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15500	https://www.youtube.com/watch?v=edNQn8jdd0s	24 divided by 4 \|\| 24÷4 Mathstoon 23400 subscribers 23 likes Description 9200 views Posted: 20 Apr 2024 Topic: How to divide 24 by 4, that is, 24 divided by 4. Answer: 24 divided by 4 is equal to 6. In other words, 24/4 =6. Subscribe for more videos and stay tuned! We post videos daily on youtube DividedBy, #Mathstoon, #mathtutorial 3 comments Transcript: in this video we will find 24 / 4 we know that this can be written as a fraction 24/ 4 now we will write 24 as 4 6 and we have four in the denominator now we will cancel the common four so it will be equal to 6 therefore 24 / 4 is equal to 6 now we will do this using the long division method we know that 4 goes into 24 by 6 times and we have 24 here and subtract this we will get zero therefore 24 / 4 is 6 and which we got here and this is our final answer thank you for watching please like share and comment on the video also do subscribe the channel
15501	https://kids.wordsmyth.net/we/?rid=35894&level=2	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| online dictionary-thesaurus \| \| \| \| \| \| --- --- \| \| \| \| \| \| --- \| Log In \| \| \| Register \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| home \| \| \| subscription \| \| \| feedback \| \| \| about us \| \| \| blog \| \| \| widget \| \| \| FAQ \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| \| \| Dictionaries \| \| \| Comprehensive \| \| Children's \| \| WILD (Illustrated) \| \| Word Exploration \| \| \| Word Explorer \| \| Browse \| \| Search Filter \| \| Reverse Search \| \| A-Z Word Parts \| \| Puzzle Solvers \| \| \| Anagram Solver \| \| Crossword Solver \| \| Teacher Tools \| \| \| Classes \| \| Students \| \| Lessons \| \| Assignments \| \| Reports \| \| Vocabulary Center \| \| \| Activities \| \| Wordlist Maker \| \| Writing Tool (Beta) \| \| Legacy activities \| \| My Wordsmyth \| \| \| Lookup History \| \| My Wordlists \| \| Legacy activities \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Word Explorer Children's Dictionary Suite \| \| \| --- \| \| \| Elementary dictionary \| \| \| Intermediate dictionary \| Help \| \| \| --- \| \| Help \| Help \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Intermediate Dictionary \| \| More results \| \| \| Show multi-word results \| \| Browse in wordlist \| \| See entries that contain "rout" \| \| \| Display options \| \| \| --- \| \| \| Show syllables \| \| \| Show Lookup History \| \| \| Double-Click Lookup \| \| \| Show Word Parts \| \| \| Show Spanish support \| \| \| Show Chinese support \| \| \| Pronunciation \| \| \| --- \| \| \| Wordsmyth \| \| \| Phonics \| \| \| IPA \| \| \| \| \| \| \| Lookup History \| \| rout1 \| \| \| \| \| --- \| \| rout1 \| \| pronunciation: raUt parts of speech: noun, verb \| \| \| --- \| \| part of speech: \| noun \| \| definition 1: \| a confused retreat of troops after they have been beaten. The battle ended in a rout of the enemy. \| \| definition 2: \| a total defeat. Our team was discouraged after the rout. synonyms: conquest \| \| related words: \| defeat, herd, upset \| \| \| \| \| part of speech: \| verb \| \| inflections: \| routs, routing, routed \| \| definition 1: \| to cause to run away or retreat. The government troops routed the rebels. similar words: chase, dispel, drive, repel, scatter, stampede \| \| definition 2: \| to defeat completely. The visitors routed the home team. synonyms: conquer, subdue similar words: beat, crush, defeat, lick \| \| related words: \| overpower, subdue, trample, upset \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Subscribe for ad-free Wordsmyth and more Learn more \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| Home \| Send Feedback \| Having a problem? \| Suggest a Word \| Privacy Policy \| Privacy Manager \| \| \| \| \| --- \| \| \| \| \| ©2025 Wordsmyth \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| online dictionary-thesaurus \| \| \| \| \| \| --- --- \| \| \| \| \| \| --- \| Log In \| \| \| Register \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| home \| \| \| subscription \| \| \| feedback \| \| \| about us \| \| \| blog \| \| \| widget \| \| \| FAQ \| \| \| \| \| \| \| \| --- --- \| \| \| \| \| \| --- \| Log In \| \| \| Register \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| home \| \| \| subscription \| \| \| feedback \| \| \| about us \| \| \| blog \| \| \| widget \| \| \| FAQ \| \| \| \| \| \| --- \| Log In \| \| \| Register \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| home \| \| \| subscription \| \| \| feedback \| \| \| about us \| \| \| blog \| \| \| widget \| \| \| FAQ \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| \| \| Dictionaries \| \| \| Comprehensive \| \| Children's \| \| WILD (Illustrated) \| \| Word Exploration \| \| \| Word Explorer \| \| Browse \| \| Search Filter \| \| Reverse Search \| \| A-Z Word Parts \| \| Puzzle Solvers \| \| \| Anagram Solver \| \| Crossword Solver \| \| Teacher Tools \| \| \| Classes \| \| Students \| \| Lessons \| \| Assignments \| \| Reports \| \| Vocabulary Center \| \| \| Activities \| \| Wordlist Maker \| \| Writing Tool (Beta) \| \| Legacy activities \| \| My Wordsmyth \| \| \| Lookup History \| \| My Wordlists \| \| Legacy activities \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Word Explorer Children's Dictionary Suite \| \| \| --- \| \| \| Elementary dictionary \| \| \| Intermediate dictionary \| Help \| \| \| --- \| \| Help \| Help \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Intermediate Dictionary \| \| More results \| \| \| Show multi-word results \| \| Browse in wordlist \| \| See entries that contain "rout" \| \| \| Display options \| \| \| --- \| \| \| Show syllables \| \| \| Show Lookup History \| \| \| Double-Click Lookup \| \| \| Show Word Parts \| \| \| Show Spanish support \| \| \| Show Chinese support \| \| \| Pronunciation \| \| \| --- \| \| \| Wordsmyth \| \| \| Phonics \| \| \| IPA \| \| \| \| \| \| \| Lookup History \| \| rout1 \| \| \| \| \| --- \| \| rout1 \| \| pronunciation: raUt parts of speech: noun, verb \| \| \| --- \| \| part of speech: \| noun \| \| definition 1: \| a confused retreat of troops after they have been beaten. The battle ended in a rout of the enemy. \| \| definition 2: \| a total defeat. Our team was discouraged after the rout. synonyms: conquest \| \| related words: \| defeat, herd, upset \| \| \| \| \| part of speech: \| verb \| \| inflections: \| routs, routing, routed \| \| definition 1: \| to cause to run away or retreat. The government troops routed the rebels. similar words: chase, dispel, drive, repel, scatter, stampede \| \| definition 2: \| to defeat completely. The visitors routed the home team. synonyms: conquer, subdue similar words: beat, crush, defeat, lick \| \| related words: \| overpower, subdue, trample, upset \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Subscribe for ad-free Wordsmyth and more Learn more \| \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| \| \| Dictionaries \| \| \| Comprehensive \| \| Children's \| \| WILD (Illustrated) \| \| Word Exploration \| \| \| Word Explorer \| \| Browse \| \| Search Filter \| \| Reverse Search \| \| A-Z Word Parts \| \| Puzzle Solvers \| \| \| Anagram Solver \| \| Crossword Solver \| \| Teacher Tools \| \| \| Classes \| \| Students \| \| Lessons \| \| Assignments \| \| Reports \| \| Vocabulary Center \| \| \| Activities \| \| Wordlist Maker \| \| Writing Tool (Beta) \| \| Legacy activities \| \| My Wordsmyth \| \| \| Lookup History \| \| My Wordlists \| \| Legacy activities \| \| \| \| \| \| \| \| \| \| \| \| Comprehensive \| \| Children's \| \| WILD (Illustrated) \| \| \| \| Word Explorer \| \| Browse \| \| Search Filter \| \| Reverse Search \| \| A-Z Word Parts \| \| \| \| Anagram Solver \| \| Crossword Solver \| \| \| \| Classes \| \| Students \| \| Lessons \| \| Assignments \| \| Reports \| \| \| \| Activities \| \| Wordlist Maker \| \| Writing Tool (Beta) \| \| Legacy activities \| \| \| \| Lookup History \| \| My Wordlists \| \| Legacy activities \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| Word Explorer Children's Dictionary Suite \| \| \| --- \| \| \| Elementary dictionary \| \| \| Intermediate dictionary \| Help \| \| \| --- \| \| Help \| Help \| \| \| \| \| \| \| \| \| \| --- \| \| \| Elementary dictionary \| \| \| Intermediate dictionary \| \| \| \| --- \| \| Help \| Help \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| Intermediate Dictionary \| \| More results \| \| \| Show multi-word results \| \| Browse in wordlist \| \| See entries that contain "rout" \| \| \| Display options \| \| \| --- \| \| \| Show syllables \| \| \| Show Lookup History \| \| \| Double-Click Lookup \| \| \| Show Word Parts \| \| \| Show Spanish support \| \| \| Show Chinese support \| \| \| Pronunciation \| \| \| --- \| \| \| Wordsmyth \| \| \| Phonics \| \| \| IPA \| \| \| \| \| \| \| Lookup History \| \| rout1 \| \| \| \| \| Show multi-word results \| \| Browse in wordlist \| \| See entries that contain "rout" \| \| \| \| --- \| \| \| Show syllables \| \| \| Show Lookup History \| \| \| Double-Click Lookup \| \| \| Show Word Parts \| \| \| Show Spanish support \| \| \| Show Chinese support \| \| \| Pronunciation \| \| \| --- \| \| \| Wordsmyth \| \| \| Phonics \| \| \| IPA \| \| \| \| \| --- \| \| \| Wordsmyth \| \| \| Phonics \| \| \| IPA \| \| \| \| Lookup History \| \| rout1 \| \| \| \| --- \| \| rout1 \| \| rout1 \| \| \| --- \| \| part of speech: \| noun \| \| definition 1: \| a confused retreat of troops after they have been beaten. The battle ended in a rout of the enemy. \| \| definition 2: \| a total defeat. Our team was discouraged after the rout. synonyms: conquest \| \| related words: \| defeat, herd, upset \| \| \| \| \| part of speech: \| verb \| \| inflections: \| routs, routing, routed \| \| definition 1: \| to cause to run away or retreat. The government troops routed the rebels. similar words: chase, dispel, drive, repel, scatter, stampede \| \| definition 2: \| to defeat completely. The visitors routed the home team. synonyms: conquer, subdue similar words: beat, crush, defeat, lick \| \| related words: \| overpower, subdue, trample, upset \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Subscribe for ad-free Wordsmyth and more Learn more \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| Home \| Send Feedback \| Having a problem? \| Suggest a Word \| Privacy Policy \| Privacy Manager \| \| \| \| \| --- \| \| \| \| \| ©2025 Wordsmyth \| \| \| \| \| --- \| \| \| \|
15502	https://www.youtube.com/watch?v=4UxiUxsiLtQ	In/Out Grouping \| LSAT Logic Games LSAT Lab 23300 subscribers 348 likes Description 25624 views Posted: 25 Jun 2021 In/Out Grouping games are a tricky game type designed to measure if-then reasoning. Properly notating conditional relationships and quickly linking them together is key to success in this game type. 0:00 Intro 0:14 Frequency 0:39 Agenda 1:02 Example Game 1:37 Gameboard 3:01 Rules 4:57 Inferences 6:53 Question 1 8:29 Question 2 10:59 Question 3 13:09 Question 4 15:10 Question 5 18:04 Closed vs. Open 19:40 Subgroups 22:00 Frames 26:21 Summary Experience the power of personalized learning when you take a free practice digital LSAT. Get our latest videos when you subscribe to the LSAT Lab YouTube channel. 111 comments Transcript: Intro (techno music) - [Matt] Hi, this is Matt at LSAT Lab and today's lesson is on in out grouping games in the logic game session. In out grouping games are really important game type. Frequency They make up 9% of all logic games. And while it's not the biggest game type out there, it's an important one that sits at that intersection between medium and hard. It's something that's gonna take a little bit of effort to work on. And some people struggle with the lack of concreteness that sometimes you might find in the game board. In today's lesson, we're going to look at the following. Agenda We're gonna start with an example game so you get a chance to see what this game type looks like. And then we're gonna talk about some of the characteristics that are associated with in out grouping games like closed versus open. So whether you know how many players go to each team or not. Whether there are different kinds of players or subgroups. And then what are the opportunities for creating frames to allow you to move faster? So let's start with this example game. Example Game What I'd like you to do is try this one first. And what I mean by try this one first, come up with a good game board, one that would allow you to create hypotheticals so that you can test cases whether things could be true or it must be true or counterexamples whether they don't have to be true in order to help you navigate the answer choices. Then notate the rules next to the game board and make any inferences that fall into place from those rules. Take a few minutes to try this on your own, hit pause, and then when you're ready to check your work against mine, hit play again. All right, welcome back. So the first thing we need to do is create a good game board Gameboard one where we can create hypotheticals to test cases, to check whether things could be true or must be true along the way. So now the game board here for in out grouping games is essentially the same game board we're gonna use for standard grouping games, which is a column chart. In this case with in out grouping, we're only going to have two columns, one for in and one for out. So it says that we're going to select at least one of the following six types of stone. So, that word select is your clue that we're looking at a game of selection or in out grouping is another way of referring to it. We can either do in and out, we can do selected, not selected. It doesn't exactly matter. What's most important is that in in out grouping games, we understand that this is a binary choice. It's not like we have team one and team two. It's that we have team one and not team one. We have onstage or not onstage. The light is on or the light is not on. When we have the logical opposite as our choices, that's a good clue that we're looking at a game of selection or in out grouping game. So we're gonna have a selected versus not selected. And then here's our cast of characters, L, M, N, O, P, and T. We'll write them down next to the game board and that's it for a good game board. We just need to have a place to put our players to create hypotheticals. It's relatively straightforward for in out grouping. The next step is to notate the rules Rules and with I out grouping games, we're looking at a lot of if then rules. So, we should expect to work with if then rules when we're working on in out grouping games. If we're not comfortable with if then rules, maybe we wanna consider whether we do the in out grouping game first or later. Right, we might need a little extra time as we struggle with the rules. So to notate the rules, we're just gonna use the language cues that create if then rules in order to organize them into if then relationships. In the first rule, the word unless is the organizing word. It introduces a necessary condition. It also implies the negation of the sufficient condition. That's a little bit trickier of a rule. This is probably the most complicated rule that we have to deal with here. But in terms of putting into notation, essentially, we're gonna put N on the right side of the arrow and we're gonna negate M and O are selected. What's the opposite of them both being selected? If either one of them is not selected, then it's not the case that they're both selected. So if either M is not selected or O is not selected, then N is not selected according to the first rule. The second rule, T is selected only if O is not selected. Only if is the language cue that tells us what introduces our, in this case, necessary condition. Only if introduces a necessary condition. The term that goes on the right side of the arrow. Okay, so we know that O is not selected, goes on the right, and T is selected, goes on the left for the rule number two. Rule number three is the same, just with L and P. So if L is selected and P is not selected, and only if is organizing that relationship. And finally, in the last rule if is organizing the if then relationship. If is introducing our sufficient condition which tells us to put N on the left side of the arrow and L on the right side of the arrow. So that's it for the rules. Next step, make inferences. Inferences On in out grouping games, inferences generally consists of connecting the if then relationships. So just like in a tree ordering game where we'd take the relative relationships and put them together and to build the tree, in an in out grouping game, we want to as much as possible connect the rules together using the common terms across the different rules. So you might look for, we might just take the first rule and just put it down somewhere. Like if M is not selected or O is not selected, then N is selected. When we're trying to represent this in kind of a bigger visual map, we're gonna go ahead and create spaces for M independently. If either M is not selected or O is not selected, then N is, we'll have spots for M and O separately. If you look at the next rule, T to O, it does connect on the front end of the first rule. So we can go and just tack that on to the front of the first rule. If you look at the next rule, L and P, there's no common term between either L or P in anything that's already together. So let's pass over the third rule for just a second and look at the fourth rule. There, N is the common term. And so if N implies that L is selected, then we can add that to the tree. And now that makes the third rule a little bit easier, if L, then not P. And so we can connect all the rules together to lead to this one kind of masterful system. You can take the contrapositive if you want. Right, so you can say that if P is selected, then L is not selected. And if L is not selected, then N is not selected. And N is not selected, then M is and O is. And if O is selected, then T is not. You can write that out as a contrapositive, if you'd like. Or you can keep it as it is here. You don't even need to connect all the rules together. Some people like to leave the rule separate. But if you can connect the rules together, I think you'll find that it's easier to make the implicit inferences or see the implicit connections across the rules. So now that we've connected the rules together, let's take a look at the questions. Question 1 So we'll let you try these questions one at a time, pause the recording, see if you can find an answer that you like and then when you're ready to check your work, hit play again and compare notes. All right, welcome back. So this is a rules question. It's just asking for an acceptable selection of stones. One that satisfies all of the rules. Probably a good idea just to take the rules and apply them one at a time and knock out wrong answer choices. So if you look at the first rule, M and O are selected, unless N is selected. So if either M is out or O is out, then N is supposed to be in, right. Well, neither of them are out in A. O is out in B, so N needs to be there and it is. M is out in C, so N needs to be there, it is. M and O, neither of them are out in D. O is out in E. So N should be there, but it's not. So that limits to answer choice E. And we're just gonna do this one rule at a time until only one edge choice remains. If we look at the second rule, T is selected only if O is not. So that means that if T is in, and O is out, anything that has both T and O, and that's answer choice C. The third rule tells us that L is selected only if P is not. So any answer choice that has L and P we can get rid of and that's answer choice A. And then the last rule, N implies L. So if we ever see an N, we need to also see an L. We see an N in both B and D. But we don't see an L and D, so we can get rid of it. And that means that B is the right answer. All right, so here's question two. Question 2 Go ahead and give this one a try on your own and when you're ready to check your work, hit play again. All right, welcome back. So this is the second question in the set. It's a global question, meaning it's something we need to be able to answer without new information. And it's asking for a pair of stones of which, at least one of them must be selected. It doesn't have to be the case that at least one of T and L is selected. Does it have to be the case that at least one of M and P is selected? Eventually we're gonna get to a pair where at least one of them has to be selected, and that'd be our answer. So if you look at answer choice A, T and L and we look at how they appear in the connected rules, there's a rule that says, if T is selected, then O is not. And if O is not, then N is selected. And if N is selected, then L is selected. So the relationship between T and L is a rule that goes from positive to positive. All right, in order for the pair to be the right answer, we need this answer choice to move from negative to positive. So a negative to positive rule is going to say that if one is out, then the other one is in. By contrapositive, if the other one is out, then the first is in. A negative to positive rule would require that at least one of them is selected. And so really what we're looking for is a pair where we can identify those stones in the tree, where a relationship is drawn between them that moves from negative to positive. From T to L in answer choice A, that's from positive to positive. And it won't require that at least one of them is selected. So let's get rid of answer choice A. Answer choice B has M and P. Now M and P, that's negative to negative. Again, we're looking from negative to positive. This doesn't quite create the same relationship that we're looking for. So let's get rid of B. Answer choice C says M and L. So if we're gonna connect M and L there, we go from a negative to a positive. So this looks a little bit better. Hold on to answer choice C. Answer choice D connects O and P. So if O is not selected, then P is not selected. That's negative to negative. We're looking for negative to positive. So let's get rid of answer choice D. Answer choice E connects T and P. Now there's positive to negative if we look at the relationship between them. And that says that most one of them is selected, but it doesn't require that either one is selected. So we'll get rid of answer choice E and that leaves us with answer choice C as the right answer. Question 3 All right, here's the next question. Go ahead and try that this one on your own. Hit pause, and when you're ready to check your work, hit play again and we'll look at it together. All right, welcome back. So this is the third question in the set, and this one is a local question, meaning it gives us a new piece of information inside of the questions that apparently has some implications and we wanna find those implications before we really look at the answer choice. So if L is not selected, we can put L out. and we look for what are the implications of that? Well, we can find L in the tree. But unfortunately, it goes from L to not P. And we don't know that L is, we know that L is not in. So we're gonna be able to employ the contrapositive of the tree in order to make inferences. And so if L is not in, and we know that N is not in. And if N is not in by contrapositive, we know that M and O are in. And if O is in, by contrapositive, we know that T is out. And so we're able to make a lot of inferences, but every stone except for P, right. At this point we have M and O are in, L, N and T are out. P can do anything it wants. So, which one of the following must be true? Let's take a look. Answer choice a says that O is not selected. We know that O is selected. So we can get rid of it. It must be false. Answer choice B, N and O are both selected. Well, no, N is not selected. So let's get rid of B. C says M and T are both selected. Well, M is selected but T is not selected. So let's get rid of C. At most, three stones are selected. I think this might be our answer because we've got M and O in, and the possibility of one more, with P. But because L, N, and T are all out, it's gonna be impossible to have more than three. So I think D is gonna be our answer. Let's hold onto it and double check E. E says that at least three stones are selected. Well, no, we've got two. We don't really know that we're gonna have to have P. We could have P, but we don't have to have P. And so we have at least two stones are selected. We could have a third, we have at most three. So answer choice D is the right answer. Question 4 Here's the next question. Give this one a try on your own. Hit pause, and when you're ready to check your work, hit play again. All right, welcome back. So this is another local question. This one is telling us that if M is not selected, we want to know what cannot be true. So if it could be true, we would get rid of it, right? But that means that there's gonna be some implications of M not being selected. We want to follow all of those implications as far as it goes, and then use that information in order to eliminate answer choices. So if M is not selected, what do we know? Well, if we look at the tree, if M is not selected, we can just follow this one forward. So we know that N is selected, which means that P is not selected. What don't we know about yet? T and O. Right, so which one of the following cannot be true? Anything about T and O could be true, except for the relationship between T and O is a positive to negative one. That means that we can't have both of them in. We could have one in, the other one in, or them both out. Right, so which one of the following can not be true? It could be true that T is selected. We can get rid of answer choice A. And it could be true that O is selected. You can get rid of answer choice B. C says that neither T nor O are selected. Well, could they both be out? Sure, they could both be out. They can't both be in, but they could both be out. So that's possible still. Get rid of C. Answer choice D, neither P nor O. Well, that's possible. If O, I mean we can do anything. We already know that P is out, and we can do pretty much anything we want with O. So, yeah, it's possible that neither P nor O are selected. Get rid of D. If you look at answer choice E, both T and P are selected. Well, T and P can't both be selected because the relationship between them is a positive to negative one, right. That's actually true under all circumstances. It cannot be the case that T and P are both selected regardless of what happens with M. So E does have to be true and that's our answer. Question 5 All right, here's the last question for this game. Hit pause and try this one on your own. And when you're ready to check your work, hit play again. All right, come back. So this is a min-max question, which are pretty common on in out grouping games. On these games, it's pretty common for them say, "Okay, well what's the minimum that could be in? What's the maximum that could be in?" If they ask you for the maximum, try putting everybody in and see what happens. What ends up happening is that you end up breaking rules, so then, you have to figure out how to conform to those rules. The rules that break are the rules that are gonna go from positive to negative when you're dealing with the maximum, right? Because the rules that prevent you from putting everybody in are positive to negative rules, right? If we look at which rules are being broken right now after putting everybody in, we know that T and O cannot both be in, right. So we need to figure out how do we get rid of one them. Let's get rid of T, for example. And then if we look at the other rules, we see that we're also still breaking the rule that says if L is selected, then P is not selected. We can't have both L and Ps. We have to get rid of one of them as well. Well, if we get rid of the P, doesn't seem like that's going to interact with any of the other rules, right? We have to get rid of one of T and O. We have to get rid of one of L and P. And if we do that, we can find a hypothetical that satisfies all of the rules, right? T is not in, so we don't have to worry about that. M and O are in, so we don't worry about them. N is in, so we better have L, and we do. L is in, so we better not have P and we don't. Right, so this hypothetical satisfies all the rules. So apparently four is the max, answer choice C. We could look at this from another perspective. So something else that's pretty common would be at the very beginning of the game, when we're kind of taking stock of our inferences, to notice that we have this rule that goes from positive to negative between T and O, and that we can't have them both in or both selected. One of them is gonna have to be out. And so we could put a placeholder on the outside if we know that either T or O is out, we can do a T slash O over there. This still allows us to put them both out, but at least one of them is for sure out. We'll do the same thing with L and P, right? You can do another placeholder for L and P in out side. Now we've got at least two stones that have to be out. Now we can do the same thing when it comes to rules that move from negative to positive as well, right. If we know that negative to positive means that at least one of them are in, right, then let's say, for example, between M and N, we know that at least one of them must be selected. So we can put a placeholder between M and N on the inside. Then we could use these placeholders in order to arrive at the same answer. Essentially, if we know that at least two of them are out, then we can tell that the maximum would be four that are in. So a couple of different ways that we could approach the question like this. But C is the correct answer. So now let's look at some of the features Closed vs. Open that are common to in out grouping games. And let's look at whether games are closed or open. Closed meaning we know exactly how many are in and how many are out. Open, like in the last game where it could change from one question to the next. So let's suppose you have a scenario that says a manager will put on sale exactly four of seven products, right? This is your standard closed scenario. We know that exactly four of the seven products are in, our column chart is simply gonna have four in and three out. Now we can add F, G, H, J, K, L, and O to the game board and we're ready to go. About 42% of in-out grouping games are closed. Meaning we know exactly how many are in and how many are out. So a large percentage of them, but not all of them, not even a majority of them. The majority of them are gonna be open, right. 58% of in out games, we won't know how many are in or how many are out. And that vagueness is one of the aspects of in out grouping games that tends to bother people. It leads to a situation where you don't have a very concrete game board, right. If you have the column chart without any of the slots, it can feel like the game board isn't very solid. So here, we have a manager who'll put on sale at least one of seven products, right. The key difference here is at least one of seven products, not exactly four. So when it's saying that there's gonna be at least a certain number, that's a good sign that we're looking at an open game. We've got our players F, G, H, J, K, L, and O. And this game board is just a little bit less clear. We won't know how many are in and how many are out. But, otherwise the game board will be identical. Subgroups Next, let's look at subgroups. What happens when you have different kinds of players? This is gonna be something that we'll see on about a third of in out grouping games. So something that shows up fairly regularly, often in connection with the game being closed, right. It's really common for us to see open games where there aren't subgroups. And closed games where there are subgroups. So those are natural pairings. So here we have a radio DJ will select exactly four songs to play during her show. Exactly two will be pop songs, F, G, H and J, and exactly two will be country songs, K, L, and M. So we're gonna choose two of each of these different kinds of songs, okay. We have pop songs and country songs. So well we know that we're gonna be selecting, that's telling us in and out. The radio DJ will select exactly four. So here it's closed. Exactly four songs, right? It's closed. So we can put four in and three out. And then for our players, we're just gonna list them separately, right? We've got the pop songs and the country songs. We'll do this just like we do subgroups in ordering games. In this case, they tell us that we're gonna have exactly two of each of the different kinds of subgroups. And so we can maybe keep track of that above, two for the pop and two for the country. That should help us make quicker deductions. So for example, one thing you can do in a situation like this is you can use the fact that the game is closed to your advantage. That closed nature of the game allows there to be fewer solutions. And because there are fewer solutions, we should get more control. We might be able to even do some upfront work. So for example, here, if we're gonna have two country songs and there are only three country songs to choose from, there's a limited number of ways in which we can make that happen. There's actually three ways which you can grab two from three, right? We can either grab K and L, K and M, or L and M, right. And so that might be a way in which we can use the fact that this has got subgroups to our advantage, to create some frames, and maybe get a headstart on the game. I would assume that with the other rules, we might be even able to fill in some of the other positions in the game board. So that brings the discussion Frames to the last topic in this lesson, which is about frames. And frames is something where gives you an opportunity to move more quickly through the questions because you do some work upfront once, and then you leverage that work as you're answering all the questions. And so it allows you to move on quickly because you're essentially not repeating your work. So what are the triggers for creating frames on in out grouping games? Well all but one scenarios definitely count. And so what we just saw wasn't all but one scenario where we were asked to choose two out of the three of the country songs. But two of three is all but one. This would also work for three out of four as well. So here, if you're choosing two out of three, remember KL, KM, LM, right, we might get a rule. We might even get a rule that says, if K is selected then M is selected. If you have two of the players connected in an if then rule within one of the subgroups, that often allows you to make a further inference. So for example, if K implies M, then K and L is not an option as a pair of two. That would satisfy the rules. We either are in a situation where we have K and M or L and M. But either way, the inference is that M is definitely selected. So when you have a rule that goes from positive to positive, it's that right side of the arrow that ends up having to be true when you're choosing all but one. We could try another example. We could look at this on the left side, and let's say we're choosing three out of four from the pop songs, right. If we were choosing three out of four, there's four ways of grabbing three from four. We can grab F, G, and H. We can grab F, G, and J. We can F, H, and J. Or G, H, and J. Four different ways of grabbing three from four. Hopefully, we won't have to use all four of those. Hopefully, there's a rule that allows us to eliminate some of them really, really quick and can help us narrow down the frames to maybe two or potentially three. So let's say we have a rule that says that if G is selected, then J is not. So it's gonna go from positive to negative. In this case, any of those scenarios where G and J are both in, we can get rid of, right. So we can get rid of the second and the fourth options. And that leaves in both of them, F and H consistently there. And so if we're gonna choose all but one, and we have I have a rule that goes from positive to negative relating two of those players within that all but one, then the others are the ones that have to be selected, right? So in this situation, if G and J, we can't have them both we're gonna have to have F and H. And so there's an inference, and we can either run this out either with G or with J. And since you can't have both, if you have G in, that's gonna imply that J's out. If you have J in, that implies that G is out. And that gives us leverage on this game because we're now starting to make, we're starting to see how the game is limited into just a couple of different solutions. And this is pretty common on closed games with subgroups. Another opportunity for creating frames on in out grouping games is based off of rules with opposite triggers. So rules that have either K on the left side and not K on the left side. Basically, the left side of two different rules need to be the logical opposite of each other. If K is selected, then G is selected. And if K is not selected, then L is selected. Because we have two rules that begin with the opposite trigger, we know something no matter what. It's going to be the case that either K is selected or the K is not selected. And that gives us an opportunity for creating frames. We can run at one frame where K is and one where K is not. Follow the inferences that follow along from that. So if K is selected, then G also is selected. Whereas if K is not selected, then L is selected. We can add those to the frames. Hopefully, the other rules would allow us to make further inferences after we got started with this as the fork in the road. And so conditionals that begin with the opposite trigger are another good opportunity for creating forks in in out grouping games. Now, you won't see that very often, but if you do, explore it. So in summary for in out grouping games, Summary the game board that we're gonna use is a standard column chart. We're gonna have one column for the inside, one column for the outside. If we don't know how many players are in and out, we call it open. If we do know how many players are in and out, we call it closed. The kinds of rules we're gonna use for in out grouping games are fixed positions where we know the player is just being assigned in or out. A fixed relationship where we know that, you know, exactly one of these is going to be selected. Or a conditional relationship and that's the one that we're gonna expect to see the most. Most of the rules on in out grouping games are if then relationships. We need to be comfortable using those conditional language cues in order to set up the if then notational form. Finally, for creating frames, we have a couple of strategies, either look for situations where you're choosing all but one of a particular subgroup. Or look for a couple of if then rules that have opposite triggers. Either of those would work. So that's it for today's lesson on in out grouping games. I invite you to check out these other lessons or visit us today at LSATlab.com.
15503	https://www.quora.com/How-do-you-calculate-planet-s-average-density-from-its-mass-and-radius	Something went wrong. Wait a moment and try again. Average Density of Planet... Space and Astronomy Planet Compositions 5 How do you calculate planet’s average density from its mass and radius? Mike Miller Studied Materials Science and Engineering · Author has 10.7K answers and 65.9M answer views · 5y How does one calculate a planet's density? How does one calculate a planet's density? Step 1: Measure the diameter and then volume of the planet. See: Calculating the Diameter of the Earth Step 2: figure out the planet’s mass. If you’re on the planet and can measure it’s surface gravity (g) directly (like with a pendulum test), then you can figure out the mass (M) with the value of surface gravity, the gravitational constant (G), and the planet’s radius (R). Solve for M. Step 3: Density is mass divided by volume. Since you’ve got mass (step 2) and volume (step 1), you can get density. Step 2 (Long Distance Version): If you can’t be on a pla How does one calculate a planet's density? Step 1: Measure the diameter and then volume of the planet. See: Calculating the Diameter of the Earth Step 2: figure out the planet’s mass. If you’re on the planet and can measure it’s surface gravity (g) directly (like with a pendulum test), then you can figure out the mass (M) with the value of surface gravity, the gravitational constant (G), and the planet’s radius (R). Solve for M. Step 3: Density is mass divided by volume. Since you’ve got mass (step 2) and volume (step 1), you can get density. Step 2 (Long Distance Version): If you can’t be on a planet’s surface and measure gravity directly, then you can calculate the planet’s mass by watching its moons. Their orbital radius (r) and orbital period (T) will tell you the mass (M) of the planet. Related questions How do you calculate the average density of Earth if it has a mass of 6.0x²⁴ kg and a radius of6.4x10⁶? Does orbital radius include the radius of a planet? How does one calculate a planet's density? How is the average density of a planet calculated if we can't directly measure it? How can I calculate the mass of any planet? Duff Howell Physics B.S. Applied Physics PhD. 25 years in high tech optical instruments. · Author has 1.7K answers and 2.5M answer views · Jun 27 Big old hint from Physics 101 to pass along to that AI you’re trying to train. Your inputs have units of…… Mass….and … anybody??? Yes, Distance. And ‘density’ has units of what? Mass and a function of distance that might be applied if you know the shape and a critical describing distance for the planet. Off you go. John Stephenson I've stared at the night sky with wonder since I was a kid. · Author has 2.4K answers and 2.8M answer views · 2y Let’s do this in SI units as it makes more sense. Find the radius in metres, if it’s in miles you’ll have to convert it, in km just multiply by 1000 Find the volume in cubic metres using V=43πr3 Divide the mass of the planet in kilograms by this volume, which will return the average density in kilograms per cubic metre. On this scale water has a density of 1,000. Alon Goldring Former Senior Laser Physicist at Lumenis · Author has 1.7K answers and 1.2M answer views · 2y To calculate the average density, you calculate the volume by: V = 4piR^3/3 ( R = radius of the planet) Then the density is given simply by M/V ( M = mass of the planet). Related questions How can we find out the density, mass, and size of a planet or star? Can a planet's average density be estimated using only its mass and radius, without considering its composition? What planet may have a density of 0.69 g cm3? How do scientists calculate the mass of Earth and other planets? If a planet doubles in mass and the density remains the same, how much does the radius increase? Sol Warda Author has 5.8K answers and 2.5M answer views · 1y Originally Answered: Can a planet's average density be estimated using only its mass and radius, without considering its composition? · Yes! What is the volume of a sphere formula? Look it up in your textbook! V ==4/3 Pi R^3, where R==Radius of the planet. Once you figure out the volume of the planet, then: Density ==Mass / Volume. Got it? Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Oldschool72 Studied Associate & Business-to-Business (Graduated 1967) · Author has 3.3K answers and 505.1K answer views · 1y Originally Answered: Can a planet's average density be estimated using only its mass and radius, without considering its composition? · Yes. Assuming the volume can be modeled as a sphere, Density = mass/volume = mass/ (4πr³/3). The density may not be constant, as in a metallic core but that is part of the mass which is known. Painterguy Lincs Lives in Lincolnshire England · Author has 4.7K answers and 973.2K answer views · Jan 17 Related How does one calculate a planet's density? Measure its mass by weighing it on a bathroom scales (assuming gravity in your bathroom is 9.8N/kg… if not, calculate N/G to find the mass), then measure its volume by submerging it in the bath and noting its displacement, then divide the mass by the volume to get the density. (If your bathroom scale isn’t big enough, propel the planet along a flat, frictionless surface using a known force and measure the acceleration, then calculate the mass by m=f/a If your bathtub isn’t big enough, submerge the planet in a fishpond to get the displacement volume. If it floats you can kill two birds with one s Measure its mass by weighing it on a bathroom scales (assuming gravity in your bathroom is 9.8N/kg… if not, calculate N/G to find the mass), then measure its volume by submerging it in the bath and noting its displacement, then divide the mass by the volume to get the density. (If your bathroom scale isn’t big enough, propel the planet along a flat, frictionless surface using a known force and measure the acceleration, then calculate the mass by m=f/a If your bathtub isn’t big enough, submerge the planet in a fishpond to get the displacement volume. If it floats you can kill two birds with one stone… measure the mass by the floating displacement, then push it under to get the volume. If the pond is still not big enough, submerge the planet in a lake to find the displacement volume. 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For the Earth as an example (which you could apply to other planets) the formula is: You have the Mass in your assumption ~= 5.98 x10^24 kg You have the Gravitational constant (it is fixed) = (6.67408 × 10^-11) m^3 kg^-1 s^-2 You need the distance from the centre of the Earth, or in other words its radius. You have the density (mass per unit volume) which in kg/m^3 is ~= 5,515.3kg/m^s. To get the radiu What do you mean by a planet’s gravity? Do you mean the speed at which an object falls due to gravity or a planet’s Gravitational Field Strength? If you mean the field strength, that’s pretty simple. For the Earth as an example (which you could apply to other planets) the formula is: You have the Mass in your assumption ~= 5.98 x10^24 kg You have the Gravitational constant (it is fixed) = (6.67408 × 10^-11) m^3 kg^-1 s^-2 You need the distance from the centre of the Earth, or in other words its radius. You have the density (mass per unit volume) which in kg/m^3 is ~= 5,515.3kg/m^s. To get the radius of the Earth from this, you first need the Volume, and from that you can get the Radius using spherical formulas (Large bodies generally coalesce into the most stable 3D shape, a sphere) as an approximation. Volume of a sphere = (4/3)πr^3 Volume is Mass/Density, which is (5.98x10^24) / (5,515.3). So you can equate these two expressions and re-arrange to find that r3=((5.98x1024)∗3)/(5,515.3∗4∗π) Or that r= 6.38x106metres(I pulled that off the internet as I was too lazy to find my scientific calculator). Then you have everything you need to find the Gravitational field strength, make sure you understand what I have done here and you should get the field strength as: The formula is very similar for acceleration due to gravity, but with a subtle difference. I’ll leave you to see that from the formula: Which is quite similarly (at sea level) about 9.8 m/s. But this number will change as you travel further from the centre of mass. It makes sense for the numbers to be the same at the surface level of large bodies, and whereas the field strength doesn’t change since its numbers are fixed, the speed due to gravity can. Sources (for further help in calculations): Gravitational Field Formula Acceleration Due to Gravity Formula Hopefully I wasn’t too sleep when I wrote this, but feel free to check my math and just make sure I haven’t been an idiot at any point! Deleted Account Lives in New York, NY · Author has 2K answers and 9.1M answer views · 8y Related How could the astronomers measure the weight of planets? The exact answer to your question is in the last paragraph. However, I need to point out that you can't really determine the weight of anything without specifying where it's being weighed. That's because weight is defined as the force applied to a body by gravity (specifically its mass times the acceleration from gravity, or Weight = F = ma, where a = g, the acceleration from gravity). This varies depending on the mass of the dominant body, i.e. the planet where something is being weighed. For example, since the moon is less massive than the Earth, it's g value is less, so I'd weigh less on th The exact answer to your question is in the last paragraph. However, I need to point out that you can't really determine the weight of anything without specifying where it's being weighed. That's because weight is defined as the force applied to a body by gravity (specifically its mass times the acceleration from gravity, or Weight = F = ma, where a = g, the acceleration from gravity). This varies depending on the mass of the dominant body, i.e. the planet where something is being weighed. For example, since the moon is less massive than the Earth, it's g value is less, so I'd weigh less on the moon than on Earth (about 84% lighter). Similarly, I'd be heavier on our most massive planet, Jupiter, where g is 250% higher. And a feather would weigh several billion tons on the “surface” of a black hole. Here is a table of gravitational force, relative to the Earth (it should be a lower case ‘g' in the Gravity column because capital ‘G' is the Universal Gravitational Constant, which is something else) Using this table, you can see that my weight on Venus is only 90% of what it is on Earth, but I would weigh 28 times more on the Sun (assuming I don't evaporate from the heat!). In other words, your question is incomplete. For example, are you asking how much the planets weigh under the gravity of the Earth? Or under the Sun? How about planets outside our solar system, which revolves around a different star? What you may have meant to ask is how to determine the mass of a planet. That said, this excellent Scientific American article should answer your question. Basically, you’d measure the gravitational effect it has based on known masses. For example, if you want to know the weight of Jupiter due to the Sun's gravity, you'd first have to know the mass of the Sun. It all starts with Newton’s Law of Universal Gravitation. First, we'd calculate the mass of the Earth. Once we know the mass of the Earth, we can calculate the mass of the Sun based on our orbit around it. For example, the Sun's gravitational pull must be equal to the centripetal force from the Earth's revolution. If not, the Earth would either fly off into space or spiral into the sun. Since we know the Sun's pull, we can calculate its mass. Using the Sun's mass for reference, we can then repeat the process for any planet, based on its distance from the Sun and its size. Refer to the above Scientific American article for all the details and calculations. Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Anonymous 5y Related How are the masses of the planets calculated? Newton's Law of Universal Gravitation. This law states that every object in the universe attracts every other object with a force directed along the line of center-to-center distance between these two objects. The force is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects. This law also maybe written as: Fg = (GMm)/r2 Where Fg is the gravitational force, G is the universal gravitational constant (G = 6.67x10-11 N m2/kg2), M and m are the masses of the objects and r is the center-to-center distance between the two objects. N Newton's Law of Universal Gravitation. This law states that every object in the universe attracts every other object with a force directed along the line of center-to-center distance between these two objects. The force is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects. This law also maybe written as: Fg = (GMm)/r2 Where Fg is the gravitational force, G is the universal gravitational constant (G = 6.67x10-11 N m2/kg2), M and m are the masses of the objects and r is the center-to-center distance between the two objects. Newton also noted that the gravitational force of attraction of the sun pulling on an orbiting body must be equal to the centripetal force. (Fc = Fg) Knowing that Fc =mac, and that centripetal acceleration is equal to (4?2r)/T2, we can equate Fc = Fg. Fc = Fg (m4?2r)/T2] = (GMm)/r2 Rearranged, we get r3/T2 = (GM)/(4?2) T is period of the satellite around central object, r is the center-to-center distance from central object. If carefully noticed, r3/T2 is part of Kepler's Third Law, and is a constant that depends only on the mass of the central object. In this way, Newton was able to provide a theoretical basis for Kepler's laws: orbiting satellites are held in place by gravity. Mass of a Planet Collectively, these laws may help us determine a number of things. One such example would be calculating the mass of a planet. Clearly it is unrealistic to visit a planet to get its measurements, so we use the measurements we can take from the Earth. Question: Uranus' moon Oberon has an orbital period of 13.5 days. Its mean distance from Uranus is 582,600 km. What is the mass of Uranus? (Note: G=6.67x 10-11) Solution: First find the orbital radius of the moon in meters. (Center to center distance) 582,600 km = 582,600,000 m Convert the time into seconds. 13.5d x 24h/d x 60m/h x 60s/m = 1,166,400 s Use the formula: r3/T2 = (GM)/(4?2) Rearrange to get mass by itself. M=(r34?2)/(GT2) Plug in the values. M=[(582,600,0003)4?2]/[(6.67x 10-11)(1,166,4002)] M=8.6 x 1025 kg Hint: Don't be afraid to use parenthesis. Richard Stratford B.Sc. in Astronomy (college major), University of St Andrews (Graduated 1973) · Author has 3.1K answers and 578.9K answer views · 3y Related How do you calculate the average density of Earth if it has a mass of 6.0x²⁴ kg and a radius of6.4x10⁶? First, you have to remember that the Earth’s radius is 6.4×10^6 metres. From this you can calculate the Earth’s volume (4/3× pi ×(6.4×10^6)³) cubic metres (m³). To help you, the volume is 1.1×10^21 m³. You then divide the Earth’s mass in kg by its volume in m³ to obtain the density in kg/m³. Allen Ries Math Major University of Alberta · Author has 25.1K answers and 9.7M answer views · 2y Related How can we find the density of an object if we know its radius and mass? density = mass/volume Since you mention that the object has a known radius we can assume the object is a sphere. The volume of a sphere is 4/3πr³. The rest is high school mathematics. Lean Consultant (2016–present) · Updated 3y Related An extrasolar planet has a mass of 4.6 Earth-masses. Assume that the higher mass compresses the planet’s interior so that its average density is 40% larger than the Earth’s average density. What is the predicted radius of the planet? It’s really fairly simple to calculate. Here is how: Let’s assume that the earth’s radius is roughly 4,000 miles. So, following the formula for calculating the volume a sphere, with a radius of 4,000 miles, earth’s volume in cubic miles is 4/3 X 3.1415 (which is Pi) X 4000 X 4000 X 4000. Let’s call this BIG number “V-earth.” Now, the extrasolar planet has a mass that is 4.6 times that of our little blue planet. To put it in another way, the mass of the new planet, therefore, is 4.6 X V-earth. But, due to its higher gravity, this planet has a density 1.4 times that of the earth. We can, therefore, It’s really fairly simple to calculate. Here is how: Let’s assume that the earth’s radius is roughly 4,000 miles. So, following the formula for calculating the volume a sphere, with a radius of 4,000 miles, earth’s volume in cubic miles is 4/3 X 3.1415 (which is Pi) X 4000 X 4000 X 4000. Let’s call this BIG number “V-earth.” Now, the extrasolar planet has a mass that is 4.6 times that of our little blue planet. To put it in another way, the mass of the new planet, therefore, is 4.6 X V-earth. But, due to its higher gravity, this planet has a density 1.4 times that of the earth. We can, therefore, safely assume that it’s volume would be (4.6 X V-earth)/1.4, which is another big number. To keep our thinking simple, let’s call it V-extra. You know that we can calculate this, right? All you need is a little calculator, or for old-timers like me, just paper and pencil. Now, if we assume the new planet’s radius to be R, it’s volume can be calculated as (4/3) X 3.1415 X R X R X R, with R being unknown. But because we KNOW the actual value of V-extra (calculated following the logic above), we can write: (4/3) X 3.1415 X R X R X R = V-extra OR, R X R X R = (V-extra)/((4/3) X 3.1415) You can now easily calculate the radius of the new planet, or R. I was crazy enough to do that. It came out to be 5,674 miles, give or take a few miles…It’s fun at times to go back to high school math….🤣 Related questions How do you calculate the average density of Earth if it has a mass of 6.0x²⁴ kg and a radius of6.4x10⁶? Does orbital radius include the radius of a planet? How does one calculate a planet's density? How is the average density of a planet calculated if we can't directly measure it? How can I calculate the mass of any planet? How can we find out the density, mass, and size of a planet or star? Can a planet's average density be estimated using only its mass and radius, without considering its composition? What planet may have a density of 0.69 g cm3? How do scientists calculate the mass of Earth and other planets? If a planet doubles in mass and the density remains the same, how much does the radius increase? Is it possible to calculate a planet's gravity based on its mass and density? Why do you think a planet’s density decreases as its distance from the sun increases? What relationships can you determine about a planet’s density? How do you calculate orbital velocity from mass and radius? Which planet has the highest average density? The mass of a planet is 100times mass of earth and its radius is 4 times the radius of earth. What is the escape velocity of the planet? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15504	https://math.stackexchange.com/questions/4090797/maximum-product-of-distances-of-a-point-and-vertices-of-equilateral-triangle	complex analysis - maximum product of distances of a point and vertices of equilateral triangle - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more maximum product of distances of a point and vertices of equilateral triangle Ask Question Asked 4 years, 5 months ago Modified2 years, 11 months ago Viewed 424 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. If A B C A B C is an equilateral triangle. P P is a variable point inside the triangle or on its sides. We need to find the maximum product of A P⋅B P⋅C P A P⋅B P⋅C P. This question is in a Complex Analysis homework. I don't know how can I use the complex analysis (maybe, maximum modulus theorem ) to find the maximum product. Updated work Let z 1 z 1, z 2 z 2, z 3 z 3 and z z correspond to the vertices A A, B B, C C, and variable point p p, respectively, then the product function is given by f(z)=\|z−z 1\|\|z−z 2\|\|z−z 3\|f(z)=\|z−z 1\|\|z−z 2\|\|z−z 3\|, clearly, (z−z 1)(z−z 2)(z−z 3)(z−z 1)(z−z 2)(z−z 3) is nonconstant holomorphic function in C C and so in T¯¯¯¯T¯, where T T represents the region inside the equilateral triangle. Then by the Maximum modulus Theorem, f f attains the maximum on the boundary ∂T∂T. To this point, is there a way to find z z explicitly that maximize f f? complex-analysis geometry maximum-principle Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Apr 6, 2021 at 3:24 math_for_evermath_for_ever asked Apr 5, 2021 at 22:08 math_for_evermath_for_ever 571 3 3 silver badges 7 7 bronze badges 4 Given one vertex and the side (base?), the third vertex may be calculated. For the purpose of finding P assume side = 2. (0,0) and (2,0) is the base. Find the third vertex and for a point (x,y) in the triangle write the equation of multiplication. Find maximum by means derivatives.Moti –Moti 2021-04-06 02:03:11 +00:00 Commented Apr 6, 2021 at 2:03 @Moti, this using Calculus, is there a way to apply the complex analysis theorems/contents?math_for_ever –math_for_ever 2021-04-06 02:35:59 +00:00 Commented Apr 6, 2021 at 2:35 Define the vertices as COMPLEX values, but you still need derivatives.Moti –Moti 2021-04-06 02:39:06 +00:00 Commented Apr 6, 2021 at 2:39 @Moti One can solve the issue in a geometric way without needing derivatives.Jean Marie –Jean Marie 2021-04-06 13:46:59 +00:00 Commented Apr 6, 2021 at 13:46 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Up to a similitude s(z)=a z+b s(z)=a z+b, we can reduce the issue to equilateral triangle T T with vertices: z 1=1,z 2=ω,z 3=ω 2=ω¯¯¯with ω=e 2 i π/3 z 1=1,z 2=ω,z 3=ω 2=ω¯with ω=e 2 i π/3 Therefore: AP.BP.CP is the absolute value of : (z−1)(z−ω)(z−ω 2)=z 3−1(z−1)(z−ω)(z−ω 2)=z 3−1 The maximum value of this holomorphic function, call it f(z)f(z), is reached for a z z belonging to the boundary (Maximum Modulus Theorem, as you have noticed). Let us parametrize the line segment from 1 1 to ω ω in the following way: z=(1−t)+t ω,0≤t≤1(1)(1)z=(1−t)+t ω,0≤t≤1 The image by f f of this line segment is the red "drop-like" cubic curve (see equation in the remark below). The two other sides z′=(1−t)ω+t ω 2,z′′=(1−t)ω 2+t z′=(1−t)ω+t ω 2,z″=(1−t)ω 2+t can be written under the form: z′=ω z,z′′=ω 2 z(1)(1)z′=ω z,z″=ω 2 z Therefore, because ω 3=1 ω 3=1 and (ω 2)3=1(ω 2)3=1, the image by f f of these two other sides obtained by taking to the third power expressions (1) gives the same curve! This is the image f(∂T)f(∂T). The image by function f f of T T itself is the inside of the drop. The sister curve described as the image of ∂T∂T by g(z):=z 3 g(z):=z 3 is magenta curve. Considering the red curve, f(z)=z 3−1 f(z)=z 3−1 reaches its maximal value for z 3 m−1=I′z m 3−1=I′, which corresponds to: z 3 m=−1 8(point I')(2)(2)z m 3=−1 8(point I') Cubic equation (2) has three solutions. As the argument of I I is π+k 2 π π+k 2 π, the arguments of solutions z m z m must be π/3+k 2 π/3 π/3+k 2 π/3. As the maximum must occur on the boundary, taking k=0,1,2 k=0,1,2 gives points J 0,J 1,J 2 J 0,J 1,J 2 resp., i.e., the midpoints of the sides of triangle T T. Fig. 1: The three maximizing solutions J 0,J 1,J 2 J 0,J 1,J 2. Remark: The parametric equations of the "drop" obtained by taking the real and imaginary parts of ((1−t)+t ω)3((1−t)+t ω)3 are (x=1−9 2 t(1−t),y=3 3√2 t(1−t)(1−2 t))(x=1−9 2 t(1−t),y=3 3 2 t(1−t)(1−2 t)) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 2, 2022 at 6:42 answered Apr 6, 2021 at 13:44 Jean MarieJean Marie 90.3k 7 7 gold badges 59 59 silver badges 132 132 bronze badges 1 I just realized that the "droplike curve" belongs to the family of (cubic) Bezier curves.Jean Marie –Jean Marie 2022-10-02 06:44:17 +00:00 Commented Oct 2, 2022 at 6:44 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 complex-analysis geometry maximum-principle 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 9Condition for 3 complex numbers to represent an equilateral triangle 14Equilateral Triangle from three complex points 14Let z 1 z 1, z 2 z 2 and z 3 z 3 be complex vertices of an equilateral triangle. 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15505	https://math.stackexchange.com/questions/2288293/is-there-a-formula-to-extract-the-real-part-of-a-complex-number	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 Is there a formula to extract the real part of a complex number? [closed] Ask Question Asked Modified 9 days ago Viewed 20k times 7 $\begingroup$ I was pondering the topic of complex numbers and how we just look at the them and extract the real and imaginary parts because it's easy given that we write it like 4+2i. I was wondering if there was some mathematical formula, in which we could put any complex number, and it would output the real (or imaginary) part of that number. EDIT: as fleablood pointed out in the comment of this post Which maybe seems circular as $ \overline z==Re(z)iIm(z)$. There's also $Re(z)=cos(arg(z))$ where $arg(z)$ is the argument angle of the complex number. Note: all of these take it as given that a complex number determined/defined by two factors and find $Re(z)$ is always a matter of taking the factor or manipulating a factor that was given. What I'm asking is, whether there's an equation or set of operations in which you could take an imaginary number and put it through some that equation (not defined by $Re(z)$ or $Im(z)$) and have it return either the real or imaginary part. Sorry for any confusion and/or ignorance on my part. complex-numbers Share edited May 20, 2017 at 1:07 Nick the coderNick the coder asked May 19, 2017 at 20:15 Nick the coderNick the coder 17911 gold badge11 silver badge44 bronze badges $\endgroup$ 3 2 $\begingroup$ Sure $Re(z)$ is the real part of complex number $z.$ A bit of a cop-out, isn't it? $\endgroup$ Doug M – Doug M 2017-05-19 20:19:18 +00:00 Commented May 19, 2017 at 20:19 $\begingroup$ Yes, it's fz=(a+bi) = a. You just see it and pull it out. That is a formula. We call it Re(z). $\endgroup$ fleablood – fleablood 2017-05-19 20:19:36 +00:00 Commented May 19, 2017 at 20:19 $\begingroup$ D'oh. I forgot the obvious sws1710 answer. Which maybe seems circular as $\overline z = Re(z) - i Im(z)$. There's also $Re(z) = cos (arg (z))$ where arg(z) is the argument angle of the complex number. Note: all of these take it as given that a complex number determined/defined by two factors and find Re(z) is always a matter of taking the factor or manipulating a factor that was given. $\endgroup$ fleablood – fleablood 2017-05-19 20:23:55 +00:00 Commented May 19, 2017 at 20:23 Add a comment \| 3 Answers 3 Reset to default 22 $\begingroup$ Let $z\in\Bbb C$. Then the real part is $\dfrac{z+\bar{z}}{2}$ and the imaginary part is $\dfrac{z-\bar{z}}{2i}$. Share edited May 19, 2017 at 22:13 answered May 19, 2017 at 20:19 szw1710szw1710 8,3701818 silver badges2727 bronze badges $\endgroup$ 0 Add a comment \| 3 $\begingroup$ If you have access to the argument $\arg(z)$ and absolute value $\|z\|$ then you can do a Taylor expansion of exponential function on the matrix $$Z_{polar} = \left[\begin{array}{rr} \log(\|z\|)&\arg(z)\ -\arg(z)&\log(\|z\|) \end{array}\right]$$ You can basically just stuff this $Z$ into the calculus formula: $$\exp(x) = \sum_{k=0}^N \frac{{x}^k}{k!}$$ For some suitable large $N$. This series converges really fast so you probably don't need very many terms. The answer for the real part will end up all along the diagonal according to the matrix representation for complex numbers: $$z = a+bi \Rightarrow M_z = \left[\begin{array}{rr}a&b\-b&a\end{array}\right]$$ Actually this answer can maybe be used as a motivation for the definition of the complex logarithm. Share edited Aug 23, 2017 at 12:01 answered May 20, 2017 at 1:05 mathreadlermathreadler 26.6k99 gold badges4040 silver badges107107 bronze badges $\endgroup$ Add a comment \| 3 $\begingroup$ If you allow taking the absolute value of z, you can use the formula: $\displaystyle{\displaylines{Re(z)=\frac{z+\frac{\|z\|^{2}}{z}}{2}}}$ Share answered Sep 18 at 22:41 Jacob HungerfordJacob Hungerford 6588 bronze badges $\endgroup$ 1 $\begingroup$ A clever way to take the conjugate of any non-zero complex number. $\endgroup$ David K – David K 2025-09-19 00:40:52 +00:00 Commented Sep 19 at 0:40 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions complex-numbers 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 Prove a complex number is real Real part of this complex number 0 Determining the condition for the real part and imaginary parts of complex function using its modulus Does imaginary part of complex number represents the meaning of down payment or stealing in real life?? 0 Show that the real part of complex number is $-1$ 4 separating real and imaginary part of complex number 2 How to find real part of $x^i$ without the "arg(x)" 2 Does the imaginary part in complex number needs to be the square root of one? Hot Network Questions Proving a certain Cantor cube is a complete metric space (by definition) - proof verification Survival analysis - is a cure model a good fit for my problem? 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15506	https://aspoonfuloflearning.com/pattern-activities-for-kindergarten/	Skip to content All Blog Posts, Math, Patterns Hands-On Pattern Activities for Kindergarten Are you looking for some hands-on activities to help your students practice patterns? This is the post for you! I’m sharing a variety of engaging pattern activities for kindergarten that you can use to enhance your math instruction with hands-on practice. Kindergarten Pattern Activities This wide variety of activities will help you introduce and practice patterns with your whole class. You’ll also find some engaging activities that you can use for math centers or small group practice. 1. Pattern Anchor Chart Activities There are many fun ways that you can incorporate anchor charts and pocket charts into your pattern practice! Your students will love to help build a pattern anchor chart as you introduce each new pattern type. You can hand out the different shape cards and students can watch for the right time to place their card in the pattern. In addition to anchor charts, students love pocket chart activities! For example, students can learn about how patterns can be represented by letters with a pocket chart activity. As you add different shape cards to the chart, students can add the corresponding letter underneath each card. This helps students see and understand the difference between the pattern labels. You can also help your students practice completing patterns with a pocket chart activity! Simply place shape cards in the pocket chart and have students decide what cards would complete the pattern. Students love to solve the mystery of the hidden shapes! 2. Pattern Mini-Books Students always enjoy being able to create their own mini books on any topic, but patterns are especially fun! These printable pattern books help students practice a variety of different pattern types. Students can practice coloring and labeling the pattern type. They can also practice continuing patterns by drawing the missing shapes. Finally, students can cut and paste small shape cards to create a pattern. 3. Pattern Books Once students have had a chance to practice individual pattern types, they can put together their own large pattern book. In this printable activity, students will focus on one pattern type at a time. They will color and identify the pattern, then write the colors that they used for each pattern. This is a fun way to practice spelling color words! 4. All About Patterns Another printable book to help students summarize the different patterns that they have learned is an “All About Patterns” mini book. To complete this book, students will cut and paste a variety of small shape cards to create different combinations of patterns. 5. Pattern Snake This cute pattern snake gives students a chance to get creative with patterns. Students will choose a variety of colored squares to create this snake craft. They can arrange and glue the squares in a pattern of their choice or you can provide them with a pattern type. These colorful snakes always look great on a bulletin board! 6. What’s the Hidden Pattern? This fun partner game is a great way to help students practice completing patterns! One student will create a pattern of their choice by placing small manipulatives on a pattern strip. Then they will use small paper cups to cover the final three objects in the pattern. Their partner (who has been covering their eyes) will then guess what is underneath each cup based on their knowledge of patterns. This is a fun way to help students identify and complete patterns! This is a great activity for math centers or morning work time! Some of your students might even enjoy pairing up to play this game as a fast finisher option. Since it’s an open-ended activity, students can play again and again until it’s time to move on to the next part of your daily routine! 7. Pattern Bracelets Once students have had plenty of practice with patterns, they can use their knowledge to create their own pattern bracelets! Students can use pony beads and pipe cleaners to create a bracelet using the pattern of their choice. The pipe cleaners add a good amount of structure to make it easier for little hands to string the beads. Students are always so excited to compare the patterns that they chose for their bracelets! When it’s time to go home for the day, your students will be eager to explain patterns to their families! 8. Pattern Musical Chairs Last, but not least, is a pattern activity that will get students moving: Pattern musical chairs! Each student starts at their seat and creates a pattern of their choice on a pattern strip. They leave the last three spaces on their pattern strip empty. Once everyone is ready, play some music and have students to walk around the room for about 10 to 15 seconds. When the music stops, students will sit at the seat closest to them, even if it’s not their seat. They will look at the pattern created at that seat and then finish the pattern. When everyone has completed the pattern in front of them, they will clear the pattern strips and create a brand new pattern. Just like the first round, they will leave the last three spaces empty. Then it’s time to start the music and play again! Printable Pattern Activities for Kindergarten All of the pattern activities mentioned above can be found in one easy-to-download resource! You’ll find all of the printable materials and instructions that you need to incorporate these hands-on pattern activities into your math practice. These activities are a great way to supplement your math instruction or provide support for students who could use additional practice with patterns. If you’d like to take a closer look at everything included in this resource, you can find it in the A Spoonful of Learning shop or on TPT. Save These Kindergarten Pattern Activities Be sure to save this post so you can come back to it whenever you need some fun pattern activities for kindergarten! Just add the pin below to your favorite teaching board on Pinterest. You’ll be able to quickly find these activities when you’re looking for a fun way to practice patterns in your classroom. PrevPrevious6 Reasons to Use Centers in Kindergarten Next6 Ways to Use Printable Kindergarten Hats for LearningNext You might also like... 10 Engaging Kindergarten Homework Ideas 5 Reasons to Use Digital Learning Activities in Kindergarten Top 5 Resources from A Spoonful of Learning Teacher Summer Prep Ideas for a Smooth School Year Facebook Instagram Pinterest-p Helpful Links Home About Blog Contact Us Literacy Math Seasonal Centers Alphabet Freebies
15507	https://en.wikipedia.org/wiki/Pseudoscalar	Jump to content Pseudoscalar Català Deutsch Français Italiano 日本語 ਪੰਜਾਬੀ Polski Русский Slovenščina Suomi Українська 中文 Edit links From Wikipedia, the free encyclopedia Scalar quantity, changing sign in mirrored coordinates \| \| \| --- \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Pseudoscalar" – news · newspapers · books · scholar · JSTOR (January 2021) (Learn how and when to remove this message) \| In linear algebra, a pseudoscalar is a quantity that behaves like a scalar, except that it changes sign under a parity inversion while a true scalar does not. A pseudoscalar, when multiplied by an ordinary vector, becomes a pseudovector (or axial vector); a similar construction creates the pseudotensor. A pseudoscalar also results from any scalar product between a pseudovector and an ordinary vector. The prototypical example of a pseudoscalar is the scalar triple product, which can be written as the scalar product between one of the vectors in the triple product and the cross product between the two other vectors, where the latter is a pseudovector. In physics [edit] In physics, a pseudoscalar denotes a physical quantity analogous to a scalar. Both are physical quantities which assume a single value which is invariant under proper rotations. However, under the parity transformation, pseudoscalars flip their signs while scalars do not. As reflections through a plane are the combination of a rotation with the parity transformation, pseudoscalars also change signs under reflections. Motivation [edit] One of the most powerful ideas in physics is that physical laws do not change when one changes the coordinate system used to describe these laws. That a pseudoscalar reverses its sign when the coordinate axes are inverted suggests that it is not the best object to describe a physical quantity. In 3D-space, quantities described by a pseudovector are antisymmetric tensors of order 2, which are invariant under inversion. The pseudovector may be a simpler representation of that quantity, but suffers from the change of sign under inversion. Similarly, in 3D-space, the Hodge dual of a scalar is equal to a constant times the 3-dimensional Levi-Civita pseudotensor (or "permutation" pseudotensor); whereas the Hodge dual of a pseudoscalar is an antisymmetric (pure) tensor of order three. The Levi-Civita pseudotensor is a completely antisymmetric pseudotensor of order 3. Since the dual of the pseudoscalar is the product of two "pseudo-quantities", the resulting tensor is a true tensor, and does not change sign upon an inversion of axes. The situation is similar to the situation for pseudovectors and antisymmetric tensors of order 2. The dual of a pseudovector is an antisymmetric tensor of order 2 (and vice versa). The tensor is an invariant physical quantity under a coordinate inversion, while the pseudovector is not invariant. The situation can be extended to any dimension. Generally in an n-dimensional space the Hodge dual of an order r tensor will be an antisymmetric pseudotensor of order (n − r) and vice versa. In particular, in the four-dimensional spacetime of special relativity, a pseudoscalar is the dual of a fourth-order tensor and is proportional to the four-dimensional Levi-Civita pseudotensor. Examples [edit] The stream function for a two-dimensional, incompressible fluid flow . Magnetic charge is a pseudoscalar as it is mathematically defined, regardless of whether it exists physically. Magnetic flux is the result of a dot product between a vector (the surface normal) and pseudovector (the magnetic field).[citation needed] Helicity is the projection (dot product) of a spin pseudovector onto the direction of momentum (a true vector). Pseudoscalar particles, i.e. particles with spin 0 and odd parity, that is, a particle with no intrinsic spin with wave function that changes sign under parity inversion. Examples are pseudoscalar mesons. In geometric algebra [edit] See also: Pseudoscalar (Clifford algebra) A pseudoscalar in a geometric algebra is a highest-grade element of the algebra. For example, in two dimensions there are two orthogonal basis vectors, , and the associated highest-grade basis element is So a pseudoscalar is a multiple of . The element squares to −1 and commutes with all even elements – behaving therefore like the imaginary scalar in the complex numbers. It is these scalar-like properties which give rise to its name. In this setting, a pseudoscalar changes sign under a parity inversion, since if is a change of basis representing an orthogonal transformation, then where the sign depends on the determinant of the transformation. Pseudoscalars in geometric algebra thus correspond to the pseudoscalars in physics. References [edit] ^ Zee, Anthony (2010). "II. Dirac and the Spinor II.1 The Dirac Equation § Parity". Quantum field theory in a nutshell (2nd ed.). Princeton University Press. p. 98. ISBN 978-0-691-14034-6. ^ Weinberg, Steven (1995). "5.5 Causal Dirac Fields §5.5.57". The quantum theory of fields. Vol. 1: Foundations. Cambridge University Press. p. 228. ISBN 9780521550017. Retrieved from " Categories: Geometric algebra Clifford algebras Linear algebra Scalars Hidden categories: CS1: long volume value Articles with short description Short description matches Wikidata Use American English from March 2019 All Wikipedia articles written in American English Articles needing additional references from January 2021 All articles needing additional references All articles with unsourced statements Articles with unsourced statements from April 2025
15508	https://pmc.ncbi.nlm.nih.gov/articles/PMC7584559/	Reliability of humeral head measurements performed using two- and three-dimensional computed tomography in patients with shoulder instability - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Int Orthop . 2020 Jul 26;44(10):2049–2056. doi: 10.1007/s00264-020-04710-x Search in PMC Search in PubMed View in NLM Catalog Add to search Reliability of humeral head measurements performed using two- and three-dimensional computed tomography in patients with shoulder instability Jakub Stefaniak Jakub Stefaniak 1 Sport Trauma and Biomechanics Unit, Department of Traumatology, Orthopaedics and Hand Surgery, Poznań University of Medical Sciences, Poznań, Poland 2 Rehasport Clinic, Poznań, Poland Find articles by Jakub Stefaniak 1,2,✉, A M Kubicka A M Kubicka 3 Institute of Zoology, Poznań University of Life Sciences, Poznań, Poland Find articles by A M Kubicka 3, A Wawrzyniak A Wawrzyniak 2 Rehasport Clinic, Poznań, Poland Find articles by A Wawrzyniak 2, L Romanowski L Romanowski 1 Sport Trauma and Biomechanics Unit, Department of Traumatology, Orthopaedics and Hand Surgery, Poznań University of Medical Sciences, Poznań, Poland Find articles by L Romanowski 1, P Lubiatowski P Lubiatowski 1 Sport Trauma and Biomechanics Unit, Department of Traumatology, Orthopaedics and Hand Surgery, Poznań University of Medical Sciences, Poznań, Poland 2 Rehasport Clinic, Poznań, Poland Find articles by P Lubiatowski 1,2 Author information Article notes Copyright and License information 1 Sport Trauma and Biomechanics Unit, Department of Traumatology, Orthopaedics and Hand Surgery, Poznań University of Medical Sciences, Poznań, Poland 2 Rehasport Clinic, Poznań, Poland 3 Institute of Zoology, Poznań University of Life Sciences, Poznań, Poland ✉ Corresponding author. Received 2019 Sep 16; Accepted 2020 Jul 6; Issue date 2020 Oct. © The Author(s) 2020 Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit PMC Copyright notice PMCID: PMC7584559 PMID: 32712787 Abstract Purpose The aim of the study was to compare two measurement methods of humeral head defects in patients with shoulder instability. Intra- and inter-observer reliability of humeral head parameters were performed with the use of 2D and 3D computed tomography. Methods The study group was composed of one hundred humeral heads measured with the use of preoperative 2D and 3D computed tomography by three independent observers (two experienced and one inexperienced). All observers repeated measurements after 1 week. The intra-class correlation coefficient (ICC) and the minimal detectable change with 95% confidence (MDC 95%) were used for statistical analysis of diagnostic agreement. Results For 3D inter-observer reliability, ICC values were “excellent” for all parameters and MDC 95% values were “excellent” or “reasonable.” All intra-observer ICC and MDC 95% values for 3D were “excellent” for experienced and inexperienced observers. For 2D-CT, ICC values were usually “good” or “moderate” with MDC 95% values higher than 10 or 30%. Conclusions Three-dimensional CT measurements are more reliable than 2D for humeral head and Hill-Sachs lesion assessment. This study showed that 2D measurements, even performed by experienced observers (orthopaedic surgeons), are burdened with errors. The 3D reconstruction decreased the risk of error by eliminating inaccuracy in setting the plane of the measurements. Keywords: Shoulder instability, Measurement reliability, Bone defects, 2D-CT, 3D-CT Introduction Shoulder instability is a common condition affecting 25 in every 100.000 people a year. It affects mainly young people, especially men (3:1) . In people under 20 years of age, the risk of recurrent instability after the first dislocation can be up to 90% . During anterior shoulder dislocation, the humeral head is displaced in front of the glenoid and its posterior surface is wedged into the anterior edge of the glenoid. The resulting bone impression, called Hill-Sachs defect, is the common diagnosis in patients with recurrent shoulder instability. The presence of a Hill-Sachs defect may predispose to a conflict between the humeral head and glenoid, and consequently to the dislocation of the shoulder joint . Diagnosis of the defects of the anterior glenoid rim and the Hill-Sachs defect is important in treatment and facilitates the selection of an appropriate treatment method [4, 5]. Recently, significant attention has been paid to the issue of coexistence of humeral bone defects with glenoid defects. Not only the presence of a humeral head defect but also its morphology, location, and its interplay with glenoid bone loss might matter from the biomechanical point of view [6, 7]. Currently, the golden standard in the treatment of patients with shoulder instability is arthroscopic Bankart repair, which in patients with normal morphology of the glenoid and humeral head proves to be highly effective and with a low risk of recurrence of instability . However, in the case of glenohumeral bone defects, the effectiveness of the soft tissue repair method drops significantly. Both glenoid and humeral head defects have been considered as the most important risk factors that should guide us in our selection of the most appropriate technique to stabilize the shoulder in case of instability [8–10]. Hence, correct tools for bone evaluation are needed. Instability severity index score (ISIS) developed by Balg and Boileau uses a quite simple way of X-ray evaluation for definition of the defects. However, Bouliane at al. found there to be very low inter-rater accuracy in the approach [11, 12]. According to Hirshmann et al. , conventional X-ray is characterized by poor reliability, two-dimensional computed tomography (2D-CT) by medium reliability, and three-dimensional computed tomography (3D-CT) by high reliability of assessment. We have been able to show in our previous study that 3D glenoid reconstruction is more reliable for glenoid bone loss assessment than 2D . Therefore, we have focused on the measurement of humeral head defects in shoulder instability. The aim The aim of this study was to assess the reliability of the most common methods to measure humeral head and its defects in patients with recurrent shoulder instability using 2D-CT and 3D-CT. We have hypothesized that 3D-CT will be more reliable than 2D-CT in assessing humeral head deficiency. In order to test the hypothesis, we have performed intra-observer and inter-observer tests by experienced and inexperienced evaluators. This study is a continuation of previous approach to evaluate the glenoid defects . Material and methods One hundred consecutive CT scans performed on 100 patients (mean age 35.5; SD 15.5; min 17; max 69) diagnosed with traumatic anterior shoulder instability were obtained from our radiology department. The scans were collected by independent orthopaedic surgeons who did not participate in the assessment of method reliability. All the patients underwent a physical examination in order to identify other shoulder pathology like rotator cuff tears, deformations, fractures, or osteoarthritis. In the next stage, a basic radiology assessment showed that 63 of the 100 CT scans displayed signs of shoulder instability (63 glenoid bone loss, of which 47 Hill-Sachs defects). Each of the CT scans was subjected to appropriate computer processing depending on the type of measurement method used: 2D-CT with multiplanar reconstruction and 3D-CT reconstruction. For both techniques, several of the most frequently measured indices of humeral head and Hill-Sachs lesion were chosen for further assessment (Table 1): circle area of humeral head, Hill-Sachs length, Hill-Sachs depth, humeral head length, humeral head height, anatomical neck width. Table 1. Measurements of humeral head \| Name of measurement \| Description \| References \| --- \| Circle area of humeral head \| A circle is fitted with its dimensions to the edge of the joint surface \| Saito et al. Cho et al. \| \| Hill-Sachs length \| The length of Hill-Sachs erosion \| Kodali et al. \| \| Hill-Sachs depth \| The length measured between the deepest point of Hill-Sachs and the border of circle area of humeral head measurement \| Kodali et al. \| \| Humeral head length \| Measurement performed from the articular surface of the humeral head to the greater tubercle, perpendicular to the long axis of the humerus \| Pearl et al. \| \| Humeral head height \| Measurement performed from the articular surface of the humeral head perpendicular to the line of the anatomical neck of the humeral head \| Pearl et al. [18, 19] \| \| Anatomical neck width \| Measurement performed along the line of the anatomical neck of the humeral head on the sagittal plane \| Pearl et al. \| Open in a new tab All related measurements were performed by three independent observers. Two observers were orthopaedic surgeons specializing in shoulder surgery. The third was an observer with no experience in medical orthopaedic treatment (not a physician). Each of them performed measurements twice with a seven day interval without prior knowledge of the results of the first measurement and the findings of the second investigator. 2D measurement method 2D-CT method was based on an analysis of two-dimensional computed tomography with multiplanar reconstruction, using the OsiriX MD 64-bit software (v.8.5). In the first stage of the assessment, the shoulder CT scans were reconstructed in the 3D Curved-MPR module. Then, this image was set in three planes: frontal, sagittal, and transverse (Fig.1) and measurements were performed (Table 1, Figs.2 and 3). Fig. 1. Open in a new tab Initial 2D-CT multiplanar reconstruction for humeral head measurements. The sagittal (c) and frontal plane (b) axes run along the long axis of the humerus; the third axis on the transverse plane (a) marks the long axis of the humeral head (Osirix MD) Fig. 2. Open in a new tab Measurements of humeral head on the transverse plane: circle area of humeral head, Hill-Sachs length, Hill-Sachs depth Fig. 3. Open in a new tab Measurements of humeral head on the sagittal plane: humeral head length, anatomical neck width, humeral head height (Osirix MD) 3D measurement method 3D-CT was based on CT analysis in three-dimensional reconstruction. In the first stage, all scans were reconstructed in three-dimensional space using the 3D Slicer software (3D Slicer ver 4.4). The program allowed conversion of a DICOM file into a Mesh file, which could then be further evaluated using the GOM Inspect (GOM, ver 8) software (Figs. 4 and 5). Fig. 4. Open in a new tab 3D-CT reconstruction of humeral head in GOM Inspect (V8) software Fig. 5. Open in a new tab Measurements of humeral head: humeral head length, anatomical neck width, humeral head height, circle area of humeral head, Hill-Sachs length (GOM Inspect (V8) software) 3D reconstruction was conducted by one observer to avoid any errors associated with the 3D reconstruction itself. Thirty randomly chosen tomograms were reconstructed with an interval of seven days. One of the CT scans in pair was marked as “model” and converted into the CAD format, while the second tomogram, called “comparative,” was converted into the Mesh format. Both CT scans as a pair were compared with each other using the GOM Inspect (V8) program. The reliability of the 3D reconstruction was positively assessed, as the average differences within pairs did not exceed 0.15 mm. Next, all measurements were performed on 3D-CT with the use of GOM Inspect program (Table 1). Statement of human and animal rights This article does not contain any studies involving human participants and animals performed by any of the authors. Statistical analysis In our study, we relied on two models of reliability testing: intra- and inter-observer reliability in which we choose three independent researchers with different levels of experience. In the process of developing of our study, we relied on existing research that concerned reliability assessment in measurement methods. Moreover, our study is the continuation of previous study, regarding the assessment of reliability of glenoid bone defects measurements on 2D and 3D CT . However, in this study, we decided to add one more observer (orthopaedic surgeon) to present reliability of measurements performed by observers with the same level of expertise. Statistical analysis was performed using Microsoft Office Excel (Microsoft ver. 16.23) and SPSS software (IBM ver. 22.0.0.1) and supported by professional statistician. Calculation of sample size for reliability tests with intra-class correlation coefficient (ICC) was performed to check whether number of patients included in the analysis allows reliable statistical analysis . In the intra-observer reliability, with the use of ICC, the sample size should not contain less than N = 13 and N drop = 15 in case of 10% dropout (expected ICC 0.92; two repetitions; the lower acceptable ICC was 0.7 and a significance level for a one-tailed test was a = 0.05). For the inter-observer reliability, the sample size should not contain less than N = 38 with N drop = 43 in case of 10% dropout (expected reliability ICC = 0.92; precision 0.05 with confidence level 95%). Intra-observer and inter-observer reliability were calculated for 100 CT scans for humeral head measurements (circle area of humeral head, humeral head length, humeral head height, anatomical neck width) and for 47 CT scans containing visible Hill-Sachs lesion (Hill-Sachs length, Hill-Sachs depth). All measurements were repeated after seven days by each observer. Reliability was calculated by means of ICC in which values can range from 0 to 1, where “0” means total non-compliance, and “1” absolute compliance of the measurement. For inter- and intra-observer reliability, ICC (2,k) (two-way random effects, absolute agreement, multiple raters/measurements) model was calculated with a 95% confidence interval (95% CI) . Compliance in the ICC range is ranked as follows: “excellent,” > 0.9; “good,” 0.75 < ICC <0.9; “moderate,” 0.5 < ICC < 0.7; and “poor,” ICC < 0.5 [21, 22]. The minimal detectable change (MDC) defined as the minimal amount of change that is required to distinguish a true performance change from a change due to variability in performance or measurement error. MDC with 95% confidence (MDC 95%) was calculated as percentages of measurement mean and showed real change and repeatability of the test. MDC 95% values lower than 30% were assessed as “reasonable” and lower than 10% as “excellent” . Results Inter-observer reliability In 2D CT method, ICC values were “excellent” for parameters circle area of humeral head and Hill-Sachs depth; “good” for Hill-Sachs length, humeral head length, and humeral head height; and “moderate” for anatomical neck width. In 3D CT method, ICC values for all parameters were “excellent.” For 3D measurements, the MDC 95% values were “excellent” for circle area of humeral head, Hill-Sachs length, Hill-Sachs-depth, humeral head length, and humeral head height (2.76–2.89) and “reasonable” for anatomical neck width (14.00). For 2D measurements, MDC 95%values were excellent for circle area of humeral head and humeral head height (5.29–9.76) and “reasonable” for Hill-Sachs depth, humeral head length, and anatomical neck width (23.94–24.73). For Hill-Sachs length measurement, MDC 95% value was higher than 30% (74.99). All measurements are presented in Table 2. Table 2. ICC values for inter-observer 2D-CT and 3D-CT measurements and statistical significance. ICC, inter-class correlation coefficient; 95% CI, 95% confidence interval \| \| N \| 2D \| 3D \| --- --- \| \| \| \| ICC \| 95%CI \| MDC% \| ICC \| 95%CI \| MDC% \| \| Circle area of humeral head \| 100 \| 0.965 \| 0.983–0.983 \| 5,29 \| 0,988 \| 0.912–0.995 \| 2.76 \| \| Hill-Sachs length \| 47 \| 0.825 \| 0.595–0.991 \| 74,99 \| 0,984 \| 0.949–0.993 \| 2.83 \| \| Hill-Sachs depth \| 47 \| 0.968 \| 0.949–0.980 \| 23,94 \| 0,990 \| 0.967–0.995 \| 2.81 \| \| Humeral head length \| 100 \| 0.786 \| 0.701–0.846 \| 12,69 \| 0,986 \| 0.963–0.993 \| 2.83 \| \| Humeral head height \| 100 \| 0.888 \| 0.844–0.920 \| 9,76 \| 0,999 \| 0.998–0.999 \| 2.89 \| \| Anatomical neck width \| 100 \| 0.735 \| 0.511–0.841 \| 24,73 \| 0,991 \| 0.985–0.994 \| 14.00 \| Open in a new tab Intra-observer reliability In 2D-CT method, ICC values for 1st experienced observer were “excellent” for Hill-Sachs length and Hill-Sachs depth, “good” for circle area of humeral head and humeral head height, and “moderate” for humeral head length and anatomical neck width. All ICC values for the first experienced observer in 3D method were “excellent.” MDC 95% values were excellent for all 3D measurements (2.78–9.46) and reasonable for all 2D parameters (11.16–22.47). For the second experienced observer in 2D-CT evaluation, ICC values were “excellent” for anatomical neck width, Hill-Sachs depth, “good” for circle area of humeral head and humeral head height, and “moderate” for Hill-Sachs length and humeral head length. All ICC values for the second experienced observer in 3D CT method were “excellent.” MDC 95% values were excellent for all 3D measurements (2.79–8.66). For 2D measurements, MDC 95% values were “excellent” for anatomical neck width and “reasonable” for circle area of humeral head, Hill-Sachs depth, humeral head height, and humeral head length (8.37–16.77). For Hill-Sachs length, MDC 95% value was higher than 30% (44.34%). For the inexperienced observer, in 2D CT method, ICC values were “good” for circle area of humeral head, Hill-Sachs depth, and humeral head height and “moderate” for Hill-Sachs length, humeral head length, and anatomical neck width. All ICC values for the in-experienced observer in 3D-CT method were “excellent.” All values of MDC 95% were “excellent” for 3D measurements (3.44–9.18). For 2D measurements, MDC 95% values were “reasonable” for circle area of humeral head, humeral head height, humeral head length, and anatomical neck width (11.87–25.21). No 2D parameters were “excellent.” Two parameters, Hill-Sachs length and Hill-Sachs depth, had MDC 95% values higher than 30–44.35 and 63.37, respectively. All results are presented in Table 3. Table 3. ICC values for 1st and 2nd experienced and in-experienced intra-observer measurements. ICC, interclass correlation coefficient; 95% CI, 95% confidence interval \| \| N \| Experienced observer 1 \| Experienced observer 2 \| In-experienced observer \| --- --- \| \| 2D \| 3D \| 2D \| 3D \| 2D \| 3D \| \| \| ICC \| 95%CI \| MDC% \| ICC \| 95%CI \| MDC% \| ICC \| 95%CI \| MDC% \| ICC \| 95%CI \| MDC% \| ICC \| 95%CI \| MDC% \| ICC \| 95%CI \| MDC% \| \| Circle area of humeral head \| 100 \| 0.867 \| 0.815–0.905 \| 11.16 \| 0.987 \| 0.982–0.991 \| 2.79 \| 0.877 \| 0.828–0.911 \| 11.17 \| 0.987 \| 0.978–0.992 \| 2.79 \| 0.876 \| 0.827–0.911 \| 11.87 \| 0.981 \| 0.973–0.986 \| 3.44 \| \| Hill-Sachs length \| 47 \| 0.978 \| 0.964–0.986 \| 11.60 \| 0.999 \| 0.998–0.999 \| 2.89 \| 0.720 \| 0.545–0.828 \| 44.34 \| 0.999 \| 0.866–0,996 \| 2.90 \| 0.724 \| 0.552–0.831 \| 44.35 \| 0.997 \| 0.995–0.998 \| 4.41 \| \| Hill-Sachs depth \| 47 \| 0.992 \| 0.987–0.995 \| 12.56 \| 0.993 \| 0.989–0.996 \| 9.46 \| 0.992 \| 0.987–0.995 \| 12.59 \| 0.996 \| 0.992–0.998 \| 8.66 \| 0.771 \| 0.628–0.859 \| 63.37 \| 0.994 \| 0.989–0.996 \| 9.18 \| \| Humeral head length \| 100 \| 0.595 \| 0.417–0.715 \| 22.47 \| 0.985 \| 0.979–0.989 \| 2.78 \| 0.750 \| 0.651–0.820 \| 16.77 \| 0.979 \| 0.968–0.986 \| 3.21 \| 0.741 \| 0.639–0.814 \| 18.19 \| 0.975 \| 0.965–0.982 \| 3.45 \| \| Humeral head height \| 100 \| 0.894 \| 0.852–0.924 \| 11.18 \| 0.990 \| 0.986–0.993 \| 2.79 \| 0.811 \| 0.736–0.864 \| 16.77 \| 0.988 \| 0.981–0.992 \| 3.08 \| 0.867 \| 0.814–0.904 \| 14.00 \| 0.976 \| 0.967–0.983 \| 4.34 \| \| Anatomical neck width \| 100 \| 0.592 \| 0.422–0.710 \| 25.17 \| 0.986 \| 0.981–0.990 \| 2.78 \| 0.925 \| 0.896–0.946 \| 8.37 \| 0.981 \| 0.971–0.987 \| 3.21 \| 0.641 \| 0.489–0.746 \| 25.21 \| 0.978 \| 0.969–0.984 \| 3.47 \| Open in a new tab Discussion Based on the results of the study, we have confirmed the hypothesis that computed tomography with 3D reconstruction is more reliable than 2D-CT for evaluation of humeral head parameters and bone defects. Furthermore, we have also shown that a 3D-CT evaluation seems to be resistant to bias resulting from the level of the researcher’s experience. In all evaluations, ICC values were “excellent” for all 3D-CT measurements. MDC 95% values for 3D measurements were “excellent” for almost all parameters (except inter-observer anatomical neck width measurement, where the MDC 95% value was “reasonable” (14.00)). For comparison, 2D measurements had usually good or moderate ICC values and “reasonable” or above 30% threshold values of MDC 95%. Bone defects on the lateral surface of the humeral head were first described in 1855 by Malgaigna , but only in 1940 did Hill and Sachs completely described and published the morphology of these defects . The incidence of the Hill-Sachs defect increases with the number of shoulder joint dislocations. After the first episode of dislocation, Hill-Sachs presence is found in about 65% of cases, and in patients with recurrent instability in almost 93% of cases [26, 27]. In our series, the incidence of humeral head lesions was 47%. The exact assessment depends on the use of an appropriate diagnostic method. The presence of Hill-Sachs bone loss is important in the case of a risk of a conflict between the humeral bone defect and the anterior glenoid rim. Therefore, an accurate assessment of the morphology of the defect (length, width, depth, and location), which is essential due to its impact on the choice of treatment method, depends on the quality of the examination methodology . The assessment of the bone defects of the anterior glenoid rim and humeral head is usually based on a two-dimensional analysis of transverse CT scans. Currently available software for computed tomography analysis provides a number of useful tools, with the help of which we can perform the necessary measurements such as a measurement of the length of a straight line joining two points, the surface area of the selected point, or volume of space. However, the analysis of a two-dimensional image of an essentially three-dimensional object leads to the risk of making a mistake resulting from image imperfections or measurement errors on the part of the person performing the measurement. Some models of two-dimensional image processing, available in commercial programs, gave the opportunity to reconstruct a stack of individual images into one three-dimensional model, the projection of which can be set in three planes (transverse, frontal, and sagittal). However, the analysis of such projections (application of measurements) is still carried out only on one plane, which will not eliminate the basic errors of the method [29, 30]. Referring to this type of “hybrid” image projection as a three-dimensional reconstruction is therefore not fully correct. The real three-dimensional method assumes the reconstruction of a virtual three-dimensional model of the tested object and gives the possibility to perform such measurements in three planes. When applying measurements within the glenoid or the humeral head, the model can be freely rotated to identify and apply the correct measurement point. This reduces the risk of error arising from faulty setting of the initial measurement projection, as in the two-dimensional method . In our previous study, we analyzed the reliability of the 2D and 3D measurement method of anterior glenoid bone loss assessment in patients with anterior shoulder instability . We have proved that ICC values for 3D-CT reconstruction were significantly more reliable for most measurements than the 2D method. Just as in this study, we have proven that the 3D method allows for more accurate measurement by researchers with different levels of experience. Similar to the measurements of the glenoid defect, different measuring methods of the Hill-Sachs bone loss were described in the literature . Kodali et al. positively assessed the reliability of the Hill-Sachs measurement by two-dimensional tomography, measuring the width and depth of the defect in three planes (sagittal, frontal, and transverse) . In contrast, the method of three-dimensional tomography was used by Cho et al. assessing the width and depth of Hill-Sachs defects and their position relative to the articular surface of the humeral head . Ho et al. assessed the reliability of 3D-CT measurements of nine anatomically shaped bone models of Hill-Sachs lesions. There was strong agreement between all raters for all measured parameters (length, width, depth) . One of the most important findings in our study was the experience of evaluation in interpretation of CT images matters if is based on 2D images. The spatial view and 3D reconstruction seem to provide more relatable tools independent of the experience of the surgeon. This aspect of measurement methods has not been, to our best knowledge, studied in shoulder imagine evaluations up until now (the exception being our previous publication on glenoid defects ). Kaup et al. evaluated the impact of radiologists’ experience in diagnostic accuracy of osteoporotic vertebral compression fractures in CT and MRI imaging . In another field of imaging, radiologists’ experience was also addressed in the assessment of salivary gland tumours with the use of CT and MRI . In both studies, higher experience resulted in greater reliability. Traditional X-rays have also been used for the evaluation of humeral head defects. They have been part of the commonly used ISIS. This score assists surgeons in identifying the risk factors for recurrence of shoulder instability following shoulder stabilization treatment. In the case of an absence of risks, arthroscopic Bankart repair has a high potential for effective treatment. Bone defects are the major criteria and misinterpretation may lead to underscoring and hence incorrect surgical planning. Burkhart et al. shows that in 67% of patients with an inverted-pear glenoid have recurrent shoulder instability after soft tissue repair and a 100% recurrence in patients with Hill-Sachs . Tauber et al. found bone defects in 57% out of 41 patients re-operated on for recurrence of instability . Finally, Boileau et al. identified risk factors for recurrence instability—attritional glenoid defect (> 25% bone loss) and Hill-Sachs with stretched anterior capsule or laxity . The inexperience of the surgeon and the case of unclear image together with low value of instruments could be some of the reasons for such weak assessments. Traditional X-ray allows us to diagnose the presence of a defect only in about 7% of cases after the first dislocation episode, in comparison, computed tomography or magnetic resonance tomography images are much more accurate and allow us to determine the presence of a defect in more than 90% of cases . Chalmers et al. report that linear measurements resulted in most aggressive recommendations of treatment . Stillwater et al. assessed that there are no significant differences between measurements performed on 3D-CT and 3D-MR postprocessed images . On the other hand, there are some studies which undermine the accuracy of 3D-CT measurements in comparison to measurements performed with the use of arthroscopy . One of the limitations of the study is that we have just focused on humeral head defects. Recently, as studied by Di Giacomo et al. and Yamamoto et al. , the importance of HSL the position (not only the size) and bipolar lesions have been found to play an important role in so called engagement. The identification of both seems to be an important factor in deciding on the choice of optimal operating technique to stabilize the shoulder. This study is a continuation of our work on glenoid evaluation. An evaluation of the interplay of bipolar lesions would exceed the scope of one research paper and is proposed as a matter for a further study. Another weakness identified in current diagnosis methods is the complexity of 3D reconstruction measurements. 3D methods of measurement with the currently available software are relatively advanced and difficult to use accurately. As a result, it may be troublesome and time consuming in everyday clinical practice. An automated process could improve the practical use applicability of CT-based image reconstruction. Such attempts have already been implanted in surgical planning for arthroplasty. Good examples of this are patients-specific instruments (PSI) software used in hip, knee, or shoulder replacement (OrthoView software etc.). To conclude, 3D-CT measurements are more reliable than 2D for humeral head and Hill-Sachs lesion assessment. This study showed that 2D measurements, even performed by experienced observers (orthopedic surgeons) are burdened with errors. The 3D reconstruction decreased the risk of error due to inaccuracy in setting the plane of the measurements and might be precise and easy to use for evaluators inexperienced in computed tomography assessment. Compliance with ethical standards Conflict of interest The authors declare that they have no conflict of interest. Footnotes Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. References 1.Zacchilli MA, Owens BD. 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[DOI] [PubMed] [Google Scholar] Articles from International Orthopaedics are provided here courtesy of Springer ACTIONS View on publisher site PDF (1.6 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Material and methods Results Discussion Compliance with ethical standards 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
15509	https://ocw.mit.edu/courses/16-50-introduction-to-propulsion-systems-spring-2012/6242f54646f5274f1b7d809400e6fe00_MIT16_50S12_lec25.pdf	Now we can write 16.50 Subjects: Velocity triangles; Compressor performance maps In the last lecture we discussed the basic mechanisms of energy exchange in compressors and drew some simple velocity triangles to show how we go from the stationary coordinate system to one in the moving blades. In more detail, the velocity triangle is: v1 = w1 tanβ1 ' v2 = w2 tanβ2 = ωr - w2 tanβ2 ' where β2 is the flow angle in the rotating coordinate system of the rotor blades. ' The advantage of the latter form for v2 is that it is β2 which is nearly set by the ' shape of the blades, i.e. β2 is nearly constant. The difference between the flow angle and the angle of the chord line at the trailing edge of the blades is called the deviation, and is usually a small number of degrees. Taking advantage of this small deviation, we can write the Euler equation as Tt 2 (! "1)MT 2 & w2 tan $2 # w1 tan $1 ( = 1 + 1 + ! "1 2 1" " Tt1 2 M1 ' %r %r ) where M1 is the flow Mach number upstream and MT is the tangential Mach number of the blade motion. Since the two flow angles expressed this way are nearly constant, the temperature ratio becomes a function primarily of two variables, the tangential Mach number MT ' and the ratio of the axial flow velocity to the blade speed. For the usual case of β2 > 0, β1 > 0 we get a functional dependence something like: 1 Lecture 25 Tt2 - 1 Tt1 MT 2 w " !r Actually, we usually plot this in terms of the pressure ratio rather than the temperature ratio, and include superimposed lines of efficiency, so that the map looks like: Here the compressor efficiency is defined as: # $1 # # $1 Tt1( pt 2 ) $ Tt1 ("Tt )Ideal pt1 ( p t 2/ p t1 ) # $1 !c = = = ("Tt )Actual Tt 2 $ Tt1 Tt 2 / Tt1 $1 2 From Kerrebrock, Jack L. Aircraft Engines and Gas Turbines. 2nd edition. MIT Press, 1992. © Massachusetts Institute of Technology. Used with permission. Since ηc<1, the total pressure rise for a given total temperature rise is less than it could be: % % ! c = [1+ " c (# c $1)] % $1 < # c % $1 ' By decreasing β1 and β2 , or even making them negative, we can increase Tt2/Tt1. What limits this increase? The answer is best given in terms of a Diffusion Factor, which describes the tendencies for the boundary layer to separate under the influence of the pressure rise in the blade passage. The critical region is the suction surface of the blade. This region feels a pressure rise due to the decrease of V’ from V’1 to V’2, (the prime is a reminder that these velocities in the relative frame) and also due to the acceleration, followed by deceleration on the suction surface. The Diffusion Factor, defined as v2 ! v1 V2 " D = 1! + ; # $ c V 1 " 2#V 1 " s is a crude, but effective way to account for these two flow deceleration components. Here c is the chord of the blades and s is their spacing (in the peripheral direction). A value of about 0.5 for D is the upper limit for good efficiency. It is conventional and useful to represent the loss in the blading in terms of a Loss Factor, defined as pt 1 " pt 2 !1 = pt 1 " p1 3 From Kerrebrock, Jack L. Aircraft Engines and Gas Turbines. 2nd edition. MIT Press, 1992. © Massachusetts Institute of Technology. Used with permission. We then find that we can correlate the losses in the form ! cos#2 " (cos#2 " )2 2$ cos#1 " .06 .04 .02 0 .1 .2 .3 .4 .5 .6 .7 Diffusion Factor, D Compressor limitations. Rotating Stall. Surge. As the incidence angles increase in the compressor a point is reached when the flow begins to separate from the blades, with serious consequences. Large incidence angles occur when the ratio (axial velocity)/(blade speed) is low, i.e., for a fixed rotational velocity, at reduced flow, and hence at increased pressure ratio. All compressors have a fairly well defined “stall line” that runs diagonally from low flow and low ! c , to high values of both. Operation must be restricted to the region below and to the right of this line. The operational line for the compressor runs roughly parallel to this stall line. As we saw in Lecture 20, The reason for the compressor to be restricted to his operation line have to do with flow continuity between it and the turbine and nozzle. It is a fortunate fact that the line thus defined tends to correspond to a constant blade incidence, which, when chosen properly, is close to that which optimizes blade performance, and so the operating line is more or less the peak compressor efficiency line. The phenomena that occur at and beyond stall are complex, dynamical and highly nonlinear. A detailed understanding of these effects has only emerged in the last two decades, and remedial measures based on this understanding are still under evaluation. Much of the pioneering work in this field has been done at MIT (see E. Greitzer, Engineering for Power, 98, 2, (1976) and J. of Fluids Engineering, 103 (1981), Paduano et al., ASME Paper 91-GT-87, 1981). A more complete exposition of the principles is given in Kerrebrock’s book, (Aircraft 4 Image by Manuel Martinez-Sanchez. Adapted from Fig. 5.12 in Kerrebrock, Jack L. Aircraft Engines and Gas Turbines. 2nd edition. MIT Press, 1992. Engines and Gas Turbines, (MIT Press, 1992, Sec. 5.7, pp. 3254-266), a short summary of which is presented below. In most axial compressors, incipient separation first leads to the Rotating Stall phenomenon: sections of the stalling rotor then operate in deep stall, with almost zero flow, while the rest carries normal or high flow per unit frontal area. These regions move backwards in the rotating frame at ~0.4 – 0.6 ωr, so that, when observed from rest they rotate forward, but only at a fraction of the rotor speed. The reason for the bimodal flow distribution is that adjacent streamtubes become unstable with respect to flow exchange: if the rotor as a whole is near stall conditions, and one streamtube loses some flow and diverts it to its neighbors, the streamtube with less flow goes into stall and loses even more flow, while the neighbors gain flow and remain stable. The reason the “stall cells” move backwards relative to the rotor is an elaboration of the same argument: if one flow passage stalls, the swirling incoming flow is re-routed such that the passage ahead of the one stalled sees a more axial flow, while the one behind sees a larger incidence angle. The stall moves to this trailing passage, and the stalled passage clears. Rotating stall, since it moves rapidly about, tends to average out and, aside from high-frequency excitation, may not be dynamically significant. On the other hand, the net compressor performance drops strongly, and in addition, it is not easy to reverse once started, except by stopping and re-starting the engine. Detecting rotating stall and instituting the appropriate control reaction is very important. If the engine controls simply detect a loss of pumping performance (pressure loss) they may react by increasing fuel flow, which combined with the reduced airflow, may lead to overheating and burnout. Under some conditions having to do mainly with the ratio of flow inertia to flow passage capacitance, the complete “pumping system” (compressor plus choked turbine nozzles) can enter a global oscillation, called Surge. Unlike rotating stall, surge involves deep oscillations or even reversals of the whole flow through the compressor, and can be mechanically destructive (certainly quite noticeable, in the form of loud, repeated bangs, accompanied by flame ejection from both ends of the engine). When stall is reached, the engine may go into either Rotating Stall or Surge. The detailed mechanisms that determine which of the phenomena will prevail are encapsulated in Greitzer’s “B parameter” !r Vp B = 2a Vc where a is the speed of sound in the burner, Vp is the “plenum” volume (mainly the burner volume) and Vc is the volume in the compressor flow passages. In a single-stage compressor, values of B below ~ 0.8 lead to rotating stall, while higher values lead to surge. For multistage (N) compressors, Bcrit. is lower by somewhere between N and N. One favorable aspect of surge (as opposed to rotating stall) is that it can usually be cleared by simply reducing fuel flow (or, in a test stand, opening the downstream throttle). 5 MIT OpenCourseWare 16.50 Introduction to Propulsion Systems Spring 2012 For information about citing these materials or our Terms of Use, visit:
15510	https://www.youtube.com/watch?v=fseuBvhJE_o	OpenStax AP Physics Chapter 16.5: Kinetic and Potential Energy for Pendulum OpenStax 9810 subscribers 321 likes Description 51131 views Posted: 1 Jul 2016 This instructional video covers Energy and the Simple Harmonic Oscillator and corresponds to Section 16.5 in OpenStax College Physics for AP® Courses. Visit the full OpenStax AP® Physics Collection at 14 comments Transcript: we did the energy of a mass on a spring let's do the energy of a mass and a string energy of a pendulum all right let's see so this one is very similar we know that the total energy has to be conserved so we would say e mechanical whatever kinetic you have plus whatever potential have you have you know that has to be constant so for the mass on a spring we just talked about those two and how the energy goes back and forth but remains constant another way you could write it as we have already in the class is Delta K plus Delta U equals Zer so let's look at the kind of problems we could think about so if you have a pendulum like this and it's going around on a curve like that it could be up high it could be down here it could be up here so in all these different places uh the Delta K is still related to 1 12 mv^ 2 and now the Delta U is actually related to MGH right so the potential energy that's driving this motion is related to gravity so mg and H or we'll call it Delta y because it's actually the penum is moving up and down a little bit as you as you release it so so one kind of problem you could say is you release from Heights H how fast at the bottom well let's see initially we could actually just Solve IT formally we could say Delta K that's K final minus K initial so let's call this initial and this final well if you released it from rest up there then it initially had zero kinetic energy and it's getting some final that you're looking for so that would be 12 m v final squar that's the final kinetic energy minus 0o Because You released it from rest and if you want a Delta U you're actually going down you're losing potential energy so you're going down to a level we could call zero so plus the final would be zero minus the initial would be mg Delta y minus the initial that has to be zero and then you see what you would expect you simply say you set the kinetic energy you're going to get equal to the potential energy that you lost mg Delta Y and then you go solve for whatever the problem ask you to solve for maybe it gave you the height and ask you to get the speed or maybe it gave you how fast it goes and ask you how high it had to be for this part it's pretty much the same as sort of a ball rolling down a hill there's not that much difference you could even get a little more complicated and say What's the difference how fast is it going if you start here and you end here right well let's see so you can see this is a little bit further along than this the the actual point it's going to get to is a little bit higher so we haven't come all the way down so those problems get tricky it turns into a bunch of sort of trigonometry and algebra what you really need to know is uh these Heights so if you really wanted to think about this height you would say okay I have a right triangle here and I have it starting at some angle and I can figure out all the properties because I know the length of the pendulum right and I can figure out if I know the length and the angle I can figure out that this is L sin Theta because that's the opposite and this is L cosine Theta because that's that side and that's because it's all way up here this side's similar you could draw a line like that and now we have a right triangle here but it might be at some different angle call it five okay you know some things you know this L right and if I tell you how far along it goes on this curve you could basically approximate that curve as this height or you could say s is Theta R but you got to think about it's actually these parts right here how high does it go this is the important part what is this length minus L when it's hanging all the way down is the change in potential energy and at this stage what is this length minus uh this length minus L to get the heights so for penum it's a little tricky to find these Heights it's not as trivial as just seeing where it is for the spring kx^ s there's a little bit of geometry to figure out how high it's gone but you can do it
15511	https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Virtual_Textbook_of_OChem_(Reusch)_UNDER_CONSTRUCTION/14%3A_Thiols_and_Sulfides	14: Thiols and Sulfides - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode Book: Virtual Textbook of OChem (Reusch) UNDER CONSTRUCTION Organic Chemistry { } { "01:_Structure_and_Bonding" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Intermolecular_Forces" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Chemical_Reactivity" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Aromaticity" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Nomenclature" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Stereoisomers_Part_I" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Stereoisomers_Part_II" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Alkanes" : "property 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18:47:51 GMT 14: Thiols and Sulfides 43971 43971 admin { } Anonymous Anonymous User 2 false false [ "article:topic-guide", "showtoc:no", "license:ccbyncnd", "licenseversion:40" ] [ "article:topic-guide", "showtoc:no", "license:ccbyncnd", "licenseversion:40" ] 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. Organic Chemistry 4. Book: Virtual Textbook of OChem (Reusch) UNDER CONSTRUCTION 5. 14: Thiols and Sulfides Expand/collapse global location Home Campus Bookshelves Bookshelves Introductory, Conceptual, and GOB Chemistry General Chemistry Organic Chemistry Supplemental Modules (Organic Chemistry) Organic Chemistry (OpenStax) Organic Chemistry (Morsch et al.) 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Organic Chemistry II (Morsch et al.) Organic Chemistry III (Morsch et al.) Understanding Organic Chemistry Through Computation (Boaz and Pearce) Organic Chemistry -Part 1 Fundamentals (Malik) Inorganic Chemistry Analytical Chemistry Physical & Theoretical Chemistry Biological Chemistry Environmental Chemistry Learning Objects 14: Thiols and Sulfides Last updated Jun 5, 2019 Save as PDF 13: Ethers 15: Benzene and Derivatives Page ID 43971 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. 1. Nucleophilicity of Sulfur Compounds 2. 2. Oxidation States of Sulfur Compounds 1. Nucleophilicity of Sulfur Compounds Sulfur analogs of alcohols are called thiols or mercaptans, and ether analogs are called sulfides. The chemical behavior of thiols and sulfides contrasts with that of alcohols and ethers in some important ways. Since hydrogen sulfide (H 2 S) is a much stronger acid than water (by more than ten million fold), we expect, and find, thiols to be stronger acids than equivalent alcohols and phenols. Thiolate conjugate bases are easily formed, and have proven to be excellent nucleophiles in S N 2 reactions of alkyl halides and tosylates. R–S(–) Na(+) + (CH 3)2 CH–Br (CH 3)2 CH–S–R + Na(+) Br(–) Although the basicity of ethers is roughly a hundred times greater than that of equivalent sulfides, the nucleophilicity of sulfur is much greater than that of oxygen, leading to a number of interesting and useful electrophilic substitutions of sulfur that are not normally observed for oxygen. Sulfides, for example, react with alkyl halides to give ternary sulfonium salts (equation # 1) in the same manner that 3º-amines are alkylated to quaternary ammonium salts. Although equivalent oxonium salts of ethers are known, they are only prepared under extreme conditions, and are exceptionally reactive. Remarkably, sulfoxides (equation # 2), sulfinate salts (# 3) and sulfite anion (# 4) also alkylate on sulfur, despite the partial negative formal charge on oxygen and partial positive charge on sulfur. 2. Oxidation States of Sulfur Compounds Oxygen assumes only two oxidation states in its organic compounds (–1 in peroxides and –2 in other compounds). Sulfur, on the other hand, is found in oxidation states ranging from –2 to +6, as shown in the following table (some simple inorganic compounds are displayed in orange). Try drawing Lewis-structures for the sulfur atoms in these compounds. If you restrict your formulas to valence shell electron octets, most of the higher oxidation states will have formal charge separation, as in equation 2 above. The formulas written here neutralize this charge separation by double bonding that expands the valence octet of sulfur. Indeed, the S=O double bonds do not consist of the customary σ & π-orbitals found in carbon double bonds. As a third row element, sulfur has five empty 3d-orbitals that may be used for p-d bonding in a fashion similar to p-p (π) bonding. In this way sulfur may expand an argon-like valence shell octet by two (e.g. sulfoxides) or four (e.g. sulfones) electrons. Sulfoxides have a fixed pyramidal shape (the sulfur non-bonding electron pair occupies one corner of a tetrahedron with sulfur at the center). Consequently, sulfoxides having two different alkyl or aryl substituents are chiral. Enantiomeric sulfoxides are stable and may be isolated. Thiols also differ dramatically from alcohols in their oxidation chemistry. Oxidation of 1º and 2º-alcohols to aldehydes and ketones changes the oxidation state of carbon but not oxygen. Oxidation of thiols and other sulfur compounds changes the oxidation state of sulfur rather than carbon. We see some representative sulfur oxidations in the following examples. In the first case, mild oxidation converts thiols to disufides. An equivalent oxidation of alcohols to peroxides is not normally observed. The reasons for this different behavior are not hard to identify. The S–S single bond is nearly twice as strong as the O–O bond in peroxides, and the O–H bond is more than 25 kcal/mole stronger than an S–H bond. Thus, thermodynamics favors disulfide formation over peroxide. Mild oxidation of disufides with chlorine gives alkylsulfenyl chlorides, but more vigorous oxidation forms sulfonic acids (2nd example). Finally, oxidation of sulfides with hydrogen peroxide (or peracids) leads first to sulfoxides and then to sulfones. The nomenclature of sulfur compounds is generally straightforward. The prefix thio denotes replacement of a functional oxygen by sulfur. Thus, -SH is a thiol and C=S a thione. The prefix thia denotes replacement of a carbon atom in a chain or ring by sulfur, although a single ether-like sulfur is usually named as a sulfide. For example, C 2 H 5 SC 3 H 7 is ethyl propyl sulfide and C 2 H 5 SCH 2 SC 3 H 7 may be named 3,5-dithiaoctane. Sulfonates are sulfonate acid esters and sultones are the equivalent of lactones. Other names are noted in the table above. 14: Thiols and Sulfides is shared under a CC BY-NC-ND 4.0 license and was authored, remixed, and/or curated by LibreTexts. Back to top 13: Ethers 15: Benzene and Derivatives Was this article helpful? Yes No Recommended articles 1: Structure and Bonding 2: Intermolecular Forces 3: Chemical Reactivity 4: Aromaticity 5: Nomenclature Article typeChapterLicenseCC BY-NC-NDLicense Version4.0Show Page TOCno on page Tags This page has no tags. © Copyright 2025 Chemistry 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 ☰ 13: Ethers 15: Benzene and Derivatives
15512	https://physics.stackexchange.com/questions/566218/i-dont-get-why-the-fx-t-asinkx-wtp-sine-wave-equation-works	homework and exercises - I don't get why the $f(x, t) = Asin(kx-wt+p)$ sine wave equation works - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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 Physics helpchat Physics 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 I don't get why the f(x,t)=A∗s i n(k x−w t+p)f(x,t)=A∗s i n(k x−w t+p) sine wave equation works [closed] Ask Question Asked 5 years, 2 months ago Modified5 years, 2 months ago Viewed 1k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Closed. This question is off-topic. It is not currently accepting answers. Homework-like questions and check-my-work questions are considered off-topic here, particularly when asking about specific computations instead of underlying physics concepts. Homework questions can be on-topic when they are useful to a broader audience. If you intend to modify your question, please read the links above carefully before editing. Note that answers with complete solutions may be deleted! Closed 5 years ago. Improve this question I really don't get how this function for calculating the Y of a sine wave based on a particular point in the X-axis and time works. f(x,t)=A sin(k x−w t+p)f(x,t)=A sin⁡(k x−w t+p) x=x= position on X-axis t=t= point in time a=a= amplitude k=k= wave number w=angular frequency p=phase I'm writing down all of the letter meanings mostly as a just in case if the letters you use are different to mine. I've noticed a lot of different letters being thrown around. this is quite simple, so as you might expect I'm new to physics. incredibly new. I just don't get how the equation works. I get why y = A sin(wt+p) works - the angular speed times the time creates the angle, plus a "headstart angle" as I like to call the phase, then you have the sin of that angle times the amplitude, to amplify it to it'll have the right 'hight'. but I just don't understand why f(x, t) = A sin(kx-wt+p) works. I can't wrap my head around it. is there any explanation you can think of? homework-and-exercises waves 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 17, 2020 at 0:13 Harish Chandra Rajpoot 2,527 22 22 gold badges 25 25 silver badges 41 41 bronze badges asked Jul 16, 2020 at 22:40 RemerazeRemeraze 125 10 10 bronze badges 2 Try plotting the function, as a function of z, for a few values of t. Use some simple values such as k=1, w=1, p=0 to begin with. First plot for t=0, then for t=0.02, then t a bit bigger, say 0.1, and hopefully you will begin to get a feel for this function. Next try a different pair of values for k and w. Always think about what you are doing of course.Andrew Steane –Andrew Steane 2020-07-16 22:56:27 +00:00 Commented Jul 16, 2020 at 22:56 I posted a (much) more detailed version of @AndrewSteane's comment, then deleted it when I saw the comment, which I think will be considerably more instructive for the OP.WillO –WillO 2020-07-16 23:06:44 +00:00 Commented Jul 16, 2020 at 23:06 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. The first thing to understand is purely mathematical. Suppose we have a function g(t)g(t). Then g(t+a)g(t+a), for some constant a a (taken to be greater than zero without loss of generality), is the function shifted a a to the left. To see this, think of it this way: the new argument is t+a t+a, so if originally I needed to be at t 0 t 0 to have the value g 0=g(t 0)g 0=g(t 0), now I must instead be at the point t 0−a t 0−a so that g(t−a+a)=g 0 g(t−a+a)=g 0. Now for your question, let us throw away the phase p p, you can see it is just a constant shift. So we are concerned with y(x,t)=A cos(k x−ω t)y(x,t)=A cos⁡(k x−ω t). To make it more physical, suppose you are looking at a fish tank and the height of a water wave is y(x,t)y(x,t), where t t is the time and x x is the position along the tank. Now, let us focus on a single point along the tank. Then y(x,t)y(x,t) for x x fixed is just a function of t t, and indeed when you look at just one point you see the height of the water oscillating in time. Now suppose we freeze time instead. Then we get a snapshot of the wave, and it will instead vary with position, since time is fixed. y(x,t)y(x,t) is then a function of x x when t t is fixed. Notice we have the term −ω t−ω t and t t is increasing. So −ω t−ω t is responsible for shifting to the right, and as we increase t t, we are moving further and further right. This gives us the direction the wave propagates along x x. 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 16, 2020 at 23:24 JamalSJamalS 19.7k 6 6 gold badges 65 65 silver badges 109 109 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. A waveform (any shape, not necessarily a periodic sine wave) that propagates in space without changing shape can always be written in the form f(x,t)=g(x−v t),f(x,t)=g(x−v t), where g(x)g(x) is a function with only one parameter and v v is the propagation velocity. You can try this out for yourself for any function (try sketching g(x)g(x)). In your case, g(x)=A sin(k x+p)g(x)=A sin⁡(k x+p) and f(x,t)=g(x−c t)f(x,t)=g(x−c t), where c c is the speed of light. In your case, c c is not written explicitly, but hidden in the relation ω=c k ω=c k. If you find sin(ω t)sin⁡(ω t) more intuitive than sin(k x)sin⁡(k x), you can do the same reasoning starting from here: F(x,t)=G(t−x v).F(x,t)=G(t−x v). 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 16, 2020 at 23:37 Han-Kwang NienhuysHan-Kwang Nienhuys 2,373 12 12 silver badges 16 16 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. It would help if you could pinpoint exactly what you don't understand about the equation, but as an overview: It doesn't really matter what kind of complicated (real) function you have inside the sine function, it will always output some value between −1−1 and 1 1. If you left the function as f(x,t)=sin(k x−w t+p)f(x,t)=sin⁡(k x−w t+p) and let time pass, at each fixed value of x x the output (height) will oscillate between −1−1 and 1 1. If you want the function to oscillate between, say, −5−5 and 5 5 you need to set A=5 A=5 in which case when the sine function is equal to 1 1 the overall function is equal to 5 5. w w is the angular frequency, if you take two wave equations, one with w=1 w=1 and second with w=10 w=10, when you allow t t to run between say 0 0 and 1 1, the second wave equation runs through 10 10 times as many cycles. This effectively determines how quickly the wave oscillates in time, hence the name. As for the phase factor, this will move the entire function along by some amount. For instance if we compare sin(x)sin⁡(x) and sin(x+π)sin⁡(x+π), we have just moved the second function over by an amount π π. The phase factors allows us to effectively choose where in the wave cycle we start at t=0 t=0. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jul 16, 2020 at 23:41 answered Jul 16, 2020 at 23:26 CharlieCharlie 7,169 1 1 gold badge 19 19 silver badges 41 41 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Starting with your general equation y=Asin(kx-wt+p), y=Asin(-wt+p) gives the values of y at any time at x=0. y = Asin(kx+p) gives the values of y at any x for t=0. In other words, if you took a photo of the wave at any particular time, there would be a variation in y with x given by the general equation with that fixed value of t. If you focused on a particular x and tracked the variation of y there, it would be given by the general equation with that fixed value of x. 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 17, 2020 at 0:17 Not_EinsteinNot_Einstein 2,826 1 1 gold badge 8 8 silver badges 11 11 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Imagine you are experimenting with a wave on a rope. Here is a nice visual: You need to describe everything about the wave that could be important to someone across the world so that they can recreate the EXACT wave in their lab. First of all it should be a sine wave shape: y(x,t)=sin(????)y(x,t)=sin⁡(????) And as you said, the wave might not have a height of one, so you multiply it by a number "A" that will give you whatever maximum height and depth you desire. y(x,t)=A sin(????)y(x,t)=A sin⁡(????) If your wave's peaks and valleys range from [−3,3][−3,3], "A" should be 3 3. Now, you need to write down the frequency of the wave in time f f. This works exactly as it does in the first equation you learned. It is convenient to multiply it by 2 π 2 π since you plan to put it inside a sine function: y(x,t)=A sin(2 π f t)y(x,t)=A sin⁡(2 π f t) Now you measure your frequency in SPACE, by taking a picture of the wave and seeing how many wavelengths fit in a meter. You get a number ν ν. It is important to understand that this is the cycles per unit distance, NOT unit time like before: cycles/meter not cycles per second. Now, you multiply it by 2 π 2 π to stuff it into a sine function: y(x,t)=A sin(2 π f t−2 π ν x)y(x,t)=A sin⁡(2 π f t−2 π ν x) We call the 2 π f 2 π f the "angular frequency in time" and shorten it to ω ω for convenience. Notice, ω ω is now in radians/second not cycles per second. We call the 2 π ν 2 π ν the "angular frequency in space" and shorten it to k k for convenience. Notice, k k is now in radians/meter not cycles per meter. We often call k k the wave number. y(x,t)=A sin(2 π f t−2 π ν x)⇒y(x,t)=A sin(ω t−k x)y(x,t)=A sin⁡(2 π f t−2 π ν x)⇒y(x,t)=A sin⁡(ω t−k x) Finally you need to communicate the "headstart angle" p p like you said. p p is the angle the wave starts at in it's cycle when your experiment begins: y(x,t)=A sin(ω t−k x+p)y(x,t)=A sin⁡(ω t−k x+p) And that's where the wave equation comes from! It contains the minimum amount of information needed to describe any regular wave. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jul 18, 2020 at 2:37 answered Jul 17, 2020 at 21:25 novawarrior77novawarrior77 504 3 3 silver badges 13 13 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions homework-and-exercises waves See similar questions with these tags. 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15513	https://calcresource.com/cross-section-circle.html	CALC RESOURCE Calculation Tools & Engineering Resources Terms of use Jump to - Calculator - Definitions x Calculation options: \| \| \| Number of digits: \| \| 10x notation for big numbers: \| Circular Section Calculator By Dr. Minas E. Lemonis, PhD - Updated: June 30, 2020 Home > Cross Sections > Circular This tool calculates the properties of a circular cross-section. Enter the circle radius R or the diameter D, below. The calculated results will have the same units as your input. Please use consistent units for any input. \| \| \| R = \| \| or... \| \| D = \| \| Geometric properties: \| \| Area = \| \| Perimeter = \| \| Properties related to any centroidal axis: \| \| I = \| \| S = \| \| Z = \| \| Rgx = \| \| Iz = \| \| ADVERTISEMENT \| Share this Table of Contents - Calculator - Definitions - Geometry - Moments of Inertia - Elastic bending and moment of inertia - Polar moment of inertia - Elastic modulus - Elastic stresses - Plastic modulus - Radius of gyration - Circular section formulas - Related pages See also Properties of a Rectangular sectionProperties of a circular tube sectionMoment of Inertia of a CircleAll Cross Section tools ADVERTISEMENT Definitions Table of contents - Geometry - Moments of Inertia - Elastic bending and moment of inertia - Polar moment of inertia - Elastic modulus - Elastic stresses - Plastic modulus - Radius of gyration - Circular section formulas - Related pages Geometry The area A and the perimeter P, of a circular cross-section, having radius R, can be found with the next two formulas: Moments of Inertia The moment of inertia (second moment of area) of a circular section around any axis passing through its centroid, is given by the following expression: where R the radius of the section. Equivalently, the moment of inertia is given, in terms of circle diameter , by the formula: Elastic bending and moment of inertia The moment of inertia (second moment or area) is used in beam theory to describe the rigidity of a beam against flexure. The bending moment M applied to a cross-section is related with its moment of inertia with the following equation: where E is the Young's modulus, a property of the material, and , the curvature of the beam due to the applied load. Therefore, it can be seen from the former equation, that when a certain bending moment M is applied to a beam cross-section, the resulting curvature is reversely proportional to the moment of inertia I. Polar moment of inertia The polar moment of inertia, describes the rigidity of a cross-section against torsional moment, likewise the planar moments of inertia described above, are related to flexural bending. The calculation of the polar moment of inertia around an axis z (perpendicular to the section), can be done with the Perpendicular Axes Theorem: where the and are the moments of inertia around axes x and y, which are mutually perpendicular with z and meet at a common origin. Since, for a circular section it is: , the above equation becomes: The dimensions of moment of inertia are . Elastic modulus The elastic section modulus of any cross section around axis x (centroidal), describes the response of the section under elastic flexural bending. It is defined as: where , the moment of inertia of the section around x axis and Y the distance from centroid of a section fiber, parallel to the axis. Typically, the most distant fibers are of particular interest. For the circle, the bigger distance from center is . Therefore, application of the above formula to a circular cross-section, provides the following expression, for the elastic modulus, around any centroidal axis: Elastic stresses If a bending moment is applied around axis x, the section will respond with normal stresses, varying linearly with the distance from the neutral axis (which under elastic regime coincides to the centroidal x-x axis). Along neutral axis the stresses are zero. Absolute maximum will occur at the most distant fiber, with magnitude given by the formula: From the last equation, the section modulus can be considered for flexural bending, a property analogous to cross-sectional A, for axial loading. For the latter, the normal stress is F/A. The dimensions of section modulus are . Plastic modulus The plastic section modulus is similar to the elastic one, but defined with the assumption of full plastic yielding of the cross section due to flexural bending. In that case the whole section is divided in two parts, one in tension and one in compression, each under uniform stress field. For materials with equal tensile and compressive yield stresses, this leads to the division of the section into two equal areas, , in tension and , in compression, separated by the neutral axis. This is a result of equilibrium of internal forces in the cross-section, under plastic bending conditions. Indeed, the internal compressive force, over the entire compressive area, would be , assuming full plastification (i.e. the material would have yielded everywhere) and that the compressive yield stress is equal to (ignoring any plastic hardening in this context). Similarly, the internal tensile force would be , using the same assumptions. Enforcing equilibrium: The axis is called plastic neutral axis, and for non-symmetric sections, is not the same with the elastic neutral axis (which again is the centroidal one). The circular section however, is a symmetrical one and therefore the plastic neutral coincides with the elastic one. In other words, the plastic neutral axes passes through the center of the circle. The plastic modulus, for flexural bending around a given axis, is given by the general formula: where , the distance of the centroid of the compressive area from the plastic neutral axis and the respective distance of the centroid of the tensile area. For the case of a circular cross-section, the plastic neutral axis passes through centroid, as already mentioned, dividing the whole area into two equal parts. The compressive area would be a semicircle, with area . Its centroid lies at a distance equal to , from the axis (check our centroids table for reference). Due to symmetry, the same applies for the tensile area, too. Therefore, the plastic section modulus, of the circular section, is found like this: which is simplified to: Radius of gyration Radius of gyration of any cross-section, relative to an axis, is given by the general formula: where I the moment of inertia of the cross-section around the same axis and A its area. The dimensions of radius of gyration are . It describes how far from centroid the area is distributed. Small radius indicates a more compact cross-section. For a circular section, substitution to the above expression gives the following radius of gyration, around any axis, through center: Circle is the shape with minimum radius of gyration, compared to any other section with the same area A. Circular section formulas The following table, includes the formulas, one can use to calculate the main mechanical properties of the circular section. \| Circular section properties \| \| Quantity \| Formula \| \| Area: \| \| \| Circumference: \| \| \| Moment of inertia: \| \| \| Elastic modulus: \| \| \| Plastic modulus: \| \| \| Radius of gyration: \| \| \| Approximations: \| \| Related pages Properties of a Rectangular sectionProperties of a circular tube sectionMoment of Inertia of a CircleAll Cross Section tools Liked this page? Share it with friends! See also Properties of a Rectangular sectionProperties of a circular tube sectionMoment of Inertia of a CircleAll Cross Section tools
15514	https://stats.stackexchange.com/questions/117741/adding-two-or-more-means-and-calculating-the-new-standard-deviation	Skip to main content Adding two (or more) means and calculating the new standard deviation Ask Question Asked Modified 12 months ago Viewed 125k times This question shows research effort; it is useful and clear 13 Save this question. Show activity on this post. There are a load of discussions online about adding means and recalculating the standard deviation, but on none have I found answer to this question. I have two sub groups with mean x1 and x2, and s1 and s2. I want to add the two means to create x1+x2=x3. How do I calculate s3. The two sub groups are additive. The exact data I have is on herbicide and fungicide use on crops on four different farms, where herbicide + fungicide = pesticide. Herbicide (kg): 2,3,1,2 Fungicide (kg): 4,7,3,1 Pesticide (kg): 6,10,4,3 Herbicide x1=2.0; s1=0.8 Fungicide x2=3.8; s2=2.5 Pesticide x3=5.8; s3=3.1 Unfortunately I don't have any of the underlying data. I have only been provided with herbicide and fungicide means and standard deviations, x1, x2, s1 and s2. Is calculating this possible? Thank you! variance standard-deviation mean Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Oct 20, 2014 at 13:22 Joseph P asked Oct 3, 2014 at 12:38 Joseph PJoseph P 14311 gold badge11 silver badge77 bronze badges 8 2 It may be true in your case that pesticide = herbicide + fungicide, but that depends on physical additivity. That's not the same question as in statistical discussions on combining means or SDs of different samples. In general, if you have two samples both measuring the same thing, the combined mean will be somewhere between the two means, not their sum. So if combining heights of men and women, the mean height for (men and women) is certainly not the mean height of mean PLUS the mean height of women. – Nick Cox Commented Oct 3, 2014 at 12:50 @NickCox In the case you mention, we're talking about subgroup means, not means of different variables. – abaumann Commented Oct 3, 2014 at 13:44 @abaumann Agreed; that's what I said: "both measuring the same thing". – Nick Cox Commented Oct 3, 2014 at 14:02 Are you conflating population notation with samples? The Greek letters are conventionally used with population quantities. – Glen_b Commented Oct 3, 2014 at 16:31 @NickCox - thank you for clarifying - yes - this adding two subgroup means. – Joseph P Commented Oct 20, 2014 at 13:01 \| Show 3 more comments 4 Answers 4 Reset to default This answer is useful 15 Save this answer. Show activity on this post. The mean E(X+Y) is equal to the sum of the means E(X) and E(Y), i.e., in your case 2+3.8=5.8. The standard deviation is the square root of the variance Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y). If you assume that the use of herbicide and fungicide are independent - a bold assertion, although I don't know much about agriculture - then this simplifies to Var(X+Y)=Var(X)+Var(Y) and allows you to calculate the standard deviation by observing that Var(X)=0.822=0.6724 Var(Y)=2.52=6.25 Which leads us to find that σ3=0.6724+6.25−−−−−−−−−−−√≈2.631 Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Oct 3, 2014 at 13:58 abaumannabaumann 2,1001515 silver badges1313 bronze badges 0 Add a comment \| This answer is useful 8 Save this answer. Show activity on this post. For combining groups, the following picture is for your reference (from: Higgins, Thomas, Chandler, Cumpston, Li, Page, Welch (eds.) (2019) Cochrane Handbook for Systematic Reviews of Interventions, of which you can browse an older version here: Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Aug 29, 2024 at 19:21 viiv 322 bronze badges answered Nov 12, 2019 at 2:26 Ching-Chang ChenChing-Chang Chen 8111 silver badge11 bronze badge 1 1 This is a more correct general answer for anyone wanting it. The other two answers assume that the groups are the same size which may not be the case in general. – adunaic Commented Sep 8, 2022 at 21:18 Add a comment \| This answer is useful 7 Save this answer. Show activity on this post. The standard deviation is calculated differently if your sample correspond to the whole population or not. In the first case (i.e. your study is conducted only on four farms), abaumann is right and the standard deviation of your pesticide sample s3 is 2.68, calculated with the formula s=1n∑i=1n(xi−x¯)2−−−−−−−−−−−−√ If instead the four farms are a random sample of a wider population (i.e. you are trying to estimate the pesticide consumption in your country using only four farms) you have to use the corrected sample standard deviation: scorr=1n−1∑i=1n(xi−x¯)2−−−−−−−−−−−−−−−√ which gives s3=3.10, as you mentioned in your question. The use of the term n−1 is called Bessel's correction. To calculate s3 you can either take s21+s22−−−−−−√, as correctly explained by abaumann, or use one of the two equation above (depends on your case) with the pesticide sample. Let's recapitulate: You have three sets of observation data: Herbicide, Fungicide and Pesticide, each with four elements. The P set is the result of the sum of H and F, so the three sets are not independent. Depending if they are the whole population or a sub-set of a bigger population, you will want to calculate the standard deviation of the sample s or the corrected sample standard deviation σ (please note these symbols are just to make things easier to follow and they are not a standard convention). Let's consider now only the first two sets, H and F. Their means and standard deviations are x1=2.00s1=0.71σ1=0.82x2=3.75s2=2.17σ2=2.50 The sum of the means x3 have standard deviations s3 and σ3 as follows: x3=x1+x2=2.00+3.75=5.75s3=s21+s22−−−−−−√=0.712+2.172−−−−−−−−−−√=2.28σ3=σ21+σ22−−−−−−√=0.822+2.502−−−−−−−−−−√=2.63 and that's the correct answer. What you are doing is calculating the mean and standard deviations of a set (the P set resulting from the sum element by element of H and F) treating it as a third, independent set. If you were given the P set without explanation, that would be the best you could do and you would get x3=5.75s3=2.68σ3=3.10. But in fact you know that P is the result of eight observation and not only four, so you can leverage this and get a lower standard deviation. The mean is independent of the number of observations hence it stays the same. In short, the two methods are not giving the same results because they are indeed calculated on different sets. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Oct 21, 2014 at 13:00 answered Oct 20, 2014 at 17:33 Joseph D'ArimatheaJoseph D'Arimathea 43833 silver badges77 bronze badges 3 @joseph-darimathea Thank you for contributing on this. abaumanns answer is 2.63 not 2.68, and I think its just coincidence that these numbers are close. Can you calculate the correct s3 of 3.10 without the underlying data by your approach (i.e. just using x1, x2, s1 and s2)? It is definitely not 0.82+2.52−−−−−−−−−√. – Joseph P Commented Oct 20, 2014 at 22:59 @GlobalSprawl I expanded the answer. In short, you can't get 3.10 from only x1,2 and s1,2. – Joseph D'Arimathea Commented Oct 21, 2014 at 10:48 @NickCox Indeed I haven't been clear, thanks for the correction. – Joseph D'Arimathea Commented Oct 21, 2014 at 13:05 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. For a generalised approach with any number of groups and group sizes you can use the following, adapted from code at ``` suppressMessages(library(dplyr)) Create raw dataframe with unequal group sizes set.seed(644) dat_raw <- tibble( group = c(rep("A", 10), rep("B",20), rep("C",30)), x = rnorm(60) ) Overall summary dat_total <- dat_raw %>% summarise( n = n(), mean = mean(x), sd = sd(x) ) dat_total > # A tibble: 1 × 3 > n mean sd > > 1 60 0.0542 0.873 Summary by group dat_grouped <- dat_raw %>% group_by(group) %>% summarise( n = n(), mean = mean(x), sd = sd(x) ) dat_grouped > # A tibble: 3 × 4 > group n mean sd > > 1 A 10 0.203 0.923 > 2 B 20 0.0917 0.997 > 3 C 30 -0.0204 0.788 Calculating overall summary from grouped data dat_total_grouped <- dat_grouped %>% mutate( ex = n mean, exx = sd^2 (n-1) + ex^2 / n ) %>% summarise(across(c(n, ex, exx), sum)) %>% mutate( mean = ex/n, sd = sqrt((exx - ex^2/n)/(n-1)) ) %>% select(n, mean, sd) dat_total_grouped > # A tibble: 1 × 3 > n mean sd > > 1 60 0.0542 0.873 Overall summary from ungroued and grouped data are equal all.equal(dat_total, dat_total_grouped) > TRUE ``` Created on 2023-09-06 with reprex v2.0.2 Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Sep 5, 2023 at 23:52 JWillimanJWilliman 14133 bronze badges 1 This would be a better answer if you explained a bit more what you're doing in this code. I realize you have some comments, but you should also explain in English what's going on. – Adrian Keister Commented Sep 6, 2023 at 0:05 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions variance standard-deviation mean See similar questions with these tags. Linked 1 Summing multiple standard deviations (repeated measures) 0 Combined standard deviation of geometric series Related 5 What is the pooled standard deviation of paired samples? 21 What does pooled variance "actually" mean? 1 Compute Standard Deviation for the Mean of Means Hot Network Questions "Intermediate value theorem" for manifolds Worried travelling to America once more after long stays in the past Justifying an 'analog horror' aesthetic in a 'future' setting What happens during the "ambassador was called by the ministry" meeting? How can a forwarding service handle certificates for me? 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15515	https://www.indeed.com/career-advice/career-development/how-to-calculate-work-efficiency	How To Calculate Work Efficiency in 6 Steps (With Examples) \| Indeed.com Skip to main content Home Company reviews Find salaries Sign in Sign in Employers / Post Job 1 new update Home Company reviews Find salaries Employers Create your resume Resume services Change country 🇺🇸 United States Help Start of main content Career Guide Finding a job Resumes & cover letters Resumes & cover letters articles Resume samples Cover letter samples Interviewing Pay & salary Career development Career development articles Starting a new job Career paths News Career development How To Calculate Work Efficiency in 6 Steps (With Examples) How To Calculate Work Efficiency in 6 Steps (With Examples) Written by Indeed Editorial Team Updated July 24, 2025 Measuring the ability of tools to do their jobs efficiently is important for professionals in industries such as engineering, manufacturing and construction. One way that professionals can measure this is work efficiency. Knowing what work efficiency is and how to calculate it can help you find the best tools for tasks you have by comparing them effectively. In this article, we discuss how to calculate work efficiency, including what it is, why it's important and six steps you can use to perform the calculation.Related:Production Efficiency Formula: What It Is and Who Uses It Related jobs on Indeed Manufacturing & utilities jobs Part-time jobs Full-time jobs Remote jobs View more jobs on Indeed What is work efficiency? Work efficiency is a way to measure the output of any machine based on its input. This can work for simple machines such as levers, pulleys, wedges, screws and inclined planes to complex machines such as automobile engines, wind turbines and bicycles. You can calculate work efficiency as a ratio and express it as a percentage of an input to a machine and what it outputs using the following formula:Efficiency = (Energy Output / Energy Input) x 100 Where: Efficiency is the overall advantage the machine provides to a particular task Output is the amount of work the machine does in Joules Input is the amount of work put into the machine in Joules 100 is to convert the ratio into a percentage When calculating work efficiency, the closer to 100% the product of the equation is, the more efficient the machine. Machines that are more efficient often are the best machines to use for work because they can help you save resources such as money, energy and the amount of work you do to gain the same profit.Related:What Is Production Efficiency? Why is calculating work efficiency important? Calculating work efficiency is important for professionals in engineering and physics because it can help them understand how energy moves through a machine and how much energy every machine loses. It is important because it can help these professionals conserve limited resources such as fossil fuels like oil or limited human resources because of operating limits in a region. Finally, making this calculation can be important for businesses that use heavy machinery because they can lower their operating costs if they use machines that need less input for the same output or more output for the same amount of input. How to calculate work efficiency Below are steps you can use to help you calculate work efficiency: 1. Collect information about input and output energy Before you can use the calculation to find what the work efficiency of a machine is, gather the amount of energy put into the machine in Joules, the unit for energy in the International System of Units, and the amount of energy the machine outputs in Joules. Machines can have more than one input of energy, so ensure you calculate each one and include it in the amount of energy that is input. All machines experience energy loss as other forms of energy. For example, hitting a nail with a hammer loses some energy to heat and some to sound.The output of a machine is almost always less than the input because all machines are subject to the law of conservation of energy, which states that energy can only change from one form to another, not be created or destroyed, so almost all percentages are less than 100%. Machines with an efficiency of 80% or higher are the most efficient, but that doesn't mean they are necessarily the best tools for a particular job. For example, using a hammer to hit a nail might have a lower percentage, but over time saves energy for the person using the hammer. 2. Fill information in for each part of the equation and simplify Once you've gathered your information about the input energy and output energy of the machine, you can use that information to fill in the parts of the equation. For example, if you find that a machine such as a lever takes 6 pounds of force from a person moving their end down 2 feet, while the other end of the lever moves a 10-pound box 1 foot in the air, your equation is:Efficiency = ((10 lbs x 1 foot of work) / (20 lbs x 1 foot of work)(6 lbs x 2 feet of work)) x 100 Which you can simplify as:Efficiency = (10 foot-pounds of work / 12 foot-pounds of work) x 100 3. Calculate energy output divided by energy input After you simplify your equation, you can perform the first part of your calculation, dividing your output energy by your input energy. Using the example from the previous step, your equation is:Efficiency = (10 foot-pounds of work / 12 foot-pounds of work) x 100 Divide 10 foot-pounds of work by 12 foot-pounds of work:10 / 12 = 0.83 Related:How To Solve Simple Equations (With Examples) 4. Multiply by 100 for your final percentage Once you have calculated the overall outcome of your output divided by your input, multiply your results by 100 to get a percentage of efficiency for your machine. Using the example above, you can use your outcome from the division, 0.83, and multiply it by 100 to get your final machine efficiency. For example:Efficiency = 0.83 x 100 Multiply the final two numbers in your equation:0.83 x 100 = 83%This means that the lever you used to move the box has an efficiency of 83%, which is relatively efficient for simple machines, such as levers and wedges. 5. Repeat steps 2 through 4 for each machine Sometimes, you want to compare the efficiency of each machine to see what works best for your particular needs. In these situations, you can repeat steps 2 through 4 of this process to calculate the efficiency of each machine. For example, instead of using a lever to lift the 10-pound box, you could use a pulley. If the pulley has an input of 5.5 pounds of force from a person over 2 feet to lift the same box one foot, then you can calculate the efficiency of the pulley as:Efficiency = (10 foot-pounds of work / (5.5 pounds of force x 2 feet)) x 100 Efficiency = (10 foot-pounds of work / 11 foot-pounds of work) x 100 Efficiency = 0.91 x 100 Efficiency = 91%Related:How to Calculate Percentage Difference 6. Compare the eventual results for each machine Once you calculate the overall efficiency for each machine, you can compare them to understand which is more efficient. Using the examples of the pulley and the lever from above, you can see that the pulley is more efficient because its percentage is 91% compared to the lever's 83%. This means that the pulley is easier to use for the person doing that job and might make their work easier over a longer time. This can also mean that exploring new ways to accomplish the same task may help you increase the overall efficiency of your production team members. Search jobs and companies hiring now Job title, keywords or company Location Search Example of calculating work efficiency Below is an example of calculating work efficiency for three simple levers with different lengths while lifting a 20-pound crate:The first step is to gather the information about each lever's input and output, like the example below: The first lever requires 6 pounds of force over four feet to lift the crate one foot in the air. The second lever requires 10 pounds of force over four feet to lift the crate two feet in the air. The third lever requires 8 pounds of force over three feet to lift the crate half a foot in the air. Next, input the information from the first lever into the equation for efficiency and simplify:Efficiency1 = ((20 pounds of force x 1 foot) / (6 pounds of force x 4 feet)) x 100 Efficiency1 = (20 foot-pounds of force) / (24 foot-pounds of force) x 100 Then, calculate output divided by input and multiply by 100:Efficiency1 = 0.83 x 100 Efficiency1 = 83%Next, repeat the steps for the other levers: Efficiency2 = ((20 pounds of force x 2 feet) / (10 pounds of force x 4 feet)) x 100 Efficiency2 = (40 foot-pounds of force) / (40 foot pounds of force) x 100 Efficiency2 = 1.00 x 100 Efficiency2 = 100% Lever three: Efficiency3 = ((20 pounds of force x 0.5 feet) / (8 pounds of force x 3 feet) x 100 Efficiency3 = ((10 foot-pounds of force) / (24 foot-pounds of force) x 100 Efficiency3 = 0.42 x 100 Efficiency3 = 42% Finally, you can compare the efficiencies of each lever. Lever one has an efficiency of 83%, lever two has an efficiency of 100% and lever three has an efficiency of 42%. In order to accomplish lifting a 20-pound crate in the air, the most efficient lever is the second one because it does the job of losing no energy. The information on this site is provided as a courtesy and for informational purposes only. Indeed is not a career or legal advisor and does not guarantee job interviews or offers Manufacturing & utilities Share: Related Articles FAQ: How Can You Work More Efficiently? How To Calculate Productivity in 4 Steps (With Methods) Effective vs. Efficient: What's the Difference? (With Tips) Explore more articles 10 Different Types of Email (And Why They're Important) How To Write Policies and Procedures in 7 Steps (With Tips) Group vs. Team: What's the Difference? (Plus an Example) How To Write an Email Asking For a Job in 7 Steps 23 Examples of Follow-Up Email Subject Lines What Does a Truck Dispatcher Do? 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15516	https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/1a592ed184fb4c444547f67c9bcdd8ec_MIT6_436JF18_lec13.pdf	MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.436J/15.085J Fall 2018 Lecture 13 MOMENT GENERATING FUNCTIONS Contents 1. Moment generating functions 2. Sum of a random number of random variables 3. Transforms associated with joint distributions Moment generating functions, and their close relatives (probability gener ating functions and characteristic functions) provide an alternative way of rep resenting a probability distribution by means of a certain function of a single variable. These functions turn out to be useful in many different ways: (a) They provide an easy way of calculating the moments of a distribution. (b) They provide some powerful tools for addressing certain counting and combinatorial problems. (c) They provide an easy way of characterizing the distribution of the sum of independent random variables. (d) They provide tools for dealing with the distribution of the sum of a random number of independent random variables. (e) They play a central role in the study of branching processes. (f) They play a key role in large deviations theory, that is, in studying the asymptotics of tail probabilities of the form P(X ≥ c), when c is a large number. (g) They provide a bridge between complex analysis and probability, so that complex analysis methods can be brought to bear on probability problems. (h) They provide powerful tools for proving limit theorems, such as laws of large numbers and the central limit theorem. 1 1 MOMENT GENERATING FUNCTIONS 1.1 Deﬁnition Deﬁnition 1. The moment generating function associated with a random variable X is a function MX : R → [0, ∞] deﬁned by sX ]. MX (s) = E[e The domain DX of MX is deﬁned as the set DX = {s \| MX (s) < ∞}. If X is a discrete random variable, with PMF pX , then X sx MX (s) = e pX (x). x If X is a continuous random variable with PDF fX , then Z MX (s) = e sxfX (x) dx. Note that this is essentially the same as the deﬁnition of the Laplace transform of a function fX , except that we are using s instead of −s in the exponent. 1.2 The domain of the moment generating function Note that 0 ∈ DX , because MX (0) = E[e0X ] = 1. For a discrete random variable that takes only a ﬁnite number of different values, we have DX = R. For example, if X takes the values 1, 2, and 3, with probabilities 1/2, 1/3, and 1/6, respectively, then 1 1 1 s 2s 3s MX (s) = e + e + e , (1) 2 3 6 which is ﬁnite for every s ∈ R. On the other hand, for the Cauchy distribution, fX (x) = 1/(ˇ(1 + x2)), for all x, it is easily seen that MX (s) = ∞, for all s 6= 0. In general, DX is an interval (possibly inﬁnite or semi-inﬁnite) that contains zero. Exercise 1. Suppose that MX (s) < ∞ for some s > 0. Show that MX (t) < ∞ for all t ∈ [0, s]. Similarly, suppose that MX (s) < ∞ for some s < 0. Show that MX (t) < ∞ for all t ∈ [s, 0]. 2 Exercise 2. Suppose that log P(X > x) lim sup , −ν < 0. x x!1 Establish that MX (s) < ∞ for all s ∈ [0, ν). 1.3 Inversion of transforms By inspection of the formula for MX (s) in Eq. (1), it is clear that the distribu tion of X is readily determined. The various powers esx indicate the possible values of the random variable X, and the associated coefﬁcients provide the corresponding probabilities. At the other extreme, if we are told that MX (s) = ∞ for every s 6= 0, this is certainly not enough information to determine the distribution of X. On this subject, there is the following fundamental result. It is intimately related to the inversion properties of Laplace transforms. Its proof requires so phisticated analytical machinery and is omitted. Theorem 1. Inversion theoremSuppose that MX (s) is ﬁnite for all s in an interval of the form [−a, a], where a is a positive number. Then, MX determines uniquely the CDF of the random variable X. In particular, if MX (s) = MY (s) < ∞, for all s ∈ [−a, a], where a is a positive number, then the random variables X and Y have the same CDF. There are explicit formulas that allow us to recover the PMF or PDF of a ran dom variable starting from the associated transform, but they are quite difﬁcult to use (e.g., involving “contour integrals”). In practice, transforms are usually inverted by “pattern matching,” based on tables of known distribution-transform pairs. 1.4 Moment generating properties There is a reason why MX is called a moment generating function. Let us con sider the derivatives of MX at zero. Assuming for a moment we can interchange the order of integration and differentiation, we obtain dMX (s) d sX ] = E[XesX ] = E[e = E[X], ds s=0 ds s=0 s=0 dmMX (s) dm sX ] sX ] = E[e = E[Xm e = E[Xm] dsm s=0 dsm s=0 s=0 3 \| \| \| \| \| \| Thus, knowledge of the transform MX allows for an easy calculation of the moments of a random variable X. Justifying the interchange of the expectation and the differentiation does require some work. The steps are outlined in the following exercise. For sim plicity, we restrict to the case of nonnegative random variables. Exercise 3. Suppose that X is a nonnegative random variable and that MX (s) < ∞ for all s ∈ (−∞, a], where a is a positive number. (a) Show that E[Xk] < ∞, for every k. (b) Show that E[XkesX ] < ∞, for every s < a. (c) Show that (ehX − 1)/h ≤ XehX . (d) Use the DCT to argue that h hX −1 i hX ] − 1 e E[e E[X] = E lim = lim . h#0 h h#0 h 1.5 The probability generating function For discrete random variables, the following probability generating function is sometimes useful. It is deﬁned by gX (s) = E[s X ], with s usually restricted to positive values. It is of course closely related to the moment generating function in that, for s > 0, we have gX (s) = MX (log s). One difference is that when X is a positive random variable, we can deﬁne gX (s), as well as its derivatives, for s = 0. So, suppose that X has a PMF pX (m), for m = 1, . . .. Then, 1 X m gX (s) = s pX (m), m=1 resulting in dm gX (s) = m! pX (m). dsm s=0 (The interchange of the summation and the differentiation needs justiﬁcation, but is indeed legitimate for small s.) Thus, we can use gX to easily recover the PMF pX , when X is a positive integer random variable. At the same time X d gX (s) = mpX (m) = E[X]. ds s=1 m≥1 4 \| \| 1.6 Examples d Example : X = Exp(λ). Then, ˆ Z 1 λ λ−s , s < λ; sxλe−λx dx = MX (s) = e ∞, otherwise. 0 d Example : X = Ge(p) ( 1 s e p X s , e < 1/(1 − p); sm 1−(1−p)es MX (s) = e p(1 − p)m−1 ∞, otherwise. m=1 In this case, we also ﬁnd gX (s) = ps/(1 − (1 − p)s), s < 1/(1 − p) and gX (s) = ∞, otherwise. d Example : X = N(0, 1). Then, Z 1 1 x2 MX (s) = √ exp(sx) exp(− )dx 2ˇ −1 2 Z 1 2 exp(s2/2) x2 + 2sx − s = √ exp(− )dx 2ˇ −1 2 = exp(s 2/2). 1.7 Properties of moment generating functions We record some useful properties of moment generating functions. Theorem 2. (a) If Y = aX + b, then MY (s) = esbMX (as). (b) If X and Y are independent, then MX+Y (s) = MX (s)MY (s). (c) Let X and Y be independent random variables. Let Z be equal to X, with probability p, and equal to Y , with probability 1 − p. Then, MZ (s) = pMX (s) + (1 − p)MY (s). Proof: For part (a), we have MX (aX + b) = E[exp(saX + sb)] = exp(sb)E[exp(saX)] = exp(sb)MX (as). 5 2 For part (b), we have MX+Y (s) = E[exp(sX + sY )] = E[exp(sX)]E[exp(sY )] = MX (s)MY (s). For part (c), by conditioning on the random choice between X and Y , we have sZ] = pE[e MZ(s) = E[e sX ] + (1 − p)E[e sY ] = pMX (s) + (1 − p)MY (s). Example : (Normal random variables) (a) Let X be a standard normal random variable, and let Y = ˙X + µ, which we know 1 to have a N(µ, ˙2) distribution. We then ﬁnd that MY (s) = exp(sµ + s2˙2). 2 d (b) Let X = N(µ1, ˙1 2) and Y = N(µ2, ˙2 2). Then, n o MX+Y (s) = exp( s(µ1 + µ2) + 1 s 2(˙1 2 + ˙2 2) . 2 d Using the inversion property of transforms, we conclude that X + Y = N(µ1 + µ2, ˙1 2 + ˙2 2), thus corroborating a result we ﬁrst obtained using convolutions. SUM OF A RANDOM NUMBER OF INDEPENDENT RANDOM VARI ABLES Let X1, X2, . . . be a sequence of i.i.d. random variables, with mean µ and vari ance ˙2 . Let N be another independent random variable that takes nonnegative PN integer values. Let Y = i=1 Xi. Let us derive the mean, variance, and mo ment generating function of Y . We have E[Y ] = E[E[Y \| N]] = E[Nµ] = E[N]E[X]. Furthermore, using the law of total variance, var(Y ) = E var(Y \| N) + var E[Y \| N] = E[N˙2] + var(Nµ) = E[N]˙2 + µ 2var(N). Finally, note that E[exp(sY ) \| N = n] = Mn (s) = exp(n log MX (s)), X 6 3 implying that 1 X MY (s) = exp(n log MX (s))P(N = n) = MN (log MX (s)). n=1 The reader is encouraged to take the derivative of the above expression, and evaluate it at s = 0, to recover the formula E[Y ] = E[N]E[X]. Example : Suppose that each Xi is exponentially distributed, with parameter λ, and that N is geometrically distributed, with parameter p ∈ (0, 1). We ﬁnd that log MX (s) e p pλ/(λ − s) λp MY (s) = = = 1 − elog MX (s)(1 − p) 1 − λ(1 − p)/(λ − s) λp − s which we recognize as a moment generating function of an exponential random variable with parameter λp. Using the inversion theorem, we conclude that Y is exponentially distributed. In view of the fact that the sum of a ﬁxed number of exponential random variables is far from exponential, this result is rather surprising. An intuitive explanation will be provided later in terms of the Poisson process. TRANSFORMS ASSOCIATED WITH JOINT DISTRIBUTIONS If two random variables X and Y are described by some joint distribution (e.g., a joint PDF), then each one is associated with a transform MX (s) or MY (s). These are the transforms of the marginal distributions and do not convey infor mation on the dependence between the two random variables. Such information is contained in a multivariate transform, which we now deﬁne. Consider n random variables X1, . . . , Xn related to the same experiment. Let s1, . . . , sn be real parameters. The associated multivariate transform is a function of these n parameters and is deﬁned by s1X1+···+snXn MX1,...,Xn (s1, . . . , sn) = E e . The inversion property of transforms discussed earlier extends to the multi variate case. That is, if Y1, . . . , Yn is another set of random variables and MX1,...,Xn (s1, . . . , sn), MY1,...,Yn (s1, . . . , sn) are the same functions of s1, . . . , sn, in a neighborhood of the origin, then the joint distribution of X1, . . . , Xn is the same as the joint distribution of Y1, . . . , Yn. 7 Remarks: (a) Consider two random variables X and Y . Their joint transform is sX tY ] = E[e MX,Y (s, t) = E[e e sX+tY ] = MZ (1), where Z = sX + tY . Thus, calculating a multivariate transform essentially amounts to calculating the univariate transform associated with a single ran dom variable that is a linear combination of the original random variables. (b) If X and Y are independent, then MX,Y (s, t) = MX (s)MY (t). 8 MIT OpenCourseWare 6.436J / 15.085J Fundamentals of Probability Fall 2018 For information about citing these materials or our Terms of Use, visit:
