has two descendants, which we specify as left and right descendants. The number of binary trees with n leaves is Cn. Figure 1 shows the correspondence with bracketed products: the tree is a “parse tree” for the product. r r r r \ \ \ \ \ \ r r r (a ◦(b ◦(c ◦d))) r r r r r r r \ \ \ \ \ (a ◦((b ◦c) ◦d)) \ \ L L L L r r r r r r r ((a ◦b) ◦(c ◦d)) \ \ \ \ \ r r r r r r r ((a ◦(b ◦c)) ◦d) \ \ \ \ \ r r r r r r r (((a ◦b) ◦c) ◦d) Figure 1: Binary trees and bracketed products Rooted plane trees The number of rooted plane trees with n edges is Cn+1. Figure 2 shows the rooted plane trees with three edges. r r r r r r @ @ r r r r r r @ @ r r r r @ @ r r r r @ @ Figure 2: Rooted plane trees 17 Dissections of polygons An n-gon can be dissected into triangles by drawing n −2 non-crossing diagonals. There are Cn−1 dissections of an n-gon. Figure 3 shows dissections of a pentagon. B B B Z Z q q q q q B B B Z Z q q q q q B B B B Z Z Z B B B Z Z q q q q q B B B Z Z q q q q q B B B B B B B Z Z q q q q q Z Z Z Figure 3: Dissections of a polygon Dyck paths A Dyck path starts at the origin and ends at (2n, 0), moving at each step to the adjacent lattice point in either the north-easterly or south-easterly direction and never going below the X-axis. (An even number of steps is required since each step either increases or decreases the Y-coordinate by 1.) Figure 4 shows the Dyck paths with n = 3. @ @ @ @ q q q q q q q q q q q q q q q q q q q q q q q q q q q q @ @q @ @ @ q q q q q q q q q q q q q q q q q q q q q q q q q q q q @ @ @@ @ q q q q q q q q q q q q q q q q q q q q q q q q q q q q @ @@ @ @ q q q q q q q q q q q q q q q q q q q q q q q q q q q q @ @ @ @ @ @ q q q q q q q q q q q q q q q q q q q q q q q q q q q q Figure 4: Dyck paths The number of Dyck paths is Cn+1, and of these, Cn never return to the X-axis before the end. I will indicate the proof since it illustrates another technique. Let Dn be the number of Dyck paths, and En the number which never return to the axis. Now a Dyck path begins by moving from (0, 0) to (1, 1) and ends by moving from (2n −1, 1) to (2n, 0); if it did not return to the axis in between, then removing these “legs” gives a shorter Dyck path. So En = Dn−1. Suppose that a Dyck path ﬁrst returns to the axis at (2k, 0). Then it is a composite of a non-returning Dyck path of length 2k with an arbitrary Dyck 18 path of length 2(n −k); so Dn = n X k=1 EkDn−k. Solving these simultaneous recurrences gives the result. Ballot numbers An election is held with two candidates A and B, each of whom receives exactly n votes. In how many ways can the votes be counted so that A is never behind in the count? It is easy to match these ballot numbers with Dyck paths. For n = 3, the ﬁve counts are AAABBB, AABABB, AABBAB, ABAABB, and ABABAB. This can be described another way. In a 2×n array, we place the numbers 1, . . . , 2n in order against the candidates who receive those votes. This gives the representations shown in Figure 5. 1 2 3 4 5 6 1 2 4 3 5 6 1 2 5 3 4 6 1 3 4 2 5 6 1 3 5 2 4 6 AAABBB AABABB AABBAB ABAABB ABABAB Figure 5: Tableaux Note that the numbers increase along each row and down each column. 3.4 Young diagrams and tableaux The ﬁve objects shown are known as Young tableaux; they arise in the rep-resentation theory of the symmetric group and much related combinatorics. A Young diagram (sometimes called a Ferrers diagram) consists of n boxes arranged in left-aligned rows, the number of boxes in each row being a non-decreasing function of the row number. This is simply a graphical representation of a partition of n: for each partition n = a1 + a2 + · · ·, with a1 ≥a2 ≥. . ., we take a1 boxes in the ﬁrst row, a2 in the second, and so on. Now a Young tableau is a ﬁlling of the boxes with the numbers 1, 2, . . . , n so that each row and each column is in increasing order. You maay like to 19 invent a ballot interpretation for the number of Young tableaux belonging to a given diagram. This combinatorics is important in describing the representation theory of the symmetric group Sn, the group of all permutations of {1, . . . , n}. It is known that the irreducible matrix representations of Sn over the complex numbers are in one-to-one correspondence with the partitions of n (that is, to the Young diagrams); the degree of a representation is equal to the number of Young tableaux belonging to the corresponding diagram. Thus, the ﬁve Young tableaux shown in the preceding section correspond to an irreducible representation of degree 5 of the group S6. There is a “hook length formula” for the number of Young tableaux cor-responding to a given diagram. The hook associated with a cell consists of that cell and all those to its right in the same row or below it in the same column. The hook length of a cell is the number of cells in its hook. Now the number of Young tableaux associated with the diagram is equal to n! divided by the product of the hook lengths of all its cells. Thus for the diagram with two rows of length 3, the formula gives 6! 4 · 3 · 2 · 3 · 2 · 1 = 5. 3.5 Wedderburn–Etherington numbers What happens if we count binary trees without the left-right distinction between the two children at each node? In other words, two binary trees will count as “the same” if a sequence of reversals of subtrees above each point converts one to the other. It can be shown that the recurrence relation for the number Wn of binary trees with this convention (the Wedderburn–Etherington numbers is Wn =            1 2 n−1 X i=1 WiWn−i if n is odd, 1 2 n−1 X i=1 WiWn−i + Wn/2 ! if n is even, and that the generating function w(x) satisﬁes w(x) = x + 1 2(w(x)2 + w(x2)). 20 This is much more diﬃcult to solve. Whereas Cn is roughly 4n (in the sense that the limit of C1/n n as n →∞is 4), Wn is roughly 2.483 . . .n in the same sense. Exercises 1 Give a counting proof of the Vandermonde convolution in the case where a and b are natural numbers. 2 Verify some of the formulae for Catalan objects in the notes, either by deriving a recurrence, or by ﬁnding bijections between the objects counted. 3 In the analysis of Dyck paths, adopt the convention that D0 = 1 and E0 = 0. Prove that, if d(x) and e(x) are the generating functions, then xd(x) = e(x), d(x) = 1 + e(x)d(x). Hence derive formulae for Dn and En. 4 Use the hook length formula to derive the formula for the Catalan number Cn. 5 Prove the recurrence relation and the equation for the generating function for the Wedderburn–Etherington numbers. 4 Unimodality It is well known that the binomial coeﬃcients increase up to halfway, and then decrease. Indeed, the shape of the bar graph of binomial coeﬃcients is well approximated by a scaled version of the “bell curve” of the normal distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 This property of binomial coeﬃcients is easily proved. Since n k + 1 = n −k k + 1 n k , the binomial coeﬃcient increases from k to k + 1, remains constant, or de-creases, according as n −k > k + 1, n −k = k + 1 or n −k < k + 1, that is, as n is greater than, equal to, or less than 2k + 1. So, if n is even, the binomial coeﬃcients increase up to k = n/2 and then decrease; if n is odd, the two middle values (k = (n −1)/2 and k = (n + 1)/2) are equal, and they increase before this point and decrease after. Other combinatorial numbers also show this unimodality property, but in cases where we don’t have a formula, we need general techniques. 4.1 Unimodality and log-concavity Given a sequence of positive numbers, say a0, a1, a2, . . . , an, we say that the sequence is unimodal if there is an index m with 0 ≤m ≤n such that a0 ≤a1 ≤· · · ≤am ≥am+1 ≥· · · ≥an. The sequence a0, a1, a2, . . . , an of positive integers is said to be log-concave if a2 k ≥ak−1ak+1 for 1 ≤k ≤n −1. The reason for the name is that the logarithms of the as are concave: setting bk = log ak, we have 2bk ≤ bk−1 + bk+1, or in other words, bk+1 −bk ≤bk −bk−1. So if we plot the points (k, bk) for 0 ≤k ≤n, then the slopes of the lines joining consecutive points decrease as k increases, so that the ﬁgure they form is concave when viewed from above. Now it is clear that a log-concave sequence is unimodal. A nice general result is: Theorem 4.1 Let A(x) = n X k=0 akxk be the generating polynomial for the numbers a0, . . . , an. Suppose that all the roots of the equation A(x) = 0 are real and negative. Then the sequence a0, . . . an is log-concave. Before we begin the proof, we note that a polynomial with all coeﬃcients positive cannot have a real non-negative root, and a polynomial all of whose roots are negative has all its coeﬃcients positive. 22 The proof is by induction: there is nothing to prove when n = 1, since any sequence of two numbers is log-concave. For n = 2, the condition for the polynomial a0 + a1x + a2x2 to have real roots is a2 1 −4a0a2 ≥0, which is stronger than log-concavity; as remarked, if the roots are real, they are negative. Now we turn to the general case. Suppose that A(x) = (x+c)B(x), where c > 0 and B(x) = bn−1xn−1 + · · · + b1x + b0. Now the polynomial B(x) has all its roots real and negative, since they are all the roots of A(x) except for −c. So the coeﬃcients are all positive, and by the inductive hypothesis, the sequence b0, . . . , bn−1 is log-concave; that is, b2 k ≥bk−1bk+1 for k = 1, . . . , n −2. Also, since A(x) = (x + c)B(x), we have a0 = cb0, an = bn−1, and ak = bk−1 + cbk for 1 ≤k ≤n −1. We ﬁrst show that bkbk−1 ≥bk+1bk−2 for 2 ≤k ≤n −2. For we have b2 kbk−1 ≥bk+1b2 k−1 ≥bk+1bkbk−2; dividing by bk gives the result. Now for 2 ≤k ≤n −2, we have a2 k −ak+1ak−1 = (bk−1 + cbk)2 −(bk + cbk+1)(bk−2 + cbk−1) = (b2 k−1 −bkbk−2) + c(bk−1bk −bk+1bk−2) + c2(b2 k −bk+1bk−1); and all three terms are non-negative since c > 0. The cases k = 1 and k = n −1 are left to the reader. 4.2 Binomial coeﬃcients and Stirling numbers For the binomial coeﬃcients, we have n X k=0 n k xk = (1 + x)n; all its roots are −1, and so the theorem shows that the binomial coeﬃcients are log-concave, and hence unimodal. 23 For the unsigned Stirling numbers of the ﬁrst kind, we have n X k=1 u(n, k)xk = x(x + 1) · · · (x + n −1), and the polynomial on the right has roots 0, −1, −2, . . . , −(n −1). We can neglect the zero root: the Stirling numbers start at k = 1 rather than zero, and dividing by x simply changes the indexing so that they start at 0. So again the Stirling numbers are log-concave and hence unimodal. The Stirling numbers of the second kind are more diﬃcult, since there is no convenient form for the generating polynomial. We start with the recurrence relation S(n, 1) = S(n, n) = 1, S(n, k) = S(n−1, k−1)+kS(n−1, k) for 1 < k < n. Let An(x) = n X k=0 S(n, k)xk. We have A0(x) = 1. For n > 0, we have A(n, 0) = 0, so zero is a root of An(x) = 0. We have to show that the other roots are real and negative. We prove this by induction: P1(x) = x has a single root at x = 0, while A2(x) = x + x2 has roots at x = 0 and x = −1; so the induction begins. From the recurrence relation, we have An(x) = n X k=1 S(n, k)xk = n X k=1 S(n −1, k −1)xk + n X k=1 kS(n −1, k)xk = x (dAn−1(x)/dx + An−1(x)) . Putting Bn(x) = An(x)ex, we see that An(x) = 0 and Bn(x) = 0 have the same roots. The identity above, multiplied by ex, gives x dBn−1(x)/dx = Bn(x). By Rolle’s Theorem, there is a root of Bn(x) between each pair of roots of Bn−1(x), and one to the left of the smallest root of Bn−1(x) (since Bn−1(x) → 0 as x →−∞); and also a a root at 0. This accounts for (n −2) + 1 + 1 roots, that is, all the roots of Bn(x). So the induction step is complete. 24 Exercises 1 Let S be a ﬁxed set of positive integers, and let rn be the number of partitions of n into distinct parts from the set S. What is the generating polynomial P rnxn? Is the sequence (rn) unimodal? 2 Let (an) be an inﬁnite sequence of positive numbers which is log-concave (that is, an−1an+1 ≤a2 n for all n ≥1). Show that the ratio an+1/an tends to a limit as n →∞. 5 q-analogues In a sense, a q-analogue of a combinatorial formula is simply another formula involving a variable q which has the property that, as q →1, the second formula becomes the ﬁrst. Of course there is more to it than that; some q-analogues are more important than others. What follows is nothing like a complete treatment; I will concentrate on a particularly important case, the Gaussian or q-binomial coeﬃcients, which are, in the above sense, q-analogues of binomial coeﬃcients. 5.1 Deﬁnition of Gaussian coeﬃcients The Gaussian (or q-binomial) coeﬃcient is deﬁned for non-negative integers n and k as n k q = (qn −1)(qn−1 −1) · · · (qn−k+1 −1) (qk −1)(qk−1 −1) · · · (q −1) . In other words, in the formula for the binomial coeﬃcient, we replace each factor r by qr −1. Note that this is zero if k > n; so we may assume that k ≤n. Now observe that lim q→1 qr −1 q −1 = r. This follows from l’Hˆ opital’s rule: both numerator and denominator tend to 0, and their derivatives are rqr−1 and 1, whose ratio tends to r. Alternatively, use the fact that qr −1 q −1 = 1 + q + · · · + qr−1, and we can now harmlessly substitute q = 1 in the right-hand side; each of the r terms becomes 1. 25 Hence if we replace each factor (qr −1) in the deﬁnition of the Gaussian coeﬃcient by (qr −1)/(q −1), then the factors (q −1) in numerator and denominator cancel, so the expression is unchanged; and now it is clear that lim q→1 n k q = n k . 5.2 Interpretations Quantum calculus The letter q stands for “quantum”, and the q-binomial coeﬃcients do play a role in “quantum calculus” similar to that of the ordi-nary binomial coeﬃcients in ordinary calculus. I will not discuss this further; see the book Quantum Calculus, by V. Kac and P. Cheung, Springer, 2002, for further details. Vector spaces over ﬁnite ﬁelds The letter q is also routinely used for the number of elements in a ﬁnite ﬁeld (which is necessarily a prime power, and indeed there is a unique ﬁnite ﬁeld of any given prime power order q – a theorem of Galois). Theorem 5.1 Let V be an n-dimensional vector space over a ﬁeld with q elements. Then the number of k-dimensional subspaces of v is n k q . Proof The proof follows the standard proof for binomial coeﬃcients count-ing subsets of a set. A k-dimensional subspace of V is speciﬁed by choosing a basis, a sequence of k linearly independent vectors. Now the number of choices of the ﬁrst vector is qn −1 (since every vector except the zero vector is eligible); the second can be chosen in qn −q ways (since the q multiples of the ﬁrst vector are now ineligible); the third in qn−q2 ways (since the q2 linear combinations of the ﬁrst two are now ruled out); and so on. In total, (qn −1)(qn −q) · · · (qn −qk−1) choices. We have to divide this by the number of k-tuples of vectors which form a basis for a given k-dimensional subspace. This number is obtained by 26 replacing n by k in the above formula, that is, (qk −1)(qk −q) · · · (qk −qk−1). Dividing, and cancelling the powers of q, gives the result. Remark Let F denote a ﬁeld of q elements. Then a set of k linearly independent vectors in F n can be represented as a k × n matrix of rank k. We may put it into reduced echelon form by elementary row operations without changing the subspace it spans; and, indeed, any subspace has a unique basis in reduced echelon form. So as a corollary we obtain Corollary 5.2 The number of k×n matrices over a ﬁeld of q elements which are in reduced echelon form is n k q . As a reminder, a matrix is in reduced echelon form if (a) the ﬁrst non-zero entry in any row is a 1 (a leading 1); (b) the leading 1s occur further to the right in successive rows; (c) all the other elements in the column of a leading 1 are 0. This has two consequences. First, it gives us another way of calculating the Gaussian coeﬃcients. For example, the 2×4 matrices in reduced echelon form are as follows, where ∗denotes any element of the ﬁeld: 1 0 ∗ ∗ 0 1 ∗ ∗ , 1 ∗ 0 ∗ 0 0 1 ∗ , 1 ∗ ∗ 0 0 0 0 1 , 0 1 0 ∗ 0 0 1 ∗ , 0 1 ∗ 0 0 0 0 1 , 0 0 1 0 0 0 0 1 . So we have 4 2 q = q4 + q3 + q2 + q2 + q + 1 = (q2 + 1)(q2 + q + 1). This expression, and the deﬁnition, are polynomials in q, which agree for ev-ery prime power q; so they coincide. In a similar way, any Gaussian coeﬃcient can be written out as a polynomial. 27 The other consequence is that algebra is not required here. Over any alphabet of size q, containing two distinguished elements 0 and 1, the number of k × n matrices in “reduced echelon form” (satisfying (a)–(c) above) with no zero rows is n k q . Lattice paths How many lattice paths are there from the origin to the point (m, n), where each step in the path moves one unit either north or east? Clearly the number is m + n m , since we must take m+n steps of which m are north and n are east, and the northward steps may occur in any m of the m + n positions. Suppose we want to count the paths by the area under the path (that is, bounded by the X-axis, the line x = m, and the path). We use a generating function approach, so that a path enclosing an area of a units contributes qa to the overall generating function. Here q is simply a formal variable; the answer is obviously a polynomial in q. Theorem 5.3 The generating function for lattice paths from (0, 0) to (m, n) by area under the path is m + n m q . We will see why in the next section. Note that, as q →1, we expect the formula to tend to m + n m . A non-commutative interpretation Let x and y be elements of a (non-commutative) algebra which satisfy yx = qxy, where the coeﬃcient q is a “scalar” and commutes with x and y. Then we have the following analogue of the Binomial Theorem (see Exercises): Theorem 5.4 (x + y)n = X k=0 n n k q x(n −k)yk. For example, (x + y)3 = xxx + xxy + xyx + yxx + xyy + yxy + yyx + yyy. 28 We can use the relation to move the y’s to the end in each term; each time we jump a y over an x we pick up a factor q. So (x + y)3 = x3 + (1 + q + q2)x2y + (1 + q + q2)xy2 + y3, in agreement with the theorem. 5.3 Combinatorial properties These properties can be proved in two ways: by using the counting inter-pretation involving subspaces of a vector space, or directly from the formula (usually easiest). The proofs are all relatively straightforward; I will just give outlines where appropriate. Proposition 5.5 n k q = n n −k q . This is straightforward from the formula. Alternatively we can invoke vector space duality: there is a bijection between subspaces of dimension k of a vector space and their annihilators (subspaces of codimension k of the dual space). Proposition 5.6 n 0 q = n n q = 1, and n k q = n −1 k −1 q + qk n −1 k q for 0 < k < n. Again, straightforward from the formula. Alternatively, consider k × n matrices in reduced echelon. If the leading 1 in the last row is in the last column, then the other entries in the last row and column are zero, and deleting them gives a (k −1)×(n−1) matrix in reduced echelon. Otherwise, the last column is arbitrary (so there are qk possibilities for it; deleting it leaves a k × (n −1) matrix in reduced echelon. Remark From this we can prove Theorem 5.3, as follows. Let Q(n, k) be the sum of the weights of lattice paths from (0, 0) to (n −k, k), where the weight of a path is qa if the area below it is a. Clearly Q(n, 0) = Q(n, n) = 1. Consider all the lattice paths from (0, 0) to (n −k, k), and divide them into two classes: those in which the last step is vertical, and those in which it is horizontal. In the ﬁrst case, the last step is from the end of a path 29 counted by Q(n−1, k −1) (ending at (n−k, k −1)), and adds no area to the path. In the second step, it is from the end of a path counted by Q(n −1, k) (ending at (n −k −1, k)), and increases the area by k, adding qkQ(n −1, k) to the sum. So the numbers Q(n, k) have the same recurrence and boundary conditions as the Gaussian coeﬃcients, and must be equal to them. From the last two results, we can deduce an alternative recurrence: n k q = qn−k n −1 k −1 q + n −1 k q . 5.4 The q-binomial theorem The q-analogue of the Binomial Theorem states: Theorem 5.7 For any positive integer n, n Y i=1 (1 + qi−1z) = n X k=0 qk(k−1)/2zk n k q . The proof is by induction on n; starting the induction at n = 1 is trivial. Suppose that the result is true for n −1. For the inductive step, we must compute n−1 X k=0 qk(k−1)/2zk n −1 k q ! 1 + qn−1z . The coeﬃcient of zk in this expression is qk(k−1)/2 n −1 k q + q(k−1)(k−2)/2+n−1 n −1 k −1 q = qk(k−1)/2 n −1 k q + qn−k n −1 k −1 q ! = qk(k−1)/2 n k q by the alternative recurrence relation. I state without proof here Heine’s formula, the q-analogue of the negative binomial theorem: n Y i=1 (1 −qi−1z)−1 = ∞ X j=0 n + j −1 j q zj. 30 5.5 Jacobi’s Triple Product Identity This is only loosely connected with the topics of this chapter, but is inter-esting in its own right. Theorem 5.8 (Jacobi’s Triple Product Identity) Y n>0 (1 + q2n−1z)(1 + q2n−1z−1)(1 −q2n) = X l∈Z ql2zl. The sharp-eyed will notice that the series on the right breaks my rules that formal Laurent series should have only ﬁnitely many negative terms. Well, this just shows that formal power series are more ﬂexible than might ﬁrst appear! You can check that the three inﬁnite products on the left contribute only ﬁnitely many terms to each power, positive or negative, of z. By replacing q by q1/2 and moving the third term in the product to the right-hand side, the identity takes the form Y n>0 (1 + qn−1/2z)(1 + qn−1/2z−1) = X l∈Z ql2/2zl ! Y n>0 (1 −qn)−1 ! , in which form we will prove it. The proof here is a remarkable argument by Richard Borcherds, and this write-up from my Combinatorics textbook. A level is a number of the form n + 1 2, where n is an integer. A state is a set of levels which contains all but ﬁnitely many negative levels and only ﬁnitely many positive levels. The state consisting of all the negative levels and no positive ones is called the vacuum. Given a state S, we deﬁne the energy of S to be X {l : l > 0, l ∈S} − X {l : l < 0, l ̸∈S}, while the particle number of S is \|{l : l > 0, l ∈S}\| −\|{l : l < 0, l ̸∈S}\|. Although it is not necessary for the proof, a word about the background is in order! Dirac showed that relativistic electrons could have negative as well as positive energy. Since they jump to a level of lower energy if possible, Dirac hypothesised that, in a vacuum, all the negative energy levels are occupied. 31 Since electrons obey the exclusion principle, this prevents further electrons from occupying these states. Electrons in negative levels are not detectable. If an electron gains enough energy to jump to a positive level, then it becomes ‘visible’; and the ‘hole’ it leaves behind behaves like a particle with the same mass but opposite charge to an electron. (A few years later, positrons were discovered ﬁlling these speciﬁcations.) If the vacuum has no net particles and zero energy, then the energy and particle number of any state should be relative to the vacuum, giving rise to the deﬁnitions given. We show that the coeﬃcient of qmzl on either side of the equation is equal to the number of states with energy m and particle number l. This will prove the identity. For the left-hand side this is straightforward. A term in the expansion of the product is obtained by selecting qn−1 2z or qn−1 2z−1 from ﬁnitely many factors. These correspond to the presence of an electron in positive level n −1 2 (contributing n −1 2 to the energy and 1 to the particle number), or a hole in negative level −(n −1 2) (contributing n −1 2 to the energy and −1 to the particle number). So the coeﬃcient of qmzl is as claimed. The right-hand side is a little harder. Consider ﬁrst the states with parti-cle number 0. Any such state can be obtained in a unique way from the vac-uum by moving the electrons in the top k negative levels up by n1, n2, . . . , nk, say, where n1 ≥n2 ≥. . . ≥nk. (The monotonicity is equivalent to the re-quirement that no electron jumps over another. The jumping process allows the possibility that some electrons jump from negative levels to higher but still negative levels, so k is not the number of occupied positive levels.) The energy of the state is thus m = n1+. . .+nk. Thus, the number of states with energy m and particle number 0 is equal to the number p(m) of partitions of m, which is the coeﬃcient of qm in P(q) = Q n>0(1 −qn)−1, as we saw in lecture 1. Now consider states with positive particle number l. There is a unique ground state, in which all negative levels and the ﬁrst l positive levels are ﬁlled; its energy is 1 2 + 3 2 + . . . + 2l −1 2 = 1 2l2, and its particle number is l. Any other state with particle number l is obtained from this one by ‘jumping’ electrons up as before; so the number of such states with energy m is p(m −1 2l2), which is the coeﬃcient of qmzl in ql2/2zlP(q), as required. 32 The argument for negative particle number is similar. Exercises 1 Prove that, for ﬁxed n, the Gaussian coeﬃcients are unimodal. 2 For ﬁxed n and k, the Gaussian coeﬃcient n k q is a polynomial in q of degree k(n −k), whose coeﬃcients a0, . . . , ak(n−k) are non-negative integers. Prove that the coeﬃcients are symmetric: that is, ai = ak(n−k)−i. Remark It is also true that the coeﬃcients are unimodal, but this is not so easy to prove. The polynomial does not have all its roots real and negative! 3 Show that, for a, b equal to 0 or 1, 2m + a 2l + b −1 =    0 if a = 0 and b = 1, m l otherwise. Remark For a more challenging exercise, ﬁnd a formula for n k ω , where ω is a primitive dth root of unity. 4 Deduce Euler’s Pentagonal Numbers Theorem from Jacobi’s Triple Prod-uct Identity. (Hint: put q = t3/2, z = −t−1/2.) 5 Consider the algebra generated by two non-commuting variables x and y satisfying the relation yx = qxy. Prove that (x + y)n = n X k=0 n k q xn−kyk. 33 6 Symmetric polynomials A symmetric polynomial in n indeterminates is one which is unchanged under any permutation of the indeterminates. The theory of symmetric polynomials goes back to Newton, but more recently has been very closely connected with the representation theory of the symmetric group, which we glanced at in Lecture 3. I will just give a few simple results here. The best reference is Ian Macdonald’s book Symmetric Functions and Hall Polynomials. 6.1 Symmetric polynomials Let x1, . . . , xn be indeterminates. If π is a permutation of {1, . . . , n}, we denote by iπ the image of i under π. Now a polynomial F(x1, . . . , xn) is a symmetric polynomial if F(x1π, . . . , xnπ) = F(x1, . . . , xn) for all π ∈Sn, where Sn is the symmetric group of degree n (the group of all polynomials of degree n). Any polynomial is a linear combination of monomials xa1 1 · · · xan n , where a1, . . . , an are non-negative integers. The degree of this monomial is a1+· · ·+ an. A polynomial is homogeneous of degree r if every monomial has degree r. Any polynomial can be written as a sum of homogeneous polynomials of degrees 1, 2, . . .. In a homogeneous symmetric polynomial of degree r, the exponents in any monomial form a partition of r into at most n parts; two monomials which give rise to the same partition are equivalent under a permutation, and so must have the same coeﬃcient. Thus, the dimension of the space of homogeneous symmetric polynomials of degree r is pn(r), the number of partitions of r with at most n parts. There are three especially important symmetric polynomials: (a) The elementary symmetric polynomial er, which is the sum of all the monomials consisting of products of r distinct indeterminates. Note that there are n r monomials in the sum. (b) The complete symmetric polynomial hr, which is the sum of all the monomials of degree r. There are n + r −1 r terms in the sum: the proof of this is given in the Appendix to these notes. 34 (c) The power sum polynomial pr, which is simply n X i=1 xr i. For example, if n = 3 and r = 2, (a) the elementary symmetric polynomial is x1x2 + x2x3 + x1x3; (b) the complete symmetric polynomial is x1x2+x2x3+x1x3+x2 1+x2 2+x2 3; (c) the power sum polynomial is x2 1 + x2 2 + x2 3. Note that er(1, . . . , n) = n r , hr(1, . . . , 1) = n + r −1 r , and pr(1, . . . , 1) = n. Also, the q-binomial theorem that we met in the last lecture shows that er(1, q, q2, . . . , qn−1) = qr(r−1)/2 n r q , and Heine’s formula shows that, similarly, hr(1, q, q2, . . . , qn−1) = n + r −1 r q . 6.2 Generating functions The best-known occurrence of the elementary symmetric polynomials is the connection with the roots of polynomials. (To avoid conﬂict with xi, the variable in a polynomial is t in this section.) The coeﬃcient of tn−r in a polynomial of degree n is (−1)rer(a1, . . . , an), where a1, . . . , an are the roots. This is because the polynomial can be written as (t −a1)(t −a2) · · · (t −an), and the term in tn−r is formed by choosing t from n −r of the factors and −ai from the remaining r. Said otherwise, and putting xi = −1/ai, this says that the generating function for the elementary symmetric polynomials is E(t) = n X r=0 er(x1, . . . , xn)tr = n Y i=1 (1 + xit), 35 with the convention that e0 = 1. In a similar way, the generating function for the complete symmetric polynomials is H(t) = X r≥0 hr(x1, . . . , xn)tr = n Y i=1 (1 −xit)−1. We also take P(t) to be the generating function for the power sum polyno-mials, with a shift: P(t) = X r≥1 pr(x1, . . . , xn)tr−1. Now we see that H(t) = E(−t)−1, so that n X r=0 (−1)r3rhn−r = 0 for n ≥1. For P(t), we have d dtH(t) = P(t)H(t), d dtE(t) = P(−t)E(t). 6.3 Functions indexed by partitions We extend the deﬁnitions of symmetric polynomials as follows. Let λ = (a1, a2, . . .) be a partition of r, a non-decreasing sequence of integers with sum r. Then, if z denotes one of the symbols e, h or p, we deﬁne zλ to be the product of zai over all the parts ai of λ; this is again a symmetric polynomial of degree r. For example, if n = 3 and λ is the partition (2, 1) of 3, we have eλ = (x1x2 + x1x3 + x2x3)(x1 + x2 + x3), pλ = (x2 1 + x2 2 + x2 3)(x1 + x2 + x3), hλ = eλ + pλ. We also deﬁne the basic polynomial mλ to be the sum of all monomials with exponents a1, a2, . . .. In the above case, mλ = x2 1x2 + x2 1x3 + x2 2x1 + x2 2x3 + x2 3x1 + x2 3x2. 36 Theorem 6.1 If n ≥r, and z is one of the symbols m, e, h, p, then any symmetric polynomial of degree r can be written uniquely as a linear com-bination of the polynomials zλ, as λ runs over all partitions. Moreover, in all cases except z = p, if the polynomial has integer coeﬃcients, then it is a linear combination with integer coeﬃcients. So the polynomials er or hr, with r ≤n, are generators of the ring of symmetric polynomials in n variables with integer coeﬃcients. For z = e, this is a version of Newton’s Theorem on symmetric polynomials (which, however, applies also to rational functions). 6.4 Appendix: Selections with repetition Theorem 6.2 The number of n-tuples of non-negative integers with sum r is n + r −1 r . The claim about the number of monomials of degree r follows immediately from this result, which should be contrasted with the fact that the number of n-tuples of zeros and ones with sum r is n r . Proof We can describe any such n-tuple in the following way. Take a line of n + r −1 boxes. Then choose n −1 boxes, and place barriers in these boxes. Let (a) a1 be the number of empty boxes before the ﬁrst barrier; (b) a2 be the number of empty boxes between the ﬁrst and second barriers; (c) . . . (d) an be the number of empty boxes after the last barrier. Then a1, . . . , an are non-negative integers with sum r. Conversely, given n non-negative integers with sum r, we can represent it with n −1 barriers in n + r −1 boxes: place the ﬁrst barrier after a1 empty boxes, the second after a2 further empty boxes, and so on. So the required number of n-tuples is equal to the number of ways to position n −1 barriers in n + r −1 boxes, which is n + r −1 n −1 = n + r −1 r , 37 as required. 7 Group actions How many ways can you colour the faces of a cube with three colours? Clearly the answer is 36 = 729. But what if we regard two colourings as the same if one can be transformed into the other by a rotation of the cube? This is typical of the problems we consider in this chapter. 7.1 The Orbit-Counting Lemma This chapter of the lectures, unlike most of the others, requires some technical background. I assume that you know the deﬁnition of a group. I will run brieﬂy through the theory of group actions, and ﬁnally reach the Orbit-Counting Lemma, which solves our introductory problem. Throughout this section, permutations act “on the right”, that is, the eﬀect of applying a permutation π to an element x of the domain is written xπ. This is not just a matter of notation; it entails the fact that the product π1π2 of two permutations is calculated by the rule “ﬁrst π1, then π2”, rather than the other way round. This ensures that x(π1π2) = (xπ1)π2 for all elements x. An action of a group G on a set X is a map associating to each group element g ∈G a permutation πg of X in such a way that the following two conditions hold: (a) πgh = πgπh for all g, h ∈G (that is, xπgh = xπgπh for all g, h ∈G and all x ∈X); (b) if 1 denotes the identity element of G, then π1 is the identity permuta-tion (that is, xπ1 = x for all x ∈X). Usually we simplify notation by not distinguishing between g and πg, writing simply xg instead of xπg. From a diﬀerent point of view, an action is a homomorphism from the group G to the symmetric group of all permutations of X. Two elements x, y ∈X are equivalent under the action if there exists an element g ∈G such that xg = y. It is routine to show that this is really an equivalence relation; its equivalence classes are called orbits, and the action is transitive if there is just one orbit. Thus we have a ﬁrst structure theorem: 38 any action can be split uniquely into transitive actions on the sets of the orbit partition of the domain. In our motivating problem, the group G of 24 rotations of the cube acts on the set X of 729 coloured cubes, and we want to count the orbits. So our immediate goal is to count the orbits in an arbitrary action. If H is a subgroup of G, then there is a natural partition of G into right cosets Hx of H, for x ∈G. Lagrange’s Theorem assures us that each coset has the same cardinality, so the number of cosets is equal to \|G\|/\|H\|. We denote the set of right cosets of H in G by cos(H, G). Now there is an action of G on the set cos(H, G): the group element g induces the permutation Hx 7→H(xg). At risk of some confusion, we write this as (Hx)g = H(xg). Now, given any transitive action of G on a set X, and x ∈X, the set {g ∈G : xg = x} is a subgroup of H, called the stabiliser of x, and denoted by StabG(x). Now there is a natural bijection between X and cos(H, G), where the element y ∈X corresponds to the set {g ∈G : xg = y} (it is easily checked that this is a coset of H). This bijection also respects the action of G: if z ∈G satisﬁes yg = z, and Hk and Hl are the cosets corresponding to y and z, then (Hk)g = (Hl). So the so-called “coset spaces” of subgroups of G give a complete list of transitive actions of G, up to a natural notion of isomorphism of actions. Note in addition that any two points in the same orbit have stabilisers of the same order. (The stabilisers are in fact conjugate subgroups of G.) In an arbitrary action of G on X, we let ﬁxX(g) denote the number of points of X which are ﬁxed by the permutation g. Now we can state the Orbit-Counting Lemma, the foundation of enumeration under group action. Theorem 7.1 Let G act on the ﬁnite set X. Then the number of G-orbits in X is equal to the average number of ﬁxed points of elements of G, that is, 1 \|G\| X g∈G ﬁxX(g). The theorem has a probabilistic interpretation. Choose a random element of G (from the uniform distribution). Then its expected number of ﬁxed points is equal to the number of orbits of G. 39 Proof Construct a bipartite graph as follows. The vertices are of two types: the elements of X, and the elements of G. There is an edge from x to g if xg = x. We count the number of edges in two diﬀerent ways. Each vertex g lies in ﬁxX(g) edges; so the number of edges is X g∈G ﬁxX(g). Now we count the other way. Take a point x ∈X. The number of edges containing it is \| StabG(x)\|. This value is the same for all the points in the orbit OG(x) containing x. So the number of edges containing points in the orbit is \| StabG(x)\| · \|OG(x)\| = \|G\|. Since each orbit contributes \|G\| edges, the number of orbits is obtained by dividing the number of edges by \|G\|, as claimed. Now consider the coloured cubes. In order to do the calculations, we need to classify the elements of the group G of rotations of the cube (a group of order 24). They are of the following types: (a) the identity; (b) “face rotations” (about an axis through two opposite face centres) through ±π/2 (six of these, two for each pair of opposite faces); (c) “face rotations” through π (three of these); (d) “edge rotations” (about an axis through two opposite edge midpoints) through π (six of these); (e) “vertex rotations” (about an axis through two opposite vertices) through ±2π/3 (eight of these, two for each pair of opposite vertices). For each type of rotation, we have to count the number of coloured cubes it ﬁxes. A cube will be ﬁxed if faces in the same cycle of the permutation have the same colour. So the answer will be 3c, where c is the number of cycles of the permutation on faces. For the ﬁve types listed above the numbers of cycles are 6 (each single face is a cycle), 3 (for the vertical axis, the top and bottom faces, and the other four in a single cycle), 4 (as the previous except that the 4-cycle splits into two 2-cycles), 3 (the faces are permuted in cycles of two), and 2 (the faces are permuted in cycles of three). So the calculation of the theorem is: 1 24(36 + 6 · 33 + 3 · 34 + 6 · 33 + 8 · 32) = 57. 40 7.2 Labelled and unlabelled Many combinatorial objects that we want to count are based on an underlying set, which we usually assume to be the set {1, 2, . . . , n}. Very often the simplest method of counting gives us the total number of objects that can be built on this set. But we may be completely uninterested in the labels 1, 2, . . . , n, and want to count two objects as being the same if there are some labellings of the underlying set that make them identical. We distinguish these two problems as counting labelled and unlabelled objects. Counting unlabelled objects is thus an orbit-counting problem: we want to know the number of orbits of the symmetric group Sn, acting on the objects in question by permuting the labels. To take an extreme case: there are n k labelled k-element subsets of an n-element set, but there is only one unlabelled subset. Here are a few more examples. Objects Labelled Unlabelled Subsets 2n n + 1 Partitions B(n) p(n) Permutations n! p(n) Linear orders n! 1 Here B(n) is the Bell number (the number of partitions of an n-set) and p(n) the number of partitions of the number n. Note that the numbers of unlabelled structures can agree and those of labelled structures disagree, or vice versa. The third entry needs a little explanation. Any permutation can be writ-ten as a product of disjoint cycles; the cycle lengths form a partition of n called the cycle structure of the permutation. Now given two permutations with the same cycle structure, we can replace the entries in one by those in the other. For example, (1)(2, 3) can be transformed into (2)(1, 3) by swap-ping the labels 1 and 2. (You might recognise this as the argument that shows that two permutations are conjugate in the symmetric group if and only if they have the same cycle structure.) In the three cases in the table, we can count the unlabelled objects di-rectly; but in more complicated cases, the Orbit-Counting Lemma is required. One example is the number of graphs on n vertices. The labelled number 41 is 2n(n−1)/2, since for each of the n(n −1)/2 pairs of vertices we can choose whether to join it by an edge or not; but the only way to calculate the nuber of unlabelled graphs is via the Orbit-Counting Lemma. 7.3 Cycle index There is a way to “mechanise” the counting in many important cases, which we now discuss. This was introduced by Redﬁeld and, independently, by P´ olya, and reﬁned by de Bruijn and others. (Incidentally, these early work-ers found the Orbit-Counting Lemma in Burnside’s group theory book, and called it “Burnside’s Lemma”, a name which is still sometimes used. How-ever, the result is due to Frobenius, and earlier to Cauchy in a special case.) The set-up is as follows. We have a set X on which a group G acts. We are going to decorate X by placing one of a set of “ﬁgures” at each point. Each ﬁgure has a weight, which is a non-negative integer. We don’t require the number of ﬁgures to be ﬁnite, but we ask that there should be only ﬁnitely many ﬁgures of any given weight. The ﬁgures can thus be counted by the ﬁgure-counting series A(x) = X n≥0 anxn, where an is the number of ﬁgures of weight n. Now one of the conﬁgurations we want to count consists of the set X with a ﬁgure at each point; this can be described by a function from X to the set of ﬁgures. Such a function f will have a weight, given by w(f) = P{w(x) : x ∈X}. There are only ﬁnitely many functions of any given weight, and the action of the group G preserves weight; so we can let bn be the number of functions of weight n, and deﬁne the function-counting series B(x) = X n≥0 bnxn. The ﬁnal ingredient is the cycle index polynomial Z(G), deﬁned as Z(G) = 1 \|G\| X g∈G sc1(g) 1 sc2(g) 2 · · · scn(g) n . Here s1, . . . , sn are indeterminates, and ci(g) is the number of cycles of length i in the cycle decomposition of g, for i = 1, . . . , n. Now the Cycle Index Theorem states: 42 Theorem 7.2 B(x) = Z(G; si ←A(xi) for i = 1, . . . , n). The notation on the right means that we substitute A(xi) for si, for i = 1, . . . , n. I won’t prove the theorem here – it follows from the Orbit-Counting Lemma with a certain amount of ingenuity – but will conclude with a simple application which doesn’t even hint at the uses of the theorem. First, let us calculate the cycle index of the rotation group of the cube. The ﬁve types of elements mentioned earlier have the following cycle struc-tures in their action on faces: (a) Identity: (1, 1, 1, 1, 1, 1) (usually abbreviated to 16). (b) Face rotations through ±π/2: 124. (c) Face rotations through π: 1222. (d) Edge rotations: 23. (e) Vertex rotations: 32. So the cycle index is Z(G) = 1 24(s6 1 + 6s2 1s4 + 3s2 1s2 2 + 6s3 2 + 8s2 3). Now any counting problem for which we can write a ﬁgure-counting series can be solved by substitution. For example: (a) Take each of the three colours to be a ﬁgure of weight 0. The ﬁgure-counting series is simply 3. We recover our earlier count. (b) Take one of the colours (say red) to have weight 1, and all the others weight 0. The ﬁgure-counting series is x + 2. So substituting xi + 2 for si gives a polynomial in which the coeﬃcient of xk is the number of types of cube which have exactly k red faces. (c) A small extension of the Cycle Index Theorem shows that, if we sub-stitute pi(x, y, z) = xi +yi +zi for si, we obtain a trivariate polynomial in which the coeﬃcient of xiyjzk is the number of cubes with i red, j blue, and k green faces. (d) The generalisation to an arbitrary number of colours is now routine. 43 Exercises 1 Perform the calcuations in the four counting problems above. 2 A necklace has ten beads, each of which is either black or white, arranged on a loop of string. A cyclic permutation of the beads counts as the same necklace. How many necklaces are there? How many are there if the necklace obtained by turning over the given one is regarded as the same? 3 Let G be a permutation group on a set X, where \|X\| = n. For 0 ≤i ≤n, let pi be the proportion of elements of G which have exactly i ﬁxed points on X, and let p(x) = P pixi be the generating function for these numbers (the probability generating function for ﬁxed points). For 0 ≤i ≤n, let Fi be the number of orbits of G in its action on the set of i-tuples of distinct elements of X, and let F(x) = X Fixi i! be the exponential generating function for these numbers. Use the Orbit-counting Lemma to show that F(x) = P(x + 1) and deduce that the proportion of ﬁxed-point-free elements in G is p0 = F(−1). Taking G to be the symmetric group Sn, show that the number of ﬁxed-point-free permutations (the derangement number) is n! n X k=0 (−1)k k! . Deduce that this number is the closest integer to n!/e. 4 Consider the set of all functions from {1, . . . , n} to {1, . . . , m}. There are mn functions in the set. Now let the symmetric group Sn act on these functions by permuting their arguments: (fπ)(x) = f(xπ−1). [Incidentally, the inverse is there to make this an action – can you see why?] 44 Show that orbits correspond to m-tuples of non-negative integers with sum n, so that the number of orbits is m + n −1 n . (See the Appendix in Lecture Notes 7.) Show that a permutation g with k cycles ﬁxes mk functions. Hence use the Orbit-Counting Lemma to show that 1 n! n X k=1 u(n, k)mk = m + n −1 n . Show that we can replace m by an indeterminate x and multiply by n! to get the identity n X k=1 u(n, k)xk = x(x + 1) · · · (x + n −1), from which some sign changes yield n X k=1 s(n, k)xk = x(x −1) · · · (x −n + 1), a formula we met in Section 1. (Here s(n, k) and u(n, k) are the signed and unsigned Stirling numbers of the ﬁrst kind.) 8 Species In this lecture I will discuss a very nice unifying principle for a number of topics in enumerative combinatorics, the theory of species, introduced by Andr´ e Joyal in 1981. Species have been used in areas ranging from inﬁnite permutation groups to statistical mechanics, and I can’t do more here than barely scratch the surface. Joyal gave a category-theoretic deﬁnition of species; I will take a more informal approach. There is a book on species, by Bergeron, Labelle and Leroux, entitled Combinatorial Species and Tree-Like Structures; but I think that Joyal’s original paper in Advances in Mathematics is hard to beat. 45 8.1 What is a species? As I said earlier, a typical combinatorial structure of the type we wish to count is often built on a ﬁnite set; we are interested in counting labelled structures (the diﬀerent structures built on a ﬁxed set) and also the unla-belled structures (essentially the isomorphism types of structures). A species is a functor F (this word is used by Joyal in its technical sense from category theory; I will be less formal but will explain what is going on) which takes an n-element set and produces the set of objects in which we are interested; it should also have the property that the functor transforms any bijection between n-element sets A and B to a bijection between the sets F(A) and F(B) of objects built on these sets. Because of this condition, we can use the standard n-element set {1, 2, . . . , n}, but don’t have to worry if during the argument we have a non-standard set (such as a proper subset of the standard set). Joyal’s intuition is that we think of a formal power series where the coef-ﬁcients are not numbers, but sets of combinatorial objects: F = X n≥0 F({1, 2, . . . , n})xn. Suitable specialisations will give us the generating functions for unlabelled and unlabelled objects. The ﬁrst specialisation is to replace the set F(A) by the sum of the cycle indices of the automorphism groups of the unlabelled structures in F(A): let us call this Z(F). This will be a formal power series in inﬁnitely many variables s1, s2, . . .. Now it turns out that the specialisations f(x) = Z(F; sn ←xn for all n), F(x) = Z(F; s1 ←x, xn ←0 for n > 1), give us, respectively, the ordinary generating function for the unlabelled structures in the species F, and the exponential generating function for the labelled structures. 8.2 Examples If this is a bit abstract, hopefully some examples will bring it back to earth. 46 Sets Let Set denote the “identity” species, where the structure on the ﬁnite set A is simply a labelling of A. Thus, for each n, there is one unlabelled srtucture, and one labelled structure. So the generating functions are set(x) = X n≥0 xn = 1 1 −x, Set(x) = X n≥0 xn n! = exp(x) respectively. The cycle index of the species Set can be computed as follows. First, Z(Sn) = 1 n! X n! 1a1 · · · nana1! · · · an! sa1 1 · · · san n , where the sum is over all partitions of n having ai parts of size i for i = 1, 2, . . . , n (the coeﬃcient is the number of permutations with this cycle struc-ture). Summing this over all n seems a formidable task, but a remarkable simpliﬁcation occurs: since n! cancels we can sum over the variables a1, . . . , an independently. We obtain Z(Set) = exp X i≥1 si i ! . Now substituting Xi for si for all i gives set(x) = exp X i≥1 xi i ! = exp(−log(1 −x)) = 1 1 −x, Set(x) = exp(x), as expected. Note that the formula for the sum of the cycle indices of the symmetric groups was known in the combinatorial enumeration community before Joyal provided it with this nice interpretation. 47 Linear orders A much easier case is the species Lin of linear (or total) orders. There are n! labelled linear orders on n points; all are isomorphic, and there are no non-trivial automorphisms, so we have Z(Lin) = X n≥0 sn 1 = 1 1 −s1 , from which the generating functions are lin(x) = Lin(x) = 1/(1 −x). 8.3 Operations on species There are three important ways that we can add two species F and G. Sum F + G is the species which constructs on the set A all the F-objects and all the G-objects (we assume these two classes to be disjoint). Clearly the cycle index and the generating functions for unlabelled and labelled objects are simply obtained by adding those for F and G. Product F · /mathbfG is the species whose objects on a set A are con-structed in the following way: partition A into two (possibly empty) parts B and C; put an F-object on B, and a G-object on C. A slightly harder calculation shows that the cycle index, and hence the generating functions for unlabelled and labelled objects, are obtained by multiplying those for F and G. Here is an example. What is Set2? Given a set A, we partition it into a subset B and its complement A \ B. So we can regard this as the species Subset. The numbers of unlabelled and labelled objects in this species on n points are n + 1 and 2n respectively, and their generating functions are (as expected) 1/(1 −x)2 and exp(2x). Substitution As with power series in general, there is a formal restriction on substitution: we can only substitute G into F provided that G(∅) = ∅. If this condition holds, then we deﬁne F[G]-objects on A as follows: partition A (into non-empty parts); put a G-structure on each part; and put a F-structure on the set of parts. The cycle index is given by substituting the cycle index of G into that of F in the following way: Z(F[G]) = Z(F : sn ←Z(G, sm ←snm)). 48 In other words, for the indeterminate sn in Z(F, we substitute the cycle index of G but in the indeterminates sn, s2n, . . . in place of s1, s2, . . .. The eﬀect on the generating functions for labelled objects is simple sub-stitution: FG = F(G(x)). For unlabelled objects it is a bit more com-plicated, we need the cycle index for F: fg(x) = Z(F; sn ←g(xn) for all n). For example, let Set∗be the species of non-empty sets. Then the e.g.f. for labelled objects is Set∗(x) = exp(x) −1. Now Set[Set∗] is the species of set partitions, where the labelled objects are counted by the Bell numbers: the exponential generaing function is thus exp(exp(x)−1), as we saw earlier. As an exercise, obtain the ordinary generating function for partitions of the integer n from this approach. Remark The fact that substituting a species into Set exponentiates the generating function for labelled structures is sometimes called the exponential principle in enumerative combinatorics. We see that substitution of species is much more general. Rooted structures This means structures where one point is distinguished. It can be shown that the eﬀect of rooting a species is to apply the operator s1 ∂ ∂s1 to the cycle index, and hence to apply the operator x d/dx to the generating function for labelled structures. I will denote the operation of rooting a species by R, and the operation of rooting and then removing the root (i.e., deleting a point) by D: this just corresponds to diﬀerentiation. There are many other nice examples, some of which are described in the exercises. 8.4 Exercises 1 Deﬁne the species Circ of circular orders and the species Perm of per-mutations, and calculate the generating functions for unlabelled and labelled objects in these species. Show that Z(Circ) = − X m≥1 φ(m) m log(1 −sm), 49 where φ is Euler’s totient function. Use the decomposition of permutations into disjoint cycles to show that Set[Circ] = Perm, and verify the appropriate identities for the generating functions. Remark It is not so easy to calculate the cycle index of Perm directly, but using the above expression it is not too hard to show that Z(Perm) = Y n≥1 (1 −sn)−1. 2 Use the fact that Catalan objects are rooted binary trees to show that the species Cat of Catalan objects satisﬁes Cat = E + Cat2, where E denotes the species of singleton sets (that is, it returns its input if this has cardinality 1, and the empty set otherwise). Show similarly that the species W of rooted binary trees without the left-right distinction (counted by Wedderburn–Etherington numbers) satisﬁes W = E + Set2[W], where Set2 is the species of 2-element sets. 3 Let F denote the species of “1-factors” or partitions of a set into subsets of size 2. Show that D(F) = E · F, F = Set[Set2]. Use each of these equations to show that the exponential generating function for labelled 1-factors is exp(x2/2). 4 Let Graph and ConnGraph be the species of graphs and connected graphs respectively. (Here, assume that a connected graph has at least one vertex.) Show that Graph = Set[ConnGraph]. (It follows from this that the e.g.f. for connected graphs is the logarithm of the e.g.f. for graphs.) 50 9 M¨ obius inversion In this section we will discuss the Inclusion-Exclusion principle, with a few applications (including a formula for the chromatic polynomial of a graph), and then consider a wide generalisation of it due to Gian-Carlo Rota, involv-ing the M¨ obius function of a partially ordered set. The q-binomial theorem gives a simple formula for the M¨ obius function of the lattice of subspaces of a vector space. 9.1 Inclusion-Exclusion The Inclusion-Exclusion Principle is one of the most familiar results in com-binatorics. For two sets A and B, it asserts simply that \|A ∪B\| = \|A\| + \|B\| −\|A ∩B\|. For the general case, we need some notation. Let A1, . . . , An be subsets of a ﬁnite set S. For any subset I of the index set {1, 2, . . . , n, we let AI = \ i∈I Ai. By convention, we take A∅= S. Theorem 9.1 The number of elements lying in none of the sets A1, . . . , An is X I⊆{1,...,n} (−1)\|I\|\|AI\|. Proof We count the contribution of each element s ∈S to the sum in the above formula. If s lies in none of the sets Ai then it is counted once in the term A∅and in none of the others. Suppose that J = {i : s ∈Ai} ̸= ∅. Then the terms to which s contributes come from sets AI with I ⊆J, and the contribution is X I⊆J (−1)\|I\| = j X k=0 j k (−1)k = (1 −1)j = 0, where j = \|J\|. □ Corollary 9.2 Suppose that the family of sets has the property that, if \|I\| = i, then \|AI\| = mi. Then the number of points lying in none of the sets is n X i=0 (−1)i n i mi. 51 9.2 Applications We begin with two standard applications of the Corollary. First, a formula for the Stirling numbers of the second kind. Theorem 9.3 The number of surjective functions from an m-set to an n-set is n X i=0 (−1)i n i (n −i)m. Proof Let S be the set of all functions from M to N, where \|M\| = m and \|N\| = n, say N = {1, . . . , n}. Let Ai be the set of functions which do not take the value i. Then a function is surjective if and only if it lies in none of the sets Ai. If \|I\| = i, then AI consists of functions which take values in the set {1, . . . , n} \ I; there are (n −i)m such functions. So the theorem follows immediately from Corollary 9.2. □ Corollary 9.4 S(m, n) = 1 n! n X i=0 (−1)i n i (n −i)m. Proof We can describe a surjective function as follows: choose a partition of the domain into n parts (we can do this in S(m, n) ways, by deﬁnition of the Stirling number); then assign each part to a point of the codomain (which can be done in n! ways). So n!S(m, n) is the number of surjective functions. □ The second application concerns derangements: these are permutations of {1, . . . , n} with no ﬁxed points. Theorem 9.5 The number of derangements of {1, . . . , n} is given by the formula dn = n! n X i=0 (−1)i i! . 52 Proof Let S be the set of all permutations, and Ai the set of permutations which ﬁx the element i ∈{1, . . . , n}. Then a permutation is a derangement if and only if it lies in no set Ai. The permutations in AI ﬁx every point in the set I, so there are (n −i)! of them if \|I\| = i. Thus Corollary 9.2 gives dn = n X i=0 (−1)i n i (n −i)! = n! n X i=0 (−1)i i! . as claimed. □ The summation here is the partial sum of the series for e−1, so dn is approximately n!/e. Indeed, it is easy to show that it is the nearest integer to n!/e. The “secretary problem” asks: a secretary puts n letters into n addressed envelopes at random: what is the probability that no letter is correctly ad-dressed? The answer is very close to 1/e, perhaps a little surprising at ﬁrst sight. For our ﬁnal application we consider graphs. A graph consists of a set V of vertices and a set E of edges, each edge being a 2-element set of vertices. Given a set of q colours, a colouring of the graph is an assignment of colours to the vertices; it is proper if the two vertices in each edge have diﬀerent colours. Theorem 9.6 For any graph G = (V, E), there is a polynomial PG(x) such that, for any natural number q, PG(q) is the number of proper colourings of G with q colours. Moreover, PG is a monic polynomial with degree n = \|V \|. This is usually proved by operations on the graph (“deletion” and “con-traction”. The Inclusion-Exclusion proof here provides a formula. Proof Let S be the set of all colourings of G with q colours. For each edge e, let Ae be the set of colourings for which the edge e is “improperly coloured”, that is, its vertices have the same colour. A colouring is proper if it lies in no set Ae. Given a set I ⊆E, how many colourings lie in AI? Consider the graph (V, I) with edge set I. A colouring in Ai assigns the same colour to all vertices in the same connected component of this graph; so \|AI\| = qc(I), where c(I) is the number of connected components of (V, I). 53 By Theorem 9.1, the number of proper colourings is X I⊆E (−1)\|I\|qc(I). It is clear that this is a polynomial in q; the leading term comes from the unique graph (V, I) with n connected components, namely I = ∅. □ This formula shows a connection between graph colouring and the Potts model in statistical mechanics, but we cannot pursue this here. 9.3 The M¨ obius function of a poset A poset, or partially ordered set, consists of a set A with a relation ≤on A which is (a) reﬂexive: a ≤a for all a ∈A; (b) antisymmetric: a ≤b and B ≤a imply a = b, for all a, b ∈A; (c) transitive: a ≤b and b ≤c imply a ≤c, for all a, b, c ∈A. An important combinatorial example consists of the case where A is the set of all subsets of a ﬁnite set S, and a ≤b means that a is a subset of b. It turns out that the Inclusion-Exclusion principle can be formulated in terms of this poset, and then generalised so as to apply to any poset. We begin with an observation which will not be proved here. Theorem 9.7 Let P = (A, ≤) be a ﬁnite poset. Then we can label the elements of A as a1, a2, . . . , an such that, if ai ≤aj, then i ≤j. This is sometimes stated “Every poset has a linear extension”. The anal-ogous result for inﬁnite posets requires a weak form of the Axiom of Choice in its proof. Now let P = (A, ≤) be a poset. We deﬁne the incidence algebra of P as follows: the elements are all functions f : A × A →R such that f(a, b) = 0 unless a ≤b. Addition and scalar multiplication are deﬁned in the obvious way, and multiplication by the rule fg(a, b) =    X a≤c≤b f(a, c)g(c, b) if a ≤b, 0 if a ̸≤b. 54 If we number the elements of A as in the preceding theorem, then we can represent a function from A × A to R by an n × n matrix; the deﬁnition of the incidence algebra shows that any function which lies in the algebra is upper triangular. The multiplication in the algebra is then just matrix multiplication, so the incidence algebra is a subalgebra of the algebra of all n × n real matrices. We now deﬁne three particular elements of the incidence algebra. (a) ι is the identity function: ι(a, b) = 1 if a = b, 0 if a ̸= b , represented by the identity matrix. (b) ζ is the zeta function: ζ(a, b) = 1 if a ≤b, 0 if a ̸≤b. (c) µ, the M¨ obius function, is the inverse of the zeta function: µζ = ζµ = ι. The zeta function is represented by an upper unitriangular matrix with integer entries; so its inverse, the M¨ obius function, is also represented by an upper unitriangular matrix with integer entries. Its deﬁnition shows that, if a < b, then X a≤c≤b µ(a, c) = 0, so that µ(a, b) = − X a≤c<b µ(a, c). This gives a recursive method for calculating the M¨ obius function, as we will see. From the deﬁnition, we immediately have the M¨ obius inversion formula: Theorem 9.8 Let P be a poset with M¨ obius function µ. Then the following are equivalent: (a) g(a, b) = P a≤c≤b f(a, c) for all a ≤b; (b) f(a, b) = P a≤c≤b g(a, c)µ(c, b) for all a ≤b. 55 9.4 Some examples The preceding remark shows that the value of µ(a, b) depends only on the structure of the interval [a, b] = {c : a ≤c ≤b}. Many important posets have a least element (which is usually called 0) and a “homogeneity property”: for any a, b with a ≤b, there is an element c such that the interval [a, b] is isomorphic to the interval [0, c]. In a poset with this property, µ(a, b) = µ(0, c), and we can regard the M¨ obius function as a one-variable function. A chain A chain, or linear order, is a poset in which every pair of elements is comparable. Any ﬁnite chain is isomorphic to {0, 1, . . . , n −1} with the usual order. Its M¨ obius function is given by µ(a, b) = ( 1 if b = a, −1 if b = a + 1, 0 otherwise. This follows immediately from the recursive method of computing µ. In this case, any interval [a, b] is isomorphic to the interval [0, b −a], so it would have suﬃced to take a = 0; but the general case is simple enough. Direct product The direct product of posets P1 = (A1, ≤1) and P2 = (A2, ≤2) has set A1 × A2 (Cartesian product), and (a1, a2) ≤(b1, b2) ⇔a1 ≤1 b1 and a2 ≤2 b2. It is easily checked that µ((a1, a2), (b1, b2)) = µ(a1, b1)µ(a2, b2). This extends in a straightforward way to the direct product of any ﬁnite number of posets. Subsets of a set The poset of all subsets of {1, 2, . . . , n} can be represented as the direct product of n copies of the 2-element chain {0, 1}; the subset a is identiﬁed with the n-tuple (a1, . . . , an), where ai = n 1 if i ∈a, 0 if i / ∈a. 56 It follows from the two preceding paragraphs that the M¨ obius function is µ(a, b) = (−1)\|b\a\| if a ⊆b, 0 if a ̸⊆b. In this case, if a ⊆b, then [a, b] is isomorphic to [∅, b \ a], and we see the homogeneity property in action. So the following are equivalent: (a) f(a) = P b≤a g(b); (b) g(a) = P b≤a f(b)(−1)\|a\b\|. With a little rearrangement, this is a generalisation of the Inclusion-Exclusion principle, with cardinality replaced by an arbitrary function (see Exercise 1). The classical M¨ obius function The classical M¨ obius function from num-ber theory is deﬁned on the natural numbers; the partial order is given by a ≤b if a divides b. Although this partial order is inﬁnite, all intervals are ﬁnite, and it has the homogeneity property: if a \| b, then the interval [a, b] is isomorphic to [1, b/a]. This poset is isomorphic to the product of chains, one for each prime power. We have µ(pa, pb) = ( 1 if b = a, −1 if b = a + 1, 0 otherwise. Hence we have the general formula: µ(m, n) = (−1)d if m \| n and n/m is a product of d distinct primes, 0 otherwise. In particular, µ(1, n) is the number-theorists’ M¨ obius function, which they write as µ(n). We have the classical M¨ obius inversion formula, the equiva-lence of the following functions f, g on N: (a) g(n) = P m\|n f(m); (b) f(n) = P m\|n f(n)µ(n/m). 57 Subspaces of a vector space For our ﬁnal example, let A be the set of all subspaces of an n-dimensional vector space over a ﬁeld of order q. If V ≤W, the structure of the interval [V, W] depends only on dim(W) −dim(V ), and so is isomorphic to [{0}, W/V ]. Recall the q-binomial theorem: n Y i=1 (1 + qi−1z) = n X k=0 qk(k−1)/2zk n k q . Putting z = −1, the left-hand side becomes 0; then we have (−1)nqn(n−1)/2 = − n−1 X k=0 (−1)kqk(k−1)/2 n k q . This shows, recursively, that if dim(V ) = n, then µ[{0}, V ] = (−1)nqn(n−1)/2. Exercises 1 Let (Ai : i = 1, . . . , n} be a family of subsets of a set X. For I ⊆ {1, . . . , n}, let • f(I) be the number of points lying in Ai for all i ∈I, and • g(I) be the number of points lying in Ai for all i ∈I and for no i / ∈I. Prove that f(I) = X J⊇I g(J), and deduce from Theorem 9.8 and the form of the M¨ obius function for the power set of a set that g(I) = X J⊃I (−1)\|J\I\|f(J). Putting I = ∅, deduce the classical form of the Inclusion–Exclusion principle. 2 There is a partial order on the set of all partitions of {1, . . . , n}, deﬁned as follows: if a and b are partitions, say that a reﬁnes b if every part of b is a union of parts of a. Can you ﬁnd the M¨ obius function of this partial order? 58 3 Prove the following “approximate version” of Inclusion-Exclusion: Let A1, . . . , An, A′ 1, . . . , A′ n be subsets of a set X. For I ⊆N = {1, . . . , n}, let aI = \ i∈I Ai , a′ I = \ i∈I A′ i . If aI = a′ I for all proper subsets I of N, then \|aN−a′ N\| ≤\|X\|/2n−1. 4 Prove that the exponential generating function for the derangement num-bers dn (Theorem 9.5) is X n≥0 dnxn n! = e−x 1 −x. Give an alternative proof of this formula, by showing that, if Derang is the species of derangements, then Perm = Set · Derang. (A set carrying a permutation is the union of the set of ﬁxed points and a set none of whose points is ﬁxed.) 5 The following problem, based on the children’s game “Screaming Toes”, was suggested to me by Julian Gilbey. n people stand in a circle. Each player looks down at someone else’s feet (i.e., not at their own feet). At a given signal, everyone looks up from the feet to the eyes of the person they were looking at. If two people make eye contact, they scream. What is the probability of at least one pair of people screaming? Prove that the required probability is ⌊n/2⌋ X k=1 (−1)k−1(n)2k (n −1)2k 2k k!, where (n)j = n(n −1) · · · (n −j + 1). 59 10 Cayley’s Theorem The course ends with four entirely diﬀerent proofs of Cayley’s theorem for the number of labelled trees on n vertices, some of which introduce new ideas. There is a direct bijective proof due to Pr¨ ufer; Joyal’s proof using species; a proof using Kirchhoﬀ’s Matrix-Tree Theorem; and a proof using Lagrange inversion. A tree is a connected graph without cycles. It is not hard to show by induction that a tree on n vertices has n−1 edges. There are 16 trees on the vertex set {1, 2, 3, 4}: four of them are “stars” in which one vertex is joined to the other three, and the other twelve are “paths”. Theorem 10.1 The number of labelled trees on the vertex set {1, . . . , n} is nn−2. 10.1 Pr¨ ufer codes We construct a bijection between the set of all trees on the vertex set {1, . . . , n} and the set of all (n −2)-tuples of elements from this set. The tuple associated with a tree is called its Pr¨ ufer code. First we describe the map from trees to Pr¨ ufer codes. Start with the empty code. Repeat the following procedure until only two vertices remain: select the leaf with smallest label; append the label of its unique neighbour to the code; and then remove the leaf and its incident edge. Next, the construction of a tree from a Pr¨ ufer code P. We use an auxiliary list L of vertices added as leaves, which is initially empty. Now, while P is not empty, we join the ﬁrst element of P to the smallest-numbered vertex v which is not in either P or L, and then add v to L and remove the ﬁrst element of P. When P is empty, two vertices have not been put into L; the ﬁnal edge of the tree joins these two vertices. I leave it as a (quite non-trivial) exercise to show that these maps are inverse bijections. This proof gives extra information: the valency of vertex i of the tree is one more than the number of occurrences of i in its Pr¨ ufer code; so the number of trees with prescribed vertex valencies can be calculated. 60 10.2 A proof using species Let Lin and Perm be the species of linear orders and permutations respec-tively. We have seen that these two species have the same counting function for labelled structures on n points (namely n!); so Lin[F] and Perm[F] will also have the same counting function for labelled structures, for any species F. Joyal takes F = RTree, the species of rooted trees (trees with a distin-guished vertex. Now Lin[RTree] consists of a linear order on a set, say {1, 2, . . . , k} with the usual order, with a rooted tree at each point. We can identify the root of the tree at point i to be i itself. What we have constructed is a tree with a distinguished path {1, 2, . . . , k}. Joyal calls such an object a vertebrate, since it has a “backbone” from the “head” 1 to the “tail” k. We get a vertebrate by taking a tree on n vertices and distinguishing two of them to be the head and the tail; in a tree there is a unique path between any two vertices. So the number of vertebrates is n2T(n), where T(n) is the number of trees. Also Perm[RTree] consists of a set of, say, k points carrying a permu-tation, with a rooted tree attached at each point. If we direct every edge of each tree towards the root, we have a picture representing what Joyal calls an endofunction, a function from {1, . . . , n} to itself. Such a function has a set of “periodic points” which return to their initial positions after ﬁnitely many steps; any other point is “transient”, and the transient points feed into periodic points in a treelike fashion. The number of endofunctions is clearly nn. So n2T(n) = nn, giving the result. 10.3 The Matrix-Tree Theorem This theorem, proved by Kirchhoﬀin the nineteenth century for analysis of electrical circuits, depends on the notion of the Laplacian matrix of a graph G = (V, E). Assuming that V = {v1, . . . , vn}, this is the n × n symmetric matrix whose (i, i) entry is the valency of vertex vi, and whose (i, j) entry for i ̸= j is −1 if {vi, vj} is an edge, and 0 otherwise. Note that the row sums of this matrix are all zero, so its determinant is zero. Recall that the (i, j) cofactor of a square matrix A is the determinant of the matrix obtained from A by deleting the ith row and the jth column, multiplied by (−1)i+j. 61 Theorem 10.2 The cofactors of the Laplacian matrix of a graph are all equal to the number of spanning trees of the graph. A tree on the vertex set {1, . . . , n} is simply a spanning tree of the com-plete graph, the graph whose edges are all pairs of vertices. The Lapla-cian matrix of the complete graph is nIn −Jn, where In and Jn denote the n × n identity and all-1 matrices. Deleting the last row and column gives nIn−1 −Jn−1. We ﬁnd the determinant of the last matrix by computing its eigenvalues. Every row and column sum is n −(n −1) = 1, so the all-1 vector is an eigenvector with eigenvalue 1. If v is a vector orthogonal to the all-1 vector, then Jn−1v = 0, so v is an eigenvector with eigenvalue n. Thus nIn−1 −Jn−1 has eigenvalues 1 (multiplicity 1) and n (multiplicity n−2); so its determinant is nn−2, which is thus the number of spanning trees. The proof of the Matrix-Tree Theorem depends on the Cauchy–Binet for-mula, a nineteenth century determinant formula which asserts the following. et A be an m × n matrix, and B an n × m matrix, where m < n. Then det(AB) = X X det(A(X)) det(B(X)), where X ranges over all m-element subsets of {1, . . . , n}. Here A(X) is the m × m matrix whose columns are the columns of A with index in X, and B(X) is the m × m matrix whose rows are the rows of B with index in X. To prove the Matrix-Tree Theorem for the graph G = (V, E) with Lapla-cian matrix L(G), choose an arbitrary orientation of the edges of G, and let M be the signed vertex-edge incidence matrix of G, with (v, e) entry +1 if v is the “head” of the arc e, −1 if v is the “tail” of e, and 0 otherwise. It is straightforward to show that MM ⊤= L(G). Let v be any vertex of G, and let N = Mv be the matrix obtained by deleting the row of M indexed by e. It can be shown that, if X is a set of n −1 edges, then det(N(X)) = n ±1 if X is the edge set of a spanning tree, 0 otherwise. By the Cauchy–Binet formula, det(NN ⊤) is equal to the number of spanning trees. But NN ⊤is the principal cofactor of L(G) obtained by deleting the row and column indexed by v. The fact that all cofactors are equal is not really necessary for us, and can be proved by elementary linear algebra. 62 10.4 Lagrange inversion Our ﬁnal approach involves another general technique, Lagrange inversion. Let G be the set of all formal power series (over the commutative ring R with identity) which have the form x+· · ·, that is, constant term is zero and coeﬃcient of x is 1. Any of these series can be substituted into any other. We make a simple observation: Proposition 10.3 The set G, with the operation of substitution, is a group. This group is sometimes called the Nottingham group, for reasons that are a little obscure. Proof Closure and the associative law are straightforward, and the formal power series x is the identity. Let f(x) = x+a2x2+a3x3+· · · be any element of G. We seek an inverse g(x) = x + b2x2 + b3x3 + · · · such that f(g(x)) = x. The coeﬃcient of xn in f(g(x)) = g(x) + a2g(x)2 + a3g(x)3 + · · · is bn + stuﬀ, where stuﬀinvolves the as and bi for i < n. Equating it to zero gives bn in terms of as and bi for i < n; so the bs can be found recursively. In a similar way, we ﬁnd a unique element h(x) ∈G for which h(f(x)) = x. Then g(x) = h(f(g(x)) = h(x), and the inverse is unique. □ The proof implicitly shows us how to ﬁnd the inverse; Lagrange inversion gives a more direct approach. Theorem 10.4 The coeﬃcient of xn in g(x) is dn−1 dxn−1 x f(x) n x=0 . n!. I will not give the proof here; it involves working with Laurent series and extending the notion of poles and the calculus of residues to formal power series. 63 Now let RTree be the species of rooted trees, as before. We clearly have the equation RTree = E · Set[RTree], where E is the species of 1-element sets; this is because a rooted tree is a (ppossibly empty) set of rooted trees all joined to a new root. Thus the exponential generating function T ∗(x) for rooted trees satisﬁes T ∗(x) = x exp(T ∗(X)). So the function T ∗(x) is the inverse (in the group G) of the function x/ exp(x). From Lagrange inversion, we ﬁnd that the coeﬃcient of xn/n! in T ∗(x) is dn−1 dxn−1 exp(nx) x=0 = nn−1. Since the number of rooted trees is n times the number of trees, we conclude that there are nn−2 trees on n vertices. 10.5 Stirling’s formula The most famous asymptotic formula in enumerative combinatorics is Stir-ling’s formula, an estimate for the factorial function. We write f ∼g to mean that f(n)/g(n) →1 as n →∞. Typically this is used with f a combinatorial counting function and g an analytic approximation to f. Stirling’s formula is an example. Theorem 10.5 n! ∼ √ 2πn n e n . It follows that, if T(n) is the number of labelled trees on n vertices, then lim n→∞ T(n) n! 1/n = e, so the exponential generating function for T(n) has radius of convergence 1/e. Using more complicated methods, Otter showed that the number of unla-belled trees on n vertices is asymptotically An−5/2cn, where A = 0.5349485 . . . and c = 2.955765 . . .. 64 Exercises 1 Calculate the chromatic polynomial of (a) the path with n vertices, (b) the cycle with n vertices. 2 A forest is a graph whose connected components are trees. Show that there is a bijection between labelled forests of rooted trees on n vertices, and labelled rooted trees on n + 1 vertices with root n + 1. Use Stirling’s formula to show that, if a forest of rooted trees on n vertices is chosen at random, then the probability that it is connected tends to the limit 1/e as n →∞. 3 Count the labelled trees in which the vertex i has valency ai for 1 ≤i ≤n, where a1, . . . , an are positive integers with sum 2n −2. 65
4559	https://artofproblemsolving.com/wiki/index.php/2024_AMC_10B_Problems/Problem_10?srsltid=AfmBOopDvF-ZQxwAfnt8_47cc8H3QW0M0OjE8Hx-V6hPoh7y4Ynucj0f	Art of Problem Solving 2024 AMC 10B Problems/Problem 10 - 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 2024 AMC 10B Problems/Problem 10 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2024 AMC 10B Problems/Problem 10 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 (Techniques) 5 Solution 4 6 Solution 5 (wlog) 7 Solution 6 (barycentrics) 8🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️) 9 Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀) 10 Video Solution 2 by SpreadTheMathLove 11 See also Problem Quadrilateral is a parallelogram, and is the midpoint of the side . Let be the intersection of lines and . What is the ratio of the area of quadrilateral to the area of ? Solution 1 Let have length and let the altitude of the parallelogram perpendicular to have length . The area of the parallelogram is and the area of equals . Thus, the area of quadrilateral is . We have from that . Also, , so the length of the altitude of from is twice that of . This means that the altitude of is , so the area of is . Then, the area of quadrilateral equals the area of minus that of , which is . Finally, the ratio of the area of to the area of triangle is , so the answer is . Solution 2 Let . Since with a scale factor of , . The scale factor of also means that , therefore since and have the same height, . Since is a parallelogram, vladimir.shelomovskii@gmail.com, vvsss Solution 3 (Techniques) We assert that is a square of side length . Notice that with a scale factor of . Since the area of is the area of is , so the area of is . Thus the area of is , and we conclude that the answer is ~Tacos_are_yummy_1 Solution 4 Let be a square with side length , to assist with calculations. We can put this on the coordinate plane with the points , , , and . We have . Therefore, the line has slope and y-intercept . The equation of the line is then . The equation of line is . The intersection is when the lines are equal to each other, so we solve the equation. , so . Therefore, plugging it into the equation, we get . Using the shoelace theorem, we get the area of to be and the area of to be , so our ratio is ~idk12345678 Solution 5 (wlog) Let be a square with side length . We see that by a Scale factor of . Let the altitude of and altitude of be and , respectively. We know that is equal to , as the height of the square is . Solving this equation, we get that This means we can also calculate the area of . Adding the area we of and we get We can then subtract this from the total area of the square: , this gives us for the area of quadrilateral Then we can compute the ratio which is equal to ~yuvag (why does the always look so bugged.) Solution 6 (barycentrics) . Since is the midpoint of , . The equation of is: The equation of is: We also know that . To find the intersection, we can solve the system of equations. Solving, we get . Therefore, . Using barycentric area formula, 🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️) Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀) ~ Pi Academy Video Solution 2 by SpreadTheMathLove See also 2024 AMC 10B (Problems • Answer Key • Resources) Preceded by Problem 9Followed by Problem 11 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 10 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " 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.