15517	https://www.quora.com/How-do-I-find-the-rank-of-a-rectangular-matrix-if-possible	How to find the rank of a rectangular matrix, if possible - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Matrix Properties Linear Algebra Course Matrices in Maths Rectangular Matrix Rank of Matrix Matrices and Vectors Mathematics , Basic Linear Algebra 5 How do I find the rank of a rectangular matrix, if possible? All related (34) Sort Recommended Assistant Bot · 1y To find the rank of a rectangular matrix, you can use several methods. The rank of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix. Here are a few common methods to determine the rank: Row Echelon Form (REF) Transform the matrix: Use Gaussian elimination to convert the matrix to row echelon form (REF). This involves using elementary row operations (swapping rows, scaling rows, adding multiples of rows to other rows). Count the non-zero rows: The rank of the matrix is equal to the number of non-zero rows in its row echelon form. Reduced Row Eche Continue Reading To find the rank of a rectangular matrix, you can use several methods. The rank of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix. Here are a few common methods to determine the rank: Row Echelon Form (REF) Transform the matrix: Use Gaussian elimination to convert the matrix to row echelon form (REF). This involves using elementary row operations (swapping rows, scaling rows, adding multiples of rows to other rows). Count the non-zero rows: The rank of the matrix is equal to the number of non-zero rows in its row echelon form. Reduced Row Echelon Form (RREF) Transform the matrix: Use Gaussian elimination to convert the matrix to reduced row echelon form (RREF). Count the leading 1s: The rank is equal to the number of leading 1s in the RREF. Determinants (for square matrices) For square matrices, if you want to find the rank using determinants: Find submatrices: Consider all square submatrices of different sizes. Calculate determinants: The rank is the largest size of any square submatrix that has a non-zero determinant. Singular Value Decomposition (SVD) Decompose the matrix: Use SVD to decompose the matrix A A into U Σ V T U Σ V T. Count non-zero singular values: The rank of the matrix is equal to the number of non-zero singular values in the diagonal matrix Σ Σ. Example Consider the matrix: A=⎛⎜⎝1 2 3 4 5 6 7 8 0⎞⎟⎠A=(1 2 3 4 5 6 7 8 0) Convert A A to REF or RREF. Suppose the REF is: ⎛⎜⎝1 2 3 0−3−6 0 0 0⎞⎟⎠(1 2 3 0−3−6 0 0 0) 3. Count non-zero rows: There are 2 non-zero rows, so the rank of matrix A A is 2. Conclusion To summarize, the rank of a rectangular matrix can be determined using methods like row echelon form, reduced row echelon form, determinants of submatrices (for square matrices), or singular value decomposition. The most common and straightforward methods are the REF and RREF approaches. Upvote · Abhinav Vadrevu Works at ITC Limited · Author has 66 answers and 211.2K answer views ·10y Convert the matrix into its RREF (Row reduced echelon form). The number of non zero rows in the RREF of the matrix is the rank of the matrix. Upvote · 9 5 Sponsored by Google Ads Google AI finds ready-to-buy shoppers this peak season. Turn browsing into sales this peak season with an AI-powered campaign from Google Ads. Sign Up 9 7 Dr. Pramod Ranjan Professor of Mathematics at Patliputra University Patna ·3y Let us consider a rectangular matrix A of order 2x3.Since the minor of order greater than 2 of matrix A can't be formed, so the highest possible rank of matrix A is 2. If there exists at least one minor of order 2 of matrix A which is non zero, then the rank of matrix A will be 2. If every minor of order 2 of matrix A is zero, then find all minors of order 1 of matrix A. If there exists at least one minor of order 1 which is non zero, then the rank of matrix A will be 1. If the matrix A is a zero matrix, then the rank of matrix A will be zero. Upvote · 9 1 Related questions More answers below Where can we find the rank of a rectangular matrix as well as a square matrix? Can a3x4 matrix have rank equal to 3? How is the rank of a square or rectangular matrix determined and can it be calculated manually? How do you find the rank of a 2 by 3 matrix? How do I find the rank of a 3x2 matrix? Ashish Jethvani Knows about Linear Algebra ·9y Number of pivot elements in matrix is rank of a matrix. Pivot elements are decided when we compute the upper triangular matrx using gauss Jordan elimination . Pivot elements are always non zero. Upvote · 9 2 Sponsored by Dell Technologies Built for what's next. Do more, faster with Dell AI PCs powered by #IntelCoreUltra processors & ready for Windows 11. Learn More 99 10 Preetham Thelluri Assistant Math Teacher (Peer Facilitator) at Alpharetta HS · Author has 174 answers and 685.9K answer views ·Updated 5y Related How can we find the rank of a 4×4 matrix? A 4 x 4 matrix can have a maximum rank of 4. A rank of a matrix is nothing but the number of linearly independent columns in the matrix. A linearly independent column is a column that cannot be expressed as a linear combination of other columns in the matrix. Let’s look at some examples (these are 3 x 3, but the same process applies for any n x n matrix): This is the 3 x 3 identity matrix. It has a rank of 3, because it has 3 linearly independent columns. No one column in this matrix can be expressed as a linear combination of the other. This is just another 3 x 3 matrix. It seems to have 3 linea Continue Reading A 4 x 4 matrix can have a maximum rank of 4. A rank of a matrix is nothing but the number of linearly independent columns in the matrix. A linearly independent column is a column that cannot be expressed as a linear combination of other columns in the matrix. Let’s look at some examples (these are 3 x 3, but the same process applies for any n x n matrix): This is the 3 x 3 identity matrix. It has a rank of 3, because it has 3 linearly independent columns. No one column in this matrix can be expressed as a linear combination of the other. This is just another 3 x 3 matrix. It seems to have 3 linearly independent columns. But is the rank 3? Not here! :D If you subtract the second column from the third column, you can see that the new matrix becomes: ⎡⎢⎣1 2 1 0 2 0 1−2 1⎤⎥⎦[1 2 1 0 2 0 1−2 1] Do you notice something? The first and third columns are the SAME! They can be expressed as linear combinations of each other! This matrix only has 2 linearly independent columns. It’s rank is 2. In all, if you want to find the rank of a matrix, just look for the number of linearly independent columns in any dimensions! Source of images: Math is Fun Hope this helped! Upvote · 99 24 9 1 Related questions More answers below Can the rank of a matrix be greater than 3? What are rectangular matrices? What is understood by the rank of the matrix? How do I find a matrix of rank one that is approximate to the square of a 22 matrix? How do you find the rank of a product of two matrices? Vijay Krishna B.Tech in Computer Science and Engineering, SRM Institute of Science and Technology (Graduated 2022) ·7y Related How can I find the rank of a matrix? First make the matrix into Echelon form. As you see in the above image this is called the echelon form A matrix A (of order m × n) is said to be in echelon form (triangular form) if (i) Every row of A which has all its entries 0 occurs below every row which has a non-zero entry. (ii) The first non-zero entry in each non-zero row is 1. (iii) The number of zeros before the first non-zero element in a row is less than the number of such zeros in the next row. By elementary operations one can easily bring the given matrix to the echelon form So to find the rank Number of non- zero rows = Rank of the matrix If Continue Reading First make the matrix into Echelon form. As you see in the above image this is called the echelon form A matrix A (of order m × n) is said to be in echelon form (triangular form) if (i) Every row of A which has all its entries 0 occurs below every row which has a non-zero entry. (ii) The first non-zero entry in each non-zero row is 1. (iii) The number of zeros before the first non-zero element in a row is less than the number of such zeros in the next row. By elementary operations one can easily bring the given matrix to the echelon form So to find the rank Number of non- zero rows = Rank of the matrix If all the element in the row is zero it is called as Zero row. For example, The number of non-zero rows = Rank of the matrix = 2. Upvote · 99 88 9 7 9 2 Sponsored by RedHat Know what your AI knows, with open source models. Your AI should keep records, not secrets. Learn More 99 36 Alexander Farrugia My PhD. thesis uses a lot of techniques from matrix theory. · Author has 3.2K answers and 27.5M answer views ·9y Related How do I find a rank of a 3x4 matrix using a minor method? Thanks for the answer request. A minor is the determinant of a square submatrix of some matrix. In order to obtain the rank of your 3×4 3×4 matrix using its minors, first obtain the determinant of each 3×3 3×3 submatrix of the 3×4 3×4 matrix. If one of these determinants is nonzero, you may stop and state that the rank of the 3×4 3×4 matrix is 3 3. However, if the determinants of these four 3×3 3×3 matrices are all zero, then the rank of the 3×4 3×4 matrix is less than 3 3. In such a case, proceed to find the determinant of each 2×2 2×2 submatrix. If one of these determinants is nonzer Continue Reading Thanks for the answer request. A minor is the determinant of a square submatrix of some matrix. In order to obtain the rank of your 3×4 3×4 matrix using its minors, first obtain the determinant of each 3×3 3×3 submatrix of the 3×4 3×4 matrix. If one of these determinants is nonzero, you may stop and state that the rank of the 3×4 3×4 matrix is 3 3. However, if the determinants of these four 3×3 3×3 matrices are all zero, then the rank of the 3×4 3×4 matrix is less than 3 3. In such a case, proceed to find the determinant of each 2×2 2×2 submatrix. If one of these determinants is nonzero, you can stop and state that the rank of the 3×4 3×4 matrix is 2 2. However, if all of these 2×2 2×2 submatrices have determinant zero, then you know that the rank of the 3×4 3×4 matrix is less than 2 2 - so it is either zero or one. In such a case, it is easy to determine whether the rank is zero or one: it is zero if the entries of the matrix are all zero, otherwise it is one. Upvote · 99 72 9 6 R.James R.McHenry Knows Finnish ·4y Related How do I find the determinant of a rectangular matrix like a 2x3 matrix? A determinant is a number that is assigned only to a square matrix. With the help of a determinant one can say something about the properties of a matrix. Here is an easy example of how to calculate the determinant of a 2 × 2 matrix: detabcd=ad-bc Example: det2-351=21--35=17th Determinant of a 3 × 3 matrix: In order not to have to remember this calculation formula, there is a calculation aid. You write the first two columns behind the matrix. Then one can apply the rule of Sarrus. Products enclosed in blue areas are added and the products in orange areas are subtracted from this. The result is exa Continue Reading A determinant is a number that is assigned only to a square matrix. With the help of a determinant one can say something about the properties of a matrix. Here is an easy example of how to calculate the determinant of a 2 × 2 matrix: detabcd=ad-bc Example: det2-351=21--35=17th Determinant of a 3 × 3 matrix: In order not to have to remember this calculation formula, there is a calculation aid. You write the first two columns behind the matrix. Then one can apply the rule of Sarrus. Products enclosed in blue areas are added and the products in orange areas are subtracted from this. The result is exactly the formula above. Here you can calculate matrices online. Upvote · 99 15 Alexandru Carausu Former Assoc. Prof. Dr. (Ret) at Technical University "Gh. Asachi" Iasi (1978–2010) · Author has 3K answers and 875.5K answer views ·10mo Related What is the rank of a matrix? Can a matrix have two different ranks? If so, how would that happen and what are some examples of this scenario? An m-by-n matrix over a numerical field F has one and only one rank : Rank A = r . But this numerical feature of a matrix has 4 (or even more) equivalent definitions. Let A = [ a_i j ] be an m-by-n matrix over a numerical field F . Def. #1 : Rank A = r = the maximum (highest) order of non-zero determinants with entries a_i j . Def. #2 : Rank A = r = the maximum (highest) order of non-singular square sub-matrices of A . Def. #3 : Rank A = r = the maximum (highest) number of linearly independent rows of A . Def. #4 : Rank A = r = the maximum (highest) number of linearly independent columns of A . Def Continue Reading An m-by-n matrix over a numerical field F has one and only one rank : Rank A = r . But this numerical feature of a matrix has 4 (or even more) equivalent definitions. Let A = [ a_i j ] be an m-by-n matrix over a numerical field F . Def. #1 : Rank A = r = the maximum (highest) order of non-zero determinants with entries a_i j . Def. #2 : Rank A = r = the maximum (highest) order of non-singular square sub-matrices of A . Def. #3 : Rank A = r = the maximum (highest) number of linearly independent rows of A . Def. #4 : Rank A = r = the maximum (highest) number of linearly independent columns of A . Def. #5 : Rank A = r = the common dimension of the subspaces spanned by rows / columns of A . I am adding a few notes for readers of this answer who are (possibly) less familiar with matrices and their properties ; by the way, just the second and third among the three (above posted) questions suggest that their sender falls in this group. Note 1. It follows from all these definitions that Rank A = r ≤ min { m , n } . If mn ==> r ≤ n . If m = n the matrix is square or order n and r ≤ n . These obvious inequalities follow from the first two definitions. Note 2. In the 3-rd and 4-th definitions, it is easy to see that all m rows of A are vectors in the real (if F = R ) Euclidean space R^n : A_i = [ a_i 1 . . . a_i j . . . a_im ] ∈ R^n , while all columns of A are vectors in the Euclidean space R^m : A^j = [ a_ 1 j . . . a_i j . . . a_m j ] ∈ R^m . These obvious properties allow to “slice” any m-by-n matrix into row of n columns, or as a column / stack or m rows : A = [ A^1 . . . A^j . . . A^n ] = [ A 1 . . . A_i . . . A^ m ]^T . (1) In the rightmost side of (1) , [ . . . ]^T denotes the transposition operator which turns an m-by-n matrix into an n-by-m matrix (and convrsely, of course) ; and a row into a column / a column into a row. This way (1) to decompose any matrix in its columns / rows offer a ground for the last two of the above definitions. Def. #6 : The row subspace of the m-by-n matrix is the set of linear combinations of the m rows in A ; I use to denote it as ROWSP_ A . The row subspace of the m-by-n matrix is the set of linear combinations of the m columns in A ; I use to denote it as COLSP_ A . Note 3. It follows from the 3-rd and 4-th definitions of the rank, and also from one of the (nine) definitions of the basis of a vector space V / of a vector subspace W ⊆subsp V that Def. #5 is just the definition of each of the two bases that (respectively) span the subspaces ROWSP A and COLSP_ A . The notion of basis is one of the most important in the LINEAR ALGEBRA. It can be found in any textbook of LinAlg, this fundamental subdomain of Mathematics. I offer its formulation given, at page 84, in the best (most complete and rigorous among the American books in this field) due to the outstanding linear-algebraic author, Professor Gilbert STRANG from the Masachussetts Institute of Technology : 2I A basis for a vector space is a set f vectors having two properties at once : (1) It is linearly independent. (2) It spans the space. Two of the nine equivalent definition of this notion are : A basis A = [ a 1 a 2 . . . a _ n ] is the least family of vectors that span the space or the largest family of linearly independent in V / in the subspace W ⊆_subsp V . Note 4. I appreciate that the second and third of the above questions were simply superfluous. [ Gilbert STRANG, Linear Algebra nd Its Applications. Harcourt Brace Jovanovich College Publishers, International Edition ; Fort Worth ★ Philadelphia ★ . . . ★ Sydney ★ Tokyo, 1988 ] Upvote · Anurag Bansode 7y Related Where can we find the rank of a rectangular matrix as well as a square matrix? The rank of matrix is the maximum order of it's non-vanishing minor. The rank of matrix can be found by reducing it to Echelon/cannonical form 1 2 -1 3 3 4 0 -1 -1 0 -2 7 Perform row operation and try to make all element but 1st element of C1 to zero. 1 2 -1 3 0 -2 3 -10 0 2 -3 10 Now move to next non-zero number in R2 and make all number below it to zero 1 2 -1 3 0 -2 3 10 0 0 0 0 Here the last row has all element as zero hence by definition the Rank (r) is 2. Or By reducing it to normal form 2 3 -1 -1 1 -1 -2 -4 3 1 3 -2 6 3 0 7 you can use column operation in this Perform suitable operatio Continue Reading The rank of matrix is the maximum order of it's non-vanishing minor. The rank of matrix can be found by reducing it to Echelon/cannonical form 1 2 -1 3 3 4 0 -1 -1 0 -2 7 Perform row operation and try to make all element but 1st element of C1 to zero. 1 2 -1 3 0 -2 3 -10 0 2 -3 10 Now move to next non-zero number in R2 and make all number below it to zero 1 2 -1 3 0 -2 3 10 0 0 0 0 Here the last row has all element as zero hence by definition the Rank (r) is 2. Or By reducing it to normal form 2 3 -1 -1 1 -1 -2 -4 3 1 3 -2 6 3 0 7 you can use column operation in this Perform suitable operation and try to make the largest Identity matrix possible in the given matrix. The largest Identity matrix here is I3 so the answer/rank is 3. I hope this satisfies your query. Upvote · 9 2 Aelson Sobral MSc in Pure Mathematics, Federal University of Sergipe (UFS) (Graduated 2020) ·3y Related How do you know if a matrix is full rank? By definition, the rank of a matrix A is the dimension of the subspace generated by its rows (or columns, since they are equal). Therefore, a matrix of full rank is the one whose rank matches the minimum between the number of rows and columns. For square matrices, you simply calculate the determinant. Nonzero determinant implies a full rank matrix. For non-square matrices, however, it gets a little bit trickier. You can calculate the dimension of the subspace generated by the rows (or columns) and check if it matches the minimum between the number of rows and columns. This method bypasses the us Continue Reading By definition, the rank of a matrix A is the dimension of the subspace generated by its rows (or columns, since they are equal). Therefore, a matrix of full rank is the one whose rank matches the minimum between the number of rows and columns. For square matrices, you simply calculate the determinant. Nonzero determinant implies a full rank matrix. For non-square matrices, however, it gets a little bit trickier. You can calculate the dimension of the subspace generated by the rows (or columns) and check if it matches the minimum between the number of rows and columns. This method bypasses the usage of determinants. If you are reluctant and still want to use determinants, then you can do the following: let’s say you have a matrix of size AxB with A < B. Then, search for all square submatrices of size A and calculate the determinant. If you can find at least one submatrix with a nonzero determinant, then your matrix is of full rank. Hope this helps! Upvote · 9 1 9 1 Charan(The Ghost) Former Decision/Data Scientist at Genpact (company) (2019–2020) ·8y Related How can I find the rank of a matrix? Look at the matrix whether it is rectangular or square matrix. For rectangular matrix, if number of rows is less than number of columns then the rank of matrix wipll be equal to number of linearly independent rows.Similarly, If number of columns is less than number of rows then rank of matrix will be equal to number of linearly independent columns. For square matrix, number of linearly independent rows or columns is called rank of matrix.please consider the following matrix:Consider the bracket as square matrix..for all the rows. Y = [1 2 3 2 3 5 3 4 7 4 5 9] Now, You can see, column 1 and 2 are indep Continue Reading Look at the matrix whether it is rectangular or square matrix. For rectangular matrix, if number of rows is less than number of columns then the rank of matrix wipll be equal to number of linearly independent rows.Similarly, If number of columns is less than number of rows then rank of matrix will be equal to number of linearly independent columns. For square matrix, number of linearly independent rows or columns is called rank of matrix.please consider the following matrix:Consider the bracket as square matrix..for all the rows. Y = [1 2 3 2 3 5 3 4 7 4 5 9] Now, You can see, column 1 and 2 are independent because they are not derived form others. but column 3 (C1 + C2) is sum of column 1 and column 2. So there are two linearly independent columns hence its rank is 2. Upvote · 99 38 9 1 Vishakh Rajendran M.S. in Aerospace and Aeronautical Engineering, Nanyang Technological University · Author has 682 answers and 3M answer views ·5y Related What is the maximum rank of a matrix? Rank is the number of independent rows or Columns in the matrix. It is to be noted that however you calculate the rank (be it from rows point of view or Columns point of view), the rank of the matrix will be the same. So we just need to find out the maximum number of independent rows (or Columns) and this would be the rank of the matrix. For a n x n matrix (square matrix), the maximum value of the rank possible = n. For a 3 x 3 matrix, the maximum rank possible = 3 (provided all the three rows are independent). But if the matrix is not symmetric, say 2 x 3 matrix. In this case, there are 2 rows Continue Reading Rank is the number of independent rows or Columns in the matrix. It is to be noted that however you calculate the rank (be it from rows point of view or Columns point of view), the rank of the matrix will be the same. So we just need to find out the maximum number of independent rows (or Columns) and this would be the rank of the matrix. For a n x n matrix (square matrix), the maximum value of the rank possible = n. For a 3 x 3 matrix, the maximum rank possible = 3 (provided all the three rows are independent). But if the matrix is not symmetric, say 2 x 3 matrix. In this case, there are 2 rows and 3 Columns. So, as I said before, rank can be Computed from rows or Columns point of view and both will basically agree with each other. So the maximum rank will be which ever is the least (As there are two rows here and not more, the rank calculated from rows point of view would be 2 if both the rows is independent). So for a m x n matrix, the maximum value of the rank that is possible = Whichever is the Least of either of m or n. Upvote · 9 9 9 1 Related questions Where can we find the rank of a rectangular matrix as well as a square matrix? Can a3x4 matrix have rank equal to 3? How is the rank of a square or rectangular matrix determined and can it be calculated manually? How do you find the rank of a 2 by 3 matrix? How do I find the rank of a 3x2 matrix? Can the rank of a matrix be greater than 3? What are rectangular matrices? What is understood by the rank of the matrix? How do I find a matrix of rank one that is approximate to the square of a 22 matrix? How do you find the rank of a product of two matrices? Why is the determinant value for a rectangular matrix not possible? How do I find a rank of a 3x4 matrix using a minor method? What is the rank of a matrix of order 1×1? Is square matrix a rectangular matrix? How do you find the rank of a matrix sizable (linear algebra, matrices, matrix rank, and math)? Related questions Where can we find the rank of a rectangular matrix as well as a square matrix? Can a3x4 matrix have rank equal to 3? How is the rank of a square or rectangular matrix determined and can it be calculated manually? How do you find the rank of a 2 by 3 matrix? How do I find the rank of a 3x2 matrix? Can the rank of a matrix be greater than 3? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15518	https://www.collinsdictionary.com/us/dictionary/english-thesaurus/alleviate	Synonyms of ALLEVIATE \| Collins American English Thesaurus - [x] - [x] TRANSLATOR LANGUAGE GAMES SCHOOLS BLOG RESOURCES More [x] English Thesaurus [x] English English Dictionary English Thesaurus English Word Lists COBUILD English Usage [x] English Grammar Easy Learning Grammar COBUILD Grammar Patterns English Conjugations English Sentences [x] English ⇄ French English-French Dictionary French-English Dictionary Easy Learning French Grammar French Pronunciation Guide French Conjugations French Sentences [x] English ⇄ German English-German Dictionary German-English Dictionary Easy Learning German Grammar German Conjugations German Sentences [x] English ⇄ Italian English-Italian Dictionary Italian-English Dictionary Easy Learning Italian Grammar Italian Conjugations Italian Sentences [x] English ⇄ Spanish English-Spanish Dictionary Spanish-English Dictionary Easy Learning Spanish Grammar Easy Learning English Grammar in Spanish Spanish Pronunciation Guide Spanish Conjugations Spanish Sentences [x] English ⇄ Portuguese English-Portuguese Dictionary Portuguese-English Dictionary Easy Learning Portuguese Grammar Portuguese Conjugations [x] English ⇄ Hindi English-Hindi Dictionary Hindi-English Dictionary [x] English ⇄ Chinese English-Simplified Dictionary Simplified-English Dictionary English-Traditional Dictionary Chinese-Traditional Dictionary [x] English ⇄ Korean English-Korean Dictionary Korean-English Dictionary [x] English ⇄ Japanese English-Japanese Dictionary Japanese-English Dictionary English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese More [x] English Italiano Português 한국어 简体中文 Deutsch Español हिंदी 日本語 English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese DefinitionsSummarySynonymsSentencesPronunciationCollocationsConjugationsGrammar Credits × Synonyms of 'alleviate' in American English alleviate (verb)in the sense of ease Synonyms ease allay lessen lighten moderate reduce relieve soothe Synonyms of 'alleviate' in British English alleviate (verb)in the sense of ease Definition to lessen (pain or suffering) A great deal can be done to alleviate back pain. Synonyms ease I made her a hot water bottle to ease the pain. reduce Consumption is being reduced by 25 per cent. relieve Drugs can relieve much of the pain. moderate They are hoping that she will be persuaded to moderate her views. smooth dull They gave him morphine to dull the pain. diminish Federalism is intended to diminish the power of the central state. soften He could not think how to soften the blow of what he had to tell her. check blunt soothe Lemon tisanes with honey can soothe sore throats. subdue He forced himself to subdue and overcome his fears. lessen The burden will lessen if you ask someone for help. lighten He felt the need to lighten the atmosphere. quell He is trying to quell fears of a looming crisis. allay He did what he could to allay the children's fears. mitigate ways of mitigating the effects of an explosion abate a government programme to abate greenhouse gas emissions slacken Inflationary pressures continued to slacken last month. assuage She was trying to assuage her guilt. quench mollify slake palliate Certain drugs can palliate the main symptoms. See examples for synonyms seeameliorate Copyright © 2016 by HarperCollins Publishers. All rights reserved. Additional synonyms in the sense of abate Definition to make or become less strong a government programme to abate greenhouse gas emissions Synonyms reduce, slow, relax, ease, relieve, moderate, weaken, dull, diminish, decrease, lessen, alleviate, quell, mitigate, attenuate, slake in the sense of allay Definition to reduce (fear, doubt, or anger) He did what he could to allay the children's fears. Synonyms reduce, quiet, relax, ease, calm, smooth, relieve, check, moderate, dull, diminish, compose, soften, blunt, soothe, subdue, lessen, alleviate, appease, quell, mitigate, assuage, pacify, mollify in the sense of assuage Definition to relieve (grief, pain, or thirst) She was trying to assuage her guilt. Synonyms relieve, ease, calm, moderate, temper, soothe, lessen, alleviate, lighten, allay, mitigate, quench, palliate You may also like English Quiz Confusables English Word lists Latest Word Submissions English Grammar Grammar Patterns Language Lover's Blog Collins Scrabble The Paul Noble Method English Quiz Confusables English Word lists Latest Word Submissions English Grammar Grammar Patterns Language Lover's Blog Collins Scrabble The Paul Noble Method English Quiz Confusables English Word lists Latest Word Submissions Browse alphabetically alleviate allergic allergic to allergy alleviate alleviating alleviation alleviative All ENGLISH synonyms that begin with 'A' 123 Wordle Helper ------------- Scrabble Tools -------------- Quick word challenge Quiz Review Question: 1 Score: 0 / 5 SYNONYMS Select the synonym for: ambassador unionpacerecklessnessminister SYNONYMS Select the synonym for: to smile to beamto maintainto educateto toughen SYNONYMS Select the synonym for: clutter pacegadgethomedisorder SYNONYMS Select the synonym for: to teach to terrifyto coachto terminateto roar SYNONYMS Select the synonym for: environment scandalcurepayterritory Your score: Check See the answer Next Next quiz Review Study guides for every stage of your learning journey Whether you're in search of a crossword puzzle, a detailed guide to tying knots, or tips on writing the perfect college essay, Harper Reference has you covered for all your study needs. 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15519	https://www.johndcook.com/blog/2019/10/02/sturm-series/	Number of real roots in an interval Skip to content MATH PROBABILITY SIGNAL PROCESSING NUMERICAL COMPUTING SEE ALL … STATS EXPERT TESTIMONY WEB ANALYTICS FORECASTING RNG TESTING SEE ALL … PRIVACY HIPAA SAFE HARBOR CRYPTOGRAPHY DIFFERENTIAL PRIVACY PRIVACY FAQ WRITING BLOG RSS FEED TWITTER SUBSTACK ARTICLES TECH NOTES ABOUT CLIENTS ENDORSEMENTS TEAM SERVICES (832) 422-8646 Contact Number of real roots in an interval Posted on 2 October 2019 by John Suppose you have a polynomial p(x) and in interval [a, b] and you want to know how many distinct real roots the polynomial has in the interval. You can answer this question using Sturm’s algorithm. Let p 0(x) = p(x) and let p 1(x) be its derivative p‘(x). Then define a series of polynomials for i ≥ 1 p i+1(x) = − p i-1(x) mod p i(x) until you reach a constant. Here f mod g means the remainder when f is divided by g. This sequence of polynomials is called the Sturm series. Count the number of sign changes in the Sturm series at a and at b. Then the difference between these two counts is the number of distinct roots of p in the interval [a, b]. Example with Mathematica As an example, let’s look at the number of real zeros for p(x) = x 5 − x − c for some constant c. We’ll let Mathematica calculate our series. p0[x_] := x^5 - x - c p1[x_] := D[p0[x], x] p2[x_] := -PolynomialRemainder[p0[x], p1[x], x] p3[x_] := -PolynomialRemainder[p1[x], p2[x], x] This works out to p 0(x) = x 5 – x – c p 1(x) = 5 x 4 – 1 p 2(x) = (4/5)x + c p 3(x) = 1 – 3125 c 4/256 Now suppose c = 3 and we want to find out how many zeros p has in the interval [0, 2]. Evaluating our series at 0 we get −3, −1, 3, −3109/16. So the pattern is − − + −, i.e. two sign changes. Evaluating our series at 2 we get 2 7, 79, 23/5, −3109/16. So the pattern is + + + −, i.e. one sign change. This says x 5 − x − 3 has one real root between 0 and 2. By the way, you can multiply the polynomials in the sequence by any positive constant you like if that makes calculations easier. This multiplies subsequent polynomials by the same amount and doesn’t change the signs. Fine print Note that the algorithm counts distinct roots; multiple roots are counted once. You can let the end points of the interval be infinite, and so count all the real roots. I first tried using Mathematica’s PolynomialMod function, but PolynomialMod[5 x^4 - 1, 4 x/5 + c] gave the unexpected result 5 x 4 − 1. That’s because PolynomialMod does not let you specify what the variable is. It assumed that 4 x/5 + c was a polynomial in c. PolynomialRemainder is explicit about the variable, and that’s why the calls to this function above have x as the last argument. More posts on function roots Few coefficients, few roots Interlaced zeros of ODEs Categories : Math Bookmark the permalink Post navigation Previous PostTotal curvature of a knot Next PostFixed points 3 thoughts on “Number of real roots in an interval” Andrew 2 October 2019 at 12:18 The “Few coefficients, few roots” link is broken. John 2 October 2019 at 12:20 Thanks, Andrew. Just fixed it. BobC 2 October 2019 at 22:20 I’m presently reviewing all of high school math in preparation of taking my CSET Math examinations on my way to becoming a teacher (as my “retirement career” from engineering). Just covered root-finding Monday evening, so this is timely (though not directly relevant to my immediate goal). Still, I thought it must have an interesting proof, and sure ‘nuf it does: Comments are closed. Search for: John D. Cook, PhD My colleagues and I have decades of consulting experience helping companies solve complex problems involving data privacy, applied math, and statistics. Let’s talk. We look forward to exploring the opportunity to help your company too. John D. Cook © All rights reserved. Search for: (832) 422-8646 EMAIL
15520	https://math.stackexchange.com/questions/2598564/given-one-side-and-the-opposite-angle-how-to-show-the-area-is-a-maximum-when-th	calculus - Given one side and the opposite angle, how to show the area is a maximum when the triangle is isosceles? - 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 Given one side and the opposite angle, how to show the area is a maximum when the triangle is isosceles? Ask Question Asked 7 years, 8 months ago Modified7 years, 8 months ago Viewed 506 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. The problem is: If one side and the opposite angle of a triangle are fixed, and the other two sides are variable, use the law of cosines to show that the area is a maximum when the triangle is isosceles. I know that, if I let c c denote the fixed side and θ θ the angle opposite to c c, then area A=1/2 a b sin θ A=1/2 a b sin⁡θ, where a a and b b are the variable sides. I've tried substituting b b by the expression for it determined by the law of cosines, but that didn't seem to get me anywhere. calculus geometry trigonometry optimization triangles Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jan 9, 2018 at 18:09 Caleb Stanford 47.2k 8 8 gold badges 76 76 silver badges 178 178 bronze badges asked Jan 9, 2018 at 17:54 SasakiSasaki 1,219 1 1 gold badge 10 10 silver badges 16 16 bronze badges Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Fix the side of length c c. Let C C be the vertex of the triangle opposite to c c. The set of points that can be chosen for C C, forming a fixed angle θ θ, forms a circle, such that c c is a chord of that circle. (Actually it forms two circles, one on each side, but we can just take one WLOG.) The area of the triangle is 1 2 c h 1 2 c h. So the question is what point C C on this circle maximizes the height h h perpendicular to c c. Somewhat clearly, that would be the point C C such that the line tangent to the circle at C C is parallel to c c. It follows from this that the triangle is isosceles. Alternate approach: If you want to make your idea work, you would be trying to maximize (1 2 sin θ)a b(1 2 sin⁡θ)a b subject to the constraint a 2+b 2−2 a b cos θ=c a 2+b 2−2 a b cos⁡θ=c, for a fixed c c and θ θ. This is a Lagrange multipliers problem, with two variables a a and b b. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 9, 2018 at 18:12 answered Jan 9, 2018 at 18:06 Caleb StanfordCaleb Stanford 47.2k 8 8 gold badges 76 76 silver badges 178 178 bronze badges 3 I'm sorry, but why does angle θ θ remain constant as we move its vertex C C along the circle? That's the only thing I didn't understand. I looked it up on Google but could only find it stated as an Euclid's theorem here: math.stackexchange.com/q/1586946. Does that theorem have a name? Or can you link me to a proof? Thank you.Sasaki –Sasaki 2018-01-10 17:13:52 +00:00 Commented Jan 10, 2018 at 17:13 1 @Sasaki Try "the inscribed angle theorem" (see Wikipedia). Angle ACB is called an "inscribed angle" -- the theorem is that the measure of an inscribed angle is half of the measure of the central angle, no matter where the point C is located on the circle. The proof is on Wikipedia, or on this ProofWiki page, or you could search "inscribed angle theorem proof" for a number of other hits.Caleb Stanford –Caleb Stanford 2018-01-10 20:10:20 +00:00 Commented Jan 10, 2018 at 20:10 Also, Jack's image is nice.Caleb Stanford –Caleb Stanford 2018-01-10 20:10:42 +00:00 Commented Jan 10, 2018 at 20:10 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. If A B A B is fixed and A C B ˆ A C B^ is fixed, the circumradius of A B C A B C is fixed as well. If C C travels on the (major, or minor) A B A B-arc of a fixed circle Γ Γ, the distance of C C from A B A B, hence the area of A B C A B C, is clearly maximized when C C lies on the perpendicular bisector of A B A B, hence when A B C A B C is isosceles with respect to the base A B A B. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 9, 2018 at 19:28 Jack D'AurizioJack D'Aurizio 372k 42 42 gold badges 419 419 silver badges 886 886 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Let a a be a fixed side and α α be a fixed angle of Δ A B C Δ A B C. Thus, in the standard notation we obtain: S Δ A B C=1 2 b c sin α=2 R 2 sin α sin β sin γ=S Δ A B C=1 2 b c sin⁡α=2 R 2 sin⁡α sin⁡β sin⁡γ= =R 2 sin α(cos(β−γ)−cos(β+γ))≤R 2 sin α(1+cos α).=R 2 sin⁡α(cos⁡(β−γ)−cos⁡(β+γ))≤R 2 sin⁡α(1+cos⁡α). The equality occurs for cos(β−γ)=1 cos⁡(β−γ)=1, which gives β=γ β=γ and we are done! Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 9, 2018 at 18:05 Michael RozenbergMichael Rozenberg 208k 31 31 gold badges 171 171 silver badges 294 294 bronze badges 2 I understand it up to the inequality sign. Why is the LHS less than or equal to the RHS? Besides, I understand that them being equal implies an isosceles triangle, but I can't quite see how that proves what is to be proven. Maybe there's some geometric interpretation to those quantities that I'm missing out on? Thanks very much.Sasaki –Sasaki 2018-01-09 22:52:54 +00:00 Commented Jan 9, 2018 at 22:52 Because cos(β−γ)≤1 cos⁡(β−γ)≤1 and −cos(β+γ)=−cos(180∘−α)=cos α−cos⁡(β+γ)=−cos⁡(180∘−α)=cos⁡α. Also, we got that R 2 sin α(1+cos α)R 2 sin⁡α(1+cos⁡α) depends on α α and a a.Michael Rozenberg –Michael Rozenberg 2018-01-10 02:21:38 +00:00 Commented Jan 10, 2018 at 2:21 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. hint Do not forget the useful equality a sin(α)=b sin(β)=c sin(γ)a sin⁡(α)=b sin⁡(β)=c sin⁡(γ) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 9, 2018 at 18:06 hamam_Abdallahhamam_Abdallah 63.8k 4 4 gold badges 29 29 silver badges 50 50 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I you use the extended version of the sine rule, with a a the side, A A the opposite angle and R R the circumradius, you find that 2 R=a sin A 2 R=a sin⁡A, so that the third vertex of the triangle lies on a fixed circle. The centre of the circle lies on the perpendicular bisector of the side you have been given (the circle passes through the two vertices you already have) and the greatest possible area is when the vertex also lies on the bisector of this side (greatest height on fixed base). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 9, 2018 at 18:12 Mark BennetMark Bennet 102k 14 14 gold badges 119 119 silver badges 232 232 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 calculus geometry trigonometry optimization triangles 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 0How can I solve a triangle knowing its area, one side and an opposite angle? 1Isosceles triangle of maximum area, given the length of a median to one of its equal sides 2maximizing the area of a triangle with given angle and opposite side 0How to find triangle with side lengths: sin θ,sin θ−−−−√sin⁡θ,sin⁡θ and sin θ−−−−√sin⁡θ 1Find the maximum area of a triangle given angle and opposite side 2If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles. 3When solving an SAS triangle, why do you have to use the law of sines to find the angle opposite of the shortest remaining side? 1Given one side and its opposite angle of a triangle, find its maximum and minimum perimeter 2If the side lengths of a triangle increase and the third side is fixed, the opposite angle decreases 1Angle between the median and altitude to one side of an isosceles triangle Hot Network Questions ICC in Hague not prosecuting an individual brought before them in a questionable manner? 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15521	https://math.stackexchange.com/questions/4980651/prove-the-alternate-segment-theorem-without-using-radius-at-right-angle-to-tange	geometry - Prove the Alternate Segment Theorem without using Radius at Right Angle to Tangent - 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 the Alternate Segment Theorem without using Radius at Right Angle to Tangent Ask Question Asked 11 months ago Modified11 months ago Viewed 217 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. The Alternate Segment Theorem, or Tangent-Chord Theorem, is formulated as follows: The angle between a tangent and a chord is equal to the angle subtended by the same chord in the alternate segment. Its proof of which I know is located here, and it uses the Right Angle to Tangent of Circle goes through Center, which I would like to avoid if possible. I would also like to avoid trigonometry, triangle similarity, as well as areas. The perfect proof would be involving the Inscribed Angle Theorem and other basics. Is there a possible proof matching my criteria? I have the following idea of a possible candidate for a proof. Construct a circle with center O O. Mark a point T T on that circle and draw tangent M N M N with T T being the tangent point. Points M M and N N should be chosen in such way that T T lies between M M and N N. Choose an arbitrary point P≢T P≢T on that same circle. Then we can assume without loss of generality that angle α=∠P T N α=∠P T N is an acute angle between tangent M N M N and chord P T P T. Construct a line parallel to M N M N through point P P. Since P P is a point on a circle not coincident with P P, there are two possible cases: The constructed line is a tangent through point P P. The constructed line is a secant through point P P. Let's examine the latter case. Then there exists another point Q≢P Q≢P where this line will intersect the circle. Then ∠T P Q=∠P T N=α∠T P Q=∠P T N=α as alternate interior angles between M N∥P Q M N∥P Q. By Inscribed Angle Theorem, we have ∠T P Q=1 2 Q T⌣∠T P Q=1 2 Q T⌣. How do I proceed from here? One idea I have is to construct a ray Q T Q T, mark a point K K on that ray so that Q T<Q K Q T<Q K, and then prove that T N T N bisects angle ∠P T K∠P T K. If so, it follows that triangle T P Q T P Q is isosceles, and P T⌣=Q T⌣P T⌣=Q T⌣, which proves the statement. geometry euclidean-geometry circles alternative-proof angle Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Oct 11, 2024 at 9:26 RusuranoRusurano asked Oct 5, 2024 at 17:38 RusuranoRusurano 1,076 6 6 silver badges 14 14 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Consider a point B′B′ on the circle near to B B, and another chord A D A D as in figure below. By the inscribed angle theorem we have: ∠B′A D=∠D B F.∠B′A D=∠D B F. In the limit B′→B B′→B line B F B F becomes tangent at B B and by the above equality the angle ∠D B F∠D B F formed by this tangent with B D B D is equal to angle ∠B A D∠B A D, as we wanted to prove. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 9, 2024 at 21:27 answered Oct 5, 2024 at 20:34 Intelligenti paucaIntelligenti pauca 56.2k 4 4 gold badges 50 50 silver badges 91 91 bronze badges 2 Are limits even defined in Euclidean geometry? The approach is interesting for sure, though.Rusurano –Rusurano 2024-10-05 21:02:14 +00:00 Commented Oct 5, 2024 at 21:02 4 @Rusurano They were widely used in geometry from XVII to early XX century.Intelligenti pauca –Intelligenti pauca 2024-10-05 21:12:20 +00:00 Commented Oct 5, 2024 at 21:12 Add a comment\| This answer is useful 0 Save this answer. +50 This answer has been awarded bounties worth 50 reputation by Rusurano Show activity on this post. In the fig., let ∠C B H=∠1∠C B H=∠1 ∠D B C=∠2∠D B C=∠2 ∠B D C=∠3∠B D C=∠3 ∠B E C=∠4∠B E C=∠4 ∠G B C=∠5∠G B C=∠5 ∠B F C=∠6∠B F C=∠6 Now, ∠1+∠2=90 0∠1+∠2=90 0 Also, ∠2+∠3=90 0∠2+∠3=90 0 ⟹∠1=∠3⟹∠1=∠3 also, ∠3=∠4∠3=∠4 ⟹∠1=∠4⟹∠1=∠4 Now, ∠1+∠5=180 0∠1+∠5=180 0 Also, P A T B P A T B is a cyclic quadrilateral. ⟹∠4+∠=180 0⟹∠4+∠=180 0⟹∠5=∠6⟹∠5=∠6 QED Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 15, 2024 at 9:10 answered Oct 13, 2024 at 13:10 KnightRiderDuttKnightRiderDutt 552 1 1 silver badge 10 10 bronze badges 2 There is no point M M in your diagram, so I have no idea what is angle ∠B A M∠B A M.Rusurano –Rusurano 2024-10-13 21:49:04 +00:00 Commented Oct 13, 2024 at 21:49 @Rusurano Notation mistake, I have corrected it KnightRiderDutt –KnightRiderDutt 2024-10-15 09:13:24 +00:00 Commented Oct 15, 2024 at 9:13 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry euclidean-geometry circles alternative-proof angle 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 3proof of the fact that the set of points equidistant from sides of an angle form a bisector of the angle 1Problem about circumscribed circle of triangle 1Two circles intersect in A,where two tangents generate different chords. Equality of two of them? 2Prove that the center of a circle within a constructed triangles lies on the angle bisector 6Congruence of constructed circle chords 3Length of a tangent segment to the vertex of the circumscribed angle 3Proving that a smaller inscribed, tangent circle to another circle makes an angle bisector of the given angle, joining the two tangent points 2Given three points on a line, construct a circle through two of these points and a certain tangent through the third point 0A proof that all segments are equal. How can this be? Hot Network Questions Riffle a list of binary functions into list of arguments to produce a result "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf Weird utility function How to fix my object in animation Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? 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15522	https://openstax.org/books/microbiology/pages/1-3-types-of-microorganisms	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in Microbiology 1.3 Types of Microorganisms Microbiology 1.3 Types of Microorganisms Search for key terms or text. Learning Objectives By the end of this section, you will be able to: List the various types of microorganisms and describe their defining characteristics Give examples of different types of cellular and viral microorganisms and infectious agents Describe the similarities and differences between archaea and bacteria Provide an overview of the field of microbiology Most microbes are unicellular and small enough that they require artificial magnification to be seen. However, there are some unicellular microbes that are visible to the naked eye, and some multicellular organisms that are microscopic. An object must measure about 100 micrometers (µm) to be visible without a microscope, but most microorganisms are many times smaller than that. For some perspective, consider that a typical animal cell measures roughly 10 µm across but is still microscopic. Bacterial cells are typically about 1 µm, and viruses can be 10 times smaller than bacteria (Figure 1.12). See Table 1.1 for units of length used in microbiology. Figure 1.12 The relative sizes of various microscopic and nonmicroscopic objects. Note that a typical virus measures about 100 nm, 10 times smaller than a typical bacterium (~1 µm), which is at least 10 times smaller than a typical plant or animal cell (~10–100 µm). An object must measure about 100 µm to be visible without a microscope. \| Units of Length Commonly Used in Microbiology \| \| \| --- \| Metric Unit \| Meaning of Prefix \| Metric Equivalent \| \| meter (m) \| — \| 1 m = 100 m \| \| decimeter (dm) \| 1/10 \| 1 dm = 0.1 m = 10−1 m \| \| centimeter (cm) \| 1/100 \| 1 cm = 0.01 m = 10−2 m \| \| millimeter (mm) \| 1/1000 \| 1 mm = 0.001 m = 10−3 m \| \| micrometer (μm) \| 1/1,000,000 \| 1 μm = 0.000001 m = 10−6 m \| \| nanometer (nm) \| 1/1,000,000,000 \| 1 nm = 0.000000001 m = 10−9 m \| Table 1.1 Microorganisms differ from each other not only in size, but also in structure, habitat, metabolism, and many other characteristics. While we typically think of microorganisms as being unicellular, there are also many multicellular organisms that are too small to be seen without a microscope. Some microbes, such as viruses, are even acellular (not composed of cells). Microorganisms are found in each of the three domains of life: Archaea, Bacteria, and Eukarya. Microbes within the domains Bacteria and Archaea are all prokaryotes (their cells lack a nucleus), whereas microbes in the domain Eukarya are eukaryotes (their cells have a nucleus). Some microorganisms, such as viruses, do not fall within any of the three domains of life. In this section, we will briefly introduce each of the broad groups of microbes. Later chapters will go into greater depth about the diverse species within each group. Link to Learning How big is a bacterium or a virus compared to other objects? Check out this interactive website to get a feel for the scale of different microorganisms. Prokaryotic Microorganisms Bacteria are found in nearly every habitat on earth, including within and on humans. Most bacteria are harmless or helpful, but some are pathogens, causing disease in humans and other animals. Bacteria are prokaryotic because their genetic material (DNA) is not housed within a true nucleus. Most bacteria have cell walls that contain peptidoglycan. Bacteria are often described in terms of their general shape. Common shapes include spherical (coccus), rod-shaped (bacillus), or curved (spirillum, spirochete, or vibrio). Figure 1.13 shows examples of these shapes. Figure 1.13 Common bacterial shapes. Note how coccobacillus is a combination of spherical (coccus) and rod-shaped (bacillus). (credit “Coccus”: modification of work by Janice Haney Carr, Centers for Disease Control and Prevention; credit “Coccobacillus”: modification of work by Janice Carr, Centers for Disease Control and Prevention; credit “Spirochete”: Centers for Disease Control and Prevention) They have a wide range of metabolic capabilities and can grow in a variety of environments, using different combinations of nutrients. Some bacteria are photosynthetic, such as oxygenic cyanobacteria and anoxygenic green sulfur and green nonsulfur bacteria; these bacteria use energy derived from sunlight, and fix carbon dioxide for growth. Other types of bacteria are nonphotosynthetic, obtaining their energy from organic or inorganic compounds in their environment. Archaea are also unicellular prokaryotic organisms. Archaea and bacteria have different evolutionary histories, as well as significant differences in genetics, metabolic pathways, and the composition of their cell walls and membranes. Unlike most bacteria, archaeal cell walls do not contain peptidoglycan, but their cell walls are often composed of a similar substance called pseudopeptidoglycan. Like bacteria, archaea are found in nearly every habitat on earth, even extreme environments that are very cold, very hot, very basic, or very acidic (Figure 1.14). Some archaea live in the human body, but none have been shown to be human pathogens. Figure 1.14 Some archaea live in extreme environments, such as the Morning Glory pool, a hot spring in Yellowstone National Park. The color differences in the pool result from the different communities of microbes that are able to thrive at various water temperatures. Check Your Understanding What are the two main types of prokaryotic organisms? Name some of the defining characteristics of each type. Eukaryotic Microorganisms The domain Eukarya contains all eukaryotes, including uni- or multicellular eukaryotes such as protists, fungi, plants, and animals. The major defining characteristic of eukaryotes is that their cells contain a nucleus. Protists Protists are an informal grouping of eukaryotes that are not plants, animals, or fungi. Some algae are protists and others are bacteria; all protozoa are examples of protists. Algae (singular: alga) are mostly made up of protists that can be either unicellular or multicellular and vary widely in size, appearance, and habitat (Figure 1.15). Algal protists are surrounded by cell walls made of cellulose, a type of carbohydrate. Algae are photosynthetic organisms that extract energy from the sun and release oxygen and carbohydrates into their environment. Cyanobacteria, a type of bacteria, is also considered an algae, but these organisms are bacterial prokaryotes and therefore have a peptidoglycan-based cell wall, unlike the cellulose-based cell wall of the algal protists. Because other organisms can use the waste products of all algae for energy, algae are important parts of many ecosystems. Many consumer products contain ingredients derived from algae, such as carrageenan or alginic acid, which are found in some brands of ice cream, salad dressing, beverages, lipstick, and toothpaste. A derivative of algae also plays a prominent role in the microbiology laboratory. Agar, a gel derived from algae, can be mixed with various nutrients and used to grow microorganisms in a Petri dish. Algae are also being developed as a possible source for biofuels. Figure 1.15 Assorted diatoms, a kind of algae, live in annual sea ice in McMurdo Sound, Antarctica. Diatoms range in size from 2 μm to 200 μm and are visualized here using light microscopy. (credit: modification of work by National Oceanic and Atmospheric Administration) Protozoa (singular: protozoan) are protists that make up the backbone of many food webs by providing nutrients for other organisms. Protozoa are very diverse. Some protozoa move with help from hair-like structures called cilia or whip-like structures called flagella. Others extend part of their cell membrane and cytoplasm to propel themselves forward. These cytoplasmic extensions are called pseudopods (“false feet”). Some protozoa are photosynthetic; others feed on organic material. Some are free-living, whereas others are parasitic, only able to survive by extracting nutrients from a host organism. Most protozoa are harmless, but some are pathogens that can cause disease in animals or humans (Figure 1.16). Figure 1.16 Giardia lamblia, an intestinal protozoan parasite that infects humans and other mammals, causing severe diarrhea. (credit: modification of work by Centers for Disease Control and Prevention) Fungi Fungi (singular: fungus) are also eukaryotes. Some multicellular fungi, such as mushrooms, resemble plants, but they are actually quite different. Fungi are not photosynthetic, and their cell walls are usually made out of chitin rather than cellulose. Unicellular fungi—yeasts—are included within the study of microbiology. There are more than 1000 known species. Yeasts are found in many different environments, from the deep sea to the human navel. Some yeasts have beneficial uses, such as causing bread to rise and beverages to ferment; but yeasts can also cause food to spoil. Some even cause diseases, such as vaginal yeast infections and oral thrush (Figure 1.17). Figure 1.17 Candida albicans is a unicellular fungus, or yeast. It is the causative agent of vaginal yeast infections as well as oral thrush, a yeast infection of the mouth that commonly afflicts infants. C. albicans has a morphology similar to that of coccus bacteria; however, yeast is a eukaryotic organism (note the nuclei) and is much larger. (credit: modification of work by Centers for Disease Control and Prevention) Other fungi of interest to microbiologists are multicellular organisms called molds. Molds are made up of long filaments that form visible colonies (Figure 1.18). Molds are found in many different environments, from soil to rotting food to dank bathroom corners. Molds play a critical role in the decomposition of dead plants and animals. Some molds can cause allergies, and others produce disease-causing metabolites called mycotoxins. Molds have been used to make pharmaceuticals, including penicillin, which is one of the most commonly prescribed antibiotics, and cyclosporine, used to prevent organ rejection following a transplant. Figure 1.18 Large colonies of microscopic fungi can often be observed with the naked eye, as seen on the surface of these moldy oranges. Check Your Understanding Name two types of protists and two types of fungi. Name some of the defining characteristics of each type. Helminths Multicellular parasitic worms called helminths are not technically microorganisms, as most are large enough to see without a microscope. However, these worms fall within the field of microbiology because diseases caused by helminths involve microscopic eggs and larvae. One example of a helminth is the guinea worm, or Dracunculus medinensis, which causes dizziness, vomiting, diarrhea, and painful ulcers on the legs and feet when the worm works its way out of the skin (Figure 1.19). Infection typically occurs after a person drinks water containing water fleas infected by guinea-worm larvae. In the mid-1980s, there were an estimated 3.5 million cases of guinea-worm disease, but the disease has been largely eradicated. In 2014, there were only 126 cases reported, thanks to the coordinated efforts of the World Health Organization (WHO) and other groups committed to improvements in drinking water sanitation.1011 Figure 1.19 (a) The beef tapeworm, Taenia saginata, infects both cattle and humans. T. saginata eggs are microscopic (around 50 µm), but adult worms like the one shown here can reach 4–10 m, taking up residence in the digestive system. (b) An adult guinea worm, Dracunculus medinensis, is removed through a lesion in the patient’s skin by winding it around a matchstick. (credit a, b: modification of work by Centers for Disease Control and Prevention) Viruses Viruses are acellular microorganisms, which means they are not composed of cells. Essentially, a virus consists of proteins and genetic material—either DNA or RNA, but never both—that are inert outside of a host organism. However, by incorporating themselves into a host cell, viruses are able to co-opt the host’s cellular mechanisms to multiply and infect other hosts. Viruses can infect all types of cells, from human cells to the cells of other microorganisms. In humans, viruses are responsible for numerous diseases, from the common cold to deadly Ebola (Figure 1.20). However, many viruses do not cause disease. Figure 1.20 (a) Members of the Coronavirus family can cause respiratory infections like the common cold, severe acute respiratory syndrome (SARS), and Middle East respiratory syndrome (MERS). Here they are viewed under a transmission electron microscope (TEM). (b) Ebolavirus, a member of the Filovirus family, as visualized using a TEM. (credit a: modification of work by Centers for Disease Control and Prevention; credit b: modification of work by Thomas W. Geisbert) Check Your Understanding Are helminths microorganisms? Explain why or why not. How are viruses different from other microorganisms? Microbiology as a Field of Study Microbiology is a broad term that encompasses the study of all different types of microorganisms. But in practice, microbiologists tend to specialize in one of several subfields. For example, bacteriology is the study of bacteria; mycology is the study of fungi; protozoology is the study of protozoa; parasitology is the study of helminths and other parasites; and virology is the study of viruses (Figure 1.21). Immunology, the study of the immune system, is often included in the study of microbiology because host–pathogen interactions are central to our understanding of infectious disease processes. Microbiologists can also specialize in certain areas of microbiology, such as clinical microbiology, environmental microbiology, applied microbiology, or food microbiology. In this textbook, we are primarily concerned with clinical applications of microbiology, but since the various subfields of microbiology are highly interrelated, we will often discuss applications that are not strictly clinical. Figure 1.21 (a) A virologist samples eggs from this nest to be tested for the influenza A virus, which causes avian flu in birds. (b) A biologist performs a procedure to identify an organism that causes ulcerations in humans (credit a: U.S. Fish and Wildlife Service; credit b: James Gathany / CDC; Public Domain) Eye on Ethics Bioethics in Microbiology In the 1940s, the U.S. government was looking for a solution to a medical problem: the prevalence of sexually transmitted diseases (STDs) among soldiers. Several now-infamous government-funded studies used human subjects to research common STDs and treatments. In one such study, American researchers intentionally exposed more than 1300 human subjects in Guatemala to syphilis, gonorrhea, and chancroid to determine the ability of penicillin and other antibiotics to combat these diseases. Subjects of the study included Guatemalan soldiers, prisoners, prostitutes, and psychiatric patients—none of whom were informed that they were taking part in the study. Researchers exposed subjects to STDs by various methods, from facilitating intercourse with infected prostitutes to inoculating subjects with the bacteria known to cause the diseases. This latter method involved making a small wound on the subject’s genitals or elsewhere on the body, and then putting bacteria directly into the wound.12 In 2011, a U.S. government commission tasked with investigating the experiment revealed that only some of the subjects were treated with penicillin, and 83 subjects died by 1953, likely as a result of the study.13 Unfortunately, this is one of many horrific examples of microbiology experiments that have violated basic ethical standards. Even if this study had led to a life-saving medical breakthrough (it did not), few would argue that its methods were ethically sound or morally justifiable. But not every case is so clear cut. Professionals working in clinical settings are frequently confronted with ethical dilemmas, such as working with patients who decline a vaccine or life-saving blood transfusion. These are just two examples of life-and-death decisions that may intersect with the religious and philosophical beliefs of both the patient and the health-care professional. No matter how noble the goal, microbiology studies and clinical practice must be guided by a certain set of ethical principles. Studies must be done with integrity. Patients and research subjects provide informed consent (not only agreeing to be treated or studied but demonstrating an understanding of the purpose of the study and any risks involved). Patients’ rights must be respected. Procedures must be approved by an institutional review board. When working with patients, accurate record-keeping, honest communication, and confidentiality are paramount. Animals used for research must be treated humanely, and all protocols must be approved by an institutional animal care and use committee. These are just a few of the ethical principles explored in the Eye on Ethics boxes throughout this book. Clinical Focus Resolution Cora’s CSF samples show no signs of inflammation or infection, as would be expected with a viral infection. However, there is a high concentration of a particular protein, 14-3-3 protein, in her CSF. An electroencephalogram (EEG) of her brain function is also abnormal. The EEG resembles that of a patient with a neurodegenerative disease like Alzheimer’s or Huntington’s, but Cora’s rapid cognitive decline is not consistent with either of these. Instead, her doctor concludes that Cora has Creutzfeldt-Jakob disease (CJD), a type of transmissible spongiform encephalopathy (TSE). CJD is an extremely rare disease, with only about 300 cases in the United States each year. It is not caused by a bacterium, fungus, or virus, but rather by prions—which do not fit neatly into any particular category of microbe. Like viruses, prions are not found on the tree of life because they are acellular. Prions are extremely small, about one-tenth the size of a typical virus. They contain no genetic material and are composed solely of a type of abnormal protein. CJD can have several different causes. It can be acquired through exposure to the brain or nervous-system tissue of an infected person or animal. Consuming meat from an infected animal is one way such exposure can occur. There have also been rare cases of exposure to CJD through contact with contaminated surgical equipment14 and from cornea and growth-hormone donors who unknowingly had CJD.1516 In rare cases, the disease results from a specific genetic mutation that can sometimes be hereditary. However, in approximately 85% of patients with CJD, the cause of the disease is spontaneous (or sporadic) and has no identifiable cause.17 Based on her symptoms and their rapid progression, Cora is diagnosed with sporadic CJD. Unfortunately for Cora, CJD is a fatal disease for which there is no approved treatment. Approximately 90% of patients die within 1 year of diagnosis.18 Her doctors focus on limiting her pain and cognitive symptoms as her disease progresses. Eight months later, Cora dies. Her CJD diagnosis is confirmed with a brain autopsy. Go back to the previous Clinical Focus box. Footnotes 10C. Greenaway “Dracunculiasis (Guinea Worm Disease).” Canadian Medical Association Journal 170 no. 4 (2004):495–500. 11World Health Organization. “Dracunculiasis (Guinea-Worm Disease).” WHO. 2015. Accessed October 2, 2015. 12Kara Rogers. “Guatemala Syphilis Experiment: American Medical Research Project”. Encylopaedia Britannica. Accessed June 24, 2015. 13Susan Donaldson James. “Syphilis Experiments Shock, But So Do Third-World Drug Trials.” ABC World News. August 30, 2011. Accessed June 24, 2015. 14Greg Botelho. “Case of Creutzfeldt-Jakob Disease Confirmed in New Hampshire.” CNN. 2013. 15P. Rudge et al. “Iatrogenic CJD Due to Pituitary-Derived Growth Hormone With Genetically Determined Incubation Times of Up to 40 Years.” Brain 138 no. 11 (2015): 3386–3399. 16J.G. Heckmann et al. “Transmission of Creutzfeldt-Jakob Disease via a Corneal Transplant.” Journal of Neurology, Neurosurgery & Psychiatry 63 no. 3 (1997): 388–390. 17National Institute of Neurological Disorders and Stroke. “Creutzfeldt-Jakob Disease Fact Sheet.” NIH. 2015. 18National Institute of Neurological Disorders and Stroke. “Creutzfeldt-Jakob Disease Fact Sheet.” NIH. 2015. Accessed June 22, 2015. 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15523	https://www.youtube.com/watch?v=eDFV0YUgmbU	MATHFLEX Ep167: Integral of (1/u)du Ryco Ricks 1260 subscribers 5 likes Description 325 views Posted: 13 Mar 2023 mathflex matholympiadquestion integralcalculus1 2 comments Transcript: so let's have this uh this or another set of integrals here so I just want to show you here that the power rule uh would not be applicable in some of the cases so let's super sample this one so if you have the integral of 1 over X DX so if you will apply power rule here so say for example this one if I will rewrite 1 over X DX so this will be X raised to negative one the X so as we have said if you will apply power rules so they develop U raised to n d n this will be U raised to n plus 1 over n plus one plus C so in this case if you have negative 1 as your exponent so if you will add one there so it would be X raised to zero over negative one plus one that's zero plus C so you have any number is zero is one over zero plus C so any number divided by zero we turn them as undefined okay so that can be possible so in other words the power rule would not be applicable on integral or the form of 1 over X DX so in short if you have one over or the integral of one over u v u so instead the answer here is long okay the absolute value of U plus C okay so in this case you will have here now line the absolute value of x plus C that's our final answer okay let's try the second one so the second one is simply in this form so you just have to take out one over three outside and then you have again 1 over x p x so it's simply the answer here is one always be long of your absolute value of x plus C okay so this is our answer again so the third one is a bit okay we put a bit Twist on it so 1 over X DX so how we can transform this one in the form of the integral of 1 over u d u okay so by substitution again so we can have like let Q is equal to two X Plus One D U that is equal to 2 DX so your Therefore your DX is d u over two okay so therefore if I'll try to substitute that one this is one half okay of your 1 over U view okay then the answer here is one over long one-half long and back substitute your U absolute value of your two X plus one plus C so this is our answer okay so this is a quick introduction of our more complicated lessons on our succeeding videos till next time guys bye thank you [Music]
15524	https://www.netlib.org/voronoi/hull.html	hull - convex hulls, Delaunay triangulations, alpha shapes A program for convex hulls Summary Source Code Examples of use Synopsis Description Bugs/Portability Author Algorithms Function call Summary =======Hull is an ANSI C program that computes the convex hull of a point set in general (but small!) dimension. The input is a list of points, and the output is a list of facets of the convex hull of the points, each facet presented as a list of its vertices. (The facets are assumed to be simplices, such as triangles in 3d; this is enforced by tiebreaking, giving a triangulation of a facet by "placing".) The program can also compute Delaunay triangulations and alpha shapes, and volumes of Voronoi regions. The program uses exact arithmetic when possible, with a moderate speed penalty. (Typically a factor of 2 or 3 for Delaunay triangulation, less for convex hulls). Output in postscript and OFF format for geomview is supported. This document is an html version of the manpage, with links to papers describing the algorithm. At long last, I've made the code a bit more portable: it compiles under gcc with the provided makefile, and I've included Visual Studio 7 solution and project files to build it on Windows (as a console application). I'm told it also compiles and runs fine under Solaris. clarkson@research.bell-labs.com. Source Code =========== Examples of use =============== Synopsis ========hull -d -f -A -aa -af -oN -ov -s -r -m -X -i -oF Description =========== The input_file (default stdin) is a sequence of points (also called sites), separated by \n; a d-dimensional point is specified as a group of d floats separated by whitespace (other than \n). The output_file (default stdout) gives d-tuples of the input points, where a point is given as an integer i if it was the i'th point in the input_file. If the convex hull lies in a flat (affine subspace) of dimension d', the output will comprise a list of d'-tuples, the vertices of the convex hull relative to that flat. The output tuples represent the facets of the convex hull of the input set. A facet which is not a simplex is output implicitly as the collection of simplices of its triangulation. The output for Delaunay triangulation includes a "point at infinity", numbered -1; facets including it correspond to facets of the convex hull of the sites. Some chatter will appear on debug_file (default stderr). The coordinates of the input points are rounded to integers. With -m , the coordinates are multiplied by multiplier before this rounding. The program attempts to use a method that gives exact answers to numerical tests; in some circumstances, this may fail, and with some warnings, the program continues, using approximate arithmetic. More detail on the options: -d compute the Delaunay triangulation of the input set. -f give the main output (convex hull or Delaunay triangulation) in output , which is by default the list of vertex numbers described above, or ps, for postscript output of planar points, or off, for OFF output of 3d points. -aa compute the alpha shape using parameter < alpha >: the output is the set of all d-tuples of sites such that there is a ball of radius < alpha > containing those sites on its bounding sphere, and containing no other sites. A Delaunay triangulation computation is implied by this option and by -A. -af output the alpha shape in format , as in option -f. -A compute the alpha shape of the input, finding the smallest alpha so that the sites are all contained in the alpha-shape. -r randomly shuffle the input points; this may speed up the program. -s randomly shuffle using < seed > for the random number generator. -oN do not produce main output (convex hull or Delaunay triangulation). If you want an alpha shape only, you need this to turn off the Delaunay output. -ov Give volumes of Voronoi regions of input sites, and in general d'-volumes of d'-dimensional Voronoi cells. Implies -d. Bugs/Portability ================ If the convex hull is a single point, the algorithm will fail to report it. All other degeneracies should be handled. Tie-breaking is done so that all reported facets are simplices. If the input points are degenerate, some hull facets may be; for example, some Delaunay simplices may have zero volume. Determining non-simplicial facets or deleting zero-volume Delaunauy simplices could be done in post-processing (not implemented). The file rand.c includes calls to pseudo-random number generators; No simplices are deleted; the only way to free storage is to free it all using free_hull_storage. Author ======Ken Clarkson,(clarkson@research.bell-labs.com), using an earlier version written by Susan Dorward, who is not to blame. Algorithms ========== Combinatorial The program implements an incremental algorithm, which can be randomized if desired; the algorithm is roughly as described in Section 3 (page 7) of K. L.Clarkson, K.Mehlhorn, and R.Seidel. Four results on randomized incremental constructions. Comp. Geom.: Theory and Applications, pages 185--121, 1993. Preliminary version in Proc. Symp. Theor. Aspects of Comp. Sci., 1992. The code is an implementation of the "all-visibilities" search procedure. The "anti-origin" point of that discussion is represented as a NULL point in the code, so that simplex s represents a facet iff s->peak.vert is NULL. Numerical The numerical code, for computing normals to facets and so on, is an implementation of K. L.Clarkson. Safe and effective determinant evaluation. In Proc. 31st IEEE Symposium on Foundations of Computer Science, pages 387--395, Pittsburgh, PA, October 1992. The conditioning procedure of that paper is used, and applied to finding a normal vector to a facet of the convex hull. The visibility test is then a dot product. Vectors a_1, a_2... are obtained from points p_0, p_1,...by translating p_0 to the origin: a_i = p_i-p_0, for i=1..cdim. Function call ============= The main function to call is simplex build_convex_hull(gsitef get_s, site_n site_numm, short dim, short vdd) returns pointer to root simplex of triangulation of convex hull (or Delaunay triangulation if vdd==1). see hull.h for gsitef typedef etc. get_s returns next site each call; hull construction stops when NULL returned; site_numm returns number of site when given site; dim dimension of point set; vdd if (vdd) then return Delaunay triangulation, otherwise convex hull;
15525	https://math.stackexchange.com/questions/3129901/show-that-the-function-fx-x-x2-1-is-bijective	real analysis - Show that the function $f(x)=x/(x^2-1)$ is bijective. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Show that the function f(x)=x/(x 2−1)f(x)=x/(x 2−1) is bijective. Ask Question Asked 6 years, 7 months ago Modified4 years, 7 months ago Viewed 8k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. I have to show that the function f:(−1,1)→R f:(−1,1)→R f(x)=x x 2−1 f(x)=x x 2−1 is bijective. I have shown that it is injective which is pretty simple. I'll write it out here for future reference. To show that f f is injective, let x 1,x 2∈(−1,1)x 1,x 2∈(−1,1). Assume that f(x 1)=f(x 2)f(x 1)=f(x 2). Then x 1 x 2 1−1=x 2 x 2 2−1 x 1 x 1 2−1=x 2 x 2 2−1 Multiplying both sides by (x 2 1−1)(x 2 2−1)(x 1 2−1)(x 2 2−1) which we know can't be 0 0. On simplifying the equation further, we get (x 1 x 2+1)(x 2−x 1)=0(x 1 x 2+1)(x 2−x 1)=0 This means either x 2 x 1=−1 x 2 x 1=−1 or x 2=x 1 x 2=x 1. It isn't possible that x 2 x 1=−1 x 2 x 1=−1 because if it were true then x 2=−1 x 1 x 2=−1 x 1 and for all the value of x 1 x 1 in the domain, x 2 x 2 will go out of the domain. Thus, it must be that x 2=x 1 x 2=x 1. This concludes the injectivity part. What I am confused about is the surjectivity part. I know that you usually invert the function and then show that for any value y y in the codomain, there exists a value x x in the domain of the function such that f(x)=y f(x)=y. I can't seem to invert this function appropriately. Any hints? real-analysis functions Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Feb 28, 2019 at 7:59 Salman QureshiSalman Qureshi 381 1 1 gold badge 4 4 silver badges 12 12 bronze badges 3 1 Make quadratic and find discriminant.. I think that's the usual way Scáthach –Scáthach 2019-02-28 08:02:58 +00:00 Commented Feb 28, 2019 at 8:02 Show that a(x 2−1)−x=0 a(x 2−1)−x=0 for some a∈(−1,1)a∈(−1,1).Wuestenfux –Wuestenfux 2019-02-28 08:03:51 +00:00 Commented Feb 28, 2019 at 8:03 One possibility would be to find the inverse. Some posts related to this: How do I find the inverse of f(x)=x x 2+1 f(x)=x x 2+1, How do you work out the inverse of functions such as f(x)=x x 2−1 f(x)=x x 2−1?Martin Sleziak –Martin Sleziak 2021-07-02 06:52:12 +00:00 Commented Jul 2, 2021 at 6:52 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. First notice that x=0⟺y=0.x=0⟺y=0. Then for \|x\|<1∧x≠0\|x\|<1∧x≠0, y=x x 2−1⟺y x 2−x−y=0.y=x x 2−1⟺y x 2−x−y=0. The discriminant is strictly positive so that there are two distinct real roots. And by Vieta, their product is −1−1, so that one lies in (−1,1)(−1,1) and the other not. Hence the function is invertible. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Feb 28, 2019 at 8:27 answered Feb 28, 2019 at 8:21 user65203 user65203 1 1 +1 for demonstrating that the solution is in (−1,+1)(−1,+1). And also, for separating x=0 x=0 as a special case.stressed out –stressed out 2019-02-28 08:29:44 +00:00 Commented Feb 28, 2019 at 8:29 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. By using the quadratic formula one has f−1(x)=1−4 x 2+1−−−−−−√2 x f−1(x)=1−4 x 2+1 2 x for the given range of x x. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Feb 28, 2019 at 8:06 Peter ForemanPeter Foreman 20.2k 2 2 gold badges 13 13 silver badges 43 43 bronze badges 4 How do you overrule the solution with ++ sign? Namely, f−1(x)=1+4 x 2+1−−−−−−√2 x f−1(x)=1+4 x 2+1 2 x stressed out –stressed out 2019-02-28 08:09:39 +00:00 Commented Feb 28, 2019 at 8:09 f(−1)=+∞⇒f−1(+∞)=−1 f(−1)=+∞⇒f−1(+∞)=−1 Peter Foreman –Peter Foreman 2019-02-28 08:11:16 +00:00 Commented Feb 28, 2019 at 8:11 1 Shouldn't we also show that −1<1−4 x 2+1√2 x<+1−1<1−4 x 2+1 2 x<+1?stressed out –stressed out 2019-02-28 08:17:50 +00:00 Commented Feb 28, 2019 at 8:17 @PeterForeman I still don't get how you overruled the other possibility.Salman Qureshi –Salman Qureshi 2019-02-28 08:39:02 +00:00 Commented Feb 28, 2019 at 8:39 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Let y y be a real number not equal to 0 0. Let f(x)=x/y+1−x 2 f(x)=x/y+1−x 2. Then f(−1)=−1/y f(−1)=−1/y and f(1)=1/y f(1)=1/y. Hence there exists x x between −1−1 and 1 1 with f(x)=0 f(x)=0. This gives x/y+1=x 2 x/y+1=x 2 or y=f(x)y=f(x). Clearly, 0=f(0)0=f(0). Hence f f is surjective. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Feb 28, 2019 at 8:12 Kavi Rama MurthyKavi Rama Murthy 361k 20 20 gold badges 102 102 silver badges 195 195 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Suppose a a is in the function's image. Let us find out conditions on this element: there exists x∈R∖{−1,1}x∈R∖{−1,1} s.t.: x x 2−1=a⟹a x 2−x−a=0⟹this quadratic's equation discriminant is non-negative:x x 2−1=a⟹a x 2−x−a=0⟹this quadratic's equation discriminant is non-negative: Δ=1+4 a 2⟹Δ≥1>0∀a∈R Δ=1+4 a 2⟹Δ≥1>0∀a∈R and you now proved your function is surjective...and you didn't have to find out what the inverse is. Last task: can you prove now that the x x that solves the above is in (−1,1)(−1,1) , as it must be? Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Feb 28, 2019 at 8:21 answered Feb 28, 2019 at 8:10 DonAntonioDonAntonio 215k 19 19 gold badges 143 143 silver badges 291 291 bronze badges 3 Shouldn't it be −1<x<+1−1<x<+1 instead of x∈R∖{−1,+1}x∈R∖{−1,+1}?stressed out –stressed out 2019-02-28 08:19:20 +00:00 Commented Feb 28, 2019 at 8:19 1 @stressedout You're completely right...but the proof of this is almost the same. Editing now, thanks.DonAntonio –DonAntonio 2019-02-28 08:21:16 +00:00 Commented Feb 28, 2019 at 8:21 Why this proves that f f is surjective? Is that because we have proved that the equation x/(x 2−1)=a x/(x 2−1)=a always have two real solutions for all a a and so f f "covers" all R R? Does the "for all a∈R a∈R" come from the fact that you've proved that the determinant is positive independently from a a so the equality holds for all a a? Thanks.ZaWarudo –ZaWarudo 2020-06-13 08:10:29 +00:00 Commented Jun 13, 2020 at 8:10 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The function f(x)=x x 2−1 f(x)=x x 2−1 is continuous and strictly (monotonically) decreasing in (−1,1)(−1,1): f′(x)=−x 2−1(x 2−1)2<0 f′(x)=−x 2−1(x 2−1)2<0 and: lim x→−1+f(x)=+∞;lim x→1−f(x)=−∞.lim x→−1+f(x)=+∞;lim x→1−f(x)=−∞. Therefore, f(x)f(x) is surjective for x∈(−1,1)x∈(−1,1). Here is the graph: Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Feb 28, 2019 at 9:58 farruhotafarruhota 32.3k 2 2 gold badges 20 20 silver badges 54 54 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Let y∈R y∈R. If y=0 y=0 then f(0)=y f(0)=y with 0∈(−1,1)0∈(−1,1). Suppose y≠0 y≠0. Let x 1,x 2 x 1,x 2 be defined by x 1=1+1+4 y 2−−−−−−√2 y x 2=1−1+4 y 2−−−−−−√2 y x 1=1+1+4 y 2 2 y x 2=1−1+4 y 2 2 y Then x 1 x 2=−1 x 1 x 2=−1, so one of them must lie in the interval [−1,1][−1,1]. Since x 1,x 2 x 1,x 2 are roots of y x 2−x−y y x 2−x−y, we see that none of them can be equal to 1 1. Hence none of them can be equal to −1−1 either. Therefore one of them must lie in (−1,1)(−1,1). Let z z be this one. Then y z 2−z−y=0 y z 2−z−y=0 and z∈(−1,1)z∈(−1,1). This implies that f(z)=z z 2−1=y f(z)=z z 2−1=y. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Feb 6, 2021 at 17:48 AlphieAlphie 5,144 2 2 gold badges 10 10 silver badges 30 30 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 real-analysis functions 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 4How to prove one-to-one and onto? 0How do you work out the inverse of functions such as f(x)=x x 2−1 f(x)=x x 2−1? 1How do I find the inverse of f(x)=x x 2+1 f(x)=x x 2+1 Related 0Is this function bijective, surjective and injective? 1Proving Hyperbolic Sine and Tan functions are bijective. 0show that f is bijective function 2Proving injectivity of a certain polynomial function 2Finding if these function is bijective within given domain/codomain 3Study the injectivity and surjectivity of the function f 0Prove or disprove: For every function f:B→C f:B→C we have that if 1≤\|C\|<\|B\|1≤\|C\|<\|B\| then f is not injective. 0Am I correct? 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15526	https://meangreenmath.com/2013/10/30/importance-of-the-base-case-in-a-proof-by-induction/	Importance of the base case in a proof by induction – Mean Green Math Skip to content Mean Green Math Explaining the "whys" of mathematics Importance of the base case in a proof by induction John QuintanillaPrecalculus, TheoremsOctober 30, 2013 September 1, 2021 5 Minutes In Precalculus, Discrete Mathematics or Real Analysis, an arithmetic series is often used as a student’s first example of a proof by mathematical induction. Recall, from Wikipedia: Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers. It is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that the given statement for any one natural number implies the given statement for the next natural number. From these two steps, mathematical induction is the rule from which we infer that the given statement is established for all natural numbers. The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps: The basis (base case): prove that the statement holds for the first natural number . Usually, or . The inductive step: prove that, if the statement holds for some natural number , then the statement holds for . The hypothesis in the inductive step that the statement holds for some is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for . As an inference rule, mathematical induction can be justified as follows. Having proven the base case and the inductive step, then any value can be obtained by performing the inductive step repeatedly. It may be helpful to think of the domino effect. Consider a half line of dominoes each standing on end, and extending infinitely to the right. Suppose that: The first domino falls right. If a (fixed but arbitrary) domino falls right, then its next neighbor also falls right. With these assumptions one can conclude (using mathematical induction) that all of the dominoes will fall right. Mathematical induction… works because is used to represent an arbitrary natural number. Then, using the inductive hypothesis, i.e. that is true, show is also true. This allows us to “carry” the fact that is true to the fact that is also true, and carry to , etc., thus proving holds for every natural number . When students first encounter mathematical induction (in either Precalculus, Discrete Mathematics, or Real Analysis), the theorems that students are asked to prove usually fall into four categories: Calculating a series (examples below). Statements concerning divisibility (for example, proving that is always a factor of ). Finding a closed-form expression for a recursively defined sequence (for example, if and if , proving that ) Statements concerning inequality (for example, proving that if ) Here’s a common first (or maybe second) example of mathematical induction applied to an arithmetic series. Theorem. Proof. Induction on . : The left-hand is simply , while the right-hand side is , which is also equal to . So the base case works. : Assume that the statement holds true for the integer . . If I replace by in the statement of the theorem, then the right-hand side becomes I find it helpful to describe this to students as my target. In other words, as I manipulate the left-hand side, my ultimate goal is to end up with this target. Once I have done that, then I have completed the proof. If I replace by in the statement of the theorem, then the left-hand side will now end on instead of : Notice that we’ve seen almost all of this before, except for the extra term . So we will substitute using the induction hypothesis, carrying the extra along for the ride. Now our task is, by hook or by crook, using whatever algebraic tricks we can think of to convert this last expression into the target. Most students are completely comfortable doing this, although they typically multiply out the term unnecessarily. Indeed, many early proofs by induction are simplified by factoring out terms whenever possible — in the example below, is factored on the third step — as opposed to multiplying them out. In my experience, proofs by induction often serve as a stringent test of students’ algebra skills as opposed to their skills in abstract reasoning. In any event, here’s the end of the proof: A fair amount of algebra was needed to prove the case. However, the first step — the base case — was especially easy. Indeed, in most proofs by induction seen by students, the base case is often quite trivial… to the point that students often wonder why the base case is needed in the first place. I first saw this next example in Calculus, by Tom M. Apostol. This next fallacious example illustrates what can happen if the base case is ignored. The statement of this “theorem” doesn’t match the formula for an arithmetic series, and so clearly something is wrong with the following “proof.” “Theorem.” “Proof.” Induction on . : Let’s just ignore the base case, it’s unimportant. : Assume that the statement holds true for the integer . . If I replace by in the statement of the theorem, then the right-hand side — my target — becomes On the left-hand side, we use the induction hypothesis: Now our task is, by hook or by crook, using whatever algebraic tricks we can think of to convert this last expression into the target. So that’s the end of the “proof.” Clearly, something went wrong with the above proof. What went wrong, obviously, is that we didn’t check the base case. If , then the left-hand side is . However, the right-hand side is . So the base case is false. So what happened? We correctly showed that, if the case is true, then the case is also true. The catch, of course, is that the case is never true. Using the domino analogy, we showed that if a domino falls, then the next domino will fall. But the first domino never falls. All this to say… yes, it’s important to check that the base case actually works. Share this: Click to share on Facebook (Opens in new window)Facebook Click to share on X (Opens in new window)X Like Loading... Related Tagged induction Published by John Quintanilla I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science. View all posts by John Quintanilla PublishedOctober 30, 2013 September 1, 2021 Post navigation Previous Post Engaging students: Computing trigonometric functions using a unit circle Next Post Graphical statistics 7 thoughts on “Importance of the base case in a proof by induction” Pingback: 100,000 page views \| Mean Green Math Pingback: 250,000 page views \| Mean Green Math Maximus Mathematicussays: September 4, 2020 at 6:17 pm A simpler fallacious example would have been n + 3 = n + 4. Reply John Quintanillasays: November 12, 2020 at 7:31 am You’re right, of course. Pedagogically, I think I prefer Apostol’s example, as the “proof” has the appearance of the types of problems that students would be expected to do for homework. But I suppose my preference is really a matter of taste. Reply James Da Silva Chensays: September 1, 2021 at 12:58 am I think you made a typo in the last three lines of your inductive step in your fallacious example. The 6 should be turned into an 8 Reply John Quintanillasays: September 1, 2021 at 8:36 am Thanks for catching that. It’s now been fixed. Reply Pingback: 500,000 page views – Mean Green Math Leave a comment Cancel reply Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. Search for: Follow Blog via Email Enter your email address to follow this blog and receive notifications of new posts by email. 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15527	https://www.ams.org/bull/2022-59-04/S0273-0979-2021-01753-8/S0273-0979-2021-01753-8.pdf	BULLETIN (New Series) OF THE AMERICAN MATHEMATICAL SOCIETY Volume 59, Number 4, October 2022, Pages 471–502 Article electronically published on October 29, 2021 HELLY-TYPE PROBLEMS IMRE B´ AR´ ANY AND GIL KALAI Abstract. In this paper we present a variety of problems in the interface between combinatorics and geometry around the theorems of Helly, Radon, Carath´ eodory, and Tverberg. Through these problems we describe the fasci-nating area of Helly-type theorems and explain some of their main themes and goals. 1. Helly, Carath´ eodory, and Radon theorems In this paper, we present a variety of problems in the interface between com-binatorics and geometry around the theorems of Helly, Radon, Carath´ eodory, and Tverberg. Helly’s theorem [Hel23] asserts that for a family {K1, K2, . . . , Kn} of convex sets in Rd, where n ≥d+1, if every d+1 of the sets have a point in common, then all of the sets have a point in common. The closely related Carath´ eodory theorem [Car07] states that for S ⊂Rd, if x ∈conv S, then x ∈conv R for some R ⊂S, \|R\| ≤d+1. The more general colorful Carath´ eodory theorem [B´ ar82] says the following. Let S1, S2, . . . , Sd+1 be d + 1 sets (or colors if you wish) in Rd. Suppose that x ∈d+1 i=1 conv Si. Then there is a transversal T = {x1, . . . , xd+1} of the system S1, . . . , Sd+1, meaning that x1 ∈S1, x2 ∈S2, . . . , xd+1 ∈Sd+1 such that x ∈ conv T. A transversal is also called a rainbow set when S1, . . . , Sd+1 are considered as colors. The uncolored version, that is, when S1 = S2 = · · · = Sd+1, is the classic result of Carath´ eodory. There is a closely related colorful version of Helly’s theorem due to Lov´ asz that appeared in [B´ ar82]. Tverberg’s theorem [Tve66] states the following. Let x1, x2, . . . , xm be points in Rd with m ≥(r −1)(d + 1) + 1. Then there is a partition S1, S2, . . . , Sr of {1, 2, . . . , m} such that r j=1 conv {xi : i ∈Sj} ̸= ∅. This was a conjecture by Birch, who also proved the planar case in a slightly diﬀerent form. The bound of (r−1)(d+1)+1 in the theorem is sharp as can easily be seen from the conﬁguration of points in a suﬃciently general position. The case r = 2 is Radon’s theorem [Rad21], another classic from 1921, which was used by Radon to prove Helly’s theorem. Helly’s original proof (published later) was based on a separation argument. Sarkaria [Sar92] gave a simple proof of Tverberg’s theorem based on the colorful Carath´ eodory theorem. Received by the editors May 12, 2021. 2020 Mathematics Subject Classiﬁcation. Primary 52A35; Secondary 52A20. Research of the ﬁrst author was partially supported by Hungarian National Research grants (no. 131529, 131696, and 133819), and research of the second author by the Israel Science Foundation (grant no. 1612/17). c ⃝2021 Imre B´ ar´ any and Gil Kalai 471 472 IMRE B´ AR´ ANY AND GIL KALAI This paper describes the fascinating area of Helly-type theorems and explains some of their main themes and goals through a large and colorful bouquet of prob-lems and conjectures. Some of these problems are very precise and clear-cut; see, for instance, Sierksma’s conjecture (Conjecture 4.1), the cascade conjecture (Con-jecture 5.1), and Problem 3.2 about volumes of intersections. Some of them are rather vague; see, for instance, Problem 2.1 about intersection patterns of Euclidean convex sets, Problem 3.9 about the mutual position of convex sets, and Problem 5.5 about topological conditions for the existence of Tverberg partitions. We hope to see the answers to many of the questions presented here in the near future. Often, results from convexity give a simple and strong manifestation of theorems from topology: Helly’s theorem manifests the nerve theorem from algebraic topology, and Radon’s theorem can be regarded as an early linear version of the Borsuk– Ulam theorem. One of our main themes is to further explore these connections to topology. Helly-type theorems also oﬀer complex and profound combinatorial connections and applications that represent a second theme of this paper. For a wider perspective and many other problems we refer the reader to survey papers by Danzer, Gr¨ unbaum, and Klee [DGK63]; Eckhoﬀ[Eck79] and [Eck93]; Tancer [Tan13]; De Loera, Goaoc, Meunier, and Mustafa [DLGMM19]; and the forthcoming book of B´ ar´ any [B´ ar]. Here is a quick summary of the paper. Section 2 deﬁnes the nerve that records the intersection pattern of convex sets in Rd, describes some of its combinatorial and topological properties, and considers various extensions of Helly’s theorem, such as the fractional Helly theorem, which asserts that if a fraction α of all sets in a family of convex sets have a nonempty intersection, then there is a point that belongs to a fraction β(α, d) of the sets in the family. Section 3 considers various reﬁnements and generalizations of Helly theorems, such as the study of dimensions of intersections of convex sets and the study of Helly-type theorems for unions of convex sets. Section 4 presents various extensions and reﬁnements of Tverberg’s theorem, starting with Sierksma’s conjecture on the number of Tverberg partitions. Section 5 studies the cascade conjecture about the dimensions of the Tverberg points and considers several connections with graph theory including a speculative connection with the four-color theorem. Section 6 deals with other Tverberg-type problems. Section 7 brings problems related to the Carath´ eodory theorem and weak-epsilon nets, and Section 8 gives a glance at common transversals; rather than piercing a family of sets by a single point or a few points, we want to stab them with a single or a few j-dimensional aﬃne spaces. Final conclusions are drawn in the last section. 2. Around Helly’s theorem 2.1. Nerves, representability, and collapsibility. We start this section with the following basic deﬁnition: for a ﬁnite collection of sets F = {K1, K2, . . . , Kn}, the nerve of F is a simplicial complex deﬁned by N(F) = {S ⊂[n] : i∈S Ki ̸= ∅}. Helly’s theorem can be seen as a statement about nerves of convex sets in Rd, and nerves come in to play in many extensions and reﬁnements of Helly’s theorem. HELLY-TYPE PROBLEMS 473 A missing face S of a simplicial complex K is a set of vertices of K that is not a face, but every proper subset of S is a face. Helly’s theorem asserts that a d-representable complex does not have a missing face with more than d+1 vertices. A simplicial complex is d-representable if it is the nerve of a family of convex sets in Rd. Problem 2.1. Explore d-representable simplicial complexes. We refer the reader to the survey on d-representable complexes by Tancer [Tan13]. Let K be a simplicial complex. A face F ∈K is free if it is contained in a unique maximal face. An elementary d-collapse step is the removal from K of a free face F with at most d vertices and all faces containing F. A simplicial complex is d-collapsible if it can be reduced to the empty complex by a sequence of elementary d-collapse steps. Wegner proved [Weg75] that every d-representable complex is d-collapsible. The converse does not hold even for d = 1: 1-representable complexes are the clique complex of interval graphs, and 1-collapsible complexes are the clique complexes of chordal graphs. Here, a clique complex of a graph G is a simplicial complex whose faces cor-respond to the sets of vertices of complete subgraphs of G. Chordal graphs are graphs with no induced cycles of length greater than 3. Intersection patterns of intervals (which are the same as 1-representable complexes) were completely char-acterized by Lekkerkerker and Boland [LB62]. They proved that interval graphs are characterized by being chordal graphs with the additional property that among every three vertices, one is a vertex or is adjacent to a vertex in any path between the other two. They also described interval graphs in terms of a list of forbidden induced subgraphs. 2.2. The upper bound theorem. For a ﬁnite collection of sets F = {K1, K2, . . . , Kn}, n ≥d + 1, in Rd, let N = N(F) be the nerve of F. We put fk(N) = \|{S ∈N : \|S\| = k + 1}\|. (The vector (f0(N), f1(N), . . . ) is called the f-vector of N, and is sometimes referred to also as the f-vector of F and written as f(F).) Helly’s theorem states that if fn−1(N) = 0, then fd(N) < n d+1 or, with the f(F) notation, fn−1(F) = 0 implies fd(F) < n d+1 A far-reaching extension of Helly’s theorem was conjectured by Katchalski and Perles and proved by Kalai [Kal84b] and Eckhoﬀ[Eck85]. Theorem 2.1 (Upper bound theorem). Let F be a family of n convex sets in Rd, and suppose that every d + r + 1 members of F have an empty intersection. Then, for k = d, . . . , d + r −1, fk(N(F)) ≤ d+r−1 j=k j −d k −d n −j + d −1 d . The theorem provides best upper bounds for fd(F), . . . , fd+r−1(F) in terms of f0(F) provided fd+r(F) = 0. The proofs rely on d-collapsibility. There is a simple case of equality: the family consists of r copies of Rd and n −r hyperplanes in general position. Theorem 2.1 is closely related to the upper bound theorem for convex polytopes of Peter McMullen [McM70]. In fact, a common proof was given by Alon and Kalai in [AK95]. 474 IMRE B´ AR´ ANY AND GIL KALAI Problem 2.2. Study cases of equality for the upper bound theorem. A place to start would be to understand 2-representable complexes K with f3(K) = 0 and f2(K) = n−1 2 . Theorem 2.1 implies the sharp version of the fractional Helly theorem of Katchal-ski and Liu [KL79]. The sharp version is due to Kalai [Kal84b]. Theorem 2.2. Let K be a d-representable complex. If fd(K) ≥α n d+1 , then dim(K) ≥βn, where β = β(d, α) = 1 −(1 −α)1/d+1. In other words, if F = {K1, . . . , Kn} is a family of convex sets in Rd (n ≥d + 1) and at least α n d+1 of the d+1 tuples in F intersect, then F contains an intersecting subfamily of size βn. This is a result of central importance around Helly’s theorem. The existence of β(α) is referred to as the fractional Helly property, and if β →1 when α →1 this is referred to as the strong fractional Helly property. We note that a complete characterization of f-vectors of d-representable com-plexes was conjectured by Eckhoﬀand proved by Kalai [Kal84a,Kal86]. 2.3. Helly numbers and Helly orders. It is useful to consider the following abstract notions of Helly numbers and Helly orders. Let F be a family of sets. The Helly number h(F) of F is the minimal positive integer h such that if a ﬁnite subfamily K ⊂F satisﬁes K′ ̸= ∅for all K′ ⊂K of cardinality ≤h, then K ̸= ∅. The Helly order ho(F) of F is the minimal positive integer h such that if a ﬁnite subfamily K ⊂F satisﬁes (1) every ﬁnite intersection of sets in K belongs to F, and (2) K′ ̸= ∅for all K′ ⊂K of cardinality ≤h, then K ̸= ∅. Of course, when we consider families of sets closed under intersection, the Helly number and the Helly order coincide. So, for example, the topological Helly theorem (discussed in Section 2.4) asserts that the Helly order of topologically trivial sets in Rd is d + 1, and Amenta’s theorem (see Section 3.3) asserts that the family of unions of k pairwise disjoint convex sets in Rd has the Helly order k(d+1). Let F be a family of sets. The fractional Helly number g(F) of F is the minimal positive integer g such that there is a function f(α) > 0, deﬁned for α > 0, with the following property: for every family K ⊂F of cardinality n, if at least α n g of the g-tuples in F intersect, then F contains an intersecting subfamily of size f(α)n. 2.4. Topological Helly theorem and Leray complexes. Helly himself proved a topological version of his theorem [Hel30]. A good cover is a family of compact subsets of Rd such that every intersection of sets in the family is either empty or topologically trivial. (By topologically trivial we mean contractible, but it is suﬃcient to assume that all homology groups vanish.) Theorem 2.3 (Topological Helly). If in a good cover of n subsets of Rd, n ≥d+1, every intersection of d + 1 sets is nonempty, then the intersection of all the sets in the family is nonempty. A simplicial complex K is d-Leray if Hi(K′) = 0 for every induced subcomplex K′ of K and for every i ≥d. The well-known nerve theorem from algebraic topology asserts that if K is a ﬁnite family of sets that form a good cover, then the nerve of K is topologically equivalent to K. (The notion of topologically equivalent corresponds to the notion of topologically trivial in the deﬁnition of good covers.) HELLY-TYPE PROBLEMS 475 It follows from the homological version of the nerve theorem that d-representable complexes are d-Leray. It is also easy to see that d-collapsible complexes are d-Leray. Remark. The nerve theorem played an important role in algebraic topology in the 1940s and 1950s, e.g., in showing that de Rham homology coincides with other notions of homology. Helly’s topological theorem is remarkable since it came ear-lier than these developments. In Section 4.2 we will mention that Radon’s theo-rem can be seen as an early incarnation of the Borsuk–Ulam theorem in topology. Topological extensions of Helly-type theorems are an important part of the the-ory. Often, such extensions are considerably more diﬃcult to prove, but in a few cases the topological proofs are the only known ones even for the geometric re-sults. There are also a few cases where natural topological extensions turned out to be incorrect. The survey paper [DLGMM19] of De Loera, Goaoc, Meunier, and Mustafa emphasizes connections with combinatorial theorems closely related to the Brouwer ﬁxed-point theorem, starting with the Sperner lemma and the Knaster– Kuratowski–Mazurkiewicz theorem. A general problem is the following. Problem 2.3. (i) Find ﬁner and ﬁner topological and combinatorial properties of d-represen-table complexes. (ii) Extend Helly-type theorems to good covers, Leray complexes, and beyond. (iii) Find weaker topological conditions that suﬃce for the topological Helly theorem to hold. There is much to say about part (ii) of Problem 2.3. In several cases the way to go about it is to extend properties of d-representable complexes to d-Leray com-plexes. We will come back to such extensions later, but we note that the upper bound theorem (Theorem 2.1) as well as the full characterization of their f-vectors, extends to d-Leray complexes; see [Kal02]. This is also closely related to Stanley’s characterization [Sta75] of f-vectors of Cohen–Macaulay complexes. Regarding part (i) of Problem 2.3, we ﬁrst note a very easy connection with embeddability: If G is a graph we denote by TG the graph where for every edge e, we add a new vertex ve that is adjacent to the endpoints of e, and remove e itself; see Figure 1. If G is not planar, e.g., when G = T5, then TG is not 2-representable. (Note however that Kn itself is 2-representable for every n.) Figure 1. TK5 476 IMRE B´ AR´ ANY AND GIL KALAI White [Whi21] deﬁned the class of d-Matouˇ sek simplicial complexes that are related to topological invariants for embeddability as follows. Let K be an abstract simplicial complex with vertices V (K) = [n] = {1, 2, . . . , n}. We deﬁne the dual simplicial complex K′, with vertices V (K′) = {J ∈K : J is inclusion maximal} and faces K′ = {α ⊆V (K′) : J∈α J ̸= ∅}. We say that K is d-Matouˇ sek, if the Z2-index of the space ˆ K = {(x, y) ∈\| \| K′ \| \| 2 : supp(x) ∩ supp(y) / ∈K} is less than d. Here supp(x) denotes the support of x in K′, which is the inclusion-minimal face of K′ containing x. It is straightforward to verify that K is d-representable iﬀthere exists a linear map f : K′ →Rd, such that for every set I ⊆V (K) not in K, we have i∈I f(αi) = ∅, where αi = {J ∈V (K′) : i ∈J}. This implies the existence of a Z2-map from ˆ K to Sd−1; thus any d-representable complex is also d-Matouˇ sek. White proved that nerves of good covers in Rd are d-Matouˇ sek, and he also showed that being 1-Matouˇ sek is equivalent to being 1-representable. Regarding part (iii), Debrunner [Deb70] showed that for the statement of the topological Helly property it suﬃces to assume that the (reduced) homology of intersections of k sets in the family 1 ≤k ≤d + 1 vanishes at and below dimension d −k. Even more general conditions were found by Montejano [Mon14]. 2.5. Conditions for the fractional Helly property. Problem 2.4. (i) Find geometric, topological, and combinatorial conditions that imply the fractional Helly property. (ii) Find geometric, topological, and combinatorial conditions that imply the strong fractional Helly property. We will mention here two conjectures regarding the fractional Helly property and two related theorems. A class of simplicial complexes is hereditary if it is closed under induced subcomplexes. Recall that for a simplicial complex K, fi(K) is the number of i-faces of K and b(K) is the sum of (reduced) Betti numbers of K. In connection with the fractional Helly theorem, Kalai and Meshulam [Kal10] formulated the following conjecture. Conjecture 2.5 (Kalai and Meshulam). Let C > 0 be a positive number. Let F be the hereditary family of simplicial complexes deﬁned by the property that for every simplicial complex K ∈F with n vertices, b(K) ≤Cnd. Then for every α > 0 there is β = β(d, C) > 0 such that K ∈F and fd(K) ≥α n d+1 imply dim(K) ≥βn. The conclusion of the conjecture is referred to as the fractional Helly property of degree d. Kalai and Meshulam further conjectured that the conclusion holds even if one replaces b(K) with \|χ(K)\|, where χ(K) is the Euler characteristic of K. When d = 0, this conjecture is about graphs and it was proved (in its strong version) in [CSSS20]; see also [SS20]. HELLY-TYPE PROBLEMS 477 Conjecture 2.6 (Kalai and Meshulam). Let U be a family of sets in Rd. Suppose that for every intersection L of m members of U, b(L) ≤γmd+1. Then U satisﬁes the fractional Helly property of order d. In some special cases the fractional Helly property has been established. For instance, Matouˇ sek [Mat04] showed that families of sets with a bounded VC dimen-sion in Rd satisfy the fractional Helly property of order d. Another case includes the so-called convex lattice sets. These are sets of the form Zd ∩C where Zd is the lattice of integer points in Rd and C is a convex set in Rd. A result of B´ ar´ any and Matouˇ sek [BM03] asserts that families of convex lattice sets in Rd satisfy the fractional Helly property of order d + 1. In both of these theorems the fractional Helly number is considerably smaller than the Helly number. For example, let F be the family of all convex lattice sets in Rd. The Helly number of F, h(F), is equal to 2d, as shown by Doignon [Doi73], while the fractional Helly number is d + 1; see [BM03]. Problem 2.7. Does the assertion of the Radon theorem imply the fractional Helly property? An aﬃrmative solution to one interpretation of this question was recently given by Holmsen and Lee [HL21], who showed that for abstract convexity spaces, the ﬁnite Radon number r implies that the fractional Helly number is bounded by some function m(r) of r. Problem 2.8. Estimate m(r). Convex sets are sets of solutions of systems of linear inequalities, and we can consider systems of polynomial inequalities of higher degrees. Conjecture 2.9. The family Bd k of sets of solutions in Rd of polynomial inequalities of degree ≤k has the fractional Helly property. It is known [Mot55] (and is an easy consequence of Helly’s theorem itself) that the class Ad k of sets in Rd of common zeroes of systems of polynomial inequalities of degree ≤k has the Helly number d+k k . And we can even ask if this formula gives the precise fractional Helly number for the class Bd k. We conclude this section by mentioning an interesting recent abstract notion of convexity described by Moran and Yehudayoﬀ[MY20], which seems relevant to various problems raised in this paper and, in particular, to Problem 2.7. In this notion of abstract convexity, which we call MY-convexity, we assume that every convex set is the intersection of half-spaces. We assume further that the VC dimension of the class of half-spaces is at most D. The class Bd k is an example of an MY-convexity space where the half-spaces are the sets of solutions of a single polynomial inequality of degree k. Problem 2.10. Consider an MY-convexity space X where the VC dimension of the class of half-spaces is at most D. (i) Does X have the fractional Helly number f(D) for some function f of D? (ii) Does X have the fractional Helly number D? 2.6. The (p, q)-property. The conclusion of Helly’s theorem is that the family is intersecting; i.e., there is a point Rd that is included in all sets in the family. 478 IMRE B´ AR´ ANY AND GIL KALAI Problem 2.11. What conditions guarantee that the family is t-pierceable, meaning that there are t points such that every set in the family contains at least one? In the language of nerves, what conditions guarantee that the set of vertices of the nerve can be expressed as the union of t faces? A family of sets has the (p, q)-property if for every p members of the family some q have a nonempty intersection. Note that here we assume p ≥q > d. (For nerves this says that every set of p vertices spans a face with q vertices, and this is closely related to Tur´ an’s problem for hypergraphs.) Hadwiger and Debrunner [HD57] introduced the (p, q)-property and proved Theorem 2.4. If a ﬁnite family of convex sets in Rd has the (p, q)-property and (d −1)p < (q −1)d, then it is p −q + 1-pierceable. A family of sets has the (p, q)r-property if in its nerve every p vertices span at least r faces with q vertices. This was introduced by Montejano and Sober´ on [MS11] and further studied by Keller and Smorodinsky [KS18]. Montejano and Sober´ on proved (among other results) Theorem 2.5. A family of convex sets in Rd with the (d + 2, d + 1)d-property is 2-pierceable. Hadwiger and Debrunner [HD57] conjectured in 1957 and Alon and Kleitman [AK92b,AK92a] proved the following important theorem. Theorem 2.6 ((p, q)-theorem). For all p ≥q > d there exists f(d, p, q) such that if a family of convex sets in Rd has the (p, q)-property, then it is f(d, p, q)-pierceable. The bound on f(d, p, q) given in [AK92b] is enormous. The ﬁrst open case is d = 2 and p = 4, q = 3. It is known that f(2, 4, 3) is between 3 and 9, the lower bound is from [KGT01], and the upper bound is a recent result of McGinnis [McG20] who brought down the upper bound of 13 of [KGT01] to 9. Substantial improvements for the general case were given by Keller, Smorodinsky, and Tardos [KST18] and by Keller and Smorodinsky [KS20]. Problem 2.12. Improve further the bounds on f(2, 4, 3) and, more generally, on f(d, p, q). Alon, Kalai, Matouˇ sek, and Meshulam [AKMM02] proved the following result that implies that the Alon–Kleitman theorem extends to good covers and Leray complexes (but with worse bounds). Theorem 2.7. For every q > d+1 there exists C(d, q) with the following property. Let F be a hereditary class of simplicial complexes satisfying the fractional Helly property of degree d. If a simplicial complex K ∈F has the property that every q vertices span a d-dimensional face, then the vertices of K can be covered by C(d, q)-faces. See Eckhoﬀ[Eck03] for a survey on (p, q)-theorems. 2.7. A Ramsey type question. Conjecture 2.13. For integers d ≥1 and r > 1 + ⌈d/2⌉, there is α = α(d, r) > 0 such that the following holds. Let F be a family of n convex sets in Rd. Then F contains nα(d,r)-sets such that either every r has a point in common or no r has a point in common. HELLY-TYPE PROBLEMS 479 There is a large literature on this and related questions starting with a theorem of Larman, Matouˇ sek, Pach, and T¨ or˝ ocsik [LMPT94] that proves the case d = 2 and r = 2. Subsequent works are [APP+05] and [FPT11]. When r = d + 1, this conjecture holds with α = 1/(d+1). This was observed by Keller and Smorodinsky (private communication) and follows from their improved (p, q)-theorems. The gen-eral phenomenon here (with several interesting manifestations) is that graphs and hypergraphs arising in geometry satisfy much stronger forms of Ramsey’s theorem than arbitrary graphs and hypergraphs. 2.8. Colorful, fractional colorful, and matroidal Helly theorems. The col-orful Helly theorem of Lov´ asz (see [B´ ar82]) asserts the following. Assume that C1, . . . , Cd+1 are ﬁnite families of convex sets in Rd with the property that if every transversal K1, . . . , Kd+1 is intersecting, then Ci ̸= ∅for some i ∈[d + 1]. Here “transversal” means that Ki ∈Ci for every i ∈[d + 1]. The colorful version implies the original one when C1 = · · · = Cd+1. The analogous colorful version of the fractional Helly theorem says that if an α fraction of all transversals of the system C1, . . . , Cd+1 is intersecting, then one of the families, say Ci, contains an intersecting subfamily of size β\|Ci\|. Here α > 0, of course, and β = β(d, α) has to be positive. Such a theorem (with β = α/(d + 1)) was proved and used ﬁrst in [ABB+09]. The dependence of β was improved by Kim [Kim17], who showed in particular that β →1 as α →1. The optimal dependence of β on α and d is a recent result of Bulavka, Goodarzi, and Tancer [BGT20]. They use Kalai’s algebraic shifting technique [Kal84b] and raise the following interesting conjecture. Conjecture 2.14. Let K be a d-Leray simplicial complex whose vertex set V is partitioned into sets V1, . . . , Vd+1, called colors, and \|Vi\| = ni for i ∈[d + 1]. Assume that K contains at least α d+1 1 ni colorful d-faces for some α > 0. Then there is i ∈[d + 1] such that the dimension of the restriction of K to Vi is at least (1 −(1 −α)1/(d+1))ni −1. Kalai and Meshulam [KM05] extended the assertion of the colorful Helly theo-rem to the topological setting and also considered a matroidal version. A matroidal complex is the complex consisting of the independent sets of a matroid. Equiva-lently, M is a matroidal complex if and only if every induced subcomplex is pure, i.e., if all its maximal faces have the same cardinality. Theorem 2.8. Let X be a d-Leray complex on the vertex set V . Suppose that M is a matroidal complex on the same vertex set V with rank function ρ. If M ⊂X, then there exists τ ∈X with ρ(V \τ) ≤d. This theorem gives the colorful Helly property when M is a transversal matroid and it suggests a general way to extend results about colorings. We will encounter this idea again in Section 4.3, where we try to move from colorful versions of Tverberg’s theorem to matroidal versions. Theorem 2.8 has interesting connections with advances in topological combinatorics related to Hall’s marriage theorem and rainbow matchings; see [AB06] and [AB09]. 480 IMRE B´ AR´ ANY AND GIL KALAI 3. More around Helly’s theorem 3.1. Dimensions of intersections: Katchalski’s theorems. Let g(d, k) be the smallest integer with the following property: for every family of n convex sets in Rd, n ≥g(d, k), such that the dimension of intersection of every g(d, k) set in the family is at least k, the dimension of intersection of all members of the family is at least k. Helly’s theorem asserts that g(d, 0) = d + 1. In 1971 Katchalski [Kat71] proved the following interesting result. Theorem 3.1. g(d, 0) = d + 1, g(d, k) = max{d + 1, 2(d −k + 1)} if 1 ≤k ≤d. Given a family K = {K1, K2, . . . , Kn} of convex sets in Rd and J ⊂[n], set K(J) = j∈J Kj and write d(J) = dim K(J). A further remarkable result of Katchalski [Kat78] “reconstructs” the dimension of the intersection: Theorem 3.2. Let K = {K1, K2, . . . , Kn} and K′ = {K′ 1, K′ 2, . . . , K′ n} be two families of compact convex sets in Rd. If dK(J) = dK′(J) for every J, \|J\| ≤d + 1, then dK(J) = dK′(J) for every J. Katchalski actually proved a stronger statement, namely, that the condition dK(J) = dK′(J) for every J with (d + 1) −⌊d/2⌋≤\|J\| ≤d + 1 suﬃces for the conclusion of Theorem 3.2. More generally he proved that for every r ≥1 if dK(J) = dK′(J) for every J with (d + r) −⌊d/(r + 1)⌋≤\|J\| ≤d + r, then dK(J) = dK′(J) for every J. Deﬁne the D-nerve of a ﬁnite set of convex sets as its nerve K where every face S ∈K is labeled by the dimension of i∈S Ki. We can regard the D-nerve as a nested collection of simplicial complexes that correspond to intersections of dimension ≥j. Problem 3.1. Explore combinatorial and topological properties of D-nerves of families of compact convex sets in Rd. 3.2. Helly with volume. Theorems about volumes of intersections are closely related to theorems about dimensions of intersections. The natural question is, given a ﬁnite family F of convex sets in Rd, what condition guarantees that the intersection F not only is nonempty but also has volume at least one, say. The ﬁrst result in this direction is in [BKP82] of B´ ar´ any, Katchalski, and Pach. Theorem 3.3 (Helly with volume). Assume that F is a ﬁnite family of convex sets in Rd, \|F\| ≥2d, such that the intersection of any 2d sets from F has volume at least one. Then vol( F) ≥d−2d2. The example of the 2d half-spaces in Rd whose intersection is the unit cube shows that the number 2d is the best possible in this theorem. In other words, 2d is the Helly number for volumes. However, the bound d−2d2 is not sharp and was improved ﬁrst by Nasz´ odi [Nas16] to (cd)−2d and later by Brazitikos [Bra17] to (cd)−1.5d. In both estimates, c > 0 is a universal constant. The following question is still open. Problem 3.2. Show that under the conditions of Theorem 3.3 vol( F) ≥(cd)−d/2 where c > 0 is a constant. A similar result was established in [BKP82] for the diameter of the intersection. The Helly number is again 2d. So if the intersection of any 2d sets from the family F HELLY-TYPE PROBLEMS 481 has diameter at least one, then diam F ≥cd−d/2. The lower bound was improved in a series of recent papers: ﬁrst by Brazitikos [Bra17] to cd−11/2, then by Ivanov and Nasz´ odi [IN21] to (2d)−3, and more recently by Almendra-Hern´ andez, Ambrus, and Kendall [AHAK21] to (2d)−2. This leads to the next problem. Problem 3.3. Assume that F is a ﬁnite family of convex sets in Rd, \|F\| ≥2d, such that the intersection of any 2d sets from F has diameter at least one. Then diam F ≥ 2/d. Recently, several further quantitative Helly-type results have appeared; see for instance [DFN21] and [DS21]. 3.3. Unions of convex sets: around the Gr¨ unbaum–Motzkin conjecture. Nina Amenta [Ame94] proved a Helly-type result on unions of disjoint convex sets. Theorem 3.4. Let F be a family of sets in Rd such that every member in F is the union of k disjoint compact convex sets. Suppose further that every intersection of members of F is also a union of k disjoint convex sets. If every k(d + 1) sets in F has a point in common, then F ̸= ∅. In the language of Section 2.3, Theorem 3.4 asserts that the Helly order of the family of disjoint unions of k compact convex sets in Rd is (d + 1)k. This was conjectured by Gr¨ unbaum and Motzkin [GM61] who proved the case k = 2, by Larman [Lar68] who proved their conjecture for k = 3, and by Amenta in its full generality. It is easy to see that this family has no ﬁnite Helly number. Kalai and Meshulam [KM08] proved that Amenta’s theorem extends topologi-cally. They consider the following setting. Let K and L be simplicial complexes with a map from V (K) to V (L) such that the inverse image of every i face in L is the union of at most k i-faces of L. If K is d-Leray, then the Leray number of L is at most dk + k −1. Eckhoﬀand Nischke [EN09] showed that Amenta’s theorem extends combinato-rially. In the setting of the previous paragraph they proved that if K has no missing face of size d + 1 or larger, then L has no missing face of size k(d + 1) or larger. 3.4. More on families of unions of convex sets. We may consider sets in Rd that can be represented as unions of k convex sets but delete the disjointness assumption. In this case Alon and Kalai [AK95] and Matouˇ sek [Mat97] proved the following result. Theorem 3.5. Assume that F is a ﬁnite family of sets in Rd such that every member in F is the union of k compact convex sets. Then F has a ﬁnite Helly order. Let us mention a recent topological Helly-type theorem by Goaoc, Pat´ ak, Pat´ a-kov´ a, Tancer, and Wagner [GPP+17] that strengthens Theorem 3.5. Theorem 3.6. For every γ > 0 there is h(γ, d) with the following property. Let U be a family of sets in Rd. Suppose that for every intersection L of some members of U and every i ≤⌈d/2⌉−1, we have bi(L) ≤γ. Then, if every h(γ, d) members of U have a point in common, then all sets in U have a point in common. We note that Theorem 3.6 implies Theorem 3.5. In fact, its proof relies on the method developed by Matouˇ sek in [Mat97]. His method, connecting topologi-cal obstructions for embeddability to Helly-type theorems, is the basis of White’s notion [Whi21] of d-Matouˇ sek complexes. 482 IMRE B´ AR´ ANY AND GIL KALAI In connection with this we mention the following curious question. Conjecture 3.4. The Helly order of families of unions of two disjoint nonempty sets in Rd is d + 1. This is known to be false if “two” is replaced by a large integer even when d = 2. We say that two compact sets intersect nicely if the long Meyer–Vietoris exact sequence splits into short exact sequences dimensionwise. Problem 3.5. Let K = {K1, K2, . . . , Kn} be a ﬁnite family of compact sets such that for every set of indices I ⊂[n], K(I) is topologically equivalent to a ﬁxed topological space Z, and for every two sets of indices I, J ⊂[n], K(I) and K(J) intersect nicely. Then K is topologically equivalent to a ﬁber bundle over N(K) with ﬁbers topologically equivalent to Z. A positive answer to Problem 3.5 would imply Conjecture 3.4 because a pair of disjoint unions of nonempty convex sets whose intersection is also a disjoint union of nonempty convex sets always intersects nicely. 3.5. A conjecture by Gao, Landberg, and Schulman. Here is an interesting Helly-type conjecture by Gao, Langberg, and Schulman [GLS08]. For a convex set K in Rd an ϵ enlargement of K is K+ϵ(K−K) (where K−K = {x−y : x, y ∈K}). Conjecture 3.6. For every d, k, and ϵ there is some h = h(d, k, ϵ) with the following property. Let F be a family of unions of k convex sets. Let Fϵ be the family obtained by enlarging all the involved convex sets by ϵ. If every h members of F have a point in common, then all members of Fϵ have a point in common. Of course, for k = 1 we can take ϵ = 0 by Helly’s theorem. 3.6. Boxes and products. Problem 3.7. Let d1, d2, . . . , dr be a partition of d. Study Helly-type theorems for families of Cartesian products K1 ×K2 ×· · ·×Kr of convex sets where dim Ki = di. The case of standard boxes, namely when d1 = d2 = · · · = dd = 1 is of special interest. Standard boxes have Helly number 2, and therefore their nerves are de-termined by their graphs. Eckhoﬀproved an upper bound theorem for standard boxes in [Eck88] and studied the extremal families in [Eck91]. It is easiest to de-scribe the families where the upper bound is attained. If fd+r = 0 (that is, the largest nonempty intersection is for d + r sets), then the family consists of r copies of Rd and roughly the same number of parallel copies of each of the d coordinate’s hyperplanes. Let K be the nerve of a family of standard boxes in Rd. Then K is a d-Leray complex and has the further property that if S is a set of vertices such that every pair of vertices in S form an edge, then S is a face of K. This property of the nerve corresponds to Helly number 2 for the original family, and we refer to it as Helly number 2. Problem 3.8. Extend Eckhoﬀ’s upper bound theorem to the class of d-Leray complexes with no missing faces of size greater than 3 (namely, those corresponding to Helly number 2). HELLY-TYPE PROBLEMS 483 3.7. Mutual position of convex sets. The study of nerves of convex sets is the study of intersection patterns of families of convex sets. When we start with a family of convex sets in Rd, we can go further and consider intersection patterns of the convex hulls of all subfamilies. (We can go even further by alternating between taking convex hulls and intersections and by considering statements regarding k-ﬂat transversals rather than plain intersections.) (a) (b) (c) (d) Figure 2. Mutual positions of three convex sets. Figure 2 shows various possible positions of three convex sets in the plane: (a) the convex hull of every two sets intersects the third set; (b) the convex hull of any two sets is disjoint from the third set, but all pairwise convex hulls have a point in common; (c) the three convex hulls of pairs of sets have no point in common; (d) the convex hull of two sets intersects the third set. Statements in this wider language can be regarded as the study of mutual positions of convex sets and they are, of course, of interest even for conﬁgurations of points, which we discuss in the next sections. Problem 3.9. Are there interesting things to say about the mutual position of convex sets? 3.8. Order types for points and sets. To conclude this section and prepare for the next, we brieﬂy mention the notion of order types (a.k.a. oriented matroids). These objects arise from conﬁgurations of points (or of hyperplanes) in real vector spaces, and can also be associated with directed graphs. Consider a sequence Y = (y1, y2, . . . , yn) of n points in Rd that aﬃnely span Rd. The order type described by Y can be seen as the set of all minimal Radon partitions. There is a more general axiomatic deﬁnition of order types that roughly requires that the restriction to every d + 3 points be an order type of d + 3 points in a real space. For general order types there is a topological representation that replaces the linear description 484 IMRE B´ AR´ ANY AND GIL KALAI of order types that correspond to point conﬁgurations. Another equivalent way to describe the order type is as follows: for every set J of subscripts i1, . . . , id+1 with 1 ≤i1 < · · · < id+1 ≤n, let sg(J, Y ) be the sign of the determinant of the (d + 1) × (d + 1) matrix (1) yi1 yi2 · · · yid+1 1 1 · · · 1 . Two sequences Y = (y1, y2, . . . , yn) and Z = (z1, z2, . . . , zn) of n points in Rd are equivalent (or have the same order type) if sg(J, Y ) = sg(J, Z) for all J ⊂[n] of size d + 1. For more on oriented matroids see [BLVS+93]. Returning to families of con-vex sets, we note that one way to record the mutual position of n convex sets K1, K2, . . . , Kn in Rd is by listing all order types of sequences y1 ∈K1, y2 ∈ K2, . . . , yn ∈Kn. Goodman and Pollack’s notion of allowable sequences for conﬁgurations [GP85] is a very useful way to study order types of planar conﬁgurations. The more general notion of interval sequences by Dhandapani, Goodman, Holmsen, and Pollack gives a way to record mutual positions of n convex planar sets [DGHP05]. 4. Around Tverberg’s theorem 4.1. Sierksma’s conjecture. Conjecture 4.1 (Sierksma’s conjecture). The number of Tverberg r-partitions of a set of (r −1)(d + 1) + 1 points in Rd is at least ((r −1)!)d. This question was raised by Sierksma [Sie79] in 1979 and not much progress has been achieved since. The best lower bound is about the square root of the conjectured one; this is a result of [Vˇ Z93] and [Hel07]. The conjecture, if true, is sharp, as shown by the example in Figure 3 for d = 2, r = 4: the vertices of the three triangles plus the point in the center is a set with ten points and 3!2 Tverberg partitions. Figure 3. Ten points with 3!2 Tverberg partitions. In Rd take analogously r −1 d-dimensional simplices with their center at the origin; their vertices together with the origin form a set of (r −1)(d + 1) + 1 points with (r −1)!d Tverberg partitions. There are further cases where equality holds, such as the one connected to the following problem raised by Perles. We need a HELLY-TYPE PROBLEMS 485 deﬁnition: a Tverberg partition S1, . . . , Sr of an m-element set X ⊂Rd is of type (a1, a2, . . . , ar) if the multisets {a1, a2, . . . , ar} and {\|S1\|, \|S2\|, . . . , \|Sr\|} coincide. Problem 4.2. Suppose that a1, a2, . . . , ar is a partition of m = (r −1)(d + 1) + 1 such that 1 ≤ai ≤d + 1 for every i. Is there a conﬁguration of m points in Rd for which all of Tverberg partitions are of type (a1, a2, . . . , ar)? This problem was raised by Perles many years ago and a positive answer was given by White [Whi17]. White’s examples provide a rich family of examples for cases of equality in Sierksma’s conjecture. An even more general family of con-structions for the equality cases, based on staircase convexity, is in the paper of Bukh, Loh, and Nivasch [BLN17]. A similar construction was given by P´ or [P´ or18] in connection with the so-called universal Tverberg partitions. Problem 4.3. Explore further examples of equality cases in Sierksma’s conjecture. 4.2. Topological Tverberg. Conjecture 4.4 (Topological Tverberg conjecture). Let f be a continuous function from the m-dimensional simplex σm to Rd. If m ≥(d + 1)(r −1), then there are r pairwise disjoint faces of σm whose images have a point in common. If f is a linear function, this conjecture reduces to Tverberg’s theorem. The case r = 2 was proved by Bajm´ oczy and B´ ar´ any [BB79] using the Borsuk–Ulam theorem. Moreover, for r = 2, one can replace the simplex by any other polytope of the same dimension. The case where r is a prime number was proved in an important paper of B´ ar´ any, Shlosman, and Sz˝ ucs [BSS81]. The prime power case was settled by ¨ Ozaydin, in an unpublished (yet available) paper [¨ Oza87]. For the prime power case, the proofs are quite diﬃcult and are based on computations of certain characteristic classes. In 2015 the topological Tverberg conjecture was disproved in a short note by Frick [Fri15]. This involves some early result on vanishing of topological obstruc-tions by ¨ Ozaydin, a theory developed by Mabillard and Wagner [MW14] extending Whitney’s trick to k-fold intersections, and a fruitful reduction by Gromov [Gro10], rediscovered and extended by Blagojevi´ c, Frick, and Ziegler [BFZ19]. Conjecture 4.5. Let f be a linear function from an m-dimensional polytope P to Rd. If m ≥(d + 1)(r −1), then there are r pairwise disjoint faces of P whose images have a point in common. Problem 4.6. Does the conclusion of the topological Tverberg conjecture hold if the images of the faces under f form a good cover (that is if all those images and all their nonempty intersections are contractible)? 4.3. Colorful Tverberg. Let C1, . . . , Cd+1 be disjoint subsets of Rd, called colors, each of cardinality at least t. A (d + 1)-subset S of Rd is said to be multicolored (or rainbow) if \|S ∩Ci\| = 1 for i = 1, . . . , d + 1. Let r be an integer, and let T(r, d) denote the smallest value t such that for every collection of colors C1, . . . , Cd+1 of size at least t each there exist r disjoint multicolored sets S1, . . . , Sr such that r i=1 conv Si ̸= ∅. The question of ﬁniteness of T(r, d) was raised in [BFL90] and proved there for the case d = 2. 486 IMRE B´ AR´ ANY AND GIL KALAI The general case was solved by an important theorem of ˇ Zivaljevi´ c and Vre´ cica [ˇ ZV92]. It asserts that T(r, d) ≤2r −1 if r is a prime, which implies that T(r, d) ≤ 4r −1 for all r and d. This theorem is one of the highlights of discrete geometry and topological combinatorics. The only known proofs of this theorem rely on topological arguments although the statement is about convex hulls, partitions, and linear algebra. The following question is a challenge for convex geometers. Problem 4.7. Find a nontopological proof of the ﬁniteness of T(r, d). B´ ar´ any and Larman [BL92] showed that T(r, 2) = r and asked the following. Conjecture 4.8 (Colorful Tverberg conjecture). T(r, d) = r. The case where r + 1 is a prime was proved by Blagojevi´ c, Matschke, and Ziegler [BMZ15]. It is a neat result by Lov´ asz from [BL92] that T(2, d) = 2 for all d. Sober´ on gives an equally neat (and very diﬀerent) proof of the same result in [Sob15]. The colorful Tverberg theorem is related to a well-known problem in discrete geometry, that of halving lines and hyperplanes. Given 2n points in general position in Rd, a halving hyperplane is a hyperplane with n points on each side. Problem 4.9. What is the maximum number H(2n, d) of partitions of a set of 2n points in Rd into equal parts via halving hyperplanes? Equivalently, what is the minimum number of non-Radon partitions with parts of equal size? A well-known conjecture, which is open even for d = 2, is that for a ﬁxed d, H(n, d) = nd−1+o(1). With the help of the colorful Tverberg theorem it was shown that H(n, d) = nd−ϵd, where ϵd is a positive constant depending on d. For d = 2 it is known that neC√nH(2, n) ≤O(n4/3). A matroid version of Tverberg’s theorem is the topic of [BKM17], which states the following. Assume that M is a matroid of rank d+1. Let b(M) denote the maximal number of disjoint bases in M. If f is a continuous map from the matroidal complex of M to Rd, then there exist t ≥1 4 b(M) independent sets σ1, . . . , σt ∈M such that t 1 f(σi) ̸= ∅. It is not clear how good this lower estimate on t is. Conjecture 4.10. In the above theorem, b(M) could be replaced by cb(M) for some absolute positive constant c. 5. The cascade conjecture and more When we have r < d + 2 points in Rd, they have a Radon partition iﬀthey are aﬃnely dependent. Are there conditions that guarantee that the existence of Tverberg partitions below the Tverberg number? In this section we will discuss the dimension of Tverberg points and the quest for conditions guaranteeing the existence of Tverberg partitions for conﬁgurations of points below the Tverberg number. 5.1. The cascade conjecture. For a set A, denote by Tr(A) the set of points in Rd that belong to the convex hull of r pairwise disjoint subsets of A. We call these points Tverberg points of order r. Let ¯ tr(A) = 1 + dim Tr(A). (Note that dim ∅= −1.) Radon’s theorem can be stated as follows: if ¯ t1(A) < \|A\|, then T2(A) ̸= ∅. There is a similar statement HELLY-TYPE PROBLEMS 487 which is still open: if ¯ t1(A) + ¯ t2(A) < \|A\|, then T3(A) ̸= ∅. We can go one step further: if ¯ t1(A) + ¯ t2(A) + ¯ t3(A) < \|A\|, then T4(A) ̸= ∅. These statements are special cases of the following. Conjecture 5.1 (Cascade conjecture). For every A ⊂Rd, r≥1 ¯ tr(A) ≥\|A\|. This is a question of Kalai from 1974 [Kal95]; see also [Kal00]. The conjecture was proved for d ≤2 by Akiva Kadari (unpublished MSc thesis in Hebrew). While this conjecture is wide open, we can ask for topological extensions of various kinds and for more general topological conditions for conﬁgurations of cardinality below the Tverberg number (r −1)(d + 1) + 1, that imply the existence of a Tverberg partition into r parts; see Problem 5.4. 5.2. Reay’s dimension conjecture. The following is a 1979 question from Reay [Rea79] where general position means weak general position; that is, no d+1 points lie in a hyperplane. Conjecture 5.2 (Reay’s conjecture). If A is a set of (d + 1)(r −1) + 1 + k points in general position in Rd, then dim Tr(A) ≥k. In particular, Reay’s conjecture asserts that a set of (d + 1)r points in general position in Rd can be partitioned into r sets of size d + 1 such that the simplices described by these sets have an interior point in common. This is easy when the points are in very general position, for instance, when they are algebraically independent. The main diﬃculty is how to use the weak general position condition. A recent result of Frick and Sober´ on [FS20] (see Section 7.1) is perhaps relevant here. While the conclusion of the cascade conjecture seems stronger than that of Reay’s dimension conjecture, it is not known how to derive it from the cascade conjecture. 5.3. Special cases of the cascade conjecture and expressing a directed graph as union of two trees. A special case of the cascade conjecture asserts that given 2d + 2 points in Rd, you can either partition them into two simplices whose interiors intersect, or you can ﬁnd a Tverberg partition into three parts. We give a reformulation based on positive hulls: given 2d nonzero vectors in Rd such that the origin is a vertex of the cone spanned by them, it is the case that either: • we can divide the points into two sets A and B so that the cones spanned by them have a d-dimensional intersection; or • we can divide them into three sets A, B, and C so that the cones spanned by them have a nontrivial intersection. Another interesting reformulation is obtained when we dualize using the Gale transform, and this has led to the problem we consider next: a very special class of conﬁgurations arising from graphs. Start from a directed graph G with n vertices and 2n −2 edges and associate with each directed edge (i, j) the vector ej −ei. This leads to the following problem. Problem 5.3. Let G be a directed graph with n vertices and 2n −2 edges. When can we divide the set of edges into two trees T1 and T2 (we disregard the orientation 488 IMRE B´ AR´ ANY AND GIL KALAI of edges) so that when we reverse the directions of all edges in T2 we get a strongly connected digraph? One of us (Kalai) conjectured that if G can be written as the union of two trees, the only additional obstruction is that there is a cut consisting only of two edges in reversed directions. Chudnovsky and Seymour found an additional necessary condition: there is no induced cycle v1, vk, . . . , v2k, v1 in G, such that each vertex ci is cubic, the edges of the cycle alternate in direction, and none of the vertices v1, . . . , v2k are sources or sinks of G. 5.4. Tverberg partitions of order 3 for conﬁgurations below the Tverberg number. Problem 5.4. When n < 2d + 3, ﬁnd conditions for the set of Radon points and the set of Radon partitions of a set X of n points in Rd that guarantee the existence of a Tverberg partition into three parts. The cascade conjecture asserts that if n = d + 2 + k and the dimension of Radon points is smaller than k, then there exists a Tverberg partition into three parts. While this is wide open, it would be interesting to propose a more general topological condition that suﬃces for the existence of a Tverberg partition into r parts. Conjecture 5.5. If the map from the Radon partitions of X to the Radon points of X is topologically degenerate (in some sense), then a Tverberg partition into three parts exists. In Problem 5.4 and Conjecture 5.5 we can relax the conclusion and can do so in various ways. For that we need a few deﬁnitions: The k-core of a ﬁnite set X in Rd is the intersection of the convex hull of all sets A ⊂X with \|X \ A\| ≤k, that is, corek X = {conv A : A ⊂X, \|X \ A\| ≤k}. The case k = 0 is the usual convex hull. The k-Radon core of a ﬁnite set X in Rd is the intersection of Radon points of all sets A ⊂X with \|X \ A\| ≤k; this is the set of points in Rd that remain Radon points of X even after we delete k points from X in all possible ways. (Clearly, the Tverberg points of order 3 are in the ﬁrst Radon core, and the points in the ﬁrst Radon core are in the 2-core.) Problem 5.6. When n < 2d + 3, ﬁnd conditions for the set of Radon points and the set of Radon partitions of a set X of n points in Rd that guarantee the following: (1) the second core of X, core2 X, is nonempty; (2) the ﬁrst Radon core of X is nonempty; (3) X admits a Reay (3,2)-partition, that is, a partition into three parts such that the convex hulls are pairwise intersecting (see Section 6.2). 5.5. Radon partitions and Radon points for conﬁgurations based on cubic graphs. Let G be a cubic graph with n vertices {v1, v2, . . . , vn}. Associate with ev-ery edge {vi, vj} in G its characteristic vector in Rd, giving a conﬁguration Conf(G) of 3n/2 points in (n −1)-dimensional space. In [Onn01] and also in personal com-munication (2011), Onn observed that the existence of a Tverberg 3-partition (or even of a Reay (3,2)-partition; see Section 6.2) is equivalent to a 3-edge coloring for G, and he concluded that deciding if a conﬁguration of 3(d + 1)/2 points in Rd (d an odd integer) admits a Tverberg partition into three parts is NP-complete. HELLY-TYPE PROBLEMS 489 The following problem is motivated by the four-color theorem. Problem 5.7. (i) Study Radon partitions and Radon points for conﬁgurations based on cubic graphs. (ii) Find conditions for the Radon points and Radon partitions of Conf(G) that guarantee a 3-edge coloring for G. It would be interesting to ﬁnd conditions for Problem 5.4 and Conjecture 5.5 that would imply the 3-edge colorability of bipartite cubic graphs and, much more ambitiously, conditions that would imply the four-color theorem, namely, the 3-edge colorability of planar cubic graphs. 6. More around Tverberg’s theorem 6.1. Eckhoﬀ’s partition conjecture. Let X be a set endowed with an abstract closure operation X →cl(X). The only requirements of the closure operation are (1) cl(cl(X)) = cl(X), and (2) A ⊂B implies cl(A) ⊂cl(B). Deﬁne tr(X) to be the largest size of a (multi)set in X that cannot be partitioned into r parts whose closures have a point in common. The following conjecture is due to Eckhoﬀ[Eck00]. Conjecture 6.1 (Eckhoﬀ’s partition conjecture). For every closure operation, tr ≤t2 · (r −1). If X is the set of subsets of Rd and cl(A) is the convex hull operation, then Radon’s theorem asserts that t2(X) = d + 1 and Eckhoﬀ’s partition conjecture implies Tverberg’s theorem. In 2010 Eckhoﬀ’s partition conjecture was refuted by Boris Bukh [Buk10]. Bukh’s beautiful paper contains several important ideas and further results. We will mention one ingredient. Recall the nerve construction for moving from a family F of n convex sets to the simplicial complex that records empty and nonempty intersections for all subfamilies G of F. Bukh studied sim-plicial complexes whose vertex sets correspond to the power set of a set of size n: starting with n points in Rd or some abstract convexity space, consider the nerve of convex hulls of all 2n subsets of these points! In Bukh’s counterexample, tr = t2 · (r −1) + 1, which is just one larger than the conjectured bound. Perhaps tr ≤t2 · (r −1) + c for some universal constant c ≥1. There is a recent and positive development about Eckhoﬀ’s conjecture. P´ alv¨ olgyi [P´ al20] has proved that tr grows linearly in r, that is, tr ≤cr where the constant c depends only on r2. Problem 6.2. Find classes of closure operations for which tr ≤t2 · (r −1). We can ask if the inequality tr ≤t2 · (r −1) holds for Moran and Yehudayoﬀ’s convexity spaces considered in Section 2.5. Bukh’s paper includes an interesting notion that extends the notion of nerves. Given a conﬁguration of points in the Euclidean space or in an abstract convexity space, we consider the nerve of convex hulls of all nonempty subsets of the points. This is a simplicial complex that we refer to as the B-nerve of the conﬁguration, 490 IMRE B´ AR´ ANY AND GIL KALAI with the additional structure that vertices are labeled by subsets, and with some additional combinatorial properties. Problem 6.3. Study properties of B-nerves of point conﬁgurations in Rd. 6.2. A conjecture by Reay. For a set X ⊂Rd, a Reay (r, j)-partition is a par-tition of X into subsets S1, S2, . . . , Sr such that j i=1 conv Ski ̸= ∅, for every 1 ≤k1 < k2 < · · · < kj ≤r. In other words, the convex hulls of any j sets of the partition intersect. Deﬁne R(d, r, j) as the smallest integer m such that every m-element set X ⊂Rd has a Reay (r, j)-partition. Reay [Rea79] conjectured that you cannot improve the value given by Tverberg’s theorem, namely, that Conjecture 6.4 (Reay’s conjecture). R(d, r, j) = (r −1)(d + 1) + 1. Micha A. Perles believes that Reay’s conjecture is false even for j = 2 and r = 3 for large dimensions but, with Moriah Sigron, he proved [PS16] the strongest positive results in the direction of Reay’s conjecture. 6.3. Two old problems and universality. Problem 6.5 (McMullen and Larman). How many points v(d) guarantee that for every set X of v(d) points in Rd there exists a partition into two parts X1 and X2 such that for every p ∈X, conv (X1\p) ∩conv (X2\p) ̸= ∅. This is a strong form of Radon’s theorem: the partition X1, X2 of X = X1 ∪X2 remains a Radon partition even after we delete any point from X. Similar questions can be asked about Tverberg partitions. Larman [Lar72] proved that v(d) ≤2d+3 and this bound is sharp for d = 1, 2, 3, 4. The lower bound v(d) ≥⌈5d 3 ⌉+ 3 is a result of Ram´ ırez Alfons´ ın [RA01]. This problem is the dual form of the original question by McMullen: What is the largest integer n = f(d) such that every set of n points in general position in Rd is projectively equivalent to the set of vertices of a convex polytope? A related problem is the following. Problem 6.6. How many points T(d; s, t) in Rd guarantee that they can be divided into two parts such that every union of s convex sets containing the ﬁrst part has a nonempty intersection with every union of t convex sets containing the second part? We explain next why R(d; s, t) is ﬁnite. This is a fairly general Ramsey-type argument and it gives us an opportunity to mention a few recent important results. The argument has two parts: (1) Prove that T(d; s, t) is ﬁnite (with good estimates) when the points are in cyclic position (to be deﬁned shortly). (2) Use the fact that for every d and n there is f(d, n) such that among every m points in general position in Rd, m > f(d, n), one can ﬁnd n points in cyclic position. The ﬁniteness of T(d; r, s) follows (with horrible bounds) from these two ingredi-ents by standard Ramsey-type results. It would be nice to understand the behavior of this function. Statement (2) is a kind of universality theorem. In a more precise form it says that for every d and n there is an integer f(d, n) such that the following holds. HELLY-TYPE PROBLEMS 491 Every sequence x1, . . . , xm in Rd in general position with m ≥f(d, n) contains a subsequence y1, . . . , yn such that all simplices of this subsequence are oriented the same way. The latter condition says, in more precise form, that for every set of subscripts i1, . . . , id+1 with 1 ≤i1 < · · · < id+1 ≤n, the sign of the determinant of the (d + 1) × (d + 1)-matrix (2) yi1 yi2 · · · yid+1 1 1 · · · 1 is the same (and diﬀerent from 0). Now a point set is cyclic if its elements can be ordered so that the simplices along this ordering have the same orientation. Statement (2) says that the property of being cyclic is universal because every long enough sequence of points in general position contains a cyclic subsequence of length n. Every ﬁnite sequence of points on the moment curve is cyclic. This shows that no other type of point sequence can be universal. Recently, a fairly good understanding of f(d, n) has been achieved in a series of papers. Theorem 6.1. f(d, n) = twrd(θ(n)). Here, twrd is the d-fold tower function. The lower bound is by Suk [Suk14] (improving earlier bounds by Conlon, Fox, Pach, Sudakov, and Suk [CFP+14]) and the upper bound comes from B´ ar´ any, Matouˇ sek, and P´ or [BMP16]. The following, somewhat vague, question emerges here naturally. Problem 6.7. Determine the universal type of n lines in R3 and in Rd. More generally, what is the universal type of n k-dimensional aﬃne ﬂats in Rd? Some preliminary results in this direction are the topic of a forthcoming paper by B´ ar´ any, Kalai, and P´ or [BKP21]. We note that the order type of a sequence of points does not determine its Tverberg partitions. Problem 6.8. Develop a notion of order type based on Tverberg partitions into at most r parts, r ≥3. Here, Perles and Sigron’s work on strong general position [PS16] and P´ or’s uni-versality theorem [P´ or18] could be relevant. 7. Carath´ eodory and weak ϵ-nets 7.1. Colorful Carath´ eodory and the Rota basis conjecture. The following question was raised in Chow’s Polymath 12 [Cho17] dedicated to Rota’s basis con-jecture. Consider the d + 1 sets (or colors if you wish) C1, C2, . . . , Cd+1 of points in Rd. Assume that each \|Ci\| = d + 1 and that the interior of each conv Ci contains the origin. Problem 7.1 (D. H. J. Polymath). Can we ﬁnd a partition of all points into d + 1 rainbow parts such that the interior of the convex hulls of the parts have a point in common. (A rainbow set is a set containing one element from each Ci.) To see the connection, ﬁrst recall Rota’s basis conjecture. Conjecture 7.2 (Rota’s basis conjecture). If B1, B2, . . . , Bn are disjoint bases in Rn (or even in an arbitrary matroid), then it is possible to ﬁnd n new disjoint bases C1, C2, . . . , Cn such that each Ci contains one element from every Bj. 492 IMRE B´ AR´ ANY AND GIL KALAI Note that Rota’s basis conjecture can be stated (over R) as follows. Consider d+1 sets (or colors) C1, C2, . . . , Cd+1 of points in Rd. Assume that each \|Ci\| = d+1 and that the interior of each conv Ci is nonempty. Then there exists a partition of all points into d + 1 rainbow parts such that the interior of the convex hulls of each part is nonempty. Returning to Problem 7.1, we note here that according to the colorful Carath´ e-odory theorem there is a rainbow set whose convex hull contains the origin. Without the words “the interiors of”, Problem 7.1 would be a special case of the colorful Tverberg conjecture (Section 4.3). A positive answer would be a strong variant of Reay’s conjecture (Section 5.2) on the dimension of Tverberg points, and, as explained before, also a strong form of Rota’s basis conjecture over the reals. A recent result of Frick and Sober´ on [FS20] is that a set of r(d + 1) points in Rd can always be partitioned into r sets, each of size d + 1, such that the convex hulls of the parts have a point in common. This theorem is related to the uncolored case of Problem 7.1 but without the word “interior”. 7.2. The complexity of the colorful Carath´ eodory theorem and of Tver-berg partitions. Problem 7.3. Consider d + 1 sets C1, C2, . . . , Cd+1 of points in Rd. Assume that each \|Ci\| = d + 1 and that each conv Ci contains the origin. Is there a polynomial-time algorithm to ﬁnd a rainbow simplex containing the origin? An interesting result in this direction is due to Meunier et al. [MMSS17]. They show that the problem lies in the intersection of complexity classes PPAD and PLS. The same applies to the analogous question about Tverberg partitions: Is there a polynomial-time algorithm to ﬁnd a Tverberg partition of an (r −1)(d + 1) + 1-element point set in Rd? There are very few geometric problems in both classes PPAD and PLS that are not known to be solvable in polynomial time. The results in [MMSS17] are the ﬁrst upper bound on the complexity of these problems. 7.3. Carath´ eodory-type theorem for cores. Recall the deﬁnition of the k-core of a ﬁnite set X in Rd from Section 5.4. The Carath´ eodory number for the k-core is the smallest integer f(d, k) with the property that a ∈corekX (where X ⊂Rd) implies the existence of Y ⊂X such that a ∈corekY and \|Y \| ≤f(d, k). So f(d, 0) = d + 1 is just the Carath´ eodory theorem. B´ ar´ any and Perles [BP90] established the ﬁniteness of f(d, k) together with some other properties of this function, for instance, that f(d, 1) = max{2(d + 1), 1 + d + ⌊d2/4⌋}, and that f(2, k) = 3(k + 1). Several questions remain open; we mention only two of them. Problem 7.4. Determine f(d, 2) and f(3, k). 7.4. The covering number theorem. Assume that X ⊂Rd is ﬁnite and \|X\| ≥ d + 1. A simplex of X is just conv Y where Y ⊂X and \|Y \| = d + 1. According to Carath´ eodory’s theorem every point in conv X is contained in a simplex of X; that is, conv X is covered by the simplices of X. Which point is covered maximally, and how many times is it covered? A famous result of Boros and F¨ uredi [BF84] says that in the planar case there is a point covered by 2 9 n 3 + O(n2) simplices (that is, triangles) of X, where n = \|X\|. This is a positive fraction of all triangles of X and the constant 2 9 is the best possible. In higher dimensions Tverberg’s theorem and the colorful Carath´ eodory theorem imply (see [B´ ar82]) the following result. HELLY-TYPE PROBLEMS 493 Theorem 7.1 (Covering number). Assume X is a set of n ≥d + 1 points in Rd. Then there is a point covered by 1 (d+1)d n d+1 simplices of X. This is again a positive fraction of all simplices of X. Deﬁne bd as the supremum of all β > 0 such for that every set X of n ≥d + 1 points in Rd there is a point covered by β n d+1 simplices of X. So bd ≥(d + 1)−d. In a remarkable paper, Gromov [Gro10] showed that bd ≥ 2d (d+1)!(d+1). Gromov’s theorem applies to continuous maps from the boundary of an (n −1)-dimensional simplex to Rd. His estimate is an exponential improvement on the previous bounds. Both Gromov’s theorem and Pach’s theorem below play an important role in the emerging theory of high-dimensional expanders [FGL+12]. From the other direction Bukh, Matouˇ sek, and Nivasch [BMN11] give an exam-ple, based on the stretched grid, that shows bd ≤ (d+1)! (d+1)d+1 . They conjecture that this is the right value of bd. Conjecture 7.5. Show that bd = (d+1)! (d+1)d+1 . More modestly, prove that bd is expo-nential in d. An interesting extension of the covering number theorem is the following result of Pach [Pac98]. Theorem 7.2 (Pach’s theorem). Assume that C1, . . . , Cd+1 are sets (colors, if you like) in Rd, each of size n. Then there is a point p ∈Rd and there are subsets Di ⊂Ci (for all i ∈[d + 1]), each of size at least c(d)n such that the convex hull of every transversal of the system D1, . . . , Dd+1 contains p. Here c(d) > 0 is a constant that depends only on d. This is a homogeneous version of the covering number theorem. It was conjec-tured in [BFL90], where case d = 2 was proved more generally even if the sets C1, C2, C3 need not have the same size. This raises the following question. Problem 7.6. Does Pach’s theorem remain true if the sets C1, . . . , Cd+1 have arbitrary sizes? We mention that Pach’s theorem does not have a topological extension, as shown in [BMNT18] and in [BH20] in a stonger form. 7.5. Weak ϵ-nets. An important application of the covering number theorem is about weak ϵ-nets. Let ϵ > 0 be ﬁxed. Given a ﬁnite set X of n ≥d + 1 points, let C be the (ﬁnite) family of sets conv Y for all Y ⊂X with \|Y \| ≥ϵn. A set F ⊂Rd is called a weak ϵ-net for X if F ∩C ̸= ∅for every C ∈C. Theorem 7.3 (Weak ϵ-net theorem). Under the above conditions there is a weak ϵ-net F for X such that \|F\| ≤ cd ϵd+1 , where cd > 0 is a constant. The upper bound on the size of F is from [AK92b] and [ABFK92] and has been improved to O(ϵ−d), disregarding some logarithmic terms. The trivial lower bound on the size of F is 1 ϵ . Bukh, Matouˇ sek, and Nivasch [BMN11] give an example (based on the stretched grid or staircase convexity) where the size of the weak ϵ-net is at least of order 1 ϵ (log 1 ϵ )d−1. So the bounds on the size of a weak ϵ-net 494 IMRE B´ AR´ ANY AND GIL KALAI are far from each other, and the general belief is that the true behavior should be slightly superlinear in 1 ϵ . Problem 7.7. Find a better upper bound for the size of a weak ϵ-net. One remarkable improvement in this direction is a result of Rubin [Rub18] who showed that in the planar case there is always a weak ϵ-net of size of order 1 ϵ1.5+δ for any δ > 0. A more recent result of Rubin [Rub21] applies in any dimension d ≥2 and gives a weak ϵ-net of size of order 1 ϵd−1/2+δ for any δ > 0. Weak ϵ-nets can be deﬁned not only for points but for k-dimensional aﬃne ﬂats in Rd. We only state the question for lines in R3 and leave the rest of the cases to our imaginary reader. Let L be a set of n lines, and let C be a ﬁnite family of convex sets in R3. Assume that every C ∈C intersects an ϵ-fraction of the lines in L, that is, \|{L ∈L : C ∩L ̸= ∅}\| ≥ϵn for every C ∈C. Conjecture 7.8 (ϵ-net of lines). Under these conditions, there is a set of lines L∗ whose size depends only on ϵ such that every C ∈C intersects some line in L∗. The set L∗can be thought of as a weak ϵ-net of lines for C. We will encounter this question again soon in connection with Problem 8.1. 8. A glance at common transversals 8.1. Transversals for intersecting families. A k-transversal of a family of con-vex sets in Rd is a k-dimensional aﬃne space that intersects every set in the fam-ily. Transversal theory deals with conditions that guarantee the existence of k-transversals. The case k = 0 is connected to Helly-type theorems, and there are some general results for hyperplane transversals, namely, k = d −1, and very few general results for 0 < k < d −1 and, in particular, for line transversals in R3. The fascinating theory geometric transversals goes beyond the scope of this paper; for surveys see Goodman, Pollack, and Wenger [GPW93], Wenger [Wen99], and Holmsen [Hol13]. We will mention only a few problems where the conditions are in terms of the intersection pattern of the sets in the family. Problem 8.1. Assume that a family C of n convex sets in R3 satisﬁes the property that any two sets in C intersect. Show that there is a line intersecting cn elements in C, where c > 0 is a universal constant. Partial results in this direction are given in [B´ ar21]. Problem 8.1 is the ﬁrst, and so far most interesting, unsolved case of a series of problems of the same type. Namely, for what numbers k, r, d is it true that, given a family C of convex sets in Rd where every k tuple is intersecting, there is an r-ﬂat intersecting a positive fraction of the sets in C? Of course, the positive fraction should depend only on k, r, and d. An interesting example satisfying the conditions is when C consists of n lines in a two-dimensional plane in R3. Then, of course, every set in C is a line transversal for all sets in C. This example shows that degenerate cases are going to make the problem diﬃcult. Figure 4 is an example of ﬁve pairwise intersecting convex sets in R3 without a common line transversal. The ﬁve sets comprise three rectangles and two triangles, all of whose vertices belong to two parallel planes H0 and H1. HELLY-TYPE PROBLEMS 495 H0 H1 Figure 4. Five sets in R3 that pairwise intersect and have no line transversal. The question comes from a paper by Mart´ ınez, Rold´ an, and Rubin [MSRPR20] and is connected to the colorful Helly theorem. They also ask the slightly more general bipartite version of the question. Problem 8.2. Assume F and G are ﬁnite families of convex sets in R3 with the property that A ∩B ̸= ∅for any two sets A ∈F and B ∈G. Show that there is a line intersecting c\|F\| elements of F or c\|G\| elements of G where c > 0 is again a universal constant. An example is two sets F and G of lines on a doubly ruled surface, which shows that degenerate cases may cause diﬃculties again. It is worth mentioning that both questions are invariant under nondegenerate aﬃne transformation. We observe here that a positive answer to Conjecture 7.8 from the last section would imply that in Problem 8.1 there is a very ﬁnite set L of lines intersecting every element of C, where by “very ﬁnite” we mean that the size of L is bounded by 1000, say, or by some other absolute constant. 9. Conclusion This paper introduces the fascinating area of Helly-type theorems, and describes some of its main themes and goals through a variety of open problems. Often, results from convexity give a simple and strong manifestation of theorems from topology: Helly’s theorem manifests the nerve theorem from algebraic topology, and Radon’s theorem can be regarded as an early “linear” appearance of the Borsuk– Ulam theorem. One of our main themes is to further explore these connections to topology. Helly-type theorems also oﬀer complex and profound combinatorial connections and applications that represent the second main theme of this paper. We note that Helly-type theorems and the interplay between convex geometry, combinatorics, and topology play an important role in the emerging theory of high-dimensional expanders. There are various related parts of this theory that we did not consider. We gave only a small taste of the theory of common transversals, we did not discuss the closely related theorems of Kirchberger and Krasnoselskiˇ i, and we did not consider the rich connections to metric geometry. For example, when you consider families of translates of a ﬁxed convex set, the theory takes interesting and surprising turns, and it has applications and connections, e.g., to the theory of Banach spaces. 496 IMRE B´ AR´ ANY AND GIL KALAI About the authors Imre B´ ar´ any is a research professor at the R´ enyi Institute of Mathematics, Bu-dapest, Hungary, and emeritus professor at University College London. 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MR1185000 502 IMRE B´ AR´ ANY AND GIL KALAI R´ enyi Institute of Mathematics, 13-15 Re´ altanoda Street, Budapest, 1053 Hungary; and, Department of Mathematics, University College London, Gower Street, London, WC1E 6BT, United Kingdom Email address: barany.imre@renyi.hu Einstein Institute of Mathematics, Hebrew University, Jerusalem 91904, Israel; and Efi Arazy School of Computer Science, IDC, Herzliya, Israel Email address: kalai@math.huji.ac.il
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Contents 1 Ellipse 1.1 Equation of ellipse 1.2 Length of latus rectum 2 Second definition of ellipse 2.1 Proof 3 The directrices 3.1 Ellipse using focus and directrix 4 General ellipse at origin 5 General Ellipse 6 An Example of the General Ellipse 7 K and "Standard Form" 8 Tangent at latus rectum 9 Reflectivity of ellipse 10 Links to related Topics Ellipse Add links Resource Discuss Read Edit Edit source View history Read Edit Edit source View history What links here Related changes Permanent link Page information Cite this page Get shortened URL Download QR code Wikimedia Projects Commons Wikibooks Wikidata Wikinews Wikipedia Wikiquote Wikisource Wikispecies Wikivoyage Wiktionary Meta-Wiki Outreach MediaWiki Wikimania Print/export Create a book Download as PDF Printable version In other projects Wikimedia Commons Wikipedia Wikiquote Wikidata item Appearance From Wikiversity \| \| \| Subject classification: this is a mathematics resource. \| \| \| \| Search for Ellipse on Wikipedia. \| Ellipse [edit \| edit source] In cartesian geometry in two dimensions the ellipse is the locus of a point that moves relative to two fixed points called The distance from one to the other is non-zero. The sum of the distances from point to foci is constant. See figure 1. The center of the ellipse is located at the origin and the foci are on the at distance from has coordinates has coordinates . Line segments By definition the length of the Each point where the curve intersects the major axis is called a are the vertices of the ellipse. Line segment perpendicular to the major axis at the midpoint of the major axis is the with length Any line segment that intersects the curve in two places is a A chord through the focus is a Each focal chord perpendicular to the major axis is a Equation of ellipse [edit \| edit source] Let point have coordinates Make appropriate substitutions, expand and the result is: or If the equation is expressed as Length of latus rectum [edit \| edit source] Length of latus rectum Second definition of ellipse [edit \| edit source] \| \| \| Ellipse is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant. Let where: is non-zero, Therefore, Let directrix have equation where At point Statement is true at point also. Section under "Proof" below proves that statement (3) is true for any point on ellipse. \| Proof [edit \| edit source] \| \| \| As expressed above in statement second definition of ellipse states that ellipse is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant. This section proves that this definition is true for any point on the ellipse. At point base Similar calculations can be used to prove the case for Focus2 and Directrix2 in which case: Therefore: where Ellipse is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant, called eccentricity \| The directrices [edit \| edit source] See Figure 1. The vertical line through with equation is a of the ellipse. Likewise for the vertical line A second definition of the ellipse: the locus of a point that moves relative to a point, the focus, and a fixed line called the directrix, so that the distance from point to focus and the distance from point to line form a constant ratio Consider the point in the figure. Let the length Consider the point : Length . Distance from center to directrix Distance from center to directrix = . . . . Distance from focus to directrix . Ellipse using focus and directrix [edit \| edit source] See Figure 2. Let the point be , the focus be and the directrix have equation where The directrix is horizontal and the major axis vertical through the origin . Expand and the result is: Compare this equation with the equation generated earlier: . When the equation is the major axis is horizontal. When the equation is the major axis is vertical. General ellipse at origin [edit \| edit source] The general ellipse allows for the major axis to have slope other than horizontal or vertical. See Figure 1. The line is the major axis of the ellipse shown in red. Points are the foci with coordinates respectively. Point is any point on the curve. By definition Length Length Make appropriate substitutions, expand and the result is: . This equation has the form where: because center is at origin, An example Let be (3,4) and The ellipse is: or Reverse-engineering ellipse at origin Given an ellipse in format calculate is non-zero. Coefficients provided could be, for example, or or where is an arbitrary constant and all groups of coefficients define the same ellipse. To produce consistent, correct values for the equations become: or: Solutions are: . This formula for is valid if both are . where You should see one positive value and one negative value for Choose the positive value and . The solutions become simpler if K == 1. if ( K != 1 ) { A ← KA; B ← KB; C ← KC; } and the solutions for become: where With values available all the familiar values of the ellipse may be calculated: Equation of major axis: in normal form. Equation of minor axis: in normal form. Equations of directrices: in normal form. An example using focus and directrix Given focus and directrix with equation , calculate equation of the ellipse in form See Figure 2. Let point be any point on the ellipse. Distance from to focus Distance from to directrix . . . . . Expand and the result is: Because the center of the ellipse is at the origin and the various lines have equations as follows. Minor axis: Other directrix: Major axis: If you reverse-engineer the ellipse using the method above, , the expression becomes , and the expression becomes . General Ellipse [edit \| edit source] The ellipse may assume any position and any orientation in Cartesian two-dimensional space. See the red curve in Figure 1. The equation of this curve may be derived as follows: Given two of length and point The expansion of this expression is somewhat complicated because it contains 5 variables The expansion may be simplified by reducing the number of variables from to , the familiar Let , the center of the ellipse, have coordinates where and Then and The expansion is: where: A different approach Begin with ellipse at origin: where: By translation of coordinates, move the ellipse so that the new center of the ellipse is: ← ← The equation above becomes: where: the same as the value of in the method above. The expansion of will show that this method produces the same results as the method above. Given foci and major axis, perhaps the simplest way to produce is to calculate and move the center from to . Center of ellipse Given we know from above that Therefore where the point is the center of the ellipse. An Example of the General Ellipse [edit \| edit source] See Figure 1. Given foci and calculate the equation of the ellipse in form Calculate the center: Equation of ellipse at origin: Move ellipse from origin to Reverse-engineering the general ellipse Given ellipse calculate the foci and the major axis. Calculate the center of the ellipse (point ): Solutions are: where In this example, If != ← ← ← The values may be calculated as in "Reverse-engineering ellipse at origin" above. Length of major axis Focus Focus Significant lines of the Ellipse The significant lines of the ellipse are: each each Consider the ellipse in Figure 2. Given foci and calculate the equations of all the significant lines. Slope of major axis Major axis has equation and it passes through Therefore, Major axis has equation: Center of ellipse Minor axis is perpendicular to major axis. Therefore, minor axis has equation: and it passes through the center Equation of minor axis (orange line through ): or or Using the equation of the minor axis, the fact that each latus rectum is parallel to the minor axis, and that the distance from minor axis to latus rectum each latus rectum has equation: Equation of latus rectum (blue line) through Equation of latus rectum (blue line) through Using the equation of the minor axis, the fact that each directrix is parallel to the minor axis, and that the distance from minor axis to directrix each directrix has equation: Equation of directrix (red line) through Equation of directrix (red line) through K and "Standard Form" [edit \| edit source] Everything about the ellipse can be derived from the last three of which are contained within: for ellipse at origin. Consider the ellipse: the green curve in Figure 1. It is tempting to say that These values satisfy However, These values for are not correct. Put the equation of the ellipse into "standard form." In this context "standard form" means that For ellipse at origin In fact ← ← ← ← The equation of the ellipse becomes: and The equation of the ellipse is in "standard form" and: The values are correct. Example 2. Consider the ellipse the red curve in Figure 1. In this example, and the equation in "standard form" is: . Example 3. Consider the ellipse the blue curve in Figure 1. The center In this example and the equation in "standard form" is: . Tangent at latus rectum [edit \| edit source] See Figure 1. The red curve is that of an ellipse at the origin with major axis vertical: The line is the directrix with equation: The green line has equation: The aim of this section is to show that the line is tangent to the ellipse at the Let the line intersect the curve. The coordinates of the point of intersection are given by: If the line is a tangent, has one value and the discriminant is or: The tangent has slope and equation: Let this line intersect the curve. The coordinates of the points of intersection are given by: Discriminant = half length of latus rectum. The tangent touches the curve where , the point where is the latus rectum. Reflectivity of ellipse [edit \| edit source] See Figure 1. The curve (red line) is an ellipse with equation: where Foci have coordinates Line is tangent to the curve at point A ray of light emanating from focus is reflected from the inside surface of the ellipse at point and passes through the other focus The aim is to prove that Point has coordinates At point Slope of line Slope of line Slope of curve at Using , if then: , , and where If you make the substitutions and expand, the result is . Therefore, angle of reflection angle of incidence and the reflected ray passes through the other focus Links to related Topics [edit \| edit source] Conic sections Quartic function#Two Conic Sections Retrieved from " Category: Mathematics Hidden category: Resources with related material at Wikipedia Add topic
15529	https://amesweb.info/Materials/Linear-Thermal-Expansion-Coefficient-Metals.aspx	Linear Thermal Expansion Coefficient for Metals AmesWeb LINEAR THERMAL EXPANSION COEFFICIENT FOR METALS Linear thermal expansion coefficients of metals including aluminum, steel, bronze, iron, brass, copper, gold, silver, invar, magnesium, nickel, titanium and zinc are given in the following thermal expansion coefficients chart. These linear thermal expansion coefficients are room temperature values of metals. Linear thermal expansion coefficient is defined as material's fractional change in length divided by the change in temperature. Coefficient of linear thermal expansion is designated by the symbol α (alpha). The SI unit of thermal expansion coefficient is (°C)-1 and U.S. customary unit is (°F)-1. Room Temperature Linear Thermal Expansion Coefficient Values for Metals Aluminum Alloys Material Coefficient of Thermal Expansion (CTE) 10-6 (°C)-1 10-6 (°F)-1 Aluminum Alloy 1100 23.6 13.1 Aluminum Alloy 2011 23.0 12.8 Aluminum Alloy 2024 22.9 12.7 Aluminum Alloy 5086 23.8 13.2 Aluminum Alloy 6061 23.6 13.1 Aluminum Alloy 7075 23.4 13.0 Aluminum Alloy 356.0 21.5 11.9 Copper-Base Alloys Material Coefficient of Thermal Expansion 10-6 (°C)-1 10-6 (°F)-1 Copper Alloy C11000 (electrolytic tough pitch)17.0 9.4 Copper Alloy C17200 (beryllium - copper)16.7 9.3 Copper Alloy C22000 (Commercial bronze, 90%)18.4 10.2 Copper Alloy C23000 (Red brass, 85 %)18.7 10.4 Copper Alloy C26000 (cartridge brass)19.9 11.1 Copper Alloy C27000 (Yellow Brass)20.3 11.3 Copper Alloy C36000 (free - cutting brass)20.5 11.4 Copper Alloy C51000 (Phosphor bronze, 5% A)17.8 9.9 Copper Alloy C62300 (Aluminum bronze, 9%)16.2 9.0 Copper Alloy C71500 (copper - nickel, 30%)16.2 9.0 Copper Alloy C93200 (bearing bronze)18.0 10.0 Cast Irons Material Coefficient of Thermal Expansion 10-6 (°C)-1 10-6 (°F)-1 Gray Irons Grade G1800 11.4 6.3 Grade G3000 11.4 6.3 Grade G4000 11.4 6.3 Ductile Irons Grade 60-40-18 11.2 6.2 Grade 80-55-06 10.6 5.9 Precious Metals Material Coefficient of Thermal Expansion 10-6 (°C)-1 10-6 (°F)-1 Gold (commercially pure)14.2 7.9 Silver (commercially pure)19.7 10.9 Steels Material Coefficient of Thermal Expansion 10-6 (°C)-1 10-6 (°F)-1 Plain Carbon and Low Alloy Steels Steel Alloy A36 11.7 6.5 Steel Alloy 1020 11.7 6.5 Steel Alloy 1040 11.3 6.3 Steel Alloy 4140 12.3 6.8 Steel Alloy 4340 12.3 6.8 Stainless Steels Stainless Alloy 304 17.2 9.6 Stainless Alloy 316 15.9 8.8 Stainless Alloy 405 10.8 6.0 Stainless Alloy 440A 10.2 5.7 Stainless Alloy 17-7PH 11.0 6.1 Titanium Alloys Material Coefficient of Thermal Expansion 10-6 (°C)-1 10-6 (°F)-1 Commercially Pure (ASTM Grade 1)8.6 4.8 Titanium Alloy Ti - 5Al - 2.5Sn 9.4 5.2 Titanium Alloy Ti - 6Al - 4V 8.6 4.8 Titanium Alloy Ti-8Mn 10.8 6.0 Various Metals Material Coefficient of Thermal Expansion 10-6 (°C)-1 10-6 (°F)-1 Magnesium Alloy AZ31B 26.0 14.4 Nickel 200 13.3 7.4 Inconel 625 12.8 7.1 Inconel 718 13.0 7.2 Monel 15.7 8.7 Monel 400 13.9 7.7 Haynes Alloy 25 12.3 6.8 Invar 1.6 0.9 Super Invar 0.72 0.40 Kovar 5.1 2.8 Chemical Lead 29.3 16.3 Antimonial lead (6 %)27.2 15.1 Tin (Commercially pure)23.8 13.2 Lead - tin solder (60Sn - 40 Pb)24.0 13.3 Zinc (Commercially pure)23.0 - 32.5 12.7 - 18.1 Reference: Callister.W.D,JR. (2007). Materials Science and Engineering: An Introduction . 7th edition. John Wiley & Sons, Inc. Oberg.E , Jones.D.J., Holbrook L.H, Ryffel H.H., (2012) . Machinery's Handbook . 29th edition. Industrial Press Inc., pp 376 - 377 About us\| Contact us\| Disclaimer \| Privacy Policy Copyright © 2013-2023
15530	https://it.wikipedia.org/wiki/Sopraelevata_Aldo_Moro	Sopraelevata Aldo Moro - Wikipedia Vai al contenuto [x] Menu principale Menu principale sposta nella barra laterale nascondi Navigazione Pagina principale Ultime modifiche Una voce a caso Nelle vicinanze Vetrina Aiuto Sportello informazioni Pagine speciali Comunità Portale Comunità Bar Il Wikipediano Contatti Ricerca Ricerca [x] Aspetto Aspetto sposta nella barra laterale nascondi Testo Piccolo Standard Grande Questa pagina utilizza sempre caratteri di piccole dimensioni Larghezza Standard Largo Il contenuto è il più ampio possibile per la finestra del browser. Colore (beta) Automatico Chiaro Scuro Questa pagina è sempre in modalità luce. Fai una donazione registrati entra [x] Strumenti personali Fai una donazione registrati entra [x] Mostra/Nascondi l'indice Indice sposta nella barra laterale nascondi Inizio 1 Storia e viabilitàAttiva/disattiva la sottosezione Storia e viabilità 1.1 Storia 1.2 Viabilità 2 Percorso 3 La costruzione 4 La manutenzione 5 Note 6 Bibliografia 7 Altri progetti 8 Collegamenti esterni Sopraelevata Aldo Moro [x] 3 lingue Deutsch Español Français Modifica collegamenti Voce Discussione [x] italiano Leggi Modifica Modifica wikitesto Cronologia [x] Strumenti Strumenti sposta nella barra laterale nascondi Azioni Leggi Modifica Modifica wikitesto Cronologia Generale Puntano qui Modifiche correlate Link permanente Informazioni pagina Cita questa voce Ottieni URL breve Scarica codice QR Carica su Commons Modifica collegamenti interlinguistici Stampa/esporta Crea un libro Scarica come PDF Versione stampabile In altri progetti Wikimedia Commons Wikiquote Elemento Wikidata Da Wikipedia, l'enciclopedia libera. Voce principale: Sistema viario di Genova. \| Sopraelevata Aldo Moro \| \| \| \| Localizzazione \| \| Stato \| Italia \| \| Regioni \| Liguria \| \| Province \| Genova \| \| Dati \| \| Classificazione \| strada sopraelevata \| \| Inizio \| SS 1 presso rotatoria Genova Foce \| \| Fine \| A7 presso casello Genova Ovest \| \| Lunghezza \| 4.51 km \| \| Data apertura \| 1961-1965 \| \| Gestore \| Città metropolitana di Genova \| \| Percorso \| \| Località servite \| Città metropolitana di Genova \| \| La strada sopraelevata in corrispondenza di Sottoripa \| \| Manuale \| La strada sopraelevata Aldo Moro, nota semplicemente come sopraelevata, è una strada urbana di scorrimento di Genova, posta a un'altezza superiore rispetto alla sede stradale ordinaria (caratteristica da cui deriva l'appellativo "sopraelevata"), che congiunge il quartiere della Foce (vicino al centro) al casello autostradale di Genova Ovest. È classificata tecnicamente come strada urbana mentre la classificazione amministrativa è quella di strada comunale. È percorsa, considerando complessivamente i due sensi, da circa 80 000 veicoli al giorno. È lunga circa 6 km (la sua lunghezza risulta maggiore se si considerano anche le rampe di accesso e uscita) ed attraversa, nella sua parte centrale, la zona del porto antico. È intitolata al politico democristianoAldo Moro, rapito ed ucciso nel 1978 dalle Brigate Rosse. Storia e viabilità [modifica \| modifica wikitesto] Storia [modifica \| modifica wikitesto] Tra la fine degli anni cinquanta e l'inizio degli anni sessanta, la diffusione della motorizzazione di massa e il notevole aumento del traffico veicolare urbano spinsero l'amministrazione genovese a definire le prime ipotesi per una strada sopraelevata destinata al traffico di attraversamento e posta a una quota più alta rispetto alla sede stradale esistente a ridosso dell'arco portuale storico. Nel febbraio 1961 si insediò la giunta comunale presieduta dal sindaco Vittorio Pertusio, che aveva la costruzione della nuova arteria tra le sue priorità. La realizzazione dell'opera era già prevista dal piano regolatore generale adottato nel 1956 e fu definitivamente approvata dal Ministero dei lavori pubblici nel 1959 con il documento di pianificazione urbanistica, definendola “strada a traffico veloce”. Il 31 marzo 1961 il consiglio comunale ratifica la decisione della giunta e decide l'affidamento della progettazione esecutiva e della costruzione alla Società C.M.F. (Società Costruzioni Metalliche Finsider), azienda del gruppo IRI. Il progetto era curato dall'ingegnereFabrizio de Miranda, con la collaborazione dell'ingegnere Ernesto Pitto. Il collaudo della struttura fu eseguito con quaranta autocarri a pieno carico, portata ritenuta necessaria per testare la resistenza dei 4.600 metri d'asfalto disposti su due carreggiate larghe in tutto sedici metri e sorrette da 210 pilastri. L'inaugurazione della strada avvenne il 25 agosto 1965. La prima auto imboccò la sopraelevata qualche giorno dopo, il 6 settembre; lo stesso giorno avvenne l'inaugurazione ufficiale, per mano del ministro delle Partecipazioni statali, il sestrese Giorgio Bo, in rappresentanza del governo Moro II. La gigantesca opera costò un miliardo e 752 milioni di lire dell'epoca, corrispondente a circa 16 500 000 euro del 2008. Il 18 marzo 1967 venne aperto al traffico il raccordo di collegamento fra la sopraelevata e l'autostrada A7 da e per Milano. Ne era stato ipotizzato anche un proseguimento a ponente, fino all'aeroporto Cristoforo Colombo. Nel 2014 è stata realizzata la strada sopraelevata Via Guido Rossa, che si collega alla sopraelevata esistente a qualche centinaio di metri di distanza e ne costituisce la prosecuzione fino all'autostrada e alla strada per l'aeroporto. Viabilità [modifica \| modifica wikitesto] Sulla sopraelevata è vietato l'accesso ai pedoni, alle biciclette, ai ciclomotori, ai veicoli a braccia e a trazione animale e ai mezzi pesanti. La strada è a due semi-carreggiate e presenta a due corsie per senso di marcia, con limiti di velocità massima di 60 km/h e minima di 40 km/h. In direzione levante la Sopraelevata ha quattro ingressi nella zona di Sampierdarena: uno direttamente dal casello autostradale di Genova Ovest, uno da via Cantore, uno da via Milano (inaugurato nel 2017) e uno da Lungomare Giuseppe Canepa nella zona di San Benigno; ha uscite in corrispondenza del Porto Antico e della zona di Portoria (tramite un tunnel che taglia in sotterranea il centro cittadino) e infine termina in prossimità della Fiera. Sotto: la sopraelevata vista da via Saffi. Si vede il cantiere della Fiera (maggio 2023), la rotonda 9 novembre 1989 che porta a viale Brigate Partigiane; in rettilineo corso Marconi; a destra piazzale John Fitzgerald Kennedy Verso ponente, sempre in prossimità della Fiera, vi sono un accesso raggiungibile da corso Quadrio e piazza Dante ed un'uscita in corrispondenza di Via di Francia (quartiere San Benigno), con una rampa che consente anche l'accesso diretto a Lungomare Canepa, mentre un'altra uscita permette di raggiungere sia Via Cantore sia il casello dell'autostrada A7 di Genova Ovest. Fino al 1990 era attiva, nella carreggiata in direzione ponente, una rampa di entrata da piazza Cavour, poi demolita per fare spazio alle nuove sistemazioni della zona per le Colombiadi del 1992 (tra cui la realizzazione di un sottopasso che corre sotto Palazzo San Giorgio). Attualmente l'abbozzo di rampa rimasto funziona da piazzola di sosta di emergenza. L'ultimo punto di rilevamento della velocità, all'altezza della foce Dopo diversi annunci, avvenuti a partire dal 2010, nel marzo 2012 sulla sopraelevata è stato attivato il sistema Tutor per il controllo della velocità. Percorso [modifica \| modifica wikitesto] Sopraelevata Aldo Moro Direzione levante Uscita Milano - Genova Ovest Casello di Genova ovest Genova via Antonio Cantore Genova Piazza Principe Dinegro Milano - Genova Ovest Casello di Genova ovest Area di servizio inversione di marcia Genova Porto Genova lungomare Canepa Savona Genova via Milano solo immissione Genova Porto Antico Piazza Cavour solo uscita Genova centro Piazza De Ferrari Piazza Corvetto solo uscita Genova Foce La Spezia Genova Brignole Sopraelevata Aldo Moro Direzione ponente Uscita Genova Foce La Spezia Genova Brignole Genova centro Piazza De Ferrari Piazza Corvetto Genova Porto Antico Piazza Cavour solo immissione svincolo dismesso nel 1990 Milano - Genova Ovest Casello di Genova ovest Genova via di Francia Genova Piazza Principe Dinegro Porto Genova lungomare Canepa Savona La costruzione [modifica \| modifica wikitesto] «60 000 m³ di demolizione, 78 000 m³ di sbancamento e di scavi, 120 fondazioni su pali, 82 normali, 18 a cassone, per uno sviluppo di 6 800 metri; 18 km di guard-rail, quasi 10 km di parapetti, migliaia di uomini impiegati, 2 anni di lavoro.» (Dal video di presentazione della sopraelevata) La strada sopraelevata sfiora Palazzo San Giorgio, al porto antico La data ufficiale dell'inizio dei lavori di costruzione della sopraelevata fu il 12 febbraio 1964, giorno della posa del primo pilone. Furono necessarie quindicimila tonnellate di lamiere d'acciaio, settantatremila di calcestruzzo, la demolizione di 300 000 metri cubi di edifici (fra cui alcune palazzine del XVII secolo adiacenti al Palazzo Millo, al porto antico) per far posto alle strutture portanti, che prevedevano 78 000 metri cubi di sbancamenti. Dopo lunghi studi di fattibilità, portati avanti dall'assessore alle Strade del Comune Ivo Lapi, della realizzazione dell'opera si occupò la CMF-Costruzioni Metalliche Finsider. La manutenzione [modifica \| modifica wikitesto] La strada sopraelevata fotografata dalla fermata della metropolitana, in via Bruno Buozzi La nuova strada richiese interventi di manutenzione piuttosto presto: nel 1977, appena dodici anni dopo l'inaugurazione, alcuni elementi della struttura in acciaio richiesero una nuova verniciatura, peraltro ampiamente prevista dai manuali di manutenzione. Nel 1988 fu eseguita la manutenzione straordinaria dei giunti di dilatazione. La strada è tuttavia rimasta sempre aperta al traffico di decine di migliaia di veicoli al giorno. Nel corso della sua storia la sopraelevata è stata oggetto di discussioni e spesso anche di polemiche legate all'impatto visivo negativo e all'inquinamento acustico e ambientale che provoca ai numerosi palazzi vicini. La sopraelevata nel tratto che corre sopra i binari della linea ferroviaria Genova-Savona Più volte è stato ipotizzato di demolire la sopraelevata e di sostituirla con un nuovo tunnel sotto il porto o, in alternativa, con un ponte in grado di scavalcare da levante a ponente, fino alla Lanterna, l'intero bacino portuale genovese. Note [modifica \| modifica wikitesto] ^Capitolato tecnico, su Comune di Genova(archiviato dall'url originale il 17 dicembre 2010). ^ Katia Bonchi, Buon compleanno sopraelevata! La strada Aldo Moro compie 50 anni. Per Tursi il futuro è il tunnel, in Genova24, 24 agosto 2015. ^articolo, in La Stampa, 6-7 settembre 1965, p.6. ^Raccordo e sopraelevata a Genova, in Le Strade, anno XLVII, n.4, Milano, Touring Club Italiano, aprile 1967, p.352. ^Piano Regolatore Generale della Città di Genova 1976 - 1980, su Comune di Genova. ^ Stefano Origone, Acceso il Tutor in Sopraelevata. Pizzicati 5 auto oltre i 100 all'ora, in la Repubblica, 6 marzo 2012. ^Sopraelevata, una strada d'acciaio, su YouTube, Archivio Nazionale Cinema Impresa. Bibliografia [modifica \| modifica wikitesto] Fabrizio de Miranda, La strada sopraelevata di Genova e sue caratteristiche di progetto, in Costruzioni Metalliche, n.5, 1965. Alberto Rogano, La Sopraelevata di Genova, in Le Strade, anno XLV, n.12, Milano, Touring Club Italiano, dicembre 1965, pp.501-509. Fabrizio de Miranda, Strada sopraelevata a Genova, in Casabella, anno XXX, n.308, Milano, Editoriale Domus, agosto 1966, pp.52-61, ISSN0008-7181(WC· ACNP). Francesco Gastaldi, La strada Sopraelevata di Genova: tra storia e attualità, in Trasporti & Cultura, n.5, 2003, pp.16-21. Mauro Marsullo, Sopraelevata di Genova – La Città Sospesa, Pedonalizzazione della Sopraelevata di Genova[collegamento interrotto], 2012. Altri progetti [modifica \| modifica wikitesto] Altri progetti Wikiquote Wikimedia Commons Wikiquote contiene citazioni sulla Sopraelevata Aldo Moro Wikimedia Commons contiene immagini o altri file sulla Sopraelevata Aldo Moro Collegamenti esterni [modifica \| modifica wikitesto] (EN) Strada sopraelevata di Genova, su Structurae. Sopraelevata, una strada d'acciaio, su YouTube, Archivio Nazionale Cinema Impresa. Informazioni sull'ipotesi del tunnel subacqueo sostitutivo della sopraelevata, su trail.liguria.it. URL consultato il 19 luglio 2008(archiviato dall'url originale il 26 settembre 2012). Portale Genova Portale Ingegneria Portale Trasporti Estratto da " Categorie: Ponti di Genova Ponti di Fabrizio de Miranda Ponti in acciaio d'Italia Strade di Genova Strade sopraelevate [altre] Categorie nascoste: Voci con modulo citazione e parametro pagina Pagine con collegamenti non funzionanti P454 letta da Wikidata Questa pagina è stata modificata per l'ultima volta il 23 ago 2025 alle 00:32. Il testo è disponibile secondo la licenza Creative Commons Attribuzione-Condividi allo stesso modo; possono applicarsi condizioni ulteriori. Vedi le condizioni d'uso per i dettagli. Informativa sulla privacy Informazioni su Wikipedia Avvertenze Codice di condotta Sviluppatori Statistiche Dichiarazione sui cookie Versione mobile Modifica impostazioni anteprima Ricerca Ricerca [x] Mostra/Nascondi l'indice Sopraelevata Aldo Moro 3 lingueAggiungi argomento
15531	https://www.ainfoinc.com/blog/blog-circular-waveguide-size.html	Home Blog Circular Waveguide Size Product was successfully added to your shopping cart. Swipe to the left Circular Waveguide Size Print By Stanley Hu 8 years ago 4851 Views \| \| \| \| \| \| \| \| --- --- --- \| BY \| IEC \| EIA WC \| Frequency Range (GHz) \| Cutoff Frequency (GHz) \| Internal Diameter \| \| \| mm \| Inch \| \| BY8 \| C8 \| 992 \| 0.803~1.100 \| 0.698 \| 251.84 \| 9.91 \| \| BY10 \| C10 \| 847 \| 0.939~1.290 \| 0.817 \| 215.14 \| 8.47 \| \| BY12 \| C12 \| 724 \| 1.100~1.510 \| 0.957 \| 183.77 \| 7.24 \| \| BY14 \| C14 \| 618 \| 1.290~1.760 \| 1.12 \| 157 \| 6.18 \| \| BY16 \| C16 \| 528 \| 1.510~2.070 \| 1.311 \| 134.11 \| 5.28 \| \| BY18 \| C18 \| 451 \| 1.760~2.420 \| 1.534 \| 114.58 \| 4.51 \| \| BY22 \| C22 \| 385 \| 2.070~2.830 \| 1.796 \| 97.87 \| 3.85 \| \| BY25 \| C25 \| 329 \| 2.420~3.310 \| 2.102 \| 83.62 \| 3.29 \| \| BY30 \| C30 \| 281 \| 2.830~3.880 \| 2.461 \| 71.42 \| 2.81 \| \| BY35 \| C35 \| 240 \| 3.310~4.540 \| 2.88 \| 61.04 \| 2.40 \| \| BY40 \| C40 \| 205 \| 3.890~5.330 \| 3.381 \| 51.99 \| 2.05 \| \| BY48 \| C48 \| 175 \| 4.540~6.230 \| 3.955 \| 44.45 \| 1.75 \| \| BY56 \| C56 \| 150 \| 5.300~7.270 \| 4.614 \| 38.1 \| 1.50 \| \| BY65 \| C65 \| 128 \| 6.210~8.510 \| 5.402 \| 32.537 \| 1.28 \| \| BY76 \| C76 \| 109 \| 7.270~9.970 \| 6.326 \| 27.788 \| 1.09 \| \| BY89 \| C89 \| 94 \| 8.490~11.600 \| 7.377 \| 23.825 \| 0.94 \| \| BY104 \| C104 \| 80 \| 9.970~13.700 \| 8.685 \| 20.244 \| 0.80 \| \| BY120 \| C120 \| 69 \| 11.600~15.900 \| 10.057 \| 17.415 \| 0.69 \| \| BY140 \| C140 \| 59 \| 13.400~18.400 \| 11.649 \| 15.088 \| 0.59 \| \| BY165 \| C165 \| 50 \| 15.900~21.800 \| 13.842 \| 12.7 \| 0.50 \| \| BY190 \| C190 \| 44 \| 18.200~24.900 \| 15.794 \| 11.125 \| 0.44 \| \| BY220 \| C220 \| 38 \| 21.200~29.100 \| 18.446 \| 9.525 \| 0.38 \| \| BY255 \| C255 \| 33 \| 24.300~33.200 \| 21.103 \| 8.331 \| 0.33 \| \| BY290 \| C290 \| 28 \| 28.300~38.800 \| 24.62 \| 7.137 \| 0.28 \| \| BY330 \| C330 \| 25 \| 31.800~43.000 \| 27.683 \| 6.35 \| 0.25 \| \| BY380 \| C380 \| 22 \| 36.400~49.800 \| 31.617 \| 5.563 \| 0.22 \| \| BY430 \| C430 \| 19 \| 42.400~58.100 \| 36.776 \| 4.775 \| 0.19 \| \| BY495 \| C495 \| 17 \| 46.300~63.500 \| 40.227 \| 4.369 \| 0.17 \| \| BY580 \| C580 \| 14 \| 56.600~77.500 \| 49.103 \| 3.581 \| 0.14 \| \| BY660 \| C660 \| 13 \| 63.500~87.200 \| 55.28 \| 3.175 \| 0.13 \| \| BY765 \| C765 \| 11 \| 72.700~99.700 \| 64.462 \| 2.769 \| 0.11 \| \| BY890 \| C890 \| 9 \| 84.800~116.000 \| 73.552 \| 2.388 \| 0.09 \| Posted in: Technical Knowledge A-INFO INC About A-INFO Careers News and Events New Products Tech. Knowledge RF Calculators Request RMA Terms & Comditions RFQ Contact Us My Account Login Create an Account My Account My Wishlist My Quote Track Your Orders Online Help Contact Information Address: US: 60 Tesla, Irvine, CA 92618, USA Phone: US: +1-949-639-9688 US: +1-949-639-9608 Fax: US: +1-949-639-9670 Email: sales@ainfoinc.com Working Days/Hours: Mon - Fri / 9:00AM - 5:30PM Customer Service Request RMA Terms & Conditions FAQ RF Calculators & Conversions Be the First to Know
15532	https://www.ck12.org/book/ck-12-elementary-and-intermediate-college-algebra/r2/section/9.8/	Graphing Reciprocal Functions in Standard Form \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up HomeAlgebraFlexBooksCK-12 Elementary and Intermediate College AlgebraCh98. Graphing Reciprocal Functions in Standard Form 9.8 Graphing Reciprocal Functions in Standard Form Difficulty Level: At Grade \| Created by: CK-12 Last Modified: Jun 30, 2017 Read Resources Details Attributions Loading... The time to empty a tank by pumping out the contents varies inversely as the rate of pumping. If the pump works at a rate of 30 gallons per hour, how much is in a tank that started with 300 gallons at 2 hours? At 5 hours? At 10 hours? Instead of having to solve three separate equations for each of these values, we can draw a graph of this situation and then read the values from the graph. I n this section, w e consider how to graph reciprocal functions in standard form. [Figure 1] Rational Functions in Standard Form A rational function is a function in the form p(x)q(x), where p(x) and q(x) are polynomials and q(x)≠0. The parent graph for some rational functions is y=1 x. Below we have a table of values for this function. \| x \| y=1 x \| --- \| \| -4 \| -1 4 \| \| -2 \| -1 2 \| \| -1 \| -1 \| \| -1 2 \| -2 \| \| 0 \| undefined \| \| 1 2 \| 2 \| \| 1 \| 1 \| \| 2 \| 1 2 \| \| 4 \| 1 4 \| Notice that the y-values are reciprocals of the x-values. Hence, this function is called the reciprocal function. The graph of the function is below. This shape is called ahyperbola. [Figure 2] The two parts of the graph are called branches and are divided by two boundary lines called asymptotes. The two asymptotes in this case are the x-axis and the y-axis. For a hyperbola, the branches are always symmetrical about the point where the asymptotes intersect. In this example, they are symmetrical about the origin. If we apply the transformation rules to this parent graph, we can express the standard form of a reciprocal function. Standard Form of a Reciprocal Function The standard form of a reciprocal function is given by the equation f(x)=a x−h+k. Below we consider how to graph reciprocal functions of the form f(x)=a x−h+k. First, we need to deal with some preliminaries. Domain of a Rational Function Recall that the domain of a function is the set of possible inputs to the function. Notice in the graph of y=1 x above that when we tried to input 0, we got an undefined result. This means we need to eliminate 0 from the domain of y=1 x. In set-builder notation, we can write this as {x∈R\|x≠0},or in interval notation we can write this as(-∞,0)∪(0,∞). In general, we need to be aware of the potential for domain problems when we are dealing with rational functions. Since they are fractions, we do not want the denominators to be equal to zero. To determine which values to eliminate from the domain of a rational function, set the denominator equal to zero and solve. The solutions are values that will make the denominator equal to zero and need to be eliminated from the domain. Example 1 Determine the domain of the rational functions below. (Note: Some of these are not in standard form. We will cover these in the next section.) a. y=3 x+4+6 b. y=3 x−1 4 x+8 c. y=4 x 2+5 x+6 d. y=1 x 2+4 Solution: a. When we set x+4=0, we find x=-4. Therefore, the domain of this function is {x∈R\|x≠-4}or (-∞,-4)∪(4,∞). b. 4 x+8=0⟶x=-2. The domain of the function is {x∈R\|x≠-2}or(-∞,-2)∪(-2,∞). c. x 2+5 x+6(x+2)(x+3)x=-2 dom y=0=0 x=-3={x∈R\|x≠-2,x≠-3}=(-∞,-3)∪(-3,-2)∪(-2,∞)Be careful to include the middle interval, which is part of the solution. d. x 2+4=0 has no real solutions. Therefore, the denominator is never zero, and the domain is all real numbers. Vertical Asymptotes in Standard Form The vertical boundary line or asymptote is caused by the value that is the domain concern, when x−h=0.Around the vertical asymptote, the y-values increase or decrease without bound. Vertical Asymptote in Standard Form There will be a vertical asymptote at x=h. Horizontal Asymptotes in Standard Form Horizontal asymptotes describe how the graph behaves for large values of \|x\|. When this is true, the 1st term in standard form is a fixed number divided by a large number, which we can estimate to be zero. Then, the only term remaining is k, so the horizontal asymptote is y=k. This is the end behavior of a rational function. The function will be close to the horizontal line, y=k. Horizontal Asymptote in Standard Form There will be a horizontal asymptote at y=k. Example 2 What are the asymptotes for f(x)=-1 x+6+9? Solution:The asymptotes are x=-6 and y=9. Graphing Rational Functions in Standard Form To Graph a Rational Function in Standard Form Determine the domain of the function. This will also be the location of the vertical asymptote. Find the horizontal asymptote. Create a table of values with some values to the right of the vertical asymptote, and some to the left of the vertical asymptote. Example 3 Graph f(x)=-2 x. Solution:First, the domain is all real numbers except x=0, so x=0 is also the vertical asymptote. The horizontal asymptote is y=0 since there is no value for k explicitly written. Let's make a table of values by choosing three values to each side of x=0. \| x \| y \| --- \| \| -4 \| 1 2 \| \| -2 \| 1 \| \| -1 \| 1 \| \| \| -2 \| \| 2 \| -1 \| \| 4 \| -1 2 \| Plotting the points and using the asymptotes, we have the graph below. [Figure 3] This video by Mathispower4u demonstrates how to determine the domain, complete a table of values, and graph a rational function. Example 4 Graph y=1 x−5+2. Solution: For y=1 x−5+2, the vertical asymptote is x=5 because that would make the denominator zero, and we cannot divide by zero. The horizontal asymptote is at y=2. Since these are asymptotes that are not also axes, we represent them with dashed lines to indicate that they are asymptotes, but not part of the graph. Let's create a table of values. x y 0 1 4 5 2 1 2 3 3 1 1 2 6 3 8 2 1 3 10 2 1 5 Drawing the asymptotes and plotting the points gives us the graph below. [Figure 4] Notice the shape and location of the branches are the same as the parent graph, just shifted to the right 5 units and up 2 units. Example 5 The time to empty a tank by pumping out the contents varies inversely as the rate of pumping. Where g is the number of gallons and t is the time in hours, how much is in a tank at 2 hours? At 5 hours? At 10 hours? Solution:We can describe this situation with the function g=361.19 t+3.2046+186.29.The graph of this function is below. [Figure 5] By finding t=2 on the t-axis, we can find the g-value, 255.688 gallons. Notice there are about 230 gallons after 5 hours, and 213 after 10 hours. This video by Mr. G demonstrates how to use transformations to graph rational functions. Example 6 Find a function that describes the graph below. [Figure 6] Solution: We know that the numerator will be negative because the branches of this hyperbola are a reflection of the branches of the parent graph. The asymptotes are x=-3 and y=-4. So far, we know y=a x+3−4. To determine a, we can use the given x-intercept. 0 4-3=a-3.75+3-4=a-0.75=a The equation is y=-3 x+3−4 How To Graph Rational Functions With a Graphing Utility To graph a rational function with a graphing utility, enter the function into "y=" and press graph. To type the fraction, you may need to highlight the numerator and use the slash "/" key to get to the denominator. Also, remember to enclose the numerator and denominator in separate parentheses. WARNING Different graphing utilities represent asymptotes differently. For instance, in Desmos, there are no asymptotes shown at all. [Figure 7] On a TI-83, the graph has a solid vertical asymptote. [Figure 8] When graphing, it is best to represent the asymptotes with dashed lines to indicate they have a role in the graph, but are not actually part of the graph (like a solid line). Summary To find the domain of a rational function, set the denominator equal to zero and solve the equation. The solutions should be eliminated from the real numbers to get the domain of the function. The vertical asymptote of a reciprocal function in standard form is x=h,and the horizontal asymptote is y=k. To graph a reciprocal function in standard form, determine the domain of the function (this will also be the location of the vertical asymptote), find the horizontal asymptote, and create a table of values with some values to the right of the vertical asymptote, and some to the left of the vertical asymptote. Review What are the asymptotes for y=2 x+8−3? What are the asymptotes for y=6−1 x−4? For problems 3-11, graph each rational function and state the equations of the asymptotes, the domain and range, and the intercepts. y=3 x y=1 x+6 y=-1 x y=-1 x+3 y=1 x+5 y=1 x−3−4 y=2 x+4−3 y=5 x+2 y=3−1 x+2 Write functions that describe the graphs below. You may assume that a = 1. 12. [Figure 9] 13. [Figure 10] Explore More Write a function that satisfies the following: a vertical asymptote at x=5 that goes through the point (6,4). Write a function that satisfies the following: a horizontal asymptote at y=-2,and a vertical asymptote at x=-3. 3.Boyle's Law states that the pressure and volume of a gas are inversely proportional. What does this mean? How does this represent an inversely proportional relationship? A bicycle pump is a great example that shows Boyle's Law. When you push down on the pump, the volume inside the bike pump decreases, and the pressure of the air increases so that it's pushed into the tire. If the volume inside of a bicycle pump is 8.2 cubic inches, and the pressure is 19.1 psi, we can find the equation that represents this situation. We know from Boyle's Law that the volume, "y", equals the constant divided by the pressure in psi. Our equation is y=156.62 x. Graph this function. What is the relationship between the pressure and volume in this situation? Answers for Review and Explore More Problems Please see the Appendix. PLIX Try these interactives that reinforce the concepts explored in this section: ck12.org/a/2187848 ck12.org/a/1824181 References "Boyle's Law," last edited May 15, 2017, Notes/Highlights \| Color \| Highlighted Text \| Notes \| \| --- --- \| \| \| Please Sign In to create your own Highlights / Notes \| Currently there are no resources to be displayed. Description This section covers graphing a rational function when it is in standard form. Learning Objectives Difficulty Level At Grade author Liz Krams contributor Lori Jordan,Kate Dirga editor Liza Cope Tags inverse,Function,Equation,denominator,LCD,reciprocal,asymptote,graph,rational,horizontal,vertical, (8 more) Subjects mathematics,Algebra Concept Nodes MAT.ALG.707.01 (Estimate Graphs of Rational Functions - Algebra) Grades Higher-Ed Standards Correlations CCSS,TX License CC BY NC Language English Date Created Jun 30, 2017 Last Modified Jun 30, 2017 Vocabulary . \| Image \| Reference \| Attributions \| --- \| \| [Figure 1] \| Credit:Todd Huffman Source: License:CC BY-NC 3.0 \| \| \| [Figure 2] \| Credit:NASA's Glenn Research Center Source: \| \| \| [Figure 3] \| Credit:NASA's Glenn Research Center Source: \| \| \| [Figure 4] \| Credit:NASA's Glenn Research Center Source: \| \| \| [Figure 5] \| Credit:NASA's Glenn Research Center Source: License:CC BY-NC 3.0 \| \| \| [Figure 6] \| Credit:NASA's Glenn Research Center Source: \| \| \| [Figure 7] \| Credit:Desmos License:CC BY-NC 3.0 \| \| \| [Figure 8] \| Credit:Desmos License:CC BY-NC 3.0 \| \| \| [Figure 9] \| Credit:NASA's Glenn Research Center Source: \| \| \| [Figure 10] \| Source: \| Show Attributions Show Details ▼ Previous Solving Rational Equations Using Cross Multiplication Next Graphing Other Rational Functions You may also like Video Another Rational Function Graph Example Reviews Was this helpful? 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15533	https://commons.wikimedi…partitions_5.svg	File:Noncrossing partitions 5.svg - Wikimedia Commons Jump to content [x] Main menu Main menu move to sidebar hide Navigate Main page Welcome Community portal Village pump Help center Language select Participate Upload file Recent changes Latest files Random file Contact us Special pages Search Search English [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 Summary 2 Licensing File:Noncrossing partitions 5.svg File Discussion [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Permanent link Page information Cite this page Get shortened URL Download QR code Concept URI Nominate for deletion Print/export Download as PDF Printable version In other projects From Wikimedia Commons, the free media repository File File history File usage on Commons File usage on other wikis Metadata Download all sizesUse this file on the webUse this file on a wikiEmail a link to this fileInformation about reusing Size of this PNG preview of this SVG file: 282 × 598 pixels. Other resolutions: 113 × 240 pixels \| 226 × 480 pixels \| 362 × 768 pixels \| 482 × 1,024 pixels \| 965 × 2,048 pixels \| 330 × 700 pixels. Original file(SVG file, nominally 330 × 700 pixels, file size: 8 KB) Open in Media Viewer File information Structured data Captions Captions Edit English Add a one-line explanation of what this file represents Summary [edit] Description Noncrossing partitions 5.svg The C 5 = 42noncrossing partitions of a 5-element set with the other 10 partitions shown below There are A111275(5) = 10 different cyclic patterns (rows in this arrangement), compare Noncrossing partition necklaces.svg Source Own work AuthorWatchduck You can name the author as "T. Piesk", "Tilman Piesk" or "Watchduck". SVG development InfoField The SVG code is valid. This vector image was created with an unknown SVG tool. Ordered like A231428 Matrices of equivalence relations Connected pentagon vertices Genji chapter symbols The 7partitions of 5 (compare this table) Compare: Partitions of a 4-element set Licensing [edit] I, the copyright holder of this work, hereby publish it under the following licenses: Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled GNU Free Documentation License. GFDL GNU Free Documentation License true true This file is licensed under the Creative CommonsAttribution 3.0 Unported license. 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15534	https://www.doubtnut.com/qna/644505246	A box contains beads of different colors There are total 200 beads in it One bead is taken at random The probability that it is blue is 0.98 Then, a) How many blue beads are there in the box? b) Some blue beads are removed from the box Now probability of blue bead becomes 0.96 So how many blue beads are removed?Solution in Malayalam Step by step video & image solution for A box contains beads of different colors There are total 200 beads in it One bead is taken at random The probability that it is blue is 0.98 Then, a) How many blue beads are there in the box? b) Some blue beads are removed from the box Now probability of blue bead becomes 0.96 So how many blue beads are removed? by Maths experts to help you in doubts & scoring excellent marks in Class 10 exams. Topper's Solved these Questions Explore 61 Videos Explore 36 Videos Similar Questions What are the coordinates of the fourth vertex of the parallelogram shown below. Write the second degree polynomials given below as the product of twofirst degree polynomials.Find also the solutions of the equation p(x)= 0 in each:- p(x)=x2−7x+12. Which gives blue bead in borax bead test ? In the Borax bead test of Co2+, the blue colour of bead is due to the formation of : Which will give borax bead test with blue bead? Blue borax bead is obtained with A chain of beads is to be prepared using 6 different red coloured beads and 3 different blue coloured beads. In how many ways can this be done so that no two blue coloured beads come together. Blue borax-bead is obtained with निम्न में से कौन - सा आयन बोरेक्स मनका परीक्षण में नीले रंग का मनका बनायेगा ? Which gives blue bead in borax bead test: A chain of beads is to be prepared using 6 different red coloured beads and 3 different blue coloured beads. In how many ways can this be done so that no two blue coloured beads come together. A box contains 6 red beads and 5 white beads. 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15535	https://www.radiologyinfo.org/en/info/galactogram	Galactography (Ductography, Galactogram) skip to main content Home Tests and Treatments Pediatric Care Screening Diseases and Conditions Safety En Español Galactography (Ductography) Rewind 10 Seconds Next Up 00:00 00:00 00:00 Chromecast Closed Captions Settings Fullscreen Galactography uses mammography and an injection of contrast material to create pictures of the inside of the breast's milk ducts. It is most commonly used when a woman has experienced a bloody or clear discharge from the breast nipple but has an otherwise normal mammogram. It's important not to squeeze the nipple prior to the exam as there may only be a small amount of fluid and it is necessary to see where it is coming from to perform the exam. Tell your doctor about any breast symptoms or problems, prior surgeries, hormone use, medical conditions, and if there's a possibility you are pregnant. Leave jewelry at home and wear loose, comfortable clothing. You may be asked to wear a gown. Don't wear deodorant, talcum powder or lotion under your arms or on your breasts as these may appear on the mammogram and interfere with correct diagnosis. What is Galactography (Ductography)? What are some common uses of the procedure? How should I prepare? What does the equipment look like? How does the procedure work? How is the procedure performed? What will I experience during and after the procedure? Who interprets the results and how do I get them? What are the benefits vs. risks? What are the limitations of Galactography? What is Galactography (Ductography)? Galactography is an x-ray examination that uses mammography, a low-dose x-ray system for examining breasts, and a contrast material to obtain pictures, called galactograms, of the inside of the breast's milk ducts. An x-ray exam helps doctors diagnose and treat medical conditions. It exposes you to a small dose of ionizing radiation to produce pictures of the inside of the body. X-rays are the oldest and most often used form of medical imaging. The breast is composed primarily of three structures: fat, lobules(that make the milk) and milk ducts (that carry the milk from the lobule to the nipple). While mammography, ultrasound and magnetic resonance imaging(MRI) are excellent ways to image the breast; they do not visualize the inside of the breast's milk ducts to the same degree as galactography. top of page What are some common uses of the procedure? The most common use of galactography is to evaluate a woman who has a bloody or clear discharge from her breast nipple and an otherwise normal mammogram. Galactography is typically NOT called for in women with the following conditions: A discharge that is milky, blue-green, green, or gray is usually not a cause for concern, especially if it comes from multiple ducts in the breast. A discharge that is from both breasts in a woman who has not had children may indicate a side effect from a drug, or may be related to a pituitary problem located in the brain. top of page How should I prepare? Very little preparation is necessary for this procedure. The only requirement is that the nipple not be squeezed prior to the exam, as sometimes there is only a small amount of fluid and it is necessary to see where that fluid is coming from to perform the exam. Tell your doctor about all the medications you take. List any allergies, especially to iodine contrast materials. Tell your doctor about recent illnesses or other medical conditions. Always inform your doctor or x-ray technologist if there is any possibility that you are pregnant. As in mammography, do not wear deodorant, talcum powder or lotion under your arms or on your breasts on the day of the exam. These can appear on the galactogram as calcium spots. In addition, before the examination you will be asked to remove all jewelry and clothing above the waist and you will be given a gown or loose-fitting material that opens in the front. top of page What does the equipment look like? This exam is performed using a mammography unit. A mammography unit is a box with a tube that produces x-rays. The unit is used exclusively for breast x-ray exams and features special accessories to limit x-ray exposure to only the breast. The unit features a device to hold and compress the breast and position it so the technologist can capture images at different angles. Other equipment needed for this exam may include small wires called dilators, a small catheter(a plastic hollow tube) and a blunt-tipped tube that is inserted into the milk duct in the nipple to inject a tiny amount of contrast material. top of page How does the procedure work? X-rays are a form of radiation like light or radio waves. X-rays pass through most objects, including the body. The technologist carefully aims the x-ray beam at the area of interest. The machine produces a small burst of radiation that passes through your body. The radiation records an image on photographic film or a special detector. Different parts of the body absorb the x-rays in varying degrees. Dense bone absorbs much of the radiation while soft tissue (muscle, fat, and organs) allow more of the x-rays to pass through them. As a result, bones appear white on the x-ray, soft tissue shows up in shades of gray, and air appears black. In galactography, a small amount of contrast material is injected into the milk duct, and a mammogram is performed so that the inside of the milk duct can be seen. If there is a filling defect (black area) in the milk duct, it often indicates a small mass. Most of these are papillomas, which are non-cancerous masses of the milk ducts. They may be pre-cancerous, and sometimes are removed. Less than 10 percent of filling defects will be cancer. The galactogram will not only find the small mass, but will also show where it is located in the breast, to help the surgeon find the area. In some cases, there are no filling defects. Rather, the ducts lead to cysts in the breast, a sign of fibrocystic change. These cysts may cause a bloody discharge, but generally are not worrisome. top of page How is the procedure performed? Your doctor will likely do this exam on an outpatient basis. The patient is seated or placed on her back with the breast exposed. The nipple is cleansed, and a tiny amount of fluid is squeezed from the nipple to identify the duct with the discharge. The milk duct may be dilated to permit a small catheter (a plastic, hollow tube) or blunt-tipped tube to be inserted into the milk duct. Occasionally a warm towel will be placed on the breast to help the milk duct become more visible and to allow easier access to the milk duct. A small amount of contrast material is then injected, and a mammogram is obtained. A second injection and mammogram may be performed. You must hold very still and may need to hold your breath for a few seconds while the technologist takes the x-ray. This helps reduce the possibility of a blurred image. The technologist will walk behind a wall or into the next room to activate the x-ray machine. When the examination is complete, the technologist may ask you to wait until the radiologist confirms they have all the necessary images. The procedure normally takes between 30 minutes to an hour. top of page What will I experience during and after the procedure? The dilation of the milk duct can sometimes be uncomfortable; however it is usually not painful. The nipple may be squeezed to identify the milk duct with the discharge. top of page Who interprets the results and how do I get them? A radiologist, a doctor trained to supervise and interpret radiology examinations, will analyze the images. The radiologist will send a signed report to your primary care or referring physician who will discuss the resultswith you. The report may be given directly to you at the time of the exam. You may need a follow-up exam. If so, your doctor will explain why. Sometimes a follow-up exam further evaluates a potential issue with more views or a special imaging technique. It may also see if there has been any change in an issue over time. Follow-up exams are often the best way to see if treatment is working or if a problem needs attention. top of page What are the benefits vs. risks? Benefits Galactography can find small cancerous and non-cancerous masses that cannot be identified in any other way so that they may be removed at an early stage. A galactogram identifies the location of the tumors in the breast for the surgeon. No radiation stays in your body after an x-ray exam. X-rays usually have no side effects in the typical diagnostic range for this exam. Risks There is always a slight chance of cancer from excessive exposure to radiation. However, given the small amount of radiation used in medical imaging, the benefit of an accurate diagnosis far outweighs the associated risk. The radiation dose for this procedure varies. See the Radiation Dosepage for more information. It is possible to injure the duct, either during the process of placing the catheter or while injecting contrast material. This almost always heals by itself. It is possible to miss the area of concern, or to put the catheter into the incorrect duct, potentially delaying diagnosis. There is always a possibility of infection of the breast, or mastitis, but this is uncommon. Women should always tell their doctor and x-ray technologist if they are pregnant. See the Radiation Safetypage for more information about pregnancy and x-rays. A Word About Minimizing Radiation Exposure Doctors take special care during x-ray exams to use the lowest radiation dose possible while producing the best images for evaluation. National and international radiology protection organizations continually review and update the technique standards radiology professionals use. Modern x-ray systems minimize stray (scatter) radiation by using controlled x-ray beams and dose control methods. This ensures that the areas of your body not being imaged receive minimal radiation exposure. top of page What are the limitations of Galactography? If there is no discharge at the time of the exam, the duct with the discharge cannot be identified, and the test will not be able to be performed. In addition, some ducts are quite small, and may not be able to be dilated. If the incorrect duct is injected, an incorrect diagnosis can be made. Not all breast tumors occur within the ducts, and a tumor may be present that is not identified on the galactogram. top of page Additional Information and Resources RTAnswers.org: Radiation Therapy for Breast Cancer(opens in a new tab) top of page This page was reviewed on October 31, 2024 Send us your feedback Did you find the information you were looking for? Yes No Please type your comment or suggestion into the text box below. Note: we are unable to answer specific questions or offer individual medical advice or opinions. Comments: Rewind 10 Seconds Next Up 00:00 00:00 01:58 Chromecast Closed Captions Settings Fullscreen Images Image Gallery Mammography procedure View full size with caption Related Articles and Media Breast Cancer Breast Cancer Treatment Breast Lumps Mammography Contrast Materials Radiation Safety Radiation Dose Images related to Galactography (Ductography) Sponsored By Please note RadiologyInfo.org is not a medical facility. Please contact your physician with specific medical questions or for a referral to a radiologist or other physician. To locate a medical imaging or radiation oncology provider in your community, you can search the ACR-accredited facilities database(opens in a new tab). This website does not provide cost information. The costs for specific medical imaging tests, treatments and procedures may vary by geographic region. Discuss the fees associated with your prescribed procedure with your doctor, the medical facility staff and/or your insurance provider to get a better understanding of the possible charges you will incur. Web page review process: This Web page is reviewed regularly by a physician with expertise in the medical area presented and is further reviewed by committees from the Radiological Society of North America (RSNA) and the American College of Radiology (ACR), comprising physicians with expertise in several radiologic areas. 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15536	https://pmc.ncbi.nlm.nih.gov/articles/PMC12148317/	Fatal Suicidal Cut-Throat Wound in a Patient With Double Valve Replacement: Interplay of Cardiac Dysfunction and Self-Harm - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Cureus . 2025 May 20;17(5):e84464. doi: 10.7759/cureus.84464 Search in PMC Search in PubMed View in NLM Catalog Add to search Fatal Suicidal Cut-Throat Wound in a Patient With Double Valve Replacement: Interplay of Cardiac Dysfunction and Self-Harm Ajay K Bhagat Ajay K Bhagat 1 Department of Forensic Medicine and Toxicology, Rajendra Institute of Medical Sciences, Ranchi, IND Find articles by Ajay K Bhagat 1, Kumar Shubhendu Kumar Shubhendu 1 Department of Forensic Medicine and Toxicology, Rajendra Institute of Medical Sciences, Ranchi, IND Find articles by Kumar Shubhendu 1, Ankur Chaudhary Ankur Chaudhary 1 Department of Forensic Medicine and Toxicology, Rajendra Institute of Medical Sciences, Ranchi, IND Find articles by Ankur Chaudhary 1, Anand Kumar Anand Kumar 1 Department of Forensic Medicine and Toxicology, Rajendra Institute of Medical Sciences, Ranchi, IND Find articles by Anand Kumar 1,✉ Editors: Alexander Muacevic, John R Adler Author information Article notes Copyright and License information 1 Department of Forensic Medicine and Toxicology, Rajendra Institute of Medical Sciences, Ranchi, IND ✉ Anand Kumar singhanand.dx@gmail.com ✉ Corresponding author. Accepted 2025 May 20; Collection date 2025 May. Copyright © 2025, Bhagat et al. This is an open access article distributed under the terms of the Creative Commons Attribution License CC-BY 4.0., which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. PMC Copyright notice PMCID: PMC12148317 PMID: 40491635 Abstract Suicidal incised neck injuries are infrequent and exhibit considerable forensic intricacies, particularly when compounded by pre-existing medical conditions.We delineate a case involving a 32-year-old male patient diagnosed with severe rheumatic heart disease and a history of double valve replacement, who succumbed to a self-inflicted incised neck wound during his hospitalization.The 7.5-cm incised wound presented two skin tags, indicative of hesitation cuts, and was devoid of any defensive injuries.The internal examination indicated an enlarged heart, artificial heart valves in situ, a pale appearance of the organs, and haemorrhagic infiltration around the wound; anatomical parts like the trachea and oesophagus were intact.The mechanism of mortality was ascribed to be haemorrhagic shock.The patient's impaired cardiac function likely contributed to the expedited fatal outcome by restricting physiological compensatory mechanisms.This case exemplifies the complex interaction between chronic illness and suicidal tendencies, underscoring the need for comprehensive forensic analysis and proactive mental health assessments in individuals with medical vulnerabilities. Keywords: cardiac valve replacement, cut-throat injury, haemorrhagic shock, medicolegal autopsy, rheumatic heart disease, suicide Introduction Self-inflicted incised wounds to the neck, commonly termed "cut-throat injuries," represent an uncommon yet striking manifestation of self-directed injury, carrying substantial clinical and forensic significance.These kinds of injuries account for a minimal share of international suicide figures, with alternative means like hanging, poisoning, self-immolation, and firearms being notably more common [1,2]. Nonetheless, their occurrence demands an exhaustive forensic analysis owing to the risk of misclassification as homicides, particularly when the characteristics of the wounds or the evidence at the scene lack clarity.The notable risk related to neck injuries stems from the collection of key anatomical parts in the cervical region, which involves the carotid arteries, jugular veins, trachea, oesophagus, and spinal cord [3,4]. Therefore, a rigorous forensic methodology is essential for the evaluation of wound morphology, directional attributes, depth, and accompanying features like hesitation marks or signs of vital reactions. The forensic differentiation between suicidal and homicidal neck injuries is often complex. Suicidal wounds typically exhibit features such as a single deep incised wound, associated hesitation cuts or "tentative wounds," a predictable wound trajectory, and an absence of defensive injuries. In contrast, homicidal neck wounds may be multiple, more forceful, and accompanied by signs of struggle or restraint, as well as injuries to other parts of the body. Additionally, in suicidal cases, the weapon is frequently found near the body, and the act is often carried out in privacy or isolation [5,6]. Psychological and psychiatric determinants are integral to understanding suicidal behaviour.Empirical investigations indicate that individuals experiencing depression, psychosis, substance dependence, or chronic medical ailments exhibit a markedly elevated risk of suicide [7,8]. Within populations afflicted by medical conditions, those diagnosed with chronic cardiac diseases, particularly those characterized by heart failure or structural valvular impairments, exhibit a significant incidence of depression and suicidal ideation.The burden of physical ailments such as dyspnoea, fatigue, impaired mobility, and the psychological turmoil linked to a terminal diagnosis engender feelings of despair, which may lead to suicidal crises in certain instances [9,10]. In these scenarios, suicidality may manifest either impulsively or as an ostensibly rational means of escaping suffering. People with earlier heart surgical procedures, particularly those involving prosthetic valves, commonly deal with ongoing complications and symptoms that can harm their general quality of life.Complications associated with the valves, requirements for anticoagulation therapy, arrhythmias, and recurrent hospital admissions, often exacerbated by financial limitations, may further impair emotional well-being.The correlation between persistent physical ailments and suicidal tendencies highlights the necessity for comprehensive care models that concurrently address both physiological and psychological health in the context of long-term disease management [11,12]. In hospital or institutional settings, suicides are particularly concerning due to questions surrounding supervision, access to means, and preventive measures. Cut-throat suicides occurring within a hospital environment are extremely rare and can raise suspicion of foul play unless carefully evaluated. These situations demand detailed investigation not only of the autopsy findings but also of the behavioural observations, psychiatric history, accessibility of instruments, and environmental controls in place at the time of the incident. This case study details the uncommon and clinically intricate circumstance of a 32-year-old male patient, who has a history of severe rheumatic heart disease and a prior double valve replacement, who succumbed to a self-inflicted incised wound to the neck while hospitalized.The case illustrates the intricate nature of suicide, notably among those enduring persistent health challenges, and underlines the need for a detailed forensic, clinical, and psychological appraisal in identifying the cause of death.This case study was presented at the 46th Annual National Conference of the Indian Academy of Forensic Medicine (Forensic Medicon 2025) on January 31, 2025. Case presentation Case background The deceased was a 32-year-old male patient admitted to the general wards, displaying indications of a lasting cough, worsening difficulty in breathing, and an inability to lie flat (orthopnoea).His medical background notably included rheumatic heart disease, characterized by serious regurgitation issues with the aortic and mitral valves, moderate tricuspid regurgitation, and compromised left ventricular function. Approximately 4.5 years prior to his demise, the patient had undergone major cardiac surgery, which entailed a double valve replacement utilizing TTK Chitra prosthetic valves (aortic 29 mm and mitral 33 mm), left atrial appendage ligation, and De Vega’s repair of the tricuspid valve. The TTK Chitra prosthetic heart valve was designed and developed by Sree Chitra Tirunal Institute for Medical Sciences and Technology (SCTIMST), Thiruvananthapuram, Kerala, India and manufactured by Heart Valve Division,TTK Healthcare Limited, Thumba, Thiruvananthapuram, Kerala, India. Notwithstanding the surgical intervention, he continued to exhibit symptoms (Figures 1, 2) and was reportedly suffering from psychological distress characterized by sensations of impending doom and despair.In the early morning hours, he was found lifeless in the hospital ward, with a kitchen knife near his body and a deep incised wound on his neck. Figure 1. Chest radiographs: posteroanterior view . Open in a new tab A. Two years after the cardiac surgery (depicting Gross cardiomegaly, sternal sutures and prosthetic cardiac valves in situ) B. Four years and five months after the cardiac surgery (depicting Gross cardiomegaly, sternal sutures and prosthetic cardiac valves in situ, along with ground-glass opacities in bilateral lung fields) Figure 2. Computed tomography pulmonary angiogram performed two months prior to the death of the deceased patient. Open in a new tab A. There was evidence of ground-glass opacity in apico-anterior segment of upper lobe of right lung, likely alveolar haemorrhages. B. Gross cardiomegaly was noted with maximally dilated left ventricle. Left atrium was also dilated. C. Prosthetic cardiac valves were seen in situ. D. There was dilatation of main pulmonary artery, as well as left and right pulmonary arteries suggesting pulmonary hypertension. No filling defect was seen. E. Segmental and sub-segmental pulmonary arteries also did not show any filling defect. There was no evidence of thromboembolism. F. Inferior vena cava was prominent. Autopsy findings External Examination An incised wound measuring 7.5 cm was noted over the upper left fronto-lateral aspect of the neck. The presence of two skin tags suggested at least three separate attempts to inflict the injury, a characteristic feature of tentative or hesitation wounds often seen in suicides (Figure 3A). Additionally, two old surgical scars were present: one measuring 18 cm over the anterior midline of the chest (consistent with sternotomy), and another smaller scar (3 cm) over the upper abdomen. Notably, no defensive wounds were observed on the upper limbs or elsewhere on the body. Figure 3. Autopsy findings in the neck . Open in a new tab A. Incised wound over the upper left fronto-lateral aspect of the neck. B. Deep incised wound involving the skin, subcutaneous tissues, and muscles. Major blood vessels were severed, but the trachea and oesophagus were spared. Internal Examination Dissection of the neck uncovered a deep incised wound affecting the integumentary layer, subcutaneous structures, and musculature.Significant vascular structures were transected, yet both the trachea and oesophagus remained intact (Figure 3B).The observation of haemorrhage and vital responses substantiated that the inflicted trauma occurred prior to death. The sternum bore evidence of a prior surgical incision fixed with steel wires.Both lungs exhibited adherence to the thoracic wall, while the pericardium demonstrated adherence to the adjacent mediastinal structures. The heart was massively enlarged, weighing approximately 1300 grams, and displayed marked hypertrophy (Figure 4A). Prosthetic valves were observed in both the mitral and aortic positions (Figure 4B). Additional findings included moderate ascites, hepatodiaphragmatic adhesions (Figure 4C), splenomegaly, and necrosis involving the mesentery, descending colon, and sigmoid colon (Figure 4D). All internal organs appeared pale, indicative of significant blood loss. Figure 4. Thoraco-abdominal autopsy findings . Open in a new tab A. Gross cardiomegaly. B. Prosthetic mitral and aortic valves. C. Hepato-diaphragmatic adhesions. D. Mesenteric ischaemic changes along with necrosis of descending colon, and sigmoid colon. Cause of Death The immediate cause of death was determined to be haemorrhagic shock due to a self-inflicted cut-throat injury using a sharp cutting cum pointed weapon. The underlying chronic cardiac disease, with severely compromised cardiac function, likely played a contributory role by hastening the fatal outcome due to reduced physiological reserve. The mechanism of death was exsanguination, leading to cardiac arrest. Forensic Interpretation The wound pattern, its location, depth, and the presence of skin tags strongly supported a self-inflicted injury. The absence of defensive wounds or signs of struggle, along with the presence of the weapon nearby and lack of evidence indicating third-party involvement, confirmed the manner of death as suicide. Psychological elements, as derived from history given by relatives portrayed a sense of impending doom and emotional burden, likely contributing to the suicidal act. The chronic and progressive nature of his heart disease, coupled with the limitations in daily functioning and persistent symptoms, appeared to have significantly impacted his mental well-being. The physical debilitation may have served as a catalyst for suicidal ideation, particularly in a hospital setting where access to means and absence of constant supervision could facilitate such actions. Discussion Suicidal incised wounds are frequently marked by significant anatomical trauma and the possibility of misidentification due to their similarity to injuries resulting from homicide.From a forensic perspective, a meticulous assessment of the characteristics of the wounds, the psychiatric background, and the surrounding circumstances is crucial in establishing the manner of death[5,6,13]. In this case, the deceased individual, a 32-year-old male patient, suffered from long-term rheumatic heart complications, highlighted by significant aortic and mitral valve leakage, complicated by prior heart operations. Individuals with long-term cardiovascular ailments typically endure compromised hemodynamic stability, diminished functional capabilities, and ongoing psychological stressors stemming from frequent hospital admissions, the burden of pharmacotherapy, and apprehension regarding potential deterioration or mortality[11,12]. These elements can profoundly influence mental well-being, rendering individuals susceptible to depressive disorders and, in extreme instances, suicidal thoughts. Research has consistently indicated a robust correlation between chronic physical ailments and an elevated risk of suicide. Notably, cardiovascular disease has been recognized as an autonomous risk factor for suicidal behaviour, particularly among younger demographics who may possess limited coping strategies or insufficient psychosocial support[14,15]. Individuals suffering from heart failure demonstrate a suicide incidence that is nearly double that of the general populace, a phenomenon attributed to perceptions of disability, persistent pain, feelings of hopelessness, and social isolation. The concept of "rational suicide," in which individuals confronting irreversible deterioration and suffering make a conscious decision to terminate their lives, may also be relevant in this context, especially when taking into account the patient’s expressed sense of doom and declining health status. From a forensic perspective, the injury characteristics support a suicidal origin. The presence of a single incised wound with two skin tags suggests multiple tentative efforts before the fatal incision, a classical feature of self-inflicted injuries. The absence of defence wounds and signs of struggle further supports the conclusion that the act was self-perpetrated. Moreover, the location (left fronto-lateral neck), directionality, and sparing of vital structures such as the trachea and oesophagus, are commonly observed in suicidal cut-throat wounds[5,6]. The presence of haemorrhage and vital reaction confirms the antemortem nature of the wound, strengthening the interpretation. Furthermore, the subject's cardiac dysfunction likely contributed to the mechanisms and speed of demise.The heart exhibited pronounced hypertrophy, suggesting considerable chronic volume overload and myocardial stress.The diminished cardiovascular reserve would have significantly restricted the physiological ability to endure blood loss.In such scenarios, even a minor haemorrhage can swiftly precipitate hypovolemic shock and mortality. The psychological component is equally critical. As per history given by the relatives, the patient's reported insomnia, breathlessness, and feelings of impending doom are consistent with depressive or anxiety-spectrum symptomatology, which are known precursors to suicidal behaviour[9,10]. In hospital settings, suicidal behaviour is often underestimated, particularly in non-psychiatric wards. Easy access to sharp instruments (like kitchen knives), lack of surveillance during early morning hours, and failure to recognize early psychological warning signs may contribute to such tragic outcomes. Hospital-based suicides, especially those involving violent methods, raise important questions about patient safety protocols. Literature underscores the need for routine psychological screening in patients with chronic and terminal illnesses, especially those demonstrating behavioural changes, withdrawal, or expressions of hopelessness. Suicide prevention in medical wards should incorporate environmental risk assessments, staff training, and multidisciplinary collaboration to ensure early detection and intervention[19,20]. This particular case underscores the critical forensic significance of reconstructing the scene and meticulous documentation.In cases involving solitary cut-throat injuries, the synthesis of clinical history, wound characteristics, autopsy results, and psychological assessments is essential for precise interpretation.This comprehensive methodology aids in ruling out foul play, validating instances of suicide, and enhancing both medicolegal determinations and the public health comprehension of suicide among medical patients. Conclusions This case exemplifies the intricate relationship between persistent physical illness and suicidal behaviour, emphasizing the necessity for a sophisticated approach to both clinical treatment and forensic analysis.The deceased, a young male patient afflicted with severe rheumatic heart disease and a history of valve replacement, epitomizes a vulnerable demographic frequently neglected in terms of psychological assistance.His choice to terminate his life using an uncommon and violent method, namely a self-inflicted cut-throat injury, highlights the profound psychological anguish that can accompany chronic cardiac ailments. From a clinical perspective, this case underscores the imperative for comprehensive patient management in contexts of chronic illness.Psychological assessment, suicide risk stratification, and the implementation of suitable safety protocols are vital components of inpatient care, particularly for individuals exhibiting emotional distress or observable behavioural alterations.The absence of these measures can result in tragic yet avoidable consequences, even in closely monitored medical settings. This case ultimately emphasizes the hidden challenges associated with chronic health issues and the urgent need to focus on both the physical and mental health elements of patient treatment.Effective suicide prevention must progress from reactive interventions to a proactive, integrated strategy that emphasizes early identification, mental health resources, and ongoing monitoring of at-risk individuals. Disclosures Human subjects: Consent for treatment and open access publication was obtained or waived by all participants in this study. Conflicts of interest: In compliance with the ICMJE uniform disclosure form, all authors declare the following: Payment/services info: All authors have declared that no financial support was received from any organization for the submitted work. Financial relationships: All authors have declared that they have no financial relationships at present or within the previous three years with any organizations that might have an interest in the submitted work. Other relationships: All authors have declared that there are no other relationships or activities that could appear to have influenced the submitted work. Author Contributions Concept and design: Anand Kumar, Kumar Shubhendu, Ankur Chaudhary, Ajay K. Bhagat Acquisition, analysis, or interpretation of data: Anand Kumar, Kumar Shubhendu, Ankur Chaudhary, Ajay K. Bhagat Drafting of the manuscript: Anand Kumar, Kumar Shubhendu, Ankur Chaudhary, Ajay K. Bhagat Critical review of the manuscript for important intellectual content: Anand Kumar, Kumar Shubhendu, Ankur Chaudhary, Ajay K. Bhagat Supervision: Kumar Shubhendu References 1.Method of suicide in the mentally ill: a national clinical survey. Hunt IM, Swinson N, Palmer B, et al. Suicide Life Threat Behav. 2010;40:22–34. doi: 10.1521/suli.2010.40.1.22. [DOI] [PubMed] [Google Scholar] 2.A review of the various suicide methods used around the world. Bidaki R, Shirani S, Shamsian M, Tezerjani EP, Heidari F, Shirani B, Aida-Farsham Aida-Farsham. International Journal of Medical Reviews. 2016;3:504–507. [Google Scholar] 3.Penetrating injuries of the neck. Albadri MLH, Albeiruty UA, Mossa AA. Iraqi Postgraduate Medical Journal. 2009;8:196–203. [Google Scholar] 4.Chana M. Hamilton Bailey’s Emergency Surgery. Vol. 336. Boca Raton, FL: CRC Press; 2025. Vascular injuries of the neck; p. 40. 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15537	https://pressbooks-dev.oer.hawaii.edu/collegephysics/chapter/14-2-temperature-change-and-heat-capacity/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 14 Heat and Transfer Methods 101 14.2 Temperature Change and Heat Capacity Summary Observe heat transfer and change in temperature and mass. Calculate final temperature after heat transfer between two objects. One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance. The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Owing to the fact that the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid). HEAT TRANSFER AND ENERGY CHANGE The quantitative relationship between heat transfer and temperature change contains all three factors: [latex]\boldsymbol{Q=mc\Delta{T}},[/latex] where[latex]\boldsymbol{Q}[/latex]is the symbol for heat transfer,[latex]\boldsymbol{m}[/latex]is the mass of the substance, and[latex]\boldsymbol{\Delta{T}}[/latex]is the change in temperature. The symbol[latex]\boldsymbol{c}[/latex]stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by[latex]\boldsymbol{1.00^{\circ}\textbf{C}}.[/latex] The specific heat cc is a property of the substance; its SI unit is[latex]\boldsymbol{\textbf{J/(kg}\cdotp\textbf{K)}}[/latex]or[latex]\boldsymbol{\textbf{J/(kg}\cdotp^{\circ}\textbf{C)}}.[/latex]Recall that the temperature change[latex]\boldsymbol{(\Delta{T})}[/latex]is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is[latex]\boldsymbol{\textbf{kcal/(kg}\cdotp^{\circ}\textbf{C)}}.[/latex] Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. In general, the specific heat also depends on the temperature. Table 1 lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. We see from this table that the specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth. Example 1: Calculating the Required Heat: Heating Water in an Aluminum Pan A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from[latex]\boldsymbol{20.0^{\circ}\textbf{C}}[/latex]to[latex]\boldsymbol{80.0^{\circ}\textbf{C}}./latex How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water? Strategy The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table 1. Solution Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature. Calculate the temperature difference: [latex]\boldsymbol{\Delta{T}=T_{\textbf{f}}-T_{\textbf{i}}=60.0^{\circ}\textbf{C}}.[/latex] 2. Calculate the mass of water. Because the density of water is[latex]\boldsymbol{1000\textbf{ kg/m}^3},[/latex]one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is[latex]\boldsymbol{m_{\textbf{w}}=0.250\textbf{ kg}}.[/latex] 3. Calculate the heat transferred to the water. Use the specific heat of water in Table 1: [latex]\boldsymbol{Q_{\textbf{w}}=m_{\textbf{w}}c_{\textbf{w}}\Delta{T}=(0.250\textbf{ kg})(4186\textbf{ J/kg}^{\circ}\textbf{C})(60.0^{\circ}\textbf{C})=62.8\textbf{ kJ}}.[/latex] 4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1: [latex]\boldsymbol{Q_{\textbf{Al}}=m_{\textbf{Al}}c_{\textbf{Al}}\Delta{T}=(0.500\textbf{ kg})(900\textbf{ J/kg}^{\circ}\textbf{C})(60.0^{\circ}\textbf{C})= 27.0\times10^4\textbf{ J}=27.0\textbf{ kJ}}.[/latex] 5. Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat: [latex]\boldsymbol{Q_{\textbf{Total}}=Q_{\textbf{W}}+Q_{\textbf{Al}}=62.8\textbf{ kJ}+ 27.0\textbf{ kJ}=89.8\textbf{ kJ}}.[/latex] Thus, the amount of heat going into heating the pan is [latex]\boldsymbol{\frac{27.0\textbf{ kJ}}{89.8\textbf{ kJ}}}[/latex][latex]\boldsymbol{\times100\%=30.1\%},[/latex] and the amount going into heating the water is [latex]\boldsymbol{\frac{62.8\textbf{ kJ}}{89.8\textbf{ kJ}}}[/latex][latex]\boldsymbol{\times100\%=69.9\%}.[/latex] Discussion In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times greater than that of aluminum. Therefore, it takes a bit more than twice the heat to achieve the given temperature change for the water as compared to the aluminum pan. Figure 2. The smoking brakes on this truck are a visible evidence of the mechanical equivalent of heat. Example 2: Calculating the Temperature Increase from the Work Done on a Substance: Truck Brakes Overheat on Downhill Runs Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. The problem is that the mass of the truck is large compared with that of the brake material absorbing the energy, and the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment. Calculate the temperature increase of 100 kg of brake material with an average specific heat of[latex]\boldsymbol{800\textbf{ J/kg}\cdotp^{\circ}\textbf{C}}[/latex]if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed. Strategy If the brakes are not applied, gravitational potential energy is converted into kinetic energy. When brakes are applied, gravitational potential energy is converted into internal energy of the brake material. We first calculate the gravitational potential energy[latex]\boldsymbol{(Mgh)}[/latex]that the entire truck loses in its descent and then find the temperature increase produced in the brake material alone. Solution Calculate the change in gravitational potential energy as the truck goes downhill [latex]\boldsymbol{Mgh=(10,000\textbf{ kg})(9.80\textbf{ m/s}^2)(75.0\textbf{ m})=7.35\times10^6\textbf{ J}}.[/latex] 2. Calculate the temperature from the heat transferred using[latex]\boldsymbol{Q=Mgh}[/latex]and where[latex]\boldsymbol{m}[/latex]is the mass of the brake material. Insert the values[latex]\boldsymbol{m=100\textbf{ kg}}[/latex]and[latex]\boldsymbol{c=800\textbf{ J/kg}\cdotp^{\circ}\textbf{C}}[/latex]to find [latex]\boldsymbol{\Delta{T}\:=}[/latex][latex]\boldsymbol{\frac{(7.35\times10^6\textbf{ J})}{(100\textbf{ kg})(800\textbf{ J/kg}^{\circ}\textbf{C})}}[/latex][latex]\boldsymbol{=92^{\circ}\textbf{C}}.[/latex] Discussion This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, so this technique is not practical. However, the same idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted by the brakes into electrical energy (battery). \| Substances \| Specific heat (c) \| --- \| \| Solids \| J/kg⋅ºC \| kcal/kg⋅ºC2 \| \| Aluminum \| 900 \| 0.215 \| \| Asbestos \| 800 \| 0.19 \| \| Concrete, granite (average) \| 840 \| 0.20 \| \| Copper \| 387 \| 0.0924 \| \| Glass \| 840 \| 0.20 \| \| Gold \| 129 \| 0.0308 \| \| Human body (average at 37 °C) \| 3500 \| 0.83 \| \| Ice (average, -50°C to 0°C) \| 2090 \| 0.50 \| \| Iron, steel \| 452 \| 0.108 \| \| Lead \| 128 \| 0.0305 \| \| Silver \| 235 \| 0.0562 \| \| Wood \| 1700 \| 0.4 \| \| Liquids \| \| \| \| Benzene \| 1740 \| 0.415 \| \| Ethanol \| 2450 \| 0.586 \| \| Glycerin \| 2410 \| 0.576 \| \| Mercury \| 139 \| 0.0333 \| \| Water (15.0 °C) \| 4186 \| 1.000 \| \| Gases 3 \| \| \| \| Air (dry) \| 721 (1015) \| 0.172 (0.242) \| \| Ammonia \| 1670 (2190) \| 0.399 (0.523) \| \| Carbon dioxide \| 638 (833) \| 0.152 (0.199) \| \| Nitrogen \| 739 (1040) \| 0.177 (0.248) \| \| Oxygen \| 651 (913) \| 0.156 (0.218) \| \| Steam (100°C) \| 1520 (2020) \| 0.363 (0.482) \| \| Table 1. Specific Heats1 of Various Substances \| Note that Example 2 is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced by a blow torch instead of mechanically. Example 3: Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan Suppose you pour 0.250 kg of[latex]\boldsymbol{20.0^{\circ}\textbf{C}}[/latex]water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of[latex]\boldsymbol{150^{\circ}\textbf{C}}.[/latex]Assume that the pan is placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium a short time later? Strategy The pan is placed on an insulated pad so that little heat transfer occurs with the surroundings. Originally the pan and water are not in thermal equilibrium: the pan is at a higher temperature than the water. Heat transfer then restores thermal equilibrium once the water and pan are in contact. Because heat transfer between the pan and water takes place rapidly, the mass of evaporated water is negligible and the magnitude of the heat lost by the pan is equal to the heat gained by the water. The exchange of heat stops once a thermal equilibrium between the pan and the water is achieved. The heat exchange can be written as[latex]\boldsymbol{\|Q_{\textbf{hot}}\|=Q_{\textbf{cold}}}.[/latex] Solution Use the equation for heat transfer[latex]\boldsymbol{Q=mc\Delta{T}}[/latex]to express the heat lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature: Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water and the final temperature: Note that[latex]\boldsymbol{Q_{\textbf{hot}}<0}[/latex]and[latex]\boldsymbol{Q_{\textbf{cold}}>0}[/latex]and that they must sum to zero because the heat lost by the hot pan must be the same as the heat gained by the cold water: [latex]\begin{array}{rcl} \boldsymbol{Q_{\textbf{cold}}+Q_{\textbf{hot}}} & \boldsymbol{=} & \boldsymbol{0,} \ \boldsymbol{Q_{\textbf{cold}}} & \boldsymbol{=} & \boldsymbol{-Q_{\textbf{hot}},} \ \boldsymbol{m_{\textbf{W}}c_{\textbf{W}}(T_{\textbf{f}}-20.0^{\circ}\textbf{C})} & \boldsymbol{=} & \boldsymbol{-m_{\textbf{Al}}c_{\textbf{Al}}(T_{\textbf{f}}-150^{\circ}\textbf{C}).} \end{array}[/latex] 4. This an equation for the unknown final temperature,[latex]\boldsymbol{T_{\textbf{f}}}[/latex] 5. Bring all terms involving[latex]\boldsymbol{T_{\textbf{f}}}[/latex]on the left hand side and all other terms on the right hand side. Solve for[latex]\boldsymbol{T_{\textbf{f}}},[/latex] and insert the numerical values: [latex]\begin{array}{lcl} \boldsymbol{T_{\textbf{f}}} & \boldsymbol{=} & \boldsymbol{\frac{(0.500\textbf{ kg})(900\textbf{ J/kg}^{\circ}\textbf{C})(150^{\circ}\textbf{C})+(0.250\textbf{ kg})(4186\textbf{ J/kg}^{\circ}\textbf{C})(20.0^{\circ}\textbf{C})}{(0.500\textbf{ kg})(900\textbf{ J/kg}^\textbf{C})+(0.250\textbf{ kg})(4186\textbf{ J/kg}^{\circ}\textbf{C})}} \ {} & \boldsymbol{=} & \boldsymbol{\frac{88430\textbf{ J}}{1496.5\textbf{ J/}^{\circ}\textbf{C}}} \ {} & \boldsymbol{=} & \boldsymbol{59.1^{\circ}\textbf{C}.} \end{array}[/latex] Discussion This is a typical calorimetry problem—two bodies at different temperatures are brought in contact with each other and exchange heat until a common temperature is reached. Why is the final temperature so much closer to[latex]\boldsymbol{20.0^{\circ}\textbf{C}}[/latex]than[latex]\boldsymbol{150^{\circ}\textbf{C}}?[/latex]The reason is that water has a greater specific heat than most common substances and thus undergoes a small temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter). TAKE-HOME EXPERIMENT: TEMPERATURE CHANGE OF LAND AND WATER What heats faster, land or water? To study differences in heat capacity: Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can achieve approximately equal masses by using[latex]\boldsymbol{50\%}[/latex]more water by volume.) Heat both (using an oven or a heat lamp) for the same amount of time. Record the final temperature of the two masses. Now bring both jars to the same temperature by heating for a longer period of time. Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes. Which sample cools off the fastest? This activity replicates the phenomena responsible for land breezes and sea breezes. Check Your Understanding 1: If 25 kJ is necessary to raise the temperature of a block from[latex]\boldsymbol{25^{\circ}\textbf{C}}[/latex]to[latex]\boldsymbol{30^{\circ}\textbf{C}},[/latex]how much heat is necessary to heat the block from[latex]\boldsymbol{45^{\circ}\textbf{C}}[/latex]to[latex]\boldsymbol{50^{\circ}\textbf{C}}?[/latex] Summary The transfer of heat[latex]\boldsymbol{Q}[/latex]that leads to a change[latex]\boldsymbol{\Delta{T}}[/latex]in the temperature of a body with mass[latex]\boldsymbol{m}[/latex]is[latex]\boldsymbol{Q=mc\Delta{T}},[/latex]where[latex]\boldsymbol{c}[/latex]is the specific heat of the material. This relationship can also be considered as the definition of specific heat. Conceptual Questions 1: What three factors affect the heat transfer that is necessary to change an object’s temperature? 2: The brakes in a car increase in temperature by[latex]\boldsymbol{\Delta{T}}[/latex]when bringing the car to rest from a speed[latex]\boldsymbol{v}.[/latex]How much greater would[latex]\boldsymbol{\Delta{T}}[/latex]be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes. Problems & Exercises 1: On a hot day, the temperature of an 80,000-L swimming pool increases by[latex]\boldsymbol{1.50^{\circ}\textbf{C}}.[/latex]What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation. 2: Show that[latex]\boldsymbol{1\textbf{ cal/g}\cdotp^{\circ}\textbf{C}=1\textbf{ kcal/kg}\cdotp^{\circ}\textbf{C}}.[/latex] 3: To sterilize a 50.0-g glass baby bottle, we must raise its temperature from[latex]\boldsymbol{22.0^{\circ}\textbf{C}}[/latex]to[latex]\boldsymbol{95.0^{\circ}\textbf{C}}.[/latex]How much heat transfer is required? 4: The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at[latex]\boldsymbol{20.0^{\circ}\textbf{C}}:/latex water; (b) concrete; (c) steel; and (d) mercury. 5: Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers. 6: A 0.250-kg block of a pure material is heated from[latex]\boldsymbol{20.0^{\circ}\textbf{C}}[/latex]to[latex]\boldsymbol{65.0^{\circ}\textbf{C}}[/latex]by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed. 7: Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water? 8: (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a[latex]\boldsymbol{54.9^{\circ}\textbf{C}}[/latex]temperature increase? (b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent. 9: Following vigorous exercise, the body temperature of an 80.0-kg person is[latex]\boldsymbol{40.0^{\circ}\textbf{C}}.[/latex]At what rate in watts must the person transfer thermal energy to reduce the the body temperature to[latex]\boldsymbol{37.0^{\circ}\textbf{C}}[/latex]in 30.0 min, assuming the body continues to produce energy at the rate of 150 W?[latex]\boldsymbol{(1\textbf{ watt}=\:1\textbf{ joule/second or }1\textbf{ W}=\:1\textbf{ J/s})}.[/latex] 10: Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails[latex]\boldsymbol{(1\textbf{ watt }=\:1\textbf{ joule/second or }1\textbf{ W}=\:1\textbf{ J/s and }1\textbf{ MW}=\:1\textbf{ megawatt})}./latex Calculate the rate of temperature increase in degrees Celsius per second[latex]\boldsymbol{(^{\circ}\textbf{C/s})}[/latex]if the mass of the reactor core is[latex]\boldsymbol{1.60\times10^5\textbf{ kg}}[/latex]and it has an average specific heat of[latex]\boldsymbol{0.3349\textbf{ kJ/kg}^{\circ}\cdotp\textbf{C}}./latex How long would it take to obtain a temperature increase of[latex]\boldsymbol{2000^{\circ}\textbf{C}},[/latex]which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the[latex]\boldsymbol{5\times10^5\textbf{-kg}}[/latex]steel containment vessel would also begin to heat up.) Figure 3. Radioactive spent-fuel pool at a nuclear power plant. Spent fuel stays hot for a long time. (credit: U.S. Department of Energy) Footnotes 1 The values for solids and liquids are at constant volume and at[latex]\boldsymbol{25^{\circ}\textbf{C}},[/latex]except as noted. 2 These values are identical in units of[latex]\boldsymbol{\textbf{cal/g}\cdotp^{\circ}\textbf{C}}.[/latex] 3 [latex]\boldsymbol{c_{\textbf{v}}}[/latex]at constant volume and at[latex]\boldsymbol{20.0^{\circ}\textbf{C}},[/latex]except as noted, and at 1.00 atm average pressure. Values in parentheses are[latex]\boldsymbol{c_{\textbf{p}}}[/latex]at a constant pressure of 1.00 atm. Glossary specific heat : the amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 ºC Solutions Check Your Understanding 1: The heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case. Problems & Exercises [latex]\boldsymbol{5.02\times10^8\textbf{ J}}[/latex] [latex]\boldsymbol{3.07\times10^3\textbf{ J}}[/latex] 617 W License College Physics chapters 1-17 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. Share This Book
15538	https://www.preprints.org/manuscript/202410.0922/v1	Loading web-font Gyre-Pagella/Size2/Regular 11 October 2024 11 October 2024 You are already at the latest version 1. Introduction 2. Key Euclidean Distance Properties for Computational Optimization 3. Key Approaches for Optimizing Euclidean Distance Computation 4. Comparative Analysis of Optimization Approaches 5. Discussion 6. Conclusion Data Availability Statement Conflicts of Interest Short Biography of Authors This version is not peer-reviewed. Optimizing Euclidean Distance Computation Rustam Mussabayev Rustam Mussabayev A peer-reviewed article of this preprint also exists. Submitted: 11 October 2024 Posted: 11 October 2024 You are already at the latest version In this paper, we present a comparative analysis of seventeen different approaches to optimizing Euclidean distance computations, a core mathematical operation that plays a critical role in a wide range of algorithms, particularly in machine learning and data analysis. The Euclidean distance, being a computational bottleneck in large-scale optimization problems, requires efficient computation techniques to improve the performance of various distance-dependent algorithms. To address this, several optimization strategies can be employed to accelerate distance computations. From spatial data structures and approximate nearest neighbor algorithms to dimensionality reduction, vectorization, and parallel computing, various approaches exist to accelerate Euclidean distance computation in different contexts. Such approaches are particularly important for speeding up key machine learning algorithms like K-means and K-nearest neighbors (KNN). By understanding the trade-offs and assessing the effectiveness, complexity, and scalability of various optimization techniques, our findings help practitioners choose the most appropriate methods for improving Euclidean distance computations in specific contexts. These optimizations enable scalable and efficient processing for modern data-driven tasks, directly leading to reduced energy consumption and a minimized environmental impact. Keywords: Euclidean distance ; optimization strategies ; K-means clustering ; K-nearest neighbors (KNN) ; vectorization ; parallelization ; triangle inequality ; spatial data structures ; block vector approximations ; approximate methods Subject: Computer Science and Mathematics - Computational Mathematics 1. Introduction Euclidean distance, a fundamental concept in geometry, is the most intuitive measure of spatial separation between points. Its computation is a core mathematical operation across various disciplines, including linear algebra, optimization, data analysis, and machine learning . In many algorithms within these fields, Euclidean distance is one of the most frequently performed computational operations and often constitutes a primary bottleneck in large-scale optimization problems, as clustering, nearest-neighbor searches, or high-dimensional data applications . Consequently, optimizing basic operations like distance computations can lead to significant improvements in the overall performance of many distance-dependent algorithms. For instance, in large datasets or real-time processing applications, reducing the overhead of distance calculations can substantially improve the speed and scalability of algorithms like K-means clustering or K-nearest neighbors (KNN) . This requires a deep understanding of both the mathematical properties of Euclidean distance and the specific characteristics of the problem at hand. To address these challenges, several optimization strategies can be employed to accelerate distance computations. In this article, we provide a comprehensive review and comparative analysis of various optimization approaches aimed at improving the efficiency of Euclidean distance computations. The primary goal of these approaches is to reduce the computational burden associated with large-scale distance calculations. Given two points and in an n-dimensional space, the Euclidean distance between them is defined as:where and are the coordinates of the points and in n-dimensional space ; and are the kth components of and , respectively; and is the norm of the vector difference between and . The Euclidean distance defines a metric space by satisfying key conditions such as non-negativity, identity of indiscernibles, symmetry, and the triangle inequality . These properties give its metric space structure, allowing for precise mathematical analysis of geometric relationships in the data. Euclidean distance is a fundamental metric for quantifying similarity and dissimilarity, playing a key role in tasks such as clustering, classification, decision making, location tracking, route planning, anomaly detection, bioinformatics, and optimization . Many machine learning algorithms, including K-Nearest Neighbors (KNN), K-Means, Hierarchical Clustering, Principal Component Analysis (PCA), Vector Quantization (VQ), Mean Shift, and DBSCAN, depend heavily on distance calculations [7,8]. Beyond machine learning, Euclidean distance is crucial in image processing for tasks like edge detection, object recognition, and image segmentation , and in geographic information systems (GIS) for route planning and spatial analysis . It is also central to anomaly detection, identifying outliers based on their distance from a central cluster, particularly useful in fraud detection and network security . In optimization problems such as the Facility Location Problem (FLP), it minimizes total travel distance, ensuring efficient facility placement, which is especially important in logistics and supply chain management [13,14]. In bioinformatics, Euclidean distance is employed to compare genetic sequences and biological samples , clustering similar gene expression patterns, identifying biomarkers, and mapping evolutionary relationships . In robotics path planning , it helps robots navigate the shortest path while avoiding obstacles, essential for precision in automated manufacturing, autonomous vehicles, and robotic surgery . In Natural Language Processing (NLP), Euclidean distance measures similarity between high-dimensional word embeddings or document vectors. These pairwise distance matrices are frequently used in graph-based methods, such as community detection, to uncover semantic clusters and build ontologies [19,20]. In numerous algorithms that employ Euclidean distance calculations, such as K-means clustering or K-nearest neighbors (KNN), computing these distances represents a significant computational burden, especially as the size of the data and the dimensionality increase . Each computation involves measuring the straight-line distance between pairs of points in a multidimensional space, which can be computationally intensive due to the necessity of performing multiple arithmetic operations. Full Euclidean distance calculations involve expensive operations like squaring differences, summing them, and taking square roots. Moreover, these distance calculations are not just performed once; they are typically executed repeatedly across many iterations of an algorithm. For instance, in each iteration of K-means, distances from each point to every cluster center must be recalculated to reassign points to the nearest cluster . Similarly, in KNN, distances are computed to all training data points to find the nearest neighbors for classification or regression . This repetitive computation can exponentially increase the time complexity and processing load, particularly with large datasets, making optimization of these distance calculations a crucial aspect of improving the efficiency and scalability of such algorithms. Optimizing the computation of Euclidean distances can markedly enhance the performance of a multitude of algorithms where it is a fundamental, repeatedly executed operation . This is particularly true for data-intensive applications in fields like machine learning, data mining, and computer vision, where used algorithms rely heavily on frequent distance calculations. By refining these calculations the overall computational demand decreases. This not only speeds up the algorithms but also reduces energy consumption and allows for processing larger datasets or achieving higher throughput in real-time applications . Such enhancements in efficiency are crucial for scaling systems to handle the growing volumes and complexities of data encountered in modern analytical tasks, thereby broadening their applicability and improving their utility in practical scenarios. Understanding these fundamental challenges and strategies is essential for researchers and practitioners working with large-scale and high-dimensional datasets. Optimizing Euclidean distance computations is not merely a technical necessity but a foundational step towards enabling more advanced data analysis and machine learning tasks in the era of big data. 2. Key Euclidean Distance Properties for Computational Optimization Here, we summarize the main mathematical properties of Euclidean distance and how they can be efficiently used for its computational optimization: Symmetry: . In pairwise distance calculations, this property lets you compute each distance once and store it in a symmetric matrix, halving the number of computations . Non-Negativity: . It provides a basic validation check during distance computations, allowing the algorithm to terminate early or avoid further incorrect computations . Identity of Indiscernibles: . This allows skipping distance calculations for identical points and filling the main diagonal of pairwise distance matrices with zeros, reducing unnecessary operations . Triangle Inequality (Subadditivity): This property states that for any points A, B, and C, the direct distance between A and C is less than or equal to the sum of the distances through an intermediate point B, i.e., . Triangle Inequality is crucial for reducing unnecessary distance calculations in clustering and nearest-neighbor algorithms. For example, in nearest-neighbor search, if and are known, you can skip calculating if the inequality shows it is not needed . This property also enables optimizations in hierarchical clustering and network-based applications by using intermediate distances to avoid redundant computations. In KD-trees, the triangle inequality helps prune unnecessary comparisons, speeding up the search process . Spatial Locality: Points closer in space have smaller Euclidean distances, while distant points have larger distances. This property enables partitioning techniques, like KD-trees or Ball Trees [29,30], to group nearby points, allowing computations to focus on local regions instead of the entire dataset. Monotonicity of the Square Root Function: The square root function is monotonically increasing, meaning that if , then . Many distance-based algorithms (e.g., KNN and K-means) only need relative distances. Skipping the square root and using squared Euclidean distance reduces computational complexity while preserving the distance order [1,31]. Dimensional Independence: Euclidean distance calculations can be performed independently for each dimension. This allows for parallelization, where each dimension’s contribution is computed separately, speeding up computations on multi-core processors . Additivity of Independent Dimensions or Subspaces: In high-dimensional spaces, Euclidean distance is the sum of squared differences across dimensions. Low-variance or threshold-exceeding dimensions can be skipped, enabling early termination and reducing computation with minimal accuracy loss . Scaling: While positive homogeneity is a property of norms, a similar scaling behavior holds for Euclidean distance under scalar multiplication. Specifically, for a scalar , the distance between two points scales as . This reflects that the distance is scaled by the absolute value of the scalar. After re-scaling, you can utilize previously computed distances and apply the scaling factor, avoiding full recomputation . Convexity: The squared Euclidean distance is a convex function of its inputs. This property is crucial for optimization, as many algorithms (e.g., gradient descent) leverage convexity to ensure efficient convergence to a global minimum with fewer iterations . While the squared Euclidean distance is convex with respect to one variable when the other is fixed, the overall optimization problem may still be non-convex. Continuity: Euclidean distance is continuous, so small changes in point positions cause small changes in distance, . This allows for distance approximations in real-time applications, enabling faster computations in dynamic environments by skipping exact recalculations . Invariance under Geometric Transformations: Euclidean distance remains unchanged under translations, rotations, or reflections. For any transformation T, , enabling optimization by skipping distance recomputation, improving performance in graphics and simulations . Minkowski Special Case: Since Euclidean distance is a special case of the Minkowski distance with , other Minkowski norms, such as the -norm () and -norm (), provide upper or lower bounds. These bounds can be used for fast filtering: if an estimate using a simpler norm is already too large, further exact calculations can be skipped, improving performance . Sparsity Robustness: Euclidean distance computations in sparse vector spaces can be significantly optimized by only computing distances on non-zero dimensions. This property is particularly useful in large-scale machine learning tasks involving high-dimensional but sparse datasets, as it reduces the number of operations . Euclidean distance forms a metric space due to its properties (symmetry, triangle inequality, non-negativity, and identity of indiscernibles), enabling optimized search algorithms such as range queries, nearest neighbor searches, and clustering by reducing redundant calculations. 3. Key Approaches for Optimizing Euclidean Distance Computation This section provides an overview of the main approaches for optimizing Euclidean distance calculations to accelerate the process. These approaches can generally be categorized into algorithm-specific methods, low-level optimizations, and hardware acceleration techniques. Additionally, hybrid approaches, which combine these methods, often yield the best results. 3.1. Algorithm-Specific Approaches Algorithm-specific approaches for optimizing Euclidean distance computation focus on modifying or designing algorithms to reduce the number of required distance calculations by leveraging the mathematical properties of Euclidean distance, thereby minimizing computational overhead while preserving accuracy in distance measurements . Squared Euclidean Distance One of the primary methods for reducing the computational load of Euclidean distance calculations involves algorithmic optimizations that minimize the number of required operations. A straightforward optimization is to eliminate the computationally expensive square root operation when only relative distances between points are needed, as in K-nearest neighbors (KNN) or K-means clustering. Since the square root is a monotonically increasing function, the ordering of distances remains unchanged, allowing the use of squared Euclidean distance instead . This simplification is particularly beneficial for large datasets, as square root operations are more computationally intensive than basic arithmetic. Additionally, using squared Euclidean distance not only enhances computational efficiency in the K-means algorithm but also improves its mathematical tractability . For instance, in clustering techniques, the Euclidean distance is integral to solving the Minimum Sum-of-Squares Clustering (MSSC) problem , where the objective is to partition a set of data points into K clusters such that the sum of the squared Euclidean distances between each point and the centroid of its assigned cluster is minimized, that is by minimizing the total within-cluster variance:where denotes the squared Euclidean distance between the data point and the centroid of cluster . From a computational perspective, the squared Euclidean distance is computed by simply eliminating the computationally expensive square root operation from the standard Euclidean distance and can be considered as one of the possible optimization approaches: Given that the square root function is monotonically increasing, minimizing the squared Euclidean distance is mathematically equivalent to minimizing the Euclidean distance : This equivalence ensures that the cluster assignments obtained by minimizing the squared distance are identical to those obtained by minimizing the unsquared distance. This elimination simplifies computations, especially for large datasets, as square root operations are more computationally intensive than basic arithmetic operations. The squared Euclidean distance is a quadratic function, which is convex with respect to each data point individually when the centroid is fixed. However, the overall objective function (2) in the MSSC algorithms is not convex with respect to both the cluster assignments and centroids simultaneously. Therefore, the K-means algorithm can converge to a local minimum, which is not necessarily the global minimum . To find solutions closer to the global minimum, more sophisticated MSSC algorithms are typically employed . Despite the non-convexity of the problem, the differentiability of the squared Euclidean distance allows for the straightforward derivation of update rules within the K-means algorithm. Specifically, the centroid of a cluster can be straightforward and efficiently computed as the mean of the data points within that cluster: The K-means algorithm iteratively assigns each data point to the cluster with the nearest centroid , and then updates each centroid by calculating the mean of all data points assigned to that cluster, i.e., , iteratively minimizing the within-cluster variance until convergence. A remarkable feature of the K-means algorithm is its ability to naturally and iteratively minimize the sum of squared Euclidean distances, , making it particularly well-suited for large-scale clustering tasks. This natural minimization occurs through the algorithm’s intrinsic process of reducing the sum of squared distances between data points and their respective cluster centroids by alternating between assignment and update steps, ultimately leading to convergence without the need for external adjustments . By utilizing the squared Euclidean distance (3), the K-means algorithm achieves computational efficiency and mathematical tractability, even though it may converge to a local minimum . This approach enables the algorithm to find meaningful cluster configurations in a computationally feasible manner, though it may not always guarantee the global optimum of the clustering objective. However, this disadvantage can be mitigated by using simple techniques like Multi-start K-means clustering, which helps to obtain near-optimal solutions. Lower Bound Techniques The block vector approximation allows the K-means algorithm to avoid unnecessary full distance calculations by providing a lower bound on the Euclidean distance between a point and a cluster center. This lower bound enables the algorithm to make early decisions about whether a cluster center can possibly be the closest center, thereby eliminating the need for a complete calculation of computationally expensive Euclidean distance in many cases. Initially, to form the block vector representation, the given a high-dimensional data point , it is subdivided into smaller sub-vectors, or blocks, of fixed sizes along the dimensions:where each represents a subset of consecutive components of . For each block, the p-norm of the sub-vector is computed, resulting in a block vector : This process reduces the dimensionality of the original vector while retaining essential information about its structure. During the K-means algorithm, each point is assigned to the closest cluster center. For a point with a known closest center , the lower bound on the distance to any other center is computed using the block vectors. By using Hölder’s inequality, the inner product is approximated by the inner product of the block vectors . If the lower bound on the distance to is greater than the actual distance to the current closest center , i.e.,then cannot be the closest center, and the algorithm skips the full Euclidean distance calculation for . Here the expression refers to the squared Euclidean norm of the vector . This results in significant computational savings, especially when there are many centers. Algorithm 1 illustrates the snippet of K-means pseudocode leveraging lower bounds from block vector approximations to accelerate clustering. \| \| \| Algorithm 1: Snippet of K-means pseudocode with Block Vectors optimization \| \| \| The core computational saving comes from the fact that block vectors are much shorter than the original vectors, and the dot product of block vectors is much cheaper than a full Euclidean distance computation. Since can be precomputed for all x, they don’t need to be recomputed in every iteration. This mechanism works particularly well when there are a large number of clusters, or when the data is high-dimensional, as the savings in computation increase in such scenarios. Triangle Inequality The triangle inequality property , , can be used to prune unnecessary distance calculations. In certain cases, if a partial distance is already larger than a known threshold, further calculation can be skipped, as the result will not affect the outcome . For example, Hamerly’s method for accelerating the K-means algorithm leverages the triangle inequality to significantly reduce the number of distance calculations required during clustering. In the naive K-means algorithm, the distance between each data point and every cluster centroid is recalculated in each iteration, resulting in a computational complexity of per iteration, where m is the number of points and k is the number of clusters. Hamerly’s method optimizes this process by maintaining upper bounds and lower bounds on the distances between each point and its closest and second closest centroids, respectively. The algorithm also calculates a threshold for each cluster, representing half the minimum distance between the centroid and any other centroid: . Using the Triangle Inequality, it can skip unnecessary distance calculations if the current bounds indicate that the point cannot be closer to another centroid than its current assignment. Mathematically, this is achieved by comparing the upper bound to the maximum of and , and only recomputing distances when necessary. Algorithm 2 presents a pseudocode snippet that demonstrates how Hamerly’s method efficiently reduces distance computations in the K-means algorithm when recalculating distances between points and centroids. In the presented pseudocode, denotes the index of the centroid (or cluster) currently assigned to data point . This approach effectively reduces the number of Euclidean distance calculations, resulting in a more efficient algorithm that converges faster than the naive K-means while maintaining the same accuracy in clustering. \| \| \| Algorithm 2: Snippet of K-means pseudocode with Triangle Inequality optimization \| \| \| Figure 1 illustrates the positions of data points and centroids on a number line, highlighting the distances and thresholds used to apply the triangle inequality in skipping distance computations during K-means clustering. In this example, the data point is closer to centroid than to centroid , and the upper bound is less than the threshold . This condition allows us to skip the computation of the distance . Figure 1. Illustration of the triangle inequality in K-means clustering to reduce computations. Since , we can skip computing . Recursive Distance Updating Another widely-used method in data analysis that heavily relies on Euclidean distance computations is agglomerative hierarchical clustering , which builds a hierarchy by repeatedly merging the closest clusters until all points form a single cluster. Starting with each point as its own cluster, the algorithm merges the pair with the smallest distance and recalculates the distances between the newly merged cluster and all remaining clusters. In its naive implementation, this requires recalculating distances from scratch after every merge, making it computationally expensive for large datasets, as the number of pairwise distance calculations increases quadratically. Ward’s method, one of the most efficient techniques for agglomerative hierarchical clustering, is an example of how to minimize Euclidean distance recalculations when solving the Minimum Sum-of-Squares Clustering (MSSC) problem. Ward’s method efficiently minimizes within-cluster variance (2) after merging clusters. It benefits from a recursive update mechanism known as the Lance–Williams formula, which optimizes the process of updating distances between the newly merged cluster and the remaining clusters. Instead of recalculating distances from scratch, the Lance–Williams formula updates the distances based on previously known distances and the sizes of the clusters involved in the merge. The Lance–Williams formula is: Here, is the distance between cluster i and the newly formed cluster z (created by merging clusters x and y); , , and are the sizes of clusters i, x, and y, respectively; , , and are the pairwise squared Euclidean distances between these clusters. Algorithm 3 presents pseudocode demonstrating how the Lance–Williams formula is used to efficiently update distances during agglomerative hierarchical clustering. \| \| \| Algorithm 3: Hierarchical clustering algorithm using Lance-Williams formula for recursive distance updating \| \| \| Using the Lance–Williams formula and the nearest-neighbor chain algorithm , agglomerative hierarchical clustering achieves a time complexity of , a significant improvement over the naive approach. Advanced Initialization One effective strategy to reduce the computational burden in distance-intensive algorithms is to employ better initialization techniques, particularly in algorithms that rely on iterative local or global search . Many such algorithms, like K-means clustering, repeatedly compute distances as they iteratively refine their solutions. Poor initialization, such as random selection of initial centroids in K-means, can lead to significantly more iterations and, consequently, a higher number of distance calculations. By contrast, advanced initialization methods can result in fewer iterations and thus fewer distance computations. In K-means, for example, the algorithm starts by randomly selecting K centroids and iteratively refines them by recalculating distances between data points and centroids in each iteration. With random initialization, the algorithm often converges slowly due to poor starting points, requiring many iterations to find optimal centroids. Each iteration involves distance computations, where n is the number of data points and K is the number of clusters. Therefore, reducing the number of iterations directly reduces the number of distance calculations. Advanced initialization techniques, such as K-means++ , provide a more strategic selection of initial centroids, leading to faster convergence. K-means++ initializes the centroids by probabilistically selecting data points that are farther apart, which ensures a better spread of centroids across the dataset. This reduces the number of iterations required for convergence compared to random initialization, ultimately reducing the number of Euclidean distance calculations performed during the algorithm. An even more sophisticated initialization technique involves using the output of Ward’s method as the initial centroids for K-means. Ward’s method is aligns well with the objective of K-means . By first applying Ward’s method and then using its cluster centroids as the starting points for K-means, the algorithm often converges in far fewer iterations than with K-means++ or random initialization. This combination can significantly reduce the number of distance computations, as the initial centroids are already near optimal. Furthermore, alternating between different MSSC clustering algorithms, such as using Ward’s method followed by K-means (i.e., using Ward’s output as the input for K-means), can often improve the clustering outcome by further minimizing the objective function (2) compared to running each algorithm in isolation, as the strengths of each method complement one another in refining the cluster assignments . Precomputing and Caching Distances In certain scenarios, such as when working with static datasets, it is possible to precompute the pairwise Euclidean distances between all points and store them in a distance matrix, which is a key component in many machine learning and data analysis tasks . This precomputation, while expensive in terms of memory, allows for instant retrieval of distances during the optimization process, thereby eliminating the need for repeated calculations. Caching strategies can also be employed when the dataset is too large to store all pairwise distances, ensuring that frequently accessed distances are quickly available. For a set of m points in n-dimensional space, the precomputed Euclidean distance matrix is an symmetric matrix, leveraging the property , where the entry represents the Euclidean distance between points and . Mathematically, is defined as: In this formulation, the symmetry of the matrix is explicitly used by calculating only for and setting for . This approach allows for the computation of only half of the distance matrix—the elements below the main diagonal (the lower triangle), significantly reducing the number of distance calculations from to for a dataset with m points. To save space by storing only the lower triangular part of a symmetric matrix , its condensed (or compact) representation can be used. The condensed matrix is a vector containing the elements of the lower triangular part of , excluding the diagonal. The index k in the condensed matrix , corresponding to the element in the full matrix , is given by:where (assuming 0-based indexing). The condensed matrix thus contains a total of elements. To reconstruct the indices in the full matrix from the index k in the condensed matrix , the following formulas can be used:where denotes the ceiling function, and denotes the floor function, which truncates the decimal part to return an integer value. The bottleneck of the K-means algorithm is the repeated distance recalculations in each iteration. This is necessary because centroids are updated as the mean values of the assigned points, which are not actual data points. The K-means algorithm can be accelerated by using a precomputed Euclidean distance matrix , where , to avoid redundant computations. An optimized alternative is the K-medoids algorithm, which improves efficiency by leveraging the precomputed distance matrix and reducing the impact of outliers by using actual data points (medoids) as cluster centers . Unlike K-means, which computes distances to centroids, K-medoids assigns points to medoids based on precomputed distances from matrix . The objective of K-medoids is to minimize the total dissimilarity between points and their medoids, defined as:where is the medoid, and is the precomputed Euclidean distance. The algorithm iteratively refines the medoids by swapping them with non-medoid points and recalculating the total cost until convergence, making it especially beneficial for large datasets where recalculating distances is expensive. In large datasets, there is a trade-off between computation time and memory usage when calculating Euclidean distances. Precomputing all pairwise distances in a matrix speeds up algorithms by avoiding redundant recalculations, but this approach demands substantial memory. Storing the full distance matrix for m points requires memory, which becomes impractical for large datasets. For example, with , it would need around 80 GB of memory. While manageable for small datasets, this approach is infeasible for larger ones. To reduce memory usage, a condensed matrix can be employed, leveraging the symmetry and zero diagonal , storing only the necessary elements. Spatial Data Structures and Approximate Methods The Euclidean distance is central in the K-nearest neighbors (KNN) algorithm, where it is employed to quantify the proximity between data points . Given a set of training data where represents the feature vectors and denotes the corresponding class labels, the KNN algorithm classifies a new data point by identifying the set of the K nearest neighbors, defined as: The algorithm then assigns the class label y to based on the majority class among its nearest neighbors:where is the indicator function. Thus, the Euclidean distance serves as a crucial metric in determining the classification of data points based on the proximity of their neighbors within the feature space. Nearest neighbor search is a common task in many machine learning algorithms . In nearest neighbor search the goal is to identify the closest point in a dataset to a query point . The naive approach involves computing the Euclidean distance from to each point in the dataset, which has a time complexity of . For large datasets and high-dimensional spaces, this approach can be prohibitively expensive. Besides the use of precomputed pairwise distance matrices, one of the most effective strategies to accelerate nearest neighbor search is the use of spatial data structures such as KD-trees and Ball Trees . These structures enable a partitioning of the dataset into hierarchical regions, allowing the algorithm to discard large portions of the dataset that are unlikely to contain the nearest neighbor, thus reducing the number of distance computations. Spatial data structures, such as KD-trees, exploit the Spatial Locality property of Euclidean distance by organizing data points in a way that nearby points are grouped together and distant points are separated. The Spatial Locality property implies that points closer in space have smaller Euclidean distances, while those farther apart have larger distances. The KD-tree algorithm operates by recursively splitting the dataset into two halves along the median of one of the dimensions at each level of the tree . Given a query point , the KD-tree can be traversed to rapidly exclude regions of the space that are farther from than the currently identified nearest neighbor. While the KD-tree is efficient in low to moderate dimensions, its performance degrades in high-dimensional spaces due to the curse of dimensionality. For example, when searching for the nearest neighbor to a query point , the KD-tree first identifies the region where resides. It then compares points within that region before expanding the search to neighboring regions only if necessary. This process avoids the need to compute distances to points in far-away regions, as the Euclidean distance between and points in those distant regions would exceed the distances already found in the current local region. This selective search process dramatically reduces the computational burden, especially in low- and moderate-dimensional spaces. Figure 2 illustrates how a KD-tree partitions the space and reduces distance computations during nearest neighbor search. The query point is located in a specific region, and the algorithm only needs to consider data points within that region and nearby regions that intersect the search radius. Pruned regions are shaded, showing how the KD-tree efficiently eliminates the need to compute distances to distant points. Building spatial data structures like KD-trees or Ball Trees can require significant memory, especially with large datasets or higher dimensions. Figure 2. KD-tree partitioning and nearest neighbor search illustrating pruned regions and search path. Dashed lines represent splitting hyperplanes at different levels of the KD-tree. Shaded areas are pruned regions that are not searched. The red star is the query point , and the circle represents the distance to the current nearest neighbor. An alternative approach is the use of Approximate Nearest Neighbors (ANN) algorithms, which trade-off exactness for speed. Techniques such as Locality-Sensitive Hashing (LSH) enable faster searches by hashing data points into buckets based on random projections, ensuring that nearby points in the original space are more likely to be hashed into the same bucket. Although ANN algorithms do not guarantee the identification of the exact nearest neighbor, they offer significant computational savings and often suffice in practical applications. Early Exit Strategies Further optimizations can be achieved by leveraging techniques like early exit strategies in algorithms that do not require exact distances for all points. For instance, the additivity property of Euclidean distance in subspaces allows it to be broken down into the sum of squared differences across individual dimensions. This allows for partial distance calculations to determine if a point can be excluded early . If the partial sum of squared differences already exceeds a known threshold, further dimensions do not need to be considered. This is an effective optimization in high-dimensional spaces, where early termination of distance computation can save substantial computational effort. Dimensionality Reduction Techniques As datasets grow in size and dimensionality, traditional methods like Euclidean distance face increasing computational challenges. The curse of dimensionality describes how the volume of space increases exponentially with the number of dimensions, causing data points to become sparse and distances less meaningful . This sparsity complicates tasks such as nearest neighbor searches, clustering, and classification, as computational costs escalate. In high-dimensional spaces, Euclidean distance often fails to effectively differentiate between nearby and distant points, prompting the need for alternative distance measures and mitigation techniques . To overcome these challenges, dimensionality reduction techniques are among the best options. These techniques aim to find a mapping , where , that preserves the essential geometric properties of the data while reducing the computational cost of distance calculations. Principal Component Analysis (PCA) is one of the most widely used linear dimensionality reduction techniques. PCA transforms the original data points into a set of orthogonal components, ranked by the amount of variance they capture from the data. By retaining only the top d components, where , PCA effectively reduces the dimensionality of the data while preserving the most significant features, allowing for faster Euclidean distance computations . Another popular method is the t-Distributed Stochastic Neighbor Embedding (t-SNE) , which is a non-linear technique designed to maintain the local structure of the data in a lower-dimensional space. While t-SNE is primarily used for visualization, it can also facilitate faster distance computations in the lower-dimensional embedding space, although with the trade-off of potentially altering global distances. Random Projection , grounded in the Johnson-Lindenstrauss Lemma, offers a probabilistic approach to dimensionality reduction. It projects the original data onto a randomly chosen lower-dimensional subspace using a random matrix , where each entry is independently drawn from a suitable distribution (e.g., Gaussian or sparse distribution): This projection approximately preserves pairwise Euclidean distances with high probability:where is a small distortion parameter. Random Projection is particularly appealing due to its computational efficiency and simplicity, as it requires no prior knowledge of the data distribution and can be implemented with minimal overhead. Dimensionality reduction techniques offer computational advantages but come with limitations. PCA, though effective for preserving variance, is a linear method and may miss complex, non-linear patterns. Its sensitivity to outliers can distort the data structure, and it may discard low-variance dimensions that are still important. t-SNE, useful for visualizing local structure, is computationally expensive and can distort global relationships, making it unsuitable for large datasets. Random Projection is efficient and easy to implement but introduces distortion in distances, potentially affecting accuracy, and its results can vary due to randomness. These techniques often involve trade-offs between efficiency, accuracy, and data-specific considerations. Approximation via Clustering and Dimensional Culling Instead of computing the exact distance matrix, one can use clustering techniques to approximate the distance matrix by grouping similar points together. For example, using cluster centroids as representatives for the points within each cluster, the distance matrix can be computed more efficiently by calculating distances between centroids rather than between individual points, which significantly reduces the number of required distance calculations . In high-dimensional spaces, not all dimensions contribute equally to the distance between points. Identifying and ignoring dimensions with low variance or limited impact on distance can reduce the number of calculations . 3.2. Low-Level Optimizations Low-level optimizations focus on improving the efficiency of Euclidean distance computations by optimizing the underlying code at a granular level. These optimizations are crucial when working with large-scale data or real-time systems where every computational cycle counts. Approximation with Lower Precision By applying scalar quantization to the individual dimensions of a vector, each component of the vector is approximated using lower precision. For scalar quantization, each component of the vector is mapped to a discrete set of values. This reduces the complexity of the distance calculation by performing operations on smaller, quantized values rather than the original high-precision numbers. For instance, instead of using floating-point arithmetic, the vector components are represented as integers, allowing faster calculations using integer arithmetic, which is typically more efficient in many hardware architectures . Scalar quantization maps each component of an n-dimensional vector to a quantized value , where is a finite set of levels. The resulting quantized value is defined as . In uniform quantization, the levels in are equally spaced. For L levels over the range , the quantization levels are given by , where and . The quantized value for is . The quantization error is the difference between and its quantized value: . Although this error accumulates in distance calculations, it allows for faster computations using lower precision. Quantization lowers the precision of vector components, introducing some error in distance calculations. The coarser the quantization, the larger the potential error. This trade-off is acceptable in cases where exact precision is less important, such as approximate nearest neighbor searches or noisy data. It is especially beneficial in large-scale or real-time applications, where faster computation and reduced memory use outweigh the accuracy loss. Loop Unrolling Loop unrolling is another low-level optimization technique that can be applied to Euclidean distance calculations. By manually unrolling loops, the number of loop control instructions (such as incrementing indices and checking loop termination conditions) is reduced, allowing the CPU to perform more useful work per cycle . For example, consider the following loop that computes the squared Euclidean distance: This loop can be unrolled by manually expanding the loop body for a fixed number of iterations: While this increases the code size, it reduces the overhead of loop control and increases instruction-level parallelism, leading to faster execution. Machine Code Optimization Another avenue for optimizing Euclidean distance computations involves low-level programming and the use of just-in-time (JIT) compilers. Implementing distance calculations in low-level languages like C or C++ allows developers to exploit the full potential of the underlying hardware, using techniques such as assembly-level optimizations, manual loop unrolling, function inlining, instruction reordering, and precise control over memory allocation and access patterns to generate more efficient code . These optimizations can minimize cache misses and take advantage of specific CPU instructions that are optimized for floating-point arithmetic. By enabling these optimizations, further performance gains can be achieved without requiring manual intervention. In environments where floating-point computation is expensive, using fixed-point arithmetic can additionally reduce the computational load. In addition to traditional low-level languages, Python developers can achieve similar performance gains by using Numba, a high-performance JIT compiler . Numba automatically translates Python functions into optimized machine code at runtime using the LLVM compiler library. This approach allows Python programs to reach speeds comparable to those of C or FORTRAN without requiring developers to leave the Python ecosystem. By simply applying a Numba decorator to a Python function, the function is automatically compiled into highly optimized machine code, often resulting in substantial speedups for numerical algorithms. This enables the efficient execution of distance calculations while retaining the flexibility and ease of use that Python offers. Vectorization Modern computing architectures offer vectorized instructions that can significantly accelerate distance calculations. Vectorization enables performing the same operation on multiple data elements simultaneously, leveraging the computational power of modern CPUs. Vectorized implementations and the use of advanced linear algebra libraries take advantage of special floating-point hardware, such as SIMD (Single Instruction, Multiple Data) operations and vector registers. By vectorizing the Euclidean distance computation, it is possible to leverage SIMD instructions to compute multiple distances in parallel within a single CPU core, thus speeding up the process . Libraries such as BLAS (Basic Linear Algebra Subprograms) and NumPy provide highly optimized routines for these vectorized operations, making them readily accessible for optimization. For instance, BLAS library implementations include a set of low-level routines for performing common linear algebra operations—like matrix multiplication, dot products, vector addition, scalar multiplication, norm computation, and sum reduction—which can be directly used to calculate Euclidean pairwise distance matrices. NumPy is particularly advantageous for vectorizing Euclidean distance computations due to its ability to efficiently handle large arrays and matrices through optimized low-level implementations . These vectorized operations are built on highly optimized C libraries, allowing NumPy to fully leverage modern CPU architectures. By utilizing vectorization, NumPy can perform element-wise arithmetic across entire arrays simultaneously, significantly reducing the time complexity compared to traditional for-loop-based approaches in Python. This efficiency is further enhanced by the SIMD capabilities of modern CPUs, which enable the same operation to be applied to multiple data points in parallel. Additionally, NumPy’s integration with highly optimized linear algebra libraries, such as BLAS and LAPACK, ensures that operations like matrix multiplication and dot products—crucial for distance calculations—are executed with minimal overhead. For example, vectorization techniques can be applied to distance calculations by expressing them in terms of matrix operations. The matrix of squared Euclidean distances between two vectors, and , is given by: Here, represents the squared norm of , is the squared norm of , and denotes the dot product between and . Instead of performing full matrix multiplication, the squared norms and are computed more efficiently using row-wise operations. This optimization avoids unnecessary calculations, as only the squared norms for each row are required, significantly reducing the computational complexity. Listing 1 provides an example of Python function for computing pairwise squared Euclidean distances between two sets of points using vectorization capabilities of NumPy library. \| \| \| Listing 1: Python code for computing pairwise squared Euclidean distances between two sets of points using vectorization capabilities of NumPy \| \| \| The vectorization capabilities of NumPy, as demonstrated in the code at Listing 1 and at Figure 3, significantly accelerate the computation of pairwise squared Euclidean distances by avoiding explicit Python loops and leveraging highly optimized matrix operations. Figure 3. Vectorized computation of pairwise squared Euclidean distances between two sets of points. In particular, the dot product between matrices and is computed using the function np.dot(X, Y.T), which takes advantage of NumPy’s efficient linear algebra routines to perform the operation in a single step across all pairs of points. Similarly, the squared norms of the rows in and are computed using SIMD element-wise multiplication operation () and sum reduction along the row dimension (axis=1), fully utilizing vectorized operations to handle entire arrays at once. The operation reshapes the 1D array into a 2D column vector, allowing NumPy to automatically expand the dimensions of and through broadcasting. In NumPy broadcasting enabling efficient element-wise operations with the 2D dot product result without explicitly copying data. This approach eliminates the need for slow, iterative computations typically found in for-loop-based implementations. Instead, all operations are executed in parallel for each row of the matrices using special vector registers within a single CPU core, significantly boosting performance through SIMD operations. The final distance matrix is assembled by combining squared norms and dot products in a single vectorized expression, further optimizing computational efficiency. Parallelization Parallelization involves distributing the workload across multiple cores or processors, enabling simultaneous execution of distance calculations . This approach is particularly advantageous when dealing with large datasets or when Euclidean distance computation needs to be performed repeatedly, as in iterative algorithms like K-means clustering or K-nearest neighbors (KNN) . By breaking down the computation into smaller tasks that can be processed concurrently, parallelization reduces the overall computation time, making it feasible to handle more extensive datasets within a reasonable timeframe. Parallelization is a powerful strategy for optimizing Euclidean distance calculations, particularly when dealing with large-scale data or computationally intensive algorithms. By effectively utilizing the available hardware resources—whether through multi-threading, multiprocessing, or distributed computing—parallelization can dramatically improve the performance and scalability of algorithms that rely on Euclidean distance measurements. The main conceptual difference between CPU and GPU parallelism lies in their design: CPUs are optimized for sequential processing with a few powerful cores, focusing on complex tasks that require strong single-thread performance, whereas GPUs are designed for massive parallelism with thousands of smaller, simpler cores that excel at handling many simultaneous tasks, making them ideal for workloads like distance computations across large datasets . Libraries and frameworks like OpenMP for CPUs, CUDA for GPUs, and Apache Spark for distributed computing facilitate parallel computation by providing interfaces to efficiently utilize multiple cores, threads, and clusters, enhancing the performance of Euclidean distance calculations on various hardware architectures . When combined with vectorization and machine code optimization, parallelization offers even greater potential for accelerating computations by leveraging modern hardware capabilities such as SIMD instructions, multi-core processors, and GPUs. By applying vectorized operations within each parallel thread or process, computational efficiency can be further enhanced, leading to significant reductions in processing time. Listing 2 provides an example of Python code for computing pairwise squared Euclidean distances between two sets of points using Numba’s JIT compilation. This approach combines automatic machine code optimization, vectorization, and parallelism across multiple CPU cores to enhance performance. \| \| \| Listing 2: Python code for computing pairwise squared Euclidean distances between two sets of points using Numba’s JIT compilation, which combines automatic machine code optimization, vectorization, and parallelism across multiple CPU cores for enhanced performance. \| \| \| In Listing 2, Numba’s JIT (Just-In-Time) compilation is applied to further accelerate the squared Euclidean distance computation compared to Listing 1. The @njit(parallel=True) decorator compiles the Python code into optimized machine code at runtime, allowing for faster execution by utilizing the CPU’s hardware capabilities. Parallelism is achieved through the prange construct, which distributes the workload of iterating over the rows of matrices X and Y across multiple CPU cores. This ensures that each core computes a portion of the pairwise distances, significantly speeding up the overall computation compared to the single-core, vectorized approach in Listing 1. This parallelism significantly accelerates computation for larger datasets, especially when multiple CPU cores are utilized, as the pairwise distance calculations can be processed concurrently. As a result, the parallelization approach is particularly advantageous for large-scale problems, offering greater performance gains than vectorization alone, particularly when dealing with high-dimensional or large datasets and using multiple CPU or GPU cores. As the problem scale and the number of cores increase, the speedup achieved by parallelization can far exceed that of non-parallel implementations. However, when applied to small datasets, parallelization can increase memory usage due to the overhead of managing multiple threads or processes, as well as potential data duplication. Furthermore, combining vectorization and parallelization can yield extra efficiency by optimizing both individual arithmetic operations and the distribution of workload across cores, resulting in even greater overall acceleration. Hardware Acceleration Leveraging specialized hardware, such as multiple core CPUs and Graphics Processing Units (GPUs), can further accelerate Euclidean distance calculations . Modern GPUs consist of thousands of cores that can execute the same operation on multiple data points simultaneously, making them well-suited for tasks like Euclidean distance calculation where the same operation (distance computation) needs to be applied across many data points. GPUs, with their massively parallel architecture, are particularly beneficial for applications involving large datasets, where the parallel computing power of GPUs can dramatically reduce computation time. Field-Programmable Gate Arrays (FPGAs) offer a different kind of hardware acceleration by allowing the user to create custom circuits that are optimized for specific tasks, such as Euclidean distance computation . Unlike GPUs, which are fixed-function devices, FPGAs can be reprogrammed to implement custom logic that executes the distance computations in an optimized fashion. FPGAs are particularly advantageous in applications where power efficiency is critical, as they can provide significant speedup with lower power consumption compared to GPUs or CPUs. Figure 4. Vectorized and parallelized computation of pairwise squared Euclidean distances using Numba’s JIT compilation and parallel processing. Figure 4. Vectorized and parallelized computation of pairwise squared Euclidean distances using Numba’s JIT compilation and parallel processing. 3.3. Hybrid Approaches In practice, combining multiple optimization strategies often yields the best results. For instance, using dimensionality reduction to lower the computational burden, followed by squared distance optimization techniques to eliminate the square root operations, and then applying vectorization and parallelization can lead to significant speedups . Similarly, integrating hardware acceleration with algorithmic optimizations—such as spatial data structures and approximate methods—can maximize performance gains. These hybrid approaches are particularly effective in large-scale applications, where combining different techniques allows for more scalable and efficient processing of complex datasets. By carefully selecting and integrating these optimization strategies, it is possible to tailor the computation of Euclidean distances to the specific needs of an application, achieving both high performance and accuracy . 4. Comparative Analysis of Optimization Approaches In this section, we compare the different optimization approaches discussed above in terms of their effectiveness, complexity, and suitability for various applications. The results of comparative analysis of optimization approaches for Euclidean distance computation are given in in Table 1. Table 1. Comparison of Optimization Approaches for Euclidean Distance Computation 4.1. Computation Speedup The primary goal of optimization is to reduce the time required to compute Euclidean distances. Techniques like squared Euclidean distance, advanced initialization, and lower-bound methods provide moderate to high speedups by eliminating redundant calculations or improving the initialization phase. Methods like early exit strategies, dimensional culling, and hardware acceleration can achieve very high speedups, particularly in large-scale or real-time applications. 4.2. Complexity Most low-level techniques, such as loop unrolling, vectorization, and advanced initialization, have low to moderate complexity (ease of implementation) and are relatively straightforward to implement. More advanced strategies, such as recursive distance updating and approximate nearest neighbors, require deeper algorithmic understanding and expertise. Techniques like machine code optimization (e.g., using Numba) are moderately complex but can be automated through libraries, while hardware acceleration and FPGA usage demand specialized knowledge. 4.3. Scalability Scalability is crucial in large-scale applications. Many of the techniques, like approximate nearest neighbors, vectorization, parallelization, and dimensional culling, scale well with large datasets. Precomputing distance matrices, while providing fast retrieval, may face scalability issues due to high memory usage in very large datasets. Parallelization techniques and spatial data structures (e.g., KD-trees) also scale well but may degrade in performance in high-dimensional data spaces. 4.4. Memory Usage Techniques such as precomputing distance matrices and clustering involve high memory usage due to storing pairwise distances or intermediate results. In contrast, methods like vectorization, early exit strategies, and loop unrolling have minimal memory requirements, making them suitable for resource-constrained environments. 4.5. Hardware Requirements Most approaches, such as advanced initialization, lower-bound techniques, and dimensionality reduction, can be implemented without specialized hardware. However, hardware acceleration approaches like GPU and FPGA acceleration require appropriate hardware resources. Similarly, vectorization benefits from SIMD-enabled CPUs, and parallelization thrives with multi-core processors or GPU support. 4.6. Accuracy Impact Accuracy impact refers to the trade-offs between computational speed and precision. Techniques like squared Euclidean distance, lower-bound methods, triangle inequality, and vectorization maintain exact accuracy while improving efficiency. In contrast, methods such as approximate nearest neighbors (ANN), dimensionality reduction, and lower precision approximations may introduce slight inaccuracies, suitable for cases where faster computation outweighs precision. Techniques like parallelization and hardware acceleration preserve accuracy when properly implemented. The choice depends on the application’s tolerance for accuracy loss versus the need for speed. 4.7. Best Use Cases The suitability of each approach depends on the specific application. Each technique has ideal use cases. Advanced initialization works best for algorithms like K-means where a good initial guess speeds up convergence. Early exit strategies are highly effective for high-dimensional nearest neighbor searches, while dimensional culling excels in high-dimensional spaces where not all dimensions contribute equally. Loop unrolling and vectorization are best for low-level numerical optimizations, and hardware acceleration is critical for real-time or large-scale processing tasks. 5. Discussion Selecting the appropriate optimization technique for Euclidean distance computations depends on the specific data characteristics and application requirements. Techniques like squared Euclidean distance, lower-bound methods, and triangle inequality are effective for algorithms such as K-means clustering and K-nearest neighbors, where exact distance calculations are crucial. These methods provide significant speedups without sacrificing accuracy, making them ideal for large datasets that demand precision. Spatial data structures like KD-trees and Ball Trees efficiently partition space to accelerate nearest neighbor searches in low to moderate-dimensional data. However, their performance diminishes in high-dimensional spaces due to the curse of dimensionality. For high-dimensional data, dimensionality reduction techniques and clustering with dimensional culling become more appropriate. Methods like Principal Component Analysis or Random Projection reduce data to a lower-dimensional space, preserving significant variance while decreasing computation time, suitable for exploratory data analysis or preprocessing in machine learning pipelines. Approximate methods, including Locality-Sensitive Hashing, are advantageous when exact nearest neighbor searches are computationally infeasible due to dataset size. They offer substantial speedups with acceptable accuracy loss, making them suitable for applications like recommendation systems or real-time search where response time is critical. Hardware acceleration techniques, such as GPU and FPGA implementations, leverage parallel processing capabilities to significantly reduce computation times while maintaining accuracy, making them effective for real-time processing and handling large-scale data in environments like deep learning and big data analytics. A critical consideration in selecting an optimization approach is balancing computation speed and potential loss of accuracy. Techniques that introduce approximations—such as approximate nearest neighbors, dimensionality reduction, and approximation with lower precision—may impact accuracy. While these methods offer high speedups, they may not be suitable for applications where precise distance calculations are essential, such as in medical imaging or security-sensitive domains. Conversely, methods like vectorization, parallelization, and machine code optimization provide significant computational benefits without compromising accuracy, making them ideal for applications requiring both high performance and precise results. When employing clustering and dimensional culling, there’s a trade-off between reducing computational load and the risk of losing important information in the omitted dimensions. This can impact accuracy, particularly if the discarded dimensions are relevant to the analysis. Assessing the significance of each dimension is crucial in the context of the specific application. The choice of optimization technique also depends on available computational resources. Hardware acceleration methods require specialized hardware like GPUs or FPGAs, which may not be feasible in all settings due to cost or infrastructure limitations. In such cases, software-based optimizations like vectorization or loop unrolling can still provide substantial speedups on standard CPUs. Parallelization offers high scalability and is effective for large datasets but may increase memory usage due to overhead from managing multiple threads or processes. Applications with limited memory resources need to consider this trade-off. For static datasets where repeated distance computations are required, precomputing and caching distances can significantly speed up retrieval times, but this approach demands high memory usage, which might not be suitable for extremely large datasets or memory-constrained environments. Ultimately, selecting the most appropriate optimization strategy requires understanding the application’s specific needs and constraints. For applications where accuracy is paramount and computational resources are ample, methods that preserve exact distances while optimizing computation—such as vectorization and parallelization—are preferable. In scenarios where computational speed is critical and some accuracy loss is acceptable, approximate methods and dimensionality reduction techniques may be more suitable. Evaluating the impact of potential inaccuracies on the application’s outcomes is essential. Combining multiple optimization approaches can yield tailored solutions that align closely with specific requirements and limitations. 6. Conclusion As data sizes and algorithmic complexity continue to grow, advancements in optimization techniques will be crucial for improving the performance of algorithms that rely on Euclidean distance calculations. Although Euclidean distance computation is simple in its formulation, it can become a significant computational bottleneck in large-scale applications. Optimizing this operation can greatly improve the overall efficiency of various analytical processes. By reducing the time and computational resources required for distance calculations, these optimizations will improve the scalability and responsiveness of data-driven applications. Additionally, increased computational efficiency directly translates to lower energy consumption, making these optimizations essential for minimizing the environmental impact of large-scale data processing. As energy efficiency is a key component of sustainable development, the reduction in power usage achieved through optimizing Euclidean distance calculations supports global efforts toward more eco-friendly and sustainable technological solutions. This paper has reviewed and compared various optimization techniques aimed at accelerating Euclidean distance computations. These techniques span algorithm-specific optimizations, low-level code optimizations, and hardware acceleration, each offering different trade-offs in terms of speed, complexity, and resource requirements. Applying these optimization techniques provides distinct advantages depending on the specific use case. Algorithm-specific methods, like squared Euclidean distance and lower-bound techniques, allow for immediate reductions in computation time without significant changes in implementation complexity, making them well-suited for standard machine learning tasks. Low-level optimizations such as vectorization, loop unrolling, and parallelization exploit modern CPU architectures for rapid computations, particularly useful in high-throughput environments. Meanwhile, hardware acceleration using GPUs or FPGAs offers the most significant speedups for real-time applications or large datasets, although they come with higher development and hardware costs. Together, these techniques allow practitioners to tailor solutions that maximize performance based on available resources, dataset size, and application demands. By comparing the effectiveness, complexity, and scalability of various optimization techniques, our findings guide practitioners in selecting the most suitable methods for improving Euclidean distance computations in their specific contexts. This comparative analysis highlights that no single optimization technique is universally superior. The best optimization strategy depends on dataset size, dimensionality, memory constraints, and the availability of hardware resources. Combining multiple techniques often offers the most efficient and scalable solution for improving Euclidean distance computations, particularly in large-scale machine learning and data analysis applications. As data sizes continue to grow and applications become more demanding, the importance of efficient Euclidean distance computation will only increase. Future research may focus on developing hybrid methods that combine multiple optimization strategies or on leveraging emerging hardware technologies, such as quantum computing, to further accelerate these computations. The continued development of efficient Euclidean distance computation techniques will remain a critical area of research, enabling advancements across numerous scientific and engineering disciplines. Funding This research was funded by the Science Committee of the Ministry of Science and Higher Education of the Republic of Kazakhstan (grant no. BR21882268). Data Availability Statement The code supporting the findings of this study is available at Conflicts of Interest The authors declare no conflicts of interest. Abbreviations The following abbreviations are used in this manuscript: \| \| \| --- \| \| ANN \| Approximate Nearest Neighbors \| \| BLAS \| Basic Linear Algebra Subprograms \| \| CPU \| Central Processing Unit \| \| CUDA \| Compute Unified Device Architecture \| \| FPGA \| Field-Programmable Gate Array \| \| GPU \| Graphics Processing Unit \| \| JIT \| Just-In-Time \| \| KNN \| K-Nearest Neighbors \| \| KD-tree \| K-Dimensional Tree \| \| LSH \| Locality-Sensitive Hashing \| \| MSSC \| Minimum Sum-of-Squares Clustering \| \| OpenMP \| Open Multi-Processing \| \| PCA \| Principal Component Analysis \| \| SIMD \| Single Instruction, Multiple Data \| \| t-SNE \| t-Distributed Stochastic Neighbor Embedding \| References Deza, M.M.; Deza, E. Encyclopedia of Distances, 4 ed.; Springer: Berlin, Heidelberg, 2016; p. 756. [Google Scholar] [CrossRef] Bottesch, T.; Bühler, T.; Kächele, M. Speeding up k-means by approximating Euclidean distances via block vectors. 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Dashed lines represent splitting hyperplanes at different levels of the KD-tree. Shaded areas are pruned regions that are not searched. The red star is the query point , and the circle represents the distance to the current nearest neighbor. Figure 3. Vectorized computation of pairwise squared Euclidean distances between two sets of points. Table 1. Comparison of Optimization Approaches for Euclidean Distance Computation \| Optimization Approach \| Computation Speedup \| Complexity \| Scalability \| Memory Usage \| Hardware Requirements \| Accuracy Impact \| Best Use Cases \| --- --- --- --- \| \| Squared Euclidean Distance \| Moderate \| Low \| High \| Minimal \| None \| None \| K-means, KNN, large datasets \| \| Lower Bound Techniques \| High \| Moderate \| High \| Low \| None \| None \| K-means clustering, nearest neighbor search \| \| Triangle Inequality \| High \| Moderate \| High \| Minimal \| None \| None \| K-means, KNN, hierarchical clustering \| \| Recursive Distance Updating \| Moderate \| High \| Low to Moderate \| Moderate \| None \| None \| Hierarchical clustering \| \| Precomputing / Caching \| High (retrieval) \| Moderate to High \| Moderate \| High \| None \| None \| Static datasets, K-medoids, DBSCAN \| \| Spatial Data Structures \| High \| Moderate \| High (low dimensions), Low in high dimensions \| Moderate \| None \| None \| Nearest neighbor search, low to moderate dimensionality \| \| Approximate Methods \| Very High \| Moderate \| Very High \| Moderate \| None \| Moderate to High \| Large-scale nearest neighbor search \| \| Dimensionality Reduction \| High \| High \| Moderate to High \| Moderate to High \| None \| Moderate to High \| High-dimensional data, exploratory data analysis \| \| Advanced Initialization \| Moderate to High \| Low \| High \| Minimal \| None \| Potential Positive Impact \| K-means, multi-start clustering, better centroid initialization \| \| Early Exit Strategies \| High \| Moderate \| High \| Minimal \| None \| Minimal \| Nearest neighbor search, high-dimensional data \| \| Clustering and Dimensional Culling \| High \| Moderate \| High \| Moderate \| None \| Moderate to High \| High-dimensional data, feature selection \| \| Approximation with Lower Precision \| High \| Low \| High \| Minimal \| None \| Low to Moderate \| Large-scale distance computations, approximate solutions \| \| Loop Unrolling \| Moderate \| Low \| Moderate \| Low \| None \| None \| Low-level optimizations, numerical computations \| \| Vectorization \| High \| Low \| High \| Low \| SIMD-enabled CPU \| None \| Any distance-intensive operation \| \| Parallelization \| Very High \| High \| Very High \| Low to Moderate \| Multi-core CPU, GPU \| None \| Large datasets, real-time processing \| \| Hardware Acceleration (GPU, FPGA) \| Very High \| High \| Very High \| Low to Moderate \| GPU, FPGA \| Potential Minimal \| High-throughput, real-time processing \| \| Machine Code Optimization \| High \| Low to Moderate \| High \| Low \| Modern CPU \| None \| C++, Assembly, Python-based applications with Numba library, dynamic environments \| Short Biography of Authors \| \| \| --- \| \| \| Rustam Mussabayev is an Associate Professor and the Head of the AI Research Lab at Satbayev University, Kazakhstan. He holds a Candidateof Engineering Sciences degree(equivalent to a PhD in Computer Science) with expertise in datascience, high-performance computing, and operations research. His research interests span a wide range of topics, including clustering, natural language processing, machine learning, and optimization. He has received numerous awards, including the StatePrize “Best Researcher 2023” of the Republic of Kazakhstan and the Best Paper Award at ACIIDS 2024. His work is widely published in top-tier journals and conferences, contributing to advancements in data analysis, algorithm development, and high-performance computing solutions. \| \| \| \| --- \| \| \| 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). 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15539	https://spinlab.wpi.edu/courses/ece504_2008/8.pdf	ECE504: Lecture 8 ECE504: Lecture 8 D. Richard Brown III Worcester Polytechnic Institute 28-Oct-2008 Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 1 / 30 ECE504: Lecture 8 Lecture 8 Major Topics We are still in Part II of ECE504: Quantitative and qualitative analysis of systems mathematical description →results about behavior of system We will cover the following qualitative properties of systems: ◮Stability ◮Reachability and controllability ◮Observability ◮Minimality Today, our focus is on analyzing the internal stability of LTI systems. We will also discuss techniques for analyzing the stability of nonlinear and time-varying systems. You should be reading Chen Chapter 5 now. Sections 5.1 and 5.3-5.5 all discuss internal stability. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 2 / 30 ECE504: Lecture 8 Stability: Some Intuition Suppose you have a CT-LTI system described by ˙ x(t) =   −1 0 0 0 1 0 0 0 1  x(t) +   1 0 0  u(t) y(t) = 1 1 −1 x(t) + 0 u(t) Is this system “stable”? Since this is an LTI system, we can compute the transfer function using our standard technique: ˆ g(s) = C(sI3 −A)−1B + D = (s −1)2 (s + 1)(s −1)2 = 1 s + 1 Is this system stable? Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 3 / 30 ECE504: Lecture 8 Internal Versus External Stability There are two types of stability that we can discuss when we use state-space descriptions of dynamic systems: 1. Internal stability (are the states blowing up?) 2. External stability (is the output blowing up?) Transfer functions can only tell you about the external stability of systems. State-space descriptions can tell you about both internal and external stability. This is another big advantage of the state-space mathematical description with respect to transfer functions. As we’ve seen, it is possible for systems to have external stability but not internal stability. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 4 / 30 ECE504: Lecture 8 Bounded Vector-Valued Functions Deﬁnition The vector-valued function z(t) : R 7→Rn is bounded if there exists some ﬁnite 0 ≤M < ∞such that ∥z(t)∥≤M for all t ∈R. Recall that ∥z(t)∥ := q z⊤(t)z(t) = q z2 1(t) + · · · + z2 n(t) is the Euclidean norm. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 5 / 30 ECE504: Lecture 8 Internal Stability of Continuous-Time Systems Deﬁnition A continuous-time system is stable if and only if, when the input u(t) ≡0 for all t ≥t0, the state x(t) is bounded for all t ≥t0 for any initial state x(t0) ∈Rn. Deﬁnition A continuous-time system is asymptotically stable if and only if it is stable and lim t→∞∥x(t)∥= 0 for any initial state x(t0) ∈Rn. Note: Our “stable” is Chen’s “marginally stable”. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 6 / 30 ECE504: Lecture 8 Continuous-Time Nonlinear System Example Suppose we had a system with scalar state dynamics ˙ x(t) = x(t)(1 −x2(t)) 1. What are the equilibria of this diﬀerential equation? 2. Is this system (internally) stable? 3. Is this system (internally) asymptotically stable? Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 7 / 30 ECE504: Lecture 8 Internal Stability of CT-LTI Systems Continuous time LTI system: ˙ x(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) Note that internal stability has nothing to do with the output equation y(t) = Cx(t) + Du(t) or the matrix B. Internal stability is all about the properties of the diﬀerential equation ˙ x(t) = Ax(t) (1) What are the equilibria of (1)? Note that it is impossible to have multiple isolated equilibria when the system is LTI. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 8 / 30 ECE504: Lecture 8 Remarks on Internal Stability ◮If you are looking at a linear system, we can focus our attention on the equilibrium point x(t) = 0. We don’t need to consider other equilibria in the nullspace of A because their behavior will be the same as the equilibrium point x(t) = 0. ◮If you are looking at a nonlinear system, the behavior at each equilibrium point may be diﬀerent. Each must be analyzed individually. ◮Our deﬁnition of asymptotic (internal) stability is appropriate for linear systems because of the implicit equilibrium at the origin. ◮Asymptotic (internal) stability may be deﬁned diﬀerently for nonlinear systems. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 9 / 30 ECE504: Lecture 8 Stability Criterion Theorem A continuous-time LTI system is (internally) stable if and only if both of the following conditions are true. 1. Re(λj) ≤0 for all j ∈{1, . . . , s} where {λ1, . . . , λs} is the set of distinct eigenvalues of A. 2. For each λj that is an eigenvalue of A such that Re(λj) = 0, the algebraic multiplicity of λj is equal to the geometric multiplicity of λj, i.e. rj = mj. Note that condition 2 is equivalent to A being “diagonalizable for all eigenvalues with zero real part”. Chen Theorem 5.4 requires each eigenvalue with zero real part to have algebraic multiplicity one. It should be clear that Chen’s condition is actually too strong. Examples... Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 10 / 30 ECE504: Lecture 8 Stability Criterion Proof Sketch Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 11 / 30 ECE504: Lecture 8 Asymptotic Stability Criterion Theorem A continuous-time LTI system is asymptotically (internally) stable if and only if Re(λj) < 0 for all j ∈{1, . . . , s} where {λ1, . . . , λs} is the set of distinct eigenvalues of A. Remarks: 1. It is easier to check asymptotic stability than stability. 2. Special name for matrices with all eigenvalues satisfying Re(λj) < 0: “Hurwitz” (the same Hurwitz as the Routh-Hurwitz stability criterion you may have seen in an undergraduate systems/controls course). 3. The proof of this theorem follows directly from the fact that lim t→0 tmeαt = 0 for any integer m if Re(α) < 0. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 12 / 30 ECE504: Lecture 8 Internal Stability of Discrete-Time Systems The deﬁnitions of internal stability and asymptotic internal stability for continuous-time systems can be easily extended to discrete-time systems: Deﬁnition A discrete-time system is stable if and only if, when the input u[k] ≡0 for all k ≥k0, the state x[k] is bounded for all k ≥k0 for any initial state x[k0] ∈Rn. Deﬁnition A discrete-time system is asymptotically stable if and only if it is stable and lim k→∞∥x[k]∥= 0 for any initial state x[k0] ∈Rn. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 13 / 30 ECE504: Lecture 8 Stability Criterion Theorem A discrete-time LTI system is (internally) stable if and only if both of the following conditions are true. 1. \|λj\| ≤1 for all j ∈{1, . . . , s} where {λ1, . . . , λs} is the set of distinct eigenvalues of A. 2. For each λj that is an eigenvalue of A such that \|λj\| = 1, the algebraic multiplicity of λj is equal to the geometric multiplicity of λj, i.e. rj = mj. The proof of this theorem is similar to the proof for the continuous-time case except that the terms of Ak look like λk j , kλk j , k2λk j , etc. ◮If \|λj\| < 1, then kmλk j is always bounded. ◮If \|λj\| = 1, then condition 2 implies that there are only λk j terms, no kmλk j terms. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 14 / 30 ECE504: Lecture 8 Discrete-Time Asymptotic Stability Criterion Theorem A discrete-time LTI system is asymptotically (internally) stable if and only if \|λj\| < 1 for all j ∈{1, . . . , s} where {λ1, . . . , λs} is the set of distinct eigenvalues of A. Remarks: 1. Special name for matrices with all eigenvalues satisfying \|λj\| < 1: “Schur” (the same Schur as the Schur-Cohn stability criterion). 2. The proof of this theorem follows directly from the fact that lim k→0 kmλk j = 0 for any integer m if \|λj\| < 1. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 15 / 30 ECE504: Lecture 8 Discrete-Time Internal Stability Examples Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 16 / 30 ECE504: Lecture 8 Lyapunov Stability ◮Lyapunov stability is a general internal stability concept that is particularly useful for nonlinear diﬀerential/diﬀerence equations. ◮The ideas of Lyapunov stability can also apply to LTI systems. The main idea: 1. Assume the input u(t) ≡0 (we are looking at internal stability). 2. Let v(x(t)) be the squared “distance” of the state x(t) from the equilibrium of interest. 3. If ˙ v(x(t)) < 0 for all x(t) not at the equilibrium of interest, then the state x(t) must be getting closer to the equilibrium. Why? 4. If the “distance” to the equilibrium is strictly decreasing for all x(t), then the system is asymptotically stable. Recall that nonlinear systems may have many diﬀerent equilibria but, in LTI systems, we only need to consider the equilibrium x(t) = 0. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 17 / 30 ECE504: Lecture 8 Lyapunov Stability t v(x(t)) Note that v(x(t)) →0 (the system is asymptotically stable) even though this example shows a case when the “distance” to the equilibrium is not a strictly decreasing function of time. Hence, ˙ v(x(t)) < 0 for all x(t) implies asymptotic stability, but the converse is not always true. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 18 / 30 ECE504: Lecture 8 An Example of Lyapunov Stability for Nonlinear Systems Let’s look again at the continuous-time nonlinear scalar diﬀerential equation ˙ x(t) = x(t)(1 −x2(t)) We know of three equilibria: x(t) ∈{−1, 0, 1}. Let’s look at the Lyapunov stability of each of these equilibria... Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 19 / 30 ECE504: Lecture 8 An Example of Lyapunov Stability for Nonlinear Systems ˙ x(t) = x(t)(1 −x2(t)) and ˙ v(x(t)) = (x(t) −ǫ)2 for ǫ ∈{−1, 0, 1} −1.5 −1 −0.5 0 0.5 1 1.5 −10 −8 −6 −4 −2 0 2 x(t) (d/dt)v(x(t)) ε = −1 ε = 0 ε = 1 None of the equilibria meet the requirements for Lyapunov stability. Why? Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 20 / 30 ECE504: Lecture 8 Positive Deﬁnite and Positive Semi-Deﬁnite Matrices Deﬁnition A matrix P ∈Rn×n is positive deﬁnite if P = P ⊤and P has all positive eigenvalues. Remarks: 1. When P = P ⊤, the matrix P is said to be symmetric. 2. Fact: All of the eigenvalues of real-valued symmetric matrices are real. 3. Note: Chen’s deﬁnition of positive deﬁniteness does not require symmetry (his theorems say “symmetric and positive deﬁnite”). Deﬁnition A matrix P ∈Rn×n is positive semi-deﬁnite if P = P ⊤and P has all non-negative eigenvalues. Obviously, any positive deﬁnite P is also positive semi-deﬁnite, but the converse is not always true. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 21 / 30 ECE504: Lecture 8 Quadratic Form In linear algebra, we call expressions like z⊤P z the “quadratic form”. ◮For any P ∈Rn×n, z ∈Rn, and any integer n ≥1, the result of the quadratic form is just a scalar. ◮If P ∈Rn×n is positive deﬁnite, then it is not too diﬃcult to show that z⊤P z > 0 for all z ∈Rn and z ̸= 0. ◮Similarly, if P ∈Rn×n is positive semi-deﬁnite, then we can show that z⊤P z ≥0 for all z ∈Rn. Relationship to Euclidean norm: ◮Recall that ∥z∥2 := z2 1 + · · · + z2 n = z⊤z = z⊤Inz. ◮Is In positive deﬁnite? ◮For positive deﬁnite P , we can deﬁne a generalized norm as ∥z∥2 P := z⊤P z. Note that, like the Euclidean norm, this generalized norm satisﬁes all of the intuitive properties of a distance measure. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 22 / 30 ECE504: Lecture 8 Lyapunov Matrix Equation: CT-LTI Systems Let P ∈Rn×n be a constant positive deﬁnite matrix and deﬁne v(x(t)) := x⊤(t)P x(t) Assuming a continuous-time LTI system, we can write ˙ v(x(t)) = d dtx⊤(t) P x(t) + x⊤(t)P d dtx(t) = d dtx(t) ⊤ P x(t) + x⊤(t)P d dtx(t) = (Ax(t))⊤P x(t) + x⊤(t)P (Ax(t)) = x⊤(t)A⊤P x(t) + x⊤(t)P Ax(t) = x⊤(t) A⊤P + P A x(t) = −x⊤(t)Qx(t) Is Q := − A⊤P + P A is symmetric? When must ˙ v(x(t)) < 0? Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 23 / 30 ECE504: Lecture 8 Lyapunov Theorem: CT-LTI Systems Recall that, for CT-LTI systems: A is Hurwitz ⇔the CT-LTI system is asymptotically stable Theorem A is Hurwitz if and only if, for any positive deﬁnite Q ∈Rn×n, there exists a unique positive deﬁnite P ∈Rn×n satisfying A⊤P + P A = −Q. The proof of this theorem can be found in the Chen textbook Section 5.4. The implied procedure: 1. You are given A ∈Rn×n. 2. You pick a particular Q ∈Rn×n that must be positive deﬁnite. A good choice is In. 3. You then solve for P ∈Rn×n. This must be done element by element for each term in P . Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 24 / 30 ECE504: Lecture 8 Lyapunov Theorem: CT-LTI Systems Example: A = −1 −1 0 −1 Clearly, this A is Hurwitz. Let’s try out the Lyapunov theorem... Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 25 / 30 ECE504: Lecture 8 Comments on the Lyapunov Theorem for CT-LTI Systems ◮In general, solving for P in the equation A⊤P + P A = −Q means that there are n2 simultaneous linear equations to solve. ◮The Lyapunov theorem has limited usefulness for low-dimension, e.g. n = 2 or n = 3, CT-LTI systems because it is usually easier to just compute the eigenvalues from the characteristic polynomial. ◮For larger dimensional problems, e.g. n ≥5, it might be easier to work with the Lyapunov theorem and the n2 linear equations. ◮The main ideas of Lyapunov stability are particularly useful in nonlinear systems, where the goal is to show that d dt∥x(t) −ǫ∥2 < 0 for all x(t) ∈Rn where ǫ is an equilibrium of the system. ◮In nonlinear systems, local Lyapunov stability is also useful, i.e. d dt∥x(t) −ǫ∥2 < 0 for all x(t) ∈Xǫ ⊂Rn. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 26 / 30 ECE504: Lecture 8 Lyapunov Stability: DT-LTI Systems In DT-LTI systems, we only need to analyze the equilibrium at the origin, so the Lyapunov idea says that we want v(x[k + 1]) −v(x[k]) < 0 where v(x[k]) = x⊤[k]P x[k] with positive deﬁnite P is a generalized squared distance as in the CT-LTI case. We can write v(x[k + 1]) −v(x[k]) < 0 ⇔ v(Ax[k]) −v(x[k]) < 0 ⇔ x⊤[k]A⊤P Ax[k] −x⊤[k]P x[k] < 0 ⇔ x⊤[k] A⊤P A −P x[k] < 0 ⇔ −x⊤[k]Qx[k] < 0 Is Q := P −A⊤P A is symmetric? When must v(x[k + 1])−v(x[k]) < 0? Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 27 / 30 ECE504: Lecture 8 Lyapunov Theorem: DT-LTI Systems Recall that, for DT-LTI systems: A is Schur ⇔the DT-LTI system is asymptotically stable Theorem A is Schur if and only if, for any positive deﬁnite Q ∈Rn×n, there exists a unique positive deﬁnite P ∈Rn×n satisfying P −A⊤P A = Q. The proof of this theorem can also be found in the Chen textbook Section 5.4. The same basic procedure as the CT-LTI case applies here (in general, you have to solve n2 simultaneous linear equations). Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 28 / 30 ECE504: Lecture 8 Stability of Time-Varying Systems ◮Example A(t) = −1 e2t 0 −1 What are the eigenvalues of A(t)? Is this system stable or asymptotically stable? ◮We can use the fundamental matrix method to compute the CT-STM: Φ(t, 0) = 1 2 2e−t et −e−t 0 2e−t Now what do you think about the internal stability of this system? ◮In general, eigenvalue analysis is not useful for time-varying systems. The concept of Lyapunov stability, however, can still be applied (but the theorems for LTI systems do not directly apply). ◮Please read Chen pp. 138-140 for a discussion of the internal stability of time-varying systems. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 29 / 30 ECE504: Lecture 8 Final Remarks 1. State-space description allows us to analyze the internal stability of dynamic systems. 2. General concepts of stability and asymptotic stability apply to linear, non-linear, time-invariant, and time-varying systems. 3. The concept of local stability applies only to nonlinear systems with distinct equilibria. 4. Several stability theorems for LTI systems. 5. Lyapunov stability analysis is a general tool that can be applied to all types of systems. Worcester Polytechnic Institute D. Richard Brown III 28-Oct-2008 30 / 30
15540	https://www.youtube.com/watch?v=jVvpQAIrJ-o	Multi-step word problem with Pythagorean theorem \| Geometry \| Khan Academy Khan Academy 9090000 subscribers 246 likes Description 24903 views Posted: 28 Jun 2020 Use the Pythagorean theorem and proportional reasoning in context. View more lessons or practice this subject at . Khan Academy is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We offer quizzes, questions, instructional videos, and articles on a range of academic subjects, including math, biology, chemistry, physics, history, economics, finance, grammar, preschool learning, and more. We provide teachers with tools and data so they can help their students develop the skills, habits, and mindsets for success in school and beyond. Khan Academy has been translated into dozens of languages, and 15 million people around the globe learn on Khan Academy every month. As a 501(c)(3) nonprofit organization, we would love your help! Donate or volunteer today! Donate here: Volunteer here: 14 comments Transcript: we're told that laney runs a string of lights from the ground straight up to a door frame that is 2.5 meters tall then they run the rest of the string in a straight line to a point on the ground that is six meters from the base of the door frame there are 10 lights per meter of the string how many total lights are on the string so pause this video and see if you can work this out all right now let's work through this together and i think this one warrants some type of a diagram so let me draw this door frame that looks like this and this door frame is 2.5 meters tall so that's its height right over there and what they're going to do what laney's doing is she is stringing up these lights so that's this yellow right over here so it goes up to the top of that door frame and then they run the rest of the light in a straight line to a point on the ground that is six meters from the base of the door frame so let me show a point that is six meters from the base of the door frame so it would look something like maybe like that so this distance right over here is six meters so they run the rest of the light from the top of the door frame to that point that's six meters away so the yellow right over here that is the light and so we need to figure out how many total lights are on the string so the way i would tackle it is first off first of all i want to figure out how how long is the total string and to figure that out i just need to figure out okay it's going to be 2.5 plus whatever the hypotenuse is of this right triangle i think it's safe to assume that this is a standard house where door frames are at a right angle to the floor and so we have to figure out the length of this hypotenuse and if we know that plus this 2.5 meters then we know how long the entire string of lights are and then we just have to really multiply it by 10 because there's 10 lights per meter of the string so let's do that so how do we figure out the hypotenuse here well of course we would use pythagorean theorem so let's call this let's call this h for hypotenuse we know that the hypotenuse squared is equal to 2.5 squared plus 6 squared so this is going to be equal to 6.2 plus 36 which is equal to 42.25 or we could say that the hypotenuse is going to be equal to the square root of 42 42.25 and i could get my calculator out at this point but i'll actually just keep using this expression to figure out the total number of lights so what's the total length of the string we have to be careful here a lot of folks would say oh i figured out the hypotenuse let me just multiply that by 10. in fact my brain almost did that just now but we got to realize that the entire string is the hypotenuse plus this 2.5 so the whole string length let me write this way string length is equal to 2.5 plus the square root of 42.25 and then we would just multiply that times 10 to get the total number of lights so now let's actually get the calculator out so we have 42.25 and then we were to take the square root of that gets us to 6.5 and then we add the other 2.5 plus 2.5 equals that that's the total length of string so the total length of string is 9 meters and there are 10 lights per meter so the number of lights number of lights is equal to 9 total meters of string times 10 lights per meter which would give us 90 lights now some of you might debate if you think really deeply about it is if you have a light right at the beginning if these were kind of set up like fence posts that maybe you could argue that there's one extra light in there but for the sake of this i kind of view this as on average there are 10 lights per meter and so if it's a 9 meter string 9 times 10 is 90 lights
15541	https://www.cff.org/intro-cf/about-cystic-fibrosis	What Is Cystic Fibrosis? Cystic fibrosis is a genetic disease that affects the lungs, pancreas, and other organs. It is progressive, meaning that it gets worse over time. There are close to 40,000 children and adults living with cystic fibrosis in the United States and an estimated 105,000 people have been diagnosed with CF across 94 countries. CF can affect people of every racial and ethnic group. There are many misconceptions about CF. Learn the facts on our page, Dispelling Misconceptions About Cystic Fibrosis. In people with CF, mutations in the cystic fibrosis transmembrane conductance regulator (CFTR) gene cause the CFTR protein to not function properly. When the protein is not working correctly, it’s unable to help move chloride — a part of salt — to the cell surface. Without the chloride to attract water to the cell surface, the mucus in various organs becomes thick and sticky. CF affects multiple organs in the body: In the lungs, the mucus clogs the airways and traps germs, like bacteria. This leads to infections, inflammation, respiratory failure, and other complications. For this reason, avoiding germs is a top concern for people with CF. In the pancreas, mucus builds up and prevents the release of the digestive enzymes that help the body absorb food and key nutrients. This results in malnutrition and poor growth. In the liver, the thick mucus can block the bile duct, causing liver disease. In men, CF can affect their ability to have children. Learn how the cystic fibrosis transmembrane conductance regulator affects the digestive system. Today, because of improved medical treatments and care, about 60% of people with CF are age 18 or older. Many people with CF can expect to live healthy, fulfilling lives into their 30s, 40s, and beyond. Symptoms of CF People with CF can have a variety of symptoms, including: Very salty-tasting skin Persistent coughing, at times with phlegm Frequent lung infections including pneumonia or bronchitis Wheezing or shortness of breath Poor growth or weight gain in spite of a good appetite Frequent greasy, bulky stools or difficulty with bowel movements Nasal polyps Chronic sinus infections Clubbing or enlargement of the fingertips and toes Rectal prolapse Male infertility Learn more about CF — from diagnosis to living with the disease as an adult — in "An Introduction to Cystic Fibrosis: For Patients and Their Families," (disponible en español) or watch the video series. Listen to CF clinicians explain: Which body parts are affected by CF Common CF symptoms How CF is treated Genetics and Diagnosis Cystic fibrosis is a genetic disease. People with CF have inherited two copies of the defective CF gene — one copy from each parent. Both parents must have at least one copy of the defective gene. People with only one copy of the defective CF gene are called carriers, but they do not have the disease. Each time two CF carriers have a child, the chances are: 25 percent (1 in 4) the child will have CF 50 percent (1 in 2) the child will be a carrier but will not have CF 25 percent (1 in 4) the child will not be a carrier and will not have CF The defective CF gene contains a slight abnormality called a mutation. There are more than 1,700 known mutations of the disease. Most genetic tests only screen for the most common CF mutations. Therefore, the test results may indicate a person who is a carrier of the CF gene is not a carrier. Diagnosing cystic fibrosis is a multistep process, and should include a: Newborn screening Sweat test Genetic or carrier test Clinical evaluation at a CF Foundation-accredited care center Although most people are diagnosed with CF by the age of 2, some are diagnosed as adults. A CF specialist can order a sweat test and recommend additional testing to confirm a CF diagnosis. Read the CF Foundation’s clinical care guidelines for diagnosing CF. “I grew up wondering why I felt sick every day. As doctors suggested unlikely diseases, such as hormonal disorders, kidney disease, lupus, and depression, I felt I was further from an answer. Then, my ENT suggested CF, a disease I had never heard of. As he described what he knew about CF, it matched all of my symptoms and promised the answer I had been looking for my whole life.” — Katie K., an adult with CF, from the Community Blog According to the Cystic Fibrosis Foundation Patient Registry, in the United States: There are close to 40,000 children and adults living with cystic fibrosis in the United States — and an estimated 105,000 people have been diagnosed with CF across 94 countries. Approximately 1,000 new cases of CF are diagnosed each year. More than 75 percent of people with CF are diagnosed by age 2. About 60% of the CF population is age 18 or older. Did you know? More than half of the cystic fibrosis population is over 18. What to Expect Cystic fibrosis is a complex disease. The types of symptoms and how severe they are can differ widely from person to person. Many factors can affect a person's health and the course the disease runs, including your age when you are diagnosed. The Outlook Tremendous advancements in specialized CF care have added years and improved the quality of the lives of people with cystic fibrosis. During the 1950s, a child with CF rarely lived long enough to attend elementary school. Today, many people with CF achieving their dreams of attending college, pursuing careers, getting married, and having kids. Watch this video to see how we "count our success in lives" as we continue writing the next chapter in CF together. Although there has been significant progress in treating this disease, there is still no cure and too many lives are cut far too short. Managing CF The types of CF symptoms and how severe they are can differ widely from person to person. Therefore, although treatment plans can contain many of the same elements, they are tailored to each person's unique needs. Erin, an adult with CF, wearing her vest. People with CF and their families have expertise in how the disease affects them and how their daily lives affect the way they approach their care. By acknowledging each other's expertise, people with CF, their families, and clinical care teams can work together to develop treatment plans that align personal life goals with health goals. “My doctor and I decided to come up with a plan that would work for me. We were able to negotiate a deal so that I was doing more treatments than I had been, but I wasn’t just sitting at home hooked up to machines.” — Betsy Sullivan, then a teenager with CF, from the CF Community Blog Each day, people with CF complete a combination of the following therapies: Airway clearance to help loosen and get rid of the thick mucus that can build up in the lungs. Inhaled medicines to open the airways or thin the mucus. These are liquid medicines that are made into a mist, or aerosol, and then inhaled through a nebulizer. They include antibiotics to fight lung infections and therapies to help keep the airways clear. Pancreatic enzyme supplement capsules to improve the absorption of vital nutrients. These supplements are taken with every meal and most snacks. People with CF also usually take multivitamins. An individualized fitness plan to help improve energy, lung function, and overall health. CFTR modulators to target the underlying defect in the CFTR protein. Because different mutations cause different defects in the protein, the medications that have been developed so far are effective only in people with specific mutations. Support From the CF Foundation The CF Foundation supports people with CF by: Accrediting more than 130 care centers. These centers are staffed by dedicated health care professionals who provide expert CF care and specialized disease management. Supporting research to discover and develop new CF treatments and maintaining a pipeline of potential therapies that target the disease from every angle. Today, the Foundation is focused on developing lifesaving new therapies for larger numbers of people with CF — including those with rare and nonsense mutations— and pursuing daring, new opportunities to one day develop a lifelong cure. Research When a group of parents started the Cystic Fibrosis Foundation in 1955, there were no treatments for cystic fibrosis. These parents set their sights high, to: Advance understanding of this little-known disease Create new treatments and specialized care for their children Find a cure In the following years, the fundraising and commitment of the CF community has enabled the Foundation to support fundamental research in the laboratory that has led to groundbreaking discoveries, including identifying the gene and protein responsible for cystic fibrosis. By expanding our knowledge of the underlying biology of the disease and its effects on the body, researchers have paved the way for creating new treatments. The Foundation's steadfast commitment to advancing CF research has helped enable more than a dozen new treatments for the disease. We have made incredible progress, including the approvals by the U.S. Food and Drug Administration (FDA) of Kalydeco(ivacaftor), Orkambi(lumacaftor/ivacaftor), Symdeko(tezacaftor/ivacaftor), Trikafta(elexacaftor/tezacaftor/ivacaftor), Cayston(aztreonam), and TOBI(tobramycin). Watch this video to see how clinical research has made a difference in the lives of people with CF. Research by dedicated scientists and clinicians from a wide range of disciplines advances our understanding of cystic fibrosis every day, helping to shape clinical care practices for people living with the disease for years to come. These include studies conducted using patient data in the CF Foundation's Patient Registry, which are helping us identify trends and track the effectiveness of treatments. The Foundation is supporting the best research here and abroad to improve the quality of life of people with CF today and increase the speed of innovative research and drug development to add tomorrows. Two major initiatives are helping with this mission. To make meaningful progress against infections, we established the five-year Infection Research Initiative to help improve the detection, diagnosis, prevention, and treatment of infections. From 2018 through 2023, we invested more than $170 million to fund research and the development of new treatments. We also conducted a comprehensive review of our research portfolio, identifying and filling gaps, evaluating our investments, and recalibrating our infection research strategy. This initiative helped us set the agenda going forward for a robust infection research program so that we can continue to meet the needs of the CF community. The second major initiative is the Path to a Cure, an ambitious research agenda to deliver treatments for the underlying cause of the disease and a cure for every person with CF.The Foundation is challenging potential collaborators to submit proposals that will accelerate the pace of progress in CF drug discovery and development and intends to allocate $500 million to the effort through 2025. The Path to a Cure centers around two core strategies to address the underlying cause of CF: restoring CFTR protein when none exists and fixing or replacing the underlying genetic mutation to address the root cause of CF. By pursuing these bold strategies and others, the CF Foundation continues to build a robust pipeline of potential new therapies that fight the disease from every angle. Learn more about the CF Foundation's key research programs: Research We Fund: See a snapshot of how the CF Foundation is funding cystic fibrosis research. CF Foundation Therapeutics Laboratory: Based in Lexington, Mass., the CF Foundation Therapeutics Laboratory identifies and tests potential groundbreaking therapies for CF, readying them for further development. Therapeutics Development Network: The Therapeutics Development Network is the largest CF clinical trials network in the world. It provides the resources and support for studies that are leading to important new therapies and better treatments. Drug Development Pipeline: Discoveries from the laboratory are being turned into potential drugs that attack both the symptoms of CF and the cause —a faulty gene that makes a defective protein. Research Centers: These CF "think tanks" are located at top universities and medical schools across North America, where scientists from many disciplines are brought together to combine their expertise to find a cure for CF.