4560	https://pubmed.ncbi.nlm.nih.gov/17437863/	The root and root canal morphology of the human mandibular first premolar: a literature review - 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. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. To: Subject: Body: Format: [x] MeSH and other data Send email Cancel Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Add to My Bibliography My Bibliography Unable to load your delegates due to an error Please try again Add Cancel Your saved search Name of saved search: Search terms: Test search terms Would you like email updates of new search results? Saved Search Alert Radio Buttons Yes No Email: (change) Frequency: Which day? Which day? Report format: Send at most: [x] Send even when there aren't any new results Optional text in email: Save Cancel Create a file for external citation management software Create file Cancel Your RSS Feed Name of RSS Feed: Number of items displayed: Create RSS Cancel RSS Link Copy Actions Cite Collections Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Permalink Permalink Copy Display options Display options Format Page navigation Title & authors Abstract Similar articles Cited by Publication types MeSH terms LinkOut - more resources Review J Endod Actions Search in PubMed Search in NLM Catalog Add to Search . 2007 May;33(5):509-16. doi: 10.1016/j.joen.2006.12.004. Epub 2007 Feb 27. The root and root canal morphology of the human mandibular first premolar: a literature review Blaine M Cleghorn1,William H Christie,Cecilia C S Dong Affiliations Expand Affiliation 1 Department of Dental Clinical Sciences, Dalhousie University, Halifax, Nova Scotia, Canada. blaine.cleghorn@dal.ca PMID: 17437863 DOI: 10.1016/j.joen.2006.12.004 Item in Clipboard Review The root and root canal morphology of the human mandibular first premolar: a literature review Blaine M Cleghorn et al. J Endod.2007 May. Show details Display options Display options Format J Endod Actions Search in PubMed Search in NLM Catalog Add to Search . 2007 May;33(5):509-16. doi: 10.1016/j.joen.2006.12.004. Epub 2007 Feb 27. Authors Blaine M Cleghorn1,William H Christie,Cecilia C S Dong Affiliation 1 Department of Dental Clinical Sciences, Dalhousie University, Halifax, Nova Scotia, Canada. blaine.cleghorn@dal.ca PMID: 17437863 DOI: 10.1016/j.joen.2006.12.004 Item in Clipboard Cite Display options Display options Format Abstract The purpose of this study was to undertake a comprehensive literature review of the root and root canal morphology of the mandibular first premolar. Published studies citing the anatomy and morphology of mandibular premolars report data for over 6,700 teeth. These studies were divided into anatomical studies reporting number of roots, number of canals, and apical morphology. Variations because of sex and ethnic background have also been reported, along with case reports of anomalies. Approximately 98% of the teeth in these studies were single-rooted. The incidence of two roots was 1.8%. Three roots when reported were found in 0.2% of the teeth studied. Four roots were rare and were found in less than 0.1% of the teeth studied. Studies of the internal canal morphology revealed that a single canal was present in 75.8% of the teeth. Two or more canals were found in 24.2% of the teeth studied. A single apical foramen was found in 78.9% of the teeth, whereas 21.1% had two or more apical foramina. The role of genetics and racial variation may result in differences of incidence of root number and canal number in human populations. The dental literature is not unique in studying ethnicity and sex variations. Higher incidences of teeth with additional canals and roots have been reported in Chinese, Australian, and sub-Sahara African populations. Physical anthropology studies seem to show the lowest incidence in Western Eurasian, Japanese and American Arctic populations. The root and root canal morphology of this tooth can be complex and requires careful evaluation prior to root canal therapy. PubMed Disclaimer Similar articles Anomalous mandibular premolars: a mandibular first premolar with three roots and a mandibular second premolar with a C-shaped canal system.Cleghorn BM, Christie WH, Dong CC.Cleghorn BM, et al.Int Endod J. 2008 Nov;41(11):1005-14. doi: 10.1111/j.1365-2591.2008.01451.x.Int Endod J. 2008.PMID: 19133090 The root and root canal morphology of the human mandibular second premolar: a literature review.Cleghorn BM, Christie WH, Dong CC.Cleghorn BM, et al.J Endod. 2007 Sep;33(9):1031-7. doi: 10.1016/j.joen.2007.03.020. Epub 2007 Jun 5.J Endod. 2007.PMID: 17931927 Review. Root form and canal morphology of mandibular premolars in a Jordanian population.Awawdeh LA, Al-Qudah AA.Awawdeh LA, et al.Int Endod J. 2008 Mar;41(3):240-8. doi: 10.1111/j.1365-2591.2007.01348.x. Epub 2007 Dec 12.Int Endod J. 2008.PMID: 18081806 The incidence of mandibular premolars with more than one root canal in a Turkish population.Kartal N, Yanikoğlu F.Kartal N, et al.J Marmara Univ Dent Fac. 1992 Sep;1(3):203-10.J Marmara Univ Dent Fac. 1992.PMID: 1308778 Root and root canal morphology of the human permanent maxillary first molar: a literature review.Cleghorn BM, Christie WH, Dong CC.Cleghorn BM, et al.J Endod. 2006 Sep;32(9):813-21. doi: 10.1016/j.joen.2006.04.014. Epub 2006 Jun 30.J Endod. 2006.PMID: 16934622 Review. See all similar articles Cited by Mandibular first premolar with three roots: a case report.Kakkar P, Singh A.Kakkar P, et al.Iran Endod J. 2012 Fall;7(4):207-10. Epub 2012 Oct 13.Iran Endod J. 2012.PMID: 23130081 Free PMC article. Root Canal Configuration of Burmese (Myanmar) Maxillary First Molar: A Micro-Computed Tomography Study.Kyaw Moe MM, Jo HJ, Ha JH, Kim SK.Kyaw Moe MM, et al.Int J Dent. 2021 Nov 30;2021:3433343. doi: 10.1155/2021/3433343. eCollection 2021.Int J Dent. 2021.PMID: 34887924 Free PMC article. Atypical anatomy of maxillary second premolar with three roots and four canals.Izaz S, Mandava P, Bolla N, Dasari B.Izaz S, et al.J Conserv Dent. 2017 Sep-Oct;20(5):370-373. doi: 10.4103/JCD.JCD_279_16.J Conserv Dent. 2017.PMID: 29386789 Free PMC article. Endodontic Management of a Two-rooted Mandibular First Premolar with Five Root Canals with Cone-beam Computed Tomography: A Case Report.Nouroloyouni A, Lotfi M, Milani AS, Nouroloyouni S.Nouroloyouni A, et al.J Dent (Shiraz). 2021 Sep;22(3):225-228. doi: 10.30476/DENTJODS.2020.83376.1049.J Dent (Shiraz). 2021.PMID: 34514072 Free PMC article. Gender difference and root canal morphology in mandibular premolars: A cone-beam computed tomography study in an Iranian population.Kazemipoor M, Hajighasemi A, Hakimian R.Kazemipoor M, et al.Contemp Clin Dent. 2015 Jul-Sep;6(3):401-4. doi: 10.4103/0976-237X.161902.Contemp Clin Dent. 2015.PMID: 26321843 Free PMC article. See all "Cited by" articles Publication types Review Actions Search in PubMed Search in MeSH Add to Search MeSH terms Age Factors Actions Search in PubMed Search in MeSH Add to Search Bicuspid / abnormalities Actions Search in PubMed Search in MeSH Add to Search Dental Pulp Cavity / abnormalities Actions Search in PubMed Search in MeSH Add to Search Ethnicity Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Mandible Actions Search in PubMed Search in MeSH Add to Search Sex Factors Actions Search in PubMed Search in MeSH Add to Search Tooth Abnormalities / ethnology Actions Search in PubMed Search in MeSH Add to Search Tooth Root / abnormalities Actions Search in PubMed Search in MeSH Add to Search LinkOut - more resources Full Text Sources ClinicalKey [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
4561	https://brainly.com/question/48055236	[FREE] How many hydrogen atoms are present in 1.00 mol of hydrogen gas, H₂? How many H₂ molecules? What is the - brainly.com 5 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 +44,5k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +10,6k Ace exams faster, with practice that adapts to you Practice Worksheets +5,9k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified How many hydrogen atoms are present in 1.00 mol of hydrogen gas, H₂? How many H₂ molecules? What is the mass of the same? a) Hydrogen atoms: 6.022×1 0 23, H₂ molecules: 6.022×1 0 23, Mass: 2.02 g b) Hydrogen atoms: 3.011×1 0 23, H₂ molecules: 6.022×1 0 23, Mass: 4.04 g c) Hydrogen atoms: 6.022×1 0 23, H₂ molecules: 3.011×1 0 23, Mass: 4.04 g d) Hydrogen atoms: 3.011×1 0 23, H₂ molecules: 3.011×1 0 23, Mass: 2.02 g 1 See answer Explain with Learning Companion NEW Asked by Paulalex7512 • 02/19/2024 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 10360349 people 10M 0.0 0 Upload your school material for a more relevant answer 1.00 mol of hydrogen gas contains 1.2044 × 1024 hydrogen atoms and 6.022 × 1023 H2 molecules, with a mass of 2.0158 g. Explanation When considering 1.00 mol of hydrogen gas (H2), we need to understand Avogadro's number, which is 6.022 × 1023, the number of units (in this case, molecules) in one mole of any substance. Since hydrogen gas exists as a diatomic molecule, H2, there are 6.022 × 1023 molecules of H2 in 1.00 mol of hydrogen gas. Each of these molecules contains two hydrogen atoms, so there are twice as many atoms as there are molecules, resulting in 2 × 6.022 × 1023 = 1.2044 × 1024 hydrogen atoms within 1.00 mol of H2 gas. Furthermore, the mass of 1 mole of H2 molecules can be determined by its molar mass, which is 2.0158 g/mol. Therefore, the mass of 1.00 mol of H2 gas is 2.0158 g. Thus, the correct answer is Hydrogen atoms: 1.2044 × 1024, H2 molecules: 6.022 × 1023, Mass: 2.0158 g. Answered by akibkhanzai12 •33.1K answers•10.4M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 10360349 people 10M 0.0 0 Introductory Chemistry Online! - Paul R. Young Chemistry - Paul Flowers Basics of General, Organic, and Biological Chemistry - David W. Ball, John W. Hill, Rhonda J. Scott Upload your school material for a more relevant answer 1.00 mol of hydrogen gas contains 1.2044×1 0 24 hydrogen atoms, 6.022×1 0 23 H₂ molecules, and has a mass of 2.0158 g. However, the options provided do not exactly match these values. Explanation To determine the number of hydrogen atoms, hydrogen molecules, and the mass of 1.00 mol of hydrogen gas (H₂), we follow these steps: Number of H₂ Molecules: According to Avogadro's number, 1 mole of any substance contains approximately 6.022×1 0 23 units. So, in 1.00 mol of H₂ gas, there are 6.022×1 0 23 molecules of H₂. Number of Hydrogen Atoms: Each H₂ molecule contains 2 hydrogen atoms. Therefore, the number of hydrogen atoms is calculated as: Number of H atoms=2×Number of H₂ molecules=2×(6.022×1 0 23)=1.2044×1 0 24 Thus, there are 1.2044×1 0 24 hydrogen atoms in 1.00 mol of H₂. Mass of H₂: The molar mass of H₂ is approximately 2.0158 g/mol. Thus, the mass of 1.00 mol of H₂ is Mass of H₂=2.0158 g Given the options provided, the correct choice would be none of the given options as they do not exactly match the calculations, but focusing on the information: Hydrogen atoms: 1.2044×1 0 24 H₂ molecules: 6.022×1 0 23 Mass: 2.0158 g. Examples & Evidence For example, if we had 2 moles of H₂, we would have 2×(6.022×1 0 23)=1.2044×1 0 24 H₂ molecules in total, which would mean 2×2=4 moles of hydrogen atoms, leading to the understanding of these quantities in larger amounts. The calculations are based on basic principles in chemistry, particularly the concept of Avogadro's number and the molecular composition of hydrogen gas. Thanks 0 0.0 (0 votes) Advertisement Paulalex7512 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers Community Answer Calculate the mass and number of molecules and the number of each constituent atom in 3.4 grams of ammonia gas (NH3). A) Mass: 3.4 g, Molecules: 6.022 x 10^23, Hydrogen atoms: 3.011 x 10^24, Nitrogen atoms: 1.505 x 10^24 B) Mass: 3.4 g, Molecules: 1.806 x 10^24, Hydrogen atoms: 9.032 x 10^24, Nitrogen atoms: 9.032 x 10^24 C) Mass: 6.8 g, Molecules: 3.011 x 10^24, Hydrogen atoms: 9.032 x 10^24, Nitrogen atoms: 9.032 x 10^24 D) Mass: 6.8 g, Molecules: 6.022 x 10^23, Hydrogen atoms: 3.011 x 10^24, Nitrogen atoms: 1.505 x 10^24 Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. Community Answer 8 For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval mc015-1. Jpg? mc015-2. Jpg mc015-3. Jpg mc015-4. Jpg mc015-5. Jpg. New questions in Chemistry What is a double bond? A. A covalent bond between 2 atoms where each atom contributes 2 electrons. B. A covalent bond between 2 atoms where each atom contributes 4 electrons. C. An ionic bond between 2 atoms where 2 electrons have been transferred. D. A covalent bond between 2 atoms where each atom contributes 1 electron. Why do noble gases rarely form bonds with other atoms? A. The noble gases are not reactive, so they don't need full valence octets. B. Gases do not form bonds with atoms that are not also gases. C. The noble gases are already stable, with full octets of valence electrons. D. The noble gases form bonds only with themselves because they have octets. A balloon that had a volume of 3.50 L at 25.0∘C is placed in a hot room at 40.0∘C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room? Use T 1V 1=T 2V 2. A. 2.19 L B. 3.33 L C. 3.68 L D. 5.60 L How are chemical bonds formed? A. Through obtaining 10 valence electrons B. Through gaining a core electron C. Through removing parts of the nucleus D. Through transferring or sharing electrons What is an ionic bond? A. A bond resulting from the transfer of 1 or more electrons B. A bond formed by sharing electrons C. A bond between 2 atoms of similar electronegativities D. A bond formed from 2 atoms of the same element Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
4562	https://www.geeksforgeeks.org/dsa/count-of-n-digit-numbers-with-all-distinct-digits/	Tutorials Courses Count of N-digit numbers with all distinct digits Given an integer N, the task is to find the count of N-digit numbers with all distinct digits. Examples: Input: N = 1 Output: 9 1, 2, 3, 4, 5, 6, 7, 8 and 9 are the 1-digit numbers with all distinct digits. Input: N = 3 Output: 648 Naive Approach: If N > 10 i.e. there will be atleast one digit which will be repeating hence for such cases the answer will be 0 else for the values of N = 1, 2, 3, ..., 9, a series will be formed as 9, 81, 648, 4536, 27216, 136080, 544320, ... whose Nth term will be 9 9! / (10 - N)!. Below is the implementation of the above approach: ```` // C++ implementation of the approach include using namespace std; // Function to return the factorial of n int factorial(int n) { if (n == 0) return 1; return n factorial(n - 1); } // Function to return the count // of n-digit numbers with // all distinct digits int countNum(int n) { if (n > 10) return 0; return (9 factorial(9) / factorial(10 - n)); } // Driver code int main() { int n = 3; cout << countNum(n); return 0; } // Java implementation of the approach class GFG { // Function to return the factorial of n static int factorial(int n) { if (n == 0) return 1; return n factorial(n - 1); } // Function to return the count // of n-digit numbers with // all distinct digits static int countNum(int n) { if (n > 10) return 0; return (9 factorial(9) / factorial(10 - n)); } // Driver code public static void main(String []args) { int n = 3; System.out.println(countNum(n)); } } // This code is contributed by Srathore Python3 implementation of the approach Function to return the factorial of n def factorial(n) : if (n == 0) : return 1; return n factorial(n - 1); Function to return the count of n-digit numbers with all distinct digits def countNum(n) : if (n > 10) : return 0; return (9 factorial(9) // factorial(10 - n)); Driver code if name == "main" : n = 3; print(countNum(n)); This code is contributed by AnkitRai01 // C# implementation of the approach using System; class GFG { // Function to return the factorial of n static int factorial(int n) { if (n == 0) return 1; return n factorial(n - 1); } // Function to return the count // of n-digit numbers with // all distinct digits static int countNum(int n) { if (n > 10) return 0; return (9 factorial(9) / factorial(10 - n)); } // Driver code public static void Main(String []args) { int n = 3; Console.WriteLine(countNum(n)); } } // This code is contributed by Princi Singh // Javascript implementation of the approach // Function to return the factorial of n function factorial(n) { if (n == 0) return 1; return n factorial(n - 1); } // Function to return the count // of n-digit numbers with // all distinct digits function countNum(n) { if (n > 10) return 0; return (9 factorial(9) / factorial(10 - n)); } // Driver code var n = 3; document.write(countNum(n)); // This code is contributed by rutvik_56. ```` Time Complexity: O(n) Auxiliary Space: O(n) Efficient Approach:We have to fill (n) places with different digit. Like we n=2 then (_ _) places to fill. first place we fill (1 to 9) any number. let we fill 9 in first place then in second place we have choice (0 to 8). So for first place we 9 choices because we can not fill 0 at first place and after that for 2nd place we 9 choice and for 3rd place we 8 choice then so on... let we take 4 digit no so ( 9choice 9 choice 8 choice 7choice) . Choices-> first place -9 second place-9 third place -8 fourth place -7 and so on....... ```` // C++ implementation of the approach include using namespace std; // Function to return the count // of n-digit numbers with // all distinct digits long long countNum(int n) { if (n > 10) return 0; long long count = 1; // Store the count long long j = 9; // take choice / take loop 1 to n and multiply with choice/ for (int i = 1; i <= n; i++) { if (i == 1) { count = count j; continue; } else { count = count j; j--; } } return count; } // Driver code int main() { int n = 3; cout << countNum(n); return 0; } // Java implementation of the approach class GFG { // Function to return the count // of n-digit numbers with // all distinct digits static long countNum(int n) { if (n > 10) return 0; long count = 1; // Store the count long j = 9; // take choice / take loop 1 to n and multiply with choice/ for (int i = 1; i <= n; i++) { if (i == 1) { count = count j; continue; } else { count = count j; j--; } } return count; } // Driver code public static void main(String[] args) { int n = 3; System.out.println(countNum(n)); } } // This code is contributed by Srathore Python implementation of the approach Function to return the count of n-digit numbers with all distinct digits def countNum(n): if n > 10: return 0 count = 1 # Store the count j = 9 # take choice # take loop 1 to n and multiply with choice for i in range(1, n+1): if i == 1: count = count j continue else: count = count j j -= 1 return count Driver code n = 3 print(countNum(n)) This code is contributed by rutikbhosale using System; class GFG { // Function to return the count // of n-digit numbers with // all distinct digits static long countNum(int n) { if (n > 10) return 0; long count = 1; // Store the count long j = 9; // take choice / take loop 1 to n and multiply with choice/ for (int i = 1; i <= n; i++) { if (i == 1) { count = count j; continue; } else { count = count j; j--; } } return count; } // Driver code static void Main(string[] args) { int n = 3; Console.WriteLine(countNum(n)); } } // Javascript implementation of the approach // Function to return the count // of n-digit numbers with // all distinct digits function countNum(n) { if (n > 10) return 0; let count = 1; // Store the count let j = 9; // take choice / take loop 1 to n and multiply with choice/ for (let i = 1; i <= n; i++) { if (i === 1) { count = j; continue; } else { count = j; j--; } } return count; } // Driver code const n = 3; console.log(countNum(n)); ```` Time Complexity: O(n) Auxiliary Space: O(1) S Explore DSA Fundamentals Logic Building Problems Analysis of Algorithms Data Structures Array Data Structure String in Data Structure Hashing in Data Structure Linked List Data Structure Stack Data Structure Queue Data Structure Tree Data Structure Graph Data Structure Trie Data Structure Algorithms Searching Algorithms Sorting Algorithms Introduction to Recursion Greedy Algorithms Graph Algorithms Dynamic Programming or DP Bitwise Algorithms Advanced Segment Tree Binary Indexed Tree or Fenwick Tree Square Root (Sqrt) Decomposition Algorithm Binary Lifting Geometry Interview Preparation Interview Corner GfG160 Practice Problem GeeksforGeeks Practice - Leading Online Coding Platform Problem of The Day - Develop the Habit of Coding Thank You! What kind of Experience do you want to share?
4563	https://cp-algorithms.com/string/lyndon_factorization.html	Skip to content Last update: June 8, 2022 Translated From: e-maxx.ru Lyndon factorization¶ Lyndon factorization¶ First let us define the notion of the Lyndon factorization. A string is called simple (or a Lyndon word), if it is strictly smaller than any of its own nontrivial suffixes. Examples of simple strings are: $a$, $b$, $ab$, $aab$, $abb$, $ababb$, $abcd$. It can be shown that a string is simple, if and only if it is strictly smaller than all its nontrivial cyclic shifts. Next, let there be a given string $s$. The Lyndon factorization of the string $s$ is a factorization $s = w_1 w_2 \dots w_k$, where all strings $w_i$ are simple, and they are in non-increasing order $w_1 \ge w_2 \ge \dots \ge w_k$. It can be shown, that for any string such a factorization exists and that it is unique. Duval algorithm¶ The Duval algorithm constructs the Lyndon factorization in $O(n)$ time using $O(1)$ additional memory. First let us introduce another notion: a string $t$ is called pre-simple, if it has the form $t = w w \dots w \overline{w}$, where $w$ is a simple string and $\overline{w}$ is a prefix of $w$ (possibly empty). A simple string is also pre-simple. The Duval algorithm is greedy. At any point during its execution, the string $s$ will actually be divided into three strings $s = s_1 s_2 s_3$, where the Lyndon factorization for $s_1$ is already found and finalized, the string $s_2$ is pre-simple (and we know the length of the simple string in it), and $s_3$ is completely untouched. In each iteration the Duval algorithm takes the first character of the string $s_3$ and tries to append it to the string $s_2$. It $s_2$ is no longer pre-simple, then the Lyndon factorization for some part of $s_2$ becomes known, and this part goes to $s_1$. Let's describe the algorithm in more detail. The pointer $i$ will always point to the beginning of the string $s_2$. The outer loop will be executed as long as $i < n$. Inside the loop we use two additional pointers, $j$ which points to the beginning of $s_3$, and $k$ which points to the current character that we are currently comparing to. We want to add the character $s[j]$ to the string $s_2$, which requires a comparison with the character $s[k]$. There can be three different cases: $s[j] = s[k]$: if this is the case, then adding the symbol $s[j]$ to $s_2$ doesn't violate its pre-simplicity. So we simply increment the pointers $j$ and $k$. $s[j] > s[k]$: here, the string $s_2 + s[j]$ becomes simple. We can increment $j$ and reset $k$ back to the beginning of $s_2$, so that the next character can be compared with the beginning of the simple word. $s[j] < s[k]$: the string $s_2 + s[j]$ is no longer pre-simple. Therefore we will split the pre-simple string $s_2$ into its simple strings and the remainder, possibly empty. The simple string will have the length $j - k$. In the next iteration we start again with the remaining $s_2$. Implementation¶ Here we present the implementation of the Duval algorithm, which will return the desired Lyndon factorization of a given string $s$. vector< string> duval(string const & s) { int n = s. size(); int i = 0; vector< string> factorization; while (i < n) { int j = i + 1, k = i; while (j < n && s[k] <= s[j]) { if (s[k] < s[j]) k = i; else k ++; j ++; } while (i <= k) { factorization. push_back(s. substr(i, j - k)); i += j - k; } } return factorization;} Complexity¶ Let us estimate the running time of this algorithm. The outer while loop does not exceed $n$ iterations, since at the end of each iteration $i$ increases. Also the second inner while loop runs in $O(n)$, since is only outputs the final factorization. So we are only interested in the first inner while loop. How many iterations does it perform in the worst case? It's easy to see that the simple words that we identify in each iteration of the outer loop are longer than the remainder that we additionally compared. Therefore also the sum of the remainders will be smaller than $n$, which means that we only perform at most $O(n)$ iterations of the first inner while loop. In fact the total number of character comparisons will not exceed $4n - 3$. Finding the smallest cyclic shift¶ Let there be a string $s$. We construct the Lyndon factorization for the string $s + s$ (in $O(n)$ time). We will look for a simple string in the factorization, which starts at a position less than $n$ (i.e. it starts in the first instance of $s$), and ends in a position greater than or equal to $n$ (i.e. in the second instance) of $s$). It is stated, that the position of the start of this simple string will be the beginning of the desired smallest cyclic shift. This can be easily verified using the definition of the Lyndon decomposition. The beginning of the simple block can be found easily - just remember the pointer $i$ at the beginning of each iteration of the outer loop, which indicated the beginning of the current pre-simple string. So we get the following implementation: string min_cyclic_string(string s) { s += s; int n = s. size(); int i = 0, ans = 0; while (i < n / 2) { ans = i; int j = i + 1, k = i; while (j < n && s[k] <= s[j]) { if (s[k] < s[j]) k = i; else k ++; j ++; } while (i <= k) i += j - k; } return s. substr(ans, n / 2);} Problems¶ UVA #719 - Glass Beads Contributors: jakobkogler (94.49%) adamant-pwn (4.72%) wikku (0.79%)
4564	https://www.bu.edu/eci/files/2020/08/Chapter-10-Appendix.pdf	OREDO'HYHORSPHQW3ROLF\&HQWHU %RVWRQ8QLYHUVLW\ %D\6WDWH5RDG %RVWRQ0$ EXHGXJGS Chapter 10: Fiscal Policy Appendices Appendices to Chapter 10 of Essentials of Economics in Context Essentials of Economics in Context – Chapter 10 Appendices [AUTHOR NAME] 1 APPENDIX A: AN ALGEBRAIC APPROACH TO THE MULTIPLIER, WITH A LUMP-SUM TAX A lump-sum tax is a tax that is simply levied on an economy as a flat amount. This amount does not change with the level of income. Suppose that a lump-sum tax is levied in an economy with a government (but no foreign sector). Consumption in this economy is: C = C + mpc Yd (the consumption function from Chapter 9, but using after-tax or disposable income in the formula). Since disposable income is: Yd = Y – T + TR we can write the consumption function as: C = C + mpc (Y – T + TR) Thus aggregate expenditure in this economy can be expressed as: AE = C + I + G = C + mpc (Y –T + TR) + I + G = (C – mpcT + mpc TR + I + G) + mpc Y By substituting this into the equation for the equilibrium condition, Y = AE, we can derive an expression for equilibrium income in terms of all the other variables in the model: Y= (C – mpcT + mpc TR + I + G) + mpc Y Y - mpc Y = (C – mpcT + mpc TR + I + G) (1-mpc) Y = (C – mpcT + mpc TR + I + G) 𝑌= 1 (1 −𝑚𝑝𝑐) (C – 𝑚𝑝𝑐T + 𝑚𝑝𝑐 𝑇𝑅 + 𝐼 + 𝐺) If autonomous consumption, investment, or government spending change, these each increase equilibrium income by mult = 1/(1 – mpc) times the amount of the original change. If the level of lump-sum taxes or transfers changes, these change Y by either negative or positive (mult)(mpc) times the amount of the original change. To see this explicitly, consider the changes that would come about in Y if there were a change in the level of the lump sum tax from T0 to a new level, T1, if everything else stays the same. We can solve for the change in Y by subtracting the old equation from the new one: Essentials of Economics in Context – Chapter 10 Appendices [AUTHOR NAME] 2 𝑌 1 = 1 (1 −𝑚𝑝𝑐) (C + 𝐼+ 𝐺 – 𝑚𝑝𝑐 𝑇 ̅1 + 𝑚𝑝𝑐 𝑇𝑅) 𝑌 0 = 1 (1 −𝑚𝑝𝑐) (C + 𝐼+ 𝐺 – 𝑚𝑝𝑐 𝑇 ̅0 + 𝑚𝑝𝑐 𝑇𝑅) 𝑌 1 − 𝑌 0 = 1 (1 −𝑚𝑝𝑐) (C −𝐶̅ + 𝐼−𝐼+ 𝐺−𝐺 – 𝑚𝑝𝑐 𝑇 ̅1 + 𝑚𝑝𝑐 𝑇 ̅0 + 𝑚𝑝𝑐 𝑇𝑅−𝑚𝑝𝑐 𝑇𝑅) But C , I, G, TR (and the mpc) are all unchanged, so most of the subtractions in parentheses come out to be 0. We are left with (taking the negative sign out in front): 𝑌 1 − 𝑌 0 = − 1 (1 −𝑚𝑝𝑐) 𝑚𝑝𝑐 (𝑇 ̅1 − 𝑇 ̅0) or ∆Y = – (mult)(mpc)∆T As explained in the text, the multiplier for a change in taxes is smaller than the multiplier for a change in government spending, because taxation affects aggregate expenditure only to the extent that people spend their tax cut or pay their increased taxes by reducing consumption. Because people may also save part of their tax cut or pay part of their increased taxes out of their savings, not all the changes in taxes will carry over to changes in aggregate expenditure. The tax multiplier has a negative sign, since a decrease in taxes increases consumption, aggregate expenditure, and income, while a tax increase decreases them. APPENDIX B: AN ALGEBRAIC APPROACH TO THE MULTIPLIER, WITH A PROPORTIONAL TAX With a proportional tax, total tax revenues are not set at a fixed level of revenues, as was the case with a lump sum tax but, rather, are a fixed proportion of total income. That is, T = tY where t is the tax rate. The equation for AE becomes AE = C + mpc (Y – tY + TR) I + G = C + mpc TR + I + G) + mpc (Y – tY) = (C + mpc TR + I + G) + mpc (1 – t) Y Substituting in the equilibrium condition, Y = AE, and solving yields: Y = (C + mpc TR + I + G) + mpc (1 – t) Y Y – mpc (1 – t) Y = C + mpc TR + I + G Essentials of Economics in Context – Chapter 10 Appendices [AUTHOR NAME] 3 (1 – mpc (1 – t)) Y = C + mpc TR + I + G 𝑌= [ 1 1 −𝑚𝑝𝑐(1 −𝑡)] (𝐶̅ + 𝑚𝑝𝑐 𝑇𝑅+ 𝐼+ 𝐺) The term in brackets is a new multiplier, for the case of a proportional tax. It is smaller than the basic (no proportional taxation) multiplier, reflecting the fact that now any change in spending has smaller feedback effects through consumption. (Some of the change in income “leaks” into taxes.) For example, if mpc = 0.8 and t = 0.2, then the new multiplier is 1/(1 – 0.64), or approximately 2.8, compared to the simple model multiplier 1/(1 – 0.8), which is 5. Changes in autonomous consumption or investment (or government spending or transfers) now have less of an effect on equilibrium income—the “automatic stabilizer” effect mentioned in the text. Is there a multiplier for the tax rate, t? That is, could we derive from the model a formula for how much equilibrium income should change with a change in the rate (rather than level) of taxes? For example, if the tax rate were to decrease from 0.2 to 0.15, could we calculate the size of the change from Y0 to Y1 illustrated in Figure 10.7? Yes, but deriving a general formula for a multiplier relating the change in Y to the change in the tax rate requires the use of calculus, which we will not pursue here. (If you are familiar with calculus, you can use the last formula above to calculate the change in Y resulting from a change in t).
4565	https://www.cliffsnotes.com/study-notes/19225119	Lit NotesStudy GuidesDocumentsQ&A Chat PDF Log In Understanding Muscle Physiology: Twitch, Summation, Tetanus & California State University, NorthridgeWe aren't endorsed by this school BIOL 282 Sep 6, 2024 Uploaded by CountMorningLemur48 Home/ Biology Date: ____February 28, 2024___________________________ Name: ___Cedric Lanier_______________________________ Lab mates names: _____Daisy, Kaivan____________________________________________________ Lab 6: Muscle Physiology II Purpose: To understand how the body decides how much force to exert to stimuli. To understand the difference between a muscle twitch, summation and tetanus. To understand how tetanus leads to fatigue. Directions: Use lecture material to answer questions IN YOUR OWN WORDS. 1. What is a muscle twitch ( 2 POINTS ). A muscle twitch is an involuntary cramp, stiffness, or tightening (contracting) of a muscle in a specific area or muscle group (e.g. Biceps, quads) that is served by a single motor nerve fiber. While they range from minor to severe, the causes can range from dehydration, electrolyte imbalance, overuse, increased demand in blood flow, and/or underlying medical conditions (e.g. arteriosclerosis, neurodegeneration). 2. Explain the difference between summation and tetanus ( 4 POINTS ). Summation is the process by which the force of a muscle contraction increases as stimuli being applied to the muscle before it has fully relaxed from the previous contraction (twitch). There are two types of summation temporal summation, which occurs when a muscle is stimulated repeatedly in quick succession before it has had a chance to relax fully and spatial summation, involving multiple motor units within a muscle being stimulated simultaneously (stronger contraction). Tetanus: Tetanus, also known as tetanic contraction or tetanic fusion, is a sustained and maximal contraction of a muscle fiber or a group of muscle fibers in response to repeated stimulation at a sufficiently high frequency. In tetanus, the muscle fibers are stimulated so rapidly that they do not have a chance to relax fully between stimuli. As a result, individual twitches blend together, leading to a smooth, sustained contraction. Tetanus occurs by increasing the frequency of stimulation beyond a certain threshold that varies by muscle type. As the frequency increases, muscle fibers cannot relax completely between stimuli, leading to fused contractions and maximal force production. Ultimately tetanus is a summation of contractions. Directions: Use Google to answer the following question IN YOUR OWN WORDS. 3. Why do our muscles fatigue faster in women than men? ( 4 POINTS ). Muscle fatigue occurs faster in women than in men due their physiological differences. One of those differences is the composition of fast twitch muscle fibers. Men are known to have a higher percentage of fast twitch muscle fibers that produce quick powerful contractions, than slow ones. Another difference is their hormonal differences of estrogen (women) and testosterone (men). Testosterone levels are higher n men and influence muscle mass, strength and recovery. While Estrogen can impact the muscles metabolism and increase their fatigue. Also, the differences of men and women muscle size, distribution of muscle mass, and structure impact fatigue rates as well. Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. Exercise Routine for Biology 282 Muscle Physiology Lab II Each student should do FOUR of the exercises . For example, one student will do one set of 12 for biceps, deltoids, abs, and quads. Start by lifting weights and getting a feeling that it is light or too hard to lift, then find the average weight between the lightest load to the hardest to lift , and use that weight for the repetitions. A team of 4 students thus will be able to fill most of the chart if they distribute exercises among themselves strategically. The figures below are to show you different exercises depending the muscle you want to use. You will need to keep a tab on your exercises (include units of measure) FILL OUT THE TABLE BELOW WITH YOUR GROUP'S INFORMATION ( 5 POINTS ). Muscle Group Name ( Lab Mate Name ) Number of repetitions Left Side / Right Side Weight Lifted Left Side / Right Side Biceps ( Cedric ) 2 sets 12/12 35 lbs./35 lbs. Biceps ( Daisy ) 2 sets 12/12 10 lbs./10 lbs. Triceps ( Kaivan ) 12/12 22 lbs./22 lbs. Triceps ( Kavian ) 12/12 22 lbs./22 lbs. Deltoid ( Cedric ) 2 sets 12/12 40 lbs./40 lbs. Deltoid ( Daisy ) 2 sets 12/12 12 lbs./12 lbs. Pectorals ( Kaivan ) 12/12 33 lbs./33 lbs. Pectorals ( Kaivan ) 12/12 33 lbs./33 lbs. Abs ( Kaivan ) 12/12 200 lbs./ (body weight) Abs ( Kaivan ) 12/12 200 lbs./ (body weight) Quads ( Cedric ) 2 sets 12/12 50 lbs./50 lbs. Quads ( Daisy ) 2 sets 12/12 40 lbs. /40 lbs. Hamstrings ( Cedric ) 2 sets 12/12 55 lbs./55 lbs. Hamstrings ( Daisy ) 2 sets12/12 20 lbs./20 lbs. Gastrocnemius ( Kaivan ) 12/12 55lbs. /55 lbs. Gastrocnemius ( Kaivan ) 12/12 55 lbs./55 lbs. Why is this page out of focus? Because this is a premium document. Subscribe to unlock this document and more. Page1of 11 Students also studied Nasm CPT Final Exam Study Guide questions and answers.docx 2023/2024 NASM CPT Final Exam Study Guide Questions And Answers (100% Correct) Graded A+ Agility - correct answer ✅ability to accelerate, decelerate, stabilize, and change direction quickly while maintaining posture Quickness - correct answer ✅the ability Chamberlain College of Nursing BIOS 255 7C Midterm 1 - AMSA Test Bank.pdf 8/24/2021 Midterm 1 Exam: Attempt review My sites / 211C-LIFESCI7C-2 / Midterm Exams / Midterm 1 Exam Summer 2021 - Session A - Week 10 \| Summer 2021 - Session C - Week 4 Summer Sessions 2021 - LIFESCI7C-2 - MULJI Started on Friday, 13 August 2021, 5:42 P University of California, Los Angeles LIFESCI 7C informational-reading-comprehension-where-are-the-stars.pdf Name _ Date _ Page 1 Informational Reading Comprehension Where Are the Stars? Read the passage and answer the questions that follow. 1 When you look up at the night sky, what do you see? If you're like 80% of people around the world, or 99% of Americans Universidad Autónoma de Yucatán EN 7TH Lab 8- Sensory Physiology.pdf Name: Valerie Guzman Date: 10-16-24 Section #: 17131 Lab 8: Sensory Physiology Purpose (1 points): The purpose of this lab is to experimentally explore and understand how different sensory systems in the body detect stimuli from the environment. We want t California State University, Northridge BIOL 282 ch 6 the muscular system key.pdf Chapter 6 The Muscular Systemn MUSCLE MOVEMENTS, TYPES, AND NAMES 14. Relative to general terminology conceming muscle activity, first fabet the following structures on Figure 6-5: insertion, origin, tendon, resting muscle, and contracting miuscle identif Memorial High School A&P 101 Quiz 2 ECE 491.docx ECE 491 Introduction to Neural Networks Quiz 2 Open books and notes Please solve each question on a separate sheet and submit your answers to Blackboard. You must solve the questions by yourself, and you should not get any help from others. 1) (35 pts) Gi University of Notre Dame Ece 491 Study your doc or PDF Upload your materials to get instant AI help Students also studied Nasm CPT Final Exam Study Guide questions and answers.docx 2023/2024 NASM CPT Final Exam Study Guide Questions And Answers (100% Correct) Graded A+ Agility - correct answer ✅ability to accelerate, decelerate, stabilize, and change direction quickly while maintaining posture Quickness - correct answer ✅the ability Chamberlain College of Nursing BIOS 2557C Midterm 1 - AMSA Test Bank.pdf 8/24/2021 Midterm 1 Exam: Attempt review My sites / 211C-LIFESCI7C-2 / Midterm Exams / Midterm 1 Exam Summer 2021 - Session A - Week 10 \| Summer 2021 - Session C - Week 4 Summer Sessions 2021 - LIFESCI7C-2 - MULJI Started on Friday, 13 August 2021, 5:42 P University of California, Los Angeles LIFESCI 7Cinformational-reading-comprehension-where-are-the-stars.pdf Name _ Date _ Page 1 Informational Reading Comprehension Where Are the Stars? Read the passage and answer the questions that follow. 1 When you look up at the night sky, what do you see? If you're like 80% of people around the world, or 99% of Americans Universidad Autónoma de Yucatán EN 7THLab 8- Sensory Physiology.pdf Name: Valerie Guzman Date: 10-16-24 Section #: 17131 Lab 8: Sensory Physiology Purpose (1 points): The purpose of this lab is to experimentally explore and understand how different sensory systems in the body detect stimuli from the environment. We want t California State University, Northridge BIOL 282ch 6 the muscular system key.pdf Chapter 6 The Muscular Systemn MUSCLE MOVEMENTS, TYPES, AND NAMES 14. Relative to general terminology conceming muscle activity, first fabet the following structures on Figure 6-5: insertion, origin, tendon, resting muscle, and contracting miuscle identif Memorial High School A&P 101Quiz 2 ECE 491.docx ECE 491 Introduction to Neural Networks Quiz 2 Open books and notes Please solve each question on a separate sheet and submit your answers to Blackboard. You must solve the questions by yourself, and you should not get any help from others. 1) (35 pts) Gi University of Notre Dame Ece 491 Other related materials S113_Module2.pdf COLLEGE DEPARTMENT 1st Semester Academic Year 2021-2022 Name: _ Course/Year/Block: _ MR. RONIE R. MARQUEZ Instructor RUBELYN M. ESPERON, PH. D. College Dean Prayer Before the Class Starts Let us put ourselves, into the holy presence of God. In the name o Tarlac State University BIO 12ART HISTORY 102 - Study Guide Introduction to Cell Biology.docx Study Guide: Introduction to Cell Biology Introduction Cell biology is a branch of biology that focuses on the structure and function of the cell, the fundamental unit of life. Understanding cell biology is crucial for comprehending various biological pro University of St. Augustine for Health Sciences HIS 102Emergency and disaster management.pptx EMERGENCY AND DISASTER MANAGEMEN T Sharon Wehrman Dr. Kelly COMM120 INTRODUCTION TO EMERGENCY AND DISASTER MANAGEMENT •The value of emergency and disaster management is to protect the communities and improve strategies regarding preparation, response, an American Public University EDMG 110Cellular Biology Study Guide.docx Cellular Biology Study Guide Table of Contents: 1. 2. 3. 4. 5. Chapter 1: Cell Structure and Function Chapter 2: Cellular Transport Chapter 3: Cellular Respiration and Photosynthesis Chapter 4: Cell Division Chapter 5: DNA and Protein Synthesis Chapter 1: Xavier University - Ateneo de Cagayan BIOLOGY 101BIOL_1105_02_homework.pdf BIOL 1105 Fall 2024 Homework Assignment #2, due Thursday September 19 at 11:59 pm Submit your answers to Parts 1-2 on Brightspace by typing or pasting your answers into the text box. Please make your answers legible and organized. Check your work before y University of Ottawa BIOL 1105microbiology week 2 critical thinking question copy.docx Week 2 Critical Thinking Questions 1. Ubiquitous and its Applicability to Bacteria and Archaea: The term Ubiquitous means "found everywhere." I would answer yes, the term "ubiquitous" is absolutely appropriate for describing bacteria and archaea. They inh Georgia Military College BIO 299-2 You might also like Microbiology Resource Announcement (MRA) Manuscript Guidelines Preparation of a Microbiology Resource Announcement (MRA) Manuscript Due Date: End of the term Length: Maximum 500 words (excluding title page, abstract, figures, acknowledgements & references) and a 50 word abstract; doubled space submitted in MS word. R University of Ottawa TMM 3300Accelerated Introduction to Neurobiology: Course Overview & Labs Syllabus: TMM 3106 Introduction to Neurobiology Winter, 2023 Course Coordinator: Stephen Gee (stevegee@uottawa.ca) Course Dates: Jan.9 - April 12 Times: Mondays 11:30-1:00 pm and Wednesdays 2:30-4:00 pm Location: 850 Peter Morand. Mondays - PMD 202A; Wedn University of Ottawa TMM 3106Module 5_ Problem Set_ Essential Human Anatomy & Physiology I w_Lab - McGuire - 2023B.pdf Module 5: Problem Set !"Due No due date !"Points 5 !"Questions 75 !"Time Limit None Instructions Anatomy of the Muscular system: Introduction & Muscles of the Head, Neck and Trunk Attempt History LATEST Attempt Time Score Attempt 1 48,350 minutes 0 out of Portage Learning BIOD 151Exploring Population Ecology: Course Overview & Objectives BIOL 3171 - Dr. Gordon Fitch Fall 2024 BIOL 3171: Population Ecology (3 credits) Course Instructor: Gordon Fitch How to address me: Dr. Fitch, Professor Fitch, Gordon Course Format: in person; lectures will usually be recorded and made available on eClass York University BIO 3171Exploring Mutations in Pigmentation Genes: Practice Questions Mutation Practice Questions BIOL 167 Living Color These questions are designed to test your knowledge of the types of mutations that can occur during DNA replication or gene expression - silent, missense, nonsense, and frameshift. The application of these Finger Lakes Community College BIO 122module 1 assignment .docx Shanice Davy Module 1 Assignment Bio 131 How does acute inflammation protect against infection? Be specific and provide details. a. Acute inflammation protects the body against infection by the speeding up the healing process and fights off the infection. SUNY Empire State College SMT 272034 You might also like Microbiology Resource Announcement (MRA) Manuscript Guidelines Preparation of a Microbiology Resource Announcement (MRA) Manuscript Due Date: End of the term Length: Maximum 500 words (excluding title page, abstract, figures, acknowledgements & references) and a 50 word abstract; doubled space submitted in MS word. R University of Ottawa TMM 3300Accelerated Introduction to Neurobiology: Course Overview & Labs Syllabus: TMM 3106 Introduction to Neurobiology Winter, 2023 Course Coordinator: Stephen Gee (stevegee@uottawa.ca) Course Dates: Jan.9 - April 12 Times: Mondays 11:30-1:00 pm and Wednesdays 2:30-4:00 pm Location: 850 Peter Morand. Mondays - PMD 202A; Wedn University of Ottawa TMM 3106Module 5_ Problem Set_ Essential Human Anatomy & Physiology I w_Lab - McGuire - 2023B.pdf Module 5: Problem Set !"Due No due date !"Points 5 !"Questions 75 !"Time Limit None Instructions Anatomy of the Muscular system: Introduction & Muscles of the Head, Neck and Trunk Attempt History LATEST Attempt Time Score Attempt 1 48,350 minutes 0 out of Portage Learning BIOD 151Exploring Population Ecology: Course Overview & Objectives BIOL 3171 - Dr. Gordon Fitch Fall 2024 BIOL 3171: Population Ecology (3 credits) Course Instructor: Gordon Fitch How to address me: Dr. Fitch, Professor Fitch, Gordon Course Format: in person; lectures will usually be recorded and made available on eClass York University BIO 3171Exploring Mutations in Pigmentation Genes: Practice Questions Mutation Practice Questions BIOL 167 Living Color These questions are designed to test your knowledge of the types of mutations that can occur during DNA replication or gene expression - silent, missense, nonsense, and frameshift. The application of these Finger Lakes Community College BIO 122module 1 assignment .docx Shanice Davy Module 1 Assignment Bio 131 How does acute inflammation protect against infection? Be specific and provide details. a. Acute inflammation protects the body against infection by the speeding up the healing process and fights off the infection. SUNY Empire State College SMT 272034 Do Not Sell or Share My Personal Information When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link. More information Manage Consent Preferences Essential Cookies Always Active Essential Cookies are required for providing you with features or services that you have requested. For example, certain Cookies enable you to log into secure areas of our Services. Share or Sale of Personal Data Always Active Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link. If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences. Advertising Cookies Always Active Advertising Cookies collect data about your online activity and identify your interests so that we can provide advertising that we believe is relevant to you. Advertising Cookies may include Retargeting Cookies. Analytics Cookies Always Active Analytics Cookies allow us to understand how visitors use our Services. They do this by collecting information about the number of visitors to the Services, what pages visitors view on our Services and how long visitors are viewing pages on the Services. Analytics Cookies also help us measure the performance of our advertising campaigns in order to help us improve our campaigns and the Services’ content for those who engage with our advertising.