15542	https://www.youtube.com/watch?v=BBnejtu6t9Y	Simplifying Powers of i (imaginary unit) GreeneMath.com 181000 subscribers 116 likes Description 12360 views Posted: 28 Aug 2019 In this lesson, we review simplifying powers of i. In order to simplify powers of i, we must know the result for raising i to the first four powers and our rules of exponents. We can use these tools to simplify any power of i. 4 comments Transcript: Intro in this lesson we want to review simplifying powers of I so over the course of the last two lessons we talked about the imaginary unit I and we talked about complex numbers in general and how to perform operations with complex numbers so one more topic that comes up when you first start looking at imaginary units you're going to think about how to simplify powers of I so if you got something like I rais to the^ 53 you are expected to write that in a simp simp form okay so it's very easy to kind of perform this simplification you just need to know your rules of exponents and you need to know the first four so I to the first Power I SAR I cubed and I to the 4th power so the first four powers of I okay with those two things the rules of exponents and the first four powers of I you can simplify any power of I that you'd want to all right so the first thing I have listed here is i to the power of Z is equal to one I just put this here for reference sake we're not going to use it in the lesson a lot of students will ask is I to the^ of 0 equal to one the answer to that is yes so any nonzero number raised to the^ of zero is always one okay when we think about the definition of I we normally get i^2 is equal to ne1 given to us so this guy right here and we know that I to the first power is just equal to I right any number raised to the power of one is just itself so that's where those two come from now when I get to I cubed I start using my rules for exponents so I break this up and I say it's equal to i^2 I okay you can say this is i to the first Power here if you want to by rules of exponents If I multiply here I stays the same and I add 2 plus 1 to get me back to three okay so I haven't done anything illegal then I take the fact that I SAR by definition is1 and I just replace it so now I have -1 I which is just negative I okay so right now we know that I the first power is just I i^2 is -1 and I cubed is I now what about I to the 4th power we're going to break this up as i^ 2 i^ 2 again rules of exponents I would stay the same 2 + 2 is 4 so nothing illegal there and again we know that I S by definition is 1 so we're just replacing it in each case1 1 is positive 1 okay so we need to write this down in your notebook or whatever you're using to follow along I of the first power is I i^ 2 is-1 I Cub is I and I to 4th power is 1 so we just need to know those four okay that's all we need so let's look at our first example Whole number exponents so suppose we see I to the power of 48 so the first thing you're going to look for if you're dealing with whole number exponents okay whole number exponents so if you have something like 48 which is a whole number the first thing you want to do is check for divisibility by four if it's divisible by four it's equal to one okay so 48 is divisible by 4 48 ID 4 is 12 so I can rewrite this using the power to power rule as I to the 4th power raised to the 12th power okay if I know that I to the 4th power by definition is one then I could say this is 1 to the 12th power which is one so if you have I raised to a whole number exponent that is divisible by 4 it simplifies to just one so if I had something like I to the^ of 60 at as an example we all know that 60 ID 4 is 15 so I could write this as I to 4th power raised to the 15th power again this is just 1 to the 15th power which is one one raised to any power is always one okay so it doesn't matter what you're raising it to once you have this I to the 4th power here being raised to something it's just one raised to whatever which is always going to give you one okay if you had something like I raised to the power of 200 as an example we know that 200 ID 4 is 50 so again I could just say that this is I to 4th power rais to the 50th power this is 1 rais to the 50th power which is just 1 okay so if you have I raised to a whole number exponent and that whole number happens to be divisible by four it's just equal to one okay you don't need to do all this you can just stop and say it's one now what if it's not divisible Not divisible by 4 by four what do we do so suppose we saw I raised to the power of 61 well what you want to do is you want to look for the next number going down that's going to be divisible by four okay so in this case we know that if you're at 61 if I go down to 60 60 divisible 4 60 ID 4 is 15 so I'm just going to break this up and say this is I to the^ of 60 let me make that a little better so again I to the^ of 60 time I to the first power now we know I to the power of 60 is what it's one right I could write it as I to the^ 4 raised to the^ 15 this is one so I can replace this right here with a one that's gone don't even think about it anymore so it's 1 time I the first Power which is just I okay very very easy process all right what about I to the power of 202 again is it divisible by four no it's not right a number for it to be divisible by four the final two digits in this case that would be two because there's a zero in front of the two has to form a number that's divisible by four two is not divisible by four right if I divide two by four I get 0 five so what I have to do is again look for a number going down that's divisible by 4 2001 wouldn't be but 200 is so I can break this up and say this is what it's I to the power of 200 time I sared okay we know that this guy is going to be a one right so we'll say 1 I SAR and we know that i^ s by definition is -1 all right what about I ra to the power of 1,3 okay we know that again I look at the final two digits of the number it's a three it's not divisible by four if that was a four it would be but we're not that high up so we need to go down to two which isn't and then to one which isn't and then to zero which is so the number 1,00 is divisible by four so I could write this as I ra to the^ of 1,00 I ra to the^ 3 we know this is one okay so 1 I cubed again we know by definition I cubed is negative I okay so this is I all right what about I raised to the power of -22 don't get scared if you see a negative exponent remember if you get a negative exponent you take the reciprocal of the base okay so it would be 1/ I and you make the exponent positive so it's to the 22nd power so again I'm just using my same process here so I have 1 over when I think about I to the^ of 20 2 is 22 divisible by 4 no is 21 no is 20 yes so I could write this as 1 over I ra power of 20 I ra^ of 2 okay so 1 over I know I raed to the^ of 20 is 1 time i^ 2 is 1 so times negative 1 so you basically have 1 over 1 1 is just 1 so 1 over -1 which is-1
15543	https://math.wvu.edu/~hlai2/Teaching/Tip-Pdf/Tip1-21.pdf	The First Derivative Test and the classiﬁcation of critical points (open inter-val min-max problems) The First Derivative Test Suppose that f(x) is continuous on an interval I and is dif-ferentiable in I except possibly a point c inside I. (i) If f′(x) > 0 on the left side of c and f′(x) < 0 on the right side of c, then f(c) is a local maximum value of f(x) on I. (ii) If f′(x) < 0 on the left side of c and f′(x) > 0 on the right side of c, then f(c) is a local minimum value of f(x) on I. Example 1 Apply the ﬁrst derivative test to classify the critical points of f(x) = 2x3 + 3x2 − 36x + 17. Solution: Since f(x) is a polynomial, f(x) is diﬀerentiable (and so continuous) in its domain (−∞, ∞). Compute f′(x) = 6x2 + 6x −36 = 6(x2 + x −6) = 6(x −2)(x + 3). Thus the critical points of f(x) are x = 2 and x = −3. These points divide the domain of f(x) into three intervals: (−∞, −3), (−3, 2) and (2, ∞). Since f′(−4) > 0, f′(0) < 0 and f′(3) > 0, we conclude that f′(x) > 0 in both (−∞, −3) and (2, ∞), and that f′(x) < 0 in (−3, 2). Therefore, f(x) is increasing in both (−∞, −3) and (2, ∞), and f(x) is decreasing in (−3, 2). By The First Derivative Test, f(−3) is a local maximum value and f(2) is a local minimum value of f(x) in its domain. Example 2 Apply the ﬁrst derivative test to classify the critical points of f(x) = x2 + 16 x . Solution: Note that f(x) is diﬀerentiable (and so continuous) in its domain which consists of two intervals (−∞, 0) and (0, ∞). Compute f′(x) = 2x −16 x2 = 2x3 −16 x2 = 2(x3 −8) x2 . Thus the only critical point of f(x) is x = 2. This point divide the domain of f(x) into three intervals: (−∞, 0), (0, 2) and (2, ∞). Since f′(−1) < 0, f′(1) < 0 and f′(3) > 0, we conclude that f′(x) < 0 in both (−∞, 0) and (0, 1), and that f′(x) > 0 in (2, ∞). Therefore, f(x) is decreasing in both (−∞, −3) and (0, 2), and f(x) is increasing in (2, ∞). By The First Derivative Test, f(2) is a local minimum value of f(x) in its domain. 1 Example 3 Apply the ﬁrst derivative test to classify the critical points of f(x) = 4 + x 2 3 . Solution: Note that f(x) diﬀerentiable (and so continuous) in its domain (infty, ∞). Compute f′(x) = 2 3x −1 3 . Thus the only critical point of f(x) is x = 0 (at which f′(x) does not exist). This point divide the domain of f(x) into two intervals: (−∞, 0) and (2, ∞). Since f′(−1) < 0, and f′(1) > 0, we conclude that f′(x) < 0 in (−∞, 0), and that f′(x) > 0 in (0, ∞). Therefore, f(x) is decreasing in (−∞, 0), and f(x) is increasing in (0, ∞). By The First Derivative Test, f(0) is a local minimum value of f(x) in its domain. Example 4 Find the point (x, y) on the graph y = 4 −x2 that is closest to the point (3, 4). Solution: By the distance formula, the distance we want to minimize of d(x) = q (x −3)2 + (y −4)2 = q (x −3)2 + (4 −x2 −4)2 = q (x −3)2 + (x2)2 = p x4 + x2 −6x + 9. This amounts to minimize the function f(x) = [d(x)]2 = x4 + x2 −6x + 9, on its domain (−infty, ∞). Compute f′(x) = 4x3 + 2x −6. Observe that f(1) = 0, and so using division, we have f′(x) = (x −1)(4x2 + 4x + 6) = 2(x −1)(2x2 + 2x + 3). Note that 2x2 + 2x −3 > 0 (for example, using quadratic formula to see that 2x2 + 2x −3 = 0 does not have real roots) for any x. Thus x = 1 is the only critical point. which divides the domain of f(x) in to these intervals: (−∞, 1) and (1, ∞). As f′(0) < 0, and f′(2) > 0. Therefore, by the ﬁrst derivative test, f(1) is the only local minimum value of f(x) in (−∞, ∞), and so f(1) = 5 is also the absolute minimum of f(x) in its domain. Therefore, the point we are looking for is (1, 3), and the shortest distance it p f(1) = √ 5. Example 5 Determine two real numbers with diﬀerence 20 and minimum possible product. Solution: Let x and y denote these two real numbers with x ≤y. Then y = 20 + x. Thus we want to minimize f(x) = xy = x(20 + x) = 20x + x2, in (−∞, ∞). Note that f′(x) = 2x+20, and so x = −10 is the only critical point. As f′(x) < 0 when x < −10 and f′(x) > 0 when x > −10. Therefore f(−10) is the only local minimum value of f(x) in (−∞, ∞), and so it is also the absolute minimum value of f(x) in its domain. Hence x = −10 and y = 10 are the two numbers we want. 2
15544	https://apps.dtic.mil/sti/tr/pdf/AD0697664.pdf	~I -14 C., 01 Sw C V ) . 0 . 4 ) C ' 0. 0) 00 cd0 ru 0.-C13 0 0 -4 C H W4'c I On Direct Methods for Solving Symmetric Systems of Linear Equations James R. Bunch ---3 irThi document SCI. Ite C & Tov , Technical Report No. 33 -3 Computer Center University of California Berkeley IThis docuinent 'iri been approved for Public rele -s -Td saile; it diztribution is untimJit,. On Direct Methods for Solving SMetric Systems of Linear Equations by James Raymond Bun:h Abstract There has been no stable direct method for solving symmetric systems of linear equations which takes advantage of the symmetry. If the system is also positive definite, then fast, stable direct methods (e.g., Cholesky and symmetric Gaussian elimination) exist which preserve the symmetry. These methods are unstable for symmetric indefinite sys-tf s. Such systems often occur in the calculation of eigenvectors. Gaussian eliminatiun with partial or complete pivoting is currently recommended for solving symmetric indefinite systems, and here symmetry is lost. We present a generalization of symmetric Gaussian elimination, called the diagonal pivoting method, in which pivots of order two as F well as one are allowed in the decomposition. We show that the diagonal pivoting method for symmetric indefinite matrices takes advantage of symmetry Lo that only n'n multiplications, at most I n3 additions, 6 3 and 2 storage locations are required to solve A x -b , where A is a non-singular symmetric matrix of order n . Furthermore, we--shwQ that the method is nearly as stable as Gaussian elimination with complete pivoting, while requiring only half the number of operations and half the storage. L.... .... .. ... ...... ....... ... .... ...... We inzlude a listing of an Algol procedure for the diagonal pivoting method, which is applicable both to symmetric definite and indefinite systems. We discuss the problem of symmetric band matrices and present an algorithm only for the tridiagonal case. Fur~ther, we discuss the problem Of 'Oquillhrattng symmetric matrices while preserving symmetry and we present a simple algorithim (and Algol procedure) for accomplishing thiti. Table of Contents Page Chapter 1 Introduction 1 1.1 Symmetric Systems of Linear Equations 1 ti.2 Our Contribution 1 1.3 Summary 2 1.4 Origin of Symmetric Indefinite Systems 3 Chapter 2 I'resentation of the Problem 5 2.1 Introduction 5 2.2 General Problem : Exact Arithmetic 5 2.3 General Problem: Failure of Previous Attempts in Finite Precision Arithmetic 6 2.4 Condition of a Matrix 7 2.5 General Case : Stable Direct Methods 7 2.6 Symmetric Case : Direct Methods 10 2.7 Symmetric Positive Definite Case 10 2.8 Symmetric Case : Failure of Cholesky and L D Lt Methods in Exact Ar.thmetic 11 2.9 Symmetric Case : Failure of L D Lt in Finite Precision Arithmetic 12 2.10 Symmetric Case : Present Situation 12 2.11 Symmetric Case : Our Problem 13 Chapter 3 : Historical Survey 14 3.1 Introduction J4 3.2 Direct Methods 14 3.3 Indirect Methods 15 Chapter 4 : Diagonal Pivoting 17 4.1 Preserving Symmetry 17 4.2 The L D Lt Decomposition 17 4.3 Orthogonal Reduction to Diagonal Form 18 4.4 Lagrange's Method of Reduction 18 4.5 Kahan's Proposal 20 4.6 Pivotal Strategy 21 4.7 Parlett's Observation 22 Chapter 5 : The Decomposition for Diagonal Pivoting 23 5.1 Definitions 23 5.2 Ti" lv'compo [ ttion 23 . Thc 'i I ivot 21 I; I Page Chapter 5 (Continued) 5.4 The 2x2 Pivot 24 5.5 Bounding v 25 5.6 The Reduced Matrices 27 5.7 Criterion for Choosing a lxI or 2x2 Pivot 27 Chapter 6 : The Complete and Partial Pivoting Strategies 30 6.1 Complete Pivoting 30 6.2 Bounding V 31 C 6.3 Operation Count for Complete Pivoting 31 6.4 Partial Pivoting for Equilibrated Matrices 33 6.5 Bounding v 33 p 6.6 Criticism of the Partial Pivoting Strategy 34 Chapter 7 : Equilibration of Symmetric Matrices 36 7.1 Introduction 36 7.2 Equilibration of General Matrices 36 7.3 Difficulties with Symmetric Matrices 37 7.4 The Obvious Attempt 38 7.5 Equilibration of Lower Triangular and Symmetric Matrices 39 7.6 The Algorithm for Null Rows in the Lower Triangle 40 7.7 Summary of Equilibration of Symmetric Matrices 41 7.8 The Algorithm for Exponent Adjustment 42 Chapter 8 : Unequilibrated Diagonal Pivoting 44 8.1 Maximal Off-diagonal Element 44 8.2 Bounding vb 45 8.3 Comments on this Strategy 46 8.4 A Partial Strategy for Equilibrated Reduced Matrices 48 8.5 k Partial Strategy for Unequilibrated Reduced Matrices 48 Chapter 9 : Operation Count 50 9.1 Solution by Diagonal Pivoting 50 9.2 Summary of the Work Required 50 9.3 Forming (1) A -M D Mt 52 9.4 Solving (2), (3), (4) 54 9.5 Total Work Required 54 9.6 Upper Bound on Mults 55 9.7 Upper Bound on Adds 55 T~~ Page Chapt?7 10 :Error Analysis for Diagonal Pivoting 56 10.1 Introduction 56 10.2 The Occurrences of Error 56 10.3 The Error Matrix E 57 10.4 Notation 57 10.5 Summary of the Error Analysis (r) 58 10.6 The Decomposition for the Reduced Matrix 59 10.7 The Error Matrix F for the Decomposition 59 10.8 The Error Matrices M 1 , 6D, M2 60 10.9 Floating Point Error Analysis 61 10.10 Floating Point Analysis for F 61 10.11 Summary of Floating Point Analysis for F 64 10.12 Comments on the Bound for F 65 10.13 Floating Point Analysis for a D 66 10.14 Floating Point For M1 and M 2 67 10.15 Floating Point Bound for iL 68 10.16 Comments on the Bound for E 70 Chapter 11 ; An A Posteriori Bound on Elemcnt Growth for 0 < a < 1 71 11.1 Introduction 71 11.2 Pivots 71 11.3 Hadamard's Inequality 72 11.4 Founding det A(k) 73 11.5 Fundamental Inequality 73 11.6 Bounding Pivot Growth 74 11.7 Bounding Element Growth 75 11.S Comments on the Bound 78 11.9 Smaller Bound on Pivot Growth 79 Chapter 12 An A Priori Bound on Element Growth for -= (i + /7)/8 82 12.1 Introduction 82 12.2 Lower Bound on c(as) h(n,a) for 0 < c I 83 12.3 Remarks 3n an Upper Bound for h(n,a) 85 12.4 An A Priori Bound for a -a0 -(1 + 1)1-8 86 80 12.5 Bound on Element Growth 91 12.6 Conjecture for Gaussian Elimination 91 12.7 Conjecture for Diagonal Pivoting 92 12.8 The Optimal Choice of ci 92 I ... I: . Page Chapter 13 : Iterative Improvement 94 13.1 The Approximate Solution 94 13.2 The Iteration 95 13.3 Convergence of the Iteration 96 Chapter 14 : Symmetric Band Matrices 98 14.1 Gaussian Elimination for Band Matrices 98 14.2 Diagonal Pivoting for Symmetric Band Matrices 99 14.3 Symmetric Tridiagonal Matrices 99 Appendix A : Miscellaneous Results A-I A.1 Vtagcnai Pivoting for Positive Definite Matrices A-1 A.2 A Results foc Symmetric Hadamard Matrices A-2 A.3 Gaussian Elimination for Tridiagonals A-3 Appendix B : Algorithm for Symmetric Equilibration B-I B.1 Discussion B-I B.2 The Algol Procedure B-i Appendix C : Algorithm for Diagonal Pivoting C-I Bibliography 4J Acknowledgtment This dimsterraLion was prepared under the direction of Profes~sor licresford N. Parlett, Chairman, Department of Computer Science, Berkeley. I am most grateful to Professor Parlett for his being willing to discuss innumerable contentions and for the go'nerous contribution of his time that he made. I wish to thank ProfessoL William Kahan for several helpful eonversations, aad I wish to thank the dissertation committee, Professor Parlett, Professor C. Keith Miller, Department of Mathe-matices, and Professor H-ugh D. McNiven, Department of Civil Engineering, toy reading the manuscript. I d160 UWAk 'Mrs. Geri Stephen for great patience and exceptional skill In typing the manuscript. Research for this dissertation was supported in part by the Office of Naval Research under the Computer Center Nonr Contract 3656(23). Chapter 1 ; Introduction 1.1 Symmitric Syatem of Linear Lquations Let us consider direct methods for solving the system of linear algebraic equations, A x - b, where A is symmetric. if A is alo positive definite, then Cholesky's method (§2.7) and :: L D Lt method (§2.6) are fast, stable, and preserve symmetry. Tf A is symmetric but indef3nite (neither positi e definite nor negative definite), Cholesky's method and the L D Lt method are unstable .nd can produce very inaccurate results (§2.8). At the present time, if A is symmetric indefinite, Gaussian elimination with partial or complete pivoting is recommended for solving the system (Fox, p. 80, 185), and thus the symmetry of A is ot no advantage. Is there an algorithm for the symmetric indefinite case wh'ch is stable, is faster than Gaussian elimination, and can take advantage of the symmetry? 1.2 Our Contribution We discuss the problem is Chapter 2 and review previous efforts in Chapter 3. In Chapters 4-6 we present a laethod, called diagonal pivoting, which fulfills the above requirements whet restricted to equilibrated matrices, In Chapter 7 we present a metbod for equilibrating symmetric matricua in a very simple manner. A variation of the diaLonal pivoting method is presented in Chapter 8i this method is applicable to unequili-brated matrices and it fulfills. the above-mentioned requirements. In Chapter 9 we show that the diagonal pivoting method is almost as fast as Cholesky or L D Lt. In Chapter 10 we perform a backwards error analysis. In Chapters 11-12 we show that the method is essentially as stable as Gaussian elimination with complete pivoting (in the sense of Wilkinson's analysis for Gauss'an elimination with completely pivoting, Wilkinson (19b1)). In Chapter 13 we show that iterative improvement is as applicable here as it Is for Gaussian elimination. In Chapter 14 we discuss the problem of symmetric band matrices. All the results proved are applicable to complex systems where A is Hermitian. 1.3 Summary Let A be an n x n matrix with max IAij 1 ii ij We want to solve A x - b. k-1 Let - C N denote C N + I C Ni where the C 0 < i < k i-O ore constants independent of n Let G.E. denote Gaussian elimin-aticn. The situation is summarized in the following table: 3 -Bound on Restric- Element tio-as Number of Number of Growth Method on A AuhLOAlications Additions Storage StaLlityl G.E. with 1 i 2in complete det A 0 -n 3 n n /nf (n) pivoting G.E, with 1 partial det A 0 3 n n2 pivoting Cholesky symmetric 1 2 Method positive n n 2 definite Diagonal symmetric, 1 n 3 1 n 2 n f (n) Pivoting det A #0 T- 4 n c" < --1 3c (ot) h (n. a) 3 n ~1l/2 n --3 Here 0 < a < I, f(n) -k i , h(n,a) is a function k-2 dependent on the pivotal strategy, and c(a) and h(n,a) are defined in 0.446 9§11.6-7. In Chapter 12 we show that c(c)h(n,) <. 3.07(n-1) < 3 v'nn for a - (I + 17)18 -c 0 1.4 Origin of Symmetric Indefinite System's The problem of indefinite systems cf linear equations is sometimes dismissed as academic by the claim that physical pr-blems always generate positive definite systems of linear equations. However, the numerical solution of natural problems often gives rise to situations which do not have a physical origin. We give two related examples. In the Rayleigh Quotient Iteration for finding eigenvalues of a positive definite matrix A (Wilkinson (1965), p. 172, and particularly £i 4 p. 629) we need to solve the systems ( A -ri ) Xi+ xi where r i = x i A x i/ x i xi Here A - r I cannot be definite because r lies between the extreme eigenvalues. In the inverse iteration method for finding the eigenvector corres-ponding to an approximation \ to an intermediate eigenvalue of a positive definite matrix A, we need to solve ( A -X I) v i+ I u, u i+ 1 - vi+ / max (v i+I ) .A - X I can be indefinite even when A is positive definite (Wilkinson (1965), pp. 618-635). I! 5 Chapter 2 Presentation of the Problem 2.1 Introduction The speed and storage capacities of current digital computers allow us to solve large systems of linear equations by direct methods. Here we shall consider direct methods for solving a system of linear equations, A x - b , whe-e A is symmetric and det A 0 0 2.2 General Problem : Exact Arithmetic First let us consider the solution, in exact arithmetic, of A x b , where A is general and det A # 0 . We know that Gaussian elimination will give the aolution provided that whenever a zero appears in the leading diagonal position we interchange that row with a lower row with non-zero leading element (such a row will exist since det A 0 0), e.g. if A = 0 and if j is the least integer for which A ji 0 , then we interchange rows I and j . Or, equivalently, there exists a permutation matrix P such that Gaussian elimination without interchanges applied to P A will give us the solution. Since we could also do the same with columns instead of with rows, there exists a permutation matrix Q such that Gaussian elimination without interchanges applied to A Q ,.,1l1 give us the solution. In matrix notation the Gaussian elimination algorithm factors A into A - L U , where L is unit lower triangular, U is upper triangular, and L and U are unique when they exist (Wilkinson (1965), p. 204). Thus " xact arithmetic there exist permutation matrices P and Q 6 such that P A = L U and A Q = L2 U2 , provided only that der AO0 2.3 General Problem : Failure of Previous Attempts in Finite Precision Arithmetic In finite precision arithmetic (Wilkinson, 1963) the above algo-ricIhm can fail if we interchange only under the condition that the ele-ment in the leading diagonal position be zero. Let A [ and b [ Then A L [J = [ [ However, if c and n are Li/: n1 -or 1/1 small enough, then in finite precision arithmetic the operation i -i/c yields -I/c . Let U and x be the matrix and vector of the values of U C c and x, respectively, computed in finite precision. Then U = and a -b 2] So xo -J.I c- I I So x-U' a - Kll) / , but x c mU co If n cE, then x i and the error in the first component of the computed solution is Id I i 7 2.4 Condition of a Matrix Wilkinson (1965, pp. 189-191) shows that the relative error in the solution of a system of linear equations is bounded in proportion to the condition number of A , K(A) -\|A\| \|A I ' > I , i.e., if e -x x c ~C then lel / Ix1 < p(A) K(A) We should not expect a small error if K(A) is large. Here A has a very satisfactory condition number. A-' [ 2I.] n Using the one-norm, 1A1 ma IAij1, we have K(A) -AI 1A- 11 = J i-I 1 [l + max(IclIn)] 11 - n,'l The computer replaces Cn - I by --I, and the computed inverse is (A-l L Then bA- b -Thus the trouble lies in the Gaussian elimination algorithmn, not in the matrix A 2.5 General Case : Stable Direct Methods (a) Direct Inversion Direct inversion (the formation of A ) of a system requires -n each of multiplications and additions (Fox, pp. 177-179). Gaussian eli-mination, however, requires only n each of multiplications and 3 additions. Thus we would prefer to use Gaussian elimination if we could obtain a satisfactory solution. 8 (b) Stability Let us attempt to solve A x -b , where A is nxn and det A # 0 . If x is the solution we obtain from the computer, we c may consider x to be the exart solution of the system (A + E) y -b We might say that the algorithm we use is stable if the elements of E are small in comparison to the corresponding elements of A . Actually the term is more often used when 1E / 1A1 is small. (Here I I is any norm.) (c) Stability for Gaussian Elimination Wilkinson (1960) showed that for Gaussian elimination we have: 1 2.01 max(1,J)2 A(n-k+l) where t is the number Eiji < .1 ~-iJ2 - a rs I of binary digits in the machine, k = min(i,j), and A(n-k+l) is the reduced matrix of order n-k+l in the elimination process. The important lesson from the above is that we must be interested in keeping the elements in the reduced matrices small. There are two well-known strategies for choosing permutation matrices P and Q such that Gaussian elimination without interchanges applied to P A Q will provide sufficiently small element growth in the reduced matrices. (d) Complete Pivoting The first strategy, called complete pivoting, requires that we bring the largest element in the reduced matrix into the leading diagonal position. This strategy is called complete since we search the entire reduced matrix. Wilkinson showed that this complete strategy gives t I I =-- --" Imom 9 1 max Ik+ < vf(k)max A where f(k) 2 k r r i ,j i ,j r-2 In words, the elements in the reduced matrices can never become too large; so this strategy is never bad. It is conjectured that the true bound is max -A n-k+J<) k where A to real (A2.4). Equivalently, the above says that there exist permutation matrices P and Q such that Gaussian elimination without interchanges applied to P A Q gives max 1E ij 1 2.01 n3 /2 f(n) 2 - t max 1A i i ,J iJ (e) Partial Pivoting The second strategy, called partial pivoting, requires that we bring the largest element in the first column of the reduced matrix into the leading diagonal position. This strategy is called partial since we search only a part of the reduced matrix. This is equivalent to the application of Gaussian elimination without interchages to P A 4h-k+lV k where P is a permutation matrix. Here max ;A- < 2 1I 0 0 0 i -1 0 0 II 1 0 1 This bound is sharp since A -. . , where A is nxn , has A(n) = n . Thus max JE -I < 2.01 n 2 - t 2 n max IA I nn i ji j and this is very weak when n > t Correspondingly, we could use a partial pivoting strategy in which we bring the largest element in the first row of the reduced matrix into I 10 the leading diagonal position. Thus there exists a permutation matrix such that Gaussian elimination without interchanges applied to A Q 2- n has an error matrix E with max tEijI < 2.01 n max l~i iji,j (f) The Error Matrix We see that the error matrix E is dependent on the decomposition L,U , the matrix A , the right hand side b , and the permutations P and Q by which we can pre- and post-multiply A , i.e. E -E(L, U, A, b, I', Q) , where L U - P A Q . (For the partial pivoting strategy on the first column of the reduced matrix, we take Q -I in the above.) 2.6 Symmetric Case : Direct Methods If A is symmetric, then we can only apply congruences to A if we want to preserve symmetry. In particular, whenever we interchange two rows, we must also interchange the corresponding columns. Thus only a diagonal element can be brought into the leading diagonal position. The cymmetric form of the Gaussian elimination decomposition L U gives t the decomposition L D Lt , where L is unit lower triangular, D is diagonal, and Lt is the transpose of L . Since we may only perform congruences on A , the error matrix is E -E(L, D, A, b, N, N t ) where tt N is a permutation matrix such that L D L N A N 2.7 Symmetric Positive Definite Case (a) Cholesky's Method The Cholesky decomposition (Wilkinson (1965), pp. 229-232) is the most well-known decomposition for a positive definite matrix A -i I t 11 (i.e. x Ax > 0 for x 0) Here A L L where L is lower triangular. No interchanges are re4uired for stability. (b) L D Lt Decomposition _ If A is positive definite, then its L D L t decomposition (the1 symmetric form of Gaussian elimination) is stable in the absence of unde.rtluw and overflow. The elements of L can be arbitrarily large, but if they do not overflow then in fact the error matrix E is as small as the error matrix for the Cholesky decomposition of A . 1/2 (Note: L -L D ) (c) Method of Congruent Transformations This method (Westlake, p. 21; De Meersman and Schctsmans, (1964)) uses decomposition (b) with interchanges. Here the largest diagonal element is brought into the leading diagonal position at each step. If A is posi-tive definite, then the elements of L are bounded by 1 and the method is stable. (d) Summary If A is positive definite, then the above three methods are stable. If A is nxn , then each method requires n2 storage positions, 1 n3 multiplications, and n 3 additions to solve A x - b 6 6 2.8 Symmetric Case : Failure of Cholesky and L D L t Methods in Exact Arithmetic If A is symmetric but indefinite, then the L D L decomposition, the method of congruent transformations, and the Cholesky decomposition fail in exact arithmetic for a matrix as simple as A [ 1 that is, there exists no permutation matrix N such :hat = - =- = 12 N A Nt has an L D L or an L Lt decomposition. Note that L Lt is always positive semi-definite. Thus Cholesky decomposition will fail in exact arithmetic whenever A is indefinite tt dnd det A ' 0 ,i.e. there exists no permutation N such that N A Nt -L Lt where L Is lower triangular. 'file L D Lt decomposition and the method of congruent transfor-mations will fail in exact arithmetic whenever all the diagonal elements in a reduced matrix are zero, i.e. there exists no permutation N such that N A Nt has an L D Lt decomposition. 2.9 Symmetric Case : Failure of L D Lt in Finite Precision Arithmetic The L D Lt decomposition in finite precision can be unstable if the diagonal elements are too small. The L D Lt decomposition on the matrix A will produce the same incorrect solution on the computer as we saw in §2.3 for Gaussian elimination without interchanges. However, here there exists no permutation matrix N such that N A Nt has a stable L D Lt decomposition. Thus the L D L t decomposition fails for symmetric indefinite matrices. 2.10 Symmetric Case : Present Situation If we ignore the symmetry of A and apply elitainption with complete or partial pivoting to A , then in general A will no longer be sym-metric after the first step of the elimtintion. We then need - n 2 1 3 storage positions in the computer and we must perform -n multi-3 plications and n additions. But we also have stability for the 3 .... . ... . . . . ... . . . ..... . .... . .. ... r 13 L U decomposition. This procedure is presently recommended for the solution of symmetric indefinite systeas of linear equations (Fox; p. 80, 185), and thus the symmetry of A is of no aavautage. 2.11 Symnetric Case Our Problem ii A is symmetric but indefinite, we would like. to find an algorithm which gives a stable decomposition when applied to N A Nt where N is a suitable permutation matrix, but which also takes advantage of the symmetry in order to require only n storage 1 n positions in the computer and to require only 6 n multiplications 1 3 and -n additions. 6 Our algorithm will fulfill all the above requirements wich the 1 3 3 exception that we will need between n and n additions. 4 t I .. .. = 1 14 Chapter 3 Historical Survey 3.1 Introduction Various methods have been proposed for symmetr;.c indefinite f MaLrices. Most of these methods have been unstable, while the stable methods have required operation counts of at least n3/3, where an operation Is defined to be a muItiplication follo-,wed by an addition. Let us look at some of these methods. 3.2 Direct Methods Usually direct muthods for symmetric indefinite systems are based t on the symmetric form of Gaussian eliminarion, L D L , which is unstdble in the absence of pivoting. The L D L method and its 'ariant, the method of congruent trans-formations (in which we use the largest diagonal elment as the pivot 23 at each step (§2.7)) require n storage locations and operations. But both methods are unstable (§2.7-2.9). These methods are of value only if It is known in advance that no element of D will vanish or be small. The Crout factorization (Hildebrand, pp. 429-435; Householder, pp. 82-83) is also a modification of L D Lt , symmecric Gaussian 2li-mination, and thus requires 2 storage locations and n-2 6 operations, but it is also unstable. These variants of L D Lt are unstz'ble. Let us consider some direct methods that are not based on L D Lt I I I I I I-I I-I-I-II i--I I Is 15 I The escalator method (Householder, pp. 78-79) uses the known solution of a subsystem as a step in solving the complete system. The problem lies In finding the solution of the subsystem. Some headway in this problem was made by Parlett and Reid (1969). They reduce a symmetric matrix to tridiagonal form by stabilized ele-mentary congruences and solve the tridiagonal system by Gauss1~n eli-mination with partial pivoting. They require n storage loca-tior and n3 operations, and the method is stable. 3 In October, 1965, W. Kahan (in correspondence with R. De Meersmans and L. Schotsmans) proposed a method for solving symmetric systems based on Lagrange's theorem on the reduction of quadratic forms to diagonal forms (§4.3 - 4.4). Kahan proposed the generalization of the idea of a pivot to include 2x2 submatrices (§4.5). Then Schotsmans (1965) prepared an algorithm in which one searches all the principal 2x2 submatrices for the one with largest determinant. . n2 ~ 1 n3 This algorithm requires - n storage, but between -n and n operations (§5.4). 3.3 Indirect Methods The Seidel iterative method (Householder, pp. 48-51, 81) and the method of relaxation (Householder, pp. 48-51, 81) require - n 2 opera-tions for each cycle. The number of cycles required depends on the matrix, the starting values, and the needed accuracy, Usually the nu-ber of cycles exceeds a , so at least n3 operations are required. : \| I . . . . I = I I " II 16 The method of steepest descent (Householder, pp. 47-51, 82) requires - 2 n? operations at each step. Again, usually at least n steps are required. So the number of operations is at least 3 2 n The congruent gradient method, also called the Stiefel-Hestenes method, is a finite iterative method designed for positive definite matrices, but it can be used for symmetric matrices (Fox, pp. 208-213; Householder, pp. 73-78, 82). It requires - 2 n2 operations at each of n steps. Once again - 2 n3 operations are required. See Reid (1967) for useful observations on this method. I r[ 17 Chapter 4 ; Diagonal Pivoting 4.1 Preservig Symmetry In order to L:ave a direct method for symmetric matrices which will preserve symmetry, we can perform only congruences on the matrix A , i.e. if we premultiply A by a non-singular matrix X then we must also postmultiply by Xt 4.2 The L D Lt Decomposition t Let us consider the L D L method in greater detail. We con-vert A to diagonal form by congruences. Let u. consider the first step of the decomposition: Let A- If a 0 ,then A L , L c B 0 B-CCtla 1C 0 1 where L and In- is the identity matrix of order n -i C/a I 1n-1] The variant of L D Lt called the method of congruent transforma-tion. (Westlake,p. 21; De Meeramans and Schotsmans) uses the largest diagonal element as pivot. This is equivalent to the L D Lt decom-position of N A Nt , where N is a permutation matrix. As we saw in §2.8 both methods are unstable for symmetric (inde-finite) systems. This instability results from our being unable to bring an off-diagonal element into the pivotal position. Since we are using only congruences, we can bring only a diagonal element into the pivotal I .. .. ..-. 18 position. When some off-diagonal element Ai , j > i is very large, we can bring Aji into the (2,1) position by congruences, but never into the (1,1) position. Thus we cannot take advantage of this valuable information. 4.3 Orthogonal Reduction to Diagonal Form t Any real quadratic form x A x of rank r can be reduced by an orthogonal transformation to a diagonal form x 2 + +X x2 11 r r where Al. ... Ar are the non-zero eigenvalues of A (Mirsky, pp. 362-363). If A is an nxn symmetric matrix with det A # 0 , then the above means that t Aff0A0 t where A = diag n 1,... , n , the Xi are the etgenvalues of A and 0 is an orthogonal matrix whose i column is an eigenvector corres-ponding to A 1 t However, this 0 A 0 decomposition involves more work than Gaussian elimination and requires the use of irrational operations. For a finite-precision algorithm we would prefer a reduction involving only rational operations. 4.4 Lagrange's Method of Reduction In 1759 Lagrange devised a method for reducing a real quadratic form of rank r by a real non-singular linear transformation to a 19 diagonal form L 2 + + x2 1 X " r r where C "...,sr are all non-zero (Mirsky, pp. 368-374), and the number of positive (and negative) squares is invariant. t This method corresponds to the L D L decomposition of a sym-metric matrix A when the L D Lt decomposition exists. Suppose A 1 1 U ... -A - O , while det A # 0 . Then the nnl t L D L decomposition for A does not exist. In this case, some A # 0 for r # s since det A 0 rs Let us assume A 11 -0 = A22 but A12 0 0 , where A A (Xl,.. .,x) x A x is a quadratic form in xl,...,x n where x - xl, ,xx n ] I In this case Lagrange proposed the following transformation: (4.4.1) x -Y + Y x -2 x- 3Y3 "Xn . Yn This maps 2 A 1 2 x I x 2 into 2 A 1 2 ( -). Thus is transformed into a quadratic form P in Yl....'y where the coeffi-2 an cients of y and y2 are non-zero. Then we can proceed with the decomposition (Mirsky, p. 371-2; Gantmacher, p. 19q). Let us consider the above transformation in matrix form. Then T y :z , where -1 0 (4.4.2) T = and In-2 is the identity matrix of order 0 1n- 2 n-2 i K! 20 t t Ct Thus x A x y (T t A T) y, and T A[ I a symmetric matrix with (Tt A T)11 ,(Tt A T) 22 0 0 Hence we have avoided the problem of zeros on the diagonal of A by use of the 2x2 matrix This procedure is also applicable to complex quadratic forms. 4.5 Kahan's Proposal Ln 1965 W. Kahan (in correspondence with R. De Meersmans and L. Schotemans) proposed that Lagrange's method could be made the basis of a stable method which preserved symmetry. Kahan adapted Lagrange's method to finite precision by observing that the use of in (4.4.2) corresponds to the use of a 2x2 submatrix as a pivot in a decomposition by linear transformations and that a 2x2 could be chosen if the diagonal elements were zero or very small (§§2.8-2.9). Suppose we used a 2x2 submatrix P as a pivot. Let us look at such a decomposition. Let A -, where A -At . det A 0 A is nxn , P is 2x2 , C is (n-2)x2 , and B is (n-2)x(n-2). Then A a L1 0 B-CP 1 Ct 1 L , where L1 -and Ik is the identity matrix of order k How do we choose whether to use a lxl or a 2x2 pivot? Can both lxl and 2x2 pivots be baL? 21 4.6 Pivotal Strategy Kahan considered two pivotal strategies. In the first, one searches the entire matrix for m c - wax [A' , AjA - A2 1 c i,j,k f A i , then interchange rows and columns 1 and i and use All as a lxl pivot. If m IA 1a A.k - A2I , then interchange rows and columns 1 with j and 2 with k , and use [ Ajj Ajk] as a 2x2 pivot. ILA k A k Since we search the entire matrix, this is called a complete pivoting strategy, In analogy with complete pivoting for Caussian Elimination. However, the searching here requires between 3 6 and -n multiplications to find mc for all steps (depending on 3 the number of lxl and 2x2 pivots used). The decomposition itself requires I n multiplications. Thus this strategy would require 6 between I n 3 and n multiplications to solve A x -b (§5.4), which is more than for Gaussian elimination. Hence Kahan rejected this strategy. Kahan considered a second pivotal strategy in which we scan only the first column and the main diagonal; this is called a partial pivoting strategy, in analogy with partial pivoting for Gaussian elimination. 1 n2 1 n2 The searching here only requires between .1 n and -n multi-4 2 plications. We take m A.max (A2 1 I A } . However, this partial pivoting strategy is unstable. I 22 At A 1' , where O<c<<l. TThus [ would be used as T h e n m p J A I A2 2 0 T h u s 2 1 ()21 8 2 a 2x2 pivot, and the reduced matrix AM is I- 1 + g - 1 C) If c is small enough, then in finite precision arithmetic the operatio 1 -2 yields -. + 8 . This can cause highly 3 C 3 3 c 3 inaccurate solutions, as in §2.3. So this partial pivoting strategy is unstable. For these reasons Kahan rejected this method for use on symmetric systems. 4.7 Parlett's Observation In 1967 B. Parlett observed that the examples for which the partial pivoting strategy was unstable were also unequilibrated. A symmetric matrix A is equilibrated if max IAijI - 1 for each row index i j (§7.1). Parlett conjecture that the partial diagonal pivoting strategy would be stable when applied to equilibrated matrices. 23 Chapter 5 The Decompoition for Diagonal Pivoting 5.1 Definitions Let A be an nxn symnetric non-singular matrix. We want to reduce A to the "diagonal" form M D Mt by congruences, where D is a block diagonal matrix, each block being of order 1 or 2, &nd H is unit lower triangular with H 0 if D 0 0 Let 0 "x IA 1 l , " max IAi , and 1 -IAn 22 -A i ,J ij1 12 5.2 The Decomposition Let A = [ [ , where C is j x (n-J), B is (n-J) x (n-J), L C B s and P is j x j , where j - I or 2. 0 01 if p-l exists, then A - L1 LL where IB-C 1 t -L and 'p. In- are the identity matrices of order p--j and n-J, respectively. Any element of C P -will be called a multiplier. 5.3 The lxl Pivot Suppose P is of order 1. (We shall not make a distinction between a matrix P of order 1 and its element, which we shall also call P .) i 24 Let us assume that we have already interchanged rowe and columns so that 1PI P i i.e. P is the maximum diagonal element. If P 1 exists (i.e. V 0), the,. let A (n - l) -B -C P-1 Ct (n-1) Then (C P-I~ A Ai P and Ai Ai~~+ (C P-i Aj+~ i 1+1 ij i+l,j4-l £ J+l,l Since I l and V - max JAiji we have the following: Lemma I: If P is of order 1 and IPI " U1 I 0 then (1) max I(c p-1 I < Po/W, i (ii) max JA (nl) I (1 + Uo/j)o i,j Thus a lxl pivot P is useful iff IFP -P' is large relative to 10 I i.e. if WI/)JO is bounded away from zero. 5.4 The 2x2 Pivot Suppose P is of order 2 and P-1 exists (i.e. v # 0). at -1 Here the (k-l) row of C P is: (Akl, Ak2) -A11 A 1 - Ar 1 (AkZ A2 2 - Ak2 A21, Ak2 A,1 - Ak. A2 1 ). (n2 -I (n-2) -let A(2) B C p Ct Then A -A -(C P C -B- CAi 2 i+2,J+2 ~ il C 1 j (C P) C £2 2j Since v -A11 A 2 - A2 1 1, lAkil and IA.k21 < 0 , and JA111, 1A 22 <111 ' we have the following: 111 _ 221 _ --Lama 2: If P is of order 2 and Idet PI " V 0 , then (1) max I (C P-1) I o 11 o + ii1)Iv (ii) max < (1 + 2o (jj + P)Iv) 'o iJ Thus a 2x2 pivot is useful iff we can bound v away from zero. In particular from §5.3, we need to have v bounded away from zero whenever p1/p0 is near zero. (Note that the use of the standard norm bound would give too crude an analysis for Lemmas I and 2.) 5.5 Bounding V We can easily bound v from above, since v -IA, A 2 2 - A:,2I.< A 2 4A 1 1 1 1A 2 V2 + M2. Thus we have: 2112~ + 222 Lea-3: IdetPI V<p2 + V2 - 0 1 This upper bound is sharp for II VO 0 l0 -L 0 0 0 -ow\n o But, as we saw in §5.4, we need a lowez bound on V which bounds I i 26 v away from zero when V 1 /M 0 s small. Clearly, such a lower bound does not exist without interchanges. Consider A - 0 0 1 with P =10 001 We shall exhibit three different pivotal strategies in §6.1, §6.4, and §8.1, which provide us with the necessary interchanges so that we have a 2x2 pivot P with: Idet PI (§6.2, §6.5, §8.2). Assuming this lower bound, we have: Lenma 4: If Idet PI _ v> I >0 then (i) max I(C P-1)ikl < P 0/(PO - U l ) for k-l,2, i,k (ii) ax I(n-2 )I.+-(ii) MAX A'( 2v10/(J 0 ij 0 0 0 1 The lower bound on V > 0 is sharp for l1 PO 0 L 0 \ K A-0 L 0 1-Thus we have a good bound on element grovth in the reduced matrix, since if Vj/ O is small then W 1 . We shall see in 1 0 1 0 Chapter 10-12 that stability follows from this. 27 5.6 The Reduced Matrices On Av(n .(n) , (n) Define A -A, )0 10 P 1 v "y1 Let A(k) be the reduced matrix of order k . Let 1 a (k) . max (k), 1 (k) . max IA(k) 1. and o(k ) . JA (k) A2(k) A (k)21 0J jj 1' 1 1 22 2 All considerations in §§5.2-5.5 hold for A(k) 5.7 Criterion for Choosing a ixl or 2x2 Pivot We must find a proper criterion for deciding whether we shall use a lxl or 2x2 pivot. In Chapter 10 we shall show that the elements of the error matrix are bounded in proportion to the elements in the reduced matrices. For stability we must ensure that the elements in the reduced matrices do not become too large. If we made our criterion to be the minimization of the number of multiplications (additions), then we would want a lxl (2x2) pivot at each step. But this would be unstable. Instead, let us aim to minimize the element growth that can take place in the transformation from one reduced matrix to the next. For further temarks see §12.6. Let F (k) be the growth factor permitted by choosing a jxj pivot for A(k) , where J-1 or 2. If the hypothesis of Lemma 4 holds (i.e. v(k) >() 2 k)2 for all A~k) ), then by Lmas 1 and 4: -4 28 F 1 + jOIl F=i + 21F-P P (k) (k). (k) F has a good bound if p 1 is not too small; while (k) (k) (k) F2 has a good bound if p IPO is not too large. Thus we are led to the following: Definition: For 0 < a < 1 , let S be the following strategy: for each reduced matrix A~k) , choose a lxi pivot iff P (k) (k) >a (and a 2x2 pivot otherwise). With S we have F(k) I < + 1a and F 2k) 1+2/(l -for all A(k) But at any stage the choice of a 2x2 pivot carries us further towards the complete reduction than does the choice of a lxl pivot. Since the growth factors from reduced matrix to reduced matrix are multiplicative, it is natural to compare the square of the growth fac-tor F( k ) permitted by choosing a lxl pivot for A(k ) with the growth factor F(k) for a 2x2 pivot. 2 Thus the problem is to find min max ((1 + I/a) 2, 1 + 2/(1 -c)} Theorem: min max (( + 1/a) 2 , I + 2/(1 - a)) -(9 + 17)18 O<a<1 and is achieved by a -a 0 (1 + r17)/8 Proof: The equation (1 + I/a) 2 -1 + 2/(1 - a) reduces to a quadratic with roots (1 +±17)/8 . Since the left side of the equation L ~. 29 is monotone decreasing, the right side is monotone increasing, and a > 0 , the minimum is given by a 0 = (1 + ,r7)/8 q.e.d. We immediately obtain the following bounds on multipliers and on elements in the reduced matrices under strategy S . Let m be any multiplier. (k) Corollary 2: Under strategy S i , if for all A K (k) >(k) 2 ( 1 1 k)2 V _ U then: Ia for a lxl pivot <mi I1 /(1 - a) for a 2x2 pivot For a a0 u (1 + /17)/8 (7i - 1)/2 < 1.562 for a lxl pivot T + 7)/4 < 2.781 for a 2x2 pivot Corollary 3: Under strategy S. with a - a0 , if (k) >(k) 2 (k) 2 Ak O >III ) ) forall A(k) ,thenfor l<i<n P M ) < P0[(9 + /J)/ 2 ] ( n - t )I2 1 U 0 (2 , 5 7 )n - i In Chapters 11-12 we will give a much better bound on the elements in the reduced matrices. The strategy S allows us to proceed in the following order: a (1) calculate V(k) and pk (k) (k) (2) if Pi a0 ji0 then we use a lxl rpivot; (3) otherwise we find a 2x2 by some strategy. Thus we earch for a 2x2 for A(k) iff (k1 1 a 0(k) 30 Chapter 6 : The Complete and Partial Pivoting Strategies 6.1 Complete Pivoting As we saw in Chapter 4, the partial pivoting strategy can be unstable when uNed on unequilibrated matric:s. The trouble lies in the fact that we do not have a lower bound on the 2x2 principal minors which bounds them away from zero when the diagonal elements are small (§5.5). Let us therefore consider a complete pivoting strategy ("complete" in the sense that we search over all the principal 2x2 minors, cf. §4.6). By interchanging rows and the corresponding columns it is possible to bring any diagonal element into the (1,1) position or any principal 2x2 submatrix into the leading 2x2 position. Let Vc max JIAi - A 21 i ,jii The complete strategy involves: (1) finding il max JAl I , 0 - max JAi I i i,j (2) choosing a Ixl or 2x2 pivot according to S §5.7); (3) for a lxl, interchanging so that IPI = l (§5.3); (4) for a 2x2, finding vc I. and interchanging so that Idet PI V c ( h wor b 2x2, firdachc rdc (§5.4). This would be repeated for each reduced matrix. i 31 6.2 Bounding V However, the result of all this work is that we do obtain a lower bound for vc in trnw of 0 and pI" K Theorem 1: V < 02 + 1" Proof: The upper bound follows from Lemma 3 of §5.5. Let IA, . Then v max IAi - AlI > IA A -A 2 Iw 0 sC rrso rs I,~ 2 A A > 2 2 since V = A2 and 0 rr as 20 01 0 rs p I =max JA iii q.e.d. 1 Thus Lemma 4 in §5.5 holds for v when pl < 1i According to S ,we choose a Ixl pivot for the reduced matrix A ( k ) if C& P > CL0 P where o 0 = (I + 11i7)18. In Chapters 10-12 we shall see that stability of this complete pivoting strategy follows. 6.3 Operation Count for Complete Pivoting Unfortunately, the operation count here is much larger than we desire. The calculation of V(k ) requires k(k-1) multiplications and additions. Let p be the number of lxl pivots used (so q -(n - p)/2 pivots of order 2 are used). Let P) denote summation over those indices k , 1 < k < n , for which A(k) uses a pivot of order j 32 The searching for all the V (k) requires: c (6.3.1) 1(') k(k -1) + 1(2) k(k -1) multiplications and additions. n 1 .lna Now (6.3.1) < I k(k - 1) -n(n + 1)(n -4) n3 with r-1 n?2 equality iff p - n , while (6.3.1) >J 2J(2j -1) T n(n + 2)(2n -1) i-I ! n3 with equality iff p - 0 (i.e. q -n/2). 6 From Chapter 9 we see that the rest of the work for solving A x - b would require . n3 multiplications, and between I n3 and 6 4 in additions. Thus the complete diagonal pivoting strategy requires between n 3 and n3 multiplications, and between _ n3 and l2 12 a additions. (To be exact, n3 + 3 1 p3 multiplications and 3 I 1-i +.1 p additions are required, where p is the number of lxi pivots used.) Examples: If A is nxn and positive definite, then p - n 0 1 0 If Ais nxn with ni even, then p -0 ii 33 6.4 Partial Pivoting for Equilibrated Matrices As we saw in §§6.2-6.3 the complete pivoting strategy is stable, but it involves more work that we are willing to perform, while in Chapter 3 we saw that a partial pivoting strategy is unstable for j unequilibrated matrices. We shall now show that a partial pivoting strategy is stable when applied to equilibrated matrices (with equilibrated reduced matrices). Let A be equilibrated, i.e. let max AiI = 0 for every i j where V > 0 (usually we normalize by taking U0 i ). Let v - max IAll A -A2 The partial strategy involves: (1) equilibrating A (thus we know 0 (2) finding p1 and choosing a lxl or 2x2 according to S (§5.7); (3) for a lxl, interchanging so that IPI Ul 05.3); (4) for a 2x2, finding v , and interchanging so that Idet P1 Vp (§5.4). 6.5 bounding V Now let us find a lower bound for Vp Theorem 2: If A is equilibrated (max lAijI -= for every i 2 2<% 2 2 4 then V 1 0 < - 2 with IAklj - p0 . If (i), then p0 1-1 and, trivially, p o 0_i - 0 . If (ii), then V > JAll Akk- i1' p? -A: P2 _O- q.ced. Thus, Lemma 4 in §5.5 holds for v if . < 0 ' According to S , we choose a lxl pivot for the equilibrated reduced matrix A(k) ci of rde k ff (k) (k) -1 /~/ of order k ifk ,where -+ Also, stability of the partial pivoting strategy for equilibrated matrices follows from Chapters 10-12. For A A (n ) , only 2(n -1) multiplications and additions are (n) (k) required to calculate v ) Thus the calculation of v k for all P p 1 2 k requires between 2 n and n(n - 1) multiplications, compared nd ~ fcrallth with between 6 3 for all the V(k) in (6.3.1). 6.6 Citticism of the Partial Pivoting Strategy The drawback to this method and the crite-.ion which we have found for the pivoting strategy is that the matrix mus. be equilibrated at the beginning and then each reduced matrix should be equilibrated. But an algorithm for equilibrating symmetric matrices has never been exhibited, i.e., for an arbitrary matrix A , "Pe seek diagonal matriceb D 1 , D2 such that D A D 2 is equilibrated, .nd if A is symmetric, 21 2 we need D -D2 in order to preserve symmetry. 1 . . . . . .. . . . . . .. . .. 2. . . . . . . 35 We resolve the above predicament by two fundamentally different approaches. In Chapter 7 we exhibit an algorithm which can equili-brate any symetric matrix in a very simple way. In §§6.1-6.3 we showed that complete diagonal pivoting avoids the problem of equili-bration and is stable, although the number of multiplications and additions required is more than we desire. But in Chapter 8 we shdll exhibit a new version of diagonal pivoting which is applicable to unequilibrated matrices; we call this unequilibrated diagonal pivoting. £his method will show that equilibration (in Wilkinson's sense) is unnecessary for this strategy and this is the algorithm that we recommend. 36 Chapter 7 : Equilibration of Symetric Matrices 7.1 Introduction Wilkinson (1961) recommends that a matrix be equilibrated before applying any algorithm for solving a system of linear equations. A matrix A is said to be equilibrated if all its rows and columns have the same length in some norm. Wilkinson's rounding error analysis for Gaussian elimination (Wilkinson, 1961) gives the most effective resultN when the matrix is equtI"brated, since a small perturbation of one row (or column) Is then of the same magnitude as that of any other row (or column). 7.2 Equilibration of General Matricee In finite precision we modify the definition of equilibration. (In this chapter we shall confine ourselves to the norm 10. - maxixil .) i A matrix A is row equilibrated if, for each row index i -I 0 < max ,AijI < WO . where a is the number base of the floating point system. A matrix A is column equilibrated If, for each columt, index j PO< max JA1jI <1O A matrix A is equilibrated if it is both row and column equili-brated. Usually we normalize by choosing pO0 -1 . We shall assume PO = 1 in this chapter. The use of 8permits a matrix to be equilibrated by changes of exponent only. I In order to row (column)-equilibrace A we seek a diagonal matrix nb~ (D) such that DA (A D ) is row (column)-equilibrated. To 1 2 is equilibrated. However, there is no unique equibrated form of a matrix for this norm (Forsythe and Moler, p. 45). The various equili-brated forms may differ greatly in their desirability for use in Gaussian elimination, since the various equilibrations cause different choices of the pivots (some are good choices, others are bad). F em ixi;) the equilibration is unique For the one-norm (Ix0I but the convergence of the algorithm is slow (Sinkhorn (1964), (1967); Siktkhorn and Knopp). 7,3 Difficulties Vith S etric Matrices -2 If A is sbymmeic, then we allow 1- instead of V0 as the lower bound it the definitions of row (and column)-equilibrated. If A is syametric, then A is row equilibrated iff A is column equilibrated 1ff A is equilibrated. In order to equilibrate a symmetric matrix and still preserve the symmetry we seek a diagonal matrix D such that D A D is equilibrat _d. Let A 1 2 -diag {1/2, D) D -diag {i, V/2} and Le A dg L 14/2] 2 D - diag {v' /4, /2 1 38 Then 1) 1 A D 1 /4 1 D2 A D2 18 ,and I 1 2 1 D3 A D3 -F are all equilibrated. Criteria for choosing pivots are based on their size. It is an open problem whether all equilibrated forms of a symmetric matrix give satisfactory pivots as judged by the usual criteria. There is no known algorithm for the symmetric case. 7.4 The Obvious Attempt Let us cc sider D A D . Let D -diag {dl,...,d n I and A be an nxn symmetric matrix. Then (DAD)ij -d i d A . Let us assume that no row of A is all zero. The method which seems the morc obvious is to equilibrated one row at a time: Let D1 diag{dl,1,...,). C Aij for 3,4 ' I Then (D AD )ij -dI Aij for i 1, J0l 2 or J'l, i#1 I All for ijml So choose d 1 = max{ 'IA 1 1 , max IA ij ) S2<j i ). Lemma 1: If T + A is row-equilibrated, then A is equilibrated. Proof: Let us consider the ith row of A . Since T + A is row-equilibrated, max Aij I = 1 . But IAjij < I for J i , since ITijI < 1 and Aji = Tij for j > i. Hence max IA ij ik each i . q.e.d. Now we shall show hcw to construct D such that D(T + A)D is row-equilibrated if T + A has no all zero row. In this case, by Lemma 1, D A D is equil!.b:'-d. In fact, it is eas- uilibrate the lower triangular part of a symmetrJc matrix and s.. preserve symmetry. Lemne 2: Let B be an nxn lower triangular matrix (i.e. B -0 for I > i ) with no all-zero row. For 1 < i < n , let di I max { -max Id Bi }. Then D B D is row equili-i 1il l i< i-l I -j brated, where D = diag dI ....d} n 40 Proof: By induction. Let d 1 , and B 11 0 by hypothesis. So Id2 BI " 1 Assume that dl,...,dl 1 > 0 have been chosen so that max Idj dk BJki - 1 for I < J < i-1 l<k~j Let d i max { , l<1max Idj BiI I By hypothesis, 1:3. <~ di > 0 . Then max Idi d ijI . j i J Hence the di exist for 1 < i < n by induction, Let D -diag {d , . . ,d n I . Then D B D is row-equilibrated. q.e.d. 7.6 The Algorithm for Null Rows in the Lower Triangle But what do we do if A 1 0 or if, for some i , Aij =0 for 1 < J < i ? Let us form D as In §7.5 with the exception that we set d i 1 1 if Aij = 0 for I < j < i . Thenfor all i,j : l(D A D) iI < 1 D A D fails to be equilibrated only if for some i (a) Aij = 0 for 1< j < i , or (b) max Idj AijI < 1 j >i If the i th row of A is not null, then the maximum in (b) is positive. For such i, define ei by e-l = max Id A I ; for all other i , let ei =1. Let E =diag {el,...,e n 41 Theorem: Let A be an nxn symmetric matrix with no null row. Let D and E be constructed as above. Let A - D E . Then A is a diagonal matrix, and 6 A A is equilibrated. Proof: If ei A I then the maximal magnitude of row i in D A D is raised to I by forming E D A D E , while in all other rows the ith element is increased in magnitude but not in excess of 1 The theorem follows from §7.5. q.e.d. 7.7 Summary of Equilibration of Symmetric Matrices If A is an nxn symmetric matrix with no null row, we can find a diagonal matrix D (in two sweeps, although only n steps) such that D A D is equilibrated (in the u-norm, \|xi. = max lxii ). i We can express D by the following algorithm: For i = l(l)n -max { /1iV, max 16 Ai ) if A 0 for _ l<ii i jAij Aij 6ii = some j ,1< J 1 if Aij - 0 for I < j <i. Then, for i - l(l)n: if A #0 for some j, < J < i i max i A ji I if Aij -0 fur 1<j< i i+l <j<nI Let D = diag (dl...,d n I . Then D A D is equilibrated. The work required to equilibrate an nxn symmetric matrix with no all zero row is: n square roots n(n+l) multiplications (n-I) 2 idul tions 42 In practice the algorithm can be expressed in a very simple manner in Algol (see kppendix B) and can be performed in only n steps. (We would actually set fi " 0 if A 0 for 1 < j < 1 and then search for max 16j AjI iff f -0 ) i+l J<n ± 7.8 The Algorithm for Exponent Adjustment In practice, we actually only require that 6-2 < max JAij I < for every i instead of max IAij I - 1 for every i so that we need lj _n only adjust the exponents of the elements of A Then our algorithm takes the following form: Let Aij = Yij B( j for j i , where -1 < Iij I <1, unless AJ4 = 0 when we take 0 = a iiJ ij Let A where C4 <1 , unless Al1 0 when we take y i i 0 ii 0 if Aij - 0 for 1 < J i For i -l(ln set 6 a x{max (a -6 )} otherwise 2X p i1 ' ma ij J i if 61 #0 For i = l(1)n: set d i max (aj - 6j) if 6, = 0 li <n i -. 43 This will give S-2 <max 16 d + A i < 1 for every ± (This equilibration can be performed very rapidly in mach rne language.) 44 Chapter 8 Unequilibrated Diagonal Pivoting 8.1 Maximal Off-diagonal Element In order to obtain a lower bound of W- 2_0 for v in Theorem 0p 1 in §6.3, we needed the fact that, due to equilibration, there existed an element of maximal absolute value in the first column. However, if P1 < P0 I then there exist integers i,j with i > j , such that jA 11 i WO We need only bring the element A up to the (2,1) position and then we will have a 2x2 pivot with a maximal off-diagonal element. We shall call this variation unequilibrated diagonal pivoting. Let U 0 . max JAiji = IArsI , where r,s are the least such inte-i,j gers. Let pI = max lA il• i Let V =A A -A' b rr ss rs This strategy involves: (I) finding u1 and the least integer k with IAkkI [ 1 (2) finding p 0 and the least integers r,s with IA rl = PO (3) choosing a lxl or 2x2 pivot according to S (§5.7); (4) for a 1xl, interchanging rows and columns I with k so that IvI = 1P (§5.3); (5) for a 2x2, interchanging rows and columns I with r and 2 with s so that Idet PI = Vb and IA2 11 -o0 (§5.4). This procedure is repeated for each reduced matrix. 45 Note that calculating Vb requires only 2 multiplications instead of n(n-l) for v and 2(n-l) for v C p Clearly from the definitions of vb, V, and V and from Lemma 3 of §5.5, we have: Lemma 1: v < V ' V < 2 + 2 -p -c -0 1" 8.2 Bounding vb Let us now bound vb from. below. From §8.1, we may assume -A 2 I and -0 b IA 11 A22 -A2 1 2 A2 00 P2 P2 < 1 2 + W Theorem 1: If IA 21 1 ' O - then 0 -1 < -- < 0 i Proof: The upper bound follows from Lemma 1. Since IA2 = 0 IA A -A2 11 -0 -A A > 2 2 211 -0 ' b 1 22 21 0 1 22 -0 1 q°e. . Here, as in §6.2 and §6.5, symmetry was used to get the lower bound on Vb By symmetry, if PO= IA 2 11 then JA 12 1 ' 00 If A were not symmetric, then p0 = 1A 21 1 does not imply that IA12 1 - P0 (in fact we could have A12 ' 0). Thus no such non-negative lower bound on the determinant of 2x2 submatrices can exist for non-symmetric matrices. Thus Theorem I impltes that Lemma 4 in 55.5 holds for vb if WI < PO According to Sa (§5.7), we would choose a lxi pivot for the reduced matrix A of order k iff p k)> W where t 0 - (l + T7)/8. in Chapters 10-12, we shall see that stability of this strategy (for unequilibrated matrices) follows from the above. -46 8.3 Comments on this Strategy From §8.1, we see that we need do no searching over the 2x2 prin-cipal minors, but we merely choose that principal minor with maximal off--diagonal element. In Chapter 9 we shall see that this searching O(1n3 and . In for the W k) requires between -I-- n3 additions and no multiplications. We used the terms "complete" and "partial" strategies in Chapter 4 to distinguish between searching over all the 2x2 principal minors and over the 2x2 principal minors with off-diagonal element in the first column. In analogy with Gaussian elimination with complete pivoting, we would like to call our strategy in §8.1 a complete pivoting strategy (k) since we search all of A for its maximal element, but we do not wish to cause confusion with the use of the word "complete" in Chapter 4 where it meant searching all the 2x2 principal minors. Now, in analogy with Gaussian elimination with partial pivoting, we ask if there could be a partial strategy where we search only the (k) first column of A for its maximal element. l Such a partial s=trategy requires at most only -n (n-i) addi-2 tions to calculate the maximal element in the first column of all the reduced matrices. If such a partial strategy were table, then this 13 3l strategy would require only a multiplictions and 6 additions to solve A x b a = At det A 0 • , de A #0 47 However, any such partial strategy is unstable for unequilibrated matrices, as the following example shows: A -1 1 where 0":aLx<1 and O<cE<<I 8.4 A Partial Strategy for Equilibrated Reduced Matrices (n (k) Clearly, if A - A and each of its reduced matrices A is A(k) (k) equilibrated (i.e. max IJA I = Pk for each i ), then the partial ij strategy, whereby we choose the 2x2 principal submatrix whose off-diagoral element is the maximal element in the first column of A(k) is stable since, by the equilibration, such a maximal element in the first column of A(k) is also a maximal element of A(k) But we would have to equilibrate A at the start, and then equi-librate each reduced matrix A(k) Let us now consider such a partial strategy when A is equilibrated, but we do not equilibrate the reduced matrices. 8.5 A Partial Strategy for Uneguilibrated Reduced Matric,=s Let A be an nxn symmetric equilibrated non-singular matrix with max %j -p0 for each i . We shall now consider the partial strategy j defined in §§8.3-8.4, but we shall not equilibrate the reduced matriceb A(k) for k < n. 48 Let A( k ) be the reduced matrix of order k ; let A ( n ) A Let -) max ( (We shall not actually calculate ij 0 (k) .(k) (k), Let)I A) say. We shall assume we have II interchanged rows and columns so that A k)I _ (k ) Then let X (k) . a JA (k), . So X(k) Ij(k) ,wie (n) (in) .1 (k) (k) We shall use a I1 pivot iff Pl > a X Let m be any multiplier (§5.1). (k) > a X W then we use A as a lxl pivot, and I 1 -- 1 Iml < x(k)/P(k) < i/a , while max IJA (k-l) ', 11 (k) + X(k)/a < I(k (j +j i/- 0 (k) () If 11(k) < a X(k) then we Interchange so that JA2kI -, ) X -(k) -I) . So v(k) >k() 22 (k) 2) X ( k) ' Then Il (k) (p(k) (k) (k) Then Im X (P + )/V and max JA (k-2)1 <(k) + 2X (k)' (P(k) + W(k) )V(k) < (I + 2/(l1 (k) As in §5.7 we would choose a - a0 -(l + v'Tl . Then wax IA (kI < (2.57)n-k . for each A(k) i But we cannot obtain a bound on A(k) as in Chapters 11-12 since we cannot express bounds on v (k) in terms of (k) 49 We leave the significance of this method (§8.5) in relation to the method in §§8.1-8.4 as an open question. Chapter 9 Operation Count 9.1 Solution by Diagonal Pivoting We now consider the amount of work required to solve A X -B by the (unequilibrated) diagonal pivoting method (Chapter 8), where A is an nxn non-singular symmetric matrix and B is an nxk matrix, i.e. there are k right hand sides. In matrix notation, we perform the following steps: (i) A= M D t t -1 (2) C =M B. (3) Y: D C (4) X= M- t Y. -t -it Here M means (M ) Let p be the number of .xl pivots used. So q = -p) pivots of order 2 are used. Let A be the reduced matrix of order i Definition: {l if A(i) uses a lxi pivot pivot [i] -2 if AW uses a 2x2 pivot L if A (i+ l) uses a 2x2 pivot We shall use the term Mults (Adds) to mean the number of multi-plications (additions). We shall count a comparison as an addition. We shall use Z (J ) to denote summation over those indices i 1 < i < n , such that pivot[i] = J1 9.2 Summary of the Work Required Stcps (1), (2), (3), (4) In 99.1 req~tire; 51 1 3 1 2 i3 Mults -+ (k + )n + (- -2k)n + (k + 3/2)p + X(2) U 1 3 2 4 22 3 n + (k + ) n+ (5- 2k)n+ -(2k 7)2 n -6 4~ 3 l Adds n3 + (k + 5/8)nW + (2 2k)n + 3/8)p + i (1-1) 4 4 4 13 1 ~-n4 +-i=p3 -3 <1n3 + (k + -) n' (k- )n l n3 For k 1 1 (i.e. one right hand side), let us compare the work required for the diagonal pivoting method (applicable to all non-singular symmetric matrices) with that required for Cholesky's method (applicable only to positive-definite matrices). Cholesky Diagonal Pivoting Exact Lower Bound Upper Bound Mults n 3 + n 3 n 2 n 3 9+ n 2 2 n + 6 2 3 6 2 3 6 4 3 12 1 3 2 7 1 3 + A3 2 5 1 3 3 2 5 Adds n + -n n -n n n + -n Root Recipro- n 0 0 cals Since the time required for a computer to perform a multiplication 1i 3 is much longer than for an addition, the requirement of -n multi-6 plications and between -3 and n3 additions is very satisfac-4 3 tory for symmetric indefinite matrices. K -52 If A is positive definite, then Cholesky's method is preferable to the diagonal pivoting method. See also Appendix A (SA.l). 9.3 Forming (k) A-MD M t t First we decompose A into A - M D M Let us consider the reduced matrix AM of order i The following chart shows our course of action. find 1() Adds- i- I find Add 1-- il P MH LL ulLs, Adds-1 yes no use lxi use 2x2 interchange inechange I I ~Mults-2 Mut -:calculate calculate det:t Ads -2I multipliers Mults, Ads 'omA(-)calculate { ults -6(-) Adds( 1 A) multipliers Adds -2(1-2) 2 (11 form A (1i-2) Mults, Adds j i-ii :- 1-2 53 Thus for A(1) Mults A. i(i+l) and Adds 1 if pivoc[i] 1 1 2 2 3 Mults -12 -3i -7 and Adds -(i-l)(j i+l) if pivotti] -2 1 Recall that there are p lxl pivots and q -(n-p) 2x2 pivots. For step (1): + 1(2) i Mults -Tjl) Ti ([+1) + ( 4i + 31 -7) (I i ( ++ (i+l) + -(i-l)i + 31 -7 t +) ) (31- 7), 1 n3 I2 19 7 1(2) (9.3.1) n + -+Zp+3 6 6- 2 13 12 1 _ n + n + 1 Adds " i 2 + (2) (ii)(3 +) S(i){43 i2 + 1) + ( 2 -i)1 + y(2) I 2 + 1 1 + _ (1_1)2 + ( -5 4 2 4 2 4) n /i)1 " ( + ( )- n-p) + I(-2 1 2) 8 1 3 5 2 1 5 1 t4l) (9.3.2) n 3 + -n 2 + .p + - z 1 (1-2) -4 8 844 8 n 5n Now let us bound 3 [42) i and 1 1 (1-2) from above. 44 [(2)i < p+2J = -2 ) + (n-p) and 54 < n 1 p2 5 (1) i (i-2) < i (1-2) -p{n 2 -(p+l)n + " p 2 + ! P i-n-p+l So we have the following upper bounds for (1): (93.3) Mults < + -n 2 2 p -6 4 3 4~ 2 1 n3 + 2 p L (I + 1 2 i(p2+P+l )n + I1 -Z+ 5 (9.3.4) Adds <1 3 n 1 p + ) n 4 8 +T1 p. 9.4 Solving (2). (3. (4) -1 For (2) C M 1 B : Multe, Adds -(1- n+!-p) k (Note that M a 0 if pivot [i] -2.) -1 For (3) Y D C : Mults 3n -2p , Adds - n -p -t For (4) X -M Y I 1n2 1 Mults, Adds n 2 -n + p) k 9.5 Total Work Required Thus steps (1), (2), (3), (4) together require: 1 32+3 1(2) Mults T1 + (k +1 + 1 -2k) n + (k + p + 3 (. 6 + ) 4P )1 3 5 2 4 7 3 2 Mults < n- + (k+ 1) + ( -2k) n + (k-) p p 1 n 1 n2 1 Mults >6 + (k + )n + (L- 2k) n. I 3 5n 3 2k k 3 i(I Adds = n + (k + g) n2 + 2k) n k ) p + 3 l i (1-2) Adds( < + (k + ) 1+ ( -kn+ 1P3 1 1 P2+ 1 3 5 2 3 7 Adds > 1- + (k + ) 2 + -2k) n. \|i 9.6 Uper Bound on Mults We would like a minimal upper bound on Mults that is independent of p , where 0 < p n Let f(x) -(k -x- By elementary calculus, f(P) < f ( " T2 (2k- 7) Thus ults < n + (k + 4 . n 9zn 2 25 (-2k) n + -- (2k-7) 2 n g + n -n + j- (wnen k i) 9.7 Uer Bound on Adds Now we would also like a minimal upper bound on Adds that is inde-peadent of p Let g(x x +( z_ x2 + ( n -n + k- ) x. i ~12 Ltgx --- 4 4 4 12 Since g'(x) > 0 on 10,n] we have (p) n n (k -n Thus 12 8 k 1 2 )n.Tu Adds < 3 n + (k + )n - (k - -) n 1 3 32 n + - n2- 2 n (where k l) 2j 56 Chapter 10 : Error Analysis for Diagonal Pivoting 10.1 Introduction Let us attempt to solve A x - b by the diagonal pivoting method. If xc is the solution we obtain from the computer, we may consider x as Lhe exact solution of the system (A + E) z -b As in §2.5, we are interested in showing that the elements of E are small in comparison to the corresponding elements of A (Wilkinson, 1960, 1961, 1963, 19b5). Such an analysis of E is called a back-ward error analysis. 10.2 The Occurrences of Error Suppose that we could perfocm the diagonal pivoting method in exact arithmetic. In order to solve A x - b , we perform the fol-lowing steps (see §9.1): (1) A - M D Mt (the decomposition) (2) c -M' b (the new right hand side) (3) y - D c (solve the lxl and 2x2 systems) (4) x - Mt y (recover the solution) However, in finite precision arithmetic, we have error at each step. Instead of decomposing A into H D Mt , we obtain M and D such that M D Mt. A + F. Instead of calculating M -] b , D- c , Mt y , we actually obtain c - (M + M 1) b , y - (D + 6) 1 C , x -(M + M2 ) - t y for some perturbations Ml , 6 D , M 2 respectively. Thus we actually perform: 57 (1) A + F MDMt (2) cm (M + M) b (3) y (D + 6 D) 1 c (4) x= (M + M 2)- t 10.3 The Error Matrix E Thus we have b - (M+M) c (M + M)(D + 6 D) y -(M + M 1)(D + 6 D)(M + M 2 )t x -(M D Mt + M1 (D + 6 D)(M + M2)t + H [6 D (M + M + D Mt 1) x . but M D Mt a A + F , while (A + E) x - b . Henc, E -F + M1 (D + 6 D)(M + M )t + M [6 D(M + M)t + D M if we can bound the elements of F , M , D , M # M 2 and 6 D 1'2' then we can bound the elements of E . We shall see that most of the error lies in F , in other words, most of the error occurs when obtaining the decomposition H 1 MD of A 10.4 Notation In the following sections, the symbol E,n will stand for any numbers with ICI < 2- in < (1.06) 2- , where t iG the num-ber of binary digits in the computer. Each occurrence of C or n in an equation should be indexed, but we shall suppress these indices for the sake of clarity. and shall denote the summation over those indices k i < k < n , with pivotik] = 1 and 2 respectively. -tt We shall write g(t) - 0 (2 - ) if lim 2 t g(t) is finite. 10.5 Summary of ihe Error Analysis In the following sections we shall show: (10.5.1) IFjI 2t ' 11 + 3.01/a] ( k) + [I + l1.02/(l-a)] (k) for I , and F - Ft (10.5.2) IEi jI s IFij + £ , where £ < 2- t ma) . (k) 2 x k max (1/a2, 1/(-,1l 2 )) Also, for a - (a 0 , we shall show: (10.5.3) IF 1 2' 1 5.71 (k) + 31.6 (k) for i J Ijj j 0 L 0 o 1 (10.5.4) 1E < 2-t max p (k) (23.54)n k 0 n where I I is the one-norm: 1E1 = max 1I j i=1 In Chapters 11 and 12, we shall show for a = a0 : (10.5.5) max IFijI < (15.8)n 2 - t rn f(n) 110 (3.07) nO 4 4 6 iJ (10.5.6) IF < (15.8)n 2 2- t n f(n) 1O (3.07) n0.446 (10.5.7) NEI < (23.54)n 2 2 - t n f(n) V' 0 (3.07)n 0 .446 For Gaussian elimination with complete pivoting (see §2.5): (10.5.8) @0< (2.01)n 2 2- t ,n f(n) 11O I where L U -A + F . For further remarks, see §10.12 and §10.16. 59 10.6 The Decomposition for the Reduced Matrix A(r) Let A(r) be the reduced matrix of order r 1 < r < n . Let a pivotir] (see §9.1). Thus a 1 1 or 2 Let A(r ) be the matrix resulting from deleting the first a B rows and columns from A(r) p(r) C(r) t Lez A (r ) where P (r) is axe c(r) Air) ] s 8 C (r) is (r-s)xs . We shall use P(r) as the sxs pivot. s s Let M(r) be the (r-s)xs matrix resulting from calculating C(r) (P nr) in finite precision. Let G be the (r-s)xs a 9 ) ~(r(r) error matrix : (r) = M(r) _ c(r) P(r)-i Let Ar) be the S S S S reduced matrix of order r-s resulting from calculating A (r) -M (r ) C Wt (r-s) in finite precision. Let G be the (r-s)×(r-s) error matrix C ( r -s ) A (r-s ) A (r) + M (r) c (r)t But M(r) (r)t - M(r) (M(r) -( r)) p(r)]t M(r) p(r) M(r)t Mr) p(r) (r)t Hence A r)+ F (-)-M~r PC Mrt + Ar) ,where _f.S SS S S S FCr-s) -G(r-s) C (r) Wt FG + M a (r) 0 (ra S S S 10.7 The Error Matrix F for the Decomposition t t From §10.6, we have A+FMDM where F F and for i > j Fij = + (M Dt , where for i > J I I 'I 60 = 2 (k) with Gij ij G(n-k) G-k,--) if pivotfn-k+l] - 1 (k) -(n-kl) C(k G(-kl if pivot[n-k+l] - 2 LI. Ui-k-I ,j-k-i 0 if pivot[n-k+11 -0 C- (Q(n-k)) 1 J-k,l if pivotin-k+l] = I and for j > k : 0 jk = (0n-kl))j-k-1, I if pivot~n-k+1] -2 (0( n - k) if pivot[n-k+l] 0 t2 )j-k-l,2 10.8 The Error Matrices 1 6 D, M 2 (M + MH) c = b From §10.6, (Mi n - J ) i if pivot[n-j+l] -1 ( 2 (n-J -I)) Mij ( 2-l if pivotin.-j+l] -2 (M2 )i-J-l,2 if pivot[n-J+l] = 0 for i > j ,with H 0 if pivot[n-j+l] = 2 J+1 ,j while Mj 1 for every j , and Mij - 0 for j > i Thus (Ml)ij = 0 for J > i and (H) = 0 if I ij 1 J+1,j pivot[n-j+1] = 2 (D + 6 D)y = c . From §10.6, D is a block diagonal matrix with blocks of order 1 and 2. The blocks are the pivots P(r) in §10.6, S L 61 s -pivotir] . Also, 6 D has the same block structure as D. (M + H2) x -y .M, has the same structure asM I2 10.9 Floating Point Error Analysis Since most computers now perform calculations In floating point arithmetic rather than in fixed point (see Wilkinson, 1965, pp. 110-188), we shall give an error analysis only in floating point. Then, from Wilkinson (1965), pp. 114-117, we have z = fX(x y) E (x y) (1 + c) , where jej < 2- t , t is the number of binary digits used by the computer, and is any arithmetic operation (+, -, x . Further, we shall assume that the computer can accumulate inner-products in floating point arithmetic. Then Ift (x 1 Y 1 + ... + X Yn) -n (XI yl + ... + xn Yn)I < ) iIy xi yi , where i~ I <3 -2t < (1.06) 2 . As in §10.4, we shall assume n is any number such that Inm < (1.06) 2-2t and we shall suppress indices. From §10.4, we may write Inl < 2-' 0 (2 - t) 10.10 Floating Point Analysis for F Consider Ar) . JA -Let i > j Case 1: s, pivot[r] I M O r and CI are (r-l)xl P A P So r-) (r), -r) C (1 + C)l (I + L) o ij 1 l j 1A _ (1 r) I PiJI (elPj I =I 62 (r-l) -(r)) C(r) (r) (2 + c) So Ei ( c (M 1 )1 (C 1 )(+ (A(r))(r) I < ( r ) 1 -IAi+1,+lLuJli 0 (C 1J)) -0 Cr-I) 2-t -1 -t r) (MI r) A (r) A(r)l (I + E) j+1,1 11 (C r) (r)-l. (1 + ) So (Q r)) -£(C (r) (r)-1 -1 <j (r () t r)r). (r) A Cr)-l I(M r))J P1 + 2- t ) I(C C 1 . 1 ) J A -A 11 -(r)/(r) < /a So 1(0 (r) < 2-/t and I(M (r)) Il< (1 + 2-:/c • Thus (10.10.1) IE-)I 2- t ( r ) + 2/c + 0( 2-t)] , and (1 .1 .2 IM r)p(r) 0 (r)t) I < (r) (1+ 2- t ) 2-t/O = (lOlO2) IM r I 1 "ij 0 1 (r) 2-t [1/a + 0(2-t)] 0 Case 2: s -pivot[r] m 2 ) A121 M2 , 0 2 and C2 are (r-2)x2 P 2j and dj ( (r)>p ) (C) r) 2 > 2 ) ,( ) 2 So A (r-2) C ( r) ( (r) (C (r)) + (1+) '+ -j -(( 2 )ij-( 2 ) i 1 dEi 63 (M~r))i2 ~~ (C ) (I + ) )(1 + ) . From §10.6, we have (r-2) (A(r) -(r) ((r) [ + C)3 1] 1j 2 ) C 2 Il ( 2 )JI () (Cr) (1+ U)2 - 2 12 ( 2 ~J2 (10.10.3) IG(r- 2 )I < 2-t (r) (, + [3 + 0 (2-t)] I(M 2 r) + [2 + 2- t] M (r-)i 2 ) A(r) (r) (r) (r) Ar) A A2 2 (1+ -A 1 A 2 (I + 0) Now M 2 )1 (r) (r) (r)' + IE) 22 A1 1 A 2 1 T(r) A(r) A(r) C(r)(r Cr) (r) (r) A A2 2 (2 + E) - A2 A2r E (2 + E) + A2 ( - A22 Al ( C ii 22i2 21 21 22 11 From §10.5, C)(r) (r) 2 Ct r) 2 t 1(0 2 )i < to 0 () + pr)) 2 -(2 + 2-) + Cr) 2 + W 1I2(r) W(r) 2- t (2 + 2 - t) 2- t p(r) (1+( 2 ) 2 -t -2 (r) _ (r) + O<{ 22 i+ 0 1r)L _ c + +--But (1 + C2 )/l + a) < 1 since O<a< 1 . Thus we have (10.10.4) (O r)) 1I 2 - (3/(l - a) + 0 (2-t)1 Similarly, we obtain (10.10.5) (O ( r) ) < 2- t {3/(l - a) + 0(2-t)) 2 12l -64 106 we recall: M (r) (r) p(r)-l + 0 (r) But Frm 106,wereal: 2 2 I 2 U2 (10.10.6) I(C~r) (.(r)-l) . (r) (r) + (r))/ (r) < 1l a Thus from (10.10.4) - (10.10.6), we have for j -1,2 (10.10.7) I(. r) < (I + 2- t (3 + 2 -t]}/(l a) From (10.10.3) and (10.10.7), we conclude: (r'2) l -t oCr)t (10.10.8) I1G 1 r 2 <1 + 5/(1 - a) + 0(2-t)} o we n (r) (r) (r)t Now we need to bound ( P 02 j From §10.6, we recall: M(r) (r) . C (r) + 0 (r) p(r) 2 2 2 2 From (10.10.4), (10.10.5), and §10.6, we have (10 .0 9) i (r) p (r) < (r) 2-t .. .. 2 2'iJ' < (1 + [3(1 + )/(! -c) + since Cr) p < C, r) From (10.10.4), (10.10.5), and (10.10.9), we conclude: (10.10.10) I(M~r) p(r) 0 r) t) I < + 0 (2 "t)} 2 p(r) 2 Ot/( 1 10.11 Summary of Floating Point Analysis for F From §10.7, (10.10.8), and (10.10.10) we have for i > J (10.11.1) IG I 2t < [1 + 2/C + 0 (2-t)] W p1 k) + [1 + 5/(l - a) + 0 ( 2-t)] Wi k) (10.11.2) I(M D Ot)I 2 t < l/A + 0(2-t) i (k) + 16/11 -a) + 0 "( 2 t 1 ----------------------------V 65 From (10.11.1) and (10.11.2), we conclude for i > j , (k) (10.11.3) IFi 1 I 2t < 11 + 3/a + O( 2 tfl , (k)O [I + 11/(1 - a) + 0 (2 -t) 1 k We shall now asp',me tha. the first 0(2- ) term in (10.11,3) is bounded by 0.01 (which is true fot t > 8 for all a ) and that the second 0(2 - t ) term in (10.11.3) is bounded by 0.02 (which is true for t > 14 when a - (0 a (I + /r7/8)). Under these assumptions, we have: (10.11.4) IF 1 I 2t < [1 + (3.01)/(J 11 (k) + 1 + (11.02)/(1 -) 1 1i]Jk while for a -a 0 , we have: (10.11.5) IF 1 I 2 ' < 5.71 + 3.16 1" w 10.12 Comments on the Bound for F From (10.11.4), (10.11.5), and §10.3, we see that we have a good bound on F (and hence on E ) only if the maximal elements (k) in the reduced matrices.( do not grow too large. From Corollary 3 of 55.7, we have for OL -C (k ) < (2.57)n-k P Thus (10.11.5) does not automatically give a good bound on F if n > t . At the present time, most computers have t 4 50 . Hence (10.11.5) cannot guarantee a good bound on F for systems of order > 50 i 66 According to Wilkinson (1965, p. 215), if L U -A + F , then max IF I < (2.01) n max k) for Gaussian elimination, where iJ k x(k) < 2 n for partial pivoting, while MIx (1) < rn f(n) )0 for complete pivoting. In Chapter 11, we shall show that for the diagonal pivoting method with 0 < a < 1, we have: max ( V<n f(n) P0 c(a) h(n,u) k 0 In Chapter 12, we shall show that for i a0 c(O O ) h(n,a ) < 3.07 (n-1)0 .446 for n > 2 Then from (10.11.5) we have for a -a 0 : IFIa < F -t rn f(n) P (3.07)n0.446 [5. 7 1p + 31.6(n-p)/2] i,j where p is the number of II pivots used. So 0 446 max IF I < 2 ((n) t1 (3.07)n [15.8n 10.0 9 p) i,j < 15.8n 2 - f(n) ii0 (3.07)n 0 4 4 6 which is within a factor 7.9(3.07)n 0 4 4 6 of the bound for max IFi I iJ for Gaussian elimination with complete pivoting. 10.13 Floating Point Analysis for a D The blocks of D are of order 1 or 2 (a) Surpose [Dii] is a block of order 1. Then so is [(6 D)il] We want to solve D i Yi .ci . but we obtain (Dii + (6 D) ii)y i due to the finite precision; pivotin-i+lJ -1 and IDiiI = (n-i+l) 67 Thus yi (c i/Di)(1 + c) and (6 D) -D C so (6 D)iil ( 2 - Ioii . But IDij- P(n-i+I) (n-i+l) Thus D)1( I < 2 t .- i+l) p 0 (b) Suppose DI+ j (n-+l) is a block of order - 2 \|Di~ D.+i+ 2. Then ptvot(n-1+l] = 2 and p(n-i+l) < (n-+l) (n-k+l) (I (1+ri)] (1++ n) V ID ii Di ~+ (+ +I +)-V~, where ni -2 - t 0(2-t) . Since le I< 2- t , we have (nk+) -de (n-i+l) 2- t [1 + 0 (2 -t)] (1 + a 2 ) pjn-i+l) 2 lu~ n -k~ l ) - det r 2 Now yi+I . [Ci+l D1 1 (I + n) - cI Di+lI (1 + I)J(l + 6)/f ( - + : ) and yi= ci D +li+l (1+n) -Ci+l D i+ I ] +) (I )/(±v (nk+1)) Since Icl < 2- t ,In < 2 - t 0 ( 2-t) and (n-i+l) < 0 (n-i+i) 0 after much manipulation, we have: 1(S )j ( D) ii+ I 1 < J 2 -1 + 0(2-t)] (n-i+1) 1(6 D) i 1(6 D) [+il 10.14 Floating Point for M1 and H2 From §10.2, (M + M1)c b and (M + M2) x -y --- - -- -- -- -- --- -- -- -68 From Wilkinson (1965, pp. 247-249), we have: I(M ),, I l(12)il < 2 - t 11 + 0(2"t)] IMii , (Ml)l = 0 = (Mt) 1 j for i < J, (M , it,.,)Il . j IMjl max (1/a, l/(I--)) (1 + 0 ( 2-t)] So for i > j . t, 2-t (10.14.1) (M)i1, M2ijI < max{lla,l/(l-a)}[l+O(2-t)13(n-2+i-j)/2 and I(M 1) 1 , I(M 2)I ' 2 t [I + 0 ( 2-t)] 10.15 Floating Point Bound for E From §§10.3, 10.11, 10.13. and 10.14, we could express Eij in terms of Fij and the elements of M, D, M I, M2, ard 6 D . But this expression is very complicated. n Let us define IAI- max I JAij I where A is nxn. (This is j i=& usually called the one-norm.) Clearly, IAijI < IAI From §10.3, we now have (10.15.1) lE jI L IFijI + E , where c -UM I ID + 6DI I(M Mt)I+ NMI ,,6DI ,(+Mt)l+ ID W1, I 12 2 2 69 IDI 3 . (1 + a) max (k) From §10.13, k 16DI i 2- t (I + ct) max p (k) 11 + 0(2"t)] k 0 FLum 910.10, IMI, IMtI < 1 + (n-1) max {I/a, 1/(1.-a)} [1 + 0(2-)] Let us assume 0(2- t) (1 + max {1/a. 1(1-ci)} 3(n + 2)) -0(2- ') i.e. let us assume n 2 2- t << 1 Then t-IM 1, IMtl < 2- 1 + 0(2-')]. 1' 2-Now we can bound c . Let 8 1 + (n-1) max {1/a, 1/(-a)) (10.15.2) c< 2- t (I + a) max (k) B(2 + a) [1 + 0(2-t)] k 0 For a -,O we have 8 < 2.781n , so ax1 (k) (10.15.3) c < 2 max k ) (7.734)n2 11 + 0 (2 -t)] k From (10.11.4), (10.15.1), and (10.15.2) we conclude: ,(k) (k) (10.)5.4) IJEI 2t < [1 + 3.0/a] Yj PO + 13 + 11.02/(1-a)] t ( o + (!+ a) max p B(2 + 1)[1 + 0(2 k From (IG.II.5. (10.15,1), and (10.15.3) we conclude for a 0 (10.15.5) 1E 1 I < 2 max p 15.80n + 7.74 nrt k Since .E\| ,i Irl + c , from (10.11.5) and (0.15.3): (10.15.6) 1E1 < (23.54)n2 ar (k) k ICi I \| It 70 10.16 Comments on the Bound for E From §10.12 and §10.15, we need to bound tile P(k) in order to have a good bound or E In Chapter 11 we shall show that max p(k) < r f(n) c 0 C(a) h(nCL) k 0 0 for the diagonal pivoting method for 0 < a < 1 In Chapter 12 we Shall silow that c(gO ) h(n'aO ) < 3.07(n-1)0 .446 Then we would have I <i: , (23.54)n 2 2 - t , n 0 (3.C7)(n-1) 0.446 For Gaussi3n elimination, in order to solve A x -b , in finite precision we actually perform: (1) L U - A + F (2) c- (L + SL) -1 b (3) x= (U+6u) c. From Wilkinson (1965, p. 215, pp. 248-252), we have (cf. §2.5): 2On max I1 and max FI 2.01n 2 - t max k) 1E1 < 2.01n 2 2- t max P(k) [1 + 0(2"t)1 k 0 For complete pivoting, max 1 O(k) < n f(n) p (Wilkinson, 1961). k 0 0 Thus our bound on 1El for diagonal pivoting differs by a factor 36(n-i)0 .446 from the bound on IE1 for Gaussian elimination with complete pivoting. I 71 Chapter U1 An A Posteriori Bound on Element Growth fuo 0 < a 1 11.1 Introduct ion We saw in Chapter 10 that the bouna on the elements of the error matrix depends on the maximum of the elementb of the reduced matvices. Let A A ( n ) be the original symnetric matrix of order n wi1h (k) det A # 0 . Let A' be the reduced matrix of order k We saw in §8.4 that max A ( k A <( 2 5 7)n-k max Aij But we ije : , "i,) shall show in §§11.2-11.7 that we cau get a much better bound by the use of Wilkinson's techniques for Gaussian elimination with complete pivoting (Wilkinson, 1961; pp. 281-285). His proof depended or, the fat that the pivots in Gaussian eli-mination with complete pivoting were maximal elements in the reduced matrices. Our pivoLs are rot necessarily maximal element;, but they are closely related to the maximal elements in the reduced matrices. We shall assume that we use strategy Sa (§5,7) for any cc (k) (k) 2 (A) 0 -I for all A (see Lemma 4, §5.5). In particular, this lower hound holds for the strate-Sies in §6.1, §6.4, .. d §8.1. 11.2 The Pivots Let A( k ) be the reduced matrix of order k -----72 Let pk) mx (A and (k) )A ij ai i • i,j i Let us assume that, whenever we shall use a 2x2, interchanges have ensured that v(k) .A(k) A(k) - A(k) 2 1 >(k) 2 (k) 2 .nsr11 22 21 -1 (In parcicular, thiF assumption holds for the strategies in §6.1-6.2, §§6.4-6.5, and §§8.1-.8.2.) From strategy S. (§5.7) and §9.1, we recall: Wk (-K i.e. i A (k) I if <if A uses a lXl pivot W ) (k) pivot(] 2O , i.e. if A k ) uses a 2x2 pivot 1 (k+l) ie. if A(k+ l ) (-0 if V j l , i.e. if A uses a 2x2 pivot. Let us now define (0c) if pivotik] -1 /- V( k ) if pivot~k] -2 Pi p. " i if pivottkl- 2 " -+7 if pivotLk] - 0 The pk will be called pivots. From the decomposition A - H D Mt we see that Idet Al -p1 .. p and Idet A(k) I -p.. .pN , ince det M - 1 11.3 hadamard's Inegality By idamard's Inequality (Gantmachcr, pp. 253-254), nI n ~ (n)2 2 1 ~ /2 nZn/2 (n).n (11.3.1) Idet-Al!( < { 11 Ai <(nv - (n o jj 0 iml j-I since p a x 140 'A€ Also (11.3.2) ldet A kk)I ~ (k))k I 73 11.4 Bounding det A(k) (1) If pivot[k] = 1 p (k) > (k) So o W (11.4.1) p1 " k det A(k), < (/P(k)k (Pk iv/a)k (2) If pivotik] - 2 v:(k) (k) If we assume , (k) > P(k) 2 -1 i(k) for all A (k) then (k) _2 (k) 2 Thus p ( k )2 < V(k)(l c 2) a -c) p 0 kV 1( (11.4.2) p1 .' Pk Idet A(k) . (k .kI(1 -c) )k 11.5 Fundamental Inelit From §11.4 (and later in §11.5), we are led to the following; Definition: SI/C, 2 if pivot[k] -1 (11.5.1) k 1/(l-ca2 ) if pivot[k] - 2 I k+l 1+1/k I if pivot[k] - 0 From (11.4.1) and (11.4.2) we have: (11.5.2) Pl "' Pk ( Pk ) k if pivot[k] # 0, 1 < k < n We would like to have a similar equation for pivot[k) = 0 for ouv" analysis in §11.6. If pivot[k] - 0 , then pivot[k+l] -2 . By (11.5.2), ,k+1. P1 Pk Pk+1 +l 6k+1 k+1 , where 8k+l 1 (I - ci and 1 k+l 1+1/k P P Since k l- by (11.5.1), I 74 P1 "'" Pk < [(k+l) 8k+l ] (k+l)/2 k ( -P ) k " Thus (11.5.3) P Pkk ) k for all k , 1< k < n 11.6 Bounding Pivot Growth Define qk = log Pk . From (11.5.3), k-- 1 (11.6.1) 1 qi < (k-) qk + 2k log (k ) k Ji Since Idet A (n) = p1 ... pn , we have: (n , n (11.6.2) log Idet A -I q i-i Dividing (1l.b.l" by k(k-1) for 2 < k < n-i and (11.6.2) by n-i and adding we have: q + q2 /2 + q3 /3 + ... qn- 2 /(n-2) + qn-1/(n-1) + qn/(n-1) nI log Idet + log (k )/(k-) k=2 + q 2/2 + q3/3 + ... + q n 1i(n - i) n1 after observing that ---I --rink r(r-l) =k-i n-i 1 n-l 1k~/(k-1) I (0) (11.6.3) q + qn/(n-l) < n lo i (k +- log Idet A k=2 From (11.5.3), Idet A(n)I < (Y/r-T- P n We define r (11.6.4) f(r) R( I ki/(k- )l 1/2 k-2 r (11.6.5) h(r,a){ 11 ak 1/(k-1)} 1 / 2 k-2 75 From (11.6.3) we now have q, + qn/(n-l) < log f(n-1) + log h(n-l,a) + n log (n n) + n q /(n-1) n2(n-1) n n With simple manipulation we have I -q < log f(n-l) + 2n- log n + log h(n-l,a) 1 1 + 2(n--l) log 8n + j log (n 8) 1 log f(n) + log h(n,cL) +1 log (n n) n Thus we conclude (11.6.6) p/P n < An f(n) IX h(n,t) Similarly, we have for 1 < k < n (11.6.7) AkPn < -k + 1 f(n - k + 1) '- l h(n- k + l,a) Pk/pnn-k+l < ( n-k+l (11.6.6) and (11.6.7) hold for all OL , 0 < a < 1 , under strategy S (§5.7) provided that v ( k) > P(k)2 k ) 2 A(k ) a _ -) for all (§55) We now have a bound on pivot growth. But we are interested in bounding element growth (§11.1). If A finishes with a 2x2 pivot we (2) are interested in bounding p2) 0 while if A finishes with a lxl pivot, we must bound (1) 11.7 Boui.dxn- 7lement Growth We shall now express VPI' P' "n Pn' and a Pn in terms of (1) (2) (n ) . .. P 0 and p .. 76 (n) u(n) (n) Let A-A ,U 0 0 U U ,W n V V III if pivotin] -I Since pivot(n] , 0 , pn i , and r if pivot[n] -2 i/a2 if pivot[n] -1 8n L1/ 1 -cL 2 ) if pivot[n] -2 But 1 < 0 always. If pivot[n] -2, then V1 < L0 , and (by _5.5) -< + _ 0- < /I -t 2 p0" Thus F0 if pivot[n] - I Pn A ' 0 if pivot[n] - 2 1/a 2 if pivot[n-i] - 1 NOw -(1 -Bn-) if pivot[n-i] -2 n-- (n On)1/(- if pivotin-I] - 0 Let us define: 1 __ _ V(1I__)I (_a } (11.7.1) rea -1x Thus we have: (11.7.2) n <m (a) V0 , and M(C) 1o if pivot[n] a 1 (11.7.3) VW9T Pn < n-_n,11 (n -i)r7-") ,+c(I-at if pivot n] -2 From (11.6.4) and (11.6.5), we have (11.7.4) (/;-)i/(nl) f(n-1) h(n-l,t) - f(n) h(n-l,x) n 77 We have two cases (§11.6) as to whether: (a) the last pivot is lxi, i.e. pivot[l] - 1 , (b) the last pivot is 2x2, i.e. pivot -2 Case (a): Let pivot[i] - 1 . Then p " () . (1) From (11.6.6) and (11.7.2) we have (11.7.5) 1.1(1) " < r f(n) p m(a) h(n,a) Case (b) : Let: pivot a 2 . Then P2 " and (2) (2) (2) (2) (2)' 41 e 1 Since we have assued v > V(2 ) > (1 Cta 2 ) P(2)'2 Hence (11.7.6) 0(2 ) -r P21V I From (11.6.7) with k -2 and from (11.7.6) we have (11.7.7) P(2) < n- f(n-1) h(n-l,ct) Y p Now we have two cases as to whether the first pivot is (i) lxl or (ii) 2x2. (i) Let pivot[n] - 1 . From (11.7.3) and (11.7.), Pa(2 < rnl f (n-1) h(n-l,ot) m(at) p0/ a" 2 But f(n-1) < f(n) and h(n-l,a) < h(n,a) . Thus (2) (11.7.8) P 0 < vn f(n) UO m(a) h(n,t)/ '-a (ii) Let pivot[n] -2 . From (11.7.3), (11.7.4), and (11.7.7), 0 (2) < ~n f (n) P h (n, a) A772( a 2) 0I 0 F _ __ __ I 78 But 2+'/(l -a 2) < m(a)/T- from (11.7,1). Hence (11.7.9) P(2) < /rn f(n) pO me(a) h(n,a)/,r-Let us now define: (11.7.10) c(a) -m(a) i o1 if pivot -2 where m(a) -max il/a, (la)/(1-a') } Then we conclude for 0 < a < 1 f pivot 2 : 2~ u ) 0 Similarly, we have for each reduced matrix A(k) (11.7.12) w (k) < VW- k+j f(n -k+j) )0 c(k,a) h(n -k+j, a) 00 1if j I where j -pivot[k] , c(k,a) -(a) x i/I 2 if j =2 and 0 < a < I Hence, for all A , we have for 0 < a < 1 (11.7.13) (k) < vr f(n) c(a) h(n,a) 11.8 Comments on the Bound The bound in (11.7.13) holds under strategy S , 0 < a < 1 , and under Lite assumption V (k) -for all A In parti-cular, (11.7.13) holds for the unequilibrated diagonal pivoting stra-tegy (9§8.1-8.2), the complete strategy (9§6.1-6.3), and the partial I 79 equilibrated strategy (9§6.4-6.5). (For the partial equilibrated stra-tegy, we assume that, given a reduced matrix A(k) with Vk) -max JA(k)a I 'Jj we equilibrated A (k) so that the maximal element in absolute value in eacit row is pok) ) Wilkinson (1961) obtains /n t(n) vO as the bound on the elements In the reduced matrices for solving A x - b , J 0 . max IAij I , by i,j Gaussian elimination with couplete pivoting. We here the extra factor c(a) h(n,) since our pivots are not necessarily maximal elements of the reduced matrices. We call the bounds in (11.7.11) - (11.7.13) a posteriori, since we cannot calculate the 8k terms of h(n,a) until we know the pnsi-tion of the blocks of order 1 and 2 in D for the decomposition A -M D Mt In Chapter 12, we shall give a bound on c(a) h(n,a) independent of the structure of D , for the value a -0 = (1 + Y17)/8 (§5.7). 11.9 Smaller Bound on Pivot Growth If H is an nxn positive-definite Hermitian matrix and Xmi n is the minimum eigenvalue of A , then (Shisha, p. 173): n n-2 (11.8.1) detR< 1 H H -(nX min • - i- ii mai n 11 i2) i-1 n~j>>I> 1 If A is an nxn non-singular real symmetric matrix and X is the minimum of the absolute values of the eigenvalues of A , then, setting H -At A in the above, we have 80 n nk 2. X2 0-2) A 2 (11.8.2) Idet Al' H ( 1 Ai) -t AI -~ 1=1 j n j>~l1 th where A is the j column of A Then Idet Al2 < (n tj2 ) C(A) , where 2(n-2) 2 7 2n (11.8.3) c(A) - 1 - X lAt A I /(n P Using an analysis similar to that in §§11.2-11.7, we obtain: if pivotil] -I 1 ) (11.8.4) < /n f(n) p0 c(a) h(n,a) T(A) {If pivot[Zi - 2 : P(2) 1 1 n (r))r(r-where T(A)2 -(A) n - £(AI and--r-2 ECA 2 "(r )2 r A(r) C(A(r) ) -2(r-2) L A r)t A(r)12J / [r V , where A -1 r 'i r t j -is the reduced matrix of order r (A = A ( n ) ) , X is the minimu- of (r) (r) the absolute values of the eigenvalues of A , and A is the th A(r) j column of A If det A # 0, then IrI > 0 for each A(r) If A(r) not Ladamard (see §12.6), then c(A(r)) < 1 . If A(r) is Hadamard, then A(r) will use a 1×i pivot and A (r- 1 ) will not be Hadamard (see Appendix A). Thus '(A) < 1 if det A # 0 . Hence (11.8.4) gives a lower bound than does (11.7.11), but (11.8.4) is so complicated we are unable to use it to advantage, and we present it merely for its academic interest. 81 Similarly, for Gaussian elimination with complete pivoting, we can obtain r- f (n) .0 T(A) as the bound on the elements in all the reduced matrices when solving A x - b , l0 . max lAijI , det A 0 0 ii If we replace te Xr in T(A) by 10 r I the singular value of A r)o minimum modulus. of_ 82 Chapter 12 An A Priori Round on Element Growth for a - a0 = (1 + Vi7)8 12.1 Introduction In (11.7.13) we obtained rn f(n) PO c(a) h(n,a) as a bound on the element growth in the reduced matrices, for any a with 0 < a < 1 From (11.5.1), (11.6.4), (11.6.5), (11.7.1), and (11.7.10) we recall; if pivot!k] -1 ak = 1/( -E2) if pivot~k] -2 ( k 2 / if pivot(kJ -0 f kl/(k-i) n 1/(k-i) H~n k i h (n, )2 11 H k , k-2 k-2 I if pivot[l] = 1 1 1 Me(a) = max --- } , and c(a) -mn(a) X p 1-ap i1 if pivot 12) 2 The term c(a) h(n,a) arose from the fact that our pivots are not necessarily the maximal elements of the reduced matrices, but can be expressed as multiples (involving a) of such maximal elements. The bound in (11.7.13) is a posteriori since we cannot calculate c(a) h(n,a) until we know the pivot selection, i.e. until we know the position of the blocks of order I and 2 '1 the block diagonal matrix D for our decomposition A "M D M t for a given value of a 0a < I. 83 We would like an a priori bound on the element growth, i.e. a bound independent of the selection of a lxl or a 2K2 pivot at each stage. We would also like the a priori bound to hold for all a 0 < a < 1. But no such bound exists, as we shall now show. Let p be the number of lxl pivotu for the deccmpc'sition (§9.1), i.e. there are p blocks of order I in D above. If p - n (e.g. for a positive definite matrix, see Apptndlx A), then h~n) -n l/ (k-1) I h(n,T) = (1/a) -w as a-1 0 . fp -0 (see §6.3), then k-2 n h(n,a) > H (i/il- 2)I(kl) w as a -I k-2 Thus we shall give art a priori bound for element growth only for the value c -a0 " (1 + vr17")/ (§5.7). 12.2 Lower Bound on c(a) h(na) for 0 < at < 1 We shall now find a lower bound on c(a) h(n,) for all , 0 < a < 1 , in order to show that the upper bound for a - a 0 that we obtain in §12.4 is a reasonable bound, i.e. that some other choice of a would not provide us with a much better upper bound on c(a) h(n,a) Since min c(a) h(n,a) > min (a) x sin h(n,a) , we need O<CL<1 0<,<l O<a< 1 only find lower bounds for the latter two minima. Lemma 1: min c(a) > 2.029 . O<OL 2.029 . q.e.d. We shall later need the following: n+l n n1 --l~ ( ) < k 1 -for n > m >1. Lemma 2: < L (r= k-m+l( k-m k 1 <k+l 1 Proof: By elementary calculus, k+- < k -< dx< . Thus <~~~~~ xn~ kx, o u-L I <f~ 1 1~ +1- <x - lo < for n,m> 1. q.e.d. Pow we cau find . lower bound for min h(n,a). Lemma 3: min h(n,C) > nlog > n0.3465 n Proof- Let o(a) - Y' 1/(k-1) . If p n , then k-2 h(n,) -(W/a)w( n) If p 0 then h(n,a) - (.// -2) w ( n) t(n) where~~~ ~ ~ ?(J_ ) 1,1 2 2 (2j1]l(j2 where L-N) 11 2j-1 Thus h(nCA) > b(,)w(n) , where b '%) - max fl/a, 11lyi-} . BuL min b(a) -V2 , and i attained by a - liV2 . From Lemma 2, O log(n) . hence min h(n.cz) > n n > n r01< < q.e.d. From Lemnas 1 and 3 we obtain a lower boun d for c(a) h(n,a) for 0 < a < 1 Theoram 1: ain c(a) h(n,a) > (2.029)n 0 . 3 4 6 0<a< 1L 85 In §12.4 we chall show that c(10) h(n,a0) < 3.07(n-1)0.446 Thus our choice of Ct O0 (95.7) will provide us with an upper bound on C() h(n,co) which does not differ much from the minimum of c(,t) h(n,a) for 0 < a < 1 12.3 Remarks on sn Upper Bound for h(na) We cannot evaluate the 0k until we knaw the pivotal selection, but if ue could bound all the Bk independent of the pivotcl selection, then we can obtain an upper bound. Theorem 2: If 8 k < c for all k > r ,where c > 0 and r > 3 is independent of n, then for 0 < a < I h(n, )ih(r--1,. ) < ¥(n-]. g where y{-(-2 " ~ n lg n-I Proof: From Lema 2 of §12.2, 1 1/(k-l) < log ( -k-r>3 1 n-1 Thus h(n,c)/h(r-l,Ci) < { i c1/ (C -I ) /2 < c k-r But xlogy = y.og x ior x,y > 0 . q.e.d. This shows that if we can bound 8k independent of the pivotal strategy for k > ' atud if we can consider the worst possible cases of pivot selection for k -2,...,r-1 and bound h(r-l,a) , then we have a bound for h(n,ci) . (In order to consider the cases for 2 <~ k < r-l , we must have r reasonably small.) In §12.4 we shall do this for a - a0 , and we shall have r -5 I . -- -. ------.-------.- ---------- -.-- ~- --86 12.4 An A Priori Bound for a a (1 = ( + Y7)/8 For a -aO max ( ras, + a;) 7(1 !/a Thus, for ri if pivotil) - 1 (12.2.1) c(a) d(a)/a0 where d(a) LI I/ a 2 if plvoti2] -2 Now we shall show that for a a0 k is maximal for lxI pivots for k > 5 . Recall §9.1 or §11.2 for the definition of pivotik] Lemma 4: For a - a 0 and k > 5 , if pivot[k] -0 , then B1/(k-l) l/k < )/(k-1) + 1/k k 6k+1 -Proof: From (11.5.1), if pivot[k] -0 , then for 0 < at < I /k+l 1/ - a2 ) and Ik k+1 1+1/k Hence 81/(k-l) 1/k -(/a2)i/(k-l) + I/k f (k+l)k+l/kk < a2 (i/a2 -1)2k kc k+1 iff (k+l) log (k+l) - k log k < log a2 + 2 k log (1/a -1) iff g(k,a) > 0 for k > k 0 for some integer k0 > 1 where g(x,a) - 2 x log (1/C&2 -1) + log a2 _ (x+l) log (x+l) + x log x -g(x,a) - 2 log (1/a 2 -1) -log (I + /x) > 0 ax iff x > l/[(1/C 2 -1)2 - i] In order to evaluate this inequality, we now fix at a0 . Let G(x) = g(x, C) 0 r 87 Thus G'(x) -x g(x'0 > 0 iff x > 0.94 . So G(x) is a monotone increasing function for x > 0.94. Now G(4) - -0.476 and G(5) - 3.698. Thus g(k, 0 ) > 0 for k > 5 q.e.d. Now we can apply Theorem 2 of §12.3 and Lemma 4 to obtain the [ following two necessary lemas. log a -log a Lemma 5: If pivot 0 2 , then h(n,ao)/h(4,aO) < 3 (n-1) 0 0k < 0.613 (n-i) 0 . 4 46 Proof: By Theorem 2 of §12.2 and Lemma 4, h(n,a0 )/h(4,±) L log a. 0 -log a0 0.446 < 3 (n-i) < 0.613 (n-) . q.e.d. Lemma 6: If pivoti5l -2 , then h(n,ao)/h(3, O) < 0.717 (n-i) 0 . 446 Proof: If pivot(5] = 2 , then pivot 0 2 , so by Theorem 2 of log (0 -log a0 §12.2 and Lemma 4, h(n,ci)/h(5,a) < 4 (n-l) Since pivot -2, / 1/3 1 /4 1/2 (4 85 00 oga 1 1 4g 0 ( 3 a4 4 1/2 < 0.717 q.e.d. 1 Now we need only to bound (2 1/2 for pivot[51, 2 and 1 1 2 3 1/2 {82 83 84 } for pivot # 2 (iL0)3/2 if pivotf2=I Lemma 7: If pivot -1 , then h(3,aO) -L 0(i.c) if pivot[2l 2 < 1.96. kI [ 88 Proof: Since pivot -1 , 0 r T and pivot 12] 0 0 0 -if pivot -1 So a 2 , and h(3,aO) < 1.96. 1-' if pivot -2q.e.d. 02 Lemma 8: If pivo13] 2 , then h(3,a) < 2.74. 3/2 Proof: Since pivot -2 , S3 -and 2 2 i q.e.d. Lemma 9: If pivot -0 , then 1/3 if pivot = 1 h (4,aO t0 x/ r 0 1 if pivot - 2 4/3 '14 Proof: Since pivot 0 , a4 ' 9 a = ) 1 0 0 and pivot 0 0 . So B2 . q.e.d. 1 if pivot = 2 Now we put Lemmas 5-9 together to get a bound on c(a0) h(n,c 0 ) which is independent of the pivotal selection. 89 0.446 Theorem 3: c(ca) h(na O ) < 3.07 (n-1) ' 3 rn for n > 2 Proof: From (12.2.1), c( 4 )- d(%o)/ 0 1 if pivo:[il -1 d ( 0 )" l//' ;- if pivoci2] -2 Now we must consider various situations. Case I: pivot[51 -2 : Then pivot 0 . By Lemma 6, 0.446 h(noa 0)/h(3,aO) < 0.717 (n-i) (a) If pivot -1: Then, by Lemma 7, h(3, O ) < 1.96 Since d( O) < A-0 , c(Q0) h(n,co ) < 2.39 (n-1)0.446 (b) if pivot 2: Then pivot[l] f1 . So d(c 0) -1 . By Lemma 8, h(3,.) < 2.74 0.446 So c(O 0 ) h(n.o 0 ) < 3.07 (n-i) Case II: pivot 0 2 : Then pivot # 0 . By Lemma 5, h(n,CO)/h(4,ot0) < 0.613 (n-1) (a) If pivot[ 4 ) -2; By Lemma 9, d(a0) h(4,ct 0) 2.58 . Thus c(a O) h(n,c 0) < 2.48 (n-1)0.446 (b) If pivot -1 Then 4- 1/a' and pivot 0 --1 90 (i) If pivoti3J = 2 Then pivotil' I , so d(aO) -1 . By Lemma 8, h(3,ca O ) < 2.74. Thus c(%o)h(n,a) < 16 0446 0446 (1/(X0)(2.74) /4 (0.613)(n-1) < 3.04 (n-1) (ii) If pivot 1 1 (1/Co)3/2 if pivot I By Lemma 7, h(3,c O ) -01 1 if pivot -2 We must use this form of Lemma 7 in order to prove the theorem. Once again we have two cases. (A) If pivot -1 Then pivotl] - 1 , and d(a) 0 1 . So c(c O h(n,a ) < (1/a017/6 (0.613)(n-1)0.446 0.446 2.17 (n-1) (B) If pivot -2: Then d(a 0 ) 1// --. So c( O h(n,CO < 00 c0% hnc 0 11/6 0.446 < .6(-)0.446 (1/0o1 l/ [I/( - a')] (0 . 613) (n-1)0"4 < 2.36 (n-1)O ' 4 Tius, in all cases, we have c(aO ) h(n,a0 ) < 3.07 (n-1) 0.446 for n > 2 Now 3.07 n0 . 446 < 3Vn/ for n > 2 q.e.d. A -1 91 12.5 Bound on Element Growth From §11.7 end §12.5, we see that the elements in all the reduced -)0.4 matrices are bounded by v'i-f(n) p10 (3.07)(n-) 46, where -max lA Ii , under strategy S with a -ao . (1 + r/17)18 for the pivotal strategies described in §6.1, §6.4, and particularly M8.. For Gaussian elimination with complete pivoting (Wilkinson, 1961; pp. 281-285). the bound on element growth is rn f(n) p.1 Thus, for a - a0 I our bound for diagonal pivoting is within a 0.446 factor 3.07(n-1) of Wilkinson's bound for Gaussian elimination with complete pivoting. 12.6 Conjecture for Gaussian Elimination It is conjectured that the best possible bound ;.s n for real matrices under Gaussian elimination with complete pivoting (fryer, p. 343). The conjecture is false for complex matrices (Tornheicr, 1965). For real matrices, the best possible bound is n for n =1,2,4 and is 2-1/4 for n-3. A matrix R is a Hadamard matrix if IJ I -1 and tha rows of H are orthogonal. Then the order of H is 2 or is divisible by 4 (Davis, p. 327). Under Gaussian elimination, 111(1)1 > n (Cryer, p, 343). Thus, a Hada-Ard matrix of order n has element growth of at least n r --92 Fl 1 If H is generated by tensor products of L J(Davis, p. 326), and if the order of H is n -2 k , hn 1111-n -2k 12.7 Conlecture for.Diagonal- Pivoin If Ai Is as in the previous paragraph, then It is symmetric and the diagonal pivoting method uses a lxi pivot at each step under stra-tegy S OLfor 0 < a <1 , and 1H11l -mn=m2 We conjecture that the optimal bound for diagonal pivotin~g is of the form n q(00) . We need a function q((%) >. 1 since L 1 has a + 1/a as its bound on element growth, where 0 < a < < Thus, for n -2 ,q(cL) -(a + l/aL)/2 , and q~CL0 1.10 Or, we could conjecture a best possible bound of the form n q(n,aL) .Then q(2,ux) -(a + 1/a)/2 and q(2,a0) 1.10 12.8 The Optimal Choice of a The optimal choice of a for 0 < a < 1 is the value which mini-mizes q(aL) in §12.7 for all n . Or, we could seek a sequence of a such that aL minimizes q(n,a) in §12.7. But we do not know q(aL) n or q(n,a) for n >2 We could choose a to minimize c(a) h(n,a) in (11.7.11). But Ai f(n) Pc(at) h(n,ax) is merely a bound on the element growth and is, by no means, the best possible bound. iI F / _ 93 Since c( 0 ) h(n O ) < 3.07 (n-1)0.446 (§12.4), while 0; 0 0.3465 min c(a) h(n,) > (2.029) n (§12.2), our choice of a -0 gives us a bound (ind -pendent of the pivotal selcctioa) for c(ctO ) h(n,(% O ) which does not differ much from a lower bound for min c(a) h(n,a) We assert, further, that q(ao) is not much greater 0<< 1 than min q(cx) . O<OL<I We note that inf q(2,) -1 -q(2,1) But this implies that -1 , so we would use a lXl iff one of the diagonal elements were maximal, i.e. if WI < 0 we must use a 2x2 and thus for n - 2 no decomposition would be performed, which is to be expected since the minimal element growth occurs when we do no decomposition at all. Thus inf q(2,ci) - 1 = q(2,1) does not provide us with any information for th the general n order case if we require a to minimize q(n,) n 94 Chapter 13 Iterative Improvement 13.1 The Approximate Solution Let us assume now that we have obtained an approximate solution z to the system Ax-b , A , det A 0 , by the method of diagonal pivoting. The approximate solution z to A x -b can be considered the exact solution to the system (A + E) z -b In exact aritunetic, the method of diagonal pivoting would per-form the following steps (see §10.2): (1) A =M D Mt (2) c= M 1 b (3) y f 1 c (4) x =M - t y However, in finite precision we have error at each step. For some error matrices F , M1 , D , and M 2 , we actually perform (see §10.2): (1) A + F - M D Mt (2) c- (M + M1) - I b (3) y- (D + 6 D) -I c (4) x ' (M + M2) - t y From §10.3, we see that z is the exact solution to (A + E)z -b where E - F + M 1 (D + 6 D)(M + M 2 ) t + M J6 D (M + M2)t + D M ] From (10.15.6), we have for = 0 (0 0 is the one-norm): a 95 iEi < (23.54) n 2-t (k) k 0 In Chapters 11 and 12 we have shown for a = a 0 : max p(k) < n f(n) o (3.07)0(n-1) 0 .46 k 00 where 110 . max IA jI . Hence, for a " 0 ij \|E < (72.3) n 2.9446 f(n) -t 13.2 The Iteration Let x, z . For m -1,2,... we obtain an improved solution x to A x -b by the following (1) r - b -A x m m (2) (A + E) d -r mm (3) xm+1 x +d We shall assume that (1) and (3) are done exactly. This is a reasonable assumption if accumulated inner products are used (see Wilkinson (1965), pp. 116-117). Thus we shall assume the only error occurred when we tried to solve A d - r but performed (2) (A + E) d -r instead. m m m m The iteration is meaningful only if the matrix A can be repre-sented exactly in the computer, i.e. if the elements of A are known exactly and no round-off occurs when the numbers are read into the computer. i i --96 13.3 Convergence of the Iteration We must now show that the x defined in §13.2 converge to the m solution xA b b s L Theorem: Let x A - 1 b Assume IA- 1 El a < 1/2 Let r b - A xj , (A + ) d=r , and x x d Then lira Ix -xl 0 Proof: (A + E) d r 1 b-Ax . So (A + E)(x -x Ax-Ax . Thus (A + E)(x -x) =E(x -x) . Since det A 0, (I + A I E)(x -x) - A E(x -x) rn r Let a IA - 1 El < 1/2 . Let T = a/(1 - a) So T < 1 Ix -xl <Ax -xl IA - 1 EII(1- IA- 1 El) T IX xl --- \| -1 -rn-TIx _ < 1 m- 1 Ix, - xl . Since T < 1 , lim Ix -x -0 rq.e.d. Thus the iterative vectors converge to the solution x -A b provided that IA-1 El < 1/2 . and the convergence in monotone in the norm, i.e. Ix.+ 1 -xl < Ix - xl. Now we must give conditions under which IA -1 El < 1/2 . We shall n n use the one-norm: 10 1 Ixii and 1A1- max ! IA 1 jl i-i I i-I 2,9446 t<1/,whr Corollary: If (72.3) n f(n) K(A) 2 < 1/2 , where K(A) A1 IA-1 1 ,then lim Ix - xl 0 where x - A- b m 97 Proof: From §13.1, \|El < (72.3) n f(n) 0 But po< IAI , so IA\|1 1 p0 < K(A) . Hence \|A- 1 El < (72.3) n2. 944 6 f(n) K(A) 2- t ard the corollary follows from the previous theorem. q.e.d. We see that the convergence of the iterates depends on the condi-tion number of A , and so we cannot expect iterative improvement to be of value for ill-conditioned matrices. For further remarks on condition numbers, see Wilkinson (1965); and on iterative improvement, see Fox (pp. 49-53, 109-113) and Moler (pp. 316-321). As we noted in §13.2 the iteration is meaningful only if no round-off occurred while reading A into the computer and if the elements of A are exact. I 98 Chapter 14 Symmetric Band Hatrices 14.1 Gaussian Elimination for Band Matrices Let A be an nxn non-singular band matrix with band widrh 2m+l << n , i.e. Aij - 0 f li-Jl > m We could store A in (2m+l)n locations rather than n2 loca-tions by ignoring Aij for li-Ji > m . If we could preserve the band structure while solving A x -b , then we would save storage and thus be able to solve band matrices of very large order in rela-tively few storage locations. If we use Gaussian elimination with complete pivoting, then we must interchange to bring the maximal element to the leading diagonal position. This could destroy the band structure and we would need n locations to store L and U in the decomposition. If we use Gaussian elimination with partial pivoting, then L is unit lower triangular with L = 0 if ji-Ji > m+l and U is upper triangular with U l 0 if li-jl > 2m+l . Thus L can be stored in mn locations and U in (2m+l)n locations. Since L and part of U can be written over A , we would nced only mn additional storage locations. Thus if A is an nxn band matrix with band width 2ar+l and if b is a vector of length n , then Gaussian elimination with partial pivoting requires only (3m+2)n storage locations to solve A x - b . Furthermore, this method requires only - (2m 2 + 4m + l)n multiplications and additions. 99 14.2 Diagonal Pivoting for Symmetric Band Matrices Let A be an nxn symmetric non-singular band matrix with band width 2m+l . If we use diagonal pivoting (see Chapters 5, 6 and 8) to solve A x - b , then interchanges can destroy the band structure I 2 and we would need 2 storage locations. We have investigated many variations of the diagonal pivoting method for the symmetric band case, but these algorithms have either been unstable or have required more storage and operations than Gaussian elimination with partial pivoting. At the present time we recommend Gaussian elimination with partial pivoting rather than the diagonal pivoting method (see Chapters 5 and 8) for symmetric band matrices, and thus we are unable to take advan-tage of the symmetry. (For the special case of symmetric tridiagonal matrices, see §14.3.) 14.3 Symmetric TridjgAonal Matrices We are able to present a stable algorithm for the symmetric tri-diagonal case which requires less storage than does Gaussian elimina-tion with partial pivoting. Let A be an nxn synmetric non-singular trLdiagonal matrix, i.e. Aij -0 if li-Jil > 1 . Let A -ai for 1 < i < n and A ii+i -bi Ai+li for 1 < i < n-1. Asume Ial < c ,while max ' a i l x Ibmil < a 2<i<n l<i . If Jail _ 2 r b , then a, # 0 and we shall use a, as a Ixl pivot. Then M 2 1 b/a A a2 - M1 , and A11 1 1 -2 -K 2 1 b A (n-1) = A Thus A (n-) is tridiagonal, IM 211 _ 2 r/ lb I ii i+1,j+1 1 S(n-) + 2 r , while A(n-l) < 8 otherwiee. If la 1 I < 2 -r b2 , then we shall use as a 2x2 II b1 a2 pivot. (Note that we make no interchanges.) Then 2 1 >b -, a2 > b2 (1 -2 r a > b2 (1 - 2-r 8) > 0 Ia I 2 1 1 1 1 21) b1 since 2 -r 8 < 1 by assumption. Now M 3 1 b b 2 /(a a 2 -b2) , M 3 2 -a 1 b 2 /(a 1 a 2 -b2) A (n-2) _ a -b and A(n-2) . A therwise. Thus A n 2) 11 3 32 2 ' ij i+2,j+2 is tridiagonal, 1M 31 1 l b2 1/[1b 1l (1 -2 r 8)] -r- (n-2)1 2-r a 1M 3 2 1 < 2 -r 81(i -2 -r 8) <All ])i 8I(1 -2 8) , while IA ( 2 ) <__ 8 otherwise. We see that the bounds on the elements of A ( n-) for a lxl and A( n - 2 ) for 2x2 are independent of the bound a on Jal F 101 Thus the pattern continues throughout all the reduced matrices. Hence we conclude ; each reduced matrix A(k) is tridiagonal, IAk < max {6 + 2r , 6/( - 2r 8)} ,while IAi'(I <B otherwise. Usually we normalize by choosing I 1 Then, Ai < max (1 + 2r 1/(1 -2r)} while IA(kI 1 otherwise, (k) r for each reduced matrix A . Since 2 >B- r > 1 . Hence A ) 1 k < I + 2r Thus given any positive integer r , we have Max max JA(k)I < 1 2 r (- 3 for r -1) k iJi A backward error analysis of this algorithm shows tht it is very stable (since the elements of all the reduced matrices are bounded by I + 2r , which takes on its minimal value 3 for r - 1 ). Thus we can decompose A - M D Mt , where D is block diagonal with blocks of order 1 and 2, and M is unit lower triangular with M - 0 if D #0 and with IM I - O if 1 > +2 il i+l'i 'ij' i >+ We shall need an n-vector array to record the pivotal selection. We set pivot[k] -1 (2) if we use a lxl (2x2) pivot for A~k) If pivotik] - 2 , we set pivot[k-1] - Mnk+3,n-k+l . Then we need only 2n storage locations to store the rest of M and D (these we write over A ). Thus we need only 3n storage locations for this algorithm. From §14.2, we see that Gaussian elimination requires 5n storage and 7n operations, and is very stable for tridiagonals (see Appendix A, §A.3). 102 We would also like the number of operations required for our algo-rithm to be less than 7n . However, if we use the algorithm in the 1 1 manner in which we have expressed it, 8;-- n - 2- p multiplications 2 2 and 5n additions are required, where p is the number of lxl 1 pivots used. Thus between 6n and 8- multiplications are required. 2 (We ignore multiplication by 2 in the count.) However, we can reduce the number of multiplications from 1 1 1 3 &2 n -2- p to 7 n -p if we implement the algorithm in the following manner: (We present the first step of the algorithm in Algol form): if b[l] -0 then M[2,1] :- 0 else begin temp :- al1]/b[l]; if abs (temp) > abs (b[lJ)x2+(-r) then begin M[2,1] :. 1/temp; a12] :- a - M[2,l]xb[l] end else begin calc :- tempxa - b[l]; M[3,1 :-- b/calc; M1[3,2] : -tempxM[3,1]; a :- a13 - M[3,2]xb end end; A backward error analysis shows that this implementation of the algorithm is also very stable (since all the reduced matrices are bounded by 1 + 2r ). 1 3 Now min (1 + 2r) -3 for rl ,while ( 71-n--Ip) -#6n r>l 2 2 as r -(since the larger r is the more likely the choice of a 103 lxi pivot becomes). But (1 + 2r) as r , and thus we would not have a good bound on the error matrix for large r . Thus in practice, we must make some reasonable choice of r > 1 so that 1+2 r is not too large but so that 71 3 is reasonably small 50 2 2j pv sraoal ml (i.e. as close to 6n as possible, and hopefully not more than 7n ). We have considered many versions of diagonal pivoting for the tridiagonal case. The minimal storage possible is 3n . The above-mentioned algorithm had the least operation count of all the versions studied. Since this algorithm requires between 6n and 7i n multiplica-tions in comparison to 7n for Gaussian elimination with partial pivoting,we can recommend this algorithm for general use only if storage of 3n rather than 5n is crucial to the user. A-1 Appendix A Miscellaneous Results A.1 Diagonal Pivoting for Positive Definite Matrices If A is an nxn symmetric positive definite matrix, then the maximal element of A is on its diagonal. So p . 0 and, according (n-1 to S , we use that r-qximl diagonal element as pivot. But A 0 -1) is also positive de ir.'te. Thus p -n (where p is the number of lxl pivots used in the decomposition.) Since V(k) = (k) for each A(k) , calculating p0 (k) is unne-0 1 03 cessary. (This calculation would require n additions for all (k) the .) Thus if we know that A is positive definite, we may (k) omit the calculation of the p k , and our method is identical to the method of congruent transformations (§2.7). If we also omit the cal-culation of p(k) and use the first diagonal element as a lx' pivot (non-zero since A is positive definite), our method is identical to L D L t (92.7). From the above we see that the L D Lt method and the method of congruent transformations (i.e. L D Lt with pivoting on the diagonal) are special cases of the c.iagonal pivoting method, and either of these may be used if A is definite. (See §§2.6 -2.11 for further remarks on this topic.) I A-2 In our algorithm for the diagonal pivoting method in Appendix C, we allow the following options: (I) If A is indefinite, then we must use diagonal pivoting. (2) If A is definite, then we may use: (a) L D with pivoting on the diagonal (by omitting the calculation of the p (k) ), or A2 (b) LBD Lt (by omitting the calculation of the P.k and the 0J A2 A Result for Symmetric Hadamard Hatricee I'An nxn real matrix H is liadamard if IIIl 1 110 for all ij iI where po> 0 , and H Ht -n pI 1. Usually we normalize by choosing ti 0 .1 . Thus all the elements of H are of the same modu-Ilus, and the columns of H re mutually orthogonal, i.e. if H is t th t te columnnof H, then H H~ -n6 whete i i if i -j 6 is the Kronecker delta. ij , If u loih o the diagonal pvoting method o symtiGa sin pemnation first step under any strategy. Then the reduced matrix H (-)has \| eawthe following re tipoperies <~~is f A sidfnie hnw yum s t di au a nitin atio th Theorem: is not Hadamard, and the angle between any two H(n-1) columns ofs H i e/3 A- 3 Proof: Let us assume 1- 1 We use H as pivot. Let (n-1). B- H Then Bij - Hi+lJ+I -Hi+l,1 H J+ l,/HI1 for 1 < ij n-I. n-I n-I BBtB Y X B Bks -(H- Hr+ll Hk+,/HIl) r s k-l ki(r+lk+l (if 6+i,k+]. -s+ll Hk+I 1 /H11 ) = H t . t ~ H t H (H /H )Ht H (Ii /H) + r+l s+1 1 +1 r+l,l 11 1 r+l s+1,1 11 tH (H H /H2)Pt H + Ht H 0 H /H2 I 1 r+l, 1 s+1,1 11 r+1 s+1 1 1 r+l-1 s+1,1 11 If r 0 s , then Ht H 0 ,40 Bt B -- n since r+1 S+1 r s t H -I for all i,j and H H1 n. ijI Thus Bt B 0 0 for r s, so H (n1) is not Hadamard. r s Further B B = n + n (H2 /H2 2n . Define r r r+1, 11I IB 12 = Bt B and cos O(r,s) B B /(IB C IB 1). Then r r r r s r s cos O(r,s) -- 1/2 , so O(r,s) ± 1- 3 for r 0 s q.e.d. A similar result holds for Hermitian Hadamard matrices. A.3 Gaussian Elimination for Tridiagonals Let T be an nxn tridiagonal matrix (i.e. T j -0 for is li-il >1x). suppose ITn[! < ci , and I~j < S otherwise. A-4 ((k) Theorem: Let T be as above. Then for any reduced matrix T~k under Gaussian elimination with partial pivoting T ( k ) is tridiagonal, 2(k)I ,(k)I IT < 2 , and I < otherwise. Proof: The situation for k - n-1 is typical. If IT111 > IT 2 1 1 , then we use T11 as the pivot. So (n-1) -'2 /T while T ( Totherwise. Thus -22 21 12 1i+l,j+l Ti Is r idiagonal, ITJ 11> < 2 and!IT '>; row ot he rwise. If ITIIl < IT 211 then we interchange te first and second rows an seT astepio. oT(n-1) .. T/ (n-1).-T T/ 21 s l2 -22/21 ' 12 -T23 T11 T2 1 ' while T(1 - T otherwdie. Thus T (n- 1) is tridiagonal, ij i+1,j+1 o IT.(n-1)I (< 28 , and IT < 18 otherwise. q.e.d. From the theorem and §2.5, we conclude that Gaussian elimination with partial pivoting is very stable for tridiagonal matrices T, since max max IT < 2 m- x ITI k ij i i B-I Appendix B Algorithm for Symmetric Equilibration B.1 Discussion The following Algol procedure will equilibrated any nxn sym-metric matrix A so that D A D is equilibrated, where D is diagonal. A is replaced by D A D , and the inverses of the diagonal elements of D are stored in the vector d (See ChapL! r 7.) B.2 The Algol Procedure procedure symequil (A, n, d); value n; array A,d; integer n; comment the symmetric matrix A of order n is equili-brated and the symmetric equilibrated matrix D A D ic -1 stored in A , where U = diag (dill,..., din]); begin integer i.j; real t; for i := ct.1 1 until n do begin iij :- sqrt (abs (A[i,i])); for 3 :- 1 step I until i-i do begin t :- abs (A[iJ]); if t > d[i] then d[(I :- t end; if d[i] 0 0 then begin for j := I step 1 unti: i do A[i,j] - Alijl/d[Li]; for j :- i step 1 until n do AIJ,i] : AiJ,i]/dti]; _i B-2 end; end; for i :- 1 stEp I until n do if d[i] -0 then begin for j :- i+l step 1 until n do begin t ;- abs (AIJ,i]); if t > d[i] then di\| :- t end; if d(i] - 0 then oto alarm; for J :- i+l step 1 until n do A[J,i] := A[J,i]/d[iJ; goto out; end; alarm: print ('this matrix has a null rows) out end; c-i C-1 Appendix C Algorithm for Diaoal Pivoting The following listing of an Algol procedure will solve A X -B by the diagonal pivoting method, where A is an nxn non-singular symmetric matrix and B is a vector of length n . The L D Lt method (symmetric Gaussian eliminatio' and the method of congruent transformations (L D Lt with pivoting on the diagonal) are special cases of the diagonal pivoting algorithm. The matrix A is assumed to be stored only in its lower triangular part. A is decomposed into A = H D Mt , where H is unit lower triangular, D is symmetric block diagonal with blocks of order 1 or 2, and M[i+l,iJ = 0 when D[i+l,i] 0 . H and D are written over the lower triangular part of A A is declared [1 n, 1 : n] and B is [1 : n . Upon exit, the solution X to A X = B is stored in B , i.e., X[i] is stored in BKil If A is indefinite, then set DEF = 0 and the general diagonal pivoting method is used. If A is (positive or negative) definite, then we may omit the calculation of the maximum off-diagonal element in the reduced matrices. If DEF = 2 then this is omitted and the algorithm is identical to L D Lt with pivoting on the diago"l. The pivoting on the diagonal C-2 may also be omitted if desired by setting DEF 1 , and then the algurithm is identical to L D Lt The algorithm,as presented below, is by no means, in its most efficient form. In particular, as written, no advantage of symmetry is taken to reduce storage. Instead of using only the lower triangular part of Ajl:n, 1:n], the algorithm should be coded so that the lower triangular part of A is stored in a one-dimensional array of length n (n+l) . Further, B[1 : n] could be replaced by B[l:n, 1:k] for solving a system with k right hand sides. C-3 pw('CF IIPF [V1 Ie (A I.p orUEF) tm)Ays lkP ,p tINTFrEP NiOEF to t ('1,IAP f;T i ';tLvFS; A X 0 R BY THF L)TAGONAL PIVOTING MFTHOD W AfP .8Tr A SYM AFTRIC MATPIX oF ORDERP N AND P IS~ A VWrT)R OF LENGTH N t C~t-rN.T A~ L5 A:Ur'ED To HE rTOPEn ONLY IN ITS LOWER lkIAl tLLAP PART, M AND D APF WRITTEN OVER A WHERIE A = 1 C m TRANSPnSE. M IS IuNIT LOWER TPIANGULAR, Al' : I' IS RLOCK DIAGONAL WITH 13LOCKS OF ORDER I OR 2. AN. "/11./ a (0 WHE~t #(KI.,OT EQUALS 0 .. 3fCfOJMFvT IF A 14; It..EFImITE9 SET DEF 0ANO THE DIAGONAL PIVOTIIc, tETHOC IS USEr., CnVMFNt\T IF jM IS (POSITIVE C10 NEGATIVE) DEFINITE, THEN SIT IF =1 AKC L C L TRANSPOSE WITHoUT PIVOTING w1Itl HE -jrEfl CR SET 0FF a 2 AND L 0 L TRANSPOSE W I c'ir CN THE DIAGONAL WILL OF USFO ., $Pl A1.1 w-nt (I. ()FT. AVF TFHP, ALPHAto $IN-TE.J- sAPpAV CIAt.oE (/,.N/) t OAPP'AYt Phw~T .91en) pLl()cFI-0pE;~ HA~.IIA, (AeKvf~..JqM1) .9 ifVAL'JF$ tI\ 99 taRPAY a o IftTFGER Kt.'qj at ~rH~~PAL$ Ml .9I trotPFI T~tCAICULATFS THF t- AXIMUM. MF THE DIAGONAL OF A , ml1 umAX A(A/,) FCR K~ OLEO$ I LEO N, AND J IS THE PFCGININ1Fnkk to 141 =Afi4(A(/KK/)) 9 J on K so FOP I K1 q TEP I UNTIL N 00 IAS(A(111T/)) tGREATFR% "I THFI PH.-IN pi au AlBqcA(/I,I/fl vg J i ENO IFr~l MAYIP~. PIWOCF[UUP. MAX ^(AvKvr9R9SH0L.ml) s VALUE'K N t mILaM1 ARPAY A to TNTEGF.R KNspoc,,L .9 PFAL ?PO.m1 .. roCtffj~iNt CALCULATE-, 100 a MAX ARS(A(/Ioj/) ) FOR I- LF'~W It. OLEO N- AtM THF INTEGERS R ANO S SUCH TpaT AES(a(/PqS/l1 z M6, IT IS ASSUMED THAT MI I PAiX (AES fA(/IT/) ) ) AND, vIa ABS( A(/LvL/) ) . iFCT~:I~T(;~It'J .9 140 9= ml P a S .m L so sF001 j STFP I UNTIL "-I n0 C-4 $L0P j ,:KSTFP 1 OUNT1L;9 N f$oO irrnpV I il STEP 1 OuNTIL N $00$ TF &Hq ( A(/y.J/1 . GPFATERO MO THNJ I4Er.TN PCO or AS~S ( At/TJ/) ) . . an I q. S or .j 0END , tpRCunuUP It.TFPCNANC-E CAtKel) . VAI.k-F KoI so ARRAY0 A .0 INTEGER Kol .. rv-~T IlNTFQC ANG-ES ROW ANDl CnLUJMN K WIT14 POW L ~t0L CO)LUM'N) I wl-EpF K CEn$ I AND A 15 THE REDUCED m!ATPIX O~F CPrFP .'I 9 S Fr:IN 0LdALj! !FPP v. INTFGFR J . rIoP , .= K+1 SlEpt I uNTIt. N 00 PC INI TFvP ow At/JoK/) so tA/JqK/) qm A(/Jl1/1 $I A(/,jei/) o3 TVMP ~jEN FnPs i tc Tol STEP I1 UNTIL K-1 000 A(/Kt../) .= TFMF Etn v, TF P . ( T l ) A / 0 / =A(/KoK/) soflA 0fi It Ti (HIpoCF ALPHA ~PTH))/8 . STAPI.. cHF,lN rwANLF (/7/) or I to GO1O$ PIVOTONE END . F~1N% !;TEc2CH4ANcE (AKtl) 'v CHANGE (/I/) or K .9 (<TC P~vCTCNF FN1t IF fMl NOT i-FSS N~O ALPHA TH.FN AFrINX lK1EPrHANG-F (AKo1) at CHANGE (/1/) 90 x g sr.C;TC' IPTvCTCNE F~) g gli-s 5 $GPIEATFQ$ I TwFNO TNTEPCMANGF (AqSvI) #9 CH.4ANCE C/I/) on R IF$ P CRtATFct$ 1.1 1i.4NO IKNTEPCHANGE (AtP9I.1) .9 2 ,-OTO PIVOTTwn ts P~vOTOfI'F IF# 011111)~ 2 6 THEN GOTn ALAPM t IenM~rT kvF U E A IEI PTVOT FOP J z: 141 $STEP' 1 UNTIL$ N 00 Cfl.,~E t /jvl/) HvSr BEEK, SET FOUAL TO TIE MULTIPLIER so gF0o, j .= 1+] fITEP I UNTIL N rO0 Ito#r z' : 1 rTEI I UNTIv_ J t'0 I(J K ) .= A 1 l c-5 srnpMFr'Tt 1 '+.. A(,e/ HAVE tiFEN~ SET To THEIR NEW VALUES . 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RC/ CtH~h'iF/I/) I) .q AVE so IF;, P~vrT(/I/l = I THEN k.FG1N rH : .= 71 TEr1 I INTIL N Cn [,(/j/) .. = kU/.j/) -A(/joT/I 4b P./I/) e lo. 1. Fkiv{0 FLSE49 pCIN S SAE . ~/Ii,).' (/I1/) . P(/ CHANflE(/I1/) as P(/ CHArFI4I/) on e SA' at FOP$ j z 1.2 iiSIEPs 1 IUNTIL N tflO I .= I+? Etn# ., IF I st\Cl GliFATER $ THEN,$ $C6oTO REPEAT at as~ srrFTg iC-4 SflL\F r. Y c Ar'fl STnPE Y IN THE VECTOR 8 . c-6 SOL VF.. 11JFS PIV~i1 (/I/) T i joffs I $(EfTvFW h. Tl4EN' COTO FiNfl x OFICF r'ryTfl SO, VE 4V~Nrs . LTM R(/HI/) o, SAVE au BUT41/I .9 DET ou PIVOT(/!4j/) . M(/1) C 1FAP(/I41.1.1/) -SAVFOA(/141.I/) )/OET go I~'/.W): SAVEA(/ItT/) -TFMPOAC/I.1,I/) )/OET 1FI #i-'iTEPgf N OTE, OGOTO0 FIN~DX FLSE r7OTO SOLVE . ()mmUF Ts I~CAw c-OtVE X am INVFRSE TRANSPOSE TImES Y wH'EPE Y IS SYI'RFr IN TWF VFCTOP 8 ANfl STrnRE X IN 8 t F t:IX CALC so W$I PlvCT(/I/ a 1 Ti.EPm FOP J T.: . STEP 1 UNTIL N son~ SAVF .= tb(/T/) ,B(/!/) az R/ CH4ANtE(/I/)I). 83(/ .=M('(I) ) SAVF . I z I-1 END FLc E 1'9LQ.INl FCR K on I-le! 00n WrR J a T.1 STEP 1 tINTIL N PO F Nfl .. OFCP$ K a~ 1-1, 1 000 EGIN$ SAVF ex P(/K/) ., R(/S(/) on RU/ CHANGE(/K/)I). BU/ C"1ANCr(/K/) oft . AVF go END$,# I =~ 1-2 tENI: $IF 7 $f,~T LFsci 1 THFNI Gr.Tfl CALC . G~OTr' OtiT ALAPPM nULTPLJT (61 #( (s SIrtrULAW MATPTX ) ~ 2. OUT .. FN[O$ PIVClT ,C Bibliography Cryer, C. W., "Pivot size in Gaussian elimination", Numerische Mathematik, 12 (1968), pp. 335-345. Davis, P. J., The Mathematics of Matrices, Blaisdell, New York (1965). Forsythe, G. and C. Moler, Computer Solution of Linear Algebraic Systems Prentice-Hall, New Jersey (1967). Fox, L., An Introduction to Numerical Linear Algebra, Oxford Uni-versity Press, London (1964). Gantmacher, F. R.. The Theory of Matrices, v. 1, Chelsea, New York (1959). Hildebrand, F. B., Introduction to Numerical Analysis, McGraw-Hill, New York (1956). Householder, A. S., The Theory of Matrices in Numerical Analysis, Blaisdell, New York (1964). Marcus, M. and M. Newman, "The permanent of a symmetric matrix", Notices Amer. Math. Soc., v. 8 (1961), p. 595. Meersman, R. de and L. Schotsmans, "Note on the inversion of symmetric matrices by the Gauss-Jordan method", I.C.C. Bulletin, 3 (1964), pp. 152-5. Mirsky, L., An Introduction to Linear Algebr, Clarendon Press, Oxford (1955). Moler, C., "Iterative refinement in floating point", Journal A.C.M., 14:2 (1967), pp. 316-321. i I I Parlett, B. N., and J. K. Reid, "On the solution of a system of linear equations whose matrix is symmetric but not definite", M.I.T. (to appear). Reid, J. K., "A note on the least square eolution of a band system of linear equations by Householder reductions", Computer Journal 10:2 (1967), pp. 188-189. Schotamans, L., "Gauss-Jordan inversion of symmetric matrices, using a 2x2 pivot submatrix", C.E.N, Report No. 621-42/65-153, Mol, Belgium (1965). Shisha, 0., editor, Inequalities, Academic Press, New York (1967). Sinkhorn, R., and P. Knopp, "Concerning non-negative matrices and doubly stochastic matrices", Pacific Journal of Math., 21:2 (1967). Sinkhorn, R., "Diagonal equivalence to matrices with prescribed row and column sums", Amer. Math. Monthly, 74:4 (1967). Sinkhorn, R., "A relationship between arbitrary positive matrices and doubly stochastic matrices", Ann. Math. Statistics, 35 (1964), pp. 876-879. Tornheim, L., "Pivot size in Gauss reduction", Tech. Report, Calif. Res. Corp., Richmond, Calif. (1964). Tornheim, L., "Maximum third pivot for Gaussian reduction", Tech. Report, Calif. Res. Corp., Richmond, Calif. (1965). Weatlake, J., Handbook of Numerical Matrix Inversion and Solution of Linear Equations, Wiley, New York (1968). Wilkin-on, J. H., "Rounding errors in algebraic processes", Infor-mation Processing, "Proceedings of the International Conference on Information Proccssing, UNESCO, Paris (1959). 1 II I I II I I II I I Wilkinson, J. H., "Error analysis of direct methods of matrix inver-sion", Journal A.C.M., 8:3 (1961), pp. 281-330. Wilkinson, J. H., Rounding errors in algebraic processes, Notes on Applied Science No. 32, Her Majesty's Stationery Office, London; Frentice-Hall, New Jersey (1963). Wilkinson, J. H., The Algebraic Eigenvalue Problem, Clarendon Press, Oxford (1965). LAI E,,'VI 'wI
15545	https://tn01917090.schoolwires.net/cms/lib/TN01917090/Centricity/Domain/1416/4th%20grade%20math%202nd%20packet.pdf	©Curriculum Associates, LLC Copying is not permitted. 231 Lesson 11 Multiply by One-Digit Numbers TRY IT Learning Target • Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. SMP 1, 2, 3, 4, 5, 6, 7 LESSON 11 You have learned how to break apart numbers to multiply and how to multiply one-digit numbers by multiples of ten. Use what you know to try to solve the problem below. What is the product of 3 and 57? Math Toolkit • base-ten blocks • counters • bowls • grid paper • sticky notes • number lines • multiplication models DISCUSS IT Ask your partner: How did you get started? Tell your partner: I started by . . . SESSION 1 Explore Multiplying by One-Digit Numbers ©Curriculum Associates, LLC Copying is not permitted. 232 Lesson 11 Multiply by One-Digit Numbers SESSION 1 LESSON 11 EXPLORE Connect It 1 LOOK BACK Explain how you found the product of 3 and 57. 2 LOOK AHEAD You can use arrays, area models, and partial products to break apart numbers to help you multiply. The array at the right uses base-ten blocks to show 3 3 157. a. Write 157 in expanded form. b. Fill in the blanks below to show how to find 3 3 157. 3 3 157 5 (3 3 ) 1 (3 3 ) 1 (3 3 ) 5 1 1 5 c. What do you notice about the number of zeros in the product of 3 and 50 and in the product of 3 and 100? How many zeros would be in the product of 3 × 1,000? Explain. 3 REFLECT How does breaking apart the multiplication problem above by place value help you solve the problem? 100 50 3 1 1 7 ©Curriculum Associates, LLC Copying is not permitted. 233 Name: Lesson 11 Multiply by One-Digit Numbers LESSON 11 SESSION 1 Prepare for Multiplying by One-Digit Numbers 2 Fill in the blanks below to show how to find 2 × 48. 2 3 48 5 (2 3 ) 1 (2 3 ) 5 1 5 1 Think about what you know about multiplication. Fill in each box. Use words, numbers, and pictures. Show as many ideas as you can. Examples Examples Examples What Is It? What I Know About It product 2 40 8 ©Curriculum Associates, LLC Copying is not permitted. 234 Lesson 11 Multiply by One-Digit Numbers LESSON 11 SESSION 1 3 Solve the problem. Show your work. What is the product of 4 and 62? Solution 4 Check your answer. Show your work. ©Curriculum Associates, LLC Copying is not permitted. 235 LESSON 11 Lesson 11 Multiply by One-Digit Numbers Read and try to solve the problem below. What is the product of 3 and 254? SESSION 2 Develop Multiplying a Three-Digit Number by a One-Digit Number TRY IT DISCUSS IT Ask your partner: Can you explain that again? Tell your partner: I agree with you about . . . because . . . Math Toolkit • base-ten blocks • grid paper • index cards • sticky notes • number lines • multiplication models ©Curriculum Associates, LLC Copying is not permitted. 236 LESSON 11 DEVELOP Lesson 11 Multiply by One-Digit Numbers Explore different ways to understand multiplying a three-digit number by a one-digit number. What is the product of 3 and 254? Model It You can use an array of base-ten blocks to help you multiply. 254 254 254 6 hundreds 15 tens 12 ones 3 3 254 5 600 1 150 1 12 Model It You can also multiply using partial products. 254 3 3 12 150 1 600 ? 3 3 4 ones 3 3 5 tens 3 3 2 hundreds The partial products are 12, 150, and 600. The product is the sum of the partial products: 12 1 150 1 600. ©Curriculum Associates, LLC Copying is not permitted. 237 Lesson 11 Multiply by One-Digit Numbers SESSION 2 CONNECT IT Now you will use the problem from the previous page to help you understand how to multiply three-digit numbers by one-digit numbers. 1 In the first Model It, what do the numbers 600, 150, and 12 in the equation below the array represent? 2 How can you find the product of 3 and 254 in the first Model It? 3 Where do you see the 6 hundreds, 15 tens, and 12 ones in the second Model It? 4 What is the sum of the partial products in the second Model It? 5 How can you use estimation to check that your answer is reasonable? 6 How do both Model Its show breaking apart a factor to multiply? 7 REFLECT Look back at your Try It, strategies by classmates, and Model Its. Which models or strategies do you like best for multiplying a three-digit number by a one-digit number? Explain. ©Curriculum Associates, LLC Copying is not permitted. 238 Lesson 11 Multiply by One-Digit Numbers LESSON 11 DEVELOP SESSION 2 Apply it Use what you just learned to solve these problems. 8 2 3 163 5 ? Show your work. Solution 9 Find the product of 5 and 738. Estimate to check that your answer is reasonable. Show your work. Solution 10 What is the product of 859 and 7? Show your work. Solution ©Curriculum Associates, LLC Copying is not permitted. 239 Name: LESSON 11 SESSION 2 Lesson 11 Multiply by One-Digit Numbers Study the Example showing one way to multiply a three-digit number by a one-digit number. Then solve problems 1–5. Example Find the product of 2 and 546. 1,000 80 12 546 546 2 3 546 5 1,000 1 80 1 12 5 1,092 1 Look at the multiplication above. Use partial products to find 2 3 546. Fill in the boxes. 3 6 ones 2 3 tens 2 3 hundreds 5 4 6 3 2 1 2 1 1 , 0 0 0 , 2 3 3 132 5 ? Show your work. Solution Practice Multiplying a Three-Digit Number by a One-Digit Number ©Curriculum Associates, LLC Copying is not permitted. 240 LESSON 11 SESSION 2 Lesson 11 Multiply by One-Digit Numbers 3 Find 6 3 915. Show your work. Solution 4 Find the product of 483 and 7. Estimate to check that your answer is reasonable. Show your work. Solution 5 There is a mistake in the multiplication shown. Explain what mistake is made. Then find the correct product. Solution 607 3 4 28 1 240 268 ©Curriculum Associates, LLC Copying is not permitted. 241 LESSON 11 SESSION 3 Lesson 11 Multiply by One-Digit Numbers Develop Multiplying a Four-Digit Number by a One-Digit Number Read and try to solve the problem below. Ezekiel has 3 building sets. Each set includes 1,125 pieces. How many pieces are in all 3 sets? TRY IT Math Toolkit • base-ten blocks • grid paper • index cards • sticky notes • number lines • multiplication models DISCUSS IT Ask your partner: Do you agree with me? Why or why not? Tell your partner: I disagree with this part because … ©Curriculum Associates, LLC Copying is not permitted. 242 LESSON 11 DEVELOP Lesson 11 Multiply by One-Digit Numbers Explore different ways to understand multiplying a four-digit number by a one-digit number. Ezekiel has 3 building sets. Each set includes 1,125 pieces. How many pieces are in all 3 sets? picture It You can use an area model to help understand the problem. 3 1,000 100 20 5 1 1 1 3 3 1,000 3 3 100 3 3 20 3 3 5 3 3 1,125 5 (3 3 1,000) 1 (3 3 100) 1 (3 3 20) 1 (3 3 5) 5 3,000 1 300 1 60 1 15 model It You can also multiply the numbers using partial products. 1,125 3 3 15 60 300 1 3,000 ? 3 3 5 ones 3 3 2 tens 3 3 1 hundred 3 3 1 thousand The partial products are 15, 60, 300, and 3,000. The product is the sum of the partial products. ©Curriculum Associates, LLC Copying is not permitted. 243 Lesson 11 Multiply by One-Digit Numbers SESSION 3 CONNECT IT Now you will use the problem from the previous page to help you understand how to multiply four-digit numbers by one-digit numbers. 1 What is the expanded form of 1,125? 1 1 1 2 How is the expanded form used in the equation in Picture It? 3 What is the sum of the numbers in the equation in Picture It and the sum of the partial products in Model It? 4 The partial products in Model It shows first multiplying the 3 by the value of the digit in the ones column. Would the product change if you first multiplied the 3 by the value of the digit in the thousands column? Explain. 5 Describe how the factor 3 is used with the factor 1,125 to find the product. 6 Explain how you multiply a four-digit number by a one-digit number. 7 REFLECT Look back at your Try It, strategies by classmates, and Picture It and Model It. Which models or strategies do you like best for multiplying a four-digit number by a one-digit number? Explain. ©Curriculum Associates, LLC Copying is not permitted. 244 Lesson 11 Multiply by One-Digit Numbers SESSION 3 LESSON 11 DEVELOP Apply it Use what you just learned to solve these problems. 8 5,342 3 4 5 ? Show your work. Solution 9 Find the product of 7 and 3,928. Estimate to check that your answer is reasonable. Show your work. Solution 10 2,041 3 6 5 ? Show your work. Solution ©Curriculum Associates, LLC Copying is not permitted. Name: Lesson 11 Multiply by One-Digit Numbers LESSON 11 SESSION 3 Study the Example showing one way to multiply a four-digit number by a one-digit number. Then solve problems 1–5. Example Jesse’s family has 4 music players. Each music player can hold 8,352 songs. What is the total number of songs all 4 music players can hold? Use an area model. 4 8,000 300 50 2 1 1 1 4 3 8,000 4 3 300 4 3 50 4 3 2 4 3 8,352 5 (4 3 8,000) 1 (4 3 300) 1 (4 3 50) 1 (4 3 2) 5 32,000 1 1,200 1 200 1 8 5 33,408 All 4 music players can hold a total of 33,408 songs. 1 Complete the multiplication to use partial products to find 4 3 8,352. 8 , 3 5 2 3 4 8 2 0 0 , 1 3 2 , 0 0 0 , 3 2 ones 4 3 5 4 3 hundreds 4 3 8 2 Show how to use partial products to find 5 3 1,643. Practice Multiplying a Four-Digit Number by a One-Digit Number 245 ©Curriculum Associates, LLC Copying is not permitted. 246 Lesson 11 Multiply by One-Digit Numbers LESSON 11 SESSION 3 3 Write 4 3 3,569 in expanded form to show the place value of each digit. Then find the product. 4 Lee earns $1,075 each month. How much does he earn in 6 months? Estimate to check that your answer is reasonable. Show your work. Solution 5 Look at Callie’s work for finding 3 3 9,423. a. Explain what Callie did wrong. b. How can using estimation show that Callie’s answer is wrong? c. What is the correct answer? 9,423 3 3 9 60 120 1 2,700 2,889 ©Curriculum Associates, LLC Copying is not permitted. 247 LESSON 11 SESSION 4 Lesson 11 Multiply by One-Digit Numbers Refine Multiplying by One-Digit Numbers Complete the Example below. Then solve problems 1–9. PAIR/SHARE How else could you solve this problem? PAIR/SHARE How can you check that your answer is reasonable? The student multiplied 6 by the value of the digit in each place in 1,785. Could you use an array to help you solve this problem? Apply it 1 Find 435 3 2. Show your work. Solution EXAMPLE An aquarium has 6 female sea turtles. Each turtle lays up to 1,785 eggs a year. Suppose each turtle lays 1,785 eggs this year. How many eggs do the turtles lay in all this year? Look at how you could show your work using an area model. 6 3 1,000 6 3 700 6 3 80 6 3 5 1,000 700 80 5 6 1 1 1 6 3 1,785 5 (6 3 1,000) 1 (6 3 700) 1 (6 3 80) 1 (6 3 5) 5 6,000 1 4,200 1 480 1 30 Solution ©Curriculum Associates, LLC Copying is not permitted. 248 LESSON 11 REFINE Lesson 11 Multiply by One-Digit Numbers 2 Find 4 3 6,309. Estimate to check that your answer is reasonable. Show your work. Solution 3 A hardware store has 147 containers of paint. Each container holds 5 gallons of paint. How many gallons of paint does the store have? A 235 B 505 C 735 D 905 Dale chose A as the correct answer. How did he get that answer? PAIR/SHARE Is your estimate close to your answer? PAIR/SHARE Does Dale’s answer make sense? How could partial products help you solve this problem? Multiply 5 by the value of the digit in each place in 147. ©Curriculum Associates, LLC Copying is not permitted. 249 Lesson 11 Multiply by One-Digit Numbers SESSION 4 4 Select all the expressions that have a product of 810. A 90 3 9 B 405 3 2 C (3 3 20) 1 (3 3 70) D (2 3 400) 3 (2 3 5) E (5 3 100) 1 (5 3 60) 1 (5 3 2) 5 Mr. Larson is planning a pizza party for 273 people. He plans on 3 slices of pizza for each person. How many slices of pizza is this in all? A 276 B 546 C 619 D 819 6 Find 2,906 3 2. • • • • • • 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 ©Curriculum Associates, LLC Copying is not permitted. 250 LESSON 11 REFINE SESSION 4 Lesson 11 Multiply by One-Digit Numbers 7 Lara says, “When you multiply a three-digit number by a one-digit number, the product is always a four-digit number.” Lara writes an equation to support her statement. Greg writes an equation to show that Lara’s statement is false. Complete the equations below to show a possible equation each person could have written. Lara’s equation: 328 3 5 Greg’s equation: 328 3 5 8 Fourth-grade students hold a recycling drive. In one week, they collect 1,238 water bottles each day. How many water bottles do the fourth graders collect that week? Estimate to check that your answer is reasonable. Show your work. [Hint: There are 7 days in one week.] Solution 9 MATH JOURNAL Explain what strategy you would use to find 357 × 8. Then use that strategy to find the product. SELF CHECK Go back to the Unit 3 Opener and see what you can check off. ©Curriculum Associates, LLC Copying is not permitted. 253 Lesson 12 Multiply by Two-Digit Numbers Learning Target • Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. SMP 1, 2, 3, 4, 5, 6, 7 LESSON 12 You have learned how to multiply two-digit numbers by one-digit numbers, how to multiply one-digit numbers by multiples of 10, and how to break apart numbers by place value to multiply. Use what you know to try to solve the problem below. What is the product of 14 and 13? TRY IT DISCUSS IT Ask your partner: Can you explain that again? Tell your partner: I am not sure how to find the answer because . . . Math Toolkit • base-ten blocks • counters • cups • paper plates • grid paper • multiplication models SESSION 1 Explore Multiplying by Two-Digit Numbers ©Curriculum Associates, LLC Copying is not permitted. 254 Lesson 12 Multiply by Two-Digit Numbers LESSON 12 EXPLORE CONNECT IT 1 LOOK BACK Explain how you found the product of 14 and 13. 2 LOOK AHEAD To multiply a two-digit number by another two-digit number, you need to understand how to multiply by multiples of 10. a. Fill in the blanks to show how to multiply by multiples of 10. Expression Think of it as . . . Think of it as . . . Product 3 3 2 3 3 2 ones 6 ones 3 3 20 3 3 2 tens tens 60 30 3 20 3 tens 3 2 tens hundreds 600 3 3 10 3 2 3 10 3 3 2 3 10 3 10 6 × b. Complete the area model. Then add the four partial products to find 25 3 32. 1 1 1 5 3 REFLECT Suppose you want to find 30 3 30. How can you use a basic fact and breaking apart numbers to find the product of these multiples of 10? 2 30 1 1 20 5 20 3 2 5 5 3 30 5 5 3 2 5 20 3 30 5 SESSION 1 ©Curriculum Associates, LLC Copying is not permitted. 255 Name: Lesson 12 Multiply by Two-Digit Numbers Prepare for Multiplying by Two-Digit Numbers LESSON 12 SESSION 1 2 Complete the area model. Then add the four partial products to find 18 × 24. 1 1 1 5 1 Think about what you know about multiplication. Fill in each box. Use words, numbers, and pictures. Show as many ideas as you can. Examples Examples Examples What Is It? What I Know About It partial products 4 1 1 10 8 20 10 3 4 5 10 3 20 5 8 3 4 5 8 3 20 5 ©Curriculum Associates, LLC Copying is not permitted. 256 Lesson 12 Multiply by Two-Digit Numbers LESSON 12 SESSION 1 3 Solve the problem. Show your work. What is the product of 16 and 12? Solution 4 Check your answer. Show your work. ©Curriculum Associates, LLC Copying is not permitted. 257 Lesson 12 Multiply by Two-Digit Numbers Develop Multiplying by Two-Digit Numbers SESSION 2 LESSON 12 Read and try to solve the problem below. Folding chairs are set up in a school auditorium for a play. There are 16 rows of chairs. Each row has 28 chairs. How many folding chairs are set up for the play? TRY IT Math Toolkit • base-ten blocks • grid paper • multiplication models DISCUSS IT Ask your partner: Why did you choose that strategy? Tell your partner: A model I used was . . . It helped me . . . ©Curriculum Associates, LLC Copying is not permitted. 258 Lesson 12 Multiply by Two-Digit Numbers LESSON 12 DEVELOP Explore different ways to understand multiplying a two-digit number by a two-digit number. Folding chairs are set up in a school auditorium for a play. There are 16 rows of chairs. Each row has 28 chairs. How many folding chairs are set up for the play? Picture It You can use an area model to multiply two-digit numbers. To solve this problem, multiply 28 by 16. 10 6 20 8 1 1 10 3 20 5 200 6 3 20 5 120 10 3 8 5 80 6 3 8 5 48 200 1 80 1 120 1 48 5 ? model It You can also multiply two-digit numbers using partial products. 28 3 16 48 120 80 1 200 ? 6 ones 3 8 ones 6 ones 3 2 tens 1 ten 3 8 ones 1 ten 3 2 tens ©Curriculum Associates, LLC Copying is not permitted. 259 Lesson 12 Multiply by Two-Digit Numbers SESSION 2 Connect It Now you will use the problem from the previous page to help you understand how to multiply a two-digit number by a two-digit number. 1 Why is the area model divided into four sections? 2 How do the four steps in the multiplication using partial products in Model It relate to the four sections in the area model in Picture It? 3 What is the sum of the partial products and also the product of 28 and 16? 4 Would the product change if 20 1 8 on the top of the area model were changed to 10 1 10 1 8? Explain. 5 How could you estimate to check the reasonableness of your answer to 28 3 16 by multiplying with easier numbers? 6 REFLECT Look back at your Try It, strategies by classmates, and Picture It and Model It. Which models or strategies do you like best for multiplying a two-digit number by a two-digit number? Explain. ©Curriculum Associates, LLC Copying is not permitted. 260 Lesson 12 Multiply by Two-Digit Numbers LESSON 12 DEVELOP Apply It Use what you just learned to solve these problems. 7 Complete the area model below. Then add the partial products to find the product of 27 and 21. Show your work. 7 20 1 1 20 1 20 3 7 5 20 3 20 5 1 3 20 5 1 3 7 5 Solution 8 Find 37 3 23. Show your work. Solution 9 Select all the expressions that have a product of 640. A 10 3 64 B 60 3 40 C 80 3 80 D (30 1 2) 3 20 E (40 3 10) 1 (40 3 6) SESSION 2 ©Curriculum Associates, LLC Copying is not permitted. 261 Name: Lesson 12 Multiply by Two-Digit Numbers Study the Example showing how to multiply a two-digit number by a two-digit number to solve a word problem. Then solve problems 1–6. Example Aaron spends 35 minutes at each guitar lesson. He has 12 guitar lessons. How many minutes does Aaron spend at his guitar lessons? Use an area model to multiply 35 by 12. 300 1 50 1 60 1 10 5 420 Aaron spends 420 minutes at his guitar lessons. 10 2 30 5 1 1 10 3 30 1 ten 3 3 tens 5 3 hundreds 300 2 3 30 2 3 3 tens 5 6 tens 60 10 3 5 1 ten 3 5 5 5 tens 50 2 3 5 5 10 1 Look at the Example above. Use partial products to multiply 35 by 12. Fill in the blanks. ones 3 ones 2 3 3 ten 3 5 ten 3 3 3 5 3 1 2 1 0 5 0 1 4 2 0 2 Show how to use an area model to multiply 71 by 48. 71 3 48 5 1 1 1 5 Practice Multiplying by Two-Digit Numbers LESSON 12 SESSION 2 ©Curriculum Associates, LLC Copying is not permitted. 262 Lesson 12 Multiply by Two-Digit Numbers LESSON 12 SESSION 2 3 Show how to use partial products to multiply 48 by 71. 48 3 71 5 4 Tell whether each equation is True or False. True False 18 3 42 5 (10 3 40) 1 (10 3 2) 1 (8 3 40) 1 (8 3 2) A B 60 3 15 5 (6 3 10) 1 (6 3 5) C D 37 3 22 5 (30 3 20) 1 (30 3 20) 1 (7 3 20) 1 (7 3 20) E F 99 3 11 5 (1 3 9) 1 (1 3 90) 1 (10 3 9) 1 (10 3 90) G H 5 Mr. Greene prepares 28 bags of glass tiles for his art class. He puts 40 glass tiles in each bag. How many glass tiles does Mr. Greene use? Estimate to check that your answer is reasonable. Show your work. Solution 6 Stephanie has 6 classes a day at school. Each class is 52 minutes long. She goes to school 5 days a week. How much time does she spend in class each week? Show two different ways to solve this problem. Show your work. Solution ©Curriculum Associates, LLC Copying is not permitted. 263 LESSON 12 Lesson 12 Multiply by Two-Digit Numbers Refine Multiplying by Two-Digit Numbers SESSION 3 Complete the Example below. Then solve problems 1–9. EXAMPLE What is the product of 73 and 58? Look at how you could show your work using partial products. 73 3 58 24 560 150 1 3,500 8 ones 3 3 ones 8 ones 3 7 tens 5 tens 3 3 ones 5 tens 3 7 tens Solution Apply it 1 Find the product of 15 and 24. Show your work. Solution The student added the partial products to find 73 3 58. Should you multiply 15 3 24 or 24 3 15? PAIR/SHARE How else could you solve this problem? PAIR/SHARE How did you decide which method to use to help you solve the problem? ©Curriculum Associates, LLC Copying is not permitted. 264 Lesson 12 Multiply by Two-Digit Numbers LESSON 12 REFINE 2 What is the product of 12 and 32? Show your work. Solution 3 A deli is preparing trays of sandwiches. There are 48 trays. Each tray has 23 sandwiches. How many sandwiches are there? A 240 B 824 C 1,104 D 1,932 Nathan chose A as the correct answer. How did he get that answer? PAIR/SHARE How could you check your answer for reasonableness? PAIR/SHARE Does Nathan’s answer make sense? How could partial products help you solve this problem? Could you use an area model to help solve the problem? ©Curriculum Associates, LLC Copying is not permitted. 265 Lesson 12 Multiply by Two-Digit Numbers SESSION 3 4 A person blinks about 16 times per minute. About how many times does a person blink in 3 hours? [Hint: 1 hour 5 60 minutes] A 48 B 96 C 960 D 2,880 5 What is the product of 47 and 91? • • • • • • 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 2 3 8 9 4 0 6 1 7 5 6 Which models below could represent the solution to the problem 45 3 15? A 10 40 5 5 B 0 30 15 45 C (4 3 1) 1 (4 3 5) 1 (5 3 1) 1 (5 3 5) D (10 3 40) 1 (10 3 5) 1 (5 3 40) 1 (5 3 5) E 0 45 90 135 180 270 360 495 585 675 630 540 405 450 315 225 ©Curriculum Associates, LLC Copying is not permitted. 266 Lesson 12 Multiply by Two-Digit Numbers LESSON 12 REFINE 7 Complete each equation below using a factor between 20 and 30 so that: • The missing factor in Ian’s equation will give the greatest possible three-digit product. • The missing factor in Tia’s equation will give the least possible four-digit product. Ian’s equation: 43 3 5 Tia’s equation: 43 3 5 8 Mo has 14 tutoring sessions. Each session is 35 minutes long. How many minutes does Mo spend in the 14 sessions? Show your work. Solution 9 MATH JOURNAL Write a word problem you can solve by multiplying 2 two-digit numbers. Solve the problem and show how to find the answer. SESSION 3 SELF CHECK Go back to the Unit 3 Opener and see what you can check off.
15546	https://sumsseriesmaths.quora.com/Two-thirds-of-a-number-is-60-What-is-one-sixth-of-the-same-number	Two-thirds of a number is 60. What is one-sixth of the same number? - Sums & Series (Maths) - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Sums & Series (Maths) Unique and Informative Mathematical Questions. Follow · 52.8K 52.8K Two-thirds of a number is 60. What is one-sixth of the same number? See parent question Answer Request Follow · 9 1 5 Answers Sort Recommended Michael Brennan B.A. in Mathematics, St. Thomas University (Graduated 1969) ·3y Given: a number (n) ⅔ n = 60 Find: ⅙ n Plan: Solve for n then multiply by ⅙ ⅔ n = 60 Multiply both sides of the equation by 3/2 (3/2)(2/3)n = (3/2)60 => n = 180/2 =>n = 90 ✅ Now (1/6)n = (1/6)90 = 15 ✅ Double Check Reasonable (2/3)90 = 60 & (1/6)90 = 15 ✅ Answer: 15 Kurt Mager Enjoys solving math problems ·3y Well, 2 n 3=1 6 2 n 3=1 6 ⟹2 n=1 2⟹2 n=1 2 ⟹n=1 4⟹n=1 4 So if two-thirds of a number is 60 60, then one-sixth of that number is 60 4 60 4 or 15.15. Anthony Hawken 3y Ratio 1/6 : 2/3 = 1/6 x 3/2 = 1/4 1/4 x 60 = 15. Dave I don't stop studying maths related topics ·3y (2/3)X=60 X=(3/2)60=180/2=90 90(1/6)=15 Ronald Deep Worked at The University of Dayton ·3y 15 View 22 other answers on parent question Related questions You invest $1,000 at a simple interest rate of 5% per year. How much interest will you earn in 3 years? What will be the total amount? What are the intercepts and then use them to graph the equation 2y=-18+9x? A can company charges a $3 flat rate in addition to $1.50 per mile the domain of this function is The range of this function is? Why don't we call irrationals indefinite precisions? After all numbers are ratios and irrational numbers implies there are numbers that are void of ratios which is an absurdity, OK? Numbers are not irrational because they cannot, not terminate. All numbers are ratios. Do you have any issues with my view of the properties of numbers? Agatha put $775 into a simple interest bearing account at a rate of 2.4% for 8 years. Her friend Dominic invested the same amount into an account that is compounded quarterly at the same rate for 6 years? Jeff has $28.75. He purchased three cookies that cost $1.50 each, five newspapers that each cost $0.50, five flowers for $1.25 each, and used the remainder of the cash on a pair of sunglasses. How much were the sunglasses? When a factory operates from 6 AM to 6 PM, its total fuel consumption varies according to the formula f(t) = 0.91 - 0.110.4 + 12, where t is the time in hours after 6 AM and f() is the number of barre? Cooper designs a photo album that he is going to give his parents as a gift. He created 15 pages of the album in June. He designed 1/10 of the album in August. He finishes the remaining 3/5 just in time for his parent's anniversary in November? Suppose a roller coaster climbs 208 feet higher than its starting point making a horizontal advance of 360 feet. When it comes down, does it make a horizontal advanced of 44 feet? What are the X and Y components of a velocity vector that has a magnitude of 62.0 m/s and an angle of 115.0 degrees? What is a much more meaningful way of saying, Pythagorean triples or Pythagoras triplets? A newspaper ad offered a set of tires at a sales price of $258.00. The regular price was $300.00, what percent of the stock price was the savings? A new long-life tire has a tread depth of 3/8 inches, instead of the more typical 7/32 inches. How much deeper is the new tire? Is the sequence 2 x 2 x 11 exponential? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15547	https://wiki.math.ntnu.no/linearmethods/basicspaces/openandclosed	Institutt for matematiske fag, NTNU Interior points, boundary points, open and closed sets Let (X,d)(X,d) be a metric space with distance d:X×X→[0,∞)d:X×X→[0,∞). A point x0∈D⊂Xx0∈D⊂X is called an interior point in D if there is a small ball centered at x0x0 that lies entirely in DD, x0 interior point def⟺∃ε>0;Bε(x0)⊂D. x0 interior point ⟺def∃ε>0;Bε(x0)⊂D. A point x0∈Xx0∈X is called a boundary point of D if any small ball centered at x0x0 has non-empty intersections with both DD and its complement, x0 boundary point def⟺∀ε>0∃x,y∈Bε(x0);x∈D,y∈X∖D. x0 boundary point ⟺def∀ε>0∃x,y∈Bε(x0);x∈D,y∈X∖D. The set of interior points in D constitutes its interior, int(D)int(D), and the set of boundary points its boundary, ∂D∂D. DD is said to be open if any point in DD is an interior point and it is closed if its boundary ∂D∂D is contained in DD; the closure of D is the union of DD and its boundary: ¯D:=D∪∂D. D¯¯¯¯:=D∪∂D. Alternative notations for the closue of DD in XX include ¯DXDX¯¯¯¯¯¯¯¯, clos(D)clos(D) and clos(D;X)clos(D;X).1) Ex. In RR with the usual distance d(x,y)=\|x−y\|d(x,y)=\|x−y\|, the interval (0,1)(0,1) is open, [0,1)[0,1) neither open nor closed, and [0,1][0,1] closed.2) The set D:={(x,y)∈R2:x>0,y≥0} D:={(x,y)∈R2:x>0,y≥0} is neither closed nor open in Euclidean space R2R2 (metric coming from a norm, e.g., d(x,y)=‖x−y‖l2=((x1−y1)2+(x2−y2)2)1/2d(x,y)=∥x−y∥l2=((x1−y1)2+(x2−y2)2)1/2), since its boundary contains both points (x,0)(x,0), x>0x>0, in DD and points (0,y)(0,y), y≥0y≥0, not in DD. The closure of D is ¯D={(x,y)∈R2:x≥0,y≥0}. D¯¯¯¯={(x,y)∈R2:x≥0,y≥0}. An entire metric space is both open and closed (its boundary is empty). In l∞l∞, B1∌(1/2,2/3,3/4,…)∈¯B1. B1∌(1/2,2/3,3/4,…)∈B1¯¯¯¯¯¯. For a general metric space, the closed ball ˜Br(x0):={x∈X:d(x,x0)≤r} B~r(x0):={x∈X:d(x,x0)≤r} may be larger than the closure of a ball, ¯Br(x0)Br(x0)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯. If we let XX be a space with the discrete metric, {d(x,x)=0,d(x,y)=1,x≠y.Then B1(x0)={x0}, so that ¯B1(x0)=¯{x0}={x0}.But ˜B1(x0)=X. ℘ (Open) balls are open Let (X,d) be a metric space, x0 a point in X, and r>0. Then Br(x0) is open in X with respect to the metric d. Proof Pick x∈Br(x0). Then d(x,x0)0;d(x,x0)<r−ε⟹d(y,x)<ε implies d(y,x0)≤d(y,x)+d(x,x0)<ε+(r−ε)=r. This means: y∈Br(x0) if y∈Bε(x), i.e. Bε(x)⊂Br(x0). 1) An alternative to this approach is to take closed sets as complements of open sets. These two definitions, however, are completely equivalent. In particular, a set is open exactly when it does not contain its boundary. 2) Equivalent norms induce the same topology on a space (i.e., the same open and closed sets). Since all norms on Rn are equivalent, it is unimportant which norm we choose.
15548	http://athometuition.com/harmonic-progressions.aspx	Online Tutoring \| Math, English, Science Tutoring \| SAT, PSAT, GMAT, TOEFL, IELTS Tutors Homework Help 105008 live tutoring sessions served! SIGN INSIGN UP 1-888-592-9928 Toggle navigation An ISO 9001:2008 Certified Company Home About Us Our Process Programs Offered How it works Pricing Contact Us Harmonic Progressions Harmonic progressions Consider the sequence 1/3, 1/6, 1/9, 1/12, . . . Let’s write the reciprocals of these terms. They are 3, 6, 9, 12, respectively, and are in arithmetic progression. Such a progression is called a harmonic progression. A progression is said to be a harmonic progression if the reciprocals of the terms form an arithmetic progression. The n th term of a harmonic progression Examples We know that a, a + d, a + 2d, . . . are in A.P. The n th term of this A.P. is a + (n - 1)d. Its reciprocal is 1/a + (n - 1)d So the n th term of a H.P. is 1/a + (n - 1)d where the first term is 1/a and d is a constant independent of n. The n th term of an harmonic progression is 1/a + (n - 1)d. There is no concise general formula for the sum of n terms in an H.P. Harmonic means If a, b, c are in H.P, b is called the harmonic mean of a and c. Now let us find the relationship. a, b, c are in an H.P. Therefore, 1/a, 1/b, 1/c are in A.P. Therefore, 2/b = 1/a + 1/c = c+a/ac, i.e., b = 2ac/(c + a) The harmonic mean of a and c is 2ac/(c + a) If a and b are positive integers, their arithmetic mean, geometric mean and harmonic mean are in a geometric progression. Let the arithmetic mean, geometric mean and harmonic mean between a and b be A, G, H, respectively. Then A =G = square root ab; H = 2ab/(a + b) A . H =(2ab/(a + b)) = ab = (square root ab) = G 2 Hence, A, G, H are in a geometric progression. Example 1 In a harmonic progression, the fourth term is 1/9 and the 13 th term is 1/27. Write the harmonic progression. Solution: If the harmonic progression is 1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d), . . . The 4 th term = 1/(a + 3d) =1/9. Therefore a + 3d = 9 —— (1) The 13 th term = 1/(a + 12d) = 1/27 therefore a + 12d = 27——— (2) Solving (1) and (2) we get a = 3 and d = 2 So the harmonic progression is 1/3, 1/5, 1/7, 1/9, . . . Example 2 Insert 4 harmonic means between 1/12 and 1/42 Solution: Let the harmonic progression be 1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d), 1/(a + 4d), 1/(a + 5d) Equating the first and last terms 1/a = 1/12 and 1/(a + 5d) = 1/42 Therefore, a = 12 and a + 5d = 42, 5d = 42 – 12 = 30, d = 6. Therefore the harmonic means are 1/18, 1/24, 1/30, 1/36. Try these questions Find the following In a H.P 4/3, 3/2, 12/7, . ., find the 4 th and 7 th terms In a HP the 3 rd term is 1/7 and the 7 th term is 1/5, then show that the 15 th term is 1 Answers Find the following In a H.P 4/3, 3/2, 12/7, . . ., Solution: find the 4 th and 7 th terms Solution: Given H.P is 4/3, 3/2, 12/7, . . . Therefore 3/4, 2/3, 7/12, . . . are in A.P. a = 3/4, d = 2/3 – 3/4 = – 1/12 General term in A.P. = Tn = a + (n – 1)d 4 th term in A.P. = t4 = 3/4 + (4 – 1)(–1/12) = 3/4 + 3(–1/12) = (3 – 1)/4 = 1/2 Therefore, 4 th item in H.P is 2 7 th term in A.P. = t7 = 3/4 + (7– 1)(–1/12) = 3/4 – 1/2 = 1/4 Therefore, 7 th item in H.P is 4. In a HP the 3 rd term is 1/7 and the 7 th term is 1/5, then show that the 15 th term is 1 Solution: The 3 rd and 7 th terms of a HP are 1/7 and 1/5 respectively. Then 3 rd and 7 th terms of the corresponding A.P. are 7 and 5 Therefore, a + 2d = 7 ——— (1) a + 6d = 5 ——— (2) On solving equation 1 and 2, we get d = –1/2 and a = 8 Therefore, 15 th term in A.P. T15 = a + 14d = 8 + 14(–1/2) = 8 – 7 = 1 Therefore 15 th term in H.P is 1 Related Topics Absolute Values Adding and Subtracting Fractions Addition of Decimals Algebra Linear Equations Algebra Quadratic Equations Algebra Simultaneous Equations Algebraic Properties Algebraic Function Analyzing and Integrating Asymptotes Bar Graphs Basics of Statistics Circular Permutations Combinations Complex Numbers Complex Numbers AddSub Complex Numbers Division ComplexNumbers Multiplication Complex Numbers Properties Composite Functions Cube and Cube Roots Data collection Add Multipli Rules Datacollection GroupedMean Datacollection Mean Datacollection Median Datacollection Mode Datacollection Probabilitybasics Datacollection Probabilityevents Dividing Rational Numbers Division of Decimals Domain of SquareRootFunction EquationsReducibleQuadratic Exponential Functions ExponentialLogarithmicFunction Factorization FactorizationAnyQPolynomial FactorizationMonicQPolynomial Fractions & Decimal Conversion Functions Geometry Basics Graph of Rational Functions Graph of SquareRootFunction GraphicalRepresentation Graph of Complex Numbers Graphs Functions HighestCommonFactor Inequalities Inverse Square Functions Inverse of a Function Justifying Solutions LeastCommonMultiple Linear Equations2Variables LinearEquations3Variables Linear Equation System Linear Equation Graphs Linear Inequalities Graphs Maxima Minima Zeros More Functions Multiplication of Decimals Multiplying Rational Numbers Multiplying Two Polynomials Other Functions Permutations Pictorial BarChart Pictorial Bivariatedata Pictorial BoxPlots Pictorial FrequencyTable Pictorial Histogram Pictorial LineCharts Pictorial PieChart Pictorial StemLeafPlot Polynomials Addition Polynomials Division Polynomials Multiplication Predicting Values Problem Solving Strategies Quadratic Equation QuadraticEquationsFormula QuadraticEquationsSolutions QuadraticInequalities Rational Behind Functions Rational Expression Rational Functions Real Numbers Reciprocal Functions Recursive Multiplication Rational Numbers Review of functions Review of Sets and Relations Rounding Numbers Scientific Notation Simple Probabilities SimplifyingRationalExpressions SolutionQuadratEquawhen Solving Fractions SolvingQuadraticEquations Square of a Binomial SquareRootFunctions SquareRootFunInequalities Squares and Square Roots Step Function Subtraction of Decimals Types of Functions Unit Conversions Measurements WordProblemsQE © 2015 Athometuition. 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15549	https://www.khanacademy.org/math/in-in-class-8-math-india-hindi/x1091119cf1369fcf:mensuration/x1091119cf1369fcf:area-of-a-trapezium/v/area-of-trapezium-hindi	Area of trapezium (Hindi) (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Florida B.E.S.T. 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Skip to lesson content Class 8 Math (India) - Hindi Course: Class 8 Math (India) - Hindi>Unit 10 Lesson 1: Area of a trapezium Area of trapezium (Hindi) Area of trapezoids Math> Class 8 Math (India) - Hindi> Mensuration> Area of a trapezium © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Area of trapezium (Hindi) Google Classroom Microsoft Teams About About this video Area of a trapezoid is found with the formula, A=(a+b)/2 x h. Learn how to use the formula to find area of trapezoids. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted seventhgrade 4 years ago Posted 4 years ago. Direct link to seventhgrade's post “how do you find the heigh...” more how do you find the height of a right triangle with only the base? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. 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15550	https://www.uclahealth.org/medical-services/surgery/endocrine-surgery/conditions-treated/thyroid/normal-thyroid-hormone-levels	English العربية 中文 Italiano 日本語 한국어 فارسی Русский Español Tiếng Việt English العربية 中文 Italiano 日本語 한국어 فارسی Русский Español Tiếng Việt 310-825-2631 Endocrine Surgery Normal Thyroid Hormone Levels Find your care We deliver effective, minimally invasive treatments in a caring environment. Call 310-267-7838 to connect with an expert in endocrine surgery. Find a Doctor Request an Appointment Our Locations What are Normal Thyroid Hormone Levels? What is thyroid hormone? Thyroid hormone is made by the thyroid gland, a butterfly-shaped endocrine gland normally located in the lower front of the neck. Thyroid hormone is released into the blood where it is carried to all the tissues in the body. It helps the body use energy, stay warm and keeps the brain, heart, muscles, and other organs working as they should. Watch Video: I Have Symptoms of Hypothyroidism but My TSH Is Normal What Does the Thyroid Gland Do? The thyroid is a small butterfly-shaped gland that sits in front of the trachea or windpipe. It is responsible for producing thyroid hormone (T4 and T3), which is involved in regulation of metabolism, bowel function, heart rate, temperature sense, menstrual regularity, and other functions. The thyroid is controlled by the pituitary gland in the brain. The pituitary gland produces Thyroid Stimulating Hormone (TSH), which controls the secretion of hormone from the thyroid gland. Thyroid hormone exists in two main forms: thyroxine (T4) and triiodothyronine (T3). T4 is the primary form of thyroid hormone circulating in the blood (about 95%). To exert its effects, T4 is converted to T3 by the removal of an iodine atom; this occurs mainly in the liver and in certain tissues where T3 acts, such as in the brain. T3 normally accounts for about 5% of thyroid hormone circulating in the blood. Most thyroid hormone in the blood is bound by protein, while only a small fraction is "free" to enter tissues and have a biologic effect. Thyroid tests may measure total (protein bound and free) or free hormone levels. Production of thyroid hormone by the thyroid gland is controlled by the pituitary, another endocrine gland in the brain. The pituitary releases Thyroid Stimulating Hormone (abbreviated TSH) into the blood to stimulate the thyroid to make more thyroid hormone. The amount of TSH that the pituitary sends into the bloodstream depends on the amount of thyroid hormone in the body. Like a thermostat, if the pituitary senses low thyroid hormone, then it produces more TSH to tell the thyroid gland to produce more. Once the T4 in the bloodstream goes above a certain level, the pituitary’s production of TSH is shut off. In this way, the pituitary senses and controls thyroid gland production of thyroid hormone. Endocrinologists use a combination of thyroid hormone and TSH testing to understand thyroid hormone levels in the body. What is a TSH test? Thyroid testsBlood tests to measure low or high TSH levels are readily available and widely used. Not all thyroid tests are useful in all situations. TSH TestThe best way to initially test thyroid function is to measure the TSH (Thyroid Stimulating Hormone) level in a blood sample. Changes in TSH can serve as an "early warning system" — often occurring before the actual level of thyroid hormones in the body becomes too high or too low. A high TSH level indicates that the thyroid gland is not making enough thyroid hormone (primary hypothyroidism). On the other hand, a low TSH level usually indicates that the thyroid is producing too much thyroid hormone (hyperthyroidism). Occasionally, a low TSH may result from an abnormality in the pituitary gland, which prevents it from making enough TSH to stimulate the thyroid (central hypothyroidism). In most healthy individuals, a normal TSH value means that the thyroid is functioning properly. What is a T4 test? T4 TestsA Total T4 test measures the bound and free thyroxine (T4) hormone in the blood. A Free T4measures what is not bound and able to freely enter and affect the body tissues. What does it mean if T4 levels are abnormal?Importantly, Total T4 levels are affected by medications and medical conditions that change thyroid hormone binding proteins. Estrogen, oral contraceptive pills, pregnancy, liver disease, and hepatitis C virus infection are common causes of increasedthyroid hormone binding proteins and will result in a high Total T4. Testosterone or androgens and anabolic steroids are common causes of decreasedthyroid hormone binding proteins and will result in a low Total T4. In some circumstances, like pregnancy, a person may have normal thyroid function but Total T4 levels outside of the normal reference range. Tests measuring free T4 – either a free T4 (FT4) or free T4 index (FTI) – may more accurately reflect how the thyroid gland is functioning in these circumstances. An endocrinologist can determine when thyroid disease is present in the context of abnormal thyroid binding proteins. What is a T3 test? T3 TestsT3 tests measure triiodothyronine (T3) levels in the blood. A Total T3 test measures the bound and free fractions of triiodothyronine. Hyperthyroid patients typically have an elevated Total T3 level. T3 tests can be used to support a diagnosis of hyperthyroidism and can determine the severity. In some thyroid diseases, the proportions of T3 and T4 in the blood change and can provide diagnostic information. A pattern of increased T3 vs T4 is characteristic of Graves’ disease. On the other hand, medications like steroids and amiodarone, and severe illness can decrease the amount of thyroid hormone the body converts from T4 to T3 (active form) resulting in a lower proportion of T3. T3 levels fall late during hypothyroidism and therefore are not routinely used to evaluate patients with underactive or surgically absent thyroid glands. Measurement of Free T3 is possible but is often not reliable and therefore may not be helpful. What is a normal thyroid (hormone) level? To assess thyroid hormone status for low or high TSH levels we use TSH and FT4 tests. The normal value for a laboratory test is determined by measuring the hormone in a large population of healthy individuals and finding the normal reference range. Normal ranges for thyroid tests may vary slightly among different laboratories, and typical ranges for common tests are given below. TSH normal values are 0.5 to 5.0 mIU/L. Pregnancy, a history of thyroid cancer, history of pituitary gland disease, and older age are some situations when TSH is optimally maintained in a different range as guided by an endocrinologist. FT4 normal values are 0.7 to 1.9ng/dL. Individuals taking medications that modify thyroid hormone metabolism and those with a history of thyroid cancer or pituitary disease may be optimally managed with a different normal FT4 range. Total T4 and Total T3 levels measure bound and free thyroid hormone in the blood. These levels are influenced by many factors that affect protein levels in the body, including medications, sex hormones, and liver disease.A normal Total T4 level in adults ranges from 5.0 to 12.0μg/dL.A normal Total T3 level in adults ranges from 80-220 ng/dL. Free T3 assays are often unreliable and not routinely used to assess thyroid function. What does it mean if my thyroid levels are abnormal? \| Lab results \| Consider... \| --- \| \| High TSH, low thyroid hormone level \| Primary hypothyroidism \| \| High TSH, normal thyroid hormone level \| Subclinical hypothyroidism \| \| Low TSH, high thyroid hormone level \| Primary hyperthyroidism \| \| Low TSH, normal thyroid hormone level \| Early or mild hyperthyroidism \| \| Low TSH, high thyroid hormone levelFollowed by…High TSH, low thyroid hormone level \| Thyroiditis (Thyroid Inflammation) \| \| Low TSH, low thyroid hormone level \| Pituitary disease \| Patterns of thyroid tests associated with thyroid disease Primary HypothyroidismA high TSH and low thyroid hormone level (e.g. low FT4) can indicate primary hypothyroidism. Primary hypothyroidism occurs when the thyroid gland makes too little thyroid hormone. Symptoms of hypothyroidism can include feeling cold, constipation, weight gain, slowed thinking, and decreased energy.Causes of primary hypothyroidism include: Autoimmune thyroid disease, including Hashimoto's thyroiditis Thyroid gland dysfunction due to a medication (e.g. amiodarone, tyrosine kinase inhibitors, or cancer immunotherapy) Removal of all or part of the thyroid gland Radiation injury to the thyroid (e.g. external beam radiation, radioactive iodine ablation treatment) Excess treatment with anti-thyroid medications (e.g. methimazole, propylthiouracil) Early or mild hypothyroidism may present as a persistently high TSH and a normal FT4 hormone level. This pattern is called subclinical hypothyroidism and your doctor may recommend treatment. Over time, untreated subclinical hypothyroidism can contribute to heart disease. It is important to remember that normal TSH levels in older individuals (ages 70 and above) are higher than the normal ranges for younger individuals. Primary HyperthyroidismA low TSH and a high thyroid hormone level (e.g. high FT4) can indicate primary hyperthyroidism. Primary hyperthyroidism occurs when the thyroid gland makes or releases too much thyroid hormone. Symptoms of hyperthyroidism can include tremors, palpitations, restlessness, feeling too warm, frequent bowel movements, disrupted sleep, and unintentional weight loss.Causes of primary hyperthyroidism include: Graves' disease Toxic or autonomously functioning thyroid nodule Multinodular goiter Thyroid inflammation (called thyroiditis) early in the course of disease Thyroid gland dysfunction due to a medication (e.g. amiodarone or cancer immunotherapy) Excess thyroid hormone therapy Early or mild hyperthyroidism may present as a persistently low TSH and a normal FT4 hormone level. This pattern is called subclinical hyperthyroidism and your doctor may recommend treatment. Over time, untreated subclinical hyperthyroidism can worsen osteoporosis and contribute to abnormal heart rhythms. ThyroiditisThyroid inflammation, also called thyroiditis, causes injury to the thyroid gland and release of thyroid hormone. Individuals with thyroiditis usually have a brief period of hyperthyroidism (low TSH and high FT4 or Total T4) followed by development of hypothyroidism (high TSH and low FT4 or Total T4) or resolution. Some forms of thyroiditis are transient, like post-partum thyroiditis or thyroiditis following an infection, and often resolve on their own without need for medication. Other forms of thyroiditis, like thyroiditis resulting from cancer immunotherapy, interferon alpha, or tyrosine kinase inhibitors, usually result in permanent hypothyroidism and require long term treatment with thyroid hormone replacement. Your endocrinologist will monitor your thyroid tests during thyroiditis and can help determine if you need short- and long-term medications to balance your thyroid function and control any symptoms. Central HypothyroidismA low TSH and a low FT4 may indicate pituitary disease. Detection of central hypothyroidism should prompt your doctor to check for problems in other pituitary hormones, an underlying cause, and you may need imaging tests to look at the pituitary gland. Central hypothyroidism is treated with thyroid hormone replacement. Importantly, adequacy of thyroid replacement in central hyperthyroidism is assessed with FT4 and Total T4 tests not TSH as in primary hyperthyroidism, and deficiency in stress hormone cortisol should be assessed before starting thyroid treatment to prevent an adrenal crisis. Causes of central hypothyroidism include pituitary gland disease, such as a pituitary mass or tumor, history of pituitary surgery or radiation, pituitary inflammation (called hypophysitis) resulting from autoimmune disease or cancer immunotherapy, and infiltrative diseases. Rare causes of abnormal thyroid functionThyroid hormone resistanceIodine induced hyperthyroidismTSH-secreting tumor (TSH-oma)Germ cell tumorsTrophoblastic diseaseInfiltrative diseases, such as systemic scleroderma, hemochromatosis, or amyloidosis. When abnormal thyroid function tests are not due to thyroid disease While blood tests to measure thyroid hormones and thyroid stimulating hormone (TSH) are widely available, it is important to remember that not all tests are useful in all circumstances and many factors including medications, supplements, and non-thyroid medical conditions can affect thyroid test results. An endocrinologist can help you make sense of thyroid test results when there is a discrepancy between your results and how you feel. A good first step is often to repeat the test and ensure there are no medications that might interfere with the test results. Below are some common reasons for mismatch between thyroid tests and thyroid disease. Non-thyroidal illnessSignificant illness, such as an infection, cancer, heart failure, or kidney disease, or recent recovery from an illness can cause transient changes in the TSH. Fasting or starvation can also cause a low TSH. An endocrinologist can help to interpret changes in thyroid function tests in these circumstances to distinguish non-thyroid illness from true thyroid dysfunction. Test interferenceBiotin, a common supplement for hair and nail growth, interferes with many thyroid function tests and can lead to inaccurate results. Endocrinologists recommend stopping biotin supplements for 3 days before having a blood test for thyroid function. Individuals who have exposure to mice, like laboratory researchers and veterinarians, may develop antibodies against mouse proteins in their blood. These antibodies cross react with reagents in multiple thyroid function tests and cause unpredictable results. A specialized assay can accurately measure thyroid hormone levels and TSH in this circumstance. I don't feel well, but my thyroid tests are normal Thyroid blood tests are generally accurate and reliable. If you do not have low or high TSH levels and your thyroid tests are normal, your symptoms may not be related to thyroid disease, and you may want to seek additional evaluation with your primary care physician. How is hypothyroidism treated? What is thyroid medication? Thyroid Hormone TreatmentLevothyroxineis thestandard of care in thyroid hormone replacement therapy and treatment of hypothyroidism. Levothyroxine (also called LT4) is equivalent to the T4 form of naturally occurring thyroid hormone and is available in generic and brand name forms. How do I take levothyroxine?To optimize absorption of your thyroid medication, it should be taken with water at a regular time each day. Multiple medications and supplements decrease absorption of thyroid hormone and should be taken 3–4 hours apart, including calcium and iron supplements, proton pump inhibitors, soy, and multivitamins with minerals. Because of the way levothyroxine is metabolized by the body, your doctor may ask you to take an extra pill or skip a pill on some days of the week. This helps us to fine tune your medication dose for your body and should be guided by an endocrinologist. For patients with celiac disease (autoimmune disease against gluten) or gluten sensitivity, a gluten free formulation of levothyroxine is available. Some individuals may have genetic variant that affects how the body converts T4 to T3 and these individuals may benefit from the addition of a small dose of triiodothyronine. Liothyronineis replacement T3 (triiodothyronine) thyroid hormone. This medication has a short half-life and is taken twice per day or in combination with levothyroxine. Liothyronine alone is not used for treatment of hypothyroidism long term. Other formulations of thyroid hormone replacement include natural or desiccated thyroid hormone extractsfrom animal sources. Natural or desiccated thyroid extract preparations have greater variability in the dose of thyroid hormone between batches and imbalanced ratios if T4 vsT3. Natural or animal sources of thyroid hormone typically contain 75% T4 and 25% T3, compared to the normal human balance of 95% T4 and 5% T3. Treatment with a correct balance of T4 and T3 is important to replicate normal thyroid function and prevent adverse effects of excess T3, including osteoporosis, heart problems, and mood and sleep disturbance. An endocrinologist can evaluate symptoms and thyroid tests to help balance thyroid hormone medications. How do I know if my thyroid dose is correct? Monitoring thyroid levels on medicationCorrect dosing of thyroid hormone is usually assessed using the same tests for diagnosis of thyroid disease, including TSH and FT4. Thyroid tests are typically checked every 4-6 weeks initially and then every 6 to 12 months once stable. In special circumstances, such as pregnancy, a history of thyroid cancer, central hypothyroidism, amiodarone therapy, or use of combination T4 and T3 thyroid hormone replacement, your endocrinologist may check different thyroid tests. Additionally, your endocrinologist will evaluate for symptoms of hyperthyroidism and hypothyroidism and perform a physical exam. Women who are pregnant and women who may become pregnant should only be treated with levothyroxine (T4). Only T4 efficiently crosses the placenta to provide thyroid hormone to the developing fetus. Thyroid hormone is critical in early pregnancy for brain development. Normal ranges for thyroid tests in pregnancy are different and change by trimester. Women with thyroid disease in pregnancy or who are considering pregnancy should be under the care of an endocrinologist to guide therapy. Individuals with a history of thyroid cancer, even if only a portion of the thyroid was removed, also have different target ranges for TSH and FT4 tests. Thyroid hormone replacement in these individuals is closely tied to ongoing thyroid cancer surveillance, monitoring of thyroid cancer tumor markers, and dynamic assessment of recurrence risk. These patients are optimally managed by a multidisciplinary team including an endocrinologist and endocrine surgeon. 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15551	https://www.quora.com/If-x-1-how-can-we-prove-1-x-n-1-nx-for-n-0	Something went wrong. Wait a moment and try again. Binomial Expansion Linear Inequalities Induction (mathematics) Algebraic Proofs Basic Algebra Mathematical Reasoning Binomial Identities 5 If x > -1, how can we prove (1+x) ^(n) ≥1+nx for n≥0? Reuven Harmelin Lecturer at Technion - Israel Institute of Technology (1982–present) · Author has 2.3K answers and 1.9M answer views · 3y This is a special case of the well-known Bernoulli’s inequality, and its proof based on the mathematical induction principle: In the verification first step you clearly have In the induction step you assume that for some natural number n the inequality holds for all real numbers x>-1. Now since we deduce the following chain: and that is enough to conclude that the inequality indeed holds for all natural numbers n and every x>-1. After proving that special case of Bernoulli’s inequality, let me demonstrate its usefulness by proving the very important Arithmetic-Geometric Means inequality for any natur This is a special case of the well-known Bernoulli’s inequality, and its proof based on the mathematical induction principle: In the verification first step you clearly have In the induction step you assume that for some natural number n the inequality holds for all real numbers x>-1. Now since we deduce the following chain: and that is enough to conclude that the inequality indeed holds for all natural numbers n and every x>-1. After proving that special case of Bernoulli’s inequality, let me demonstrate its usefulness by proving the very important Arithmetic-Geometric Means inequality for any natural number n>1 and every list of n positive real numbers For n=2 it readily follows from the the next line So we continue by induction, assuming that the A-G Means inequality really holds for some natural number n>1 and we would like to deduce from that that this inequality holds also the consecutive natural number n+1. First observe that the induction hypothesis yields On the other hand we have and since we may apply Bernoulli’s inequality and obtain Take the root of order n+1 from both sides and you are done. Related questions How do I prove that √ 1 − x 1 + x < ln ( 1 + x ) arcsin x < 1 if 0 < x < 1 ? Given the series x 1 x 2 + x 2 x 3 + . . . + x n − 1 x n + x n x 1 = n , ∀ n ∈ N ∗ . How can I prove that x n + 1 = x n ? Is \| x − 1 \| + \| x \| = 1 equivalent to 4 equations + ( x − 1 ) + x = 1 , − ( x − 1 ) − x = 1 , + ( x − 1 ) − x = 1 , − ( x − 1 ) + x = 1 or equivalent to 2 equations + ( x − 1 ) + x = 1 , − ( x − 1 ) − x = 1 ? How do you show that if a0 then (a^2+b^2) /2≥ ((a+b) /2) ^2? How can I simplify 1 1 + x m − n + 1 1 + x n − m ? Mohammad Afzaal Butt B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views · 3y We can prove the required result by mathematical induction. The result is true for n = 0, since (1+x)n=(1+x)0=1 The result is also true for n = 1, since (1+x)1=1+x Let the result be true for n = k, that is (1+x)k≥1+xk We need to prove that the result is also true for n = k + 1, that is (1+x)k+1≥1+x(k+1) by our assumption (1+x)k≥1+xk multiplying both sides by (1 + x) (1+x)k+1≥(1+x)(1+xk) ⟹(1+x)k+1≥(1+xk)+x(1+xk) ⟹(1+x)k+1≥1+x(k+1)+x2k2 ⟹(1+x)k+1≥1+x(k+1)∵x2k We can prove the required result by mathematical induction. The result is true for n = 0, since (1+x)n=(1+x)0=1 The result is also true for n = 1, since (1+x)1=1+x Let the result be true for n = k, that is (1+x)k≥1+xk We need to prove that the result is also true for n = k + 1, that is (1+x)k+1≥1+x(k+1) by our assumption (1+x)k≥1+xk multiplying both sides by (1 + x) (1+x)k+1≥(1+x)(1+xk) ⟹(1+x)k+1≥(1+xk)+x(1+xk) ⟹(1+x)k+1≥1+x(k+1)+x2k2 ⟹(1+x)k+1≥1+x(k+1)∵x2k2>0 ∴ the result is also true for n = k + 1. Hence by the principle of mathematical induction, the result is true for all n∈N+{0} Simon Tsai Lives in Taiwan · Author has 4.9K answers and 2M answer views · Updated 3y For any x∈R and t≥1, ex≥(1+xt)t≥1+x The second inequality proves part of your claim: Let x↦tx and see that it’s true. Now, we are proving the second inequality: y=(1+xt)t Take the logarithms of both sides and differentiate: ∂ln(y)∂t=ln(1+xt)−xt+x Recall the integral definition of logarithm: \displaystyle \ln \left( 1 + w \right) = \int_{0}^{w} \frac {1}{1 + For any x∈R and t≥1, ex≥(1+xt)t≥1+x The second inequality proves part of your claim: Let x↦tx and see that it’s true. Now, we are proving the second inequality: y=(1+xt)t Take the logarithms of both sides and differentiate: ∂ln(y)∂t=ln(1+xt)−xt+x Recall the integral definition of logarithm: ln(1+w)=∫w011+u∂u≥w1+w Letting w=x/t completes the proof. Francesco Amato Studied at Polytechnic University of Turin (Graduated 1977) · Author has 4.5K answers and 1M answer views · 3y Let’s start with n=0, and then n=1,2,3 (1+x)0=1+nx\|n=0=1 (1+x)1=1+nx\|n=1=1+x (1+x)2=1+2x+x2≥1+nx\|n=2=1+2x since x2≥0 for all x∈R (1+x)3=1+3x+3x2+x3≥1+nx\|n=3=1+3x but if 3x2+x3≥0→x≥−3 Generally we can expand the function f(x)=(1+x)n in McLaurin series f(x)=f(0)+x⋅f′(0)+x2⋅f′′(0)/2!+x3⋅f′′′(0)/3!+…xn⋅f(n)(0)/n!+… f(0)=1; f′(0)=n(1+x)n−1\|x=0=n; f′′(0)=n(n−1)(1+x)n−2\|x=0=n(n−1); f′′′(0)=n(n−1)(n−2) (1+x)n=1+nx+n(n−1)x2/2+n(n−1)(n−2)x3/6+…+n!(n−r)!r!xr+… T Let’s start with n=0, and then n=1,2,3 (1+x)0=1+nx\|n=0=1 (1+x)1=1+nx\|n=1=1+x (1+x)2=1+2x+x2≥1+nx\|n=2=1+2x since x2≥0 for all x∈R (1+x)3=1+3x+3x2+x3≥1+nx\|n=3=1+3x but if 3x2+x3≥0→x≥−3 Generally we can expand the function f(x)=(1+x)n in McLaurin series f(x)=f(0)+x⋅f′(0)+x2⋅f′′(0)/2!+x3⋅f′′′(0)/3!+…xn⋅f(n)(0)/n!+… f(0)=1; f′(0)=n(1+x)n−1\|x=0=n; f′′(0)=n(n−1)(1+x)n−2\|x=0=n(n−1); f′′′(0)=n(n−1)(n−2) (1+x)n=1+nx+n(n−1)x2/2+n(n−1)(n−2)x3/6+…+n!(n−r)!r!xr+… This series, if n≥0 is convergent for x2≤1, that is −1≤x≤1, but when n is a positive integer, the expression consists of a finite numbers of terms, alternately positive and negative, and the relation (1+x)n≥1+nx should be verified for each n and some restriction set on the allowed x values. 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How do I prove that ∀m,n∈in such that n≥m²+m we have n! >m^(n+1)? My teachers even struggled. How do I prove that √ 1 − x 1 + x < ln ( 1 + x ) arcsin x < 1 if 0 < x < 1 ? Let n ≥ 3 be an integer, and suppose x 1 , x 2 , ⋯ , x n are positive real numbers, such that x 1 + x 2 + ⋯ + x n = 1. How do I show that x 1 − x 2 1 + x 1 − x 3 2 ⋯ + x 1 − x n n − 1 + x 1 − x 1 n < 2 ? Carlo Knows Italian · Author has 1.4K answers and 550.8K answer views · 3y Mathematical induction is used to prove Bernoulli's inequality. Base case. If x > -1 and n=0 then 0 = 0 Induction step. Inductive hypothesis. (1+x)n≥1+nx is true for n≥0. Thesis. (1+x)n+1≥1+(n+1)˙x will be true. (1+x)n+1=(1+x)⋅(1+x)n=(1+x)⋅(1+nx)=1+nx+x+nx2=1+(n+1)˙x note. we use the inductive hypothesis in the second step we use nx² ≥ 0 in the fourth step Brian Constantinescu Author has 176 answers and 728K answer views · Updated 6y Related If x i ∈ R , 0 ≤ x 1 ≤ . . . ≤ x n , n ≥ 2 and 1 1 + x 1 + . . . + 1 1 + x n = 1 , prove that √ x 1 + . . . + √ x n ≥ ( n − 1 ) ( 1 √ x 1 + . . . + 1 √ x n ) I started running into problems when I considered the properties of the HM-GM inequality, which got me pretty irked. I think I saw this problem in a book before, which is why I’m trying to pull through and find an answer. We are given the following fact: n∑i=111+xi=1 Let’s consider another fact simultaneously, that 0≤x1≤…≤xn∀n≥2 A simple solution to the previous summation for any n that satisfies the condition n≥2 is for all x1,x2…xn=n−1. The solution is true because the x_ I started running into problems when I considered the properties of the HM-GM inequality, which got me pretty irked. I think I saw this problem in a book before, which is why I’m trying to pull through and find an answer. We are given the following fact: n∑i=111+xi=1 Let’s consider another fact simultaneously, that 0≤x1≤…≤xn∀n≥2 A simple solution to the previous summation for any n that satisfies the condition n≥2 is for all x1,x2…xn=n−1. The solution is true because the xi are allowed to equal each other. This simple observation means that the following sum is also true for our restricted xi: n∑i=1xi1+xi=n−1 In fact, as Jan van Delden pointed out in the comments (thankfully), this equation is true for any xi, which can be easily seen if we consider another way to rewrite n−1: n−1=n∑i=11−n∑i=111+xi =n∑i=1(1−11+xi)=n∑i=1xi1+xi Now that we’ve got this piece figured out, let’s look at what we’re trying to prove. We need to show that n∑i=1√xi−(n−1)n∑i=11√xi≥0 ⟹n∑i=111+xin∑i=1√xi−n∑i=1xi1+xin∑i=11√xi≥0 We’ve got some summation multiplication happening over here, which can be written more compactly as: ∑(i,j)∈{1,…,n}2√xi1+xj−∑(i,j)∈{1,…,n}2xj(1+xj)√xi≥0 ∑(i,j)∈{1,…,n}2xi−xj(1+xj)√xi≥0 The problem then becomes to prove that this sum evaluates to something positive. That’s another hiccup. We have to come up with a relation between xi and xj which might shed some light on this. So check it out, 11+xi+11+xj=2+xi+xj1+xi+xj+xixj≤1 This would imply that xixj≥1, and this is the key! Since the question also states that xi≥0, then our newfound discovery also proves that xj≥0. This will be a little important laters. Back to our previous sum. There’s no terms like xixj in it yet, but do not fret! Really, we are only concerned with pairs of indices where i>j, and if we consider this condition, then the previous sum would be equivalent to this: ∑i>j(√xi−√xj)2(√xi+√xj)(√xixj−1)√xixj(1+xi)(1+xj)≥0 Let’s dissect this sum a little bit. The numerator has a term that’s squared, which is always positive. It also has a term which is the sum of principle square roots of positive numbers, which is again always positive. Moreover, the last term is √xixj−1 which we know must be greater than or equal to zero. The denominator has two terms which are just a positive number plus one, which is always positive. Lastly, another term which is just √xixj, which we know is positive. Therefore, we see that the sum adds up only positive values or zero, which means that it must be greater than or equal to zero. And with that, we conclude the answer. Let me know how bad I messed up on this one, because I’m not very good with inequality questions, and have a great day! Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Bernard Montaron PhD in Mathematics & Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (Graduated 1980) · Author has 3.2K answers and 2.1M answer views · 2y Related How do you prove that for any positive integer n and real number x > -1, the following inequality holds: (1+x) ^n ≥ 1 + nx? This is true for n=1. Consider now n>1. The derivative of the function f(x)=(1+x)n−(1+nx) is f′(x)=n((1+x)n−1−1). It is such that f′(0)=0 and for n>1 f′(x)<0 in the interval −1<x<0. This implies that f(x)≥f(0) in this interval, i.e. (1+x)n≥1+nx, because f(0)=0. And if x>0, i.e. x+1>1, we have f′(x)=n((1+x)n−1−1)>0 which implies that f(x) is growing above f(0)=0, i.e. it stays positive. So, again, (1+x)n≥1+nx Voilà! Amitabha Tripathi have more than a working knowledge of Z · Upvoted by Shubhankar Datta , Master of Science Mathematics, Jadavpur University (2022) and Michael Jørgensen , PhD in mathematics · Author has 4.7K answers and 13.9M answer views · 7y Related How do I prove that [ n x ] = [ x ] + ∑ n − 1 1 [ x + 1 / n ] ? For x∈R and n∈N, prove that ⌊nx⌋=⌊x⌋+n−1∑i=1⌊x+in⌋=n−1∑i=0⌊x+in⌋. Write x=⌊x⌋+{x}, where 0≤{x}<1. Then there is a unique k∈{0,1,2,…,n−1} for which kn≤{x}<k+1n. Fix i∈{0,1,2,…,n−1}. Then ⌊x+in⌋=⌊x⌋+⌊{x}+in⌋ = \begin{cases} \lfloor x \rfloor & \mbox{if}\: 0 \le i \le n-k- For x∈R and n∈N, prove that ⌊nx⌋=⌊x⌋+n−1∑i=1⌊x+in⌋=n−1∑i=0⌊x+in⌋. Write x=⌊x⌋+{x}, where 0≤{x}<1. Then there is a unique k∈{0,1,2,…,n−1} for which kn≤{x}<k+1n. Fix i∈{0,1,2,…,n−1}. Then ⌊x+in⌋=⌊x⌋+⌊{x}+in⌋ ={⌊x⌋if0≤i≤n−k−1;⌊x⌋+1ifn−k≤i≤n−1. Therefore n−1∑i=0⌊x+in⌋=n−k−1∑i=0⌊x⌋+n−1∑i=n−k(⌊x⌋+1) =n−1∑i=0⌊x⌋+n−1∑i=n−k1 =n⌊x⌋+k =⌊nx⌋, since nx=n⌊x⌋+n{x} and k≤n{x}<k+1. ■ Sponsored by CDW Corporation How can AI help your teams make faster decisions? CDW’s AI solutions offer retrieval-augmented generation (RAG) to expedite info with stronger insights. Doug Dillon Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views · 6y Related If x i ∈ R , 0 ≤ x 1 ≤ . . . ≤ x n , n ≥ 2 and 1 1 + x 1 + . . . + 1 1 + x n = 1 , prove that √ x 1 + . . . + √ x n ≥ ( n − 1 ) ( 1 √ x 1 + . . . + 1 √ x n ) Thomas Bell Ph.D. in Mathematics, University of Oregon (Graduated 2013) · Author has 3.9K answers and 686.8K answer views · 2y Related How do you prove that for any positive integer n and real number x > -1, the following inequality holds: (1+x) ^n ≥ 1 + nx? This is equality for n=1. Then use induction on n by multiplying both sides of the inductive hypothesis by 1+x. The right side is what you want with an extra nx^2 term, which is greater than or equal to zero. Awnon Bhowmik Studied at University of Dhaka · Author has 3.7K answers and 11.2M answer views · 9y Related How do I prove that if n 1 and n ∈ Z , x n − 1 = ( x − 1 ) ( 1 + x + x 2 + . . . + x n − 1 ) ? So many people have asked me this question that I was compelled to post this question myself and answer it. Here goes…. Consider the series 1+x+x2+x3+…..+xn−1 This is what we call a geometric series. Let us find the sum. We know that, for a geometric series Sn=arn−1r−1, for r>1, we will consider the common ratio to be greater than 1 here. Using this result for our series 1+x+x2+x3+…..+xn−1=(1)(xn−1)x−1 ⟹1+x+x2+x3+…..+xn−1=xn−1x−1,x≠1 [which is obvious, if the common ratio is 1, then we do not have a geometric series, rather a repeating numbe So many people have asked me this question that I was compelled to post this question myself and answer it. Here goes…. Consider the series 1+x+x2+x3+…..+xn−1 This is what we call a geometric series. Let us find the sum. We know that, for a geometric series Sn=arn−1r−1, for r>1, we will consider the common ratio to be greater than 1 here. Using this result for our series 1+x+x2+x3+…..+xn−1=(1)(xn−1)x−1 ⟹1+x+x2+x3+…..+xn−1=xn−1x−1,x≠1 [which is obvious, if the common ratio is 1, then we do not have a geometric series, rather a repeating number series] Cross multiplying xn−1=(x−1)(1+x+x2+…..+xn−1) [This thing doesn’t work for n<1 since it does not have a closed form for n<1] Tons of ways to skin a cat, just choose one. If you take it literally and do try to skin a cat, I will call the PETA on you, I love cats. Thank you. Mike Hirschhorn Honorary Associate Professor of Mathematics at UNSW · Upvoted by Bernard Montaron , PhD Mathematics & Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (1980) · Author has 8.1K answers and 2.7M answer views · 2y Related How do you prove that for any positive integer n and real number x > -1, the following inequality holds: (1+x) ^n ≥ 1 + nx? By induction on n . Abdelhadi Nakhal Lives in Morocco · Author has 1.3K answers and 435.7K answer views · 6y Related How can one prove [ x ] + [ x + 1 n ] + [ x + 2 n ] + … + [ x + n − 1 n ] = [ n x ] ? So we have to prove that any real number x and any natural number n≥1: ⌊x⌋+⌊x+1n⌋+⋯+⌊x+kn⌋+⋯+⌊x+n−1n⌋=⌊nx⌋ x=⌊x⌋+y for some y with 0≤y<1 It is easy to show that there exist a unique (p,α) such that: y=pn+α with p≤(n−1) and α<1n (p is an integer and α a real) x=⌊x⌋+pn+α ⇒nx=n⌊x⌋+p+nα if we apply floor function to both sides and taking into account that nα<1 we o So we have to prove that any real number x and any natural number n≥1: ⌊x⌋+⌊x+1n⌋+⋯+⌊x+kn⌋+⋯+⌊x+n−1n⌋=⌊nx⌋ x=⌊x⌋+y for some y with 0≤y<1 It is easy to show that there exist a unique (p,α) such that: y=pn+α with p≤(n−1) and α<1n (p is an integer and α a real) x=⌊x⌋+pn+α ⇒nx=n⌊x⌋+p+nα if we apply floor function to both sides and taking into account that nα<1 we obtain: ⇒⌊nx⌋=n⌊x⌋+p Now it suffices to prove that: ⌊x⌋+⌊x+1n⌋+⋯+⌊x+kn⌋+⋯+⌊x+n−1n⌋=n⌊x⌋+p We have: ⌊x⌋≤x+kn=⌊x⌋+p+kn+α<⌊x⌋+1 whenver p+k≤n−1⇒k≤n−p−1 ⌊x⌋+1<x+kn<⌊x⌋+2 whenever k≥n−p ∑n−10⌊x+kn⌋=∑n−p−10⌊x+kn⌋+∑n−1n−p⌊x+kn⌋ =(n−p)⌊x⌋+p(⌊x⌋+1)=n⌊x⌋+p Proof finished Related questions Let a, b, c in R, such that a x 2 + b x + c ≥ 0 for every x≥0. How do I prove that a 1998 + b 1997 + c 1996 ≥ 0 ? For all n≥1, let a n = ∑ n − 1 k = 1 sin ( ( 2 k − 1 ) π 2 n ) cos 2 ( ( k − 1 ) π 2 n ) cos 2 ( k π 2 n ) . What is ? How can you show that n+5/n²+3 ≤6/n for all n≥1? How can I prove (a^n - b^n) /(a-b) =a^(n-1) +a^(n-2) b+⋯+ab^(n-2) +b^(n-1) for all n≥2,a≠b? [How can one prove [ x ] + [ x + 1 n ] + [ x + 2 n ] + … + [ x + n − 1 n ] = [ n x ] ?]( Would you prove that for every n≥0 (a^ (2n+1) +b^ (2n+1)) /(a+b) =a^(2)-a^ (2n-1) b +…-ab^ (2n-1) +b^(2), with a,b and n ∈ N and a > b ≥ 0? How can I prove unbounded? Let K= {x: Ax≤b, x≥0}, b≥0. If there is non-zero solution x≥0 of Ax≤0, show that K is unbounded. (A is mxn, x is nx1 matrix, b is mx1 matrix)? Can you explain the solution? How do I prove that ∀m,n∈in such that n≥m²+m we have n! >m^(n+1)? My teachers even struggled. How do I prove that √ 1 − x 1 + x < ln ( 1 + x ) arcsin x < 1 if 0 < x < 1 ? Let [math]n\ge 3[/math] be an integer, and suppose [math]x_1,x_2,\cdots ,x_n[/math] are positive real numbers, such that [math]x_1+x_2+\cdots +x_n=1.[/math] How do I show that [math]x_1^{1-x_2}+x_2^{1-x_3}\cdots+x_{n-1}^{1-x_n}+x_n^{1-x_1}<2[/math] ? How do I prove that [math] [nx] = [x] + \sum_{1}^{n-1}[x + 1/n] [/math] ? How do I find the value of [math]\sum^{2025}_{n = 0}\int^1_0\frac{1}{2(x + n + 1)\sqrt{(x + n)(x + n + 1)}}\,dx[/math] ? [What is minimum of [math]\sum_{i = 1}^{n}[/math] [math]\frac{1}{x_{i}(1+x_{i+1})}[/math] when [math]\prod_{i=1}^{n} x_{i} = 1[/math] ( [math]x_{n+1} = x_{1}[/math] and all [math]x_{i}[/math] 's are positive) ?]( If [math]f(x) =\frac{\sin{x}}{x}[/math] , how do you prove that [math]\|f^{(n)}(x)\|\leq \frac{1}{n+1}[/math] ? How can I prove that ∫ x 0 t 2 n + 1 1 − t 2 = ln 1 √ 1 − x 2 − n ∑ k = 1 x 2 k 2 k for − 1 < x < 1 ? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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It shows their types, their formulas for area and perimeter, and their properties. Updated: 11/21/2023 Create an account to begin studying today Used by over 30 million students worldwide Create an account Characteristics of a Trapezoid ------------------------------ A trapezoid (or trapezium) is a 2D four-sided polygon (quadrilateral) that has at least two parallel sides. A quadrilateral with two pairs of parallel sides, for example, is usually called a parallelogram. Therefore, by definition, a parallelogram has the characteristics of a trapezoid. With all that said, it is possible to find educational literature excluding parallelograms from the trapezoid category, which includes only quadrilaterals that specifically have one pair of parallel sides. Figure 1: Trapezoids. Figure 2: Trapezoids with only one pair of parallel sides. Figure 2 shows the most popular shapes called trapezoids. For those, the parallel sides are called bases, and the remaining two sides are called legs. 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Take QuizWatch Next Lesson Replay Just checking in. Are you still watching? Yes! Keep playing. Your next lesson will play in 10 seconds 0:02 Definition of a Trapezoid 0:26 Special Trapezoids 1:28 Properties of a Trapezoid 2:26 Practice 3:30 Lesson Summary View Video Only Save Timeline 198K views Video Quiz Course Video Only 198K views What Is the Formula for a Trapezoid? ------------------------------------ The next subsections respond to the question: what is the formula for a trapezoid? This includes equations to find perimeter and area that can be applied to all trapezoids. Trapezoid Formula: Perimeter The perimeter of a trapezoid formula follows the logic of perimeter calculations for other polygons. It is simply the sum of the length of its sides. Therefore: Trapezoid perimeter = base + base 2 + leg + leg 2 The sides are usually labeled as a, b, c, and d, transforming the formula into: P=a+b+c+d Where P stands for perimeter and will be measured by the same unit of length used to measure each of the sides. Figure 3: trapezoid with sides labeled as a, b, c, and d. Example 1: Calculate the perimeter of the trapezoid shown in figure 4. Figure 4: trapezoid for example 1. Solution: P=a+b+c+d P=7+12+9+9=37 c m Trapezoid Formula: Area The area of a trapezoid is calculated by the formula: A=(b+B)∗h 2 Where B is the longer base, b is the shorter base, and h is the height of the trapezoid. The unit to measure area will be the square of the unit used to measure the length of the sides. For a trapezoid with sides labeled as a, b, c, and d, the formula becomes: A=(a+b)∗h 2 Figure 5: Area of a trapezoid. Example 2 Calculate the area of the trapezoid showed in figure 6. Figure 6: trapezoid for example 2. Solution: A=(a+b)∗h 2 A=(7+12)∗9 2=85.5 c m 2 Types of Trapezoids: Non-Parallelogram Trapezoids ------------------------------------------------- The next sections will present different types of trapezoids that are non-parallelograms. The three types that will be introduced are: right trapezoids, scalene trapezoids, and isosceles trapezoids. What Is a Right Trapezoid? -------------------------- A right trapezoid is one that has at least two right angles. Figure 7 shows an example of what is a right trapezoid. Figure 7: Right trapezoid. All rectangles are right trapezoids by definition. They have at least two parallel sides and at least two right angles. Indeed, rectangles have two pairs of parallel sides and four right angles. Nevertheless, it is not possible for a trapezoid (as the ones studied here) to have an odd number of right angles. Area of a Right Trapezoid Although the previously presented area formula serves to find the area of a right trapezoid. There is another method to get to its value. It requires that the polygon be divided into two parts: a rectangle and a right triangle. Then, one just needs to calculate the area of each and add them up. Figure 8: area of rectangle + area of right triangle = area of trapezoid. Example 3 Find the area of the right trapezoid of figure 9. Figure 9: Trapezoid for example 3. Solution: Area of trapezoid formula: A=(a+b)∗h 2=(8+10)∗6 2=54 c m 2 Area of triangle + Area of rectangle: A (triangle) = B∗h 2=2∗6 2=6 c m 2 A (rectangle) = Side Side = 8∗6=48 c m 2 A of trapezoid = 54 c m 2 What Is a Scalene Trapezoid? ---------------------------- The characteristics of a trapezoid categorized as scalene are simple: none of its four sides have a length equal to the other. Figure 10: Scalene trapezoid. What Is an Isosceles Trapezoid? ------------------------------- Isosceles trapezoids can be visually identified by their symmetry. Figure 11: Isosceles trapezoid symmetry. Its main properties are: Its legs are congruent. Its lower base angles are congruent. Its upper base angles are congruent. Any lower base angle is supplementary to any upper base angle. Its diagonals are congruent. Figure 12: Trapezoid angles and legs. The gridlines in figure 12 helps one to observe that the angles at vertices B and C are congruent, as are the angles at vertices A and D. The pairs of angles of vertex locations AB, AC, DB, and DC are supplementary. Finally, the legs A B― and C D― are congruent. Figure 13: Trapezoid diagonals. The diagonal B D― and A C― are congruent. Lesson Summary -------------- This lesson described quadrilaterals called trapezoids. They are characterized by having at least one pair of parallel sides, which, by definition, include parallelograms. However, parallelograms are often excluded from the study of trapezoids for one of the two reasons: parallelograms are studied separately; some sources consider trapezoids only quadrilaterals that have specifically one pair of parallel sides. The perimeter of a trapezoid is the sum of the length of its four sides. The area of a trapezoid is given by the formula A=(a+b)∗h 2. Although these equations work for all trapezoids, the ones that have at least two right angles (right trapezoids) can have their area calculated by an alternative method as well. They can be divided into one right triangle and one rectangle. Adding up the areas of these two polygons results in the trapezoid area. The other two types of trapezoids are the scalene and the isosceles. Scalene trapezoids are ones with no sides of equal length. Isosceles trapezoids have more specific properties: their legs are congruent, their adjacent angles connected by the bases are congruent, their adjacent angles connected by the legs are supplementary, and their diagonals are congruent. Video Transcript Definition of a Trapezoid Image of a trapezoid A trapezoid is a 2-dimensional geometric figure with four sides, at least one set of which are parallel. The parallel sides are called the bases, while the other sides are called the legs. The term 'trapezium,' from which we got our word trapezoid has been in use in the English language since the 1500s and is from the Latin meaning 'little table.' Special Trapezoids There are a few special trapezoids that are worth mentioning. In an isosceles trapezoid, the legs have the same length and the base angles have the same measure. Isosceles trapezoid In a right trapezoid, two adjacent angles are right angles. Right trapezoid If the trapezoid has no sides of equal measure, it is called a scalene trapezoid. Scalene trapezoid A parallelogram is a trapezoid with two sets of parallel sides. Parallelogram There is actually some controversy over whether a parallelogram is a trapezoid. One group states that the definition of a trapezoid is having only one set of parallel sides, which would exclude the parallelogram because it has two sets of parallel sides. The other, more mainstream group, states that the definition of a trapezoid is having at least one set of parallel sides, which includes the parallelogram. For our discussions, because it is the more widely accepted view, we will consider a parallelogram to be a trapezoid. Properties of a Trapezoid The formula for the perimeter of a trapezoid is P = (a + b + c + d). To find the perimeter of a trapezoid, just add the lengths of all four sides together. The formula for the area of a trapezoid is A = (1/2)(h)(a + b), where: h = height (This is the perpendicular height, not the length of the legs.) a = the short base b = the long base Dimensions of a trapezoid An isosceles trapezoid has special properties that do not apply to any of the other trapezoids: Opposite sides of an isosceles trapezoid are the same length (congruent). The angles on either side of the bases are the same size or measure (also congruent). The diagonals are congruent. Adjacent angles (next to each other) along the sides are supplementary. This means that their measures add up to 180 degrees. Practice Let's try a couple of practice problems to test your newfound trapezoid knowledge. Feel free to pause the video at any point to work through the problems yourself. 1.) Find the perimeter and area of the following trapezoid: Trapezoid for problem 1 To find the perimeter, simply add all four sides together. P = 12mm + 14mm + 18mm + 13mm = 57mm To find the area, use the formula A = (1/2)(h)(a + b). A = (1/2)(11mm)(12mm + 18mm) = 165mm^2 2.) Find the area of the following trapezoid: Trapezoid for problem 2 Again, use the area formula A = (1/2)(h)(a + b). A = (1/2)(6ft)(9ft + 4ft) A = 39ft^2 Lesson Summary A trapezoid is a 2-dimensional figure with four sides. In order for it to be classified as a trapezoid, it must have at least one set of parallel sides. Trapezoids play a key role in architecture and also can be found in numerous everyday items. Take a look at the glass you are drinking from at your next meal. From the side, it's probably shaped like a trapezoid. Learning Outcomes Review the video lesson and its corresponding transcript so that you can: Define a trapezoid and identify its properties Illustrate several special trapezoids Point out the properties of isosceles trapezoids Find the perimeter and area of a trapezoid Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a Member Already a member? Log In Back Resources created by teachers for teachers Over 30,000 video lessons& teaching resources‐all in one place. Video lessons Quizzes & Worksheets Classroom Integration Lesson Plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Back FAQ Do trapezoids have congruent diagonals? Isosceles trapezoids do have congruent diagonals. They also have congruent legs. Their adjacent angles connected by the bases are congruent, and their adjacent angles connected by the legs are supplementary. What is the equation of a trapezoid? To calculate the perimeter of a trapezoid, one can simply add up its four sides. To calculate the are, one must add the bases, multiply by the height, and divide it by two. What are properties of a trapezoid? Trapezoids are quadrilaterals that have at least two parallel sides. Some sources consider trapezoids only the figures with one pair of parallel sides, excluding the parallelogram from the trapezoid category. Are trapezoids isosceles? Isosceles trapezoids are one type of trapezoids. Their properties are: their legs are congruent, their adjacent angles connected by the bases are congruent, their adjacent angles connected by the legs are supplementary, and their diagonals are congruent. More Trapezoid Lessons Trapezoid Lesson Plan Area of a Trapezoid Lesson Plan Median of a Trapezoid \| Formula & Examples How to Find the Altitude of a Trapezoid How to Find the Perimeter of an Isosceles Trapezoid Isosceles Trapezoid Properties & Formula \| What is an Isosceles Trapezoid? 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Overview of Math Concepts Less Than Symbol \| Overview, Application & Examples 4:10 2D Shapes Overview, Names & Examples 4:35 Trapezoid \| Characteristics, Properties & Formulas 3:58 5:56 Next Lesson Surface Area \| Definition, Formula & Examples Parentheses in Math \| Definition, Uses & Examples 3:58 Universal Set Symbol, Definition & Examples 6:03 Complement of a Set \| Overview & Examples 5:59 Zero Exponent Rule \| Overview & Examples 4:32 Quotient of Powers Definition, Properties & Examples 4:58 Simplest Form \| Definition, Overview & Examples 5:49 Slope \| Definition, Formula & Examples 7:10 Skewed Distribution \| Definition, Types & Examples 5:09 Change of Base Formula \| Logarithms, Examples & Proof 4:54 Transformation in Math \| Definition, Types & Examples 6:27 Translation in Math \| Definition, Rules & Examples 4:23 Fixed Interval Reinforcement \| Definition, Schedule & Examples 4:00 Scatterplot & Correlation \| Overview, Graphs & Examples 7:48 Dilation in Math \| Definition, Formula & Examples 5:30 Simplified Fractions \| Definition, Steps & Examples 4:44 More Trapezoid Lessons How to Find the Perimeter of an Isosceles Trapezoid Isosceles Trapezoid Properties & Formula \| What is an Isosceles Trapezoid? Midsegment of a Trapezoid \| Theorem, Formula & Examples Perimeter & Area of a Trapezoid \| Overview & Formula Proofs & Angles of an Isosceles Trapezoid \| Overview & Diagram Trapezoid \| Definition, Types & Attributes Trapezoid Courses Geometry Supplemental Math: Study Aid View course Geometry Big Ideas Math Geometry: Online Textbook Help View course Geometry Discovering Geometry An Investigative Approach: Online Help View course Geometry Explorations in Core Math - Geometry: Online Textbook Help View course Geometry Holt McDougal Larson Geometry: Online Textbook Help View course Like this lessonShare Explore our library of over 88,000 lessons Search Browse Browse by subject College Courses Business English Foreign Language History Humanities Math Science Social Science See All College Courses High School Courses AP Common Core GED High School See All High School Courses Other Courses College & Career Guidance Courses College Placement Exams Entrance Exams General Test Prep K-8 Courses Skills Courses Teacher Certification Exams See All Other Courses Upgrade to enroll× Upgrade to Premium to enroll in Supplemental Math: Study Aid Enrolling in a course lets you earn progress by passing quizzes and exams. 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15553	https://turnawoodbowl.com/drying-green-wood-bowls-6-methods-success/	Drying Green Wood Bowls – 6 Methods For Success - Turn A Wood Bowl Skip to content Recommended Equipment Woodturning Tools Chucks and Faceplates Lathe Accessories Safety Equipment Sharpening Equipment Sanding Equipment Bowl Coring Equipment Finishing Supplies Green Wood Equipment Photo Product Shoot Gear Quick List Reference Articles About Search Courses Shop Cart My Account Login Recommended Equipment Woodturning Tools Chucks and Faceplates Lathe Accessories Safety Equipment Sharpening Equipment Sanding Equipment Bowl Coring Equipment Finishing Supplies Green Wood Equipment Photo Product Shoot Gear Quick List Reference Articles About Search Courses Shop Cart My Account Login Recommended Equipment Woodturning Tools Chucks and Faceplates Lathe Accessories Safety Equipment Sharpening Equipment Sanding Equipment Bowl Coring Equipment Finishing Supplies Green Wood Equipment Photo Product Shoot Gear Quick List Reference Articles About Search Courses Shop Cart My Account Login Recommended Equipment Woodturning Tools Chucks and Faceplates Lathe Accessories Safety Equipment Sharpening Equipment Sanding Equipment Bowl Coring Equipment Finishing Supplies Green Wood Equipment Photo Product Shoot Gear Quick List Reference Articles About Search Courses Shop Cart My Account Login Drying Green Wood Bowls – 6 Methods For Success How do I start drying green wood bowls? There are many approaches for drying green wood bowls once they have been turned or roughed out, including using paper bags, a microwave, kiln, denatured alcohol, even desiccant drying beads, or just simply waiting. Inside Drying Green Wood Bowls In a dream world, we could find a freshly cut tree, process the green wood into bowl blanks, turn a bowl, then set that bowl on the shelf and it will always look like it just came off the lathe. Unfortunately, that is not the nature of wood. Fresh cut or green wood contains moisture. For that matter, even logs cuts years ago usually hold moisture to some degree. This moisture inside wood can remain trapped or escape over time, based on many different factors. If not managed, uneven moisture movement causes stresses in the green wood and will likely change the overall shape of a turned bowl, or cause structural failure in the form of cracks, checks, and splits. As we have covered in Turning Green Wood Bowls, the term “drying green wood,” is really not a thing. A more accurate term is “equalizing wood,” but since “drying” is what most people think and say, we’ll use the term “drying” to keep this simple. Ultimately, we really want to control our green wood bowl turning to fit our specific needs. Once-turned and twice-turned bowls have different characteristics. The quicker and more abrupt the drying process happens for once-turned bowls, the more chances there are for cracks and unpleasant results. Waiting for full twice-turned bowls to dry is boring and forces us to wait to have the beautiful final turned bowl we are imagining. With the following methods, we can manage the even moisture equilibrium within our drying green wood bowls. If you’d like to learn all about the twice-turning process, be sure to read this article. Green Wood Off the Lathe It is possible to take a freshly turned green wood bowl from the lathe and place it on your shelf and it will dry fine without problems. However, this is the exception and will be highly dependent on several factors. Drying Green Wood Bowls, Dynamic Contributing Factors Include tree species time since cut relative humidity temperature air flow light sources heat sources bowl wall thickness bowl shape Enemies To more closely control and ease the drying process we need to be aware of our enemies. The two biggest enemies are unevenness and environment. If the wood is turned uneven, with thick and thin areas, it is almost impossible to dry these areas smoothly without tension and cracks forming. Controlling the environment is critical to ease the process of drying green wood bowls. Specifically, humidity, heat, moving air, and light are the elements working against us obtaining a crack-free final bowl. We must protect our new green wood bowl turning from these mean evil forces. The best way to accomplish this is to turn an even bowl shape and protect that turned piece afterward from the natural forces working in opposition. Drying Green Wood Bowls Elements Nature Drying Green Wood Bowls Slow or Quick When drying green wood bowls, we have two different routes to take. The first route is to turn the wood once and ease the drying process, so the wood does not change too much or too quickly. We want to avoid causing cracks and splits. With once-turned bowls, because the bowl walls are usually thin and will quickly leach out moisture and potentially dry uneven, we are trying to slow and even out the drying time. The other route to take is to twice-turn the green wood, which we’ll discuss more in a moment. The first turning is deliberately oversized and designed to take the stress of change during drying. While we still want a smooth process to a “dry” bowl with the twice-turned method, we are also usually trying to speed this method because the wood is thicker and will take much longer to dry naturally. Drying Green Wood Bowls Once-Turned Drying Green Wood Bowls Once Turned Thin Wall A once-turned bowl is a piece of green wood turned into a bowl at final shape and wall thickness. After the turning, the initial form of the bowl is finished at that point. Now, nature takes over. With a once-turned bowl, depending on wall thickness, the piece will dry, move and transform into its finished form as is dries. If the bowl is turned thin, it may shift, warp, and buckle much more noticeably. This can be a desirable or not-so-desirable effect. Either way, a thin-walled bowl will dry quickly, usually within a few days, and reach its finished look in a short time. Very thin walled bowls can dry in a matter of hours. It is not recommended to seal the end grain of once turned bowls. Since it won’t be turned a second time, removing the sealer can be difficult or nearly impossible without damaging the finished bowl. Sealing end grain is more important with twice-turned bowls, which we will cover in a moment. Instead of sealing the end grain of once turned bowls, we need to slow the drying process by controlling the bowl’s environment. Using the paper bag method (below) and/or spraying the end grain periodically with water to keep the wood evenly moist are two such approaches. Once-turned bowls tend to have a very organic, natural look to them with offset curves and rims that may rise and lower from side to side. This may or may not be the look you’d like to achieve for the final bowl. FREE eBook from Turn A Wood Bowl.com Sign Up for this FREE eBook and Weekly Wood Bowl Turning TIPS! Bowl Turning Experience? [x] Haven't Turned A Bowl Yet [x] A Few Bowls [x] Lost Count After 50 Bowls Send My Free eBook Now We won't send you spam. Unsubscribe at any time. Built with Kit Drying Twice-Turned Green Bowls As the name implies, twice-turned green bowls will be turned twice. Why you might ask? To obtain a more accurate round shape for the final bowl. The first turning of a twice-turned bowl will be allowed to thoroughly dry. In the process of drying the bowl will usually pull outward at the pith, contract on the sides and essentially stretch. With the walls of a twice-turned bowl made thick, 10 percent of the overall diameter, there is wiggle room to “correct” this movement. I will also carefully make my tenon a bit larger so that it will still be held properly by the four jaw chuck when it’s trued up after drying. Basically, we let the first turned bowl do all the warping, moving, and shaping, then, once dry, turn it a second time to the final shape and final wall thickness. Because the wood, after the first turning and some time, is dry or equalized, the amount of movement after the second turning is minimal. All of the methods below will work for twice-turned bowls. Because of the thick walls of the first turned bowl, it’s a great idea to seal the bowl’s end grain with Anchorseal before starting the drying process. Drying Green Wood Bowls Sealing End Grain Sealing the end grain of the bowl slows the amount of moisture leaking from the open end grain. The goal is to have a more even moisture loss from both the side and end grain. Evenness is the key. Final twice-turned bowls usually look perfectly round with clean circular rims and fresh, crisp lines. Drying Green Wood Bowls Twice Turned Example DRY GREEN WOOD BOWLS – Six Methods Method One – Hope and A Prayer Take a piece of green wood and turn it into a finish shaped bowl or a rough-turned bowl. See, Turning Green Wood Bowls – The Process, for more about twice, or rough-turning. Set the turned bowl aside. Again, remember every wood species is different, and all the external forces acting on that bowl are always present. Some woods are super cooperative, and you will have no trouble, while others species seem to crack when you look at them wrong. The finished turned bowl needs to be reasonably uniform in thickness throughout. If, for example, the walls near the rim are thin and the base is thick, more moisture will remain in the base after the thin walls have dried and this can cause uneven stress to crack the bowl. Drying Green Wood Bowls Even Wall Thickness For a final turned bowl, think how the environment will affect the equalizing wood. A low, cool, dark, sealed cupboard will be a much better storage location than a countertop, in the sunlight, near a breezy open door or heat source. Set the final turned bowl aside and hope for the best. In a few days or weeks, the results will be revealed. I’ve done this many times with success and several times with failure. I have turned green wood and done nothing else to it afterward, with decent results. The bowls that come to mind were turned from two-year-old cut hickory, so, while green, the wood was much drier than when first cut with the chainsaw. By the way, there are many aspects to making your chainsaw blade sharp at all times. Check out this article for all the tips to perfectlysharpen your chainsaw. Another contributing factor is the nature of the hickory species, which is a solid and stout structure. Also, the hickory bowls were turned with walls about 1/4” to 3/8” thick, not thin and not too thick. Even though they dried fine without cracks, there was some movement, noticeably waves along the rims. Method Two – Paper Bags To The Rescue Yes, brown grocery bags work well to allow the wet green wood bowl to release moisture slowly and evenly over time. Because the paper bag breathes, it will gradually release excess moisture and progressively dry, exactly what we need. The paper bag with wet fresh shavings makes a controlled micro-climate for our wet green wood bowl. Simply collect some wet shaving from the shop floor, preferably the ones from the same bowl, and line a paper grocery bag. Place the bowl or roughed blank inside, cover the bowl with more wet shavings and fold over the bag top and loosely tape the bag shut. Put this is a cool dark area that has little air movement. A cabinet or cupboard works well. This is an important note. Don’t place a green wood bowl in a plastic sealed bag. In moist locations with humidity, plastic bags will hold in moisture and most likely cause mold, bacteria and rot to form, not things we want for our beautiful bowls. The exception to the plastic bag rule is severely dry, arid locations, such as the desert. In these locations, plastic bags are used to try to retain moisture in the green wood bowl longer and slow the evaporation process. Everything is relevant! For the rest of us, that live in areas where there is a fair amount of relative humidity in the air on a regular basis, paper bags breathe and help regulate the passage of moisture very effectively. The moist, fresh shavings will slow drying but because they are surrounding the bowl, they aid in regulating or buffering the amount and speed at which the moisture leaves the green wood bowl. After a week or so, it is a good idea to move the green wood bowl turning to a new fresh paper bag. You can check the bowl’s dryness when you move it as well. The fresh paper will prevent excess moisture build-up and reduce bacteria or mold formation. Approximately every week or so, (days if the piece is thin) check the piece and see if it is drying evenly without cracks. Remove the shavings if they are dried up. Return the turned bowl to a new fresh grocery bag without shavings, close the top, and return it to a quiet area free from too much air movement. Check the moisture content of the wood as we will discuss below and once it equalizes the process is complete. Drying Green Wood Bowls Paper Bag Technique Method Three: Kiln Drying If you have access to a dryer or kiln, this can be a great way to dry twice-turned thicker green wood bowls quickly. However, again, I suggest experimenting with this process too. Every species will act differently. One bowl might dry beautifully in a dryer, while another may disintegrate. A homemade kiln can be made out of an old refrigerator, an insulated cabinet, or a box made from foam house insulation. Just cut a vent hole in the old refrigerator base and one in the top and place a light fixture with an incandescent light bulb inside. The heat from the light will slowly dry the wood bowl blanks. A systematic approach for advancing lightbulb sizes can be used to increase the temperature slowly and gradually over the course of several weeks by using progressively larger bulbs. Start with a 40w bulb, then a 60w and finally a 100w bulb. Check the moisture content of the wood with a moisture meter or weigh the pieces to determine when they have equalized. Be careful, however. Keep in mind that quick, sudden, unstable changes in the wood structure are what will result in the most problems. So easy does it and the more even everything is along the way, the better. This method is best suited for drying rough-turned bowls that will later be twice turned to their final shape and form. A kiln or drier can take the drying time for thicker twice-turned bowls from months or even years to only a few weeks. Drying Green Wood Bowls Custom Dryer Kiln Method Four: Microwave Green Bowl Drying Microwave drying a wood bowl blank can be done with short times in the microwave at full power, around 30 seconds, followed by cooling time, around 30 minutes. Repeat the process weighing the wood bowl after each session until the weight of the bowl stabilizes. If your bowl or roughed blank will fit into the microwave and if your spouse approves or is unaware you may want to try this technique. A garage sale microwave, just for drying green wood in your shop might be a better way to avoid domestic disputes. Just saying. 😉 Microwave the turned bowl on high for a very short period, 20-30 seconds. After each session, remove the piece and set it out to cool for at least a half hour. I place the blank on the counter in the kitchen and prop it up on a side so that most of the bowl has good air exposure. The heated bowl will lose moisture through evaporation, so it’s essential that all surfaces of the bowl breathe. After at least a half hour of cooling, I pop it back in the microwave for another 20-30 second dose. Repeat this process eight to ten times until the moisture content equalizes and the bowl is dry. Don’t be mistaken, this isn’t as fast as a microwaved meal, but still much quicker than nature ever imagined. If you’re microwave drying a thick rough turned bowl, it may take a few more trips to the microwave until the wood equalizes, then you’re ready to do the final turning. Microwave drying green wood bowls aren’t as fast as making a bag of popcorn, but it sure beats waiting years. Drying Green Wood Bowls Microwave Process Method Five: Drying Green Wood Bowls in Denatured Alcohol Soaking rough turned or finished bowls in denatured alcohol creates a chemical process that bonds alcohol to the internal water in the bowl’s wood cells. During the drying process, this mixture evaporates from the wood very quickly. This process requires that the bowl is completely submerged in denatured alcohol. A resealable plastic container just big enough to hold the bowl is ideal. High quality denatured alcohol can be purchased in gallon containers, and enough is needed to completely cover the whole bowl. Soak the bowl for approximately 24 hours before removing the bowl from the denatured alcohol. It’s not a bad idea to let the excess denatured alcohol drip back into the container so it can be used again later. After the excess denatured alcohol is no longer dripping from the bowl, place the bowl in a paper grocery bag and press the paper around the bowl snuggly. In my research, I found some people take the time to wrap the bowl in brown craft paper, taping it securely shut with tape, like a wrapped gift. The time it takes the denatured alcohol and water mixture to escape the wood cells in the bowl will vary but can take from one to three weeks. Like, everything we’ve discussed so far, all the various factors will contribute in different ways to affect the drying time. Once the wood does not smell like denatured alcohol, the process should be complete. Use the techniques described below to determine dryness or the equilibrium of the moisture content in the wood bowl. While I have not tried this technique yet, it appears to be a viable green wood drying solution for both finished once-turned bowls and roughed twice-turned bowls. Method Six: Desiccant Green Wood Bowl Drying When I first learned of this process, I had to stop and try to figure out how to pronounce the word, “desiccant.” Desiccant defined, is a hygroscopic substance used as a drying agent. Think of the little silica gel packets that come in some product packaging to keep the product dry. Yes, you can purchase large amounts of desiccant and use it in the process of drying green wood bowls. I have a link in my Green Wood Resource Guide for buying desiccant in volume. No need to try to collect 10,000 little packets from various product packaging. Here’s the cool thing about desiccant. Desiccant drying beads are color-coded, and they change color when they are saturated with moisture. And the best part about using desiccant is that it’s reusable. Just spread the moist beads out on a cookie sheet and bake them in the oven at 250° until they are dry and ready to use again. Desiccant needs to be sealed and not exposed to air ever. If the desiccant beads are left exposed to air, they will suck up the humidity in the air until they become saturated again. To use the desiccant beads to dry a green wood bowl, place a layer of desiccant in a sealable plastic bag and then place the green bowl inside the bag. Completely cover the bowl with desiccant until the wood bowl is not visible, then tightly seal the plastic bag. Depending on the size, thickness, and shape of the green wood bowl, the desiccant can dry the wood in as little as 24 hours. Check the bowl for dryness and see if the beads have changed color to indicated moisture saturation. If the beads are wet and the bowl is not entirely equalized, bake the beads until dry and do it again. Drying Green Wood Bowls With Desiccant Managing Twice-Turned Green Wood Bowls If the particular wood species you are using is prone to cracking, it is a good idea to seal the end grain immediately after turning the twice-turned rough green wood bowl. Anchorseal is a breathable sealer that allows moisture to slowly escape and it aids in preventing cracking end grain. For the bowls I twice-turn, I seal the end grain liberally inside and out on the roughed bowls. As the wood drys, it will lose moisture more evenly from the end and unsealed side grain. Without the Anchorseal end-grain sealing application, the green wood pours the moisture primarily out the end grain with very little escaping from the side grain. This unevenness, like unevenly turned bowl walls, causes unbalanced stress and cracking of the turned bowl. Think of the cells and grain of the wood like drinking straws. The end grain is the open end of the straws, and moisture quickly leaves these areas unevenly and causes stress cracks. Later, when the rough-turned bowl has dried or equalized, it can be turned a second time to the final shape and thickness. At that point, because the green wood has undergone the drying and movement process already, it will stay very close to the final turned appearance with little or no further shapeshifting. Measuring Green Wood Bowl Dryness So how do we know when the wood is “dry,” (a.k.a. equalized)? We need to measure and check. When you’ve worked with a particular wood long enough, you can sense when the wood is dry and stable. However, that can be an elusive and time-consuming sense to acquire. There are two ways to measure for dryness. A standard wood moisture meter can be used to measure the moisture content in your bowl. The only problem with this method is it can leave small pinprick holes in the bowl surface. Also, readings taken from different areas of the bowl can reveal very different results. I’ve found that I don’t always get a confident answer with moisture meters. However, I do use a moisture meter to determine very wet wood compared to more dry wood. For more precision, I use the following technique to measure green wood equilibrium. Weighing Dryness The method I find most useful for measuring the dryness of a green wood bowl turning is weighing the bowl. Yes, weighing the bowl will give you the best results because you actually measure the amount of water in the bowl. I find weighing in grams is the easiest way to get a precise, and easily comparable weight reading each time. Use a small digital scale, like this one, which has a gram reading and weigh the bowl. On a slip of paper, write the date and the weight in grams. Come back a few days later and repeat the process. Here’s the cool simple conclusion to this process. When the weight stabilizes, the bowl’s moisture content has equalized for its environment. It’s really that simple. Drying Green Wood Bowls Weighing Moisture Content Equilibrium Can Go Both Ways Believe it or not, I had a twice-turned bowl roughed out and placed in a dryer. I kinda forgot about it until one day I decided to pull it out and turn it for the final finished shape. Everything was fine with the wood, no cracks formed and it turned very well. With the piece complete, I decided to weigh the bowl to measure how much moisture was still inside. As the days went by and I noted the progress, I was quite surprised. This bowl was not losing weight, but instead GAINING weight. What had happened? Well, in the time I forgot the roughed bowl in the dryer, the green wood went past the point of equilibrium, and the wood cells held less moisture than the relative surrounding air humidity. When the bowl was removed from the dryer and returned to the relatively moist air it began absorbing that moisture and gained weight, just like a dry kitchen sponge soaking up water. The moisture weight gain was minimal and did not change the bowl shape or structure enough to notice, but it was a great example of how the wood is never truly dry and always evolving. Experiment I hope this article gives you new ideas for how to work with green wood. There are many advantages to working with green wood, but most importantly, remember to experiment. With all the possible variables it’s impossible to know precisely what will work and what will not. Without committing too much time, simply turn a few pieces and try drying them a couple different ways. Take notes about your drying processes so you can recall what works. Once you’ve discovered a process for a particular species of wood, forge ahead and start creating beautiful green wood turned bowls with your new knowledge! Drying Green Wood Bowls Conclusion Green wood is fantastic to turn, low on flying dust, easy on tools, and usually very affordable, as in FREE. Awareness of the changes that take place once the green wood is turned is the key. It is important to realize a transformation is taking place from a living structure to a more static artistic form. The transition between these states must be acknowledged and respected. With a bit of experimenting and knowledge about how wet green wood dries, or equalizes to the surrounding ambient conditions, we can more reliably make fantastic green wood bowls. While it might seem a bit frustrating, while striving for the best looking bowls possible, cracks and mishaps are going to happen. Learn from them. Remember what works and what doesn’t work, then play and have a blast turning and drying green wood bowls over and over. Here are other green wood articles to consider: • TWICE TURNING WOOD BOWLS – HOW TO STEP BY STEP • BANDSAW BASICS FOR WOOD BOWL TURNERS • GREEN WOOD BOWL BLANK MAKING PROCESS • BOWL TURNING GRAIN ORIENTATION – WOOD BLANK DIRECTION Happy Turning, Kent 70 Responses Lang Kinlawsays: July 1, 2024 at Should I use anchorseal (or latex paint) on rough cut bowls that I plan to put in a microwave? I have used the microwave method with some success but oak and maple still seem to crack some using this method. Especially the oak. The maple cracks were small and I was able to turn the cracks out when I finished them, but the I had to turn the oak down so much to get the cracks out that it made a completely different bowl and much smaller than I had planned. Reply Kent Wsays: July 11, 2024 at Lang, Good question. No, when microwave drying I do not seal the end grain because I want the moisture to escape as soon as possible. If cracks start to form I might take a break and glue them to reduce their spread, but that’s all I’d do. Happy Turning! Kent Reply Bill Pattersonsays: May 3, 2024 at Good day Kent, Using a moister meter, what percentages are we looking for before we start turning whether we are turning a 2nd turn or for a one and done? Reply 1. Kent Wsays: May 17, 2024 at Bill, Good question. I usually don’t use a moisture meter because they rarely are accurate with thick timber. We cover this more in the Tree to Bowl course. All the best to you and Happy Turning! Kent Reply Rick Ludemansays: December 16, 2023 at Would you recommend (or not) drying your piece before you turn it? That is just after you have cut your log into “blank” pieces and cut into circular pieces to turn. I am considering using an infrared.heater, that is one that turns red hot to heat and dry the wood “blanks” before turning. Reply Kent Wsays: December 19, 2023 at Rick, There’s no need to dry the wood before turning. You can turn dripping wet wood, medium wet wood, somewhat dry wood and very dry wood. All wood moisture types can be turned at any time. Also, attempting to dry the wood with a heater will most likely result in severely cracked wood. We cover all the details to this topic thoroughly in the Tree to Bowl – Understanding Green Wood course It is truly worth every cent. This is a topic that is rarely covered well and needs to be understood. All the best to you and Happy Turning! Kent Reply Alex Hendersonsays: February 28, 2023 at I recently turned an emerging bowl from a 7″ walnut log. The log was much wetter than I thought. What method of drying would you recommend based on the half the bowl having a thin wall and the other half being half of the log. I applied some finish and so far so good. The other half log I put in a paper bag with shavings from the first one but being 3.5″ thick would mean 3.5 years before I could turn it. Thanks for your great articles and videos I have learned a lot. Cheers Alex Reply 1. Kent Wsays: March 4, 2023 at Alex, Good question. Unfortunately, there are many elements to consider. I recommend checking out my Tree to Bowl online Course. We cover all this info and much more there. Happy Turning! Kent Reply Reiner Jakelsays: February 13, 2023 at Hi Kent, love your channel and have learned so much over the past 4 years of turning. I read all the comments about the different methods to dry or stabilize green wood bowls. The one method I did not see is using a food dehydrator for this purpose. .I stumbled across this in a you tube video by the Wood Whirler about a year ago. Jumped right in and bought a dehydrator and went for it. I have turned 10 green bowls. Nine of them were finished turned with 2 coats of sanding sealer before placing them in the dehydrator. Ran them at 131 degrees for between 40 and 70 hours and none of them developed any cracks. I just did a twice turned bowl where I lowered the temp to 100 degrees thinking being thicker slower would be better. I gave a presentation to my club ( Southern Utah Wood Turners ) which sparked a lot of interest. I would love your feedback on this. Thanks Reiner Reply 1. Kent Wsays: February 22, 2023 at Reiner, Thank you for writing and sharing! Sounds fascinating. How big are that food dehydrator and your bowl blanks? I’m guessing your environment is helping a bit with drying too. Thanks again and Happy Turning! Kent Reply Shan Andersonsays: May 5, 2022 at I am attempting to make a dryer box using an old kitchen proofer. I have disconnected the heating element, as this would result in a fast spinning meter, and probably way too high temp. I left the fan connected, and the back of the proofer has a vent where air travels up through a register and is displaced evenly inside the proofer. My plan is to install a 60w light bulb, and install some insulation board on the wall interior. There is a crack at the door seal, which I think is enough for a decent vent path for air flow. I have two questions: 1: should I apply achorseal to the end grain of the bowl(s) after the first turning prior to putting them in the dryer box? 2: Prior to making the second turn on the bowl(s), how do I re-shape the tenon if needed? (jam chuck and tail stock center). By the way, I’m very new and really have learned a lot from your videos. Reply Kent Wsays: May 7, 2022 at Hello Shan, Your dryer sounds great. And yes you only need temps around 80-100° F to dry the blanks. Seal the end-grain of the roughed bowls first. And I have several videos about twice-turning. Here’s one to view All the best to you and Happy Turning! Kent Reply 1. Denny Fastsays: August 17, 2022 at Can a person use salt to dry a green turned bowl? You can buy salt at a feed store reasonable . A 50 lb bag. I have not done it yet but have thought a lot about it. Covering a turned bowl like you wood with decadent. Reply 1. Kent Wsays: August 21, 2022 at Denny, I have never tried it, but it sounds possible. I say give it a try. Experiment! All the best to you and Happy Turning! Kent Reply Brian Beningersays: April 18, 2021 at Hi Kent, I’ve had my lathe for about 5 months & first want to thank you for your on-line content – my go-to source for all things turning. My first bowl projects were from various dry hardwoods & now I’m playing with turning green wood. I cut a lot of firewood every year & look forward to the day I can make some nice bowls from the tree crotches instead of fighting with them on the splitter – a win/win. You mention above that you coat the end grain of rough green wood bowls with Anchorseal (I use Lee Valley’s log sealer – looks like the same stuff). Prior to reading this I rough turned a green wood bowl & coated the whole thing with log sealer. I also coated all cut surfaces on several blanks I made. Is it a mistake to coat side grain too? Now I’m thinking maybe this will slow the drying process unnecessarily. Reply 1. Kent Wsays: April 21, 2021 at Brian, Great question. In general, the side grain will release moisture much slower and it doesn’t need to be coated. But that said, it depends on the species and conditions. Experiement with a batch and see if there’s a difference with coating side grain or not. All the best to you and Happy Turning! Kent Reply 2. Mike Denneysays: June 30, 2022 at I simply put my once-turned live edge bowls on the shelf in my shop and the vast majority do not crack. I used to seal them but it did not reduce the cracking significantly, it added work, and I simply filled the ones that cracked with turquoise and could raise my asking price for that piece. Humidity level in my shop is 40-50%. If I want to dry them faster, I microwave them until they are very hot, but I let them cool in the microwave or put them in a plastic bag until cool. They lose a lot of moisture but don’t crack because they’re in a very high humidity environment. I don’t dry them all the way, but this reduced the total drying time quite a bit. Reply Forrest Colliersays: March 21, 2021 at Ken – Very helpful. I’ve been woodworking for many years, but I”ve only been into green wood twice turning about a year and a half now. I’m doing all of the things you mention but still have a problem with cracking. I notice you seal just the end grain, i’ve been sealing the whole thing – might that be my problem? Also – the 1/10 rule for wall thickness gives a very thick wall on big bowls – should there be a limit there, say not thicker than 1.25″? Your content is very helpful, thanks for sharing! Reply 1. Kent Wsays: March 27, 2021 at Yes, try sealing just the end grain areas inside and out. Moisture escapes slowly from the side grain, if it too is sealed almost no moisture will escape from the side and then there is uneven moisture loss. That is what causes cracking. By sealing the end grain moisture is lost more evenly throughout the blank. The 10 percent rule is to assure there is enough material remaining after the piece distorts to turn a “true” round wall. If you limit this there will be issues. For example, a 20″ bowl roughed with only a 1.25″ wall thickness can distort and move more than an inch+, depending on species and moisture. When that roughed dry bowl is returned to the lathe, the send turning could blow right through the sidewalls because they have moved farther inward than the end grain. The 10% rule is the way to go, regardless of the size of the bowl. Happy Turning! Reply Brian Sacksays: December 26, 2020 at Hi, Kent, Great video on microwave drying. I tried it recently with a small piece of peach that resulted in a 4″ diameter bowl that I left the edge thickness about 15% of the diameter (followed your example of a turned base above the dovetail, which worked well). I put it in the microwave (1200 watts) for 15 seconds and it was rather warm when I took it out. Weighed it, waited an hour, did it again, and it came out with a number of cracks. Further research indicated that cracks result from too much heat, so I’ll turn another one the same size and try 15 seconds with 30-60 minutes between cycles and see how that works. Two quick questions: 1) Is there a way to tell or guess at how much time should be used for each microwave cycle? Some guys suggest using 50% power; most suggest longer times. Obviously, too little time won’t dry the bowl; too much cracks it. Equally obvious, both the wood and the size of the bowl impact the drying cycle. Perhaps I should start with very short times and work up based on how warm the bowl is at the end of each cycle? 2) Have you ever used a product called Cedarshield, (it apparently used to be called Cedar Treat or Turner’s Choice)? Supposedly, this treatment reacts with the moisture in the wood and turns it into a stable gel and prevents cracks. It is marketed by a company called Cedarcide, designed for any type of wood and allows the use of just about any finish once it dries in about 72 hours. It is reportedly used by a commercial company in Hawaii who struggled for years with turned bowls that would crack when moved to a much less humid location in the mainland. 3) How do I join? I was unable to find a place on your website to establish an account, as the login requires an email address and a password that obviously has to be established before logging on. Thanks. Reply 1. Kent Wsays: December 27, 2020 at Brian, It sounds like you are doing to the right thing by paying attention to the process and keeping notes. Remember every wood is different as well as conditions. 1) If the wood feels hot, I’d say that’s too much. At most it should feel warm. 2) I have not heard of this product, but I’ll check it out. 3) You can join one of my eCourses here> Or you can sign up for weekly Turning Tips here> All the best to you! Happy Turning, Kent Reply 1. Ed Wishartsays: May 7, 2023 at I have been carving pictures in burls for a number of years. My first one was a small burl about 14 inches x 8 inches, a green red cedar burl from a growing tree, obviously very wet. An older gentleman came and told me a secret, his, for removing water from wood so I tried his method and it worked. I was making a head stone for my daughters grave and so time was critical. the process is this and I have used it many times. set your piece of wood down so it is as flat as you can, take some diesel and rub it into the wood and let it sit overnight, about 15 hours. Take your tool for evening the outside of the piece and work the surface until you have removed the immediate surface of the wood. Repeat this process for a few days and you will find the water inside has disappeared. I had to do this every day for 7 days to dry that cedar block and it never even checked let alone cracked. This process has never let me down yet after 40 years. There is no odor from the diesel and it makes the wood easier to work. I encased the head stone on fiber glass to preserve it. Don’t take my word for it, experiment with a scrap piece of wood and be amazed at what happens. Reply 1. Kent Wsays: May 11, 2023 at Ed, Thank you for writing and sharing! I’m sorry about your daughter. Hopefully the wood carving/drying process was a bit of therapy that aided you during that time. This sounds like a very interesting technique. I might give it a try sometime. Thanks for sharing! All the best to you and Happy Turning! Kent Reply Douglas wayne Campbellsays: December 13, 2020 at I am just beginning to turn green wood mainly maple after several years of turning segmented and solid dry wood bowls and found this information very useful , many thanks for the tips. Reply Kent Wsays: December 13, 2020 at My pleasure. I’m glad it was helpful. All the best to you! Happy Turning! Reply John Harrissays: November 5, 2020 at I will be trying some of these tips out. Cheers Reply Kent Wsays: November 5, 2020 at Thanks, John All the best to you. Happy Turning, Kent Reply Trevor Boydsays: October 7, 2020 at Hi Kent, I live down under in Australia and am a relatively new wood turner (March 2020) and because of Covid19 am self taught. I love reading your posts and have gained an enormous amount of information on how to turn wood. Thank you. I am fortunate that I have access to quite a bit of free wood in log form and have been reading your articles on preparing green wood bowl blanks. My question is; Does it make any difference to the drying time, if I remove the bark completely or should it be left on. I have also started using a “Woodcut” Max3 bowl saver, so should I cut multiple bowls in the green form and bag dry, or leave until the initial blank is dryer? Reply 1. Kent Wsays: October 7, 2020 at Great question. The bark is a tricky thing. If it is firmly holding onto the wood, you can leave it. If it is loose and breaking off, take it all off. I have found that loose bark invites more insects and critters to hide and it also holds moisture. It really depends on the tree species. Thanks for asking. Reply Mike Larasays: August 6, 2020 at Thanks for the great info, Kent. For the microwave drying technique, do you also anchor seal the end grain of those bowls prior to microwaving? Or do you just use anchor seal on the end grain for air drying? Reply 1. Kent Wsays: August 13, 2020 at Mike, That’s a good question. No, I would not use Anchorseal if using the microwave technique. You’ll just be cooking the Anchorseal and that’s probably not a good idea. Happy Turning, Kent Reply BGsays: July 23, 2020 at Kent, Thanks again for your breadth of knowledge and outstanding formats of education. I recently stumbled on your YouTube videos. Very nice. Let’s back up the crack prevention effort a couple steps: In the back yard is a black walnut tree with an 18 inch diameter at the bottom. The top is dead and it has been deemed by the spouse and our tree guy to be a risk for falling. As it would fall on my shop so it is coming down sometime but can safely wait until winter. I recall reading somewhere that bark stays on better when trees are felled in the winter. I also stumbled on some info that suggested cracking is worse if the tree is dropped in the summer. What is the best time of year to drop the walnut tree to limit checking? (We live in Delaware.) Once it is down, what should I do with it? There is a bunch of information for plankers out there and not much for bowlers. I do not have a kiln or drying box. I do not have the time to first turn all the wood right after it comes down though I suspect least cracking would be first turn early, anchorseal and second turn when dry stable. Do I cut logs at diameter plus two inches on each end, anchor seal and stack? Or cut the log, cut out the pith, seal and stack? Or cut the log, cut out the pith, turn the bowl blank and seal? Thanks! Reply 1. Kent Wsays: July 27, 2020 at Hello BG, I’d say just send me an invite and I’ll be glad to take care of that tree for you. LOL I’m just kidding…sorta. LOL Winter is a great time to drop the tree as there will be less stress and moisture lost from the hot sun. Once it is down, follow the examples in this article. Cut the logs about 1.25 to 1.5 longer than they are wide. So an 18″ diameter section, cut it about 24″-ish inches long. If you can, cut the pith sections out. This requires cutting with the grain straight down through the log. The larger sections will require a couple of inches, so make two cuts around the pith. There is an added huge benefit, you will get quarter sawn sections which can be amazing. See this video Depending on how much moisture is in the tree, you may find it best to wait a few months before turning. Too much moisture can also make a dramatic change after a bowl has been turned. I’ve found it’s good to let a tree mellow for a bit. That being said, I’d still be turning pieces at different time intervals to see how it’s performing. Try to get the logs cut in half at a minimum and then seal the ends. That relieves tremendous cracking pressure. If you can section out the quarter-sawn pieces and remove the pith completely, all the better. Best of luck with it! Happy Turning, Kent Reply 1. BGsays: July 28, 2020 at Awesome. So helpful. Knowledge is addictive and I continue to be thankful that you share yours in such an informative fashion. Reply 1. Kent Wsays: July 28, 2020 at BG, Awesome! I’m so happy this is helpful for you! Happy Turning, Kent Reply Kevin Hsays: July 19, 2020 at Great site, Kent. I turned 3 bowl blanks for the twice turned method from very freshly cut wet oak, stored them using method 2 after coating end grain with anchorseal back in April. I just checked them and all 3 have cracks 🙁 I had wrapped them up using packing paper, after covering them with the wet shavings. They are in a basement with humidity no lower than 45%. Any ideas why they still cracked? I tried to make sure the wall thickness was even. Thanks again for your great contributions to bowl turning. Reply 1. Kent Wsays: July 22, 2020 at Hello Kevin, I feel your frustration. It’s hard to say what makes some wood crack more than others. I can tell you I also had a very fresh and very wet black cherry tree that acted similarly. What I discovered, the hard way was to let the logs rest for a few weeks. The wood I worked quickly, in its very wet state, cracked like crazy. After a bit of time passed and the logs “mellowed” they turned more controllably and without as much cracking. The lesson I learned was to reduce the extremes. Instead of going from very wet to rough turned, it was better to go from very wet, to cut logs (with some time), to cut blanks (some more time), to roughed bowls. Think of releasing the moisture from the log like landing an airplane. We want to bring the moisture down slow and controlled, not all at once and immediate. Let me know if that makes any sense. Happy Turning, Kent Reply Danilo Benvenutosays: May 27, 2020 at Me gusta mucho leer estos artículos, son muy completos en datos empíricos. Gracias por compartir toda tu experiencia. saludos desde Uruguay. “I really like reading these articles, they are very complete in empirical data. Thanks for sharing all your experience. Greetings from Uruguay.” Reply 1. Kent Wsays: June 4, 2020 at Danilo, Thank you for writing and greetings to Uruguay! I wish you all the Best! Happy Turning, Kent Reply Chasesays: May 19, 2020 at Hi Kent, I love your site. Thank you for all the great info on here! Like Val above I started turning my first piece and mistakenly stored it in a plastic bag. Now it’s moldy! I read your response to Val, started storing it in a paper bag, and at started spraying it with a bleach/water mix But to no avail! How can I fix this? If I turn it off and then seal it, can the mold still grow? Thanks for your help and God bless! Reply Kent Wsays: May 19, 2020 at Chase, Thanks for writing and for your kind words. If you still have plenty of material to turn, yes, turn off the moldy area. If the piece is complete, you can apply an oil finish, like Tried and True Danish Oil, which is pure linseed oil. This will prevent mold, and also slow the drying process which helps to prevent cracking. Place it in a dry paper bag with no shavings (if you seal with the oil) and check the bag day-to-day to see if it’s moist. If moist, switch it out for a dry bag. And, yes, the wood will dry with the oil applied. It’s almost as if the water and oil change places. I know that’s not exactly what happens, but you get the idea. The moisture still escapes even with the oil finish. Let me know if that works for you. Happy Turning and All the Best, Kent Reply 1. Chasesays: May 20, 2020 at Thanks Kent. I will let you know! Also, is it ok to carve on a finished bowl? Do you have to reapply oil after? Reply 1. Kent Wsays: May 20, 2020 at Chase, Thanks for the question. Yes, you will probably want to add oil to the freshly carved areas when they are complete. Happy Turning, Kent Reply BGsays: April 13, 2020 at Kent–Quick question about use of Anchor Seal. Do you use it on the end grain after you have turned bowl blanks on a band saw? I just made a bunch of blanks and I know many will not become bowls for weeks or months. Thanks for all the great info on your site. Reply Kent Wsays: April 13, 2020 at BG, Yes, if you know you won’t be turning them soon cover the end grain with Anchor Seal. All the best to you! Happy Turning, Kent Reply Jim Bustillosays: April 9, 2020 at Kent, Thank you for the article on the various methods for green bowl drying. I’m new to turning and am wondering why you didn’t include a seventh method — vacuum drying at lower temperatures than kiln drying. Is using a vacuum to reduce the required drying temperature a good idea? Regards, Jim Reply 1. Kent Wsays: April 13, 2020 at Hello Jim, Thanks for writing. I’m not familiar with the vacuum and low temp process. It sounds like it might require a good deal of space and time. Have you tried this? Thanks, Kent Reply 2. Bill Wellssays: June 29, 2020 at Jim, I have used vacuum drying for wood items often and with success, but not for wood bowls. However, I am going to try it! My immediate problem is finding a large enough chamber to hold a decent size bowl. Also, there may still be the issue of uneven drying. I can assure you that it will speed up drying. Reply Trent Jesseesays: March 14, 2020 at This is seriously the best most concise and clear guide I’ve come across on the web about drying green wood to turn into bowls. Thank you so much! I learned a ton. Just turned two bowls last week and trying the microwave method right now. About to try the dessicant method tomorrow. The first bowl out of a silver Maple log I cut out the pith and turned but it was already cracked a little before turning it. The second bowl out of green River Birch turned green very nicely but has now started to get some slight checking before I started today doing the microwave drying method. So we’ll see. Any ideas on how to fix little cracks in bowls to still use them? Epoxy? CA Glue mixed with it’s own saw dust possibly? Thanks again! Reply 1. Kent Wsays: March 19, 2020 at Hello Trent, Thank you so much for your kind words. It means a lot to me to know this content helps you! OK, here’s how I fix wood cracks. If the cracks are hairline thin cracks, I use this technique. For larger cracks, I will use this trick to epoxy fill, and sometimes I’ll even use turquoise as a filler. Check out those articles and let me know what you think. Happy Turning, Kent Reply Barry Kellersays: November 23, 2019 at Hi Kent. Just split the log and am roughing out my first greenwood bowl tonight when I happened onto your site. Great timing! Thanks for the info. I’m sure it varies, but approximately how long will a bowl (maybe an inch to inch and a quarter thick) sealed with Anchorseal take to dry? Reply 1. Kent Wsays: November 23, 2019 at Hi Barry, So glad you found this site. Welcome! Well, as you guessed there are many things to take into account to determine how long a piece will dry or equalize. In general it takes a wood bowl blank an inch thick about a year to equalize (dry) and stop noticeably moving. If it’s really wet, you will want to help make that process more gradual with the paper bag technique. If the tree has been down for some time and the green wood is just slightly moist, placing it in a cool cabinet may work just fine. One big thing to do, is experiment with a few pieces and see what works best. Every tree and every tree species is different, add to that your local climate and the condition the timber is in when you acquired it and things can get more complicated. Just pay attention and you will start to see certain woods responding in predictable ways. So glad you’re here. Enjoy! Happy Turning, Kent Reply Val Desjardinssays: September 1, 2019 at I use yellow carpenter’s glue to seal end grains on both my rough turned bowls and on the end of all the logs I drag home. I apply 1 coat, let it dry for one day and then apply a second coat. It is the best stuff I have ever tried and I have tried a lot of stuff. I also use to seal the end of green sawn lumber. I also use my microwave oven to dry rough turned bowls. On them I do not use the yellow glue, it can make a mess in the microwave. I cook the bowl for 1 minute on high heat and then pack it in a paper bag stuffed with the shavings I turned off. I let it cool for a couple of hours and repeat the process until it is dried. I weigh the bowl before every burn and record the weigh directly on the bowl with a pencil. That way I don’t get the bowls mixed up and I can see how little it loose weigh after about 18 burns. I don’t have to mess with my wife, I have my own microwave. I picked it up at a yard sale for 5$, been using it for 5 years. Reply 1. Kent Wsays: September 1, 2019 at Val, Thanks for writing and sharing your information! I’m sure the glue works well for sealing the ends. Does it ever work too well and cause mold in the wood? And do you dilute the wood glue to make it thinner for easier spreading? Good tips about microwaving. Marking each bowl with a pencil is a great tip and getting your own “shop” microwave is probably the best tip, at least for keeping your spouse happy. 😉 Thank you for sharing! Happy Turning, Kent Reply Alan machardysays: February 28, 2019 at Having tried most methods with a fairly high failure rate I now wait for my wife to go to her Yoga session and slip my green bowl in the microwave. I use 45 second times with long in between periods for cooling. I also use desiccant beads from compressor dryers instead of silica due to cost. So far the micro wave seems to be working. Reply Kent Wsays: February 28, 2019 at Alan, Thank goodness for Yoga. Watch out for Laurel Oak! I just got some from a local downed tree and it smells like old feet. Might make the microwave suspicious! 😉 I’m glad you’ve found a method that works for you. What are compressor dryers? I’m not familiar with that term. Thanks for leaving a comment. Happy Turning, Kent Reply Karl Kochsays: February 12, 2019 at Hi Kent, Excellent article, thanks! I now turn green wood almost exclusively and prefer to rough turn and then finish turn. I frequently use a coring system so I end up with quite a few rough turned bowls at the same time. Similar to your denatured alcohol method, I stabilize my bowls in a 50/50 mix of cheap dishwashing liquid and water in 30 gallon trash cans. The denatured alcohol proved to be too expensive and, after a couple of uses it becomes saturated with water and no longer effective. I then do my second turning immediately or transfer the rough cores to paper bags of brown paper until I can get to them. I rarely have any cracking and warpage is kept well within the 10% thickness. Although I use Anchorseal on the end-grain of the log segments in my woodpile that are waiting to be cut up and turned, it never occurred to me to use it on the end-grain portions my rough bowls. That is brilliant, thank you! I am going to try it on the next batch that I process. Thank you for sharing your experience. Karl Reply 1. Kent Wsays: February 12, 2019 at Karl, Thank you for leaving a comment and thanks for sharing your 50/50 solution for stabilizing bowl blanks. I have a couple of questions. How long do you typically leave the bowls in the solution? 50/50 sounds like a LOT of dish soap. What type of soap do you use and about how much do you mix up at a time? Great information! Thanks again. Happy Turning, Kent Reply Gary Anzalonesays: February 9, 2019 at I’m a beginner and this is very helpful information. Thanks! Reply Kent Wsays: February 10, 2019 at Hey Gary, You’re welcome. Thanks for the compliment! Happy Turning, Kent Reply Roger Behnkesays: January 29, 2019 at Thank you for this article. I’m new to green wood turning and have been reading about drying green wood only to end up confused by all of the different methods and opinions. Your article quickly cleared my confusion. Reply Kent Wsays: January 29, 2019 at Roger, Thank you for reading my article and sharing your findings. What other wood bowl turning topics are you interested in or finding difficult to get good information about? I’m always looking for viewer-driven questions and issues to help create the most helpful, relevant and useful articles. Thanks and Happy Turning, Kent Reply 1. Mikesays: May 8, 2020 at I am new to wood turning. I have 2 blow that have molded on me. How can I kill it? Reply 1. Kent Wsays: May 8, 2020 at Mike, Thanks for the question. Be sure to place the bowl in a ventilated area and not in plastic. Paper bags work well. If it gets wet, change it out with a dry paper bag. 1-gallon water to one cap full of bleach solution sprayed on the wood surface should stop the mold. Let me know if this works for you. Happy Turning, Kent Reply 2. Davesays: October 2, 2020 at Hi Kent, Great read and learned loads from reading your article. Can I ask your opinion on green burl, I came across a large fallen branch that was full of burls ( beech I think ) I have cut it all up and got enough for around 30 bowls from it so now I need to work out what’s the best thing to do with them. I have rough turned 3 of them leaving a 1 inch wall on them and they are quite wet so my question is – do I turn all 30 of them ASAP and wrap them all in paper and hope for the best or do I let the blanks dry some more before I turn them. Many thanks , Dave. Reply 1. Kent Wsays: October 7, 2020 at Dave, Great question! Please keep in mind every tree is different. I think you can try experimenting first before turning all the pieces. Burl is very random and not uniform like straight growing wood fibers. So, burl doesn’t deform like regular wood. I’d try microwave drying a roughed out blank or two and see what they do. Check this out In general, you want to turn cut wood as soon as possible as decay, dryness, and other factors work to deteriorate the material the longer it sits. But with burl, you may have a bit more time to work with. Let us know how it goes! Happy Turning, Kent Reply bgalinatsays: December 28, 2018 at Kent, Great information. A book that may be worthy of the gifts for woodturners is, “Moulthrop, A Legacy in Wood,” by Kevin Wallace. In it there is a description of their drying technique. This involves a formula of polyethylene glycol (PEG 1000) and water. After soaking they place the vessels in a room with a dehumidifier. Granted not something the novice need try early on, however, I am very interested in your thoughts on their method. Enjoy, BG Reply 1. Kent Wsays: December 28, 2018 at BG, Thanks for the heads up on this book. I’ll check it out. Like I try to stress throughout my site, I think we each need to do what works best for us. If you can dry something effectively by placing it on a shelf and waiting, great. However, it is nice to know as many different ways as possible and then find what works best for you. Thanks for the input! Reply Kevin Hubersays: October 31, 2018 at Hi Kent, Have you tried vacuum drying wood blanks? I’d like to try it, but I’m not sure I should invest in the equipment, if I’m not sure if it will work. Simplified, Vuccum canister & Vaccum pump, APX 1 week per inch Do you have any advice? Reply 1. Kent Wsays: October 31, 2018 at Kevin, This is a new one to me. I have not heard of that technique, but I will be investigating it for you. Stay tuned. Reply Leave a Reply Cancel reply Your email address will not be published.Required fields are marked Comment Name Email Website Search Here Search for: Hi, I’m Kent Hi! I’m Kent, a husband, dad, papa, graphic designer, photographer, artist, traveler, birder, dark chocolate lover and I’m addicted to turning wood bowls! Learn more about me, see the online courses I made for you, and join our group on Facebook. Ready for your wood bowl adventure? Click here to Get Started Turn A Wood Bowl is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to amazon.com. Subscribe for Weekly Tips Each week I will send you an email with tips, insights, and ideas to help you improve your wood bowl turning skills. Bowl Turning Experience? [x] Haven't Turned A Bowl Yet [x] I've Turned A Few Bowls [x] Lost Count After 50 Bowls I Want Bowl Turning Tips Every Week, Please! Built with Kit SHOP CART SUPPORT YouTube ChannelInstagramFacebook Group My Wood Bowls — Etsy Shop T-Shirts and Merch Privacy PolicyTerms and ConditionsArtwork and Content Usage The Ward Design Group – St Augustine Website Design
15554	https://www.khanacademy.org/math/ap-statistics/analyzing-categorical-ap	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.
15555	https://www.andrew.cmu.edu/user/ramesh/teaching/course/48-175/lectures/3.LinesInDescriptiveGeometry.pdf	3 Lines in 3-D Descriptive Geometry 3.1 DEPICTING A LINE SEGMENT We have previously seen that to depict a line segment in an orthographic view we simply project its end points in the view and join the projected end points to form a segment. In two adjacent views the end points of the segment lie on projection lines that are perpendicular to the folding line. 3-1 A line segment in two adjacent views 3.1.1 Auxiliary view of a line segment On occasions, it is useful to consider an auxiliary view of a line segment. The following illustrates how the construction shown in the last chapter (see Figure 2.38) can be used to solve certain basic problems, which, in turn, may be part of a larger problem. The front top At Bt Af Bf 88 construction takes advantage of the fact that a parallel projection between planes maps segments on segments and preserves endpoints. Observe the transfer distance “rule” being applied to construct the auxiliary view. 3-2 Auxiliary view of a line segment 3.2 POINTS ON A LINE SEGMENT The basic problem considered here is to visualize a point on a segment in two adjacent views. There are two cases to consider: when the views of the segments are perpendicular to the folding line and when they are not, as illustrated by the following construction. Construction 3-1 Adjacent views of a point on a segment Given a segment in two adjacent views, top and front, and the view of a point, X, on the segment in one view, say top view, construct the view of X in front view, Xf. There are two cases to consider. If the views of the segment are not perpendicular to the folding line t \| f, Xf can immediately be projected from Xt (see Figure 3-3a). Otherwise, if the views are perpendicular, we go through the following steps: 1. Use the auxiliary view construction (Figure 3-2) to project the end-points of the segment into a view, a, adjacent to t (see Figure 3-3b) and connect them to find the view of the segment. dA dB dA dB aux f front top Ba Aa At Bt Af Bf 89 2. Project Xt on the segment. 3. The distance of Xa from folding line t \| a, dX, is also the distance of Xf from folding line t \| f and serves to locate that point in f. Try to visualize why this construction works in space! a. b. 3-3 Constructing adjacent views of a point on a segment 3.3 TRUE LENGTH OF A SEGMENT The true length (TL) of a segment is the distance between its end-points. Finding the true length of a segment is a basic problem in descriptive geometry. The constructions that can be used to solve this problem are based—among other things—on the property that a parallel projection between two coplanar lines preserves distances. There are some facts that we can rely upon. When a line segment in space is oriented so that it is parallel to a given projection plane, it is seen in its true length in the projection on to that projection plane. a. f t Xf At Bt Af Xt Bf dX dX Transfer distance dx in the auxiliary view, a, is transferred back into the front view, f, to obtain the point X in the front view b. t a t f Xf Xa Xt Ba Aa At Bt Bf Af 90 3-4 Line segments parallel to projection planes are seen in true length Segments parallel to one of the projection planes is seen in an adjacent view as a line parallel to the folding line. 3-5 Line segments seen in TL TL in projection Projection plane #3 Projection plane #1 Projection plane #2 TL in projection TL in projection Line in space B3 B2 A A3 B A2 parallel Line AB is parallel to the frontal projection plane True length A2B2 seen when viewing the frontal projection plane 2 1 B A A1 B1 A2 B2 parallel 2 1 Frontal plane seen as an edge when viewing the horizontal plane A1 B1 A2 B2 91 3-5(continued) Line segments seen in TL Based on this observation, when applied to orthographic projections, this means that in order to find the TL of a segment, we must find its view in a picture plane containing the segment parallel to it (because the projection lines through the segment define a plane containing the segment and its image on parallel lines). Given the standard assumption that a segment is given in two adjacent views, top and front, the problem is easily solved if • The segment appears parallel to the folding line in, at least, one view, say top (see Figure 3-6a). In this case, the segment is parallel to the picture plane of front and appears in TL in front (the segment can be parallel to both views). • The segment appears as a point (point view of a segment, PV) in one view, say top (see Figure 3-6b). In this case, it is on a projection line, and any picture plane parallel to that line, including front, will show the segment in TL. TL parallel Line AB is parallel to the horizontal projection plane 1 2 True length A2B2 seen when viewing the horizontal projection plane B A A1 B2 A2 B1 parallel Horizontal plane seen as an edge when viewing the frontal plane 1 2 A2 B2 A1 B1 92 a. b. 3-6 Line segment seen in TL in an adjacent view There is a special case which occurs when the line segment is perpendicular to the folding line, in which case it can be constructed to appear in TL in an auxiliary view, say a side elevation. 3-7 Line segment seen in TL in an auxiliary view Are there any other special cases? TL a. front top Bt At Af Bf TL b. front top Af Bf PV (point view of AtBt) TL dA dB TL dB dA Edge view of auxiliarly projection plane #3 when viewing the frontal plane #2 2 3 1 B3 A3 B3 A3 A B B1 B1 A1 A2 B2 A1 A2 B2 93 3.3.1 General case of the true length of a line segment A segment that neither appears in point view nor is not parallel to the folding line in any view is called oblique or inclined. Its true length can be determined with the help of an auxiliary view as the following construction shows. 3-8 An oblique segment Construction 3-2 True length of an oblique segment - auxiliary view method Given two adjacent views of an oblique segment, determine the TL of the segment. Suppose the given adjacent views are numbered 1 and 2. There are three steps. 1. Select a view, say 1, and draw a folding line, 1 \| 3, parallel to the segment for an auxiliary view 3. 2. Project the endpoints of the segment into the auxiliary view (see Figure 3-2). 3. Connect the projected endpoints. The resulting view shows the segment in TL. The construction is illustrated in Figure 3-9(left). Figure 3-9(right) shows the same construction using an auxiliary 4 and the folding line 2 \| 4. Auxiliary plane #3 is parallel to AB A3 B3 B A B1 A1 A2 B2 94 3-9 True length of a segment 3.3.2 Worked example – True length of a chimney tie We show below a practical application of Construction 3-2. Figure 3-10 shows the top and front views, T and F, of a roof with a chimney and two ties, s and t, that connect the chimney to the ridge of the roof. Both views are easy to construct once we select the points where the ties meet the ridge in the top view and the points where they meet the chimney in the front view. But neither view shows either tie in true length (why?). dA dB TL dB dA Edge view of auxiliary inclined projection plane #3 seen in view #1 3 1 2 1 B3 A3 A1 B1 A2 B2 TL dB dA dA dB Edge view of auxiliary inclined projection plane #4 seen in view #2 4 2 2 1 B4 A4 A1 B1 A2 B2 95 3-10 Problem – What is the length of the chimney tie? Figure 3-11 shows how an auxiliary view, A, can be used to show one tie, s, in TL. 3-11 Length of the chimney tie s t s t F T s t s t s d d this view shows s in true length Q P T A F T 96 The figure shows the entire auxiliary view of the roof assembly, which can be constructed entirely by the constructions introduced so far with the exception of the bottom side edge of the chimney. Its front end point, P, is easy to construct from the given views. But its rear end point, Q, is not that obvious because it is not given in the front view, and we can consequently not find immediately the appropriate transfer distance. Standard constructions to solve this problem will be introduced in subsequent chapters. But a little thought will solve this problem based on what we know so far about parallel projections. I leave this as an exercise to the reader. Use the construction hint illustrated in the figure. 3.3.3 Distance between two points Construction 3-2 can also be used to determine the distance between two points in space: it is the true length of the segment connecting the two points. 3.4 SUCCESSIVE AUXILIARY VIEWS To solve certain basic problems in descriptive geometry, a second auxiliary view may be needed. This kind of view is constructed from a preceding auxiliary view by repeating the construction in Figure 3-2 (on page 88) for all points of interest. Successive auxiliary views are normally identified by indices 1, 2, ... that show the order of construction. 3.4.1 Point view of a line 3-12 Point view of a line segment With a line of sight perpendicular to an auxiliary elevation that is parallel to AB, the projection shows the true slope of AB (since horizontal plane is shown in edge view) Auxiliary plane #4 in which line AB is seen as a point. Plane #4 is perpendicular to AB (and therefore is also perpendicular to A3B3 which is a true length projection of AB) Auxiliary plane #3 is parallel to AB A3 B3 B A B1 A1 A2 B2 A4,B4 97 The next construction demonstrates an important application of successive auxiliary views that generates the point view of a line or segment; that is, it finds a picture plane to which the segment is perpendicular so that it belongs to the line family that generates an orthographic projection into the picture plane. See Figure 3-12. Construction 3-3 Point view of a line Given an oblique segment in two adjacent views, find a point view of the segment. Suppose the adjacent views are numbered 1 and 2. There are three steps: 1. Apply Construction 3-2 (on page 93) to obtain a primary auxiliary view 3 showing the segment in TL. 2. Place folding line 3 \| 4 in view 3 perpendicular to the segment to define an auxiliary view 4. 3. Project any point of the segment into view 4. This is the point view of the entire segment. 3-13 Constructing the point view of a segment dB1 dA1 TL dA2 dA1 dA2 dB1 The d's represent transfer distances measured from the respective folding line to the point. Line AB seen in true length in view #3. Note that all projectors are perpendicular to their respective folding lines. Point view of line AB seen in view #4. 4 3 3 2 1 2 A,B4 B3 A3 A1 B1 B2 A2 98 Note that if the segment is shown to be parallel to folding line 1 \| 2 in one of the given views, the construction can be shortened because the segment appears in TL in the other view. Here, only one auxiliary view is needed to obtain the point view. A point view of a segment is also the point view of the line to which it belongs. Therefore Construction 3-3 can also be used to obtain the point view of a line; any convenient distinct two points on the line can serve as the basis for the construction. 3.5 ORTHOGONAL LINES The basic constructions introduced thus far can serve to solve seemingly more intricate problems in descriptive geometry. One type of problem considers two lines that are orthogonal; that is, they are either parallel or perpendicular. Orthogonality is important in many applications, as an attribute to be either tested for or desired; it also plays a crucial role in finding various types of distances. The following sections introduce constructions that address these issues. 3.5.1 Parallel Lines We know (from Property 2-5 on page 58) that a parallel projection between planes preserves parallelism between lines. Since two parallel lines define a plane, the orthographic projection of the lines into a picture plane can be considered a parallel projection between planes and thus preserves their parallelism except when the lines are projection lines themselves, in which case they project into points, or when they are projected by the same lines, in which case their images coincide. That is: • If two lines are truly parallel, they are parallel in every view, except when they coincide or appear in point view. • Furthermore, if two lines are not parallel in a particular view and neither coincide nor appear as points in that view, they cannot be truly parallel. Choosing views wisely It should be noted that single pairs of adjacent views do not always readily help in interpreting an object. In many cases, additional views are needed. Fortunately, these additional views can, almost always, be produced by purely two -dimensional constructions, from the two adjacent views that are given. What am I looking at? ? Elevation Plan 99 Which is it? The converse unfortunately is not always true: two lines that are parallel in a particular view or coincide might not be truly parallel. Figure 3-14 illustrates this with an assembly of two wedge-shaped objects that contain, among others, two segments, l and m, that are not parallel, but appear parallel in the top and front views of the assembly. Such situations are not infrequent in architectural and engineering applications. 3-14 Non-parallel lines that appear in adjacent views as parallel 3.5.2 Testing for parallel lines True parallelism between two lines that appear parallel in two adjacent views can always be established by the following construction. Construction 3-4 Testing for parallel lines Given two lines shown parallel in two adjacent views, determine whether the lines are truly parallel. There are two cases to consider. • The lines are not perpendicular to the folding line. See Figures 3-15a) through 3-15c). In this case, the lines are truly parallel. 100 • The lines are perpendicular to the folding line. This case is also illustrated in Figure 3-15. Construct an auxiliary view not parallel to the folding line. The lines are truly parallel if and only if they are also parallel in this view. In the example shown in Figure 3-15d) the lines are not parallel, and in Figure 3-15e) the lines are parallel. a) b) c) d) e) 3-15 Testing for parallel lines Case (a), (b), (c) and (e): lines are parallel; Case (d): lines not parallel When two lines appear both in point view, they belong to the family of projection lines and thus are truly parallel. This leaves the case of two lines that coincide in a view. If they coincide also in the other view, they must be identical or be perpendicular to the folding line, in which case an auxiliary view as in Figure 3-15 decides the case. If the lines do not coincide in the adjacent view, they are truly parallel if and only if they appear as points or parallel in that view. This is illustrated in Figure 3-16. a) 2 1 b) 1 2 c) 1 2 ma m l l m la d) 2 a 1 2 m m l l la ma e) a 2 1 2 101 3-16 Lines seen simultaneously in point view are parallel 3.5.3 Distance between parallel lines The distance between two parallel lines is the distance between the intersection points of the lines with any line perpendicular to them. The following construction allows you to find this distance. Construction 3-5 Distance between two parallel lines Given two adjacent views of two parallel lines, find the distance between the two lines. There are two steps: 1. Use two successive auxiliary views as in Figure 3-13 to show one of the lines in point view. This will also show the other line in point view. (This step can be shortened or omitted if the lines are parallel to the folding line in one or both of the given views; see the following example.) TL TL 5 4 Lines AB and CD seen in true length and perpendicular to the folding line in view #5 Lines AB and CD seen as points in view #4 Lines AB and CD seen in true length in view #3 4 3 3 2 1 2 A5 C5 B5 D5 A4,B4 C,D4 B3 A3 C3 D3 D1 C2 D2 A2 B2 A1 B1 C1 102 2. Measure the distance between the two point views, which is also the distance between the lines (see Figure 3-17). 3-17 Distance between parallel lines 3.5.4 Worked example – A practical application Figure 3-18 illustrates a practical application of Construction 3-5. It shows the top and front view of a small polygonal balcony as it may emerge during the design of a building. The architect plans to use three pairs of parallel tubes as railing and has drawn them in both views. For further detailing, the architect is interested in the distance between the tubes in any pair, measured for example between their center lines. Neither view shows this distance in TL for any pair because no pair is perpendicular to the folding line, and their center line consequently do not appear in point view in any of the views. Note that the right-most pair appears in TL in both views. We thus can skip generating the first auxiliary in step 1 and proceed immediately with the construction of an auxiliary view using a folding line perpendicular to the pair in the front view; this view shows the desired distance as illustrated in the figure. Note also that the top view l m l m l3 m3 View #3 shows lines l and m in true length distance View #4 shows lines l and m in point view, the distance between them giving the required result 4 3 3 1 2 1 m2 l2 103 shows every pair in TL (why?) and therefore, it could also be used in different auxiliary views to construct point views for each pair of center lines. 3-18 Distance between (center-lines) of parallel railings—a practical example This example is meant not only to demonstrate the application of Construction 3-5, but also to make a more general point. The constructions based on auxiliary views introduced in this and the preceding chapter can be used flexibly to answer questions about the geometry of an evolving design as the design process unfolds. It is often sufficient to produce auxiliary views only of a portion of the design, which can often be done on-the-fly in some convenient region of the drawing sheet. The starting point in every case is the following: Select an appropriate folding line (or picture plane). Furthermore, you are urged: To pay particular attention to the way in which the constructions depend on properly selected folding lines. The general principles that guide these selections are the most important aspects to assimilate. 104 3.6 PERPENDICULAR LINES Two intersecting lines are perpendicular whenever rays pointing away from their point of intersection form a right angle. We call two lines perpendicular in a particular view if both appear as lines that form a right angle, or if one appears in point view and the other one in line view. Given this notion, two perpendicular lines appear perpendicular in any view showing at least one line in TL. 3-19 Perpendicularity can be verified if one line is shown in TL Note that in this view, the other line might appear in point view. Figure 3-20 illustrates this by showing two perpendicular rods in space one in TL and the other in PV. Notice that in the view in which one rod is shown in TL, the other rod is seen at right angles to it. Observe that we have ignored the thickness of the rods and treat them as lines. 3-20 Two perpendicular rods with at least one shown in TL and the other in PV TL 90° 90° projection plane p parallel to CD A C Cp D Bp Ap Dp B 105 3-20continued) Two perpendicular rods with at least one shown in TL and the other in PV The converse is also true: If in a view two intersecting lines are perpendicular and at least one line appears in TL, the lines are truly perpendicular. Again, the other line might be shown in point view. It should be easy to test whether two lines given in two adjacent views are perpendicular, by using Construction 3-2 (on page 93) to construct an auxiliary view in which at least one of the lines appears in TL. The lines are truly perpendicular if and only if they are perpendicular in this view. See Figure 3-21. 106 3-21 Two intersecting perpendicular lines However as Figure 3-22 shows, the same conditions also hold for two non-intersecting (skew) lines that are in directions at right angles to one another. 3-22 Two non-intersecting (skews) lines in directions at right angles to one another TL TL TL TL 90° 90° Line CD appears as a point in view #4, and AB is seen in true length. This view proves AB and CD to be perpendicular. Since CD appears in true length in view #3, a right angle between AB and CD is seen. 5 4 4 3 3 1 1 2 D5 C5 B5 A5 C4,D4 B4 A4 C2 C3 D3 A3 B3 A1 B1 D1 C1 A2 B2 D2 TL TL TL TL ang = 90.00° ang = 90.00° Vew #5 shows that two non-intersecting skew lines AB and CD in true length. Vew #4 shows that lines AB and CD do not intersect but are perpendicular as AB is seen in true length and CD in point view. 5 4 4 3 3 1 1 2 D5 C5 C,D4 B5 A5 B4 A4 C2 C3 D3 A3 B3 A1 B1 D1 C1 A2 B2 D2 107 Figure 3-21 motivates the following construction. Construction 3-6 A line perpendicular to given line Given a line and a point in two adjacent views, find the line through the perpendicular to the segment through the point. Let l and O be the given line and point respectively in adjacent views, 1 and 2. 3-23 Line perpendicular to line – Problem configuration 3-24 Constructing a line through a point perpendicular to a given line l l Find the foot of the perpendicular from O to l, using an auxiliary view 1 2 O O l TL 90° In views t and f, X is the foot of the perpendicular from O to l, determined from the auxiliary view a In view a, OX is constructed to be perpendicular to l a 1 2 1 X X Xa Oa O O 108 There are three steps: 1. Use Construction 3-2 (on page 93) to show l in TL in an auxiliary view a. 2. In a, draw a line through O perpendicular to l. Call the intersection point X. This segment defines the desired line in a. 3. Project back into the other views. The construction is illustrated in Figure 3-24. 3.6.1 Shortest distance between a point and a line The shortest distance between a point and a line lies on a line perpendicular to the given line passing through the point. If we have a point, O, line l, and line m, through O perpendicular to l, the shortest distance between O and l is the distance between O and the intersection point between m and l. Construction 3-6 can be extended to find the shortest distance between a point and a line. Construction 3-7 Shortest distance between a point and a line Let l and O be the given line and point respectively in adjacent views, 1 and 2. We look to find the true distance between O and l. 3-25 Shortest distance between a point and a line l TL 3 View #4 shows line AB in point view from which the shortest distance between O and AB can be easily determined shortest distance PV 4 3 2 2 1 A,B4 O4 B3 O3 A3 A1 B1 O B2 O2 A2 109 There are two steps: 1. Construct in a second auxiliary view, 4, the point view of l. 2. Project O into view 4. The distance between O4 and the point view shows the true distance between O and l. The construction is illustrated in Figure 3-25. 3.7 SPECIFYING LINE SEGMENTS IN SPACE Line segments can be specified in space whenever we know or can establish the position of either a) two points on the segment, or b) a point on the segment and the angular position of the line with respect to the frame or system of reference. 3.7.1 Two points specifying a segment Mostly, for this course, and for descriptive geometry, in general, folding or reference lines serve as the frames of reference. Therefore, when applying orthographic projection concepts, we locate lines by perpendicular distances from horizontal and vertical projection planes. That is, distances of points below the horizontal plane are seen in the vertical projection view and the distances behind the vertical plane are in the seen the horizontal projection view. 3-26 Specifying a segment by distances of its end points behind and below the projection planes S U R T L M Edge view of the horizontal and proﬁle projection planes seen in view #2 Edge view of the frontal projection plane seen in view #1 3 2 2 1 A1 A2 B1 B2 110 3.7.2 One point and an angle specifying a segment Alternatively, a line segment is specified by a point, an angle of inclination with respect to the horizontal plane, and a direction, termed bearing, relative to a compass reading. BEARING The bearing is always seen in a horizontal projection plane, typically the top view, relative to the compass North. Observe that the bearing of a line has no relationship to the angle of incline. 3-27 The bearing of a line is seen in the top view Bearing always measured from a compass direction (typically north or south) to a compass direction through a certain angle. Here the bearing reads 60° from north towards west Position of line AB in the frontal projection has no effect on the bearing of line AB (Two positions AB and A'2B'2 are shown) 2 1 N 60° W E W A N S B B2 A2 B'2 A'2 111 ANGLE OF INCLINATION OR SLOPE ANGLE The angle of inclination of a line segment is the angle it makes to any horizontal plane. It is the slope angle between the line and the horizontal projection plane and is seen only when — The line is in true length and the horizontal plane is seen in edge view. By applying Construction 3-2 (on page 93) to the line in top view, we obtain an auxiliary view in which the line is seen in TL and the horizontal plane is seen in edge view. 3-28 Constructing the angle of inclination of a line Edge of the hrizontal rojection plane Observer simultaneously sees the true length of AB and edge view of the horizontal projection plane in order to see the true slope angle of AB Slope angle in degrees A B A3B3 = true length of AB True slope angle 1 3 B3 A3 A1 B1 A2 B2 112 GRADE The slope angle also described as a percentage grade is given as: The ratio of a vertical distance for a given horizontal distance 3-29 The grade of a line WORKED EXAMPLE – Adjacent views from a line specification Given a point, the bearing, angle of inclination and true length of a line, construct the top and front views of the line The data supplied includes the top and front projections of the given point, A, bearing N30°E, a downward slope 45° and true length = 1.5” of the line. The following are the steps in the construction. 1. Establish a top view 1 by drawing a line from A, with the supplied bearing. Assume North points upwards. 2. Choose the point A in front view 2 arbitrarily vert. disp. horiz. disp. Grade = Edge of the hrizontal rojection plane Observer simultaneously sees the true length of AB and edge view of the horizontal projection plane in order to see the true grade of AB vertical displacement ----------------------------------horizontal displacement A B 2 1 N 30° E A A2 113 3. Construct an auxiliary view 3 using a folding line 3\|1 parallel to the top view of the given line. 4. Project A1 to A3 using the transfer distance from the front view 2. 5. Draw a line from A3 with given downward slope and measure off the supplied true length to construct point B3 6. Project B3 to meet the line in top view at B1. A1B1 is the required top view. 7. Project B1 to the front view and measure off the transfer distance from the auxiliary view 3 to get B2. A2B2 is the required front view. 3-30 Top and front views of a line given a point, its bearing, downward slope and true length TL A3B3 = 1.50 in. 3 1 slope angle = 45° 2 1 N 30° E A3 A1 A2 B3 TL A3B3 = 1.50 in. 3 1 slope angle = 45° 2 1 N 30° E B2 B1 A3 A1 A2 B3 114 3.7.3 Traces of a line These are points in which a line, extended if necessary, intersects the horizontal and vertical planes. The trace on the horizontal plane is called the horizontal trace, HT, and that on the vertical plane, the vertical trace, VT. See Figure 3-31. 3-31 Traces of a linegth. It should be clear that: • A line may have zero, one or two traces. • The traces of a line specify its direction without specifying its position or length VT HT A B B A HT VT B A B A HT VT A B B A HT VT B A B A 115 • A trace of a line can be used to determine the true length and angle of inclination of the line to both the horizontal and vertical planes as the construction in Figure 3-32. 3-32 True length and angles of inclination of a line using its traces 3.7.4 Specifying lines on quad coordinate paper Practical problems in descriptive geometry require methods for specifying or laying out lines with a degree of precision. It is impractical to draw lines that are, say 20 feet long. Clearly, lines are represented according to a scale. It is convenient to use coordinate dimensions. Conventionally coordinate dimensions are specified in inches and drawings are made to full scale irrespective of the scale of the problem. It is common to use coordinate paper, also known as quad paper, which is 8½" x 11" divided into ¼" squares, and is very convenient for solving practical problems using coordinate dimensions. Sometimes a working area of 8" x 10" is employed to allow for the axis to be shown. θ φ zA zB TL zB yB yA zA TL yD yB θ angle of inclination to the horizontal plane φ : angle of inclination to the vertical plane VT HT A B B A 116 3-33 Using quad coordinate paper: A(2, X, 9), P(2, 3, 7½) and B (3, 3, X) The origin is always assumed to be the lower left hand corner of the working area. The top and front views of a point are plotted on the sheet by three coordinate dimensions, which are always given in the same order. For some problems some of the coordinate values are unknown and may be omitted in the any accompanying data. If a coordinate is unknown and the complete location of a point is part of the problem, the letter X is introduced in the data The first is the distance from the left border of the working area to the line joining the top and front views. 10 9 8 7 6 5 4 3 2 1 2 4 6 8 Lower border Left border Front view Top view P P B A 117 The second is the distance from the lower border of the working area to the front view. The third is the distance from the lower border of the working area to the top view. Thus, P: 2, 3, 7 ½ - also written P(2, 3, 7 ½) - represents a point 2 inches to the right of the left border. The front view is 3 inches above the lower border and the top view is 7 ½ inches above the lower border. For some problems some of these coordinate values are unknown and are omitted in the any accompanying data. If a coordinate is unknown or if the complete location of a point is part of the problem, the letter X is introduced in the data. That is, when a position is not specified because it is either not needed or must be found during the course of a solution this is indicated by an “X.” Thus, A: 2, X, 9 specifies a point in top view only and B: 3, 3, X specifies a point in front view only. Hence, a point is represented as the distance in inches (from the left border, in front view, in top view). See Figure 3-33. Additionally, on rare occasions, we may want to increase precision in which case we will also use the center of the ¼" square or the mid-point of a grid square. In the drawing shown in Figure 3-34 three kinds of points are specified: i) at a grid point; ii) at the center of a grid cell; or iii) mid-way between two grid lines. The diagram also show how these points might be reproduced using simple construction lines between grid points. It is usual to ensure that the construction lines are not made visible. 3-34 More on using quad coordinate paper 118 WORKED EXAMPLE – Views of a truncated pyramid On quad paper, line A: (2, 2, 6), D: (2, 2, 9) is a diagonal of a horizontal hexagonal base of a right pyramid. The vertex is 3” above the base. The pyramid is truncated by a plane that passes through points P: (1, 4 1/2, X) and Q: (4, 1 1/2, X) and projects edgewise in the front view. Draw the top and front views of the truncated pyramid. We use the construction shown here to produce a hexagon given its diameter (that is, two opposite points). The construction of the truncated pyramid is shown in Figure 3-35. Notice that we need the sides of the pyramid to determine where the truncating plane meets them. 3-35 Constructing a truncated hexagonal pyramid on quad paper 10 9 8 7 6 5 4 3 2 1 2 4 6 8 A,D D A P Q 119 3.7.5 Typical problems involving lines Problem (Structural Framework) Figure 3-36 shows the top and front elevation views of a structural framework between two neighboring buildings to scale. The problem is to determine the true length of structural members AB and CD and the percentage grade of member BC. 3-36 Problem configuration: A structural framework To solve this problem we construct an auxiliary view by drawing a folding line T \| 1 parallel to AB and CD in plan view and projecting onto view 1. The auxiliary view gives the true length and grade as required. 7'-0" 7'-0" 7'-0" 4'-0" 10'-0" 8'-0" 4'-0" F T A C B D B,D A,C 120 3-37 Solution: Take an auxiliary view to find member lengths and percentage grade Problem (Mine reclamation - locating a new tunnel [line]) Pittsburgh has a number of old mines and consequently, mine tunnels, which are being reclaimed to farm mushrooms. Consider two such mine tunnels AB and AC, which start at a common point A. Tunnel AB is 110' long bearing N 40º E on a downward slope of 18º. Tunnel AC is 160' long bearing S 42º E on a downward slope of 24º. Suppose a new tunnel is dug between points B and C. What would be its length, bearing, and percent grade? 8'-0" 4'-0" BC=7'-5" CD = 13'-7" AB = 11'-5" T 1 Grade = 18.5% A D B C B,D A,C 121 We can solve this problem from plan view drawing it according to some scale (which may be provided). We start by locating A in plan and draw indefinite lines with the given bearings. Next, we construct an auxiliary view 1 showing AB in true length and locate B by projecting back into the plan view. See figure below. We repeat this step for C by creating an auxiliary view 2 in which AC is seen in true length and then locating C by projecting back into the top view. To determine the true length of BC we construct another auxiliary view 3 and using the transfer distances for B and C in views 1 and 2, we obtain the true length of BC. The bearing and percent grade are then easily determined. The construction is shown in Figure 3-38. N 40º E S 42º E A N 40º E S 42º E AB=110'-0" Folding line T \| 1 is parallel to the line bearing N 40º E AB is seen in true length and slope in view 1 T 1 18º B B A A 122 3-38 Solving the tunnel location problem Problem (Locating a point) Sometimes a problem may appear to contain more information than seems necessary. Here is an example. Let ABC be a triangular planar surface with B 25' west 20' south of A and at the same elevation. C is 12' west 20' south and 15' above A. Locate a point X on the triangle 5' above and 10' south of A. Determine the true distance from A to X. The construction is given in Figure 3-39 originally drawn to a scale, 1" = 5', on quad paper. We draw the triangular plane ABC in both top and front elevation views as shown. In front elevation, draw a line 5' above A. This meets sides BC and AC at points 1 and 2. X must lie on this line. Project 1 and 2 to top view. N 40º E S 42º E AB=110'-0" AC=160'-0" S 5º22' E dB dC dB dC BC=203'-0" percent grade = 15.6 3 T Folding line T \| 2 is parallel to the line bearing S 42º E Folding line T \| 1 is parallel to the line bearing N 40º E T 2 AC is seen in true length and slope in view 2 24º AB is seen in true length and slope in view 1 T 1 18º C B C C B B A A A 123 In top view, X lies on the line 12. In top view, draw a line 10' south of A. X must lie on this line. Therefore, X is the intersection of this line and 12. Project X into the elevation. The last step is to take an auxiliary view parallel to AX and project A and X to determine the true distance between them. 3-39 Solving the problem true distance = 13'-10" d d scale 1" = 5' line 10' South of A line 5' above A A X X X 1 2 2 1 C A B A C B
15556	https://infectagentscancer.biomedcentral.com/articles/10.1186/s13027-023-00540-9	Risk of cervical intraepithelial neoplasia grade 3 or more diagnoses for human papillomavirus16/18-positive women by cytology and co-infection status \| Infectious Agents and Cancer \| Full Text Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Advertisement Search Explore journals Get published About BMC My account Search all BMC articles Search Infectious Agents and Cancer Home About Articles Submission Guidelines Submit manuscript Risk of cervical intraepithelial neoplasia grade 3 or more diagnoses for human papillomavirus16/18-positive women by cytology and co-infection status Download PDF Download ePub Download PDF Download ePub Research Open access Published: 09 October 2023 Risk of cervical intraepithelial neoplasia grade 3 or more diagnoses for human papillomavirus16/18-positive women by cytology and co-infection status Mengyin Ao1,2, Xiaoxi Yao1,2, Danxi Zheng1,2, Xuesai Gu3& … Mingrong Xi1,2 Show authors Infectious Agents and Cancervolume 18, Article number:57 (2023) Cite this article 2364 Accesses 10 Citations 1 Altmetric Metrics details Abstract Background Human papillomavirus (HPV) 16 and 18 cause approximately 70% of cervical cancer cases. The aim of this study was to evaluate whether co-infected with other HPV genotypes will affect the risk of cervical carcinogenesis in HPV16/18 positive-women. Methods In this cross-sectional study, cervical cytology and histological classifications from women who tested positive for HPV 16/18 and underwent colposcopy within 6 months, between January 2010 and May 2021 were obtained from West China Second University Hospital of Sichuan University. Main outcomes and measures Immediate risk of cervical intraepithelial neoplasia grade 3 or more diagnoses (CIN 3+). Results A total of 7940 HPV 16/18-positive women were included, with a median age of 40 years (range 25–84 years). Among them, 2710 (34.1%) were infected with multiple genotypes, 6533 (82.28%) had cytology results and 2116 (26.65%) women were diagnosed with CIN 3+. The effects of HPV 16/18 coinfecting with other HPV on CIN3 + risk varied with specific HPV genotypes. After adjusting for cofactors, compared to single HPV 16 infection, the CIN 3 + risk was significantly reduced in women infected with HPV 16 + other high-risk HPV (hrHPV) [odds ratio (OR) = 0.621, 95% confidence interval (CI) 0.511–0.755], HPV 16 + low-risk HPV (lrHPV) (OR = 0.620, 95% CI 0.436–0.883), and HPV 16 + lrHPVs + other hrHPVs (OR = 0.248, 95% CI 0.157–0.391). The prevalence of CIN 3 + was associated with increased severity of cytologic abnormalities in HPV 16/18-positive women and peaked at cytology HSIL + (89.9% and 82.3%), which held a substantially greater risk than that of NILM (OR = 65.466, 95% CI 50.234–85.316). Conclusions In this cross-sectional study of HPV 16/18-positive women, the effects of multiple infection were likely complicated and varied with specific HPV genotypes. The coinfection of HPV 16 and other genotypes of HPV except HPV 18 was associated with decreased CIN 3 + risk. Cytologic results were informative when HPV 16/18 was positive. It might be reasonable to recommend expedited treatment for patients with HPV 16/18 positive and HSIL + cytology in the Chinese population. Background Persistent human papillomavirus (HPV) infection is the main etiological factor for cervical cancer. HPV 16 and 18 cause approximately 70% of cervical cancer cases [18 .")]. However, it is controversial whether co-infection status will influence the risk of cervical carcinogenesis in HPV 16/18-positive women. Since many studies have recommended HPV-based testing as primary screening method [2:102–31. ."),3:46.e1-46.e11. ."),4:263–70. ."),5:321–46. .")], research on this topic is essential. In large studies of populations from China, the United States, Norway and Italy, the prevalence of multiple infection is 4.6–8.2%, accounting for 20.4–43.9% of HPV-positive women 63 ."),[7:75–83. ."),8:29–34. ."),9 genotypes: prevalence and impact on the risk of precancerous epithelial lesions. J Med Virol. 2009;81(4):703–12. ."),10:405–8. .")]. A better understanding of interactions among different HPV genotypes contributes a lot to the risk assessment of women with multiple HPV infection and might provide information for the development of the next generation of HPV vaccine via predicting the possible genotypes that interact with those targeted by current vaccines. However, there is no consensus on the effects of multiple HPV infection. Some studies showed that multiple infection increased the risk of cervical intraepithelial neoplasia grade 3 or more severe diagnoses (CIN 3 +) [9 genotypes: prevalence and impact on the risk of precancerous epithelial lesions. J Med Virol. 2009;81(4):703–12. ."),10:405–8. ."),11:534–9. .")], and it was reported as the result of additive or synergistic effects [6:436–43. ."), 8:29–34. ."), 12:910–20. .")]. Conversely, Ping Xu et al. [6:436–43. .")] found that HPV 16 co-infected with other high risk HPV (hrHPV) was associated with a lower risk of CIN 3+. Besides, most studies did not distinguish HPV genotypes except for HPV 16 and 18, and thus the influence of specific HPV remains unclear. The CIN 3 + risk of HPV 16/18-positive women with different cytologic results has been well assessed from the general screening population in the United States 130 ."),[14:433–42. ."),15:47–54. .")]. The American Society of Colposcopy and Cervical Pathology published a risk-based management consensus guideline in 2019, which suggested that the combination of HPV 16/18 genotyping and cytology permitted more precise management [2:102–31. .")]. Nevertheless, limited data on the Chinese population is available [6:436–43. ."), 16, 17:650–7. .")]. Regardless of cytology results, colposcopy rather than expedited treatment remains the primary recommendations to HPV 16/18-positive women in China. It has been demonstrated that the prevalence of HPV genotypes in China is different from that in Western countries [18:692–705. ."), 19:1037–43. .")]. Therefore, it is questionable whether the recommendations based on risk estimates of American women can be applied to the Chinese population. In light of these facts, the study aimed to evaluate the effects of co-infection status and estimate the CIN 3 + risk for different cytology results in HPV 16/18-positive women. Methods Study design and participant enrollment The Ethics Committee of West China Second University Hospital of Sichuan University (WCSUH) approved the protocol before the initiation of this investigation. Informed consent was waived because the study was an anonymous analysis of retrospective data. The study manuscript follows the Strengthening the Reporting of Observational Studies in Epidemiology (STROBE) reporting guidelines for cross-sectional studies. A HPV genotyping examination with or without cytology was performed by gynecologists at WCSUH. The selection of cytology was determined by determined by the preference of clinicians and the specific condition of patients. Women who tested positive for HPV-16/18 were referred to colposcopy and biopsied if necessary. We recruited patients who tested positive for HPV 16/18 and underwent colposcopy within 6 months, between January 2010 and May 2021 at the outpatient department of WCSUH. The inclusion criteria were as follows: 25 years old or older; non-pregnant. Patients were excluded if they were immunocompromised or had a history of total hysterectomy or pelvic radiotherapy. We collected the following data from medical records: age, vaginal bleeding or not, results of HPV genotyping, cytology that obtained at the time of HPV genotyping, and pathological diagnoses. If patients had several HPV 16/18-positive results available, we collected the relevant information when the patients were positive for HPV 16/18 for the first time. HPV genotyping HPV genotyping was performed by one of the following two methods: (1) the HPV 23 Genotyping Assay (YanengBIO, Shenzhen, China), which tests for 17 hrHPV types (HPV 16, 18, 31, 33, 35, 39, 45, 51, 52, 53, 56, 58, 59, 66, 68, 73, 82) and 6 low-risk HPV (lrHPV) types (HPV 6, 11, 42, 43, 81, 83) or (2) the Tellgenplex HPV27 Genotyping Assay (Tellgen, Shanghai, China), which tests for 17 hrHPV types (HPV 16, 18, 26, 31, 33, 35, 39, 45, 51, 52, 53, 56, 58, 59, 66, 68, 82) and 10 lrHPV types (HPV 6, 11, 40, 42, 43, 44, 55, 61, 81, 83) 191 ."),[20:472–9. ."),21:e50. ."),22:e28302. ."),23]. Both detection assays have been validated and approved by the CFDA. The method for each patient was chosen according to the preference of the patient and gynecologist. Testing procedures followed kit protocols provided by the manufacturers. Cytology Cytology tests were performed using one of the following methods: the ThinPrep Pap test (histologic, Bedford, MA), or the SurePath Pap test (BD Diagnosis, Franklin Lakes, NJ). Samples were obtained by gynecologists and slides were prepared according to the manufacturers’ specifications. Cytology results were classified by WCSUH pathologists according to the 2001 Bethesda System terminology, including atypical squamous cells of undetermined significance (ASC-US), low-grade squamous intraepithelial lesion (LSIL), atypical squamous cells cannot exclude high-grade squamous intraepithelial lesion (ASC-H), atypical glandular cells (AGC, without subdivision), and high-grade intraepithelial lesion or worse (HSIL+). If two types of cytologic abnormalities (e.g. ASC-US and AGC-not otherwise specified, AGC-NOS) were detected in a sample, we would count both in during statistical analyses. Colposcopy-directed biopsy and histopathologic examination All suspicious lesions under colposcopy were biopsied. Endocervical curettage (ECC) was performed when the cervical squamous column junction was unsatisfied. Histological results were obtained from colposcopy-directed biopsy, ECC, loop electrosurgical excision procedure (LEEP), cold knife conization (CKC), and hysterectomy. All pathological diagnoses were categorized into normal (including cervicitis), CIN 1, CIN 2, CIN 3, adenocarcinoma in situ (AIS), and cervical cancer. The diagnoses were based on the worst results if there were a range of severities. Some women with adequate colposcopy assessment and normal findings might not be biopsied and classified as normal. Statistical analyses Histological results were classified as ≤ CIN 2 (including normal, CIN 1, and CIN 2) and CIN 3 + (including CIN 3, AIS, and cervical cancer). Variables were analyzed with (\chi^{2}) test, Fisher’s exact test, and Kruskal–Wallis test, as appropriate. Logistics regressions were conducted to explore the CIN3 + risk of different HPV infection patterns and cytology results. A two-tailed P value < 0.05 was considered statistically significant. Statistical analyses were conducted using IBM SPSS statistics software (version 21.0; IBM Corp, Armonk, NY). Results A total of 7940 HPV 16/18-positive women were included with a median age was 40 years (range, 25–84 years); 6211 (78.22%) and 1895 (23.87%) were positive for HPV 16 and 18, respectively. Among them, 34.1% (2710/7940) were infected with multiple genotypes, 82.28% (6533/7940) had cytology results and 26.65% (2116/7940) women were diagnosed with CIN 3 + (Additional file 1: Table S1). The prevalence of CIN 3 + in HPV 16-positive women (1899/6211, 30.57%) was significantly higher than that of HPV 18 (260/1895, 13.72%) (P< 0.001). The prevalence of CIN 3 + increased with age and differed significantly (Additional file 1: Fig. S1) (P< 0.001). The proportion of CIN 3 + in women with vaginal bleeding (316/588, 53.74%) was significantly higher than that in women without (1800/7352, 24.48%) (P< 0.001). Approximately 33.02% (2051/6211) of HPV 16-positive women had multiple infections, and that of HPV 18-positive women was 43.64% (827/1895). HPV 52, 58, 53, 81, 51, and 18 were common genotypes coinfecting with HPV 16, accounting for 22.09%, 16.43%, 11.21%, 10.24%, 9.36%, and 8.09%, respectively. HPV 18 often coinfected with HPV 16, 52, 58,53, 58, and 56, accounting for 17.91%, 16,72%, 11.43%, 9.17%, 8.95%, and 8.09%, respectively. Compared to single HPV 16 infection (Fig.1A), HPV 16 + 33 was significantly associated with an increased risk of CIN 3 + (odds ratio [OR] = 1.722, 95% confidence interval [CI] 1.102–2.691). In contrast, HPV 16 + 39 (OR = 0.467, 95% CI 0.241–0.905), HPV 16 + 59 (OR = 0.301, 95% CI 0.127–0.712), HPV 16 + 42 (OR = 0.426, 95% CI 0.228–0.799), HPV 16 + 81 (OR = 0.575, 95% CI 0.348–0.950), HPV 16 + 51 (OR = 0.428, 95% CI 0.248–0.739), HPV 16 + 53 (OR = 0.592, 95% CI 0.358–0.981), and HPV 16 + 66 (OR = 0.378, 95% CI 0.157–0.909) were significantly involved with a lower risk of CIN 3+. Compared to single HPV 18 infection (Fig.1B), HPV 18 + 33 (OR = 4.112, 95% CI 1.443–11.720), HPV 18 + 35 (OR = 4.486, 95% CI 1.775–11.335), HPV 18 + 58 (OR = 2.110, 95% CI 1.098–4.054), and HPV 18 + 16 (OR = 3.028, 95% CI 1.851–4.952) were related to a higher risk of CIN 3+. Fig. 1 Log odds ratio for CIN 3 + in coinfection of HPV 16 (A) or HPV 18 (B) and other HPV genotypes. The analyses took single HPV 16 or HPV 18 infection as a reference, respectively. The vertical solid line represents the null log odds ratio of 0.HPV, human papillomavirus; CIN 3+, cervical intraepithelial neoplasia grade 3 or more severe diagnoses Full size image According to the presence of other hrHPV and lrHPV, the included women were classified into 12 groups (Table 1). The diagnoses were significantly different in HPV 16-positive groups (P< 0.001), while there was no significant difference in HPV 18-positive groups (P = 0.266) and HPV 16 + 18-positive groups (P = 0.068). Notably, the prevalence of CIN 3 + in single HPV 16 (1408/4160, 33.85%), single HPV 18 (149/1068, 13.95%), and HPV 16 + HPV 18 (27/82, 32.93%) were the highest among HPV 16, HPV 18 and HPV 16 + 18 positive groups, respectively. Table 1 Distribution of HPV 16/18 infection patterns among the study population (n = 7940) Full size table The cytology results of 6533 included women were available (Table 2). Most (2932, 44.88%) had NILM cytology, 1514 (23.17%) had ASC-US, 906 (13.87%) had LSIL, 334 (5.11%) had ASC-H, 151 (2.31%) had AGC and 770 (11.79%) had HSIL+. As expected, the prevalence of CIN 3 + generally increased with the severity of cytologic abnormalities in HPV 16-positive women: NILM (11.4%), ASC-US (23.2%), LSIL (28.8%), ASC-H (72.3%), AGC (66.0%) and HSIL + (89.9%). Results in HPV 18-positive women showed a similar trend, and it was noteworthy that HPV 18-positive women with AGC had a relatively high prevalence of CIN 3 + (72.0%). Table 2 CIN 3 + risk of different cytology results in HPV 16/18-positive women (n = 6533) Full size table To further investigate the association between the abovementioned factors and CIN 3 + risk, we conducted two logistic regression analyses which took single HPV 16 infection (Table 3) and single HPV 18 infection as reference (Additional file 1: Table S2), respectively. Older age women (OR = 1.017, 95% CI 1.011–1.024) and those with vaginal bleeding (OR = 2.674, 95% CI 2.113–3.384) were at higher risk of CIN 3 + independently. HPV 16/18-positive women with cytologic abnormalities also had a higher risk of CIN 3+. Of note, HSIL + (OR = 65.466, 95% CI 50.234–85.316), ASC-H (OR = 17.339, 95% CI 13.223–22.736), and AGC (OR = 14.963, 95% CI 9.562–24.413) were associated with a substantially greater risk of CIN 3 + than that of NILM. As demonstrated in the logistics regression analysis using single HPV 16 infection as reference (Table 3), HPV 16 + other hrHPVs (OR = 0.621, 95% CI 0.511–0.755), HPV 16 + lrHPVs (OR = 0.620, 95% CI 0.436–0.883), and HPV 16 + lrHPVs + other hrHPVs (OR = 0.248, 95% CI 0.157–0.391) were associated with decreased risk of CIN 3 + . Meanwhile, regardless of the HPV infection patterns of HPV 18 infected women, the CIN 3 + risk of HPV 18 infection was lower than that of HPV 16 infection: single HPV 18 (OR = 0.327, 95% CI 0.255–0.420), HPV 18 + other hrHPVs (OR = 0.229, 95% CI 0.153–0.342), HPV 18 + lrHPVs (OR = 0.315, 95% CI 0.149–0.665) and HPV 18 + lrHPVs + other hrHPVs (OR = 0.210, 95% CI 0.094–0.467). However, there was no significant difference in CIN 3 + risk between single HPV 16 infection and HPV 16 + HPV 18 (OR = 0.747, 95% CI 0.364–1.534), HPV 16 + HPV 18 + other hrHPVs (OR = 0.710, 95% CI 0.322–1.562), or HPV 16 + HPV 18 + lrHPVs + other hrHPVs (OR = 0.370, 95% CI 0.076–1.801). The results of logistic regression analyses using single HPV 18 and HPV 16 infection as references were identical except for HPV infection patterns. Compared to single HPV 18 infection, the CIN 3 + risk did not differ significantly in HPV 18 + other hrHPVs (OR = 0.699, 95% CI 0.443–1.102), HPV 18 + lrHPVs (OR = 0.962, 95% CI 0.442–2.096) and HPV 18 + lrHPVs + other hrHPVs (OR = 0.641, 95% CI 0.280–1.468) (Additional file 1: Table S2). Table 3 Binary logistic regression analysis of factors influencing the risk of CIN 3 + in HPV 16/18-positive women Full size table Discussion This study is one of the largest research on HPV 16/18-positive women. Research on the effects of co-infection status might be helpful to risk stratification of HPV 16/18 infected women and more detailed interventions can be accessible in the future. Data on the CIN 3 + risk of different cytology results in a large population may provide evidence for improving the screening strategies of China. The proportion of multiple infection in HPV 16/18 infected cases was similar to a large population study in which 26.5% of HPV-positive women had multiple infection [246 .")]. In HPV 16/18 infected women, multiple infection is a common phenomenon. Our data suggested that for HPV 16/18 positive women, regardless of co-infection status, the immediate CIN 3 + risk was greater than the US benchmark (greater than or equal to 4%) for referral to colposcopy [2:102–31. .")]. Thus, it is reasonable for the referral to colposcopy in the Chinese HPV 16/18 positive population. There is no consensus on the association between multiple HPV infection and CIN 3 + risk. Previously it was assumed that each HPV genotype contributes to the risk of cervical precancer or cancer independently 250 ."), [26:E946–53. .")]. However, recent findings demonstrated that both antagonistic and synergistic interactions might exist among different HPV genotypes [6:436–43. ."), 12:910–20. ."), 27:391–8. ."),28:1458–64. ."),29:68–72. ."),30:855–64. .")]. Chaturvedi et al. [12:910–20. .")] reported that multiple HPV infection within the A9 species or oncogenic types increased the risk of cervical diseases in contrast to single HPV infection. Likewise, our results demonstrated that the co-infection of two genotypes belonging to A9 species, HPV 16 and HPV 33 was significantly associated with an increased risk of CIN 3+. Consistent with a large population-based study [6:436–43. .")], women with HPV 16 + 18 and either none, other hrHPVs, or lrHPVs were at a higher risk of CIN 3 + than single HPV 18 infection, whereas single HPV 16 infection and HPV 16 + HPV 18 had a similar risk of CIN 3 + in our study. Therefore, there might be synergistic interactions between HPV 16/18 and specific HPV genotypes. Meanwhile, we cannot exclude the possibility that the risk of multiple HPV infection was similar to the sum of estimated risk from individual genotypes, because the study did not include single infections of other HPV genotypes except HPV 16/18. We also observed antagonistic interactions between HPV 16 and other HPV genotypes, including HPV 39, 59, 42, 81, 51, 53, and 66. Furthermore, through multivariate analysis adjusted for several cofactors, we found that single HPV 16 infection was associated with a greater risk of CIN 3 + than those who were simultaneously infected by lrHPVs, and/or hrHPVs except for HPV 18. These findings were in agreement with those of Wheeler et al. [31:1291–9. .")], Wu et al. [6:436–43. .")], and Sundström et al. The mechanism of antagonistic interactions among HPV genotypes remain unclear. It was proposed that intergenotypic competition might interfere with the progression to CIN3 + in women with multiple HPV infections, which perhaps involved the key stages of HPV infection process, such as binding receptors, utilization of host cell organelles, synthesis of viral DNA, and insertion of viral DNA into the host genome [27:391–8. .")]. Another possible mechanism is relevant to the immune response, especially cross-protection [6:436–43. ."), 27:391–8. ."), 33:451–62. .")]. Concurrent infection of multiple HPV genotypes may induce a more effective immune response than a single HPV infection [27:391–8. .")], and perhaps the infection of a less carcinogenic or non-carcinogenic HPV genotype triggers the immune response in advance, thereby reducing the pathogenicity of subsequent infection of HPV 16/18 [6:436–43. .")]. Overall, the effects of multiple HPV infection might depend on specific HPV combinations and further research is needed to elucidate the mechanisms. Generally, our data demonstrated that cytologic results were informative when HPV 16/18 was positive. It is noteworthy that HPV 16/18-positive women with cytology HSIL + were at a very high risk of CIN 3 + (over 80%), showing an approximately 65 folds higher risk than cytology NILM. By contrast, Risk estimates from a subset of women in the Kaiser Permanente Northern California screening program identified HPV16 and 18 with HSIL + had immediate CIN 3 + risk of 60% and 30%, respectively [130 .")]. The possible explanation for the variation is that most of the Chinese population, especially those in rural areas, do not have access to regular screening, leading to a relatively higher risk of CIN 3 + at the first visit. [34:212–22. ."),35:e0233986. ."),36:2089–100. .")] Consequently, it might be reasonable to recommend treatment without biopsy for non-pregnant women 25 years or older with positive HPV 16/18 and cytology HSIL + in China. Unlike HSIL+, cytology ASC-H in HPV 16/18-positive women predicted different risks of CIN 3+: 72.3% for HPV 16 and 34.7% for HPV 18. Accordingly, HPV 16/18-positive with ASC-H might not require the same intensive management as HSIL+, especially HPV 18. Colposcopy might still be preferred for HPV 16/18-positive women with ASC-H cytology. Another important finding is that almost half of the included women had HPV 16/18 and cytology NILM, among which CIN 3 + was identified in 11.4% of HPV 16-positive women while 4.8% of HPV 18-positive cases. These results were consistent with previous studies, showing that even if the cytology was NILM, HPV 16/18-positive women still conferred a relatively high risk of CIN 3+. [13:144–7. ."), 24:391–8. .")] Therefore, this study supported the strategy of HPV-16/18 genotyping for women with HPV-positive and cytology-negative cytology in the Chinese population. Limitations The study had some limitations. First, the research was conducted in a single institution retrospectively. Hence the conclusions may not be generalized to the whole Chinese population without caution. Second, the number of cases was relatively low for coinfection of HPV 16/18 and several HPV genotypes. Third, medical records were incomplete for some patients, and there were no records of past history, smoking history, age at first intercourse, number of sexual partners, especially HPV vaccination status. However, a nationwide survey demonstrated that the HPV vaccine uptake rate of females was only 3% in mainland China [379 .")], hence it is reasonable to infer that most of the included women were not vaccinated against HPV. Fourth, two types of HPV genotyping tests may result in bias, but they have been well verified and reached a high agreement [19:1037–43. ."),20:472–9. ."),21:e50. ."),22:e28302. ."),23]. Conclusion In this cross-sectional study of HPV 16/18-positive women, the effects of multiple infection were likely complicated and varied with specific HPV genotypes. Generally, HPV16 co-infected with other genotypes of HPV except HPV18 was associated with decreased risk of CIN 3 + independently. More attention should be paid to the effects of multiple infection. Besides, Cytologic results were informative when HPV 16/18 was positive. Cytologic abnormalities in HPV 16/18 infected women were associated with a greater risk of CIN 3+, especially ASC-H and HSIL+. In the case of HPV 16/18 infection and HSIL + in the Chinese population, expedited treatment might be acceptable. References de Sanjose S, Quint W, Alemany L, et al. Human papillomavirus genotype attribution in invasive cervical cancer: a retrospective cross-sectional worldwide study. Lancet Oncol. 2010;11(11):1048–56. 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A nationwide post-marketing survey of knowledge, attitude and practice toward human papillomavirus vaccine in general population: implications for vaccine roll-out in mainland China. Vaccine. 2021;39(1):35–44. ArticlePubMedGoogle Scholar Download references Funding This study was supported by the National Natural Science Foundation of China (NSFC) (No. 82272710) and the Health Care Scientific and Technology Project of Sichuan Province (2022-1701). Author information Authors and Affiliations Department of Gynecology and Obstetrics, West China Second University Hospital, Sichuan University, Number 20, Third Section of People’s South Road, Chengdu, 610000, China Mengyin Ao,Xiaoxi Yao,Danxi Zheng&Mingrong Xi Key Laboratory of Birth Defects and Related Diseases of Women and Children, Sichuan University, Chengdu, Sichuan, China Mengyin Ao,Xiaoxi Yao,Danxi Zheng&Mingrong Xi Department of Information Management, West China Second University Hospital, Sichuan University, Number 20, Third Section of People’s South Road, Chengdu, 610000, China Xuesai Gu Authors 1. Mengyin AoView author publications Search author on:PubMedGoogle Scholar 2. Xiaoxi YaoView author publications Search author on:PubMedGoogle Scholar 3. Danxi ZhengView author publications Search author on:PubMedGoogle Scholar 4. Xuesai GuView author publications Search author on:PubMedGoogle Scholar 5. Mingrong XiView author publications Search author on:PubMedGoogle Scholar Contributions Study concept and design: MA and MX; data extraction: MA, XG, and XY; statistical analysis: MA; manuscript preparation: MA; manuscript editing and review: MX, DZ. All authors approved the final manuscript. Corresponding author Correspondence to Mingrong Xi. Ethics declarations Competing interests All the authors declare that they have no conflicts of interest. Additional information Publisher's Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Supplementary Information Additional file1 . Table S1: Patients characteristics. Table S2: Binary logistic regression analysis of HPV 16/18 infection patterns, adjusted by age, vaginal bleeding, and cytology results (referring to Single HPV 18 infection). Figure S1: Prevalence of CIN 3+ in HPV 16/18-positive women stratified by age (n = 7940). 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Reprints and permissions About this article Cite this article Ao, M., Yao, X., Zheng, D. et al. Risk of cervical intraepithelial neoplasia grade 3 or more diagnoses for human papillomavirus16/18-positive women by cytology and co-infection status. Infect Agents Cancer18, 57 (2023). Download citation Received: 25 July 2023 Accepted: 28 September 2023 Published: 09 October 2023 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords HPV 16 HPV 18 Cervical cancer screening CIN 3 Multiple HPV infection Download PDF Download ePub Associated content Collection New insights in pathogen-related cancers Advertisement Infectious Agents and Cancer ISSN: 1750-9378 Contact us General enquiries: journalsubmissions@springernature.com 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. © 2025 BioMed Central Ltd unless otherwise stated. Part of Springer Nature.