4566	https://www.quora.com/How-do-you-solve-for-k-T-2%CF%80-m-k	How to solve for 'k? T = 2π √(m/k)' - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Simple Harmonic Oscillato... Variables and Constants Period (frequency) Physics Oscillations Spring Constant Algebraic Equations Harmonic Motion 5 How do you solve for "k? T = 2π √(m/k)"? All related (36) Sort Recommended Assistant Bot · 1y To solve for in the equation follow these steps: Isolate the square root: Start by dividing both sides by : Square both sides: To eliminate the square root, square both sides of the equation: Cross-multiply: Rearranging gives: Simplify: This can be simplified to: Thus, the formula for is: Upvote · 9 2 Related questions More answers below What does k represent in this formula t=2pi((m/k)/^1/2? How do I solve 1/(a-b) +1/(a+b)? How can l Derive the equation T=2π√ [(m+MS/3) /k]? How do you get the value of ? How would you solve (a+b) ^4? Donald Hartig PhD in Mathematics, University of California, Santa Barbara (Graduated 1970) · Author has 7.4K answers and 2.8M answer views ·6y Upvote · 9 8 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 Continue Reading 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: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. 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! Upvote · 999 201 99 34 9 3 Lindsey Philpott Former Engineer at Self-Employment (1991–2017) · Author has 515 answers and 689.3K answer views ·6y In any equation requiring one value to be the equivalent of the other values, you need to isolate that value on one side of the equation. Here the value “k” is shown within parentheses and the parentheses are shown to be under a square root sign. My first action would be to square both sides of the equation, resulting in the expression T2 = (2Pi)2 X (m/k). Then I would multiply both sides by the value “k” to start the isolation process. Having done this we would end up with an expression on the left side of kT2 = [RHSide]. Then divide both sides by T2 to isolate the value of “k” on the lef Continue Reading In any equation requiring one value to be the equivalent of the other values, you need to isolate that value on one side of the equation. Here the value “k” is shown within parentheses and the parentheses are shown to be under a square root sign. My first action would be to square both sides of the equation, resulting in the expression T2 = (2Pi)2 X (m/k). Then I would multiply both sides by the value “k” to start the isolation process. Having done this we would end up with an expression on the left side of kT2 = [RHSide]. Then divide both sides by T2 to isolate the value of “k” on the left and you would have an expression of k on the left by itself and all other values on the right side. Put in the other values numerically and solve for “k” - QED! Upvote · 9 3 9 1 Sanjiv Soni Former Pvt Tutor · Author has 9.7K answers and 1.9M answer views ·Apr 6 T=2π√(m/k) Square both the sides T^2=4π^2(m/k) Multiply by k on both the sides KT^2=4π^2m K=4π^2m/T^2 K=(2π/T)^2m Upvote · Related questions More answers below Given that c=2πr and a=πr^2, how would you show that a=(c^2) /(4π)? How will you solve for k if k^3 + k = 30? How do you solve for 2(m-v) (2m+3v)? What value of k will make 64 1/k = 2? How can you calculate ? Marco Biagini BS in Mathematics, Eidgenössische Technische Hochschule (Graduated 1985) ·6y Solve for k: T = 2 π sqrt(m/k) T = 2 sqrt(m/k) π is equivalent to 2 sqrt(m/k) π = T: 2 π sqrt(m/k) = T Divide both sides by 2 π: sqrt(m/k) = T/(2 π) Raise both sides to the power of two: m/k = T^2/(4 π^2) Take the reciprocal of both sides: k/m = (4 π^2)/T^2 Multiply both sides by m: Answer: k = (4 π^2 m)/T^2 Upvote · 99 14 Sherlock Bradshaw 5y Related What is the proof of T=2π√m/k? The elastic force exerts a force with opposite direction to the displacement (hence the minus sign in the ode). Then we solve the second order differential equation and find that x(t) is a sin function. Now, we want the period T, which means that the angle inside the sine function must be 2π, so we multiply sqrt(k/m) by a value equal to its opposite times 2π, and this value is a time value, the period. I hope that this was helpful. Continue Reading The elastic force exerts a force with opposite direction to the displacement (hence the minus sign in the ode). Then we solve the second order differential equation and find that x(t) is a sin function. Now, we want the period T, which means that the angle inside the sine function must be 2π, so we multiply sqrt(k/m) by a value equal to its opposite times 2π, and this value is a time value, the period. I hope that this was helpful. Upvote · Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 618 Rambabu Dontu PhD in Physics, Tata Institute of Fundamental Research (Graduated 1986) · Author has 3.6K answers and 1.3M answer views ·Updated 1y Related How can l Derive the equation T=2π√ [(m+MS/3) /k]? It is not clear what the context is and what the symbols are. I suppose this refers to the oscillations of a block of mass of m attached to a spring of mass M fixed at one end. Let x be the displacement of the mass m from its equilibrium position. The potential energy of the spring is kx²/2 The velocity of the mass m is v=dx/dt. The kinetic energy of the mass m is (m/2)(dx/dt)² = mv²/2. Let us now find the KE of the spring itself. The spring is attached to a post at one end, where the displacement and velocity are zero. The displacement as well as velocity of a point on the spring are proportional Continue Reading It is not clear what the context is and what the symbols are. I suppose this refers to the oscillations of a block of mass of m attached to a spring of mass M fixed at one end. Let x be the displacement of the mass m from its equilibrium position. The potential energy of the spring is kx²/2 The velocity of the mass m is v=dx/dt. The kinetic energy of the mass m is (m/2)(dx/dt)² = mv²/2. Let us now find the KE of the spring itself. The spring is attached to a post at one end, where the displacement and velocity are zero. The displacement as well as velocity of a point on the spring are proportional to the distance y from the fixed end. The velocity is (y/L)v = uv where u= y/L. A small portion dy of the spring has mass dM = (M/L) dy = M du Its kinetic energy is (1/2)(dM)u²v² = (1/2)Mv² u²du. The total KE of the spring = (Mv²/2) S u² du The integral, taken from 0 to 1, is S u²du = 1/3 K.E of spring = (1/2)(M/3)v² By conservation of energy, mv²/2+(M/3)v²/2 +kx²/2 = constant (m+M/3)(dx/dt)²/2 +kx²/2 =constant Differntiating w.r.t t, (m+M/3) (dx/dt)(d²x/dt²) + kx(dx/dt)= 0 d²x/dt² = - {k/(m+M/3)} x This is the equation of a simple harmonic oscillator with w² =k/(m+M/3) The period of oscillation is T = (2pi)/w = (2pi)[(m+M/3)/k]^½ Upvote · 9 1 NICHOLAS GALLAGHER MS in Mathematics, Fairfield University (Graduated 2020) · Upvoted by Fabio García , MSc Mathematics, CIMAT (2018) and Justin Rising , PhD in statistics ·8y Related How can I solve ? You cant. cannot be solved. No, not because it is irrational. No, not because we can’t simplify it further. Because it is an expression, and unlike equations, expressions cannot be solved because they do not represent an equivalence relationship [to some unknown value which we can solve for]. Expressions simply express a value. You have expressed a value, that value is an exact number which is known. The exact number represented by your expression is [equal to] . No more, no less, but exactly . If, on the other hand, your goal was to determine a simpler expression which expresses the Continue Reading You cant. cannot be solved. No, not because it is irrational. No, not because we can’t simplify it further. Because it is an expression, and unlike equations, expressions cannot be solved because they do not represent an equivalence relationship [to some unknown value which we can solve for]. Expressions simply express a value. You have expressed a value, that value is an exact number which is known. The exact number represented by your expression is [equal to] . No more, no less, but exactly . If, on the other hand, your goal was to determine a simpler expression which expresses the same value as , then I would argue that there is no simpler way of expressing this number- but that is my opinion and is strictly based on representations which are known to me at this time. Fun question, thank you! Upvote · 99 21 Sponsored by LPU Online Career Ka Turning Point with LPU Online. 100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support. Apply Now 999 262 John Ball Studied Mathematics (Graduated 1990) · Author has 62 answers and 14.6K answer views ·6y T = 2π √(m/k) square both sides of the equation T^2 = 4pi^2 (m/k) k T^2 = 4pi^2 m k = 4pi^2 m / T^2 Upvote · 9 2 Michael Lamar PhD in Applied Mathematics · Upvoted by Justin Rising , PhD in statistics · Author has 3.7K answers and 17.5M answer views ·7y Related How can I solve ? I like Samuel S. Watson's answer but he opted to use the less useful definition of for and . It’s far more useful to define this operation as: with the function defined as the everywhere convergent series and the function defined by the definite integral So . One advantage of this approach to defining exponentiation is that it can easily be extended to the domain of complex numbers. Another advantage i Continue Reading I like Samuel S. Watson's answer but he opted to use the less useful definition of for and . It’s far more useful to define this operation as: with the function defined as the everywhere convergent series and the function defined by the definite integral So . One advantage of this approach to defining exponentiation is that it can easily be extended to the domain of complex numbers. Another advantage is that rational approximations of some desired precision can be obtained at far less computational cost with this approach compared to the continuous extension of the definition for rational that is based on repeated multiplication and roots described in Mr. Watson’s answer. The disadvantage of what I described is that the approach described by Mr. Watson is far more intuitive. What else could you possibly mean by than the number that is the limit of the sequence of for some sequence of rational numbers, , that converges to ? Thankfully the two definitions are provably equivalent (on the domains for which each makes sense). Proving that equivalence is not simple, but once it is established, you are free to use whichever works better for your needs. I think you’ll find that the approach I described is far more useful for practical problems than the alternative. Upvote · 9 6 9 1 Promoted by Betterbuck Anthony Madden Writer for Betterbuck ·Updated Aug 15 What are the weirdest mistakes people make on the internet right now? Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr Continue Reading Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice. More often than not, it saves even more than 50% on load times - here’s a test I ran: Using an ad blocker saved a whopping 6.5+ seconds of load time. Here’s a link to Total Adblock, if you’re interested. Not getting paid for your screentime Apps like Freecash will pay you to test new games on your phone. Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website): You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS). If you're scrolling on your phone anyway, why not get paid for it? I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool). Upvote · 999 557 99 60 9 4 Robert Parry Studied at University of the Witwatersrand · Author has 108 answers and 65.4K answer views ·8y Related Can you put the equation T=2pi ((m/k) ^1/2) in terms of m? Upvote · 9 6 Anderson McCammont Studied at Loughborough University ·Updated 7y Related Can you put the equation T=2pi ((m/k) ^1/2) in terms of m? Upvote · 9 2 Steve Johnson BS in Physics, North Dakota State University (Graduated 1966) · Author has 5.7K answers and 2.3M answer views ·4y Related How do we derive T = 2π √(l/g)? Wikipedia has a thorough discussion about that. Pendulum (mathematics) - Wikipedia Caution, it is complicated. Upvote · 9 2 Related questions What does k represent in this formula t=2pi((m/k)/^1/2? How do I solve 1/(a-b) +1/(a+b)? How can l Derive the equation T=2π√ [(m+MS/3) /k]? How do you get the value of ? How would you solve (a+b) ^4? Given that c=2πr and a=πr^2, how would you show that a=(c^2) /(4π)? How will you solve for k if k^3 + k = 30? How do you solve for 2(m-v) (2m+3v)? What value of k will make 64 1/k = 2? How can you calculate ? How do I prove that ? How can we solve ? How do you solve (A+B) ^2? How do I prove that ? How do I prove that ? Related questions What does k represent in this formula t=2pi((m/k)/^1/2? How do I solve 1/(a-b) +1/(a+b)? How can l Derive the equation T=2π√ [(m+MS/3) /k]? How do you get the value of ? How would you solve (a+b) ^4? Given that c=2πr and a=πr^2, how would you show that a=(c^2) /(4π)? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Performance Cookies Always Active These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Functional Cookies Always Active These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Targeting Cookies Always Active These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices
4567	https://www.reddit.com/r/AskEngineers/comments/1tbmi5/how_do_you_calculate_the_true_crosssectional_area/	How do you calculate the True Cross-Sectional Area of Steel Wire Rope? (X-Post from r/AskEngineers) : r/AskEngineers Skip to main contentHow do you calculate the True Cross-Sectional Area of Steel Wire Rope? (X-Post from r/AskEngineers) : r/AskEngineers Open menu Open navigationGo to Reddit Home r/AskEngineers A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to AskEngineers r/AskEngineers•12 yr. ago LeftTurnDown How do you calculate the True Cross-Sectional Area of Steel Wire Rope? (X-Post from r/AskEngineers) Hey r/AskEngineers, I'm currently working on a suspended pedestrian trail bridge project in Mozambique, and am in the process of trying to determine the characteristics of cable I'll need to use. Thus far, I've been following design guidelines provided by a bridge building organization that's supporting me along the way. Recently, I hit a snag though. Determination of the bridge cable configuration (ie. how many decking cables will be used and what the diameter will be) depends on the calculated breaking strength (in pounds) of the available options. All of the options that I've found thus far (for example: provide a tensile strength (in the provided example, it's listed as Grade and is 1770 N/mm...I think this is an error though and it is supposed to read N/mm 2). I believe that I can use this value in N/mm 2 to calculate the breaking strength by multiply it by the wire's cross-sectional area (mm 2) and then converting the resultant product from Newtons into pounds. Problem is, finding the exact cross-sectional area is proving difficult. The fact that steel wire is composed of various strands means that there is void space within the cable and therefore I cannot just use the basic formula for area (=pir 2). I was wondering if anyone has any suggestions on how to best estimate or calculate this area. Perhaps there is a proven method for finding the True Cross-Sectional Area of a steel wire rope in the industry? Thank you in advanced! EDIT: Sorry about the Title, it's not actually cross-posted anywhere... Read more Share Related Answers Section Related Answers Innovative mechanical design trends 2024 Career paths for multidisciplinary engineers Emerging materials in mechanical engineering Chemical process optimization methods Computer engineering for embedded systems New to Reddit? Create your account and connect with a world of communities. 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 December 20, 2013 Reddit reReddit: Top posts of December 2013 Reddit reReddit: Top posts of 2013 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
4568	https://www.iejme.com/download/process-of-mathematical-representation-translation-from-verbal-into-graphic.pdf	CORRESPONDENCE Dwi Rahmawati dwirahmawati1083@gmail.com © 2017 Dwi Rahmawati et al. Open Access terms of the Creative Commons Attribution 4.0 International License apply. The license permits unrestricted use, distribution, and reproduction in any medium, on the condition that users give exact credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if they made any changes. ( Introduction Representation is one of the five standards of mathematical processes besides problem solving, reasoning, connections and communication stated in Principles and standards for school mathematics (NCTM, 2000). While in Process of Mathematical Representation Translation from Verbal into Graphic Dwi Rahmawatia,b, Purwantoa, Subanjia, Erry Hidayantoa, Rahmad Bustanul Anwarb a State University of Malang, INDONESIA; b University of Muhammadiyah Metro, INDONESIA; ABSTRACT The ability to do translation from one form of representation to another representation form is a fundamental ability to build a conceptual and mathematical thinking. Related to the important of translation process, this study aimed to investigate the process of mathematical representation translation from verbal to graph. This research was a qualitative research. Collecting data was done through the assignment sheet and interviews after the subjects completed the task given. The result showed that the students were able to do the process of translation from verbal representation to graph well at every stage of the translation. The translation process was done in four stages: unpacking the source, preliminary coordination, constructing the targets, and determining equivalence. The translation process of verbal to graph representations required more than one translation process. This process through the intermediary of some other representations like symbolic, schematic, equations, numerical. In general, students do the same activity except at preliminary coordination activity. Preliminary coordination activity can be done in two ways, namely students determined the formula of the relationship between distance and time, and by connecting the distance between the two cars and the increasing time. The more the time increased, the distance the two cars decreased. KEYWORDS ARTICLE HISTORY translation process; verbal representation, grahical representation Received 26 September 2016 Revised 13 November 2016 Accepted 11 February 2017 INTERNATIONAL ELECTRONIC JOURNAL OF MATHEMATICS EDUCATION e-ISSN: 1306-3030. 2017, VOL. 12, NO. 3, 367-381 OPEN ACCESS 368 DWI RAHMAWATI ET AL. Indonesia the idea of representation has also been included in the objectives of learning mathematics at schools. Representation as one of the mathematical process standards indicates that representation is an important process that the representation in learning to develop and optimize students' thinking skills, which is a process of construction and abstraction of mathematical knowledge. To think mathematically and communicate the mathematical ideas, it needs to represent in various forms of representations such as verbal representation, images, numeric, symbols, algebra, tables and graphs. The use of various types of representation and translation between representations is very important in expressing mathematical ideas (Bal, 2015). The use of various representations have an important role in learning mathematics (Cai & Lester, 2005), especially in helping to understand and interpret the mathematical concepts in a variety of representations (Pape & Tchoshanov, 2001). This is in line with Tripathi (2008) that says in mathematics, the use of multiple representations is an important tool that allows students to understand mathematical concepts. However, generally in mathematics, representation is only as complementary in solving mathematical problems. This leads to a very limited ability representation. The weakness of mathematical representation capability is due to the difficulty in bridging the representations and change from one representation to another representation (Yerushalmy, 1997). Thus, the use of representation deserves serious attention in learning. Some experts have devided representation in a different way (Bruner, 1966; Lesh et al, 1987; Goldin, 2002; Cai & Lester, 2005; Gagatsis & Elia, 2004; Villeges et al, 2009). From some classifications, representations covered in this study are verbal and graphic representations. Representation as a standard process, NCTM determines the standard of representation that can be mastered in learning is being able to select, implement and conduct translation inter mathematical representations to solve problems (NCTM, 2000). Fluency in doing translation from one form of representation to another representation form is a fundamental ability to be possessed to build a conceptual and mathematical thinking. NCTM (2000) identifies the ability to translate inter mathematical representations as a critical ability in learning and math activities. If students can do the translation inter forms of representation, it means students have access to the representation forms. The more complete the translation ability of the students have, the stronger the understanding of mathematical ideas they have. Through translation capability inter representation done by the students, teachers can see the understanding of the concept. Therefore it is very important for teachers to know the process of translation inter representations done by the students. Teachers need to understand how students perform translation inter representations to be able to track the location and type of errors made by the students when doing translation. Mistakes made by students can be a source of information to improve learning. It is also very important for students as pre service teachers to be able to do the translation inter representations. Janvier (1987) suggests that translation as a process involving transfer / change from one representation form to other forms of representation. While Bosse et al (2012) state that translation as a cognitive process in transforming information from one representation form to another representation form. Translation activities from one representation to another representation are INT ELECT J MATH ED 369 very important in the learning process (Duval, 2006). The translation ability inter representation of graphs, tables, symbolic and verbal is essential in understanding and communicating on mathematical concepts (Bosse, 2011). However, the reality shows that the ability of translation inter verbal representations, tables, graphs and symbolic in mathematical relation is still low (Gagatsis & Shiakalli, 2004). Students have a difficulty in determining the symbolic representation that is appropriate with the relationship represented in graphs or tables (McCoy, 1996). The previous research linked to mathematical representation translation has been done at all levels of education, both at the secondary and higher education. Ipek & Okumus (2012) who did a research on prospective primary school teachers in the use of representation in the problem solving process state that prospective teachers are not able to do the translation inter representations. Celik & Arslan (2012) conducted a study on pre service teachers of primary school to determine the ability of students’ translation among verbal representations, tables, graphs and physical context. In the research they found that the students have not been able to do the translation in the types of representation, particularly that of the physical context to the graph. Students are more successful in determining the correct graph rather than constructing a graph. While Bal (2015) in a study conducted on 134 pre service teachers of primary school in state university in Turkish found that students are usually successful in the transformation inter representation, but do not succeed in the transformation particularly verbal representation to another representation. The success of students in solving problems is most in symbolic representation and the least is in verbal representation. The students state that the different forms of representation help to understand the problem better and produce some solutions. Bal suggests that the next study should examine the process of translation between representation and various forms of representation in problem solving. Bosse et al (2014), in their study on students of 15 to 17 years old found that there are four activities done by the students in doing translation from the graph to the symbolic. The activities are unpacking the source, preliminary coordination, constructing the targets, and determining equivalence. Bosse suggests that further studies, about mathematical representation translation besides from the graph to the symbolic, are also needed in completing the research to examine in more detail in the translation process. Related to the previous study, the result of this study will describe the process of translation from verbal representation to grahical representation. The results of this study are expected to be used to reflect and evaluate the translation process in the forms of representations as the consideration in arranging appropriate learning. Research Method This study aimed to reveal the translation process from verbal representation to the graph. To reveal the translation process of the research subjects, this research was conducted by asking the research subjects to resolve the problem in writing and expressing verbally. Translation process was seen from the behavior subject to resolve problems based on translational activity (Bosse, 2014). In this study, the researcher was as a means to collect data, analyze data descriptively in the form of behavior words experienced by the 370 DWI RAHMAWATI ET AL. subject. According to Cresswell (2012) this research was qualitative research. The subjects in the study were students of mathematics education. Subjects were asked to solve the problem. The researcher chose two from ten students as subjects. The research subject was selected by taking into account the work result and communication skill. The instrument of this study was the researcher guided by the assignment sheet. The researcher acted as the instrument that collected data, analyzed the data, interpreted the data and reported the results. The work sheet Instruments used was the development of assignment sheet (Çelik & Arslan, 2012). The assignment sheet is as follows: Car A goes from town X to town Y at a steady pace. Car B goes from town Y to town X at a constant speed. Both cars go at the same speed, time and route. Draw a graph showing the relationship between the distance of the two cars on the travel time! Give an explanation for your answer! The data in this study were words and note of the behavior of the research subjects when solving a given problem. Data were obtained from the written answers and field notes when the subject solved the problem. Field notes were the notes of behavior and verbal statement recorded when the subject resolved a given problem. The result of the interview was in the form of clarification of the subject’s written answer. Based on the research objective, qualitative data analysis technique was used in this study. The researcher analyzed every word, action and the student’s written responses related to problems solving produced by the subject. The data analysis included the steps of transcribing the data collected, analyzing and reducing the data, compiling every piece of data in units which were then encoded, analyzing the translation process and making conclusions. Analyzing data was conducted during the research when data retrieval simultaneously or after retrieval of data. Results and Dıscussıon The research subject was given a problem presented in verbal form. Subject was asked to perform the translation from mathematical representation verbal to graph. In general, based on the translation activity proposed by Bosse (2014), the research subject did the process of mathematical representations translation from verbal to graphics through four activities. The activities were unpacking the source, preliminary coordination, constructing the targets, and determining equivalence. The process of translation in four activities undertaken by research subject is as follows: Subject S1 The first activity was unpacking the source. After facing with the problem, S1 could instantly know the important words related the problem. He was able to understand the essential ideas in a verbal representation problem. It indicated that S1 was able to perform unpacking the source activities well. S1 behavior when performing unpacking the source was he read the problem given repeatedly and marked the words that were considered important by underlining the words on a given problem. S1 revealed the verbal representations given by identifying the components in verbal representations namely the same time, the distance of the two cars, mileage car, same speed, constant speed, distance of city X & Y and direction. Next the components of the verbal representations were translated into the components forming graphic INT ELECT J MATH ED 371 images. Here are the excerpts of interviews related to the activity of unpacking the source: P: What information do you get from the problem? S1: here there are car A and B running at the same speed and constant, V A = V B = V, then car A and B go in the opposite direction like this (pointing to the scheme that has been made), car A and B run in the same time so that the travel time is the same. P: Why do you make this schema? S1: to help me describe the event ma’am. P: Okay. Then is there information that you do not understand? S1: there is no ma’am. P: Do you understand the question on the problem? Explain, please. S1: Yes ma'am, here I was told to draw a graph of distance between the two cars towards the time. P: Is the information enough to answer the questions? S1: Yes ma’am. Interview result showed that S1 was able to recognize all of the information in the sentence that could be used to solve the problem, identify the key words that declared the information in the source representation and understand the questions. It means that S1 had conducted unpacking the source activities well to solve the problem. To obtain a graph component, first S1 did the coordination preliminary activities. The second activity was done by S1 smoothly. S1 was able to connect the literature to formulate a target representation and transfer the content of the source representation into the target representation. S1 was able to connect the information that had been identified with the understood concept. It showed that S1 was able to choose any elements that need to be exemplified and select the symbol used. S1 changed the verbal component in the form of speed, distance and time through analogy with formal symbols. For verbal component within the two cars, S1 expressed it by symbol x. The verbal component of constant speed and distance of city X-Y was expressed as V and y. While the verbal component in the form of directions, S1 changed it by making schema that indicated the travel direction of both cars from city A and B. Based on the schemas that had been made, S1 determined the interval. The problem solving of S1 when creating the schema can be seen below : Figure 1. Answer S1 while Unpacking The Source by Creating a Schema S1 then set up information that might be used to construct the target representations. S1 connected inter verbal component that had been declared symbolically and schematically to form the mathematics equation by performing a series of algebraic operations to obtain the relationship between the distance of the two cars towards the time. S1 associated it with the formula of distance, speed and time he had. It showed S1 was able to create networks and the idea relationship of inter representations, such as statements (Duval, 2006). S1’s ability in integrating ideas inter these representations was helpful in the translation stage. By forming this mathematical equation, S1 could earn some coordinate points and the declivity of the graph, so it would help in making the 372 DWI RAHMAWATI ET AL. graphic images in the next stage of the translation. The mathematical equation formed by S1 is 𝑥= 2𝑡.𝑣; 𝑡>1 2 y 𝑉 𝑦−2𝑡.𝑣; 0≤𝑡≤1 2 y 𝑉. Here is the excerpts of the interview: P: How will you draw a graph using the information in the problem you get? S1: First of all let's say, S A and S B are the mileage of car A and B. t A and t B are the travel time of car A and B. y is the distance of city X and Y, and x is the distance of the two cars. After that I count the distance of two cars: x = y- (S A + S B) = y - (tV + tV) = y - 2t.V. It is because the travel time and speed of both cars are the same. Well, because the two cars run in the opposite directions (pointing to the schema), then the car will met. So after meeting, the distance is x = 2TV. P: When will the two cars meet? S1: At t = 1 2 𝑦 𝑉, so 𝑥= 2𝑡.𝑣; 𝑡>1 2 𝑦 𝑉 𝑦−2𝑡.𝑣; 0≤𝑡≤1 2 𝑦 𝑉 P: Do you think there is another way to draw graphs? S1: mmm ... there is no ma'am. The result of S1 when doing presuppose (Figure 1) and algebraic operations (Figure 2) preliminary coordination activity is as follows: Figure 2. Answer S1 while Preliminary Coordination by Doing Presuppose Figure 3. Answer S1 while Preliminary Coordination by Algebraic Operations In the third activity, S1 did constructing the target well. After S1 symbolizes the verbal component of the distance of both cars with x, S1 determined x as the X-axis that was as component of the graph while the verbal component of the time represented by symbol t became the graph component namely Y-axis. Then by linking T, V and Y based on the concept already understood, S1 determined the range members namely 0 and 1 2 y 𝑉. For the graph components as domain member, S1 acquired it from the verbal component of distance of XY city by turning it into symbol y first. Domain members obtained were 0 and y. By using the equations that had been obtained, S1 determined the coordinate points which were the component maker of graphic images. S1 substituted t as the range members that fulfilled the equation. He obtained coordinates of point (0, 1 2 y 𝑉) and (y,0). Then, based on the equation and interval obtained in the preceding stage, S1 determined the graph gradient namely where the graph had a declivity to the right or left. Based on the graphs maker components obtained, S1 made a sketch of each graph maker component. First of all S1 sketched X axis as x, then the Y-axis as t. Then S1 put the domain member on the X axis and the range member INT ELECT J MATH ED 373 on Y axis with the right scale. S1 then described each pair of points obtained and then draw a line connecting the pair of points on the Cartesian coordinate plane. Here are the excerpts of interview: P: What do you do to draw a graph that corresponds to the problem after you obtain the formula? S1: The graphics is made by determining sb-X as the distance of the two cars nemely x, sb-Y as the travel time namely t, then for example for t = 0 then x = y, and for x = 0 then 𝑡= 1 2 𝑦 𝑉. Then I draw it like this (pointing to the picture). Here is the result of S1 problem resolution when doing constructing the targets: Figure 4. Answer S1 when Doing Constructing the Target In the fourth activity namely determining equivalence, S1 checked all the steps that had been done and recalculated the answers that had been obtained. S1 discovered again if there was no error in the operation or not. Then he rechecked the graph suitability with the information existed in the verbal representation (source). The excerpts of the interview are as follows: P: Are you sure with your answer? S1: yes, I have rechecked it, Ma’am. P: How do you check that the graph you are getting correct? Explain, please. S1: this is at t = 0, then the distance between the two cars are the same as the distance of city X and Y namely y, the two cars meet at t = 1 2 𝑦 𝑉 so that the distance is 0. Here is the structure of the translation process done by S1 based on the activity of translation: 374 DWI RAHMAWATI ET AL. Figure 5. Structure of Translation Process by S1 Table 1. Description of the coding on the process of verbal representations translation to the graph by S1 Term Code Verbal representation Verbal Reading matter Bc Identify Idf Presuppose Mis Connect Hub Decide variables sb-X and sb-Y Sb substitute the value Subs Determine the gradient Gr Check the suitability of the graph Kes A car speed of A equal to the speed of car B Kec sm The travel time of car A equal to travel time of car B Wkt sm Car A runs from city X to Y, car B runs from city Y to X Skm Speed of Car A and B constant Kec tetap Speed of car A, the speed of car B, travel time of car A, travel time of car B, distance between car A and B, the distance between city X and Y, the speed of the car, car travel time V A, V B, t A, t B, x, y, v, t Connect the dots Hub ttk Determine the interval Int Determine domain member Dm Determine range members Rn Determine the point coordinates Kt Sketch Skt Graphic representation Grafik Based on the structure process of the verbal representations translation to the graph above, S1 unpacked source in the form of verbal representations given by identifying the components of the same time, the distance of the two cars, mileage car, same speed, constant speed data, the distance of city X & Y and direction. Each component of verbal was translated into components of graph maker to illustrate the graph that corresponded to the verbal representation. The components forming the graphic of the translation result of verbal component made by S1 in the form of the X axis, Y axis, point coordinates, a member of a domain, range members and gradients. To make the process of translation from verbal problem into a graph, S1 required intermediaries of some representations such as symbolic, equations, schema and numeric. S1 drew the graph by connecting all the components to make graphics that had been obtained. S1 created cartesian coordinate with x as the X-axis and t as Y axis. Then S1 put the domain members and range on the coordinate axes. Next, S1 made a point for each pair of points obtained and then drew a line connecting the points couple with due regard to the gradient. Subject S2 Based on the answers of S2 when solving problems, it showed that S2 was able to resolve the problem properly. For the first activity, S2 did unpacking the source by understanding the problem well, finding things that were considered important in a given problem. S2 had been able to recognize all the information in the sentence that could be used to solve the problem, identify key words that declared that the information on the source representation and understand the questions. S2 revealed the verbal problem by identifying verbal components that INT ELECT J MATH ED 375 existed on the source problem. Some verbal components were time, the same time, the distance of the two cars, car mileage, same speed, constant-speed, the distance of city X & Y and direction. These verbal components which would then be translated into a graph would be the component forming graph, which would be used in creating graphic images. S2 had been able to identify all the components needed. It meant S2 had conducted unpacking the source well to solve the problem. Here are the excerpts of the interview: P: What information do you get from the problem? S2: The speed of car A and B is same and constant, V A = V B. Car A and B run at the same time, car A runs from city X and Y, car B runs from town Y to X like this (showing a picture ) P: Are you familiar with the information? S2: yes P: Do you understand the question on the matter? Explain, please. S2: yes, it is asked the graph relationship between the distance of the two cars towards the time. P: Is there enough information to answer questions? S2: Yes, there is. In the second activity, S2 was able to perform preliminary coordination activity by linking literature to formulate a target representation and transfer the content of the source representation into the target representation. S2 was able to choose any elements needed to be exemplified and select the symbol used. It showed S2 performed analogy of verbal component in the form of distance of city X & Y, time, speed, distance between the two cars, car mileage by conducting symbolization S, t, v, S i, S Ai, S Bi formally. It meant he was able to connect information that had been identified with the concept understood. For the verbal component of directions, S2 made the schema related to travel direction of the two cars. The result of S2 completion while doing analogy (Figure 5) and making the schema (Figure 6) related to the travel direction is as follows: Figure 6. Answer S2 while Preliminary Coordination by Doing presuppose Figure 7. Answer S2 while Preliminary Coordination by Making Schema S2 connected the time change towards the distance of the two cars where the speed of both cars was constant. S2 performed a series of algebraic operations to determine S i and defining a relationship between S 0, S 1, S 2, S 3 etc. S2 created patterns between the distance of the two cars if t increased more. S2 started at t = 0, the result was S 0 = S that was when t = 0 the result was the distance of the two cars equal to the distance of the city X and Y. Then S2 replaced t with 1,2,3 and so on. Thus the result was pattern that the more the time increased, the smaller the distance of the two cars. The distance the two 376 DWI RAHMAWATI ET AL. cars would be 0 when two cars met, then it would increase gradually. Here is the excerpt of the interview: P: How will you draw a graph using the information in the questions you get? S2: by exemplifying: S i is the distance between the car A and B for t = i, S Ai is the distance traveled by car A for t = i, S Bi is the distance traveled by car B for t = i. The distance between city X and Y is S. Both cars run at the same time, then t A and t B are equal because V A = V B and t A = t B then S A = S B. For determining S i (distance of the two cars) for some value of t (travel time). For t = 0, then S 0 = S - (S A0 B0 + S) = S For t = 1, S 1 = S - (S A1 + S B1) For t = 2 then S 2 = S - (S A2 + S B2) For t = 3 then S 3 = S - (S + S A3-B3) For t = n, n is the time when S i = 0, that is S + S Ai Bi = S P: When t = n? S2: When car A and B meet, when running halfway distance of city X and Y P: How is the relationship between S 0, S1, S2 and so on? S2: S0> S1> S2> ...>Sn = 0 P: Why is that? S2: because S Ai = S Bi and S A1 S1> S2> ...>Sn Here is the result of the problem completion by S2 at the time of preliminary coordination: Figure 8. Answer S2 while Preliminary Coordination by Algebraic Operations For the third activity, S2 did constructing the target by specifying Si as the X-axis and t as Y-axis to state time. From the calculation of Si S2, it was stated that S0, S1, S2, S3 and so on were the members of the range in the graph. Similarly to the graphics component in the form of domain member, expressed numerically 0, 1, 2, and so on as the domain members. Based on the pattern obtained, S2 determined the points coordinate as the components forming graph. They were (0, S0), (1, S1), (2, S2), (3, S3) and so on. Based on the relationship of S0> S1> S2> S3> ...> Sn and Sn <Sn + 1 <Sn + 2 <..., S2 stated verbally that INT ELECT J MATH ED 377 graph of t = 0 moved from the top down, then it moved from the bottom up upward. This verbal statement was a graphic component of gradient. After the components forming the graph were obtained, S2 made Cartesian coordinate with t as X-axis and S i as Y-axis. Then he put the domain member and range at the correct scale. S2 marked points on the coordinate plane which was the location of points coordinate obtained. Furthermore S2 drew the graph by connecting the points that had been put up and took notice of the movement. S2 drew graphs with pictures starting from t = 0 (sb-small X), moved from the top down with the declivity to the left, then the graph moved up to the top with a declivity to right. Here's the interview: P: What do you do to draw a graph that corresponds to the problem? S2: First I determine that sb-X states the travel time, sb-Y represents the distance of the two cars in Cartesian coordinates. Then I draw a graph starting from t = 0 (sb-small X), moving from the top down and moving bottom up as like this (pointing to the picture) P: Have you had improved the graph that you have created? Explain, please. S2: yes ma'am, I forgot that car A and B after meeting are still running so it will repel each other so that the distance of both enlarge like this (pointing to the picture) The fourth activity is determining equivalence, S1 checked all the steps that had been done and recalculated the answer. S1 searched if there was an error in the operation or not. Then he checked the suitability of graph with the information in verbal representation (source). After rechecking, S2 improved the graphics for he forgot that the cars still ran after the meeting, so the graph changed. The excerpts of the interview are as follows: P: Are you sure with your answer? S2: yes. P: How do you check that you are getting the correct graph? Explain, please. S2: I reread from the top ma’am Here is the result of S2’s problem solving when performing determining equivalence: Figure 9. Answer S2 when Perfoming Determining Equivalence Here is the structure of the translation process done by S2 based on the translation activity: 378 DWI RAHMAWATI ET AL. Figure 10. Structure of Translation Process by S2 Table 2. Coding Description on the translation process of verbal representations to the graph by S2 Term Code Verbal representation Verba l Reading matter Bc Identifying Idf Exemplifying Mis Connecting Hub Calculating Hit Making a schema of car A runs from city X to Y, car B runs from city Y to X Skm Determining domain member Dm Determining range member Rn Determining coordinate points Kt Determining the gradient Gr Define variable sb-X and sb-Y Sb Sketching Skt Connecting the coordinate points Hub ttk Checking the suitability of the graph Kes Speed of car A equal to the speed of car B Kec sm Traveling time Wkt Travel time of car A equals the travel time of car B Wkt sm Speed of Car A and B is constant Kec ttp The distance between city X and Y, speed of car A , speed of car B, travel time, the distance between car A and B when t = i, mileage of car A when t = i, mileage of car B when t = i and speed S, V A, V B, t, S i, S Ai, S Bi, v Mileage of car A equals the mileage of car B for each t = i S Ai = S Bi Graphic representation Grafik Translation process of verbal representation to graph done by S2 through four translation activities (Bosse et al,2014). S2 identified the component of INT ELECT J MATH ED 379 verbal representation from the problem given namely time, the same time, the distance of the two cars, car mileage, the same speed, constant speed, distance of city X & Y and directions. S2 translated the verbal components forming graphic image in the form of X-axis, Y-axis, domain members, range members, point coordinates and gradient. In doing translation, S2 required intermediaries of some representations like symbolic, schematic, numerical and verbal. Based on the answers, subject S1 and S2 both written and verbal, the results showed that subjects could perform translation process of verbal representation to graph well through four translation activities as stated by Bosse (2014) ie unpacking the source, preliminary coordination, constructing the targets, and determining equivalence. Preliminary coordination could be done by determining the formula of the relationship between distance and time of the event given just as done by subject S1. S2 did it by connecting the distance between the two cars and the increasing time. The more the time increased (t) the less the distance of the two cars are. For other activities in general S1 and S2 performed the same activity. Subject S1 and S2 performed the translation process of verbal to graph through intermediaries of some representations. It meant that the translation process of verbal representations to graph needed more than one translation process or in other words translation indirectly (Janvier, 1987) or globally (Duval, 2006). Based on the translation activity, subject showed that the translation process of each subject was different. Translation process has different complexity depending on the activity (Bosse et al, 2011). The process of translation from verbal is the most complex translation process (Bal, 2015; Bosse et al, 2011), therefore it is very important for pre service teachers and teachers know the activity of translation, especially from the verbal to the graph. By knowing the translation process inter representations, it is expected that future pre service teachers and future teachers can improve the ability of students in translation process effectively. Conclusıons and Suggestıons Based on the research that has been done it can be concluded that the students are able to make the process of translation from verbal representations to graph well. Translation process is done in four stages: unpacking the source, preliminary coordination, constructing the targets, and determining equivalence. The subjects did translation in every verbal representation component to graph representation component. The process of translation is from verbal to graph through the intermediary of other representations. Intermediary representation used by each subject is different, so the complexity level of each subject translation process varies depending on the activity performed. The activity of preliminary coordination can be done in two ways, namely students determine the formula of relationship between distance and time of the event given, and by connecting the distance between the two cars and the increasing time. The more the time (t) increases, the less the distance of the two cars is. For the other activities in general students do the same activity. This research is limited only on the verbal representations to the graph. The reverse process, namely the graph to the verbal has not been seen yet. And it does not give attention to the characteristics of the subject deeply as the level of ability that might affect the representation translation process. The limitation in this study has the potential to be studied in further research. 380 DWI RAHMAWATI ET AL. Disclosure statement No potential conflict of interest was reported by the authors. Notes on contributors Dwi Rahmawati – Department of Mathematics Education Doctoral, State University of Malang, Malang, Indonesia; Department of Mathematics Education, University of Muhammadiyah Metro, Metro, Indonesia Purwanto – Department of Mathematics Education, State University of Malang, Malang, Indonesia Subanji – Department of Mathematics Education, State University of Malang, Malang, Indonesia Erry Hidayanto – Department of Mathematics Education, State University of Malang, Malang, Indonesia Rahmad Bustanul Anwar – Departmen of Mathematics Education, University of Muhammadiyah Metro, Metro, Indonesia References Bal, A. P. (2015). Skills Of Using And Transform Multiple Representations Of The Prospective Teachers. Journal of Mathematical Behavior, 197(Hal.), 582-588. Bosse, M. J., Gyamfi, K. A& Chandler, K. (2011). Translation among Mathematical Representation: Teacher Belief and Practices. (Online), ( Bosse, M. J., Gyamfi, K. A& Chandler, K. (2012). Lost in Translation: Examining Translation Errors Assosiated with Mathematical Representation. School science and Mathematics, 112(3),159-170 Bosse, M. J., Gyamfi, K. A& Chandler, K. (2014). Students Differented Translation Processes, (Online),( Bruner, J. (1966). Towards a theory of instruction. Cambridge, MA: Harvard University Press. Cai, J., & Lester, F. K. 2005. Solution representations and pedagogical representations in Chinese and U. S. classrooms. Journal of Mathematical Behavior, 24, 221-237. Celik, D.& Arslan, A. S. (2012). The Analysis of Teacher Candidats Translating skill in Multiple Representations, (Online), ( ). Creswell, J.W. (2012). Educational Research. Pearson. Duval, R. 2006. The cognitive analysis of problems of comprehension in the learning of mathematics. Mediterranean Journal for Research in Mathematics Education, 1(2), 1-16. Gagatsis, A. & Elia, I. (2004). The Effects of Different Modes of Representation on Mathematical Problem Solving. Proceedings of The 28th Conference of The International Group for The PMA, 2, 447-454. Gagatsis, A., & Shiakalli, M. (2004). Ability to translate from one representation of the concept of function to another and mathematical problem solving. Educational Psychology, 24(5), 645-657. INT ELECT J MATH ED 381 Goldin, G. A. (2002). Representational systems, learning, and problem solving in mathematics. Journal of Mathematical Behavior, 17(2), 137-165. İpek, A. S., & Okumuş, S. (2012). İlköğretim matematik öğretmen adaylarının matematiksel problem çözmede kullandıkları temsiller. Gaziantep Üniversitesi Sosyal Bilimler Dergis, 11(3), 681 -700. Janvier, C. (1987). Translation Processes in Mathematics Education. Dalam Janvier(Ed). Problems of Representation in the Teaching and Learning of Mathematics, 27-32, Hillsdale, NJ: Lawrence Erlbaum Associates. Lesh, R., Post, T., & Behr, M. (1987). Representations and translations among representations in mathematics learning and problem solving. New Jersey: Lawrence Erlbaum Associates. McCoy,L.P., et. Al. (1996). Using Multiple Representation to Communicate: an Algebra Challenge. Reston. VA: NCTM National Council of Teachers of Mathematics. (2000). Principles and Standards for School Mathematics. Reston, VA: NCTM. National Council of Teachers of Mathematics. 2000. Principles and Standards for School Mathematics. Reston, VA: NCTM. Pape, S.J.& Tchoshanov, M.A. (2001). The Role of Representation(s) in Developing Mathematical Understanding. Theory into Practice, 40(2), 118-125. Tripathi, P. N. (2008). Developing mathematical understanding through multiple representations. Mathematics Teaching in Middle School, 13(89), 438-445. Villegas, J. L., Castro, E., & Gutierrez, J. (2009). Representation in problem solving: A case study with optimization problems. Electronic Journal Of Research In Educational Psychology, 7(1), 279-308. Yerushalmy, M. (1997). Designing Representations: Reasoning about Functions of Two Variables. Journal for Research in Mathematics Education, 27 (4), 431-4.
4569	https://math.stackexchange.com/questions/104786/contest-problem-on-domain-and-range-of-square-root-function	algebra precalculus - Contest problem on domain and range of square root function - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Contest problem on domain and range of square root function Ask Question Asked 13 years, 8 months ago Modified13 years, 8 months ago Viewed 579 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I have no clue how to do this problem: Let f(x)=√a x 2+b x f(x)=a x 2+b x−−−−−−−√. For how many real values a a is there at least one positive real value of b b for which the domain of f f and the range of f f are the same set? The answer is two but what is the complete solution? algebra-precalculus contest-math Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Feb 2, 2012 at 1:29 Austin Mohr 26.3k 4 4 gold badges 75 75 silver badges 129 129 bronze badges asked Feb 2, 2012 at 1:11 VictorVictor 8,528 7 7 gold badges 58 58 silver badges 115 115 bronze badges 3 1 Have you tried just computing all the cases? Note that a=0 a=0 is one of the cases that works ; in that case since b>0 b>0 the image and the range are the non-negative real numbers.Patrick Da Silva –Patrick Da Silva 2012-02-02 01:25:22 +00:00 Commented Feb 2, 2012 at 1:25 @PatrickDaSilva - just try to figure out the cases cause me headache Victor –Victor 2012-02-02 01:29:08 +00:00 Commented Feb 2, 2012 at 1:29 There is a smart way to figure out all the cases but to get rid of some very quickly ; look at Lopsy's answer.Patrick Da Silva –Patrick Da Silva 2012-02-02 01:32:25 +00:00 Commented Feb 2, 2012 at 1:32 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. First off, a x 2+b x a x 2+b x is a quadratic, and the domain of f f is the region where the quadratic is non-negative. If this set includes any negative numbers, it's game over: f(x)f(x) can never be negative, so the range won't match the domain. Therefore, a x 2+b x<0 a x 2+b x<0 when x<0 x<0. It immediately follows that a a is negative or zero. If a a is zero, we win: setting b=1 b=1 (or your favorite positive number) works. So a=0 a=0 is one solution to the problem. Otherwise, a a is negative. Solving the quadratic, the solutions are 0 0 and −b/a−b/a, so the domain of f f is [0,−b/a][0,−b/a]. This has to include only positive numbers, so b must be positive. Now, the quadratic ranges from f(0)f(0) to its maximum f(−b/2 a)f(−b/2 a), since −b/2 a−b/2 a is right between the zeroes, so the range of the function f f is [0,b√−1/4 a][0,b−1/4 a−−−−−√]. Therefore, since the range and domain are equal, −b/a=b√−1/4 a→−1/a=√−1/4 a−b/a=b−1/4 a−−−−−√→−1/a=−1/4 a−−−−−√, and the unique solution of this latter equation is a=−4 a=−4. In summary, there are two solutions, a=0 a=0 and a=−4 a=−4. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 2, 2012 at 1:33 Patrick Da Silva 42.7k 6 6 gold badges 93 93 silver badges 145 145 bronze badges answered Feb 2, 2012 at 1:29 LopsyLopsy 4,620 23 23 silver badges 28 28 bronze badges 3 2 I just added some math typesetting. Very good answer ; you thought it through properly.Patrick Da Silva –Patrick Da Silva 2012-02-02 01:34:12 +00:00 Commented Feb 2, 2012 at 1:34 lopsy- i think you make some mistake on the sentence that after the first paragraph accoding to this: artofproblemsolving.com/Wiki/index.php/2003_AMC_12A_Problems/…Victor –Victor 2012-02-02 04:02:33 +00:00 Commented Feb 2, 2012 at 4:02 What exactly is the mistake?Lopsy –Lopsy 2012-02-02 12:29:15 +00:00 Commented Feb 2, 2012 at 12:29 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 algebra-precalculus contest-math See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 2domain of square root 2Domain and Range of f(x) 11How to find the range of the function: f(x)=√x−1+2√3−x f(x)=x−1−−−−−√+2 3−x−−−−−√ 2Constants for the Domain and Range of a Function 3Find the domain and range of the following functions. 1Finding domain and range of relations 3Possible range of a function 1what is the domain of 1√x−[x]1 x−[x]√ where [x] denotes the greatest integer function and find the range . 2Finding the domain and range of a composite function Hot Network Questions How long would it take for me to get all the items in Bongo Cat? alignment in a table with custom separator What's the expectation around asking to be invited to invitation-only workshops? Gluteus medius inactivity while riding Origin of Australian slang exclamation "struth" meaning greatly surprised Why are LDS temple garments secret? Riffle a list of binary functions into list of arguments to produce a result Spectral Leakage & Phase Discontinuites Identifying a thriller where a man is trapped in a telephone box by a sniper Find non-trivial improvement after submitting How to locate a leak in an irrigation system? What can be said? Matthew 24:5 Many will come in my name! Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done? I have a lot of PTO to take, which will make the deadline impossible Bypassing C64's PETSCII to screen code mapping How can the problem of a warlock with two spell slots be solved? Is existence always locational? Can you formalize the definition of infinitely divisible in FOL? Numbers Interpreted in Smallest Valid Base Does the curvature engine's wake really last forever? A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? Where is the first repetition in the cumulative hierarchy up to elementary equivalence? more hot questions Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4570	https://www.sciencedirect.com/topics/mathematics/cylinder-set	Skip to Main content My account Sign in Cylinder Set In subject area:Mathematics Cylinder sets refer to specific subsets of a measurable space, characterized by their structure that allows for the representation of sequences or paths within that space. They are defined in relation to measures and possess properties that enable analysis of their behavior under various conditions, such as bounded distortion. AI generated definition based on: Handbook of Dynamical Systems, 2006 How useful is this definition? Add to Mendeley Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Gaussian Processes 2018, Stochastic Analysis of Mixed Fractional Gaussian ProcessesYuliya Mishura, Mounir Zili 1.3.5.1Cylinder sets Consider a probability space , a non-empty set , the set of the real-valued functions on and the set of continuous real-valued functions on . Definition 1.12 A subset A of (respectively, ) is called a cylinder set in (in), if there exist an integer , for j = 1, … , n and a Borel subset such that The smallest σ-field containing all cylinder sets of (respectively, ) is denoted by (respectively, ). View chapterExplore book Read full chapter URL: Book2018, Stochastic Analysis of Mixed Fractional Gaussian ProcessesYuliya Mishura, Mounir Zili Chapter Gaussian Processes 2018, Stochastic Analysis of Mixed Fractional Gaussian ProcessesYuliya Mishura, Mounir Zili 1.3.5Equivalence and singularity of measures related to Gaussian processes 1.3.5.1Cylinder sets Consider a probability space , a non-empty set , the set of the real-valued functions on and the set of continuous real-valued functions on . Definition 1.12 A subset A of (respectively, ) is called a cylinder set in (in), if there exist an integer , for j = 1, … , n and a Borel subset such that The smallest σ-field containing all cylinder sets of (respectively, ) is denoted by (respectively, ). 1.3.5.2Probability measure induced by a stochastic process Definition 1.13 Consider a stochastic process defined on the probability space . We introduce a set function on a measurable space such that for any t1, …, and for any Borel subset . Remark 1.13 It is possible to prove that is indeed a probability measure on a measurable space , see, for example, [MIS 17]. The probability measure is called the measure induced by the process X. 1.3.5.3Gaussian measures: entropy and entropy distance for two measures Definition 1.14 A probability measure on is said to be Gaussian if it is induced by a Gaussian process . If, in particular, the measure is supported by the space of continuous functions on , then is called a Gaussian measure on . Definition 1.15 Consider two probability measures , k = 1, 2, on a measurable space , and let be a class of all possible finite partitions of Ω. Then, the value is called the relative entropy of the probability measure relative to . Note that in the formula for , we assume that 0 log 0 = 0 and log 0 = −∞. Definition 1.16 The entropy distance between two measures and is defined by Remark 1.14 Because the function F(x) = x log x is convex, we can apply Jensen’s inequality to conclude that This means that the relative entropy is non-negative. Remark 1.15 Recall the notions of absolutely continuous, equivalent and singular probability measures. We say that the probability measure is absolutely continuous w.r.t. the measure , both defined on a probability space (denoted by ), if for any set such that we have that . If , then there exists a non-negative random variable (Radon–Nikodym derivative, Radon–Nikodym density) such that for any set We say that the probability measures and are equivalent, , if both and . In this case, there exist both and , both of which are positive a.s., and moreover, We say that the probability measures and are singular if there exists such a set that and (in this case, obviously, there exists a set such that and , therefore the definition of singularity is symmetric). If two measures and are singular, then obviously . However, we cannot state that if , then the measures and are singular. Conversely, they can even be equivalent. To clarify this situation, let us take any positive random variable Z such that . Then, Z can be considered as a Radon–Nikodym derivative, for some probability measure . Assume that , and we immediately get an example of equivalent probability measures with infinite relative entropy and entropy distance. Lemma 1.9 Let and be two probability measures on , and let there exist some ε0 ∈ (0, 1] such that for any ε ≤ ε0, in turn, there exists an event Aε such that and . Then, the measures and are singular. Proof Put , and denote Bk = Aεk. Consider the event Then, on the one hand, for any n ≥ 1 whence it follows immediately that On the other hand, the events ∪k = n∞Bk are decreasing, therefore and a lemma is proved. The proof of the following lemmas can be found, for example, in [HID 93], Chapter 6, section 2. Lemma 1.10 1) : If a probability measure is absolutely continuous with respect to a probabilitty measure , then the entropy can be expressed by using the Radon–Nikodym density as follows 2) : If is not absolutely continuous with respect to , then . Lemma 1.11 Let {Xk, 1 ≤ k ≤ n} be a finite number of random variables on a measurable space . Assume that Xk are independent and Gaussian with respect to both probability measures and , and let where and Vari are the mean and variance with respect to , i = 1, 2, respectively. Let be the sub-σ-algebra generated by the sequence {Xi, 1 ≤ i ≤ n}. Then, the restrictions and of and , respectively, to are equivalent probability measures and the following relations hold: i) ii) iii) iv) v) Now, we consider a process that is Gaussian with respect to both probability measures and . To be specific, we take the basic measurable space to be on which two Gaussian measures and are given. Let be the sub-σ-field of generated by {Xt, }, where is a finite subset of . The proof of the following lemma can be found, for example, in [HID 93], Chapter 6, section 2. Lemma 1.12 Suppose that an increasing sequence {, n ≥ 1} of sub-σ-fields converges to so that . Let and be the restrictions of the given probability measures and to , respectively. If then is absolutely continuous with respect to and Definition 1.17 Let a process be Gaussian with respect to the measure . We say that the process X is non-degenerate w.r.t. this measure if, for any n ≥ 1 and any t1, t2, …, covariance matrix (Cov(Xti, Xtj), 1 ≤ i, j ≤ n) is positive definite. The next result can be proved similarly to lemma 1.12. Lemma 1.13 Entropy distance between two measures and equals where , Δ is the class of all finite subsets of , and (respectively ) are the restrictions of (respectively ) to . 1.3.5.4Equivalency and singularity of Gaussian measures From now on, we assume that the set is infinite. Definition 1.18 Let a process be Gaussian with respect to the measure . We say that the process X is non-degenerate w.r.t. this measure if for any n ≥ 1 and any t1, t2, …, covariance matrix (Cov(Xti, Xtj), 1 ≤ i, j ≤ n) is positive definite. In the opposite case, we say that the process X is degenerate w.r.t. the measure . Now our goal is to establish singularity conditions for two Gaussian measures corresponding to the same process. Consider the process and assume that it is Gaussian with respect to two probability measures and . Take any finite number of points t1, t2, … , tn from and consider vector X = (Xt1, Xt2, …, Xtn) which is a Gaussian vector with respect to both measures. Denote KX,1 and KX,2 its covariance matrices w.r.t. and , respectively. According to theorem A.8, there exists a non-degenerate matrix S such that both STKX,1S and STKX,2S are diagonal matrices. Moreover, according to lemma 1.8, STKX,1S and STKX,2S are covariance matrices of the vector SX with respect to the corresponding measures. In this connection, let us consider two cases. i) : Let there exist such a vector X that matrices STKX,1S and STKX,2S have different numbers of non-zero diagonal elements. This means that vector SX has a different number of independent components w.r.t. measures and . Let them be 1 ≤ k1 ≤ k2 ≤ n. Then, there exists a random vector Y = (Y1, Y2, …, Yk2) whose components are taken from SX and non-random non-zero vector λ = (λ1, λ2, …, λk2) such that the linear combination ∑k = 1k2λkYk has a non-degenerate Gaussian distribution w.r.t. and a degenerate distribution w.r.t. , i.e. and This means that the measures and are singular. As was mentioned before, in this case, their entropy distance . Of course, this is also true in the case when the process X is non-degenerate w.r.t. one of the measures and degenerate w.r.t. another. ii) : Let, for any vector X, matrices STKX,1S and STKX,2S have the same number of non-zero diagonal elements. This means that vector SX has the same number of independent components w.r.t. both measures. Assume that the process X is degenerate in the sense that the number of such components is restricted by some number k. Consider any vector X with k linearly independent components. Then, both matrices KX,1 and KX,2 are symmetric and positive definite; therefore, we can apply theorem A.9 and conclude that the restrictions of the measures and on this vector are both non-degenerate. Then, we can apply lemma 1.11 and conclude that the measures and , being concentrated on the vector X, are equivalent. From now on, we can exclude both cases (i) and (ii) considered in detail above and concentrate on non-degenerate Gaussian processes. Recall that this means that any vector (Xt1, …, Xtn) consists of linearly independent components and consequently has a non-degenerate Gaussian distribution w.r.t. two probability measures, and , on . The idea of the next statement was taken from [HID 93], Chapter 6; however, we present this proof with slight modifications, for the reader’s benefit. Theorem 1.6 Let the process be Gaussian with respect to two measures and on and non-degenerate w.r.t. both measures. If the entropy distance is infinite, then and are singular. Proof Without loss of generality, assume that . Indeed, if this is not the case, then we can take in place of Xt. Then, the covariance matrices are the same, the linear independence of the components is preserved and further proof is carried out in the same way as without this condition. For any finite subset , denote as before the covariance matrices of the vector (Xt1, …, Xtn) with respect to and , respectively. According to our supposition, both covariance matrices are positive definite. Then, it follows from theorem A.9 that there exists an invertible matrix C = (Cij)1 ≤ i, j ≤ n such that CTKX,1C and CTKX,2C are simultaneously diagonal and one of them, say CTKX,2C, is the unit matrix. Set Then, Y = {Yk, 1 ≤ k ≤ n} is a Gaussian sequence consisting of independent, w.r.t. both measures and , random variables, with Set . Denoting by and the restrictions of and to , respectively, and , we obtain from lemmas 1.10 and 1.11 that [1.8] where , . To prove that the measures and are singular, we consider two cases. i) Let there exist two positive constants c1 and c2 such that for any finite subset of . It follows from equality [1.8] that there exist positive constants a and b such that [1.9] [1.10] and [1.11] hold. Denote by A′ the event in defined by [1.12] It follows from Chebyshev’s inequality that with , where in the last inequality we used equations [1.9], [1.10] and [1.11]. Moreover, it follows from equation [1.8] that Thus Again, it follows from Chebyshev’s inequality together with relations [1.9], [1.10] and [1.11] that with the same constant with . From the assertion and lemma 1.13, it follows that for any ϵ > 0, there exist a finite number of points t1, … , tn and an event Aϵ′, given by equation [1.12], such that [1.13] and [1.14] Inequalities [1.13] and [1.14], together with lemma 1.9, imply that and are singular. ii) Let, there exist for any ϵ > 0, both a finite subset of and a number k ∈ {1, … , n} such that σk < ϵ or . Let us consider the latter case. For the event belonging to , we obtain the following bounds and Letting ϵ → 0, we see that and . Again, applying lemma 1.9, we get that and are singular. Theorem 1.7 Two Gaussian measures on a measurable space (Ω, ), induced by the same Gaussian process, are either equivalent or singular. Proof Consider two Gaussian measures and on a measurable space (Ω, ), and let be the entropy distance between them. If , then it follows from definition 1.16 that and . Applying lemma 1.12, we see that and are equivalent. Conversely, if , then it follows from theorem 1.6 that the measures and are singular. Remark 1.16 Theorem 1.7 is very well known as the Feldman–Hajek dichotomy for Gaussian measures, see papers [FEL 58] and [HÁJ 58]. Another proof of this dichotomy, which is obtained with the help of martingale methods, is in [ENG 80]. View chapterExplore book Read full chapter URL: Book2018, Stochastic Analysis of Mixed Fractional Gaussian ProcessesYuliya Mishura, Mounir Zili Chapter Handbook of Dynamical Systems 2006, Handbook of Dynamical SystemsStefano Luzzatto 3.1.1The measure of cylinder sets A straightforward but remarkable consequence of the bounded distortion property is that the measure of cylinder sets tends to zero uniformly. Lemma 3.1 Notice that in the one-dimensional case, the measure of an interval coincides with its diameter and so this implies in particular that the diameter of cylinder sets tends to zero, implying the essential uniqueness of the symbolic representation of itineraries. Proof It is sufficient to show that there exists a constant τ ∈ (0,1) such that for every n ⩾ 0 and every ω(n) ⊂ ω(n–1) we have (3) Applying this inequality recursively then implies \|ω(n) \| ⩽ τ \| ω(n−1) \| ⩽ τ2 \|ω(n−2) \| ⩽ · · · ⩽ τn\|ω0 \| ⩽ τn\|Δ\|. To verify (3) we shall show that (4) To prove (4) let first of all Then from the definition of cylinder sets we have that Fn (ω(n–1)) = Δ and that , and therefore \|Fn (ω(n))\| ⩽ δ or, equivalently, \|Fn (ω(n–1) \ ω(n))\| ⩾ \|Δ\| – δ > 0. Thus, using the bounded distortion property we have and (4) follows choosing . The next property actually follows only from the conclusions of Lemma 3.1 rather than from the bounded distortion property itself. It essentially says that it is possible to “zoom in” to any given set of positive measure. Lemma 3.2 For any ε > 0 and any Borel set A with \|A\| > 0 there exists n ⩾ 1 and such that Proof Fix some ε > 0. Suppose first of all that A is compact. Then, using the properties of Lebesgue measure it is possible to show that for any η > 0 there exists an integer n ⩾ 1 and a collection such that and \|ωη\| ⩽ \|A\|+ η. Now suppose by contradiction that \|ω(n) ∩ A\| ⩽ (1 – ε)\|ω(n) \| for every ω(n) ∈ ωη for any given η > 0. Using that fact that the ω(n) ∈ ωn are disjoint and thus Σ\|ω(n)\| = \|ωη\|, this implies that Since η can be chosen arbitrarily small after fixing ε this gives a contradiction. If A is not compact we can approximate if from below in measure by compact sets and repeat essentially the same argument. View chapterExplore book Read full chapter URL: Handbook2006, Handbook of Dynamical SystemsStefano Luzzatto Chapter Nonstandard Analysis and Measure Theory 2002, Handbook of Measure TheoryPeter A. Loeb PROOF Let B be any measurable set in the continuum product σ-algebra on Ω. The set B is in the σ-algebra generated by a countable number of cylinder sets. Therefore, B is determined on a countable index set C in the sense that for any α and β in ω, if α(t) = β(t) for all t ∈ C, then α ∈ B if and only if β ∈ B. Now suppose that B contains Mh. Take any ω ∈ Ω. Define ω' so that it agrees with ω on C and with h on T\C. Since ω'(t) = h(t) except for countably many t ∈ T, we have ω' ∈ Mh, and thus ω' ∈ B. On the other hand, since ω and ω' agree on C, the property of C implies that ω ∈ B also. This means that B = Ω, and hence the outer measure P(Mh) of Mh is one. View chapterExplore book Read full chapter URL: Book2002, Handbook of Measure TheoryPeter A. Loeb Chapter Handbook of Dynamical Systems 2006, Handbook of Dynamical SystemsStefano Luzzatto Lemma 3.1 Notice that in the one-dimensional case, the measure of an interval coincides with its diameter and so this implies in particular that the diameter of cylinder sets tends to zero, implying the essential uniqueness of the symbolic representation of itineraries. Proof It is sufficient to show that there exists a constant τ ∈ (0,1) such that for every n ⩾ 0 and every ω(n) ⊂ ω(n–1) we have (3) Applying this inequality recursively then implies \|ω(n) \| ⩽ τ \| ω(n−1) \| ⩽ τ2 \|ω(n−2) \| ⩽ · · · ⩽ τn\|ω0 \| ⩽ τn\|Δ\|. To verify (3) we shall show that (4) To prove (4) let first of all Then from the definition of cylinder sets we have that Fn (ω(n–1)) = Δ and that , and therefore \|Fn (ω(n))\| ⩽ δ or, equivalently, \|Fn (ω(n–1) \ ω(n))\| ⩾ \|Δ\| – δ > 0. Thus, using the bounded distortion property we have and (4) follows choosing . The next property actually follows only from the conclusions of Lemma 3.1 rather than from the bounded distortion property itself. It essentially says that it is possible to “zoom in” to any given set of positive measure. View chapterExplore book Read full chapter URL: Handbook2006, Handbook of Dynamical SystemsStefano Luzzatto Chapter Dimension theory 2016, Fractal Functions, Fractal Surfaces, and Wavelets (Second Edition)Peter R. Massopust 5The box dimension of projections In this section the box dimension of a self-affine set in is related to the box dimension of the orthogonal projection onto a coordinate axis. For this purpose, let (x, y) be a Cartesian coordinate system of ; let XA = (A1, …, AN) be the attractor of the recurrent IFS (X, w, P), where X is—without loss of generality—the unit square in , and the maps are affine and of the form (3.48) with 0 < \|ai\|, \|bi\| < 1, . Furthermore, it is assumed that the connection matrix C = (cij) associated with the recurrent IFS (X, w, P) is irreducible (ie, cij = 0 or 1). For a set , denote the orthogonal projection of E onto the y-axis by Ey. Note that (XA)y is the attractor of the recurrent IFS (Xy, wy, P) with given by It follows from a result in Ref. that exists and equals . The next theorem relates the box dimension of (XA)y to that of A. Theorem 64 Let (X, w, P) be a recurrent IFS as defined earlier. Assume w satisfies the OSC Let dbe the unique positive number such that and let d be determined by the formulawhere . If (3.49) then . Proof Since and , it follows that Furthermore, the irreducibility of the connection matrix C implies that each component Ai contains a nonsingular affine image of Aj, and thus , . Now let 0 < ε < 1 be given. Denote by Σε the set of all finite codes i(n) = i(n)(ε) such that (3.50) (3.51) and i = i−1 ∈Σε if . It is easy to see that Σε generates a partition of the code space Σ into cylinder sets i(n). To simplify the notation, letandNote that the codes are “time reversed.” For , let and note that . Let E be a bounded set in , and let be a class of covers such that each consists of (ε × ε)-squares with sides parallel to the coordinate axis. Denote by the cardinality of a minimal cover from , and by the minimum number of compact intervals of length ε needed to cover Ey. The proof is based on the following geometrical observations: Let ε > 0 and let i(n) ∈Σε. Then 1. : and wi(n)X is a rectangle of width ai(n) and height bi(n); 2. : since ai(n) ≤ ε, any cover of Ai(n) that is in may be arranged in such a way that each (ε × ε)-square meets both vertical sides of wi(n)X. (This is possible since each (ε × ε)-square is wider than .) If Q is such a square, then wi(n)−1Q is a rectangle of height ε/bi(n) that meets the lines x = 0 and x = 1. Hence iff . This now defines a one-to-one correspondence between covers of Ai(n) in and covers of consisting of compact intervals of length ε/bi(n). Consequently, To establish the dimension result, the cardinality of Σε is needed. For this purpose the following probability measure on Σε is introduced. Let v = (v1, …, vN)t be a right positive eigenvector of and define a row-stochastic irreducible matrix M = (mij) by Let δ = (δ1, …, δN)t be the unique stationary distribution associated with the matrix M. In other words, Mδ = δ and . Let μM denote the probability measure on Σ generated by M with initial distribution δ. Then for any i(n) ∈Σε, one hasEqs. (3.50), (3.51) now imply thatfor some positive constants c1 and c2. Consequently,The remainder of the proof involves estimates on certain sums. To obtain these estimates a second row-stochastic matrix is now introduced. For 0 ≤ β ≤ 1, let α be the unique positive number such that (3.52) and let u be a positive left eigenvector of with eigenvalue 1. Let be defined byGiven the initial distribution , the induced probability measure satisfiesHence (3.53) Now all the tools are available to prove the dimension result. First it is shown that d is an upper bound for . Note that, since Σε generates a partition of Σ, ThusLet β > dy and α be as in Eq. (3.52). Then it follows from the Perron-Frobenius theorem that . As β > dy, there exists a constant c > 0 with the property thatfor all . Let . Thenfor all , whereas if , then ε/bi(n) ≤ 1 and thus for some c > 0,ThereforeThe last inequality follows from the fact that ai(n) ≤ ε for i(n) ∈Σε. Since and since by Eq. (3.53)the preceding inequality becomesLetting , yields . Finally it is established that d is also a lower bound for . Note that the OSC implies that for all i(n), j(n) ∈Σε with i(n)≠j(n). Thus any (ε × ε)-square in a cover belonging to the class meets at most L = 2a−1 + 4 of the rectangles . Hence The last equality holds since for . Choosing β < dy and proceeding as above yields the result.If (X, w, P) is an IFS—that is, C = (1)—Eq. (3.49) reduces to with . Hence the following corollary holds. Corollary 4 Let (X, w) be an IFS with attractor A and maps of the form Eq. (3.48). Suppose that the OSC and hold. If d is the unique positive solution ofthen . View chapterExplore book Read full chapter URL: Book2016, Fractal Functions, Fractal Surfaces, and Wavelets (Second Edition)Peter R. Massopust Chapter Handbook of Dynamical Systems 2006, Handbook of Dynamical SystemsE. Glasner, B. Weiss Definition 9.3 A set B ∈ χ is uniform if A partition is uniform if, for all N, every set in is uniform. The connection between uniform sets, partitions and unique ergodicity lies in Proposition 8.1. It follows easily from that proposition that if is a uniform partition, say into the sets {P1, P2, …, Pa}, and we denote by also the mapping that assigns to x ∈ X, the index 1 ⩽ i ⩽ a such that x ∈ Pi, then we can map X to by: Pushing forward the measure μ by π, gives π ∘ μ and the closed support of this measure will be a closed shift invariant subset, say . Now the indicator functions of finite cylinder sets span the continuous functions on E, and the fact that is a uniform partition and Proposition 8.1 combine to establish that (E, shift) is uniquely ergodic. This will not be a model for (X, χ, μ, T) unless modulo null sets, but in any case this does give a model for a non-trivial factor of X. Our strategy for proving Theorem 9.2 is to first construct a single non-trivial uniform partition. Then this partition will be refined more and more via uniform partitions until we generate the entire σ-algebra χ. Along the way we will be showing how one can prove a relative version of the basic Jewett–Krieger theorem. Our main tool is the use of Rohlin towers. These are sets B ∈ χ such that for some N, B, T B, …, TN−1 B are disjoint while fill up most of the space. Actually we need Kakutani–Rohlin towers, which are like Rohlin towers but fill up the whole space. If the transformation does not have rational numbers in its point spectrum this is not possible with a single height, but two heights that are relatively prime, like N and N + 1 are certainly possible. Here is one way of doing this. The ergodicity of (X, χ, μ, T) with μ non-atomic easily yields, for any n, the existence of a positive measure set B, such that With N given, choose n ⩾ 10 · N2 and find B that satisfies the above. It follows that the return time is greater than 10 · N2 on B. Let Since ℓ is large (if Bℓ is non-empty) one can write ℓ as a positive combination of N and N × 1, say Now divide the column of sets {Ti Bℓ 0 ⩽ i < ℓ} into uℓ-blocks of size N and υℓ-blocks of size N + 1 and mark the first layer of each of these blocks as belonging to C. Taking the union of these marked levels (Ti Bℓ for suitably chosen i) over the various columns gives us a set C such that rC takes only two values—either N or N + 1 as required. It will be important for us to have at our disposal K–R towers like this such that the columns of say the second K–R tower are composed of entire subcolumns of the earlier one. More precisely we want the base C2 to be a subset of C1—the base of the first tower. Although we are not sure that this can be done with just two column heights we can guarantee a bound on the number of columns that depends only on the maximum height of the first tower. Let us define formally: View chapterExplore book Read full chapter URL: Handbook2006, Handbook of Dynamical SystemsE. Glasner, B. Weiss Chapter Handbook of Dynamical Systems 2006, Handbook of Dynamical SystemsStefano Luzzatto Definition 10 F : Δ → Δ has the (volume) bounded distortion property if there exists a constant such that for all n ⩾ 1 and any measurable subset we have (2) This means that the relative measure of subsets of a cylinder set of any level n are preserved up to some factor under iteration by fn. A crucial observation here is that the constant is independent of n. Thus in some sense the geometrical structure of any subset of Δ reoccurs at every scale inside each partition element of up to some bounded distortion factor. This is in principle a very strong condition but we shall see below that it is possible to verify it in many situations. We shall discuss in the next section some techniques for verifying this condition in practice. First of all we state the first result of this section. View chapterExplore book Read full chapter URL: Handbook2006, Handbook of Dynamical SystemsStefano Luzzatto Chapter Functional Integration and Quantum Physics 1979, Pure and Applied Mathematics 2Construction of Gaussian Processes We begin with some probabilistic notions; see Breiman or Feller for further discussion. A probability measure space is a triple (X, ℱ, μ) of a set X, a σ-field ℱ of subsets of X, and a positive measure μ on X with μ(X) = 1. (Such a measure is called a probability measure.) A real-valued measurable function on X is called a random variable. Given a random variable f, the measure dμf on (−∞, ∞) defined by μf(A) = μ(f−1[A]) is called the probability distribution of f (to be distinguished from the “distribution function,” F(t) = μf(−∞ t] which is common in the literature and which we shall not use). Given n random variables f1,…, fn we define f1 ⊗ … ⊗ fn: X → ℝn by (f1, ⊗ … ⊗ fn)(x) = (f1(x), f2(x), …, fn(x)) and the joint probability distribution μf1…., fn(A) ≡ μ([f1 ⊗ … ⊗fn]−1[A]), a probability measure on ℝn. We use E(A) for μ(A), E(f) for ∫ f dμ, and E(f; A) to denote the integral of f over the set A. There is a reason for introducing the term “random variable” for the equivalent “measurable function,” namely, an implied change of viewpoint. For suppose that one has a “random function on (−∞, ∞),” i.e., for each t ∈ ℝ a random variable ft (with some measurability in t); a functional analyst would most naturally think about ft(·) for each t or perhaps the function ft(x) of two variables. The probabilist's language leads to the consideration of the functions f. (x) for each x. As is usual, one does not distinguish random variables which are equal almost everywhere. Typically, one goes even further and, at least formally, removes the points of X from consideration: two sets in ℱ, A and B, are called equivalent if and only if μ(A Δ B) = 0 where A Δ B = (A\B)∪(B\A). The equivalence classes are called events and the family of events, ℱ/ℒμ, inherits the notions of intersection and union. A random variable is, more properly, a map (f−1) from ℬ, the Borel subsets of ℝ, to ℱ/ℒμ preserving countable unions and intersections and μf is the composition of f−1: ℬ → ℱ/ℒμ and μ: ℱ/ℒμ → [0, ∞). An isomorphism of two probability measure spaces (X, ℱ, μ) and (X′, ℱ′, μ′) is a map T: ℱ/ℒμ → ℱ′/ℒμ, which is a bijection respecting countable unions and intersections and with μ′ T = μ. Random variables f on X and f' on X' are said to correspond under T if and only if T f−1 = (f')−1. Having pointed out the need for the above general abstract framework, we will usually be colloquial and talk about points, about random variables as functions, etc. Let ℒ be an index set for a family of random variables. If I ⊂ ℒ is a set with n ≡ #(I) < ∞, we have a joint probability distribution μI on ℝn for {fα}α∈I. The measures μI are consistent in the sense that if I ⊂ I', then μI, can be obtained from μI' by “integrating out” the variables in I'I; i.e., if {f1,…fn} ≡ {fα}α∈I and {f1,…, fm} = {fα}α∈I' (m ≥ n), then μI(A) = μI'(A × ℝm-n). A fundamental result in probability theory is: Theorem 2.1 (Kolmogorov's theorem) Let ℒ be a countable set and let a probability measure μI, on ℝ#(I) be given for each finite set I ⊂ ℒ so that the family of μI's is consistent. Then, there is a probability measure space (X, ℱ, μ) and random variables {fα}α∈ℒ so that μI is the joint probability distribution of {fα}α∈I. Moreover, this space is unique in the sense that if (X′, ℱ′, μ′) and {f'α}α∈ℒ also have these properties and if ℱ (and respectively, ℱ′) is the smallest σ-field with respect to which the fα (respectively, f′α) are measurable, then there is an isomorphism of the probability measure spaces under which each fα corresponds to f'α. Proof The existence and uniqueness aspects are quite distinct. To prove existence we will take X = ℒ where = ℝ ∪ {∞} is the one-point compactification of ℝ. Since X is compact, we can use the Reisz–Markov theorem to construct Baire measures (this finesses the proof of countable additivity, or more properly, places it on the shoulders of the proof of the Reisz–Markov theorem; see, e.g., Berberian 11 for this proof). Let Cfin(X) denote the family of continuous functions of finitely many coordinates {xα}α∈I (as I runs through all finite subsets) and let if f is a function of {xα}α∈I. ℓ is a well-defined positive linear functional on Cfin(X) because of the consistency conditions and clearly \|ℒ(f)\| ≤ \|\|f\|\|∞. By the Stone–Weierstrass theorem, Cfin(X) is dense in C(X), so ℒ extends uniquely to a positive linear functional on C(X). Therefore, by the Reisz–Markov theorem, there is a Baire probability measure, μ, on X with Define fα(x) = xα if xα ≠ ∞ and zero if xα = ∞. Then clearly the dμI are the joint distribution of the {fα}α∈I. This proves existence. To prove uniqueness, we first show that the condition that the fα's generate ℱ implies that the bounded functions of finitely many fα's are dense in L2(X, dμ). For let ℋ be the closure of these functions in L2 and let χA in ℋ be a characteristic function of some set A. Then we claim that there are An whose characteristic functions, χn, depend on only finitely many fα's with χn χA. For let gn χA and by passing to a subsequence, suppose that gn → χA pointwise almost everywhere. Let . Then χn → χA pointwise almost everywhere and so in L2 by the dominated convergence theorem. Since χn → χA, the set of A's with χA in ℋ is closed under finite intersections since this is true of the generating sets. Clearly it is closed under complements and it is closed under countable unions by the monotone convergence theorem. Thus, the set of A with χA in ℋ is a σ-field, and so it is ℱ. Thus ℋ = L2(X, dμ). Given two models (X, ℱ, μ), {fα} and (X', ℱ′, μ) {f'α}, define U: L2(X, dμ) → L2(X′, dμ′) by U[F(fα)] = F(f′ α), which is well defined and unitary and extends to L2 by the above density result. It is not hard to see that U takes characteristic functions into characteristic functions (use χn → χA) so that T defined by U(χA) = χT(A) defines a map of ℱ/ℒμ to ℱ′/ℒμ′. This T is easily seen to be an isomorphism under which fα corresponds to f′ α. (X, ℱ, μ) is called a model for the μI. If we take ℒ = {0, 1, 2,…}, then by the above ℝℒ can be taken as a model. (In the above, ℒ was used a priori, but one notes that μ{x\|some xα = . It is often useful to know if some nice subset of ℝℒ has measure one. Two particularly useful subsets are and ′ defined as follows: for m ∈ ℤ, ≡ ∩m ′m with the Fréchet topology induced by the \|\|·\|\|m and ′ ≡ ∪m m. ′ is the topological dual of if x ∈ ′ is associated to the linear functional The cylinder sets σ-field on ′ is the smallest σ-field with respect to which the functionals x → Lx(y) are measurable for each y ∈ . It is easy to see that this is identical to the σ-field generated by viewing ′ as a subset of ℝℒ with its natural σ-field. Given a probability measure, μ, on ′ with the cylinder set σ-field, we can define the function Φ on by (2.1) called the Fourier transform of μ. Often this is called the characteristic function of μ. To avoid confusion with the characteristic function of a set, we only use the former name in this book. Φ clearly has the following three properties [(c) follows from the dominated convergence theorem and the continuity of Lx(·)]: (a) : Φ(0) = 1. (b) : Φ is positive definite; i.e., given z1,…, zn ∈ ℂ and y1,…, yn ∈ , (c) : Φ is continuous when is given its Fréchet topology. The following infinite-dimensional analog of Bochner's theorem (see for a proof of Bochner's theorem) is a special case of a theorem of Minlos . The beautiful proof we use we learned from van Hemmen (although the literature on Minlos' theorem is so large, it may well be older): Theorem 2.2 (Minlos’ theorem for ′) A necessary and sufficient condition for a function Φ on to be the Fourier transform of a probability measure on ′ is that it obey (a)–(c). Proof Necessity has already been discussed, so suppose that (a)–(c) hold. By Bochner's theorem, for any finite I, there is a measure μI on ℝI with By the uniqueness part of Bochner's theorem, the μ's are consistent, so by our proof of Kolmogorov's theorem, there is a measure μ on ℝℒ so that (2.1) holds for all y ∈ with yα nonzero for only finitely many α's. Each m and thus ′ is measurable in ℝℒ. If we show that μ(′) = 1, then μ may be restricted to ′ and (2.1) will extend to all of . The proof is thus reduced to using (c) to show that μ(′) = 1. Given ɛ, we can find m and δ so that \|\|y\|\|m ≤ δ implies that \|Φ(y) − 1\| ≤ ɛ. We claim that (2.2) for all y ∈ . For (2.2) holds if \|\|y\|\|2m ≤ δ2, since \|1 – Φ(y)\| ≤ ɛ in that case, and it holds if \|\|y\|\|2m > δ2, since Re Φ(y) ≥ − 1 for all y [we use here the fact that condition (b) implies that \|Φ(y)\|≤ Φ (0)]. Fix a sequence {qn} and for α and N let dσα,N be the measure on ℝN+1: Notice that (for all integrals over ℝN+1): (2.3a) (2.3b) Integrating (2.2) with respect to dσα,N and using (2.3), we find that (2.4) Choose qn so that and take N → ∞, using the monotone convergence theorem: Now take α to zero, using the monotone convergence theorem again to obtain Choosing qn = (1 + n2)−m−1, we see that so that, if we take ɛ to zero, we see that μ(′) = 1. The above theorem only depends on the structure of and ′ as a particular topological vector space and its dual. But is topologically isomorphic to (ℝ) under the Hermite expansion f ∈ → {fn} with where Φn(x) = (2nn!)-1/2(-1)nϕ-1/4ex2/2(d/dx)ne−x2, the nth harmonic oscillator wave function (see or , Appendix to Section V.3]) and also to (ℝv) for each v. As a result, Theorem 2.2 immediately extends to the following. Definition cylinder set measure on ′(ℝv) is a measure on the σ-field generated by the functions T → T(ϕ) as ϕ runs through all of (ℝv). Theorem 2.3 (Minlos‘ theorem for ′) A necessary and sufficient condition for a function Φ(·) on (ℝv) to be the Fourier transform of a cylinder set probability measure on ′(ℝ) is that Φ(0) = 1, Φ be positive definite, and Φ be continuous in the Fréchet topology on (ℝv). Remark The above proof shows a little more; namely, if Φ is continuous in the norm associated to sm, then μ is concentrated on any set of the form (ɛ > 0), For example, these methods imply that if ω(t) is a Wiener path and Φ ∈ C∞0, then Φω is in the domain of \|d/dx\|1/2−ɛ This is just short of implying continuity (any function in \|d/dx\|1/2+ɛ is continuous), so we will need a subtler argument to get continuity (see Section 5). In the above, we considered countably many random variables, but much of the discussion is applicable to arbitrary families, e.g., families indexed by t ∈ (a, b) ⊂ ℝ (“stochastic processes”). In particular, Kolomogorov's theorem, in the form we give it (and its proof) extends to arbitrary families. We can use ℝℒ for X; however, the use of ℝℒ does not extend since {x\|xα = ∞ for some α} is no longer even measurable! There is one subtlety (“the problem of versions”) associated with this extension that we shall discuss now and henceforth generally brush under the rug. The subtlety can be illustrated in the following trivial example: Example Let (X, ℱ, μ) be [0, 1] with its Borel sets and Lebesgue measure. For each t ∈ [0, 1], let q(t) be the random variable on X which is identically one and let (t) be defined by Then for each fixed t, q and are equal almost everywhere so that, from the Kolmogorov theorem point of view, q and are identical families. But notice that for every x ∈ X, the map t ↦ q(t)(x) is continuous while t ↦ (t)(x) is discontinuous! This example illustrates dramatically that if is a family of random variables on (X, ℱ, μ) and ℱ is the minimal σ-field, then {x\|t → q(t)(x) is continuous} may not be measurable. This example is especially disturbing because one of the most celebrated results in the development of the Wiener process is that the “paths,” q(t)(x) are continuous in t for almost every x. One can, for example, define the Wiener process for t ≤ 1 a priori on [0, 1] with the Baire field, in which case it does not strictly make sense to say that t → q(t) is pointwise almost everywhere continuous. (It does make sense to say that t → q(t) is continuous in L2-norm and this will be trivial to prove.) There are a variety of ways around this problem: (a) : (The one we will use) Suppose we show that for almost every x ∈ X, there is a Cx so that (2.5) for some fixed δ and all rational t, s in [0, 1]. Since only countably many t and s are involved, (2.5) is “version independent.” Choose any explicit X for the process and consider the q(t) for t rational and let X0 = {x ∈ X\|(2.5) holds}. For x ∈ X0, q(t)(x) defined for t rational extends to a unique continuous function (t)(x) for all t ∈ [0, 1] and thereby, we can define random variables (t) on X0 and extend them to be zero on X\X0. Clearly t → (t)(x) is continuous for each x ∈ X. Moreover, the joint distributions of (q(t1),…, q(tn)) agree with those for ( (t1),…, (tn)). This is obvious for ti rational and extends to all t by the pointwise continuity of and the L2 continuity of q. We thus have a “version of q′ with continuous ”paths.“ This point of view is further discussed in . (b) : One directly constructs a probability measure on the space of continuous functions and defines this to be the Wiener process; the above philosophy is not then directly relevant. One disadvantage of this method is that a detailed proof of countable additivity can get somewhat more involved than in our discussion. This procedure is used in ; see also Section 17. (c) : Use Minlos' theorem to construct a measure on ′(ℝ) so that the joint distribution of T(f1),…,T(fn) agrees with that of ƒ q(t)f1(t) dt,…, ƒ q(t) fn(t) dt. Then find in ′ a measurable family, F, of distributions equal (as distributions) to continuous functions with μ(′F) = 0. The Lévy 168–Ciesielski proof (described, e.g., in ) can be interpreted in this way. (d) : Use the compact model of [0, 1] but use the Borel field rather than the Baire field (see, e.g., for the distinction) in which case the Hölder continuous functions are Borel measurable. The construction we used in Theorem 2.2 only defines a Baire measure which will have many Borel extensions, but precisely one regular Borel extension. In this extension, the Hölder continuous functions have measure one. This is a point of view advocated by Nelson . While the words leading to the final result are different in the above arguments, the end definition of Wiener measure as a measure on the space of continuous functions is the same. A particularly important class of abstract processes are the Gaussian processes. It is easiest to describe them by using the Fourier transforms of the joint probability distributions; i.e., One reason for this is that dμf1,…,fn is consistent with dμf1,…,fn (m ≥ n) if and only if We call f a Gaussian random variable of variance a if and only if We call f1,…,fn jointly Gaussian with covariance {aij} (aij = aji) if and only if (2.6) aij is only dependent on fi and fj, since If {aij}1≤i≤j≤n is a nonsingular matrix with inverse b, then (2.6′) Given an n × n real symmetric matrix aij, it will be the covariance of some jointly Gaussian random variables if and only if a is positive semidefinite. For, in that case, (2.6) defines a positive definite function and so, by Bochner's theorem, a measure via Fourier transformation. Occasionally (but not here), one discusses Gaussian random variables with mean mi and covariance aij in which case those described above are Gaussian random variables of mean zero. The right-hand side of (2.6) is then replaced by Throughout this book, the phrase “f and g are Gaussian random variables” will be used to indicate they are jointly Gaussian with mean zero. Theorem 2.3A Let ℋ be a separable real Hilbert space. Then there exists a probability measure space (X, ℱ, μ) and for each v ∈ ℋ a random variable, Φ(v), so that v ↦ Φ(v) is linear and so that for any v1,…,vn ∈ ℋ, (Φ(v1),…, Φ(vn)) are jointly Gaussian with covariance vi, vj (·,· ≡ inner product on ℋ). The same kind of uniqueness as occurs in the Kolmogorov theorem holds here. Remark We will call {Φ(v)} the Gaussian process with covariance ·,·. Proof Pick v1,…,vn,… an orthonormal basis for ℋ. Let c(t) be defined for any sequence t1,…,tn,… eventually zero, by By the above remarks on when is positive definite and on consistency, we can apply Kolmogorov's theorem to construct (X, ℱ, μ) and (Φ1,…, Φn,…, jointly Gaussian with covariance δij (this is somewhat circumlocutory, one can just take X = ℝ∞, Φi = xi, and directly). Now given a finite sum . Then ƒ\|Φ(v)\|2 dμ = \|\|v\|\|2, so, by continuity, Φ extends from finite sums to a map from ℋ to L2(X, dμ). It is easy to use continuity to see that so the Φ's are jointly Gaussian with the proper covariance. Uniqueness of the process restricted to finite sums of the vi, follows from the uniqueness part of Kolmogorov's theorem and general uniqueness follows from the L2 continuity deduced above. Corollary 2.4 Let c(t, s) be a jointly continuous real-valued function on K × K where K is a separable topological space. Suppose that for any Φ1,…, tn ∈ K, c(ti, tj) is a positive semidefinite matrix. Then, there exists an essentially unique measure space (X, ℱ, μ) and a random variable q(t) for each t ∈ K so that the q(t) are jointly Gaussian with covariance c. Proof For each t ∈ K, introduce a formal symbol δt and consider the vector space of finite sums ∑ aiδti. Define an inner product on this space by By hypothesis, this is a positive semidefinite inner product, so by the usual procedure of quotienting by null vectors and completing we can form a Hilbert space, ℋ. The separability of K and continuity of c imply that ℋ is separable. Let ϕ be the Gaussian process with covariance (·, ·)ℋ and let q(t) = ϕ(δ1). Warning: q(t) is not, in general, linear in t. We want to give one final result in the abstract theory of Gaussian processes, a result due to Feldman [80a] and Hajek [121a]. It is essentially equivalent to a result of Shale about the implementability of Bogoliubov (≡ symplectic) automorphisms in the theory of the free Bose field. Most field theorists are unaware of the work of Feldman-Hajek and most probabilists of the work of Shale! Theorem 2.5 Let ℋ be a real Hilbert space and let A be a bounded positive invertible operator on ℋ. A necessary and sufficient condition for there to exist a single measure space (X, ℱ) with functions {Φ(v)}v∈ℋ and two mutually absolutely continuous probability measures dμ and dv so that the Φ(v) are jointly Gaussian on (X, ℱ, μ) with covariance (·, ·)ℋ and jointly Gaussian on (X, ℱ, v) with covariance (·, A ·)ℋ is that A − 1 be Hilbert–Schmidt. A proof may be found, for example, in . It is quite easy to prove; indeed, we will essentially prove it as Lemma 18.6. The useful direction is that A − 1 Hilbert–Schmidt implies mutual absolute continuity. The idea is to choose a basis for ℋ with AΦn = αnΦn. Then formally and . The condition can be used to show that converges in L2(X, dμ). Of course, the square of the limit turns out to be dv/dμ. There is an extensive probabilistic literature on properties of Gaussian processes; see, e.g., [61a, 82b, 82c, 85a, 180, 195a]. View chapterExplore book Read full chapter URL: Book series1979, Pure and Applied Mathematics Review article Elicitation of multivariate prior distributions: A nonparametric Bayesian approach 2010, Journal of Statistical Planning and InferenceFernando A. MoalaAnthony O’Hagan In general, is a joint probability, which may also not be easy for the expert to specify. However, if A is a cylinder set, is a marginal probability, which will usually be easier to specify. Marginal probabilities may also be specified implicitly as marginal quantiles; for instance, by specifying that u is the upper quartile of her distribution for , the expert implies that , where . We will illustrate in Section 5 how in practice we can obtain good results by eliciting marginal probabilities supplemented by a small number of joint probabilities. It is also possible to incorporate elicited values for means or other moments, and also modes, within our framework (Oakley and O’Hagan, 2007), but for simplicity we assume that the elicited summaries from the expert comprise a data set where is the expert's assessment of the chance that . View article Read full article URL: Journal2010, Journal of Statistical Planning and InferenceFernando A. MoalaAnthony O’Hagan Related terms: Brownian Motion Probability Theory Bernoulli Shift Generated -Field Markov Partition Probability Measure Marginals Preserving System Proposition View all Topics
4571	https://artofproblemsolving.com/wiki/index.php/Diophantine_equation?srsltid=AfmBOooiIyEvyAdjltJ3pPDnFUsZ3Ftu6KPQ_c7AT0lh-FdWWn_gHulN	Art of Problem Solving Diophantine equation - 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 Diophantine equation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Diophantine equation A Diophantine equation is an equation relating integer (or sometimes natural number or whole number) quanitites. Finding the solution or solutions to a Diophantine equation is closely tied to modular arithmetic and number theory. Often, when a Diophantine equation has infinitely many solutions, parametric form is used to express the relation between the variables of the equation. Diophantine equations are named for the ancient Greek/Alexandrian mathematician Diophantus. Contents [hide] 1 Linear Combination 2 Pythagorean Triples 2.1 Method of Pythagoras 2.2 Method of Plato 2.3 Babylonian Method 3 Sum of Fourth Powers 4 Pell Equations 5 Methods of Solving 5.1 Coordinate Plane 5.2 Modular Arithmetic 5.3 Induction 5.4 General Solutions 6 Fermat's Last Theorem 7 Problems 7.1 Introductory 7.2 Intermediate 7.3 Olympiad 8 References 9 See also Linear Combination A Diophantine equation in the form is known as a linear combination. If two relatively prime integers and are written in this form with , the equation will have an infinite number of solutions. More generally, there will always be an infinite number of solutions when . If , then there are no solutions to the equation. To see why, consider the equation . is a divisor of the LHS (also notice that must always be an integer). However, will never be a multiple of , hence, no solutions exist. Now consider the case where . Thus, . If and are relatively prime, then all solutions are obviously in the form for all integers . If they are not, we simply divide them by their greatest common divisor. See also: Bézout's identity. Pythagorean Triples Main article: Pythagorean triple A Pythagorean triple is a set of three integers that satisfy the Pythagorean Theorem, . There are three main methods of finding Pythagorean triples: Method of Pythagoras If is an odd number, then is a Pythagorean triple. Method of Plato If , is a Pythagorean triple. Babylonian Method For any (), we have is a Pythagorean triple. Sum of Fourth Powers An equation of form has no integer solutions, as follows: We assume that the equation does have integer solutions, and consider the solution which minimizes . Let this solution be . If then their GCD must satsify . The solution would then be a solution less than , which contradicts our assumption. Thus, this equation has no integer solutions. If , we then proceed with casework, in . Note that every square, and therefore every fourth power, is either or . The proof of this is fairly simple, and you can show it yourself. Case 1: This would imply , a contradiction. Case 2: This would imply , a contradiction since we assumed . Case 3: , and We also know that squares are either or . Thus, all fourth powers are either or . By similar approach, we show that: , so . This is a contradiction, as implies is odd, and implies is even. QED [Oops, this doesn't work. 21 (or ) are equal to and not even...] Pell Equations Main article: Pell equation A Pell equation is a type of Diophantine equation in the form for natural number. The solutions to the Pell equation when is not a perfect square are connected to the continued fraction expansion of . If is the period of the continued fraction and is the th convergent, all solutions to the Pell equation are in the form for positive integer . Methods of Solving Coordinate Plane Note that any linear combination can be transformed into the linear equation , which is just the slope-intercept equation for a line. The solutions to the diophantine equation correspond to lattice points that lie on the line. For example, consider the equation or . One solution is (0,1). If you graph the line, it's easy to see that the line intersects a lattice point as x and y increase or decrease by the same multiple of and , respectively (wording?). Hence, the solutions to the equation may be written parametrically (if we think of as a "starting point"). Modular Arithmetic Sometimes, modular arithmetic can be used to prove that no solutions to a given Diophantine equation exist. Specifically, if we show that the equation in question is never true mod , for some integer , then we have shown that the equation is false. However, this technique cannot be used to show that solutions to a Diophantine equation do exist. Induction Sometimes, when a few solutions have been found, induction can be used to find a family of solutions. Techniques such as infinite Descent can also show that no solutions to a particular equation exist, or that no solutions outside of a particular family exist. General Solutions It is natural to ask whether there is a general solution for Diophantine equations, i.e., an algorithm that will find the solutions for any given Diophantine equations. This is known as Hilbert's tenth problem. The answer, however, is no. Fermat's Last Theorem Main article: Fermat's Last Theorem is known as Fermat's Last Theorem for the condition . In the 1600s, Fermat, as he was working through a book on Diophantine Equations, wrote a comment in the margins to the effect of "I have a truly marvelous proof of this proposition which this margin is too narrow to contain." Fermat actually made many conjectures and proposed plenty of "theorems," but wasn't one to write down the proofs or much other than scribbled comments. After he died, all his conjectures were re-proven (either false or true) except for Fermat's Last Theorem. After over 350 years of failing to be proven, the theorem was finally proven by Andrew Wiles after he spent over 7 years working on the 200-page proof, and another year fixing an error in the original proof. Problems Introductory Two farmers agree that pigs are worth dollars and that goats are worth dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way? (Source) Intermediate Let be a polynomial with integer coefficients that satisfies and Given that has two distinct integer solutions and find the product . (Source) Olympiad Determine the maximum value of , where and are integers satisfying and . (Source) Solve in integers the equation . References Proof of Fermat's Last Theorem See also Number Theory Pell equation Retrieved from " Category: Number theory 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.