15557	https://dlmf.nist.gov/10.17	DLMF: §10.17 Asymptotic Expansions for Large Argument ‣ Bessel and Hankel Functions ‣ Chapter 10 Bessel Functions DLMF Index Notations Search Help? Citing Customize Annotate UnAnnotate About the Project 10 Bessel FunctionsBessel and Hankel Functions10.16 Relations to Other Functions10.18 Modulus and Phase Functions §10.17 Asymptotic Expansions for Large Argument ⓘ Keywords:Bessel functions, asymptotic expansions for large argumentPermalink: also:Annotations for Ch.10 Contents §10.17(i) Hankel’s Expansions §10.17(ii) Asymptotic Expansions of Derivatives §10.17(iii) Error Bounds for Real Argument and Order §10.17(iv) Error Bounds for Complex Argument and Order §10.17(v) Exponentially-Improved Expansions §10.17(i) Hankel’s Expansions ⓘ Keywords:Hankel functions, Hankel’s expansions, asymptotic expansions for large argument, for Bessel and Hankel functionsNotes:See Olver (1997b, pp.237–242).Referenced by:§10.17(ii), §10.17(ii), §10.40(i), §10.67(i), §10.74(i), §10.74(i), §10.74(ii)Permalink: also:Annotations for §10.17 and Ch.10 Define a 0⁡(ν)=1, 10.17.1 a k⁡(ν)=(4⁢ν 2−1 2)⁢(4⁢ν 2−3 2)⁢⋯⁢(4⁢ν 2−(2⁢k−1)2)k!⁢8 k=(1 2−ν)k⁢(1 2+ν)k(−2)k⁢k!, k≥1, ⓘ Defines:a k⁡(ν): polynomial coefficient (locally)Symbols:(a)n: Pochhammer’s symbol (or shifted factorial), !: factorial (as in n!), k: nonnegative integer and ν: complex parameterReferenced by:§10.49(i), Erratum (V1.1.3) for AdditionsPermalink: pMML, pngAddition (effective with 1.1.3): An alternative Pochhammer symbol representation was added. See also:Annotations for §10.17(i), §10.17 and Ch.10 10.17.2 ω=z−1 2⁢ν⁢π−1 4⁢π, ⓘ Defines:ω (locally)Symbols:π: the ratio of the circumference of a circle to its diameter, z: complex variable and ν: complex parameterPermalink: pMML, pngSee also:Annotations for §10.17(i), §10.17 and Ch.10 and let δ denote an arbitrary small positive constant. Then as z→∞, with ν fixed, 10.17.3 J ν⁡(z)∼(2 π⁢z)1 2⁢(cos⁡ω⁢∑k=0∞(−1)k⁢a 2⁢k⁡(ν)z 2⁢k−sin⁡ω⁢∑k=0∞(−1)k⁢a 2⁢k+1⁡(ν)z 2⁢k+1), \|ph⁡z\|≤π−δ, ⓘ Symbols:J ν⁡(z): Bessel function of the first kind, ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, cos⁡z: cosine function, ph: phase, sin⁡z: sine function, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, a k⁡(ν): polynomial coefficient and ωA&S Ref:9.2.5 Referenced by:§10.17(i), §10.17(iii), §10.17(iv), §10.18(iii), §10.24, §10.49(i), §10.61(v), §10.7(ii)Permalink: pMML, pngSee also:Annotations for §10.17(i), §10.17 and Ch.10 10.17.4 Y ν⁡(z)∼(2 π⁢z)1 2⁢(sin⁡ω⁢∑k=0∞(−1)k⁢a 2⁢k⁡(ν)z 2⁢k+cos⁡ω⁢∑k=0∞(−1)k⁢a 2⁢k+1⁡(ν)z 2⁢k+1), \|ph⁡z\|≤π−δ, ⓘ Symbols:Y ν⁡(z): Bessel function of the second kind, ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, cos⁡z: cosine function, ph: phase, sin⁡z: sine function, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, a k⁡(ν): polynomial coefficient and ωA&S Ref:9.2.6 Referenced by:§10.17(iii), §10.17(iv), §10.18(iii), §10.24, §10.7(ii), §11.6(i)Permalink: pMML, pngSee also:Annotations for §10.17(i), §10.17 and Ch.10 10.17.5 H ν(1)⁡(z)∼(2 π⁢z)1 2⁢e i⁢ω⁢∑k=0∞i k⁢a k⁡(ν)z k, −π+δ≤ph⁡z≤2⁢π−δ, ⓘ Symbols:H ν(1)⁡(z): Bessel function of the third kind (or Hankel function), ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, i: imaginary unit, ph: phase, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, a k⁡(ν): polynomial coefficient and ωA&S Ref:9.2.7 Referenced by:§10.17(i), §10.17(iii), §10.17(iv), §10.41(v)Permalink: pMML, pngSee also:Annotations for §10.17(i), §10.17 and Ch.10 10.17.6 H ν(2)⁡(z)∼(2 π⁢z)1 2⁢e−i⁢ω⁢∑k=0∞(−i)k⁢a k⁡(ν)z k, −2⁢π+δ≤ph⁡z≤π−δ, ⓘ Symbols:H ν(2)⁡(z): Bessel function of the third kind (or Hankel function), ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, i: imaginary unit, ph: phase, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, a k⁡(ν): polynomial coefficient and ωA&S Ref:9.2.8 Referenced by:§10.17(i), §10.17(iii), §10.17(iv), §10.41(v), §10.49(i), §10.61(v)Permalink: pMML, pngSee also:Annotations for §10.17(i), §10.17 and Ch.10 where the branch of z 1 2 is determined by 10.17.7 z 1 2=exp⁡(1 2⁢ln⁡\|z\|+1 2⁢i⁢ph⁡z). ⓘ Symbols:exp⁡z: exponential function, i: imaginary unit, ln⁡z: principal branch of logarithm function, ph: phase and z: complex variablePermalink: pMML, pngSee also:Annotations for §10.17(i), §10.17 and Ch.10 Corresponding expansions for other ranges of ph⁡z can be obtained by combining (10.17.3), (10.17.5), (10.17.6) with the continuation formulas (10.11.1), (10.11.3), (10.11.4) (or (10.11.7), (10.11.8)), and also the connection formula given by the second of (10.4.4). §10.17(ii) Asymptotic Expansions of Derivatives ⓘ Keywords:Bessel functions, Hankel functions, asymptotic expansions for large argument, derivativesNotes:These results follow by differentiation of the corresponding expansions in §10.17(i); compare §2.1(iii).Referenced by:§10.40(i), §10.67(i)Permalink: also:Annotations for §10.17 and Ch.10 We continue to use the notation of §10.17(i). Also, b 0⁡(ν)=1, b 1⁡(ν)=(4⁢ν 2+3)/8, and for k≥2, 10.17.8 b k⁡(ν)=((4⁢ν 2−1 2)⁢(4⁢ν 2−3 2)⁢⋯⁢(4⁢ν 2−(2⁢k−3)2))⁢(4⁢ν 2+4⁢k 2−1)k!⁢8 k. ⓘ Defines:b k⁡(ν): polynomial coefficient (locally)Symbols:!: factorial (as in n!), k: nonnegative integer and ν: complex parameterPermalink: pMML, pngSee also:Annotations for §10.17(ii), §10.17 and Ch.10 Then as z→∞ with ν fixed, 10.17.9 J ν′⁡(z)∼−(2 π⁢z)1 2⁢(sin⁡ω⁢∑k=0∞(−1)k⁢b 2⁢k⁡(ν)z 2⁢k+cos⁡ω⁢∑k=0∞(−1)k⁢b 2⁢k+1⁡(ν)z 2⁢k+1), \|ph⁡z\|≤π−δ, ⓘ Symbols:J ν⁡(z): Bessel function of the first kind, ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, cos⁡z: cosine function, ph: phase, sin⁡z: sine function, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, ω and b k⁡(ν): polynomial coefficientA&S Ref:9.2.11 Permalink: pMML, pngSee also:Annotations for §10.17(ii), §10.17 and Ch.10 10.17.10 Y ν′⁡(z)∼(2 π⁢z)1 2⁢(cos⁡ω⁢∑k=0∞(−1)k⁢b 2⁢k⁡(ν)z 2⁢k−sin⁡ω⁢∑k=0∞(−1)k⁢b 2⁢k+1⁡(ν)z 2⁢k+1), \|ph⁡z\|≤π−δ, ⓘ Symbols:Y ν⁡(z): Bessel function of the second kind, ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, cos⁡z: cosine function, ph: phase, sin⁡z: sine function, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, ω and b k⁡(ν): polynomial coefficientA&S Ref:9.2.12 Permalink: pMML, pngSee also:Annotations for §10.17(ii), §10.17 and Ch.10 10.17.11 H ν(1)′⁡(z)∼i⁢(2 π⁢z)1 2⁢e i⁢ω⁢∑k=0∞i k⁢b k⁡(ν)z k, −π+δ≤ph⁡z≤2⁢π−δ, ⓘ Symbols:H ν(1)⁡(z): Bessel function of the third kind (or Hankel function), ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, i: imaginary unit, ph: phase, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, ω and b k⁡(ν): polynomial coefficientA&S Ref:9.2.13 Referenced by:§10.18(iii)Permalink: pMML, pngSee also:Annotations for §10.17(ii), §10.17 and Ch.10 10.17.12 H ν(2)′⁡(z)∼−i⁢(2 π⁢z)1 2⁢e−i⁢ω⁢∑k=0∞(−i)k⁢b k⁡(ν)z k, −2⁢π+δ≤ph⁡z≤π−δ. ⓘ Symbols:H ν(2)⁡(z): Bessel function of the third kind (or Hankel function), ∼: Poincaré asymptotic expansion, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, i: imaginary unit, ph: phase, k: nonnegative integer, z: complex variable, ν: complex parameter, δ: small positive constant, ω and b k⁡(ν): polynomial coefficientA&S Ref:9.2.14 Referenced by:§10.18(iii)Permalink: pMML, pngSee also:Annotations for §10.17(ii), §10.17 and Ch.10 §10.17(iii) Error Bounds for Real Argument and Order ⓘ Keywords:Bessel functions, Hankel functions, asymptotic expansions for large argument, error boundsNotes:See Watson (1944, pp.205–206) and Olver (1997b, pp.268–269).Permalink: also:Annotations for §10.17 and Ch.10 In the expansions (10.17.3) and (10.17.4) assume that ν≥0 and z>0. Then the remainder associated with the sum ∑k=0 ℓ−1(−1)k⁢a 2⁢k⁡(ν)⁢z−2⁢k does not exceed the first neglected term in absolute value and has the same sign provided that ℓ≥max⁡(1 2⁢ν−1 4,1). Similarly for ∑k=0 ℓ−1(−1)k⁢a 2⁢k+1⁡(ν)⁢z−2⁢k−1, provided that ℓ≥max⁡(1 2⁢ν−3 4,1). In the expansions (10.17.5) and (10.17.6) assume that ν>−1 2 and z>0. If these expansions are terminated when k=ℓ−1, then the remainder term is bounded in absolute value by the first neglected term, provided that ℓ≥max⁡(ν−1 2,1). §10.17(iv) Error Bounds for Complex Argument and Order ⓘ Notes:See Olver (1997b, pp.266–267).Referenced by:§10.49(i), §10.74(i)Permalink: also:Annotations for §10.17 and Ch.10 For (10.17.5) and (10.17.6) write 10.17.13 H ν(1)⁡(z)H ν(2)⁡(z)}=(2 π⁢z)1 2⁢e±i⁢ω⁢(∑k=0 ℓ−1(±i)k⁢a k⁡(ν)z k+R ℓ±⁡(ν,z)), ℓ=1,2,…. ⓘ Symbols:H ν(1)⁡(z): Bessel function of the third kind (or Hankel function), H ν(2)⁡(z): Bessel function of the third kind (or Hankel function), π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, i: imaginary unit, k: nonnegative integer, z: complex variable, ν: complex parameter, a k⁡(ν): polynomial coefficient, ω and R ℓ±⁡(ν,z): remainderReferenced by:§10.17(iv), §10.17(v)Permalink: pMML, pngSee also:Annotations for §10.17(iv), §10.17 and Ch.10 Then 10.17.14\|R ℓ±⁡(ν,z)\|≤2⁢\|a ℓ⁡(ν)\|⁢𝒱 z,±i⁢∞⁡(t−ℓ)⁢exp⁡(\|ν 2−1 4\|⁢𝒱 z,±i⁢∞⁡(t−1)), ⓘ Defines:R ℓ±⁡(ν,z): remainder (locally)Symbols:exp⁡z: exponential function, i: imaginary unit, 𝒱 a,b⁡(f): total variation, z: complex variable, ν: complex parameter and a k⁡(ν): polynomial coefficientReferenced by:§10.17(iv), Erratum (V1.0.10) for Equation (10.17.14)Permalink: pMML, pngErrata (effective with 1.0.10): Originally the factor 𝒱 z,±i⁢∞⁡(t−1) in the argument to the exponential was written incorrectly as 𝒱 z,±i⁢∞⁡(t−ℓ). Reported 2014-09-27 by Gergő Nemes See also:Annotations for §10.17(iv), §10.17 and Ch.10 where 𝒱 denotes the variational operator (2.3.6), and the paths of variation are subject to the condition that \|ℑ⁡t\| changes monotonically. Bounds for 𝒱 z,i⁢∞⁡(t−ℓ) are given by 10.17.15 𝒱 z,i⁢∞⁡(t−ℓ)≤{\|z\|−ℓ,0≤ph⁡z≤π,χ⁡(ℓ)⁢\|z\|−ℓ,−1 2⁢π≤ph⁡z≤0 or π≤ph⁡z≤3 2⁢π,2⁢χ⁡(ℓ)⁢\|ℑ⁡z\|−ℓ,−π<ph⁡z≤−1 2⁢π or 3 2⁢π≤ph⁡z<2⁢π, ⓘ Symbols:π: the ratio of the circumference of a circle to its diameter, ℑ⁡: imaginary part, i: imaginary unit, ph: phase, 𝒱 a,b⁡(f): total variation, z: complex variable and χ⁡(ℓ)Referenced by:§10.17(iv)Permalink: pMML, pngSee also:Annotations for §10.17(iv), §10.17 and Ch.10 where χ⁡(ℓ)=π 1 2⁢Γ⁡(1 2⁢ℓ+1)/Γ⁡(1 2⁢ℓ+1 2); see §9.7(i). The bounds (10.17.15) also apply to 𝒱 z,−i⁢∞⁡(t−ℓ) in the conjugate sectors. Corresponding error bounds for (10.17.3) and (10.17.4) are obtainable by combining (10.17.13) and (10.17.14) with (10.4.4). §10.17(v) Exponentially-Improved Expansions ⓘ Keywords:Bessel functions, Hankel functions, asymptotic expansions for large argument, error bounds, exponentially-improvedNotes:See Olver (1991b, Theorem 1) or Olver (1993a, Theorem 1.1), and (10.16.6).Referenced by:§10.74(i)Permalink: also:Annotations for §10.17 and Ch.10 As in §9.7(v) denote 10.17.16 G p⁡(z)=e z 2⁢π⁢Γ⁡(p)⁢Γ⁡(1−p,z), ⓘ Symbols:Γ⁡(z): gamma function, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, Γ⁡(a,z): incomplete gamma function, G p⁡(z): rescaled terminant function and z: complex variableReferenced by:§10.40(iv)Permalink: pMML, pngSee also:Annotations for §10.17(v), §10.17 and Ch.10 where Γ⁡(1−p,z) is the incomplete gamma function (§8.2(i)). Then in (10.17.13) as z→∞ with \|ℓ−2⁢\|z\|\| bounded and m (≥0) fixed, 10.17.17 R ℓ±⁡(ν,z)=(−1)ℓ⁢2⁢cos⁡(ν⁢π)⁢(∑k=0 m−1(±i)k⁢a k⁡(ν)z k⁢G ℓ−k⁡(∓2⁢i⁢z)+R m,ℓ±⁡(ν,z)), ⓘ Symbols:π: the ratio of the circumference of a circle to its diameter, cos⁡z: cosine function, i: imaginary unit, G p⁡(z): rescaled terminant function, m: integer, k: nonnegative integer, z: complex variable, ν: complex parameter, a k⁡(ν): polynomial coefficient and R ℓ±⁡(ν,z): remainderPermalink: pMML, pngSee also:Annotations for §10.17(v), §10.17 and Ch.10 where 10.17.18 R m,ℓ±⁡(ν,z)=O⁡(e−2⁢\|z\|⁢z−m), \|ph⁡(z⁢e∓1 2⁢π⁢i)\|≤π. ⓘ Defines:R ℓ±⁡(ν,z): remainder (locally)Symbols:O⁡(x): order not exceeding, π: the ratio of the circumference of a circle to its diameter, e: base of natural logarithm, i: imaginary unit, ph: phase, m: integer, z: complex variable and ν: complex parameterPermalink: pMML, pngSee also:Annotations for §10.17(v), §10.17 and Ch.10 For higher re-expansions of the remainder terms see Olde Daalhuis and Olver (1995a) and Olde Daalhuis (1995, 1996). 10.16 Relations to Other Functions10.18 Modulus and Phase Functions © 2010–2025 NIST / Disclaimer / Feedback; Version 1.2.4; Release date 2025-03-15. 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15558	https://math.stackexchange.com/questions/1195891/help-deriving-the-midpoint-formula	linear algebra - Help Deriving the Midpoint Formula - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Help Deriving the Midpoint Formula Ask Question Asked 10 years, 6 months ago Modified10 years, 6 months ago Viewed 425 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. One of the problems in one of the packets that I'm going through to review for a pre-test for an independent-study calculus class has asked me to derive the midpoint formula. I've gotten to the point where I have the following equation: 2 x 2 x m−2 x m x 1+2 y 2 y m−2 y m y 1=x 2 2−x 2 1+y 2 2−y 2 1 2 x 2 x m−2 x m x 1+2 y 2 y m−2 y m y 1=x 2 2−x 1 2+y 2 2−y 1 2 Would it be mathematically correct to split this into the following two equations…: {2 x 2 x m−2 x m x 1=x 2 2−x 2 1 2 y 2 y 2−2 y m y 1=y 2 2−y 2 1{2 x 2 x m−2 x m x 1=x 2 2−x 1 2 2 y 2 y 2−2 y m y 1=y 2 2−y 1 2 …and treat them as a system of equations? If so, then how would I go about doing this? linear-algebra analytic-geometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Mar 18, 2015 at 20:24 RandomDSdevelRandomDSdevel asked Mar 18, 2015 at 18:35 RandomDSdevelRandomDSdevel 345 2 2 silver badges 14 14 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. You can not split this equation. That would be the same as saying a+b=c+d a+b=c+d can be split into a=c a=c and b=d b=d. If the points are P 1,P 2,P 1,P 2, and P m P m, for P m P m to be the midpoint between P 1 P 1 and P 2 P 2, then one way of writing this is that P m P m must be on the line between P 1 P 1 and P 2 P 2 (i.e., P m=t P 1+(1−t)P 2 P m=t P 1+(1−t)P 2 for 0≤t≤1 0≤t≤1) and P m P m must be at the halfway position (i.e., t=1 2 t=1 2). Another way is that \|P 1−P m\|=\|P 2−P m\|\|P 1−P m\|=\|P 2−P m\| and \|P 1−P m\|=\|P 2−P 1\|/2\|P 1−P m\|=\|P 2−P 1\|/2 . This way gives you two equations for the two unknowns of the x and y components of P m P m. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 18, 2015 at 21:11 marty cohenmarty cohen 111k 10 10 gold badges 88 88 silver badges 186 186 bronze badges 2 Ah, of course! I'm a nincompoop for not realizing I would need a second equation when I started doing my derivation from the first of the two equations you listed in your second derivation method!RandomDSdevel –RandomDSdevel 2015-03-18 22:16:32 +00:00 Commented Mar 18, 2015 at 22:16 Glad to be of service.marty cohen –marty cohen 2015-03-19 04:26:30 +00:00 Commented Mar 19, 2015 at 4:26 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra analytic-geometry See similar questions with these tags. 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15559	https://math.stackexchange.com/questions/1633619/can-you-find-the-maximum-or-minimum-of-an-equation-without-calculus	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 Can you find the maximum or minimum of an equation without calculus? [closed] Ask Question Asked Modified 9 years, 8 months ago Viewed 26k times 3 $\begingroup$ Without using calculus is it possible to find provably and exactly the maximum value or the minimum value of a quadratic equation $$ y:=ax^2+bx+c $$ (and also without completing the square)? I'd love to know the answer. analysis Share edited Feb 1, 2016 at 13:45 Karlie KlossKarlie Kloss asked Jan 30, 2016 at 20:34 Karlie KlossKarlie Kloss 13511 gold badge11 silver badge99 bronze badges $\endgroup$ 8 $\begingroup$ Also here is meta.math.stackexchange.com/questions/5020/… for formatting your functions properly. $\endgroup$ Arbuja – Arbuja 2016-01-30 20:35:54 +00:00 Commented Jan 30, 2016 at 20:35 $\begingroup$ Odd question... No calculus, no completion of the square... I guess asking the teacher should work. $\endgroup$ user228113 – user228113 2016-01-30 21:16:30 +00:00 Commented Jan 30, 2016 at 21:16 2 $\begingroup$ Without completing the square, or without calculus? $\endgroup$ Mark Bennet – Mark Bennet 2016-01-30 21:16:36 +00:00 Commented Jan 30, 2016 at 21:16 $\begingroup$ This is like asking how to win a martial arts tournament while unconscious. $\endgroup$ Matt Samuel – Matt Samuel 2016-01-31 00:13:29 +00:00 Commented Jan 31, 2016 at 0:13 $\begingroup$ That said, I would guess the ancient Greeks knew how to do this, and I think completing the square was discovered less than a thousand years ago. $\endgroup$ Matt Samuel – Matt Samuel 2016-01-31 00:33:17 +00:00 Commented Jan 31, 2016 at 0:33 \| Show 3 more comments 4 Answers 4 Reset to default 10 $\begingroup$ The vertex of $y=x^2$ is $(0, 0)$. The vertex of $y = Ax^2$ is $(0, 0)$. The vertex of $y = A(x - k)^2$ is just shifted right $k$, so it is $(k, 0)$. The vertex of $y = A(x - k)^2 + j$ is just shifted up $j$, so it is $(k, j)$. Multiply that out, you get $y = Ax^2 - 2Akx + Ak^2 + j$. So you get $$b = -2ak \tag{1}$$ $$c = ak^2 + j \tag{2}$$ Solve (1) for $k$ and plug it into (2), then solve for $j$,you get: $$k = \frac{-b}{2a}$$ $$c = a\left(\frac{-b}{2a}\right)^2 + j \implies j = \frac{4ac - b^2}{4a}$$ So the vertex occurs at $(j, k) = \left(\frac{-b}{2a}, \frac{4ac - b^2}{4a}\right)$. Share answered Jan 31, 2016 at 6:27 DanielVDanielV 24.6k55 gold badges4141 silver badges7070 bronze badges $\endgroup$ 2 $\begingroup$ So it works out the values in the shifts of the maxima or minima at (0,0) , in the specific quadratic, to deduce the actual maxima or minima in any quadratic. I think this is a good answer to the question I asked. $\endgroup$ Karlie Kloss – Karlie Kloss 2016-02-07 13:22:15 +00:00 Commented Feb 7, 2016 at 13:22 $\begingroup$ This is one of the best answer I have come across $\endgroup$ Kirthi Raman – Kirthi Raman 2016-10-15 22:44:59 +00:00 Commented Oct 15, 2016 at 22:44 Add a comment \| 2 $\begingroup$ We assume (for the sake of discovery; for this purpose it is good enough if this is just an inspired guess) that the curve $y = ax^2 + bx + c$ is symmetric around a vertical axis. It's obvious this is true when $b = 0$, and if we have plotted $y = ax^2 + bx + c$ for various other values of $a$, $b$, and $c$, we may observe enough appearance of symmetry to suppose that it might be true in general. So it's reasonable to say: supposing it were true, what would that tell us about the minimum/maximum value of the polynomial? We find the points on this curve of the form $(x,c)$ as follows: \begin{align} y &= c. \ c &= ax^2 + bx + c. \ 0 &= ax^2 + bx = (ax + b)x. \end{align} Hence if $(x,c)$ is on the curve, then either $ax + b = 0$ or $x = 0$. Setting $x_1 = -\dfrac ba$ and $x_2 = 0$, we can plug in these two values for $x$ and confirm that indeed the two points $\left(-\frac ba, c\right)$ and $(0, c)$ are on the curve. Using the assumption that the curve is symmetric around a vertical axis, the vertical axis would have to be halfway between $\left(-\frac ba, c\right)$ and $(0, c)$, that is, it is the line $x = -\dfrac b{2a}$. If there is a global maximum or minimum, it is a reasonable guess that it would be on this line, so let's see what we have at $x_0 = -\dfrac b{2a}$. Plugging this into the equation and doing the algebra to find the point $(x_0, y_0)$ on the curve, \begin{align} y_0 &= a\left(-\frac b{2a}\right)^2 + b\left(-\frac b{2a}\right) + c \ &= c - \frac{b^2}{4a}. \end{align} So that's our candidate for the maximum or minimum value. To prove this is correct, consider any value of $x$ other than $-\dfrac b{2a}$. Any such value can be expressed by its difference from $-\dfrac b{2a}$, that is, we let $$ x = -\frac b{2a} + t$$ where $t \neq 0$. Now plug this value into the equation and do the algebra: \begin{align} y &= a\left(-\frac b{2a} + t\right)^2 + b\left(-\frac b{2a} + t\right) + c \[.5ex] &= at^2 + c - \frac{b^2}{4a}. \tag 1 \end{align} If $a$ is positive, $at^2$ is positive, hence $y > c - \dfrac{b^2}{4a} = y_0$ for every point $(x,y)$ on the curve such that $x \neq x_0$, and therefore $y_0 = c - \dfrac{b^2}{4a}$ is a minimum. But if $a$ is negative, $at^2$ is negative, and similar reasoning says that $y_0 = c - \dfrac{b^2}{4a}$ is a maximum. Note that the proof made no assumption about the symmetry of the curve. On the contrary, the equation $y = at^2 + c - \dfrac{b^2}{4a}$ can be used to prove that the curve is symmetric. If we take this a little further, we can even derive the standard quadratic formula from it. The roots of the equation $y = ax^2 + bx + c$ are the values of $x$ such that $y = 0$. But as we know from Equation $(1)$, above, if we make the substitution $x = -\dfrac b{2a} + t$, that means \begin{align} 0 = y &= ax^2 + bx + c \ &= at^2 + c - \frac{b^2}{4a}. \end{align} A little algebra (isolate the $at^2$ term on one side and divide by $a$) gives us $$ t^2 = \frac{b^2}{4a^2} - \frac ca. \tag 2 $$ In general, if $p^2 = q$ then $p = \pm \sqrt q$, so Equation $(2)$ tells us that \begin{align} t &= \pm \sqrt{\frac{b^2}{4a^2} - \frac ca} \ &= \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}\ &= \pm \frac{\sqrt{b^2 - 4ac}}{\lvert 2a \rvert}\ &= \pm \frac{\sqrt{b^2 - 4ac}}{2a}, \end{align} and recalling that we set $x = -\dfrac b{2a} + t$, \begin{align} x &= -\frac b{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \ &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}, \end{align} which is precisely the usual quadratic formula. Is the reasoning above actually just an example of "completing the square," or is it sufficiently different from the usual method of "completing the square" that it can be considered a different method? The equation $x = -\dfrac b{2a} + t$ is equivalent to $t = x + \dfrac b{2a}$; the method of completing the square involves expanding $\left(x + \dfrac b{2a}\right)^2$; and in fact we do see $t^2$ figuring prominently in the equations above. Certainly we could be inspired to try completing the square after noticing how neatly the equation $ax^2 + bx + c = at^2 + c - \dfrac{b^2}{4a}$ simplified the problem; but we never actually expanded the binomial $\left(x + \dfrac b{2a}\right)^2$, and we never subtracted the original polynomial from it to find the amount we needed to "complete" the square. Instead, the quantity $c - \dfrac{b^2}{4a}$ just "appeared" in the original equation as the result of a direct substitution. I think that may be about as different from "completing the square" as a purely algebraic method can get. Share edited Feb 2, 2016 at 2:05 answered Feb 1, 2016 at 15:29 David KDavid K 110k88 gold badges9191 silver badges242242 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ One approach for finding the maximum value of $y$ for $y=ax^2+bx+c$ would be to see how large $y$ can be before the equation has no solution for $x$. First rearrange the equation into a standard form: $ax^2+bx+c-y=0$ Now solving for $x$ in terms of $y$ using the quadratic formula gives: $x= \frac{-b\pm \sqrt{b^2-4a(c-y)}}{2a}$ This will have a solution as long as $b^2-4a(c-y) \geq 0$ You can rearrange this inequality to get the maximum value of $y$ in terms of $a,b,c$. See if you get the same answer as the calculus approach gives. Remember that $a$ must be negative in order for there to be a maximum. Share answered Jan 30, 2016 at 20:42 HughHugh 2,3511616 silver badges1515 bronze badges $\endgroup$ 4 $\begingroup$ Yes a variation of this idea can be used to find the minimum too $\endgroup$ Karlie Kloss – Karlie Kloss 2016-01-30 20:59:43 +00:00 Commented Jan 30, 2016 at 20:59 5 $\begingroup$ @Karlie Kloss Technically speaking this solution is also not without completion of squares because you are still using the quadratic formula and how do you get that??? $\endgroup$ Anurag A – Anurag A 2016-01-30 21:09:49 +00:00 Commented Jan 30, 2016 at 21:09 $\begingroup$ @KarlieKloss Just because you don't see something spelled out in its full detail doesn't mean it is "not used." The question then is, what is the proof of the quadratic formula that does not use any form of completing the square? $\endgroup$ David K – David K 2016-02-01 14:58:32 +00:00 Commented Feb 1, 2016 at 14:58 $\begingroup$ Yes, t think now that is a better question to ask. So this method answers the question if there is a proof of the quadratic formula that does not use any form of completing the square. $\endgroup$ Karlie Kloss – Karlie Kloss 2016-02-01 16:46:22 +00:00 Commented Feb 1, 2016 at 16:46 Add a comment \| 0 $\begingroup$ This is almost the same as completing the square but .. for giggles. If $a = 0$ we know $y = xb + c$ will get "extreme" and "extreme" positive and negative values of $x$ so no max or minimum is possible. So if $ax^2 + bx + c = a(x^2 + x b/a)+c := a(x^2 + b'x) + c$ So finding the max/min is simply a matter of finding the max/min of $x^2 + b'x$ and multiplying by $a$ and adding $c$. In other words .... wolog $a = 1$ and $c = 0$. Now we know $x^2 + bx$ has only a min as $x^2$ is positive and as $\|x\|$ increases the $x^2$ term "overpowers" the $bx$ term. Or if $x > \|b\|/2$ then $(x+ h)^2 + b(x + h) = x^2 + bx +h(2x + b) + h^2 > 0$ so the expression has no max value. So we want to find the minimum of $x^ + b'x = x(x + b)$. Let $y := x - b'/2$ then $x(x + b')=(y -b'/2)(y + b'/2)= y^2 - (b'^2/4)$. As $y^2 \ge 0$ the min will occur when $y = 0$ or in other words, $x= b'/2 = b/2a$ So the max/min of $ax^2 + bx + c$ occurs at $x = b/2a$ and the max/min value is $b^2/4 + b^2/2a + c$ Okay, that really was the same thing as completing the square but it didn't feel like it so what the @@@@. Share edited Feb 2, 2016 at 3:08 answered Feb 2, 2016 at 2:54 fleabloodfleablood 132k55 gold badges5252 silver badges142142 bronze badges $\endgroup$ Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions analysis 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 Why can ALL quadratic equations be solved by the quadratic formula? 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15560	http://galileo.phys.virginia.edu/classes/241w.rmm5a.fall04/manual/Lab01.pdf	5 ____________ University of Virginia Physics Department PHYS 241W, Fall 2004 Name _____ Date _ Partners____ Lab 1 – ELECTROSTATICS OBJECTIVES • To understand the difference between conducting and insulating materials. • To observe the effects of charge polarization in conductors and insulators • To understand and demonstrate two ways to charge an object: conduction and induction • To determine the polarity of charge on a charged object based on macroscopic observations OVERVIEW Source: Fishbane, Paul M., Stephen Gasiorowicz, and Stephen T. Thornton. Physics for Scientists and Engineers, 3rd Edition. Prentice-Hall, Inc. Upper Saddle River, NJ. 2005. pp. 609-617 Electrostatics1 is the study of charges, which are not in motion and the interactions between them. Most of the phenomena we observe in the study of electrostatics arise from ionization. Although atoms are electrically neutral, their valence electrons are sometimes easily transferred; when an atom gains or loses electrons, the resulting imbalance of charge is referred to as ionization. Positive ions are atoms which have lost electrons, while negative ions are atoms which have gained electrons. A material with easily detached valence electrons (which can then move through the material easily) is referred to as a conductor; conversely, a material with strongly attached electrons is called an insulator. Objects can be charged by making contact with another charged object, a phenomena known as charging by conduction. When objects are charged without coming into contact with a charge source, the process is known as charging by induction; this process primarily works with conductors. One method of charging involves moving a charged object to the vicinity of two uncharged conductors in contact with each other, as shown in Figure 1. An induced charge flows to one conductor, leaving the other conductor oppositely charged. When separated the conductors have equal, but opposite, charges. As always, it is important to remember for these experiments that charge is conserved and that like charges repel while opposite charges attract. 1 Adapted from Dr. Richard Lindgren, Charlene Wyrick, Karyn Traphagen, and Lynn Lucarto. The Shocking Truth: Lessons in Electrostatics. 2000. 6 Lab 1 - Electrostatics ____________ University of Virginia Physics Department PHYS 241W, Fall 2004 INTRODUCTION TO THE APPARATUS The electroscope is an instrument that detects the presence of charge on an object, either through actual contact (conduction) or through induction. When the electroscope itself is charged, its two conductive components (which vary from electroscope to electroscope) will acquire like charge and deflect from the vertical position of gravitational equilibrium. Thus a rod is proven to possess a charge when contact between the rod and the electroscope transfers charge to the previously neutral electroscope. A charged object brought in the vicinity of the charged electroscope will change the angle of deflection, indicating the presence of charge via induction (or induced polarization). This process is explained in greater detail in the experiments. Step 1 Step 2 Step 3 Step 4 Figure 1: Charging by Induction UVa Electroscope Insulating base Metal Tube Brass Support Steel Pin Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 7 The UVA electroscopes have been designed with the following considerations: • The base of the electroscope is constructed out of acrylic, an insulator, to minimize charge leakage to the table (or to the object on which the base sits). • The tube is made of metal and is suspended slightly above its center of gravity so that it will quickly reach stable equilibrium when no charge is present. • The steel pin, which supports the tube, has been fitted inside the tube to minimize charge leakage. • The support structure is made of brass, a conductor. The rounded edges of the support minimize charge leakage into the atmosphere (sharp points on conductors tend to leak charge away easily). If a charged rod is brought into contact with the upper lip of the support structure, charge will distribute across the brass and the aluminum, causing the tube to deflect from the brass structure. The following hints will optimize your use of the electroscope: • Insulators (e.g. Teflon and acrylic) do not transfer charge easily to other objects, so draw or scrape the rod across the brass lip in order to transfer charge to the electroscope. • Oils transferred from your hand to the rod may adversely affect experimental results; you may need to clean the rod with alcohol to ensure that it is as clean as possible. • Handle only the marked end of the Teflon rod and charge the unmarked end to minimize the effects of hand oils. • When using the electroscope to detect the presence of charge, it may be necessary to bring a charged rod near the tube in a back-and-forth rhythmic motion to cause movement of the tube. • Humidity increases the charge leakage into the atmosphere. Ideally, humidity should be below 50%. Winter is thus the best time in Virginia for static electricity demonstrations, but we use air conditioners and dehumidifiers to help improve the situation. THE TRIBOELECTRIC SERIES Materials possess various tendencies to acquire or lose electrons; the ordering of these tendencies is referred to as the triboelectric series. When you use the silk to charge the Teflon rod, you are engaging in a process known as triboelectric charging. Teflon’s electrical nature dictates that it will acquire a negative net charge, as it has a tendency to take electrons from the silk. Glass, on the other hand, has a tendency to acquire a positive net charge. The list below orders a 8 Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 number of common materials by their electrical nature2. Note that human hands have a strong tendency to gain positive charge. We will use Teflon and silk. Human hands Asbestos Rabbit Fur Glass, Mica Human Hair Nylon, Wool Lead Silk Aluminum Paper Cotton Steel Wood Amber Hard Rubber Mylar Nickel, Copper Silver, Brass Gold, Platinum Polyester, Celluloid Saran Wrap Polyurethane Polyethylene Polypropylene Vinyl, Silicon Teflon Silicon Rubber Figure 2 2 This Triboelectric series was adapted from Allen, Ryne C., Desco Industries Inc. (DII), December 2000. Downloaded from July 31, 2002. WARNING: One should use the triboelectric series such as the one illustrated above with caution. The ability of a material to acquire or lose electrons is related to several physical factors, including the material’s conductivity and surface finish as well as the microscopic surface area contact between the two charging elements. Also, many materials are not pure, and impurities in the material may change its electrical nature. We have found tables with acrylic being positive and negative. The acrylic rods we have are more positive than negative. But one should not be surprised if acrylic, for example, gains negative instead of positive charge. Tendency to gain POSITIVE charge Tendency to gain NEGATIVE charge ZERO Lab 1 - Electrostatics ____________ University of Virginia Physics Department PHYS 241W, Fall 2004 9 Question 1: If you rub a glass rod with silk, what is the polarity of the charge on each object? What about Teflon and silk? Gold and lead? Use the Triboelectric series above. Glass and silk: Glass_ Silk_ Teflon and silk: Teflon_ Silk___ Gold and lead: Gold_ Lead_ INVESTIGATION 1: CONDUCTORS AND INSULATORS The purpose of the experiment is to observe the electrical properties associated with an insulator and a conductor. Recall that, as stated in the Overview, electrons can move freely through conductors; they are not free to move within insulators. The following equipment is needed for this investigation: • Spinner • Alcohol • Silk • Paper towels (one roll per room) • Teflon rods (2) • Brass rod • Wooden rod • Electroscope Activity 1-1: Teflon In the following activity, you will observe the interaction between two negatively charged Teflon rods, one held in your hand and one placed on a spinner. You will charge only one end of each Teflon rod. Question 1-1: Why is it possible to charge only one end of the Teflon rod (as opposed to charging the entire rod)? Is this true for all materials? The “spinner” consists of a metal pin attached to a plastic acrylic base and a second piece of plastic which rotates on the metal pin. It thus rotates very freely. The rod is positioned on this second piece of plastic. 10 Lab 1 - Electrostatics ____________ University of Virginia Physics Department PHYS 241W, Fall 2004 Prediction 1-1: Predict what will happen if you hold the charged end of the Teflon rod in your hand along the charged end of the Teflon rod on the spinner without touching(Figure 3)? Do not yet try it. What about along the uncharged end of the Teflon rod on the spinner? Figure 3 1. Charge one end of the first Teflon rod by striking it on the silk and place this Teflon rod on the spinner. Now charge one end of the second Teflon rod. 2. Use the spinner and the Teflon rods to test your predictions: After charging both rods, hold one Teflon rod in your hand parallel to the Teflon rod on the spinner as shown in Figure 3 in order to ensure the greatest possible interaction between the two. If nothing happens, you may need to recharge the Teflon rods. MAKE SURE THAT YOU CHARGE THE SAME END OF THE TEFLON RODS THAT YOU CHARGED BEFORE. Otherwise, it will be necessary for you to discharge the rods to produce accurate results. You can discharge the rods by rubbing the charged end with your hand to allow excess negative charge to flow away. The Charging Process: When you strike the silk with the Teflon rod, a charge transfer occurs between the two materials. As Teflon attracts negative charge, the Teflon rod attracts the loose electrons from the silk’s surface and becomes negatively charged. Because charge is conserved, the silk is left positively charged. Transfer of electrons is responsible for charging; atoms do not transfer protons. Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 11 Question 1-2: What were your results? Was there any interaction with the uncharged end of the rod on the spinner? If so, discuss. (You need to hold the rods very close together in this case.) Activity 1-2: Movement of Charge in Metals How the Electroscope Works: Metals contain valence electrons which are not tightly bound to atoms and which are able to move through the conductor. When you charge the electroscope, excess charge distributes itself evenly across the conducting parts of the electroscope (the brass support and the metal tube). This distribution occurs as a result of the force of repulsion between like charges, which attempt to move as far away from each other as possible. After the charge reaches static equilibrium, the tube should be deflected at a constant angle. 1. Use the charged Teflon rod to charge the electroscope by rubbing the top of the brass support with the end of the Teflon rod that is charged. You may need to charge the rod and rub the support a few times until the tube deflects a little. 2. Recharge the Teflon rod and bring it slowly near the lower end of the tube. At first you should notice repulsion between the tube and the charged Teflon rod; as the rod slowly approaches the tube, it should then begin to attract. Electroscope Insulating base Metal Tube Brass Support 12 Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 Question 1-3: Explain your two observations above in terms of metal’s conducting properties and the charge distribution. What happens to the charge distribution in the electroscope when the negatively charged Teflon rod is nearby? Activity 1-3: Using Conductors to Discharge the Electroscope When you provide the electrons in a charged object a route of escape through contact with an uncharged conducting object, electrons will flow from the charged object to the uncharged conducting object until electrical balance is achieved. 1. To demonstrate this gradual discharging, touch the metal pin of the spinner to the top of the electroscope for a few seconds and then remove it. The electroscope’s deflection angle should have decreased. 2. Discharge the pin by touching it with your finger and repeat the process described in Step 1. Watch the electroscope gradually discharge. Note: It is necessary that you discharge the pin, as it will reach a saturation point and no longer be able to drain charge from the electroscope. You will learn more about this phenomenon when you study capacitors. When you bring the metal parts of the electroscope into contact with a conducting element, which offers a pathway from the charged metal to the ground, you are performing an operation known as “grounding” or total discharging. As the spinner’s metal pin does not provide a pathway for charges to the ground, it does not completely “ground” the electroscope. Prediction 1-2: Predict what would happen if you touched the top of the charged electroscope with your finger? What does this indicate about the conducting properties of the human body? Lab 1 - Electrostatics ____________ University of Virginia Physics Department PHYS 241W, Fall 2004 13 3. Touch the top of the brass support with your finger and observe the results. Question 1-4: Were your observations consistent with your prediction? If not, explain. Prediction 1-3: What would happen if you touched the top of the charged electroscope with a brass rod held in your hand? An acrylic rod? A wooden rod? Justify your prediction on the basis of the properties of conductors and insulators. 4. Touch the charged electroscope with each of the rods mentioned in Prediction 1-3 and observe and record the results. Question 1-5: Explain your observations in terms of each rod’s conducting properties. INVESTIGATION 2: CHARGE POLARIZATION An object is said to be neutral if it contains the same number of positive and negative charges. A neutral object can, however, produce some of the same phenomena as a charged object as a result of a process known as polarization. We already know that opposite charges attract. If we recall that charges are somewhat free to move within an object, we should not be surprised that a positively charged object will induce a charge alignment in a neutral object so that the object’s electrons are as near to the 14 Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 positively charged object as possible. As a result, the neutral object will appear to react to an electric force as though it were charged. When you charge by induction, you are exploiting polarization. Recall that in a conductor, electrons are free to move. Polarization in a conductor, then, is a result of a movement of electrons to one side of an object. Electrons are not free to move in an insulator; this does not mean, however, that an insulator does not experience polarization. Polarization in an insulator is a result of an alignment of the charge within each individual molecule (Figure 4). A positive charge in the vicinity of an atom, for example, will attract the atom’s electron cloud and thus shift the cloud toward the positive charge. This results in a reorientation of the charges of the atom. In a neutral object, atoms closest to the charged object will experience the greatest amount of stretching. Polarization does not create a permanent charge; it is instead a temporary effect caused by the proximity of a charged object. The polarized object acquires no net charge. Question 2-1: What charge will the left and right sides of the insulating rod tend to have in Figure 4? Figure 4 In addition to the equipment used previously, you will need for this investigation: Insulating rod Charged rod Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 15 • Acrylic (insulating) rod Activity 2-1: Charge Polarization in an Insulator 1. Place the acrylic rod on the spinner. Prediction 2-1: Predict what will happen if you bring a charged Teflon rod near the acrylic rod? Will it matter which end of the acrylic rod you place the charged rod near? Will it matter which side of the acrylic rod you choose? 2. Test your predictions using a charged Teflon rod. Question 2-2: Was your prediction correct? If not, explain. In Activity 1-2 you charged the electroscope with the charged Teflon rod until the tube deflected just a little. Then you used the charged Teflon rod to come near the lower end of the tube, but didn’t touch it. At first, when you approached 5-10 cm away, the tube repelled. As you came closer, the tube was attracted. If you don’t recall this experiment, you may want to repeat it now. Question 2-3: In terms of polarization, what causes the attraction of the tube to the Teflon rod? Question 2-4: Why does the tube initially repel and then attract? 16 Lab 1 - Electrostatics ____________ University of Virginia Physics Department PHYS 241W, Fall 2004 Activity 2-2: Charge Polarization in a Conductor 1. Charge the electroscope using the Teflon rod and silk. 2. Bring a charged Teflon rod near the bottom of the brass support next to the tube, but not touching. You should observe an increased deflection angle between the tube and the brass support. Question 2-5: In terms of polarization, what causes this increased deflection? Question 2-6: What charge sign do the elements of the electroscope have? Indicate in the diagram below the sign of any charged element of the electroscope. INVESTIGATION 3: CHARGING BY CONDUCTION AND INDUCTION When you charge the electroscope by touching the negatively charged rod to the top of the brass support, you are charging via conduction, which requires contact. The electroscope acquires the same charge as the charged rod; the negatively charged rod (which Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 17 has taken electrons from the silk) distributes electrons on the electroscope. Charging via conduction necessitates that the charged object physically touch the object to be charged, and both objects will have the same sign of charge. Charging via induction occurs without contact between the charged object and the object to be charged. The charged object (which we will refer to as an inducer) is brought near the neutral object, inducing polarization. Suppose the inducer has a negative charge. If you place your finger on the polarized object, your finger will drain electrons from the object, thus creating a net positive charge (Figure 5). When your finger and then the inducer are both removed, this net positive charge remains. The newly charged object is thus of the opposite polarity as the initially charged inducer. Figure 5 You do not need any new equipment for this investigation. Activity 3-1: Charging by Induction 1. Ground the electroscope to neutralize it. You can do this by touching the top of the brass support with your finger, thereby providing a pathway for the charge to the ground. 2. Have your partner hold his/her finger to the top of the brass support. 3. While the finger is on the brass support, bring the charged Teflon rod to the base of the brass support without touching it until you see a deflection of the tube. 18 Lab 1 - Electrostatics ___________ University of Virginia Physics Department PHYS 241W, Fall 2004 4. Have your partner remove his/her finger, then remove the Teflon rod. 5. Observe the deflection angle of the tube. Question 3-1: What is the sign of the charge on the electroscope? Explain what has happened. Question 3-2: How can you verify the sign of charge on the electroscope? Describe. Now try it, and explain your results.
15561	https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/11%3A_Transition_Metal_Complexes/11.03%3A_Electron_Counting	Skip to main content 11.3: Electron Counting Last updated : Oct 4, 2022 Save as PDF 11.2: Building Blocks- Metal Ions and Ligands in Transition Metal Complexes 11.4: Chelation Page ID : 189677 Chris Schaller College of Saint Benedict/Saint John's University ( \newcommand{\kernel}{\mathrm{null}\,}) Transition metals bind to ligands in order to edge closer to electronic saturation. In other words, the electrons donated by the ligands to form the bonds to the metal help that metal get to the next noble gas configuration. It does not always get there, and sometimes it overshoots, but that's a rough guide. Consider the coordination complex below. [Co(NH3)6]Cl3 contains a cobalt complex ion and three chloride counterions; the counterions just there to balance the charge. We could think of that complex as being assembled from a cobalt ion and six ammonia molecules. The six ammonia molecules each lend a lone pair to help the cobalt towards its octet. If we go backwards, and disassemble the complex into its components, we see that each of those ammonia molecules was neutral. There is no formal charge on any of those nitrogens. If they are all neutral, but the complex has a positive charge, then where did the positive charge come from? The answer has to be the cobalt ion. The cobalt has a 3+ charge (or "oxidation state"). Knowing that, we can figure out how many electrons cobalt has in its valence shell within the complex. In the periodic table, cobalt has nine valence electrons. However, if the overall complex has a charge of 3+, then the cobalt also had a charge of 3+. It has lost three electrons, so it only has six left. Once it forms the complex, each ammonia donates a pair of electrons; that's twelve total. The valence shell around cobalt includes its own d electrons plus that twelve donated by the ligands, for a total of eighteen. We can summarise in a table: \| \| \| --- \| \| metal valence e- (or d e-) \| 9 e- \| \| charge on complex \| 3+ \| \| charge on ligands \| 0 \| \| charge on metal \| 3+ \| \| metal ion d e- \| 6 e- \| \| e- donated by ligands \| 6 x 2 = 12 e- \| \| total \| 18 e- \| [Co(NH3)5Cl]Cl2 is a very similar example. This time, when we dissect the complex, we find a chlorine that would have a formal charge of 1- (we always assume the ligands all had octets before they joined the metal). The overall complex has a 2- charge; taking into account that there is a charge of -1 on the chloride within that complex, then the cobalt ion must have a charge of 3+ in order to arrive at that overall 2+ charge. Again,we can use a table to count the electrons on cobalt in the complex. \| \| \| --- \| \| metal valence e- (or d e-) \| 9 e- \| \| charge on complex \| 2+ \| \| charge on ligands \| 1- \| \| charge on metal \| 3+ \| \| metal ion d e- \| 6 e- \| \| e- donated by ligands \| 12 e- \| \| total \| 18 e- \| Exercise Confirm the electron count on the metal is 18 electrons in each of the following complexes. a) Pd(PPh3)4 b) Cr(CO)6 c) [Cu(NCCH3)4]ClO4 d) Fe(CO)5 e) K4[Fe(CN)6] Answer a: : \| \| \| --- \| \| Pd valence e- \| 10 e- \| \| charge on complex \| 0 \| \| charge on ligands \| 0 \| \| charge on Pd \| 0 \| \| revised Pd e- \| 10 e- \| \| e- donated by ligands \| 4 x 2 = 8 e- \| \| total \| 18 e- \| Answer b: : \| \| \| --- \| \| Cr valence e- \| 6 e- \| \| charge on complex \| 0 \| \| charge on ligands \| 0 \| \| charge on Cr \| 0 \| \| revised Cr e- \| 6 e- \| \| e- donated by ligands \| 6 x 2 = 12 e- \| \| total \| 18 e- \| Answer c: : \| \| \| --- \| \| Cu valence e- \| 11 e- \| \| charge on complex \| +1 \| \| charge on ligands \| 0 \| \| charge on Cu \| +1 \| \| revised Cu e- \| 10 e- \| \| e- donated by ligands \| 4 x 2 = 8 e- \| \| total \| 18 e- \| Answer d: : \| \| \| --- \| \| Fe valence e- \| 8 e- \| \| charge on complex \| 0 \| \| charge on ligands \| 0 \| \| charge on Fe \| 0 \| \| revised Fe e- \| 8 e- \| \| e- donated by ligands \| 5 x 2 = 10 e- \| \| total \| 18 e- \| Answer e: : \| \| \| --- \| \| Fe valence e- \| 8 e- \| \| charge on complex \| -4 \| \| charge on ligands \| -6 \| \| charge on Fe \| +2 \| \| revised Fe e- \| 6 e- \| \| e- donated by ligands \| 6 x 2 = 12 e- \| \| total \| 18 e- \| Exercise Determine the electron count on the metal in each of the following complexes. a) RhClH2(PPh3)2 b) Ni(OH2)42 c) Cu(NH3)42 d) K2[PtCl6] Answer a: Answer b: Answer c: Answer d: On the next page, we will see how some ligands can bind to a metal more than once. That helps them hold on more tightly. 11.2: Building Blocks- Metal Ions and Ligands in Transition Metal Complexes 11.4: Chelation
15562	https://undergroundmathematics.org/divisibility-and-induction/factorial-fun/solution	Solution \| Factorial fun \| Divisibility & Induction \| Underground Mathematics underground mathematics Map Search Browse User More Home How-to guide Underground hub About and contact Your mathematical classroom Exit fullscreen mode ### Divisibility & Induction Food for thought Factorial fun Add to your resource collection Printable/supporting materialsFullscreen modeTeacher notes Problem Solution Solution We denote the product of the first 20 20 natural numbers by 20!20! and call this 20 20 factorial. What is the highest power of 5 5 which is a divisor of 20 20 factorial? We have 20!=20×19×18×⋯×3×2×1 20!=20×19×18×⋯×3×2×1. Since 5 5 is a prime number, we can just add the highest power of 5 5 dividing each of the numbers 1 1, 2 2, 3 3, …, 20 20. This is 0 0 for each number not divisible by 5 5. There are only four numbers in this range (5 5, 10 10, 15 15, 20 20) that are divisible by 5 1 5 1, and none of them is divisible by 5 2 5 2. So the highest power of 5 5 dividing 20!20! is 5 4 5 4. Just how many factors does 20!20! have altogether? The first thing we need to do is find the prime factorisation of 20!20!. As 20!=20×19×18×⋯×3×2×1 20!=20×19×18×⋯×3×2×1, we can can do this by finding the prime factorisation of each of the numbers 2 2, 3 3, …, 20 20 (ignoring 1 1 since it won’t contribute any primes) and multiplying them together. \| Number \| Prime factorisation \| --- \| \| 2 2 \| 2 2 \| \| 3 3 \| 3 3 \| \| 4 4 \| 2 2 2 2 \| \| 5 5 \| 5 5 \| \| 6 6 \| 2×3 2×3 \| \| 7 7 \| 7 7 \| \| 8 8 \| 2 3 2 3 \| \| 9 9 \| 3 2 3 2 \| \| 10 10 \| 2×5 2×5 \| \| 11 11 \| 11 11 \| \| 12 12 \| 2 2×3 2 2×3 \| \| 13 13 \| 13 13 \| \| 14 14 \| 2×7 2×7 \| \| 15 15 \| 3×5 3×5 \| \| 16 16 \| 2 4 2 4 \| \| 17 17 \| 17 17 \| \| 18 18 \| 2×3 2 2×3 2 \| \| 19 19 \| 19 19 \| \| 20 20 \| 2 2×5 2 2×5 \| Multiplying the prime factorisations in the right-hand column together and simplifying, we get 20!=2 18×3 8×5 4×7 2×11×13×17×19.20!=2 18×3 8×5 4×7 2×11×13×17×19. Now we can use this to calculate the number of divisors of 20!20!. Each divisor will have a unique prime factorisation, which must be ‘contained’ within the prime factorisation of 20!20!. Let m m be a divisor of 20!20!. Then there are nineteen possible values for the highest power of 2 2 dividing m m (0 0, 1 1, …, 18 18). Similarly, there are nine possible values for the highest power of 3 3 dividing m m (0 0, 1 1, …, 8 8). Continuing in this way for all the prime factors of 20!20!, we can calculate that there are 19×9×5×3×2×2×2×2=41040 19×9×5×3×2×2×2×2=41040 divisors of 20!20!. Show that the highest power of p p that divides 500!500!, where p p is a prime number and p t<500<p t+1 p t<500<p t+1, is ⌊500/p⌋+⌊500/p 2⌋+⋯+⌊500/p t⌋,⌊500/p⌋+⌊500/p 2⌋+⋯+⌊500/p t⌋, where ⌊x⌋⌊x⌋ (the floor of x x) means to round x x down to the nearest integer. (For example, ⌊3⌋=3⌊3⌋=3, ⌊4.7⌋=4⌊4.7⌋=4, ⌊−2.7⌋=−3⌊−2.7⌋=−3, and so on.) We can see that ⌊500/p⌋⌊500/p⌋ is the number of multiples of p p that are less than or equal to 500 500. For example, if p p goes into 500 500 “seven and a bit” times, this means that p p, 2 p 2 p, …, 7 p 7 p are less than 500 500 but 8 p 8 p is greater than 500 500. Similarly, ⌊500/p 2⌋⌊500/p 2⌋ is the number of multiples of p 2 p 2 that are less than or equal to 500 500, and so on. Now, we can work out the highest power of p p that divides 500!500! by considering the number of multiples of p p, p 2 p 2, …, p t p t less than 500 500. We need to count all the multiples of p p. We need to count the multiples of p 2 p 2 twice, since they contribute 2 2 to the exponent, but we have already counted them once as they are also multiples of p p so we need to count them just once more. Similarly, we need to count the multiples of p 3 p 3 three times in total, but we have already counted them twice (once in the multiples of p p and once in the multiples of p 2 p 2), so we need to count them just once more. And so on for the remaining powers. So the highest power of p p that divides 500!500! has exponent ⌊500 p⌋+⌊500 p 2⌋+⋯+⌊500 p t⌋⌊500 p⌋+⌊500 p 2⌋+⋯+⌊500 p t⌋ Note that we had to assume that p p was prime. What would have gone wrong if it had not been? Answer Consider the case that p=10 p=10, for example. Then in addition to all of the numbers which are multiples of 10 10 or 10 2 10 2 (which give ⌊500/10⌋+⌊500/10 2⌋=50+5=55⌊500/10⌋+⌊500/10 2⌋=50+5=55 factors of 10 10), there are further factors of 10 10 from other multiples of 2 2 and 5 5 such as 2×5 2×5, 15×18 15×18, and so on. The same sort of problem will occur for many composite values of p p. How many factors does n!n! have? We can generalise the above result beyond the case of 500!500!. The highest power of p p that divides n!n!, where p t≤n<p t+1 p t≤n<p t+1, is equal to ⌊n p⌋+⌊n p 2⌋+⋯+⌊n p t⌋,⌊n p⌋+⌊n p 2⌋+⋯+⌊n p t⌋, by exactly the same reasoning as in (b). We can use this information to find the prime factorisation of n!n!. Let P P be the largest prime with P≤n P≤n. Also, for any number m m, let t m t m be the integer such that m t m≤n<m t m+1 m t m≤n<m t m+1. Using the information from part (b), we can then calculate the prime factorisation of n!n!: we have n!=2(⌊n/2⌋+⌊n/2 2⌋+⋯+⌊n/2 t 2⌋)×3(⌊n/3⌋+⌊n/3 2⌋+⋯+⌊n/3 t 3⌋)×⋯×P(⌊n/P⌋+⌊n/P 2⌋+⋯+⌊n/P t P⌋).n!=2(⌊n/2⌋+⌊n/2 2⌋+⋯+⌊n/2 t 2⌋)×3(⌊n/3⌋+⌊n/3 2⌋+⋯+⌊n/3 t 3⌋)×⋯×P(⌊n/P⌋+⌊n/P 2⌋+⋯+⌊n/P t P⌋). Then we can use the same reasoning as at the end of part (a) to calculate the number of factors of n!n!: we get (1+⌊n 2⌋+⌊n 2 2⌋+⋯+⌊n 2 t 2⌋)×(1+⌊n 3⌋+⌊n 3 2⌋+⋯+⌊n 3 t 3⌋)×⋯×(1+⌊n P⌋+⌊n P 2⌋+⋯+⌊n P t P⌋).(1+⌊n 2⌋+⌊n 2 2⌋+⋯+⌊n 2 t 2⌋)×(1+⌊n 3⌋+⌊n 3 2⌋+⋯+⌊n 3 t 3⌋)×⋯×(1+⌊n P⌋+⌊n P 2⌋+⋯+⌊n P t P⌋). We can check that this expression works for the example of 20!20!: (1+⌊20 2⌋+⌊20 4⌋+⌊20 8⌋+⌊20 16⌋)×(1+⌊20 3⌋+⌊20 9⌋)×(1+⌊20 5⌋)×(1+⌊20 7⌋)×(1+⌊20 11⌋)×(1+⌊20 13⌋)×(1+⌊20 17⌋)×(1+⌊20 19⌋)=(10+5+2+1+1)×(6+2+1)×(4+1)×(2+1)×(1+1)×(1+1)×(1+1)×(1+1)=19×9×5×3×2×2×2×2=41040.(1+⌊20 2⌋+⌊20 4⌋+⌊20 8⌋+⌊20 16⌋)×(1+⌊20 3⌋+⌊20 9⌋)×(1+⌊20 5⌋)×(1+⌊20 7⌋)×(1+⌊20 11⌋)×(1+⌊20 13⌋)×(1+⌊20 17⌋)×(1+⌊20 19⌋)=(10+5+2+1+1)×(6+2+1)×(4+1)×(2+1)×(1+1)×(1+1)×(1+1)×(1+1)=19×9×5×3×2×2×2×2=41040. This is the same answer as in part (a), suggesting that this formula works! Previous Based on this NRICH resource, used with permission. The NRICH resource remains Copyright University of Cambridge, All rights reserved. Last updated 20-Jul-17 Tags Generalising and specialising Station Divisibility & Induction Lines Number Add to your collection Add the current resource to your resource collection EmailTwitter TermsCookiesPrivacy underground mathematics Rich resources for teaching A level mathematics Copyright © and Database Right 2013-2025 University of Cambridge All rights reserved
15563	https://www.sciencedirect.com/science/article/pii/S2468602619300075	Skip to article My account Sign in View PDF JSES Open Access Volume 3, Issue 2, July 2019, Pages 65-69 Innervation of the subscapularis: an anatomic study Author links open overlay panel, , , , rights and content Under a Creative Commons license Open access Background Successful healing of the subscapularis during anatomic total shoulder arthroplasty surgery is critical to optimize functional outcomes and avoid complications. The purpose of this study was to examine the upper and lower subscapularis nerve insertion in relation to the musculotendinous junction to estimate the risk of nerve injury. Our hypothesis was that arm position changes the risks to these nerves when exposing the anterior glenoid. Methods Twenty cadaveric shoulders were dissected, and the subscapular nerves were identified from the posterior cord of the brachial plexus to the muscle insertion. The nerve length from the origin to the muscle insertion and the distance to the myotendinous junction were measured in various shoulder positions including neutral, external, and internal rotation. Results The mean length of the upper subscapular nerve was 51.4 ± 12.8 mm; that of the lower subscapular nerve was 50.5 ± 14 mm. The mean distance from the insertion of the upper subscapular nerve to the myotendinous junction 53.0 ± 14.7 mm with external rotation, 38.5 ± 9.7 mm with neutral rotation, and 30.0 ± 9.2 mm with internal rotation. The mean distance from the lower subscapular nerve to the myotendinous junction was 44.5 ± 13.8 mm with external rotation, 31.9 ± 9.3 mm with neutral rotation, and 25.4 ± 8.8 mm with internal rotation. The internally rotated position placed these nerves closest to the glenohumeral joint. Conclusion The upper and lower subscapular nerves insert in the muscle belly close to the myotendinous junction, putting them at risk of iatrogenic injury. Care must be taken to avoid damage with retractor placement in the anterior glenoid neck as these nerves are at risk of compression or torsional injury. Level of evidence Anatomy Study Cadaveric Dissection Keywords Subscapularis shoulder arthroplasty anatomy cadaveric study deltopectoral approach innervation subscapularis Cited by (0) : Institutional review board approval was not required. © 2019 The Author(s). Published by Elsevier Inc. on behalf of American Shoulder and Elbow Surgeons.
15564	https://wayground.com/en-us/product-rule-worksheets	50+ product rule worksheets on Quizizz \| Free & Printable Flashcards For Teachers Schools and Districts en-us Enter Code... Login Signup Enter Code Explore our full library of resources Including flashcards, videos, quizzes, and reading comprehension materials to enhance your students’ learning. Browse all resources Browse all resources Free Printable product rule worksheets Math teachers, discover a valuable collection of free printable product rule worksheets to help students master the concept of multiplying exponential expressions. 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15565	https://www.youtube.com/watch?v=d7C75Y2rjbg	Physics Lecture Chapter 4: Motion in 2 and 3 Dimensions Dot Physics 51700 subscribers 402 likes Description 24189 views Posted: 4 Jan 2023 Here is my lecture review of Halliday Resnik and Walker Fundamentals of Physics (9th Edition). Chapter 4: Motion in 2 and 3 Dimensions. In this chapter: - kinematic equations in vector form - derivatives of vectors - projectile motion - derivation of centripetal acceleration - relative velocity Here is the playlist with all of the lectures 17 comments Transcript: hello I am continuing my uh lecture series uh based on holiday resident and Walker 9th Edition College physics chapter four that's where we're at where chapter four Motion in two and three dimensions so just a quick recap because I think it's important to look back at this for a second in the last chapter and the link to all of my lectures for this textbook are down below I'll put a link to the uh the playlist we had the kinematic equations so these kinematic equations Define the position and velocity and acceleration time and how they relate so we have uh the final position initial position I put this as the x0 we just have V 0 before because now we're gonna deal with two Dimensions um a x is the acceleration the X Direction and T is the time we also had the following uh v x was the derivative of x with respect to time and ax was the derivative of v x with respect to time but those are all in one dimension after that we did that was actually chapter two chapter three we we looked at this idea of okay well we can express uh two Dimensions or more as a vector R and one of the ways to this is the position Vector R which determines the position we could just say uh R is equal to some x coordinate in the X Direction plus some y coordinate in the y direction plus Z coordinate in the Z Direction now the book holiday resident and Walker writes it like this I hat J hat z k hat I I don't I like the XYZ better because it makes more sense these X hat y hat Z hats are vectors with a magnitude of one this points in the X Direction so this is really just the x value times the direction Vector X and the Y value times a y Vector so that's how we can find the position that way uh so we had a whole bunch of things uh you could write also the velocity as V I'm going to use the X hat notation v x X hat plus v y hat plus v z z hat so they have an X Y and Z component in the uh for velocity also okay so um if you remember one way there's a couple ways to derive this equation um and it used the definition the the non-calculus way I don't know which way you want to do it uh the non-calculus way was to use the definition of average velocity so it turns out that we can write all these definitions and all these kinematic equations as vectors also I'm not going to read derive them because the derivation is essentially the same but I will say that we could say V average as instead of Delta X over delta T I'm going to say Delta r over delta T and that's equal to V 0 plus v vector vector over 2. so it looks the same way except when we have vectors we have vectors so this would be Delta R is really just R2 I'm going to I'm going to do they use that I'm going to use two and one I like that better R2 minus R1 over delta T and let's just write this out so you can see that it's the same thing right if I write R2 and R1 as X1 y1 Z1 X2 y2z2 this is going to be uh X to X hat plus Y2 y hat plus Z2 Z hat that's R2 minus X1 X hat plus y1 y hat plus Z1 Z hat all that divided by delta T so if you remember when we have vectors and we want to add or subtract vectors we add or subtract the components so if I do that I have an X hat here and an X hat there so it's going to be X2 minus X1 so I'm going to get x2 minus X1 X hat and then I'm going to get Y2 minus y1 this is going to be plus Y2 minus y1 y hat and then I'm going to get the Z's plus Z2 minus Z1 Z hat and all of that's divided by delta T now if I have this is a vector on the top and I divide by the scalar delta T I have to divide each component by delta T so I can actually write this as X2 minus X1 over Delta t x hat plus Y2 that's a y minus y1 over delta T y hat plus Z2 minus Z1 Z hat over delta T and what do we have here well this is going to be VX this is going to be VX is this X hat this is average X right the Del it says Delta X over delta T this is v y hat plus v z Z hat and this is this and this is that so we actually get if we just break it into components I get VX is Delta X over Delta t v y is Delta y over delta T and VZ is Delta 0 is delta T and so I can't break it up into components and so it is equivalent to the same thing so anything that you can do with this you can write as vectors so that means that these well not that one that one's a little bit more difficult these two can be written as the following vector equation so we we have the following Vector equations I'm going to start from the top let's rewrite V average is Delta R over delta T so you'll notice that's a vector that's a vector that's a scalar so this all works the instantaneous is going to be the derivative of r with respect to T so how do I do a derivative of a vector well let's just write this as D DT and they're going to write r as x x hat plus y y hat plus z z hat so when I take the derivative of a vector I have to take the derivative of each component so this is actually just going to be DX DT X hat plus d y d t y hat hat plus d z DT Z hat so what's DX DT well that's VX VX X hat what's dydt why that's v y y hat and what's DZ DT is v z z hat so I this is true right now I have that is v v is v x x at v y y hat and so forth and then we can do the same thing for the acceleration it's DV DT and then we can do the two kinematic equations I can say our final I'll do it in the format the book uses r0 plus v 0 t plus one half a t squared so notice vector vector vector vector everything has to be a vector um and so T is a scalar so I can multiply by uh that my thing froze up there it goes okay uh and then we have the other equation V equals V 0 plus a T now you can't write the final one because it deals with v 0 squared which that's not a vector uh and then Delta R uh it depends on uh other things too so it's a little bit more difficult to write that but these two are very useful okay so we still have displacement uh Delta R we still position velocity and I'm just looking at the the summary here the acceleration oh so a would be uh we could write this as a x x hat plus a y y hat plus a z z hat that's nice okay uh average acceleration um projectile motion this is a big deal projectile motion so this is the motion of an object with only the gravitational force so if I take and I'll give an example like this here's some ledge and I launch a ball with an initial velocity v0 and you can put it at some angle Theta and it's going to go up and back down projectile motion is the motion that we have when an object has an acceleration of 0 x hat minus 9.8 meters per second squared y hat plus 0 Z hat so this is the X and Y directions so that's true and something that happens if that's true then my kinematic equations ever break down the kinematic equations um I'll write it this way first I have R equals r0 Plus v0 t plus a one-half t squared now let's write this in terms just the X components if I just take the X components of all these out I get x equals x 0 plus now what's the X component of the initial velocity I can actually find that right from this triangle if I know the initial velocity magnitude is V 0 and it's at some angle Theta then this would be V 0 cosine Theta T and then the last term is one half a t squared well what's a it's zero the X component for the acceleration zero so that's it now let's do the same thing for the y direction in the y direction I have y y zero now I have the Y velocity initial velocity which is right here that's the opposite side of that triangle so that's going to be equal to plus v 0 sine Theta T and then I'm going to have the accelerations negative 9.8 we call that we say G is 9.8 meters per second squared this is going to be minus one half g t squared so you get these two equations one for the X Direction and one for the Y and and this is the key right there to projectile motion the other key is this because this motion up here has no y's in it and this one has no x's in it then the T here has to be the same as the T here so when you have a motion you can break this into a one-dimensional x motion of an object moving at a constant speed a one-dimensional y motion of an object going up and down and those two motions take the same time so the trick is to solve almost always solve this one for time and then use that up there to solve for uh position now you be careful if I have an initial y velocity this is going to have a t and a t squared term in it so it you might have to use a quadratic equation to solve for time and you'll get two times uh and if I launch this up here like this I'd get two times one would be a positive the time it takes to get over here and one would be negative so if you went back in time to when y was equal to zero it'd get over there there's a whole bunch of stuff to project on motion but you know this is a chapter summary really um if you need lots of problems you can do that there's okay I'm looking right here they have uh the trajectory equation I think that's a little complicated for introductory physics um I wouldn't worry about that the range equation I I think is a terrible terrible terrible equation uh you'll see it though R equals V 0 squared over G times sine of 2 Theta this is the range equation be very careful one you can derive this up here if you take this situation and you start with some velocity at some angle at y equals zero and you end at y equals zero you get this now there is a double angle formula that you have to use to get this but it's not a big deal it only works and into the same location if you start higher up and then lower down this doesn't work if you start low and end high that doesn't work okay so I I really don't think that is a big deal okay this one is a big deal uniform circular motion so suppose I have an object moving in a circle that's a pretty good Circle right there so uniform Circle motion circular motion means that we have some object moving uh with a constant magnitude velocity and some circle of radius r now if you look at the definition of acceleration it's DV DT now if the magnitude is constant it doesn't mean that the derivative is common is zero because in fact as you see this one as it when it's over here it'll be going down so the X and Y components of this this is the X Direction This is the Y do change so even though it's moving with a constant speed it does have an acceleration and I'll go ahead and tell you the answer because that's the important part the acceleration is going to the magnitude I'll write it as a magnitude is V squared over r the direction of this acceleration uh I'll put it I'll put over here because I'm going to use that picture in a second if that's V the direction of the acceleration is towards the center of the circle and that's why this is called centripetal acceleration tall acceleration this literally means Center pointing Center pointing acceleration don't confuse that with this Central Centra Fugal acceleration or Force either way is fine Excel aeration it's hard to spell when you're on YouTube this is Center fleeing so this is the force that pushes away from the center of the circle this is actually uh what we call a fictitious Force it's not normally used in introductory physics courses I just want to tell you because I think you're grown up and you can handle it or a teenager you know whatever you are I think you can handle if you're in this course you can handle it this is a fake Force so what this does is if I want to assume my frame of reference is accelerating I need to add this Force to make physics work again so we tend to use this centripetal acceleration in physics just because it causes fewer problems okay so let's see if we can derive this value right here um so I'm going to say that I have this r and Theta okay so the object is at some position Theta and the radius of the circles are and that's a constant magnitude velocity so let's see if we can write the velocity as a vector well let me take this blue marker right here I need the X and Y components well if I draw this actually I don't need that one I want here is my my why and here's my X right so it's going that much in the at this instant it's going that way in the y direction and that way in the X Direction and you'll notice that this angle Theta is the same as that so just imagine that this whole thing went back down to here then that angle Theta would go to zero that angle Theta would go to zero so I can write this as the following Vector so the X component is going to be the magnitude of the Velocity V but it's in the negative X direction right it's moving this way so it's going to be negative V and then I'm going to say sine Theta X hat because that's the opposite side of this right triangle and then the X component the Y component is going to be V cosine Theta y hat so that's my uh Vector acceleration vector I can get sine and cosine of theta by looking at this triangle since that's the same triangle I'm going to write this Theta r X Y so if this triangle right here has the same angle remember sine is just a ratio of sides this has the same as the right triangle with the same angle it has to have the same ratio of sides so if I take the sine of theta right here I could take the sine of theta down here and sine of theta I'll write it right here is equal to opposite of hypotenuse that's going to be equal to Y over R and then cosine Theta is going to be adjacent over hypotenuse that's going to be X over r so I can put those in up here instead of sine and cosine and I get the velocity is equal to negative V over r y x hat plus v over r x y hat but now I and this is a constant right so I can actually V and R constant V is a magnitude to the velocity we said it's moving at a constant velocity so this is going to be equal to V over R times negative y x hat plus X y hat just a note if this derivation is like blowing your mind it's fine uh I don't normally see students having to use this um just as long as you know this is true you don't always have to derive it but now I know the expression for the velocity I can derive an expression for the acceleration so let's just recopy that down because it's important so I'm going to say V equals oh yeah V over r negative y x hat plus X Y hat now I'm going to take the derivative I'm going to say the acceleration is DV DT and what is constant and what's not well V is constant and R is constant the same size circle it's moving at the same speed so that's a constant so when I take the derivative I just have to take the derivative of the stuff inside of here so this is going to be let's write it out V over R negative d y d y d t X hat plus d x DT y hat well what is DX DT that's the x velocity that's the Y velocity I'm sorry and that's the x velocity so this is actually going to be equal to V over R negative v x no v y x-hat plus v x y hat and that's a vector but I want to find the magnitude of this Vector a magnitude is going to be uh I could break this into two pieces but I'll I'll just do it all in one time I can factor out that all I need to do is find the magnitude of this part right because that's just a scalar anyway so this is going to be equal to V over R times the square root of the X component which is going to be v y squared plus the square root of the X component which is the Y component which is the x squared but if you see right here I have v y squared plus v x squared square rooted that's just the magnitude of the velocity of V so this is going to be V over R times V or V squared over r okay so now we need to get uh the the direction the direction of this Vector so let's go ahead and go back a second to this definition of velocity so I'm going to write this V equals negative V sine Theta X hat plus v cosine Theta y hat so if I put that in up here I can say a equals V over R I'm going to factor out the the V so V squared over R times negative sine Theta X hat plus cosine Theta y hat and this is what we would say this is a unit vector actually that's not right I did something wrong let me see I've ordered on the board over here with one second so this is maybe I should write this as [Music] um okay so then if I draw this so I'm right here Theta I'm right there what direction is the acceleration well it's going to be negative sine Theta so sine Theta is this side so it's going to be down no it's going to be back and up okay I'm not going to worry about the direction I messed up on the direction um that's fine I'm cool with that hope you're cool with that I had it and I just forgot how to I don't want to make it all messy and stuff like that okay now there is one other thing in here that I don't think is well it's kind of a big deal and that's relative velocity so imagine that there is a car a train car and it's moving with the velocity V train with respect to the ground and now I have a person inside there that's moving this way with the velocity of the person with respect to the train and then I have a person over here and they want to see that person moving because there's a window in this for some reason I can say the following the velocity of the person with respect to the ground is equal to the velocity of the person with respect to the train plus the velocity of the train with respect to the ground and this is your relative velocity formula so one of the truth the key things here is always to use a notation of the object with respect to the reference frame object reference frame object reference frame and then if you line these up so these two match in your in your equation then you get these two I know it's kind of stupid but that is true and you can see that it makes sense here right if this is moving at 10 meters per second and the person's moving at 5 meters per second then this would be um we can actually write it out velocity of the plane with respect to the train is 10 x hat meters per second velocity of the person with respect to the the train with respect to the ground I'm sorry that's this one the TG v p t is equal to five x hat meters per second and then if I add those two together I get 15 meters per second extraction so it works but here's the awesome thing it doesn't even have to be lined up right so if I have a river like this is a common one here's a river flowing velocity of the water with respect to the ground and then I have a boat that aims directly across velocity of the boat with respect to the water then I can get the velocity of the boat with spec to the ground velocity of the boat ground is velocity of the boat with respect to the water plus the velocity of the water with respect to the ground and if you write these as vectors as a vector equation everything will works and it works out just fine so it's not too bad um okay so that's it for chapter four and you know just a quick reminder there is a lot of stuff here okay uh and I fly through this uh you you what you need to do is you need to read through the chapter or study this stuff practice problems work through the practice problems do the homework uh don't feel like you can just watch this YouTube video and get it there are maybe some people that can do that but that's not how most people learn you have to really uh just seeing it is not just getting it okay so make sure you you work on that if you have questions post the questions down below I'll try to answer them if I can so I hope that helps and I'll see you in the next chapter
15566	https://www.expii.com/t/intro-to-scalene-isosceles-and-equilateral-triangles-980	Expii Intro to Scalene, Isosceles, and Equilateral Triangles - Expii An equilateral triangle has sides (and angles) that are all equal. An isosceles triangle has two sides (and angles) that are identical. In a scalene triangle, all the sides and angles are unique!. Explanations (4) Daniel Liu Text 10 Triangles can be classified into three types: Equilateral, isosceles, and scalene. An Equilateral triangle is just what it sounds like: a triangle with all its sides equal. Because all of its sides are equal, it follows that all of its angles are also equal. A second type of triangle is the isosceles triangle. This type of triangle has at least two sides equal; thus, an equilateral triangle is also an isosceles. An isosceles triangle also has the property that the two angles opposite the two equal sides are also equal. The last type of triangle is the scalene triangle. This triangle has no three sides equal; in other words, all the sides are of different lengths. As you can guess, all the angles of this triangle also are distinct. Report Share 10 Like Related Lessons Perpendicular Bisectors of a Triangle Meet Isosceles and Equilateral Symmetry and Special Lines Angle Bisector Theorem Proving a Triangle Is Isosceles View All Related Lessons Jesse Zhang Text 2 All triangles can be categorized into two mutually exclusive classes of triangles: isosceles and scalene. An isosceles triangles has at least two sides which are equal in length, and a scalene triangle has no two sides which are equal. You may notices that because of this, isosceles triangles always have at least one line of symmetry while scalene triangles do not have any. An equilateral triangle is a special case of an isosceles triangle that has all three sides equal in length. As a result, all equilateral triangles have three lines of symmetry and three angles equal to 60∘. Report Share 2 Like Freya Edholm Text 2 Here is some extra practice on identifying different types of triangles based on their sides. Be warned, this worksheet uses variables instead of numbers; just assume each variable is distinct. GLHF! Report Share 2 Like Muhammad Hamza Waseem Text 1 Triangles are usually divided into three types based on the the lengths of their sides: 1)Equilateral 2)Isosceles 3)Scalene An equilateral triangle is the one having all the three sides equal in length. It follows that all of its angles are equal in measure. An isosceles triangle is one having two of its sides equal in length. It follows that the two angles opposite to the two equal sides are equal in measure. An scalene triangle is the one having all the sides different in length. It follows that all of its angles are unequal in measure. Report Share 1 Like You've reached the end How can we improve? General Bug Feature Send Feedback
15567	https://web.math.princeton.edu/~js129/PDFs/teaching/analysis_1_fall_2014/Recitation_Week_3.pdf	Analysis 1 Recitation Session of Week 3 October 3, 2014 1 Homework Sheet Number 3 1.1 Inverse Image • For a map f : X →Y , and a subset B ⊂Y , we have deﬁned a new set, a subset of X, denoted byf −1 (B) which is deﬁned as f −1 (B) := { x ∈X \| f (x) ∈B } . • The inverse image possess many nice properties with set operations which are either only partially true or completely true for the image of A ⊂X under f: f (A) := { y ∈Y \| ∃x ∈A : f (x) = y } . • Here is an example: Claim: f S j∈I Aj = S j∈I f (Aj) Proof: – Claim 1: f S j∈I Aj ⊆S j∈I f (Aj) Proof: 1. Let y0 ∈f S j∈I Aj be given. 2. ⇐ ⇒y0 ∈ n y ∈Y h ∃x ∈S j∈I Aj i : f (x) = y o 3. ⇐ ⇒ ∃x ∈S j∈I Aj : f (x) = y0 4. ⇐ ⇒(∃j0 ∈I : ∃x ∈Aj0) : f (x) = y0. 5. ⇐ ⇒∃j0 ∈I : (∃x ∈Aj0 : f (x) = y0) (basic result in formal logic). 6. ⇐ ⇒∃j0 ∈I : (y0 ∈{ y ∈Y \| [∃x ∈Aj0] : f (x) = y }) 7. ⇐ ⇒y0 ∈S j∈I { y ∈Y \| [∃x ∈Aj] : f (x) = y } 8. ⇐ ⇒y0 ∈S j∈I f (Aj) 9. Thus we have shown that an arbitrary element of f S j∈I Aj is also an element of S j∈I f (Aj), which means that any element of f S j∈I Aj is also an element of S j∈I f (Aj). – Claim 2: f S j∈I Aj ⊇S j∈I f (Aj) Proof: 1 ∗Let y0 ∈S j∈I f (Aj) be given. ∗Because all the implications above were ⇐ ⇒(two-sided), we can use the procedure above to conclude that y0 ∈f S j∈I Aj . ∗Thus we have shown that an arbitrary element of S j∈I f (Aj) is also an element of f S j∈I Aj , which means that any element of S j∈I f (Aj) is also an element of f S j∈I Aj . ■ • Here is another example: Claim: B1 ⊆B2 = ⇒f −1 (B1) ⊆f −1 (B2) for any two subset B1 and B2 of Y . Proof: – Let x0 ∈f −1 (B1) be given. – ⇐ ⇒x0 ∈{ x ∈X \| f (x) ∈B1 } – ⇐ ⇒f (x0) ∈B1 – But B1 ⊆B2 according to our assumption, so that f (x0) ∈B2. – f (x0) ∈B2 ⇐ ⇒x0 ∈{ x ∈X \| f (x) ∈B2 } ⇐ ⇒x0 ∈f −1 (B2). – Because x0 was arbitrary, we conclude that f −1 (B1) ⊆f −1 (B2). ■ • Claim: A1 ⊆A2 = ⇒f (A1) ⊆f (A2). Proof: – Let y0 ∈f (A1) be given. – That means that ∃x ∈A1 such that f (x) = y0. – But A1 ⊆A2, so that x ∈A2. – That means that ∃x ∈A2 such that f (x) = y0. – That is, y0 ∈f (A2). ■ • Claim: f (A1) ⊆f (A2) = ⇒A1 ⊆A2 is not in general true. Proof: – To prove this we merely have to give one counterexample. – Take X = {1, 2} and Y = {3}. – Then there can be only one map f : X →Y , namely, f (x) = 3. – Deﬁne A1 := {1} and A2 := {2}. Then clearly f (A1) = f (A2) = {3}, yet, A1 ⊈A2. ■ • Here is an example of why inverse images are easier to “work with” than images: Claim: f (A1 ∩A2) ⊆f (A1) ∩f (A2). Proof: 2 – A1 ∩A2 ⊆A1, so that f (A1 ∩A2) ⊆f (A1). – A1 ∩A2 ⊆A2, so that f (A1 ∩A2) ⊆f (A2). – Given these two conclusion, we must have f (A1 ∩A2) ⊆f (A1 ∩A2). ■ • Claim: f (A1) ∩f (A2) ⊆f (A1 ∩A2) is not in general true. Proof: – To prove this we merely have to give one counterexample. – Take X = {1, 2} and Y = {3}. – Then there can be only one map f : X →Y , namely, f (x) = 3. – Deﬁne A1 := {1} and A2 := {2}. – Then f (A1) ∩f (A2) = {3} yet A1 ∩A2 = ∅and so f (∅) = ∅. – It is clear that {3} ⊈∅. ■ 1.2 Function Composition 1.2.1 Deﬁnition • Given two functions f : X →Y and g : Y →Z, we may deﬁne a new function g ◦f : X →Z by the following rule: x 7→g (f (x)). • To verify that this deﬁnition makes sense, we must make sure that the function is well-deﬁned, that is, that if x1 = x2 then (g ◦f) (x1) = (g ◦f) (x2). This is of course true because f and g are separately well deﬁned (as you can well verify). 1.2.2 Example with Surjectivity • Claim: Let f1 : X1 →X2, f2 : X2 →X3, . . . , fn : Xn →Xn+1. If fi is surjective ∀i ∈{1, 2, . . . , n} then fn ◦fn−1 ◦· · · ◦f1 : X1 → Xn+1 is surjective. Proof: – Let α ∈Xn+1 be given. – Our goal is to show that ∃x ∈X1 such that (fn ◦fn−1 ◦· · · ◦f1) (x) = α. – Because fn is surjective, ∃αn ∈Xn such that fn (αn) = α. – Because fn−1 is surjective, ∃αn−1 ∈Xn−1 such that fn−1 (αn−1) = αn. That means that ∃αn−1 ∈Xn−1 such that fn (fn−1 (αn−1)) = α. – We can continue in this fashion until we reach that there must ∃α1 ∈X1 such that f1 (α1) = α2, that is, ∃α1 ∈X1 such that fn (fn−1 (. . . (f2 (f1 (α1))))) = α. ■ 3 1.2.3 Example with Injectivity • Claim: Given two functions f : X →Y and g : Y →Z, if g ◦f is injective, g may in general, be not-injective (even though f is always injective in that scenario). Proof: – First we show that f is injective: ∗Assume that for given (x1, x2) ∈X2 we have f (x1) = f (x2). ∗To prove the assertion that f is injective we must show that x1 = x2. ∗Because f (x1) = f (x2), then necessarily as g is well deﬁned we have g (f (x1)) = g (f (x2)), that is, (g ◦f) (x1) = (g ◦f) (x2). ∗But g ◦f is injective, which means that x1 = x2. – Suﬃce to give one counterexample to prove this claim: ∗Let X = [0, ∞), Y = R and Z = R. ∗Deﬁne f : X →Y by x 7→x and g : Y →Z by y →y2. ∗Then clearly f is injective (as it is just the identity map) and as we’ve seen before g is not injective. ∗However, g ◦f is injective as you can well verify explicitly: · Assume for two x1 and x2 in X ((x1, x2) ∈X2) we have that (g ◦f) (x1) = (g ◦f) (x2). · That means that g (f (x1)) = g (f (x2)). · Because f is just the identity map, that means g (x1) = g (x2), or x1 2 = x2 2. · But x1 and x2 are positive numbers as we’ve deﬁned X, so that x1 = x2. ∗Conclusion: Even though g is not injective, g\|f(X) : f (X) →Z (the restriction of g to the image of f) is going to be injective, and this is the intuitive piece we were missing in this theorem. 1.3 Cardinalities 1.3.1 The Set of Finite Subsets of R Has the Same Cardinality as R • Deﬁne S := { A ⊂R \| \|A\| ∈N } • Claim: \|R\| = \|S\| Proof: – Deﬁne Sn := { A ⊂R \| \|A\| = n } for all n ∈N. – Then clearly, S = S n∈N Sn. – Claim: \|R\| ≤\|S\|. Proof: ∗S1 = { A ⊂R \| \|A\| = 1 } = { {r} ⊂R \| r ∈R } so that \|S1\| = \|R\| and thus as S1 ⊆S, \|S\| ≥\|R\|. – Claim: \|R\| ≥\|S\|. Proof: ∗Let A ∈Sn be given. That is, A is a set of real numbers with n elements. ∗Then we can write these elements as a1, a2, . . . , an and WLOG can assume that a1 < a2 < · · · < an. ∗Deﬁne a map f : Sn →Rn by A 7→(a1, a2, . . . , an). (That is, by the order of the elements). 4 ∗Claim: f is well deﬁned. Proof: · Clearly by ordering the elements we remove any ambiguity about the representation of the sets: if two sets are equal they will necessarily lead to the same n-tuplet in Rn. ∗Claim: f is injective. Proof: · By the same argumentation as above it is clear that if two n-tuplets are equal in the image of f they must have come from the same ﬁnite subset of R. ∗Because f is injective, we necessarily have \|Sn\| ≤\|Rn\|. ∗Claim: \|Rn\| = \|R\|. · Claim: \|(0, 1)\| = \|R\|. Proof: the function tan πx −π 2 from (0, 1) to R is a bijection. (Proof as homework.) · Claim: \|(0, 1)\| = (0, 1)2 . Proof: 1. For every real number between 0 and 1, pick one particular convention for a decimal representation. (For example, pick 0.01 always instead of 0.009999999999 . . . ) 2. Deﬁne a map g : (0, 1)2 →(0, 1) by the relation (0.a1a2a3 . . . , 0.b1b2b3 . . . ) 7→0.a1b1a2b2a3b3 . . . . 3. Claim: g is injective. Proof: If g ((0.a1a2a3 . . . , 0.b1b2b3 . . . )) = g ((0.c1c2c3 . . . , 0.d1d2d3 . . . )) then 0.a1b1a2b2a3b3 · · · = 0.c1d1c2d2c3d3 . . . , in which case it is clear that ai = ci and bi = di for all i, and so really the two pairs are equal. 4. Claim: g is surjective. Proof: Given a random number 0.x1x2x3 · · · ∈(0, 1), Take the pair (0.x1x3x5 . . . , 0.x2x4x6 . . . ) which will map to it. · Returning to the actual proof that \|Rn\| = \|R\|, we perform induction on n. · The induction basis is n = 2. For this case, we have that R2 = \|R × R\| = \|(0, 1) × (0, 1)\| = \|(0, 1)\| = \|R\| · Assume that \|Rn\| = \|R\| for some n ∈{2, 3, . . . }. · Then Rn+1 = \|Rn × R\| = \|R × R\| = \|R\| · Thus by the principle of induction we have that \|Rn\| = \|R\| for all n ∈{2, 3, . . . }. Of course the statement is true for n = 1! ∗Thus we have \|Sn\| ≤\|R\|. ∗But \|S\| = S n∈N Sn , and all the Sn’s are disjoint. Because each has \|Sn\| ≤\|R\|, we may estimate the cardinality of S n∈N Sn ∗ ≤ S n∈N {n} × R = \|N × R\| ≤\|R × R\| = \|R\|, where in ∗we used the fact that all the Sn’s are disjoint, so we can say the cardinality of the union is necessarily smaller than the cardinality of \|N\| disjoint copies of R. – Because \|S\| ≤\|R\| and \|S\| ≥\|R\|, using the Schroeder-Bernstein theorem we conclude that \|S\| = \|R\|. 5 1.3.2 \|N\| is the Smallest Possible Inﬁnite Cardinality • Claim: For any inﬁnite set X, ∃an injection N →X (which shows that \|N\| ≤\|X\|. Proof: – Because X is inﬁnite, it is not empty, so ∃x0 ∈X. Deﬁne f (0) := x0. – Because X is inﬁnite, X\ {x0} is not empty, so ∃x1 ∈X. Deﬁne f (1) := x1. – Because X is inﬁnite, X\ {x0, x1} is not empty, so ∃x2 ∈X. Deﬁne f (2) := x2. – ... – Because X is inﬁnite, X\ {x0, x1, x2, . . . , xn} is not empty, so ∃xn+1 ∈X. Deﬁne f (n + 1) := xn+1. This is possible due to the axiom of dependent choice which says that you can make countably many choices, each depending on the previous ones you made. In fact we can prove that DC stems from AC: – Claim: AC implies DC (axiom of choice implies dependent axiom of choice) Proof: ∗Deﬁne for each A ⊂X, R (A) := { B ⊂X \| A ⊂B ∧\|B\| = \|A\| + 1 }. ∗By assumption we assume R (A) ̸= ∅∀A ⊂X. This amounts to the information that X is inﬁnite. ∗Using the axiom of choice, ∃f : P (X) →P (X) such that f (A) ∈R (A) ∀A ⊂X. ∗So for any A ⊂X, the sequence (An)n∈N := f n (A), where f n denotes composition of f n times on itself, has An ∈R (An−1) for all n ∈N. – Clearly f is injective because we always pick our next value from the complement of the previous values. 1.4 Speciﬁc Tips 1.4.1 Question 4 • Typo in question: You will get an injective map from N →X which shows that any inﬁnite set has \|N\| ≤\|X\|, that is, \|N\| is the smallest cardinality. Point of the exercise is (as far as I can see) to repeat the proof above but without the full axiom of choice, rather, instead, just the countable axiom of choice. • (a): Use induction to prove the statement for all n ∈N. Use the fact that X is inﬁnite, so that in particular it is not empty and has as many elements as you want, even after you remove a ﬁnite number from it. • (b): Use the axiom of choice (the inﬁnite product of nonempty sets is nonempty) to ascertain that there is such a map, because the product Q k∈N A ⊂X \|A\| = 2k is non-empty (because the sets in the product, A ⊂X \|A\| = 2k , are not empty using (a)). But now we have: Y k∈N A ⊂X \|A\| = 2k ≡ A ⊂X \|A\| = 20 × A ⊂X \|A\| = 21 × A ⊂X \|A\| = 22 × . . . = n (A, B, C, . . . ) ∈P (X)N \|A\| = 1, \|B\| = 2, \|C\| = 4, . . . o We can identify any inﬁnite product (a, b, c, . . . ) ∈XN as a map A : N →X (don’t confuse A as a set or Aas a map–sorry for the duplicity). A map is just a rule that associates with every number n ∈N an element of our set X, and this is exactly what an inﬁnite product does as well: given a number n ∈N, we have associated with it an element of X: for n = 0 we have associated a, for n = 1 we have associated b and so on. Thus n (A, B, C, . . . ) ∈P (X)N \|A\| = 1, \|B\| = 2, \|C\| = 4, . . . o = A : N →P (X) \|A (k)\| = 2k∀k ∈N 6 • (c): Deﬁne S (k) := A (k) \ Sk−1 j=1 A (k) . Make estimates on the size of Sk−1 j=1 A (k) to show that S (k) is nonempty. Then using the axiom of choice we know that Q k∈N S (k) is not empty, and an element in this product is just a function f : N →X. (think why and use the same reasoning as above). 1.4.2 Question 5 • (a): Show that the In’s build nested intervals, and as seen on the lecture, nested intervals correspond to unique real numbers. As a result, g is (well) deﬁned exactly as that one unique number which is in the intersection of all nested intervals. • (b): Pick a random real number x ∈[0, 1]. Show that we can deﬁne a binary sequence that will give rise to nested intervals that “close in” on this number. This will show surjectivity. To deﬁne the sequence, think about how these intervals “close in” on a number: for example, for the ﬁrst number in the sequence pick a0 := 0. For the second, a1, if x ∈(0, 1 2] then pick a1 := 0 and otherwise pick a1 := 1. How to generalize this? Is this choice always canonical? (this will answer injectivity) 1.4.3 Question 6 • (c) and (d): Use this “trick”: a + ib c + id = a + ib c + id c −id c −id \| {z } 1 = (a + ib) (c −id) c2 −d2 = ac + bd + i (bc −ad) c2 −d2 = ac + bd c2 −d2 + ibc −ad c2 −d2 • (e) and (f): Plug in z = a + ib where (a, b) ∈R2 and ﬁnd a and b. Make the necessary computations to get a complex equation of the form: a + ib = c + id. This will give you then two real equations for two real variables, a and b (which correspond to one complex variable, z). 1.4.4 Question 7 • Translate the conditions into conditions on two real variables a and b, and then sketch the solutions on R2. • (e): What is the imaginary part of the absolute value of a complex number? What kind of values can the absolute value take anyway? 2 Homework Sheet Number 1 • Common mistakes: – Review induction!! In particular, what’s “n = n + 1”? Make sure you understand: (a) How to formulate a proof by induction and (b) Why is the formulation the way it is (stems from a particular logic). – Set builder notation: distinction between sets and formulas. 7 – How to prove that two sets are equal (especially not using Venn diagrams). 1. Question 1 (a) Deﬁne a statement A (n) for every n ∈N, such that A (n) is true iﬀPn k=1 k2 = 1 6 [n (n + 1) (2n + 1)]. Claim: A (n) ∀n ∈N Proof: i. Claim: A (1). Proof: • A (1) says that P1 k=1 k2 = 1 6 [1 (1 + 1) (2 · 1 + 1)] which is of course true. ii. Claim: [A (n) = ⇒A (n + 1)] ∀n ∈N. Proof: • Let n ∈N be given, and assume that A (n). • A (n + 1) says that Pn+1 k=1 k2 = 1 6 [(n + 1) (n + 2) (2 (n + 1) + 1)]. • Start from the left hand side of the equation: Pn+1 k=1 k2 = Pn k=1 k2 + (n + 1)2 = 1 6 [n (n + 1) (2n + 1)] + (n + 1)2 = (n + 1) 1 6 [n (2n + 1)] + (n + 1) = 1 6 (n + 1) [n (2n + 1) + 6 (n + 1)] = 1 6 (n + 1) 2n2 + 7n + 6 = 1 6 [(n + 1) (n + 2) (2n + 3)] = 1 6 [(n + 1) (n + 2) (2 (n + 1) + 1)] • Thus we have reached the right hand side of the desired equation. ■ 2. 3. 4. 5. Question 5 (a) (b) { x ∈R \| \|x −2\| ≥\|x\| −2 } =? • The inequality: \|x −2\| ≥\|x\| −2 is equivalent to \|x −2\| + 2 ≥\|x\|. • Case 1: x ≥2, then we have x −2 + 2 ≥x which is always true. • Case 2: 0 ≤x ≤2, then we have −x + 2 + 2 ≥x which means 2x ≤4 or x ≤2 which we anyway have. 8 • Case 3: x < 0, then we have −x + 2 + 2 ≥−x which is always true. • General conclusion from this exercise: whenever you are a bit confused about many absolute values, just separate into all possible cases. (c) (d) 1 ≤\|x\| + \|y\| ≤2. • Compare with 1 ≤x2 + y2 ≤2. • In general, separate into four diﬀerent cases: i. x ≥0 and y ≥0, then we have 1 ≤x + y ≤2, which means 1 −x ≤y ≤2 −x. ii. Like that all four possible cases. 6. 7. Question 7 • Let L be the set of all books. Thus \|L\| is the number of all books in the library. • ∀B ∈L, \|B\| is the number of words in a book B. • Claim: There is at least one empty book. Proof: – Assume otherwise. That is, there is no empty book. – We know that the number of books, \|L\|, is larger than the sum of words in all books, P B∈L \|B\|. That is, \|L\| > P B∈L \|B\|. – If no book is empty, then \|B\| ≥1 for all B ∈L. Then P B∈L \|B\| > P B∈L 1 = \|L\|. – Thus we ﬁnd that \|L\| > \|L\|, a contradiction. – Thus, there must be at least one empty book in the library. • Claim: There must be less than three books in the library. Proof: – We know that there are no two books in the library with the same number of words. – Thus, order all the books by word number, say, 0, 3, 50, 3000, 3500, . . . . – The only thing we know is that ∗this sequence starts with zero, ∗that there are \|L\| numbers in it, and ∗that each item in the sequence must be bigger by at least one than its predecessor. – Using these constraints we can think of the “worst case” scenario, meaning, the case in which there is the minimal number of words in every book: 0, 1, 2, . . . . Any actual sequence of number of words of books in the library will be bigger than that. – Thus, P B∈L \|B\| ≥P\|L\|−1 k=0 k = 1 2 (\|L\| −1) \|L\|. – But we know that \|L\| > P B∈L \|B\| so that \|L\| > 1 2 (\|L\| −1) \|L\|. – This equation may only be fulﬁlled when 0 < \|L\| < 3. – Thus we can only have either one or two books in the library. – But one of the two books has to be empty, so if there are two books, the next one must have exactly one word to maintain the equation: \|L\| \|{z} 2 > \|B1\| \|{z} 0 + \|B2\|. 9
15568	https://mathoverflow.net/questions/8407/projective-closure-of-affine-curve	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 Projective closure of affine curve Ask Question Asked Modified 1 year, 7 months ago Viewed 9k times 7 $\begingroup$ Is there a generalized method to find the projective closure of an affine curve? For example, I read that the projective closure of $y^2 = x^3x+1$ in $\mathbb{P}^2$ is $y^2z = x^3xz^2+z^3$. If I want to find the the closure of another affine curve, what method should I employ? I can't seem to find an adequate description in any text. Thanks ag.algebraic-geometry Share Improve this question edited Dec 10, 2015 at 23:11 BlackAdder 52144 silver badges1414 bronze badges asked Dec 10, 2009 at 1:35 ElijahElijah 12322 silver badges55 bronze badges $\endgroup$ Add a comment \| 2 Answers 2 Reset to default 32 $\begingroup$ When the curve is a plane curve of degree $d$, the formula is simple (and in fact, this works for any hypersurface) you take $f(x,y)=0$ and replace it by $z^d f(x/z,y/z)=0$. This will be homogeneous of degree $d$, and when $z\neq 0$, you recover your curve. Now, if you have a more general affine variety, given by $\langle f_1,\ldots,f_k\rangle$ in $k[x_1,\ldots,x_n]$, then you first compute a Groebner Basis, which you can learn how to do using Cox, Little and O'Shea's "Ideals, Varieties and Algorithms" (assumes no background whatsoever) and then you do this same trick with each polynomial in the Groebner basis. Why can't you just use any basis? Look at the twisted cubic. $t\mapsto (t,t^2,t^3)$. This is cut out, in affine space, by $y-x^2$ and $z-x^3$, though $y-x^2$ and $xy-z$ are better to use. But still, $yw-x^2$ and $xy-zw$ don't give the twisted cubic! Bezout tells us that they give something of degree four containing the cubic, and it's not a hypersurface, so you get the twisted cubic plus a line. To get the cubic itself, you have to use a third quadratic polynomial that you get in the ideal, $y^2-xz$, which is not in the ideal given by the homogenized generators. These three, however, form a Groebner basis for the ideal, and then homogenization gives the homogeneous ideal of the projective twisted cubic. Share Improve this answer edited Feb 17, 2024 at 11:04 The Amplitwist 1,66144 gold badges1414 silver badges2828 bronze badges answered Dec 10, 2009 at 1:54 Charles SiegelCharles Siegel 16.2k88 gold badges9191 silver badges137137 bronze badges $\endgroup$ 1 $\begingroup$ I would like to mention that the Groebner basis must be w.r.t. a degree-compatible term order. In particular, with lex term order $z > y > x$, the basis $y - x^2$, $z - x^3$ would be a Groebner basis, but the term order is not degree-compatible. $\endgroup$ Simon Pohmann – Simon Pohmann 2023-01-22 11:08:50 +00:00 Commented Jan 22, 2023 at 11:08 Add a comment \| 7 $\begingroup$ Let me add a comment with some geometric flavor to Charles's clean answer. Once the field is algebraic closed, then the affine curve $C\subset \mathbb{A}^2$ is never compact. However it admits an embedding into the projective space (remember that we know that $\mathbb{A}^2\subset \mathbb{P}^2$) where we can look for a compactification. The construction above, geometrically speaking, takes a hyperplane in $\mathbb{P}^2$ and compactifies with it. In other words, there is a hyperplane (divisor) $H=[z=0]\subset \mathbb{P}^2$ which tells you which points to add to your curve $C$, in order to get it compact: let's denote it by $\tilde{C}$. At the end of the day, you will have that $\tilde{C}\cap H$ are the points that you have added to $C$. In a more general setting (related to the Groebner basis), if you have a manifold $M$, you may ask yourself that it would be desirable for a compactification of it to be algebraic. Now, as far as I now, there are two main obstructions for a compactification to be algebraic. There may not be a "nice" compactification and it may not have enough meromorphic functions. By "nice" here, I mean a compactification given by divisors (with normal crossings usually is asked), and by "enough meromorphic functions", there may not be ample line bundles. Questions on -When such obstructions do not cause troubles?- are big theorems in algebraic geometry. Not referring to self-confidence issues... "here, positivity helps a lot!" Share Improve this answer answered Dec 10, 2009 at 9:49 Csar Lozano HuertaCsar Lozano Huerta 2,18222 gold badges2525 silver badges3030 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 ag.algebraic-geometry See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Related 4 Finding divisors on a curve 24 Is every smooth affine curve isomorphic to a smooth affine plane curve? Projective to Affine? Automorphism group of an affine curve Global Affine Flag Variety and Affine Flag Variety Affine Paved Affine Varieties; not Affine Space Presentation for a Finite Etale Cover of an (Affine) Elliptic Curve Question feed
15569	https://www.fishbase.se/summary/Monodactylus-argenteus.html	This website uses cookies to enhance your browsing experience and ensure the functionality of our site. For more detailed information about the types of cookies we use and how we protect your privacy, please visit our Privacy Information page.. Accept All Accept Necessary Cookies Only Cookie Settings × Cookie Settings This website uses different types of cookies to enhance your experience. Please select your preferences below: Strictly Necessary- [x] These cookies are essential for the website to function properly. They include session cookies, which help maintain your session while you navigate the site, as well as cookies that remember your language preferences and other essential functionalities. Without these cookies, certain features of the website cannot be provided. Performance- [x] These cookies help us understand how visitors interact with our website by collecting and reporting information anonymously. For example, we use Google Analytics to generate web statistics, which helps us improve our website's performance and user experience. These cookies may track information such as the pages visited, time spent on the site, and any errors encountered. Save and Close Monodactylus argenteus, Silver moony : fisheries, aquarium You can sponsor this page Common name (e.g. trout) Genus + Species (e.g. Gadus morhua) About this page More Info Plus d'info Mais info Languages Arabic Bahasa/Malay Bangla Chinese(Si) Chinese(Tr) Deutsch English Español Farsi Français Greek Hindi Italiano Japanese Lao Nederlands Português(Br) Português(Pt) Russian Swedish Thai Vietnamese User feedbacks Comments & Corrections Fish Forum Guest Book Facebook Citation Uploads Attach website Upload photo Upload video Upload references Fish Watcher Related species Species in Monodactylus Species in Monodactylidae Classification - Monodactylidae Eupercaria/misc Teleostei Chordata Animalia Monodactylusargenteus (Linnaeus, 1758) Silver moony Add your observation in Fish Watcher Native range \| All suitable habitat \| Point map \| Year 2050 This map was computer-generated and has not yet been reviewed. Monodactylus argenteusAquaMaps Data sources: GBIFOBIS Upload your photosandvideos Pictures \| Videos \|Stamps, coins, misc. \| Google image Monodactylus argenteus Picture by Ramani Shirantha Classification / Names Common names \| Synonyms \| Catalog of Fishes(genus, species) \| ITIS \| CoL \| WoRMS \| Cloffa Teleostei (teleosts) >Eupercaria/misc (Various families in series Eupercaria) >Monodactylidae (Moonyfishes or fingerfishes) Etymology: Monodactylus:Greek, monos = one + Greek, daktylos = finger (Ref. 45335). More on author: Linnaeus. Environment: milieu / climate zone / depth range / distribution range Ecology Marine; freshwater; brackish; pelagic-neritic; depth range 0 - 12 m (Ref. 86942). Tropical; 24°C - 28°C (Ref. 1672); 31°N - 42°S, 19°E - 169°W Distribution Countries \| FAO areas \| Ecosystems \| Occurrences \| Point map \| Introductions \| Faunafri Indo-West Pacific: Red Sea and east coast of Africa south to the Breë River (Ref. 7293, 52193); also the Persian Gulf (Ref. 96822) to Samoa, north to the Yaeyamas, south to New Caledonia and Australia (Ref. 4959). Known from the freshwater tidal zone of the Mekong delta (Ref. 12693). Length at first maturity / Size / Weight / Age Maturity: L m13.0 range ? - ? cm Max length : 27.0 cm SL male/unsexed; (Ref. 44894); common length : 12.0 cm TL male/unsexed; (Ref. 5450) Short description Identification keys \| Morphology \| Morphometrics Dorsalspines (total): 7 - 8; Dorsalsoft rays (total): 27 - 31; Analspines: 3; Analsoft rays: 27 - 32. Body shape(shape guide): short and / or deep;Cross section: compressed. Biology Glossary (e.g. epibenthic) Found in bays, mangrove estuaries, tidal creeks, and lower reaches of freshwater streams (Ref. 2847, 44894, 48636); occasionally in silty coastal reefs (Ref. 9710). It feeds on plankton and detritus (Ref. 5213, 44894). Commonly seen in large schools (Ref. 44894, 52193); small juveniles either solitary or in small aggregations (Ref. 48635). Usually only juveniles enter freshwater (Ref. 52193). Highly territorial (Ref. 9710). Caught with throw nets (Ref. 30573). Life cycle and mating behavior Maturity \| Reproduction \| Spawning \| Eggs \| Fecundity \| Larvae Main reference Upload your references \| References \| Coordinator \| Collaborators Heemstra, P.C., 1984. Monodactylidae. In W. Fischer and G. Bianchi (eds.) FAO species identification sheets for fishery purposes. Western Indian Ocean (Fishing Area 51). Vol. 3. FAO, Rome. [pag. var.]. (Ref. 7293) IUCN Red List Status (Ref. 130435: Version 2025-1) Least Concern (LC); Date assessed: 28 June 2016 CITES Not Evaluated CMS (Ref. 116361) Not Evaluated Threat to humans Harmless Human uses Fisheries: minor commercial; aquarium: commercial FAO - Publication: search \| FishSource \| Sea Around Us More information Trophic ecology Food items (preys) Diet composition Food consumption Food rations Predators Ecology Ecology Home ranges Population dynamics Growth parameters Max. ages / sizes Length-weight rel. Length-length rel. Length-frequencies Mass conversion Recruitment Abundance Life cycle Reproduction Maturity Maturity/Gills rel. Fecundity Spawning Spawning aggregations Eggs Egg development Larvae Larval dynamics Distribution Countries FAO areas Ecosystems Occurrences Introductions BRUVS - Videos Anatomy Gill area Brain Otolith Physiology Body composition Nutrients Oxygen consumption Swimming type Swimming speed Visual pigments Fish sound Diseases & Parasites Toxicity (LC50s) Genetics Genome Genetics Heterozygosity Heritability Human related Aquaculture systems Aquaculture profiles Strains Ciguatera cases Stamps, coins, misc. Outreach Collaborators Taxonomy Common names Synonyms Morphology Morphometrics Pictures References References Tools Bio-Quiz \| E-book \| Field guide \| Identification keys \| Length-frequency wizard \| Life-history tool \| Point map \| Catch-MSY \| Special reports Check for Aquarium maintenance \| Check for Species Fact Sheets \| Check for Aquaculture Fact Sheets Download XML Summary page \| Point data \| Common names \| Photos Internet sources AFORO (otoliths) \| Aquatic Commons \| BHL \| Cloffa \| Websites from users \| Check FishWatcher \| CISTI \| Catalog of Fishes: genus, species \| DiscoverLife \| ECOTOX \| FAO - Publication: search \| Faunafri \| Fishipedia \| Fishtrace \| GloBI \| Google Books \| Google Scholar \| Google \| IGFA World Record \| National databases \| OneZoom \| Open Tree of Life \| Otolith Atlas of Taiwan Fishes \| Public aquariums \| PubMed \| Reef Life Survey \| Socotra Atlas \| TreeBase \| Tree of Life \| Wikipedia: Go, Search \| Zoological Record Estimates based on models Preferred temperature (Ref. 123201): 22.9 - 28.8, mean 27.6 °C (based on 638 cells). Phylogenetic diversity index (Ref. 82804):PD 50 = 0.6250 [Uniqueness, from 0.5 = low to 2.0 = high]. Bayesian length-weight: a=0.02818 (0.01685 - 0.04715), b=3.00 (2.85 - 3.15), in cm total length, based on LWR estimates for this species & (Sub)family-body (Ref. 93245). Trophic level (Ref. 69278):3.1 ±0.26 se; based on food items. Fishing Vulnerability (Ref. 59153):Low vulnerability (23 of 100). 🛈 Price category (Ref. 80766):Unknown. Nutrients (Ref. 124155):Calcium = 187 [94, 330] mg/100g; Iron = 1.44 [0.85, 2.58] mg/100g; Protein = 18.3 [17.2, 19.3] %; Omega3 = 0.172 [0.089, 0.312] g/100g; Selenium = 32.5 [15.1, 68.3] μg/100g; VitaminA = 14.6 [4.7, 46.6] μg/100g; Zinc = 1.89 [1.28, 2.77] mg/100g (wet weight); Random Species Back to Search Comments & Corrections Back to Top Accessed through: Not available FishBase mirror site : localhost Page last modified by : mrius-barile - 20 July 2016 Total processing time for the page : 0.0328 seconds
15570	https://wongchemistry.weebly.com/uploads/5/1/3/6/5136424/14_solubility_equilibria.pdf	APChemistry Solubility Equilibria AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product. © 2008 by René McCormick. All rights reserved. SOLUBILITY EQUILIBRIA (Ksp, THE SOLUBILITY PRODUCT) ¾ Saturated solutions of salts are another type of chemical equilibria. Remember those solubility rules? The fine print said that “soluble” is defined as 3.0 g salt dissolving in 100 g water. What if 2.9 g dissolves? We call it “insoluble”, but a good bit does actually dissolve. So, “insoluble” is not an absolute term. Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution. ♦ When the solid is first added to water, no ions are initially present. ♦ As dissolution proceeds, the concentration of ions increases until equilibrium is established. This occurs when the solution is saturated. The equilibrium constant, the Ksp, is no more than the product of the ions in solution. (Remember, solids do not appear in equilibrium expressions.) For a saturated solution of AgCl, the equation would be: AgCl(s) R Ag+(aq) + Cl−(aq) The solubility product expression would be: Ksp = [Ag+][Cl−] The AgCl(s) is left out since solids are left out of equilibrium expressions (constant concentrations). Solubility Equilibrium 2 ¾ You can find loads of Ksp values on tables contained in any text. Look up a table of Ksp values in your text and write the Ksp expression and value for the following salts: CaF2(s) R Ca+2 + 2 F− Ksp = Ag2SO4(s) R 2 Ag+ + SO4 2− Ksp = Bi2S3(s) R 2 Bi+3 + 3 S2− Ksp = DETERMINING Ksp FROM EXPERIMENTAL MEASUREMENTS ¾ In practice, Ksp is determined by careful laboratory measurements using various spectroscopic methods. Remember STOICHIOMETRY!! Example: Lead(II) chloride dissolves to a slight extent in water according to the equation: PbCl2 R Pb+2 + 2 Cl− • Calculate the Ksp if the lead ion concentration has been found to be 1.62 × 10−2M. • If the lead ion concentration is “x” then chloride’s concentration is “2x”. Also note that the “molar solubility” is equal to “x” since the coefficient on the salt is “1” (think stoichiometry!). So. . . . Ksp = (1.62 × 10−2)(3.24 × 10−2)2 = 1.70 × 10−5 Exercise 1 Calculating Ksp from Solubility I Copper(I) bromide has a measured molar solubility of 2.0 × 10−4 mol/L at 25°C. Calculate its Ksp value. Ksp = 4.0 × 10−8 Solubility Equilibrium 3 Exercise 2 Calculating Ksp from Solubility II Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a molar solubility of 1.0 × 10−15 mol/L at 25°C. Ksp = 1.1 × 10−73 ESTIMATING SALT SOLUBILITY FROM Ksp ¾ Example: The Ksp for CaCO3 is 3.8 × 10−9 @ 25°C. Calculate the molar solubility of calcium carbonate in pure water in a) moles per liter & b) grams per liter: The relative solubilities can be deduced by comparing values of Ksp BUT, BE CAREFUL! These comparisons can only be made for salts having the same ION:ION ratio. Please don’t forget solubility changes with temperature! Some substances become less soluble in cold while some become more soluble! Exercise 3 Calculating Solubility from Ksp The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4 × 10−7 at 25°C. Calculate its solubility at 25°C. = 3.3 × 10−3 mol/L Solubility Equilibrium 4 SOLUBILITY AND THE COMMON ION EFFECT The presence of a common ion will cause the equilibrium to shift so that even less of the substance with the smaller Ksp value will dissolve. ¾ Experiment shows that the solubility of any salt is always less in the presence of a “common ion”. WHY? ¾ LeChatelier’s Principle, that’s why! Be reasonable and use approximations when you can!! ¾ pH can also affect solubility. Evaluate the equation to see who would want to “react” with the addition of acid or base. ¾ Would magnesium hydroxide be more soluble in an acid or a base? Why? Mg(OH)2(s) R Mg2+ (aq) + 2 OH− (aq) (milk of magnesia) Exercise 4 Solubility and Common Ions Calculate the solubility of solid CaF2 (Ksp = 4.0 × 10−11) in a 0.025 M NaF solution. = 6.4 × 10−8 mol/L Ksp AND THE REACTION QUOTIENT, Q ¾ With some knowledge of the reaction quotient, we can decide whether a ppt will form at the conditions given AND calculate the concentrations of ions required to begin the ppt. of an insoluble salt 1. Q < Ksp, the system is not at equil. (unsaturated) 2. Q = Ksp, the system is at equil. (saturated) 3. Q > Ksp, the system is not at equil. (supersaturated) Precipitates form when the solution is supersaturated!!! Solubility Equilibrium 5 ¾ Precipitation of insoluble salts Metal-bearing ores often contain the metal in the form of an insoluble salt, and, to complicate matters, the ores often contain several such metal salts. Dissolve the metal salts to obtain the metal ion, concentrate in some manner, and ppt. selectively only one type of metal ion as an insoluble salt. Exercise 5 Determining Precipitation Conditions A solution is prepared by adding 750.0 mL of 4.00 × 10−3 M Ce(NO3)3 to 300.0 mL of 2.00 × 10−2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 × 10−10) precipitate from this solution? yes Exercise 6 Precipitation A solution is prepared by mixing 150.0 mL of 1.00 × 10−2 M Mg(NO3)2 and 250.0 mL of 1.00 × 10−1 M NaF. Calculate the concentrations of Mg2+ and F− at equilibrium with solid MgF2 (Ksp = 6.4 × 10−9). [Mg2+] = 2.1 × 10−6 M [F−] = 5.50 × 10−2 M Solubility Equilibrium 6 QUALITATIVE ANALYSIS (a favorite lab question topic on the AP Exam!) ¾ A Qualitative Analysis Scheme introduces you to the basic chemistry of various ions ¾ It also illustrates how the principles of chemical equilibria can be applied. ¾ Objective: Separate the following metal ions: silver, lead, cadmium and nickel Adding HCl causes silver and lead to ppt. (Aren’t you glad you now know your solubility rules?) while nickel and cadmium will stay in solution. Separate nickel and cadmium by filtration. Add HOT water to the ppts and filter while HOT and the lead(II) chloride will redissolve. Separate the chloride ppts by filtration. Separating Cd and Ni is more subtle. We use their sulfide Ksp values to determine which sulfide precipitates first! Exercise 7 Selective Precipitation A solution contains 1.0 × 10−4 M Cu+ and 2.0 × 10−3 M Pb2+. If a source of I− is added gradually to this solution, will PbI2 (Ksp = 1.4 × 10−8) or CuI (Ksp = 5.3 × 10−12) precipitate first? Specify the concentration of I− necessary to begin precipitation of each salt. CuI will precipitate first Concentration in excess of 5.3 × 10−8 M required Solubility Equilibrium 7 ACID−BASE AND PPT. EQUILIBRIA OF PRACTICAL SIGNIFICANCE ¾ SOLUBILITY OF SALTS IN WATER AND ACIDS the solubility of PbS in water: PbS (s) R Pb+2 + S2− Ksp = 8.4 × 10−28 the hydrolysis of the S2− ion in water S2− + H2O R HS− +OH− Kb = 0.077 ¾ Overall process: PbS + H2O R Pb+2 + HS− + OH− Ktotal = Ksp × Kb = 6.5 × 10−29 May not seem like much but it can increase the environmental lead concentration by a factor of about 10,000 over the solubility of PbS calculated from simply Ksp! Any salt containing an anion that is the conjugate base of a weak acid will dissolve in water to a greater extent than given by the Ksp. This means salts of sulfate, phosphate, acetate, carbonate, and cyanide, as well as sulfide can be affected. If a strong acid is added to water-insoluble salts such as ZnS or CaCO3, then hydroxide ions from the anion hydrolysis is removed by the formation of water. This shifts the anion hydrolysis further to the right; the weak acid is formed and the salt dissolves. Carbonates and many metal sulfides along with metal hydroxides are generally soluble in strong acids. The only exceptions are sulfides of mercury, copper, cadmium and a few others. Insoluble inorganic salts containing anions derived from weak acids tend to be soluble in solutions of strong acids. Salts are not soluble in strong acid if the anion is the conjugate base of a strong acid!!
15571	https://user.phil.hhu.de/~petersen/Riga/print_Folien_Riga_NLT.pdf	CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Introduction to the Theory of Formal Languages Wiebke Petersen Heinrich-Heine-Universität Düsseldorf Institute of Language and Information Computational Linguistics www.phil-fak.uni-duesseldorf.de/~petersen/ Riga, 2006 Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Outline 1 Introduction to Computational Linguistics 2 Preliminaries alphabets and words operations on words formal languages 3 Chomsky hierarchy describing formal languages formal grammars Chomsky-hierarchy 4 Regular languages regular expressions right-linear grammars ﬁnite-state automata closure properties and pumping lemma 5 Context-free languages context-free grammars pumping lemma and closure properties pushdown automaton Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages alphabets and words alphabets and words Deﬁnition alphabet Σ: nonempty, ﬁnite set of symbols word: a ﬁnite string x1 . . . xn of symbols. length of a word \|w\|: number of symbols of a word w (example: \|abbaca\| = 6) empty word ϵ: the word of length 0 Σ∗is the set of all words over Σ Σ+ is the set of all nonempty words over Σ (Σ+ = Σ∗\ {ϵ}) Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages alphabets and words Substrings Preﬁx, sufﬁx, inﬁx Given words w, v ∈Σ∗: preﬁx: w is a preﬁx of v iff there exists a word u ∈Σ∗with v = wu. sufﬁx: w is a sufﬁx of v iff there exists a word u ∈Σ∗with v = uw. inﬁx: w is an inﬁx of v iff there exist words u1, u2 ∈Σ∗ with v = u1wu2. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages operations on words Concatenation Deﬁnition The concatenation of two words w = a1a2 . . . an and v = b1b2 . . . bm with n, m ≥0 is w ◦v = a1 . . . anb1 . . . bm Sometimes we write uv instead of u ◦v. w ◦ϵ = ϵ ◦w = w neutral element u ◦(v ◦w) = (u ◦v) ◦w associativity (Σ∗, ◦) is a semi-group with neutral element (monoid). Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages operations on words Exponents and reversals Exponents wn: w concatenated n-times with itself. w0 = ϵ w∗= S n≥0 wn ϵ ∈w∗for any word w Reversals The reversal of a word w is denoted w R (example: (abcd)R = dcba. A word w with w = w R is called a palindrome. (madam, mum, otto, anna,. . . ) Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages formal languages Formal language Deﬁnition A formal language L is a set of words over an alphabet Σ, i.e. L ⊆Σ∗. Examples: language Lpal of the palindromes in English Lpal = {mum, madam, . . . } language LMors of the letters of the latin alphabet encoded in the Morse code: LMors = {·−, −· ··, . . . , −−··} the empty set the set of words of length 13 over the alphabet {a, b, c} English? Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages formal languages Operations on formal languages Deﬁnition If L ⊆Σ∗and K ⊆Σ∗are two formal languages over an alphabet Σ, then K ∪L, K ∩L, K \ L are languages over Σ too. The concatenation of two formal languages K and L is K ◦L := {v ◦w ∈Σ∗\|v ∈K, w ∈L} Ln = L ◦L ◦L . . . ◦L \| {z } n-times L∗:= S n≥0 Ln. Note: {ϵ} ∈L∗for any language L. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages formal languages Operations on formal languages Example K = {abb, a} and L = {bbb, ab} K ◦L = {abbbbb, abbab, abbb, aab} and L ◦K = {bbbabb, bbba, ababb, aba} K ◦∅= ∅ K ◦{ϵ} = K K 2 = {abbabb, abba, aabb, aa} Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages describing formal languages Enumerating all elements of a language Peter says that Mary is fallen of the tree. Oskar says that Peter says that Mary is fallen of the tree. Lisa says that Oskar says that Peter says that Mary is fallen of the tree. . . . The set of strings of a natural language is inﬁnite. The enumeration does not gather generalizations. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages describing formal languages Grammars Grammar A formal grammar is a generating device which can generate (and analyze) strings/words. Grammars are ﬁnite rule systems. The set of all strings generated by a grammar is a formal language (= generated language). S → NP VP VP → V NP → D N D → the N → cat V → sleeps Generates: the cat sleeps Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages describing formal languages Automata Automaton An automaton is a recognizing device which accepts strings/words. The set of all strings accepted by an automaton is a formal language (=accepted language). accepts: L(ab⋆a) Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages describing formal languages Formal grammar Deﬁnition A formal grammar is a 4-tupel G = (N, T, S, P) with an alphabet of terminals T, an alphabet of nonterminals N with N ∩T = ∅, a start symbol S ∈N, a ﬁnite set of rules/productions P ⊆{⟨α, β⟩\| α, β ∈(N ∪T)∗and α ̸∈T ∗}. Instead of ⟨α, β⟩we write also α →β. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages formal grammars Formal grammar Vocabulary Let G = (N, T, S, P) be a grammar and v, w ∈(T ∪N)∗: v is directly derived from w (or w directly generates v), w →v if w = w1αw2 and v = w1βw2 such that ⟨α, β⟩∈P. v is derived from w (or w generates v), w →∗v if there exists w0, w1, . . . wk ∈(T ∪N)∗(k ≥0) such that w = w0, wk = v and wi−1 →wi for all k ≥i ≥0. →∗denotes the reﬂexive transitive closure of → L(G) = {w ∈T ∗\|S →∗w} is the formal language generated by the grammar G. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages formal grammars Example G1 = ⟨{S,NP ,VP ,N,V,D,N,EN}, {the, cat, peter, chases}, S, P⟩ P =    S → NP VP VP → V NP NP → D N NP → EN D → the N → cat EN → peter V → chases    L(G1) = the cat chases peter peter chases the cat peter chases peter the cat chases the cat “the cat chases peter” can be derived from S by: S →NP VP →NP V NP →NP V EN →NP V peter →NP chases peter →D N chases peter →D cat chases peter →the cat chases peter Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages formal grammars Derivation tree S NP D the N cat VP V chases NP EN peter One derivation determines one derivation tree, but the same derivation tree can result from different derivations. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages formal grammars Not all formal languages are derivable from a formal grammar The set of all formal languages over an alphabet Σ = {a} is POW(Σ∗); hence, the set is uncountably inﬁnite. The set of grammars generating formal languages over Σ with ﬁnite sets of productions is countably inﬁnite. Hence, the set of formal languages generated by a formal grammar is a strict subset of the set of all formal languages. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Chomsky-hierarchy Chomsky-hierarchy The Chomsky-hierarchy is a hierarchy over structure conditions on the productions. Constraining the structure of the productions results in a restricted set of languages. The language classes correspond to conditions on the right- and left-hand sides of the productions. The Chomsky-hierarchy reﬂects a special form of complexity, other criteria are possible and result in different hierarchies. Linguists beneﬁt from the rule-focussed deﬁnition of the Chomsky-hierarchy. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Chomsky-hierarchy Noam Chomsky Noam Chomsky (∗7.12.1928, Philadelphia) Noam Chomsky, Three Models for the Description of Language, (1956). Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Chomsky-hierarchy Chomsky-hierarchy A grammar (N, T, S, P) is a Type 3) regular grammar (REG): iff every production is of the form A →βB or A →β with A, B ∈N and β ∈T ∗(right-linear grammar); or iff every production is of the form A →Bβ or A →β with A, B ∈N and β ∈T ∗(left-linear grammar). Type 2) context-free grammar (CF): iff every production is of the form A →β with A ∈N and β ∈(N ∪T)∗. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Chomsky-hierarchy Chomsky-hierarchy (cont.) A grammar (N, T, S, P) is a Type 1) context-sensitive grammar (CS): iff every production is of the form γAδ →γβδ with γ, δ, β ∈(N ∪T)∗, A ∈N and β ̸= ϵ; or of the form S →ϵ, in which case S does not occur on any right-hand side of a production. Type 0) phrase-structure grammar (recursively enumerable grammar) (RE): iff every production is of the form α →β with α ∈(N ∪T)∗\ T ∗and β ∈(N ∪T)∗. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Chomsky-hierarchy Main theorem L(REG) ⊂L(CG) ⊂L(CS) ⊂L(RE) L(RE) L(CS) L(CG) L(REG) Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages regular expressions Regular expressions RE: syntax The set of regular expressions REΣ over an alphabet Σ = {a1, . . . , an} is deﬁned by: ∅is a regular expression. ϵ is a regular expression. a1, . . . , an are regular expressions If a and b are regular expressions over Σ then (a + b) (a • b) (a⋆) are regular expressions too. (The brackets are frequently omitted w.r.t. the following dominance scheme: ⋆dominates • dominates +) Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages regular expressions Regular expressions RE: semantics Each regular expression r over an alphabet Σ denotes a formal language L(r) ⊆Σ∗. Regular languages are those formal languages which can be expressed by a regular expression. The denotation function L is deﬁned inductively: L(∅) = ∅, L(ϵ) = {ϵ}, L(ai) = {ai} L(a + b) = L(a) ∪L(b) L(a • b) = L(a) ◦L(b) L(a⋆) = L(a)∗ Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages right-linear grammars Type3-languages Deﬁnition A grammar (N, T, S, P) is right-linear iff all productions are of the form: A →w or A →wB with A, B ∈N and w ∈T ∗. A language generated by a right-linear grammar is said to be a right-linear language. Proposition If L is a formal language, the following statements are equivalent: 1 L is right-linear 2 L is regular 3 (L is left-linear) Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages right-linear grammars Proof: Each regular language is right-linear Σ = {a1, . . . , an} 1 ∅is generated by ({S}, Σ, S, {}), 2 {ϵ} is generated by ({S}, Σ, S, {S →ϵ}), 3 {ai} is generated by ({S}, Σ, S, {S →ai}), 4 If L1, L2 are regular languages with generating right-linear grammars (N1, T1, S1, P1), (N2, T2, S2, P2), then L1 ∪L2 is generated by (N1 ⊎N2, T1 ∪T2, S, P1 ∪⊎P2 ∪{S →S1, S →S2}), 5 L1 ◦L2 is generated by (N1 ⊎N2, T1 ∪T2, S1, P′ 1 ∪⊎P2) (P′ 1 is obtained from P1 if all rules of the form A →w (w ∈T ∗) are replaced by A →wS2), 6 L∗ 1 is generated by (N1, Σ, S1, P′ 1 ∪{S1 →ϵ}) (P′ 1 is obtained from P1 if all rules of the form A →w (w ∈T ∗) are replaced by A →wS1). Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Deterministic ﬁnite-state automaton (DFSA) Deﬁnition A deterministic ﬁnite-state automaton is a tuple ⟨Q, Σ, δ, q0, F⟩with: 1 a ﬁnite, non-empty set of states Q 2 an alphabet Σ with Q ∩Σ = ∅ 3 a partial transition function δ : Q × Σ →Q 4 an initial state q0 ∈Q and 5 a set of ﬁnal states F ⊆Q. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Language accepted by an automaton Deﬁnition A situation of a ﬁnite-state automaton ⟨Q, Σ, δ, q0, F⟩is a triple (x, q, y) with x, y ∈Σ∗and q ∈Q. Situation (x, q, y) produces situation (x ′, q′, y′) in one step if there exists an a ∈Σ such that x ′ = xa, y = ay′ and δ(q, a) = q′, we write (x, q, y) ⊢(x ′, q′, y′) ((x, q, y) ⊢∗(x′, q′, y′) as usual). Deﬁnition A word w ∈Σ∗gets accepted by an automaton ⟨Q, Σ, δ, q0, F⟩ if (ϵ, q0, w) ⊢∗(w, qn, ϵ) with qn ∈F. An automaton accepts a language iff it accepts every word of the language. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Example accepts L(ab⋆a) Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Nondeterministic ﬁnite-state automaton NDFSA Deﬁnition A nondeterministic ﬁnite-state automaton is a tuple ⟨Q, Σ, ∆, q0, F⟩with: 1 a ﬁnite non-empty set of states Q 2 an alphabet Σ with Q ∩Σ = ∅ 3 a transition relation ∆⊆Q × Σ × Q 4 an initial state q0 ∈Q and 5 a set of ﬁnal states F ⊆Q. Theorem A language L can be accepted by a DFSA iff L can be accepted by a NFSA. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Example DEA / NDEA The language L(ab⋆+ ac⋆) gets accepted by Note: Even automatons with ϵ-transitions accept the same languages like NDEA’s. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Complete deterministic ﬁnite-state automata Complete deterministic ﬁnite-state automata have a total transition function: Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Finite-state automatons accept regular languages Theorem (Kleene) Every language accepted by a DFSA is regular and every regular language is accepted by some DFSA. proof idea: Each regular language is accepted by a NDFSA: 0 q 0 q 0 q 1 q a Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Proof of Kleene’s theorem (cont.) If R1 and R2 are two regular expressions such that the languages L(R1) and L(R2) are accepted by the automatons A1 and A2 respectively, then L(R1 + R2) is accepted by: Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Proof of Kleene’s theorem (cont.) L(R1 • R2) is accepted by: Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages ﬁnite-state automata Proof of Kleene’s theorem (cont.) L(R∗ 1) is accepted by: Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages closure properties and pumping lemma Closure properties of regular languages Type3 Type2 Type1 Type0 union + ✓ + + + intersection + -+ + complement + -+ -concatenation + ✓ + + + Kleene’s star + ✓ + + + intersection with a regular language + + + + complement: construct complementary DFSA intersection: implied by de Morgan Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages closure properties and pumping lemma Pumping lemma for regular languages Lemma (Pumping-Lemma) If L is an inﬁnite regular language over Σ, then there exists words u, v, w ∈Σ∗such that v ̸= ϵ and uviw ∈L for any i ≥0. proof sketch: Any regular language is accepted by a DFSA with a ﬁnite number n of states. Any inﬁnite language contains a word z which is longer than n (\|z\| ≥n). While reading in z, the DFSA passes at least one state qj twice. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages closure properties and pumping lemma Pumping lemma for regular languages (cont.) Lemma (Pumping-Lemma) If L is an inﬁnite regular language over Σ, then there exists words u, v, w ∈Σ∗such that v ̸= ϵ and uviw ∈L for any i ≥0. proof sketch: Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages closure properties and pumping lemma L = {anbn : n ≥0} is not regular L = {anbn : n ≥0} is inﬁnite. Suppose L is regular. Then there exists u, v, w ∈{a, b}∗, v ̸= ϵ with uvnw ∈L for any n ≥0. We have to consider 3 cases for v. 1 v consists of a’s and b’s. 2 v consists only of a’s. 3 v consists only of b’s. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages closure properties and pumping lemma Exercises Are the following languages regular? 1 L1 = {w ∈{a, b}∗: w contains an even number of b′s}. 2 L2 = {w ∈{a, b}∗: w contains as many b′s as a′s}. 3 L3 = {wwR ∈{a, b}∗: wwR is a palindrome over {a, b}∗}. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages closure properties and pumping lemma Intuitive rules for regular languages L is regular if it is possible to check the membership of a word simply by reading it symbol for symbol while using only a ﬁnite stack. Finite-state automatons are too weak for: counting in N (“same number as”); recognizing a pattern of arbitrary length (“palindrome”); expressions with brackets of arbitrary depth. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages closure properties and pumping lemma Summary: regular languages regular expression regular grammar ﬁnite-state automaton regular language speciﬁes generates accepts equivalent equivalent Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages context-free grammars Context-free language Deﬁnition A grammar (N, T, S, P) is context-free if all production rules are of the form: A →α, with A ∈N and α ∈(T ∪N)∗. A language generated by a context-free grammar is said to be context-free. Proposition The set of context-free languages is a strict superset of the set of regular languages. Proof: Each regular language is per deﬁnition context-free. L(anbn) is context-free but not regular (S →aSb, S →ϵ). Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages context-free grammars Examples of context-free languages L1 = {wwR : w ∈{a, b}∗} L2 = {aibj : i ≥j} L3 = {w ∈{a, b}∗: more a′s than b′s} L4 = {w ∈{a, b}∗: number of a′s equals number of b′s}    S → aB A → a B → b S → bA A → aS B → bS A → bAA B → aBB    Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages context-free grammars Ambiguous grammars and ambiguous languages Deﬁnition Given a context-free grammar G: A derivation which always replaces the left furthest nonterminal symbol is called left-derivation Deﬁnition A context-free grammar G is ambiguous iff there exists a w ∈L(G) with more than one left-derivation, S →∗w. Deﬁnition A context-free language L is ambiguous iff each context-free grammar G with L(G) = L is ambiguous. Left-derivations and derivation trees determine each other! Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages context-free grammars Example of an ambiguous grammar G = (N, T, NP, P) with N = {D, N, P, NP, PP}, T = {the, cat, hat, in}, P =    NP → D N D → the N → hat NP → NP PP N → cat P → in PP → P NP    NP NP NP D the N cat PP P in NP D the N hat PP P in NP D the N hat NP NP D the N cat PP P in NP NP D the N hat PP P in NP D the N hat Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages context-free grammars Chomsky Normal Form Deﬁnition A grammar is in Chomsky Normal Form (CNF) if all production rules are of the form 1 A →a 2 A →BC with A, B, C ∈T and a ∈Σ (and if necessary S →ϵ in which case S may not occur in any right-hand side of a rule). Proposition Each context-free language is generated by a grammar in CNF. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages pumping lemma and closure properties Pumping lemma for context-free languages Lemma (pumping lemma) For each context-free language L there exists a p ∈N such that for any z ∈L: if \|z\| > p, then z may be written as z = uvwxy with u, v, w, x, y ∈T ∗, \|vwx\| ≤p, vx ̸= ϵ and uviwxiy ∈L for any i ≥0. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages pumping lemma and closure properties Pumping lemma: proof sketch S A A x v y w u . . . . . . S A A x v y w u . . . . . . A . . . v x \|vwx\| ≤p, vx ̸= ϵ and uv iwxiy ∈L for any i ≥0. Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages pumping lemma and closure properties Existence of non context-free languages L1 = {anbncn} L2 = {anbmcndm} L1 = {ww : w ∈{a, b}∗} Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages pumping lemma and closure properties Closure properties of context-free languages Type3 Type2 Type1 Type0 union + + + + intersection + -+ + complement + -+ -concatenation + + + + Kleene’s star + + + + intersection with a regular language + + + + union: G = (N1 ⊎N2 ∪{S}, T1 ∪T2, S, P) with P = P1 ∪⊎P2 ∪{S →S1, S →S2} intersection: L1 = {anbnak}, L2 = {anbkak}, but L1 ∩L2 = {anbnan} complement: de Morgan concatenation: G = (N1 ⊎N2 ∪{S}, T1 ∪T2, S, P) with P = P1 ∪⊎P2 ∪{S →S1S2} Kleene’s star: G = (N1 ∪{S}, T1, S, P) with P = P1 ∪{S →S1S, S →ϵ} Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages pumping lemma and closure properties decision problems Given: grammars G = (N, Σ, S, P), G′ = (N′, Σ, S′, P′), and a word w ∈Σ∗ word problem Is w derivable from G ? emptiness problem Does G generate a nonempty language? equivalence problem Do G and G′ generate the same language (L(G) = L(G′))? Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages pumping lemma and closure properties Results for the decision problems Type3 Type2 Type1 Type0 word problem D D D U emptiness problem D D U U equivalence problem D U U U D: decidable; U: undecidable Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Chomsky-hierarchy (1956) Type 3: REG ﬁnite-state automaton WP: linear Type 2: CF pushdown-automaton WP: cubic Type 1: CS linearly restricted automaton WP: exponential Type 0: RE Turing machine WP: not decid-able Formal Language Theory Wiebke Petersen CL Preliminaries Chomsky hierarchy Regular languages Context-free languages Literature Beesley & Karttunen (2003) Finite State Morphology. CSLI. Hopcroft, Motwani & Ullman (2001) Introduction to Automata and Language Theory. Addison-Wesley, 2nd edition. Partee, ter Meulen & Wall (1990) Mathematical Methods in Linguistics. Kluwer Academic Publishers. Sipser (2005) Introduction to the Theory of Computation. Thomson Course Technology, 2nd edition. Sudkamp (1996) Languages and Machines: An Introduction to the Theory of Computer Science. Addison Wesley, 2nd edition. Formal Language Theory Wiebke Petersen
15572	https://eurjbreasthealth.com/articles/american-joint-committee-on-cancers-staging-system-for-breast-cancer-eighth-edition-summary-for-clinicians/ejbh.galenos.2021.2021-4-3	American Joint Committee on Cancer’s Staging System for Breast Cancer, Eighth Edition: Summary for Clinicians - European Journal of Breast Health Abstracting and Indexing Turkish Federation of Breast Diseases Societies E-ISSN: 2587-0831ISSN: 2587-0831 About Us Aims and Scope Editorial Board Instructions to Authors Instructions to Authors Manuscript Preparation Ethical Policy EJBH - Study Group Contact Us American Joint Committee on Cancer’s Staging System for Breast Cancer, Eighth Edition: Summary for Clinicians PDF Cite Share Request Review VOLUME: 17 ISSUE: 3 P: 234 - 238 July 2021 American Joint Committee on Cancer’s Staging System for Breast Cancer, Eighth Edition: Summary for Clinicians Eur J Breast Health 2021;17(3):234-238 DOI: 10.4274/ejbh.galenos.2021.2021-4-3 Haoling Zhu Başak E. Doğan Author Information Copyright & Licence Information Galenos & Eur J Breast Health Disclaimer Department of Radiology UT Southwestern Medical Center, Dallas, Texas No information available. No information available Received Date: 06.04.2021 Accepted Date: 06.06.2021 Publish Date: 24.06.2021 PDF Cite Share Request ABSTRACT Introduction Conclusion References ABSTRACT Breast cancer is commonly staged using the American Joint Committee on Cancer (AJCC) staging system. The 7 th edition of the AJCC Staging Manual, was a purely anatomic staging method, which uses primary tumor size (T), nodal involvement (N), and metastasis (M) based on clinical and pathological evaluations. Advancements in tumor biology and prognostic biological markers, such as estrogen receptor (ER)/progesterone receptor (PR), HER2/neu, and Ki-67, have allowed clinicians to understand why similarly staged patients had significantly different outcomes. The most recent update to the staging system integrates molecular markers with disease extent for more optimal estimation of prognosis. This change improves the prognosis of breast cancer patients and better informs physicians in the planning of treatments. This review summarizes the changes in the AJCC Staging Manual, 8 th edition and their impact on practicing radiologists in breast cancer management. Keywords: AJCC, biomarkers, Breast cancer, Breast imaging, cancer staging Key Points •Until the implementation of the 8th edition of the AJCC Staging Manual, a purely anatomic staging method, which uses primary tumor (T) size, nodal (N) involvement, and metastasis (M) based on clinical and pathological evaluations, was employed. •Advancements in tumor biology and prognostic biological markers, such as estrogen receptor (ER)/progesterone receptor (PR), HER2/neu, and Ki-67, have allowed clinicians to understand why similarly staged patients had significantly different outcomes. •The most recent update to the staging system integrates molecular markers with disease extent for more optimal estimation of prognosis. Introduction Breast cancer is the most diagnosed cancer among women in the United States (US). It is also the leading cause of cancer-related deaths following lung cancer. In 2020, 276,480 new cases of breast cancer were projected to be diagnosed among American women along with 48,530 new cases of non-invasive breast cancer. One out of eight American women is projected to develop invasive breast cancer at some point in her life (1). In 1959, the American Joint Committee for Cancer Staging and End-Results Reporting, now the American Joint Committee on Cancer (AJCC), standardized the tumor, node, and metastasis (TNM) cancer staging system. The first edition of the AJCC Staging Manual, published in 1977, allowed clinicians to standardize treatment and evaluate treatment results between different institutions (2, 3). Since then, the manual has been periodically updated to reflect clinical and technological advancements in the field. Until the implementation of the 8 th edition of the AJCC Staging Manual in 2018, a purely anatomic staging method, which uses primary tumor (T) size, nodal (N) involvement, and metastasis (M) based on clinical and pathological evaluations, was employed. Advancements in tumor biology and prognostic biological markers [estrogen receptor (ER) and progesterone receptor (PR), HER2/neu, and Ki-67] have allowed clinicians to understand why similarly staged patients had significantly different outcomes. The most recent update to the staging system integrates anatomic staging with prognostic staging, which uses tumor grade, hormone receptors and oncogene expression, and multigene testing (4). Incorporating the prognostic stage into the breast cancer staging system has allowed physicians to individualize the patient prognosis, leading to a more optimal estimation of prognosis. This article highlights key changes from the 7th edition to the 8th edition of the AJCC breast cancer staging system, with multimodality imaging demonstration of its application. We will review the anatomic TNM staging categories: clinical staging, pathologic staging, post-therapy or post-neoadjuvant therapy staging, and restaging. Furthermore, we will summarize the prognostic staging (both clinical and pathological), the implementation of gene assays, and how they are integrated in different scenarios. Other changes in the AJCC Staging Manual, 8 th Edition, will also be discussed, including re-categorization of lobular carcinoma in situ (LCIS) and the clinical N stage based on physical examination and imaging studies (4). Anatomic staging Anatomic TNM staging is further categorized into: clinical staging, which relies on physical examination, imaging, and biopsy of the affected areas; pathologic staging, which is determined after a patient has had surgery to remove the primary tumor and regional lymph nodes; post-neoadjuvant therapy staging, which determines how much cancer remains after a patient completes preoperative systemic or radiation therapy, and may incorporate both clinical and pathologic staging; and restaging, which is performed if a cancer recurs after treatment and determines the extent of disease recurrence. Since this review is primarily for radiologists, we will spend most of this section discussing clinical anatomic staging. Importantly, imaging findings are considered relevant to staging if obtained within 4 months of diagnosis or completion of surgery, whichever confers a longer time period, provided the disease has not worsened. Clinical anatomic staging Clinical staging of the primary T begins with the measurement of the tumor size based on physical examination and imaging (4). The staging categories range from Tis to T4 and are the same for both clinical and pathological staging of the primary tumor, where the prefixes “c” and “p” indicate clinical stage and pathologic stage, respectively. Tis refers to ductal carcinoma in situ with no invasive cancer. T1–T4 refers to the tumor size, ranging from 2 cm to >5 cm, and the involvement of chest wall and/or skin (ulceration or macroscopic nodules), respectively. The T4 category is further subdivided into T4a–T4d, where T4a indicates chest wall involvement; T4b skin indicates involvement through ulceration, ipsilateral macroscopic satellite nodules, and/or skin edema (e.g., peau d’orange), which does not meet the criteria for inflammatory carcinoma; T4c is indicated when both T4a and T4b are present; and T4d inflammatory carcinoma. Lastly, LCIS is no longer staged via TNM in the Tis category, as it is now viewed as benign; however, it carries a risk of future malignancy. Similar to primary tumor staging, clinical staging of regional axillary lymph nodes (N) should begin with the prefix “c.” The staging categories range from N0 to N3, where N0 indicates no regional lymph node metastases as revealed by imaging or clinical examination; N1 indicates metastases to movable ipsilateral level I-II [where level I nodes are lateral to the lateral border of the pectoralis minor muscle and level II nodes are between the medial and lateral borders of the pectoralis minor and also include the interpectoral (Rotter’s) lymph nodes] axillary lymph nodes; N2 indicates metastases to ipsilateral level I-II axillary nodes that are clinically fixed or matted, or metastases to ipsilateral internal mammary nodes without axillary lymph node involvement (N2a and N2b, respectively); N3 indicates metastases to level III (ipsilateral infraclavicular) lymph nodes, ipsilateral internal mammary lymph nodes with level I-II axillary node metastases, or metastases to ipsilateral supraclavicular lymph nodes (N3a, N3b, and N3c, respectively). Moreover, for most patients, category cNX (suggesting regional lymph nodes cannot be assessed) is considered invalid and should be listed as cN0, unless the patient has been previously subjected to axillary dissection. Assessment of the metastases (M) stage involves categorizing patients into M0 or M1 using clinical examination. The M0 category indicates no clinical or radiographic evidence of distant metastases; however, this stage is designated cM0 since category pM0 is invalid (4). The designation pM1 may be used for patients with histologically proven metastases with at least 1 tumor deposit >0.2 mm. We also have cM0 (+), which indicates no clinical or radiographic evidence of distant metastases in the presence of histologically detected tumor deposits that are <2 mm in circulating blood, bone marrow, or other non-regional nodal tissues, without symptoms of metastases. The designation M1 indicates distant metastases, where cM1 is detected clinically or radiographically and pM1 is histologically detected with at least 1 tumor deposit >0.2 mm. Finally, if a patient is assigned the M1 category, he/she is categorized as stage IV of the disease. The patient remains in stage IV regardless of response to any preoperative systemic therapy; however, if the patient has not received preoperative therapy, the stage should be updated if postoperative imaging within 4 months of diagnosis reveals distant metastases. Prognostic staging Reflected in the 8 th edition of the AJCC staging system, prognostic staging incorporates tumor grade, hormone receptors and oncogene status [estrogen receptors (ER), progesterone receptors (PR), and human epidermal growth factor receptor 2 (HER2)], and multigene panel results, in addition to the anatomic staging system discussed above. Tumor grade The AJCC 8 th edition manual highlights that, along with other factors, such as proliferative index, hormone receptor expression, and gene expression profiles, grade is a key assessment of tumor differentiation, which in turn is an important tool for prognosis. Tumor grade used by the manual for staging is defined by the histologic grading system of Scarff, Bloom, and Richardson and standardized by the Nottingham group (4). Regardless of hormonal therapy or chemotherapy, high-grade or poorly differentiated tumors have a worse prognosis than low-grade or well-differentiated tumors. Furthermore, the Survey, Epidemiology, and End Results Program of the National Cancer Institute revealed that histologic grade is a valuable prognostic factor, regardless of tumor size or number of positive lymph nodes (5). Hormone receptors and HER2 According to the AJCC 8 th edition manual, hormone receptor and HER2 status need to be determined for all invasive breast cancers. Previous studies have demonstrated that ER- and PR- positive tumors can be effectively treated with selective ER modulators (SERMs), such as tamoxifen, to slow or stop tumor progression. Moreover, the higher the hormone receptor expression, the more effective the treatment becomes (6, 7). While ER- and PR-positive tumors are most likely to respond to SERMs, the response rate is lower for ER-positive and PR-negative, ER-negative and PR-positive, and ER-negative and PR-negative tumors, in descending order (6, 7, 8). Gene amplification or protein overexpression of the oncogene HER2 in untreated patients, whether node-positive or node-negative, has been associated with a worse prognosis (9, 10). Since HER2 positivity is associated with poor differentiation, it is more likely to be observed in high-grade invasive ductal carcinoma than invasive lobular carcinoma (11). Moreover, it is also associated with higher cell proliferation rates and hormone receptor negativity (12, 13, 14). The emergence of anti-HER2 targeted therapies has drastically improved the prognosis of patients with HER2-positive breast cancer. Particularly, the use of the monoclonal antibody trastuzumab in combination with a chemotherapeutic regimen significantly improves the disease-free and overall survival of these women (15). Biological subtypes In addition to tumor grade, hormone receptor, and oncogene expression, breast cancers vary widely on a genetic basis and this variation plays a significant role in prognosis. Four subtypes of breast cancer have been identified by genomic analysis: Luminal A, Luminal B, HER2-like, and Basal-like (16, 17). Categorizing cancers in such manner guides clinicians in both prognosis and treatment (18). Particularly, luminal A-type tumors are deemed to have a favorable prognosis since they are typically low-grade invasive ductal carcinomas (not otherwise specified type) or special types, including tubular, cribiform, or mucinous carcinomas. This subtype is responsive to endocrine therapy but has a poor response to traditional chemotherapy. In contrast, luminal B-type tumors are typically poorly differentiated and respond better to traditional chemotherapy than endocrine therapy. Although HER2-like tumors previously had the worst prognosis among all subtypes, the introduction of anti-HER2 therapy has drastically improved prognosis of patients with these cancers. Finally, basal-like tumors, which usually have a triple-negative (ER-negative, PR-negative, and HER2-negative) phenotype, have the worst prognosis and are the most challenging to treat with adjuvant therapy (4, 16, 17). Multigene panels Multigene panels can be used to obtain expression levels of multiple genes in breast cancer tissue. Many of these panels are useful prognostic tools; particularly, the Oncotype DX Breast Recurrence Score (Genomic Health, Redwood City, Calif), which measures 21 genes to predict likelihood of recurrence, has been incorporated into the updated staging system (19, 20). Despite this, one disadvantage of multigene panels is the substantial cost currently associated with their use. Thus, the AJCC 8 th edition manual states that hormone receptor and HER2 expression should be tested before obtaining a multigene panel and that the panel should be used only for certain subsets of cancers. Specifically, smaller (T1-T2) node-negative, hormone receptor-positive, HER2-negative tumors, and multigene panels may be incorporated into prognostic staging. When these tumors have an Oncotype DX score of <11, they can be considered stage IA, which can result in a downstage (4). Implementing the new staging system and challenges Upon the implementation of the AJCC 8 th edition staging system, many patients are restaged to better reflect their prognosis (Figures 1, 2, 3). A previous analysis was conducted to compare the 7 th edition to the revised 8 th edition staging system using the data obtained from 501,451 women in the National Cancer Database diagnosed from 2004 to 2014, excluding patients who underwent neoadjuvant chemotherapy. It was found that 23% of patients in stages I–III were downstaged and 19% of patients in stages I–III were upstaged (3). Figure 1 Figure 2 Figure 3 Several studies involving large cohorts have revealed that these changes to patients’ stages reflect more refined stratification and prediction of disease outcome. One study including 3327 patients who underwent surgery as an initial intervention at the University of Texas MD Anderson Cancer Center database revealed that the incorporation of grade, ER, and HER2 status into AJCC prognostic staging provided a more refined stratification in terms of patients’ disease-specific survival (21). Furthermore, an analysis of 54,727 patients in the California Cancer Registry revealed that the AJCC 8 th edition prognostic stage provided more accurate prognostic information than the anatomic stage alone (22). Despite the benefits associated with incorporating prognostic markers into breast cancer staging, there are challenges to the implementation of this new staging system. In many parts of the world, biological markers and multigene panels are not routinely available (23). Also, even in other parts of the world where biomarkers are more accessible, such as Europe, barriers in policy, reimbursement, and regulation have also delayed the widespread adoption of prognostic testing compared to the US (24). Therefore, the continued use of anatomic TNM staging in these regions emphasizes both its relevance and consistent usage. Conclusion The 8 th edition of the AJCC Cancer Staging Manual incorporates validated prognostic molecular biomarkers with standard tumor (T), node (N), and metastasis (M) anatomic categories. Given the large number of possible combinations of T, N, and M categories combined with grade, ER, PR, and HER2 status, integrating prognostic staging into multidisciplinary breast cancer care will be more complicated than with anatomic staging. Despite these challenges, prognostic staging facilitates more refined and accurate stratification of patients regarding survival outcomes than anatomic staging alone, thereby ultimately allowing clinicians to better serve their patients. Although prognostic staging is preferred for patient care, anatomic staging is retained as a key component of cancer care in regions of the world where biomarker tests are not routinely available (4). References 1 Breastcancer.org. U.S. Breast Cancer Statistics. 2021. Last Accessed Date: 28.02.2021. Available from: 2 Amin MB, Greene FL, Edge SB, Compton CC, Gershenwald JE, Brookland RK, et al. The Eighth Edition AJCC Cancer Staging Manual: continuing to build a bridge from a population-based to a more “personalized” approach to cancer staging. CA Cancer J Clin 2017; 67: 93-99. (PMID: 28094848) PubMed 3 Plichta JK, Campbell BM, Mittendorf EA, Hwang ES. Anatomy and breast cancer staging: is it still relevant? Surg Oncol Clin N Am 2018; 27: 51-67. (PMID: 29132565) PubMed 4 Hortobagyi GN, Connolly JL, D’Orsi CJ, Edge SB, Mittendorf EA, et al. Breast. In: Amin MB, Edge S, Greene F, et al, eds; American Joint Committee on Cancer. AJCC cancer staging manual. 8th ed. New York, NY: Springer, 2017. 5 Schwartz AM, Henson DE, Chen D, Rajamarthandan S. Histologic grade remains a prognostic factor for breast cancer regardless of the number of positive lymph nodes and tumor size: a study of 161 708 cases of breast cancer from the SEER Program. Arch Pathol Lab Med 2014; 138: 1048-1052. (PMID: 25076293) PubMed 6 Davies C, Godwin J, Gray R, Clarke M, Cutter D, Darby S, et al. Early Breast Cancer Trialists' Collaborative Group (EBCTCG). Relevance of breast cancer hormone receptors and other factors to the efficacy of adjuvant tamoxifen: patient-level meta-analysis of randomised trials. Lancet 2011; 378: 771-784. (PMID: 21802721) PubMed 7 Barnes DM, Harris WH, Smith P, Millis RR, Rubens RD. Immunohistochemical determination of oestrogen receptor: comparison of different methods of assessment of staining and correlation with clinical outcome of breast cancer patients. Br J Cancer 1996; 74: 1445-1451. (PMID: 8912543) PubMed 8 Hammond ME, Hayes DF, Dowsett M, Allred DC, Hagerty KL, Badve S; American Society of Clinical Oncology; College of American Pathologists. American Society of Clinical Oncology/College of American Pathologists guideline recommendations for immunohistochemical testing of estrogen and progesterone receptors in breast cancer (unabridged version). Arch Pathol Lab Med 2010; 134: e48-e72. doi: 10.5858/134.7.e48. (PMID: 20586616) PubMed 9 Slamon DJ, Clark GM, Wong SG, Levin WJ, Ullrich A, McGuire WL. Human breast cancer: correlation of relapse and survival with amplification of the HER-2/neu oncogene. Science 1987; 235: 177-182. (PMID: 3798106) PubMed 10 Ross JS, Slodkowska EA, Symmans WF, Pusztai L, Ravdin PM, Hortobagyi GN. The HER-2 receptor and breast cancer: ten years of targeted anti-HER-2 therapy and personalized medicine. Oncologist 2009; 14: 320-368. (PMID: 19346299) PubMed 11 Rosenthal SI, Depowski PL, Sheehan CE, Ross JS. Comparison of HER-2/neu oncogene amplification detected by fluorescence in situ hybridization in lobular and ductal breast cancer. Appl Immunohistochem Mol Morphol 2002; 10: 40-46. (PMID: 11893034) PubMed 12 Eccles SA. The role of c-erbB-2/HER2/neu in breast cancer progression and metastasis. J Mammary Gland Biol Neoplasia 2001; 6: 393-406. doi: (PMID: 12013529) PubMed 13 Piccart M, Lohrisch C, Di Leo A, Larsimont D. The predictive value of HER2 in breast cancer. Oncology 2001;61(Suppl 2):73-82. (PMID: 11694791) PubMed 14 Yarden Y. Biology of HER2 and its importance in breast cancer. Oncology 2001;61(Suppl 2):1-13. (PMID: 11694782) PubMed 15 Slamon D, Eiermann W, Robert N, Pienkowski T, Martin M, Press M, et al; Breast Cancer International Research Group. Adjuvant trastuzumab in HER2-positive breast cancer. N Engl J Med 2011; 365: 1273-1283. (PMID: 21991949) PubMed 16 Konecny G, Pauletti G, Pegram M, Untch M, Dandekar S, Aguilar Z, et al. Quantitative association between HER-2/neu and steroid hormone receptors in hormone receptor-positive primary breast cancer. J Natl Cancer Inst 2003; 95: 142-153. (PMID: 12529347) PubMed 17 Eiermann W, Rezai M, Kümmel S, Kühn T, Warm M, Friedrichs K, et al. The 21-gene recurrence score assay impacts adjuvant therapy recommendations for ER-positive, node-negative and node-positive early breast cancer resulting in a risk-adapted change in chemotherapy use. Ann Oncol 2013; 24: 618-624. (PMID: 23136233) PubMed 18 Coates AS, Winer EP, Goldhirsch A, Gelber RD, Gnant M, Piccart-Gebhart M, et al; Panel Members. Tailoring therapies--improving the management of early breast cancer: St Gallen International Expert Consensus on the Primary Therapy of Early Breast Cancer 2015. Ann Oncol 2015; 26: 1533-1546. (PMID: 25939896) PubMed 19 Xin L, Liu YH, Martin TA, Jiang WG. The era of multigene panels comes? The clinical utility of oncotype dx and mammaPrint. World J Oncol 2017; 8: 34-40. (PMID: 29147432) PubMed 20 Goncalves R, Bose R. Using multigene tests to select treatment for early-stage breast cancer. J Natl Compr Canc Netw 2013; 11: 174-182; quiz 182. (PMID: 23411384) PubMed 21 Mittendorf EA, Chavez-MacGregor M, Vila J, Yi M, Lichtensztajn DY, Clarke CA, et al. Bioscore: a staging system for breast cancer patients that reflects the prognostic significance of underlying tumor biology. Ann Surg Oncol 2017; 24: 3502-3509. (PMID: 28726077) PubMed 22 Weiss A, Chavez-MacGregor M, Lichtensztajn DY, Yi M, Tadros A, Hortobagyi GN, et al. Validation study of the American Joint Committee on cancer eighth edition prognostic stage compared with the anatomic stage in breast cancer. JAMA Oncol 2018; 4: 203-209. (PMID: 29222540) PubMed 23 Giuliano AE, Connolly JL, Edge SB, Mittendorf EA, Rugo HS, Solin LJ, et al. Breast cancer-major changes in the American Joint Committee on Cancer eighth edition cancer staging manual. CA Cancer J Clin 2017; 67: 290-303. Erratum in: CA Cancer J Clin 2017; 67: 345. (PMID: 28294295) PubMed 24 Horgan D, Ciliberto G, Conte P, Baldwin D, Seijo L, Montuenga LM, et al. Bringing greater accuracy to europe's healthcare systems: the unexploited potential of biomarker testing in oncology. Biomed Hub 2020; 5: 182-223. (PMID: 33564664) PubMed About Us Aims and Scope Editorial Board Instructions to Authors Contact Us 2024 ©️ Galenos Publishing House
15573	https://www.youtube.com/watch?v=8HxvX-203Ws	Prove that in triangle ABC,. Area(ABC)=abc/4R Doctor of Mathematics 24200 subscribers 62 likes Description 4030 views Posted: 19 Feb 2022 Prove that in triangle ABC,. Area(ABC)=abc/4R Solution of triangle 15 comments Transcript: this problem is related to documentary um name of chapter is solution of triangle prove that in a triangle abc area of triangle is equal to abc upon 4 r we know that area of any triangle we can find it is as um [Music] name but as is to abc and for finding the area of triangle abc we have to find the base and altitude we know that for angle a in front side is is denoted by small aim and also from for angle c its opposite side is a small c and for this angle opposite side is small b we know it properly um if we want to area of triangle a abc which is equal to 1 by 2 base into altitude based into altitude you can take a any base now we i take a visa base and so perpendicular from c to a b is its altitude i write this point to d which is perpendicular from c to on a b where to reduce c d um because a b is equal to small c so i write 1 by 2 you know c into a small c here and this is capital c this point c and c is small c is the distance from a point to b we have to now find the value of c d that is altitude and this value we can find by using ah definition of sine this is equal to area of triangle c i write it it is as equation number one and for finding the value of c d we have to go in triangle c d a which is a right angle triangle at right angle at d in triangle um c d a i apply sine a we know that sine is ratio of two sides in which in numerator is opposite side of this angle which is called perpendicular perpendicular cd upon hypotenuse which is opposite to the 90 degree angle because which is here ac so we can find by this distance cd which we will use in equation number one cd upon ac but here we are saying that ac is a distance which is denoted by small b so we can on multiplying b to right hand side on right hand side we get b sine a which is equal to c d so we put the value of c d in equation number one and we get putting this value this value of cd in equation 1 we get area of a triangle a b c is equal to 1 by 2 i write it from equation number 1 by 2 into c and value of c d which we find here b sine a so we can write it as 1 by 2 bc sine a but we know that uh by sine rule [Music] by law of sine science in drawing for triangle abc a upon sine a equal to b upon sine b it is formula beta should we must be learned by heart um this is a small c with resistance from a point to b point and it is angle c between side uh ac and bc this value this ratio all ratio equal to 2r uh we have um to find the value of sine a from here i write it it as a equation number two and i will write put the value of sine a from this law of science so first i need sine only so i write a upon sine first equal to ah fourth because all are equal to each other so i write first equal equal to four take only these 2 and find the value by transferring these two terms and factors so a upon 2 are equal to sine a by transferring these terms and putting this value of sine a in equation number two putting this value of sine a in equation number two putting this value this value in equation number 2 we get area of triangle abc equal to 1 by 2 bc i write it from equation number two area of triangle absolute 1 by 2 bc into sine a i write a value of sine sine a from here which we find here e upon 2r therefore we have to find uh area of triangle a bc is equal to uh a abc abc upon 2 into 22 equal to 4 r which was to be proved uh prove which was to be proved quite intimate random thank you very much indeed for listening to little
15574	https://math.stackexchange.com/questions/524890/proving-na-%E2%88%AA-b-na-nb-%E2%88%92-na-%E2%88%A9-b	elementary set theory - Proving $N(A ∪ B) = N(A) + N(B) − N(A ∩ B)$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Proving N(A∪B)=N(A)+N(B)−N(A∩B)N(A∪B)=N(A)+N(B)−N(A∩B) Ask Question Asked 11 years, 11 months ago Modified11 years, 11 months ago Viewed 632 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Let A A and B B be finite sets. N(A) is the number of elements in the set A. I need to prove that: N(A∪B)=N(A)+N(B)−N(A∩B)N(A∪B)=N(A)+N(B)−N(A∩B) Using the following representation of A ∪ B as a union of disjoint sets: A∪B=(A−(A∩B))∪B A∪B=(A−(A∩B))∪B I'm not sure where I would start with this... any help is appreciated! elementary-set-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Oct 13, 2013 at 18:44 Brian M. Scott 633k 57 57 gold badges 824 824 silver badges 1.4k 1.4k bronze badges asked Oct 13, 2013 at 18:41 Eric JEric J 11 1 1 silver badge 1 1 bronze badge 2 You were missing a required pair of parentheses in the second displayed line; I’ve added them.Brian M. Scott –Brian M. Scott 2013-10-13 18:44:36 +00:00 Commented Oct 13, 2013 at 18:44 @Newb Did you even read the post? =)Pedro –Pedro♦ 2013-10-13 18:49:28 +00:00 Commented Oct 13, 2013 at 18:49 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. HINTS: I don’t want to say very much, since this is an extremely straightforward problem, but these two questions might get you thinking in the right direction: If the finite sets X X and Y Y are disjoint, what is N(X∪Y)N(X∪Y) in terms of N(X)N(X) and N(Y)N(Y)? If X X is finite, and Y⊆X Y⊆X, what is N(X∖Y)N(X∖Y) in terms of N(X)N(X) and N(Y)N(Y)? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 13, 2013 at 18:46 Brian M. ScottBrian M. Scott 633k 57 57 gold badges 824 824 silver badges 1.4k 1.4k bronze badges 11 I've never studied sets in my life and need to learn them now. I know the logical solution to this. Since obviously A union B counts the intersection of A and B twice. However, I am not sure how to use this fact: A∪B=(A−(A∩B))∪B Eric J –Eric J 2013-10-13 21:10:35 +00:00 Commented Oct 13, 2013 at 21:10 @Eric: Forget proving anything for a moment; can you simply answer the two questions that I asked?Brian M. Scott –Brian M. Scott 2013-10-13 21:13:10 +00:00 Commented Oct 13, 2013 at 21:13 If they are disjoint N(X)+N(Y) and N(X\Y) is N(X)-N(Y)?Eric J –Eric J 2013-10-13 21:19:16 +00:00 Commented Oct 13, 2013 at 21:19 @Eric: Yes, that’s exactly right. And I’m fairly sure that you’re not expected to prove those two facts, unless your course has dealt pretty rigorously with the notion of finite cardinalities. Can you see how to use those facts to calculate N(A∖(A∩B)N(A∖(A∩B) and then the desired result?Brian M. Scott –Brian M. Scott 2013-10-13 21:21:35 +00:00 Commented Oct 13, 2013 at 21:21 So from (A−(A∩B))∪B; is it not just (A−(A∩B))+B which is A + B - (A∩B) ?Eric J –Eric J 2013-10-13 21:26:02 +00:00 Commented Oct 13, 2013 at 21:26 \|Show 6 more comments 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 See similar questions with these tags. 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15575	https://www.youtube.com/watch?v=nimoS70zuP0	How to Find Profit Percentage Easy Trick - Profit Percentage Formula sam tube 138000 subscribers 1801 likes Description 151686 views Posted: 12 Jan 2022 How to Calculate Profit Percentage Easy Trick - Profit Percentage Formula In this video i am showing you how to calculate the profit percentage. You can use the formula which is profit divided by cost price , multiply with 100. Profit percentage are important in every business. profitpercentage#profitlosspercent#pofit WANNA HELP OUT THE CHANNEL? Please use this affiliate Amazon link to purchase anything (at no cost to you): DISCLAIMER: The description contains affiliate links, which means that if you click on one of the product links, I will receive a small commission at NO COST to you. This helps support the channel and allows me to continue to make videos like this. Thanks for your support! 72 comments Transcript: hello guys i hope you are doing really really well in today's video i will show you how to find out profit percentage so guys let's say we have a shirt and the cost price of the shirt is 20 rupees okay cost price of shirt equals to 20 rupees or 20 dollars whatever you want okay and the selling price okay the selling price of shirt is 30 rupees okay guys and now we want to calculate how much will be the profit okay so the profit is selling price minus cost price okay so sp it means selling price cp it means cost price so we already have both so when we do selling price minus cost price so selling price is 30 and the cost price is 20 so 30 minus 20 will be the 10 so 10 rupees is our profit okay now there is one more deal so this is the first thing the second deal is let's say cost price of a trouser is equal to 40 rupees okay and the selling price of that trouser is 50 rupees okay so now again we want to calculate how much is the profit in this deal so profit again as i mentioned profit is equal to sp which is the selling price minus cp which is the cost price so selling price is 50 cost price is 40 50 minus 40 is again 10 rupees so now guys you can see there are two deals okay so this is the first deal and this is the second deal so in both of the days you see the profit is same okay which is 10 rupees but guys in both deal one of the deal make bigger profit okay so how you can calculate which deal make the bigger profit so you have to use the profit percentage formula okay so now the profit percentage come into play so the profit percentage equals to profit percentage equal to profit okay divide it with cost price multiply with 100 because we have to calculate the percentage okay so this is the formula to calculate the profit percentage so now we will put the values for uh both of the deals and we will uh figure it out and we will know that which deal makes the bigger profit so for the first deal if we put the values over here so in the profit in the first deal is 10 rupees so we put 10 over here and we divide it with the cost price so the cost price in the first deal was as you can see 20 okay so we will put here 20 we multiply it with 100 okay so when we calculate it the answer will be 50 okay so this is for the first deal 50 so i write here firstly okay so guys for the second we will again put the formula for profit percentage so the profit percentage as i already mentioned is equal to profit divide it with cost price multiply with 100 okay so in the second deal we again made a profit of 10 rupees okay and the cost price in second deal was 40 okay and so and then we multiplied with 100 okay so the answer will be 25 okay so this is for the second deal for the trouser okay and the first deal i already mentioned the profit was 50 okay guys so we we know that with this calculation that in the first deal we make the bigger profit by using the profit percentage so you can see over here we make bigger profit in the first deal in which we and the cost price of the shirt was 20 we sell it for 30 the profit was 10 but in the first deal we make a bigger profit so this is how you can calculate the profit percentage so guys this is for today's video i hope you liked the video i hope you enjoyed the video so please do subscribe to my channel thank you very much
15576	https://www.sketchy.com/mcat-lessons/ohms-law	Opens in a new window Opens an external website Opens an external website in a new window We use cookies to provide necessary functionality and improve your experience. By remaining on this website you indicate your consent. Cookie Policy Try for FreeLogin GET 20% OFF SKETCHY MCAT WITH CODE REG20 \| REGISTRATION DAY SALE MCAT Curriculum / Physics / Circuits / Ohm's Law Ohm's Law Tags: voltage current resistance Physics Ohm's Law quantifies the relationship between voltage, current, and resistance using the equation V = IR. With this knowledge, you can understand how resistors behave in electrical circuits. Resistors can be connected in series or parallel configurations, directly impacting their behavior and power dissipation. In series resistors, the resistor with the largest resistance experiences the highest voltage drop and dissipates the most power. For parallel resistors, the resistor with the smallest resistance experiences the highest current and dissipates the most power. Power in a circuit defines the rate at which energy is converted between forms, with an equation of P = IV. To simplify circuits with resistors in series or parallel, one can use equivalent resistance equations. For a series connection, equivalent resistance is the sum of the individual resistances (Req = R1 + R2 +...+ Rn). For a parallel connection, the inverse of the equivalent resistance is equal to the sum of the inverse of the individual resistances (1⁄Req = 1⁄R1 + 1⁄R2 +...+ 1⁄Rn). To measure current and voltage in a circuit, an ammeter (in series) and a voltmeter (in parallel) are used, respectively. Lesson Outline Introduction Voltage, Current, and Resistance Ohm's Law: V = IR Ohm's Law and Power Power: P = IV, measured in watts Relationship with energy: Power = Energy / Time (joules per second, J/s) Resistor Configurations Series Resistors Connected one after another along a single wire Same current through each resistor Largest resistor has highest voltage drop and dissipates most power Equivalent resistance: sum of individual resistances (Req = R1 + R2 +...+ Rn) Ammeter: device connected in series to measure current Parallel Resistors Connected with different branches from a single point Same voltage across each resistor Smallest resistor has largest current and dissipates most power Equivalent resistance: inverse of the sum of inverse of individual resistances (1⁄Req = 1⁄R1 + 1⁄R2 +...+ 1⁄Rn) Voltmeter: device connected in parallel to measure voltage Don't stop here! Get access to 29 more Physics lessons & 8 more full MCAT courses with one subscription! Try 7 Days Free FAQs What is Ohm's Law and how does it relate to voltage, current, and resistance? Ohm's Law is an essential principle in electronics that defines the relationship between the voltage, current, and resistance in a simple circuit. It states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. Mathematically, it can be represented as V = IR, where V is the voltage, I is the current, and R is the resistance. How is power calculated in a circuit using Ohm's Law? Using Ohm's Law, power in a circuit can be calculated in terms of voltage, current, and resistance. The formula for power is P = IV, where P is the power, I is the current, and V is the voltage. By substituting Ohm's Law, V = IR, into the power formula, we can find two alternative formulas: P = I2R or P = V2/R, where P is the power, I is the current, R is the resistance, and V is the voltage. How do resistors and equivalent resistance impact Ohm's Law in circuits? Resistors are passive components used in circuits to limit or control the flow of current and influence the voltage distribution by providing a specific resistance value. The equivalent resistance of a circuit, which is either the sum of individual resistances in a series configuration or the inverse sum of reciprocal resistances in a parallel configuration, is used in Ohm's Law to calculate the overall current and voltage distribution. Depending on the arrangement of resistors and the equivalent resistance value, Ohm's Law can be applied to analyze and design intricate circuits. How do you measure current and voltage in a circuit using an ammeter and a voltmeter? An ammeter is an instrument used to measure the current in a circuit, while a voltmeter measures voltage. To measure the current, the ammeter is connected in series with the component or section of the circuit whose current you want to measure. It must be turned off before making the connection to avoid potential damage. For voltage measurement, a voltmeter is connected in parallel with the component or section of the circuit you want to measure the voltage across. Both devices help to verify the accuracy of Ohm's Law calculations and determine the circuit parameters. How do you use Ohm's Law to determine an unknown circuit parameter (voltage, current, or resistance)? When analyzing a circuit, you may need to determine an unknown parameter based on given values of voltage, current, or resistance. You can use Ohm's Law to calculate the missing value. For example, if you know the voltage and resistance but not the current, you can rearrange Ohm's Law (V = IR) to solve for the current by dividing both sides by the resistance: I = V/R. Similarly, with the current and resistance values, you can find the voltage using V = IR, and with the voltage and current values, you can find the resistance using R = V/I. This helps to understand the behavior of a circuit and troubleshoot or optimize its performance. Programs MedicalMCATPANP
15577	https://www.ncbi.nlm.nih.gov/books/NBK560653/	An official website of the United States government 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 in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Cefotaxime Inderbir S. Padda; Shivaraj Nagalli. Author Information and Affiliations Authors Inderbir S. Padda1; Shivaraj Nagalli2. Affiliations 1 Richmond University Medical Center/Mount Sinai 2 Yuma Regional Medical Center Last Update: July 10, 2023. Continuing Education Activity Cefotaxime is a medication used to manage and treat cervicitis/urethritis and pneumonia. Cefotaxime is a beta-lactam antibiotic classified as a third-generation cephalosporin. Its broad-spectrum antibacterial activity is useful in treating the susceptible strains of bacteria affecting the lower respiratory tract, genito-urinary tract, central nervous system, intra-abdominal infections, bone and joint infections, skin infections, gynecologic infections, and septicemia. This activity outlines the indications, mechanism of action, and contraindications for cefotaxime is a valuable agent in treating and managing bacterial infections. This activity will highlight the mechanism of action, adverse event profile, resistance, and other key factors pertinent to healthcare team members in the care of patients with pneumonia and urethritis/cervicitis and related conditions. Objectives: Describe the pathophysiology of cefotaxime resistance. Identify the most common adverse events associated with cefotaxime therapy. Outline the mechanism of action of cefotaxime. Review some interprofessional team strategies for improving care coordination and communication to advance cefotaxime and improve outcomes. Access free multiple choice questions on this topic. Indications Cefotaxime is a beta-lactam antibiotic classified as a third-generation cephalosporin, which was first synthesized in 1976 and is FDA approved to treat gram-positive, gram-negative, and anaerobic bacteria. Its broad-spectrum antibacterial activity is useful in treating the susceptible strains of bacteria affecting the lower respiratory tract, genito-urinary tract, central nervous system, intra-abdominal infections, bone and joint infections, skin infections, gynecologic infections, and septicemia. Cefotaxime may also be used prophylactically prior to surgery to prevent surgical infections. Among the susceptible strains, Enterobacteriaceae is particularly sensitive to cefotaxime and may treat multi-drug resistant strains of Enterobacteriaceae. Although it has a broad spectrum of bactericidal activity, it is not as effective against pseudomonas aeruginosa infections compared to other third-generation antibiotics and is not recommended as monotherapy. Intramuscular treatment with cefotaxime for sexually transmitted infections with Neisseria gonorrhoeae has shown a positive outcome in both men and women. Cefotaxime has beneficial therapeutic activity treating pneumonia affecting the lower respiratory tract primarily caused by Gram-negative bacilli, and the bactericidal agent has shown significant efficacy in treating complicated infections affecting the urinary tract. Compared with the other cephalosporins, a favorable characteristic of cefotaxime is that it does not cause a notable occurrence of coagulopathies and pseudocholelithiasis. Trials comparing cefotaxime with the third-generation cephalosporin ceftriaxone have exhibited similar clinical efficiency. Clinical trials have also shown 75% to 100% resolution in hospitalized patients with moderate to severe infections. Cefotaxime may also be interchangeable with ceftriaxone as off-label use for the treatment of endocarditis by Haemophilus parainfluenzae, H. aphrophilus, Actinobacillus actinomycetemcomitans, Cardiobacterium hominis, Eikenella corrodens, and Kingella kingae(HACEK) organisms. Cefotaxime can readily cross the blood-brain barrier when administered intravenously and may treat gram-negative infections resistant to previous generations of cephalosporins. Susceptible Organisms Gram-positive bacteria Enterococcus spp Staphylococcus aureus Staphylococcus epidermidis Streptococcus pneumoniae Streptococcus pyogenes Streptococcus viridans spp Anaerobic bacteria Bacteroides spp. Clostridium spp Fusobacterium spp Peptococcus spp Peptostreptococcus spp Gram-negative bacteria Acinetobacter spp. Citrobacter spp Enterobacter spp Escherichia coli Haemophilus influenzae Haemophilus parainfluenzae Klebsiella spp. Morganella morganii Neisseria gonorrhoeae Neisseria meningitidis Proteus mirabilis Proteus vulgaris Providencia rettgeri Providencia stuartii Serratia marcescens Mechanism of Action Cefotaxime is a bactericidal agent that exerts its mechanism of action by binding penicillin-binding proteins (PBPs) via beta-lactam rings and inhibiting the definitive activity of transpeptidation in peptidoglycan cell wall synthesis of susceptible bacterial organisms. Its action demonstrates a great affinity for PBP Ib and PBP III cell wall proteins. The inability to form a bacterial cell wall further causes the autolysis of the bacteria. Similarly to other third-generation cephalosporins, its broad spectrum action makes it efficacious against gram-positive and gram-negative bacteria. Resistance Beta-lactamases can cause hydrolysis to cefotaxime, further hindering its bactericidal effects. Although susceptive, cefotaxime is quite durable against the activity of most β-lactamases. Metabolism Once administered, cefotaxime undergoes metabolism within the liver, and the majority of it is excreted renally. In the liver, cefotaxime converts to desacetylcefotaxime, which is further converted to desacetylcefotaxime lactone and then to M metabolites. More than 80% is recovered in the urine, with one-third being in the form of desacetylcefotaxime (des-CTX). Although desacetylcefotaxime (des-CTX) is the major metabolite of cefotaxime, its activity is eight-fold weaker than cefotaxime. Administration Cefotaxime is available and distributed in powder form and as a premixed solution for intramuscular and intervenous administration. The powder form is available in 500 mg, 1 g, 2 g, and 10 g vials. The premixed solution is available as 1g and 2g for injection. Gonococcal Infections Urethritis (Males): 0.5 g intramuscular injection (can be administered as a single dose) Cervicitis (Females): 0.5 g intramuscular injection (can be administered as a single dose) Rectal infection (Males): 1 g intramuscular injection (can be administered as a single dose) Rectal infection( Females): 0.5 g intramuscular injection (can be administered as a single dose) Cefotaxime has no coverage for Chlamydia trachomatis, and treatment should be added if this organism is suspected. Septicemia 2 g I.V. every 6 to 8 hours. (Daily dose of 6 to 8 grams) Spontaneous Bacterial Peritonitis (SBP) Cefotaxime is the drug of choice in patients with SBP due to its ability to achieve excellent levels in the blood and ascitic fluid. The typical dose in SBP would be 2 g intravenous every 8 hours. Uncomplicated Infections 1 g intramuscular or IV every 12 hours. (Daily dose of 2 grams) Moderate to Severe Infections 1 to 2 g intramuscular or IV every 8 hours. (Daily dose of 3 to 6 grams) Life-threatening Infections 2 g intramuscular or IV every 4 hours. (Daily dose of 12 grams) Cesarean Section First dose: 1 g IV (Umbilical cord should be clamped) Second dose: 1 g intramuscular or IV (six hours after the first dose) Third dose: 1 g intramuscular or IV (twelve hours after the first dose) Surgery Prophylaxis 1 g intramuscular or IV 30 minutes before surgery. Neonates (age 0 to 4 weeks) (Age 0 to 1 week) 50 mg/kg per dose IV every 12 hours (Age 1 to 4 weeks) 50 mg/kg per dose IV every 8 hours Infants and Children (age 1 month to 12 years old) 50 to 180 mg/kg intramuscular or IV every 6 to 8 hours (for individuals with body weight <50kg) 1 to 2 grams intramuscular or IV every 8 hours. (for individuals with body weight >50kg) Individuals with a body weight>50 kg should follow adult dosing. The daily dosage should not exceed 12 grams for infants and children. Adverse Effects Local reaction: pain, swelling Hypersensitivity: rash, pruritis, anaphylaxis Gastrointestinal effects: nausea, vomiting, diarrhea Pseudomembranous Colitis Headache Elevation in liver enzymes Elevation in BUN and creatinine Hematologic: Neutropenia, leukopenia, agranulocytosis Local reactions such as pain, swelling, and rash are the most common adverse effects following cefotaxime administration. Like other cephalosporins, cefotaxime does not cause disulfiram-like reactions. Cefotaxime used concurrently with nephrotoxic agents may promote nephrotoxic effects on the kidney, and such use requires caution. Patients with hypersensitivity to the cephalosporin or penicillin group may result in an anaphylactic reaction and are manageable with epinephrine, antihistamines, vasopressors, or corticosteroids. Contraindications Hypersensitivity to cefotaxime is an absolute contraindication to its use. Patients with known allergies to penicillin or other cephalosporins should also avoid cefotaxime. Monitoring Cefotaxime administration and dosing require adjusting in geriatric populations, patients with decreased renal function, and hepatic dysfunction. Renal function and liver enzymes require routine monitoring. The half-life of cefotaxime is generally one hour, and severe kidney dysfunction may prolong the half-life of cefotaxime and its metabolite desacetylcefotaxime. CBC should also be monitored with cefotaxime use as there are reports of hematologic changes such as neutropenia, leukopenia, and agranulocytosis. Cefotaxime, like other cephalosporins, may also cause a false positive direct coombs test. Cefotaxime is an FDA Pregnancy Category B drug. Cefotaxime use in pregnancy has not been studied clearly and should be used cautiously. Cefotaxime is reported to cross the placenta during pregnancy. It is also present in low concentrations in breast milk during lactation. Toxicity Cefotaxime is metabolized by the liver and excreted through the kidneys, and dysfunctions may result in decreased drug clearance leading to increased plasma concentrations. About 50 to 60% of the agent is excreted unchanged, and 15 to 20% is excreted as a desacetyl metabolite desacetylcefotaxime. Toxicity may result in convulsions, dyspnea, hypothermia, cyanosis, reversible encephalopathy, and death. Mortality has occurred with dosages of 6000 mg/kg/day. Treatment for cefotaxime toxicity requires supportive management. Enhancing Healthcare Team Outcomes Cefotaxime is a broad-spectrum antibiotic that is FDA-approved and indicated to treat gram-positive, gram-negative, and anaerobic organisms of susceptible strains causing pneumonia, urinary tract infections, cervicitis, endometritis, urethritis, and sepsis. The care for patients suffering from infectious diseases prompts critical care from an interprofessional team of healthcare professionals, as preventable contagious disorders can lead to medication resistance, complications, and mortality. These healthcare professionals include a primary care physician, an internist, an infectious disease specialist, critical care, a gynecologist, a nurse, and a pharmacist. Primary care clinicians, internists, and specialists should educate the patients about the consequences of non-compliance with therapy for the full duration and how resistance to treatment can further cause complications and result in mortality. The primary care physician should routinely monitor renal function, liver enzymes, and CBC as cefotaxime is metabolized and cleared in the liver and kidneys, respectively, and has also been shown to cause hematologic adverse effects. Cefotaxime should be renally dosed in patients with compromised renal function, such as CKD or ESRD, and patients receiving hemodialysis. Patients developing diarrhea while receiving treatment with antibiotics should be assessed for Clostridium difficile infection. Colonic flora is changed when receiving treatment with antibiotics, making it susceptible to Clostridium difficile infection resulting in mild to severe forms of diarrhea. Diagnostics and treatment focused on Clostridium difficile, electrolyte, and volume depletion should be initiated, and discontinuing management with cefotaxime should be considered. Counseling and careful monitoring are necessary during pregnancy, as clinical studies during its use in pregnancy are limited, and cefotaxime FDA pregnancy category B. Cefotaxime is reported to also be present in breastmilk in low amounts, and infants should be monitored accordingly. Physicians should be up to date with the newly FDA-approved cefotaxime indications dosing, and their effects in the event drug resistance does develop. During the treatment of gonorrhea causing urethritis or cervicitis, treatment for chlamydia should be added as cefotaxime does not have coverage for this organism. An interprofessional healthcare team approach to antimicrobial care with cefotaxime involving collaborative interventions and communication is key to building patient rapport and developing a therapeutic alliance so the patients comply with therapy adequately to eradicate the bacteria and prevent further spread. Continued communication and teamwork between healthcare professionals will improve antimicrobial stewardship, improve patient outcomes, limit microbial resistance, and lower the incidence of multidrug-resistant organisms. Review Questions Access free multiple choice questions on this topic. Comment on this article. References 1. : Dudley MN, Barriere SL. Cefotaxime: microbiology, pharmacology, and clinical use. Clin Pharm. 1982 Mar-Apr;1(2):114-24. [PubMed: 6309465] 2. : Todd PA, Brogden RN. Cefotaxime. An update of its pharmacology and therapeutic use. Drugs. 1990 Oct;40(4):608-51. [PubMed: 2083516] 3. : Carmine AA, Brogden RN, Heel RC, Speight TM, Avery GS. Cefotaxime. A review of its antibacterial activity, pharmacological properties and therapeutic use. Drugs. 1983 Mar;25(3):223-89. [PubMed: 6303743] 4. : Plosker GL, Foster RH, Benfield P. Cefotaxime. A pharmacoeconomic review of its use in the treatment of infections. Pharmacoeconomics. 1998 Jan;13(1 Pt 1):91-106. [PubMed: 10175990] 5. : Brogden RN, Spencer CM. Cefotaxime. A reappraisal of its antibacterial activity and pharmacokinetic properties, and a review of its therapeutic efficacy when administered twice daily for the treatment of mild to moderate infections. Drugs. 1997 Mar;53(3):483-510. [PubMed: 9074846] 6. : Baddour LM, Wilson WR, Bayer AS, Fowler VG, Bolger AF, Levison ME, Ferrieri P, Gerber MA, Tani LY, Gewitz MH, Tong DC, Steckelberg JM, Baltimore RS, Shulman ST, Burns JC, Falace DA, Newburger JW, Pallasch TJ, Takahashi M, Taubert KA., Committee on Rheumatic Fever, Endocarditis, and Kawasaki Disease. Council on Cardiovascular Disease in the Young. Councils on Clinical Cardiology, Stroke, and Cardiovascular Surgery and Anesthesia. American Heart Association. Infectious Diseases Society of America. Infective endocarditis: diagnosis, antimicrobial therapy, and management of complications: a statement for healthcare professionals from the Committee on Rheumatic Fever, Endocarditis, and Kawasaki Disease, Council on Cardiovascular Disease in the Young, and the Councils on Clinical Cardiology, Stroke, and Cardiovascular Surgery and Anesthesia, American Heart Association: endorsed by the Infectious Diseases Society of America. Circulation. 2005 Jun 14;111(23):e394-434. [PubMed: 15956145] 7. : Bui T, Patel P, Preuss CV. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Feb 17, 2024. Cephalosporins. [PubMed: 31855361] 8. : Lassman HB, Coombes JD. Metabolism of cefotaxime: a review. Diagn Microbiol Infect Dis. 1984 Jun;2(3 Suppl):3S-12S. [PubMed: 6086216] 9. : LeFrock JL, Prince RA, Leff RD. Mechanism of action, antimicrobial activity, pharmacology, adverse effects, and clinical efficacy of cefotaxime. Pharmacotherapy. 1982 Jul-Aug;2(4):174-84. [PubMed: 6302641] 10. : Coombes JD. Metabolism of cefotaxime in animals and humans. Rev Infect Dis. 1982 Sep-Oct;4 Suppl:S325-32. [PubMed: 6294781] 11. : Chin NX, Neu HC. Cefotaxime and desacetylcefotaxime: an example of advantageous antimicrobial metabolism. Diagn Microbiol Infect Dis. 1984 Jun;2(3 Suppl):21S-31S. [PubMed: 6086215] 12. : Drugs and Lactation Database (LactMed®) [Internet]. National Institute of Child Health and Human Development; Bethesda (MD): Aug 15, 2023. Cefotaxime. [PubMed: 30000425] : Disclosure: Inderbir Padda declares no relevant financial relationships with ineligible companies. : Disclosure: Shivaraj Nagalli declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK560653PMID: 32809488 Share Views PubReader Print View Cite this Page Padda IS, Nagalli S. Cefotaxime. [Updated 2023 Jul 10]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. In this Page Continuing Education Activity Indications Mechanism of Action Administration Adverse Effects Contraindications Monitoring Toxicity Enhancing Healthcare Team Outcomes Review Questions References Related information PubMed Links to PubMed Recent Activity Clear)Turn Off)Turn On) Cefotaxime - StatPearls Cefotaxime - StatPearls Your browsing activity is empty. 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15578	https://flexbooks.ck12.org/cbook/ck-12-cbse-chemistry-class-12/section/4.1/primary/lesson/electronic-configurations-and-physical-properties-of-the-d-block-elements/	Skip to content Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 4.1 General Properties of the d-Block Elements Written by:Divya Ladha Fact-checked by:The CK-12 Editorial Team Last Modified: Jul 04, 2025 Use of Transition Metals in Coloured Glass Transition metals are used in the manufacturing of coloured glass. Transition metal ions such as cobalt (Co²⁺), chromium (Cr³⁺), and iron (Fe³⁺) are commonly used as pigments because of their ability to absorb specific wavelengths of light due to d-d electronic transitions. For example, cobalt gives glass a deep blue colour, while chromium imparts green or red shades depending on its oxidation state. These vivid colours are not only aesthetically pleasing but also stable at high temperatures, making them ideal for stained glass windows, decorative pottery, and even traffic light lenses. This application directly arises from the unique electronic configurations of transition metals, which enable them to exhibit a wide range of colours. Introduction to d-Block Elements The periodic table is systematically organised into blocks based on the type of atomic orbitals being filled. The d-block of the periodic table encompasses elements from Group 3 to Group 12, where the d-orbitals are progressively filled across each of the four long periods. These elements are commonly referred to as transition metals. Historically, this name was derived from the observation that their chemical properties were "transitional" or intermediate between those of the highly reactive s-block elements (alkali and alkaline earth metals) and the diverse p-block elements. However, the International Union of Pure and Applied Chemistry (IUPAC) has provided a more precise and chemically rigorous definition: transition metals are defined as metals which possess an incomplete d-subshell either in their neutral atomic state or in any of their common and stable ionic states. This definition distinguishes them from certain elements in Group 12, namely zinc (Zn), cadmium (Cd), and mercury (Hg). These elements, although located in the d-block, have a completely filled d10 electronic configuration in their ground state (n-1)d10 ns2 as well as in their common and stable oxidation states (e.g., Zn2+ has a 3d10 configuration). Therefore, according to the strict IUPAC definition, they are not considered true transition metals. Nevertheless, their chemical behaviour often shows similarities to that of transition metals, and they typically mark the end of their respective d-series, leading to their chemistry being studied alongside that of the transition metals. Series of Transition Metals The elements of the d-block are organised into four distinct series, corresponding to the principal quantum number (n) of the d-orbital being filled: 3d series: This series begins with scandium (Sc, Z=21) and ends with zinc (Zn, Z=30). These are some of the most commonly encountered transition metals, including iron, copper, and chromium. 4d series: This series includes elements from yttrium (Y, Z=39) to cadmium (Cd, Z=48). Examples include molybdenum and silver. 5d series: This series is unique as it includes lanthanum (La, Z=57), followed by elements from hafnium (Hf, Z=72) to mercury (Hg, Z=80). Notable examples are gold and platinum. 6d series: This series starts with actinium (Ac, Z=89) and includes elements from rutherfordium (Rf, Z=104) to copernicium (Cn, Z=112). These are mostly synthetic and highly radioactive elements. Electronic Configuration of d-Block Elements The general electronic configuration of the outer orbitals for d-block elements is (n-1)d1-10ns1-2. In this notation, (n-1) refers to the penultimate (inner) d-orbitals, which can hold one to ten electrons, while the outermost @$\begin{align}ns\end{align}@$ orbital typically contains one or two electrons. It's crucial to remember that when d-block elements form ions, the electrons from the @$\begin{align}ns\end{align}@$ orbital are removed before those from the (n-1)d orbitals. However, this general rule has its exceptions. These deviations arise due to the very small energy difference between the (n-1)d and ns orbitals. This minimal energy gap allows for an electron to move from the ns orbital to the (n-1)d orbital, leading to more stable electronic configurations that are either half-filled (@$\begin{align}d^5\end{align}@$ ) or completely filled @$\begin{align}(d^{10})\end{align}@$. This enhanced stability is a consequence of factors such as increased exchange energy (resulting in more possible pairs of parallel spins in degenerate orbitals) and symmetrical distribution of electron density. Chromium (Cr) has atomic number (Z) of 24. Based on Aufbau principle, the expected electronic configuration should be @$\begin{align}[Ar]3d^4 4s^2\end{align}@$. However, its actual electronic configuration is @$\begin{align}[Ar]3d^5 4s^1\end{align}@$. An electron from the 4s orbital jumps to the 3d orbital to achieve a more stable half-filled 3d subshell @$\begin{align}(3d^5)\end{align}@$. This configuration, with five parallel spins, maximises exchange energy and thus enhances stability. Copper (Cu) has atomic number of 29. Its expected electronic configuration as per the Aufbau principle is [Ar]3d94s2; however copper exhibits a configuration of [Ar]3d104s1. Here, an electron from the 4s orbital moves to the 3d orbital to achieve a highly stable completely filled 3d subshell (3d10). Palladium (Pd), Z = 46 (from 4d series) is an even more striking exception. Its expected configuration would be [Kr]4d8 5s2. However, its actual configuration is 4d10 5s0. Both 5s electrons migrate to the 4d orbital, resulting in a completely filled 4d subshell, demonstrating the profound stability associated with a d10 configuration even at the expense of an empty outermost s-orbital. The presence of these partly filled d-orbitals (or the potential for them to be partly filled in ionic states) fundamentally distinguishes transition elements from non-transition elements and gives rise to their unique set of characteristics: Variable Oxidation States: The availability of electrons in both the (n-1)d and ns orbitals, coupled with the small energy difference between them, allows for a range of electrons to be lost or shared in chemical bonding. Formation of Coloured Ions: The presence of unpaired d-electrons enables d-d electronic transitions. When white light falls on these ions, specific wavelengths are absorbed to excite an electron from one d-orbital to another d-orbital within the same subshell. The remaining unabsorbed light is transmitted, resulting in the observed colour. For example, Cu2+ ions (with a 3d9 configuration) typically form blue solutions due to the absorption of red-orange light. Tendency to Form Complexes: The vacant or partly filled d-orbitals are ideal for accepting lone pairs of electrons from ligands (molecules or ions that can donate electron pairs), leading to the formation of a vast array of complex (coordination) compounds. For example, Fe3+ ions form a blood-red complex with thiocyanate ions, [Fe(SCN)6]3-. Catalytic Properties: Many transition metals and their compounds act as excellent catalysts in various chemical reactions, both in homogeneous and heterogeneous phases. This is attributed to their ability to exhibit multiple oxidation states and form unstable intermediate compounds, which provide alternative reaction pathways with lower activation energies. A common example is Vanadium(V) oxide (V2O5) used in the Contact Process for sulfuric acid manufacturing. Paramagnetic Behaviour: The presence of unpaired electrons in the d-orbitals leads to paramagnetic behaviour, meaning they are weakly attracted to a magnetic field. The strength of paramagnetism increases with the number of unpaired electrons. For instance, Ti3+ (3d1) is paramagnetic, while Ti4+ (3d0) is diamagnetic. General Properties of Transition Elements (d-Block) Transition elements exhibit a remarkable array of physical and chemical properties that are largely influenced by their electronic configurations and the nature of their metallic bonding. Most transition elements display characteristic metallic properties: High Tensile Strength, Ductility, and Malleability: Their strong metallic bonding, involving delocalised electrons shared across many atoms, contributes to these properties. This allows metals like silver (Ag) and copper (Cu) to be drawn into wires (ductility) or hammered into sheets (malleability). High Thermal and Electrical Conductivity: The presence of highly mobile, delocalised valence electrons allows for efficient conduction of heat and electricity. Copper (Cu), with its excellent conductivity, is widely used in electrical wiring. Metallic Lustre: The interaction of light with the delocalised electrons gives them a characteristic shiny appearance. High Melting and Boiling Points: With the notable exceptions of Zn, Cd, and Hg (due to their completely filled d-orbitals, which reduces the number of available electrons for strong metallic bonding), transition metals are very hard and possess exceptionally high melting and boiling points. This characteristic is primarily attributed to the involvement of a greater number of electrons from both the (n-1)d and ns orbitals in forming strong interatomic metallic bonds. For example, Tungsten (W) from the 5d series has the highest melting point of all metals (around 3422 °C), making it ideal for high-temperature applications like light bulb filaments and arc welding electrodes. The trend in melting points, as shown in the Figure below, generally rises to a maximum around the middle of each series (typically at the d5 configuration, such as Cr, Mo, W) because this configuration allows for the maximum number of unpaired electrons to participate in metallic bonding. Subsequently, the melting points decrease as the d-orbitals become more filled and the number of unpaired electrons reduces. High Enthalpies of Atomisation: This property quantifies the energy required to break one mole of metallic bonds to form gaseous atoms. Transition metals exhibit significantly high enthalpies of atomisation. This directly reflects the strength of their metallic bonding. The greater the number of unpaired d-electrons available for bonding, the stronger the interatomic forces. For example, molybdenum (Mo), with its [Kr]4d5 5s1 electronic configuration, has a very high enthalpy of atomisation, contributing to its exceptional strength and use in high-strength steels and superalloys for aircraft components. Variation in Atomic and Ionic Radius Across a given series (horizontal row) of transition elements, there is a progressive, albeit slight, decrease in atomic and ionic radii with increasing atomic number as shown in the Figure below. This trend is primarily due to the increasing nuclear charge (one proton added for each successive element) that pulls the electrons closer to the nucleus. While new electrons are added to the (n-1)d orbitals, the shielding effect of these d-electrons is not as effective in compensating for the increased nuclear charge as compared to electrons in s or p orbitals. Consequently, the effective nuclear charge experienced by the outermost electrons increases gradually, leading to a slight contraction of the atomic and ionic radii. An intriguing and highly significant phenomenon occurs when comparing the atomic radii of elements from different series: There is an expected increase in atomic radii when moving from the 3d to the 4d series, as a new principal energy shell is added. However, the atomic radii of the 5d series elements are virtually the same as those of the corresponding 4d series elements. Reference image: This unexpected similarity in size between the 4d and 5d series is attributed to Lanthanoid Contraction. This effect arises because the 4f orbitals are filled before the 5d series begins (from Cerium to Lutetium). The 4f electrons provide very poor shielding of the nuclear charge due to their diffuse shapes and limited penetrability towards the nucleus. As a result, the increasing nuclear charge along the lanthanoid series (from Z=58 to Z=71) causes a significant and cumulative contraction in the size of the atoms. This "lanthanoid contraction" largely compensates for the expected increase in atomic size when moving from the 4d to the 5d series. For example, zirconium (Zr) (4d series) and hafnium (Hf) (5d series) have nearly identical atomic radii (Zr: 160 pm, Hf: 159 pm). This remarkable similarity in size leads to very similar physical and chemical properties between corresponding elements of the 4d and 5d series, making their separation in geological samples and industrial processes quite challenging. This contraction affects all subsequent elements in the periodic table, including the 6d series, making gold (Au) and silver (Ag) exhibit similar noble characteristics, and influencing the chemistry of post-lanthanoid elements. Ionisation Enthalpies The ionisation enthalpy (IE) is the energy required to remove an electron from a gaseous atom. It generally increases across each series of transition elements from left to right. This is primarily due to the increasing effective nuclear charge that pulls the electrons more strongly. However, there can be certain irregularities in the ionisation enthapies such as vanadium first ionisation enthalpy is less than titanium. However, a key difference from p-block elements is that the increase in successive ionisation enthalpies for transition elements is not as steep. This is because the added d-electrons effectively shield the outer ns electrons from the increasing nuclear charge to some extent, leading to a less rapid increase in ionisation energies. It's crucial to remember that when d-block elements form ions, the electrons from the ns orbital are removed before those from the (n-1)d orbitals. For example, when iron (Fe, 3d6 4s2) forms Fe2+ ions, the two 4s electrons are removed first, resulting in a 3d6 configuration. The Table below depicts the first three ionization enthalpies of 3d, 4d and 5d series elements. \| Element \| Series \| IE₁ (kJ/mol) \| IE₂ (kJ/mol) \| IE₃ (kJ/mol) \| --- --- \| Sc \| 3d \| 631 \| 1235 \| 2388 \| \| Ti \| 3d \| 658 \| 1309 \| 2652 \| \| V \| 3d \| 650 \| 1414 \| 2830 \| \| Cr \| 3d \| 653 \| 1590 \| 2987 \| \| Mn \| 3d \| 717 \| 1509 \| 3248 \| \| Fe \| 3d \| 762 \| 1561 \| 2957 \| \| Co \| 3d \| 760 \| 1648 \| 3232 \| \| Ni \| 3d \| 737 \| 1753 \| 3395 \| \| Cu \| 3d \| 745 \| 1958 \| 3555 \| \| Zn \| 3d \| 906 \| 1733 \| 3833 \| \| Y \| 4d \| 600 \| 1180 \| 1980 \| \| Zr \| 4d \| 640 \| 1270 \| 2210 \| \| Nb \| 4d \| 652 \| 1380 \| 2416 \| \| Mo \| 4d \| 684 \| 1560 \| 2618 \| \| Tc \| 4d \| 702 \| 1470 \| 2850 \| \| Ru \| 4d \| 710 \| 1600 \| 2820 \| \| Rh \| 4d \| 720 \| 1740 \| 2990 \| \| Pd \| 4d \| 804 \| 1870 \| 3177 \| \| Ag \| 4d \| 731 \| 2070 \| 3360 \| \| Cd \| 4d \| 868 \| 1631 \| 3616 \| \| Hf \| 5d \| 658 \| 1440 \| 2250 \| \| Ta \| 5d \| 761 \| 1500 \| 2650 \| \| W \| 5d \| 770 \| 1700 \| 2700 \| \| Re \| 5d \| 760 \| 1600 \| 2730 \| \| Os \| 5d \| 840 \| 1700 \| 2800 \| \| Ir \| 5d \| 880 \| 1800 \| 2900 \| \| Pt \| 5d \| 870 \| 1791 \| 2699 \| \| Au \| 5d \| 890 \| 1980 \| 3554 \| \| Hg \| 5d \| 1007 \| 1810 \| 3300 \| There are also some interesting irregularities in the trends of ionisation enthalpies due to the stability of half-filled and completely filled d-orbitals: The first ionisation enthalpy generally increases. However, the second ionisation enthalpy for chromium (Cr) is higher than that of manganese (Mn). Cr+ has a 3d5 configuration, which is highly stable due to its half-filled nature. Removing another electron from this stable 3d5 configuration to form Cr2+ is energetically unfavourable. In contrast, Mn+ has a 3d5 4s1 configuration; removing the 4s electron to form Mn2+ results in a stable 3d5 configuration, making this step relatively easier for Mn. Similarly, the third ionisation enthalpy of manganese (Mn) is higher than that of Iron (Fe). Mn2+ has a stable 3d5 configuration, making the removal of a third electron very difficult. Whereas Fe2+ has a 3d6 configuration, and losing one electron to form Fe3+ results in a stable 3d5 configuration. This gain in stability drives the process, making the third ionisation enthalpy of iron comparatively lower. These anomalies highlight the importance of achieving stable electronic configurations in determining the energy requirements for ionization. Oxidation States One of the most defining characteristics of transition elements is their ability to exhibit a wide variety of oxidation states in their compounds. This versatility arises from two primary factors: Small Energy Difference: The energies of the (n-1)d and ns orbitals are very close, allowing electrons from both orbitals to participate in chemical bonding. This means a variable number of electrons can be lost or shared. Unpaired Electrons: The presence of multiple unpaired electrons in the d-orbitals facilitates the formation of various bonds, leading to different oxidation states. The element showing the largest number of oxidation states is typically found in or near the middle of the series, where the number of unpaired d-electrons is maximised. Manganese (Mn), for example, is the most versatile, displaying oxidation states ranging from +2 (e.g., MnSO4) to +7 (e.g., in the intensely purple potassium permanganate, KMnO4). At the extreme ends of the series, the number of oxidation states is more limited. Scandium (Sc), with its [Ar]3d1 4s2 configuration, almost exclusively exhibits a +3 oxidation state (e.g., ScCl3), as all three valence electrons are lost. Similarly, Zinc (Zn), with its 3d10 4s2 configuration, primarily exhibits a +2 oxidation state (e.g., ZnCl2), as only the two 4s electrons are involved, leaving the d10 subshell intact. The maximum oxidation state generally increases with the atomic number up to the middle of the series and then tends to decrease. For example, titanium (Ti) commonly exhibits +4 (e.g., TiO2), vanadium (V) up to +5 (e.g., V2O5), chromium (Cr) up to +6 (e.g., K2Cr2O7), and manganese (Mn) up to +7 (e.g., KMnO4). Beyond manganese, the stability of higher oxidation states generally decreases, with iron primarily showing +2 and +3, and copper +1 and +2. A key distinction from non-transition elements is that the oxidation states of transition elements frequently differ by unity (e.g., vanadium exhibits +2, +3, +4, and +5 oxidation states). In contrast, p-block elements often show oxidation states differing by units of two (e.g., Carbon in organic compounds often shows -4, -2, 0, +2, +4). Interestingly, for heavier members of a group in the d-block, higher oxidation states tend to be more stable, which is the opposite of the trend observed in the p-block (due to the "inert pair effect" in p-block). For instance, in Group 6, while chromium(VI) in the form of dichromate (Cr2O72-) in acidic medium is a strong oxidising agent (meaning it readily gets reduced). Molybdenum(VI) (MoO3) and tungsten(VI) (WO3) are significantly more stable and less oxidising. This increased stability in heavier elements is partly attributed to relativistic effects and better orbital overlap with larger atoms. Low oxidation states (like 0 or +1) are observed in complex compounds where the metal is bonded to ligands that are strong π-acceptors. These ligands can accept electron density from the filled metal d-orbitals into their empty π antibonding orbitals, forming a synergistic bond (back-bonding) that stabilises the low oxidation state. A classic example is iron pentacarbonyl (Fe(CO)5), where iron is in a 0 oxidation state. Another example is K4[Fe(CN)6], where the oxidation state of Fe is +2. Trends in Stability of Higher Oxidation States The stability of higher oxidation states in transition metals is profoundly influenced by the nature of the bonding element. The elements like fluorine and oxygen are particularly effective in stabilising high oxidation states. Halides: Fluorine, plays a critical role in stabilising the highest oxidation states of transition metals. This is due to either the high lattice energy for ionic fluorides (e.g., CoF3) or high bond enthalpy terms for covalent fluorides (e.g., VF5 and CrF6). For example, while manganese does not form a simple halide with a +7 oxidation state, the oxofluoride MnO3F is known. For vanadium (V), VF5 is a known stable halide, but other vanadium(V) halides (e.g., VCl5) are less stable or do not exist, readily undergoing hydrolysis to form oxohalides like VOCl3. Another interesting trend is the instability of lower oxidation state halides with heavier halogens; for example, CuI is stable, but CuI2 does not exist. Instead, Cu2+ readily oxidises I- to I2: @$\begin{align}2Cu^{2+}(aq) + 4I^-(aq) \rightarrow Cu_2I_2(s) + I_2(s)\end{align}@$ This demonstrates the oxidising power of Cu2+ towards larger, more easily oxidizable iodide ions. Many Cu+ compounds are unstable in aqueous solution and undergo disproportionation (@$\begin{align}2Cu^+ \rightarrow Cu^{2+} + Cu\end{align}@$ ), highlighting the greater stability of Cu2+ in aqueous solution due to its much more negative hydration enthalpy. Oxides: Oxygen is even more effective than fluorine in stabilising the highest oxidation states of transition metals. This superiority is primarily due to oxygen's ability to form multiple bonds (double or triple bonds) with metal atoms. These multiple bonds increase the stability of the compound by enhancing bond strength. The highest oxidation number in oxides typically coincides with the group number and is attained from Sc to Mn (e.g., Sc2O3 (+3), TiO2 (+4), V2O5 (+5), CrO3 (+6), Mn2O7 (+7)). Beyond group 7, the highest stable oxides show a decrease in oxidation state; for instance, for iron, the highest common oxide is Fe2O3. Besides simple oxides, oxygen also stabilises high oxidation states in oxocations (e.g., VO2+ where vanadium is in +5 state, and TiO2+ where titanium is in +4 state) and oxoanions. For example, the tetrahedral permanganate ion ([MnO4]-) and chromate ion ([CrO4]2-), both involve oxygen stabilising the highest oxidation states of manganese (+7) and chromium (+6) respectively. These compounds are powerful oxidising agents due to the high oxidation state of the central metal atom. Electrode Potentials and Stability The standard electrode potential (E°) values for transition metals provide crucial insights into their reactivity, their tendency to undergo oxidation or reduction, and the relative stability of their various oxidation states in aqueous solutions. Generally, the E° (M2+\|M) values tend to become less negative across a series from left to right. This indicates a decreasing tendency of the element to lose electrons and form M2+ ions, suggesting a relative increase in stability of the elemental form or a higher energy cost for ionisation. Copper (Cu) has a positive E°value (+0.34 V) for the Cu2+\|Cu couple. This positive value is unique among the 3d transition metals and has significant implications: It means that copper does not readily liberate hydrogen gas from dilute acids. This is because, for a metal to displace hydrogen from an acid, its standard electrode potential must be negative (i.e., it must be more easily oxidised than hydrogen). The positive E° value for copper implies that the overall energy change for the oxidation of Cu(s) to Cu2+(aq) is unfavourable. This high energy cost is primarily due to the high enthalpy of atomisation (energy required to convert solid metal to gaseous atoms) and the high sum of the first and second ionisation enthalpies of copper. These energy inputs are not sufficiently compensated by the hydration enthalpy of the Cu2+ ions (energy released when gaseous ions are solvated by water molecules). Therefore, copper only reacts with oxidising acids like nitric acid or hot concentrated sulphuric acid, where the acid itself gets reduced, rather than hydrogen. The general trend of less negative E° values across the series is related to the overall increase in the sum of the first and second ionisation enthalpies. However, some elements show deviations from this trend: The E° values for manganese (Mn) and zinc (Zn) are more negative than expected. For Mn, this is linked to the extra stability of the half-filled d5 configuration in Mn2+. For Zn, it's due to the high stability of the completely filled d10 configuration in Zn2+. The E° value for nickel (Ni) is slightly more negative than expected, which is attributed to its exceptionally high negative hydration enthalpy for Ni2+, indicating strong solvation of the ion in water. Examining the E°(M3+\|M2+) values (which indicate the ease of reduction from the +3 to the +2 oxidation state) reveals further insights into relative stabilities: The value for Sc3+\|Sc2+ is very low (meaning Sc3+ is very stable), reflecting the high stability of Sc3+ which has a noble gas configuration ([Ar]). The value for Zn3+\|Zn2+ is very high (meaning Zn2+ is extremely stable and difficult to oxidise further), due to the removal of an electron from the stable d10 configuration of Zn2+. The E°(Mn3+\|Mn2+) value is quite high (+1.57 V), indicating that Mn3+ is a strong oxidising agent and readily reduces to Mn2+. This is because Mn2+ possesses the highly stable half-filled d5 configuration, making this reduction very favourable. In contrast, the E°(Fe3+\|Fe2+) value is comparatively low (+0.77 V). This shows that Fe3+ is also reasonably stable (due to its d5 configuration) and does not have as strong an oxidising tendency as Mn3+. \| \| \| Summary of General Properties of the d-Block Elements \| \| The d-block elements (Groups 3–12) are transition metals where electrons progressively fill the (n−1)d orbitals. According to IUPAC, only those with incomplete d-subshells in atoms or stable ions are classified as true transition metals, excluding Zn, Cd, and Hg, despite their location in the d-block. These elements are grouped into four series based on the principal quantum number: 3d (Sc–Zn), 4d (Y–Cd), 5d (La–Hg), and 6d (Ac–Cn), with increasing complexity and radioactivity across the series. Their general electronic configuration is (n−1)d1-10ns1-2, but anomalies like Cr ([Ar]3d54s1) and Cu ([Ar]3d104s1) arise due to extra stability from half-filled or fully filled d-subshells. The major transition metal properties include, variable oxidation states, coloured ions, complex formation, catalytic ability, and paramagnetism, stem from the presence of partially filled d-orbitals and low energy gap between ns and (n−1)d orbitals. Physically, transition metals are strong, malleable, and ductile with high melting points and electrical conductivity, owing to delocalised electrons and strong metallic bonding. Atomic and ionic radii decrease slightly across a series due to poor shielding by d-electrons. Lanthanoid contraction causes the 5d series to have atomic sizes nearly identical to 4d elements, e.g., Zr ≈ Hf. Ionisation enthalpies increase gradually across each series but exhibit irregularities due to variations in electronic configurations. Transition metals show a wide range of oxidation states due to close ns and (n−1)d energy levels. Their stability depends on electron configuration and ligand type, with high oxidation states stabilised by electronegative elements like oxygen and fluorine. \| Review Questions on General Properties of the d-Block Elements Why are Zn, Cd, and Hg not considered true transition metals according to IUPAC, even though they belong to the d-block? What is the general electronic configuration of d-block elements, and why do elements like Cr and Cu deviate from this configuration? How does the partially filled d-orbital in transition metals lead to the formation of coloured compounds? Explain with an example. Define lanthanoid contraction and explain how it affects the atomic radii of 5d transition series compared to 4d. Why do transition metals exhibit variable oxidation states, and which series element shows the maximum number of oxidation states? What is the reason behind the high melting and boiling points of most transition metals? Name an exception and justify. Explain why Fe2+ is more stable than Fe3+ in aqueous solution, whereas Mn3+ is less stable than Mn2+. Which electrons are lost first when a transition metal atom is ionised — ns or (n−1)d? Give a reason for your answer. Why is the third ionisation enthalpy of Mn higher than that of Fe, even though Fe lies to the right of Mn in the periodic table? \| Image \| Reference \| Attributions \| --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Adaptive Practice Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)42/ 100 This lesson has been added to your library. No Results Found Your search did not match anything in . \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for: Go to the Table of Contents
15579	https://en.wikipedia.org/wiki/Circular_segment	Circular segment - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 FormulaeToggle Formulae subsection 1.1 Radius and central angle 1.2 Chord length and height 1.3 Arc length and area 1.4 Other properties 2 Applications 3 See also 4 References 5 External links [x] Toggle the table of contents Circular segment [x] 36 languages العربية Azərbaycanca Беларуская Català Čeština Dansk Deutsch Ελληνικά Español Esperanto فارسی Français 한국어 Հայերեն Bahasa Indonesia Italiano Lietuvių Македонски മലയാളം Nederlands 日本語 Norsk bokmål Norsk nynorsk ភាសាខ្មែរ Polski Português Română Русский Slovenščina کوردی Svenska தமிழ் ไทย Українська Tiếng Việt 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia Area bounded by a circular arc and a straight line A circular segment (in green) is enclosed between a secant/chord (the dashed line) and the arc whose endpoints equal the chord's (the arc shown above the green area). In geometry, a circular segment or disk segment (symbol: ⌓) is a region of a disk which is "cut off" from the rest of the disk by a straight line. The complete line is known as a secant, and the section inside the disk as a chord. More formally, a circular segment is a plane region bounded by a circular arc (of less than π radians by convention) and the circular chord connecting its endpoints. Formulae [edit] Let R be the radius of the arc which forms part of the perimeter of the segment, θ the central angle subtending the arc in radians, c the chord length, s the arc length, h the sagitta (height) of the segment, d the apothem of the segment, and a the area of the segment. Usually, chord length and height are given or measured, and sometimes the arc length as part of the perimeter, and the unknowns are area and sometimes arc length. These can't be calculated simply from chord length and height, so two intermediate quantities, the radius and central angle are usually calculated first. Radius and central angle [edit] The radius is: R=h 2+c 2 8 h{\displaystyle R={\tfrac {h}{2}}+{\tfrac {c^{2}}{8h}}} The central angle is θ=2 arcsin⁡c 2 R{\displaystyle \theta =2\arcsin {\tfrac {c}{2R}}} Chord length and height [edit] The chord length and height can be back-computed from radius and central angle by: The chord length is c=2 R sin⁡θ 2=R 2(1−cos⁡θ){\displaystyle c=2R\sin {\tfrac {\theta }{2}}=R{\sqrt {2(1-\cos \theta )}}}c=2 R 2−(R−h)2=2 2 R h−h 2{\displaystyle c=2{\sqrt {R^{2}-(R-h)^{2}}}=2{\sqrt {2Rh-h^{2}}}} The sagitta is h=R−R 2−c 2 4=R(1−cos⁡θ 2)=R(1−1+cos⁡θ 2)=c 2 tan⁡θ 4{\displaystyle h=R-{\sqrt {R^{2}-{\frac {c^{2}}{4}}}}=R(1-\cos {\tfrac {\theta }{2}})=R\left(1-{\sqrt {\tfrac {1+\cos \theta }{2}}}\right)={\frac {c}{2}}\tan {\frac {\theta }{4}}} The apothem is d=R−h=R 2−c 2 4=R cos⁡θ 2{\displaystyle d=R-h={\sqrt {R^{2}-{\frac {c^{2}}{4}}}}=R\cos {\tfrac {\theta }{2}}} Arc length and area [edit] The arc length, from the familiar geometry of a circle, is s=θ R{\displaystyle s={\theta }R} The area a{\displaystyle a} of the circular segment is equal to the area of the circular sector minus the area of the triangular portion (using the double angle formula to get an equation in terms of θ{\displaystyle \theta }): a=R 2 2(θ−sin⁡θ){\displaystyle a={\tfrac {R^{2}}{2}}\left(\theta -\sin \theta \right)} In terms of c and R, a=R 2 2(2 arcsin⁡c 2 R−sin⁡(2 arcsin⁡c 2 R))=R 2(arcsin⁡c 2 R−c 2 R 1−(c 2 R)2){\displaystyle a={\tfrac {R^{2}}{2}}\left(2\arcsin {\tfrac {c}{2R}}-\sin \left(2\arcsin {\tfrac {c}{2R}}\right)\right)=R^{2}\left(\arcsin {\frac {c}{2R}}-{\frac {c}{2R}}{\sqrt {1-\left({\frac {c}{2R}}\right)^{2}}}\right)} In terms of R and h, a=R 2 arccos⁡(1−h R)−(R−h)R 2−(R−h)2{\displaystyle a=R^{2}\arccos \left(1-{\frac {h}{R}}\right)-\left(R-h\right){\sqrt {R^{2}-\left(R-h\right)^{2}}}} In terms of c and h, a=(c 2+4 h 2 8 h)2 arccos⁡(c 2−4 h 2 c 2+4 h 2)−c 16 h(c 2−4 h 2){\displaystyle a=\left({\frac {c^{2}+4h^{2}}{8h}}\right)^{2}\arccos \left({\frac {c^{2}-4h^{2}}{c^{2}+4h^{2}}}\right)-{\frac {c}{16h}}(c^{2}-4h^{2})} What can be stated is that as the central angle gets smaller (or alternately the radius gets larger), the area a rapidly and asymptotically approaches 2 3 c⋅h{\displaystyle {\tfrac {2}{3}}c\cdot h}. If θ≪1{\displaystyle \theta \ll 1}, a=2 3 c⋅h{\displaystyle a={\tfrac {2}{3}}c\cdot h} is a substantially good approximation. If c{\displaystyle c} is held constant, and the radius is allowed to vary, then we have∂a∂s=R{\displaystyle {\frac {\partial a}{\partial s}}=R} As the central angle approaches π, the area of the segment is converging to the area of a semicircle, π R 2 2{\displaystyle {\tfrac {\pi R^{2}}{2}}}, so a good approximation is a delta offset from the latter area: a≈π R 2 2−(R+c 2)(R−h){\displaystyle a\approx {\tfrac {\pi R^{2}}{2}}-(R+{\tfrac {c}{2}})(R-h)} for h>.75 R As an example, the area is one quarter the circle when θ ~ 2.31 radians (132.3°) corresponding to a height of ~59.6% and a chord length of ~183% of the radius.[clarification needed] Other properties [edit] The perimeter p is the arclength plus the chord length: p=c+s=c+θ R{\displaystyle p=c+s=c+\theta R} Proportion of the whole area of the circle: a A=θ−sin⁡θ 2 π{\displaystyle {\frac {a}{A}}={\frac {\theta -\sin \theta }{2\pi }}} Applications [edit] The area formula can be used in calculating the volume of a partially-filled cylindrical tank lying horizontally. In the design of windows or doors with rounded tops, c and h may be the only known values and can be used to calculate R for the draftsman's compass setting. One can reconstruct the full dimensions of a complete circular object from fragments by measuring the arc length and the chord length of the fragment. To check hole positions on a circular pattern. Especially useful for quality checking on machined products. For calculating the area or locating the centroid of a planar shape that contains circular segments. See also [edit] Chord (geometry) Spherical cap Circular sector References [edit] ^Mathematics distinguishes when necessary between the words circle and disk: a disk is a plane area having a circle as its boundary, while a circle is the closed curve forming the boundary itself. ^These terms refer to a line which intersects a curve. In this case, the curve is the circle forming the disk's boundary. ^The fundamental relationship between R{\displaystyle R}, c{\displaystyle c}, and h{\displaystyle h} derivable directly from the Pythagorean theorem among R{\displaystyle R}, c/2{\displaystyle c/2}, and R−h{\displaystyle R-h} as components of a right triangle is: R 2=(c 2)2+(R−h)2{\displaystyle R^{2}=({\tfrac {c}{2}})^{2}+(R-h)^{2}} which may be solved for R{\displaystyle R}, c{\displaystyle c}, or h{\displaystyle h} as required. Weisstein, Eric W."Circular segment". MathWorld. External links [edit] Definition of a circular segment With interactive animation Formulae for area of a circular segment With interactive animation Retrieved from " Category: Circles Hidden categories: Articles with short description Short description is different from Wikidata Wikipedia articles needing clarification from December 2021 This page was last edited on 18 August 2025, at 02:46(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Search Search [x] Toggle the table of contents Circular segment 36 languagesAdd topic
15580	https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratic-formula-a1/a/quadratic-formula-explained-article	Published Time: Wed, 20 Aug 2025 22:53:54 GMT Quadratic formula explained (article) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Math: Florida B.E.S.T. Grade 3 math (FL B.E.S.T.) Grade 4 math (FL B.E.S.T.) Grade 5 math (FL B.E.S.T.) Grade 6 math (FL B.E.S.T.) Grade 7 math (FL B.E.S.T.) Grade 8 math (FL B.E.S.T.) Algebra 1 (FL B.E.S.T.) Geometry (FL B.E.S.T.) Algebra 2 (FL B.E.S.T.) 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15581	https://fiveable.me/key-terms/ap-stats/law-of-large-numbers	Law of Large Numbers - (AP Statistics) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms AP Statistics Law of Large Numbers 📊ap statistics review key term - Law of Large Numbers Citation: MLA Definition The Law of Large Numbers states that as the number of trials in a probability experiment increases, the experimental probability of an event will converge to the theoretical probability of that event. This concept highlights the stability of long-term results as the sample size grows larger, emphasizing the importance of using larger samples for more accurate predictions. 5 Must Know Facts For Your Next Test The Law of Large Numbers can be divided into two types: the Weak Law and the Strong Law, with both demonstrating convergence but under different conditions. This law is crucial in fields like insurance and finance, where predicting outcomes accurately over many trials is essential for risk assessment. The law assures us that while short-term results may vary widely, long-term averages become more stable and predictable as sample size increases. In simulations, applying the Law of Large Numbers allows for better estimation of probabilities by running more trials. It is important to note that while larger samples provide better estimates, they do not guarantee outcomes for individual trials. Review Questions How does the Law of Large Numbers influence decision-making in fields such as insurance or finance? The Law of Large Numbers plays a crucial role in decision-making within insurance and finance by allowing these industries to predict outcomes with greater accuracy over many trials. For example, insurers rely on large data sets to determine risk and set premiums, ensuring that their long-term results align closely with expected probabilities. This law gives confidence that as more policies are written, the actual loss ratio will converge to the predicted loss ratio, thus guiding financial strategies effectively. Discuss how the Law of Large Numbers applies to simulations used in estimating probabilities. In simulations aimed at estimating probabilities, the Law of Large Numbers ensures that as the number of simulated trials increases, the estimated probabilities will get closer to the theoretical probabilities. For instance, if you simulate flipping a coin many times, you will observe that the proportion of heads and tails approaches 50% as you increase the number of flips. This principle underlines the importance of running sufficient trials in simulations to achieve reliable results and enhance predictive accuracy. Evaluate the implications of the Law of Large Numbers on interpreting statistical data and making predictions based on smaller samples. The implications of the Law of Large Numbers emphasize caution when interpreting statistical data derived from smaller samples. Small samples can produce results that are significantly different from true probabilities due to random variation. However, as sample sizes increase, one can expect those results to stabilize and become more representative. This understanding is vital when making predictions or generalizations about populations based on limited data, underscoring the need for larger sample sizes to ensure accuracy in statistical analysis. Related terms Probability: The measure of the likelihood that an event will occur, ranging from 0 (impossible) to 1 (certain). Random Variable: A variable whose values are determined by the outcomes of a random phenomenon, used to quantify uncertain events. Sample Size: The number of observations or trials included in a statistical sample, which affects the reliability and accuracy of statistical estimates. 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15582	https://www.cuemath.com/questions/which-are-factors-of-x2-4x-5/	Which are factors of x2 - 4x - 5? Check all that apply. (x - 5), (x - 4),(x - 2), (x + 1),(x + 5) Solution: Let us factorise the polynomial to find the value of x by splitting the middle term. Step 1: Identify the values of a, b and c. In the above equation, a is coefficient of x2 = 1, b is the coefficient of x which is - 4 and c is the constant term = - 5. Step 2: Multiply a and c and find the factors that add up to b. 1 × (- 5) = - 5 ⇒ - 5 and 1 are the factors that add up to b. Step 3: Split bx into two terms. x2 - 5x + 1x - 5 = 0 Step 4: Take out the common factors by grouping. x(x - 5) + 1 (x - 5) = 0 (x - 5) (x + 1) = 0 By putting the factors equal to zero we get two values of x x - 5 = 0 and x + 1 = 0 x = 5 and x = - 1 Thus, the two values that satisfy the equation are 5 and - 1. Which are factors of x2 - 4x - 5? Check all that apply. (x - 5), (x - 4),(x - 2), (x + 1),(x + 5) Summary: (x - 5) ( x + 1) are the solutions that apply to the equation x2 - 4x - 5.
15583	https://www.youtube.com/watch?v=QIjgOB9Gf9Q	Solving Recurrences using Substitution Method (Solved Problem 6) Neso Academy 3020000 subscribers 57 likes Description 4409 views Posted: 13 Dec 2024 Algorithms: Solving Recurrences using Substitution Method (Solved Problem 6) Topics discussed: 1. A Solved Problem on Solving Recurrence Relations using the Substitution Method. Algorithm Playlist: Follow Neso Academy on Instagram: @nesoacademy ( Contribute: Memberships: Discord: WhatsApp: Books: Website ► App ► Facebook ► Twitter [X] ► Music: Axol x Alex Skrindo - You [NCS Release] AlgorithmsByNeso #Algorithms #SubstitutionMethod 7 comments Transcript: [Music] let's solve problem number six on recurrences using substitution here is the problem solve the following recurence relation tnal to squ < TK of n tunk of n + n if n is greater than 2 if n is equal to 2 then TN is equal to 2 this is the recurrence relation and we need to solve this recurence relation this time the algorithm is not given we do not have to write the recurence relation and then solve it using the substitution method this time we have the recurence relation and we need to solve this recurrence relation using the substitution method so now let's try to solve this recurence relation using the substitution method we know according to the substitution method we need to start from the recursive case that is TN = to square < TK of n tunk of n + n so let's start from TN = to S < TK of n tunk of n + n this is the big problem to solve the small problem is TN = to 2 if n is equal to 2 if n is 2 then TN is a constant and this constant is two this is the small problem and hence this represents the base case we are starting from the recursive case according to the substitution method now let's substitute tunk of n by squ otk of otk n tunk of < TK of n+ squ < TK of n we know we can represent square root of n as n^ 1X 2 so we can replace < TK of n by n^ 1X 2 here also we can replace root of n by n^ 1 by 2 so TN is same as n ^ 1X 2 t n ^ 1X 2 + n now in place of T n^ 1X 2 we can write n^ 1X 4 because square root of root of n is same as n^ 1X 2 1X 2 1X 2 1X 2 is 1x 4 so we will get n^ 1X 4 and here also within parenthesis we will get n^ 1X 4 because n will be replaced by square root of n this is squ root of squ root of n again this is n^ 1x 2 1X 2 so we will get n^ 1X 4 here so this is T n^ 1X 4 then here we will get square root of n or we can write it as n^ 1 by 2 so clearly T squ root of n or t n^ 1X 2 is same as n^ 1X 4 T n^ 1x4 4 + n^ 1X 2 so this is the new TN so obtained here we have written n^ 1X 2 as it is this is square root of n we are writing it as it is and then we have tqu < TK of N and then + n according to this expression tunk of n is replaced by n^ 1X 4 T n^ 1X 4 + n^ 1X 2 now we know this is the new new TN in terms of T n^ 1X 4 now we need to multiply n^ 1X 2 by n^ 1X 4 T n^ 1X 4 and we need to multiply n^ 1X 2 by n power 1X 2 now let's open these brackets and let's multiply n^ 1X 2 by these two terms here we will get n^ 1X 2 n^ 1X 4 TN n^ 1X 4 what do we get from here what is the multiplication of n^ 1x 2 and n^ 1X 4 we know when the bases are same then the powers will add so we need to add 1X 2 and 1X 4 1X 2 + 1X 4 is 3x 4 so we can write n^ 3x 4 here so we'll get n^ 3x 4 tn^ 1X 4 and then we need to multiply n^ 1X 2 by n^ 1X 2 we know here the bases are same so the powers will add and 1X 2 + 1X 2 is 1 only so we will get n^ 1 here this means we will get n here now we have n + n and n + n is equal to 2 N so we will get 2 N here so this expression is same as n^ 3x 4 t n ^ 1X 4 + 2 N now let's substitute tn^ 1X 4 by n^ 1X 8 because square root of n^ 1x 4 is same as n^ 1X 4 1X 2 so in the power 1X 4 is Multiplied to 1x 2 we will get 1X 8 as the power of n so we will get n^ 1X 8 t n^ 1x8 + n^ 1X 4 because n should be replaced by n^ 1X 4 so tn^ 1X 4 is same as n^ 1X 8 T n^ 1X 8 + n^ 1X 4 so new TN is equal to n^ 3x 4 tn^ 1X 4 + 2 N this is TN ^ 1X 4 which is same as n^ 1X 8 T n^ 1X 8 + n^ 1X 4 now we need to open these brackets and we need to multiply n^ 3x 4 by n^ 1X 8 and n^ 3x 4 by n^ 1X 4 what is the multiplication of these two terms here we know the bases are same so we need to add the powers 3x 4 + 1 by 8 is same as 7x 8 we can follow the same process which we have followed for 1X 2 + 1X 4 in the same way we can add 3x 4 and 1X 8 we will get 7x 8 so we will get n^ 7x 8 here and n^ 7x 8 will be multiplied to T n^ 1X 8 and here we have n^ 1X 4 we now need to multiply n^ 1X 4 by I n^ 3x 4 what is the addition of 3x4 and 1X 4 we need to add these two Powers because the bases are same 3x 4 + 1X 4 is same as 4X 4 and 4x 4 is equal to 1 so we'll get n Only and n + 2 N is 3 n so the new expression so obtained is n^ 7 by 8 tn^ 1X 8 + 3 n this is the value of TN now we can see the pattern here we have n^ 1X 2 here we know square root is same as 1X 2 so it is n^ 1x 2 and this is n^ 3x 4 and here we have n^ 7 by 8 we can observe in the power of n the denominator is always the powers of two here in the power the denominator is 2 which is 2 power 1 here we have 4 which is 2^ 2 here we have 8 which is 2^ 3 so it is clear that the denominators are always 2 power something and what about the numerator here we can observe the numerators are always one less than the denominators here we have three because we have four here here we have seven because we have eight here here we have one because we have two in the denominator so clearly if the denominator is 2 power some k then the numerator must be 2 power K minus 1 so this part is clear now what about TN power something here we can observe we have tn^ 1X 4 these two denominators are same here also these two denominators are same so it is clear it is T n^ 1 by 2^ K as we are assuming the denominator of the power of n is 2 power K so here also we will have 2 power K now what about the remaining part here we have + 3 n in this case we have + 2 N in this case we have plus n so it is clearly K n because here we have 1 by 2^ 3 here we have 1 by 2^ 2 we know we are representing the power of 2 as K and here we have have 2^ 3 because the power of two is three here also we have three here we have 2^ 2 because the power of 2 is 2 here we have 2 here we have 1 by 2 as the power of n and this means the power of two is 1 that is why we have one here so it is clear this is K n so if you proceed in this way then the generalized expression is in this form TN is n^ 2^ K - 1 / 2^ K T n^ 1 by 2^ k + K n this is the equation so obtained now we need to represent K in terms of n because n represents the size of the input and eventually we want to represent TN in terms of the ASM totic notation where we need to represent it in terms of the input size hence it it is important to represent K in terms of the input size also we need to eliminate TN power 1X 2^ k for this let's assume n^ 1X 2^ K is equal to 2 this means the base case is reached then only n^ 1X 2 power K will be replaced by 2 here we can observe if n is 2 then we will get T2 here here in place of n we have n ^ 1X 2^ K so we need to assume n^ 1X 2^ K as 2 we will get T2 here and T2 is equal to 2 So eventually tn^ 1X 2^ K will be replaced by a constant and the constant in this case is 2 so let us assume n^ 1X 2^ K is equal to 2 now from this we can easily find the value of K in order to find the value of K we need to bring K to the base and for this we know what we need to do we need to take log on both sides here we have the constant two so we will take log base 2 on both sides here we will get log n^ 1X 2^ K base 2 and in the right hand side we will get log 2 base 2 so the left side is log n^ 1X 2^ K base 2 and the right hand side is log 2 base 2 what is log 2 base 2 log 2 base 2 is 1 what about log n^ 1 by 2^ K base 2 we know the property of logarithm log a power B base C is same as B log a base C so 1X 2^ K comes in front of log n base 2 so we are getting 1X 2^ K log and base 2 in the left hand side and one in the right hand side of the equation now how do we solve this it is simple to solve this type of equation we need to multiply 2^ K on both sides in this way we can remove this denominator we will get log n base 2 in the left hand side and 2 power K in the right hand side this is the equation so obtained log n base 2 is same as 2^ K now we know we want to find the value of K and K is in the power of 2 we need to bring k a to the base for this we will take log once again and this time also we will take log base 2 on both sides because here the constant is two so let's take log base 2 on both sides after applying log base 2 on both sides we will get log n base 2 log base 2 in the left hand side and log 2^ K Bas 2 in the right hand side so this is the equation so obtained we know log to ^ K base 2 is same as K log 2 base 2 which is equal to K that's why we're getting K here and here in the left hand side we have log n base 2 log base 2 so this is the value of K in terms of n now here we can observe that 2^ K can be replaced by log in base 2 directly so we can replace 2^ K by log n base 2 we will get n^ log n base 2 - 1 divide by log in base 2 now we can replace n^ 1 by 2^ K by 2 we will get T2 here and T2 is equal to 2 we already know this from this recurence relation so we can replace tn^ 1 by 2^ K by 2 so this is the expression so obtained we are getting n^ log n base 2 - 1 / log n base 2 2 and here we have K n we know what's the value of k k is equal to log n base 2 log base 2 so let's replace K by log n base 2 log base 2 here we are getting n log n base 2 log base 2 now let's focus on this fraction here we have log n base 2 - 1 divide by log n base 2 we can rewrite this as 1 - 1 by log n n base 2 so in place of log n base 2 - 1 IDE log n base 2 we can write 1 - 1 by log n base 2 now we can rewrite this as n^ 1 / n^ 1 by log in base 2 because the sign here is minus if you have plus here then it will be n^ 1 n^ 1 by log n base 2 but as we have minus here we will get n^ 1 IDE by n^ 1 by log n base 2 so this is the fraction so obtained now the interesting part here we have n^ 1 by log n base 2 we know n^ 1 by 2^ K is equal to 2 and what is 2^ k 2^ k is log n base 2 so we can replace 2^ K by log n base 2 we will get n^ 1 by log n base 2 = 2 here also we have n^ 1 by log n base 2 so we can replace n^ 1 by log n base 2 by 2 so we will get n by 2 2 here we can easily cancel 2x 2 we will get n here so the expression so obtained is n + n log n base 2 log base 2 now what do you think out of these two which is the dominating term here we have n 1 one is the constant here we have n login base 2 log base 2 login base 2 log base 2 is the logarithmic function we know the growth rate of the logarithmic function is greater than the growth rate of the constant function in both these terms n is common but this is n 1 and this is n log n base 2 log base 2 here n is multiplied by the constant function and here n is multiplied by the logarithmic function clearly the growth rate of n the logarithmic function is greater than the growth rate of n the constant function therefore this is the leading term here and hence TN is equal to P of n log log n TN is represented based on the leading term which is n log log n there is no need to write the bases we can write the asymptotic notation like this in the ASM totic notation we do not have to mention the bases and the constants that's why we have n log log n here so TN is Big go of n log log n now we have understood how to solve these type of complex recurence relations and that two using the substitution method so with this we are done with this topic and we are done with this presentation okay friends this is it for now thank you for watching this presentation I will see you in the next one [Applause] [Music]
15584	https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/What_is_the_pKa_of_water	Skip to main content What is the pKa of water? Last updated : Apr 9, 2025 Save as PDF The Use of Curly Arrows Hydrocarbons Page ID : 35871 ( \newcommand{\kernel}{\mathrm{null}\,}) In most general chemistry textbooks, the pKa of water at 25 ºC is listed as 14.0. In many organic chemistry textbooks and some biochemistry texts, however, the pKa of water at 25ºC is listed as 15.7. This module describes the derivation of these two values and describes why the value of 15.7 should not be used. Much of the information in this module comes from three papers with more comprehensive coverage of the topic.1,2,18 Two Values for the Same Equilibrium Constant? The value of 14.0 for the pKw of water at 25ºC has been experimentally measured.3,4 This value can also be obtained by the same process used to calculate the pKa of all other water-soluble compounds that can act as acids in aqueous solution: from the analysis of thermodynamic or electrochemical data for these aqueous solutions. The origin of the value of 15.7 for the pKa of water can be tracked back at least as far as 1928, when Johannes Brønsted postulated that there could be two different values for the pKa of water.5 One value was the “conventional”, thermodynamically-sound equilibrium constant, in which the activity of the water is used, and assigned its correct value of unity (1) because it is the solvent in a dilute solution. Brønsted called the second constant the “rational” acidity constant (which we distinguish as Ka,H2O), in which he included a molar concentration of water. But Brønsted stated that this molar concentration “does not possess an explicit value” and thus it is not possible to calculate a value for Ka,H2O or pKa,H2O. By 1937, Brønsted, seemed to have given this problem further thought and gave the “conventional” values precedence over the untenable “rational” values. Specifically, in his 1937 textbook Physical Chemistry,6 Brønsted defined an “autoprotolysis” constant in water, K(H2O) = [H+][OH-] which is not only equal to our modern autoionization constant, Kw, but also, by analogy to his discussion of the ionization of dilute acids, equal to Ka,H2O. Mathematically, Brønsted stated here that pKw = pKa,H2O. In addition, Brønsted included only the conventional values in a table in this textbook that lists the pKa values of several well-known acids. Thus, nine years after presenting both the “rational” and “conventional” values, Brønsted seems to have set aside the “rational” values in favor of the activity-based, thermodynamically correct “conventional” values. Nevertheless, the erroneous 15.7 value, as well as the erroneous -1.7 value for the pKaof the aqueous proton (H3O+ or H+), was used by many organic chemists who employed the "Brønsted equation" in their research into the relationship between acid-base strength and catalytic power, even though these erroneous values led to poor correlations when plotted along with other aqueous acids.7,8,9,19 A second common source of the incorrect 15.7 value is the erroneous explanation of why the concentration of water, as the solvent, does not appear in the law of mass action for a reaction. This misunderstanding, and the corresponding miscalculation, will be fully described and remedied later in this module. A third, deliberately developed, source of the value of 15.7 for the pKa of water was a theoretical proposal made by physical organic chemists who were under the false impression that solvents would not appreciably affect the relative strength of acids. These chemists experimentally determined the pKa of methanol dissolved in water at 25°C to be 15.5.10 They knew that previous research had shown that when 2-propanol was used as the solvent, methanol was roughly 3 times more acidic than water.11 They also knew that the commonly accepted pKa of pure water was 14.0. Thus, as it stood, the relative acidity of methanol and water dissolved in 2-propanol (methanol > water) differed from the relative acidity of water and methanol dissolved in water (water > methanol). In an attempt to make their experimental data conform to the incorrect assumption that the solvent identity does not affect relative acid strength, the chemists who measured the pKa of methanol mathematically manipulated the pKa of water so that the relative acidities in the two different solvents would agree. It will be shown below that this mathematical manipulation is incorrect and unnecessary because solvent is known to affect relative acid strength.12,13 The pKa of the Na+ Ion: Experimental and Anecdotal Proof that the pKa of Water cannot be 15.7 It is common practice to ignore the change in pH caused by the addition of NaCl to aqueous solutions. This practice is valid for Cl- ions (pKb≈ 20)a because it is a much weaker base than water is. But is this practice valid for the Na+ ion, which has a pKa = 13.9?b If we were to take the pKa of water to be 15.7, the Na+ ion would be over 60 times as strong an acid as water itself is. If that were the case, a 0.1 M solution of NaCl would have a decrease in pH of about 0.5 pH units below the pH of pure water. Ignoring the change in pH caused by the addition of Na+ ions would not be valid in this scenario. However, if we take the pKa of water to be 14.0, then the Na+ ion is only 1.3 times as strong an acid as water itself is, and the decrease in pH caused by the added Na+ can be ignored in most situations. This is the common experience of everyone who has measured the pH of a solution of NaCl and the expectation of anyone who has made a solution of NaCl. a A. Trummal, L. Lipping, I. Kaljurand, I. A. Koppel, and I. Leito, J. Phys. Chem. A 2016, 120, 20, 3663–3669 b R.M. Smith, A.E. Martell, and R.J. Motekaitis, NIST Critical Stability Constants of Metal Complexes Database 46 (Gaithersburg, MD: NIST, 2001) Thermodynamically Correct Derivation of Ka and pKa Thermodynamic equilibrium constants (Keq) are defined based on the activity of each of the substances involved in the chemical reaction.14 The activity of a species is a measure of the “effective concentration” of that species that results from the attractive and repulsive interactions of particles in non-ideal mixtures. Thus, for the reaction: bB+cC−⇀↽−dD+eE The law of mass action is written as Keq=adD·aeEabB·acC where adD indicates activity of species D raised to the d power. In an ideal dilute solution, the solutes are treated with Henry’s law so that activities can be approximated with concentrations. It is most appropriate to approximate the activity of a solute in a dilute solution with the molality of that solute (m = moles of solute per kilogram of solvent). However, for an ideal dilute aqueous solution at 25 ºC, the molality of a solute is often approximated with the molarity of that solute (M = moles of solute per Liter of solution) because the density of water at 25 ºC is 0.997 kilograms per Liter: aD≈molalityD≈[D] where [D] indicates the concentration of the species D in units of molarity. The solvent in an ideal dilute solution is treated with Raoult’s law so that the solvent can be approximated as the pure liquid. When the solvent is approximated as a pure liquid, its activity is approximated as unity: asolvent≈1 The concentration of the solvent does not explicitly appear in the equilibrium expression (Equation 2). The approximation being made is that intermolecular forces are negligible. If the solution is non-ideal, then the effects of the intermolecular forces must be accounted for by employing activity coefficients as correction factors. The use of activity coefficients is the thermodynamically valid method to describe nonideal solutions, and is unrelated to the method of involving the solvent concentration in the calculations described below. Thus, for a reaction in which water is a participant and also the solvent, the activity of water is not “left out” of the law of mass action. Rather, it remains in the law of mass action, but it is assumed to have a value of unity (the number 1), and so its value has no effect on the value of Keq. This process is demonstrated in Example 1. Example 1: Dissociation of Hypochlorous Acid Consider the acid dissociation of hypochlorous acid HClO(aq)+H2O(l)−⇀↽−ClO−(aq)+H3O+(aq) with the following equilibrium constant Ka=aClO−·aH3O+aHClO·aH2O≈[ClO−][H3O+][HClO]{1}=[ClO−][H3O+][HClO] Using values of ΔGof at 25ºC15 for the species involved, and the Equation ΔGº=−RTlnK, it can be shown that for Equation 3: ΔGoreaction={ΔGofClO−(aq)+ΔGofH3O+(aq)}−{ΔGofHClO(aq)+ΔGofH2O(l)}={−36.8kJ+−237.1kJ}−{−79.9kJ+−237.1kJ}=43.1kJ thus Ka=2.8×10−8 or pKa=7.55. Some chemists prefer to use the short hand notation of H+ rather than H3O+ to represent the proton exchanged during the acid/base reaction, resulting in the reaction equation: HClO(aq)−⇀↽−ClO−(aq)+H+(aq) for which K′a=aClO−·aH+aHClO≈[ClO−][H+][HClO] Using values of ΔGof at 25ºC15 for the species involved, and the Equation ΔGo=−RTlnK, one can show that for Equation 4: ΔGoreaction={ΔGofClO(aq)−+ΔGofH(aq)+}−{ΔGofHClO(aq)}={−36.8kJ+0kJ}−{−79.9kJ}=43.1kJ thus K′a=Ka=2.8×10−8 or pKa=7.55. As expected, the value of Ka equals the value of K′a and thus does not depend on the manner in which the exchanged proton is represented. A similar discussion applies to the autoprotolysis of water, in which one water molecule acts as an acid, and the other water molecule acts as a base: H2O(l)+H2O(l)−⇀↽−OH−(aq)+H3O+(aq) or H2O(l)−⇀↽−OH−(aq)+H+(aq) resulting in the following equilibrium constant: Keq=aOH−·aH3O+a2H2O≈[OH−][H3O+]12=[OH−][H3O+] or K′eq=aOH−·aH+aH2O≈[OH−][H+]1=[OH−][H+] Using values of ΔGof at 25ºC15 for the species involved, and the Equation ΔGo=−RTlnK, one can show that for reaction 5: ΔGoreaction={ΔGofOH−(aq)+ΔGofH3O+(aq)}−{2×ΔGofH2O(l)}={−157.2kJ+−237.1kJ}−{2(−237.1kJ)}=79.9kJ thus Keq=1×10−14=Ka or pKa=14.0. Using values of ΔGof at 25ºC15 for the species involved, and the Equation ΔGo=−RTlnK, one can show that for Equation 6: ΔGoreaction={ΔGofOH−(aq)+ΔGofH+(aq)}−{ΔGofH2O(l)}={−157.2kJ+0kJ}−{−237.1kJ}=79.9kJ thus K′eq=1×10−14=Keq=Ka or pKa=14.0. In addition, it is possible to show that for both reactions 5 and 6, Keq=Ka=1×10−14 using values of Eo at 25ºC and the Equation nFEo=RTlnK.1 It is a standard and accepted practice to calculate Ka values in both of these ways, and experimentally measured values (or extrapolated estimates)9 confirm these calculated values. Once again, the value of Keq (and thus Ka) does not depend on the manner in which the exchanged proton is represented. It will be shown below that this equivalency of equations is not possible if the pKa of water is set as 15.7. Non-Thermodynamic Derivations Some general chemistry textbooks list the thermodynamically correct values of the Ka for species in aqueous solutions, but use an incorrect explanation as to why water does not appear in the law of mass action equation. These textbooks state that the concentration of water is constant and then claim that the constant value of the concentration has been included in the value of Ka. Other texts explicitly state that the concentration of water is 55.33 Molar at 25ºC (1.000L of water at a density of 0.9970 g/mL, and a molar mass of 18.02 grams/mole), and then explicitly multiply Keq by 55.33 to obtain what they claim is a different constant, Ka. As pointed out above, the concentration of water does not have a place in the determination of the value of Ka for water, or for the K of any reaction involving water that occurs in aqueous solution. Thus, although the correct Ka values are shown in these texts, the assumptions that led to the simplification of the law of mass action are lost, especially the assumption that the activity of the water, as solvent, is unity. The mathematical and chemical rationales presented by the chemists attempting to reconcile the relative acidity of substances in water and in non-aqueous solvents contain similar mistakes. The correct and commonly accepted ways of writing the equations for the autoprotolysis of water are Equations 5 or 6. Instead, Ballinger and Long claim that is possible to differentiate in solution between the water molecules that are acting as solvent molecules and the water molecules that are acting as solute molecules in the role of Brønsted acids.10 This claim suggests the following equations for the autoprotolysis of water: H2O(aq)+H2O(l)−⇀↽−OH−(aq)+H3O+(aq) or H2O(aq)−⇀↽−OH−(aq)+H+(aq) Note the difference in the phases of water between Equations 7 or 8 with Equations 5 or 6. Ballinger and Long make the commonly accepted assumption that the solvent water molecules be treated as the solvent part of an ideal dilute solution, with an activity of 1. They then assume that the solute water molecules be treated as the solute part of an ideal solution whose activity is approximated as the numerical value of concentration of water, 55.33. With these assumptions, the law of mass action for Equations 7 would be expressed as: K′a=aOH−·aH3O+aH2O(aq)·aH2O(l)≈[OH−][H3O+][H2O(aq)]·1=[OH−][H3O+][H2O(aq)] and the law of mass action for Equation 8 would be expressed as: K″a=aOH−·aH3O+aH2O(aq)≈[OH−][H3O+][H2O(aq)] At 25 ºC, the product of [OH−(aq)][H3O+(aq)]=1×10−14 and [H2O(aq)]=55.33, thus K′a=K″a≈[OH−(aq)][H3O+(aq)][H2O(aq)]=Keq55.33=1×10−1455.33=2×10−16=K∗a Therefore pK∗a=15.7 The assumptions made in this non-thermodynamic derivation require the acceptance of the following conundrums: The idea that the few water molecules acting as the solute have a concentration of 55.33 M and that the water molecules acting as the solvent are pure liquid implies that we have two sets of water molecules (solute ones and solvent ones) both with a concentration of 55.33 M. This is false. The sum of the two water concentrations would have to be 55.33 M: [H2O(l)]+[H2O(aq)]=55.33MAdditionally, setting the activity of the “solute” water molecules as 55.33 and the activity of the “solvent” water molecules as 1 implies that one set of water molecules is 55.33 times more reactive than the other set. This is an absurd assertion. Activity is not merely a convention. It is a description of the potential for a chemical species to react.16 Regarding item 1, some chemists argue that the acid dissociation constant for a mixture of labeled "solute" water acting as an acid in unlabeled water solvent should be calculated as (H3O+)(OH−)(mole fraction labeled"solute"water)(molarity labeled"solute"water)=(1.0x10−7)(1.0x10−7)(mole fraction labeled"solute"water)(molarity labeled"solute"water)If we are to entertain the premise of "solute" water molecules, then it is important to realize that both the "solute" and the solvent water molecules in the autoionization reaction are water molecules, making the solution an ideal solution. In an ideal solution in which the "solute" and the solvent particles experience identical intermolecular forces, both "solute" and solvent molecules obey Raoult's Law.17 Therefore, the activity coefficients of the "solute" and the solvent are both 1. In such cases, the mole fraction equals the activity. Thus, it is incorrect to use the concentration of the "solute" to replace the activity of the "solute" in the calculation of the equilibrium constant. Instead, the mole fraction of the "solute" molecules must be used. (H3O+)(OH−)(mole fraction labeled"solute"water)(mole fraction labeled"solute"water)=(1.0x10−7)(1.0x10−7)(mole fraction labeled"solute"water)(mole fraction labeled"solute"water)=1.0x10−14 As stated above, it is common practice to represent the autoprotolysis of water as Equations 7 or 8. The fact that these two reactions are equivalent is absolutely dependent upon the fact that all of the water molecules are identical. Note: Regarding the Erroneous claim that Ka = [H2O]Keq Many documents make the erroneous claim that the Ka for any acid is equal to [H2O]Keq. Not only is this statement incorrect because of its improper use of [H2O], but it is also not the Equation that Ballinger and Long derived (incorrectly!) in their attempt to make the pKa of water equal 15.7. Notice that in equation10, K∗a=Keq55.33. Rearranging this Equation gives K∗a[55.33]=Keq, not K∗a=[55.33]Keq. The first instance of the use of Ka=[H2O]Keq has not yet been tracked down, but it must have been made by someone who not only misunderstood thermodynamics, but also misread the Ballinger and Long paper. Notation Issues The following discussion shows that keeping track of the phase notation of H2O(l) versus H2O(aq) allows one to show that the H2O(aq) notation is untenable. Using the Correct Notation The process used to prove the equivalency of Equations 5 and 6 employs a basic rule of equilibrium chemistry; if two reactions are added together to make a new reaction, the equilibrium constant of the new reaction is the product of the equilibrium constants of the two original reactions. Applying that rule to convert from Equation 5 to Equation 6 requires a third equation: H3O+(aq)−⇀↽−H2O(l)+H+(aq) Equation 13 is developed from Equation 5 as follows. One of the H2O(l) molecules in Equation 5 acts as an acid, and its conjugate base is OH-. The other H2O(l) molecule must then act as a base, and its conjugate acid is H3O+. (There is no chemical difference between the two H2O(l) molecules in the equation. The likelihood of a particular water molecule’s acting as an acid or as a base is governed by the statistics of random events.) The value of K for Equation 13 can be obtained from the equation Ka×Kb=1×10−14 which holds for conjugate acid/base pairs in water at 25ºC. Because Kb = 1.0 x 10-14 for H2O, the Ka = 1 for H3O+ . It is also true that pKa + pKb = 14.0 for conjugate acid/base pairs in water. Thus, because the pKb of H2O is 14.0, the pKa of H3O+ is 0.0 The conversion of Equation 5 to Equation 6 is then: \| \| \| \| --- \| H2O(l)+H2O(l)⇌OH−(aq)+H3O+(aq) +H3O+(aq)⇌H2O(l)+H+(aq) \| K6=1×10−14 K14=1 \| pK = 14.0 pK = 0.0 \| \| H2O(l)⇌OH−(aq)+H+(aq) \| K7=K6·K14=1×10−14 \| pK = 14.0 \| Equations 5 and 6 are balanced, result in equivalent products, and because K6=K7, the two equations have the same value of Keq. These are equivalent equations. Using the Incorrect Notation Attempting this same conversion with Equations 7 and 8 leads to at least two impossible scenarios caused by the differentiation between H2O(l) and H2O(aq). To obtain an Equation equivalent to Equation 13, conjugate acid/base pairs must be assigned in Equation 7: H2O(aq)⏟acid+H2O(l)⏟base⇌OH−(aq)⏟conjugate base+H3O+(aq)⏟conjugate acid For example, if H2O(aq) is the acid, then its conjugate base is OH−. This means that H2O(l) and H3O+ must be a conjugate pair. There is no conjugate acid related to H2O(aq) in this reaction. Without a conjugate acid for H2O(aq), the conversion from 7 to 8 cannot be carried out. One possible solution is to obtain a conjugate acid for H2O(aq) by proposing that the very active H2O(aq) molecules are also able to act as Brønsted bases. These H2O(aq) molecules would thus have H3O+ as their conjugate acid. The equation for this reaction would be: H3O+(aq)−⇀↽−H2O(aq)+H+(aq) Although this proposal is not unreasonable, it is still a part of an impossible scenario because the conversion of Equation 7 to Equation 8 would be: H2O(aq)+H2O(l)−⇀↽−OH−(aq)+H3O+(aq) +H3O+(aq)−⇀↽−H2O(aq)+H+(aq) H2O(l)−⇀↽−OH−(aq)+H+(aq) The resulting Equation is not 8, but 6. This is a glaring error that clearly and simply shows that Ballinger and Long’s proposal is unsupportable. A second method for obtaining a conjugate acid for H2O(aq) is commonly found in most of the textbooks espousing Ballinger and Long’s value of pKa. In this method, Equation 13 is accepted as the Equation describing H3O+ as the conjugate acid of H2O(aq), with complete disregard for the phase designation of the H2O. Nevertheless, using the fact that pKa+pKb=14.0 in water at room temperature, and the claim that because the pKa of water is 15.7, the pKb of water must also be 15.7, the pKa of H3O+ is calculated to be -1.7. To clarify, the Equation 13 is (incorrectly!) assigned a pKa = -1.7, with a corresponding value of 50 for the Ka. The resulting combination of 7 and 13, with their (incorrectly!) assigned K values givess: \| \| \| \| --- \| H2O(aq)+H2O(l)⇌OH−(aq)+H3O+(aq) +H3O+(aq)⇌H2O(l)+H+(aq) \| K8=2×10−16 K14=50 \| pK = 15.7 pK = -1.7 \| \| H2O(aq)⇌OH−(aq)+H+(aq) \| K9=K8·K14=1×10−14 \| pK = 14.0 \| In this case, the two equations are balanced and they result in the equivalent products, but because K8 does not equal K9, the two equations do not have the same value of Keq. These are not equivalent equations. This is a second error that clearly and simply shows that Ballinger and Long’s proposal is unsupportable. For Further Consideration: Dissociation Constant of Methanol The great irony behind all of this discussion is that the Ballinger and Long10 paper not only presents data that clearly show that methanol is a weaker acid than water in aqueous solution, but also employs the value of 14.0 for the pKa of water to calculate the pKa of methanol. The authors determined that for the reaction CH3OH(aq)+OH−(aq)−⇀↽−CH3O−(aq)+H2O(l) the equilibrium constant, K, has a value of 0.029 at 25.0oC. This value clearly shows that water is the stronger acid than methanol, because the methanol is unlikely to protonate the hydroxide ion (the conjugate base of water). Instead, the opposite is true; water is likely to protonate the methoxide ion (the conjugate base of the methanol.) To obtain the direct reaction between CH3OH and H2O, CH3OH(aq)+H2O(l)−⇀↽−CH3O−(aq)+H3O+(aq) along with the value of the equilibrium constant for this reaction, Ballinger and Long add reactions 18 and 5 to obtain 19: \| \| \| \| --- \| CH3OH(aq)+OH−(aq)⇌CH3O−(aq)+H2O(l) +H2O(l)+H2O(l)⇌OH−(aq)+H3O+(aq) \| K18=0.029 K6=1×10−14 \| pK = 1.54 pK = 14.0 \| \| CH3OH(aq)+H2O(l)⇌CH3O−(aq)+H3O+(aq) \| K19=K18·K6=2.9×10−16 \| pK = 15.54 \| Thus, there are two more pieces of evidence in the paper that seems to have started the discussion, that the entire discussion concerning the relative strength of methanol and water as acids in aqueous solution has been completely misunderstood. All experimental data show that water has a pKw = pKa = 14.0 at 25.0oC, and all experimental data show that methanol is a weaker acid than water in aqueous solution. There is absolutely no experimental evidence to contradict either of these two statements. Conclusions There would be little or no confusion in this matter if it were made clear from the outset that solvents can affect the relative acidity of compounds.7,8 A straightforward method of clarifying the effect of solvent on relative acid strengths would be to present a separate table for each solvent. There would also be less confusion in this matter if it were made clear that the solvent does belong in the law of mass action in its complete form, but that the solvent does not usually appear in the final form of the law of mass action because the solvent in an ideal dilute solution has an activity of unity. It is incorrect and misleading to present the value of 15.7 for the pKaof water, yet this value has entered the fields of organic chemistry and biochemistry. The proposed value of 1.8 x 10-16 for the Ka of water cannot be justified with thermodynamic data, nor are there any experimental data to support this value. In fact, 1.8 x 10-16 is a hypothetical value that was arrived at using specious arguments in order to justify an incorrect assumption that the relative strengths of acids were not affected by changes in solvent. There is no reason to use this value. The Ka of water at 25 ºC is 1×10−14 and the pKa is 14.0. References Meister, E.C.; Willeke, M.; Angst, W.; Togni, A.; Walde, P. Helv. Chim. Acta 2014, 97, 1. Silverstein, T.P.; Heller, S.T. J. Chem. Educ., 2017, 94 (6), pp 690–695. Marshall, W. L.; Franck, E. U. J. Phys. Chem. Ref. Data, 1981, 10, 295. Bandura, A. V.; Lvov, S. N. J. Phys. Chem. Ref. Data, 2006, 35, 15. Bronsted, J. Acid and Basic Catalysis. Chem. Rev. 1928, 5 (3), 231–338. J.N. Brønsted, Physical Chemistry, William Heinemann Ltd., London, 1937. Available at Bell, R.P. Rates and Equilibria in the Ionisation of C-H Bonds. Trans. Farad. Soc. 1943, 39, 253-259. Bell, R.P. The Brönsted Equation—Its First Half-Century. In: Chapman, N.B., Shorter, J. (eds) Correlation Analysis in Chemistry. Springer, Boston, MA. 1978 A. J. Kresge, The Brønsted relation – recent developments. Chem. Soc. Rev., 1973, 2, 475-503. Ballinger, P.; Long, F.A. J. Am. Chem. Soc. 1960, 82, 795. Hine, J.; Hine, M. J. Am. Chem. Soc. 1952, 74, 5266. Cox, B.G. Acids and Bases: Solvent Effects on Acid–Base Strength, Oxford University Press, 2013. (Note: Even though the author clearly describes solvent effects, he, too, follows the incorrect convention of including concentrations in calculating K values.) Heller, S.T., Silverstein, T.P. pKa values in the undergraduate curriculum: introducing pKa values measured in DMSO to illustrate solvent effects. ChemTexts 6, 15 (2020). Harris, D. C. Quantitative Chemical Analysis, 7th ed. New York: W. H. Freeman and Co., 2007, pages 141-146. The NBS Tables of Chemical Thermodynamic Properties, J. Phys. Chem. Ref. Data, 1982, 11, Supplement No. 2. (as referenced in Noggle, J. H. Physical Chemistry 3rd ed. New York: HarperCollins, 1996) Neils, T.L.; Silverstein, T.P.; and Schaertel, S. J. Chem. Educ., 2023 100 (4), 1676-1679. Atkins, P. and dePaula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10thed. New York: W. H. Freeman and Co., 2014, pages 187-190. Neils, T.L.; Schaertel, S.; and Silverstein, T.P. Helvetica Chimica Acta, 2024, "The pKa of Water and the Fundamental Laws Describing Solution Equilibria: An Appeal for a Consistent Thermodynamic Pedagogy", Silverstein, T. P.; Neils, T. European Journal of Organic Chemistry, February, 2025 28(9), "A Meta‐Analysis of Published Brønsted Plots Supports Brønsted's “Conventional” pKa Values: 14.00 for Water and 0.00 for H(aq)." Contributors Tom Neils (Grand Rapids Community College) Stephanie Schaertel (Grand Valley State University) The Use of Curly Arrows
15585	https://matheducators.stackexchange.com/questions/27577/why-do-some-pre-calculus-text-allow-r0-in-polar-coordinates	precalculus - Why do some (pre-) calculus text allow $r<0$ in polar coordinates? - Mathematics Educators Stack Exchange Join Mathematics Educators By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Why do some (pre-) calculus text allow r<0 r<0 in polar coordinates? Ask Question Asked 1 year, 6 months ago Modified1 year, 6 months ago Viewed 404 times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. Form this question, I was surprised to learn that it is common for calculus textbooks in the US to allow r<0 r<0 when discussing polar coordinates. This answer by Dan Fox summarizes some mathematical and pedagogical drawbacks of this. Let me outline them and add some more: If we restrict to r≥0 r≥0, then (r,θ)(r,θ) have a natural geometric meaning - the distance to the origin and the polar angle, the latter defined up to 2 π 2 π by necessity. Allowing r<0 r<0 makes it much messier and necessarily de-emphasizes the important geometric picture. This aligns well with the view that coordinates are functions on the manifold, or, if you wish, that they are numbers assigned to physical points. Then, equations in polar coordinates, such as r 2=sin θ r 2=sin⁡θ, simply describe the set of points which satisfy the equation when fed into it (with some choice of a branch for θ θ), just as it is the case with equations in Carthesian coordinates. An alternative view, as expressed in comments, is that polar coordinates is a map P:(r,θ)↦(r cos θ,r sin θ)P:(r,θ)↦(r cos⁡θ,r sin⁡θ), and the equation denotes an image of the corresponding set in (r,θ)(r,θ) plane. Apart from a need to re-define the meaning of an equation, in practice one is never interested in an image of a given set under P P - rather, one is given a set V V in the physical plane and looks for a description of that set in polar coordinates, i.e., a set U U such that P(U)=V P(U)=V, e.g., to integrate in polar coordinates. In practice, this is nearly always done with the restriction r≥0 r≥0, especially since bijectivity on U U is desirable. So, the question is: what are the benefits of allowing r<0 r<0? If it is in the textbooks by historical reasons, what was the historical point of view that lead to it? What are some applications of polar coordinates where this is useful? precalculus vector-calculus Share Share a link to this question Copy linkCC BY-SA 4.0 Follow Follow this question to receive notifications asked Mar 8, 2024 at 10:12 Kostya_IKostya_I 1,471 8 8 silver badges 13 13 bronze badges 3 3 This is similar to allowing θ θ outside [0,2 π][0,2 π].Gerald Edgar –Gerald Edgar 2024-03-08 13:14:01 +00:00 Commented Mar 8, 2024 at 13:14 1 The answers so far do not seem to adequately address the pedagogical side of this question. Precalc students don't need to care about branch cuts and so on. To me, the benefit of keeping r>0 r>0 is that it gives students a useful geometric lodestar: r r is the distance from the origin. If we throw this away, we should get something very good in return.user1149748 –user1149748 2024-03-08 21:24:29 +00:00 Commented Mar 8, 2024 at 21:24 Also: polar coordinates are useful in situations with rotational symmetry, or that are close enough to rotationally symmetric to make polar coordinates a better choice than Euclidean. Conveniently describing a line through the origin, or a circle not centered at the origin, does not strike me as convincing because polar coordinates aren't made for these tasks.user1149748 –user1149748 2024-03-08 21:28:55 +00:00 Commented Mar 8, 2024 at 21:28 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 11 Save this answer. Show activity on this post. There is still a simple geometric meaning for polar coordinates when we allow for negative values of r r. (It's how I was taught polar coordinates.) Given an ordered pair (r,θ)(r,θ), Start at the origin facing the polar axis. Rotate by the directed angle θ θ. (Counterclockwise if θ>0 θ>0 and clockwise if θ<0 θ<0.) Travel the directed distance r r. (Forwards if r>0 r>0 and backwards if r<0 r<0.) This does not fit the notion of a coordinate chart, which should assign to each point a unique list of coordinates. But it makes perfect sense as a coordinate transformation, which assigns to each list of coordinates a point. If we restrict P P to the domain [0,∞)×[0,2 π)[0,∞)×[0,2 π), then we have an (almost) bijection that can be used to define a coordinate chart. But it is sometimes more convenient to restrict to R×(−π 2,π 2]R×(−π 2,π 2]. And when we don't need a coordinate chart and bijectivity isn't important, why not allow for the domain to be R 2 R 2? I'm not sure what you mean by needing to "re-define the meaning of an equation." I was taught that an equation is an open statement of equality involving some number of free variables. The solutions of an equation, given an ordering of the variables, are lists of values that satisfy the equation. To produce a graph of an equation, we interpret the solutions as coordinates, then map coordinates to points. In addition to the examples of circles, spirals, hyperbolas, and roses others have already given, there are many more interesting and beautiful curves generated by simple polar equations that rely on negative values of r r and/or values of θ θ outside of [0,2 π)[0,2 π). For example, the graph of r=1 2+cos(θ)r=1 2+cos⁡(θ) is a limaçon with an inner loop. Here's another situation where it makes perfect sense to use negative values of r r. Consider a particle traveling down the y y-axis whose position in Cartesian coordinates is described by r C(t)=(0,1−t)r C(t)=(0,1−t). Its position in polar coordinates can be described simply as r P(t)=(1−t,π 2)r P(t)=(1−t,π 2). This matches our sense that the particle never "changes direction," even while passing through the origin. But if negative values of r r are not allowed, the parametrization must be discontinuous at t=1 t=1: r P(t)=(1−t,π 2)r P(t)=(1−t,π 2) for t≤1 t≤1 and r P(t)=(t−1,3 π 2)r P(t)=(t−1,3 π 2) for t>1 t>1. The same applies to almost any other smooth path through the origin. Share Share a link to this answer Copy linkCC BY-SA 4.0 Follow Follow this answer to receive notifications edited Mar 8, 2024 at 14:57 answered Mar 8, 2024 at 14:44 Justin HancockJustin Hancock 3,365 8 8 silver badges 26 26 bronze badges 2 I think you answer sheds some light on the question "how did it get into textbooks": it came from the times when "function" meant an analytic function and "curve" meant an (entire) analytic branch; so restriction to r>0 r>0 would sometimes mean that we abruptly cut such a branch and "lose a part of it". Since this point of view is entirely abandoned elsewhere in curricula now, I don't see a good reason to stick to it here either.Kostya_I –Kostya_I 2024-03-08 22:27:43 +00:00 Commented Mar 8, 2024 at 22:27 In usual professional mathematical usage "coordinates" mean a bijective map (with some further restrictions depending on the category in which one works). The idea is that coordinates should uniquely specify a point on the space under consideration. There is nothing wrong with allowing negative r, but it seems to me an error to refer to such as a "coordinate".Dan Fox –Dan Fox 2024-03-20 07:07:22 +00:00 Commented Mar 20, 2024 at 7:07 Add a comment\| This answer is useful 8 Save this answer. Show activity on this post. It would be nice if we distinguished between "polar coordinates" and "polar parameterizations". Let Ω⊂R 2 Ω⊂R 2 be an open set which does not contain any loop around the origin. Then a choice of polar coordinates for Ω Ω is a choice of smooth section of P P. Such a section always exists and, if Ω Ω is connected, the section is completely determined up to a sign of r r and a multiple of 2 π 2 π for θ θ. You are correct that we never have polar coordinates where r r changes sign. On the other hand, we can describe some subsets S x,y⊂R 2 S x,y⊂R 2 conveniently using polar parameterizations. A polar parameterization is just describing S x,y S x,y as S x,y=P(S r,θ)S x,y=P(S r,θ) for some S r,θ⊂R 2 S r,θ⊂R 2. Note that polar coordinates are polar parameterizations, but not all polar parameterizations are polar coordinates! When we write "polar equations" are are actually talking about polar parameterizations, not polar coordinates. Let's look at the equation r=cos(θ)r=cos⁡(θ), with θ∈[0,π]θ∈[0,π] to describe a circle. Note that the restriction of P P from the graph of r=cos(θ)r=cos⁡(θ), with θ∈[0,π]θ∈[0,π], is smooth through r=0 r=0! This is as nice as it gets. In general if f:(θ 1,θ 2)→R f:(θ 1,θ 2)→R is any smooth function, the map θ↦(f(θ),θ)↦P(f(θ),θ)θ↦(f(θ),θ)↦P(f(θ),θ) will also be smooth. When we write r=f(θ)r=f(θ) for θ∈(θ 1,θ 2)θ∈(θ 1,θ 2) we are actually thinking about a parameterized curve in the (x,y)(x,y) plane. We are not (as you claim in the OP) graphing the equation r=f(θ)r=f(θ) where (r,θ)(r,θ) are polar coordinates for some region containing the circle (in fact, no such polar coordinates exist since no region containing the origin has a smooth section). Usually you will not want to include a point where r=0 r=0 even in a parameterization of a 2D region, since P P is locally "two to one" on any punctured neighborhood of a point (0,θ 0)(0,θ 0). However, there are even circumstances when this is natural. When I want to find the area of the disk bounded by r=cos(θ)r=cos⁡(θ), it is natural to use an integral from θ=0 θ=0 to θ=π θ=π. It is still one to one and nice enough that you can write the whole area of the disk as ∫π 0 cos 2(θ)d θ∫0 π cos 2⁡(θ)d θ with no issue (breaking it up from 0 0 to π 2 π 2 and π 2 π 2 to π π gives two nice non-overlapping pieces). I wouldn't usually think about the pre-image when I write down this integral: I just think about cutting the disk into small pieces. While I agree that for many purposes it is nice to restrict to r>0 r>0 it is not always needed. I want to illustrate how a region can have polar coordinates without restricting θ∈[0,2 π)θ∈[0,2 π). For example, P P restricts to a diffeomorphism between the rectangle (0.5,3)×(0,6 π)(0.5,3)×(0,6 π) in the (r,θ)(r,θ) plane and the region 0.5+θ<r<3+θ 0.5+θ<r<3+θ with θ∈(0,6 π)θ∈(0,6 π). I would like to be able to work with this region without breaking it up into chunks which are artificially restricted to only go from θ=0 θ=0 to θ=2 π θ=2 π. I will also note that contending with these kinds of issues is bread and butter stuff in complex analysis. EDIT: One more note. The OP says that: This aligns well with the view that coordinates are functions on the manifold, or, if you wish, that they are numbers assigned to physical points. Then, equations in polar coordinates, such as r 2=sin(θ)r 2=sin⁡(θ), simply describe the set of points which satisfy the equation when fed into it (with some choice of a branch for θ θ ), just as it is the case with equations in Cartesian coordinates. This point of view has a few serious drawbacks. First of all θ θ is not a coordinate map. It is locally but not globally. Hence the need for a choice of the branch. More seriously, this interpretation of a "polar equation" never admits r=0 r=0 as a solution. Let's be explicit and attempt to define r(x,y)=x 2+y 2−−−−−−√r(x,y)=x 2+y 2 θ(x,y)=atan2(x,y)θ(x,y)=atan2⁡(x,y) See this definition of atan2(x,y)atan2⁡(x,y). Then according to OP definition a solution of r 2=sin(θ)r 2=sin⁡(θ) is a point (x,y)(x,y) satisfying x 2+y 2=sin(atan2(x,y))x 2+y 2=sin⁡(atan2⁡(x,y)) Since atan2(0,0)atan2⁡(0,0) we must admit that (0,0)(0,0) is not a solution of r 2=sin(θ)r 2=sin⁡(θ) under the definition of the OP. This is not to say that this is a poor definition: it is good for some purposes and not for others. Share Share a link to this answer Copy linkCC BY-SA 4.0 Follow Follow this answer to receive notifications edited Mar 8, 2024 at 17:51 answered Mar 8, 2024 at 13:36 Steven GubkinSteven Gubkin 26.6k 4 4 gold badges 66 66 silver badges 113 113 bronze badges 8 1 I agree that not restricting θ θ to [0,2 π)[0,2 π) is fundamental. But in the first example, what are the advantages compared to taking θ∈[−π/2,π/2)θ∈[−π/2,π/2) (or, if you wish, θ∈[0,π/2)∪[3 π 2,2 π)θ∈[0,π/2)∪[3 π 2,2 π)?)Kostya_I –Kostya_I 2024-03-08 15:17:12 +00:00 Commented Mar 8, 2024 at 15:17 2 I am saying that, for curves, "r=f(θ)r=f(θ)" is a convenient shorthand for θ↦(f(θ)cos(θ),f(θ)sin(θ))θ↦(f(θ)cos⁡(θ),f(θ)sin⁡(θ)): a perfectly respectable parameterized curve no matter the sign of f(θ)f(θ).Steven Gubkin –Steven Gubkin 2024-03-08 15:19:49 +00:00 Commented Mar 8, 2024 at 15:19 If you really want to be consistent and say that "coordinates are functions on the manifold", then you have to deal with the issue of branch cuts. My definition of "a choice of polar coordinates" above is the only rigorous fix I am aware of. Restricting P P to r>0 r>0 and θ∈[0,2 π)θ∈[0,2 π) is not really "coordinates" by that definition either. It only gives us coordinates for Ω=R 2−{(x,0):x≥0}Ω=R 2−{(x,0):x≥0}.Steven Gubkin –Steven Gubkin 2024-03-08 15:25:45 +00:00 Commented Mar 8, 2024 at 15:25 If you want a looser definition of coordinates, so that for you r:R 2−{0,0}→(0,∞)r:R 2−{0,0}→(0,∞) and θ:R 2→S 1 θ:R 2→S 1 (as you seem to imply when you say that θ θ is only defined up to 2 π 2 π), then your "polar coordinates" do not function at the origin. It would not be legit to ever use r=0 r=0 or describe the point (0,0)(0,0) with these coordinates. So the circle centered at (0.5,0)(0.5,0) of radius 0.5 0.5 is simply not expressible using your coordinates.Steven Gubkin –Steven Gubkin 2024-03-08 15:34:02 +00:00 Commented Mar 8, 2024 at 15:34 I don't know why you are dwelling on the origin so much: I'm happy to exclude the origin; polar coordinates are badly behaved there either way. I see no natural reason why the curve r=sin θ r=sin⁡θ must contain the origin but the logarithmic spiral r=e θ r=e θ must not.Kostya_I –Kostya_I 2024-03-08 22:10:55 +00:00 Commented Mar 8, 2024 at 22:10 \|Show 3 more comments This answer is useful 4 Save this answer. Show activity on this post. Allowing r<0 r<0 certainly does not de-emphasize the geometric picture. The image below is the graph of r=sin(3 θ).r=sin⁡(3 θ). As θ θ goes from 0 0 to π/3,π/3,3 θ 3 θ goes from 0 0 to 1 1 and back down to 0,0,r r goes from 0 0 to 1 1 and then back down to 0,0, and you see the part of this figure that is in the first quadrant. It lies below the line θ=π/3,θ=π/3, the 60∘60∘ ray. As θ θ goes on from π/3 π/3 to 2 π/3,2 π/3,r r goes from 0 0 to −1−1 and back up to 0,0, giving us the part of this picture that is below the x x-axis. The third leaf, in the second quadrant, has θ θ going from 2 π/3 2 π/3 to π.π. Then as θ θ goes from π π to 4 π/3,4 π/3,r r is negative again and we traverse the first leaf, in the first quadrant, for the second time. And so on. Negative values of r r have a natural geometric meaning, just as positive values do. Share Share a link to this answer Copy linkCC BY-SA 4.0 Follow Follow this answer to receive notifications answered Mar 8, 2024 at 15:30 Michael HardyMichael Hardy 2,128 13 13 silver badges 26 26 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Polar curves are a very old subject. This article, published in 1949 in the American Mathematical Monthly, suggests that the subject goes back to Newton. Newton as an Originator of Polar Coordinates. In this framework, we represent curves in the plane by the scheme R=f(θ)R=f(θ). There is simplicity and elegance that would not be present if we decreed that the polar variable R R must be positive. A simple example is the curve R=cos(θ)R=cos⁡(θ). If there is a rule forbidding R<0 R<0 then we are stuck with a semicircle. But without the restriction, the graph is the entire circle, traversed for θ∈[0,π]θ∈[0,π]. Another example is the equation of a conic in polar coordinates. A very general form is R=l 1−e cos(θ)R=l 1−e cos⁡(θ) where e e is the eccentricity. The equation unifies conic sections. But alas, if we had to live under a regime where R<0 R<0 is forbidden, the hyperbolas would be missing one of their branches. It is also amusing to think about what would happen to curves like the four-leaf rose R=cos(2 θ)R=cos⁡(2 θ). I suppose we would have to rename it, since it would have just two lobes. Even then it is awkward because you have to wait and look the other way while θ θ progresses through the intervals where cos(2 θ)<0 cos⁡(2 θ)<0. Share Share a link to this answer Copy linkCC BY-SA 4.0 Follow Follow this answer to receive notifications answered Mar 8, 2024 at 13:53 user52817user52817 12.1k 21 21 silver badges 50 50 bronze badges 4 1 Sorry, I don't follow. You don't need R<0 R<0 for the equation "R=cos θ R=cos⁡θ" to describe the full circle; just use the range −π 2≤θ<π 2−π 2≤θ<π 2, which is also geometrically the range of polar rays that do intersect the circle. In other two examples, taking the absolute value, or squaring both sides, restores all the desired branches.Kostya_I –Kostya_I 2024-03-08 14:58:47 +00:00 Commented Mar 8, 2024 at 14:58 2 @Kostya_I The fact that you have to futz around with domain restrictions is unnecessary overhead, resulting from requiring r>0 r>0.user52817 –user52817 2024-03-08 15:54:49 +00:00 Commented Mar 8, 2024 at 15:54 1 if you prefer [0,2 π)[0,2 π) as the domain, the equation r=cos θ r=cos⁡θ still describes the full circle. A semi-circle only entered the picture because you futzed around with the domain.Kostya_I –Kostya_I 2024-03-08 16:09:34 +00:00 Commented Mar 8, 2024 at 16:09 @Kostya_I You want the equation r=cos(θ)r=cos⁡(θ) for θ∈[−π 2,π 2]θ∈[−π 2,π 2] to mean "the set of all allowable point (x,y)(x,y) where the equation r(x,y)=cos(θ(x,y))r(x,y)=cos⁡(θ(x,y)) is satisfied, given my definition of the functions r r and θ θ". You must then admit that (x,y)=(0,0)(x,y)=(0,0) is not in the solution set, as θ θ is undefined at (x,y)=(0,0)(x,y)=(0,0). It only makes sense to include (0,0)(0,0) in the solution set if r=cos(θ)r=cos⁡(θ) is interpreted as a parametric curve. When you do this, there is no need to restrict r>0 r>0: it is artificial.Steven Gubkin –Steven Gubkin 2024-03-08 17:06:06 +00:00 Commented Mar 8, 2024 at 17:06 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. There is a clear interpretation of negative angles, and we allow angles to be negative. There is a clear interpretation of negative radii, so it's only natural (or at least "consistent") to do the same and allow radii to be negative. Sure, there are some more advanced contexts where that causes an issue. But we can resolve those issues when we get to them (by introducing the restriction that radii are to be non-negative in those contexts). Share Share a link to this answer Copy linkCC BY-SA 4.0 Follow Follow this answer to receive notifications edited Mar 9, 2024 at 3:22 answered Mar 8, 2024 at 15:24 Justin SkycakJustin Skycak 13.1k 5 5 gold badges 42 42 silver badges 63 63 bronze badges 2 3 Given the very different sorts of roles played by angles and radii in polar coordinates, it makes no sense to say that it's only natural to do the same things with one that one does with another. There are reasons to allow radii to be negative, but that can't be it.Michael Hardy –Michael Hardy 2024-03-08 17:23:48 +00:00 Commented Mar 8, 2024 at 17:23 @MichaelHardy okay, then replace the word "natural" with "consistent".Justin Skycak –Justin Skycak 2024-03-08 20:17:30 +00:00 Commented Mar 8, 2024 at 20:17 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 precalculus vector-calculus See similar questions with these tags. 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15586	https://allanbickle.wordpress.com/wp-content/uploads/2016/05/2tonejcpaper.pdf	2-Tone Coloring of Joins and Products of Graphs Allan Bickle Department of Mathematics Western Michigan University 1903 W. Michigan Kalamazoo, MI 49008 email: allan.e.bickle@wmich.edu Abstract A 2-tone coloring of a graph assigns two distinct colors (a label) to each vertex with the restriction that adjacent vertices have no common colors, and vertices at distance two have at most one common color. The 2-tone chromatic number of a graph is the minimum number of colors in any 2-tone coloring. When considering the 2-tone chromatic number of the join of two graphs, we nd that each part of the join cannot repeat any label. Adding this restriction leads to the denition of pair coloring. We determine basic bounds and exact values of the pair chromatic number for specic classes of graphs. We also determine the 2-tone chromatic number for Cartesian prod-ucts of some graphs. These results show some unexpected connections to Latin squares and perfect matchings. 1 Introduction Classical vertex coloring of graphs has been generalized in many dierent ways. (See , for basic concepts and results.) A set coloring assigns a set of k colors to each vertex so that adjacent vertices have no common colors . A distance k coloring requires that vertices within distance k receive distinct colors. A related notion is an L (2, 1) labeling, in which every vertex receives a nonnegative integer label, with vertices at distance one having labels at least two apart and vertices at distance two having labels at least one apart . These denitions have been generalized to dene a coloring that assigns k colors to each vertex with restrictions on which sets may appear within distance k of each other that are stronger the smaller the distance. 1 Denition 1. Let G be a graph, k, t ∈N, [k] = {1, 2, ..., k}, and let Pt([k]) denote the set of t-element subsets of [k]. A function f : V (G) →Pt([k]) is called a proper t-tone k-coloring (or sometimes just a t-tone coloring) of G if \|f(u) ∩f(v)\| < d(u, v) for all distinct vertices u and v of G. A graph is t-tone k-colorable if it has a proper t-tone k-coloring. The t-tone chromatic number of G, denoted by τt(G), is the smallest positive integer k for which G has a proper t-tone k-coloring. This denition was suggested by Gary Chartrand. It was rst studied in a class project of Nicole Fonger, Josh Goss, Ben Phillips, and Chris Segroves . It was explored in more detail in a paper by Ben Phillips and this author . A paper by Dan Cranston, Jaehoon Kim, and William Kinnersley provides some improved upper bounds. A paper by Deepak Bal, Patrick Bennett, An-drzej Dudek, and Alan Frieze determines the 2-tone chromatic number of the random graph. Note that for t = 1, τ1 (G) = χ (G), the usual chromatic number of a graph G. This paper shall be solely concerned with the 2-tone chromatic number. We shall often call f(v) the label associated with the vertex v of the coloring f, and the elements of f(v) will be called colors. Thus, in a 2-tone coloring, each vertex has a label of 2 distinct colors. Adjacent vertices have no common colors, and vertices distance two apart have at most one common color. When the context is clear, the label {a, b} will be denoted ab. Vertices distance two apart are called second-neighbors. A 2-chord of a cycle is a pair of vertices of the cycle at distance two apart. Some basic results are immediate. The 2-tone chromatic number exists for all graphs. If H is a subgraph of G then τ2(H) ≤τ2(G). We have τ2(Kn) = 2n. If G has components Gi, then τ2(G) = max τ2(Gi). Several concepts related to proper coloring will be useful. The k-core of a graph G is the maximal induced subgraph H ⊆G such that δ(G) ≥k, if it exists. The core number of a vertex is the largest value for k such that v is in the k-core. Let D (G) be the degeneracy of G, the maximum k such that G has a subgraph H with δ (H) ≥k. A graph is k-degenerate if it has degeneracy at most k. It is immediate from the denition that D (G) ≤△(G). A deletion sequence of a graph is a sequence of vertices formed by successively deleting a vertex of minimum degree. A construction sequence is the reversal of a deletion sequence. (See , for background on these concepts.) The 2-tone chromatic number has been determined for many basic classes of graphs in and . For complete multipartite graphs, τ2 (Ka1,...,ar) = r X i=1 1 + √1 + 8ai 2 . Hence for the nontrivial star, τ2 (K1,s) = l 5+√1+8s 2 m . For a nontrivial tree T 2 Figure 1: A 2-tone 5-coloring of the Petersen graph. with maximum degree △, τ2 (T) = l 5+√1+8△ 2 m . Consider forming a graph whose vertices are the ten possible labels for a 2-tone 5-coloring and all possible edges are added. This graph is the Petersen graph. In fact, the Petersen graph can be dened with just this labeling. Thus its 2-tone chromatic number is ve, so any subgraph of the Petersen graph is 2-tone 5-colorable. (See Figure 1.) This result can also be seen by noting that the Petersen graph is the complement of the line graph of K5. Let Ln = L (Kn), the complement of the line graph of Kn. Thus L5 is the Petersen graph. For n ≥4, we have τ2 (Ln) = n . For the cycle Cn, τ2 (Cn) = 6 n = 3, 4, 7 5 else . If τ2 (G) = k, we call a 2-tone k-coloring of G a minimum coloring. The minimum colorings of Cn are unique up to isomorphism of colors and graphs for n = 3, 4, 5, 6, 8, 9. That is, any 2-tone k-coloring with k = τ2 (Cn) can be converted to any other by renaming of the colors and taking an automorphism . The theta graph θi,j,k is formed by taking paths of lengths i, j, k and identifying them at their endvertices. It necessarily contains three cycles of lengths a = i + j, b = i + k, c = j + k. We will use (i, j, k) for θi,j,k. The theta graph θ1,2,2 has τ2 (θ1,2,2) = 7, and (3, 3, 3), (3, 3, 5), (3, 3, 6), (4, 4, 4), (4, 4, 5), and (3, 3, 9), all have 2-tone chromatic number 6. For all other theta graphs, τ2 (θi,j,k) = max {τ2 (Ca) , τ2 (Cb) , τ2 (Cc)}, where a = i + j, b = i + k, c = j + k . 3 A graph G is 2-tone k-critical if τ2 (G) = k and for any proper subgraph H of G, τ2 (H) < k. The following theorem proves useful in several later theorems. Theorem 2. Let G be a nonempty graph with u, v ∈V (G) not adjacent and set e = uv. Then τ2 (G + e) −τ2 (G) ≤1. 2 Upper Bounds Upper bounds for for 2-tone and t-tone coloring were studied in 2011 by Cranston, Kim, and Kinnersley . We consider some slight improvements on their bounds. Theorem 3. Let G be a graph with degeneracy k and maximum degree ∆= ∆(G). Then τ2 (G) ≤2k + 1+√ 9+8(2∆k−∆−k2) 2 . Proof. Color G with a construction sequence. When colored, a vertex v has j ≤k neighbors already colored, which excludes 2j colors. Each of the j neighbors has at most ∆−1 second-neighbors. Then v has at most ∆−j uncolored neighbors, each of which has at most k−1 colored second-neighbors. Thus there are at most j (∆−1) + (∆−j) (k −1) second-neighbors already colored, which is maximized when j = k. Thus we need r extra colors, where r 2 ≥2∆k −∆−k2 + 1 ≥j (∆−1) + (∆−j) (k −1) + 1. Solving, we nd r ≥ 1+√ 9+8(2∆k−∆−k2) 2 . Theorem 4. Let G be a graph with maximum degree ∆≥2. Then τ2 (G) ≤ 2∆−1 + 1+√ 1+8∆(∆−1) 2 . Proof. We may assume G is connected and regular. Now G −e is ∆−1-degenerate. As in the previous proof, color G−e using a construction sequence. Each vertex has at most ∆−1 neighbors which exclude at most 2 (∆−1) colors. There are at most (∆−1)2+(∆−2) = ∆(∆−1)−1 second-neighbors already colored, so we need r extra colors, where r 2 ≥∆(∆−1). Solving, we nd r ≥ 1+√ 1+8∆(∆−1) 2 . Lastly, adding e back requires at most one more color by Theorem 2. Note that proved the upper bound τ2 (G) ≤2∆+ 1+√ 9+8∆(∆−1) 2 ≤ 2 + √ 2 ∆. The bound in the previous theorem is at least one better. In , 2-tone 5-colorable cubic graphs are characterized. It was conjectured that 4 Conjecture 5. Let G be a cubic graph. Then a. τ2 (G) ≤8; b. τ2 (G) ≤7 when G does not contain K4; c. τ2 (G) ≤6 when G does not contain K4 −e. Part a was proved in . Theorem 6. If G is cubic, τ2 (G) ≤8. It is not hard to check that part c holds for Cn × K2 and Mobius ladders. Proposition 7. Let n ≥3. We have τ2 (Cn × K2) = 6. Proof. Since Ci × K2 contains C4, τ2 (Cn × K2) ≥6. The tables below repre-sent colorings of Ci × K2, for i = 3, 4, 5. Each cell represents a vertex, and neighboring cells and cells on opposite ends of a row are adjacent. The rst two tables can be concatenated to build products of longer cycles and K2. 25 34 16 14 26 35 25 13 24 16 14 26 15 23 23 16 24 15 46 14 25 13 26 35 A Mobius ladder can be formed by adding edges between opposite vertices of an even cycle. It can also be considered as a product of a cycle and K2 with a 'twist' in it, hence the name. Proposition 8. Let G be a Mobius ladder. Then τ2 (G) = 6. Proof. Any Mobius ladder contains C4. Consider the following Mobius ladders, where the left and right edges are identied with a 'twist'. In the latter two, the columns with parentheses can be successively inserted one at a time. 12 46 23 45 13 56 12 (35) 46 (15) 23 (16) 45 (26) 13 (24) 56 (34) 12 (35) 46 15 26 45 16 23 45 (26) 13 24 35 12 34 56 Larger Mobius ladders can be constructed by concatenating smaller ones an odd number of times to maintain the 'twist'. Part c was refuted by demonstrating that it fails for the Heawood graph, which does not contain K4−e (and indeed has girth 6). Their proof uses a clever but somewhat involved contradiction. The following shorter proof analyzes the maximal independent sets of the Heawood graph. Theorem 9. The Heawood Graph is not 2-tone 6-colorable. Proof. Recall that the Heawood Graph G is the incidence graph for the Fano Plane, or equivalently, the Steiner Triple System of order 7 (STS(7)). Hence it is bipartite and any two vertices in the same partite set have exactly one common neighbor. 5 Figure 2: The cubic 2-tone 7-critical graph of order 10 Consider the maximal independent sets of G. Each partite set is a maxi-mum independent set of size seven. A vertex in one partite set is adjacent to three in the other, so there can be four vertices from the other partite set in the maximal independent set. Call an independent set with four vertices in one partite set and one in the other a 1-4-set. Two vertices in one partite set have a total of ve distinct neighbors in the other set. Hence there can be an independent set with two vertices in each partite set. Suppose that G is 2-tone 6-colorable. If there is a color class of size seven, then it is one of the partite sets, and each vertex requires a distinct color for its other color. If there is a color class of size six contained in one of the partite sets, then there are at most two color classes with two vertices in this partite set, and at least four more colors are needed. Hence each color class has size at most ve. But then the six color classes must have sizes 5,5,5,5,4,4. Restricted to one partite set, they have sizes 5, 4, 2, or 1. Hence there are at least two with size at least four. Now any two color classes can overlap on at most one vertex of a partite set, so there are two with size four, which must 1-4-sets. Now WLOG any 1-4-set contains four vertices that do not contain a triangle of the STS(7). But no other 1-4-set can contain only one vertex of this set, since then it would contain a triangle. This is a contradiction. A 2-tone 7-coloring of the Heawood graph appears in , verifying that it has 2-tone chromatic number 7. After learning of the falsication of part c, this author checked all 21 cubic graphs of order 10 and found that one of them also violates part c. Proposition 10. Let G be the graph formed by starting with two copies of K2,3 and adding a matching between the vertices of degree two in the two K2,3's (see Figure 2). Then G is 2-tone 7-critical. Proof. Suppose G has a 2-tone 6-coloring. Now K2,3 is uniquely 6-colorable, and each partite set requires three distinct colors. Let A and B be the partite 6 sets of one K2,3 and C and D be the partite sets of the other K2,3, with A and C being those that have three vertices, and hence are joined by the matching. If there is a color in common between A and C, then it must appear on two vertices in each set, and hence on adjacent vertices. If they have no common colors, then B and C use the same three colors. But then they have two common labels at distance two apart. G has only two edge orbits. Considering G −e for one edge of each type, 2-tone 6-colorings are easily obtained. Hence G is 7-critical. The existence of a cubic 7-critical graph is of interest, especially since no cubic graph is 6-critical . It is also interesting that both counterexamples are bipartite. 3 Joins and Pair Coloring For the 2-tone chromatic number of a join of graphs, we have the following partial results. Proposition 11. For the join G + H, τ2 (G + H) ≥τ2 (G) + τ2 (H) . If G and H have diameter at most 2, then this is an equality. Proof. The inequality follows since no common color can be used in both fac-tors G and H of the join. If G and H both have diameter at most 2, then so does G + H. Therefore combining minimum colorings for G and H creates no conict, so the bound is achieved. The bound may not be exact because vertices at distance greater than two in G will have distance two in G+H. The converse is false, as for example the wheel W6 = C6 +K1 achieves the bound even though the 6-cycle has diameter 3. Thus for any factor in a join, the vertices must have a 2-tone coloring with the additional restriction that each label must be distinct. This motivates the following denition. Denition 12. A pair k-coloring of a graph G is a 2-tone k-coloring in which every label is distinct. A graph is pair k-colorable if it has a pair k-coloring. The pair chromatic number of G, pc (G), is the smallest k for which it has a pair k-coloring. Some results on the pair chromatic number are immediate. We have pc (G) ≥ τ2 (G), and if diam (G) ≤2, then this is an equality. Hence it is an equality for almost all graphs. If H is a subgraph of G, then pc (H) ≤pc (G). It is not 7 dicult to show that pc (G + e) −pc (G) ≤1. It is also straightforward to see that pc (G + H) = pc (G) + pc (H). A graph G is pair k-colorable if and only if it is contained in Lk. Thus if n > k 2 , pc (G) > k. Equivalently, pc (G) ≥ 1+√1+8n 2 . Thus pc Kn = l 1+√1+8n 2 m . This also implies that given n1 = n (G) and n2 = n (H), τ2 (G + H) ≥τ2 (Kn1,n2) = X i=1,2 1 + √1 + 8ni 2 . since G + H contains Kn1,n2 as a subgraph. This bound appears to be good for sparse graphs, but it is unclear exactly when it is an equality. Theorem 13. Let G have degeneracy k ≤n −1. Then 1 + √1 + 8n 2 ≤pc (G) ≤2k + & 1 + p 1 + 8 (n −k) 2 ' . Proof. The lower bound has already been justied. Color G with a construction sequence. Each vertex v has j ≤k neighbors which exclude at most 2j colors. There are at most n−j−1 labels that have already been used on non-neighbors of v. Thus we need r extra colors, where r 2 ≥n −j. Solving, we nd r ≥ 1+√ 1+8(n−j) 2 . Thus we need at most 2j + 1+√ 1+8(n−j) 2 colors to label v, which is maximized when j = k. The upper bound is attained for the graph Kk + Kn−k. Since forests are exactly the 1-degenerate graphs, we have the following corollary. Corollary 14. For a forest F, l 1+√1+8n 2 m ≤pc (F) ≤2 + 1+√ 1+8(n−1) 2 . Thus there are usually three possible values for the pair chromatic number of a forest, but there are only two for n = r 2 +1, r ≥2. Note that stars attain the upper bound. Characterizing the trees that attain the upper bound may be possible, but distinguishing between the other two values appears dicult. Conjecture 15. Let F be a forest with order n and let r be the smallest integer such that n ≤ r 2 + 1. Then pc (F) = 2 + 1+√ 1+8(n−1) 2 if and only if ∆(F) ≥ r−1 2 + 1. The reverse direction of this conjecture is obvious, but the forward direction seems dicult. If there is a counterexample, it cannot have maximum degree r−1 2 . 8 Proposition 16. Let F be a forest with order n and let r be the smallest integer such that n ≤ r 2 + 1. If ∆(F) = r−1 2 , then pc (F) ≤1 + 1+√ 1+8(n−1) 2 . Proof. The result is easily checked for 1 ≤n ≤7. Let F be a forest with order n ≥8 and let r ≥5 be the smallest integer such that n ≤ r 2 +1. Add edges if necessary to form a tree T with the same maximum degree. Let v be a vertex with degree r−1 2 , which WLOG receives label 12. Then its r−1 2 neighbors must receive all possible labels from {3, 4, ..., r + 1}. Now F has at most r −1 vertices remaining, and 2 (r −1) labels left. Label the remaining vertices using a construction sequence. If the ith vertex labeled is adjacent to a neighbor of v, then there are at least 2 (r −1) −(i −1) −4 = 2r −i −5 ≥2r −(r −1) −5 = r −4 > 0 labels remaining. If the ith vertex labeled is not adjacent to a neighbor of v, then its neighbor u must have the color 1 or 2, and either excludes r−1 possible labels. The other color on u excludes one more label. Since u uses one of the labels already excluded, the preceding i−1 vertices exclude at most i−2 labels. Thus there are at least 2 (r −1) −(r −1) −1 −(i −2) = r −i ≥r −(r −1) = 1 labels remaining. Thus r + 1 colors suce to label T. We now present two variations on Theorem 13. Corollary 17. Let G have degeneracy k ≤n −1. Then 1 + √1 + 8n 2 ≤pc (G) ≤max 0≤j≤k ( 2j + & 1 + p 1 + 8 (nj −j) 2 ') . Proof. Color G with a construction sequence. Let the k-core of G have order nk. If vertex v is the ith vertex colored and j is the core number of v, then neighbors of v exclude at most 2j colors. There are at most nj −j −1 labels that have already been used on non-neighbors of v. Thus we need r extra colors, where r 2 ≥nj −j. Solving, we nd r ≥ 1+√ 1+8(nj−j) 2 . Thus we need at most 2j+ 1+√ 1+8(nj−j) 2 colors to label v, and we must take the maximum over all j. This may be an improvement when D (G) < ∆(G). For regular graphs, the following corollary is an improvement. Corollary 18. Let G be a connected graph with maximum degree ∆≥1. Then l 1+√1+8n 2 m ≤pc (G) ≤2∆−1 + 1+√ 1+8(n−∆+1) 2 . Proof. Since G is connected, G −e is ∆−1-degenerate. By Theorem 13, pc (G −e) ≤2 (∆−1) + 1+√ 1+8(n−(∆−1)) 2 . Lastly, adding e back requires at most one more color. 9 Denition 19. A graph G is pair k-critical if for any proper subgraph H of G, pc (H) < pc (G) = k. For small values of k, it is possible to list all such graphs, since these are also the critical forbidden subgraphs of Lk. k Pair k-Critical Graphs 2 K1 3 K2 4 K4, K2 5 K7, P3 It is considerably more dicult to determine all pair 6-critical graphs, which are the critical forbidden subgraphs of L5 (the Petersen graph). We consider this problem below. Note that for any k, there is a nite number of pair k-critical graphs. Proposition 20. Any graph G has nitely many critical forbidden subgraphs. Proof. Let G have order n. Then Kn+1 is a critical forbidden subgraph of G, so any other critical forbidden subgraph must have order at most n. There are nitely many such graphs, some subset of which are not subgraphs of G. Some subset of these are critical. Aside from the connection to 2-tone coloring, there is another reason why pair colorings are interesting. Consider labeling the vertices of a complete graph Kn with 1 to n. Then each edge can be labeled with the pair of labels of its vertices. Each possible label occurs exactly once. Thus a pair k-coloring of a graph corresponds to a (usually dierent) edge-induced subgraph of Kk. Thus we can transform a question on pair k-coloring of a disconnected graph into a question on packing (or decomposition) of a complete graph. As packings have been widely studied, results on them can be applied to pair coloring. Some examples of graphs and corresponding subgraphs for their minimum colorings are given in the following table. Graph Corresponding Subgraph Kn nK2 C4 2P3 C5 C5 C6 K2,3 C8 W4 C9 K5 −e P4 P5 Ln Kn K( r 2),( s 2) Kr ∪Ks 10 For example, consider a union of complete graphs. Each clique must have no common colors on its vertices. Thus its coloring corresponds to a matching in the complete graph. We employ the following lemma. Lemma 21. Let a1, ..., ak be integers with 1 ≤a1 ≤... ≤ak ≤ n 2 and P ai ≤ n 2 . Then Kn has a packing with matchings of sizes ai. Proof. It is well-known that there is a decomposition D of Kn into n or n −1 matchings of size n 2 . Form a packing B of Kn with matchings Bi of sizes 0 ≤b1 ≤... ≤bk ≤ n 2 and P ai = P bi from D by adding the necessary number of empty matchings and deleting the necessary number of edges. If ai ̸= bi for some i, there must be integers i < j such that bi < ai ≤aj < bj. Form a subgraph H by merging matchings Bi and Bj together. Each component of H must be a path or even cycle. Since the sizes of Bi and Bj were unequal, H must have a component path of odd length. Swap the edges on this path between Bi and Bj. This modied packing has k matchings whose sizes are closer to the the integers a1, ..., ak than before. (That is, the number P \|bi −ai\|has been reduced by 2.) We can apply this process repeatedly until we produce a packing with matchings of sizes ai. Theorem 22. Let a1, ..., ak be integers with 1 ≤a1 ≤... ≤ak and n = P ai. Then pc ∪ i Kai = max n 2ak, l 1+√1+8n 2 mo . Proof. Both lower bounds are immediate. Let N = max n 2ak, l 1+√1+8n 2 mo . By the lemma, KN can be packed with matchings of sizes a1, ..., ak. Hence a pair N-coloring of ∪ i Kai exists. Corollary 23. Let G be a disconnected graph with components Gi with orders a1, ..., ak, 1 ≤a1 ≤... ≤ak. Then max n pc (Gi) , l 1+√1+8n 2 mo ≤pc (G) ≤ max n 2ak, l 1+√1+8n 2 mo . Hence graphs for which the lower inequality is strict are of interest. Such a graph must be disconnected with pair chromatic number at least 6. Four such critical graphs are listed in the table below. They have pair chromatic number at least 6 since (1) none of these graphs are subgraphs of the Petersen graph and (2) the the corresponding decompositions do not pack K5. Graph Corresponding Decomposition T3−3 ∪P3 {K1 + 2K2, P3 ∪K2} C6 ∪2K2 {K2,3, 2K2, 2K2} C5 ∪K1,3 {C5, K3 ∪K2} 2K1,3 {K3 ∪K2, K3 ∪K2} 11 Now we can determine all pair 6-critical graphs. We dene some notation. Let Ta−b−...−c be a tree (caterpillar) with a spine having vertices of degree a, b, ..., c. Similarly, TPa−Pb−...−Pc is formed by appending paths of lengths a, b, ..., c to a path. Theorem 24. There are exactly 11 pair 6-critical graphs, namely K11, K1,4, 2K1,3, T3−3 ∪P3, T3−P4−3, K3, C4, C7, C10, C6 ∪2K2, C5 ∪K1,3. Proof. [sketch] We consider L5 (the Petersen graph). Certainly K11 is pair 6-critical, so any other such graph has order at most 10. Now K1,4 is pair 6-critical, so any other such graph has maximum degree at most 3. Consider forests. We have 2K1,3 6-critical since any two vertices of the Petersen graph have distance at most two apart. Since the Petersen graph has a Hamiltonian path, no path is 6-critical. Checking cases shows that no spider (tree with exactly one vertex of degree three) is 6-critical. Checking cases when there are exactly two vertices of degree three at distance two apart does not nd any 6-critical graph. When there are two adjacent vertices of degree 3, we nd T3−3 ∪P3 is 6-critical. Otherwise, checking cases shows that T3−P4−3 is the only other 6-critical forest. Certainly the only 6-critical cycles are K3, C4, C7, C10. Now consider graphs containing a single cycle. The cycle C9 is not the basis for any 6-critical unicyclic graph. Checking cases for C8 does not produce any new 6-critical graphs. For C6, we rst nd the disconnected graph C6 ∪2K2. Checking cases for connected 6-critical unicyclic graphs does not nd any more. Starting with C5, we see C5 ∪K1,3 is a disconnected 6-critical graph. Checking cases when we consider appending one, two, three, four, or ve trees does not produce any new 6-critical graphs. There are two 2-tone 6-critical theta graphs (θ3,3,3 and θ3,3,5) with at most order 10, but both contain 2K1,3. Checking cases shows that no graph formed by appending trees to a theta graph is 6-critical. Consider adding another path to a theta graph. There are two possibilities. One produces two disjoint cycles, the other produces a subdivided K4. Note rst that any 6-critical graph can be produced by adding an edge between nonadjacent vertices of the Petersen graph and then deleting some number of edges. It is easily seen that this number must be at least four since adding the edge creates a 3-cycle, two 4-cycles, several 7-cycles, and two vertices of degree 4 that must be disrupted. Now two disjoint cycles must be 5-cycles, but checking cases does not produce any new 6-critical graphs. Checking cases organized by length of the longest cycle in a subdivided K4 also does not produce any new 6-critical graphs. Finally consider disconnected graphs that are not forests and not uni-cyclic. Begin by deleting a subgraph of small order and then considering what other component could be added to make the graph not pair 5-chromatic, and whether the resulting graph is 6-critical. Deleting one, two, or three vertices 12 produces nothing new. Deleting four or more vertices eliminates all but at most one cycle. This exhausts the search. We can determine the pair chromatic number of cycles. Theorem 25. We have pc (Cn) =      5 n = 5, 6, 8, 9 6 n = 3, 4, 7, 10 −15 l 1+√1+8n 2 m n ≥11 . Proof. There are two relevant lower bounds to consider. First, pc (Cn) ≥ τ2 (Cn). This is exact for 3 ≤n ≤9 since the unique minimum colorings for all but C7 do not repeat a pair and C7 has a minimum coloring that does not repeat a pair . The second lower bound requires that every vertex of the cycle have a distinct pair, so pc (Cn) ≥ l 1+√1+8n 2 m . We now consider explicit colorings of cycles where no pair is repeated. First consider the following broken cycle. −12−(56)−34−25−(36)−14−23−45−(26)−13−(46 −15)−24−16−35− Without the pairs in parentheses, we have a ten-cycle. The pairs in paren-theses can be 'inserted' into the cycle, preserving the necessary properties. That is, we can subdivide some edges and assign previously unused pairs to the new vertices. The two pairs in a single set of parentheses must be inserted at the same time. This provides constructions up to the 15-cycle, for which all pairs formed from six colors are used. We next use induction on r to prove the existence of constructions for larger values of n. Assume that for r ≥6, there exists a 2-tone coloring of the cycle with r 2 vertices using r colors, so that each possible pair is used exactly once. We want to insert new pairs of colors in between some of the existing pairs. Allowing color r + 1 adds r new pairs to insert. We model this situation with a bipartite graph as follows. One partite set is the r new pairs to be added. The other partite set is the r 2 possible locations for insertion. An edge joins two vertices if the particular pair can be inserted in the particular location. We seek a maximum matching in this bipartite graph. Since each pair is distinct, and each pair uses the new color r + 1, a pair can be inserted as long as the other color is not one of the four used on the vertices between which the new vertex will be inserted. Thus each vertex in the location partite set has degree r −4. Each existing color is used r −1 times times on the cycle. Since the vertices on which it is used form an independent set, each excludes two locations, leaving r 2 −2 (r −1) = 1 2 (r −1) (r −4) valid locations, so this is the degree of the vertices in this partite set. 13 Consider a subset S of new pairs of order s, and let its neighborhood N (S) have order n. Then s 1 2 (r −1) (r −4) ≤n (r −4), so n ≥s. Thus the bipartite graph satises Hall's condition, so it has a maximum matching. Thus the new pairs can be successively inserted up to a cycle of length r+1 2 . By induction, we have constructions for all n ≥15. Our constructions achieve one of the lower bounds in all but the case n = 10, for which our construction is one larger. The bound cannot be achieved in this case since the Petersen graph is non-hamiltonian. A corollary on paths follows immediately. Corollary 26. For n ≥3, pc (Pn) = ( 5 3 ≤n ≤10 l 1+√1+8n 2 m n ≥11 . Proof. The path requires at least ve colors, and enough so that every vertex has a distinct pair. For n ≥11, breaking the cycle constructed in the proof of the theorem for wheels yields the appropriate construction. For smaller values, the appropriate construction exists because the Petersen graph has a hamiltonian path. Corollary 27. For the wheel Wn = Cn + K1 and the fan Fn = Pn + K1, n ≥3, we have τ2 (Wn) =      7 n = 5, 6, 8, 9 8 n = 3, 4, 7, 10 −15 l 5+√1+8n 2 m n ≥11 . τ2 (Fn) = ( 7 3 ≤n ≤10 l 5+√1+8n 2 m n ≥11 . 4 2-tone Coloring of Cartesian Products We now consider 2-tone coloring of cartesian products of graphs. We rst consider upper bounds. Note that if Gi has maximum degree ∆i and degen-eracy ki, then ∆(G1 × G2) = ∆1 + ∆2 and D (G1 × G2) = k1 + k2 . We apply the technique of . Theorem 28. Let Gi have degeneracy ki = D (Gi) and maximum degree ∆i = ∆i (G). Further, let k = k1 + k2 and M = 2∆1k1 −∆1 −k2 1 + 2∆2k2 −∆2 −k2 2 + k1∆2. Then τ2 (G1 × G2) ≤2k + l 1+√9+8M 2 m . 14 Proof. Number the vertices of Gi in increasing order according to a construc-tion sequence. Number the vertices of G1 × G2 lexicographically and arrange them in a grid. Consider a construction sequence of increasing lexicographic order. Color G1 × G2 with this construction sequence, so each vertex has at most k neighbors which exclude at most 2k colors. When colored, a vertex v has at most 2∆1k1 −∆1 −k2 1 + 2∆2k2 −∆2 −k2 2 + k1∆2 = M second-neighbors already colored by the proof of Theorem 3. The rst term refers to second-neighbors in the same row, the second to second-neighbors in the same column, and the third to second-neighbors whose row and column are both dierent from v's. Thus we need r extra colors, where r 2 ≥M + 1. Solving, we nd r ≥1+√9+8M 2 . Theorem 29. Let graphs G1 and G2 have maximum degrees ∆1 ≥∆2 ≥1, ∆1 ≥2, and let ∆= ∆(G1 × G2) = ∆1 + ∆2. Then τ2 (G1 × G2) ≤2∆−1 + & 1 + p 17 + 8 (∆2 −∆−∆2 −∆1∆2) 2 ' . Proof. We may assume G1 ×G2 is connected and regular. Number the vertices of Gi in increasing order according to a construction sequence. Number the vertices of G1 × G2 lexicographically and arrange them in a grid. Consider a construction sequence of increasing lexicographic order. Delete the edge e between the last two vertices of this sequence. Now G1 × G2 −e is ∆−1-degenerate. Coloring it using a construction sequence, each vertex has at most ∆−1 neighbors which exclude at most 2 (∆−1) colors. The largest number of second-neighbors already colored occurs on the last two vertices colored. They have at most ∆1 (∆1 −1) + (∆2 −1)2 + ∆1∆2 = (∆1 + ∆2)2 −(∆1 + 2∆2) − ∆1∆2 + 1 = ∆2 −∆−∆2 −∆1∆2 + 1 second-neighbors already colored. The rst term refers to second-neighbors in the same row, the second to second-neighbors in the same column, and the third to second-neighbors which have neither row or column in common. Thus we need r extra colors, where r 2 ≥ ∆2 −∆−∆2 −∆1∆2 + 2. Solving, we nd r ≥ 1+√ 17+8(∆2−∆−∆2−∆1∆2) 2 . Lastly, adding e back requires at most one more color by Theorem 2. Note that proved the upper bound τ2 (G) ≤2∆+ 1+√ 9+8∆(∆−1) 2 ≤ 2 + √ 2 ∆. The improvement on this bound is near greatest when ∆1∆2 is largest (for xed ∆), namely when ∆1 = ∆2 = 1 2∆. Then ∆2 −∆−∆2 − ∆1∆2 = 3 4∆2 −3 2∆= 3 4 (∆−1)2 −3 4, so       1 + r 17 + 8 3 4 (∆−1)2 −3 4 2       =     1 + q 11 + 6 (∆−1)2 2     ≤ &√ 6∆ 2 ' 15 = √ 1.5∆ and τ2 (G1 × G2) ≤2∆−1 + √ 1.5∆ = 2 + √ 1.5 ∆ −1. We now consider specic classes of graph products. Theorem 30. Let m ≤n. Then τ2 (Km × Kn) = 6 m = n = 2 2n else . Proof. Let G = Km×Kn, m ≤n. Now Kn ⊆G, so τ2 (G) ≥2n. Now certainly τ2 (K2 × K2) = 6, and τ2 (K1 × K2) = 4. Furthermore, we see G ⊆Kn × Kn, so the proof reduces to this case. Now it is well-known that there exist at least two mutually orthogonal Latin squares for all n except 2 and 6. We construct a 2-tone coloring of Kn ×Kn by using numbers 1 to n for the rst Latin square and n + 1 to 2n for the second Latin square. Now juxtapose them into a Graeco-Latin square. This can be viewed as a 2-tone coloring of Kn × Kn, where each cell is a vertex and each pair of vertices in the same row or column are adjacent. This leaves us to nd a 2-tone coloring of K6 × K6 with 12 colors. The following table represents this coloring, completing the proof. AB 37 28 59 46 10 90 68 35 7A 1B 24 78 2B 40 13 5A 69 56 0A 17 48 29 3B 34 15 9B 26 70 8A 12 49 6A 0B 38 57 Note that a 2-tone coloring exists in this last case even though a Graeco-Latin square does not because we have 66 labels to choose from, rather than 36. This result yields bounds on the 2-tone chromatic number of a cartesian product of graphs. Corollary 31. Let G and H be nontrivial graphs, not both K2, with orders r and s, respectively. Then max {τ2 (G) , τ2 (H) , 6} ≤τ2 (G × H) ≤max {2r, 2s} . We can also consider products of paths. Proposition 32. For the grid Pm × Pn, m, n ≥2, we have τ2 (Pm × Pn) = 6. Proof. Since the grid contains a C4, τ2 (Pm × Pn) ≥6.. Tile the grid with the following block, where cells represent vertices and each pair of neighboring cells are adjacent. This denes a 2-tone coloring. 36 15 24 25 34 16 14 26 35 16 The grid can be wrapped around on itself to form a product of cycles. The same coloring shows the following. Corollary 33. Let i, j be positive integers. We have τ2 (C3i × C3j) = 6. These problems suggest the problem of determining τ2 (Ci × Cj) for all i and j. We have the following partial results. Theorem 34. We have τ2 (C3 × Ci) = 6 i = 3, 6, 8, 9, 11, 12, i ≥14 7 i = 4, 5, 7 with i = 10, 13 undecided between 6 and 7. Proof. We have already seen the result for C3 × C3i. The following blocks of length 3 and 8 agree on two consecutive columns, so they can be concatenated to obtain the other values. 14 56 23 25 34 16 36 12 45 14 56 24 15 36 14 56 23 25 34 16 23 45 26 13 46 36 12 35 46 12 35 24 15 For i = 4, 5, 7, 2-tone 7-colorings are easily determined, and the rst is included in the following theorem. Suppose that G = C3 × C4 has a 2-tone 6-coloring. Then each color is used four times, since every color appears once in each column. Then each color appears twice in some row, necessarily distance two apart. Thus there is a pair of columns distance two apart with three such pairs of colors in the same rows. Clearly no two can be in the same row. This trio of colors must be rotated in the column in between. The trio of the other three colors must be rotated to all three positions over these three columns. But this leads to a contradiction. For G = C3 × C5, each color is used ve times, and so occurs twice in two rows distance two apart. Thus there are twelve such pairs, so of the ve pairs of columns distance two apart, one must have at least three such pairs of colors in the same rows. A contradiction follows as before. For G = C3 × C7, it is possible to show by an exhaustive search that there is no 6-coloring. Corollary 35. For j = 3, 6, 8, 9, 11, 12, and j ≥14, τ2 (C3i × Cj) = 6. Lemma 36. If a product of cycles has a 2-tone 6-coloring, then no vertex has both its two neighbors in its row sharing a common color and its two neighbors in its column sharing a common color. Proof. Let v be a vertex in a product of cycles G. Suppose to the contrary that G has a 2-tone 6-coloring and v has both its two neighbors in its row sharing a common color and its two neighbors in its column sharing a common color. Let u and w be neighbors of v that are not in the same row or column, and let 17 their mutual neighbor other than v be x. Then u, v, w, and x induce C4, so u and w have a common color. Thus every pair of neighbors of v has a color in common, and clearly no two neighbors of v have the same label. Form a graph H whose vertices are the four colors not used on v, and edges represent the labels of the neighbors of v. Then H has size four, and every pair of edges is adjacent. But this is impossible. Theorem 37. Let i ≥3. Then τ2 (C4 × Ci) = 7. Proof. Consider a given row of G = C4 × Ci. It is possible for a row to be 2-tone colored with six colors so that every vertex has the property that each two neighbors share no common colors if and only if its length is a multiple of three. First suppose i ̸= 0 mod 3 and G has a 2-tone 6-coloring. Now every 4-cycle 2-tone colored with six colors has every pair of nonadjacent vertices having one common color. Thus G has some vertex v so that its pairs of neighbors in both its row and column each have a color in common. But by the previous lemma, this is impossible. Now suppose i = 0 mod 3, and G has a 2-tone 6-coloring. By the lemma, each row must have the property that each two neighbors share no common colors. Then each six colors appear in every three consecutive vertices of a row, and two rows distance two apart must have a color repeated in each of the columns. Then a contradiction follows as in the case of the graph G = C3×C4. Thus τ2 (C4 × Ci) ≥7. We now show that equality can be achieved. The following tables corre-spond to colorings of C4 × C3, C4 × C4, and C4 × C5. Each entry represents a vertex and cells are adjacent if they are neighbors in a row or column, or on opposite ends of a row or column. The rst two colorings can be concate-nated to form products of C4 and larger cycles since they agree on the rst two columns. 15 46 23 37 25 14 12 36 57 34 27 16 15 46 12 47 37 25 34 26 12 36 17 45 34 27 35 16 15 46 12 47 36 37 25 34 26 14 12 36 17 45 67 34 27 35 16 25 Lemma 38. Let n ≥5, n ̸= 0 mod 3. Then any 2-tone 6-coloring of Cn uses at least four 2-chords. Proof. Case 1. Let n = 3r + 1. Then Cn needs 6r + 2 colors with repetition. A color class using no 2-chords can use at most r vertices. A color class with r + 1 vertices uses at least two 2-chords. Four classes of r vertices and two of r + 1 produce 6r + 2 colors. Using a color class of r + k colors would require at least 3k −1 2-chords, so there is no advantage to using a class of more than r + 1 vertices. 18 Case 2. Let n = 3r + 2. Then Cn needs 6r + 4 colors with repetition. A color class using no 2-chords can use at most r vertices. A color class with r + 1 vertices uses at least one 2-chord. Two classes of r vertices and four of r + 1 produce 6r + 4 colors. Using a color class of r + k colors would require at least 3k −2 2-chords, so there is no advantage to using a class of more than r + 1 vertices. In both cases, at least four 2-chords are required. Proposition 39. Let (i, j) = (5, n), 5 ≤n ≤19, n ̸= 0 mod 3, or (7, 7) or (7, 8). Then τ2 (Ci × Cj) ≥7. Proof. Each 2-chord corresponds to the vertex between its ends. In each case, we show that if the graph has a 2-tone 6-coloring, more 2-chords are required than there are vertices in the product of cycles. Then by the pigeonhole princi-ple, some vertex must use 2-chords in both its row and column. But by Lemma 36, this is impossible, so the 2-tone chromatic number is at least 7. For (i, j) = (5, n), 5 ≤n ≤19, n ̸= 0 mod 3, we have order 5n and at least 4n + 4 · 5 > 5n 2-chords. For (i, j) = (7, 7), we have order 49 and at least 4 · 7 + 4 · 7 = 56 2-chords. For (i, j) = (7, 8), we have order 56 and at least 4 · 7 + 4 · 8 = 60 2-chords. For hypercubes, we have τ2 (Q1) = 4, τ2 (Q2) = τ2 (Q3) = 6, and τ2 (Q4) = 7 since Q1 = K2, Q2 = C4, Q3 = C4 × K2, and Q4 = C4 × C4. It was shown in by Josh Goss that the 5- and 6-cubes can be 8-colored. Hence 7 ≤τ2 (Q5) ≤τ2 (Q6) ≤8. Acknowledgments Thanks to Ben Phillips for introducing me to the topic of t-tone coloring, and for his advice and assistance along the way. References D. Bal, P. Bennett, A. Dudek, and A. Frieze, The t-tone chromatic number of random graphs. To appear in Graphs and Combinatorics. A. Bickle, The k-Cores of a Graph. Ph.D. Dissertation, Western Michigan University (2010). A. Bickle, Cores and Shells of Graphs. Mathematica Bohemica. 138 1 (2013), 43-59. A. Bickle and B. Phillips, t-Tone Colorings of Graphs. Submitted. 19 B. Bollobas and A. Thomason, Set colourings of graphs. Discrete Math. (25) 1979 21-26. G. Chartrand and P. Zhang, Chromatic Graph Theory. CRC Press, (2009) D. Cranston, J. Kim, and W. Kinnersley, New results in t-tone coloring of graphs. Electronic Journal of Combinatorics 20 (2) (2013), #P17 N. Fonger, J. Goss, B. Phillips, and C. Segroves, Math 6450: Final Report. nalReport2.pdf. (2009) J. R. Griggs and R. K. Yeh, Labelling Graphs with a condition at distance 2. SIAM J. Discrete Math. 5 (1992) 586-595. D. West, Introduction to Graph Theory, (2nd ed.). Prentice Hall, (2001). 20
15587	https://www.teacherspayteachers.com/browse/free?search=printable%20periodic%20table%20of%20elements	Printable Periodic Table of Elements Matter Atoms & Periodic Table of Elements Digital + Printable Notebook Setup Free Printable Periodic Table of Elements: 4 Versions with IUPAC Spelling Get more with resources under $5 Printable Periodic Table of Elements and Chemistry Reference Sheet Printable Periodic Table of Elements Printable Periodic Table of Elements Flash Cards Printable Periodic Table of Elements Worksheet 5th grade Science Reading Passage Periodic Table of Elements-Printable The Periodic Table of Elements Printable Tables Learning Tools Printable Periodic Table of the Elements Free: Periodic Table of Elements Color by Category \| print and digital Matter Atoms Periodic Table of Elements Activities \| Physical Science Notebook Matter Unit Word Search Puzzle Coloring Activity - Periodic Table of Elements FREE Chemistry Basics Series The Periodic Table of the Elements Periodic Table of Elements in Post-it Note Design FREE Periodic Table of Elements Matching Worksheet (Science Worksheet Activity) Periodic Table of Elements - FREE SAMPLE Creating Words From the Periodic Table of Elements Periodic Table of the Elements Periodic Table Printable - With Element Square Example Matter Unit - Atoms and Periodic Table of Elements Notebook Set Up Periodic Table of Elements Reading Passage and Color by Number Activity Printable Periodic Table Reference Sheet Free: Periodic Table of Elements Color by Category \| Print activity \| Chemistry Periodic Table Printable - Symbol, Name, Atomic Number, Atomic Mass Simplified Periodic Table of Elements with rounded atomic mass Periodic Table of Elements Word Search Activity - Physical Science Matter Unit Physical Science Lab and Projects Supply List Outline Printable Periodic Table of Elements \| Simple Student-Friendly Version \| PDF
15588	https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence?srsltid=AfmBOopWW84sGdE1Ya0exctn5fdoNvxQ4cfpOr03egKwjjxzcoa--6IJ	Art of Problem Solving Arithmetic sequence - 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 Arithmetic sequence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Arithmetic sequence In algebra, an arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence. For example, is an arithmetic sequence with common difference and is an arithmetic sequence with common difference ; however, and are not arithmetic sequences, as the difference between consecutive terms varies. More formally, the sequence is an arithmetic progression if and only if . A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in arithmetic progression if and only if . Contents [hide] 1 Properties 2 Sum 3 Problems 3.1 Introductory problems 3.2 Intermediate problems 4 See Also Properties Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let be the first term, be the th term, and be the common difference of any arithmetic sequence; then, . A common lemma is that given the th term and th term of an arithmetic sequence, the common difference is equal to . Proof: Let the sequence have first term and common difference . Then using the above result, as desired. Another common lemma is that a sequence is in arithmetic progression if and only if is the arithmetic mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of arithmetic sequences. Sum An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value. The first is that if an arithmetic series has first term , last term , and total terms, then its value is equal to . Proof: Let the series be equal to , and let its common difference be . Then, we can write in two ways: Adding these two equations cancels all terms involving ; and so , as required. The second is that if an arithmetic series has first term , common difference , and terms, it has value . Proof: The final term has value . Then by the above formula, the series has value This completes the proof. Problems Here are some problems with solutions that utilize arithmetic sequences and series. Introductory problems 2005 AMC 10A Problem 17 2006 AMC 10A Problem 19 2012 AIME I Problems/Problem 2 2004 AMC 10B Problems/Problem 10 2006 AMC 10A, Problem 9 2006 AMC 12A, Problem 12 Intermediate problems 2003 AIME I, Problem 2 Find the roots of the polynomial , given that the roots form an arithmetic progression. See Also Geometric sequence Harmonic sequence Sequence Series Retrieved from " Categories: Algebra Sequences and series Definition 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.