4572	https://math.stackexchange.com/questions/4744521/finding-the-farthest-point-on-ellipse-from-origin	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 Finding the farthest point on ellipse from origin? [closed] Ask Question Asked Modified 2 years, 2 months ago Viewed 705 times 2 $\begingroup$ I need to find the farthest point of the ellipse from the origin, I thought it should be on the $x$-axis but I don't know how to prove it. Maybe a point has a greater distance due to the Pythagorean theorem, for example, one value can cause a bigger distance because of its y value, etc. I thought that I have to calculate it from the formula but couldn't understand how. This is the formula of my ellipse: $$ \dfrac{\left(x-3\right)^2}{25}+\frac{y^2}{16}=1 $$ This is the original image of the question, I need to get the farther point on the ellipse from the center of the moon (0,0) (origin). geometry algebra-precalculus Share edited Jul 29, 2023 at 20:10 basilinniabasilinnia asked Jul 29, 2023 at 17:54 basilinniabasilinnia 10377 bronze badges $\endgroup$ 9 3 $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ Another User – Another User 2023-07-29 17:59:30 +00:00 Commented Jul 29, 2023 at 17:59 1 $\begingroup$ Hint. Your guess is a reasonable one. What is that point's distance from the origin? If it is $r$, then the equation of the circle at that distance is $x^2+y^2 = r^2$. Can you tell if that circle intersects the ellipse in more than that one point? ETA: The first comment is well-taken. What are you studying? What do you know how to do? What is the distance formula? Etc. $\endgroup$ Brian Tung – Brian Tung 2023-07-29 17:59:49 +00:00 Commented Jul 29, 2023 at 17:59 $\begingroup$ farthest points on an ellipse are just the points on major axis, look at auxillary circle for example; the proof is that each point on ellipse can be written as (acosx,bsinx) so we wanna maximise $a^2cos^2x + b^2sin^2x=a^2-(a^2-b^2)sin^2x$ which because $a>b$ is clearly minimised for x=0 or 180 in which case points are $(\pm a,b)$ $\endgroup$ Aditya_math – Aditya_math 2023-07-29 18:00:55 +00:00 Commented Jul 29, 2023 at 18:00 $\begingroup$ have you tried drawing the ellipse ? once the picture is drawn the answer is not very difficult $\endgroup$ Thomas – Thomas 2023-07-29 18:05:30 +00:00 Commented Jul 29, 2023 at 18:05 $\begingroup$ @Aditya_math: "farthest points on an ellipse are just the points on major axis"; that's not necessarily true for an ellipse that is not centered at the origin. For example, if the $25$ is replaced by $4$, the furthest point is not a point on either the major or the minor axis. $\endgroup$ Brian Tung – Brian Tung 2023-07-29 18:06:07 +00:00 Commented Jul 29, 2023 at 18:06 \| Show 4 more comments 4 Answers 4 Reset to default 4 $\begingroup$ This might be an overkill for a problem in 2D with a simple geometry, where a graphical solution is possible, but one may use the Lagrangian approach here. So you are trying to maximize the distance from origin, i.e. $f(x,y) = x^2 + y^2$, where $(x,y)$ are supposed to lie on the ellipse, i.e. $g(x,y) = (x-3)^2/25 + y^2/16 - 1 = 0$. We form the Lagrangian $L(x,y,\lambda) = f(x,y)+\lambda g(x,y)$. Constraint qualifications are satisfied for this problem, so the points of maxima must be critical points of the Lagrange function. Therefore $$ \begin{aligned} \partial_x L &=& 2x + 2\lambda(x-3)/25 &= 0,\ \partial_y L &=& 2y + 2\lambda y/16 &= 0,\ \partial_\lambda L &=& (x-3)^2 + y^2/16 -1 &= 0. \end{aligned} $$ If $\lambda=0$ then also $(x,y)=0$ from the first two equations, which does not satisfy the third one. We can isolate $x$ in the first equation and $y$ in the third: $x = \frac{6\lambda/25}{2+2\lambda/25}$, assuming $\lambda/25 \neq -1$; $y=0$, assuming $\lambda/16 \neq -1$. This is then substituted into the third equation. When $\lambda/16 \neq -1$ we have $y=0$ from the second equation, $x-3 = \pm 5$ from the third, so $x=8$ or $x = -2$. The corresponding values for $\lambda$ can be computed from the first equation. This gives us two critical points: $(x,y,\lambda) = (8,0,-40)$, and $(x,y,\lambda)=(-2,0,-10)$. When $\lambda/16 = -1$, i.e., $\lambda = -16$, we get $x = -16/3$ from the first equation, and subsequently $y^2<0$ from the third, which is not feasible. All in all, out of the two critical points, $(x,y)=(8,0)$ is furthest from the origin (point of global maximum). Here is the picture Share edited Jul 29, 2023 at 18:56 answered Jul 29, 2023 at 18:40 rafexiaprafexiap 69033 silver badges88 bronze badges $\endgroup$ Add a comment \| 4 $\begingroup$ The points on your ellipse can be parameterized by $x=3 + 5\cos\theta$ and $y=4\sin \theta$. The squared distance from the origin is $$ \begin{eqnarray} (3 + 5\cos\theta)^2 + (4\sin\theta)^2&=&9+30\cos\theta+25\cos^2\theta+16\sin^2\theta\ &=&25+30\cos\theta+9\cos^2\theta\ &=&(5+3\cos\theta)^2, \end{eqnarray} $$ which is clearly maximized when $\cos\theta=+1$. The corresponding point is $(x,y)=(3+5\cos\theta,4\sin\theta)=(8,0)$. Share answered Jul 29, 2023 at 18:52 mjqxxxxmjqxxxx 43.4k33 gold badges6464 silver badges115115 bronze badges $\endgroup$ Add a comment \| 2 $\begingroup$ Parametrization $$ x=3 + 5\cos\theta, ~y= 4\sin \theta $$ Distance squared from origin is $$ \rho^2= x^2+y^2 =(3 + 5\cos\theta)^2 $$ Extreme $\rho$ s at $ \cos \theta = \pm 1, ~\theta =(0, \pi)$ which occur at $$ y=0,~x= 3\pm 5 = (8,-2) $$ on $x$-axis. Share answered Jul 29, 2023 at 20:39 NarasimhamNarasimham 42.5k77 gold badges4646 silver badges112112 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ What you're looking for is optima of $f(x,y)=x^2+y^2$ under the constraint of the equation of the ellipse, that we note $g(x,y)=0$. Optima are reached at points where the gradients of both functions $f$ and $g$ are collinear. Doing the algebra, one finds the equation $y\cdot(3x+16)=0$. So indeed, you have optima for $y=0$ but also for $x=-16/3$. However, the second case yields non real complex solutions. The farthest point from the origin on the ellipse is thus (8,0). Share answered Jul 29, 2023 at 18:07 Eric BrierEric Brier 55122 silver badges77 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry algebra-precalculus 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 Related Calculating Distance of a Point from an Ellipse Border 1 Finding a polar equation for an ellipse 1 Calculate tangent point on ellipse Calculating the distance from a point on an ellipse and focal points 0 length of an arm from a point along major axis of ellipse 0 Calculate Distance from Point to Ellipse edge 0 Angle of an ellipse point shifted up in respect to a similar at origin 0 Finding the coordinates of the farthest point from a point inside an ellipse without knowing a or b? Hot Network Questions Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Traversing a curve by portions of its arclength What is the meaning of 率 in this report? Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Is it safe to route top layer traces under header pins, SMD IC? Interpret G-code How to locate a leak in an irrigation system? I'm having a hard time intuiting throttle position to engine rpm consistency between gears -- why do cars behave in this observed way? Why do universities push for high impact journal publications? how do I remove a item from the applications menu Gluteus medius inactivity while riding How to use \zcref to get black text Equation? Why is the definite article used in “Mi deporte favorito es el fútbol”? Identifying a movie where a man relives the same day Who is the target audience of Netanyahu's speech at the United Nations? The rule of necessitation seems utterly unreasonable A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man Bypassing C64's PETSCII to screen code mapping How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? Can I use the TEA1733AT for a 150-watt load despite datasheet saying 75 W? How to home-make rubber feet stoppers for table legs? My dissertation is wrong, but I already defended. How to remedy? Can a GeoTIFF have 2 separate NoData values? "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf more hot questions
4573	https://www.labcorp.com/content/dam/labcorp/drupal/metachromatic%20leukodystrophy.pdf	InheritestSM is a service mark of Laboratory Corporation of America® Holdings. ©2012 Laboratory Corporation of America® Holdings. All rights reserved. What is Metachromatic Leukodystrophy? Metachromatic leukodystrophy (MLD) is an inherited disease with a variable age of onset and is characterized by the progressive loss of motor skills and intellectual function. It involves defects in the enzyme called arylsulfatase A (ARSA), which breaks down molecules called sulfatides (a type of lipid found in cell membranes), particularly within the nervous system. Symptoms associated with MLD are due to the build-up of sulfatides that destroy myelin (the protective material that surrounds nerve cells). Metachromatic leukodystrophy is also known as arylsulfatase A deficiency.1 What are the symptoms of Metachromatic Leukodystrophy and what treatment is available? Metachromatic leukodystrophy is a progressive disease that occurs in three types, based primarily on the age of onset.2 Type (rate of progression) Typical Age of Onset Initial Symptoms Age at Death Late-infantile (steady progression) 1-2 years of age Clumsiness, frequent falls, toe walking, & slurred speech ~3.5 years after onset of symptoms, occasionally second decade of life Juvenile (slower progression) 4-14 years of age Decline in school performance & behavioral problems ~10-20+ years after onset of symptoms Adult (periods of relative stability between periods of decline) After puberty (~14 years), but may not occur until the 40s-50s Decline in school or work performance, personality changes, substance abuse issues, emotional changes ~20-30 years after onset of symptoms Shared symptoms of MLD include 2: • Hypotonia (low muscle tone) that progresses with age • Loss of coordination • Slurred speech • Pain in arms and legs • Peripheral neuropathy (loss of sensation in hands and feet) • Weak bladder • Seizures • Vision and hearing loss • Nerve damage • Eventual wheelchair dependence InheritestSM is a service mark of Laboratory Corporation of America® Holdings. ©2012 Laboratory Corporation of America® Holdings. All rights reserved. There is no cure for MLD. Treatment includes supportive care for symptoms and may include seizure management and physical therapy. Additional therapies, including stem cell transplant, may be considered.2 How is Metachromatic Leukodystrophy inherited? Metachromatic leukodystrophy is an autosomal recessive disease caused by mutations in the ARSA gene.1 An individual who inherits one copy of an ARSA gene mutation is a “carrier” and is not expected to have related health problems. An individual who inherits two mutations in this gene, one from each parent, is expected to be affected with MLD. If both members of a couple are carriers, the risk for an affected child is 25% in each pregnancy; therefore, it is especially important that the reproductive partner of a carrier be offered testing. Who is at risk for Metachromatic Leukodystrophy? Metachromatic leukodystrophy can occur in individuals of all races and ethnicities. In the Japanese population, the incidence is approximately 1 in 70,0003, with a calculated carrier frequency of 1 in 132. In Europe, the incidence is estimated to be 1 in 80,0004, with a calculated carrier frequency of 1 in 141. Having a relative who is a carrier or who is affected can increase an individual’s risk of being a carrier. Consultation with a genetics health professional may be helpful in determining carrier risk and appropriate testing. What does a positive test result mean? If a gene mutation is identified, an individual should speak to a physician or genetics health professional about the implications of the result and appropriate testing for the reproductive partner and at-risk family members. What does a negative test result mean? A negative result reduces, but does not eliminate, the possibility that an individual carries a gene mutation. The likelihood of being a carrier is also influenced by family history, medical symptoms, and other relevant test results. Where can I get more information? • MLD Foundation: • The Evanosky Foundation: • United Leukodystrophy Foundation: References 1. Metachromatic leukodystrophy. Genetics Home Reference Available at: Accessed: Feb 20, 2012 2. Fluharty AL. Arylsulfatase A Deficiency. GeneReviews. Available at Accessed: Feb 20, 2012 3. Eto Y et al Molecular characteristics in Japanese patients with lipidosis: Novel mutations in metachromatic leukodystrophy and Gaucher disease Mol Cell Biochem 1993; 119:179-184. 4. Lugowska A et al Population Carrier Rates of Pathogenic ARSA Gene Mutations: Is Metachromatic Leukodystrophy Underdiagnosed? PLoS ONE 2011; 6(6):e20218.
4574	https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Book%3A_Electric_Circuits_IV_-_Digital_Circuitry_(Kuphaldt)/07%3A_Boolean_Algebra	7: Boolean Algebra - Workforce LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode Book: Electric Circuits IV - Digital Circuitry (Kuphaldt) Electronics Technology { "7.01:Introduction_to_Boolean_Algebra" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_Boolean_Arithmetic" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Boolean_Algebraic_Identities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Boolean_Algebraic_Properties" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.05:_Boolean_Rules_for_Simplification" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.06:_Circuit_Simplification_Examples" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.07:_The_Exclusive-OR_Function-_The_XOR_Gate" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.08:_DeMorgans_Theorems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.09:_Converting_Truth_Tables_into_Boolean_Expressions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Numeration_Systems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Binary_Arithmetic" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Logic_Gates" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Switches" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Electromechanical_Relays" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Ladder_Logic" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Boolean_Algebra" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Karnaugh_Mapping" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Combinational_Logic_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Multivibrators" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Sequential_Circuits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Shift_Registers" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Digital-Analog_Conversion" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:_Digital_Communication" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15:_Digital_Storage(Memory)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "16:_Principles_Of_Digital_Computing" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "Book:Electric_Circuits_IV-Digital_Circuitry(Kuphaldt)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electrical_Fundamentals_Competency_(Industry_Training_Authority_of_BC)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electric_Circuits_III_-Semiconductors(Kuphaldt)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electric_Circuits_II_-Alternating_Current(Kuphaldt)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electric_Circuits_I_-Direct_Current(Kuphaldt)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electric_Circuits_VI_-Experiments(Kuphaldt)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Electric_Circuits_V_-References(Kuphaldt)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Hydraulics_and_Electrical_Control_of_Hydraulic_Systems_(Pytel)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Trigonometry_and_Single_Phase_AC_Generation_for_Electricians_(Flinn)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Troubleshooting_Motors_and_Controls_(Dickson-Self)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sat, 20 Mar 2021 03:11:19 GMT 7: Boolean Algebra 937 937 admin { } Anonymous Anonymous User 2 false false [ "article:topic-guide", "authorname:tkuphaldt", "Boolean Algebra", "license:gnufdl", "licenseversion:13", "source@ ] [ "article:topic-guide", "authorname:tkuphaldt", "Boolean Algebra", "license:gnufdl", "licenseversion:13", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Electronics Technology 4. Book: Electric Circuits IV - Digital Circuitry (Kuphaldt) 5. 7: Boolean Algebra Expand/collapse global location Book: Electric Circuits IV - Digital Circuitry (Kuphaldt) Front Matter 1: Numeration Systems 2: Binary Arithmetic 3: Logic Gates 4: Switches 5: Electromechanical Relays 6: Ladder Logic 7: Boolean Algebra 8: Karnaugh Mapping 9: Combinational Logic Functions 10: Multivibrators 11: Sequential Circuits 12: Shift Registers 13: Digital-Analog Conversion 14: Digital Communication 15: Digital Storage (Memory) 16: Principles Of Digital Computing Back Matter 7: Boolean Algebra Last updated Mar 20, 2021 Save as PDF 6.6: Programmable Logic Controllers (PLC) 7.1: Introduction to Boolean Algebra Page ID 937 Tony R. Kuphaldt Schweitzer Engineering Laboratories via All About Circuits ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents No headers All arithmetic operations performed with Boolean quantities have but one of two possible outcomes: either 1 or 0. There is no such thing as “2” or “-1” or “1/2” in the Boolean world. It is a world in which all other possibilities are invalid by fiat. As one might guess, this is not the kind of math you want to use when balancing a checkbook or calculating current through a resistor. However, Claude Shannon of MIT fame recognized how Boolean algebra could be applied to on-and-off circuits, where all signals are characterized as either “high” (1) or “low” (0). 7.1: Introduction to Boolean Algebra 7.2: Boolean ArithmeticIn Boolean mathematics, addition is equivalent to the OR logic function, multiplication is equivalent to the AND logic function, and complementation is equivalent to the NOT logic function. 7.3: Boolean Algebraic IdentitiesIn mathematics, an identity is a statement true for all possible values of its variable or variables. The algebraic identity of x + 0 = x tells us that anything (x) added to zero equals the original “anything,” no matter what value that “anything” (x) may be. Like ordinary algebra, Boolean algebra has its own unique identities based on the bivalent states of Boolean variables. 7.4: Boolean Algebraic PropertiesThe commutative, associative, and distributive properties apply to Boolean algebra. 7.5: Boolean Rules for SimplificationBoolean algebra finds its most practical use in the simplification of logic circuits. If we translate a logic circuit’s function into symbolic (Boolean) form, and apply certain algebraic rules to the resulting equation to reduce the number of terms and/or arithmetic operations, the simplified equation may be translated back into circuit form for a logic circuit performing the same function with fewer components. 7.6: Circuit Simplification Examples 7.7: The Exclusive-OR Function - The XOR GateOne element conspicuously missing from the set of Boolean operations is that of Exclusive-OR, often represented as XOR. Whereas the OR function is equivalent to Boolean addition, the AND function to Boolean multiplication, and the NOT function (inverter) to Boolean complementation, there is no direct Boolean equivalent for Exclusive-OR. This hasn’t stopped people from developing a symbol to represent this logic gate, though. 7.8: DeMorgan’s TheoremsA mathematician named DeMorgan developed a pair of important rules regarding group complementation in Boolean algebra. By group complementation, I’m referring to the complement of a group of terms, represented by a long bar over more than one variable. 7.9: Converting Truth Tables into Boolean ExpressionsIn designing digital circuits, the designer often begins with a truth table describing what the circuit should do. The design task is largely to determine what type of circuit will perform the function described in the truth table. There are procedural techniques available and Boolean algebra proves its utility in a most dramatic way. This page titled 7: Boolean Algebra is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Tony R. Kuphaldt (All About Circuits) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 6.6: Programmable Logic Controllers (PLC) 7.1: Introduction to Boolean Algebra Was this article helpful? Yes No Recommended articles 7.7: The Exclusive-OR Function - The XOR GateOne element conspicuously missing from the set of Boolean operations is that of Exclusive-OR, often represented as XOR. Whereas the OR function is equ... 7.8: DeMorgan’s TheoremsA mathematician named DeMorgan developed a pair of important rules regarding group complementation in Boolean algebra. By group complementation, I’m r... 7.9: Converting Truth Tables into Boolean ExpressionsIn designing digital circuits, the designer often begins with a truth table describing what the circuit should do. The design task is largely to deter... 7.1: Introduction to Boolean Algebra 7.2: Boolean ArithmeticIn Boolean mathematics, addition is equivalent to the OR logic function, multiplication is equivalent to the AND logic function, and complementation i... Article typeChapterAuthorTony R. KuphaldtLicenseGNU FDLLicense Version1.3 Tags Boolean Algebra source@ © Copyright 2025 Workforce LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 6.6: Programmable Logic Controllers (PLC) 7.1: Introduction to Boolean Algebra
4575	https://www.mecapedia.uji.es/pages/criterio_de_von_Mises.html	MECAPEDIA • Criterio de von Mises Conceptos Elementos Fenómenos Lecciones Leyes Modelos Parámetros Procedimientos Sistemas Todo Página aleatoria Animaciones Nomenclatura Catálogos Estándares Calculadores Contactar Inicio Modelos Criterio de von Mises Criterio de von Mises Modelo 29/07/2022 Marta C. Mora El criterio de von Mises, también llamado criterio de la máxima energía de distorsión, es un criterio de resistencia estática, aplicado a materiales dúctiles, según el cual, el material no fluirá en el punto analizado siempre que la energía de distorsión por unidad de volumen en el punto no supere la energía de distorsión por unidad de volumen que se da en el momento de la fluencia en el ensayo de tracción. El criterio de resistencia se escribe matemáticamente como: 1 2((σ 1−σ 2)2+(σ 1−σ 3)2+(σ 2−σ 3)2)<S y donde S y>0 es el límite de fluencia a tracción. El coeficiente de seguridad en el punto analizado, de acuerdo con el criterio de von Mises, se obtiene de: n s=S y 1 2((σ 1−σ 2)2+(σ 1−σ 3)2+(σ 2−σ 3)2) expresión que es válida para cualquier signo de las tensiones principales. En el caso del estado tensional biaxial el criterio de von Mises puede representarse gráficamente en un diagrama σ A−σ B como donde éstas representan las dos tensiones principales no nulas, como se indica en la figura. La zona sombreada representa la zona segura, para la cual el material no fluye de acuerdo con dicho criterio. Criterio de von Mises Ver también: Criterio de RankineCriterio de TrescaCriterio de Coulomb-MohrCriterio de Mohr modificadoCriterio de resistencia estáticaTensiones principalesEnsayo de tracción Área de Ingeniería Mecánica. Universitat Jaume I - Sobre los autores La Universitat Jaume I no supervisa la veracidad ni la precisión del contenido proporcionado aquí y no es responsable en ningún caso del mismo. Los únicos responsables son los autores materiales de este contenido. Esta web utiliza cookies propias y de terceros para ofrecer una mejor experiencia. Al continuar la navegación entendemos que acepta nuestra política de cookies.
4576	https://devforum.roblox.com/t/issues-calculating-the-angle-between-two-vectors/1885690	Issues calculating the angle between two vectors I want to get the angle between two vectors. I found this website. On that website I found this equation α = arccos[(a · b) / (\|a\| \|b\|)] I tried implementing this in code with the following print(math.deg(0-math.cos(vectorA:Dot(vectorB) / ( vectorA.Magnitude vectorB.Magnitude)))) but I have a problem. One I couldn’t find arccos or inverse cosine so please correct me if I’m wrong but I believe 0-math.cos() could be used as a substitute I also tried math.cos()^-1 but this I don’t believe worked. Secondly when I input the values of vectors A & B into the website which has a calculator for want I want todo I get a different answer to theirs. print(math.deg(0-math.cos(vectorA:Dot(vectorB) / ( vectorA.Magnitude vectorB.Magnitude)))) One I couldn’t find arccos or inverse cosine math.acos() exists??? math.acos() I believe 0-math.cos() could be used as a substitute I also tried math.cos()^-1 That’s not how it works. It’s the inverse operation of the function, not the additive/multiplicative inverse. If x gives you y, the inverse of the function is y → x. x y y x I get a different answer to theirs Either because of your faulty “substitute”, or you weren’t converting between radians and degrees properly. Also, when used with unit vectors, you can disregard the division and it just becomes math.deg(math.acos(vec1:Dot(vec2))) math.deg(math.acos(vec1:Dot(vec2))) your correct with acos however just for your information math.deg(math.acos(vec1:Dot(vec2))) seems to be giving me a lot of nan not all the time but a lot of the time when I use acos in my equation it doesn’t do this math.deg(math.acos(vectorA:Dot(vectorB) / ( vectorA.Magnitude vectorB.Magnitude))) regardless ty math.deg(math.acos(vec1:Dot(vec2))) math.deg(math.acos(vectorA:Dot(vectorB) / ( vectorA.Magnitude vectorB.Magnitude))) seems to be giving me a lot of nan That’s an issue with floating point precision i think, just clamp the number with math.clamp between the interval [-1, 1] before using arccosine on it math.clamp Powered by Discourse, best viewed with JavaScript enabled
4577	https://www.machinelearningmastery.com/a-gentle-introduction-to-hessian-matrices/	Navigation A Gentle Introduction To Hessian Matrices Hessian matrices belong to a class of mathematical structures that involve second order derivatives. They are often used in machine learning and data science algorithms for optimizing a function of interest. In this tutorial, you will discover Hessian matrices, their corresponding discriminants, and their significance. All concepts are illustrated via an example. After completing this tutorial, you will know: Hessian matrices Discriminants computed via Hessian matrices What information is contained in the discriminant Let’s get started. A Gentle Introduction to Hessian Matrices. Photo by Beenish Fatima, some rights reserved. Tutorial Overview This tutorial is divided into three parts; they are: Definition of a function’s Hessian matrix and the corresponding discriminant Example of computing the Hessian matrix, and the discriminant What the Hessian and discriminant tell us about the function of interest Prerequisites For this tutorial, we assume that you already know: Derivative of functions Function of several variables, partial derivatives and gradient vectors Higher order derivatives You can review these concepts by clicking on the links given above. What Is A Hessian Matrix? The Hessian matrix is a matrix of second order partial derivatives. Suppose we have a function f of n variables, i.e., 𝑓:𝑅𝑛→𝑅 The Hessian of f is given by the following matrix on the left. The Hessian for a function of two variables is also shown below on the right. Hessian a function of n variables (left). Hessian of f(x,y) (right) We already know from our tutorial on gradient vectors that the gradient is a vector of first order partial derivatives. The Hessian is similarly, a matrix of second order partial derivatives formed from all pairs of variables in the domain of f. Want to Get Started With Calculus for Machine Learning? Take my free 7-day email crash course now (with sample code). Click to sign-up and also get a free PDF Ebook version of the course. What Is The Discriminant? The determinant of the Hessian is also called the discriminant of f. For a two variable function f(x, y), it is given by: Discriminant of f(x, y) Examples of Hessian Matrices And Discriminants Suppose we have the following function: g(x, y) = x^3 + 2y^2 + 3xy^2 Then the Hessian H_g and the discriminant D_g are given by: Hessian and discriminant of g(x, y) = x^3 + 2y^2 + 3xy^2 Let’s evaluate the discriminant at different points: D_g(0, 0) = 0 D_g(1, 0) = 36 + 24 = 60 D_g(0, 1) = -36 D_g(-1, 0) = 12 What Do The Hessian And Discriminant Signify? The Hessian and the corresponding discriminant are used to determine the local extreme points of a function. Evaluating them helps in the understanding of a function of several variables. Here are some important rules for a point (a,b) where the discriminant is D(a, b): The function f has a local minimum if f_xx(a, b) > 0 and the discriminant D(a,b) > 0 The function f has a local maximum if f_xx(a, b) < 0 and the discriminant D(a,b) > 0 The function f has a saddle point if D(a, b) < 0 We cannot draw any conclusions if D(a, b) = 0 and need more tests Example: g(x, y) For the function g(x,y): We cannot draw any conclusions for the point (0, 0) f_xx(1, 0) = 6 > 0 and D_g(1, 0) = 60 > 0, hence (1, 0) is a local minimum The point (0,1) is a saddle point as D_g(0, 1) < 0 f_xx(-1,0) = -6 < 0 and D_g(-1, 0) = 12 > 0, hence (-1, 0) is a local maximum The figure below shows a graph of the function g(x, y) and its corresponding contours. Graph of g(x,y) and contours of g(x,y) Why Is The Hessian Matrix Important In Machine Learning? The Hessian matrix plays an important role in many machine learning algorithms, which involve optimizing a given function. While it may be expensive to compute, it holds some key information about the function being optimized. It can help determine the saddle points, and the local extremum of a function. It is used extensively in training neural networks and deep learning architectures. Extensions This section lists some ideas for extending the tutorial that you may wish to explore. Optimization Eigen values of the Hessian matrix Inverse of Hessian matrix and neural network training If you explore any of these extensions, I’d love to know. Post your findings in the comments below. Further Reading This section provides more resources on the topic if you are looking to go deeper. Tutorials Derivatives Gradient descent for machine learning What is gradient in machine learning Partial derivatives and gradient vectors Higher order derivatives How to choose an optimization algorithm Resources Additional resources on Calculus Books for Machine Learning Books Thomas’ Calculus, 14th edition, 2017. (based on the original works of George B. Thomas, revised by Joel Hass, Christopher Heil, Maurice Weir) Calculus, 3rd Edition, 2017. (Gilbert Strang) Calculus, 8th edition, 2015. (James Stewart) Summary In this tutorial, you discovered what are Hessian matrices. Specifically, you learned: Hessian matrix Discriminant of a function Do you have any questions? Ask your questions in the comments below and I will do my best to answer. Get a Handle on Calculus for Machine Learning! Feel Smarter with Calculus Concepts ...by getting a better sense on the calculus symbols and terms Discover how in my new Ebook: Calculus for Machine Learning It provides self-study tutorials with full working code on: differntiation, gradient, Lagrangian mutiplier approach, Jacobian matrix, and much more... Bring Just Enough Calculus Knowledge to Your Machine Learning Projects See What's Inside More On This Topic A Gentle Introduction to Sparse Matrices for Machine… Introduction to Matrices and Matrix Arithmetic for… A Complete Guide to Matrices for Machine Learning… Gentle Introduction to Predictive Modeling A Gentle Introduction to XGBoost for Applied Machine… A Gentle Introduction to the Gradient Boosting… discriminant, Hessian, local extrema, local maximum, local minimum, saddle points, second order derivatives A Gentle Introduction to the Jacobian A Gentle Introduction to the Laplacian 13 Responses to A Gentle Introduction To Hessian Matrices Jack August 5, 2021 at 12:24 pm # Hi dear Mehreen Saeed I have a suggestion， could you combine the application with courses by using Python，and also show some real problem by using machine learning，because we need use the technology to solve real problems in real time，on the other hand , CV also can post on it like custom object-tracking detection，thank you so much！ Reply Mehreen Saeed August 5, 2021 at 10:23 pm # Sure, such articles will follow. For this article we focus on learning the math before implementation in Python. It is always best to understand how things work and then go on the implementation part 🙂 Reply 2. Stephen Mark August 5, 2021 at 8:04 pm # Thank you for bothering to write this article. Reply Mehreen Saeed August 5, 2021 at 10:24 pm # 🙂 Reply 3. Douglas August 7, 2021 at 3:24 am # Small comment: your rule about the discriminant being either positive or negative only applies to 2 x 2 matrices. For higher dimensional matrices, the general rule is that the Hessian must be either positive definite or negative definite to determine extrema. Of course, for symmetric 2 x 2 matrices, the determinant being positive guarantees that the two eigenvalues are positive; so while you say that works for 2×2 matrices, I do not believe it works in general. Never-the-less, you say “Evaluating them helps in the understanding of a function of several variables.” so this suggests what you are saying generalizes. Now, I could be wrong, but if so, please let me know. Reply 4. Tarique Ahmad August 7, 2021 at 4:05 am # An step by step solution using S-Math studio may be more illustrative. Reply 5. Max August 7, 2021 at 4:22 am # 100% Eigenvalues and the Hessian. Reply 6. JG August 7, 2021 at 4:26 am # Nice post! Thank you! Normally on machine learning problems we do not know the explicit function y =f(x1,…xn). And we solve it…it does not give us this explicit output-input dependency … So with the exception of testing a known function , trough a discrete DATASET production and later on with some genetic algorithms or similar methods to try to identify maximum and minimum of de dataset that we can check directly via Hessian component and discriminant analysis … I do not see how are Hessian matrix used in machine learning algorithms ? regards Reply 7. Jim Winburn August 7, 2021 at 7:21 am # Thank you, Mehreen. Your explanations are excellent! Python code examples would be really helpful! Reply 8. vandana kushwaha September 7, 2021 at 4:01 pm # Hi Mehreen, Can you tell us how to find the Hessian matrix for a mutli-function with multi-variable? Like fx(theta1, theta2)=0 and fy(theta1, theta2)=0. I know how to get Jacobian matrix for fx and fy but how to get hessian for both?? Reply Adrian Tam September 8, 2021 at 1:20 am # Hessian matrix is for scalar-valued functions. If you treat fx and fy as two separate functions, then you can get each a Hessian. If you consider [fx,fy] as a vector (hence a vector-valued function), Jacobian is all you can get. Reply 9. Ken March 17, 2023 at 5:44 pm # What I don’t understand is: why can’t you figure out if a point is a maximum, minimum, or saddle point by looking at only fxx and fyy? Why do you need to compute D = fxx + fyy – (fxy)^2? It seems straight forward enough to see that if both fxx and fyy are negative then you have a maximum, if both are positive then you have a minimum, and if one is positive and one is negative then you have a saddle point. Can you think of an example where both fxx and fyy have the same sign and yet the surface has a saddle? Maybe I am not understanding the geometric interpretation of fxy if there is one… Is there? Thank you! Best article I’ve found yet on this topic Reply James Carmichael March 18, 2023 at 8:34 am # Hi Ken…If you have a surface already you can state that there appears to be a maximum or minimum via visual inspection. The equation proves it mathematically. This is important because other code cannot “visually inspect” a curve as we can as humans. So if other decisions need to be made based upon finding a maximum and/or minimum the must be a mathematical basis. Reply Leave a Reply Click here to cancel reply.
4578	https://brainly.com/question/5471599	Published Time: Tue, 03 Sep 2024 23:38:32 GMT [FREE] Solve the equation using the quadratic formula. Then solve the equation by factoring to check your - brainly.com Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Try advanced features for free Join for free Log in / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Solve the equation using the quadratic formula. Then solve the equation by factoring to check your solutions. x 2−2 x+1=0 2 See answers See Expert answer Asked by olgarpt • 09/24/2017 0:02 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer 5.0 1 Upload your school material for a more relevant answer using the formula x = [-(-2) +/- sqrt((-2)^2 - 4 1 1)] / 2 = 2 +/- sqrt 0 -------------- = 2/2 = 1 2 The solution is x = 1 Duplicity 2 Factoring we have (x - 1)(x - 1) = 0 which is x = 1 Duplicity 2 Answered by Brainly User Thanks 1 5.0 (4 votes) Textbook &Expert-Verified⬈(opens in a new tab) 5.0 1 Calculus-Based Physics - Jeffrey W. Schnick Chemistry - Paul Flowers Physics for AP® Courses 2e - Kenneth Podolak, Henry Smith Upload your school material for a more relevant answer The equation x 2−2 x+1=0 can be solved using the quadratic formula to find x=1 with a multiplicity of 2. Factoring the equation also confirms the solution as (x−1)(x−1)=0. Both methods give the same result, confirming that x=1 is the solution. Explanation To solve the equation x 2−2 x+1=0 using the quadratic formula, we first identify the coefficients from the standard form of a quadratic equation a x 2+b x+c=0. Here, we have: a=1 b=−2 c=1 The quadratic formula is given by: x=2 a−b±b 2−4 a c Substituting the values into the formula: x=2⋅1−(−2)±(−2)2−4⋅1⋅1 This simplifies to: x=2 2±4−4 x=2 2±0 x=2 2±0 Thus, we find: x=2 2=1 This means that the solution is x=1 with a multiplicity of 2, since it is a repeated root. Next, we can verify our solution by factoring the quadratic equation. The original equation can be rewritten as: (x−1)(x−1)=0 This confirms that: x−1=0→x=1 Thus, both methods (quadratic formula and factoring) yield the same solution of x=1. Examples & Evidence For example, if you had a quadratic equation like x 2−4 x+4=0, you would similarly use the quadratic formula to find that it also factors to (x−2)(x−2)=0 and has a repeated root of x=2. Using the quadratic formula allows us to handle any quadratic equation and confirms the solutions are valid. Factoring provides a quick way to verify the results, proving that both approaches are accurate. Thanks 1 5.0 (4 votes) Advertisement Community Answer This answer helped 43944400 people 43M 2.0 1 The solutions to the given quadratic equation are x=1 and x=1. The given quadratic equation is x²-2x+1=0. The roots of a quadratic equation ax² + bx + c = 0 are given by x = [-b ± √(b² - 4ac)]/2a. Here, a=1, b=-2 and c=1 Now, x = [2 ± √(4 - 4)]/2 x=2±0/2 x=1 With factoring, we get x²-2x+1=0 x²-x-x+1=0 x(x-1)-1(x-1)=0 (x-1)(x-1)=0 (x-1)=0 x=1 Therefore, the solutions to the given quadratic equation are x=1 and x=1. To learn more about the quadratic formula visit: brainly.com/question/11540485. SPJ6 Answered by bhoopendrasisodiya34 •13.6K answers•43.9M people helped Thanks 1 2.0 (1 vote) Advertisement ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics If you determine that the x value of a system of linear equations is 1, what should you do next? A. You can stop because you know there is no solution. B. Substitute 1 for y in one of the equations, and solve for x. C. Substitute 1 for x in one of the equations, and solve for y. D. You can stop because you know that there is an infinite number of solutions. Find the sum 6 y a+7 y a. The number of boys and girls in different grades of a school are shown in the table below. Pick the appropriate scatter plot for the data. \| Grades \| Boys \| Girls \| --- \| 8th \| 20 \| 25 \| \| 9th \| 30 \| 28 \| \| 10th \| 40 \| 35 \| \| 11th \| 60 \| 45 \| \| 12th \| 80 \| 65 \| Find the middle term in the expansion of (2 x−y)4 and simplify your answer. 5=2.23606797749… Previous questionNext question What’s your question? Community Brainly Community Brainly for Schools & Teachers Brainly for Parents Brainly Scholarships Honor Code Community Guidelines Insights: The Brainly Blog Become a Volunteer Company Homework Questions & Answers Textbook Solutions Online Tutoring Careers Advertise with us Terms of Use Copyright Policy Privacy Policy Cookie Preferences Help Signup Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Why are you reporting this ad? Please make a selection. Played audio Had inappropriate content Covered the page Other Additional Information Please help us by describing the ad. Only 500 characters are allowed. REPORT AD Thank you for letting us know. This ad has already been reported. Powered by × Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
4579	https://pubmed.ncbi.nlm.nih.gov/16260547/	ACOG Committee Opinion #323: Elective coincidental appendectomy - 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. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. To: Subject: Body: Format: [x] MeSH and other data Send email Cancel Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Add to My Bibliography My Bibliography Unable to load your delegates due to an error Please try again Add Cancel Your saved search Name of saved search: Search terms: Test search terms Would you like email updates of new search results? Saved Search Alert Radio Buttons Yes No Email: (change) Frequency: Which day? Which day? Report format: Send at most: [x] Send even when there aren't any new results Optional text in email: Save Cancel Create a file for external citation management software Create file Cancel Your RSS Feed Name of RSS Feed: Number of items displayed: Create RSS Cancel RSS Link Copy Full text links Wolters Kluwer Full text links Actions Cite Collections Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Permalink Permalink Copy Display options Display options Format Page navigation Title & authors Similar articles Cited by MeSH terms LinkOut - more resources Obstet Gynecol Actions Search in PubMed Search in NLM Catalog Add to Search . 2005 Nov;106(5 Pt 1):1141-2. doi: 10.1097/00006250-200511000-00060. ACOG Committee Opinion #323: Elective coincidental appendectomy ACOG Committtee on Gynecologic Practice PMID: 16260547 DOI: 10.1097/00006250-200511000-00060 Item in Clipboard ACOG Committee Opinion #323: Elective coincidental appendectomy ACOG Committtee on Gynecologic Practice. Obstet Gynecol.2005 Nov. Show details Display options Display options Format Obstet Gynecol Actions Search in PubMed Search in NLM Catalog Add to Search . 2005 Nov;106(5 Pt 1):1141-2. doi: 10.1097/00006250-200511000-00060. Author ACOG Committtee on Gynecologic Practice PMID: 16260547 DOI: 10.1097/00006250-200511000-00060 Item in Clipboard Full text links Cite Display options Display options Format No abstract available PubMed Disclaimer Similar articles Re: Elective Female Genital Cosmetic Surgery: ACOG Committee Opinion Summary, Number 795.Seftel AD.Seftel AD.J Urol. 2020 Jul;204(1):170. doi: 10.1097/JU.0000000000001047.04. Epub 2020 Apr 15.J Urol. 2020.PMID: 32293946 No abstract available. [Pros and cons of elective appendectomy in gynecologic laparotomy].Grys E, Pawlaczyk M, Paczkowska A, Bembnista M.Grys E, et al.Ginekol Pol. 1996 Jul;67(7):361-5.Ginekol Pol. 1996.PMID: 9138999 Review.Polish. ACOG committee opinion. Length of hospital stay for gynecologic procedures. Number 191, October 1997 (Replaces no. 134, March 1994). Committee on Gynecologic Practice. American College of Obstetricians and Gynecologists.[No authors listed][No authors listed]Int J Gynaecol Obstet. 1998 Feb;60(2):189.Int J Gynaecol Obstet. 1998.PMID: 9509962 No abstract available. ACOG committee opinion. Statement on surgical assistants. Number 240, August 2000. American College of Obstetricians and Gynecologists. Committee on gynecologic practice. Committee on obstetric practice.[No authors listed][No authors listed]Int J Gynaecol Obstet. 2001 Dec;75(3):341.Int J Gynaecol Obstet. 2001.PMID: 11794299 No abstract available. Ambulatory gynaecological surgery: risk and assessment.Pasternak LR, Johns A.Pasternak LR, et al.Best Pract Res Clin Obstet Gynaecol. 2005 Aug;19(5):663-79. doi: 10.1016/j.bpobgyn.2005.06.003. Epub 2005 Jul 11.Best Pract Res Clin Obstet Gynaecol. 2005.PMID: 16011908 Review. See all similar articles Cited by Unindicated hysterectomies in India: the aftermath.Davis AA.Davis AA.BMJ Case Rep. 2019 Dec 17;12(12):e230129. doi: 10.1136/bcr-2019-230129.BMJ Case Rep. 2019.PMID: 31852688 Free PMC article. Fallopian Tube Ligation or Salpingectomy as Means for Reducing Risk of Ovarian Cancer.Szender JB, Lele SB.Szender JB, et al.AMA J Ethics. 2015 Sep 1;17(9):843-8. doi: 10.1001/journalofethics.2015.17.9.stas1-1509.AMA J Ethics. 2015.PMID: 26390206 Free PMC article.No abstract available. Histopathological features of incidental appendectomy specimens obtained from living liver donors.Akbulut S, Koç C, Sarıcı B, Özcan M, Şamdancı E, Yılmaz S.Akbulut S, et al.Turk J Gastroenterol. 2020 Mar;31(3):257-263. doi: 10.5152/tjg.2020.19010.Turk J Gastroenterol. 2020.PMID: 32343238 Free PMC article. Cost-effectiveness of prophylactic appendectomy: a Markov model.Newhall K, Albright B, Tosteson A, Ozanne E, Trus T, Goodney PP.Newhall K, et al.Surg Endosc. 2017 Sep;31(9):3596-3604. doi: 10.1007/s00464-016-5391-y. Epub 2017 Jan 11.Surg Endosc. 2017.PMID: 28078461 Natural orifice-assisted laparoscopic appendectomy.Nezhat C, Datta MS, Defazio A, Nezhat F, Nezhat C.Nezhat C, et al.JSLS. 2009 Jan-Mar;13(1):14-8.JSLS. 2009.PMID: 19366534 Free PMC article. See all "Cited by" articles MeSH terms Appendectomy / adverse effects Actions Search in PubMed Search in MeSH Add to Search Elective Surgical Procedures / adverse effects Actions Search in PubMed Search in MeSH Add to Search Gynecologic Surgical Procedures Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Patient Selection Actions Search in PubMed Search in MeSH Add to Search Risk Assessment Actions Search in PubMed Search in MeSH Add to Search LinkOut - more resources Full Text Sources Ovid Technologies, Inc. Wolters Kluwer Full text links[x] Wolters Kluwer [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
4580	https://math.stackexchange.com/questions/931117/prove-that-the-power-set-of-an-n-element-set-contains-2n-elements	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 Prove that the power set of an $n$-element set contains $2^n$ elements Ask Question Asked Modified 4 years ago Viewed 110k times 37 $\begingroup$ Theorem. Let $X$ denote an arbitrary set such that $\|X\|=n$. Then $\|\mathcal P(X)\|=2^n$. Proof. The proof is by induction on the numbers of elements of $X$. For the base case, suppose $\|X\|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $\|\mathcal P(X)\|=1=2^0$. Now, the induction step. Suppose $\|X\|=n$; by the induction hypothesis, we know that $\|\mathcal P(X)\|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup{a}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup{a}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $\|\mathcal P(Y)\|=2^n+2^n=2^{n+1}$. $\square$ Image that replaced text. From the above explanation, I don't understand why the set that contains ${a}$ will contain $2^{\|n\|}$ elements when it should clearly be $2^{\|1\|}$. The construction of a new set $S$ is the union of the old set with cardinality $n$ and a new element ${a}$, therefore the set that does not contain ${a}$ still has cardinality $n$ and the set that contains ${a}$ is just ${a}$, one element. Can someone please elucidate? elementary-set-theory induction Share edited Sep 28, 2021 at 21:50 Archer 6,61966 gold badges4646 silver badges9494 bronze badges asked Sep 14, 2014 at 16:36 FraïsséFraïssé 11.7k77 gold badges7373 silver badges163163 bronze badges $\endgroup$ 1 2 $\begingroup$ It is the sets that contain $a$. $\endgroup$ André Nicolas – André Nicolas 2014-09-14 16:45:13 +00:00 Commented Sep 14, 2014 at 16:45 Add a comment \| 6 Answers 6 Reset to default 34 $\begingroup$ Here is an another, combinatorial proof. Let $X$ be a set with $n$ elements. To form a subset of $X$, we go over each element of $X$ and exercise a choice of whether or not to include in the subset. Every sequence of choices gives a different subset. Since for every element, there are 2 choices, for $n$ elements, there are $2 \times 2 \times ...$ $n$ times, choices. Therefore, there are $2^n$ distinct subsets of $X$. Share answered Sep 14, 2014 at 16:56 Amey JoshiAmey Joshi 1,12488 silver badges1414 bronze badges $\endgroup$ 1 3 $\begingroup$ this explanation was very clear to understand $\endgroup$ Sujal Mandal – Sujal Mandal 2018-07-23 04:54:32 +00:00 Commented Jul 23, 2018 at 4:54 Add a comment \| 26 $\begingroup$ You misread the proof. Since set $X$ has $n$ elements, the induction hypothesis tells us that $\|\mathcal{P}(X)\| = 2^n$. The set $Y = X \cup {a}$ has $n + 1$ elements. It subsets are either subsets of $X$, of which there are $2^n$ by the induction hypothesis, or the union of a subset $Z$ of $X$ with ${a}$. By the induction hypothesis, there are $2^n$ subsets $Z$ of $X$. Hence, there are $2^n$ subsets of the form $Z \cup {a}$ of the set $Y$. Hence, $Y$ has $2^n$ subsets that do not contain $a$ and $2^n$ subsets that do contain $a$ for a total of $2^n + 2^n = 2 \cdot 2^n = 2^{n + 1}$ subsets of $Y$, which is what the author wants to show. Share answered Sep 14, 2014 at 16:46 N. F. TaussigN. F. Taussig 79.3k1414 gold badges6262 silver badges7777 bronze badges $\endgroup$ 4 1 $\begingroup$ Sorry, you lost me on the fourth sentence. Can you show me explicitly what is $Z$? $\endgroup$ Fraïssé – Fraïssé 2014-09-14 17:00:18 +00:00 Commented Sep 14, 2014 at 17:00 $\begingroup$ Oh I think I see it now, could you elaborate whether if I take the union of two sets say {A,B} U {C}, do I get {A, B, C} or {A, B, {A,C}, {B,C}}? $\endgroup$ Fraïssé – Fraïssé 2014-09-14 17:11:18 +00:00 Commented Sep 14, 2014 at 17:11 1 $\begingroup$ @ Aåkon I think it's the former. He introduced $Z$ ($2^{n}$ times) to show that $\mid \mathcal{P}(Y) \setminus \mathcal{P}(X) \mid = \mid \mathcal{P}(X)\mid$ $\endgroup$ Chandran Goodchild – Chandran Goodchild 2019-02-09 13:02:58 +00:00 Commented Feb 9, 2019 at 13:02 $\begingroup$ @N.F .Taussig: There are a confusing point in this proof. You state that $Y = X \cup {a }$ has as its subsets either subsets of $X$, or unions of ${a}$ with subsets $Z$ of $X$. But what is the proof of this assumption? $\endgroup$ Edward.Lin – Edward.Lin 2022-02-25 18:56:07 +00:00 Commented Feb 25, 2022 at 18:56 Add a comment \| 11 $\begingroup$ Suppose you've already shown that $X={1,2}$ has $2^2=4$ subsets, namely ${\cal P}(X)={\emptyset,{1},{2},X}$. Now you add a new element $a=3$ to get $Y=X\cup {a}={1,2,3}$. The four subsets of $X$ are also subsets of $Y$, but you get new subsets - those which contain $a$, e.g. ${1,3}$. But each of these is one of the ones you already had together with the new $a$, e.g. ${1,3}$ is ${1} \cup {3}$. So, the new ones are $\emptyset \cup {3}$, ${1} \cup {3}$, ${2} \cup {3}$, and $X \cup {3}$ and those are exactly as much as you already had - four of them. Which implies that ${\cal P}(Y)$ has twice as much elements (the old ones and the new ones) as ${\cal P}(X)$, so $2\times2^2=2^3=8$. $$ \begin{array}{\|c\|c\|} \hline \text{Subsets of }X&\text{New subsets}\ \hline \emptyset&{3}\ \hline {1}&{1,3}\ \hline {2}&{2,3}\ \hline X&Y\ \hline \end{array} $$ Share edited Sep 14, 2014 at 16:55 answered Sep 14, 2014 at 16:46 FrunobulaxFrunobulax 6,9142828 silver badges4444 bronze badges $\endgroup$ Add a comment \| 4 $\begingroup$ Alternative proof: Represent a subset of $X$ as a binary number such that its $k^{th}$ bit is set if and only if the $k^{th}$ element is taken. E.g., ${a,b,c}\to111$, ${a,c}\to101$, ${c}\to001$, ... You can convince yourself that the two representations are equivalent, and that the second includes all binary numbers from $0$ to $2^{\|X\|}-1$. Note that if you add an element, the subset of $X$ with signature $m$ generates two new subsets, with signatures $2m$ and $2m+1$. Share edited Feb 9, 2021 at 8:39 answered Feb 8, 2021 at 17:40 user65203user65203 $\endgroup$ Add a comment \| 2 $\begingroup$ Here is another one that uses the Binomial Theorem. Take, for example, the set $A = {1, 2, 3, 4}$ Then, $P(A) = {, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {2, 3}, {2, 4}, {3, 4}, {4, 1}, {1, 2, 3}, {2, 3, 4}, {3, 4, 1}, {4, 1, 2}, {1, 2, 3, 4}}$ In $P(A)$, see that the number of subsets with cardinality 0 is 1, there are 4 subsets with cardinality 1, 6 with cardinality 2, then 4 again with 3 and then 1 more with cardinality 4. $\pmb{1-4-6-4-1}$; $16$ total. Note that these are the coefficients of the terms when you expand the binomial theorem (binomial coefficients) with $n=4$. Now we want the sum of these numbers. This is can be easily achieved by replacing $x=y=1$ in the binomial formula where $n$ is the cardinality of set A and the formula follows from there. Why the pattern appears is very simple. Let $n$ be $\|A\|$. You want to fill subsets of cardinalities $0, 1, \cdots,\ n$. You have n elements to choose from. So to fill the first subset with cardinality 0, you use $\binom{n}{0}$, for the second with cardinality 1 we use $\binom{n}{1}$ and so on. Share answered Jan 8, 2021 at 10:12 RiddhimanRiddhiman 3344 bronze badges $\endgroup$ 0 Add a comment \| 1 $\begingroup$ I will prove the claim by induction. First suppose $\|A\|=1$. That is, $A={a_1}$. Thus, $P(A) ={ {a}, { \emptyset } }$. So, it can be concluded that $\|A\|=2^1=2$. My inductive hypothesis is that $\|A\|=n$ and $\|P(A)\|=2^n$. Now let's consider $A'$, where $A'$ is $A$ with one new element added, call it $a_{n+1}$. That is to say $A'= {a_1,a_2,a_3,\dots,a_{n+1} }$. When considering $P(A')$ two types of sets arise, those containing $a_{n+1}$ and those that do not. Allow $B_i$ to represent any particular member of $P(A)$. Notice that $\|P(A')\|=\|P(A)\|+\|B_k \cup{a_{n+1}}\|$, for all $k\in \mathbb{N}$ satisfying $1\leq k\leq 2^n$. Therefore $\|P(A')\|=2^n+2^n=2\cdot 2^n=2^{n+1}$. $\square$ Share edited Apr 5, 2020 at 19:26 user26857 53.4k1414 gold badges7676 silver badges166166 bronze badges answered Mar 31, 2020 at 14:53 Lucas WillhelmLucas Willhelm 19911 silver badge44 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 elementary-set-theory induction 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 1 Use Mathematical Induction to show that for any set $S$, If $\vert S\vert=n$, then $\vert\mathcal P(S)\vert=2^n$ It is based on set theory. Here I wan to know that how can I derive the modulus of P(A) is equal to 2¿. Find Possible Combination with set of options without repetitions. Related 2 Prove that if a set A of natural numbers contains $n_0$ and whenever A contains k it also contains k+1. 3 Power set of a finite set is finite proof, no cardinality of power set 1 Show that if $S$ is a finite set with n elements, then $S$ has $2^n$ subsets by using mathematical induction 0 Summation of the products of the elements of subsets 0 Set of subsets of elements of $n$ sets that contain at least one element of each set 0 Trying to prove every nonempty finite set of Z has a smallest element by induction 1 Set Theoretic Notation Pertaining to the Power Set Hot Network Questions Checking model assumptions at cluster level vs global level? в ответе meaning in context Separating trefoil knot on torus I have a lot of PTO to take, which will make the deadline impossible how do I remove a item from the applications menu Traversing a curve by portions of its arclength Passengers on a flight vote on the destination, "It's democracy!" How to use cursed items without upsetting the player? Suspicious of theorem 36.2 in Munkres “Analysis on Manifolds” How to home-make rubber feet stoppers for table legs? How to locate a leak in an irrigation system? Is this commentary on the Greek of Mark 1:19-20 accurate? How can I get Remote Desktop (RD) to scale properly AND set maximum windowed size? What meal can come next? Do sum of natural numbers and sum of their squares represent uniquely the summands? Implications of using a stream cipher as KDF How do I disable shadow visibility in the EEVEE material settings in Blender versions 4.2 and above? Lingering odor presumably from bad chicken Determine which are P-cores/E-cores (Intel CPU) Origin of Australian slang exclamation "struth" meaning greatly surprised How many stars is possible to obtain in your savefile? Numbers Interpreted in Smallest Valid Base What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? Change default Firefox open file directory more hot questions Question feed
4581	https://www.intechopen.com/chapters/80577	Open Access is an initiative that aims to make scientific research freely available to all. To date our community has made over 100 million downloads. It’s based on principles of collaboration, unobstructed discovery, and, most importantly, scientific progression. As PhD students, we found it difficult to access the research we needed, so we decided to create a new Open Access publisher that levels the playing field for scientists across the world. How? By making research easy to access, and puts the academic needs of the researchers before the business interests of publishers. We are a community of more than 103,000 authors and editors from 3,291 institutions spanning 160 countries, including Nobel Prize winners and some of the world’s most-cited researchers. Publishing on IntechOpen allows authors to earn citations and find new collaborators, meaning more people see your work not only from your own field of study, but from other related fields too. Brief introduction to this section that descibes Open Access especially from an IntechOpen perspective Want to get in touch? Contact our London head office or media team here Our team is growing all the time, so we’re always on the lookout for smart people who want to help us reshape the world of scientific publishing. Home > Books > Keratinocyte Biology - Structure and Function in the Epidermis Open access peer-reviewed chapter Keratinocyte Stem Cells: Role in Aging Written By Submitted: 26 November 2021 Reviewed: 21 January 2022 Published: 27 February 2022 DOI: 10.5772/intechopen.102805 Cite this chapter There are two ways to cite this chapter: From the Edited Volume Keratinocyte Biology - Structure and Function in the Epidermis Edited by Mayumi Komine Chapter metrics overview 891 Chapter Downloads Cite this chapter There are two ways to cite this chapter: Impact of this chapter Total Chapter Downloads on intechopen.com Total Chapter Views on intechopen.com Citations Abstract Stem cells located in the skin are responsible for continual regeneration, wound healing, and differentiation of different cell lineages of the skin. The three main locations of skin stem cells are the epidermis, dermis, and hair follicles. The keratinocyte stem cells are located in the epidermal basal layer (the interfollicular stem cells), hair follicle bulge region (the hair follicle stem cells), and sebaceous glands (the sebaceous gland stem cells) and are responsible for the epidermal proliferation, differentiation, and apoptosis. The interfollicular (IF) stem cells are responsible for epidermis regeneration by proliferating basal cells that attach to the underlying basement membrane and with time they exit from the cell cycle, start terminal differentiation, and move upward to form the spinous, the granular, and the stratum corneum layers. The hair follicle (HF) stem cells are responsible for hair regeneration and these stem cells undergo a cycle consists three stages; growth cycles (anagen), degeneration (catagen), and relative resting phase (telogen). The sebaceous gland (SG) stem cells located in between the hair follicle bulge and the gland and are responsible for producing the entire sebaceous gland which secretes oils to moisture our skin. The role of epidermal stem cells is extremely crucial because they produce enormous numbers of keratinocytes over a lifetime to maintain epidermal homeostasis. However, the age-associated changes in the skin; for example; alopecia, reduced hair density, gray or thin hair, reduced wound healing capacity are related to skin stem cells’ decline functionality with age. Keywords Author Information Tuba Musarrat Ansary Koji Kamiya Mamitaro Ohtsuki Address all correspondence to: tuba2020@jichi.ac.jp 1. Introduction Skin, the largest organ in the human body gives a protective barrier against the harmful events of the environment, for example, radiation, germs, temperature, and toxic substances. Moreover, the skin also protects our body from excessive dehydration and works as a permeability barrier. To support the above-mentioned roles and repair skin injury and wounds, the skin needs to regenerate and proliferate with the help of skin stem cells. Broadly the skin can be divided into three parts: Epidermis, the outermost layer, is mainly composed of keratinocytes and is known as the squamous stratified epithelium. The dermis, the middle layer, is separated by the basement membrane from the epidermis and is mainly composed of the extracellular matrix of tough collagen fibers, blood vessels, and nerves. The hypodermis, the third layer, mainly consists of fibroblasts, adipose tissues, and connective tissues. Adult skin development involves a multi-stage process that involved cells from diverse embryonic origins. Following gastrulation, the neuroectoderm cells stimulate the formation of the nervous system and skin epidermis. The neural and epidermal fate of these cells is dependent on different signaling pathways, for example, Wnt signaling, fibroblast growth factor (FGF), and bone morphogenetic protein (BMP) signaling, and Notch signaling pathway . During development, the ectoderm-derived cells are become the epidermal basal layer and are responsible for all epidermal structures, for example, the hair follicles, sebaceous glands, and sweat glands . A complex and multiple embryonic origins contribute to the dermis formation at different regions of the body. In a broad perspective, the mesoderm-derived cells are responsible for the dermis of the ventral and flank regions of the body and neural-crest-derived cells are responsible for the head dermis . The mesoderm-derived cells are also responsible for the development of the adipose tissues in the hypodermis . In everyday life skin has to perform many functions which are essential for our survival, for example, to protect from physical and mechanical injuries, harmful radiations, and after injury, it needs to form new cells to repair. In this regard, the skin stem cells (SCs) in the epidermis play the most essential roles to maintain skin renewal throughout life and repairing wounds after injury. In this review, we attempt to clarify the 1. Types of stem cells located in the epidermis, 2. The location, function, and markers to identify the epidermal SCs, 3. Role of chronological and photoaging on epidermal stem cells. 2. Epidermal stem cell The epidermis is mainly consisting of five sub-layers with distinct characteristics although they function together to maintain tissue homeostasis and regeneration (Figure 1). The innermost or the deepest layer of the epidermis is the Basal layer, which is a single layer of proliferative basal cells that attach to a basement membrane (BM) by hemidesmosomes. These cells can continually divide and after some period lose attachment from the BM, get pushed by new cells and a program of differentiation has triggered. Above the basal layer, the Squamous cell layer contains the mature basal cells which are now known as squamous cells. This is the thickest layer of the epidermis and spiny projections that are attached to the surrounding cells by desmosomes. The keratinocytes are then got bigger and flatter and push towards two thin layers of the epidermis; the Stratum granulosum and the Stratum lucidum. The stratum corneum is the uppermost layer of the epidermis which contains the dead keratinocytes also referred to as anucleate squamous cells. The skin appendages include the sweat glands, sebaceous gland, nails, hair shafts, and hair follicles, and have both epidermal and dermal components which help the skin to complete its function . The epidermal SCs reside in the basal layer are responsible for the maintenance of the epidermal stratified epithelium by self-renewal, wound repair, and differentiation capabilities. The epidermal proliferative unit (EPU) consists of basal cells that are responsible for the maintenance of the cornified layers by self-renewing and producing stem cells and non-stem cells . There are several techniques available to understand and locate epidermal stem cells, for example, lineage tracing and genetic fate mapping. In the case of lineage tracing, an individual cell is labeled and the location, status, and the number of descendants from that cell can be identified by that label . The genetic fate mapping technique involves marking an individual cell or a group of cells in the embryo and the trace the final position of all descendant cells until the completion of the development . The epidermal SCs are also found in skin appendages, for example, the hair follicle bulge and the sebaceous glands, and have the potency to differentiate into different lineages that are present in adult skin . The epidermal SCs niche can be classified into three groups according to their location; (1) The interfollicular (IF) stem cell located in the basal layer, (2) The hair follicle (HF) stem cell located in the bulge, and (3) The sebaceous gland (SG) stem cell located in between the bulge and hair shaft (Figure 2) [9, 10]. Figure 1. The five main layers of epidermis. The basal cells can divide themselves and move upwards. As they move to the next layer they become flatter and start losing nuclei. The five main layers of epidermis. The basal cells can divide themselves and move upwards. As they move to the next layer they become flatter and start losing nuclei. Figure 2. Location of epidermal stem cells. The interfollicular (IF) stem cells located in the basal layer of the epidermis. Hair follicle (HF) stem cells located in the bulge area of the hair shaft. Melanocyte stem cells are found in the bulge and also in the hair matrix. The sebaceous gland (SG) stem cell located in the sebaceous gland. Location of epidermal stem cells. The interfollicular (IF) stem cells located in the basal layer of the epidermis. Hair follicle (HF) stem cells located in the bulge area of the hair shaft. Melanocyte stem cells are found in the bulge and also in the hair matrix. The sebaceous gland (SG) stem cell located in the sebaceous gland. 2.1 The interfollicular (IF) stem cell The IF stem cells are found along with the basement membrane which is a specialized thin layer of extracellular matrices. The IF stem cells help to maintain the epidermis regeneration by self-renewal as well as produce progenitor cells named transit-amplifying (TA) cells which divide a limited number of times and then undergo terminal differentiation and give rise to flattened and dead keratinocytes in the cornified layer . Apart from the multipotency these IF stem cells also show other characteristics, such as plasticity, which means these cells can lose their original identity and differentiate into other lineages . Depending on the proximity from the wound and the specific stem cell niche origin, the epidermal stem cells participate in wound healing and tissue regeneration. By performing genetic fate mapping analysis, one report demonstrated that during initial wound healing there is an abundance of long-lived IF stem cells recruitment which promote re-epithelialization, and as the wound repairing the HF-derived stem cells increased in the wound area . 2.1.1 IF stem cell markers There are several markers that can be used to identify IF stem cells (Figure 3). Integrins are the heterodimeric cell-surface receptors that consist of α and β subunits and are responsible for cell adhesion, proliferation, and migration . Adherence of the IF stem cells with the BM and extracellular matrixes is regulated by the integrins . Several types of integrins are expressed in the epidermis; α2β1 (receptor for Collagen), α3β1 (receptor for laminin 5), α6β4 (receptor for laminin), α5β1 (receptor for fibronectin), [16, 17]. Among all integrins, the α6 or CD49f is the most used marker for epidermal stem cells . The well-recognized marker for IF stem cells are the high α6 and the low transferrin receptor CD71 (α6-bright/CD71-dim) . As there is a positive correlation between the IF stem cell proliferation and adherence, the proliferative IF stem cells can be distinguished from the low-adhered TA cells with the higher β1 integrins expression . During terminal differentiation, the IF stem cells express involucrin, a differentiation marker, and filaggrin, an intermediate filament (IF)-associated protein [21, 22]. Figure 3. Markers of epidermal stem cells. Markers of epidermal stem cells. 2.2 The hair follicle (HF) stem cell The HF is one of the mini-organs in our body which go through life-long cyclic regeneration and involution [23, 24, 25]. The HF is located in attachment with the sebaceous gland and arrector pili muscle and it has two main segments; an epithelium made of keratinocytes and a dermal papilla (DP) made of mesenchymal cells [26, 27]. The cyclic regeneration of HF is mainly consisting of these phases; an active growth phase (anagen), a regression or involution phase (catagen), and a relative rest phase (telogen) and after the hair is shed a new hair cycle begins [28, 29]. The upper HF does not cycle visibly and is mainly divided into 2 segments; the infundibulum and the isthmus and the lower HF which consistently regenerates within the hair cycle divided into the hair bulb and the suprabulbar region [30, 31]. The infundibulum is the uppermost segment of the follicle which is funnel-shaped and begins from the epidermis surface and extends to the sebaceous gland opening . The isthmus is the lower part of the upper HF and is located between the sebaceous gland and the bulge . The bulb is the cyclic portion and the base of the HF which regenerates in every hair cycle and includes dermal papilla and HF matrix . The suprabulbar region includes three parts; outer root sheath, inner root sheath, and the hair shaft and it lies between the hair bulb and the isthmus . Bulge is the region where the HF stem cells are located and which lies between the sebaceous gland and the arrector pili muscle. These quiescent and long-lived stem cells have the potential to generate all epithelial lineages of the skin, including HF and hair [34, 35, 36]. The HF stem cells contribute to wound healing by recruiting multipotent SCs and life-long HF regeneration by providing new cells. They are normally known as quiescent, slow-cycling, and label-retaining cells. Another type of stem cell that resides in the bulge of the HF is the long-lived neural-crest cell-derived melanocyte stem cell, which performs a crucial role in hair pigmentation maintenance . Generally, melanoblasts, the immature progenitors of melanocytes, proliferate and differentiate into melanocytes in the epidermis and migrate to the hair follicle and divided into two categories; the hair matrix melanocytes responsible for pigmenting the original hair and the bulge melanocyte stem cells which are responsible for the following hair cycle pigmentation . The regeneration of the follicular Melanocyte stem cells is synchronized with the HF cycle. During the anagen phase, the melanogenically active Melanocyte stem cells reside in the hair matrix proliferate and differentiate into Melanocyte progenitors to produce melanin and transfer to the neighboring keratinocytes and serve as a reservoir of the pigmentary unit for eye, hair, and skin . In the catagen phase, the differentiated Melanocyte stem cells die because of the high apoptosis rate . Melanin not only gives the essential pigmentation but also protects our skin from harmful UV radiation as the melanin granules work like a UV absorbent. To identify the lineage of the Melanocyte stem cell, a transgenic mouse model has been developed. The undifferentiated Melanocyte stem cells reside in the bulge express Dopachrome tautomerase (DCT) and tyrosinase-related proteins 1 (TRP-1) and serve as an early marker of Melanocyte stem cell. Nishimura et al. developed a transgenic mouse by using a lacz reporter manipulated by the DCT promoter and which enables people to find out the DCT positive melanoblasts . However, both progenitor and mature melanocyte stem cells express DCT, so it cannot be regarded as a specific marker for the Melanocyte stem cell . The CD34 can also use to identify the Melanocyte stem cells, for example, one paper reported that CD34 negative Melanocyte expressed high DCT, KIT (KIT Proto-Oncogene, Receptor Tyrosine Kinase), Tyr (tyrosinase), Tyrp1, Pmel17 (premelanosome protein), and MITF (Melanocyte Inducing Transcription Factor) . Sox10 (Sry-related high-mobility-group box 10) can also be used as a marker for Melanocyte stem cells as this transcription factor plays an important role during the differentiation of the neural crest cell to Melanocyte stem cell [43, 44]. 2.2.1 Location There are some differences of opinion regarding the location of the stem cells and between the species. Some reports demonstrated that the germinative cells located in the lower area of the bulb are the HF stem cells as they have the differentiation ability [45, 46], however, several reports have challenged this idea and showed that HF stem cells are located in the bulge which is the upper and permanent portion of the HF. Several lines of experiments using pulse-chase experiment, in-vitro analysis, lineage analysis, seminal experiments, and BrdU- labeling experiments have proven the HF stem cells residing in the bulge [35, 46, 47, 48, 49, 50, 51]. The in-vitro clonal analysis has shown that 95% of multipotent stem cells reside in the bulge and the rest of 5% are in the bulb, which is known as matrix cells or transit-amplifying (TA) cells [2, 52, 53]. 2.2.2 Major functions The HF stem cells located in the HF bulge area are label-retaining slow-cycling cells that perform several functions; for example, hair regeneration, reepithelization after a wound. HF stem cells play an important role in the generation of all layers of HF and hair regeneration by fueling the hair cycle [46, 48, 49, 54]. In general, when an anagen phase starts, the HF stem cells become activated and an HF will grow and regenerate and push the club hair above. During the anagen phase, the stem cells from the bulge area and the hair germ cells are activated by the mesenchymal cells from the DP, start proliferating in descending order, move to the bulb area, and create an outer root sheath (ORS) . Throughout the anagen, the matrix or transit-amplifying (TA) cells originated from the bulge stem cells, move upward, and start to differentiate into follicle cells [49, 56]. By performing lineage tracing and double pulse-chase experiments one report confirmed that these TA cells then return to the bulge niche and lose the stemness property . The catagen phase started when the matrix TA cells are exhausted and undergo apoptosis . During catagen, apoptosis causes a huge decline in TA cell number, regression of approximately two-thirds of the hair follicles and only long-lived stem cells survive [57, 58]. After catagen, the resting phase or telogen phase will start. The telogen phase includes quiescent HF stem cells and shedding of the old HFs and this phase becomes longer progressively throughout life [57, 58]. In response to the signals from the DP, a new anagen phase started after the telogen phase and a new hair cycle begins [30, 59]. In addition to hair regeneration, the HF stem cells also play an important role in wound healing and re-epithelization. HF stem cells have the potential to differentiate into multiple lineages; for example, keratinocytes, smooth muscle cells, glial cells, neurons, and melanocytes and promote angiogenesis [60, 61, 62, 63]. Many reports perform in-vitro and ex-vivo analysis using rodent and human samples showed that the HF stem cell can differentiate into audiogenic, osteogenic lineages as well as illustrate similar properties as bone-marrow-derived mesenchymal stem cells [64, 65]. Because of this property, the HF stem cells are regarded as one of the powerful stem cell candidates for cell therapy in the case of cutaneous wound healing and tissue regeneration . In clinical studies, using graft transplantation from the scalp in patients with leg ulcers showed better therapeutic potential compared to the non-hairy grafts [67, 68, 69, 70]. Performing double-label analysis and lineage tracing in the wound-repair model in animal reports showed that HF stem cells rapidly mobilize to the epidermis after injury and participate in epidermal repair by proliferating TA cells [36, 49, 71]. Using HF patch transplantation assay it has been demonstrated that HF stem cells contribute to generating new follicles in wounded mouse skin areas . A complete reduction of HF stem cells in transgenic mice displayed a delay in wound healing after a full-thickness wound in the dorsal area . Similarly, a delay in re-epithelization after the wound is observed in Edaraddcr/cr mice, that have a mutation in HF development . Additionally, it has been shown that administration of HF stem cell in the wound area accelerates the healing process [75, 76]. 2.2.3 HF stem cell markers Several signaling pathways are important for the regulation and initiation of the anagen phase in the quiescent, slow-cycling label-retaining bulge stem cells . Wnt/β catenin pathway plays an essential role in HF stem cells activation, maintenance, and differentiation . The importance of Wnt/β catenin signaling in HF development is further proven by the report that showed complete HF follicle loss in a transgenic mouse with ectopic expression of Wnt inhibitor (Dickkopf 1) . The fibroblast growth factor (FGF) signaling plays a crucial role in HF stem cell differentiation, hair cycle clock regulation, and angiogenesis [80, 81]. Sonic hedgehog (Shh) signaling expressed in the HF matrix is crucial for HF regeneration and neogenesis [82, 83]. Bone morphogenetic proteins (BMP) are also essential for HF regeneration, activation, quiescence, and TA cell differentiation and are expressed in the matrix [77, 84]. There are primarily four techniques to study skin stem cells; for example, label retention, clonogenic assays, skin reconstitution, and genetic lineage tracing . Several markers have been identified to locate bulge and non-bulge stem cells in murine and human skin. Among the epithelial stem or progenitor markers, the most widely used marker for murine bulge stem cells are Keratin 15 (K15) and Clusters of differentiation 34 (CD34) [86, 87, 88]. In the case of human bulge stem cells, the most used markers are K15, Keratin 19 (K19), and Clusters of differentiation 200 (CD200) [89, 90]. The leucine-rich G protein–coupled receptor 5 (Lgr5), a Wnt target gene label the mouse lower bulge stem cells during the telogen phase and lower ORS HF during the anagen phase . Several transcriptional factors are used to mark HF stem cells; such as Lim-homeodomain transcription factor Lhx2, SRY (Sex-determining region Y)-box 9 (Sox-9), transcription factor 3 (TCF-3), cytoplasmic 1 (NFATC1) (Figure 3) [92, 93, 94]. 2.3 The sebaceous gland (SG) stem cell Among other appendages in the skin, the sebaceous glands produce sebocytes and sebum to keep the lipid homeostasis and plays important role in barrier functions . Unlike HF’s cyclic growth, the SG has a continuous growth similar to the epidermis, and SG is typically found in association with the HF or as a modified version found independently in eyelids . The resident stem cells in SG proliferate in the basal layer of the SG, differentiate into sebocytes and gather sebum, then move upwards and rupture the content inside into the pilosebaceous canal . The specific markers that are used to identify the SG stem cells are K5, K14, K79, Leucine Rich Repeats, and Immunoglobulin Like Domains 1 (LRIG1), leucine-rich repeat-containing G protein-coupled receptor 1 (LGR6), B lymphocyte-induced maturation protein 1 (Blimp1) [98, 99, 100, 101]. 3. Epidermal stem cells and aging Skin stem cells, residing in a protective niche, maintain the skin homeostasis by self-renewal and terminal differentiation. Unlike other somatic stem cells, the skin stem cells are quite resistant to aging as the number and self-renewal capacity of the stem cell do not reduce with age . In general, stem cells stay at a quiescent state for a long time in their niche, and upon activation by numerous intrinsic and extrinsic factors these stem cells can exit this quiescent stage and differentiate into multiple lineages. Stem cell exhaustion is a state where the stem cells fail to renew themselves and thereby decrease in number which is mainly caused by aging. Several reports compared IF stem cells between young and old mice showed no difference in the number of stem cells, telomere length, gene expression related to aging, an abundance of K15 positive HF stem cells, and multipotency [103, 104]. On the contrary, by performing colony-forming essays in human keratinocytes, one report demonstrated that the cells from the aged donor have retarded colony-forming ability . As we age there is an increase in senescent cells accumulation and DNA damage resulting in a decline in stem cells’ function to produce new progenitor and effector cells [106, 107]. In this notion, Ultraviolet radiation (UVR) plays a crucial role in DNA damage in stem cells that ultimately lead to photoaging . 3.1 Photoaging and epidermal stem cell The major characteristics of aged HF stem cells are imbalance in the phases of the hair cycle, stem cell exhaustion, and loss of hair (alopecia) and the appearance of the hair becoming dry, gray, or thin [109, 110]. A proper balance between the proliferation and quiescence state of the hair cycle is a prerequisite for HF stem cell lifespan and self-renewal. In this regard, the competitive balance between Wnt and BMP pathways is essential for HF homeostasis and cycle activation. During the regression phase, there is a decreased expression of Wnt and increased expression of BMP pathways which cause inhibition in keratinocyte proliferation and differentiation . Specifically, the Wnt10 activated the anagen phase and BMP6 is the inhibitor of the hair cycle . One report showed that persistent expression of Wnt1 causes mice HF to retain in the growth phase, initiate cellular senescence, and finally cause stem cell exhaustion and premature hair loss . Increased Wnt signaling pathway, specially Wnt10 and β-catenin expression is observed in C57BL6/J mice exposed to UVR which causes HF miniaturization and gray hair . UVR exposure can cause p53, a checkpoint protein, overactivation through DNA damage which is also associated with decreased stem cell renewal capacity, stem cell exhaustion, and premature aging . The stem cell niche or microenvironment homeostasis is maintained by the interaction among mesenchymal cells, integrins expressed by the stem cells, and the extracellular matrix. It has been reported that UVR increased the expression of c-Myc, a transcriptional factor, which reduces the β1 integrin expression and thus impair β1 integrin-initiated adherence to the niche and migration [115, 116, 117]. Reactive oxidative species (ROS) induced by the UVR also cause a decrease in stem cell renewal capacity, senescence, and exhaustion [118, 119]. Another indication of stem cell aging is telomere shortening which resulted in hair loss and impaired stem cell proliferation and ultimately premature aging . One report demonstrated chronic UV exposure to transgenic mice causes DNA damage and telomere shortening by modulating telomerase activity [121, 122]. A major hallmark of photoaged skin is altered wound repair capacity. Mitogen-activated protein kinase (MAPK) plays an essential role in cutaneous wound healing and an in-vivo study one report showed that chronic UV irradiation cause MAPK downregulation [123, 124]. 3.2 Chronological aging and epidermal stem cell A progressive decline in skin regeneration, repair, and homeostasis, thin hair, loss of hair, wrinkle, thin dermis, and epidermis, etc. are associated with accelerated aging. The major characteristic of the aged hair follicles is the hair density reduction and the resting period of the hair cycle increase. One report compared HF stem cell functions between 2 month and 24-month-old mice and found that the old mice HF showed a longer telogen phase, defective proliferation, and shorter hair growth phase . Loss of HF stem cells is also associated with age-related hair shaft miniaturization . Another typically aged phenotype related to hair is hair thinning, graying, or hair loss. There are several genetic mutation mice models that depicts accelerated aging phenotypes, for example, gray, brittle, and fragile hair and alopecia as a result of genetic instability . It has been reported that DNA damage causes downregulation of Collagen 17 (Col17) expression in HF stem cells and these defected stem cells start to differentiate terminally and pushed themselves upward and eliminate . As it is well known that Col17 is a crucial component in HF homeostasis and Col17 deficit can cause premature aging phenotype in hair, for example, alopecia or atrophy in HF . In comparison to HF stem cells the role of aging on IF stem cells are not yet clarified. Some reports showed increased proliferation in epidermal stem cells and others confirmed the proliferation decreased as organisms aged. There are several proteins (Heat shock cognet 71, Stress protein 70, Myc associated protein, Cyclin D1, Glucose related protein) that expressed similarly in epidermal stem cells of adult and aged human skin and epidermal stem cells collected from the neonatal and aged mice have the same plasticity when injected into the blastocysts . In normal physiological conditions, melanocytes, differentiated from the Melanocyte stem cell in the hair matrix, produce melanin during the anagen phase and transfer it to the neighboring keratinocytes. Following hair cycles, the TA melanocytes produced from the melanocyte stem cells are responsible for producing melanin and pigmenting new hair follicles. Several external or internal factors can cause aging-associated modifications in Melanocyte stem cells which cause gray hair, one of the most evident signs of aging. The genotoxic stress caused by radiation results in differentiation of the Mc stem cells in the niche and thereby diminish their self-renewal ability and which leads to hair pigmentation impairment in the following hair cycle . Aging, itself is associated with Melanocyte stem cell reduction. An age-associated decline in Melanocyte stem cells measured by immunofluorescence with Kit antibody was observed in aged mice . Another paper confirmed that the number of melanoblasts in the niche decreased with aging as well as the incidence of pigmented melanoblasts which means the ectopic differentiation increased with aging . It can be speculated that due to the lack of enough progenitor cells the melanin production may be hampered and cause hair graying. 4. Conclusions Skin stem cells participate in wound healing and maintain skin integrity and homeostasis by self-renewal and producing progenitor cells. Unlike other stem cells, epidermal stem cells maintain an appropriate number throughout life and showed quite a resistance against aging. However, as we age the increased amount of DNA damage response and senescence can affect the epidermal stem cell’s functions; for example, self-renewal capacity, increase exhaustion, mobility to the wound area or reduction in the number and that lead to skin aging phenotypes, for example, premature hair loss, gray or thin hair, reduced wound healing capacity (Figure 4). Figure 4. Major effects of chronological and photo aging on different skin stem cells. Major effects of chronological and photo aging on different skin stem cells. Acknowledgments I would like to express my sincere gratitude and acknowledgment to my supervisor and mentor Dr. Mayumi Komine for the guidance and supervision which help me to complete this project. Conflict of interest The author declares no conflict of interest. References Sections Cite this chapter There are two ways to cite this chapter: Next chapter Cytokeratins of Tumorigenic and Highly Malignant Respiratory Tract Epithelial Cells By Carol A. Heckman 168 downloads \| 1 cites Written By Submitted: 26 November 2021 Reviewed: 21 January 2022 Published: 27 February 2022 Cite this chapter There are two ways to cite this chapter: © 2022 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Continue reading from the same book Keratinocyte Biology Edited by Mayumi Komine Published: 28 September 2022 IntechOpen Limited 167-169 Great Portland Street, London, W1W 5PF, UNITED KINGDOM Phone: +44 20 8089 5702 © 2025 IntechOpen. All rights reserved. IntechOpen Author/Editor? To get your discount, log in. Your cart Discounts available on purchase of multiple copies. View rates \| \| \| --- \| \| Subtotal \| €0.00 \| Local taxes (VAT) are calculated in later steps, if applicable. Support: orders@intechopen.com
4582	https://mrbrownswhs.wordpress.com/2015/10/31/thursday-october-29-and-friday-october-30-diction-connotation-imagery-and-tone/	Thursday, October 29 and Friday October 30: Diction, Connotation, Imagery and Tone \| Mr. Brown's Blog Mr. Brown's Blog Search Primary MenuSkip to content Home English II English II Pre-AP Search for: English II Pre-AP Thursday, October 29 and Friday October 30: Diction, Connotation, Imagery and Tone October 31, 2015Mr. BrownLeave a comment Toward the end of this week, we discussed five important literary terms and concepts—diction, denotation, connotation, imagery and tone—in preparation for the upcoming literary analysis essay on “The Ones Who Walk Away from Omelas.” Here are brief definitions of each term, followed by more thorough explanations and recaps of how we studied them. Diction: an author’s specific word choices Denotation: the (neutral, literal) dictionary definition of a word Connotation: the associated (non-dictionary) meanings of a word (can be as simple as “positive” or “negative” or more complex) Imagery: details an author includes to create sensory “images” for the reader (usually sight but also sound, smell, taste or touch) Tone: the emotion or attitude of an author or narrator conveyed through the text Diction Authors use specific word choices to achieve the effects they want. What we, as readers, respond to when we encounter strong or particular word choices is theconnotations. We began class on Thursday with a few examples in which the literal meaning or referent of different words was the same, but the connotations were quite different. Examples: (a) That girl is slender. (b) That girl is scrawny. (c) That girl is skinny. (a) “I think we should go out sometime soon,” he said. (b) “I’d like to take you on a date sometime soon,” he said. (a) “Are those your children?” (b) “Are those your little ones?” (c) “Are those your brats?” (a) Her values are old-fashioned. (b) Her values are traditional. In the examples above, the grouped sentences sound different and imply different attitudes because of specific word choices, even though literally they mean more or less the same thing. The word choices have different connotations. Also, consider the verbs trotted, pranced, bolted and bounded. All refer to motion faster than a walking pace. Think of the differences created by plugging each of these words into the sentence “The dog ___ to the door.” Saying that the dog trotted or pranced to the door might imply a lighthearted state, while saying that the dog bolted or bounded to the door could imply excitement or aggression.Bolted can connote either excitement or anxiety. Advertisement Barbara Kingsolver wrote, “Art is the antidote that can call us back from the edge of numbness, restoring the ability to feel for another.” Imagine substituting gift for antidote. The difference is that while a gift is something nice and appreciated, an antidote is something urgently needed as protection or salvation. Using this specific medical term adds a connotation of urgency, suggesting that emotional numbness toward others is like a disease for which art is the cure. Imagery Any details that appeal to one or more of the five senses create imagery in writing. Consider the passage below from When I Was Puerto Rican by Esmeralda Santiago. Details that create imagery are underlined. A ripe guava is yellow, although some varieties have a pink tinge. Their skin is thick, firm, and sweet. Its heart is bright pink and almost solid with seeds. The most delicious part of the guava surrounds the tiny seeds. If you don’t know how to eat a guava, the seeds end up in the crevices between your teeth. When you bite into a ripe guava, your teeth must grip the bumpy surface and sink into the thick edible skin without hitting the center. . . . A green guava is sour and hard. You bite into it at its widest point, because it’s easier to grasp with your teeth. You hear the skin, meat, and seeds crunching inside your head, while the inside of your mouth explodes in little spurts of sour. I want a ripe guava—RIGHT NOW! Don’t you? Or at the very least, doesn’t the narrator sound like she knows what she’s talking about? The passage is ripe with sensory details (hehe, get it?) creating images that are visual(e.g.,yellow,pink tinge), auditory (e.g.,crunching inside your head), gustatory (e.g.,sweet,spurts of sour), and tactile (e.g.,seeds end up in the crevices between your teeth,grip the bumpy surface). Tone Tone is a complex concept. It can be defined asthe attitude of an author/narrator conveyed through a text, but also as the approach an author takes to a text. For example, you might say that the narrator of When I Was Puerto Rican tells about guavas with the appreciative tone of a true connoisseur.Appreciative describes her attitude, and she approaches the subject as a connoisseur, or expert, describing the correct and incorrect techniques of eating a guava and leaving utterly no doubt that she has a lot of practice. Advertisement Another way to think about tone ishow the “voice” of the text sounds. One of the ways that you will know you are developing into a more mature and sophisticated reader is when you begin to hear a voice in your head as you read. Nope, that’s not schizophrenia. That’s your awareness of voice and tone. It’s a wonderfully pleasurable sense to develop as a reader. An important guideline is to remember thattone is always directed toward someone or something. Santiago’s tone is directed toward guavas; Kingsolver’s, toward art. The reason we discuss diction and imagery is thatdiction and imagery play important roles in shaping the tone of writing. There are other literary devices that develop tone as well, but these are the ones we are focusing on. Consider this excerpt from “Racism” (1964), by Ayn Rand. Important word choices are underlined. Racism is the lowest, most crudely primitive form of collectivism. It is the notion of ascribing moral, social or political significance to a man’s genetic lineage—the notion that a man’s intellectual and characterological traits are produced and transmitted by his internal body chemistry. Which means, in practice, that a man is to be judged, not by his own character and actions, but by the characters and actions of a collective of ancestors. Racism claims that the content of a man’s mind . . . is inherited. . . . This is the caveman’s version of the doctrine of innate ideas . . . which has been thoroughly refuted by philosophy and science. Racism is a doctrine of, by and for brutes. It is a barnyard or stock-farm version of collectivism, appropriate to a mentality that differentiates between various breeds of animals, but not between animals and men. Now, look at this excerpt from “The Letter from Birmingham Jail” (), by Dr. Martin Luther King, Jr. Important word choices and images are underlined. Perhaps it is easy for those who have never felt the stinging darts of segregation to say, “Wait.” But when you have seen vicious mobs lynch your mothers and fathers at will and drown your sisters and brothers at whim; when you have seen hate-filled policemen curse, kick and even kill your black brothers and sisters; when you see the vast majority of your twenty million Negro brothers smothering in an airtight cage of poverty in the midst of an affluent society; when you suddenly find your tongue twisted and your speech stammering as you seek to explain to your six year old daughter why she can’t go to the public amusement park that has just been advertised on television, and see tears welling up in her eyes when she is told that Funtown is closed to colored children, and see ominous clouds of inferiority beginning to form in her little mental sky, and see her beginning to distort her personality by developing an unconscious bitterness toward white people; when you have to concoct an answer for a five year old son who is asking: “Daddy, why do white people treat colored people so mean?”; when you take a cross-country drive and find it necessary to sleep night after night in the uncomfortable corners of your automobile because no motel will accept you; when you are humiliated day in and day out by nagging signs reading “white” and “colored”; when your first name becomes “nigger,” your middle name becomes “boy” (however old you are) and your last name becomes “John,” and your wife and mother are never given the respected title “Mrs.”; when you are harried by day and haunted by night by the fact that you are a Negro, living constantly at tiptoe stance, never quite knowing what to expect next, and are plagued with inner fears and outer resentments; when you are forever fighting a degenerating sense of “nobodiness”—then you will understand why we find it difficult to wait. There comes a time when the cup of endurance runs over, and men are no longer willing to be plunged into the abyss of despair. Advertisement Both authors are opposed to racism, but the tones are very different. Rand uses words such as “lowest,” “crudely primitive,” “caveman,” and “brutes” to create a tone of contempt for racism. Her writing treats racism as something stupid and savage, a lower way of thinking for intellectually under-developed people, something worthy of scorn. You can almost imagine her wrinkling her nose in disgust. King, on the other hand, uses words and images such as “stinging darts,” “tears welling up in her eyes,” “ominous clouds of inferiority,” “humiliated,” “nagging,” “degenerating,” and “abyss of despair” to create a tone of agony and outrage toward racism. In King’s view, racism is something excruciatingly painful, something that harasses the soul until it just can’t take any more abuse. Thus, you see that two writers can take quite different tones toward the same subject, even when they basically agree about it. On Friday, students received a Tone Vocabulary Listwith adjectives commonly used to describe tone in writing. I want to stress that this is NOT an exhaustive or comprehensive list. Any word or words that accurately describes the attitude or approach of an author or narrator describes the tone of a piece of writing. Students completed in-class practice on diction, connotation, imagery and toneusing two passages from The Hunger Games. For this practice exercise, students focused on the tone of Katniss Everdeen, the first-person narrator of the novel, toward various people and situations. Advertisement Share this: Click to share on X (Opens in new window)X Click to share on Facebook (Opens in new window)Facebook Like Loading... Related This Week in Review (11/2–6): Gathering Grounds for the Literary Analysis EssayNovember 8, 2015 In "English II Pre-AP" The End of the SemesterDecember 2, 2015 In "English II Pre-AP" Critiquing Your Own Rough DraftNovember 14, 2015 In "English II Pre-AP" Post navigation Previous Post Tuesday, October 27: Allegory, continuedNext Post This Week in Review (11/2–6): Gathering Grounds for the Literary Analysis Essay Leave a comment Cancel reply Δ "If you haven't surprised yourself, you haven't written." ~Eudora Welty Blog at WordPress.com. Comment Reblog SubscribeSubscribed Mr. Brown's Blog Sign me up Already have a WordPress.com account? Log in now. Mr. Brown's Blog SubscribeSubscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Privacy & Cookies: This site uses cookies. By continuing to use this website, you agree to their use. To find out more, including how to control cookies, see here: Cookie Policy %d Design a site like this with WordPress.com Get started Advertisement
4583	https://communities.springernature.com/posts/facts-myths-and-sandwiches-anions-of-18-annulene	Facts, Myths and Sandwiches: Anions of Annulene \| Research Communities by Springer Nature Skip to main content Share your thoughts about the Research Communities in our survey. Research Communities by Springer NatureSearch Clear Search Register Sign In Home Activity Feed ECR Hub NEW! Collection Hubs Health & Clinical Research Biomedical Research General & Internal Medicine Healthcare & Nursing Paediatrics, Reproductive Medicine & Geriatrics Pharmacy & Pharmacology Public Health Surgery Humanities & Social Sciences Arts & Humanities Behavioural Sciences & Psychology Business & Management Economics Education Law, Politics & International Studies Philosophy & Religion Social Sciences Life Sciences Agricultural & Food Science Anatomy & Physiology Bioengineering & Biotechnology Cancer Cell & Molecular Biology Ecology & Evolution Genetics & Genomics Immunology Microbiology Neuroscience Plant Science Zoology & Veterinary Science Mathematics, Physical & Applied Sciences Astronomy Chemistry Civil Engineering Computational Sciences Earth & Environment Electrical & Electronic Engineering Materials Mathematical & Computational Engineering applications Mathematics Mechanical Engineering Physics Statistics Protocols & Methods Research Data Sustainability Help and Support How do I create a post? Communities Guidelines Find a Journal RegisterSign In Behind the Paper Facts, Myths and Sandwiches: Anions of Annulene Published in Chemistry Mar 06, 2024 Wojciech Stawski and Harry L. Anderson 2 contributors Like Liked by Joan Serrano-Plana and 2 others Close Share this post Choose a social network to share with, or copy the URL to share elsewhere Share with... X (Twitter)FacebookLinkedInWhatsAppEmail ...or copy the link communities.springernature.com Facts, Myths and Sandwiches: Anions of Annulene This is a representation of how your post may appear on social media. The actual post will vary between social networks Explore the Research nature.com The anti-aromatic dianion and aromatic tetraanion of annulene Nature Chemistry - A previous investigation of the anti-aromatic dianion of annulene concluded that it consists of a mixture of two isomers. Now it has been shown that this dianion exists as a… A plethora of beautiful new aromatic compounds, with increasing intricacy and complexity, are reported every year. On the other hand, when one looks back at the seminal work in the field, one sometimes discovers surprisingly simple systems that remain unexplored and mysterious, even for decades. This happened to us when considering the pioneering papers of Franz Sondheimer and co-workers, describing chemistry of annulenes.1,2 [n]Annulenes are fully pi-conjugated polyenes, where the number n indicates ring size, for example: annulene is (CH)18. The gargantuan efforts of Sondheimer’s team in the 1960s verified the validity of Hückel’s rule: annulenes are aromatic if they contain 4 n+2 electrons in the circuit, and anti-aromatic when there are 4 n electrons (where n is a positive integer). Annulene is one of the iconic molecules of organic chemistry, yet surprisingly underexplored. Perhaps the myth of its instability and the nontrivial synthesis are what hindered researchers from further exploration. In 1973, Oth, Woo and Sondheimer reported that reduction of annulene with potassium metal leads to an anti-aromatic dianion, with enormous differences between the magnetic shielding inside and outside of the ring (ca. 30 ppm by 1 H NMR) reflecting its strong anti-aromatic character.3 Based only on the symmetry and integration of the poorly resolved spectrum, they concluded that annulene dianion exists in two structural forms (Fig. 1). While the first geometry seems reasonable, the second one looks strange, potentially with substantial steric clashes between the inner protons, so we decided to re-investigate this archetypal compound. Figure 1. Geometry of previously postulated species involved in reduction of annulene with potassium metal together with reported 1 H NMR chemical shift values.3 The easiest way to get unequivocal structural information would be to grow single crystals and determine the geometry from X-ray diffraction. That’s why my PhD supervisor, Harry, contacted Marina Petrukhina — her group in Albany (NY, US) has outstanding expertise in crystallization and coordination chemistry of highly air-sensitive anions. I synthesised annulene, travelled with it to Albany, and got paired up with a PhD student from Marina’s laboratory, Yikun Zhu. Our collaboration and mutual understanding made the research efficient and genuinely exciting. We worked together on the alkali metal reduction — at first, very cautiously, because according to Sondheimer, the anion was expected to decompose above 0 °C. Another difficulty was stopping the reaction at an appropriate stage — the original work did not even mention the beautiful colour changes accompanying the reduction, so we were not sure what to expect, and when to stop the reaction: is it monoanion? Dianion? Or did it already decompose? Fortunately, all the stars aligned for us, and we managed to get suitable crystals (using lithium metal as a reducing agent) in only two weeks (Fig. 2). Figure 2. Flame-sealed crystallization ampules (left) and crystals of tetra-anion under microscope (right). Fortunately, Marina’s group was awarded extra synchrotron beamtime at Argonne National Laboratory, so analysis of our precious material became possible even on very small crystals. Zheng Wei, the departmental crystallographer, took our sample to Chicago for a night shift. I remember being woken up by a few notifications on my phone and reading the exciting news: X-ray diffraction analysis indicated not only that instead of dianion we crystallised an unexpected tetraanion, but also that it has a totally different geometry to that assigned 50 years ago and, on the top of that, it forms a lithium-intercalated sandwich (Fig. 3,4)! Up to this point, the only hydrocarbon capable of forming similar sandwiched complexes was corannulene.4 This was the first of many Eureka! moments during this project. Figure 3. Geometry of the tetra-anion and crystal structure of the self-assembled sandwiched complex. Color code: blue - Li, grey - C, red - O. Hydrogen atoms were removed for clarity. Figure 4. 3D-printed model of the annulene sandwich, together with lithium metal in oil and yellow solution of annulene, in front of an X-ray diffractometer. Nuclear magnetic resonance (NMR) experiments in solution indicated that lithium metal yields an anti-aromatic di-anion and aromatic tetra-anion; with potassium, the reaction does not reach the tetra-anionic stage. Although we did not manage to force the di-anion to crystallize, the symmetry of the NMR signals clearly indicated that its shape is identical to that of the tetra-anion. Contrary to the original paper, we found that these anions are stable at room temperature, provided they are protected from oxygen and moisture. After two months in Albany, I came back to Oxford, synthesised more material and performed a series of variable temperature NMR experiments. They showed that the dianion is flexible and undergoes dynamic conformational changes in solution, indicated by well-resolved multiplets recorded only at or below –70 °C. In contrast,the tetra-anion is rigid, and its NMR spectra do not exhibit significant temperature-dependence, remaining sharp even at 55 °C. The last (but not least) eureka moment came at the final stage of the project. I still had a few milligrams of material left and realized that due to the resemblance between annulene and corannulene tetra-anions (both having five internal bays coordinating lithium cations in their sandwich complexes) it would be interesting to mix them together and see, whether there is any trace of a heteroleptic sandwich. Looking at the NMR spectrum made me absolutely stunned — it suggested selective formation of the mixed compound! I ran directly into Harry’s office with printed spectra, and we discussed the results. In less than a week, crystals were formed, and I was able to measure them and solve the structure, getting tangible evidence for formation of a mixed sandwich (Fig. 5). Figure 5. Geometry of corannulene and annulene tetra-anions and fragment of a crystal structure of the heteroleptic sandwich. The reasons for the drastic change in geometry during the reduction were unclear until our postdoc, Igor Rončević, performed a series of state-of-the-art theoretical calculations. According to the picture drawn from the relative energies and electronic structure of the different possible geometries, the observed change of shape minimizes the repulsion of the negative charge accumulated on the apical positions of the ion. Our research demonstrates that the annulene dianion exists as a single isomer, in contrast with previous reports. The reduction process results in an unusual change of geometry when going from the neutral molecule to the anti-aromatic dianion and (previously unknown) aromatic tetraanion. Self-assembly of the tetra-anion with lithium cations leads to a metallocene-like, diannulene sandwich complex and a heteroleptic (mixed) sandwich with a corannulene tetra-anion. This research re-writes history and shows potential of using higher annulenes as multicoordinating ligands. References: Sondheimer, F. The annulenes. P. Roy. Soc. A - Math. Phy. 297, 173–204 (1967). Sondheimer, F., Wolovsky, R. & Amiel, Y. Unsaturated macrocyclic compounds. XXIII. The synthesis of the fully conjugated macrocyclic polyenes cycloöctadecanonaene (annulene), cyclotetracosadodecaene (annulene), and cyclotriacontapentadecaene (annulene). J. Am. Chem. Soc. 84, 274–284 (1962). Oth, J. F. M., Woo, E. P. & Sondheimer, F. Unsaturated macrocyclic compounds. LXXXIX. Dianion of annulene.J. Am. Chem. Soc. 95, 7337–7345 (1973). 4.Zabula, A. V., Filatov, A. S., Spisak, S. N., Rogachev, A. Yu. & Petrukhina, M. A. Main group metal sandwich: Five lithium cations jammed between two corannulene tetraanion decks. Science333, 1008–1011 (2011). +1 Close Contributors Author Wojciech Stawski DPhil student, University of Oxford Follow Contributing author Harry L. Anderson Professor of Chemistry, University of Oxford Follow Multiple Contributors Wojciech Stawski and Harry L. Anderson View all Close Contributors Author Wojciech Stawski DPhil student, University of Oxford Follow Contributing author Harry L. Anderson Professor of Chemistry, University of Oxford Follow Please sign in or register for FREE If you are a registered user on Research Communities by Springer Nature, please sign in Sign InRegister Follow the Topic Organic Chemistry Physical Sciences > Chemistry >Organic Chemistry Inorganic Chemistry Physical Sciences > Chemistry >Inorganic Chemistry Organometallic Chemistry Physical Sciences > Chemistry > Organic Chemistry >Organometallic Chemistry Coordination Chemistry Physical Sciences > Chemistry > Inorganic Chemistry > Organometallic Chemistry >Coordination Chemistry Nature Chemistry#### Nature Chemistry A monthly journal dedicated to publishing high-quality papers that describe the most significant and cutting-edge research in all areas of chemistry, reflecting the traditional core subjects of analytical, inorganic, organic and physical chemistry. More about the journal Latest Content Behind the Paper, From the Editors Radiative Cooling Materials for Extreme Environmental Applications Behind the Paper, From the Editors Comprehensive Understanding of Closed Pores in Hard Carbon Anode for High‑Energy Sodium‑Ion Batteries Behind the Paper, From the Editors Scalable and Sustainable Chitosan/Carbon Nanotubes Composite Protective Layer for Dendrite‑Free and Long‑Cycling Aqueous Zinc‑Metal Batteries Real-world evaluation of cinacalcet on hard outcomes in hemodialysis patients in Saudi Arabia Behind the Paper, Empower Your Research, ECR Hub The graphic design of predatory papers This community is not edited and does not necessarily reflect the views of Springer Nature. Springer Nature makes no representations, warranties or guarantees, whether express or implied, that the content on this community is accurate, complete or up to date, and to the fullest extent permitted by law all liability is excluded. Website terms of use Online privacy notice Cookie policy Report content Manage Cookies Copyright © 2025 Springer Nature All rights reserved. Built with Zapnito Research Communities by Springer Nature Cookies We use cookies to ensure the functionality of our website, to personalize content and advertising, to provide social media features, and to analyze our traffic. If you allow us to do so, we also inform our social media, advertising and analysis partners about your use of our website. You can decide for yourself which categories you want to deny or allow. Please note that based on your settings not all functionalities of the site are available. Further information can be found in our privacy policy. Close Cookie Control Customise your preferences for any tracking technology The following allows you to customize your consent preferences for any tracking technology used to help us achieve the features and activities described below. To learn more about how these trackers help us and how they work, refer to the cookie policy. You may review and change your preferences at any time. See the full cookie policySee the full privacy policy Reject all Accept all [x] Strictly Necessary More information These trackers are used for activities that are strictly necessary to operate or deliver the service you requested from us and, therefore, do not require you to consent. [x] Advertising More information These trackers help us to deliver personalized marketing content and to operate, serve and track ads. [x] Social & Targeting More information These trackers help us to deliver personalized marketing content to you based on your behaviour and to operate, serve and track social advertising. [x] Measurement More information These trackers help us to measure traffic and analyze your behaviour with the goal of improving our service. [x] Experience Enhancements More information These trackers help us to provide a personalized user experience by improving the quality of your preference management options, and by enabling the interaction with external networks and platforms. Save and continue → Learn more and customise Reject all Accept all
4584	https://pmc.ncbi.nlm.nih.gov/articles/PMC4381887/	Lymphogranuloma venereum: diagnostic and treatment challenges - 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 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 Infect Drug Resist . 2015 Mar 27;8:39–47. doi: 10.2147/IDR.S57540 Search in PMC Search in PubMed View in NLM Catalog Add to search Lymphogranuloma venereum: diagnostic and treatment challenges Romana Ceovic Romana Ceovic 1 Department of Dermatology and Venereology, University Hospital Center Zagreb and School of Medicine, Zagreb, Croatia Find articles by Romana Ceovic 1,✉, Sandra Jerkovic Gulin Sandra Jerkovic Gulin 2 Department of Dermatology and Venereology, General Hospital Sibenik, Sibenik, Croatia Find articles by Sandra Jerkovic Gulin 2 Author information Article notes Copyright and License information 1 Department of Dermatology and Venereology, University Hospital Center Zagreb and School of Medicine, Zagreb, Croatia 2 Department of Dermatology and Venereology, General Hospital Sibenik, Sibenik, Croatia ✉ Correspondence: Romana Ceovic, Department of Dermatology and Venereology, School of Medicine, University Hospital Center Zagreb, 10000 Zagreb, Salata 4, Croatia, Tel +385 1236 8917, Fax +385 1237 9925, Email romana.ceovic@zg.htnet.hr Collection date 2015. © 2015 Ceovic and Jerkovic Gulin. This work is published by Dove Medical Press Limited, and licensed under Creative Commons Attribution – Non Commercial (unported, v3.0) License The full terms of the License are available at Non-commercial uses of the work are permitted without any further permission from Dove Medical Press Limited, provided the work is properly attributed. PMC Copyright notice PMCID: PMC4381887 PMID: 25870512 Abstract Lymphogranuloma venereum is a sexually transmitted disease caused by L1, L2, and L3 serovars of Chlamydia trachomatis. In the last 10 years outbreaks have appeared in North America, Europe, and Australia in the form of proctitis among men who have sex with men. Three stages of disease have been described. The disease in primary stage may go undetected when only a painless papule, pustule, or ulceration appears. The diagnosis is difficult to establish on clinical grounds alone and frequently relies upon either serologic testing, culture, or more recently, nucleic acid amplification testing of direct specimens. A proper treatment regimen cures the infection and prevents further damage to tissues. Lymphogranuloma venereum causes potentially severe infections with possibly irreversible sequels if adequate treatment is not begun promptly. Early and accurate diagnosis is essential. Doxycycline is the drug of choice. Pregnant and lactating women should be treated with erythromycin or azithromycin. Patient must be followed up during the treatment, until disease signs and symptoms have resolved. Repeated testing for syphilis, hepatitis B and C, and HIV to detect early infection should be performed. Keywords: sexually transmitted disease, Chlamydia trachomatis, early and accurate diagnosis Introduction Lymphogranuloma venereum (LGV) is a sexually transmitted disease (STD) caused by L1, L2, and L3 serovars of Chlamydia trachomatis that primarily infects the lymphatics and can be transmitted through unprotected vaginal, anal, or oral sexual contact. C. trachomatis is the most common cause of bacterial STDs in both men and women.1C. trachomatis is an obligate intracellular gram-negative bacterium that during its developmental cycle alternates between two forms: the infectious elementary body (EB) and the noninfectious replicating form, the reticulate body. The attachment process of the elementary body to a host cell is the event most crucial to a successful infection. Elementary bodies attach to columnar epithelial cells followed by endocytosis and inhibition of lysosomal fusion. A number of chlamydial ligands have been identified and characterized. These include the major outer membrane protein (MOMP) as well as the cysteine-rich OmcB (Omp2) protein, hsp70, the polymorphic outer membrane proteins, and the thermolabile 34-kDa membrane protein. In addition to these proposed ligands, there is a considerable amount of experimental evidence to suggest that the glycosaminoglycan, heparan sulfate, is involved in the chlamydial attachment-infectivity process. C. trachomatis does not produce heparan sulfate. Heparan sulfate acts as a host cell receptor for MOMP.2 The disease was described in 1833 by Wallace, but it was defined as a clinical and pathological entity in 1913 by Durand, Nicolas, and Favre. LGV is a subtype of genital ulcer diseases that includes other STDs such as Herpes simplex virus-2 (HSV-2), syphilis, and chancroid. LGV synonyms include lymphopathia venerea, tropical bubo, climatic bubo, strumous bubo, poradenitis inguinales, Durand-Nicolas-Favre disease, and lymphogranuloma inguinale. LGV is endemic in tropical and subtropical areas of the world (certain areas of Africa, Southeast Asia, India, the Caribbean, and South America). The incidence has been low in developed world, but in the last 10 years outbreaks have appeared in North America, Europe, and Australia in the form of proctitis among men who have sex with men (MSM).1,3–5 Three stages of disease have been described. The disease in primary stage may go undetected when only a painless papule, pustule, or ulceration appears.6 The diagnosis is difficult to establish on clinical grounds alone and frequently relies upon either serologic testing, culture, or more recently, nucleic acid amplification testing (NAAT) of direct specimens.7–9 LGV in MSM is associated with a high rate of coinfection with gonorrhea, syphilis, Herpes simplex, hepatitis C, and/or HIV coinfection. When LGV is suspected, diagnostic tests for other potentially coexisting STDs must be performed. Differential diagnoses for LGV include cat scratch disease, chancroid, syphilis, HSV-2, and granuloma inguinale. Epidemiology LGV probably affects both sexes equally, although it is more commonly reported in men because early manifestations of LGV are more apparent in men. Men typically present with the acute form of the disease, whereas women often present when they develop complications from later stages of the disease. LGV may appear at any age, but the highest incidence is between 15 and 40 years (sexually active population). Most cases in Europe and North America have been identified among white, frequently HIV-positive MSM patients presenting with proctitis. Since 2003, there have been a series of LGV outbreaks reported across Europe.10–13 Since formal surveillance was launched in 2004, the United Kingdom has seen the highest number of confirmed MSM cases globally.13 Between April 2003 and June 2012, more than 2,000 cases of LGV had been confirmed in the United Kingdom.12 A large increase in diagnosed cases occurred in late 2009, peaking at 150 cases per quarter in mid-2010; since then, UK rates have remained steady at approximately 80 cases per quarter until 2012. Seventy-seven percent of cases have been diagnosed in London, Brighton, and Manchester. In one report from the United Kingdom (outbreak of 327 new cases of LGV), 76% MSM were also HIV positive, 39% had a diagnosis of another STD, and approximately 19% were also infected with hepatitis C virus. This is a major public health concern because enhanced shedding of HIV during clinical proctitis could increase the risk of HIV transmission to uninfected men.14 Before the outbreaks in MSM, LGV was primarily endemic in heterosexuals in areas of East and West Africa, India, parts of Southeast Asia, and the Caribbean where it is manifested as the classic form of disease with genital ulcers and lymphadenopathy (without proctitis). In these regions, a survey from Madagascar showed that C. trachomatis is not the main cause of genital ulcers. In that survey, 76% of 196 patients with genital ulcers had chlamydial antibodies, but only 8% of patients had LGV confirmed by multiplex polymerase chain reaction (PCR).15,16 From the beginning of the year 2007 until the end of 2011, 146 cases of LGV were notified in Barcelona.17 There are new cases in Finland,18 Czech Republic,19 and France where the first case of C. trachomatis L2b proctitis was described in a woman.20 Authors from the Netherlands reported a case study of a female patient with bubonic LGV caused by serovar L2b, which was probably contracted via her bisexual male partner.21 There are also complicated cases such as LGV infection mimicking deep-vein thrombosis22 and reactive arthritis associated with LGV.23 LGV among MSM in Europe is caused in the majority of cases by the C. trachomatis serovar L2b, which shows a high degree of clonal relatedness as found by Multi Locus Sequence Typing in a molecular epidemiological study. This is in contrast to the strains circulating among MSM in the United States, which show more molecular diversity. Based on these findings, it is now speculated that the LGV epidemic among MSM in Europe caused by the L2b variant may have been imported to Europe from the United States by the end of the previous century via the highly internationalized network of sexual contacts among MSM.4,24 The same serotype of C. trachomatis was identified in rectal swabs between 1979 and 1985 from patients infected with HIV in San Francisco and between 2000 and 2005 in 92 MSM patients with LGV in Amsterdam.24 A small number of cases of L serovar LGV have been reported in heterosexuals in the United Kingdom and Europe (including seven women in the United Kingdom since 2004), but these appear linked to bisexual male partners or sexual contact with those returning from endemic regions.13,14 Prior to 2003, most cases in Europe were imported via travelers, sailors, and soldiers.4,25 In the United States, the true incidence is unknown because national reporting of LGV ended in 1995. Between November 2004 and January 2006, LGV was identified in 180 people, with 27 people identified as being obtained from homosexual males. A study published in 2011 reporting LGV surveillance data from multiple sites in the United States found that less than 1% of the samples obtained from rectal swabs of MSM that were positive for C. trachomatis tested positive for LGV.26 Pathophysiology C. trachomatis is divided into 15 serovars (A, B, Ba, C–K, and L1–L3) based on analysis of the MOMP. Serovars L1–L3 cause LGV. The L2 serovar can be further separated into L2, L2′, L2a, or L2b according to minor differences in their component amino acids. LGV is predominantly a disease of the lymphatic tissue.6 In contrast to serovars A–K, which remain confined to the mucosa, the LGV serovars C. trachomatis induce a lymphoproliferative reaction, it gains entrance through skin breaks and abrasions or it crosses the epithelial cells of mucous membranes. LGV serovars and other strains of C. trachomatis appear to bind to epithelial cells via heparan sulfate receptors27 and then it travels via lymphatics to multiply within mononuclear phagocytes in regional lymph nodes. After lymphangitis, areas of necrosis occur within the nodes, followed by the formation of abscess.27 In one study of 12 patients, lymph node macrophages contained organisms that stained black with Warthin–Starry stain,28 the organisms were clustered within vacuoles and both reticulate and elementary bodies were visible by electron microscopy. PCR testing of the lymph nodes for LGV serovars of C. trachomatis was positive in 9/12 cases. The pathologic findings are compatible with LGV or Bartonella henselae (ie, “cat scratch” disease), but are not diagnostic of either entity.29 The lymph node reaction may take several weeks to develop and can result in substantial inflammation and subsequent fibrosis.6 In a small number of cases, dissemination and systemic disease can occur. Neither the degree of infectiousness nor the reservoir of disease has been accurately defined, but heterosexual transmission has been attributed largely to asymptomatic female carriers and in the MSM population; asymptomatic rectal infection and/or penile infection is the likely source of onward transmission.30 Clinical features First stage (primary LGV) The incubation period lasts 3–30 days, after which a primary lesion occurs in the form of a small painless papule, pustule, nodule, shallow erosion, or herpetiform ulcer (Table 1). The initial lesions may be differentiated from the more common herpetic lesions by the lack of pain associated with the lesion. Differentiation from a syphilitic chancre requires serologic testing. The primary lesion of LGV is most commonly located on the coronal sulcus of men and on the posterior vaginal wall, fourchette (known as the frenulum of labia minora/posterior commissure of the labia minora) or vulva and on the cervix of women. The lesion usually heals within 1 week and may go unnoticed in the urethra, vagina, or rectum. Mucopurulent discharge from urethra, cervix, or rectum may be present regarding the inoculation site. Some ulcers in the recent MSM outbreak have been described as indurated and of variable tenderness; their duration has been as long as several weeks.31 Extra-genital lesions have been reported such as ulcers and fissures in the perianal area in MSM,32 the lip or oral cavity (tonsil), and extra-genital lymph nodes.33,34 Table 1. Clinical features of lymphogranuloma venereum \| Men \| Women \| :--- \| \| First stage \| First stage \| \| – Painless papule/pustule/nodule/erosion/ulcer on penis/anus/ lip and oral cavity (tonsil) \| – Painless papule/pustule/nodule/erosion/ulcer on vulva/posterior vaginal wall/cervix/anus/ lip and oral cavity \| \| – Proctitis (symptoms: rectal pain, anorectal bleeding, mucoid and/or hemopurulent rectal discharge, tenesmus, constipation) \| – Proctitis (symptoms: rectal pain, anorectal bleeding, mucoid and/or hemopurulent rectal discharge, tenesmus, constipation) \| \| Second stage \| Second stage \| \| – Lymphadenitis \| – Lymphadenitis \| \| – Intra-abdominal or retroperitoneal lymphadenopathy \| – Intra-abdominal or retroperitoneal lymphadenopathy \| \| – Inguino and/or femoral lymph adenopathy (typically unilateral, heterosexuals) \| – Inguino and/or femoral lymph adenopathy (typically unilateral, only 20% of women) \| \| – Bubo formation (fluctuant and suppurative lymph nodes that may rupture) \| – Bubo formation (fluctuant and suppurative lymph nodes that may rupture) \| \| – Fever/arthritis/pneumonitis/perihepatitis/abnormal hepatic enzymes (systemic spread) \| – Fever/arthritis/pneumonitis/perihepatitis/abnormal hepatic enzymes (systemic spread) \| \| Third stage \| Third stage \| \| – Genito-anorectal syndrome (more often in women) \| – Genito-anorectal syndrome (more often in women) \| \| – Chronic proctitis \| – Chronic proctitis \| \| – Fistulae \| – Fistulae \| \| – Strictures \| – Strictures \| \| – Stenosis of rectum \| – Stenosis of rectum \| \| – Genital lymphedema (elephantiasis, “saxophone penis”) \| – Genital lymphedema (elephantiasis) \| \| – “Lymphorroids” \| – “Lymphorroids” \| \| \| – Scarring of vulva (esthiomene) \| Open in a new tab Note: Rare cases. The new clinical picture is mostly seen among the MSM. Hemorrhagic proctitis in MSM is the primary manifestation of infection following direct transmission to the rectal mucosa. It might also occur in women with rectal exposure.35 In the recent outbreaks of LGV in MSM in Western Europe, approximately 96% of all cases presented with proctitis; symptoms included rectal pain, anorectal bleeding, mucoid and/or hemopurulent rectal discharge, tenesmus, constipation, and other symptoms of lower gastrointestinal inflammation. However, patients may present to gastroenterologists or colorectal surgeons for persisting symptoms.36 Endoscopic features are nonspecific, with a wide range of differential diagnoses including Crohn’s disease, lymphoma, anorectal carcinoma, and other sexually transmitted ulcerative infections (eg, syphilis, herpes).37,38 LGV proctitis mimics chronic inflammatory bowel disease, both clinically and in the pathological substrate. These cases may present with an incomplete or undisclosed history of proctosigmoiditis, without the characteristic adenopathy syndrome. During the initial evaluation and colonoscopy, there is a strong clinical and endoscopic suspicion of inflammatory bowel disease by virtue of presentation and endoscopic and histological findings. The diagnosis of IBD is subsequently modified to LGV proctosigmoiditis when one or more of the following transpire: 1) there is a failure of response to inflammatory bowel disease therapy; 2) additional components of history are identified (MSM/travel); 3) initially performed chlamydia antibody test shows seroconversion in the follow-up sample; and 4) response to antibiotics effective against chlamydia.39 Some patients reported systemic symptoms (fever and malaise). Genital ulcers and inguinal symptoms were less common.36 Several cases of pharyngeal MSM LGV infection have been reported recently. LGV serovars can cause symptomatic ulceration and pharyngitis but also asymptomatic carriage at this site.33 Secondary lesions (secondary LGV) Secondary LGV begins within 2–6 weeks after the onset of primary lesion. Depending on the site of inoculation, LGV can cause inguinal syndrome (after primary lesion of the anterior vulva, penis, or urethra) or anorectal syndrome (usually after primary lesion of the posterior vulva, vagina, or anus). Inguinal syndrome presents with painful inflammation of the inguinal (superficial and deep) lymph nodes and occurs mostly in men (occurs in only 20% of women with LGV). It is the most common clinical manifestation of genital LGV among heterosexuals. In two-thirds of cases, this produces unilateral enlargement, inflammation, suppuration, and abscesses. The disease process may involve one or many lymph nodes, and if adjacent to one another, they may coalesce. The central areas of such lymph nodes may then undergo necrosis. Fluctuant and suppurative lymph nodes then develop, causing the classic “bubo” of LGV. These “buboes” may rupture in one-third of patients, which may lead to sinus tract formation. In other cases, they may develop into hard, nonsuppurative masses. When both inguinal and femoral lymph nodes are involved, they can be separated by the inguinal (Poupart’s) ligament, “Groove sign”. This sign is pathognomonic of LGV, but occurs in only 15%–20% of cases.40 Women often have primary involvement of upper vagina, cervix, or posterior urethra; however, if they are receptive to anal sex they may have primary involvement of the rectum. These regions drain to the deep iliac or perirectal nodes and cause intra-abdominal or retroperitoneal lymphadenopathy that may lead to symptoms of lower abdominal pain or low-back pain.41 The systemic spread of LGV C. trachomatis may be associated with low-grade fever, chills, malaise, myalgias, and arthralgias. In addition, systemic spread occasionally results in arthritis, pneumonitis, abnormal hepatic enzymes, and perihepatitis. Rare systemic complications include cardiac involvement, aseptic meningitis, and ocular inflammatory disease.40 Reactive arthritis in MSM following LGV proctitis has been reported in several cases in recent years.42 In the rare pharyngeal syndrome affecting the mouth and throat, cervical lymphadenopathy and buboes can occur.18 Tertiary-stage/genito-anorectal syndrome/anogenitorectal syndrome/third-stage LGV This stage manifests predominantly in women, but also in homosexual men, because of the location of the involved lymphatics. It is characterized by a chronic inflammatory response and the destruction of tissue, which is followed by the formation of perirectal abscess, fistulas, strictures, and stenosis of rectum. Lymphorroids are hemorrhoid-like swellings of obstructed perirectal, and intestinal lymphatics may also occur. If it is not treated, chronic progressive lymphangitis leads to chronic edema and sclerosing fibrosis. This results in strictures and fistulas that can cause elephantiasis of the genitals, esthiomene (chronic ulcerative disease of vulva leading to disfiguring fibrosis and scarring), and frozen pelvis syndrome.9,41 Penile and scrotal edema and distortion have been termed “saxophone penis”.43 LGV proctitis can lead not only to rectal stricture but also to development of mega colon.44 Within the current MSM outbreak, tertiary complications of anorectal LGV such as strictures and fistulas have been observed rarely.45 Finally, regarding clinical features, it has to be noted that in the present LGV epidemic among MSM, UK cohorts showed almost all LGV to be symptomatic,3,14 but Dutch studies had a significant proportion of asymptomatic infection detected.46,47 In a study conducted in the Netherlands that included LGV among MSM diagnosed in 2003–2004, only one patient, with onset of illness in April 2003, had symptoms usually associated with LGV (ie, inguinal adenopathy and a painful genital ulcer); all other patients had gastrointestinal symptoms (eg, bloody proctitis with a purulent or mucous anal discharge and constipation).45 Diagnosis Modern techniques now rely on nucleic acid amplification tests (NAATs) in well-equipped laboratories.35 The assays have high sensitivity and specificity. C. trachomatis is an intracellular organism, so samples should contain cellular material. For the detection of LGV serovars of C. trachomatis, different DNA samples can be used: 1) primary anogenital lesion swab (ulcer base exudate), 2) rectal mucosa swab (when anorectal LGV is suspected), or 3) enlarged or fluctuant lymph nodes or buboes aspirate (when inguinal LGV is suspected).9,41 After topical disinfection, a 21-gauge needle should be inserted into the lymph node through healthy adjacent tissue and the pus aspirated into a syringe; a small volume (<0.5 mL) of saline solution may be injected and re-aspirated for nonfluctuant nodes. Urethral swab or first-catch urine specimen can be used as a sample when urethritis and/or inguinal lymphadenopathy is present and LGV is suspected as the cause. Urine specimen usually shows negative PCR results in case of anorectal manifestation of LGV.35 A two-step procedure is usually followed. The first step includes C. trachomatis NAAT test (test only confirms presence of C. trachomatis, and does not allow serogroup identification). NAATs are not approved for testing samples from extra-genital sites, but studies have demonstrated high sensitivity and specificity in rectal chlamydial infections.9,41,47–49C. trachomatis NAATs can be performed using one of three methods: PCR, strand displacement amplification, or transcription-mediated amplification. The second step is performed only if the first step test detects C. trachomatis in the sample. The second step diagnostic test is LGV biovar–specific DNA NAAT from the same sample used in first step test. Two tests for the second step are available: a real-time PCR-based test (detects all C. trachomatis LGV biovar strains)9,41,50 and a real-time quadriplex PCR-based assay. Both PCR techniques are reliable; the assay correctly identified 100% of non-LGV chlamydial specimens, 100% with no chlamydial infection, and 96.36% LGV specimens.47,51 For patients with genital disease, genital and lymph node specimens (lesion swabs and bubo aspirate) may be tested for C. trachomatis by 1) culture or 2) direct immunofluorescence (relatively insensitive). It is not always possible to culture the organism. Culture at primary stage is not often possible because the primary lesion often goes unnoticed. Culture on cyclohexamide-treated McCoy cells has a sensitivity of 75%–85% at best when the specimen is from LGV-suspected lesions and less for bubo aspirates.35,52 This method requires a lot of time, work, and money, and the availability is questionable. If NAATs are not available, then Chlamydia genus-specific serological assays can be used. Four techniques are available: 1) the complement fixation, 2) the single L-type immunofluorescence test, 3) the micro-immunofluorescence test, and 4) anti-MOMP IgA assay (UK). With appropriate clinical presentation, a complement fixation antibody titer of higher than 1:64 is considered diagnostic. Titers that are greater than 1:256 are highly suggestive. In addition, a four-fold increase in the complement fixation titer of blood samples taken 2 weeks apart is indicative.2 A four-fold rise in antibody or single-point titers of >1/64 and >128 for the micro-immunofluorescence test has been considered positive.35,53,54 When clinical features are suggestive of LGV, high Ig anti-MOMP antibodies titer supports the diagnosis. Recent study from the Netherlands showed the anti-MOMP IgA serological assay to be the most useful for rectal LGV infection. The test has 75.5% sensitivity and 74.3% specificity in asymptomatic MSM with rectal C. trachomatis, and 85.7% sensitivity and 84.2% specificity in symptomatic MSM. The IgA anti-MOMP assay can identify a considerable proportion of the (asymptomatic) anal LGV infections correctly. Yet, biovar L-specific NAAT are still the preferred diagnostic tests in clinical settings.35,55,56 It is important to stress that neither a low titer excludes LGV nor does a high titer in patient without symptoms confirm LGV.48,49 The identification of rectal polymorphonuclear leucocytes from rectal swabs is predictive of LGV proctitis, especially in HIV-positive MSM. In a study of 87 MSM with confirmed LGV, majority of samples (Gram-stained anorectal swabs) had more than 10 white blood cells (WBCs) per high-power field.35,47 Histology of the lymph nodes is not specific: follicular hyperplasia, abscesses, cryptitis, and crypt abscesses without distortion of crypt architecture (last two were the most common findings in 12 anorectal biopsies from LGV in MSM).30,49 Management A proper treatment regimen cures the infection and prevents further damage to tissues. LGV causes potentially severe infections with possibly irreversible sequels if adequate treatment is not begun promptly. Early and accurate diagnosis is essential.4,9,41 The management of the ongoing epidemic LGV in industrialized Western countries caused by the C. trachomatis variant L2b still needs improvements in diagnosis, therapy, and prevention. Therefore, the rapid C. trachomatis variant L2b-specific PCR has been developed to circumvent laborious ompA gene sequencing.57C. trachomatis should be treated with antibacterial drugs reaching high intracellular concentrations. In general, intracellular-acting agents such as doxycycline, erythromycin, and azithromycin and certain quinolones are advised.58 Doxycycline is the drug of choice for patients who are not pregnant.9,35,41 Pregnant and lactating women should be treated with erythromycin or azithromycin. Only one case of clinical failure with extended doxycycline therapy has been reported in an HIV-negative MSM with anorectal LGV. He subsequently responded to treatment with moxifloxacin 400 mg daily for 10 days; no isolate was available for resistance testing.35,59 Three MSMs who developed inguinal LGV failed the recommended 21-day doxycycline treatment. A recent paper suggests that inguinal LGV may require prolonged courses of doxycycline, exceeding the currently advised 21-day regimen.3 Current STD treatment guidelines for MSM in the United States recommend treatment of rectal chlamydia with a single 1-g dose of azithromycin, but there are increasing concerns about its effectiveness with treatment failures reported.59–61 A recent retrospective study conducted in the United States compared azithromycin and doxycycline for the treatment of rectal chlamydial infection; persistent/recurrent infection was higher among men treated with azithromycin (22%) compared with doxycycline (8%). These data suggest that doxycycline may be more effective than azythromycin in the treatment of rectal chlamydial infections.61 Factors that may be contributing to treatment failure for rectal chlamydia include: pharmacokinetic properties of azithromycin and doxycycline in rectal tissue, the ability of chlamydia to transform into a persistent state that is less responsive to antimicrobial therapy, the impact of the rectal microbiome on chlamydia, and heterotypic resistance.60 Recommended treatment regimens for LGV are listed in Table 2. As adjunctive therapy, the aspiration of fluctuant buboes is recommended for pain relief and prevention of rupture or chronic sinus formation, in contrast to surgical incision of buboes due to potential complications. The pharynx is a reservoir for chlamydia and LGV and may play a role in ongoing transmission. Although spontaneous clearance may occur in untreated patients with pharyngeal chlamydia, in high-risk STD clinic patients, testing the pharynx for chlamydia should be considered.56 Table 2. Recommended treatment regimens for lymphogranuloma venereum \| Drug \| Regimen \| Mechanism/possible side effects \| Comment \| :--- :--- \| \| Doxycycline \| 100 mg twice daily for 21 days, orally \| – Inhibits protein synthesis by binding to 30S ribosomal subunits of susceptible bacteria – Dyspepsia; nausea; diarrhea; photosensitivity; darkening of skin, nails, eyes, teeth, gums, or scars; esophageal ulceration; Fanconi syndrome (nephrotoxicity); steatosis (hepatotoxicity); headache and vision problems (secondary intracranial hypertension-pseudotumor cerebri)62–65 \| – First choice, recommended by Centers for Disease Control – Contraindicated in pregnancy and breastfeeding – Antacids that contain aluminum, calcium, or magnesium or any product that contains iron such as vitamin or mineral supplements should not be taken \| \| Erythromycin \| 500 mg four times daily for 21 days, orally \| – Inhibits bacterial growth by blocking dissociation of peptidyl tRNA from ribosomes – Diarrhea, stomach pain, nausea, vomiting66,67 \| Second choice; recommended by Centers for Disease Control \| \| Azithromycin \| 1 g stat, orally 1g weekly for 3 weeks, orally \| – Inhibits bacterial protein synthesis by binding to the 50S ribosomal subunit of the bacterial 70S ribosome – The same as for erythromycin \| Should be considered as second choice, but evidence is lacking to recommend this drug currently \| \| Tetracycline \| 500 mg four times daily for 21 days, orally \| – The same as for doxycycline – The same side effects listed earlier for doxycycline \| The same as for doxycycline \| \| Minocycline \| 300 mg loading dose, followed by 200 mg twice daily for 21 days, orally \| – The same as for doxycycline – Vertigo, dizziness, ataxia, tinnitus, and the side effects listed earlier for doxycycline68,69 \| The same as for doxycycline \| \| Moxifloxacin \| 400 mg once daily for 21 days, orally \| – Blocks DNA gyrase enzyme (it is responsible for production and repair of bacterial DNA) and it leads to bacteria death – Nausea, dizziness, diarrhea, QT prolongation, and photosensitivity \| – Administration should be separated from aluminum- and magnesium-containing antacids, sucralfate, and multivitamins because they can lower absorption of moxifloxacin and reduce its effectiveness – Should be used with caution with warfarin (increases risk for bleeding) and sotalol (abnormal heart rhythm) \| Open in a new tab A recent study suggests that HIV seropositivity was the strongest risk factor for LGV and that proctoscopic findings and elevated WBC counts in anorectal smear specimens were the only clinically relevant predictors for LGV infection. For all MSM reporting receptive anorectal intercourse, rectal chlamydia screening is recommended. If chlamydia test results are not available, for MSM reporting receptive anorectal intercourse who have proctitis noted by proctoscopic examination and a WBC count of >10 cells/high-power field in an anorectal Gram stain specimen or who have proctitis noted by proctoscopic examination and HIV seropositivity, treatment with doxycycline (100 mg twice per day) is advised until chlamydia test results are available, with a minimum duration of treatment of 7 days. If the anorectal chlamydia test result is negative, no treatment should be administered or doxycycline treatment should be stopped after a minimum of 7 days. If the anorectal chlamydia test result is positive and LGV testing is available, doxycycline treatment (100 mg twice per day) should be started or continued until LGV is rejected. A 7-day course of doxycycline is effective for treatment of a non-LGV chlamydia. Therapy should be continued until 21 days after confirmation of LGV. If the anorectal chlamydia test result is positive and if LGV testing is unavailable, doxycycline treatment (100 mg twice per day for up to 21 days) should be started or continued for MSM who meet one of the following criteria: proctitis noted during initial proctoscopic examination, >10 WBCs/high-power field in the initial Gram-stained anorectal smear specimen, or HIV seropositivity.9 Persons who have had sexual contact with a patient who has LGV within 4 weeks before the onset of the patient’s symptoms, or the last 3 months if asymptomatic LGV is detected, must be tested for C. trachomatis/LGV and start treatment (doxycycline 100 mg twice daily for 21 day). Patients must be followed up during the treatment, until disease signs and symptoms have resolved. Repeated testing for syphilis, hepatitis B and C, and HIV to detect early infection should be performed. A test of cure for LGV is not considered necessary if the recommended 21-day course of doxycycline is completed.9,41 Health promotion/prevention The outbreaks of LGV among MSM in developed countries support the need for careful screening of these patients. A routine screening for rectal chlamydia is recommended in asymptomatic men at risk, as this may represent an important reservoir for the onward transmission of infection. Treatment of this group of patients is essential in the attempt to eradicate the disease. Patients should get information about safer sex behavior. Condom use may reduce the risk of LGV transmission, but uncovered ulcerated areas remain a problem. With early and accurate diagnosis and appropriate antibiotic therapy, the prognosis is excellent, but reinfection and relapses may occur.70 Footnotes Disclosure The authors have no conflict of interest related to this paper. References 1.Beigi RH. Lymphogranuloma venereum. In: Beigi RH, editor. Sexually Transmitted Diseases. West Sussex, UK: John Wiley & Sons, LTD; 2012. pp. 49–52. [Google Scholar] 2.Taraktchoglou M, Pacey AA, Turnbull JE, Eley A. Infectivity of Chlamydia trachomatis serovar LGV but not E is dependent on host cell heparansulfate. Infect Immun. 2001;69(2):968–976. doi: 10.1128/IAI.69.2.968-976.2001. [DOI] [PMC free article] [PubMed] [Google Scholar] 3.Oud EV, de Vrieze NH, de Meij A, de Vries HJ. Pitfalls in the diagnosis and management of inguinal lymphogranuloma venereum: important lessons from a case series. Sex Transm Infect. 2014;90(4):279–282. doi: 10.1136/sextrans-2013-051427. [DOI] [PubMed] [Google Scholar] 4.de Vries HJ, Zingoni A, Kreuter A, Moi H, White JA. 2013 European guideline on the management of lymphogranuloma venereum. J Eur Acad Dermatol Venereol. 2015;29(1):1–6. doi: 10.1111/jdv.12461. [DOI] [PubMed] [Google Scholar] 5.de Vrieze NH, de Vries HJ. Lymphogranuloma venereum among men who have sex with men. An epidemiological and clinical review. Expert Rev Anti Infect Ther. 2014;12(6):697–704. doi: 10.1586/14787210.2014.901169. [DOI] [PubMed] [Google Scholar] 6.Mabey D, Peeling RW. Lymphogranuloma venereum. Sex Transm Infect. 2002;78(2):90–92. doi: 10.1136/sti.78.2.90. [DOI] [PMC free article] [PubMed] [Google Scholar] 7.Behets FM, Brathwaite AR, Hylton-Kong T, et al. Genital ulcers: etiology, clinical diagnosis, and associated human immunodeficiency virus infection in Kingston, Jamaica. Clin Infect Dis. 1999;28(5):1086–1090. doi: 10.1086/514751. [DOI] [PubMed] [Google Scholar] 8.Ndinya-Achola JO, Kihara AN, Fisher LD, et al. Presumptive specific clinical diagnosis of genital ulcer disease (GUD) in a primary health care setting in Nairobi. Int J STD AIDS. 1996;7(3):201–205. doi: 10.1258/0956462961917627. [DOI] [PubMed] [Google Scholar] 9.Van der Bij AK, Spaargaren J, Morré SA, et al. Diagnostic and clinical implications of anorectal lymphogranuloma venereum in men who have sex with men: a retrospective case-control study. Clin Infect Dis. 2006;42(2):186–194. doi: 10.1086/498904. [DOI] [PubMed] [Google Scholar] 10.Ward H, Alexander S, Carder C, et al. The prevalence of Lymphogranuloma venereum (LGV) infection in men who have sex with men: results of a multi-centre case finding study. Sex Transm Infect. 2009;85(3):173–175. doi: 10.1136/sti.2008.035311. [DOI] [PMC free article] [PubMed] [Google Scholar] 11.Hughes G, Alexander S, Simms I, et al. LGV Incident Group Lymphogranuloma venereum diagnoses among men who have sex with men in the UK: interpreting a cross-sectional study using an epidemic phase-specific framework. Sex Transm Infect. 2013;89(7):542–547. doi: 10.1136/sextrans-2013-051051. [DOI] [PubMed] [Google Scholar] 12.de Vries HJ, van der Bij AK, Fennema JS, et al. Lymphogranuloma venereum proctitis in men who have sex with men is associated with anal enema use and high-risk behavior. Sex Transm Dis. 2008;35(2):203–208. doi: 10.1097/OLQ.0b013e31815abb08. [DOI] [PubMed] [Google Scholar] 13.Savage EJ, van de Laar MJ, Gallay A, et al. European Surveillance of Sexually Transmitted Infections (ESSTI) network Lymphogranuloma venereum in Europe, 2003–2008. Euro Surveill. 2009;14(48):19428. doi: 10.2807/ese.14.48.19428-en. [DOI] [PubMed] [Google Scholar] 14.Ward H, Martin I, Macdonald N, et al. Lymphogranuloma venereum in the United Kingdom. Clin Infect Dis. 2007;44(1):26–32. doi: 10.1086/509922. [DOI] [PubMed] [Google Scholar] 15.Behets FM, Andriamiadana J, Randrianasolo D, et al. Chancroid, primary syphilis, genital herpes, and lymphogranuloma venereum in Antananarivo, Madagascar. J Infect Dis. 1999;180(4):1382–1385. doi: 10.1086/315005. [DOI] [PubMed] [Google Scholar] 16.Brathwaite AR, Figueroa JP, Ward E. A comparison of prevalence rates of genital ulcers among persons attending a sexually transmitted disease clinic in Jamaica. West Indian Med J. 1997;46(3):67–71. [PubMed] [Google Scholar] 17.Vargas-Leguas H, Garcia de Olalla P, Arando M, et al. Lymphogranuloma venereum: a hidden emerging problem, Barcelona 2011. Euro Surveill. 2012;17(2):20057. [PubMed] [Google Scholar] 18.Korhonen S, Hiltunen-Back E, Puolakkainen M. Genotyping of Chlamydia trachomatis in rectal and pharyngeal specimens: identification of LGV genotypes in Finland. Sex Transm Infect. 2012;88:465–469. doi: 10.1136/sextrans-2011-050458. [DOI] [PubMed] [Google Scholar] 19.Vanousova D, Zakoucka H, Jilich D, et al. First detection of Chlamydia trachomatis LGV biovar in the Czech Republic, 2010–2011. Euro Surveill. 2012;17(2):20055. [PubMed] [Google Scholar] 20.Peuchant O, Baldit C, Le Roy C, et al. First case of Chlamydia trachomatis L2b proctitis in a woman. Clin Microbiol Infect. 2011;17(12):21–23. doi: 10.1111/j.1469-0691.2011.03661.x. [DOI] [PubMed] [Google Scholar] 21.Verweij SP, Ouburg S, de Vries H, et al. The first case record of a female patient with bubonic lymphogranuloma venereum (LGV), serovariant L2b. Sex Trans Infect. 2012;88(5):346–347. doi: 10.1136/sextrans-2011-050298. [DOI] [PubMed] [Google Scholar] 22.Kennedy JE, Higgins SP. Complicated lymphogranuloma venereum infection mimicking deep vein thrombosis in an HIV-positive man. Int J STD AIDS. 2012;23(3):219–220. doi: 10.1258/ijsa.2011.011217. [DOI] [PubMed] [Google Scholar] 23.Pendle S, Growers A. Reactive arthritis associated with proctitis due to Chlamydia trachomatis serovar L2b. Sex Transm Dis. 2012;39(1):79–80. doi: 10.1097/OLQ.0b013e318235b256. [DOI] [PubMed] [Google Scholar] 24.Schachter J. Confirming positive results of nucleic acid amplification tests (NAATs) for Chlamydia trachomatis: all NAATs are not created equal. J Clin Microbiol. 2005;43(3):1372–1373. doi: 10.1128/JCM.43.3.1372-1373.2005. [DOI] [PMC free article] [PubMed] [Google Scholar] 25.Basta-Juzbasic A, Ceovic R. Chancroid, lymphogranuloma venereum, granuloma inguinale, genital herpes simplex infection, and molluscum contagiosum. Clin Dermatol. 2014;32(2):290–298. doi: 10.1016/j.clindermatol.2013.08.013. [DOI] [PubMed] [Google Scholar] 26.Hardick J, Quinn N, Eshelman S, et al. The HPTN 061 Study Team Multi-site screening for lymphogranuloma venereum (LGV) in the USA. Sex Transm Infect. 2011;87(1):60–64. [Google Scholar] 27.Chen JC, Stephens RS. Trachoma and LGV biovars of Chlamydia trachomatis share the same glycosaminoglycan-dependent mechanism for infection of eukaryotic cells. Mol Microbiol. 1994;11(3):501–507. doi: 10.1111/j.1365-2958.1994.tb00331.x. [DOI] [PubMed] [Google Scholar] 28.Hadfield TL, Lamy Y, Wear DJ. Demonstration of Chlamydia trachomatis in inguinal lymphadenitis of lymphogranuloma venereum: a light microscopy, electron microscopy and polymerase chain reaction study. Mod Pathol. 1995;8(9):924–929. [PubMed] [Google Scholar] 29.Jones RB, Batteiger BE. Chlamydia trachomatis (trachoma, perinatal infections, lymphogranuloma venereum, and other genital infections) In: Mandell GL, Bennett JE, Dolin R, editors. Principles and Practice of Infectious Diseases. Fifth ed. Philadelphia: Churchill Livingstone; 2000. p. 1989. [Google Scholar] 30.Jebbari H, Alexander S, Ward H, et al. UK LGV Incident Group Update on lymphogranuloma venereum in the UK. Sex Transm Infect. 2007;83(4):324–326. doi: 10.1136/sti.2007.026740. [DOI] [PMC free article] [PubMed] [Google Scholar] 31.Sethi G, Allason-Jones E, Richens J, et al. Lymphogranuloma venereum presenting as genital ulceration and inguinal syndrome in men who have sex with men in London, UK. Sex Transm Infect. 2009;85(3):165–170. doi: 10.1136/sti.2008.034348. [DOI] [PubMed] [Google Scholar] 32.Singhrao T, Higham E, French P. Lymphogranuloma venereum presenting as perianal ulceration: an emerging clinical presentation? Sex Transm Infect. 2011;87(2):123–124. doi: 10.1136/sti.2010.047019. [DOI] [PubMed] [Google Scholar] 33.Dosekun O, Edmonds S, Stockwell S, French P, White JA. Lymphogranuloma venereum detected from pharynx in four London men who have sex with men. Int J STD AIDS. 2013;24(6):495–496. doi: 10.1177/0956462412472830. [DOI] [PubMed] [Google Scholar] 34.Andrada MT, Dhar JK, Wilde H. Oral lymphogranuloma venereum and cervical lymphadenopathy. Mil Med. 1974;139(2):99–101. [PubMed] [Google Scholar] 35.White J, O’Farrell N, Daniels D. 2013 UK national guideline for the management of lymphogranuloma venereum. Int J STD AIDS. 2013;24(8):593–601. doi: 10.1177/0956462413482811. [DOI] [PubMed] [Google Scholar] 36.Nieuwenhuis RF, Ossewaarde JM, Götz HM, et al. Resurgence of lymphogranuloma venereum in Western Europe: an outbreak of Chlamydia trachomatis serovar L2 proctitis in The Netherlands among men who have sex with men. Clin Infect Dis. 2004;39(7):996–1003. doi: 10.1086/423966. [DOI] [PubMed] [Google Scholar] 37.Ahdoot A, Kotler DP, Suh JS, Kutler C, Flamholz R. Lymphogranuloma venereum in human immunodeficiency virus-infected individuals in New York City. J Clin Gastroenterol. 2006;40:385–390. doi: 10.1097/00004836-200605000-00005. [DOI] [PubMed] [Google Scholar] 38.Tinmouth J, Rachlis A, Wesson T, Hsieh E. Lymphogranuloma venereum in North America: case reports and an update for gastroenterologists. Clin Gasteroenterol Hepatol. 2006;4:469–473. doi: 10.1016/j.cgh.2005.12.006. [DOI] [PubMed] [Google Scholar] 39.Gallegos M, Bradly D, Jakate S, Keshavarzian A. Lymphogranuloma venereum proctosigmoiditis is a mimicker of inflammatory bowel disease. World J Gastroenterol. 2012;18(25):3317–3321. doi: 10.3748/wjg.v18.i25.3317. [DOI] [PMC free article] [PubMed] [Google Scholar] 40.Roest RW, Van der Meijden WI. European guideline for the management of tropical genito-ulcerative diseases. Int J STD AIDS. 2001;12(3):78–83. doi: 10.1258/0956462011923994. [DOI] [PubMed] [Google Scholar] 41.de Vries HJ, Zingoni A, White JA, Ross JD, Kreuter A. 2013 European guideline on the management of proctitis, proctocolitis and enteritis caused by sexually transmissible pathogens. Int J STD AIDS. 2013;25(7):465–474. doi: 10.1177/0956462413516100. [DOI] [PubMed] [Google Scholar] 42.Kober C, Richardson D, Bell C, Walker-Bone K. Acute seronegative polyarthritis associated with lymphogranuloma venereum infection in a patient with prevalent HIV infection. Int J STD AIDS. 2011;22(1):59–60. doi: 10.1258/ijsa.2010.010262. [DOI] [PubMed] [Google Scholar] 43.Koley S, Mandal RK. Saxophone penis after unilateral inguinal bubo of lymphogranuloma venereum. Indian J Sex Transm Dis. 2013;34(2):149–151. doi: 10.4103/2589-0557.120575. [DOI] [PMC free article] [PubMed] [Google Scholar] 44.Pinsk I, Saloojee N, Friedlich M. Lymphogranuloma venereum as a cause of rectal stricture. Can J Surg. 2007;50(6):31–32. [PMC free article] [PubMed] [Google Scholar] 45.Centers for Disease Control and Prevention (CDC) Lymphogranuloma venereum among men who have sex with men – Netherlands, 2003–2004. MMWR Morb Mortal Wkly Rep. 2004;53(42):985–988. [PubMed] [Google Scholar] 46.Spaargaren J, Fennema HS, Morré SA, de Vries HJ, Coutinho RA. New lymphogranuloma venereum Chlamydia trachomatis variant. Amsterdam. Emerg Infect Dis. 2005;11(7):1090–1092. doi: 10.3201/eid1107.040883. [DOI] [PMC free article] [PubMed] [Google Scholar] 47.Morton AN, Fairley CK, Zaia AM, Chen MY. Anorectal lymphogranuloma venereum in a Melbourne man. Sex Health. 2006;3(3):189–190. doi: 10.1071/sh06029. [DOI] [PubMed] [Google Scholar] 48.Cusini M, Boneschi V, Arancio L, et al. Lymphogranuloma venereum: the Italian experience. Sex Transm Infect. 2009;85(3):171–172. doi: 10.1136/sti.2008.032862. [DOI] [PubMed] [Google Scholar] 49.Bachmann LH, Johnson RE, Cheng H, et al. Nucleic acid amplification tests for diagnosis of Neisseria gonorrhoeae and Chlamydia trachomatis rectal infections. J Clin Microbiol. 2010;48(5):1827–1832. doi: 10.1128/JCM.02398-09. [DOI] [PMC free article] [PubMed] [Google Scholar] 50.Morré SA, Spaargaren J, Fennema JS, de Vries HJ, Coutinho RA, Peña AS. Real-time polymerase chain reaction to diagnose lymphogranuloma venereum. Emerg Infect Dis. 2005;11(8):1311–1312. doi: 10.3201/eid1108.050535. [DOI] [PMC free article] [PubMed] [Google Scholar] 51.Lister NA, Tabrizi SN, Fairley CK, Garland S. Validation of roche COBAS amplicor assay for detection of Chlamydia trachomatis in rectal and pharyngeal specimens by an omp1 PCR assay. J Clin Microbiol. 2004;42(1):239–241. doi: 10.1128/JCM.42.1.239-241.2004. [DOI] [PMC free article] [PubMed] [Google Scholar] 52.Stamm WE. Lymphogranuloma venereum. In: Holmes KK, Sparling PF, Stamm WE, et al., editors. Sexually Transmitted Diseases. 4th ed. New York: Mc-Graw Hill; 2008. pp. 595–606. [Google Scholar] 53.Centers for Disease Control and Prevention 2010 Sexually transmitted diseases treatment guidelines. MMWR Morb Mortal Wkly Rep. 2010;59(RR–12):1–110. [Google Scholar] 54.Bauwens JE, Orlander H, Gomez MP, et al. Epidemic lymphogranuloma venereum during epidemics of crack cocaine use and HIV infection in the Bahamas. Sex Transm Dis. 2002;29(5):255–258. doi: 10.1097/00007435-200205000-00001. [DOI] [PubMed] [Google Scholar] 55.de Vries HJ, Smelov V, Ouburg S, et al. Anal lymphogranuloma venereum infection screening with IgA anti-Chlamydia trachomatis specific major outer membrane protein serology. Sex Transm Dis. 2010;37(12):789–795. doi: 10.1097/OLQ.0b013e3181e50671. [DOI] [PubMed] [Google Scholar] 56.van Rooijen MS, Schim van der Loeff MF, Morré SA, van Dam AP, Speksnijder AG, de Vries HJ. Spontaneous pharyngeal Chlamydia trachomatis RNA clearance. A cross-sectional study followed by a cohort study of untreated STI clinic patients in Amsterdam, The Netherlands. Sex Transm Infect. 2014 doi: 10.1136/sextrans-2014-051633. [DOI] [PubMed] [Google Scholar] 57.Verweij SP, Catsburg A, Ouburg S, et al. Lymphogranuloma venereum variant L2b-specific polymerase chain reaction: insertion used to close an epidemiological gap. Clin Microbiol Infect. 2011;17(11):1727–1730. doi: 10.1111/j.1469-0691.2011.03481.x. [DOI] [PubMed] [Google Scholar] 58.Stock I, Henrichfreise B. Infections with Chlamidia trachomatis. Med Monatsschr Pharm. 2012;35:209–222. [PubMed] [Google Scholar] 59.Méchaï F, de Barbeyrac B, Aoun O, Mérens A, Imbert P, Rapp C. Doxycycline failure in lymphogranuloma venereum. Sex Transm Infect. 2010;86(3):278–279. doi: 10.1136/sti.2009.042093. [DOI] [PubMed] [Google Scholar] 60.Hocking JS, Kong FY, Timms P, Huston WM, Tabrizi SN. Treatment of rectal chlamydia infection may be more complicated than we originally thought. J Antimicrob Chemother. 2014 doi: 10.1093/jac/dku493. [DOI] [PubMed] [Google Scholar] 61.Khosropour CM, Dombrowski JC, Barbee LA, Manhart LE, Golden MR. Comparing azithromycin and doxycycline for the treatment of rectal chlamydial infection: a retrospective cohort study. Sex Transm Dis. 2014;41(2):79–85. doi: 10.1097/OLQ.0000000000000088. [DOI] [PMC free article] [PubMed] [Google Scholar] 62.Smith K, Leyden JJ. Safety of doxycycline and minocycline: a systematic review. Clin Ther. 2005;27(9):132–142. doi: 10.1016/j.clinthera.2005.09.005. [DOI] [PubMed] [Google Scholar] 63.Akcam M, Artan R, Akcam FZ, Yilmaz A. Nail discoloration induced by doxycycline. Pediatr Infect Dis J. 2005;24(9):845–846. doi: 10.1097/01.inf.0000177283.71692.56. [DOI] [PubMed] [Google Scholar] 64.Foster JA, Sylvia LM. Doxycycline-induced esophageal ulceration. Ann Pharmacother. 1994;28(10):1185–1187. [PubMed] [Google Scholar] 65.Tabibian JH, Gutierrez MA. Doxycycline-induced pseudotumor cerebri. South Med J. 2009;102(3):310–311. doi: 10.1097/SMJ.0b013e31818f98f0. [DOI] [PubMed] [Google Scholar] 66.Periti P, Mazzei T, Mini E, Novelli A. Adverse effects of macrolide antibacterials. Drug Saf. 1993;9(5):346–364. doi: 10.2165/00002018-199309050-00004. [DOI] [PubMed] [Google Scholar] 67.Oberg KC, Bauman JL. QT interval prolongation and torsades de pointes due to erythromycin lactobionate. Pharmacotherapy. 1995;15(6):687–692. [PubMed] [Google Scholar] 68.WeeseMayer DE, Yang RJ, Mayer JR, Zaparackas Z. Minocycline and Pseudotumor cerebri: the well-known but well-kept secret. Pediatrics. 2001;108(2):519–520. doi: 10.1542/peds.108.2.519. [DOI] [PubMed] [Google Scholar] 69.Macneil M, Haase DA, Tremaine R, Marrie TJ. Fever, lymphadenopathy, eosinophilia, lymphocytosis, hepatitis, and dermatitis: a severe adverse reaction to minocycline. J Am Acad Dermatol. 1997;36(2):347–350. doi: 10.1016/s0190-9622(97)80414-8. [DOI] [PubMed] [Google Scholar] 70.Annan NT, Sullivan AK, Nori A, et al. Rectal chlamydia – a reservoir of undiagnosed infection in men who have sex with men. Sex Transm Infect. 2009;85(3):176–179. doi: 10.1136/sti.2008.031773. [DOI] [PubMed] [Google Scholar] Articles from Infection and Drug Resistance are provided here courtesy of Dove Press ACTIONS View on publisher site PDF (224.9 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases 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
4585	https://math.stackexchange.com/questions/503744/prove-that-a-triangle-is-right-angled	trigonometry - Prove that a triangle is right angled ... - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Prove that a triangle is right angled ... Ask Question Asked 12 years ago Modified12 years ago Viewed 3k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. ... if one side is double the other and the angle opposite these sides differ by π 3 π 3 It would be better if this was proven using only algebraic trigonometric identities(sine/cosine rules, etc.) Let A,B,C be the angles and a,b,c be the respective sides opposite to them. b=2 a\|B−A\|=π 3 sin(B−A)=√3 2 sin B cos A−cos B sin A=√3 2 b k b 2+c 2−a 2 2 b c−a k a 2+c 2−b 2 2 a c=√3 2 1 k c(b 2−a 2)=√3 2 3 a 2 k c=√3 2 b\|B−A\|sin(B−A)sin B cos A−cos B sin A b k b 2+c 2−a 2 2 b c−a k a 2+c 2−b 2 2 a c 1 k c(b 2−a 2)3 a 2 k c=2 a=π 3=3–√2=3–√2=3–√2=3–√2=3–√2 And I'm stuck at that. Another approach a sin A=b sin B a sin A=2 a sin(π 3+A)√3 2 cos A−1 2 sin A=2 sin A tan A=√3 5 a sin A a sin A 3–√2 cos A−1 2 sin A tan A=b sin B=2 a sin(π 3+A)=2 sin A=3–√5 That doesn't lead anywhere too trigonometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Sep 24, 2013 at 17:06 user80551user80551 asked Sep 24, 2013 at 16:56 user80551user80551 569 1 1 gold badge 9 9 silver badges 18 18 bronze badges 3 If B B is the bigger angle, b b should be 2 a 2 a, rather than a=2 b a=2 b.Zvi Rosen –Zvi Rosen 2013-09-24 17:02:42 +00:00 Commented Sep 24, 2013 at 17:02 Oh right, corrected user80551 –user80551 2013-09-24 17:07:05 +00:00 Commented Sep 24, 2013 at 17:07 What is the angle "opposite to two sides" in a triangle?!DonAntonio –DonAntonio 2013-09-24 17:21:55 +00:00 Commented Sep 24, 2013 at 17:21 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. You have a mistake computing sin(π 3+A)sin(π 3+A), the correct formula is sin(x+y)=sin x cos y+cos x sin y, sin(x+y)=sin x cos y+cos x sin y, which leads here to 1 2 sin A+√3 2 cos A=2 sin A, 1 2 sin A+3–√2 cos A=2 sin A, and further to the well-known tan A=1√3, tan A=1 3–√, whence A=π 6 A=π 6. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 24, 2013 at 17:24 Daniel FischerDaniel Fischer 212k 19 19 gold badges 307 307 silver badges 438 438 bronze badges 0 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Visual aid for in case you're interested in developing a synthetic proof: Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 24, 2013 at 17:31 David HDavid H 32.7k 3 3 gold badges 84 84 silver badges 143 143 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let Δ A B C Δ A B C be our triangle, with sides \|A B\|=x,\|B C\|=2 x\|A B\|=x,\|B C\|=2 x , and ∠A=β,∠C=α,β−α=π 3 ∠A=β,∠C=α,β−α=π 3 Apply now the Law of Sines: \|A B\|cos α=\|B C\|sin β⟺x sin α=2 x sin(α+π 3)=2 x 1 2 sin α+√3 2 cos α⟹ \|A B\|cos α=\|B C\|sin β⟺x sin α=2 x sin(α+π 3)=2 x 1 2 sin α+3√2 cos α⟹ ⟹x sin α=4 x sin α+√3 cos α⟹sin α+√3 cos α=4 sin α⟹ ⟹tan α=1√3⟹α=π 6 and thus β=α+π 3=π 6+π 3=π 2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 24, 2013 at 17:31 DonAntonioDonAntonio 215k 19 19 gold badges 143 143 silver badges 291 291 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 trigonometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 12geometric meaning of a trigonometric identity 4What is a rational trigonometric function? Is cos x rational? 5How do I deal with sines or cosines greater than 1? 0Proving that a triangle is isosceles 1Is there an auxiliary formula for cosine and tan? 1Proving (b+c)cos B+C 2=a cos B−C 2 for any △A B C 4Finding all sides and angles of a triangle Hot Network Questions Gluteus medius inactivity while riding Another way to draw RegionDifference of a cylinder and Cuboid Is encrypting the login keyring necessary if you have full disk encryption? Why include unadjusted estimates in a study when reporting adjusted estimates? Origin of Australian slang exclamation "struth" meaning greatly surprised How many color maps are there in PBR texturing besides Color Map, Roughness Map, Displacement Map, and Ambient Occlusion Map in Blender? Drawing the structure of a matrix Program that allocates time to tasks based on priority Are there any legal strategies to modify the nature and rules of the UN Security Council in the face of one, or more SC vetos? How to fix my object in animation What were "milk bars" in 1920s Japan? Numbers Interpreted in Smallest Valid Base How do you emphasize the verb "to be" with do/does? Is existence always locational? My dissertation is wrong, but I already defended. How to remedy? How to convert this extremely large group in GAP into a permutation group. Cannot build the font table of Miama via nfssfont.tex If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church? Matthew 24:5 Many will come in my name! Fundamentally Speaking, is Western Mindfulness a Zazen or Insight Meditation Based Practice? What happens if you miss cruise ship deadline at private island? Overfilled my oil What's the expectation around asking to be invited to invitation-only workshops? Storing a session token in localstorage Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
4586	https://www.youtube.com/watch?v=9-kZBR6-c2A	Solve Non-Linear Systems of Equations - Substitution Method \| Eat Pi Eat Pi 18700 subscribers 317 likes Description 23276 views Posted: 5 Oct 2022 In this video, I teach you how to solve nonlinear systems of equations using the substitution method. The problems either have one, two, or no solutions and I do an example of each one. 0:00 - System with 2 Solutions 3:47 - System with 1 Solutions 5:10 - System with No Solutions If you have any questions, please leave them in the comment section below! Also, if you find the videos helpful, please like, share, and subscribe! 13 comments Transcript: System with 2 Solutions what's up you freaking Junior says so in this video I'm going to teach you how to solve a non-linear system of equations using the substitution method right there's basically three different cases you're either going to get two solutions one solution or no Solutions all right so I'm going to do an example of each one so let's do this one first so this one is y is equal to x squared plus x minus 1 and Y is equal to negative 2x plus 3. all right so as you can see it's saying that Y is equal to all this crap over here and Y is also equal to all this crap over here so the first thing we can do is set this crap equal to this scrap right here so that's what we're going to do we're going to say x squared plus or sorry plus x minus 1 is equal to this one over here negative 2x plus 3. okay now the second step is just getting all your terms on one side and the reason we have to do that is because we're going to have to factor okay so the first thing I'm going to do here is ADD 2x to both sides and then I'm also going to add or subtract 3 from this side right subtract 3 so then we subtract 3 from this side right so then on this side these terms cancel out right they just go to zero and then on this side we're going to have x squared plus 3x minus 4 is equal to zero right since we don't have anything on that side anymore okay so this is what we're going to have to factor so we have again x squared plus 3x minus four Okay so we're going to factor this now we have an x squared here so we can break that down into x times x and then to figure out the two numbers that go right here they have to multiply to negative four okay they have to multiply to negative 4. so two numbers now multiply to negative four but they have to add up to positive three the middle number okay so in this case we would have to use a positive 4 and a negative one right because 4 times negative 1 is equal to negative 4 and 4 plus negative 1 is equal to positive three so positive four and negative 1. and again this is all equal to zero okay and then the last thing we have to do here is just set each of our parentheses equal to zero so we have X plus 4 is equal to 0 and x minus 1 is equal to zero right so then on this top one our answer would be X is equal to negative four and then on this bottom one our answer would be X is equal to positive one okay so you get two answers right basically one for each set of parentheses okay so since we got two answers for X we're gonna basically get two answers for y and the way that we find y now is just by plugging these back into one of these original equations and I'm going to pick this one right here because that one looks like it's a lot easier to work with Okay so this one is an already over here Y is equal to negative two X plus three okay so let's plug in our first X term right here so we're going to get Y is equal to negative two times negative four plus 3 and that's equal to let's see negative 2 times negative 4 is positive eight so eight plus three is eleven okay and then for our other one we're going to get Y is equal to negative 2 times positive one plus three okay so negative two times one is negative two and negative two plus three is equal to positive one okay and I'm going to move these down just to kind of match them up a little bit more clearly so first we plugged in X is equal to negative four and we got Y is equal to eleven okay so that's our first answer our first pair so at negative four eleven and then the second one we did was X is equal to 1 and then we got Y is equal to one also all right so that's our second solution or a second pair so that one's at one comma one okay so those are your two solutions to this system of equations System with 1 Solutions okay here's the next one so we have y minus 5 is equal to negative x squared and Y is equal to 5. okay so one thing we can do here is well the second equation it's basically telling us exactly what Y is equal to right it's equal to five so we can just plug in a five for y right there okay so then we'll plug into 5 for y so we're gonna get five minus five is equal to negative x squared okay 5 minus five is zero so zero is equal to negative x squared okay now uh to get rid of the negative sign we can divide both sides by negative one so zero divided by negative one is zero so we get zero is equal to then those basically cancel out so we get x squared right so we get x squared is equal to zero now the only number we could possibly plug in for X right here would be zero right because zero squared is equal to zero so here we have one solution for X we have X is equal to zero okay so we know what our x coordinate is and then we can plug it in back into one of these equations over here but if you notice something it already tells us exactly what Y is equal to right it says Y is equal to five so we know that Y is equal to five okay so then our answer our solution our ordered pair would be zero comma five right so then this problem would only have one solution System with No Solutions okay and lastly here we have y is equal to Negative X plus seven and Y is equal to negative x squared minus two x minus one okay so again it's saying that Y is equal to all this over here and Y is equal to all this over here so we can just set all of this equal to all of this all right so that's what we're going to do so we're going to say Negative X plus seven is equal to negative x squared minus two x minus one okay and then you just want to again get everything on one side so we'll move it all to the right so here we're going to add X and we're going to subtract 7 right so then on this side we're going to add X and then subtract seven okay so then over here those cancel out this whole side goes to 0. all right so we're going to get 0 is equal to negative x squared minus x minus eight okay now I don't like this negative x squared over here so what I'm going to do is I'm going to divide this whole side by negative 1 and again what we did one side we do to the other right so then 0 divided by negative 1 is just 0 and then here all of these signs are going to flip right because we're dividing by negative one so then we're going to have positive x squared plus X plus eight okay that's a little cleaner to work with right so now we just have to factor this okay so here we have an x squared so again we can break that into x times x and then we have a a positive 8 here right so what two numbers can I multiply together to get positive 8 but they have to have a difference of basically a positive one right here right the middle number well in this case there is no solution right because the only ways we can multiply two numbers to get eight would be one times eight and two times four right so do either of these have a difference or do they add up to I should say to positive one no because 1 plus 8 is equal to nine and two plus four is equal to six right so neither of these pairs have a difference of positive one so that means we can't actually Factor this okay and since we can't Factor it it's not factorable that means there is no solution for this problem so if you found the video helpful definitely leave a thumbs up down below and if you have any other questions or want to see any other examples just let me know in the comments section below
4587	https://pubmed.ncbi.nlm.nih.gov/20306635/	Interventions for erythema nodosum leprosum. A Cochrane review - 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. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. To: Subject: Body: Format: [x] MeSH and other data Send email Cancel Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Add to My Bibliography My Bibliography Unable to load your delegates due to an error Please try again Add Cancel Your saved search Name of saved search: Search terms: Test search terms Would you like email updates of new search results? Saved Search Alert Radio Buttons Yes No Email: (change) Frequency: Which day? Which day? Report format: Send at most: [x] Send even when there aren't any new results Optional text in email: Save Cancel Create a file for external citation management software Create file Cancel Your RSS Feed Name of RSS Feed: Number of items displayed: Create RSS Cancel RSS Link Copy Actions Cite Collections Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Permalink Permalink Copy Display options Display options Format Page navigation Title & authors Abstract Similar articles Cited by Publication types MeSH terms Substances Related information Lepr Rev Actions Search in PubMed Search in NLM Catalog Add to Search . 2009 Dec;80(4):355-72. Interventions for erythema nodosum leprosum. A Cochrane review Natasja H J Van Veen1,Diana N J Lockwood,Wim H Van Brakel,J Ramirez Jr,Jan Hendrik Richardus Affiliations Expand Affiliation 1 Department of Public Health, Erasmus MC, University Medical Center Rotterdam, The Netherlands. PMID: 20306635 Item in Clipboard Interventions for erythema nodosum leprosum. A Cochrane review Natasja H J Van Veen et al. Lepr Rev.2009 Dec. Show details Display options Display options Format Lepr Rev Actions Search in PubMed Search in NLM Catalog Add to Search . 2009 Dec;80(4):355-72. Authors Natasja H J Van Veen1,Diana N J Lockwood,Wim H Van Brakel,J Ramirez Jr,Jan Hendrik Richardus Affiliation 1 Department of Public Health, Erasmus MC, University Medical Center Rotterdam, The Netherlands. PMID: 20306635 Item in Clipboard Cite Display options Display options Format Abstract Introduction: Treatment for erythema nodosum leprosum (ENL), an immunological complication of leprosy, is diverse. We undertook a systematic review as it was not clear which treatments were most beneficial. Methods: We did a systematic search to identify randomised controlled trials (RCTs) comparing treatment with placebo, no treatment or another therapy. Two authors assessed quality and checked data. Results: We included 13 studies involving 445 participants. These trials assessed: betamethasone, thalidomide, pentoxifylline, clofazimine, indomethacin and levamisole. The quality of the trials was generally poor and no results could be pooled due to the treatments being so heterogeneous. Treatment with thalidomide showed a significant benefit compared to aspirin (RR 2.43; 95% CI 1.28 to 4.59). Clofazimine treatment was superior to prednisolone (more treatment successes; RR 3.67; 95% CI 136 to 9.91) and thalidomide (fewer recurrences; RR 0.08; 95% CI 0.01, 0-56). Minor adverse events were significantly lower in participants on a low dose thalidomide regimen compared to a high dose thalidomide regimen (RR 0.46; 95% CI 0.23 to 0.93). Significantly more minor adverse events were reported in participants taking clofazimine compared with prednisolone (RR 1.92; 95% CI 1.10 to 3.35). None of the studies assessed quality of life or economic outcomes. Conclusion: There is some evidence of benefit for thalidomide and clofazimine, but generally we did not find clear benefits for interventions in the management of ENL. This does not mean they do not work because the studies were small and poorly reported. Larger studies using clear definitions and internationally recognised scales are urgently required. PubMed Disclaimer Similar articles Interventions for erythema nodosum leprosum.Van Veen NH, Lockwood DN, van Brakel WH, Ramirez J Jr, Richardus JH.Van Veen NH, et al.Cochrane Database Syst Rev. 2009 Jul 8;2009(3):CD006949. doi: 10.1002/14651858.CD006949.pub2.Cochrane Database Syst Rev. 2009.PMID: 19588412 Free PMC article. Systemic pharmacological treatments for chronic plaque psoriasis: a network meta-analysis.Sbidian E, Chaimani A, Garcia-Doval I, Doney L, Dressler C, Hua C, Hughes C, Naldi L, Afach S, Le Cleach L.Sbidian E, et al.Cochrane Database Syst Rev. 2021 Apr 19;4(4):CD011535. doi: 10.1002/14651858.CD011535.pub4.Cochrane Database Syst Rev. 2021.Update in: Cochrane Database Syst Rev. 2022 May 23;5:CD011535. doi: 10.1002/14651858.CD011535.pub5.PMID: 33871055 Free PMC article.Updated. Drugs for discoid lupus erythematosus.Jessop S, Whitelaw DA, Grainge MJ, Jayasekera P.Jessop S, et al.Cochrane Database Syst Rev. 2017 May 5;5(5):CD002954. doi: 10.1002/14651858.CD002954.pub3.Cochrane Database Syst Rev. 2017.PMID: 28476075 Free PMC article. Interventions for Old World cutaneous leishmaniasis.Heras-Mosteiro J, Monge-Maillo B, Pinart M, Lopez Pereira P, Reveiz L, Garcia-Carrasco E, Campuzano Cuadrado P, Royuela A, Mendez Roman I, López-Vélez R.Heras-Mosteiro J, et al.Cochrane Database Syst Rev. 2017 Dec 1;12(12):CD005067. doi: 10.1002/14651858.CD005067.pub5.Cochrane Database Syst Rev. 2017.PMID: 29192424 Free PMC article. Interventions for Old World cutaneous leishmaniasis.Heras-Mosteiro J, Monge-Maillo B, Pinart M, Lopez Pereira P, Reveiz L, Garcia-Carrasco E, Campuzano Cuadrado P, Royuela A, Mendez Roman I, López-Vélez R.Heras-Mosteiro J, et al.Cochrane Database Syst Rev. 2017 Nov 17;11(11):CD005067. doi: 10.1002/14651858.CD005067.pub4.Cochrane Database Syst Rev. 2017.Update in: Cochrane Database Syst Rev. 2017 Dec 01;12:CD005067. doi: 10.1002/14651858.CD005067.pub5.PMID: 29149474 Free PMC article.Updated. See all similar articles Cited by Serological responses to prednisolone treatment in leprosy reactions: study of TNF-α, antibodies to phenolic glycolipid-1, lipoarabinomanan, ceramide and S100-B.Raju R, Suneetha S, Jadhav RS, Chaduvula M, Atkinson S, Jain S, Visser LH, Das L, Panhalkar R, Shinde V, Reddy PP, Barkataki P, Lockwood DNj, Van Brakel WH, Suneetha LM.Raju R, et al.Lipids Health Dis. 2014 Jul 28;13:119. doi: 10.1186/1476-511X-13-119.Lipids Health Dis. 2014.PMID: 25070345 Free PMC article. Imatinib Triggers Phagolysosome Acidification and Antimicrobial Activity against Mycobacterium bovis Bacille Calmette-Guérin in Glucocorticoid-Treated Human Macrophages.Steiger J, Stephan A, Inkeles MS, Realegeno S, Bruns H, Kröll P, de Castro Kroner J, Sommer A, Batinica M, Pitzler L, Kalscheuer R, Hartmann P, Plum G, Stenger S, Pellegrini M, Brachvogel B, Modlin RL, Fabri M.Steiger J, et al.J Immunol. 2016 Jul 1;197(1):222-32. doi: 10.4049/jimmunol.1502407. Epub 2016 May 27.J Immunol. 2016.PMID: 27233968 Free PMC article. Leprosy--evolution of the path to eradication.Dogra S, Narang T, Kumar B.Dogra S, et al.Indian J Med Res. 2013 Jan;137(1):15-35.Indian J Med Res. 2013.Retraction in: Indian J Med Res. 2013 Dec;138(6):1038.PMID: 23481049 Free PMC article.Retracted. Methotrexate and prednisolone study in erythema nodosum leprosum (MaPs in ENL) protocol: a double-blind randomised clinical trial.de Barros B, Lambert SM, Shah M, Pai VV, Darlong J, Rozario BJ, Alinda MD, Sales AM, Doni S, Hagge DA, Shrestha D, Listiawan MY, Yitaye AM, Nery JAC, Neupane KD, Dias VLA, Butlin CR, Nicholls PG, Lockwood D, Walker SL.de Barros B, et al.BMJ Open. 2020 Nov 17;10(11):e037700. doi: 10.1136/bmjopen-2020-037700.BMJ Open. 2020.PMID: 33203627 Free PMC article.Clinical Trial. COVID-19 vaccination and leprosy-A UK hospital-based retrospective cohort study.de Barros B, Pierce R, Sprenger C, Ong ELH, Walker SL.de Barros B, et al.PLoS Negl Trop Dis. 2023 Aug 4;17(8):e0011493. doi: 10.1371/journal.pntd.0011493. eCollection 2023 Aug.PLoS Negl Trop Dis. 2023.PMID: 37540711 Free PMC article. See all "Cited by" articles Publication types Research Support, Non-U.S. Gov't Actions Search in PubMed Search in MeSH Add to Search Systematic Review Actions Search in PubMed Search in MeSH Add to Search MeSH terms Aspirin / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Clofazimine / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Erythema Nodosum / drug therapy Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Leprostatic Agents / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Leprosy, Lepromatous / drug therapy Actions Search in PubMed Search in MeSH Add to Search Prednisolone / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Randomized Controlled Trials as Topic Actions Search in PubMed Search in MeSH Add to Search Thalidomide / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Treatment Outcome Actions Search in PubMed Search in MeSH Add to Search Substances Leprostatic Agents Actions Search in PubMed Search in MeSH Add to Search Thalidomide Actions Search in PubMed Search in MeSH Add to Search Prednisolone Actions Search in PubMed Search in MeSH Add to Search Clofazimine Actions Search in PubMed Search in MeSH Add to Search Aspirin Actions Search in PubMed Search in MeSH Add to Search Related information Cited in Books MedGen PubChem Compound PubChem Compound (MeSH Keyword) PubChem Substance [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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
4588	https://www.youtube.com/watch?v=7-vNSuzBcp8	Combining Ratios & Three Part Ratios (Part 2) \| Grade 5+ Series \| GCSE Maths Tutor The GCSE Maths Tutor 306000 subscribers 1304 likes Description 84098 views Posted: 21 Feb 2020 A video revising the techniques and strategies for combining 3 part ratios (Part 2) - Higher and Foundation. This video is part of the Ratio & Proportion module in GCSE maths, see my other videos below to continue with the series. Revision videos: The Entire GCSE in Only 2 Hours!! Higher and Foundation Revision \| GCSE Maths Tutor - Everything for a Grade 6-9 in your GCSE Maths Exam! Higher Maths Exam Revision \| GCSE Maths Tutor - All The Topics You NEED To Know For The 9-1 Maths GCSE 2020 (Higher) - More videos in the series: All of Ratio in 30 Minutes!! - Part 1 - Sharing in a Ratio Problems - Part 2 - Combining Ratio Problems - Part 3 - Ratios of Ratios (Higher Only) - Part 4 - Equating Ratios (Higher Only) - Proportion Series Inverse Proportion in Context - Direct and Inverse Proportion (Higher Only) - Similar Shapes with Lengths - Similar Shapes with Area and Volume (Higher Only) - These are the calculators that I recommend 💎 Casio fx-83GTX Scientific Calculator, Black - Casio Fx-991ex Scientific Calculator Fx 991 Ex - New + 552 Function - Classwiz - These are the new revision guides I recommend 💡 Higher - New Grade 9-1 GCSE Maths Edexcel Student Book - Higher (with Online Edition) (CGP GCSE Maths 9-1 Revision) - Foundation - New Grade 9-1 GCSE Maths Edexcel Student Book - Foundation (with Online Edition) (CGP GCSE Maths 9-1 Revision) - 🔹 Watch the video 🔹 Make notes 🔹 Practice the questions 🔹 Mark your work 🔹 Review everything you have learnt Please like ✅ please comment ✅ please subscribe ✅ Enjoy the video! Follow me here: Instagram: Twitter: Music: Over N Over Musician: Not The King.👑 94 comments Transcript: Oh Oh [Music] [Music] okay so this video we're gonna have like a combining ratios or sometimes also referred to as three-part ratios we don't look at how we approach these sorts of questions particularly some of the problems later on as well so grab a piece of paper but grab a pair make some notes we're gonna kick off for this question to start with so it says given that a 2b is in the ratio 4 to 5 and B to C is in the ratio 3 to 2 find the ratio a 2 B to C and there's our three-part ratio but we're starting off with two two-part ratios so that this so can only happen when we've got like what I call a crossover within those ratios so in that first one that we've got a 2 B I'm going to highlight the be there because in the next one we've also got B to C so that one we've also got a B now in those ratios the B is represented by the 5 and it's represented by the 3 in the second one so as you can see both those numbers are different now if they were the same we could just combine this straight away but that's not a problem because we can make them the same quite nice and easy and it does sake if you're uncertain its simplest form or we'll discuss that in a second now forward if I just write the ratio a 2 B 2 C and let's just see what we have at the moment so at the moment we've got 4 2 5 so I'll put that underneath the a in the B we've got that in the first on now the next one we've got 3 2 2 4 v 2 C now I can't just write the 3 over the top of the 5 so I'm just gonna write it on the next level down or the next layer down however you want to look at it I'm just gonna write that there 3 2 2 now what I want to do is I want to make that B number the same in both so as long as it's the same in both I can I can ignore that it's two different ratios and I can almost sort of slop them together or squish them together however you want to sort of visualize that but I'm just going to think about what number I could make that now 5 and 3 the lowest common multiple of 5 and 3 and we can work out a multiple quite nice and easy just by doing 3 times 5 and that does actually give us the lowest common multiple in this case is 15 so I'm gonna make that number 15 in the middle now in order to do that if I want to change a ratio I can do it just like I can change a fraction and I can make equivalent fractions I kind of make an equivalent ratio it's gonna make the numbers bigger so I'm gonna get the ratio on the top here the 4 to 5 I'm gonna times that ratio by 3 that's going to turn the 4 into a 12 and a 5 into a 50 that's we're looking for that beam in Bisbee 15 so that falls going to become 12 and the 5 is going to become 15 there we go and that's that first ratio changed now I also want to change the one below because I want that 3 to be a 15 so the ratio below I'm gonna have to times up one by five and if we times that one by 5 so 3 becomes 15 so I don't need to write that down that's what we're aiming for and the 2 becomes 10 there we go and there's our ratio 12 to 15 to 10 now we need to have a look at this point does it simplify now the one that I've written out doesn't actually simplify now the reason that mine isn't simplifies I found the lowest common multiple of 3 and 5 but obviously he could have found a different multiple he could have gone for 30 or 60 or anything else that sort of 3 and 5 goes into in which case you would have had to simplify at this point but because I've gone for the lowest common multiple there and I've got the 15 it actually just means that this ratio doesn't simplify now doesn't mean it won't always if I find the lowest common multiple but in this particular circumstance it hasn't so there's my final answer 12 to 15 to 10 right let's have a look at another question with some slightly different wording okay so this question says the ratio of fish to dogs is 3 to 2 and the ratio of dogs to cuts is 5 to 3 find the ratio of cats to fish right so cats to fish obviously we can't find straight away because fishes in our thirst ratio in cats is in our second one so until we've got this all combined we're not going to be able to actually compare those two so we need to make a three part ratio here so says find the ratio of cats to fish and give your answer in its simplest form so let's get rid of that let's just remember what we're looking for cats to fish right so at the moment what's our crossover our crossovers dogs so we have dogs in one represented by the two dogs in the other represented by the 5 so if we write ourselves a three-part ratio here let's go fish two dogs two cats and I always try and make sure that crossover number is in the middle there or that you can do this if it's not I just quite like it visually when it's in the middle so at the moment we've got fish two dogs three two two and we've got dogs two cats five two three which I'll put on the level below again so I'm going to look at these now two and five now the lowest common multiple of two and five is ten so let's make that middle number 10 so I'll times the top one by five that's going to turn that two into a 10 but it also turns that 3 into a 15 got 15 to 10 now for that ratio the one below to make that a 10 we got after times that one by 2 that's going to make the 5 or 10 like we're looking for and it makes the 3 become a 6 and there we go there's our final answer and again because I found the lowest common multiple has actually resulted in me just having a fully simplified ratio there my final step but there we go we need to actually find the ratio in the question here so it said find the ratio of cats to fish now at the moment let's have a look cats is here 6 so cats let's just label this to the side cuts the fish cats is 6 let's just make sure we get that in the right order and fish is on the left there 15 and there we go 6 to 15 now hopefully you've spotted here this is not in its simplest form so 6 to 15 can actually simplify because both of those numbers there divided by 3 so we can simplify this down so divide 6 by 3 we get 2 and if we divide 15 by 3 we get 5 so there's our final answer for the full question there 2 to 5 right okay here's a couple for you to have a go out then okay so it's 2 questions they're very similar to the ones we've just done so have a go see if you can see if you can complete these pause the video there and whenever go at the answers in a sec right ok so for this first one then e 2 B 2 C and the numbers we've got at the moment we've got 3 to 7 and we've got 2 to 5 now the lowest common multiple of 2 & 7 is 14 it's all times the top 1 by 2 and the bottom one by 7 so we'll get 6 we'll get 14 in the middle and we'll get 35 on the right there so 6 to 14 to 35 let's just double check that we've times the top 1 by 2 6 and 14 the bottom 1 by 7 14 and 35 perfect I pulled that as a final answer and it's fully simplified again as we found the lowest common multiple over to the next one fish two dogs two cats so we've got fish dogs and cats and we've got four to three on the first ratio and five to two on the next one so we've got a three and a five again so again we're going to times this to make 15 it's all times a topping by 5 the bottom one by the and that's going to give us 20 to 15 to 6 and then again we need to look at our last step here so it says find the ratio of fish to cats so fish we've got there is 20 and cats we've got there is 6 so that would be 20 to 6 and again it wants in its simplest form so we'll divide both sides there by 2 and we'll get 10 to 3 and there we go and there's our final answer for that particular question all right ok so that's how we're going to approach three part ratios let's type okay our next bit okay so this is where we're going to apply some of our 3-part ratios onto some further problems we're going to look at sharing in a ratio as well which again is something I discussed in the previous video which is part 1 linked in the description so obviously check that one out if you haven't already but we're gonna have a look at actually applying these combined ratios when we've got some further problems to actually analyze here so it says Annabeth and Charlotte Sara share 138 pounds the amount Anna and Beth get is in the ratio 4 to 5 and the amount Beth and Charlotte get is in the ratio two to one how much does Anna get okay so we've got three people we've got two separate ratios so this is our little hint that we're going to make this three part ratio so we're going to make that to start with and then we'll have a look at actually answering the question so we've got a 2 B 2 C and at the moment we've got another and Beth sharing in the ratio 4 to 5 and Beth and Charlotte sharing in the ratio two to one and there we go so we can line that all up just like before so let's make that 10 in the middle sometimes the top them by two in the bottom one by five and that's going to give us 8 and 10 on our top ratio and 10 and 5 on our bottom ratio and there we go there's our three part ratio done now we can actually have a look at the question because it says that they share 138 pounds how much does Anna get so all we're doing is we're sharing in a ratio just like we've done before so we just have a look at all these three parts which is you know they're all sharing it it doesn't say that there's a difference between them it doesn't say that one of them's getting 130 a so it's all of them there so if we have those all together 8 plus 10 plus 5 is 23 so we've got 23 parts that it says it's equal to 138 pounds and just like before when we looked at ratios we're looking at the value of one part so to get the value of one part we need to divide both sides here by 23 so 138 divided by 23 gives us the value of 6 so one part is equal to 6 pounds now we don't need to go ahead and find out how much everyone gets because we only really care about what anna gets there so anna in our new ratio that we're using is the 8 so Anna's getting these 8 parts and 8 times 6 pounds it gives us our value for an hour 8 times 6 is 48 pounds so Anna gets 48 pounds and that is our final answer there and we'll just highlight that offers a fine well so there we go 48 pounds right there we go so just thinking about obviously applying some sharing in a ratio as well as this 3-part combined ratio we can have a look at one more question where the wording is slightly trickier and here it is okay so it says Dillon Emily and Frank share some money the amount of money Dillon gets to the amount of money Emily gets is in the ratio 4 to 3 the amount of money that Dillon gets to the amount that Frank gets is in the ratio 5 to 2 given that Emily gets 20 one-pound more than Frank work out how much Emily gets so we've got one of these questions here where it's saying somebody gets more than the other so if we can get a three-part ratio put together we'll have a look at how many parts more this person gets so I start putting this together we've got Dillon Emily and Frank now it says Nico be care for the wording here it says Dillon to Emily is 4 to 3 so 4 it's a 3 and then it says Dillon to Frank okay says the amount Dillon gets we're gonna highlight this the amount Dillon gets the amount Frank gets is in the ratio 5 to 2 so that's not going to line up perfectly in the middle here just because of the way the wording was given so what's that Dillon is 5 Frank is 2 there we go so it's not quite lined up as nicely so I'm gonna do is I'm just gonna rearrange these I'm gonna write it in different way don't have to do this but I just prefer that number being in the middle although it really doesn't need to be you don't need to worry about it so I'm just gonna rearrange this I want D in the middle so I'm gonna have Emily - Dillon - Frank yeah and honestly you really don't have to do this but for the first one there I've got three two forks I'm now swapping it around the other way which may just slightly confuse things if you're not happy doing that but I prefer that number be in the middle so I'm just going to do it like that all right there we go so in fact I might just do it both sides he could decide for yourself which one you prefer I'm gonna time to the top one by five and the bottom one by four that's gonna make 20 in the middle so we'll get 15 20 and then the bottom one that two becomes eight again so I could have done that on the side over here on this one times five times four and we'd have got 22 15 to 8 so it really doesn't matter obviously which way you do it I just really like that number being in the middle but you obviously don't have to go about doing that now it says here that given that Emily gets 20 one-pound more than Frank so Emily and Frank are these two so emily is getting an extra seven parts Emily gets 15 Frank gets 8 so that's plus 7 parts so those extra 7 parts are equal to 21 pounds it's obviously from there we can work out the value of one part we can do 21 divided by 7 and that equals 3 pounds so each part is worth 3 pounds now it says in the question obviously work out how much Emily gets now it doesn't matter which ratio we use here which side you've done it as long as you make sure you use this Emily number which is 15 so we can finish that off because emily is gonna get let's just write this here 15 lots of the 3 pounds so 15 times 3 and that equals 45 pounds that Emily gets obviously just thinking about how we've got to they're finding that gap between the two so there we go she gets 45 pounds just making sure that you find out the value of one part again just like we have in the previous video looking obviously some of these ratios but there we go that's how to approach a couple of these questions here where you've got some split ratios that you need to combine first right okay so here's call for you to have a go up right okay so there's two questions for you to go out so pause the video there have a go we'll go over the ants in a sec all right okay so this first one we have James le and Alicia there we go and it says that they share 120 pounds it says the amount James and Ellie get is in the ratio three to two I'm the amount Elian Alyssa get is in the ratio four to five how does Eddie get now you might not have spotted this when you're having a go but actually we don't have to change one of the ratios here look because I could open just times the top 1 by 2 and that's going to turn that 2 4 le into a 4 so that would just make it 6 2 4 and the other one stays as a 5 now you could have obviously change them all if you wanted you could have made it 8 in the middle there but obviously it doesn't matter you're gonna get the same answer either way so let's have a look at what we get when we just approach it like this so going from here it says they all share 120 so it's all of these numbers here and they add up to 15 so we've got 15 parts that equal 120 pounds and obviously we want to know the value of one part there we go this will divide by 15 and 120 divided by 15 is 8 so each part is worth 8 pounds so to finish this off how much does Ellie get well Ellie here is the 4 so if we do for lots of the 8 pounds there we go we get 32 pounds for the amount that Ellie gets there we go and we'll just highlight that off done right onto our next one so we've got an abrino and Charlie so a B and C and the amount of money the Bruno gets is in the ratio so Hannah to Bruno is 5 to 3 there we go and then Bruno - charlie is in the ratio of 5 to 2 that's quite nice that lines up nicely on this question there we go so given that Anna gets 57 pound more than Charlie work out how much Bruno gets so I make it's 57 pounds more than charlie and how much Bruno gets that's what we're gonna work out right so well let's create outs our 3-part ratio then so we'll make it 15 in the middle against all times up by 5 on this 1 by 3 and that's going to be 25 for a 15 for B and the times on the bottom 1 by 3 gives us 6 at the end there right so there is our 3-part ratio let's have a look so it says Anna gets 57 pounds more than Charlie so Anna and Charlie what's the gap between them okay so that is a gap of 19 there we go so we've got 19 parts I have not got much space there let's just write this at the top so that's plus 19 that's right this appear so we get nineteen parts equals fifty seven pounds so divide that by nineteen and we get one part equals three pounds there we go so one part equals three now what it says is how much does Bruno get while Bruno here is our fifteen so Bruno gets fifteen lots of three so fifteen times three pounds there you go equals 45 maybe R so Bruno gets forty five pounds and that's the final answer there for our second question all right let's have a look at something slightly different then okay so there's a lot more words here and this is a good question because there's actually two different ways of approaching this so I'm going to approach it in a way that we've been doing all video but then I might discuss another way that we can actually potentially have a look at this question so it says in a pack of pens the number of red pens are the number of blue pens are in the ratio three to five and the number of blue and two green is in the ratio three to seven says there are 27 red pens in the pack how many green pens are in the pack so let's have a look we've got red to blue to green and at the moment we've got three two five four red to blue and we've got three to seven four blue to green so again that's just approach this like normal make our three part ratio it's all times this one by three this one by five and let's see what we get we get nine to fifteen and on the bottom there we get thirty five and there we go there's our ratio for the three part ratio for these pens it says there are 27 red pens so let's have a look underneath red then we know that this number here is 27 so the nine parts equals 27 so just like it did before if you get this sorry canvas think alright well what are times nine by to get 27 and that numbers three okay obviously we can do the working out to the side 27 divided by nine three and that just means we just need to times all of these by three to get our full ratio so 15 times three would be 45 and 35 times three gives us a hundred and five there we go one zero five so it says how many green pens are in the pack or green pens is that last one now 105 so that's our final answer 105 green pens so there we go that's quite nice and easy just doing that using a three-part ratio obviously just allocating that number to the one part in that three part ratio and just obviously thinking about it like that now you could actually approach this question in a slightly different way because it does say there are 27 red pens so before actually going about changing any of this so if we look at the original ratio look if there were 27 red pens and how do we get that 3 to become 27 we were times it by 9 and we can do that for the 5 as well so we know that there's 45 blue pens and that's right from the start there we go and that matches our 45 in our final answer okay just think about what we did there we just times both those numbers in the ratio by 5 by by 9 sorry so if we know this 45 blue pens we can also apply that for this ratio here so if that one's 45 then what are we times 3 by 2 get 45 over times by 15 so you can also just times 7 by 15 and that would give you 105 there we go so you can't just do it straight from the ratios up there although that is very is very unique to this one type of question but obviously just a different way that you could think about it if you have stuck on a question like this potentially you could always think about it in that way as well but it does only work for these sort of for in a unique way for these questions but here we go here's one for you to go out and here it is okay so here we go here's your question so pause the video there have a go we'll go over the answers in a sec okay so this question says in a village the number of houses - number of flats are in the ratio 7 to 4 and then flats the bungalow so we've got houses 2 flats 2 bungalows and the first ones in the ratio 7 to 4 and their flats the bungalows is in the ratio 8 to 5 so I've got a similar scenario here that where you only actually had to change the one so four just times the top 1 by 2 that's going to make that crossover number there become an 8 so we get 14 to 8 so 5 and again you could have changed them both if you wanted you'll still get to the final answer you might get some not-so-nice decimals along the way but there we go it says now that there are let's have a look what's it say it says that there are 50 bungalows so that's 50 under here under the bungalows let's just rewrite that last fives not very good there we go so we've got 50 bungalows and we have to think is how do we get from 5 to 50 we time by 10:00 there we go so we just need to times all of these by 10 and then we'll answer the question so 8 times 10 in the middle is 80 14 times 10 is hundred and 40 and what's the question asking for it says how many houses are there what houses is right there 140 and there is our final answer 140 houses right okay so before we finish I got one more type of question for us to have a look at it's just something a little bit different just to finish off with something that's just a different way of thinking to some of these questions but we'll have a look at that question now before we finish this up okay so this question here looks like a three-part ratio we've got the points a B C and D line order on a straight line we've got a to be B to DS 2 to 5 AC C D is 4 to 7 and we want to find the ratio a to B B to C C to D so it looks like and it's going to be a three part ratio the way we look at it but it looks like we're gonna have to try and find some sort of crossover now if you have a look a 2b is not in either of the other ratio B to D isn't also not in either of the other ratio so we've not got a crossover so we have to think about this slightly differently now it just say that they lie on a straight line so if we just imagine this line that's just imagine here it is here's a line we've got a b c d let's just label this up so we've got a b c and d now it says the ratio a to B B to D is 2 to 5 so that's this bit here I'm gonna do this in a different color we've got a to be there and that's 2 and then B to D to the end 5 then we've got the other ratio that's given to us we've got a to C which is 4 and we've got C to D which is 7 right so we haven't got a crossover but we have got totals and what we're gonna have a look at is if we have something like this sort of scenario where we can draw a little picture we can actually just combine the both these ratios together by looking at them if the totals add up to the same what I mean by that look is 2 & 5 gives us a total over here of 7 the other one 4 & 7 over here gives us a total of 11 now if they both have the same total this would make this problem a lot easier so let's think what's a lowest common multiple of 7 and 11 all that is 70 so to make these numbers out of 77 I would have two times the top one by 11 and the bottom one by seven and that would give me new numbers here on my number line so if we change this look we've got 22 number times up by 11 and here we've got 55 and on the bottom there 4 times 7 is 28 and 7 times 7 is 49 now they have the same totals we can pretty much just get the readings from our picture so A to B we've got appear 22 and that's the first part in our three-part ratio there so let's have a look we've got a to B which is 22 now we're looking at B to C B to C we can't find because it's crossed over from the other two so there's no way we can get that at the moment but then C to D we've got that that's given to us down here C to D is that 49 so we've got 49 at the end there we're just missing the one in the middle and here's a bit of an idea of how do we actually find that number in the middle now we know that the total over here had to add up to 77 that's what we did we made these both add up to 77 so this total that we're looking at has to add up to 77 okay just like those other two these have thought up to 77 so what if we got at the moment we've got 22 and 49 nice add those together see what we've got so far 49 and 22 make 71 so to get from 71 to 77 we would have to add in an extra six and there we go and there's our three-part ratio here from a diagram so something a little bit different okay do have to draw a little picture for this I think it's probably quite difficult unless you draw a picture but there's just something a little different just to finish this off when we're looking at this combining ratios and rather than looking at a crossover we're looking at the total between them so that we're able to combine them together and have a look at these sorts of missing distances in the middle okay so I'm only going to do one of these questions I've got one for you to have a go up now so here it is right so here's your last question so pause the video there have a go from over the unseen a sec right okay so ABC and D lie on a straight line so let's draw it out again a b c d there we go we'll go a to b c and d and put these ratios on so we've got a to be it's three B to D it's five and then underneath we've got a to C is 5 and C to D is six now let's have a look what we got as the total sin so three and five had to make eight and five and six to make it 11 so we've got a very similar scenario here we're gonna make the lowest common multiple which is 88 it's all times the top numbers by 11 the bottom numbers by 8 and let's see what our new totals are here so we've got 33 and the times up by 11 and we've got 55 here on the bottom all the other times up by 8 so 5 times 8 is 40 and 6 times 8 is 48 there we go alright so we have a total here what's the total of these so they're both add up to 88 now that may 8 times 11 is 88 and 11 times 8 is 88 we've got the same total on both so we should be able to make a fair comparison here so let's have a look quick or a to B B to D sorry B to C and then C to D now we've got one of the we've got what this one up here we've got 33 that's form A to B so we can put 33 in there we go we've also got just like before we've got the 48 down here from C to D so that's 48 and they stop those together and see what we miss him so we've got 48 33 that adds up to 81 there we go so we are missing 7 to get that 81 up to the 88 so there's our final answer there's our three-part ratio and there is just something a little different to be thinking about with some of these or nasty or three-part ratio questions but there we go have deals helpful if it was useful please like please come and please subscribe and I will see for the next one [Music] [Music]
4589	https://www.youtube.com/watch?v=7W6RNETxVbQ	Factor x^2 - 9 using the Difference of Squares Formula The Math Sorcerer 1220000 subscribers 68 likes Description 8361 views Posted: 20 Jun 2023 Factor x^2 - 9 using the Difference of Squares Formula Useful Math Supplies My Recording Gear (these are my affiliate links) Math, Physics, and Computer Science Books Epic Math Book List Pre-algebra, Algebra, and Geometry College Algebra, Precalculus, and Trigonometry Probability and Statistics Discrete Mathematics Proof Writing Calculus Differential Equations Books Partial Differential Equations Books Linear Algebra Abstract Algebra Books Real Analysis/Advanced Calculus Complex Analysis Number Theory Graph Theory Topology Graduate Level Books Computer Science Physics These are my affiliate links. As an Amazon Associate I earn from qualifying purchases. If you enjoyed this video please consider liking, sharing, and subscribing. Udemy Courses Via My Website: Free Homework Help : My FaceBook Page: My Instagram: My TikTok: There are several ways that you can help support my channel:) Consider becoming a member of the channel: My GoFundMe Page: My Patreon Page: Donate via PayPal: Udemy Courses(Please Use These Links If You Sign Up!) Abstract Algebra Course Advanced Calculus Course Calculus 1 Course Calculus 2 Course Calculus 3 Course Calculus 1 Lectures with Assignments and a Final Exam Calculus Integration Insanity Differential Equations Course Differential Equations Lectures Course (Includes Assignments + Final Exam) College Algebra Course How to Write Proofs with Sets Course How to Write Proofs with Functions Course Trigonometry 1 Course Trigonometry 2 Course Statistics with StatCrunch Course Math Graduate Programs, Applying, Advice, Motivation Daily Devotionals for Motivation with The Math Sorcerer Thank you:) 2 comments Transcript: hello in this video we're going to factor X squared minus 9. let's go ahead and work through it solution so this is an example of something called the difference of squares there's a formula that says if you have a squared minus B squared this is equal to a minus B times a plus b so in this particular example all we have to do is write this in the form that allows us to use this formula so let's write it right below the formula so we have x squared minus and then we can write the 9 as 3 squared then you can see here just by matching that a is equal to X and B is equal to three so this is equal to parentheses x minus three times and then X plus three directly applying the formula and so this would be the full factorization of x squared minus nine hope this video has been helpful to someone out there good luck
4590	https://en.wikipedia.org/wiki/Carbon-13_nuclear_magnetic_resonance	Jump to content Search Contents (Top) 1 Receptivity 2 Chemical shifts 2.1 Coupling constants 3 Implementation 3.1 Sensitivity 3.2 Coupling modes 4 Distortionless enhancement by polarization transfer spectra 5 Attached proton test spectra 6 See also 7 References 8 External links Carbon-13 nuclear magnetic resonance Čeština Dansk Deutsch Euskara Français 日本語 Português Slovenčina Svenska Українська 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Analytical method \| \| \| This article may be too technical for most readers to understand. Please help improve it to make it understandable to non-experts, without removing the technical details. (February 2024) (Learn how and when to remove this message) \| Carbon-13 (C13) nuclear magnetic resonance (most commonly known as carbon-13 NMR spectroscopy or 13C NMR spectroscopy or sometimes simply referred to as carbon NMR) is the application of nuclear magnetic resonance (NMR) spectroscopy to carbon. It is analogous to proton NMR (1H NMR) and allows the identification of carbon atoms in an organic molecule just as proton NMR identifies hydrogen atoms. 13C NMR detects only the 13C isotope. The main carbon isotope, 12C does not produce an NMR signal. Although about 1 million times less sensitive than 1H NMR spectroscopy, 13C NMR spectroscopy is widely used for characterizing organic and organometallic compounds, primarily because 1H-decoupled 13C-NMR spectra are simpler, have a greater sensitivity to differences in the chemical structure, and thus are better suited for identifying molecules in complex mixtures. At the same time, such spectra lack quantitative information about the atomic ratios of different types of carbon nuclei, because the nuclear Overhauser effect used in 1H-decoupled 13C-NMR spectroscopy enhances the signals from carbon atoms with a larger number of hydrogen atoms attached to them more than from carbon atoms with a smaller number of H's, and because full relaxation of 13C nuclei is usually not attained (for the sake of reducing the experiment time), and the nuclei with shorter relaxation times produce more intense signals. The major isotope of carbon, the 12C isotope, has a spin quantum number of zero, so is not magnetically active and therefore not detectable by NMR. 13C, with a spin quantum number of 1/2, is less abundant (1.1%), whereas other popular nuclei are 100% abundant, e.g. 1H, 19F, 31P. Receptivity [edit] 13C NMR spectroscopy is much less sensitive (ca. by 4 orders of magnitude) to carbon than 1H NMR spectroscopy is to hydrogen, because of the lower abundance (1.1%) of 13C compared to 1H (>99%), and because of a lower(0.702 vs. 2.8) nuclear magnetic moment. Stated equivalently, the gyromagnetic ratio (6.728284 107 rad T−1 s−1) is only 1/4th that of 1H. On the other hand, the sensitivity of 13C NMR spectroscopy benefits to some extent from nuclear Overhauser effect, which enhances signal intensity for non-quaternary 13C atoms. Chemical shifts [edit] The disadvantages in "receptivity" are compensated by the high sensitivity of 13C NMR signals to the chemical environment of the nucleus, i.e. the chemical shift "dispersion" is great, covering nearly 250 ppm. This dispersion reflects the fact that non-1H nuclei are strongly influenced by excited states ("paramagnetic" contribution to shielding tensor. This paramagnetic contribution is unrelated to paramagnetism). For example, most 1H NMR signals for most organic compounds are within 15 ppm. The chemical shift reference standard for 13C is the carbons in tetramethylsilane (TMS), whose chemical shift is set as 0.0 ppm at every temperature. Typical chemical shifts in 13C-NMR Coupling constants [edit] Homonuclear 13C-13C coupling is normally only observed in samples that are enriched with 13C. The range for one-bond 1J(13C,13C) is 50–130 Hz. Two-bond 2J(13C,13C) are near 10 Hz. The trends in J(1H,13C) and J(13C,13C) are similar, except that J(1H,13C are smaller owing to the modest value of the 13C nuclear magnetic moment. Values for 1J(1H,13C) range from 125 to 250 Hz. Values for 2J(1H,13C) are near 5 Hz and often are negative. Implementation [edit] Sensitivity [edit] As a consequence of low receptivity, 13C NMR spectroscopy suffers from complications not encountered in proton NMR spectroscopy. Many measures can be implemented to compensate for the low receptivity of this nucleus. For example, high field magnets with wider internal bores are capable of accepting larger sample tubes (typically 10 mm in diameter for 13C NMR versus 5 mm for 1H NMR). Relaxation reagents allow more rapid pulsing. A typical relaxation agent is chromium(III) acetylacetonate. For a typical sample, recording a 13C NMR spectrum may require several hours, compared to 15–30 minutes for 1H NMR. The nuclear dipole is weaker, the difference in energy between alpha and beta states is one-quarter that of proton NMR, and the Boltzmann population difference is correspondingly less. One final measure to compensate for low receptivity is isotopic enrichment. Some NMR probes, called cryoprobes, offer 20x signal enhancement relative to ordinary NMR probes. In cryoprobes, the RF generating and receiving electronics are maintained at ~ 25K using helium gas, which greatly enhances their sensitivity. The trade-off is that cryoprobes are costly. Coupling modes [edit] Another potential complication results from the presence of large one bond J-coupling constants between carbon and hydrogen (typically from 100 to 250 Hz). While potentially informative, these couplings can complicate the spectra and reduce sensitivity. For these reasons, 13C-NMR spectra are usually recorded with proton NMR decoupling. Couplings between carbons can be ignored due to the low natural abundance of 13C. Hence in contrast to typical proton NMR spectra, which show multiplets for each proton position, carbon NMR spectra show a single peak for each chemically non-equivalent carbon atom. In further contrast to 1H NMR, the intensities of the signals are often not proportional to the number of equivalent 13C atoms. Instead, signal intensity is strongly influenced by (and proportional to) the number of surrounding spins (typically 1H). Integrations are more quantitative if the delay times are long, i.e. if the delay times greatly exceed relaxation times. The most common modes of recording 13C spectra are proton-noise decoupling (also known as noise-, proton-, or broadband- decoupling), off-resonance decoupling, and gated decoupling. These modes are meant to address the large J values for 13C - H (110–320 Hz), 13C - C - H (5–60 Hz), and 13C - C - C - H (5–25 Hz) which otherwise make completely proton coupled 13C spectra difficult to interpret. With proton-noise decoupling, in which most spectra are run, a noise decoupler strongly irradiates the sample with a broad (approximately 1000 Hz) range of radio frequencies covering the range (such as 100 MHz for a 23,486 gauss field) at which protons change their nuclear spin. The rapid changes in proton spin create an effective heteronuclear decoupling, increasing carbon signal strength on account of the nuclear Overhauser effect (NOE) and simplifying the spectrum so that each non-equivalent carbon produces a singlet peak. Both the atoms, carbon and hydrogen exhibit spins and are NMR active. The nuclear Overhauser Effect is in general, showing up when one of two different types of atoms is irradiated while the NMR spectrum of the other type is determined. If the absorption intensities of the observed (i.e., non-irradiated) atom change, enhancement occurs. The effect can be either positive or negative, depending on which atom types are involved. The relative intensities are unreliable because some carbons have a larger spin-lattice relaxation time and others have weaker NOE enhancement. In gated decoupling, the noise decoupler is gated on early in the free induction delay but gated off for the pulse delay. This largely prevents NOE enhancement, allowing the strength of individual 13C peaks to be meaningfully compared by integration, at a cost of half to two-thirds of the overall sensitivity. With off-resonance decoupling, the noise decoupler irradiates the sample at 1000–2000 Hz upfield or 2000–3000 Hz downfield of the proton resonance frequency. This retains couplings between protons immediately adjacent to 13C atoms but most often removes the others, allowing narrow multiplets to be visualized with one extra peak per bound proton (unless bound methylene protons are non-equivalent, in which case a pair of doublets may be observed). Distortionless enhancement by polarization transfer spectra [edit] Distortionless enhancement by polarization transfer (DEPT) is an NMR method used for determining the presence of primary, secondary and tertiary carbon atoms. The DEPT experiment differentiates between CH, CH2 and CH3 groups by variation of the selection angle parameter (the tip angle of the final 1H pulse): 135° angle gives all CH and CH3 in a phase opposite to CH2; 90° angle gives only CH groups, the others being suppressed; 45° angle gives all carbons with attached protons (regardless of number) in phase. Signals from quaternary carbons and other carbons with no attached protons are always absent (due to the lack of attached protons). The polarization transfer from 1H to 13C has the secondary advantage of increasing the sensitivity over the normal 13C spectrum (which has a modest enhancement from the nuclear overhauser effect (NOE) due to the 1H decoupling). Attached proton test spectra [edit] Another useful way of determining how many protons a carbon in a molecule is bonded to is to use an attached proton test (APT), which distinguishes between carbon atoms with even or odd number of attached hydrogens. A proper spin-echo sequence is able to distinguish between S, I2S and I1S, I3S spin systems: the first will appear as positive peaks in the spectrum, while the latter as negative peaks (pointing downwards), while retaining relative simplicity in the spectrum since it is still broadband proton decoupled. Even though this technique does not distinguish fully between CHn groups, it is so easy and reliable that it is frequently employed as a first attempt to assign peaks in the spectrum and elucidate the structure. Additionally, signals from quaternary carbons and other carbons with no attached protons are still detectable, so in many cases an additional conventional 13C spectrum is not required, which is an advantage over DEPT. It is, however, sometimes possible that a CH and CH2 signal have coincidentally equivalent chemical shifts resulting in annulment in the APT spectrum due to the opposite phases. For this reason the conventional 13C{1H} spectrum or HSQC are occasionally also acquired. See also [edit] Nuclear magnetic resonance Hyperpolarized carbon-13 MRI Triple-resonance nuclear magnetic resonance spectroscopy References [edit] ^ Brian E. Mann, Brian F. Taylor (1981). ¹³C NMR data for organometallic compounds. Academic Press. ISBN 9780124691506. ^ R. M. Silverstein; G. C. Bassler; T. C. Morrill (1991). Spectrometric Identification of Organic Compounds. Wiley. ISBN 9780471634041. ^ Peter Atkins (2009). Physical Chemistry (5 ed.). Freeman. ^ "The Theory of NMR - Chemical Shift". Archived from the original on 2015-01-23. Retrieved 2014-01-23. ^ Caytan E, Remaud GS, Tenailleau E, Akoka S (2007). "Precise and Accurate Quantitative 13C NMR with Reduced Experimental Time". Talanta. 71 (3): 1016–1021. doi:10.1016/j.talanta.2006.05.075. PMID 19071407. ^ "Measuring 13C NMR Spectra". University of Wisconsin. ^ Molinski, Tadeusz F. (2010). "NMR of natural products at the 'nanomole-scale'". Natural Product Reports. 27 (3): 321–9. doi:10.1039/B920545B. PMID 20179874. ^ "Introduction to Carbon NMR". University of Puget Sound. ^ a b c d Lal Dhar Singh Yadav (2013-08-13). Organic Spectroscopy. Springer. pp. 197–199. ISBN 9781402025754. ^ Pavia, Donald L., ed. (2009). Introduction to spectroscopy (4th ed.). Belmont, CA: Brooks/Cole, Cengage Learning. ISBN 978-0-495-11478-9. ^ Doddrell, D.M.; Pegg, D.T.; Bendall, M.R. (1982). "Distortionless enhancement of NMR signals by polarization transfer". J. Magn. Reson. 48 (2): 323–327. Bibcode:1982JMagR..48..323D. doi:10.1016/0022-2364(82)90286-4. ^ Keeler, James (2010). Understanding NMR Spectroscopy (2nd ed.). John Wiley & Sons. p. 457. ISBN 978-0-470-74608-0. External links [edit] Carbon NMR spectra, where there are three spectra of ethyl phthalate, ethyl ester of orthophthalic acid: completely coupled, completely decoupled and off-resonance decoupled (in this order). For an extended tabulation of 13C shifts and coupling constants. \| v t e NMR spectroscopy by isotope \| \| Isotope \| 1H 2H 3He 11B 13C 15N 17O 19F 29Si 31P 51V 57Fe 59Co 77Se 195Pt 199Hg 207Pb \| Retrieved from " Categories: Nuclear magnetic resonance Carbon Hidden categories: Pages using the EasyTimeline extension Articles with short description Short description matches Wikidata Wikipedia articles that are too technical from February 2024 All articles that are too technical Carbon-13 nuclear magnetic resonance Add topic
4591	https://catdir.loc.gov/catdir/toc/fy0805/2008924835.html	Table of contents for Library of Congress control number 2008924835 Table of contents for Differential equations with boundary-value problems / Dennis G. Zill. Bibliographic record and links to related information available from the Library of Congress catalog Note: Electronic data is machine generated. May be incomplete or contain other coding. INTRODUCTION TO DIFFERENTIAL EQUATIONS 1 1.1 Definitions and Terminology 2 1.2 Initial-Value Problems 13 1.3 Differential Equations as Mathematical Models 19 CHAPTER 1 IN REVIEW 32 FIRST-ORDER DIFFERENTIAL EQUATIONS 34 2.1 Solution Curves Without a Solution 35 2.1.1 Direction Fields 35 2.1.2 Autonomous First-Order DEs 37 2.2 Separable Variables 44 2.3 Linear Equations 53 2.4 Exact Equations 62 2.5 Solutions by Substitutions 70 2.6 A Numerical Method 75 CHAPTER 2 IN REVIEW 80 SMODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS 82 3.1 Linear Models 83 3.2 Nonlinear Models 94 3.3 Modeling with Systems of First-Order DEs 105 CHAPTER 3 IN REVIEW 113 1 1 i 1 HIGHER-ORDER DIFFERENTIAL EQUATIONS 117 4.1 Preliminary Theory-Linear Equations 118 4.1.1 Initial-Value and Boundary-Value Problems 118 4.1.2 Homogeneous Equations 120 4.1.3 Nonhomogeneous Equations 125 4.2 Reduction of Order 130 4.3 Homogeneous Linear Equations with Constant Coefficients 133 4.4 Undetermined Coefficients-Superposition Approach 140 4.5 Undetermined Coefficients-Annihilator Approach 150 4.6 Variation of Parameters 157 4.7 Cauchy-Euler Equation 162 4.8 Solving Systems of Linear DEs by Elimination 169 4.9 Nonlinear Differential Equations 174 CHAPTER 4 IN REVIEW 178 MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS 181 5.1 Linear Models: Initial-Value Problems 182 5.1.1 Spring/Mass Systems: Free Undamped Motion 182 5.1.2 Spring/Mass Systems: Free Damped Motion 186 5.1.3 Spring/Mass Systems: Driven Motion 189 5.1.4 Series Circuit Analogue 192 5.2 Linear Models: Boundary-Value Problems 199 5.3 Nonlinear Models 207 CHAPTER 5 IN REVIEW 216 SSE R I E S S O L U T I O N S O F L I N E A R E Q U A T I O N S 2 1 9 6.1 Solutions About Ordinary Points 220 6.1.1 Review of Power Series 220 6.1.2 Power Series Solutions 223 6.2 Solutions About Singular Points 231 6.3 Special Functions 241 6.3.1 Bessel's Equation 241 6.3.2 Legendre's Equation 248 CHAPTER 6 IN REVIEW 253 7 THE LAPLACE TRANSFORM 255 7.1 Definition of the Laplace Transform 256 7.2 Inverse Transforms and Transforms of Derivatives 262 7.2.1 Inverse Transforms 262 7.2.2 Transforms of Derivatives 265 7.3 Operational Properties I 270 7.3.1 Translation on the s-Axis 271 7.3.2 Translation on the t-Axis 274 7.4 Operational Properties II 282 7.4.1 Derivatives of a Transform 282 7.4.2 Transforms of Integrals 283 7.4.3 Transform of a Periodic Function 287 7.5 The Dirac Delta Function 292 7.6 Systems of Linear Differential Equations 295 CHAPTER 7 IN REVIEW 300 8 SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS 303 8.1 Preliminary Theory-Linear Systems 304 8.2 Homogeneous Linear Systems 311 8.2.1 Distinct Real Eigenvalues 312 8.2.2 Repeated Eigenvalues 315 8.2.3 Complex Eigenvalues 320 8.3 Nonhomogeneous Linear Systems 326 8.3.1 Undetermined Coefficients 326 8.3.2 Variation of Parameters 329 8.4 Matrix Exponential 334 CHAPTER 8 IN REVIEW 337 9 NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 339 9.1 Euler Methods and Error Analysis 340 9.2 Runge-Kutta Methods 345 9.3 Multistep Methods 350 9.4 Higher-Order Equations and Systems 353 9.5 Second-Order Boundary-Value Problems 358 CHAPTER 9 IN REVIEW 362 '10 PLANE AUTONOMOUS SYSTEMS 363 10.1 Autonomous Systems 364 10.2 Stability of Linear Systems 370 10.3 Linearization and Local Stability 378 10.4 Autonomous Systems as Mathematical Models 388 CHAPTER 10 IN REVIEW 395 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 397 11.1 Orthogonal Functions 398 11.2 Fourier Series 403 11.3 Fourier Cosine and Sine Series 408 11.4 Sturm-Liouville Problem 416 11.5 Bessel and Legendre Series 423 11.5.1 Fourier-Bessel Series 424 11.5.2 Fourier-Legendre Series 427 CHAPTER 11 IN REVIEW 430 _2 BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES 432 12.1 Separable Partial Differential Equations 433 12.2 Classical PDEs and Boundary-Value Problems 437 12.3 Heat Equation 443 12.4 Wave Equation 445 12.5 Laplace's Equation 450 12.6 Nonhomogeneous Boundary-Value Problems 455 12.7 Orthogonal Series Expansions 461 12.8 Higher-Dimensional Problems 466 CHAPTER 12 IN REVIEW 469 1 3 BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS 471 13.1 Polar Coordinates 472 13.2 Polar and Cylindrical Coordinates 477 13.3 Spherical Coordinates 483 CHAPTER 13 IN REVIEW 486 14 INTEGRAL TRANSFORMS 488 14.1 Error Function 489 14.2 Laplace Transform 490 14.3 Fourier Integral 498 14.4 Fourier Transforms 504 CHAPTER 14 IN REVIEW 510 115 NUMERICAL SOLUTIONS OF PARTIAL DIFFERENTIAL EQUATIONS 511 15.1 Laplace's Equation 512 15.2 Heat Equation 517 15.3 Wave Equation 522 CHAPTER 15 IN REVIEW 526 APPENDICES I Gamma Function APP- II Matrices APP-3 III Laplace Transforms APP-21 Answers for Selected Odd-Numbered Problems ANS-1 Library of Congress subject headings for this publication: Differential equations Textbooks, Boundary value problems Textbooks
4592	https://www.nature.com/articles/nrmicro1892	Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. View all journals Search Log in Explore content About the journal Publish with us Subscribe Sign up for alerts RSS feed Review Article Published: The ecology and biotechnology of sulphate-reducing bacteria Gerard Muyzer1 & Alfons J. M. Stams1 Nature Reviews Microbiology volume 6, pages 441454 (2008)Cite this article 29k Accesses 2037 Citations 50 Altmetric Metrics details Key Points Sulphate reducers have the unique ability to respire using sulphate as an electron acceptor. Sulphate-reducing bacteria and archaea are ubiquitous in the environment. The sulphate reducers that have been isolated and described thus far can be divided into seven phylogenetic lineages, five within the Bacteria and two within the Archaea. Most sulphate reducers belong to approximately 23 genera, with the Gram-negative bacteria belonging to the Deltaproteobacteria and the Gram-positive bacteria within the Clostridia. Sulphate reducers can thrive in a broad range of environmental conditions. They have been detected in shallow marine and freshwater sediments and in deep subsurface environments, such as oil wells, hydrothermal vents and mud volcanoes. They occur in environments with extremely low or high pH, extremely low or high temperature and high or low salt concentrations. They are also present in living organisms, such as ruminants and in the human intestinal tract. In marine worms, they form an intimate relationship with aerobic sulphide-oxidizing bacteria. Sulphate reducers play a key part in the carbon and sulphur cycles. They are extremely versatile with respect to the electron donors and electron acceptors that are used for growth. They can grow in a sulphate-dependent manner using hydrogen and a wide range of organic compounds. However, polymeric compounds (polysaccharides and proteins) are not typically used by sulphate reducers. Although named after their ability to respire using sulphate, other inorganic compounds can also be used as electron acceptors. Some sulphate reducers can even respire with oxygen. In environmental biotechnology, sulphate reducers have an important role. One unwanted effect is the production of hydrogen sulphide in anaerobic digesters that are used for waste and waste-water treatment and its role in the corrosion of iron and steel. However, sulphate reducers are also beneficially used in biotechnological processes to remove heavy metals and oxidized sulphur compounds from gas and water. In nature, sulphate reducers live in close vicinity with other microorganisms, which results in metabolic interactions with other anaerobes. In the presence of sulphate, they compete with methanogens and acetogens for common substrates, such as hydrogen and acetate. In the absence of sulphate, however, sulphate reducers grow acetogenically in syntrophy with methanogens. The complete genomes of different sulphate reducers have been, or are currently being, sequenced. Comparative analysis of these genome sequences will provide important information on their carbon and sulphur metabolism and open up the possibility for functional genomics. Abstract Sulphate-reducing bacteria (SRB) are anaerobic microorganisms that use sulphate as a terminal electron acceptor in, for example, the degradation of organic compounds. They are ubiquitous in anoxic habitats, where they have an important role in both the sulphur and carbon cycles. SRB can cause a serious problem for industries, such as the offshore oil industry, because of the production of sulphide, which is highly reactive, corrosive and toxic. However, these organisms can also be beneficial by removing sulphate and heavy metals from waste streams. Although SRB have been studied for more than a century, it is only with the recent emergence of new molecular biological and genomic techniques that we have begun to obtain detailed information on their way of life. Access through your institution Buy or subscribe This is a preview of subscription content, access via your institution Access options Access through your institution Subscribe to this journal Receive 12 print issues and online access $259.00 per year only $21.58 per issue Learn more Buy this article Purchase on SpringerLink Instant access to full article PDF Buy now Prices may be subject to local taxes which are calculated during checkout Similar content being viewed by others Transcriptome-wide marker gene expression analysis of stress-responsive sulfate-reducing bacteria Article Open access 27 September 2023 Oxygen respiration and polysaccharide degradation by a sulfate-reducing acidobacterium Article Open access 10 October 2023 Active microbial sulfate reduction in fluids of serpentinizing peridotites of the continental subsurface Article Open access 10 May 2021 References JÃ¸rgensen, B. B. Mineralization of organic matter in the seabed the role of sulphate reduction. Nature 296, 643645 (1982). Article Google Scholar 2. Rabus, R., Hansen, T. A. & Widdel, F. in The Prokaryotes (eds Dworkin, M., Schleifer, K.-H. & Stackebrandt, E.) 659768 (Springer Verlag, New York, 2006). An excellent overview of the physiology, biochemistry and molecular biology of sulphate- and sulphur-reducing prokaryotes. Book Google Scholar 3. Widdel, F. Anaerober Abbau von FettsÃ¤uren und BenzoesÃ¤ure durch neu Isolierte Arten Sulfat-reduzierender Bakterien. Thesis, GÃ¶ttingen Univ. (1980). Google Scholar 4. Brandis-Heep, A., Gebhardt, N. A., Thauer, R. K., Widdel, F. & Pfennig, N. Anaerobic acetate oxidation to CO2 by Desulfobacter postgatei. I. Demonstration of all enzymes required for the operation of the citric acid cycle. Arch. Microbiol. 136, 222229 (1983). Article CAS Google Scholar 5. Schauder, R., Eikmanns, B., Thauer, T. K., Widdel, F. & Fuchs, G. Acetate oxidation to CO2 in anaerobic bacteria via a novel pathway not involving reactions of the citric acid cycle. Arch. Microbiol. 145, 162172 (1986). Article CAS Google Scholar 6. Oude Elferink, S. J. W. H., Akkermans-van Vliet, W. M., Bogte, J. J. & Stams, A. J. M. Desulfobacca acetoxidans gen. nov. sp. nov., a novel acetate-degrading sulphate reducer isolated from sulfidogenic sludge. Int. J. Syst. Bacteriol. 49, 345350 (1999). Article PubMed Google Scholar 7. Ollivier, B., Cord-Ruwisch, R., Hatchikian, E. C. & Garcia, J.-L. Characterization of Desulfovibrio fructosovorans sp. nov. Arch. Microbiol. 149, 447450 (1988). Article CAS Google Scholar 8. Sass, A., Rutters, H., Cypionka, H. & Sass, H. Desulfobulbus mediterraneus sp. nov., a sulphate-reducing bacterium growing on mono- and disaccharides. Arch. Microbiol. 177, 468474 (2002). Article CAS PubMed Google Scholar 9. Baena, S., Fardeau, M.-L., Labat, M., Ollivier, B., Garcia, J.-L. & Patel, B. K. C. Desulfovibrio aminophilus sp. nov., a novel amino acid degrading and sulphate reducing bacterium from an anaerobic dairy wastewater lagoon. J. Syst. Appl. Microbiol. 21, 498504 (1998). Article CAS Google Scholar 10. Stams, A. J. M., Hansen, T. A. & Skyring, G. W. Utilization of amino acids as energy substrates by two marine Desulfovibrio strains. FEMS Microbiol. Ecol. 31, 1115 (1985). Article CAS Google Scholar Nanninga, H. J. & Gottschal, J. C. Properties of Desulfovibrio carbinolicus sp. nov. and other sulfate reducing bacteria isolated from an anaerobic purification plant. Appl. Environ. Microbiol. 53, 802809 (1987). CAS PubMed PubMed Central Google Scholar 12. Nazina, T. N., Ivanova, A. E., Kanchaveli, L. P. & Rozanova, E. P. A new sporeforming thermophilic methylotrophic sulphate-reducing bacterium, Desulfotomaculum kuznetsovii sp. nov. Mikrobiologiia 57, 823827 (1987). Google Scholar 13. Parshina, S. N. et al. Desulfotomaculum carboxydovorans sp. nov., a novel sulphate-reducing bacterium capable of growth at 100% CO. Int. J. Syst. Evol. Microbiol. 55, 21592165 (2005). Article CAS PubMed Google Scholar 14. Henstra, A. M., Dijkema, C. & Stams, A. J. M. Archaeoglobus fulgidus couples CO oxidation to sulphate reduction and acetogenesis with transient formate accumulation. Environ. Microbiol. 9, 18361841 (2007). Described growth of the sulphate-reducing archaeon A. fulgidus on carbon monoxide, both in the presence and absence of sulphate. Article CAS PubMed Google Scholar 15. Tanimoto, Y. & Bak, F. Anaerobic degradation of methylmercaptan and dimethyl sulfide by newly isolated thermophilic sulfate-reducing bacteria. Appl. Environ. Microbiol. 60, 24502455 (1994). CAS PubMed PubMed Central Google Scholar 16. Bak, F. & Pfennig, N. Chemolithotrophic growth of Desulfovibrio sulfodismutans sp. nov. by disproportionation of inorganic compounds. Arch. Microbiol. 147, 184189 (1987). Showed for the first time that some sulphate reducers can grow by dismutation of the inorganic sulphur compounds sulphite and thiosulphate. Article CAS Google Scholar 17. Bottcher, M. E., Thamdrup, B., Gehre, M. & Theune, A. S34/S32 and O18/O16 fractionation during sulphur disproportionation by Desulfobulbus propionicus. Geomicrobiol. J. 22, 219226 (2005). Article CAS Google Scholar 18. Rabus, R., Nordhaus, R., Ludwig, W. & Widdel, F. Complete oxidation of toluene under strictly anoxic conditions by a new sulfate-reducing bacterium. Appl. Environ. Microbiol. 59, 14441451 (1993). CAS PubMed PubMed Central Google Scholar 19. Harms, G. et al. Anaerobic oxidation of o-xylene, m-xylene, and homologous alkylbenzenes by new types of sulfate-reducing bacteria. Appl. Environ. Microbiol. 65, 9991004 (1999). CAS PubMed PubMed Central Google Scholar 20. Morasch, B., Schink, B., Tebbe, C. C. & Meckenstock, R. U. Degradation of o-xylene and m-xylene by a novel sulphate-reducer belonging to the genus Desulfotomaculum. Arch. Microbiol. 181, 407417 (2004). Article CAS PubMed Google Scholar 21. Aeckersberg, F., Rainey, F. A. & Widdel, F. Growth, natural relationships, cellular fatty acids and metabolic adaptation of sulfate-reducing bacteria that utilize long-chain alkanes under anoxic conditions. Arch. Microbiol. 170, 361369 (1998). Article CAS PubMed Google Scholar 22. So, C. M. & Young, L. Y. Isolation and characterization of a sulfate-reducing bacterium that anaerobically degrades alkanes. Appl. Environ. Microbiol. 65, 29692976 (1999). CAS PubMed PubMed Central Google Scholar 23. Davidova, I. A., Duncan, K. E., Choi, O. K. & Suflita, J. M. Desulfoglaeba alkanexedens gen. nov., sp. nov., an n-alkane-degrading, sulphate-reducing bacterium. Int. J. Syst. Evol. Microbiol. 56, 27372742 (2006). Article CAS PubMed Google Scholar 24. Cravo-Laureau, C., Matheron, R., Cayol, J.-L., Joulian, C. & Hirschler-RÃ©a, A. Desulfatibacillum aliphaticivorans gen. nov., spec. nov., and n-alkane- and n-alkene-degrading, sulphate-reducing bacterium. Int. J. Syst. Evol. Microbiol. 54, 7783 (2004). Article CAS PubMed Google Scholar 25. Grossi, V. et al. Anaerobic 1-alkene metabolism by the alkane- and alkene-degrading sulfate-reducer Desulfatibacillum aliphaticivorans strain CV2803T. Appl. Environ. Microbiol. 73, 78827890 (2007). Article CAS PubMed PubMed Central Google Scholar 26. Kniemeyer, O. et al. Anaerobic oxidation of short-chain hydrocarbons by marine sulphate-reducing bacteria. Nature 449, 898901 (2007). This interesting paper describes, for the first time, the anaerobic oxidation of the short-chain hydrocarbons ethane, propane and butane by SRB. Article CAS PubMed Google Scholar 27. Reeburgh, W. S. Methane consumption in Cariaco trench waters and sediments. Earth Planet. Sci. Lett. 28, 337344 (1976). Article CAS Google Scholar 28. Hoehler, T. M., Alperin, M. J., Albert, D. B. & Martens, C. S. Field and laboratory studies of methane oxidation in an anoxic sediment evidence for a methanogensulphate reducer consortium. Global Biogeochem. Cycles 8, 451463 (1994). Article CAS Google Scholar 29. Boetius, A. et al. A marine microbial consortium apparently mediating anaerobic oxidation of methane. Nature 407, 623626 (2000). Showed that anaerobic methane oxidation is mediated by a syntrophic consortium of archaea and SRB. Article CAS PubMed Google Scholar 30. Orphan, V. J. et al. Comparative analysis of methane-oxidizing archaea and sulfate-reducing bacteria in anoxic marine sediments. Appl. Environ. Microbiol. 67, 19221934 (2001). Article CAS PubMed PubMed Central Google Scholar 31. Nauhaus, K., Albrecht, M., Elvert, M., Boetius, A. & Widdel F. In vitro cell growth of marine archaealbacterial consortia during anaerobic oxidation of methane with sulphate. Environ. Microbiol. 9, 187196 (2007). Article CAS PubMed Google Scholar 32. Wilms, R., Sass, H., Kopke, B., Cypionka, H. & Engelen, B. Methane and sulphate profiles within the subsurface of a tidal flat are reflected by the distribution of sulphate-reducing bacteria and methanogenic archaea. FEMS Microbiol. Ecol. 59, 611621 (2007). Article CAS PubMed Google Scholar 33. Nauhaus, K., Boetius, A., KrÃ¼ger, M. & Widdel, F. In vitro demonstration of anaerobic oxidation of methane coupled to sulphate reduction in sediment from a marine gas hydrate area. Environ. Microbiol. 4, 296305 (2002). Article CAS PubMed Google Scholar 34. Moran, J. J. et al. Methyl sulfides as intermediates in the anaerobic oxidation of methane. Environ. Microbiol. 10, 162173 (2008). CAS PubMed Google Scholar 35. Heijs, S. K., Haese, R. R., van der Wielen, P. W., Forney, L. J. & van Elsas, J. D. Use of 16S rRNA gene based clone libraries to assess microbial communities potentially involved in anaerobic methane oxidation in a Mediterranean cold seep. Microb. Ecol. 53, 384398 (2007). Article CAS PubMed PubMed Central Google Scholar 36. LÃ¶sekann, T. et al. Diversity and abundance of aerobic and anaerobic methane oxidizers at the Haakon Mosby Mud Volcano, Barents Sea. Appl. Environ. Microbiol. 73, 33483362 (2007). Article PubMed PubMed Central CAS Google Scholar 37. Leloup, J. et al. Diversity and abundance of sulphate-reducing microorganisms in the sulphate and methane zones of a marine sediment, Black Sea. Environ. Microbiol. 9, 131142 (2007). Article CAS PubMed Google Scholar 38. Fitz, R. M. & Cypionka, H. Formation of thiosulphate and trithionate during sulphite reduction by washed cells of Desulfovibrio desulphuricans. Arch. Microbiol. 154, 400406 (1990). Article CAS Google Scholar 39. Broco, M., Rousset, M., Oliveira, S. & Rodrigues-Pousada, C. Deletion of flavoredoxin gene in Desulfvibrio gigas reveals its participation in thiosulphate reduction. FEBS Lett. 579, 48034807 (2005). Article CAS PubMed Google Scholar 40. Badziong, W. & Thauer, R. K. Growth yields and growth rates of Desulfovibrio vulgaris (Marburg) growing on hydrogen plus sulphate and hydrogen plus thiosulphate as sole energy sources. Arch. Microbiol. 117, 209214 (1978). Article CAS PubMed Google Scholar 41. Thauer, R. K., Stackebrandt, E. & Hamilton, W. A. in Sulphate-Reducing Bacteria: Environmental and Engineered Systems (eds Barton, L. L. & Hamilton, W. A.) 137 (Cambridge Univ. Press, 2007). Book Google Scholar 42. Odom, J. M. & Peck, H. D. Hydrogen cycling as a general mechanism for energy coupling in the sulphate-reducing bacteria, Desulfovibrio sp. FEMS Microbiol. Lett. 12, 4750 (1981). Article CAS Google Scholar 43. Dalsgaard, T. & Bak, F. Nitrate reduction in a sulfate-reducing bacterium, Desulfovibrio desulfuricans, isolated from rice paddy soil: sulphide inhibition, kinetics, and regulation. Appl. Environ. Microbiol. 60, 291297 (1994). CAS PubMed PubMed Central Google Scholar 44. LÃ³pez-CortÃ©s, A., Fardeau, M. L., Fauque, G., Joulian, C. & Ollivier, B. Reclassification of the sulphate- and nitrate-reducing bacterium Desulfovibrio vulgaris subsp. oxamicus as Desulfovibrio oxamicus sp. nov., comb. nov. Int. J. Syst. Evol. Microbiol. 56, 14951499 (2006). Article PubMed CAS Google Scholar 45. Keith, S. M. & Herbert, R. A. Dissimilatory nitrate reduction by a strain of Desulfovibrio desulfuricans. FEMS Microbiol. Lett. 18, 5559 (1983). Article CAS Google Scholar 46. Moura, I., Bursakov, S., Costa, C. & Moura, J. J. G. Nitrate and nitrite utilization in sulphate-reducing bacteria. Anaerobe 3, 279290 (1997). Article CAS PubMed Google Scholar 47. Lovley, D. R., Roden, E. E., Phillips, E. J. P. & Woodward, J. C. Enzymatic iron and uranium reduction by sulphate-reducing bacteria. Mar. Geol. 113, 4153 (1993). Article CAS Google Scholar 48. Park, H. S., Lin, S. & Voordouw, G. Ferric iron reduction by Desulfovibrio vulgaris Hildenborough wild type and energy metabolism mutants. Antonie van Leeuwenhoek 93, 7985 (2007). Article PubMed CAS Google Scholar 49. Lovley, D. R. & Phillips, E. J. Reduction of uranium by Desulfovibrio desulfuricans. Appl. Environ. Microbiol. 58, 850856 (1992). CAS PubMed PubMed Central Google Scholar 50. Lloyd, J. R., Ridley, J., Khizniak, T., Lyalikova, N. N. & Macaskie, L. E. Reduction of technetium by Desulfovibrio desulfuricans: biocatalyst characterization and use in a flowthrough bioreactor. Appl. Environ. Microbiol. 65, 26912696 (1999). CAS PubMed PubMed Central Google Scholar 51. Tucker, M. D., Barton, L. L. & Thompson, B. M. Reduction of Cr, Mo, Se and U by Desulfovibrio desulfuricans immobilized in polyacrylamide gels. J. Ind. Microbiol. Biotechnol. 20, 1319 (1998). Article CAS PubMed Google Scholar 52. Lovley, D. R. & Phillips, E. J. Reduction of chromate by Desulfovibrio vulgaris and its c 3 cytochrome. Appl. Environ. Microbiol. 60, 726728 (1994). CAS PubMed PubMed Central Google Scholar 53. Macy, J. M., Santini, J. M., Pauling, B. V., O'Neill, A. H. & Sly, L. I. Two new arsenate/sulphate-reducing bacteria: mechanisms of arsenate reduction. Arch. Microbiol. 173, 4957 (2000). Article CAS PubMed Google Scholar 54. Jonkers, H. M., van der Maarel, M. J. E. C., van Gemerden, H. & Hansen, T. A. Dimethylsulfoxide reduction by marine sulphate-reducing bacteria. FEMS Microbiol. Lett. 136, 283287 (1996). Article CAS Google Scholar 55. Lie, T. J., Pitta, T., Leadbetter, E. R., Godchaux, W. 3rd & Leadbetter, J. R. Sulfonates: novel electron acceptors in anaerobic respiration. Arch. Microbiol. 166, 204210 (1996). Article CAS PubMed Google Scholar 56. Dolfing, J. & Tiedje, J. M. Kinetics of two complementary hydrogen sink reactions in a defined 3-chlorobenzoate degrading methanogenic co-culture. FEMS Microbiol. Ecol. 86, 2532 (1991). Article CAS Google Scholar 57. DeWeerd, K. A. A., Mandelco, L., Tanner, R. S., Woese, C. R. & Suflita, J. M. Desulfomonile tiedjei gen. nov., sp. nov., a novel, anaerobic, dehalogenating, sulphate-reducing bacterium. Arch. Microbiol. 154, 2330 (1990). Article CAS Google Scholar 58. Bryant, M. P., Campbell, L. L., Reddy, C. A. & Crabill, M. R. Growth of Desulfovibrio in lactate or ethanol media low in sulfate in association with H2-utilizing methanogenic bacteria. Appl. Environ. Microbiol. 33, 11621169 (1977). The role of SRB in sulphate-depleted methanogenic environments became apparent. CAS PubMed PubMed Central Google Scholar 59. Chartrain, M. & Zeikus, J. G. Microbial ecophysiology of whey biomethanation: characterization of bacterial trophic populations and prevalent species in continuous culture. Appl. Environ. Microbiol. 51, 188196 (1986). CAS PubMed PubMed Central Google Scholar 60. Schink, B. & Stams, A. J. M. in The Prokaryotes (eds Dworkin, M., Schleifer, K.-H. & Stackebrandt, E.) 309335 (Springer Verlag, New York, 2006). Book Google Scholar 61. Plugge, C. M., Balk, M. & Stams, A. J. M. Desulfotomaculum thermobenzoicum subsp. thermosyntrophicum subsp. nov., a thermophilic, syntrophic, propionate-oxidizing, spore-forming bacterium. Int. J. Syst. Evol. Microbiol. 52, 391399 (2002). Article CAS PubMed Google Scholar 62. Boone, D. R. & Bryant, M. P. Propionate-degrading bacterium, Syntrophobacter wolinii sp. nov. gen. nov., from methanogenic ecosystems. Appl. Environ. Microbiol. 40, 626632 (1980). CAS PubMed PubMed Central Google Scholar 63. Wallrabenstein, C., Hauschild, E. & Schink, B. Pure culture and cytological properties of Syntrophobacter wolinii. FEMS Microbiol. Lett. 123, 249254 (1994). Article CAS Google Scholar 64. Harmsen, H., Wullings, B., Akkermans, A. D. L., Ludwig, W. & Stams, A. J. M. Phylogenetic analysis of Syntrophobacter wolinii reveals a relationship with sulphate-reducing bacteria. Arch. Microbiol. 160, 238240 (1993). CAS PubMed Google Scholar 65. Stams, A. J. M., Oude Elferink, S. J. W. H. & Westermann, P. Metabolic interactions between methanogenic consortia and anaerobic respiring bacteria. Adv. Biochem. Eng. Biotechnol. 81, 3156 (2003). CAS PubMed Google Scholar 66. Brysch, K., Schneider, C., Fuchs, G. & Widdel, F. Lithoautotrophic growth of sulphate-reducing bacteria, and description of Desulfobacterium autotrophicum gen. nov., sp. nov. Arch. Microbiol. 148, 264274 (1987). Article CAS Google Scholar 67. Weijma, J. et al. Competition for H2 between sulphate reducers, methanogens and homoacetogens in a gas-lift reactor. Water Sci. Technol. 45, 7580 (2002). Article CAS PubMed Google Scholar 68. SchÃ¶nheit, P., Kristjansson, J. K. & Thauer, R. K. Kinetic mechanism for the ability of sulphate reducers to out-compete methanogens for acetate. Arch. Microbiol. 132, 285288 (1982). Article Google Scholar 69. Oude Elferink, S. J. W. H., Visser, A., Hulshoff-Pol, L. W. & Stams, A. J. M. Sulphate reduction in methanogenic bioreactors. FEMS Microbiol. Rev. 15, 119136 (1994). CAS Google Scholar 70. Omil, F., Lens, P., Visser, A., Hulshoff Pol, L. W. & Lettinga, G. Long-term competition between sulphate reducing and methanogenic bacteria in UASB reactors treating volatile fatty acids. Biotechnol. Bioeng. 57, 676685 (1998). Article CAS PubMed Google Scholar 71. Laanbroek, H. J., Geerligs, H. J., Sijtsma, L. & Veldkamp, H. Competition for sulfate and ethanol among Desulfobacter, Desulfobulbus, and Desulfovibrio species isolated from intertidal sediments. Appl. Environ. Microbiol. 47, 329334 (1984). CAS PubMed PubMed Central Google Scholar 72. Parkes, R. J. in Ecology of Microbial Communities (eds Fletcher, M., Gray, T. R. & Jones, J. G.) 147177 (Cambridge Univ. Press, 1987). Google Scholar 73. Mori, K., Kim, H., Kakegawa, T. & Hanada, S. A novel lineage of sulphate-reducing microorganisms: Thermodesulfobiaceae fam. nov., Thermodesulfobium narugense, gen. nov., sp. nov. a new thermophilic isolate from a hot spring. Extremophiles 7, 283290 (2003). Article CAS PubMed Google Scholar 74. Itoh, T., Suzuki, K-I. & Nakase, T. Thermocladium modestius gen. nov., sp. nov. a new genus of rod-shaped, extremely thermophilic crenarchaeote. Int. J. Syst. Bacteriol. 48, 879887 (1998). Article PubMed Google Scholar 75. Itoh, T., Suzuki, K-I., Sanches, P. C. & Nakase, T. Caldivirga maquilingensis gen. nov., sp. nov. a new genus of rod-shaped crenarchaeote isolated from a hot spring in the Philippines. Int. J. Syst. Bacteriol. 49, 11571163 (1999). Article CAS PubMed Google Scholar 76. Daly, K., Sharp, R. J. & McCarthy, A. J. Development of oligonucleotide probes and PCR primers for detecting phylogenetic subgroups of sulfate-reducing bacteria. Microbiology 146, 16931705 (2000). Article CAS PubMed Google Scholar 77. Wagner, M., Roger, A. J., Flax, J. L., Brusseau, G. A. & Stahl, D. A. Phylogeny of dissimilatory sulfite reductases supports an early origin of sulfate respiration. J. Bacteriol. 180, 29752982 (1998). CAS PubMed PubMed Central Google Scholar 78. Meyer, B. & Kuever, J. Phylogeny of the alpha and beta subunits of the dissimilatory adenosine-5²-phosphosulfate (APS) reductase from sulfate-reducing prokaryotes origin and evolution of the dissimilatory sulfate-reduction pathway. Microbiology 153, 20262044 (2007). Article CAS PubMed Google Scholar 79. Dhillon, A., Teske, A., Dillon, J., Stahl, D. A. & Sogin, M. L. Molecular characterization of sulfate-reducing bacteria in the Guaymas Basin. Appl. Environ. Microbiol. 69, 27652772. Article CAS Google Scholar 80. Dar, S. A., Kuenen, J. G. & Muyzer, G. Nested PCR-denaturing gradient gel electrophoresis approach to determine the diversity of sulfate-reducing bacteria in complex microbial communities. Appl. Environ. Microbiol. 71, 23252330 (2005). Article CAS PubMed PubMed Central Google Scholar 81. Dar, S. A., Yao, L., van Dongen, U., Kuenen, J. G. & Muyzer, G. Analysis of diversity and activity of sulfate-reducing bacterial communities in sulfidogenic bioreactors using 16S rRNA and dsrB genes as molecular markers. Appl. Environ. Microbiol. 73, 594604 (2007). Article CAS PubMed Google Scholar 82. Geets, J. et al. DsrB gene-based DGGE for community and diversity surveys of sulphate-reducing bacteria. J. Microbiol. Methods 66, 194205 (2006). Article CAS PubMed Google Scholar 83. Minz, D. et al. Diversity of sulfate-reducing bacteria in oxic and anoxic regions of a microbial mat characterized by comparative analysis of dissimilatory sulfite reductase genes. Appl. Environ. Microbiol. 65, 46664671 (1999). CAS PubMed PubMed Central Google Scholar 84. Meyer, B. & Kuever, J. Molecular analysis of the diversity of sulfate-reducing and sulfur-oxidizing prokaryotes in the environment, using the aprA as functional marker gene. Appl. Environ. Microbiol. 73, 76647679 (2007). Article CAS PubMed PubMed Central Google Scholar 85. Loy, A. et al. Oligonucleotide microarray for 16S rRNA gene-based detection of all recognized lineages of sulfate-reducing prokaryotes in the environment. Appl. Environ. Microbiol. 68, 50645081 (2002). Article CAS PubMed PubMed Central Google Scholar 86. Loy, A. et al. Microarray and functional gene analyses of sulfate-reducing prokaryotes in low sulfate acidic fens reveal co-occurrence of recognized genera and novel lineages. Appl. Environ. Microbiol. 70, 69987009 (2004). Article CAS PubMed PubMed Central Google Scholar 87. Stubner, S. Enumeration of 16S rDNA of Desulfotomaculum lineage 1 in rice fields soil by real-time PCR with SybrGreen detection. J. Microbiol. Methods 50, 155164 (2002). Article CAS PubMed Google Scholar 88. Stubner, S. Quantification of Gram-negative sulphate-reducing bacteria in rice field soil by 16S rRNA gene-targeted real-time PCR. J. Microbiol. Methods 57, 219230 (2004). Article CAS PubMed Google Scholar 89. Foti, M. et al. Diversity, activity, and abundance of sulfate-reducing bacteria in saline and hypersaline soda lakes. Appl. Environ. Microbiol. 73, 20932100 (2007). Article CAS PubMed PubMed Central Google Scholar 90. Ben-Dov, E., Brenner, A. & Kushmaro, A. Quantification of sulfate-reducing bacteria in industrial wastewater by real-time polymerase chain reaction (PCR) using dsrA and apsA genes. Microb. Ecol. 54, 439451 (2007). Article CAS PubMed Google Scholar 91. Neretin, L. N. et al. Quantification of dissimilatory (bi)sulphite reductase gene expression in Desulfobacterium autotrophicum using real-time RT-PCR. Environ. Microbiol. 5, 660671 (2003). Article CAS PubMed Google Scholar 92. LÃ¼ckner, S. et al. Improved 16S rRNA-targeted probe set for analysis of sulfate-reducing bacteria by fluorescence in situ hybridization. J. Microbiol. Methods 69, 523528 (2007). Article CAS Google Scholar 93. Stahl, D. A., Loy, A. & Wagner, M. in Sulphate-Reducing Bacteria: Environmental and Engineered Systems (eds Barton, L. L. & Hamilton, W. A.) 167183 (Cambridge Univ. Press, 2007). Google Scholar 94. Mussmann, M., Ishii, K., Rabus, R. & Amann, R. Diversity and vertical distribution of cultured and uncultured Deltaproteobacteria in an intertidal mud flat of the Wadden Sea. Environ. Microbiol. 7, 405418 (2005). Article PubMed Google Scholar 95. Ito, T. et al. Phylogenetic identification and substrate uptake patterns of sulfate-reducing bacteria inhabiting an oxicanoxic sewer biofilm determined by combining microautoradiography and fluorescence in situ hybridization. Appl. Environ. Microbiol. 68, 356364 (2002). Article CAS PubMed PubMed Central Google Scholar 96. Boschker, H. T. S. et al. Direct linking of microbial populations to specific biogeochemical processes by 13C-labelling of biomarkers. Nature 392, 801804 (1998). Article CAS Google Scholar 97. Webster, G. et al. A comparison of stable isotope probing of DNA and phospholipids fatty acids to study prokaryotic functional diversity in sulfate-reducing marine sediment enrichment slurries. Environ. Microbiol. 8, 15751589 (2006). Article CAS PubMed Google Scholar 98. Ramsing, N. B., KÃ¼hl, M. & JÃ¸rgensen, B. B. Distribution of sulfate-reducing bacteria, O2, and H2S in photosynthetic biofilms determined by oligonucleotide probes and microelectrodes. Appl. Environ. Microbiol. 59, 38403849 (1993). CAS PubMed PubMed Central Google Scholar 99. Wawer, C., Jetten, M. S. & Muyzer, G. Genetic diversity and expression of the NiFe hydrogenase large-subunit gene of Desulfovibrio spp. in environmental samples. Appl. Environ. Microbiol. 61, 43604369 (1997). Google Scholar 100. Ravenschlag, K., Sahm, K., Knoblauch, C., JÃ¸rgensen, B. B. & Amann, R. Community structure, cellular rRNA content, and activity of sulfate-reducing bacteria in marine Arctic sediments. Appl. Environ. Microbiol. 66, 35923602 (2000). Article CAS PubMed PubMed Central Google Scholar 101. Jeanthon, C. et al. Thermodesulfobacterium hydrogeniphilum sp. nov., a thermophilic, chemolithoautotrophic sulfate-reducing bacterium isolated from a deep-sea hydrothermal vent at Guaymas Basin and emendation of the genus Thermodesulfobacterium. Int. J. Syst. Evol. Microbiol. 52, 765772 (2002). CAS PubMed Google Scholar 102. Knittel, K. et al. Activity, distribution, and diversity of sulfate reducers and other bacteria in sediments above gas hydrate (Cascadia Margin, Oregon). Geomicrobiol. J. 20, 269294 (2003). Article CAS Google Scholar 103. Stadnitskaia, A. et al. Biomarker and 16S rDNA evidence for anaerobic oxidation of methane and related carbonate precipitation in deep-sea mud volcanoes of the Sorokin Trough, Black Sea. Mar. Geol. 217, 6796 (2005). Article CAS Google Scholar 104. Rissati, J. B., Capman, W. C. & Stahl, D. A. Community structure of a microbial mat: the phylogenetic dimension. Proc. Natl Acad. Sci. USA 91, 1017310177 (1994). Article Google Scholar 105. Sen, A. M. Acidophilic Sulphate Reducing Bacteria: Candidates for Bioremediation of Acid Mine Drainage Pollution. Thesis, Univ. Wales (2001). Google Scholar 106. Nilsen, R. K., Beeder, J., Thostenson, T. & Torsvik, T. Distribution of thermophilic marine sulfate reducers in North Sea oil field waters and oil reservoirs. Appl. Environ. Microbiol. 62, 17931798 (1996). CAS PubMed PubMed Central Google Scholar 107. Kovacik, W. P. Jr. Molecular analysis of deep subsurface Cretaceous rock indicates abundant Fe(III)- and S°-reducing bacteria in a sulfate-rich environment. Environ. Microbiol. 8, 141155 (2006). Article CAS PubMed Google Scholar 108. Sass, H., Wieringa, E., Cypionka, H., Babenzien, H. D. & Overmann, J. High genetic and physiological diversity of sulfate-reducing bacteria isolated from an oligotrophic lake sediment. Arch. Microbiol. 170, 243251 (1998). Article CAS PubMed Google Scholar 109. Hines, M. E. et al. Molecular phylogenetic and biogeochemical studies of sulfate-reducing bacteria in the rhizosphere of Spartina alterniflora. Appl. Environ. Microbiol. 65, 22092216 (1999). CAS PubMed PubMed Central Google Scholar 110. Bahr, M. et al. Molecular chacterization of sulfate-reducing bacteria in a New England salt marsh. Environ. Microbiol. 7, 11751185 (2005). Article CAS PubMed Google Scholar 111. Dubilier, N. et al. Endosymbiontic sulphate-reducing and sulphide-oxidizing bacteria in an oligochaete worm. Nature 411, 298302 (2001). Article CAS PubMed Google Scholar 112. Woyke, T. et al. Symbiosis insights through metagenomic analysis of a microbial consortium. Nature 443, 950955 (2006). Intriguing paper on the symbiosis of four bacteria, two sulphate-reducing and two sulphur-oxidizing, in a gutless marine worm. Article CAS PubMed Google Scholar 113. Mattorano, D. A. & Merinar, T. Respiratory protection on offshore drilling rigs. Appl. Occup. Environ. Hyg. 14, 141148 (1999). Article CAS PubMed Google Scholar 114. Kaster, K. M., Grigoriyan, A., Jenneman, G. & Voordouw, G. Effect of nitrate and nitrite on sulphide production by two thermophilic, sulphate-reducing enrichments from an oil field in the North Sea. Appl. Microbiol. Biotechnol. 75, 195203 (2007). Article CAS PubMed Google Scholar 115. Hubert, C. & Voordouw, G. Oil field souring control by nitrate-reducing Sulphurospirillum spp. that outcompete sulfate-reducing bacteria for organic electron donors. Appl. Environ. Microbiol. 73, 26442652 (2007). Article CAS PubMed PubMed Central Google Scholar 116. Hulshoff-Pol, L. W., Lens, P. N. L., Stams, A. J. M. & Lettinga, G. Anaerobic treatment of sulphate-rich wastewaters. Biodegradation 9, 213224 (1998). Article CAS PubMed Google Scholar 117. Lens, P. N. L., Vallero, M. & Esposito, R. in Sulphate-Reducing Bacteria: Environmental and Engineered Systems (eds Barton L. L. & Hamilton, W. A.) 283404 (Cambridge Univ. Press, 2007). Google Scholar 118. Kaufmann, E. N., Little, M. H. & Selvaraj, P. T. A biological process for the reclamation of flue gas desulfurization using mixed sulfate reducing bacteria with inexpensive carbon sources. Appl. Biochem. Biotechnol. 63, 677693 (1996). Google Scholar 119. Koschorreck, M. et al. Processes at the sediment water interface after addition of organic matter and lime to an Acid Mine Pit Lake mesocosm. Environ. Sci. Technol. 41, 16081614 (2007). Article CAS PubMed Google Scholar 120. Van Houten, B. H. G. W. et al. Occurrence of methanogenesis during the start-up of a full-scale synthesis gas-fed reactor treating sulphate and metal-rich wastewater. Water Res. 40, 553560 (2006). Article CAS PubMed Google Scholar 121. Dar, S. A., Stams, A. J., Kuenen, J. G. & Muyzer, G. Co-existence of physiologically similar sulphate-reducing bacteria in a full-scale sulfidogenic bioreactor fed with a single organic electron donor. Appl. Microbiol. Biotechnol. 75, 14631472 (2007). Article CAS PubMed PubMed Central Google Scholar 122. van Houten, B. H. G. W. et al. Desulfovibrio paquesii sp. nov., a hydrogenotrophic sulfate-reducing bacterium isolated from a full-scale synthesis gas fed bioreactor treating zinc and sulfate-rich wastewater. Int. J. Syst. Evol. Microbiol. (in the press). 123. Buisman, C. J. N., Geraats, B. G., Ijspeert, P. & Lettinga, G. Optimisation of sulphur production in a biotechnological sulphide-removing reactor. Biotechnol. Bioeng. 35, 5056 (1990). Article CAS PubMed Google Scholar 124. Janssen, A. J. H., Ruitenberg, R. & Buisman, C. J. N. Industrial applications of new sulphur biotechnology. Water Sci. Technol. 44, 8590 (2001). Article CAS PubMed Google Scholar 125. Rao, A. G., Ravichandra, P., Joseph, J., Jetty, A. & Sarma, P. N. Microbial conversion of sulphur dioxide in flue gas to sulphide using bulk drug industry wastewater as an organic source by mixed cultures of sulphate reducing bacteria. J. Hazard Mater. 147, 718725 (2007). Article CAS PubMed Google Scholar 126. Weijma, J., Stams, A. J. M., Hulshoff Pol, L. W. & Lettinga, G. Thermophilic sulphate reduction and methanogenesis with methanol in a high rate anaerobic reactor. Biotechnol. Bioeng. 67, 354363 (2000). Article CAS PubMed Google Scholar 127. Ingham, C. J. et al. The micro-Petri dish, a million-well growth chip for the culture and high-throughput screening of microorganisms. Proc. Natl Acad. Sci. USA 104, 1821718222 (2007). Article CAS PubMed PubMed Central Google Scholar 128. Wagner, M., Nielsen, P. H., Loy, A., Nielsen, J. L. & Daims, H. Linking microbial community structure with functions: fluorescence in situ hybridizationmicroautoradiography and isotope arrays. Curr. Opin. Biotechnol. 17, 8391 (2006). Article CAS PubMed Google Scholar 129. Dumont, M. G. & Murrell, J. C. Stable isotope probing linking microbial identity to function. Nature Rev. Microbiol. 3, 499504 (2005). An excellent overview of the use of SIP in microbial ecology. Article CAS Google Scholar 130. Li, T. et al. Simultaneous analysis of microbial identity and function using NanoSIMS. Environ. Microbiol. 10, 580588 (2008). Article CAS PubMed PubMed Central Google Scholar 131. Coleman, J. R., Culley, D. E., Chrisler, W. B. & Brockman, F. J. mRNA-targeted fluorescent in situ hybridization (FISH) of Gram-negative bacteria without template amplification or tyramide signal amplification. J. Microbiol. Methods 71, 246255 (2007). This paper describes SIMSISH, a novel approach to identify active community members at a single-cell level. Article CAS PubMed Google Scholar 132. Klenk, H.-P. et al. The complete genome sequence of the hyperthermophilic, sulphate-reducing archaeon Archaeoglobus fulgidus. Nature 390, 364370 (1997). Article CAS PubMed Google Scholar 133. Heidelberg, J. F. et al. The genome sequence of the anaerobic, sulphate-reducing bacterium Desulfovibrio vulgaris Hildenborough. Nature Biotechnol. 22, 554559 (2004). Article CAS Google Scholar 134. Rabus, R. et al. The genome of Desulfotalea psychrophila, a sulphate-reducing bacterium from permanently cold Arctic sediments. Environ. Microbiol. 6, 887902 (2004). Article CAS PubMed Google Scholar 135. Mukhopadhyay, A. et al. Salt stress in Desulfovibrio vulgaris Hildenborough: an integrated genomics approach. J. Bacteriol. 188, 40684078 (2006). Article CAS PubMed PubMed Central Google Scholar 136. Fournier, M. et al. Response of the anaerobe Desulfovibrio vulgaris Hildenborough to oxidative conditions: proteome and transcript analysis. Biochimie 88, 8594 (2006). Article CAS PubMed Google Scholar 137. Beijerinck, W. M. Ãber Spirillum desulphuricans als Ursache von Sulfatreduktion. Zentralb. Bakteriol. Parasitk. Infekt. Abt. II 1, 4959 (1895). Google Scholar 138. JÃ¸rgensen, B. B. The sulphur cycle of a coastal marine sediment (Limfjorden, Denmark). Limnol. Oceanogr. 22, 814832 (1978). Article Google Scholar 139. Sass, A. & Cypionka, H. in Sulphate-Reducing Bacteria: Environmental and Engineered Systems (eds Barton, L. L. & Hamilton, W. A.) 167183 (Cambridge Univ. Press, 2007). Book Google Scholar 140. Mogensen, G. L., Kjeldsen, K. U. & Ingvorsen, K. Desulfovibrio aerotolerans sp. nov., an oxygen tolerant sulphate-reducing bacterium isolated from activated sludge. Anaerobe 11, 339349 (2005). Article CAS PubMed Google Scholar 141. Kjeldsen, K. U., Joulian, C. & Ingvorsen, K. Oxygen tolerance of sulphate-reducing bacteria in activated sludge. Environ. Sci. Technol. 38, 20382043 (2004). Article CAS PubMed Google Scholar 142. Dilling, W. & Cypionka, H. Aerobic respiration in sulphate-reducing bacteria. Arch. Microbiol. 71, 123128 (1990). Showed for the first time that some sulphate reducers can use molecular oxygen as a terminal electron acceptor. CAS Google Scholar 143. Sigalevich, P., Meshorer, E., Helman, Y. & Cohen, Y. Transition from anaerobic to aerobic growth conditions for the sulfate-reducing bacterium Desulfovibrio oxyclinae results in flocculation. Appl. Environ. Microbiol. 66, 50055012 (2000). Article CAS PubMed PubMed Central Google Scholar 144. Beech, I. B. & Sunner, J. A. in Sulphate-Reducing Bacteria: Environmental and engineered systems (eds Barton L. L. & Hamilton, W. A.) 459482 (Cambridge Univ. Press, 2007). Book Google Scholar 145. Lovley, D. R. & Klug, M. J. Sulfate reducers can outcompete methanogens at freshwater sulfate concentrations. Appl. Environ. Microbiol. 45, 187192 (1983). CAS PubMed PubMed Central Google Scholar 146. Ingvorsen, K. & JÃ¸rgensen, B. B. Kinetics of sulphate uptake by freshwater and marine species of Desulfovibrio. Arch. Microbiol. 139, 6166 (1984). Article CAS Google Scholar 147. Coetser, S. E. & Cloete, T. E. Biofouling and biocorrosion in industrial water systems. Crit. Rev. Microbiol. 31, 213232 (2005). Article CAS PubMed Google Scholar 148. Madigan, M. T. & Martinko, J. M. Brock Biology of Microorganisms 11th edn (Pearson Education, London, 2006). Google Scholar 149. Pruesse, E. et al. SILVA: a comprehensive online resource for quality checked and aligned ribosomal RNA sequence data compatible with ARB. Nucleic Acids Res. 35, 71887196 (2007). Article CAS PubMed PubMed Central Google Scholar 150. Ludwig, W. et al. ARB: a software environment for sequence data. Nucleic Acids Res. 32, 13631371 (2004). Article CAS PubMed PubMed Central Google Scholar 151. Thauer, R. K., Jungermann, K. & Decker, K. Energy conservation in chemotrophic anaerobic bacteria. Bacteriol. Rev. 41, 100180 (1977). An excellent and still appreciated review on the metabolism of chemotrophic microorganisms. CAS PubMed PubMed Central Google Scholar Download references Acknowledgements We thank the three anonymous reviewers for their constructive comments. We are grateful to L. Robertson, curator of the Beijerinck Museum, for allowing the reproduction of Vibrio desulfuricans. We acknowledge the long-lasting collaboration on sulphur biotechnology between Wageningen University and the Delft University of Technology. A. Janssen and D. Sorokin are thanked for creative discussions. We thank Paques (Balk, The Netherlands) and Shell Global Solutions International B.V. (Amsterdam, The Netherlands) for advice and financial support. Our research was supported by the Netherlands Organization for Scientific Research, division for Earth and Life Sciences and division for Technical Sciences, and the Technology Programme of the Ministry of Economic Affairs. Author information Author notes Laboratory of Microbiology, Wageningen University, Dreijenplein 10, 6703 HB, Wageningen, The Netherlands. Authors and Affiliations Department of Biotechnology, Delft University of Technology, Julianalaan 67, Delft, 2628 BC, The Netherlands Gerard Muyzer & Alfons J. M. Stams Authors Gerard Muyzer View author publications Search author on:PubMed Google Scholar 2. Alfons J. M. Stams View author publications Search author on:PubMed Google Scholar Corresponding author Correspondence to Gerard Muyzer. Related links Related links FURTHER INFORMATION Gerard Muyzer's homepage Genomes OnLine Database Paques ProbeBase The Integrated Microbial Genomes (IMG) Database The ARB Project SILVA rRNA database project Glossary Chemolithotropic : Metabolism of an organism that obtains energy from inorganic compounds and carbon from carbon dioxide. Citric acid cycle : A cyclic series of reactions that result in the conversion of acetate to carbon dioxide and NADH. Acetyl-CoA pathway : A pathway of autotrophic carbon dioxide fixation and acetate oxidation in obligate anaerobes. Dismutation : The splitting of a chemical compound into two new compounds, one that is more oxidized and one that is more reduced than the original compound. Syntrophic : Growth of two or more organisms that depend on each other for their growth. Substrate-level phosphorylation : Synthesis of high-energy phosphate bonds through the reaction of inorganic phosphate with an activated organic substrate. Homoacetogen : A bacterium that produces acetate as the sole product from sugar fermentation or from hydrogen and carbon dioxide. Acetoclastic methanogen : A methanogen that uses acetate as a substrate to produce methane and carbon dioxide. Phospholipid fatty acid : A key component of the cellular membrane of living cells that can be used to identify specific groups of microorganisms and to monitor their physiological state. CARD-FISH : Fluorescence in situ hybridization with horseradish peroxidase-labelled oligonucleotide probes and fluorochrome-labelled tyramides. The tyramides are deposited at the hybridization site, resulting in enhanced fluorescence intensity. Microautoradiography : A photographic technique to visualize the uptake of radioactive substrates by single cells. Stable isotope probing : A technique to identify microorganisms in environmental samples that have taken up a stable isotope-labelled substrate. Niche differentiation : The tendency for coexisting species to differ in their use of resources. Acid-mine drainage site : Acid water that contains H2SO4 derived from microbial oxidation of sulphidic minerals. Rights and permissions Reprints and permissions About this article Cite this article Muyzer, G., Stams, A. The ecology and biotechnology of sulphate-reducing bacteria. Nat Rev Microbiol 6, 441454 (2008). Download citation Published: Issue Date: DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative This article is cited by Hydrogen storage and geo-methanation in a depleted underground hydrocarbon reservoir Cathrine Hellerschmied Johanna Schritter Andreas P. Loibner Nature Energy (2024) ### Cobalt speciation and cycling in Linsley Pond, Connecticut, USA Zhemin Xuan Gaboury Benoit Aquatic Sciences (2024) ### Use of nitrate, sulphate, and iron (III) as electron acceptors to improve the anaerobic degradation of linear alkylbenzene sulfonate: effects on removal potential and microbiota diversification Joelithon L. Costa Luiz Galdino Silva Mario Takayuki Kato Environmental Science and Pollution Research (2024) ### Chemical characterization of archaeological marine wooden piles from the ancient harbor Theodosius in Istanbul Oktay Gonultas Mualla Balaban-Ucar European Journal of Wood and Wood Products (2024) ### Geochemical transformations of sulfur and their role in the formation of different types and subtypes of saline lakes in Southeastern Transbaikalia S. V. Borzenko I. A. Fedorov Applied Water Science (2024) Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Sign up for the Nature Briefing newsletter what matters in science, free to your inbox daily. Get the most important science stories of the day, free in your inbox. Sign up for Nature Briefing
4593	https://www.timeofcare.com/the-difference-between-lethargy-obtundation-stupor-and-coma/	Home Pharmacy Mnemonics Resources Select Page The Difference Between Lethargy, Obtundation, Stupor, and Coma PSYCHIATRY There is a spectrum of impaired consciousness that goes from full arousal to complete unresponsiveness. Coma, which is a state of unarousable unresponsiveness is the worst degree of impairment of a patient’s arousal and consciousness. Words like lethargy, obtunded, and stupor all describe various degrees to which a patient’s arousal is impaired. However, these terms are imprecise. In a clinical setting, it is more useful to describe the patient’s responses to specific stimuli. What is the difference between being lethargic, obtunded, stuporous, or in a coma? \| \| \| --- \| \| Level of Consciousness \| Description \| \| Clouding of consciousness \| The patient has a very mild form of altered mental status in which the patient has inattention and reduced wakefulness. \| \| Confusional State \| The patient has a more profound deficit than clouding of consciousness that includes disorientation, bewilderment, and difficulty following commands. \| \| Lethargy \| The patient has severe drowsiness. He/she can be aroused by moderate stimuli, but then drifts back to sleep. \| \| Obtundation \| “is a state similar to lethargy in which the patient has a lessened interest in the environment, slowed responses to stimulation, and tends to sleep more than normal with drowsiness in between sleep states.” \| \| Stupor \| Being in stupor means that “only vigorous and repeated stimuli will arouse the patient, and when left undisturbed, the patient will immediately lapse back to the unresponsive state.” \| \| Coma \| “Coma is a state of unarousable unresponsiveness.” \| Reference Tindall SC. Level of Consciousness. In: Walker HK, Hall WD, Hurst JW, editors. Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition. Boston: Butterworths; 1990. Chapter 57. Available from: print Related Well Child Check Well Adult Exam Low Income ACID-BASE DERMATOLOGY Forms & Charts History Templates PE Templates Prevention ECG Tips Guidelines Quick Physical Exam A good Assessment & Plan MV Tips EBM OB/GYN GERIATRICS PSYCHIATRY RADIOLOGY DIFFERENTIAL DIAGNOSIS Clinic Quick Reference Hospital Quick Reference InfoGraphics Book Chapter in Time of Care Notes Coding Guide Office Management
4594	https://www.chemicalaid.com/tools/equationbalancer.php?equation=Ba%28OH%292+%2B+Ca%28OH%292+%3D+BaCaO2+%2B+H2O&hl=en	Ba(OH)2 + Ca(OH)2 = BaCaO2 + H2O - Chemical Equation Balancer ChemicalAid Calculators Chemical Equation Balancer Molar Mass Calculator Empirical Formula Calculator Significant Figures Calculator Reaction Stoichiometry Calculator Limiting Reagent Calculator Oxidation Number Calculator Net Ionic Equation Calculator Redox Reaction Calculator Solubility Calculator All Calculators Elements Periodic Table Periodic Trends Element Charts Future Elements Info Chemical Formulas Reference Tables Chemistry Books Question Solutions Chemistry Quizzes Chemistry Jokes Words from Elements Help Chat Forums 🌎 EN English 👤 Login Ba(OH)2 + Ca(OH)2 = BaCaO2 + H2O - Chemical Equation Balancer Balanced Chemical Equation Ba(OH)2 + Ca(OH)2 → BaCaO 2 + 2 H 2 O Reaction Information Disclaimer Ba(OH)2+Ca(OH)2=BaCaO 2+H 2 O Reaction Type Double Displacement (Metathesis) Reactants Barium Hydroxide - Ba(OH)2 Ba(Oh)2 Barium Hydroxide Lime Caustic Baryta Ba(OH)2 Molar MassBa(OH)2 Oxidation NumberLewis Structure Calcium Hydroxide - Ca(OH)2 Hydralime Ca(Oh)2 Hydrated Lime Caustic Lime Calcium Hydrate Slaked Lime Ca(OH)2 Molar MassCa(OH)2 Oxidation NumberLewis Structure Products BaCaO 2 Water - H 2 O Hydroxic Acid H₂O [Oh2] Aqua Pure Water Hydrogen Oxide Oxidane Dihydrogen Oxide H2O Molar MassH2O Oxidation NumberLewis Structure Calculate Reaction Stoichiometry Calculate Limiting Reagent Warning: One of the compounds in Ba(OH)2 + Ca(OH)2 = BaCaO2 + H2O is unrecognized. Verify 'BaCaO2' is entered correctly. Chemical Equation Balancer 🛠️ Balance Equation ➜ Instructions To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above. Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. Replace immutable groups in compounds to avoid ambiguity. For example, C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but XC2H5 + O2 = XOH + CO2 + H2O will. Compound states [like (s) (aq) or (g)] are not required. You can use parenthesis () or brackets []. How To Balance Equations Balance any equation or reaction using this chemical equation balancer! Find out what type of reaction occured. Read our article on how to balance chemical equations or ask for help in our chat. Examples Ba(OH)2 + Ca(OH)2 = Ba(OH)2 + Ca Ba(OH)2 + Ca(OH)2 = CaBaO2 + H2O CUSO4 + NaOH = Na2SO4 + CU(OH)2 NH4Br + Ca(NO2)2 = CaBr2 + N2 + H2O NH3 + Na = H2 + Na(NH2) Al + NaOH + H2O = NaAl(OH)4 + H2 N2O5 + Ba(OH)2 = N2 + BaO5 + H2O CH3CHO + C2H6 = (CH3)2CO + H2O AgClO4 + Cl2 = AgCl + Cl2O + O2 CuSO4 + Na2S2O3 = Cu2SO4 + Na2S4O6 + Na2SO4 H2 + BaSO4 = H2SO4 + Ba C3Cl3N3O3 + Au + NH4Cl + H3PO4 = HAuCl4 + C3N3O3H3 + (NH4)3PO4 Recently Balanced Equations Algebraic Method Using Inspection Balance Ba(OH)2 + Ca(OH)2 = BaCaO2 + H2O Using the Algebraic Method Step 1: Label Each Compound With a Variable Label each compound (reactant or product) in the equation with a variable to represent the unknown coefficients. a Ba(OH)2 + b Ca(OH)2 = c BaCaO 2 + d H 2 O Step 2: Create a System of Equations Create an equation for each element (Ba, O, H, Ca) where each term represents the number of atoms of the element in each reactant or product. Ba: 1 a + 0b = 1 c + 0d O: 2 a + 2 b = 2 c + 1 d H: 2 a + 2 b = 0c + 2 d Ca: 0a + 1 b = 1 c + 0d Step 3: Solve For All Variables Use substitution, Gaussian elimination, or a calculator to solve for each variable. Using Substitution or Elimination 1a - 1c = 0 2a + 2b - 2c - 1d = 0 2a + 2b - 2d = 0 1b - 1c = 0 1a = 1c 2a + 2b = 2c + 1d 2a + 2b = 2d 1b = 1c Using Linear Systems / Algebra Use your graphing calculator's rref() function (or an online rref calculator) to convert the following matrix into reduced row-echelon-form: [ 1 0 -1 0 0] [ 2 2 -2 -1 0] [ 2 2 0 -2 0] [ 0 1 -1 0 0] The resulting matrix can be used to determine the coefficients. In the case of a single solution, the last column of the matrix will contain the coefficients. Convert to RREF and Solve Step-by-Step Simplify the result to get the lowest, whole integer values. a = 1 (Ba(OH)2) b = 1 (Ca(OH)2) c = 1 (BaCaO2) d = 2 (H2O) Step 4: Substitute Coefficients and Verify Result Count the number of atoms of each element on each side of the equation and verify that all elements and electrons (if there are charges/ions) are balanced. Ba(OH)2 + Ca(OH)2 = BaCaO 2 + 2 H 2 O \| \| Reactants \| Products \| \| --- \| \| Ba \| 1 \| 1 \| ✔️ \| \| O \| 4 \| 4 \| ✔️ \| \| H \| 4 \| 4 \| ✔️ \| \| Ca \| 1 \| 1 \| ✔️ \| Since there is an equal number of each element in the reactants and products of Ba(OH)2 + Ca(OH)2 = BaCaO2 + 2H2O, the equation is balanced. Balance Ba(OH)2 + Ca(OH)2 = BaCaO2 + H2O Using Inspection Step 1: Count the number of each element on the left and right hand sides \| \| Reactants (Left Hand Side) \| Products (Right Hand Side) \| \| --- --- \| \| \| Reactants \| Products \| \| \| \| Ba(OH)2 \| Ca(OH)2 \| Total \| BaCaO 2 \| H 2 O \| Total \| \| \| Ba \| 1 \| \| 1 \| 1 \| \| 1 \| ✔️ \| \| O \| 2 \| 2 \| 4 \| 2 \| 1 \| 3 \| ❌ \| \| H \| 2 \| 2 \| 4 \| \| 2 \| 2 \| ❌ \| \| Ca \| \| 1 \| 1 \| 1 \| \| 1 \| ✔️ \| Step 2: Multiply coefficients for compounds to balance out each element For each element that is not equal, try to balance it by adding more of it to the side with less. Sometimes there may be multiple compounds with that element on one side, so you'll need to use your best judgement and be prepared to go back and try the other options. O is not balanced. Add 1 molecule of H 2 O to the product (right-hand) side to balance Oxygen: Ba(OH)2 + Ca(OH)2 = BaCaO 2 + 2 H 2 O \| \| Reactants \| Products \| \| --- --- \| \| Ba \| 1 \| 1 \| ✔️ \| \| O \| 4 \| 4 \| ✔️ \| \| H \| 4 \| 4 \| ✔️ \| \| Ca \| 1 \| 1 \| ✔️ \| Step 3: Verify that the equation is balanced Since there are an equal number of atoms of each element on both sides, the equation is balanced. Ba(OH)2 + Ca(OH)2 = BaCaO 2 + 2 H 2 O Practice Balancing Ba(OH)2 -+ + Ca(OH)2 -+ = BaCaO 2 -+ + H 2 O -+ \| \| Reactants \| Products \| \| --- --- \| \| Ba \| 1 \| 1 \| ✔️ \| \| O \| 4 \| 3 \| ❌ \| \| H \| 4 \| 2 \| ❌ \| \| Ca \| 1 \| 1 \| ✔️ \| Hint: The sum of the balanced coefficients is 5. See Step-By-Step Solution Calculators Equations & Reactions Chemical Equation Balancer Reaction Stoichiometry Calculator Limiting Reagent Calculator Ionic Equation Calculator Redox Calculator Compounds & Formulas Empirical Formula Calculator Molar Mass Calculator Oxidation Number Calculator Solubility Calculator Lewis Structure Calculator Bond Polarity Calculator Calculations Significant Figures Calculator Chemistry Equation Calculators Ideal Gas Law Unit Converter Copyright © 2008-2025 About Us · Privacy Policy · Terms of Service 🇺🇸🌎 English Languages 🇺🇸English 🇪🇸español 🇫🇷français 🇩🇪Deutsch 🇵🇹português 🇮🇹italiano 🇷🇺русский 🇵🇱polski 🇨🇿čeština 🇻🇳Tiếng Việt 🇮🇩Indonesia 🇰🇷한국어 🇯🇵日本語 🇨🇳中文 🇸🇦العربية 🇹🇷Türkçe ✕ Do not sell or share my personal information. You have chosen to opt-out of the sale or sharing of your information from this site and any of its affiliates. To opt back in please click the "Customize my ad experience" link. This site collects information through the use of cookies and other tracking tools. Cookies and these tools do not contain any information that personally identifies a user, but personal information that would be stored about you may be linked to the information stored in and obtained from them. This information would be used and shared for Analytics, Ad Serving, Interest Based Advertising, among other purposes. For more information please visit this site's Privacy Policy. CANCEL CONTINUE Your Use of Our Content ✕ The content we make available on this website [and through our other channels] (the “Service”) was created, developed, compiled, prepared, revised, selected, and/or arranged by us, using our own methods and judgment, and through the expenditure of substantial time and effort. This Service and the content we make available are proprietary, and are protected by these Terms of Service (which is a contract between us and you), copyright laws, and other intellectual property laws and treaties. This Service is also protected as a collective work or compilation under U.S. copyright and other laws and treaties. We provide it for your personal, non-commercial use only. You may not use, and may not authorize any third party to use, this Service or any content we make available on this Service in any manner that (i) is a source of or substitute for the Service or the content; (ii) affects our ability to earn money in connection with the Service or the content; or (iii) competes with the Service we provide. These restrictions apply to any robot, spider, scraper, web crawler, or other automated means or any similar manual process, or any software used to access the Service. You further agree not to violate the restrictions in any robot exclusion headers of this Service, if any, or bypass or circumvent other measures employed to prevent or limit access to the Service by automated means. Information from your device can be used to personalize your ad experience. Do not sell or share my personal information. Terms of Content Use A Raptive Partner Site
4595	https://www.alphadictionary.com/goodword/word/alacrity	alacrity - Good Word Word of the Day alphaDictionary Free English On-line Dictionary Â Top Pages Rebel-Yankee Test The 100 Funniest Words in English The 100 Most Beautiful Words in English The 100 Most Interesting Words in English Good Word Today's Good Word Good Word Archive Good Word Dictionary Sign Up! Podcasts GW Office Dr. Goodword Best Words Grammar and Style Russian Grammar Word Fun Fun & Games Language Fun Laughing Stock Agora Donate! Word Blog Search Website: Share this page Share Â • alacrity • Printable Version Pronunciation:ê-læ-krê-tee• Hear it! Part of Speech: Noun, mass (no plural) Meaning: Willing promptness in responding, cheerful briskness. Notes: Today is 'two-for-one-day' at alphaDictionary: we are talking about both alacrity and its near synonym, celerity. Both these words refer to quickness or speed. Celerity is a neutral term for swiftness though the cleanness in its sound further implies a certain smoothness. Alacrity refers only to human quickness for it also connotes willingness if not eagerness to move swiftly. This word's adjective is an oddball: alacritous may be used with or without the [t], i.e. alacrious, though the latter is a bit dated. In Play: Today's word refers to a willing, eager speed: "The people of this nation responded to the plight of the victims of hurricane Katrina in New Orleans with an alacrity unknown to the government." Use this word to measure the zeal with which someone responds to a challenge: "Jessie Skape cleaned the restrooms of the restaurant where he worked with little alacrity." Word History: Today's Good Word comes from Latin alacritas, the noun from the adjective alacer "quick, eager, lively" + -itas, a noun suffix. It shares a source with Old English ellen "zeal, courage", which didn't make it down to us, but did in Modern German as eilen "to hurry". Latin alacer came to Italian as allegro "happy, cheerful", which English-speaking musicians borrowed, too, in the sense of "brisk, lively (with alacrity)". (We are happy that Tim Ward brought up today's Good Word in the Agora with such alacrity as we could not ignore.) P.S. - Register for the Daily Good Word E-Mail! - You can get our daily Good Word sent directly to you via e-mail in either HTML or Text format. Go to our Registration Page to sign up today! Come visit our website at for more Good Words and other language resources! Next Word: alarum Previous Word: akimbo [ A ] Home â€¢Alpha Agora â€¢Language Blog Laughing Stock â€¢Privacy Policy â€¢About Us â€¢Contact us Â©2004-2024 Robert Beard All Rights Reserved.
4596	https://reference.wolfram.com/language/ref/ModularInverse.html	ModularInverse—Wolfram Documentation Products Wolfram\|One The definitive Wolfram Language and notebook experience Mathematica The original technical computing environment Wolfram Notebook Assistant + LLM Kit All-in-one AI assistance for your Wolfram experience System Modeler Wolfram Player Finance Platform Wolfram Engine Enterprise Private Cloud Application Server Wolfram\|Alpha Notebook Edition Wolfram Cloud App Wolfram Player App More mobile apps Core Technologies of Wolfram Products Wolfram Language Computable Data Wolfram Notebooks AI & Linguistic Understanding Deployment Options Wolfram Cloud wolframscript Wolfram Engine Community Edition Wolfram LLM API WSTPServer Wolfram\|Alpha APIs From the Community Function Repository Community Paclet Repository Example Repository Neural Net Repository Prompt Repository Wolfram Demonstrations Data Repository Group & Organizational Licensing All Products Consulting & Solutions We deliver solutions for the AI era—combining symbolic computation, data-driven insights and deep technical expertise Data & Computational Intelligence Model-Based Design Algorithm Development Wolfram\|Alpha for Business Blockchain Technology Education Technology Quantum Computation WolframConsulting.com Wolfram Solutions Data Science Artificial Intelligence Biosciences Healthcare Intelligence Sustainable Energy Control Systems Enterprise Wolfram\|Alpha Blockchain Labs More Wolfram Solutions Wolfram Solutions For Education Research Universities Colleges & Teaching Universities Junior & Community Colleges High Schools Educational Technology Computer-Based Math More Solutions for Education Contact Us Learning& Support Get Started Wolfram Language Introduction Fast Intro for Programmers Fast Intro for Math Students Wolfram Language Documentation More Learning Highlighted Core Areas Demonstrations YouTube Daily Study Groups Wolfram Schools and Programs Books Grow Your Skills Wolfram U Courses in computing, science, life and more Community Learn, solve problems and share ideas. Blog News, views and insights from Wolfram Resources for Software Developers Tech Support Contact Us Support FAQs Support FAQs Contact Us Company About Wolfram Career Center All Sites & Resources Connect & Follow Contact Us Work with Us Student Ambassador Initiative Wolfram for Startups Student Opportunities Jobs Using Wolfram Language Educational Programs for Adults Summer School Winter School Educational Programs for Youth Middle School Camp High School Research Program Computational Adventures Read Stephen Wolfram's Writings Wolfram Blog Wolfram Tech \| Books Wolfram Media Complex Systems Educational Resources Wolfram MathWorld Wolfram in STEM/STEAM Wolfram Challenges Wolfram Problem Generator Wolfram Initiatives Wolfram Science Wolfram Foundation History of Mathematics Project Events Stephen Wolfram Livestreams Online & In-Person Events Contact Us Connect & Follow Wolfram\|Alpha Your Account User Portal Wolfram Cloud Products Wolfram\|One Mathematica Wolfram Notebook Assistant + LLM Kit System Modeler Wolfram Player Finance Platform Wolfram\|Alpha Notebook Edition Wolfram Engine Enterprise Private Cloud Application Server Wolfram Cloud App Wolfram Player App More mobile apps Core Technologies Wolfram Language Computable Data Wolfram Notebooks AI & Linguistic Understanding Deployment Options Wolfram Cloud wolframscript Wolfram Engine Community Edition Wolfram LLM API WSTPServer Wolfram\|Alpha APIs From the Community Function Repository Community Paclet Repository Example Repository Neural Net Repository Prompt Repository Wolfram Demonstrations Data Repository Group &Organizational Licensing All Products Consulting & Solutions We deliver solutions for the AI era—combining symbolic computation, data-driven insights and deep technical expertise WolframConsulting.com Wolfram Solutions Data Science Artificial Intelligence Biosciences Healthcare Intelligence Sustainable Energy Control Systems Enterprise Wolfram\|Alpha Blockchain Labs More Wolfram Solutions Wolfram Solutions For Education Research Universities Colleges & Teaching Universities Junior & Community Colleges High Schools Educational Technology Computer-Based Math More Solutions for Education Contact Us Learning & Support Get Started Wolfram Language Introduction Fast Intro for Programmers Fast Intro for Math Students Wolfram Language Documentation Grow Your Skills Wolfram U Courses in computing, science, life and more Community Learn, solve problems and share ideas. Blog News, views and insights from Wolfram Resources for Software Developers Tech Support Contact Us Support FAQs More Learning Highlighted Core Areas Demonstrations YouTube Daily Study Groups Wolfram Schools and Programs Books Support FAQs Contact Us Company About Wolfram Career Center All Sites & Resources Connect & Follow Contact Us Work with Us Student Ambassador Initiative Wolfram for Startups Student Opportunities Jobs Using Wolfram Language Educational Programs for Adults Summer School Winter School Educational Programs for Youth Middle School Camp High School Research Program Computational Adventures Read Stephen Wolfram's Writings Wolfram Blog Wolfram Tech \| Books Wolfram Media Complex Systems Educational Resources Wolfram MathWorld Wolfram in STEM/STEAM Wolfram Challenges Wolfram Problem Generator Wolfram Initiatives Wolfram Science Wolfram Foundation History of Mathematics Project Events Stephen Wolfram Livestreams Online & In-Person Events Contact Us Connect & Follow Wolfram\|Alpha Wolfram Cloud Your Account User Portal Wolfram Language& System Documentation Center ModularInverse See Also PowerMod Mod Power PowerModList ExtendedGCD PolynomialExtendedGCD MultiplicativeOrder EulerPhi PrimitiveRoot CoprimeQ Related Guides Number Theory Number Theoretic Functions Numerical Functions Integer Functions See Also PowerMod Mod Power PowerModList ExtendedGCD PolynomialExtendedGCD MultiplicativeOrder EulerPhi PrimitiveRoot CoprimeQ Related Guides Number Theory Number Theoretic Functions Numerical Functions Integer Functions ModularInverse[k,n] gives the modular inverse of k modulo n. Details Examples Basic Examples Scope Numerical Evaluation Applications Basic Applications Number Theory Properties & Relations Possible Issues Interactive Examples Neat Examples See Also Related Guides Related Links History Cite this Page BUILT-IN SYMBOL See Also PowerMod Mod Power PowerModList ExtendedGCD PolynomialExtendedGCD MultiplicativeOrder EulerPhi PrimitiveRoot CoprimeQ Related Guides Number Theory Number Theoretic Functions Numerical Functions Integer Functions See Also PowerMod Mod Power PowerModList ExtendedGCD PolynomialExtendedGCD MultiplicativeOrder EulerPhi PrimitiveRoot CoprimeQ Related Guides Number Theory Number Theoretic Functions Numerical Functions Integer Functions ModularInverse ✖ ModularInverse ModularInverse[k,n] ✖ ModularInverse[k,n] gives the modular inverse of k modulo n. Details ModularInverse is also known as modular multiplicative inverse. Integer mathematical function, suitable for both symbolic and numerical manipulation. Typically used in modular arithmetic and cryptography. ModularInverse[k,n] gives the number r such that the remainder of the division of r k by n is equal to 1. If k and n are not coprime, no modular inverse exists and ModularInverse[k,n] remains unevaluated. Examples open all close all Basic Examples(2)Summary of the most common use cases Compute the inverse of 3 modulo 5 and check the result: In:=1 ✖ Out=1 Plot the sequence with a fixed modulus: In:=1 ✖ Out=1 Scope(2)Survey of the scope of standard use cases Numerical Evaluation(2) Compute using integers: In:=1 ✖ Out=1 In:=2 ✖ Out=2 Gaussian integers: In:=3 ✖ Out=3 Compute using large integers: In:=1 ✖ Out=1 Applications(4)Sample problems that can be solved with this function Basic Applications(2) Two numbers are modular inverses of each other if their product is 1: In:=1 ✖ In:=2 ✖ In:=3 ✖ Out=3 Modular computation of a matrix inverse: In:=1 ✖ First compute the matrix adjoint: In:=2 ✖ Then compute the modular inverse of a matrix: In:=3 ✖ Out=3 Check that the inverse gives the correct result: In:=4 ✖ Out=4 Number Theory(2) Build an RSA-like toy encryption scheme. Start with the modulus: In:=1 ✖ Out=1 Find the universal exponent of the multiplication group modulo n: In:=2 ✖ Out=2 Private key: In:=3 ✖ Out=3 Public key: In:=4 ✖ Out=4 Encrypt a message: In:=5 ✖ Out=5 Decrypt it: In:=6 ✖ Out=6 Create a random number generator that uses the current time as a seed: In:=1 ✖ Choose modulus parameters: In:=2 ✖ Compute 20 random numbers between 0 and 1: In:=3 ✖ Out=3 Properties & Relations(6)Properties of the function, and connections to other functions ModularInverse is a periodic function: In:=1 ✖ Out=1 ExtendedGCD returns modular inverses: In:=1 ✖ Out=1 In:=2 ✖ Out=2 Compute using PowerMod: In:=1 ✖ Out=1 The results have the same sign as the modulus: In:=1 ✖ Out=1 In:=2 ✖ Out=2 If and are coprime, then is invertible modulo : In:=1 ✖ Out=1 In:=2 ✖ Out=2 Computing ModularInverse twice yields the original argument: In:=1 ✖ Out=1 Possible Issues(1)Common pitfalls and unexpected behavior For nonzero integers k and n, ModularInverse[k,n] exists if and only if k and n are coprime: In:=1 ✖ Out=1 In:=2 ✖ Out=2 In:=3 ✖ Out=3 However, 10 and 22 are not coprime: In:=4 ✖ Out=4 In:=5 ✖ Out=5 In:=6 ✖ Out=6 Interactive Examples(1)Examples with interactive outputs Visualize inverses modulo varying prime numbers: In:=1 ✖ Out=1 Neat Examples(2)Surprising or curious use cases Visualize when a number is invertible modulo 12: In:=1 ✖ Out=1 Modular inverses of sums of two squares: In:=1 ✖ Out=1 See Also PowerModModPowerPowerModListExtendedGCDPolynomialExtendedGCDMultiplicativeOrderEulerPhiPrimitiveRootCoprimeQ Function Repository: FractionMod Related Guides ▪ Number Theory ▪ Number Theoretic Functions ▪ Numerical Functions ▪ Integer Functions Related Links MathWorld History Introduced in 2017 (11.1) Cite this as:Wolfram Research (2017), ModularInverse, Wolfram Language function, ✖ Wolfram Research (2017), ModularInverse, Wolfram Language function, Text Wolfram Research (2017), ModularInverse, Wolfram Language function, ✖ Wolfram Research (2017), ModularInverse, Wolfram Language function, CMS Wolfram Language. 2017. "ModularInverse." Wolfram Language & System Documentation Center. Wolfram Research. ✖ Wolfram Language. 2017. "ModularInverse." Wolfram Language & System Documentation Center. Wolfram Research. APA Wolfram Language. (2017). ModularInverse. Wolfram Language & System Documentation Center. Retrieved from ✖ Wolfram Language. (2017). ModularInverse. Wolfram Language & System Documentation Center. Retrieved from BibTeX @misc{reference.wolfram_2025_modularinverse, author="Wolfram Research", title="{ModularInverse}", year="2017", howpublished="\url{ note=[Accessed: 28-September-2025]} ✖ @misc{reference.wolfram_2025_modularinverse, author="Wolfram Research", title="{ModularInverse}", year="2017", howpublished="\url{ note=[Accessed: 28-September-2025]} BibLaTeX @online{reference.wolfram_2025_modularinverse, organization={Wolfram Research}, title={ModularInverse}, year={2017}, url={ note=[Accessed: 28-September-2025]} ✖ @online{reference.wolfram_2025_modularinverse, organization={Wolfram Research}, title={ModularInverse}, year={2017}, url={ note=[Accessed: 28-September-2025]} Give FeedbackTop Introduction for Programmers Introductory Book Wolfram Function Repository\|Wolfram Data Repository\|Wolfram Data Drop\|Wolfram Language Products Top Products Wolfram\|One Mathematica Notebook Assistant + LLM Kit System Modeler Wolfram\|Alpha Notebook Edition Wolfram\|Alpha Pro Mobile Apps Wolfram Player Wolfram Engine Volume & Site Licensing Server Deployment Options Consulting Wolfram Consulting Repositories Data Repository Function Repository Community Paclet Repository Neural Net Repository Prompt Repository Wolfram Language Example Repository Notebook Archive Wolfram GitHub Learning Wolfram U Wolfram Language Documentation Webinars & Training Educational Programs Wolfram Language Introduction Fast Introduction for Programmers Fast Introduction for Math Students Books Wolfram Community Wolfram Blog Public Resources Wolfram\|Alpha Wolfram Problem Generator Wolfram Challenges Computer-Based Math Computational Thinking Computational Adventures Demonstrations Project Wolfram Data Drop MathWorld Wolfram Science Wolfram Media Publishing For Customers Online Store Product Downloads User Portal Your Account Support FAQ Contact Support Company About Wolfram Careers Contact Events Wolfram CommunityWolfram Blog Legal&Privacy Policy WolframAlpha.com\|WolframCloud.com © 2025Wolfram © 2025Wolfram\|Legal&Privacy Policy\| English English 日本語简体中文 Find out if you already have access to Wolfram tech through your organization ×
4597	https://www.youtube.com/watch?v=gVgJxk8JsDU	The Mole 4 - Converting Mass to Atoms: Practice Problems Beals Science 55400 subscribers 168 likes Description 16327 views Posted: 28 Apr 2016 Craig Beals from Beals Science explains how to convert between mass in grams to atoms using the mole. The video includes molar mass, Avogadro's number, converting and mole practice problems. To learn more visit: For the previous video: Molar Mass, click here: For the next video: Mole conversions of compounds, click here: (coming soon!) 15 comments Transcript: Intro hey welcome back it's Greg Beals and we've been working our way through the mole here and we're finally up to the point where we can start doing some converting we're gonna start with converting mass to atoms and atoms to mass now in the last video we talked a lot about molar mass so if you don't understand the concepts of molar mass make sure you take one step back I'll put the link down here below me somewhere and down in the description so that you can play catch up if you need to but when it comes to converting mass Conversion Factors to atoms there's a couple of conversion factors that you really need to take note of and those are the following the molar mass which I just spoke about the molar mass is the mass grams per mole so typically that ends up being one mole and whatever the mass of that is and we need to use our periodic table to find that for example if I was going to do one mole of hydrogen the elemental hydrogen it would be one point zero one grams now I can also put that as one mole of hydrogen over one point zero one grams of hydrogen I should probably put my hydrogen up here I apologize for the poor handwriting but that's molar mass those are the two conversion factors we're gonna need the other one here is of course when we're talking about the mole is Avogadro's number okay so I've Oh God Rose number which is the mole or represented as a mole so if we have one mole of hydrogen we'll go right back to hydrogen we have 6.02 times 10 to the 23rd atoms of hydrogen and likewise we have 6.02 times 10 to the 23rd atoms of hydrogen in one mole of hydrogen so you're gonna see these conversion factors all the way through here let's Dimensional Analysis start with our first example I'm gonna refer back to this all the time and you've seen me do it the whole way through if you've been watching the videos these are the easiest ways to work through these problems and we call this dimensional analysis first part says write down what you know and what you know is what's given in the problem in this case it's 45 grams of mercury mercury is Hg in dimensional analysis we need to get rid of this grams of mercury because we want to know how many atoms so right now we're stuck in grams of mercury the only way to get into atoms is to get rid of those units of grams of mercury so I'm gonna put grams of mercury down here and then the next part is a tricky part figuring out well what do I put on top of the next conversion factor grams can only turn into one thing it can only turn into moles this is why we're learning about moles because once we turn things into moles we can turn it into anything we want so I can put moles of mercury up here now remember molar mass mole and mass mole and mass this grams represents mass mole and mass molar mass is the molar mass it is the mass of one mole of mercury where do I get that from my periodic table I find period I find mercury on the periodic table it's number 80 wait down here hard to see that on yours but I can see it on here very quickly very clearly it's 200 point five nine grams so the molar mass is 200 point five nine grams per mole boy I made a mess these are gonna cancel and I'm left with moles of mercury I don't want to be in moles of mercury when I get done I want to be in atoms of mercury so I need another conversion factor right here and I'm trying to get rid of moles of mercury so I'm gonna put moles of mercury down here it's the easiest thing to do just so they'll cancel out I can turn moles into anything I want that's the beautiful thing about moles I'm trying to get my answer over here in atoms of mercury and I just said I can turn moles into anything I want just about so I can turn that into atoms of mercury now remember when we're going from moles to atoms there's only one conversion factor that conversion factors Avogadro's number 6.02 times 10 to the 23rd anytime you see atoms like this anytime you see atoms you're gonna have Avogadro's number in the front of it now let's take care of some accounting you get rid of the moles of mercury these cancel because they're on opposite sides of the line the only units I have left right here are atoms of mercury well that's what I want my answer in so this has got to be correct the only thing I have to do now we've done number two is and one is solve I'm gonna get out my trusty calculator for that and here's the calculator I'm gonna clear that off from some math we were doing before I'll walk you through this let's take 45 for the 45 grams times 1 but I'm gonna skip that step because I don't need to do it I'll go straight to divided by 200 point 5 9 I'm gonna hit enter just so that the calculator can keep track of what's going on let me move this out of the way so we can see now I've got an answer for all of this and I need to times that by Avogadro's number 6.02 times 10 to the 23rd so the calculator is keeping track all I have to do is hit the x button and now here's the part that people mess up all the time they try to get fancy don't get fancy we're trying to put this number in right here 6 let me move this 6.02 e e the e e means times 10 to the so ee the last part is 23 I hit enter here's my answer I've got 1 point 3 5 don't forget to look way over here this e 23 tells me times 10 to the 23 I'm gonna move this down and write it right up here we've got 1 point 3 5 times 10 to the 23rd equals 1 point 3 5 times 10 to the 23rd and what are my only units left atoms of mercury so there you go there's our first answer nicely done let's try another one how many atoms of Practice Problem 2 sulfur in 2.1 2 grams of sulfur let's go right back to our rules we write down what we know and what we know is what we're given or the one that has a number and that's 2.1 two grams of sulfur there we go we're trying to turn that over here into atoms of sulfur so we start off with what we know which is 2.1 two grams of sulfur and dimensional analysis i need a conversion factor that has grams of sulfur down here on the bottom grams of sulfur cuz i need these two to cancel out that steps the easiest step I don't want my answer in grams of sulfur so I need to have grams of sulfur down here the tricky part is remembering what can I turn that into up here well grams can only turn into one thing they can turn into moles that's why we're learning about moles so grams can turn into moles so I'm gonna go ahead and put mole up here moles of sulfur and then whenever I see moles and grams grams is mass moles and mass moles and mass molar mass that's the molar mass so the mass of one mole is how many ever grams or the atomic mass units of sulfur which in this case is 30 2.07 thirty two point zero seven take care of these units grams of sulfur is going to cancel right here if I was to stop and get my calculator my answer would be in moles of sulfur but I wanted in atoms of sulfur so I need another conversion factor here in my dimensional analysis I have moles of sulfur here I'm gonna put moles of sulfur down here I can turn moles into anything I want that's a great thing about moles so if I've got it in moles I can turn it into atoms of sulfur let's go ahead and put those units here if you remember from the last slide we said that one mole of sulfur is equal to abogados number of sulfur anytime I have atoms here I've got to have Avogadro's number which is 6.02 times 10 to the 23rd atoms of sulfur cancel some units just so we make sure we're on the right track good the only thing I have left is atoms of sulfur now right down at this part we set up our units we already did that we just need to solve it the way we solve it is to get out our calculator here's the calculator with everything cleaned up and I'm gonna work my way right down through here I've got 2 point 1 2 times 1 but I don't need to do the times 1 so I'm just gonna do directly divided by 32 point 0 7 divided by 32 point 0 7 I'm gonna hit energies because this calculator doesn't keep track in sequence unless I use parentheses but I don't need to the next thing is I have 2 times that whole thing by Avogadro's number 6.02 times 10 to the 23rd I hit the x button 6.02 e e means times 10 to that then I just hit the 23 you can see this e means times 10 to the and I hit enter my answer now is going to be three point nine eight times 10 to the 22 3 point 9 8 times 10 to the 22 the only units I have left right here are atoms of sulfur so I can put atoms of sulfur right here that's my answer I've got two examples to take this the other direction so we're starting with the number of atoms and we're going to turn it to the mass nothing changes though we write down what we know and what we know is three point five zero times 10 to the 15 three point five zero times 10 to the 15th atoms of helium and as always in dimensional analysis when I'm converting moles I need a conversion factor with atoms of helium on the bottom so that they cancel atoms of helium I can only turn atoms into one thing I can only turn atoms into moles of helium well how do I do that I have moles and I have atoms if you remember from before one mole is 6.02 times 10 to the 23rd atoms of helium Avogadro's number the 6.02 times 10 to the 23rd is always gonna be with that unit of atoms I like to cancel things out right away just to see where I'm at and make sure I'm on the right track right now if I stopped and started doing my math that would be in moles of helium but I don't want to be in moles I want to be in grams or the mass need another conversion factor to get myself there I can put moles of helium down here moles of helium and I can turn moles into anything I want which means I can turn it into grams or the mass grams of helium how do I do that I have moles and mass moles and mass moles and mass that's the molar mass it comes from the periodic table I find helium right up here helium has got a mass of 4.00 that's what we're gonna use for point zero zero grams of helium in one mole of helium these units cancel I'm in grams of helium and that's what I want my answer in all I have to do now is get out the calculator we type this in three point five zero times ten to the that's the EE button 15 then I need to take that times one which I'm gonna skip so I'm just going to go straight to divided by Avogadro's number divided by six point zero to EE means times 10 to the 23rd I hit enter and here's what I get but I still have some more work to do I need to take this whole number times four grams times four equals this is going to be scientific notation we need to move this one two three four five six seven eight so we'll have two point three three we're gonna round this up two point three three times ten to the negative eight here we go two point three three times ten to the negative eight grams of helium in the next series of videos we're gonna convert moles of a compound into mass of that compound and actually vice versa we'll take the mass and we'll turn that into moles we've got a couple more steps but eventually we're gonna end up where we can take mass of something and turn it into mass of something else by going through this concept of moles to get there I will put the link to the next video right down here and you get the link to the last couple videos popping up on the screen here or you can come on over to Beal science com I've got a whole series of videos that's gonna help you out with chemistry help you out with biology or if you just like to watch stuff blow up and burn and bubble like I do I've got a bunch of videos of that to keep on learning
4598	https://artofproblemsolving.com/wiki/index.php/Miquel%27s_point?srsltid=AfmBOooJqE3VhZ6KXHdgi4eGcBzsQMmvhqVIk1qLeA1ZNFQhPZT6iVBy	Art of Problem Solving Miquel's point - 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 Miquel's point Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Miquel's point Contents 1 Miquel and Steiner's quadrilateral theorem 2 Circle of circumcenters 3 Triangle of circumcenters 4 Analogue of Miquel's point 5 Six circles crossing point Miquel and Steiner's quadrilateral theorem Let four lines made four triangles of a complete quadrilateral. In the diagram these are Prove that the circumcircles of all four triangles meet at a single point. Proof Let circumcircle of circle cross the circumcircle of circle at point Let cross second time in the point is cyclic is cyclic is cyclic is cyclic and circumcircle of contain the point Similarly circumcircle of contain the point as desired. vladimir.shelomovskii@gmail.com, vvsss Circle of circumcenters Let four lines made four triangles of a complete quadrilateral. In the diagram these are Prove that the circumcenters of all four triangles and point are concyclic. Proof Let and be the circumcircles of and respectively. In In is the common chord of and Similarly, is the common chord of and Similarly, is the common chord of and points and are concyclic as desired. vladimir.shelomovskii@gmail.com, vvsss Triangle of circumcenters Let four lines made four triangles of a complete quadrilateral. In the diagram these are Let points and be the circumcenters of and respectively. Prove that and perspector of these triangles point is the second (different from ) point of intersection where is circumcircle of and is circumcircle of Proof Quadrungle is cyclic Spiral similarity sentered at point with rotation angle and the coefficient of homothety mapping to , to , to are triangles in double perspective at point These triangles are in triple perspective are concurrent at the point The rotation angle to is for sides and or angle between and which is is cyclic is cyclic. Therefore is cyclic as desired. Similarly, one can prove that Double perspective triangles vladimir.shelomovskii@gmail.com, vvsss Analogue of Miquel's point Let inscribed quadrilateral and points be given. Prove that points and are concyclic. Proof Corollary The points and are concyclic. The points and are concyclic. vladimir.shelomovskii@gmail.com, vvsss Six circles crossing point Let point point be given. Denote tangent to tangent to Prove that the circles and have the common point. Proof Let points and are concyclic, Similarly is the Miquel point of quadrungle is tangent to Similarly, is tangent to vladimir.shelomovskii@gmail.com, vvsss Retrieved from " 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.
4599	https://banotes.org/macroeconomics-i/determining-equilibrium-output-open-economy/	Determining Equilibrium Output in an Open Economy Framework • BA Notes Anthropology Anthropology & Research Methods Anthropology of Indigenous People Applied Anthropology Archaeological Anthropology Biological Anthropology Public Health & Epidemiology Social & Cultural Anthropology Social Policies & Administration Tourism Anthropology Economics Indian Economy-I Indian Economy-II Microeconomics-I Microeconomics-II Macroeconomics-I Macroeconomics-II History India History of India Earliest Times to 300 C.E. History of India C.300 to 1206 History of India C.1206-1707 History of India 1707-1950 Traditions of History Writing in India Others History of China C. 1840-1978 History of Modern East Asia – Japan (C.1868–1945) European History 1789-1945 History of Environment Political Science Comparative Government and Politics Conflict Resolution & Peace Building Democracy and Development in Northeast India Gandhi & Contemporary World India’s Foreign Policy in a Globalising World Indian Government and Politics Introduction of International Relations Introduction to Political Theory Introduction to South Asia State Politics in India Public Administration Administrative System at State & District Levels Administrative System at Union Level Administrative System in BRICS Administrative Thinkers Disaster Management E-Governance Governance – Issues and Challenges Logistics Management Organisational Behaviour Perspectives on Public Administration Right to Information Stress & Time Management Contact Macroeconomics-I Determining Equilibrium Output in an Open Economy Framework December 8, 2023 When studying macroeconomics, one of the most crucial concepts to understand is how nations determine their economic equilibrium when engaged with the global marketplace. In an open economy, where countries trade goods and services internationally, the traditional closed-economy models must be expanded to account for these external interactions. The equilibrium output in an open economy framework is determined by considering not just domestic consumption, investment, and government spending, but also the critical component of net exports. Table of Contents The transition from closed to open economy models Components of the open economy equilibrium The role of net exports Import function and the marginal propensity to import The open economy multiplier Numerical example of the open economy multiplier Determining equilibrium in an open economy External trade and the equilibrium adjustment process Initial impact phase Multiplier effect phase Balance of payments adjustment Exchange rates and their impact on equilibrium output Currency depreciation effects Currency appreciation effects Policy implications in an open economy Fiscal policy considerations Monetary policy and interest rates Trade policy implications Global interdependence and equilibrium output Spillover effects Coordinated policy responses Modern challenges to open economy equilibrium Global value chains Digital trade and services Financial integration The transition from closed to open economy models 🔗 In a closed economy, the equilibrium output is determined by the familiar equation: Y = C + I + G, where aggregate demand consists of consumption (C), investment (I), and government spending (G). However, this simplified model doesn’t reflect the reality of today’s interconnected global economy. When we introduce international trade, we must expand our model to include exports (X) and imports (M), resulting in the open economy equation: Y = C + I + G + (X – M) The term (X – M) represents net exports, which can be positive (trade surplus) or negative (trade deficit). This addition fundamentally changes how we analyze and understand economic equilibrium. Components of the open economy equilibrium 🔗 The role of net exports 🔗 Net exports function as an injection or leakage in the circular flow of income, depending on whether they are positive or negative: When X > M: A trade surplus exists, which acts as an injection into the economy, boosting aggregate demand and likely increasing equilibrium output. When X < M: A trade deficit exists, serving as a leakage from the domestic economy, which can reduce aggregate demand and potentially lower equilibrium output. The significance of net exports becomes apparent when we consider that they represent demand for domestically produced goods and services from foreign markets, minus domestic demand for foreign goods and services. Import function and the marginal propensity to import 🔗 To properly model an open economy, we need to understand how imports relate to income. The import function is typically expressed as: M = mY Where ‘m’ represents the marginal propensity to import (MPI), which indicates the fraction of an additional unit of income that will be spent on imported goods. For example, if MPI = 0.2, then for every additional $100 of income, $20 will be spent on imports. The import function can be more realistically represented as: M = M₀ + mY Where M₀ represents autonomous imports that occur regardless of income level. The open economy multiplier 🔗 The introduction of international trade modifies the simple Keynesian multiplier. In a closed economy, the multiplier is calculated as: Multiplier = 1 / (1 – MPC) Where MPC is the marginal propensity to consume. In an open economy, the multiplier becomes: Multiplier = 1 / (1 – MPC + MPI) This formula reveals an important insight: the open economy multiplier is smaller than the closed economy multiplier. This occurs because some of the increased spending leaks out of the domestic economy through imports. Numerical example of the open economy multiplier 🔗 Consider an economy with an MPC of 0.8 and an MPI of 0.2: Closed economy multiplier: 1 / (1 – 0.8) = 5 Open economy multiplier: 1 / (1 – 0.8 + 0.2) = 1 / 0.4 = 2.5 This example demonstrates that the same increase in autonomous spending (like government expenditure) would generate only half as much additional output in an open economy compared to a closed one. Determining equilibrium in an open economy 🔗 To find the equilibrium output in an open economy, we need to solve for the level of income where aggregate demand equals aggregate supply: Y = C + I + G + (X – M) Substituting the consumption function (C = C₀ + cY) and import function (M = M₀ + mY): Y = C₀ + cY + I + G + X – M₀ – mY Rearranging to solve for Y: Y – cY + mY = C₀ + I + G + X – M₀ Y(1 – c + m) = C₀ + I + G + X – M₀ Y = (C₀ + I + G + X – M₀) / (1 – c + m) This equation gives us the equilibrium output level in an open economy. External trade and the equilibrium adjustment process 🔗 When external trade conditions change, the economy undergoes an adjustment process to reach a new equilibrium. This process involves several phases: Initial impact phase 🔗 When there’s a change in exports or autonomous imports, there’s an immediate impact on aggregate demand. For instance, an increase in exports directly increases aggregate demand, shifting the AD curve rightward. Multiplier effect phase 🔗 The initial change triggers a multiplier effect. As income rises due to increased exports, consumption increases (based on the MPC), but some income also leaks out through imports (based on the MPI). The final change in equilibrium output will be: ΔY = (Initial change in exports or autonomous imports) × (Open economy multiplier) Balance of payments adjustment 🔗 Changes in the equilibrium output affect the balance of payments. Higher domestic income typically leads to increased imports, which can partially offset the initial improvement in net exports. This feedback mechanism is crucial for understanding the long-term equilibrium in an open economy. Exchange rates and their impact on equilibrium output 🔗 Exchange rates play a pivotal role in determining equilibrium output in an open economy by influencing both exports and imports: Currency depreciation effects 🔗 When a domestic currency depreciates (loses value relative to foreign currencies): Exports become cheaper: Foreign buyers can purchase more domestic goods with the same amount of their currency, potentially increasing export demand. Imports become more expensive: Domestic consumers need more of their currency to purchase the same foreign goods, potentially decreasing import demand. If these price effects are sufficiently strong (meeting the Marshall-Lerner condition), currency depreciation can improve net exports and increase equilibrium output. Currency appreciation effects 🔗 Conversely, when a domestic currency appreciates: Exports become more expensive: Potentially reducing foreign demand for domestic goods. Imports become cheaper: Potentially increasing domestic demand for foreign goods. This typically leads to a deterioration in net exports and may reduce equilibrium output, all else being equal. Policy implications in an open economy 🔗 Understanding equilibrium output determination in an open economy has significant implications for policy formulation: Fiscal policy considerations 🔗 The effectiveness of fiscal policy is reduced in an open economy due to the smaller multiplier. When government increases spending, some of the stimulative effect “leaks” abroad through increased imports. This phenomenon is often called the “import leakage effect.” For small, highly open economies with large MPIs, fiscal policy might be less effective than in larger, more closed economies. Monetary policy and interest rates 🔗 Monetary policy in an open economy must consider international capital flows. Lower interest rates might stimulate domestic investment but could also lead to capital outflows and currency depreciation. The combination of these effects makes monetary policy in an open economy more complex but potentially more powerful, as it works through both interest rate and exchange rate channels. Trade policy implications 🔗 Trade policies like tariffs and quotas directly affect the external sector component of aggregate demand. However, their overall impact on equilibrium output depends on how trading partners respond and whether there are retaliatory measures. Generally, protectionist policies can distort resource allocation and may reduce overall economic efficiency, even if they temporarily boost net exports. Global interdependence and equilibrium output 🔗 In today’s globalized world, no economy exists in isolation. The equilibrium output of one country is inherently linked to economic conditions and policies in other countries. Spillover effects 🔗 Economic growth or contraction in major economies creates spillover effects for their trading partners. For example, a recession in the United States typically reduces demand for Chinese exports, affecting China’s equilibrium output. Coordinated policy responses 🔗 Given this interdependence, coordinated policy responses (like those seen during the 2008 global financial crisis) can be more effective than unilateral actions. When major economies coordinate expansionary policies, they can avoid the “beggar-thy-neighbor” dynamics that might otherwise emerge. This global interdependence highlights why international organizations like the IMF, World Bank, and WTO play important roles in fostering economic stability and growth. Modern challenges to open economy equilibrium 🔗 Several contemporary challenges complicate the determination of equilibrium output in open economies: Global value chains 🔗 The rise of global value chains means that products are no longer made entirely in one country. This blurs the distinction between “domestic” and “foreign” goods, complicating the analysis of how trade affects equilibrium output. Digital trade and services 🔗 Traditional models of open economy equilibrium focused primarily on trade in goods. However, digital trade and services now constitute a growing portion of international trade, and they may respond differently to income and price changes than physical goods. Financial integration 🔗 Increased financial integration means that capital flows can rapidly affect exchange rates and interest rates, creating additional channels through which external factors influence domestic equilibrium output. What do you think? How might increasing protectionism and trade tensions globally affect the equilibrium output determination of developing economies that rely heavily on exports? And considering the digital transformation of international trade, how might the traditional models of open economy equilibrium need to evolve to remain relevant in the 21st century? How useful was this post? Click on a star to rate it! Submit Rating Average rating 0 / 5. Vote count: 0 No votes so far! Be the first to rate this post. We are sorry that this post was not useful for you! Let us improve this post! Tell us how we can improve this post? Submit Feedback PDF 📄 ←Previous: Transformations in the Rural Landscape of Meiji Japan: Tenancy, Agriculture, and Resistance Next: Prehistory and Its Significance in Archaeological Anthropology→ Comments 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. Macroeconomics-I 1 Issues and Concepts Why Study Macroeconomics? Certain Concepts Production Possibility Curve Importance of Economic Growth Inflation and Unemployment Business Cycle 2 National Income Accounting Circular Flow of Income National Income and Related Concepts Measurement of Related Aggregates 3 Measuring Economic Performance Methods of Measuring National Income Measures of Aggregates: Saving and Wealth Real and Nominal GDP Limitations of GDP Balance of Payments 4 Classical and Keynesian Systems The Classical Approach Output and Employment in the Classical System Aggregate Supply Function The Keynesian Approach 5 Keynesian Model of Income Determination Equilibrium and Aggregate Demand Consumption Function Relationship between Consumption and Aggregate Demand Formula for Equilibrium Output Concept of Multiplier Investment Multiplier Limitations of Multiplier 6 Fiscal Policy in Keynesian Model The Government Sector Government Spending and the Multiplier Automatic Stabilizers Effect of Change in Government Spending and Tax Rate Government Budget 7 External Sector Types of Flows in an Open Economy Gross Domestic Product (GDP) and Gross National Product (GNP) Balance of Trade Invisibles Current and Capital Accounts Net Exports Function Equilibrium Output in Open Economy 8 Functions of Money Functions of Money Measures of Money Supply Hot Money Credit Creation by Banking System 9 Demand for Money Quantity Theory of Money: Fisher’S Approach Quantity Theory of Money: Cambridge Approach Keynesian Theory of Demand For Money Determination of Equilibrium Interest Rate 10 Monetary Policy Objectives Of Monetary Policy Instruments Of Monetary Policy Monetary Policy In India Quantitative Easing Report a missing Topic/Subject Share the details and we will consider adding that Send Email For Other Courses Hospitality Institute 🌃 The MBA Institute 🎓 Psychology Town 🧠 BA LLB Notes ⚖️ Journalism & Mass Comm. 📰 BA Notes 📚 Philosophy Institute 💭 Sociology Institute 👥 Teachers Institute 🍎 The Tourism Institute 🌍 Nursing Institute 🏥 B.Com Institute 📊 Explore Anthropology 🧬 BBA in Retail 🛍️ Public Administration 🏛️ Political Science 🗳️ Food Safety Institute 🧪 Agriculture Institute 🌱 Gender Studies 👫 Environmental Studies (EVS) 🌎 Event Management (EVM) 🎊 Urban Studies 🌆 CSR Education 🏭 Population & Family Health Studies 👨‍👩‍👧‍👦 Disaster Management 🔥 Distance Education 💻 Social Work 💞 Value Education 🌟 NGO Management 🤲 Tribal Studies 🏹 DELED Institute 📚 Adult Education 📖 Creative Writing ✍️ Library & Information Science 📓 Climate Change 🌎 Sustainability Science ♻️ Diet & Nutrition 🥗 Facility & Services Mgt. 🚧 Copyright © 2025 BA Notes All Rights Reserved Share This Facebook WhatsApp LinkedIn Copy Share on Mastodon Enter your Mastodon instance URL (optional) Share

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.