15589	https://discretemathproject.net/curriculum/curriculum/dmps.html	Discrete Math Pre-Collegiate (DMPS) Discrete Math Problem Solving Home Curriculum Overview DM Pre Collegiate DM Problem Solving Course Adoption Students/Parents Counselors/Admin Teachers Professional Learning Blog Resources About Us Leadership Team Our Story Gallery Contact Sample Lessons Trading Places The Legend of The Golden Disks ngMaki It Dciuflfti to daeR Counting FUNdamentals If these samples have sparked your interest, contact us to discuss using DMPC in your school or district. Course Description Length of course: Full Year Subject area: Mathematics (‘C’) / Advanced Mathematics Prerequisites: IM 2 or Algebra I and Geometry Grade levels: 11th, 12th In this student-centered course, students engage in problem solving experiences that allow for productive struggle and mathematical knowledge construction. The goal of the course is to advance students’ thinking as defined by the Standards for Mathematical Practice, while Discrete Math content is used as a context in which students develop these ways of thinking. Each lesson presents students with phenomena from within Introductory: Game Theory, Graph Theory (Connectivity and Traceability), Combinatorics, Cryptography, and the Study of Sequences and Series (i.e., Functions defined explicitly or recursively over the set of natural numbers). From these experiences students will engage in many forms of mathematical reasoning, including inductive and deductive reasoning, reasoning with recursion, definitional and structural reasoning. Definitions, concepts and skills are necessitated through problem-solving. Students are responsible for defining, modeling, pattern generalizing, conjecturing and justifying. They are expected to collaborate and communicate mathematical ideas regularly in verbal and written forms. See below for the list of topics of study and sample lessons for this curriculum. PoWer Up This introductory module poses up to four problems teachers can use to establish the tone and the expectations for problem solving, critical thinking, group collaboration, student agency and writing in mathematics. Each lesson introduces students to the mathematical domains they will study during the year: Games, Graph Theory, Iteration/Recursion, Cryptography and Counting. Shall We Play a Game? This module largely explores a class of combinatorial games called two-player impartial games, where the same set of moves are available to both players, regardless of whose move it is, each game has a finite number of possible moves, and games must end with a win (or loss). No ties or draws are possible. Students will repeatedly explore, observe, model, conjecture, generalize and prove. Exploration: These games have been chosen to provide students ample opportunities to search for and make use of structures within and between games. Students will develop iterative and recursive thinking as they play games with smaller numbers of objects. Throughout this module, students will experience the problem-solving approaches of starting small, building up to larger cases systematically and/or working backwards. When students create valuable representations or models of the game, we expect that they will search for opportunities to share them with the class at a time that doesn’t spoil the fun of discovery for the others. Exploration continues as students play more games and begin to search for commonalities between games and between winning strategies. Conjecture and Proof: As they play this type of game, students find winning strategies through pattern recognition. Stating a winning strategy is a form of generalizing. The role of the teacher is to foster students’ curiosity about whether or not a pattern always holds and push them to explain how the rules of the game causes their conjectures to hold. Oh The Places You’ll Go! This module is an introduction to vertex-edge graphs and their properties. Through a process of guided reinvention, students (re)create the concept of graph in three canonical forms; as a drawing, a table and a list (commonly referred to as vertex-edge, adjacency matrix and set). Students will learn that (and how) graphs can be useful tools for tracking relationships among objects or people and for modeling a wide variety of situations. The main focus of study is the mathematical structures of undirected and unweighted planar graphs for modeling pairwise relations. Using graphs as a context, the overarching goal is to advance students’ ability to reason deductively (SMP3), communicate ideas (SMP6), and make connections (SMP7). In keeping with our assumptions about teaching and learning the module appeals to situations students can imagine, asking them to model problems arising in everyday life, society, and the workplace. In this way, we aim at necessitating graphs as a tool for problem-solving. Students will learn what a graph is through creating and interpreting representations of relationships. Another learning goal entails developing a heightened awareness of the difference between what one chooses to represent and how one chooses to represent it. By the end of the module, students will have had substantial experiences solving problems using graphs. Patterns, Patterns, Patterns... This module engages students in pattern generalizing, leveraging numerical and geometric situations. Students are introduced to the idea of a recurrence relation through the Tower of Hanoi problem, which is easy to solve for smaller, more concrete cases but requires recursive reasoning to reach the answer for more general cases. We ask students to justify why their recursive relations hold for all cases by explaining what it is about the rules of the game that causes the result. The module is themed by introducing problems which can be modeled by a Fibonacci sequence. Students will search for, generalize, and justify patterns in the Fibonacci sequence. In addition, they will construct and apply recursive patterns including the Golden Ratio and the Golden Rectangle found in nature, architecture, and other real-life contexts. This module also focuses on arithmetic and geometric sequences and series, building on work from the previous lessons. We want students to recognize that arithmetic and geometric sequences and series are particular kinds of recurrence relations. Additionally, we approach this from a geometric angle, using geometric problems to approach arithmetic series and questions stemming from fractals – geometric objects defined recursively – to approach geometric series. Again, we ask students the familiar question, what is it about the way the structure is defined that causes your conjecture to hold for all cases? YhpargotpyrC This module aims at engaging students in exploring and working with a selection of mathematical concepts utilized in the field of cryptography that are accessible for high school students who have completed algebra. It provides the perfect stage to engage students in thinking about math in a way that will require them to explore and uncover structures that must be precisely communicated in order to generate and decrypt coded communication. In this module, students will be introduced to various transposition and substitution ciphers. These encryption systems will necessitate reasoning with functions and the development of basic modular arithmetic. Students will explore how language constructs such as word length and letter frequency play a role in cryptography. Furthermore, this module provides a natural setting to advance students’ counting strategies, particularly the Fundamental Counting Principle, permutations and combinations. Counting Is FUNdaMENTAL This unit focuses on arithmetic and geometric sequences and series, building on the work from the Iteration and Recursion unit. We want to recognize that arithmetic and geometric sequences and series are particular kinds of recurrence relations. In particular, we want to motivate the fact that sequences defined by recurrence relations often have closed-form formulas, and arithmetic and geometric sequences and series are particularly nice in this regard. Students will experience iteration as a powerful tool to use for coming up with an explicit formula for a recurrence relation. Additionally, we approach this from a geometric angle, using geometric problems to approach arithmetic series and questions stemming from fractals – geometric objects defined recursively – to approach geometric series. Home Curriculum DM Problem Solving
15590	https://www.quora.com/What-is-the-general-formula-to-find-the-minimum-and-maximum-numbers-where-the-sum-of-the-digits-of-a-number-is-k	Something went wrong. Wait a moment and try again. Minimum Value Digit Sum Formulas in Maths Mathematical Equations Number Theory Maximum Value Math Science 5 What is the general formula to find the minimum and maximum numbers where the sum of the digits of a number is k? Anita S Vasu Principal Engineer at Amazon (company) (2016–present) · Author has 352 answers and 826.6K answer views · 8y Largest number assuming zero is not permitted is easy it will be just 11111.. k times. Smallest non negative number is exhaust with 9 until it is no longer possible, and then the remainder, and then reverse the whole thing. (a): K = 15 Largest number (without zero): 111,111,111,111,111 Smallest number (non negative): 69 (b) K = 22 Largest number (without zero): 1,111,111,111,111,111,111,111 Smallest number (non negative): 499 if you really need: Largest(k)=10k−19 Quotient, q=⌊k/9⌋ Reminder, r=k−9q Smallest(k)=(r+1 Largest number assuming zero is not permitted is easy it will be just 11111.. k times. Smallest non negative number is exhaust with 9 until it is no longer possible, and then the remainder, and then reverse the whole thing. Example (a): K = 15 Largest number (without zero): 111,111,111,111,111 Smallest number (non negative): 69 (b) K = 22 Largest number (without zero): 1,111,111,111,111,111,111,111 Smallest number (non negative): 499 Equation if you really need: Largest(k)=10k−19 Quotient, q=⌊k/9⌋ Reminder, r=k−9q Smallest(k)=(r+1)10q−1 Related questions What is the general term for the sum of the digits? What is the sum of all 3 digit numbers? How many numbers are between 1 to 10? How many eight-digit numbers have a sum of 4? What is the formula to find the number of digits in a number raised to a power? Mark Gritter recreational mathematician · Author has 5.7K answers and 11.8M answer views · 8y The maximum is easy: just make our number extend as far as possible to the left. How many digits is that? Well, we get furthest if every digit is 1. (I assume 0 is not allowed because then the question is not well-formed.) For example, the highest number with digit-sum 3 is 111. The highest number with digit-sum 17 is 11,111,111,111,111,111. For general k, this number is ∑k−1i=010i which is just a geometric series, so we get 19(10k−1). To get the lowest number, we have to shorten the number as much as possible, while shifting digits rightwards to lower place values. T The maximum is easy: just make our number extend as far as possible to the left. How many digits is that? Well, we get furthest if every digit is 1. (I assume 0 is not allowed because then the question is not well-formed.) For example, the highest number with digit-sum 3 is 111. The highest number with digit-sum 17 is 11,111,111,111,111,111. For general k, this number is ∑k−1i=010i which is just a geometric series, so we get 19(10k−1). To get the lowest number, we have to shorten the number as much as possible, while shifting digits rightwards to lower place values. This is a little more complex, but not much. We write as many 9s as are needed, and then the remaining digit. For example, the lowest number with digit-sum 60 is 6999999. For general k, this number is 10⌊k/9⌋(kmod9)+⌊k/9⌋−1∑i=09×10i Note that if k is exactly a multiple of 9, then the first term is zero, as desired. The second term is a geometric series so we can rewrite it like we did above, getting 10⌊k/9⌋(kmod9)+10⌊k/9⌋−1 =10⌊k/9⌋(kmod9+1)−1 Check: for k=60 we have 106(6+1)−1=6999999. Arnab Sen Former Computer science at Indian Institute of Science, Bangalore (IISc) (2013–2015) · Author has 103 answers and 263.4K answer views · 8y Originally Answered: What is the general formula to find the minimum and maximum number whose digit sum is k? · General formula is: maximum number would be infinity as you can add as many zeros you want anywhere without affecting the digit sum. If for specific case we assume no zero is allowed, then our objective would be to have maximum number of digits. So the maximum number is = 111111….k times. Similarly for minimum number, our focus would be to use least number of digits. So we repeatedly minus from k , 9 as many time as possible and at the end take the remaining digit and at sort them in ascending order to finally get the number. For example, if k=35, then we get, 35=9+9+9+8 so the lowest number=8999. Mark Gritter PhD dropout in Computer Science, Stanford University (Graduated 2006) · Author has 5.7K answers and 11.8M answer views · 8y Originally Answered: What is the general formula to find the minimum and maximum numbers where the string length of digit sum is k? · Recursively apply the formulas given in What is the general formula to find the minimum and maximum numbers where the sum of the digits of a number is k? under the same assumption that zeros are not permitted. If the string length of a digit sum is k, then the maximum possible value is 10k−1. (That is, k repeated 9’s.) Plug this into the formula for maximum possible number with digit-sum and we get 19(1010k−1−1) For example, the largest number whose digit sum has 3 digits, will have digit sum 999, so it consists of 999 1’s. 11111111111111111111111111111111111111111111 Recursively apply the formulas given in What is the general formula to find the minimum and maximum numbers where the sum of the digits of a number is k? under the same assumption that zeros are not permitted. If the string length of a digit sum is k, then the maximum possible value is 10k−1. (That is, k repeated 9’s.) Plug this into the formula for maximum possible number with digit-sum and we get 19(1010k−1−1) For example, the largest number whose digit sum has 3 digits, will have digit sum 999, so it consists of 999 1’s. 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 It should be obvious from the construction that no smaller 3-digit digit-sum produces a larger number. If the string length of a digit sum is k, its smallest possible value is (1/9)(10k−1), that is, k repeated 1’s. Substituting this into the formula from the other answer gives us 10⌊(10k−1)/81⌋(19(10k−1)mod9+1)−1 We know that (10k−1)/9 consists of k 1’s, so modulo 9 it is just k. (10 mod 9 = 1, so the power of 10 doesn’t make any difference.) 10⌊(10k−1)/81⌋(kmod9+1)−1 Again, the construction makes it clear that we can’t do better by picking some other, larger, digit-sum. (If you believe the formulas in the other answer then just note that they are monotone increasing. ) The smallest number whose digit-sum is 3 digits will have digit-sum 111, which means the number is 3999999999999. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. 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Related questions What is the sum of digits in the number 4444 4444 ? What is the sum of all three digit odd numbers? What is the sum of all even 2-digit numbers? How many 4-digit number(s) is/are there in which the sum of digit is 9? Is there a three digit number where the sum of its digits results in another three digit number? Assistant Bot · Sep 7 For a nonnegative integer k, consider decimal integers whose digit sum equals k. Minimum (smallest positive integer) with digit-sum k Represent k in base 9 with quotient and remainder: q = k div 9, r = k mod 9. The minimal positive decimal integer with digit-sum k is if r > 0: the digits are [r] followed by q copies of 9, i.e. number = r · 10^{q} + 999…9 (q nines). if r = 0: the digits are 1 followed by (q−1) copies of 9, i.e. number = 1 · 10^{q−1} + 999…9 (q−1 nines). Closed-form description: Let q = ⌊k/9⌋, r = k − 9q. Then if r > 0: Min(k) = r × 10^{q} + (10^{q} − 1). if r = 0 (k>0): Min(k) = 10^{q−1 For a nonnegative integer k, consider decimal integers whose digit sum equals k. Minimum (smallest positive integer) with digit-sum k Represent k in base 9 with quotient and remainder: q = k div 9, r = k mod 9. The minimal positive decimal integer with digit-sum k is if r > 0: the digits are [r] followed by q copies of 9, i.e. number = r · 10^{q} + 999…9 (q nines). if r = 0: the digits are 1 followed by (q−1) copies of 9, i.e. number = 1 · 10^{q−1} + 999…9 (q−1 nines). Closed-form description: Let q = ⌊k/9⌋, r = k − 9q. Then if r > 0: Min(k) = r × 10^{q} + (10^{q} − 1). if r = 0 (k>0): Min(k) = 10^{q−1} + (10^{q−1} − 1) = 10^{q−1} + 10^{q−1} − 1 = 2·10^{q−1} − 1 (equivalently the digit string 1 followed by q−1 nines). For k = 0: Min(0) = 0. Maximum (largest integer with a fixed number of digits n and digit-sum k) Without a fixed digit-length, there is no finite maximum: for any integer with digit-sum k you can prepend arbitrarily many zeros (which do not change value) only if leading zeros allowed; to increase numeric value while preserving digit-sum you can add blocks of digits that sum to 0 impossible except zeros — numeric value cannot be increased indefinitely while keeping digit-sum fixed and no leading zeros allowed. Therefore the set of positive integers with digit-sum k has no finite upper bound if leading zeros are allowed to create more digits? Clarification: to maximize numeric value with fixed digit-sum you must fix the number of digits n. Maximum with a fixed number of digits n (n≥1): Put as large digits as possible from left to right: let remaining sum s0 = k. For position i from 1 to n set digit di = min(9, s_{i−1}); update s_i = s_{i−1} − di. If after n digits s_n ≠ 0 there is no solution for that n. Constructed number = concatenation d1 d2 … dn (largest numerically for that n). In particular, to get the absolute maximum for a given k and a constrained number of digits n = ⌈k/9⌉ (smallest n that can accommodate sum k), use the pattern: Let q = ⌊k/9⌋, r = k − 9q. With n = q + (r>0 ? 1 : 0), the maximal n-digit number with sum k is: if r > 0: digits = [9 repeated q times] with a remaining r placed at the leftmost position of the n-digit number: number = r × 10^{q} + (10^{q} − 1) (which equals the same numeric string as the minimum construction but with digits arranged differently depending on interpretation of digit positions). if r = 0: digits = [9 repeated q times] (n = q), number = 10^{q} − 1. Practical summary: Smallest integer with digit-sum k: let q = ⌊k/9⌋, r = k − 9q. If r>0 then Min = r·10^{q} + (10^{q} − 1). If r=0 and k>0 then Min = 10^{q−1} + (10^{q−1} − 1) = 2·10^{q−1} − 1. Min(0)=0. Largest n-digit integer with digit-sum k: greedily fill digits left-to-right with 9s as described; no finite global maximum exists unless the number of digits is fixed. Doug Dillon Ph.D. Mathematics · Author has 12.4K answers and 11.4M answer views · 8y Originally Answered: What is the general formula to find the minimum and maximum numbers where the string length of digit sum is k? · Not sure there is. Perhaps to find the minimum number. Suppose k=1, the the minimum number is 1. There is no maximum number since 1 followed by as many 0’s as you want will produce a digit sum of 1. This analysis is valid for all 1 digit digit-sums. For a two digit sum, 47 for example, 999992 will produce the digit sum as will 888887 as will 7777775. Of these none are the smallest but perhaps 299999 is. Again there is no largest since we can just add 0’s to any of them Pat Devlin Former Aerospace Engineer, part time physicist · Author has 434 answers and 281.3K answer views · 8y Originally Answered: What is the general formula to find the minimum and maximum number whose digit sum is k? · I'm not sure that there is a general formula however imagine that the digit sum is just 1 then the largest number possible is unbounded i.e. 1000000000000000000…. = + infinity The minimum number considering negative integers is -1000000000000……= There are also non integers and complex numbers My guess is that there is no such formula Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Bernard Montaron PhD in Mathematics & Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (Graduated 1980) · Upvoted by Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 3.2K answers and 2.1M answer views · Updated 1y Related What’s the sum S ( k ) = ∑ n ∈ A k 1 n where A k is the set of all positive integers with digit sum equal to k 0 (in base 10)? What’s the minimum value? Obviously, S(1)=∑n≥0110n=109 Then S(2)=12∑n≥010−n+∑n≥1∑n−1k=0110n+10k=59+109∑n≥1110n+1 This sum converges quickly towards 0.668800203353559689093351448689… Following the same idea we obtain S(3)=13∑n≥010−n+109(112+121)+109(1102+1111+1201) +109(11002+11011+11101+12001)+… This sum converges qui Obviously, S(1)=∑n≥0110n=109 Then S(2)=12∑n≥010−n+∑n≥1∑n−1k=0110n+10k=59+109∑n≥1110n+1 This sum converges quickly towards 0.668800203353559689093351448689… Following the same idea we obtain S(3)=13∑n≥010−n+109(112+121)+109(1102+1111+1201) +109(11002+11011+11101+12001)+… This sum converges quickly to 0.546633210598567… Then we have (see the note at the end) S(4)=0.503712… S(5)=0.493577… S(9)>S(8)>S(7)>S(6)>1/2 Therefore, one could think that the minimum value is reached at k=5, but that may not be the case, as it could just be a ‘local’ minimum. S(9)=0.587513… is a local maximum. Then at k=10 the sequence decreases to a value close to S(7)=0.521206…. It then continues to decrease down to S(12)=0.50147... to then increase up to S(17)=0.5179... and decrease again after that. Here is the graph : S(25)≈0.5125 also is a local maximum. It would be interesting to know if for all k>5 it is true that S(k)>1/2. Note: Here is an example of effective upper bound applied to the computation of S(4) : The sequence of integers to be considered is 4, 13, 22, 31, 103, 112, 121, 202, 211, 301, 1003, 1012, 1021, 1102, 1111, 1201, 2002, 2011, 2101, 3001, etc. For numbers with at least 5 digits, there is one number with k digits starting with 3, it’s 30…01, there are k−1 numbers starting with 2, and 1+2(k−2)+(k−22) starting with 1. Therefore, the sum of reciprocals of these numbers with k digits is bounded by 110k−1(13+k−12+2k−3+(k−22))=3k2−16×10k−1 One can calculate the partial sum of these integers having at most 8 digits (with the multiplication factor 10/9 to account for numbers ending with zeros) S4=109(14+113+122+131+…+130000001)=0.50371159694… And the sum of the following terms is bounded by 109∑k≥93k2−16×10k−1=0.00000051112… Finally, S(4)<0.50371210807… For S(5), integers to consider are 5, 14, 23, 32, 41, 104, 113, 122, 131, 203, 212, 221, 302, 311, 401, 1004, 1013, 1022, 1031, 1103, 1112, 1121, 1202, 1211, 1301, 2003, 2012, 2021, 2102, 2111, 2201, 3002, 3011, 3101, 4001, 10004, 10013, 10022, 10031, 10103, 10112, 10121, 10202, 10211, 10301, 11003, 11012, 11021, 11102, 11111, 11201, 12002, 12011, 12101, 13001, 20003, 20012, 20021, 20102, 20111, 20201, 21002, 21011, 21101, 22001, 30002, 30011, 30101, 31001, 40001, etc. For numbers with al least 6 digits, there is only one number with k digits starting with 4, it’s 40…01, there are k−1 numbers starting with 3, there are 1+2(k−2)+(k−22)=k(k−1)2 numbers starting with 2, and 1+3(k−2)+3(k−22)+(k−23)=k(k2−1)6 starting with 1. Therefore, the sum of reciprocals of these numbers with k digits is bounded by 110k−1(14+k−13+k(k−1)4+k(k2−1)6)=2k3+3k2−k−112×10k−1 For example, the partial sum of reciprocals of numbers with up to 6 digits (with the multiplication factor 10/9 to account for numbers ending with zeros) is S5=109(15+114+123+132+141+…+1400001)=0.49349154… And the sum of the following terms is bounded by 109∑k≥72k3+3k2−k−112×10k−1=0.00008937… Finally, S(5)<0.49358091… Jan van Delden MSc Math and still interested · Upvoted by Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 4.8K answers and 6.5M answer views · 1y Related What’s the sum S ( k ) = ∑ n ∈ A k 1 n where A k is the set of all positive integers with digit sum equal to k 0 (in base 10)? What’s the minimum value? This is a rather difficult question, especially if we want to study the behaviour of [math]S(k)[/math] for larger values of [math]k[/math] . One reason is that it is hard to find an exact value for [math]S(k)[/math] , if [math]k>1[/math] . Another reason is that you’d better keep track of all ordered partitions of [math]k[/math] where we limit the members of such a partition to the digits that are available to us. In the answer given by Bernard Montaron he writes down the partitions of [math]k[/math] in a specific order in order to be able sum the associated [math]n\in A_k[/math] in groups. Lets repeat that, but I’ll add a bit of explanation, since it went a bit too fast for my taste [math]\[/math] This is a rather difficult question, especially if we want to study the behaviour of [math]S(k)[/math] for larger values of [math]k[/math]. One reason is that it is hard to find an exact value for [math]S(k)[/math], if [math]k>1[/math]. Another reason is that you’d better keep track of all ordered partitions of [math]k[/math] where we limit the members of such a partition to the digits that are available to us. In the answer given by Bernard Montaron he writes down the partitions of [math]k[/math] in a specific order in order to be able sum the associated [math]n\in A_k[/math] in groups. Lets repeat that, but I’ll add a bit of explanation, since it went a bit too fast for my taste. He splits the problem into two problems. One Find all numbers [math]n[/math] with [math]m[/math] digits that don’t end with a [math]0[/math] and add the contribution for all numbers of the form [math]n\cdot 10^j[/math] in one go. But that’s easy. [math]\displaystyle \sum_{j=0}^{\infty} \dfrac{1}{n\cdot 10^j}=\dfrac{1}{n} \sum_{j=0}^{\infty} \left(\dfrac{1}{10}\right)^j=\dfrac{1}{n} \dfrac{1}{1-\frac{1}{10}}=\dfrac{10}{9} \dfrac{1}{n}[/math] since we have a geometric series. Two Find these numbers [math]n[/math] with [math]m[/math] digits that don’t end with a [math]0[/math], by building these one digit a time. And here the fun starts, certainly if you wish to write a routine. A definition [math]T_{m,k}[/math]: the set of numbers [math]n_{m,k}[/math] with [math]m[/math] digits with a sum of digits equal to [math]k[/math] Given [math]T_{m,k}[/math] we may describe [math]T_{m+1,k}[/math] by either using all ordered partitions of [math]k[/math] having [math]m+1[/math] digits We’ll only have to include this step if [math]m+1\le k[/math], since these ordered partitions don’t include a [math]0[/math]. The maximal value of [math]m+1[/math] is obtained by the partition [math]k=1+1+\ldots+1[/math]. If [math]9s+1 \le k < 9s+10[/math] then we know that we need at least [math]s[/math] digits. all [math]n_{m+1,k}[/math] where we add one zero in between the first and last position at all available positions, for all [math]n_{m,k}\in T_{m,k}[/math] (and must make sure that we dispose of options that are otherwise present twice). We may distribute the numbers associated with the ordered partitions of [math]3[/math] to the [math]T_{m,3}[/math] as follows [math]T_{1,3}={3},\;T_{2,3}={12,21},\;T_{3,3}={111}[/math] The set [math]T_{1,3}[/math] is complete. We can’t add a zero in front and step one will correct for zeroes at the back, so [math]T_{2,3}[/math] is complete as well. But we can add one zero to the members in [math]T_{2,3}[/math] to find [math]T_{3,3}={102,111,201}[/math] To find [math]T_{4,3}[/math] we may add zeroes ‘in the middle’ for any of the members of [math]T_{3,3}[/math] to find [math]T_{4,3}={1002,1011,1101,2001}[/math] [math]T_{5,3}={10002,10011,10101,11001,20001}[/math] [math]T_{6,3}={100002,100011,100101,101001,110001,200001}[/math] We find [math]S(3)=\dfrac{10}{9} \displaystyle \sum_{m\ge 1} \sum_{n \in T_{m,3}} \dfrac{1}{n}[/math] or [math]S(3)=\dfrac{10}{9}\Big[\overbrace{\dfrac{1}{3}}^{m=1}+\overbrace{\left(\dfrac{1}{12}+\dfrac{1}{21}\right)}^{m=2}+\overbrace{\left(\dfrac{1}{102}+\dfrac{1}{111}+\dfrac{1}{201}\right)}^{m=3}+\ldots\Big][/math] Finding the exact value of [math]S(k)[/math] is difficult. The case [math]k=1[/math] is straightforward, but even [math]k=2[/math] constitutes a problem, since we have [math]S(2)=\dfrac{10}{9} \Big[\dfrac{1}{2}+\left(\dfrac{1}{11}+\dfrac{1}{101}+\dfrac{1}{1001}+\ldots\right)\Big]=\dfrac{5}{9}+\dfrac{10}{9}\displaystyle \sum_{m\ge 1} \dfrac{1}{10^m+1}[/math] It certainly converges, since we have [math]\displaystyle \sum_{m\ge 1} \dfrac{1}{10^m+1}<\sum_{m\ge 1} \dfrac{1}{10^m}=\dfrac{1}{9}[/math] so [math]S(2)<\dfrac{55}{81}\approx 0.6790[/math] Proving convergence here benefitted from the fact that we could describe the numbers in [math]T_{m,2}[/math] exactly. For larger [math]k[/math] we have to be more inventive. Even though the numbers in [math]T_{m,k}[/math] grow in size with increasing [math]m[/math] we need some upperbound on the size of these sets. For [math]k=3[/math] this is easy, one may show that [math]\text{#}T_{m,3}=m[/math]. We obtain [math]S(3) < \dfrac{10}{9}\displaystyle \sum_{m\ge 1} \dfrac{m}{10^m}=\dfrac{100}{81}[/math] we could improve upon this bound by realising that one of the terms in [math]T_{m,k},m\ge 2[/math] starts with a [math]2[/math]. Use that [math]j<10^m,\; d\cdot 10^m+j>d\cdot 10^m[/math] and find [math]S(3) < \dfrac{10}{9}\displaystyle \left(\dfrac{1}{3}+\sum_{m\ge 2} \dfrac{m-1}{10^{m-1}}+\dfrac{1}{2}\sum_{m\ge 2} \dfrac{1}{10^{m-1}}\right)=\dfrac{415}{729}\approx 0.5693[/math] For [math]k\le 9[/math] we’re sure that we don’t run into ordered partitions consisting of a number larger than [math]9[/math], if we only consider the elements of [math]T_{m,k}[/math] that don’t contain a [math]0[/math] then for [math]k\le 9,\; m\le k[/math]: [math]\text{#}T_{m,k}=\displaystyle \binom{k-1}{m-1}[/math] which you can find by considering [math]k[/math] [math]1[/math]’s in a row and inserting [math]m-1[/math] vertical bars to split these up in an ordered partition, for instance, we find [math]T_{2,4}[/math] by inserting one vertical bar to [math]1\;1\;1\;1[/math] in three ways [math]1\|1\;1\;1,\; 1\;1\|1\;1,\;1\;1\;1\|1[/math]. Add the numbers that are separated by these bars and find the splits [math]1\|3,\;2\|2,\;3\|1[/math] leading to [math]T_{2,4}={13,22,31}[/math]. Next, we need to know how large [math]T_{m,k}[/math] actually is once we add members by adding zeroes to [math]T_{m-1,k}[/math], where we should follow the chain starting at [math]k=2[/math], as demonstrated for [math]k=3[/math] at the beginning. We probably would like a tight bound since the binomial itself can get pretty big. In order to prove convergence one should probably consider subsets of [math]T_{m,k}[/math] based on their first digit (as I demonstrated in the simple cases [math]k\in {2,3}[/math]). If we order the numbers in the sets [math]T_{m,k}[/math] based on their first digit [math]d[/math] then we need to be able to count [math]a_{m,d,k}[/math], the number of elements of [math]T_{m,k}[/math] that start with [math]d[/math] and may bound [math]S(k)<\dfrac{10}{9}\displaystyle \sum_{d=1}^{\text{min}(9,k)} \dfrac{1}{d} \sum_{m\ge 1} \dfrac{a_{m,d,k}}{10^{m-1}}[/math] Note that [math]a_{1,9,k}=0[/math] if [math]k>9[/math]. The good news is that for [math]k\ge 10[/math] a lot of ordered partitions are no longer admissible. The bad news is that this requires even more accurate counting, if we are to correct for this. Without feeling the need to delve deeper into this convergence issue, we could just let it rip, like Bernard did. I was in luck because Maple calls such an ordered partition (without zeroes) a composition. I decided to halt the routine if the contribution of [math]T_{m,k}[/math] proved to be less than [math]5\cdot 10^{-7}[/math], hoping to retrieve [math]6[/math] significant digits. For [math]k=9[/math] I found that [math]\text{#}T_{11,9}=31824[/math] and [math]S(9)=0.587514[/math]. For [math]k=10[/math] I found that [math]\text{#}T_{10,10}=75573[/math] and [math]S(10)=0.4875973[/math]. If [math]m\ge 10,\;T_{m,10}\le T_{10,10}\cdot 2^{m-10}[/math], which I believe will hold true, then the exact value of [math]S(10)[/math] can be estimated, a bound for the surplus will be (assuming this factor [math]2[/math] is distributed evenly over the subsets having a fixed starting digit [math]d[/math]) [math]\displaystyle 5\cdot 10^{-7} \sum_{k=0}^{\infty} \left(\dfrac{2}{10}\right)^k=6.25\cdot 10^{-7}[/math] where [math]5\cdot 10^{-7}[/math] was the contribution of [math]T_{10,10}[/math] to the sum. I’ll have to double check later, since it would imply that [math]S(10)<S(5)[/math]. I think that a claim that [math]S(30)<S(5)[/math] could do with a well optimized routine, mine certainly isn’t. Sponsored by CDW Corporation How can AI help your teams make faster decisions? CDW’s AI solutions offer retrieval-augmented generation (RAG) to expedite info with stronger insights. Krishna Pandit Studied Education & Games at Little Flower High School · 3y Related What is the sum of the largest 3-digit number and the smallest 4-digit number? Ellis Cave 40+ years as an Electrical Engineer · Author has 7.9K answers and 4.3M answer views · 2y Related If a 3-digit number is k times larger than its digits' sum, what is the minimum value for k? Brute force solution, using the J programming language: If k is an integer: Find all digit sums of three digit integers Divide each each 3-digit integer by its digit sum Extract all integer ratios List the smallest integer ratio: <./(#~(=>.))d=.n%+/"1 s=.sep n=.100 to 999 The minimum value for k as an integer is 11: The integer is 198: 198 = 1+9+8 = 18, k18 = 1118=198 If k doesn’t have to be an integer: ]sm=.<./d=.n%+/"1 s=.sep n=.100 to 999 The minimum value for k if it doesn’t have to be an integer is 10.4737. The integer is 199: 199 = 1+9+9 = 19, k19 = 10.473719 = 199 Shubham Patil Engineer at Marquardt India (2019–present) · Author has 277 answers and 925.3K answer views · 7y Related What is the sum of the largest 2 digit number and the smallest 3 digit number? Considering only positive numbers.. Largest 2 digit number=99 Smallest 3 digit number=100 Sum =100+99=199 Considering negative too.. Largest 2 digit number=99 Smallest 3 digit number=-999 Mikael Segercrantz Studied Computer Engineering at Helsinki Metropolia University of Applied Sciences (Graduated 2020) · Author has 294 answers and 325.6K answer views · 3y Related What is the sum of the greatest one digit number and the smallest three digit number? First, let’s consider only positive integers here, when the greatest one digit number would be 9 and the smalles three digit number 100, their sum is 109. But 100 is not the smallest three digit number, that (still considering only integers) honor belongs to -999, which summed with the greatest one digit number (still the same 9) has the sum of -990. If not considering the negative numbers or integers, three digits is what eg 0.01 has, which summed with (again the greatest one digit number is 9) the 9, is 9.01. There may be other cases, too, but I’m kind of tired and don’t want to spend time tryi First, let’s consider only positive integers here, when the greatest one digit number would be 9 and the smalles three digit number 100, their sum is 109. But 100 is not the smallest three digit number, that (still considering only integers) honor belongs to -999, which summed with the greatest one digit number (still the same 9) has the sum of -990. If not considering the negative numbers or integers, three digits is what eg 0.01 has, which summed with (again the greatest one digit number is 9) the 9, is 9.01. There may be other cases, too, but I’m kind of tired and don’t want to spend time trying to think of any other cases. Related questions What is the general term for the sum of the digits? What is the sum of all 3 digit numbers? How many numbers are between 1 to 10? How many eight-digit numbers have a sum of 4? What is the formula to find the number of digits in a number raised to a power? What is the sum of digits in the number 4444 4444 ? What is the sum of all three digit odd numbers? What is the sum of all even 2-digit numbers? How many 4-digit number(s) is/are there in which the sum of digit is 9? Is there a three digit number where the sum of its digits results in another three digit number? What is the maximum digit number? What is the sum of the digits of all integers 1-1000 inclusive? What is the sum of two digit odd numbers? Can you add 3 odd numbers to get 30? What is the general formula to find the minimum and maximum numbers which are perfect nth power of a number whose digit count is k? What is the relationship between base of number systems (g), n, and k in this? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15591	https://www.quora.com/Can-you-explain-what-an-open-and-closed-interval-is-on-the-real-number-line-R	Something went wrong. Wait a moment and try again. Interval of Interest The Real Numbers Closed Integral Number Line Limits and Intervals Real Number System Open Interval Closed Intervals 5 Can you explain what an open and closed interval is on the real number line (R)? Andrew Droffner Studied Mathematics at Rutgers University (Graduated 1995) · Author has 8.8K answers and 5.7M answer views · 2y The real number line R is a complete ordered field. The ordered part means inequalities make sense over R, e.g. 0<x≤5,x∈R. Open and closed intervals are just another way to write inequalities. The example inequality above is a half-open interval, (0,5],x∈R. An open side has a parenthesis (UK bracket) and the number is outside the interval. The closed side includes the number. Assume x∈R in each case. The interval on the left represents the inequality on the right. (0,5]⟺0<x≤5 half-open interval (−π,42)⟺−π<x<42 full open interval [− The real number line R is a complete ordered field. The ordered part means inequalities make sense over R, e.g. 0<x≤5,x∈R. Open and closed intervals are just another way to write inequalities. The example inequality above is a half-open interval, (0,5],x∈R. An open side has a parenthesis (UK bracket) and the number is outside the interval. The closed side includes the number. Examples Assume x∈R in each case. The interval on the left represents the inequality on the right. (0,5]⟺0<x≤5 half-open interval (−π,42)⟺−π<x<42 full open interval [−2π,3.140]⟺−2π≤x≤3.140 closed interval Sanjay Chakradeo Engineer, interested in Basic Mathematics. · Author has 3.5K answers and 4.3M answer views · 3y Originally Answered: What is the difference between an open interval and a closed interval on a number line or real line? · I shall explain with examples. (-1,3) indicates the open interval from -1 to 3. It includes all the points between -1 & 3, but excluding the end points -1 & 3. It is indicated on a number line as small hollow circles at number points -1 & -3 & dark line segment between these 2 points. It is represented as {x: x is real, -1<x<3}. [-2,4] indicates the closed interval from -2 to 4. It includes both the I shall explain with examples. (-1,3) indicates the open interval from -1 to 3. It includes all the points between -1 & 3, but excluding the end points -1 & 3. It is indicated on a number line as small hollow circles at number points -1 & -3 & dark line segment between these 2 points. It is represented as {x: x is real, -1<x<3}. [-2,4] indicates the closed interval from -2 to 4. It includes both the points -2 & 4 plus all the points between -2 & 4. It is indicated on number line as solid circles at -2 & 4 and dark line between these 2 points. It is represented as { x: x is real, -2 <= x <= 4}. The interval can be open at one end & closed at other end or vice a versa. For example the interval (-3,1] is open at lower end (-3) & closed at... William La Chenal one-time analytical chemist.Now into science of homeopathy. · Author has 2.6K answers and 5.7M answer views · 3y Originally Answered: What is the difference between an open interval and a closed interval on a number line or real line? · A closed set contains its boundaries Notation, square brackets [x,y] An open set does not contain its own boundaries Notation, round brackets (x,y) If this was a human student, i would tell them to look into the separate topic of “study methods”, thus to improve their chances. But the robot is just annoying. More inanity from Don’t block it. Report it. Downvote it.. Related questions What is the difference between an open interval and a closed interval on a number line or real line? What is the definition of an open interval on a number line? What is the definition of a closed interval on a number line? What is Jungkook's real phone number? Why is a closed interval called a closed interval and an open interval called an open interval? Is [0, 1] an open or closed interval? Assistant Bot · 1y Certainly! In the context of the real number line (R), intervals are used to describe a set of numbers that fall within certain bounds. There are two main types of intervals: open intervals and closed intervals. Open Interval An open interval does not include its endpoints. It is denoted as (a,b), where a and b are real numbers, and a<b. The set includes all numbers x such that: a<x<b Example: The interval (2,5) includes all numbers greater than 2 and less than 5, but does not include 2 and 5 themselves. Closed Interval A closed interval includes its endpoints. It is denoted as [a, Certainly! In the context of the real number line (R), intervals are used to describe a set of numbers that fall within certain bounds. There are two main types of intervals: open intervals and closed intervals. Open Interval An open interval does not include its endpoints. It is denoted as (a,b), where a and b are real numbers, and a<b. The set includes all numbers x such that: a<x<b Example: The interval (2,5) includes all numbers greater than 2 and less than 5, but does not include 2 and 5 themselves. Closed Interval A closed interval includes its endpoints. It is denoted as [a,b]. The set includes all numbers x such that: a≤x≤b Example: The interval [2,5] includes all numbers from 2 to 5, including both 2 and 5. Summary Open Interval (a,b): Excludes endpoints a and b. Closed Interval [a,b]: Includes endpoints a and b. Half-Open (or Half-Closed) Intervals There are also half-open intervals, which include one endpoint but not the other: - Half-Open Interval (a,b]: Includes b but not a. - Half-Open Interval [a,b): Includes a but not b. These definitions are fundamental in calculus and real analysis, especially when discussing limits, continuity, and integration. Ajay Sreenivas Former Aerospace Engineer/ Staff Consultant at Ball Aerospace (1980–2010) · Author has 5.3K answers and 1.7M answer views · 2y closed interval consists of all the points from a to b including the end points a and b. open interval consists of all points from a to b, but not a and b. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. 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If this means absolutely nothing to you, I recommend looking into complex analysis Related questions What is the difference between a close interval, an open interval, and a clopen interval? Can you explain the difference between an open interval (a, b) and closed interval [a, b] on a number line diagram? What does it mean to be open or closed with respect to a given interval on the real line R? Is 0 a number? [Does ]0,180 mean open interval or closed interval? Robert Sheraw Author has 1.3K answers and 215K answer views · 2y An interval is a segment of the real number line. A closed interval includes the endpoints and an open interval does not. A closed interval is denoted [a, b} and an open interval (a, b). Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Scott Birrell PhD from Newcastle-upon-Tyne University · Author has 141 answers and 184.3K answer views · 2y An open interval doesn't contain the endpoints and a closed interval contains both endpoints. Cyril Anderson Did my undergrad in physics, math · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2.7K answers and 9.5M answer views · 10y Related Why is a closed interval called a closed interval and an open interval called an open interval? An open set, you can identify a boundary made up of the points that are as "close" to the set as possible without being in it. But when you look at the points that are in the set, you can't identify any outermost points within the set. Say you have an interval (0,1) in the real line. If you look at the left hand side, suppose you have a point 0.000...01 with a million zeroes after the decimal point. Well, you can always add another zero. Million and one, million and two, billion, trillion, it never ends. There is no firm, hard line where the set ends, within the set. There is that greatest low An open set, you can identify a boundary made up of the points that are as "close" to the set as possible without being in it. But when you look at the points that are in the set, you can't identify any outermost points within the set. Say you have an interval (0,1) in the real line. If you look at the left hand side, suppose you have a point 0.000...01 with a million zeroes after the decimal point. Well, you can always add another zero. Million and one, million and two, billion, trillion, it never ends. There is no firm, hard line where the set ends, within the set. There is that greatest lower bound, 0, but it's not in the set. In an open set, every point within has some room around it on every side that is completely within the set (an "epsilon ball"). An open set doesn't have an edge; it's fuzzy. A closed set, on the other hand, you can identify outermost points, and for those boundary points, on one side of them are other points inside the set, and on the other side are points outside the set. Closed sets are closed in another way as well, as I recall. If you have a convergent sequence of points within the set, the limit will also be within the set. So it's "closed" in that way as well. Sponsored by CDW Corporation How well do your distributed teams work together? Enable seamless and secure communication for everyone with Cisco collaboration solutions from CDW. Choudhary Deepak Lives in Jhunjhunu, Rajasthan, India (2016–present) · 8y Originally Answered: What is the difference between open interval and closed interval? · In open interval end points are not included and in closed interval end points are included. Nikhil Panikkar Communications and Signal Processing Engineer · Author has 1.1K answers and 2.8M answer views · 5y Related Why is differentiation defined on an open interval and continuity on a closed interval? Differentiation is a stronger condition than continuity. Differentiability implies continuity but the reverse is not true. Generally, for an interior point, two-sided limits are needed for both continuity and differentiability. But at the endpoints of an interval, we work with one-sided limits for both continuity and differentiability. Some functions are continuous at the endpoints of an interval, but not differentiable at its endpoints, while some are both continuous and differentiable at the endpoints. Relaxing the condition for differentiability allows for the applicability of certain theorem Differentiation is a stronger condition than continuity. Differentiability implies continuity but the reverse is not true. Generally, for an interior point, two-sided limits are needed for both continuity and differentiability. But at the endpoints of an interval, we work with one-sided limits for both continuity and differentiability. Some functions are continuous at the endpoints of an interval, but not differentiable at its endpoints, while some are both continuous and differentiable at the endpoints. Relaxing the condition for differentiability allows for the applicability of certain theorems to be extended to a greater range of functions. As an example consider the function f(x)=x2. This function is both continuous and differentiable in the closed interval [0,1]. But consider the function g(x)=√x. This function is continuous in [0,1], but not differentiable in it. The derivative does not exist at x=0. Now if we were to insist on differentiability in the closed interval, then we wouldn’t be able to apply Rolle’s theorem to g(x), while we would be able to do the same for f(x). Relaxing the criterion of differentiability on the closed interval, allows us to apply Rolle’s theorem to the square root function. Why is this so? This is because Rolle’s theorem doesn’t need the derivative to be defined on the end points. It only cares about the average value between them. So while it needs the function to be defined at the endpoints of the interval, it does not care about whether it is differentiable at those endpoints. Similar considerations apply for the Mean Value theorem(which is the generalization of Rolle’s theorem). Similarly, the Intermediate Value theorem would not work if the function were not defined on the endpoints of the interval. However, as it doesn’t involve any derivatives, it doesn’t care if the function is differentiable at those endpoints. That said, there are some theorems which require the function to be differentiable in a closed interval. Darboux's theorem (which is the Intermediate Value theorem applied to derivatives) requires the function to be differentiable at the endpoints. Matthew Bond Statistics B.A., University of Florida. Math Ph.D., Michigan State (2011) · Upvoted by Justin Rising , PhD in statistics · Author has 359 answers and 1.2M answer views · Updated 2y Related Why is differentiation defined on an open interval and continuity on a closed interval? Both continuity and differentiability are properties defined for individual points. But there are also one-sided definitions as well. Since we have notions of one-sided limits, we can also have notions of one-sided derivatives or continuity. For example, f(x)=\|x\| at the origin has slope 1 as you approach from the right, and slope -1 as you approach from the left. It’s also possible to have two-sided differentiability fail or succeed on the endpoint of an interval of differentiable points. It is often reasonable to interpret differentiability and/or continuity as a one-sided definition when the Both continuity and differentiability are properties defined for individual points. But there are also one-sided definitions as well. Since we have notions of one-sided limits, we can also have notions of one-sided derivatives or continuity. For example, f(x)=\|x\| at the origin has slope 1 as you approach from the right, and slope -1 as you approach from the left. It’s also possible to have two-sided differentiability fail or succeed on the endpoint of an interval of differentiable points. It is often reasonable to interpret differentiability and/or continuity as a one-sided definition when the function is only assumed to exist on a specific closed interval. However, one of the most important theorems of Calculus is the Mean Value Theorem. It says that if a function f is differentiable on (a,b) and also continuous at the endpoints, then for some point c in the interior (a,b), the instantaneous rate of change f′(c) equals the average rate of change, f(b)−f(a)b−a. The endpoint continuity is necessary, or else silly examples like f(x)=x, x≠1, f(1)=0 would give a counterexample on [0,1]. The endpoints need to be in the “right place” or else the interior rate of change will have no relation whatsoever to where you end up. But one-sided differentiability at the endpoints is not needed. For example, f(x)=xsin(1/x),x≠0 and f(0)=0 satisfies the MVT on [0,1π] quite easily, but the one-sided derivative at 0 can be shown to be undefined, with the secant lines oscillating with slopes bouncing toward ±∞. You would need to read a proof to grasp why this works out just fine. Continuity in the interior automatically follows from the assumption of differentiability on (a,b), but books often choose to state the condition as continuity on [a,b], even though only the endpoints remain to be checked (in the sense of two one-sided limits from the direction of the interior of the interval). A lot of results require conditions like this, simply because so many results directly build on the Mean Value Theorem in the first place. The derivative does the “steering”, and continuity at the endpoint just guarantees that you “stop where you expect to stop” and also guarantees that you weren’t going toward a vertical asymptote. There’s nothing theoretically wrong with a function being differentiable on [a,b], but theorems involving this property are less common and less likely to be useful than the MVT. Alex Eustis Ph.D. in Mathematics, University of California, San Diego (Graduated 2013) · Upvoted by Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) · Author has 4.6K answers and 23.8M answer views · 1y Related Can an open interval be similar to a closed interval? So, I'm going to assume intervals in R (with the usual topology), along with the usual definition of similarity meaning: two sets are similar if there is a scale transformation x→ax+b,a,b∈R,a≠0 that maps one onto the other. It's easy to prove that every such transformation is a homeomorphism of R with itself. In particular, it maps open sets to open sets and closed sets to closed sets. Therefore, the only way an open set can be similar to a closed set is when both sets are clopen (closed and open simultaneously). Furthermore, the only clopen subsets of R are the empty set and th So, I'm going to assume intervals in R (with the usual topology), along with the usual definition of similarity meaning: two sets are similar if there is a scale transformation x→ax+b,a,b∈R,a≠0 that maps one onto the other. It's easy to prove that every such transformation is a homeomorphism of R with itself. In particular, it maps open sets to open sets and closed sets to closed sets. Therefore, the only way an open set can be similar to a closed set is when both sets are clopen (closed and open simultaneously). Furthermore, the only clopen subsets of R are the empty set and the full space. (That is, R is connected.) Conclusion: an open set in R can only be similar to a closed set if both sets are empty, or both are R. Whether or not ∅,R count as “intervals” is just a matter of definition. Related questions What is the difference between an open interval and a closed interval on a number line or real line? What is the definition of an open interval on a number line? What is the definition of a closed interval on a number line? What is Jungkook's real phone number? Why is a closed interval called a closed interval and an open interval called an open interval? Is [0, 1] an open or closed interval? What is the difference between a close interval, an open interval, and a clopen interval? Can you explain the difference between an open interval (a, b) and closed interval [a, b] on a number line diagram? What does it mean to be open or closed with respect to a given interval on the real line R? Is 0 a number? [Does ]0,180 mean open interval or closed interval? What is the definition of a real number? What is the definition of an open interval? What is the definition of a closed interval? How do you show intervals on a number line? Can you explain what you mean by representation of real numbers on a number line? [If an open interval contains all non negative real numbers, is it written as 0,ꝏ) or [0,ꝏ] , what is your reason? What is the least real number contained in an open interval (0,1)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15592	https://www.quora.com/What-is-the-criteria-for-a-function-y-f-x-to-have-local-maxima-and-local-minima-at-a-point	Something went wrong. Wait a moment and try again. Local Maxima Second Derivative Test Functions (general) Critical Point Derivatives and Different... Maxima and Minima One Var... Calculus (Mathematics) Local Minima 5 What is the criteria for a function y=f(x) to have local maxima and local minima at a point? B. S. Thomson Lived in Vancouver, BC · Author has 1.2K answers and 2.9M answer views · Mar 5 Q1. How would you write d y / d x in terms of d y / d t and d x / d t ? You are asking naively about the chain rule. Get a textbook and study it. Ignore most Quora answers since they won't give you an adequate account. You need to know the exact conditions under which you can use such a formula. That is covered in all calculus textbooks with illustrative examples. Q2. How would you define increasing and decreasing function? In the calculus the phrase is "increasing [decreasing] on an interval ." But you need to get a textbook and study it since the terminology can be confusing: increasing, strictly increasing, Q1. How would you write d y / d x in terms of d y / d t and d x / d t ? You are asking naively about the chain rule. Get a textbook and study it. Ignore most Quora answers since they won't give you an adequate account. You need to know the exact conditions under which you can use such a formula. That is covered in all calculus textbooks with illustrative examples. Q2. How would you define increasing and decreasing function? In the calculus the phrase is "increasing [decreasing] on an interval ." But you need to get a textbook and study it since the terminology can be confusing: increasing, strictly increasing, nondecreasing, monotonic, etc. Make sure you know which is being used. Then you will find that your textbook is focused almost to an obsession about providing you with an understanding of increasing functions and tools to solve problems about that. Q3. What is the necessary condition that y = f ( x ) has local maxima and local minima at a point? In mathematics we would not say the necessary condition since there is hardly ever a unique one apart from just restating the definition. So do you really, I mean really, want a necessary condition for a local maximum/minimum? I doubt it. If so, here is one: A necessary condition for a function to have a local extremum at a point is that the point belongs to the domain of the function. If a poster mentions the second derivative here...well that is not necessary. A first derivative ...not a necessary condition. Continuity also not necessary. I think the best advice is to get a textbook and study it. The first lesson is that there is no necessary condition which will help solve any standard calculus problem. Maybe you would like a few sufficient conditions. This is a popular topic in all calculus textbooks and all of the answers to your questions will be found there. Related questions What are the criteria for a function to have local maxima and local minima? Is it possible that a function may have local maxima or local minima at non-stationary or at non-critical point? Is the following statement correct: “It is impossible for a function f ( x , y ) to have two local minima and no local maxima”? Why while computing maxima and minima we put f '(x) is equal to zero? How do you solve f ( x y + 1 ) − f ( x + y ) = f ( x ) f ( y ) for real x, y, and f ( − 1 ) ≠ 0 ? Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 6y I hope you are expecting an answer in plain English because that is what I am giving you. A function or a CURVE has a TURNING point sometimes called a STATIONARY point when the GRADIENT is ZERO. BUT there are 3 types of turning points. A curve has a local MAXIMUM at a point if the gradient on its left is positive I hope you are expecting an answer in plain English because that is what I am giving you. A function or a CURVE has a TURNING point sometimes called a STATIONARY point when the GRADIENT is ZERO. BUT there are 3 types of turning points. A curve has a local MAXIMUM at a point if the gradient on its left is positive and the gradient on its right is negative. A curve has a local MINIMUM at a point if the g... Stephen Kazoullis I studied mathematics at the University of Queensland to third year level . · Author has 6.3K answers and 8.4M answer views · Updated 2y For maxima and minima, a function must be continuous and the first derivative is equal to zero at a some point. For a minimum, the second derivative must be positive at that point, and for a maximum it mus t be negative. For a PHI, the first derivative must be tested before and after the point where f’(x) =0. Senia Sheydvasser PhD in Mathematics · Upvoted by David Joyce , Professor of Mathematics at Clark University · Author has 2.5K answers and 39.9M answer views · 5y Related Is the following statement correct: “It is impossible for a function f ( x , y ) to have two local minima and no local maxima”? As written, the statement isn’t really well-posed, since it doesn’t specify what the domain of this function is. Given the tags and the words “local minima”, I assume that we are talking about functions f:R×R→R. Even with such a restriction, the statement is clearly false. Consider the following counter-example: f(x,y)={0if x=0, y=1x2+y2otherwise. Now, f(x,y) clearly has a local minimum at (x,y)=(0,0). More interestingly, it also has a local minimum at (x,y)=(0,1) As written, the statement isn’t really well-posed, since it doesn’t specify what the domain of this function is. Given the tags and the words “local minima”, I assume that we are talking about functions f:R×R→R. Even with such a restriction, the statement is clearly false. Consider the following counter-example: f(x,y)={0if x=0, y=1x2+y2otherwise. Now, f(x,y) clearly has a local minimum at (x,y)=(0,0). More interestingly, it also has a local minimum at (x,y)=(0,1). It doesn’t have any other local minima, and certainly doesn’t have any local maxima. You might be wondering whether you can request that f be continuous (and perhaps even differentiable, smooth, or analytic) and still have the statement be false. The answer is “yes”—indeed, Sergei Michailov's solution almost works. You can fix it just by tilting the starting plane. Just so, below I have drawn the plot of f(x,y)=y−4(11+(x−2)2+y2+11+(x+2)2+y2). Since this is a smooth (in fact, analytic) function, we know that we can find local minima and maxima by determining where the gradient of this function is zero. In this case, the gradient will be some rational function, which of course can only be zero at finitely many points, which are easy enough to enumerate—for this function, there are precisely four, which I have conveniently added to the plot. Two of these points are obviously local minima. I claim that the remaining two are neither local minima nor maxima. The way to see this is to compute the second derivatives in both the x and y directions at those points—you will see that the second derivative is positive in one direction, and negative in the other, demonstrating that locally these points look like saddles. And there you go: here is an example of an analytic function that has two local minima and no local maxima. Sponsored by Grubhub For Merchants Ready to reach Amazon Prime customers? Reach more customers through our Amazon partnership. Prime members receive free Grubhub+. Related questions How can I find maxima and minima of any given function? What is the maxima and minima at the points where the function is not differentiable? What are the local maxima and local minima (x-1) (x-2) (x-3)? How do I find local maxima and minima? Does a function need to be continuous to have local maxima or minima? Assistant Bot · Sep 6 Critical criteria for local extrema (local maxima or minima) of a real-valued function y = f(x) at a point x = a, presented as precise tests and useful caveats. Definitions Local (relative) maximum at a: there exists a neighborhood (a − δ, a + δ) such that for all x in that neighborhood, f(x) ≤ f(a), with strict inequality for a strict/local strong maximum. Local minimum at a: there exists a neighborhood such that for all x in that neighborhood, f(x) ≥ f(a). Necessary conditions If f is differentiable at a and a is a local extremum, then f′(a) = 0. If f is not differentiable at a, a can still be an Critical criteria for local extrema (local maxima or minima) of a real-valued function y = f(x) at a point x = a, presented as precise tests and useful caveats. Definitions Local (relative) maximum at a: there exists a neighborhood (a − δ, a + δ) such that for all x in that neighborhood, f(x) ≤ f(a), with strict inequality for a strict/local strong maximum. Local minimum at a: there exists a neighborhood such that for all x in that neighborhood, f(x) ≥ f(a). Necessary conditions If f is differentiable at a and a is a local extremum, then f′(a) = 0. If f is not differentiable at a, a can still be an extremum (e.g., cusp). So differentiability is not required. Sufficient tests (when differentiability or higher derivatives exist) First derivative test (most robust for one-dimensional problems): If f′ changes sign from positive to negative at a (f′(x) > 0 for x < a near a, f′(x) < 0 for x > a), then f has a local maximum at a. If f′ changes sign from negative to positive at a, then f has a local minimum at a. If f′ does not change sign (stays positive or stays negative), then a is not a local extremum (it may be a saddle/monotone point). Second derivative test (convenient when f is twice differentiable at a and f′(a) = 0): If f″(a) > 0 → local minimum at a (concave up). If f″(a) < 0 → local maximum at a (concave down). If f″(a) = 0 → test is inconclusive; need higher-derivative test or first-derivative sign analysis. Higher-derivative test (if f is sufficiently smooth and lowest nonzero derivative at order n, with n ≥ 2): If f′(a) = f″(a) = … = f^(n−1)(a) = 0 and f^(n)(a) ≠ 0: If n is even and f^(n)(a) > 0 → local minimum. If n is even and f^(n)(a) < 0 → local maximum. If n is odd → no extremum (inflection-type stationary point). Alternative sufficient criterion (second-order Taylor / quadratic form) If f is C^2 near a, write f(x) ≈ f(a) + ½ f″(a)(x − a)^2 + higher terms. Positive quadratic coefficient implies local min, negative implies local max. Edge cases and practical notes Boundary points: extremum at an endpoint of domain requires one-sided neighborhood comparisons: check values on the domain side only. Saddle points / horizontal inflection: stationary points with no sign change of f′ (or odd-order nonzero derivative) are not extrema. Nondifferentiable extrema: evaluate one-sided derivatives or compare nearby function values directly. For multivariable functions, replace f′ with gradient and f″ with Hessian: gradient = 0 is necessary; positive-definite Hessian → local minimum, negative-definite → local maximum, indefinite → saddle. Summary checklist to test a point a for being a local extremum If a is interior: check differentiability. If differentiable, compute f′(a). If f′(a) ≠ 0 → no extremum. If f′(a) = 0 → proceed. Use first-derivative sign changes or second-derivative test. If inconclusive, use higher-derivative test or compare f(x) values in a small neighborhood. For endpoints use one-sided checks. This set of necessary and sufficient criteria covers standard cases encountered in calculus and optimization. Sambhuti Vijay Studied at DAV Public School, Unit-8, Bhubaneswar · Author has 115 answers and 13.1K answer views · Mar 5 Originally Answered: How would you write dy/dx in terms of dy/dt and dx/dt? How would you define increasing and decreasing function? What is the necessary condition that y=f(x) has local maxima and local minima at a point? · PART 1 To express dydx in terms of dydt & dxdt - Using the chain rule, if y and x are both functions of t, then: dydx=dydtdxdt provided that dxdt≠0. PART II Definition of Increasing and Decreasing function A function f(x) is said to be increasing on an interval I if: f′(x)>0for all x∈I. This means that as x increases, f(x) also increases. A function f(x) is said to be decreasing on an interval I if: f'(x) < 0 \quad \text{for PART 1 To express dydx in terms of dydt & dxdt - Using the chain rule, if y and x are both functions of t, then: dydx=dydtdxdt provided that dxdt≠0. PART II Definition of Increasing and Decreasing function A function f(x) is said to be increasing on an interval I if: f′(x)>0for all x∈I. This means that as x increases, f(x) also increases. A function f(x) is said to be decreasing on an interval I if: f′(x)<0for all x∈I. This means that as x increases, f(x) decreases. PART III A function y=f(x) has a local maximum or local minimum at a point x=c if: f′(c)=0 This means that the derivative must be zero at x=c, which indicates a critical point. However, this condition alone can not guarantee a local max or min; further tests (like the second derivative test) are required to confirm whether it is a maximum or minimum. If f′′(c)>0 then it is a point of minima. If f′′(c)<0 then it is a point of maxima. Note : f′(x) is the first derivative of x whilef′′(x) is the second derivative of x. Deb P. Choudhury Former Professor at University of Allahabad · Author has 10K answers and 8M answer views · Updated 4y If the point is 'a’, then it is a local maximum (resp.minimum) point of f(x), if f is defined in a neighborhood of a and there is a real k>0, such that for every x with a-k <x < a and a < x < a+k, f(a) > f(x) (resp.f(a) < f(x) ) holds. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Nikhil Panikkar Communications and Signal Processing Engineer · Author has 1.1K answers and 2.8M answer views · Updated 5y Related Is it possible that a function may have local maxima or local minima at non-stationary or at non-critical point? Yes, it is possible. Remember that the critical point test works by checking for points where the function’s derivative changes its sign(i.e the derivative goes from being increasing to decreasing or vice versa). But this does not work for functions that are strictly monotonic. Strictly monotonic functions don’t have a global maxima or minima, but will have a local maxima or minima at the endpoints of a closed interval. However, because the function is strictly monotonic, the first derivative retains its sign at the endpoints and can’t be detected at by the critical point test. As an example cons Yes, it is possible. Remember that the critical point test works by checking for points where the function’s derivative changes its sign(i.e the derivative goes from being increasing to decreasing or vice versa). But this does not work for functions that are strictly monotonic. Strictly monotonic functions don’t have a global maxima or minima, but will have a local maxima or minima at the endpoints of a closed interval. However, because the function is strictly monotonic, the first derivative retains its sign at the endpoints and can’t be detected at by the critical point test. As an example consider the function f(x)=x over the interval [−2,2]. This is a strictly monotonically increasing function, and does not have a global maxima or minima. It has a local maximum at 2 and a local minimum at −2. But these points can’t be detected by the critical function test, because the derivative retains its sign around these points. Even functions that are not strictly monotonic, might have regions where are strictly monotonic. For example, consider the function f(x)=x2 on [−2,2]. This function is not strictly monotonic as f(−x)=f(x). But the function is strictly monotonically increasing on the positive x-axis and strictly monotonically decreasing on the negative x-axis. As you can see, the function does have local extrema at 0(minima) and 2 and −2(maxima). While the minima at 0 is detected by the critical point test, the maxima at 2 and −2 are not, as the derivative retains its sign at these points. A second problem arises when functions are continuous, but not differentiable at some points in their domain. The critical point test, which relies on the derivative being defined at a particular point is useless here. As an example, consider the function f(x)=\|x\| on [−2,2] . This function while being continuous at 0, doesn’t have a derivative at that point because the differential coefficient has different signs on either side of 0. So the critical point test doesn’t give this point, even though the function has a local minimum at the point. A third problem arises when in addition to the first derivative being zero, the second derivative is also zero. The point is then not a point of maxima or minima, but one of inflection i.e the function transitions from concave to convex or vice versa at this point. As an example consider the function f(x)=x3 on [−2,2]. This function has a critical point at x=0, but it is not an extemum. The extrema occur at the endpoints of the interval −2 and 2, while at the point 0, the function transitions from being concave to convex. Remember that the Weierstrass Extreme Value Theorem guarantees that any function, which is continuous in a closed interval, will have both a maxima and a minima in the interval. So whenever the critical point test fails to give points of extrema, one should check for: Regions of strictly monotonic increase or decrease. Points where the function is not differentiable. Points of inflection. Chaula Dixit MSc in Mathematics, Christ (deemed university) (Graduated 2020) · 7y Related How can I find maxima and minima of any given function? For any function of one variable: f(x) Step 1- Find f'(x) Step 2- Find ‘a' for which f'(a)=0 (a is called critical point) Step 3- Find f”(x) Step 4- Substitute a in f”(x) If f”(x) > 0, then f has minimum value f(a) If f”(x) < 0, then f has maximum value f(a) For any function of two variables: f(x,y) Step 1- Find fx (partial derivative w.r.t. x) & fy (partial derivative w.r.t. y) Step 2- Solve for fx = 0 and fy = 0, call it (a,b) Step 3- Find fxx, fyy & fxy at (a,b) Step 4- Find value of equation: t = fxxfyy - (fxy)^2 If t < 0, (a,b) is called a saddle point If t > 0, f has min or max at (a,b) (fxx & fyy > For any function of one variable: f(x) Step 1- Find f'(x) Step 2- Find ‘a' for which f'(a)=0 (a is called critical point) Step 3- Find f”(x) Step 4- Substitute a in f”(x) If f”(x) > 0, then f has minimum value f(a) If f”(x) < 0, then f has maximum value f(a) For any function of two variables: f(x,y) Step 1- Find fx (partial derivative w.r.t. x) & fy (partial derivative w.r.t. y) Step 2- Solve for fx = 0 and fy = 0, call it (a,b) Step 3- Find fxx, fyy & fxy at (a,b) Step 4- Find value of equation: t = fxxfyy - (fxy)^2 If t < 0, (a,b) is called a saddle point If t > 0, f has min or max at (a,b) (fxx & fyy > 0, f is minimum fxx & fyy < 0, f is maximum) Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Maurice Bourdin Web Programmer (2019–present) · Author has 3.1K answers and 1.9M answer views · 8y Related What are the criteria for a function to have local maxima and local minima? If the function is differentiable and the derivative is continuous, then local extrema can be found when the derivative reaches zero and then changes sign (minimum if - before and + after, maximum if + before and - after) If the function is continuous, and differentiable almost everywhere, see when the derivative changes sign , even if it is not defined at the very point you are looking at. for piecewise continuous functions, points of discontinuity should be considered as well. Ted Horton 25+ years experience teaching physics and math. · Author has 401 answers and 688.5K answer views · 5y Related Where can I locate all relative maxima, relative minima, and saddle points, if any, for the function f (x, y) = ysinx? Quick edit 9:05 am: See after graph at end. Note that the domain of your function covers all real numbers x and y. First read my answer to: Ted Horton's answer to How do you minimize a function with two variables? That answer really covers the general idea for minima, maxima and saddle points. In particular you need to find the quantity D that is described there, which is used in a sort-of “second derivative test”. If you are working these kinds of problems finding D is crucial. Also, I am going to assume you are familiar with the notations: fx , fy , fxx , fyy , fxy , and fyx for fi Quick edit 9:05 am: See after graph at end. Note that the domain of your function covers all real numbers x and y. First read my answer to: Ted Horton's answer to How do you minimize a function with two variables? That answer really covers the general idea for minima, maxima and saddle points. In particular you need to find the quantity D that is described there, which is used in a sort-of “second derivative test”. If you are working these kinds of problems finding D is crucial. Also, I am going to assume you are familiar with the notations: fx , fy , fxx , fyy , fxy , and fyx for first and second order partial derivatives. We need all these derivatives so let’s calculate them for your function: fx=y⋅cos(x) fy=sin(x) fxx=−y⋅sin(x) fyy=0 fxy=fyx=cos(x) Critical points can only occur when the first partial derivatives are zero simultaneously. Start with that, and it gives us two equations in x and y. Hopefully we can solve for all (x,y) values where these 2 equations are simultaneously satisfied: i. fx=y⋅cos(x)=0 ii. fy=sin(x)=0 Notice that the second equation (the y derivative) makes our solutions very restrictive. It only works if sin(x)=0, which means we must choose x-values: x=k⋅π where k is any integer. This implies something very important for the first equation, the x derivative: The factor cos(x) is never 0 for these x values. The equations must be solved simultaneously, so the first equation tells us that the only y- value possible for critical points is: y=0 In general we can describe our critical points as all (x,y) such that: Pc=(xc,yc)=(k⋅π,0) As was already stated, k is any integer. For the “second derivative test”, we need the quantity “D” that I mentioned, and we also need fxx . We should evaluate these for any possible critical point. D is defined as: D=fxx⋅fyy−(fxy)2 D=0−cos2(x) D=−cos2(x) The cosine is squared which means with the minus sign that D is always negative . You can put in your critical values of x if you like, the fact is D is still negative: D=−cos2(k⋅π) Cosine of k⋅π is +1 for even integers k, and it is -1 for odd integers k. But we square that, so we always are left with a +1 for cosine squared, and an overall minus sign from subtracting: D=−1 , for all critical points. We also need fxx : fxx=−y⋅sin(x) It does not matter what x is, we require that y=0 at the critical points. Therefore fxx is always 0 at the critical points: fxx=0 for all critical points. Now let’s make a general list of the conditions for the test: Local minimum: D>0 , and fxx>0 Local maximum: D>0 , and fxx<0 Saddle point: D<0 If D=0, you are stuck, unless you know how to do the “third derivative test”. In your case we can conclude that. (1). Your function does not have any local maxima. (2). Your function does not have any local minima. and (3). All of the critical points are saddle points. Weird, huh? Here is a graph (courtesy of Wolfram Alpha): This may help you see why there are only saddle points. The lines that run along the peaks and valleys of the ripples can be described by: z=±y Quick edit 9:05 am: Also notice that all of the saddle points lie at specific x-values along the x-axis (the “valley” that runs along the middle). The x-axis is y=0, one of our conditions. Mohammad Afzaal Butt B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views · 5y Related What is the local maxima minima of function y=x^4-4x^3+4x^2? y=x4−4x3+4x2 y′=4x3−12x2+8x y′′=12x2−24x+8 In order to find local maxima and minima we equatey′=0 ⟹4x3−12x2+8x=x(4x2−12x+8)=0 ⟹x=0∧4x2−12x+8=0 4x2−12x+8=0⟹x2−3x+2=0 ⟹x2−2x−x+2=0 ⟹x(x−2)−1(x−2)=0 ⟹(x−2)(x−1)=0 ⟹x=2∧x=1 [math]\text{When x = 1} \quad y’’ \lt 0 \therefore\,\,\text{ x = 1 is maxima}[/math] [math]\text{When x = 2} \quad y’’ \gt 0 \t[/math] [math]y = x^4 - 4 x^3 + 4 x^2[/math] [math]y’ = 4 x^3 - 12 x^2 + 8 x[/math] [math]y’’ = 12 x^2 - 24 x + 8[/math] [math]\text{In order to find local maxima and minima we equate}\,\,y’ = 0[/math] [math]\implies 4 x^3 - 12 x^2 + 8 x = x (4 x^2 - 12 x + 8) = 0[/math] [math]\implies x = 0\quad \wedge \quad 4 x^2 - 12 x + 8 = 0[/math] [math]4 x^2 - 12 x + 8 = 0 \implies x^2 - 3 x + 2 = 0[/math] [math]\implies x^2 - 2 x - x + 2 = 0[/math] [math]\implies x (x - 2) -1(x - 2) = 0[/math] [math]\implies (x - 2) (x - 1) = 0[/math] [math]\implies x = 2\quad \wedge \quad x = 1[/math] [math]\text{When x = 0} \quad y’’ \gt 0 \therefore\,\,\text{ x = 0 is minima}[/math] [math]\text{When x = 1} \quad y’’ \lt 0 \therefore\,\,\text{ x = 1 is maxima}[/math] [math]\text{When x = 2} \quad y’’ \gt 0 \therefore\,\,\text{ x = 2 is minima}[/math] Ajesh K C Studied at Mar Athanasius College of Engineering, Kothamangalam · Author has 1.3K answers and 3.3M answer views · 5y Related What is the local maxima minima of function y=x^4-4x^3+4x^2? [math]y =x^4-4x^3+4x^2[/math] Differentiating with respect to [math]x[/math] [math]y' = 4x^3-12x^2+8x[/math] [math]y' = 4x(x^2-3x+2)[/math] [math]y' = 0[/math] [math]{\implies}[/math] [math]4x(x-1)(x-2) = 0[/math] [math]x = 0,1,2[/math] [math]y'' = 12x^2-24x+8[/math] [math]y''(0) = 8 >0 [/math] [math]{\implies}[/math] minimum at [math]x= 0[/math] [math]y''(1) = -4 <0 [/math] [math]{\implies}[/math] maximum at [math]x= 1[/math] [math]y''(2) = 8 >0 [/math] [math]{\implies}[/math] minimum at [math]x= 2[/math] Local maxima at [math]x =1[/math] and maximum value [math]= 1[/math] [math]y =x^4-4x^3+4x^2[/math] Differentiating with respect to [math]x[/math] [math]y' = 4x^3-12x^2+8x[/math] [math]y' = 4x(x^2-3x+2)[/math] [math]y'=4x(x-1)(x-2)[/math] [math]y' = 0[/math] [math]{\implies}[/math] [math]4x(x-1)(x-2) = 0[/math] [math]x = 0,1,2[/math] Now [math]y'' = 12x^2-24x+8[/math] [math]y''(0) = 8 >0 [/math] [math]{\implies}[/math] minimum at [math]x= 0[/math] [math]y''(1) = -4 <0 [/math] [math]{\implies}[/math] maximum at [math]x= 1[/math] [math]y''(2) = 8 >0 [/math] [math]{\implies}[/math] minimum at [math]x= 2[/math] So Local maxima at [math]x =1[/math] and maximum value [math]= 1[/math] Related questions What are the criteria for a function to have local maxima and local minima? Is it possible that a function may have local maxima or local minima at non-stationary or at non-critical point? Is the following statement correct: “It is impossible for a function f ( x , y ) to have two local minima and no local maxima”? Why while computing maxima and minima we put f '(x) is equal to zero? How do you solve f ( x y + 1 ) − f ( x + y ) = f ( x ) f ( y ) for real x, y, and f ( − 1 ) ≠ 0 ? How can I find maxima and minima of any given function? What is the maxima and minima at the points where the function is not differentiable? What are the local maxima and local minima (x-1) (x-2) (x-3)? How do I find local maxima and minima? Does a function need to be continuous to have local maxima or minima? What is the maxima and minima of the function f(x,y) = x^3-2y^2-2y^4+3x^2y? How do you find all the local maxima, local minima, and saddle points of the function f ( x , y ) = 3 x 2 − 2 x 3 − 3 y 2 + 6 x y ? What are some examples for functions with (second derivative) f''(x) =0 and has maxima/minima at that point? What is the definition of a function that does not have any local maxima or minima? Can any function have local maxima and local minima equal valued? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15593	https://connect.learnpad.com/content/profile.cfm?id=11652	KS1 Bitesize - Content - ClassConnect English (US) ManageDashboardContentHelp Sign InRegister Info Error KS1 Bitesize Science, Mathematics, English KS1 5 - 7 2MB Free KS1 Bitesize Literacy KS1 Bitesize Maths KS1 Bitesize Science Scan QrKey Description The KS1 Bitesize profile is a collection of activities taken from the BBC Bitesize website that are appropriate for supporting teaching in Literacy, Maths and Science at Key Stage 1. The literacy category includes sentence structure, punctuation and world level activities. Children can rehearse skills including spelling, phonics, alphabetical order, pronouns, connectives (joining words), synonyms, sentence structure and punctuation. The maths category includes activities related to calculating, the number system, handling data, shape and space, measures; and using and applying. Children can find out about length and weight, addition and subtraction, times tables, division, ordering and sequences, place value, money, time, shapes, collecting data and bar charts. The science category includes activities linked to life processes and living things, materials and their properties, and physical properties. Children think about how to identify living things, health and growth, variation, types of plants and animals, the properties of materials and how they change; electricity, rocks, light and sound. Contents KS1 Bitesize Literacy River Rhyming Alphabetical Adventure Pirate Spelling Travelling Pronouns Balloon Joining Words Matching Word Golf Alien Punctuation Kung-Fu Sentences Deep Sea Phonics KS1 Bitesize Maths Harbour Measurements Number Pyramid Camel Times Tables Division Mine Forest Ordering Adventure Sequences Safari Units Igloo Shopping Clockworks Shape Lab Animal Island Data KS1 Bitesize Science Forest Life Trail Animal Hospital Garden Guru Variation Detective Enviro Spotter Desert Island Materials Kitchen Materials Electro-matic Factory Rock and Roll Spooky Mansion Play Your Sounds Right You Are Here ClassConnect Content KS1 Bitesize Privacy Notice \| © 2025 Avantis Systems Limited Language Selection Close Select Language Loading... Initial language selection is based on your web browser preferences. Change Language Close
15594	https://pubs.usgs.gov/unnumbered/70039582/report.pdf	Map Scales 1:63,360 scale Bench Lake U.S. Department of the Interior U.S. Geological Survey Earth Science Information Center (ESIC) To be most useful, a map must show locations and distances accurately on a sheet of paper of convenient size. This means that everything included in the map ground area, distance, rivers, lakes, roads, and so on must be shown proportionately smaller than it really is. The proportion chosen for a particular map is its scale. Large Is Small Simply defined, scale is the relationship between distance on the map and distance on the ground. A map scale might be given in a drawing (a graphic scale), but it usually is given as a fraction or a ratio-1/10,000 or 1:10,000. These "representative fraction" scales mean that one unit of measurement on the map 1 inch or 1 centimeter represents 10,000 of the same units on the ground. If the scale were 1:63,360, for instance, then 1 inch on the map would represent 63,360 inches or 1 mile on the ground (63,360 inches divided by 12 inches = 5,280 feet or 1 mile). The first number (map distance) is always 1. The second number (ground distance) is different for each scale; the larger this second number is, the smaller the scale of the map. "The larger the number, the smaller the scale" sounds confusing, but it is easy to understand. A map of an area 100 miles long by 100 miles wide drawn at a scale of 1:63,360 would be more than 8 feet square! To make this map a more convenient size, either the scale used or the amount of area included must be reduced. If the scale is reduced to 1:316,800, then 1 inch on the map represents 5 miles on the ground, and an area 100 miles square can be mapped on a sheet less than 2 feet square (100 miles at 5 miles/inch equals 20 inches, or 1.66 feet). On the other hand, if the original 1:63,360 scale is used but the mapped area is reduced to 20 miles square, the resulting map will also be less than 2 feet square. Such maps would be much handier. But would they be more useful? In the small-scale map (1:316,800), there is less room; therefore, everything must be drawn smaller, and some landmarks must be left out altogether. On the other hand, the larger scale map (1:63,360) permits more detail, but it also covers much less ground. Many areas have been mapped at different scales. When choosing a map that is, when choosing a scale the most important consideration is its intended use. A town engineer, for instance, may need a very detailed map in order to precisely locate house lots, power and water lines, and streets and alleys in a community. A commonly used scale for this purpose is 1:600 (1 inch on the map represents 50 feet on the ground). This scale is so large that many features such as buildings, roads, railroad tracks that are usually represented on smaller scale maps by symbols can be drawn to scale. 1:500,000 scale _ 1:20,000 scale 1:24,000 scale 1:100,000 scale 1:250,000 scale U.S. Geological Survey Scales The U.S. Geological Survey publishes maps at various scales. The scale used for most U.S. topographic mapping is 1:24,000. Maps published at this scale cover 7.5 minutes of latitude and 7.5 minutes of longitude; they are commonly called "7.5-minute quadrangle" maps. Map coverage for the United States has been completed at this scale, except for Puerto Rico, which is mapped at 1:20,000 and 1:30,000, and a few States that have been For more information contact any Earth Science Information Center (ESIC) or call 1-800-USA-MAPS or the following office: mapped at 1:25,000. Most of Alaska has been mapped at 1:63,360, with some populated areas also mapped at 1:24,000 and 1:25,000. The 1:24,000 scale is fairly large. A map at this scale provides detailed information about the natural and manmade features of an area, including the locations of important buildings and most campgrounds, caves, ski lifts, watermills, and even drive-in theaters. Footbridges, drawbridges, fence lines, private roads, and changes in the number of lanes in a road are also shown at this scale. They would be omitted, usually, from maps in the 1:50,000 to 1:100,000 scale range; these maps cover more area while retaining a reasonable level of detail. Maps at these scales most often use the 15-minute or 30-by-60 minute quadrangle formats. Small-scale maps (1:250,000 and smaller) show large areas on a single map sheet, but details are limited to major features boundaries, State parks, airports, major roads, and railroads. USGS Topographic Maps Scale 1 :20,000 1 :24,000 1 :25,000 1 :50,000 1 :62,500 1:63,360 1:100,000 1:100,000 1:125,000 1:250,000 1:250,000 1:500,000 1:500,000 1:1,000,000 Series Puerto Rico 7.5 minute 7.5 minute 7.5X15 minute Intermediate 15 minute Alaska 1:63,360 Intermediate Intermediate 30 minute United States Antarctica Antarctica State maps United States 1 inch represents 1 ,667 feet (about) 2,000 teet 2,083 feet (about) .Smile (about) 1 mile (about) 1 mile 1.6 miles (about) 1.6 miles (about) 2 miles (about) 4 miles (about) 4 miles (about) 8 miles (about) 8 miles (about) 16 miles (about) 1 centimeter represents 200 meters 240 meters 250 meters 500 meters 625 meters 634 meters (about) 1 kilometer 1 kilometer 1.25 kilometers 2.5 kilometers 2.5 kilometers 5 kilometers 5 kilometers 10 kilometers Standard quadrangle size (latitude-longitude) 7.5 X7.5min. 7.5 X7.5min. 7.5 x 15min. NA 15 x 15min. 15 x 20 to 36 min. 30 x 60 min. NA 30 x 30 min. 1° x 2° or 3° 1° x 3° to 15° 2° x 7.5° NA 4° x 6° Quadrangle area (square miles) 71 49 to 70 98 to 140 county 1 97 to 282 207 to 281 1,568 to 2,240 county 788 to 1,128 4,580 to 8,669 4,089 to 8,336 28,1 74 to 30,462 NA 73,73410102,759 1992
15595	https://reference.medscape.com/calculator/508/panic-disorder-severity-scale-pdss	For You News & Perspective Tools & Reference Edition English Medscape Editions English Deutsch Español Français Português UK About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Register Log In EN Medscape Editions English Deutsch Español Français Português UK X Univadis from Medscape About You Professional Information Newsletters & Alerts Your Watch List Formulary Plan Manager Log Out Register Log In close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel Tools & Reference > Calculators CalculatorAboutReferences CalculatorAboutReferences Panic Disorder Severity Scale (PDSS) 7 question scale to grade panic disorder severity Questions 1.How many panic and limited symptoms attacks did you have during the week?2.If you had any panic attacks during the past week, how distressing (uncomfortable, frightening) were they while they were happening?3.During the past week, how much have you worried or felt anxious about when your next panic attack would occur or about fears related to the attacks?4.During the past week, were there any places or situations you avoided, or felt afraid of, because of fear of having a panic attack?5.During the past week, were there any activities that you avoided, or felt afraid of, because they caused physical sensations like those you feel during panic attacks or that you were afraid might trigger a panic attack?6.During the past week, how much did the above symptoms altogether interfere with your ability to work or carry out your responsibilities at school, or home?7.During the past week, how much did panic and limited symptom attacks or worry about attacks interfere with your social life? About The Panic Disorder Severity Scale (PDSS) is a brief, clinician rating scale that was developed in 1997 with the promise of becoming a standard global rating scale for panic disorder. The PDSS has become a simple, efficient way for clinicians to rate severity and treatment progress in patients with established diagnoses of panic disorder. The PDSS, is modeled after the Yale-Brown Obsessive Compulsive Scale so, it contains items that assess the severity of seven dimensions of panic disorder and associated symptoms: the frequency of panic attacks, distress during panic attacks, anticipatory anxiety, agoraphobic fear and avoidance, interoceptive fear and avoidance, impairment of or interference in work functioning; and impairment of or interference in social functioning. In the first study of its performance, this scale demonstrated adequate internal consistency and reliability, excellent inter-rater reliability, good discriminant validity and sensitivity to change. A replication study with a new set of patients confirmed its reliability and convergent and discriminant validity, and added information about cut-off scores to discriminate patients with/without current panic disorder and severity. It’s useful for a clinical as it can gauge response and remission of treatment. The scale has gained wide acceptance and has been translated into Spanish, Portuguese, Italian, Hungarian, Finnish, Serbo-Croatian, Japanese, Korean, Turkish, with satisfactory reliability and validity, comparable to the original English version. References Shear MK, Brown TA, Barlow DH, et al. Multicenter collaborative panic disorder severity scale. American Journal of Psychiatry 1997, 154 (11): 1571-5 Furukawa TA, Katherine shear M, Barlow DH, et al. Evidence-based guidelines for interpretation of the Panic Disorder Severity Scale. Depression and Anxiety 2009, 26 (10): 922-9 Houck PR, Spiegel DA, Shear MK, Rucci P. Reliability of the self-report version of the panic disorder severity scale. Depression and Anxiety 2002, 15 (4): 183-5 The Panic Disorder Severity Scale (PDSS) calculator is created by QxMD. Default Units 1. How many panic and limited symptoms attacks did you have during the week? 0/7 completed About The Panic Disorder Severity Scale (PDSS) is a brief, clinician rating scale that was developed in 1997 with the promise of becoming a standard global rating scale for panic disorder. The PDSS has become a simple, efficient way for clinicians to rate severity and treatment progress in patients with established diagnoses of panic disorder. The PDSS, is modeled after the Yale-Brown Obsessive Compulsive Scale so, it contains items that assess the severity of seven dimensions of panic disorder and associated symptoms: the frequency of panic attacks, distress during panic attacks, anticipatory anxiety, agoraphobic fear and avoidance, interoceptive fear and avoidance, impairment of or interference in work functioning; and impairment of or interference in social functioning. In the first study of its performance, this scale demonstrated adequate internal consistency and reliability, excellent inter-rater reliability, good discriminant validity and sensitivity to change. A replication study with a new set of patients confirmed its reliability and convergent and discriminant validity, and added information about cut-off scores to discriminate patients with/without current panic disorder and severity. It’s useful for a clinical as it can gauge response and remission of treatment. The scale has gained wide acceptance and has been translated into Spanish, Portuguese, Italian, Hungarian, Finnish, Serbo-Croatian, Japanese, Korean, Turkish, with satisfactory reliability and validity, comparable to the original English version. References Shear MK, Brown TA, Barlow DH, et al. Multicenter collaborative panic disorder severity scale. American Journal of Psychiatry 1997, 154 (11): 1571-5 Furukawa TA, Katherine shear M, Barlow DH, et al. Evidence-based guidelines for interpretation of the Panic Disorder Severity Scale. Depression and Anxiety 2009, 26 (10): 922-9 Houck PR, Spiegel DA, Shear MK, Rucci P. Reliability of the self-report version of the panic disorder severity scale. Depression and Anxiety 2002, 15 (4): 183-5 The Panic Disorder Severity Scale (PDSS) calculator is created by QxMD. Legal Notices and Disclaimer © 2020 QxMD Software Inc., all rights reserved. No part of this service may be reproduced in any way without express written consent of QxMD. 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15596	https://wendy-heller.com/wp-content/uploads/2016/01/winged_feet_and_mute_eloquence_dance_in.pdf	Cambridge University Press is collaborating with JSTOR to digitize, preserve and extend access to Cambridge Opera Journal. Winged Feet and Mute Eloquence: Dance in Seventeenth-Century Venetian Opera Author(s): Irene Alm, Wendy Heller and Rebecca Harris-Warrick Source: Cambridge Opera Journal, Vol. 15, No. 3 (Nov., 2003), pp. 216-280 Published by: Cambridge University Press Stable URL: Accessed: 05-06-2015 15:05 UTC REFERENCES Linked references are available on JSTOR for this article: You may need to log in to JSTOR to access the linked references. Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact support@jstor.org. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Cambridge Opera Journal, 15, 3, 216-280 ( 2003 Cambridge University Press DOL 10.1017/S0954586703001733 Winged feet and mute eloquence: dance in seventeenth-century Venetian opera IRENE ALM (edited by Wendy Heller and Rebecca Harris-Warrick) Abstract: This article shows how central dance was to the experience of opera in seventeenth-century Venice. The first part provides an introduction to the use of dance in Venetian opera and the primary sources - libretti, scores, treatises, and various eyewitness reports. The second section summarizes the extraordinary variety of subjects and style of the dances. A third section treats the musical sources, describing stylistic features of the dance music, as well as providing important insights as to how to identify which vocal or instrumental excerpts would likely have been danced. Coming on to the stage, the dancer honours the public He endeavours to tell stories with his skilful hands. And now, when the pleasing retinue pours out sweet songs, Which the singer echoes, he demonstrates by dancing; He fights, plays, loves, revels as Bacchus, turns, stands, With illustration he gracefully completes the performance. The man has as many languages as limbs; Wondrous is the Art Which makes fingers silently speak.' Of all the arts, dance is the most ephemeral. Like sculpture, dance exists in three-dimensional space, but it is also kinaesthetic. And while music also moves through time, dance - for the vast majority of its history - has lacked any form of notation or written score. Time has erased or buried so much direct knowledge, thus historians of dance are left with only scattered fragments of documentation - names of dancers, of steps, of ballets - and the feeble power of words to capture and record the movements of bodies on the stage. As Nino Pirrotta so eloquently wrote regarding the commedia dell'arte: But it often happens in the history of music that the more widely diffused and popular are the facts the historian wishes to examine, the fewer precise elements of knowledge are available to him. In this case at the time of its performance everyone knew the music performed and the ways and means of its execution, but time has swallowed and buried this direct knowledge and has left us only scattered and second- or third-hand documents. We need to gather them together and laboriously interpret them to recover a pale image of a 1 'Bellissima e la descrizione fatta da un Poeta antico dell'azioni d'un Saltatore. ["Very beautiful is the description written by a poet of antiquity of the movements of a dancer":] Ingressus scenam, populos saltator adorat. / Solerti tendit prodere gesta manu. / Nam cum grata cohors diffundit cantica dulcis, / Quae resonat Cantor, motibus ipse probat; / Pugnat, ludit, amat, Baccatur, vertitur, astat, / Illustrans verum cuncta decore replet. / Tot linguae quot membra viro; Mirabilis est Ars / Quae facit articulos ore silente loqui.' Andrea Perrucci, Dell'arte rappresentativa premeditata, ed all'improviso (Naples, 1699), 184. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions CAMBRIDGE JOURNAL 4?,7m.? hz,?,7 lana This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 217 reality that in its own time must have imposed itself with the most obvious power of suggestion.2 Pirrotta's words could equally well describe the task of studying, reconstructing, and writing much of the history of dance.3 Moreover, in Western culture that which goes unrecorded has often been dismissed as of little value. Perhaps because of this, the history of dance has been a latecomer to academic and scholarly studies. Compounding the problem is a centuries-old Western tradition of viewing dance at best with suspicion and at worst as an immoral and even dangerous activity. Choreographers and dancers were rarely given the same status and respect as other artists and musicians. The treatises of the great fifteenth-century Italian dancing masters all begin with a defence of the place dance held among the arts and sciences - in essence a plea for respect. Of the many Italian cities and courts producing opera during the seventeenth century, I have chosen Venice as the focus of this study for a number of reasons. Principal among them is that, following the opening of the first commercial theatre there n 1637, an explosion of operatic activity established Venice as the leading producer of operas during the remainder of the seventeenth century. Moreover, these operas were exported to cities and courts throughout Italy and Europe. The central role of Venice in shaping and expanding this new musical-dramatic repertoire, as well as the substantial documentation of the operas performed there, make it an ideal place to begin a study of Italian theatrical dance during this period. Understanding the function and style of dance in Venetian opera is fundamental to future studies of theatrical choreography in other Italian and European cities. My aim is to provide a foundation for further research through a thorough study of the Venetian ballo.4 Even a cursory glance through the hundreds of libretti for Venetian operas shows that balli were indeed a standard feature of productions during the seventeenth century. The quantity of these dances and the diversity of their subjects provide undeniable evidence that ballet was not created and developed solely in France, but in fact has a rich history in Venice and throughout Italy. To dismiss the Venetian balli as a marginal element of Venetian opera simply because they are different from the well-documented and better-known French dances, perpetuates the false notion cultivated by French writers of the seventeenth and eighteenth centuries that ballet 2 Nino Pirrotta, 'Commedia dell'Arte and Opera', Musical Quarterly, 41 (1955), 170; reprinted in Music and Culture in Italy from the Middle Ages to the Baroque (Cambridge, MA, 1984), 344. 3 On the problems of reconstructing dance - not just steps but also style - see Shirley Wynne, 'Reviving the Gesture Sign: Bringing the Dance Back Alive', in The Stage and the Page: London's 'Whole Show' in the Eighteenth-Centu~y Theatre, ed. Geo. Winchester Stone, Jr. (Berkeley and Los Angeles, 1981), 193-208, and by the same author, 'Baroque Manners and Passions in Modern Performance', in Opera &e Vivaldi, ed. Michael Collins and Elise K. Kirk (Austin, 1984), 170-78. 4 The term 'ballo' means dance in a general sense, but is also specifically used for the majority of theatrical dances in Venetian operas (rather than 'balletto' or 'danza'). I have not translated 'ballo' as 'ballet', since that term has specific connotations associated with French dance and with later styles of theatrical dancing. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 218 Irene Alm is entirely a French art. It also ignores the considerable influence of Italian theatrical dance on eighteenth- and nineteenth-century European ballet.5 There are fundamental reasons why we know so much less about the Venetian balli during this period. These have to do with the status and position of the dancers, and the ways the opera houses were managed. Dancing in the Venetian opera house was a professional activity, and thus there was no need for the kinds of treatises and dance manuals that instructed the nobility of late Renaissance Italy or eighteenth- century France. Choreographies for the operas were probably never notated, but were most likely created and taught during rehearsals, then memorized by the corps of dancers - a practice that continues in most theatres to this day. In this sort of oral tradition (perhaps more aptly thought of as a 'physical' or 'corporeal' tradition), the repertoire, techniques, and styles were passed directly from one generation to the next. Information about the Venetian balli must thus be gleaned from a variety of sources, none of which provides the level of detail that we might desire. No dance treatise deals directly with the Venetian repertoire; the writings on theatrical aesthetics by Doni and other theorists speak only obliquely of operatic balli, focusing much of their attention on the inheritance from the ancients. Diaries, newsletters, and chronicles provide tantalizing - but often frustratingly brief - glimpses of operatic spectacle including dance. The choreographers for Venetian opera did not write treatises, but they nonetheless left a fascinating trail of evidence that also helps us to reconstruct the history of dance in Venetian opera. This is most evident in the career of Giovanni Battista Balbi (ft. 1636-57), who was involved with Venetian opera from its inception. Best known as an impresario and producer, Balbi was responsible for producing Venetian operas in such cities as Naples and Paris, and he collaborated closely with the composer Francesco Cavalli and stage designer Giacomo Torelli. The style of dancing that developed under his direction was an essential part of the production process and was widely imitated. His choreographies for the Paris production of Francesco Sacrati's Lafintapazza (1645), immortalized in a set of engravings by Valerio Spado, were particularly fanciful, incorporating monkeys, bears, ostriches, parrots, along with Indians and Turks. In the ballo for Turks, for example, the exoticism was heightened by the addition of dancing bears to the final part of the dance (see Fig. 1). A handful of other choreographers, such as Giovanni Battista Martini and Olivieri Vigasio, are identified in the surviving account books, which also provide information about 5 This skewed view of dance history has been perpetuated even by Italian historians. Lorenzo Tozzi, for example, begins his study of the eighteenth-century Italian choreographer and dancer Gasparo Angiolini with a chapter on ballet from 1650 to 1750. Although he notes the number of prominent Italian dancers in France during this period, he discusses the rise of professional theatrical dancing only in France, with no mention of the hundreds of Venetian operas, which included balli danced by professionals. Lorenzo Tozzi, 'Breve excursus storico sulle condizioni del balletto tra il 1650 e il 1750', 1i balletto pantomimo del settecento: Gaspare Angiolini (L'Aquila, 1972), 47-53. [Ed. note: For a more balanced view of Italian dance history, see Kathleen Kuzmick Hansell, 'Theatrical Ballet and Italian Opera', in Opera on Stage, vol. 5 of The History of Italian Opera, ed. Lorenzo Bianconi and Giorgio Pestelli (Chicago, 2002), 177-308.] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 219 Fig. 1: Conclusion of the ballo for the Turks, with dancing bears, one of the eighteen engravings by Valerio Spada for the balli by Balbi in La finta pazza, as it was performed in Paris in 1645. (Balletti d'invenzione nella Finta Pazza di Giovanbattista Balbi [n.p., c. 1658].) their relative salaries, the dances, theatres, and operas with which they are associated.6 Unlike Balbi, who was involved in all aspects of production and even wrote the dedication for the Venetian and Neapolitan editions of Veremonda Amazzone di Aragona (1652/3), these choreographers were rarely identified in the libretti. Libretti provide the most important evidence for the extraordinary variety in subject, style, and tone that characterize the Venetian balli. Even the texts of arias and choruses may tell us something about how bodies moved on the stage. And whereas the surviving scores only inconsistently offer up music for the dance, what does exist is richly varied. Considered together, all of these sources provide us with insight into the special union of arts represented by dance on the Venetian opera stage. Indeed, an anonymous keyboard treatise ostensibly from 1664, describes the 6 Ed. note: The account books were discovered and first discussed by Beth L. and Jonathan Glixon, 'Marco Faustini and Venetian Opera Production in the 1650s: Recent Archival Discoveries', Journal of Musicology, 10 (1992), 48-73. They will be explored further in their forthcoming book on opera production in mid-seventeenth-century Venice, which will include further information on the choreographers. On the choreographers, see also Alm's 'Theatrical Dance in Seventeenth-Century Venetian Opera', Ph.D. diss. (University of California at Los Angeles, 1993), chapter 5, and her article 'Balbi, Giovanni Battista', The New Grove Dictionary of Music and Musicians, 2nd edn, ed. Stanley Sadie and John Tyrrell (London, 2000). On the engravings associated with Balbi and the Torelli production, see her 'Giovanni Battista Balbi, "Veneziano Ballerino celebre"', in Giacomo Torelli: L'invenzione scenica nell'Europa barocca, ed. Francesco Milesi (Fano, 2000), 214-26. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 220 Irene Alm connection of dance to the other arts, a connection mirrored in the variety of sources needed to recover its history. From what has been said above of poetry, of music, and of dance, it is evident that all three of these fine arts have a common origin, which is the imitation of beautiful nature, and they have a common goal, which is to communicate to others the ideas and the sentiments of our spirit and our heart. Therefore the peak of their graces remains in their coming together. The masters can, however, separate these three arts, but only to cultivate and polish each one particularly with more care. They still should never lose sight of the first law of nature, nor believe that one can exist well without the others. Nature and taste require that all three be always brought together. In all things there must be a common centre, and point of return, to which the most distant parts aspire. If it is poetry that produces spectacles, it is what must rule at the centre, not music and dance. They must show off more rigorously the ideas of sentiments already expressed in its verses. Therefore music and dance must enhance poetry, not obscure it. And this is the case with opera.7 Sources for the dance Libretti As I noted above, the history of seventeenth-century Italian opera is to a great extent literary; libretti serve as the principal source of information about this repertoire.8 Due in part to the eighteenth-century passion for collecting, libretti survive for nearly all of the more than three hundred operas produced between 1637 and 1700 in Venice, with the vast majority printed rather than manuscript.9 Yet, surprisingly, the copious information on dance contained in the libretti has been largely overlooked, and scholars have instead based their views of dance on the music (or lack of music) in the scores, which represent only about one-third of the operas produced during this period. The libretti tell a very different story from the scores. The early librettists felt a need to defend their forays into opera, arguing for the validity of this new genre, which - although based on the principles of classical tragedy - aimed to attract and 7 Precetti ragionati per apprendere l'accompagnamento del basso sopra gli strumenti da tasto come il gravicembalo il cembalo etc. Venezia MDCLXIIII. I-Vnm, Cod. It. IV 739 (= 10269), fol. 6". [Ed. note: We are grateful to Lorenzo Bianconi for pointing out that the treatise is actually an ingenious forgery by a late eighteenth or early nineteenth-century music historian who supplemented his considerable knowledge of seventeenth-century practices with extensive borrowings from a 1775 keyboard treatise by Vicenzo Manfredini. See Tharald Borgir, The Performance of the Basso Continuo in Italian Baroque Music (Ann Arbor, 1987), 138-40.] 8 Bianconi and Walker point to this literary tradition and its effect on the historiography of opera in their introduction to 'Production, Consumption and Political Function of Seventeenth-Century Opera', Early Music History, 4 (1984), 210-15. 9 The collection at UCLA, which includes 470 seventeenth-century and 816 eighteenth-century Venetian libretti, has been used as the principal source for this study. All references to libretti are based on the copies in this collection, unless otherwise noted. For further information on this collection, see my Catalog of Venetian Librettos at the University of California, Los Angeles (Berkeley and Los Angeles, 1993). The principal libretto collections in Venice have also been consulted: three at the Biblioteca Marciana (the Collezione Groppo, 1637-1796; the Collezione Zeno, 1637-1750; and the Collezione Rossi, 1637-1836), the Cicogna collection at Casa Goldoni, and the Rolandi collection (not exclusively Venetian libretti) at the Fondazione Giorgio Cini. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 221 entertain contemporary audiences.10 Many of the first Venetian libretti and scenarios include lengthy prefatory essays in which the poets cite Greek and Roman sources, attempting to justify the genre or defending their concessions to modern taste - and discussing their use of dance."1 For example, the scenario for Le nozze d'Enea con Lavinia (1641) includes a preface in the form of a letter from the anonymous author to some of his friends, explaining his use of dance in place of choruses at the ends of acts.12 The chorus then was an integral part of the ancient tragedies, entering not only as a character, but singing principally between acts with gestures and leaps, and with those so-to-speak moans and howls. But in the modern [tragedies] it is a less considerable part, being seen in some to do little more than separate the acts. As I have introduced even more choruses within the same acts, I therefore did not make use of them at their end; for since the entire tragedy is sung, also singing the chorus [at the end] would prove to be too tedious; thus to better satisfy the audience by means of variety, balli have been introduced, derived in some way from the plot, just as the ancient choruses danced to song in tetrametre, a verse most appropriate to movements of the body.13 The phrase 'derived in some way from the plot' is particularly important; the librettist's aim for integration, or unity of action, is explicitly stated. In his preface to Venere gelosa (1643), the librettist Niccol6 Enea Bartolini also cites numerous classical authors to show that song and dance were essential parts of drama. Bartolini accordingly includes dancing in the final scene of Act I and in the penultimate scenes of Acts II and II1.14 Each of the three balli is accompanied by alternating solo and choral singing: Niso and a chorus of Nymphs; Trulla and a chorus of toys ('trastulli'); and a solo satyr with a chorus of satyrs. The following year, both the libretto and scenario for L'Ulisse errante (1644) contain an extensive discussion of history and aesthetics in an essay addressed to the opera's dedicatee, Michel'Angelo Torcigliani, by 'Assicurato, Academico Incognito', the academic name of the librettist Badoaro. This essay includes a briefly 10 Ellen Rosand presents an extensive discussion of the attempts by librettists to justify their works by drawing upon classical principles. See especially chapter 2 of her Opera in Seventeenth-Century Venice (Berkeley, 1991). 11 Scenarios are scene-by-scene summaries, either printed separately or included in the libretto itself. On the history of the scenario in Venice, see Ellen Rosand, 'The Opera Scenario, 1638-1655: A Preliminary Survey', In Cantu et in Sermone: For Nino Pirrotta on his 80th Birthday (Florence, 1989), 335-46. 12 The libretto was apparently never printed, but a number of manuscript copies from the seventeenth and eighteenth centuries survive. According to the manuscript libretto at UCLA, Le nozze d'Enea con Lavinia was written by Giacomo Badoaro, although his authorship has been seriously questioned by Thomas Walker; see 'Gli errori di Minerva al tavolino: Osservazioni sulla cronologia delle prime opere veneziane', in Venezia e il melodramma nel seicento, ed. Maria Teresa Muraro (Florence, 1976), 11-12. The printed scenario, which bears the title Argomento et scenario delle noze d'Enea in Lavinia, is not found in the collection of libretti at UCLA. I would like to thank Ellen Rosand for sending me copies of the relevant sections from the scenario. See her comments on the extant examples, in 'The Opera Scenario', 344. 13 Argomento et scenario delle nozze d'Enea in Lavinia, 21. 14 Two additional balli for Venere gelosa are documented in the Apparati sceniciper lo Teatro Novissimo di Venetia nell'anno 1644 d'inventione e cura di lacomo Torelli da Fano (Venice, 1644). This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 222 Irene Alm summarized history of tragedy, and also mentions the practice of substituting balli for choruses: The precepts of poetry are not permanent, because the changes of centuries give birth to the diversity of composition; so that although in its earliest days, Tragedy was recited by the poet alone, his face tinted with the dregs of crushed grapes, later characters and masks were introduced, then they added choruses, music, instruments, scene changes, [and] balli in place of choruses - and perhaps in the future as times change our descendants will see new forms introduced.15 But most importantly, the descriptions of the balli in the libretti provide an invaluable documentation about the use of dance in Venetian opera. Table 1 (see Appendix), organized by theatre and year, shows the number of balli in each opera. Table 2 (see Appendix) analyzes by decade the percentage of seventeenth-century Venetian operas that include balli. Approximately 660 balli appear in the 346 seventeenth-century operas. Most of these operas are in three acts, and incorporate two balli, one to end each of the first two acts.16 Yet libretti, often printed in haste, do not always tell the full story about balli. Supplemental information can sometimes be found in scenarios, aggiunte (printed leaflets of additions and other changes or corrections), or second editions.17 In the case of Giovanni Faustini's L'Eritrea performed at S. Apollinare in 1652, for example, no balli are listed in the libretto, but two are specified in the separately printed scenario and are confirmed in the account books.18 Occasionally, second editions of libretti give new information about dances or clarify their location. For example, the first edition of I re infante (1683) lists three balli,19 but only gives the location of the third, in the opera's finale. According to the second edition, issued midway through its run, the first ballo, danced by pages with torches, occurs at the end of Act I; the second, changed to a 'popular battle' in the 15 Badoaro, L'Ulisse errante, 11. 16 Five-act structure was used in only a half-dozen operas between 1640 and 1644, but was revived at the end of the century. Seven operas (8.2 per cent) during the mid-1690s and another fifteen operas (15.8 per cent) between 1701 and 1710 are in five acts (see the operas indicated by asterisk in Table I). The librettist Girolamo Frigimelica Roberti favoured this structure, and typically closes each act with an elaborate choral scene with balli or separates the acts with intramezzi that include dance. Although a five-act structure easily allowed for the inclusion of more balli, high numbers of balli are also found in three-act operas. Several of the earliest three-act operas have three or four balli; many operas from the 1680s and 1690s have four or five; and, in fact, the two operas that include six balli are in three acts. 17 The statistics on balli in the tables reflect only those scenarios, aggiunte, and second editions found in the collection of libretti at UCLA. 18 Ed. note: On the account books, see Alm's 'Theatrical Dance in Seventeenth-Century Venetian Opera', chapter 4. Since scenarios do not survive in large numbers - Rosand ('The Opera Scenario', see n.11l) lists nineteen scenarios between the years 1639 and 1655 - it is difficult to speculate as to whether any of the other operas lacking mention of balli in their libretti actually included dances in performance. 19 'Ballo di Paggi con Torci', 'Ballo di maschere con archi', and 'Ballo di Damme e fanciulli'. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 223 style of the traditional guerra de' pugni, takes place at the end of Act II1.20 In other cases the subject of a ballo is altered more radically. In La schiavafortunata (1674) by Moniglia and Corradi, for example, the ending of Act II changes from a ballo for naiads who emerge from tigers to a less spectacular, more commonplace dance by satyrs and shepherdesses - a change that may have been made for financial reasons. In addition to financial exigencies, these editions may reflect last-minute changes made after the libretto went to press, perhaps stemming from production problems with sets, costumes, or personnel. In some second editions, dances were added or cut. Other sources confirm that the statistics based on libretti may provide at best a conservative estimate of the number of balli performed in Venetian operas. For example, Niccol6 Enea Bartolini's libretto for Venere gelosa (1643) lists three balli, performed in the final scene of Act I and the penultimate scenes of Acts II and III. The set of engravings entitled Apparati sceniciper lo Teatro Novissimo by the renowned stage designer Giacomo Torelli documents two additional balli for this opera.21 The libretto includes a prologue sung by Flora who emerges from the earth and is carried through the air by Zephyrs. The Apparati scenici, however, provides more detail: 'The Chorus of Nymphs, who with movement of their feet, now fast and now slow, performed a most beautiful dance to the singular delight of the spectators. Thus began the performance of the opera.'22 Dances may also have been routinely performed whenever the sung verses mention dance, even if the standard phrase 'Segue il ballo' was not printed in the libretto. If this is so, then the statistics cited in Tables 1 and 2 would be even higher. By the 1660s and 1670s, the placement of balli at the ends of Acts I and II had become a standard feature of Venetian opera.23 About a quarter of the balli that end internal acts actually take place within the closing scene; in the remainder, the ballo is literally the last event of the act, with the ubiquitous 'segue il ballo' printed immediately before or after 'Fine dell'Atto'. By contrast, the majority of balli that serve as finales to whole operas take place earlier within the closing scene (in 63.6 per cent, or twenty-one of the thirty-three finales). Five of the twenty-five balli 20 'Trombe al suono de le quali invezze di ballo segue alla vista di Ergisto Battaglia Popolare.' The guerra de' pugni, also called the forze d'Ercole, was a traditional Venetian entertainment in which mock battles were fought by residents of two different sestieri, or neighbourhoods, on the bridges of the city. Chassebras de Cramailles gives a detailed account of I1 ri infante in the Mercure galant (March 1683), 256-71. He later (297-300) describes the additions, which included six elephants bearing a machine, the 140 people engaged in the mock battle, and a giant turtle that breaks into sixty or seventy pieces used by the soldiers as shields. In the next issue (April 1683, 71-8), he explains that during the last days of Carnival the Grimani had the Castellani and the Nicolotti stage the battle on the bridge in earnest (resulting in bloodshed), in order to show the foreigners how the guerra de' pugni was fought. 21 The numbers given in the tables are based solely on the collection at UCLA and therefore do not include these extra balli. For further information on Torelli and reproductions of some of these designs, see Per Bjurstrom, Giacomo Torelli and Baroque Stage Design (Stockholm, 1961). See also Rosand, Opera in Seventeenth-Centu~y Venice, 91-109. 22 Apparati scenici, 8. [Ed. note: See also Wendy Heller, 'Dancing Desire on the Venetian Stage' in this issue.] 23 For tables detailing the frequency and location of the balli by decade, see Alm, 'Theatrical Dance' (see n.6), 277-78. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 224 Irene Alm labelled intermedi also occur within a closing scene rather than in independent intermedi.24 For example, in the libretto for Amore innamorato (1642) by Giovanni Battista Fusconi and Pietro Michiel, the final scenes of Acts I to IV contain balli; the scenario states, however, that these scenes serve as intermedi for the opera. Thus, the hundreds of seventeenth-century libretti reveal the indisputable and ubiquitous presence of balli in Venetian opera. From the premiere of Andromeda in 1637 through the hundreds of operas that followed in Venetian theatres over the course of the century, theatrical dance held an important place among the ingredients essential to creating the Baroque spectacle. Moreover, the extensiveness of the literary record makes it possible to chart the use and placement of theatrical dances over the course of the century. Scores In contrast to the virtually complete documentation for libretti, scores survive for only about one-third of seventeenth-century Venetian operas. Whereas the libretti were usually printed in quantity and sold to the public, the scores for these operas were never printed - all of the surviving scores are in manuscript. Occasionally a noble patron must have commissioned a score, for some of the extant manuscripts are clearly presentation copies, but commercial opera in Venice had little need for the elegant documentation associated with court opera. Many of the manuscripts are obviously working scores, with passages glued or sewn in, and indications of transpositions and cuts. Opera scores usually became the property of the theatre management, and once a score no longer had commercial value, there was little reason to preserve it. Occasionally scores were adapted for revivals in Venice or in other cities; thus, while some of the surviving scores closely match Venetian libretti, others show varying degrees of revision. The largest group of these scores belonged to a single collector, Marco Contarini, and is now housed at the Biblioteca Marciana in Venice.25 Among this relatively small number of surviving scores, somewhat less than half include music unambiguously intended for dancing (see Tables 3 and 4 below). However, often the verses that introduce the dancing are set in the score, so originally the ballo must have followed,26 and still other scores include vocal music that may have been danced. Nonetheless, the apparent lack of dance music in some of these scores has spawned two assumptions on the part of many scholars: first, 24 The seventeenth-century intermedio or intermezzo is quite different from the eighteenth-century comic intermezzo, which, with its independent plot, was an outgrowth primarily of the comic scenes rather than of the intermedi and dances of seventeenth-century opera. See Charles E. Troy, The Comic Intermezzo: A Study in the History of Eighteenth-Century Italian Opera (Ann Arbor, MI, 1979), especially chapter 1, 'Origins and Early Stages'. 25 This collection is discussed by Giovanni Morelli and Thomas Walker, 'Migliori plettri', in Aurelio Aureli and Francesco Lucio, II Medoro, ed. Morelli and Walker, in Drammaturgia Musicale Veneta 4 (Milan, 1986), CXLI-CXLVII. 26 Scores with verses introducing balli include Artemesia (1656), II Medoro (1658), L'Ofeo (1673), and many others. [Ed. note: For a complete concordance of identifiable dances in Venetian operas, listing all mentions of dance in either score or libretto, see Aim, 'Theatrical Dance' (n.6), Appendix III, 320-63.] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 225 that the balli were not actually performed or were optional; and second, that another, 'inferior', composer wrote the dance music. There is considerable evidence to contradict the first premise: account books and reviews of performances confirm that balli were regularly performed.27 The absence of dance music in many manuscripts most likely stems from the practical circumstances of rehearsals, which at least some of the time took place apart from the opera and thus would necessitate separate scores.28 Moreover, letters and contracts between agents, impresarios, composers, and singers reveal that Venetian operas were often prepared under severe time constraints, so that rehearsals usually began before the score was completed.29 Working scores often lack not only the dance music, but also instrumental obbligati, ritornelli, and even texts for arias. If the copyist did not have access to the separate folios of dance music, it could not be included, even in presentation scores. The second notion - that another composer (usually characterized as 'lesser' or 'second rate') might have been responsible for the dance music -is also often contradicted by the surviving evidence. For example, many of these scores were copied by several different people, yet in only one, the Venetian copy of Cavalli's Xerse (1654), are the dances in a different hand than the music immediately preceding it. These particular bass lines seem to have been hastily copied into the full score, perhaps as cues, but that fact alone does not provide sufficient evidence to determine whether the dances were by Cavalli or by someone else. Furthermore, many of the balli in the earliest scores for Venetian operas are woven into the fabric of a scene and are without question by the principal composer of the opera.30 The only scores in which ballo music is known to be by a different composer are those used for productions of Venetian operas in other cities. Since Venice provided a substantial portion of the repertoire for many other Italian cities and courts during the seventeenth century, a different composer might adapt the opera for its new context. For example, the poet Giovanni Filippo Apolloni and the composer Alessandro Stradella supplied prologues and intermezzi for several out-of-town productions of Venetian operas, including revivals of Cavalli's Giasone (as 1/ novello Giasone) and Scipione africano at the Teatro Tordinona in Rome in 1671.31 In both of these, Stradella's dance music is strikingly different, both in terms of length and elaborate repetition schemes, and he also relies upon standard social dances which, 27 Ed. note: See Alm, 'Theatrical Dance', chapter 4, especially Tables Va and Vb, 281-83, which show expense records for the dances in Venetian opera houses. 28 Ed. note: Alm cites a description of a dance rehearsal held at a private home, reported in the travel memoirs of Abbh Antonio Olivieri. See Alm, 'Theatrical Dance', 149-53. 29 See, for example, Bruno Brunelli, 'L'impressario in angustie', Rivista italiana del dramma, 3 (1941), 311-41, and Carl Schmidt, 'An Episode in the History of Venetian Opera: The Tito Commission (1665-66)', Journal of the American Musicological Society, 31 (1978), 422-66. 30 See, for example, the extended dance scenes in Cavalli's Le nozge di Teti e di Peleo (1639) discussed below (265-68). 31 See Owen Jander, 'The Prologues and Intermezzos of Alessandro Stradella', Analecta musicologica, 7 (1969), 87-111. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 226 Irene Aim as we shall see below, were used infrequently in Venetian opera.32 Similarly, it was Lully who supplied ballets for Cavalli's operas Xerse and Ercole amante when they were performed in Paris in 1660 and 1662. It seems dangerous to extrapolate from practices in other cities or courts to the commercial opera houses in Venice. For example, although there was a tradition of hiring a ballet composer at the courts in Turin and in Vienna during the seventeenth century, in Venice no payments to a composer for balli have been found among the surviving account books.33 Nor can eighteenth-century Venetian practice be used as a basis for seventeenth-century practice in the same city, since the role of dance in opera changed substantially after the turn of the century. Balli became more independent in the eighteenth century, with plots entirely separate from the opera.34 In the seventeenth century, however, when the balli were more closely integrated with the operas, it cannot be assumed that the dance music was by a separate composer. Since the surviving scores served a variety of purposes (some being presentation scores, others working scores, others for non-Venetian revivals) and they represent less than half of the repertoire, it is impossible to determine whether certain 32 The score for the Venetian production of Giasone (1649) does not contain instrumental music for the ballo for spirits that ends the spectacular incantation scene in Act I, or for the dance for sailors at the end of Act II. For the Roman production, however, Stradella omitted a full intermezzo after Medea's powerful invocation of the spirits at the close of Act I in Giasone, replacing it instead with two alternating dances: 'Sarabande / Balletto for the Furies / When the Balletto has finished, da capo to the Sarabande two more times, and then the Balletto another time, and then da capo to the Sarabande until they have mounted the horses' (I-Sc L.V.33, fol. 102r-). After Act II, an intermedio for Satiro and Amore ends with a 'Balletto d'Amorini' in three sections, 'Balletto / Sarabanda / Presto', i.e., essentially a suite of three dances: a promenade-style entry in common time, a sarabande, and a gigue. Stradella used a similar approach to dance music for the Roman revival of Scipione affricano in 1671. His first intermedio contains a dance for the cyclops, consisting of a 'Presto' in compound metre, a 'Balletto' in common time, an 'Adagio' in common time, and a da capo of the 'Presto'. The second intermedio includes a ballo with sections for Spanish, French, German, and Italian dancers, in various dance rhythms, followed by a sarabande with sections in alternating tempos. The score for the Roman production also contains new music for the games of the gladiators in Act I, scene 2. Once again this takes the form of a suite. 33 For Turin and the La Pierre dynasty of ballet composers, see Marie-Thbrbse Bouquet, Musique et musiciens a Turin de 1648 a 1775 (Turin, 1969), the same author's II teatro di corte dalle origini al 1788 (Turin, 1976), and Mercedes Viale Ferrero, 'Repliche a Torino di alcuni melodrammi veneziani e loro caratteristiche', in Venezia e il melodramma nel seicento, ed. Maria Teresa Muraro (Florence, 1976), 145-72. In Vienna, Johann Heinrich Schmelzer and his eldest son Andreas Anton Schmelzer served as ballet composers for Leopold I, and collaborated on many of the Minato-Draghi operas. See Egon Wellesz, Die Ballett-Suiten von Johann Heinrich und Anton Andreas Schmelzer (Vienna, 1914) and Paul Nettl, 'Die Wiener Tanzkomposition in der zweiten Hialfte des siebzehnten Jahrhunderts', Studien zur Musikwissenschaft, 8 (1921), 45-175. Among other ballet composers in Vienna are Wolfgang Ebner, J. J. Hoffer, and Nicola Matteis. See Herbert Seifert, Die Oper amnt Wiener Kaiserhof inm 17. Jahrhundert (Tutzing, 1985), and Andrew D. McCredie, 'Nicola Matteis, the Younger: Caldara's Collaborator and Ballet Composer in the Service of the Emperor, Charles VI', in Antonio Caldara: Essays on his Life and Times, ed. Brian W. Pritchard (Aldershot, 1987), 153-82. 34 Ed. note: see Hansell, 'Theatrical Ballet' (n.5). This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 227 composers were more concerned with dance music than others. Indeed, it was probably the librettist, or perhaps the impresario, who determined the number, placement, and subjects of the balli. Yet several composers are well represented among these scores, and a few general observations about composers and dance music can be made in a brief chronological survey of the sources. 1637-1660 Twenty-five scores survive from these first decades of Venetian opera, in contrast to libretti for seventy-six works. Nineteen of these scores are by Francesco Cavalli, who dominated the early decades of commercial opera.35 More of his operas (twenty-two) were performed in Venice during the 1640s and 1650s than of any other composer and scores survive for all but three of them.36 The remaining scores are by Claudio Monteverdi,37 Francesco Sacrati,38 Antonio Cesti, Pietro Andrea Ziani, and Francesco Lucio. Five of these are for works that apparently did not use balli (none is listed in the libretti), and another five omit or change the verses that would have introduced the dances. Seven contain music clearly intended for the balli, and eight others have verses introducing the dances or choruses that may have been danced (see Table 3 for details). Cavalli's first opera, Le nozze di Teti e di Peleo (1639) contains a substantial amount of dance music ranging from short instrumental pieces to extensive choreographed scenes involving solo singers, chorus, and instruments. The thorough integration of the dances into several scenes of this opera sets it apart from the works that follow; in fact, danced scenes of this length do not reappear until the ballroom scenes of the 1670s and 1680s. By contrast, Cavalli's Gli amori d'Apollo e di Dafne (1640) and La virtZ de' strali d'Amore (1642) contain only brief instrumental dance music and one longer danced chorus. Cesti's Alessandro vincitor di se stesso (1651) also includes a more extensive chorus to accompany a ballo. 35 Many of Cavalli's operas were also central to the dissemination of Venetian opera throughout Italy, enjoying numerous performances in other cities. On Cavalli, see Lorenzo Bianconi, 'Caletti, (Caletti-Bruni), Pietro Francesco, detto Cavalli', in Dizionario biografico degli italiani (Rome, 1973), vol. 16, 686-96; Jane Glover, Cavalli (New York, 1978); and Thomas Walker and Irene Alm, 'Cavalli, Francesco', in The New Grove Dictionary of Music and Musicians, 2nd edn, ed. Sadie and Tyrrell. 36 The Contarini collection contains copies of all of Cavalli's extant operas, and additional copies of a half-dozen operas survive elsewhere. Amore innamorato (1642), I Titone (1645), and Antioco (1659) lack scores. Cavalli's last two operas Coriolano for Piacenza (May 1669) and Massenzio (composed for S. Salvatore 1673, but never performed), are also lost, and ten operas considered doubtful attributions lack scores. 37 Unfortunately, no score survives for Le nozze d'Enea con Lavinia (1641), the opera by Monteverdi with the greatest amount of dance indicated in the libretto. Of his two Venetian operas that do survive, 1 ritorno d'Ulisse in patria (1640) contains only one ballo, which the score omits, and L'incoronatione di Poppea (1643) contains no balli in either libretto or score. 38 The score for Sacrati's La finta paza should soon be available in a facsimile edition with critical commentary by Lorenzo Bianconi, as volume 1 of the series Drammaturgia musicale veneta. I would like to thank Ellen Rosand for allowing me to consult her copyflow of the score. There are substantial differences with the Venetian libretto and neither dance scene is included. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 228 Irene Alm Although many of the scores from the 1650s lack dance music, Marco Faustini's account books for S. Apollinare and S. Cassiano show that balli were indeed performed in several of Cavalli's operas, as well as in Pietro Andrea Ziani's Lefortune di Rodope e Damira (1657)." The scores for Cavalli's La Rosinda (1651), Veremonda Amazzone di Aragona (1652/3), and Elena (1660) contain choruses that may have accompanied the balli, or may have simply introduced instrumental dance music. Instruments could have repeated the choral music, or contrasting dance music may have followed. Since these balli all occur at the ends of acts, it would have been easy to keep instrumental dance music in a separate score. In general, the scores from these first two and a half decades of Venetian opera show a trend away from Cavalli's early use of elaborate choral dance music, reminiscent of the late Renaissance intermedi. As balli become a conventional means of closing the first two acts, shorter instrumental dance pieces, occasionally introduced by a chorus, are favoured. 1661-1680 Libretti show the 1660s and 1670s to have been decades of conformity in many ways for Venetian opera. Of the one hundred operas from these two decades, ninety-eight contain balli (usually to close Acts I and II). Scores survive for sixty-three of these one hundred operas, including works not only by Cavalli and Cesti, but also Pietro Andrea Ziani, Antonio Sartorio, and Giovanni Antonio Boretti, several of whom became 'house composers' at particular theatres.40 Of the sixty-three scores, twenty-seven contain ballo music and another thirty-two include some or all of the verses that introduce the balli. Approximately two-thirds of the balli in these scores are instrumental pieces, with binary form and duple metre increasingly favoured in the 1670s. A few choral accompaniments for balli still appear and some pieces alternate solo and chorus, but in the vocal pieces connected with balli there is a greater emphasis on solo voices.41 The first social dance scenes also appear during this period. In Seleuco (1666) and Galieno (1676), these scenes are composed as conversations during a ball, with instruments supplying continuous dance music in the background. 1681-1690 After two decades of relative conformity, composers began to explore new styles during the 1680s. Balli remained a mainstay of productions, however, and are " See Beth L. and Jonathan Glixon, 'Marco Faustini and Venetian Opera Production in the 1650s' (n.6). 40 Antonio Sartorio dominated at S. Salvatore, Carlo Pallavicino at the two Grimani theatres (SS. Giovanni e Paolo, and, after it opened in 1678, S. Giovanni Grisostomo), and Domenico Freschi at S. Angelo (which opened in 1677). Boretti and Pietro Andrea Ziani continued to be active during the first part of the decade, and newcomers included Giovanni Legrenzi and Ziani's nephew Marc'Antonio. 41 For example, in L'Argia (1669), Alceo sings an aria accompanied by instruments that is danced by a chorus of slaves, and in Massenzio (1673) a chorus of sailors dances to an aria sung by one of the sailors. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 229 indicated in the libretti for seventy of the eighty-five operas during this decade. Scores survive for only seventeen of these works: five are by Domenico Gabrielli, three by Pallavicino and three by Freschi, with a number of other composers represented by one score apiece. Nine scores contain ballo music, even though for two of these - Rodoaldo re d'Italia (1685) and Le generose gare tra Cesare e Pompeo (1686), both by Gabrielli - the libretti fail to mention balli. These dances and several others from this decade are notated only by a bass line. Ballroom scenes were extremely popular, appearing in five scores. Most of these scenes weave together both vocal and instrumental music to accompany the dance. At the end of the decade, Giuseppe Felice Tosi's Amulio e Numitore (1689) is exceptionally rich in instrumental dance music. 1691-1700 During the final decade of the century, libretti for eighty of the eighty-seven operas mention balli. Only thirteen operas have extant scores and nine of these contain dance music.42 Ten of the scores are by Carlo Francesco Pollarolo, who dominated Venetian opera in the 1690s, much as Cavalli had earlier in the century.43 The balli in his works are also significantly longer, especially those for the intramezzi and the final choral scenes of Frigimelica Roberti's neo-classical libretti, Ilpastor d'Anfriso (1695) and Rosimonda (autumn 1695).44 Their interweaving of vocal music with instrumental pieces may exhibit French influence. Surprisingly, there are no extant scores by Pollarolo's leading rival, Marc'Antonio Ziani, who wrote eighteen operas during this decade, principally for S. Angelo and S. Salvatore.45 The majority of Giacomo Perti's operas were written for cities other than Venice, but scores for two of his Venetian operas from this decade survive, Furio Camillo (1692) and Nerone fatto Cesare (1693), and both contain balli. Although Perti's instrumental dance music is not nearly as elaborate as Pollarolo's balli, French influence is also seen in his use of a 'Borea' (bourree) and a 'Rigadon' (rigaudon) in Neronefatto Cesare. Choreographic sources The late sixteenth and early seventeenth centuries saw a flowering of Italian treatises on dance, the most important of which were those by Fabrizio Caroso and Cesare 42 Two other surviving scores do not match Venetian productions. 43 There are twenty-six operas for Venetian theatres attributed solely to Pollarolo, as well as two collaborations. Pollarolo has been credited with expanding the dimensions of the aria and increasing the use of instrumental accompaniment. On Pollarolo, see Olga Termini, 'Carlo Francesco Pollarolo: His Life, Time, and Music with Emphasis on the Operas', Ph.D. diss. (University of Southern California, 1970), and by the same author 'Carlo Francesco Pollarolo: Follower or Leader in Venetian Opera?' Studi musicali, 7 (1979), 223-72 and 'Pollarolo, Carlo Francesco', New Grove Dictionary of Music and Musicians, 2nd edn, ed. Sadie and Tyrrell. 44 See examples 88-97 in Alm, 'Theatrical Dance'. Not all of Pollarolo's extant scores for libretti by Frigimelica Roberti include the intermedio music: Ottone (1694) and Irene (1695) lack these pieces. There is no evidence, however, that they were by another composer. 45 Ziani's first seven operas (to 1685) survive, but full scores for all the operas that he composed from that time until his move to Vienna in 1700 are lost. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 230 Irene Alm Negri.46 However, these manuals are predominantly concerned with social dances performed by the nobility, despite the presence of plots or dramatic subjects in a few.47 Caroso's Nobilta di dame (1600) was reprinted at least as late as 1630, and some of the dances for which he left choreographies may have been used in Venetian balli, perhaps as late as mid-century. Nonetheless, already in 1620 Alessandri da Narni wrote, 'I think that if Prospero Lutij, Fabritio Caroso, and Cesare Negri were alive at this time, they would not use many of the dances, passeggi, and variations that we see in their writings. Therefore it does not seem right to me, that we imitate them.'48 Atypically for the period, the anonymous treatise II corago, likely written between 1628 and 1637, does deal with dance in the context of theatrical music. The corago was roughly equivalent to a combination stage manager and director, and the treatise provides much valuable information about opera, devoting several chapters to dance - although, once again, in a courtly context.49 Overall, then, we are left with a situation in which the best-known manuals from the earlier part of the century have only limited applicability to Venetian opera, and are followed by nearly a century of virtual silence from Italian choreographers and ballet masters. 46 Ed. note: For further discussion of the various treatises that inform our understanding of Venetian dance, see Alm, 'Theatrical Dance', chapter 1, 1-34. Both Caroso's Nobilta di dame and the earlier treatise on which it was based, II ballarino, are available in facsimile: II ballarino (Venice, 1581; rpt. New York, 1967) and Nobilta di dame (Venice, 1600, 1605; rpt. Bologna, 1970). See also Julia Sutton's translation and edition, Nobilta di Dame: A Treatise on Courtly Dance, Together with the Choreography and Music of 49 Dances (Oxford, 1986; rpt. New York, 1995). Negri's Le gratie d'amore (Milan, 1602), is available in two facsimile reprints (Bologna, 1969, and New York, 1969). In addition to the reissue as Nuove invenzioni di balli (Milan, 1604), there is a 1630 manuscript translation into Spanish by Don Balthasar Carlos for Sefior Conde, Duke of St. Lucar (E-Mn, MS 14085). See also Yvonne Kendall, 'Le gratie d'amore 1602 by Cesare Negri: Translation and Commentary', DMA diss. (Stanford University, 1985). Giovanni Battista Doni's Trattato della musica scenica (1640; published in Lyra Barberina II [Florence, 1768; rpt. Bologna, 1974]), while not a dance manual, is also important to an understanding of theatrical dance during the first decades of the seventeenth century and the relationship of modern practice to the writings of the ancients. See especially chapters XXXI-XL, and in the 'Appendice a Trattati di Musica: Musica Scenica, Parte I', chapters XIX-XXVI. For a more extensive discussion of the classical dances, see Irene Alm, 'Humanism and Theatrical Dance in Early Opera', Musica Disciplina, 49 (1995), 79-93. 47 Some of Negri's choreographies designed for spectacles are analyzed by Pamela Jones, 'Spectacle in Milan: Cesare Negri's Torch Dances', Early Music, 14 (1986), 182-96. 48 Alessandri da Narni, Discoroso sopra il ballo (1620), 54; the Italian is transcribed in Julia Sutton's preface to her edition of Caroso's Nobilta di dame, 19. One must keep in mind that these treatises were addressed - for the most part - to noble amateurs, albeit skilled and practised dancers, who performed before an audience of their peers. 49 Ed. note: I1 corago was perhaps written by Pierfrancesco Rinuccini (1592-1657), son of the librettist Ottavio Rinuccini. The term corago itself is derived from the Greek Xoprlyoo (choregus, or one who leads the chorus); regarding the duties of the position, see the introduction to the modern edition by Paolo Fabbri and Angelo Pompilio, I corago o vero alcune osservaTZioni per metter bene in scena le composiTzioni drammatiche (Florence, 1983), 8-10, as well as Roger Savage and Matteo Sansone, 'll Corago and the Staging of Early Opera: Four Chapters from an Anonymous Treatise circa 1630', Ear~y Music, 17 (1989), 495-99. Chapter XVII addresses the role of the corago in arranging the choreography for the chorus, with more specific information on dances in chapter XVIII, with reference to the practice of the ancients. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 231 By the early eighteenth century, however, some writers had begun to provide more detailed descriptions of balli. In 1704 Vincenzo Coronelli published a surprisingly technical summary of current dances. Yet our balli have diverse names, like the Moresca, the Passo e mezzo, the Cinquepassi, the Sette passi, the Saltarello, the Gagliarda, the Corrente, etc., and those from various places like the Francese, the Spagnuola, the Fiorentina, the Berganmesca, the Pavana, the Veneziana, the Furlana, the Siciliana, the Romana, the Canaria, etc. In our balli, three things especially converge, which are Passo, Salto, and Capriola: the Passo is divided into several types, which are coupi or step, balanced step, joined step, split, beaten, etc. The Salto is divided into the leap forward, backward, turning, to the side, and many others. The Capriola is of various types, which are the simple caper, and cross cut, half caper, the caper cut up to the number eight, so that among the professors of ballo, they speak of doing a quadruple, quintuple, or sextuple caper, and it is used by Ballerini with other terms, and especially with French terms, from which nation for the most part are wont to come new styles and inventions of balli to our Italy.50 Coronelli's nod to French influence at the end of this passage is doubtless a reflection of the flood of French treatises and choreographies in print since 1700, beginning with Feuillet's Choregraphie.51 Of the Italian dance types he mentions, several of which date at least as far back as Caroso and Negri, only the corrente, canario, and passo e mezzo appear in libretto verses, and these only occasionally. (As we shall see, standard social dances such as the sarabanda and ciaccona appear only in ballroom scenes.) However, as the prominence given to the capriola in this passage suggests, Italian dancing was known for its athleticism. Italians had used the capriola - a jump in which the dancer crossed or beat his legs in the air and which existed in many variants (see Fig. 2) - in theatrical dances at least as early as 1637, as shown in Stefano della Bella's engravings of Le nozze degli dei, a favola with music performed in Florence for the wedding of Ferdinando II de' Medici and Vittoria della Rovere.52 Capriole also figure prominently in Gregorio Lambranzi's Neue und curieuse theatralische Tantzschul (Nuremberg, 1716), the only substantial source for Italian theatrical dance practices published in the early eighteenth century.53 The work consists of 101 engravings illustrating one or more dancers on stage; each plate includes a melody at the top of the page and a caption at the bottom describing the 50 Vincenzo Coronelli, Biblioteca universale sacro-profana antico-moderna, vol. 5 (Venice, 1704), col. 225. 51 Ed. note: For information regarding the choreographies in Feuillet notation, see Meredith Ellis Little and Carol G. Marsh, La Danse Noble: An Inventory of Dances and Sources (Williamstown, 1992). 52 Della Bella's engravings show Giulio Parigi's designs for Le nozze degli dei, which had music by five composers to a libretto by C. Coppola. The engraving with the capriole is reproduced in a number of sources including Cesare Molinari, Le nozge degli ddi: un saggio sulgrande spettacolo italiano nel seicento (Rome, 1968), plate 66. 53 The original publication (Niirnberg, 1716) is in both German and Italian. It has been translated from the German by Friderica Derra de Moroda as New and Curious School of Theatrical Dancing, ed. Cyril W. Beaumont (New York, 1966). Derra de Moroda has also published a facsimile of Lambranzi's original drawings, New and Curious School of Theatrical Dancing by Gregorio Lambranzi. A Facsimile of the Original in the Bavarian State Library (New York, 1972). This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 232 Irene Aim ASSAI BEN BALLA.A CVI FORTVNA SVONA Gefuu. 6 barro, o iro salleu in alto Chi concorde a rumO pa.si hd /a Fortuna, Che .aro. in tempo e la caden(.,e t'Palto. Fig. 2: Dancer performing a capriola. (Mitelli, Proverbi figurati, 1678.) subject, action and style of movement. It includes dances by peasants, a drunken couple, a satyr, buffoons, Turks (see Fig. 3), Moors, and prisoners in chains, among many other character types. The connection of Lambranzi's style with that used in the Venetian opera houses is only speculative. However, a number of Lambranzi's figures are so strikingly similar to the subjects and descriptions of the balli in the libretti, that it seems reasonable to bear Lambranzi's illustrations in mind as we read the descriptions of dancing in the literary sources. For it turns out that the libretti themselves provide usable choreographic information - not just about styles of dancing, but even about specific movements. Many of the choreographic descriptions found in libretti illustrate the same kind of athleticism suggested by Lambranzi - what we might characterize as the 'bravura' style: leaps and jumps (salti), speed, turns, and agility. Salti appear in all styles of dances, performed by every type of character. For example, in Act V of L'Adone This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 233 $fAk& mWernacX Aw, f tzuztre~z wit Ae~rknrw ;Sr mil, - •,Aaties hauz Slg a; ri m4 bi y p Fig. 3: "Four Turks enter, one after the other, and dance with joined hands as shown; backwards, forwards, and to right and left, with ballonnes and other suitable pas. The air is played three times." (Lambranzi, Neue und curieuse theatralische Tantz-Schul , Book II, plate 38.) (1640), four dancers costumed as a lion, a tiger, a bear, and a boar enter leaping to begin the ballo. A very different setting is the sombre, penitential dance of suffering ('Ballo di Sofferenza') in La finta pazza (1641). Here Giunone sings, 'Begin the dance, strong boys', and Minerva chimes in, 'Yes, yes, begin the Greek custom, and This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 234 Irene Alm while your foot leaps, sound the whip.'54 Two operas, L'Orontea (1666) and L'Alciade (1667), even have balli specifically designated for saltatori, probably indicating tumblers or acrobats, rather than simply jumpers. Moreover, contrary to frequent assumptions about dance in this period, salti were not exclusively performed by male dancers. For example, the stage directions for L'Andromeda (1637) tell us that the 'leaping labyrinth' was danced by six ladies. In Gli amord d'Apollo e di Dafne (1640), the chorus sings 'Dance and leap / Women and men.'"5 The libretti use the term girl in two ways in regard to the balli. Galatea tells the nymphs in L'Ulisse errante (1644) to dance with flickering feet, leap, and turn, thus suggesting that they turn or twirl in one place.56 In Martio Coriolano (1683), Vendetta (Vengeance) instructs her ministers to do 'a hundred plus a hundred' turns in their ballo and 'in front of a dying Rome trace garlands in my hair'. In ballroom scenes, the terms gimi or in gii were also used to mean circling or taking a turn around the room. In Falaride tiranno d'Agrigento (1684), the stage directions for Act I, scenes 7 to 9 read 'here the ballo begins ... The ballo circles through the porticos ... The dance circles around and they exit from the porticos.'57 Carole, round dances or carols, are found in Venetian operas throughout the seventeenth century, beginning with L'Andromeda (1637). Round dances are among the oldest choreographies depicted in art, seen, for example, in figures on ancient Greek vases.58 Notably, no specific steps are prescribed for carols, and they are not mentioned in any of the treatises dealing with social dances. Nonetheless, the libretti provide invaluable information about the various woven or braided patterns combined with the round dances. In Act I, scene 9 of La Rosinda (1651), for example, a chorus of six goblins ('spiritelli') sings 'Let's weave carols to rejoice' ('Carole al giubilo tessiamo'). Many choreographers also made skilful use of woven patterns in other contexts. The intrecci in Act II, scene 2 of Amore inamorato (1686) must have made a striking effect, as two groups of twelve cupids with lit torches form braided figures. Another important source of choreographic information is the operatic character who acts as a sort of dancing master, giving advice on the ballo. For example, in Gli amon d'Apollo e di Dafne (1640) Act I, scene 4, Dafne reminds her dancers to stay in parallel lines and not to make a false step: Seguite pur l'incominciato ballo Giolive ninfe, allegri pastorelli, Facciano i piedi vostri i paralleli A' chi la su non pon mai piede in fallo. 54 Giunone: 'Si cominci la danza, Fortissimi garzoni' and Minerva: 'Si cominci, si, si, la Greca usanza, E mentre salta il pid la sferza suoni.' ss Act I, scene 4: 'Danzino e saltino / Femine & huomini.' 56 'Ballate, danzate / Col tremulo pie / Saltate, girate / Ch'il Cieco non v'.' 57 'Qui pincipia il ballo ... Gira il Ballo na sottoportici ... Gira la Danza, & escono da i sotto portici.' 58 See also the fourteenth-century frescoes by Ambrogio Lorenzetti in the Palazzo Pubblico, Siena. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 235 [By all means continue the ballo begun, joyful nymphs, happy shepherdesses; Let your feet make the parallels, upon which don't ever make a false step.] Grace, lightness, and speed are three qualities mentioned often in the verses that introduce or accompany balli. In the finale of I re' infante (1683) Venus commands her followers to sparkle, celebrate, and perform graceful dances with their feet.59 In Act II, scene 10 of Venere gelosa (1643), Trulla's dancing is described as 'lighter than a bubble' ('piuf leggiero d'una galla'). Nearly all references to the speed of movements use terms such as 'veloce' (fast or rapid), 'snello' (quick or agile), 'rapido' (swift or fast), 'agile' (agile or nimble), and 'volante' (flying). The dancers in L'Eurip (1649) race with the breeze: Lascivo, e snello II pie festeggi Ii pie gareggi Col venticello. Leggiere a prova Danza formate Compagne amate Leggiadra e nova. [Lascivious and nimble, the foot celebrates, the foot races with the breeze. Lightly in contest, beloved companions, perform a graceful, new dance.] Eyewitness accounts Eyewitness accounts by viewers who attended performances in Venice also provide information about choreography, especially when read in conjunction with the libretti. Visitors to Venice were often overwhelmed by the sheer number of operas or enthralled with the stage machinery and the singers; a few writers, some local and some foreign, comment on the balli. Robert Bargrave, an English merchant, had a passionate interest in music and wrote detailed and vivid descriptions of church music, opera, and the other musical experiences that he enjoyed during his travels.60 Venetian opera captivated him, and he listed among its delights the 'most exquisite Anticks and Masking Dances'.61 The 1656 Carnival season hosted only two operas, both by Cavalli; Bargrave must have seen Artemisia at SS. Giovanni e Paolo and L'Erismena at S. Apollinare.62 The two dances in Artemesia both involve comic characters: at the end of Act I, eight archers come to the rescue of the old nurse Erisbe, and at the close of Act II, Niso is mocked by eight pages. Erismena includes 59 'Miei seguaci voi brillate / Festeggiate / Danze leggiadre ora col pie formate.' 60 'A Relation of sundry voyages and Journeys made by me Robert Bargrave', Bodleian Library, Oxford: GB-Ob, MS. Rawlinson C.799. Bargrave even composed songs and choreographed dances for an English wedding in Constantinople (although it was called off at the last moment). See Michael Tilmouth, 'Music on the Travels of an English Merchant: Robert Bargrave (1628-61)', Music and Letters, 53 (1972), 143-59. 61 Quoted in Tilmouth, 'Music on the Travels of an English Merchant', 156. 62 The chronology of operas at SS. Giovanni e Paolo during the 1650s is particularly murky. It is possible that Cavalli's La Statira principessa di Persia was performed in 1655/6 and that Artemisia was performed in 1656/7. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 236 Irene Alm a ballo for prisoners who free their feet from chains and celebrate their freedom with a dance, weaving the chains together, and another ballo for Moorish men and women. The Pallade Veneta, a monthly news-sheet that circulated in manuscript, some- times contained reviews that provide valuable details about choreography.63 The January 1687 issue, for example, contains Francesco Coli's review of Elmiro re' di Corinto, which had opened at S. Giovanni Grisostomo on 26 December 1686. The libretto calls for two balli and a combattimento; the unusually large company of twenty-four dancers was apparently all male. The ballo for soldiers (Act II, scene 13) is described in the libretto as follows: 'Twelve followers of Pace [Peace] descend; fighting with the twelve followers of Sdegno [Scorn], they form a pleasing pattern that serves as the ballo.' Coli's review vividly expands upon this brief stage direction: In this very noble scene one sees a monster of immense size and frightful appearance, representing a flying toad on which Sdegno rides, singing an arietta all in a rage. Behind the machine of the terrifying monster one sees in the air among dense white clouds Pace followed by a large chorus of her ministers nicely arranged in that cloudy heaven, a scene in effect of supreme joy, and then Pace and Sdegno quarrel and dispute together in song. The monster representing Sdegno spews forth from its enormous and fiery mouth a great many furious and terrible men, and they pour out, having been restricted in a small space, when the impetuous followers of Pace descend in a number equal to that disgorged by Sdegno who, with various patterns and well-measured turns, have a scattered fight of fine and gallant forces, and the followers of Sdegno seeming to surrender, Pace remains victorious.64 Reviews of Venetian opera also appeared from time to time in the French monthly, the Mercure galant. The first of these articles appeared in the August 1677 issue65 and discussed all seven of the operas presented during the 1677 Carnival season in some detail: for I/Nicomede in Bitinia, the dances are the main focus of the review. The first act finished with a ballet of stonecutters. They each held their hammers and chisels, and made their movements in rhythm around a statue of Nicomede, which they seemed to complete while dancing; but all this in a manner so well planned, that one could see nothing more precise. An entree of peasants and farmhands with their shovels and their hoes finished the following act; and the second scene of the third act was agreeably interrupted by a dance by several heroes, who remembering their former loves, each took the end of one of the various coloured ribbons that were hanging from the branches of a tall myrtle in the middle of the theatre. There was nothing so amusing as to see them tangling and 63 For a history of this publication and a list of extant copies, see Eleanor Selfridge-Field, Pallade Veneta: Writings on Music in Venetian Society 1650-1750 (Venice, 1985). 64 For the complete text of the review, see Selfridge-Field, Pallade Veneta, 134-42; the passage quoted here may be found on 140. 65 Pagination varies among extant copies of the Mercure galant. In the August 1677 issue that I consulted, the discussion of Venetian opera appears on pages 72-105. Selfridge-Field cites pages 37-53; she quotes part of the review, but omits all the relevant dance passages (Pallade Veneta, 338-40). This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 237 disentangling each other, which they did in different manners and always with a skill that brought acclamation from everyone.66 The libretto for I1Nicomede in Bitinia was issued in two editions, both of which list the first two dances, but not the third - further evidence that libretti provide only a conservative gauge of the role of dance in Venetian opera.67 If it were not for this journalistic account, we would not even know that the third ballo had existed, let alone something about its choreography. Moreover, accounts such as these can be read in conjunction with other sources to help imagine how such scenes may have looked. Gregorio Lambranzi, whose writing we considered above, explained exactly how to choreograph a dance such as the one in the first ballo (see Fig. 4): Here is a wooden statue which has been covered with pieces of stone, made to adhere by means of plaster, so that it appears shapeless. It is set upon the stage. Then enter two sculptors who chisel the statue as they dance, so that the pieces of stone fall off and the mass is transformed into a statue. The pas (steps) can be arranged at pleasure.68 Subject and style The sheer variety of subjects used in Venetian balli makes it impossible to treat them all in detail. The imagination of librettists and choreographers led them to create all sorts of situations, actions, and roles for dancers, and the freedom of style and technique allowed for endless variety. Some subjects had been used in theatrical dance in Italy for at least two centuries: Bacchantes, nymphs, satyrs, hunters with bears, lions or other wild animals, fire-breathing statues, pages with torches, soldiers in combat and battle scenes, madmen, Amazons, Moors, and Turks. Some of these became incorporated into the conventions of seventeenth-century Venetian opera, while others were uniquely tailored to a single opera's plot. An important way in which the various subjects were expressed to the audience was through pantomime, one of the most distinctive features of Venetian theatrical dancing. Venetian opera was by no means the first place in which pantomime appeared in Italian choreography; expressive gesture and mime are mentioned in descriptions of danced entertainments as early as the fifteenth century, and were features of many sixteenth- and early seventeenth-century pastorales, intermedi, and balli.69 Pantomime was also an essential element of the commedia dell'arte; although commedia characters do not appear in Venetian operas or balli, their miming was seen in the other theatres and in squares throughout Carnival, such that the public readily absorbed their vocabulary of gestures. 66 Mercure galant (August 1677), 93-5. 67 The libretti list a 'ballo by stonecutters with hammers and chisels around the statue' and a 'ballo by restorers with shovels and hoes'. 68 Trans. F. Derra de Moroda (see n.53), Part Two, 3. " Mime was used, for example, in the choreographies for Giovanni Battista Guarini's Il pastor fido (1584; published Venice 1590), Emilio de Cavalieri's La rappresentatione diAnima, e di Corpo (1600), Marco da Gagliano's Dafne (1607), and many of Claudio Monteverdi's balli and combattimenti. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 238 Irene Alm P k a? {ang4 U )kidem teize. 22d rgo,, ndet ?Iz a Bru/a ie a Schlbiap fantPEfndQ) an ',Afcltak~ e ri g-cakknch 3 07-; Bgryas kanJ~ 72.nack Adida m1irv,: rywrdie ra mahl /?,I/ Fig. 4: Two sculptors chiselling a statue as they dance. (Lambranzi, Neue und curieuse, Book II, plate 24.) Pantomimed balli in seventeenth-century Venetian operas were, admittedly, on a much smaller scale than mid-eighteenth-century pantomime ballets by choreogra- phers such as Franz Hilverding, Gasparo Angiolini, and Jean-Georges Noverre, yet they provide significant evidence that the ballet en action also had historical roots in This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 239 Italy as well as France. Several seventeenth- and eighteenth-century theatrical treatises specifically mention pantomime among the many aspects of Italian chore- ography that they trace back to Greek and Roman sources. Doni, for example, drew upon Lucian in discussing the use of gesture in ancient dance, and Perrucci began chapter XI of Dell'arte rappresentativa (1699), 'Del Gestire conveniente al Rappresen- tante', by tracing the classical art of gesture in English, French, and Italian theatre.70 In Venetian opera, pantomime could be both comic and tragic, and might explore ideas, emotions, actions, or sentiments not readily expressed with the voice. A particularly vivid example of tragic pantomime comes at the close of the second act of II Romolo e'l Remo (1645). Six matrons from Alba dance a ballo grave as they mourn for their husbands and relatives, and weep amidst the spoils of war. They carry an urn with the ashes, a flask to collect their tears, a purse to pay Charon for the crossing, and lamps for the eternal flame as used in antiquity. The libretto describes the lamenting gestures of the women, whereas the scenario tells of sighs of pain;71 all of this would have been represented very differently through song. Pantomime was also used routinely in balli by soldiers, gladiators, and fencers, as well as to depict other moments of violence. Act I of L'Adelaide (1672), for example, ends with a ballo based on two miners attempting to molest Adelaide and the ensuing brawl: Here Adelaide leaves the mine, re-climbing the stone stairs by which she had descended, to return to the mouth of the exit. While two workers want to follow her to molest her, they are held back by the other fellows who, angered, begin to argue among themselves, forming a curious ballo in the form of a fight for the end of Act I. When this has finished, the dust already clinging to them in their niches catches on fire, and noisily exploding, a great deal of marble is demolished in more parts of the mine, with the ruin of some workers, changing the scene at the same time. Smoking and drinking are among the ballo subjects that commonly involved pantomime. Pipes and tobacco are used with mime in Sardanapalo (1679), Olimpia vendicata (1682), and I Pertinace (1689). Pantomimed drunkenness was used for comic effect,72 and bacchanalian dances naturally mimed drinking (see Fig. 5). The stage directions for the opening scene of Amage regina de Samarti (1694) are quite specific: 70 Giovanni Battista Doni, Trattato della musica scenica (1640), in Lyra Barberina II, 93 and Perrucci, Dell'arte rappresentativa, 110-111. [Ed. note: For further information regarding the theoretical underpinnings of pantomime, see Irene Alm, 'Pantomime in Seventeenth-Century Venetian Theatrical Dance', in Creature di Prometeo: Il ballo teatrale. Dal divertimento al dramma. Studi offerti a Aurel M. Milloss, ed. Giovanni Morelli (Florence, 1996), 87-102.] 71 The libretto states 'Ballo di sei Matrone Albane, che piangano i mariri [sic] morti nell'uccisione del Tiranno. Havranno le ceneri in un vaso, un'ampolla da raccoglier le lagrime, una borsa per pagare il passo a Charonte, ed alcune lucerne per formarne i lume eterni secondo l'uso degli antichi, con due trofei d'armi intorno a i quali si aggireranno con gesti lamentevoli.' Four strophes of verse follow. The scenario states 'Ballo grave con cenni dogliosi di Matrone Albane, che piangano intorno ad alcuni trofiei d'arme la morete de' loro Mariti e Parenti seguita nella mischia precedente, ed hanno il vaso delle ceneri in mano, ed alcune ampolette da raccoglier le lacrime, una borsa di monete per pagar il passo A Caronte, ed alcuni lumi eterni com'era costume degli antichi.' 72 For example, the revival of L'Orontea (1683) has a 'Ballo di Ubriachi'. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 240 Irene Alm ?5t4 l9rL an )ZIZ emn 7t t%7lc??7JcImo, Fig. 5: 'Two persons carry in Bacchus and hold him up until the air ends. Then they set him down and dance the following beautiful dance.' (Lambranzi, Neue and cunrieuse, Book II, plate 20.) The triumph of Bacchus is presented by Ulderico in a garden of the kingdom. ... Royal guards garlanded with flowers. The curtain having risen, four bacchantes dancing a ballo will lead the aforementioned company. . . . Niso goes around offering sweet drinks to the bacchantes, who two by two, after drinking, will enter into the ballo. .... The orchestra This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 241 b'Jrdsitv #=Wnt 0;1FCO 1c~F~ zF~rmov AlAsiraoa ~"?~?~? d~L~31h~06~~ ~ PIL ~g~ L~~?lp ~T~~ba d I . xorat 1;?~?v/,~??~7mnt~F Fig. 6: 'Ballo di Filosofi' from Giulio Cesare trionfante, Act II, scene 1. (Fold-out engraving in the libretto, Venice, 1682.) answers the music of the above verses . . . In the same prescribed order he offers the drink to the other two, and repeats the preceding verses ... after which the ballo ends. Props were also used in pantomimic dance to clarify characterizations or specific situations in a ballo. Soldiers, hunters, and gladiators carry a variety of weapons including arrows, spears, and swords.73 Peasants and workers use scythes, hoes, and shovels.74 Prisoners dance in chains, as in Diocletiano (1675), or celebrate their freedom by carrying the chains, as in L'Erismena (1656). The scholars of Archimedes hold instruments of geometry while they dance in Marcello in Siracusa (1670), and seven philosophers each with a symbolic gesture or prop (such as Diogenes with his lantern) are portrayed in the first ballo of Giulio Cesare trionfante (1682) (see Fig. 6). The extraordinary variety of subjects and characters in the balli was in large part a result of librettists' efforts to link the balli to the plot or subject of the opera, even though they did not necessarily play a critical role in advancing the action. Librettists devised many motivations for balli. One of the chief reasons was allegrezza, or joy: weddings, military victories, the rescue of prisoners, the arrival of good news - all 73 Among numerous examples are a 'Ballo d'Arcieri con Archi, e Frezze [freccie]' in Ilperfetto Ibraim gran visir di Costantinopoli (1679) and a 'Ballo di Soldati, con Lancie, e Spade' in L'amazzone corsara, overo L'Alvilda regina de Goti (1686). 74 For example, the two balli in L'Adone (1676) are 'Ballo di Satiri, e Villanelle con Bastoni, e Falci' and 'Ballo di Resauratori con Zappe, e Badili'. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 242 Irene Alm of these and more were occasions for dance. In addition, mythological and supernatural characters were prone to dance, and comic characters (buffoons, court jesters, and madmen) could always provide the opportunity for a ballo. For example, in Le fortune di Rodope e Damira (1657), the librettist Aureli uses a dance by pa?zi (madmen) to expand on the idea of Damira's madness. These madmen interrupt a marital spat between Bato and Nerina, and Bato then becomes involved in their dance.75 In some cases, the principals of the opera converse at a ball, or are entertained by giochi (games) danced by gladiators or swordsmen. In other instances the connections are somewhat less obvious, though no less intriguing. Contemporary artistic or philosophical interests might have been reflected in the dancing painters, scholars, sculptors, or philosophers. There were also balli for farmers and peasants, miners and gardeners, sailors and fishermen, guards and prisoners, eunuchs and slaves.76 Despite the remarkable variety and individuality of the Venetian balli, there are many topics to which librettists and choreographers frequently return. These are summarized in the section below. The mythological, allegorical, and pastoral As a number of commentators have noted, the problem of introducing song and dance into drama may well have been of concern to the earliest opera librettists.77 One immediate solution was to give this heightened speech (song) and movement (dance) to higher beings, differentiating them from mere mortals who merely spoke (recitative). The notion of dance having heavenly origins was most famously celebrated in the Florentine intermedi for Lapellegrina (1589). Opera's very origins are tied to Greek and Roman myths, and characters from these stories populate the balli as well. In any number of operas, gods descend from the heavens to dance and are accompanied by various Arcadian creatures: Cupids dance for Venus; the Muses dance for Apollo; fauns and satyrs celebrate bacchanalian revels.78 The enduring popularity of pastoral themes is seen in the many balli danced by nymphs and shepherds.79 As it happens, many of the balli that refer to carols are danced by characters of this type. In Act I, scene 4 of Gli amori d'Apollo e di Dafne, Dafne invites 75 Jane Glover states that Aureli 'dramatically prepares' for balli whereas Faustini does not, and cites this as one of her examples; see 'The Teatro Sant'Apollinare and the Development of Seventeenth-Century Venetian Opera', Ph.D. diss. (Oxford University, 1975), 266. Yet many of the balli in Faustini's libretti either involve or are introduced by one of the opera's characters. In fact, Faustini's Elena (1660) has a dance by freed slaves similar to the dance by freed prisoners in Aureli's L'Erismena (1656) cited by Glover. See also Faustini's La virtr de' strali d'Amore (1642), L'Euripo (1649), La Rosinda (1651), La Calisto (1652), and L'Eupatra (1655). 7 Lists of these and other roles can be found in the 'Index of Balli', in Aim, Catalog of Venetian Librettos, 984-97. n Ed. note: On the question of operatic verisimilitude, see Nino Pirotta, 'Early Opera and Aria', Music and Theatre from Poliziano to Monteverdi (Cambridge, 1982), esp. 275-80, and Rosand, Opera in Seventeenth-Century Venice, 40-45. 78 Mythological characters listed in balli include the gods Amore, Apollo, Aurora, Bacco, Cibele, Diana, Imeneo, Marte, and Pallade. Dancers, however, usually played lesser mythological figures such as 'amori', 'ciclopi', 'driadi', 'fauni', or 'zeffiri'. SSee 'ninfe', 'pastorelle', and 'pastori' in the 'Index of Balli', Aim, Catalog of Venetian Librettos. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 243 a chorus of nymphs to celebrate her happiness 'con danze e carole', and in Act II, scene 17 of Circe abbandonata da Ulisse (autumn 1697), Bleso introduces a ballo for shepherdesses and gardeners, singing 'Charming gardeners / Come, come graceful friends / Celebrate / Dance carols'.80 Carols were typically joyful dances, and even the soldiers danced them in Xerse (1654), when Eumene calls for celebrations with the happy carols of victory following the second battle. Dancers also portray followers of allegorical figures, such as suspicions ('sos- petti'), the followers of Gelosia, in Aureli's Perseo (1665), and the ministers of Discordia in Corradi's La divisione del mondo (1675).81 Other allegorical subjects include the four elements, the twelve months, the twenty-four rays of the sun, and the four parts of the world. The supernatural Dances by spirits, goblins, elves, ghosts, demons, furies, and phantoms are among those most frequently found throughout the seventeenth century. Not only did these balli allow for special effects and spectacular staging, they were easily introduced through types of scenes that had become conventions in Venetian opera.82 Sleep scenes, for example, involved phantoms and spirits who appeared in dreams. In Francesco Piccoli's L'incostanza trionfante overo, 1l Theseo (1658), Act II closes with a scene for Anthiope. As she falls asleep, Morfeo (Morpheus, the god of dreams and son of Sleep) sings an introduction to the ballo, which is danced by various phantoms who represent impending events, showing Anthiope her future through her dreams. In Aureli's Teseo tra le rivali (1685), Act III, scene 1, Fedra falls asleep and dreams of a fantastic monster and phantoms.83 A more pleasant dream is danced by a chorus of heroes in Act II, scene 19 of Ercole in Tebe (1671) by Moniglia and Aureli.84 The finale of Frigimelica Roberti's Ilpastore d'Anfriso (1695) is an elaborate allegorical scene for La Notte (Night) who appears with 'Sogni e Fantasme' (Dreams and Phantoms) who play instruments, sing, and dance. Another convention was the infernal invocation, in which spirits were conjured up to assist in carrying out a plan or to thwart someone's actions. The most famous scene of this type occurs in Cicognini's and Cavalli's Giasone (1649), which enjoyed 80 'Giardiniere vezzosette / SU', si amiche leggiadrette / Festeggiate / Carolate.' 81 The allegorical figures who routinely have followers include Allegrezza, Capriccio, Costanza, Inganno, Inventione, Pace, Paura, Riso, Sdegno, and Virtui. See 'seguaci di' (followers of) in the "Index of Balli" ', Alm, Catalog of Venetian Librettos. 82 For further discussion on the use of conventions in Venetian opera, see Rosand, Opera in Seventeenth-Centuly Venice, chapter 11. 83 'Qui al suono di grave sinfonia Fedra s'addormenta. E mentre ella dorme le apparir in sogno fantastico Mostro con alquanti varij fantasmi che formano il Ballo, qual terminato spariscono i fantasmi col mostro e Fedra si risveglia.' 84 'Mentre Megara dorme gli apparisce in sogno Ercole assiso in Trono con Pelio superato B suoi piedi. Vede quel vittorioso Heroe coronato dalla Fama d'alloro, la qual poscia alzando il volo, e suonando laurea sua Tromba chiama un Choro d'Eroi, quali compariscono ad'inchinar Ercole in forma di Ballo, qual terminato sparisce il sogno, e Megara si sveglia.' This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 244 Irene Aim a remarkable popularity during the seventeenth century."5 At the close of Act I, Medea calls the spirits to her chamber of spells and commands them to assist her. Following the chorus of spirits, Volano (a demon) sings, and after Medea's final lines, the spirits dance. Further information on at least one version of the choreography is given in three Milanese libretti from around 1660.86 In these, Medea has additional text before the 'Ballo di Spiriti', telling the spirits how she wants them to look and dance: 'not in the shape of frightful ghosts, / but with a ridiculous and charming appearance, / press the ground, draw out / affectionate dances, joking among us.'87 In this version she seems to encourage a more grotesque and comic dance, rather than a sinister or frightening one. This caution may have been intended to avoid anything that seemed too close to actually invoking infernal powers, thereby attracting the attention of the Inquisition (a particular concern in a conservative city such as Milan).88 Comic mishaps involving magic also conjured up dances of spirits and demons, who frightened or even tormented an unfortunate character. Another supernatural theme used in balli was that of statues who come to life and dance. At the end of Act II of Aureli's L'Erismena (1656), Clerio is alone on stage, and his curiosity leads him to open a book given to him by a court sorcerer to deliver to his master, Idraspe. No sooner has he opened it than several statues in the gallery begin to move. Terrified at the sight, he runs off, and the statues join in a dance to conclude the act.89 Animals Hunt scenes were popular in Venetian operas and were one way to introduce animals into balli. Bears, wild boar, and stags were pursued by hunters. Stagehands or extras may have portrayed the animals in some cases, but in other instances dancers took these roles, as in Paolo Vendramino's L'Adone (1639), where the ballo is danced by a lion, a tiger, a bear and a wild boar, or Faustini's La Calisto (1652), where six bears dance at the end of Act I (cf. Fig. 1).90 Children may have danced 85 The enduring popularity of Giasone is especially notable in view of the fact that the opera 'industry' of the time produced works designed to be extremely successful for a season, but ephemeral. Many sources for Giasone survive (numerous libretti and scores from various revivals in Venice and elsewhere), and these still await a comprehensive study. 86 I wish to thank Martin Morell for sending me this information. One undated libretto is now at the Bologna Conservatory (I-Bc no. 6519) and two libretti are at the Milan Conservatory, one undated (I-Mc, Y.104) and one with a handwritten date of 1660 (I-Mc, Y.105). 87 'E non in forme / Di larve spaventose, / Ma in sembianze ridicole, e vezzose / Premete il suol, trahete / Scherzando tra di noi danze amorose.' 88 Venice was somewhat more liberal; the characters in Domenico Balbi's II sfortunato patiente (1667) form a circle to invoke devils who dance. In 1686, however, Noris's II demone amante, overo Giugurta was banned in Venice (Act I ended with a 'Ballo d'Ombre'). The manuscript mercuri (news-sheet) dated 12 January 1685 [M.V.] reported that the authorities had suspended performances and chastised the authors for including scenes inimicable to the Catholic religion. I-Vnm Cod. It. VI 463 (= 12107). 89 This dance is described in the scenario; a different ballo appears in the libretto. 9o Notably, the account book lists expenses for the bear costumes. See Alm, Theatrical Dance, table Va. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 245 the parts of baboons and monkeys in operas such as La Bradamante (1650) by Pietro Paolo Bissari. Among the more unusual animals that dancers portrayed are the frogs in Antonio Arcoleo's La Rosaura at S. Angelo in 1689. Ostriches were surprisingly popular and appear in two operas at S. Giovanni Grisostomo in 1681, Antioco il Grande by Frisari and IR Creso by Corradi, as well as in Fulgenzio Mattia Gualazzi's La schiavitzifortunata at S. Angelo in 1695. The seventeenth-century chronicler of opera, Cristoforo Ivanovich testifies that Venetian theatres also used both live and mechanical animals: 'Thus there are real elephants, live camels, great chariots drawn by wild beasts and by horses: also flying horses, horses that dance, the most superb machines, presented in air, on land, and on the sea with extravagant artifice and praiseworthy invention.'91 Although in some operas the elephants and camels were stage machinery, at least one opera at S. Giovanni Grisostomo, Licinio imperatore (1684), actually featured two live camels that had been captured during the Turkish siege of Vienna.92 Horses were more often seen on stage. For example, in a review of Totila, performed at SS. Giovanni e Paolo in 1677, a reporter for the Mercure galant writes, 'This Act [Act I] finished with a dance of cavaliers mounted on real horses'. The bears at the close of the second act of Totila were more likely dancers or extras in costume, and the reporter simply states 'An entree of soldiers attacked by two bears finished the act'. The horses returned for the battle scene in the third act: 'Some carts loaded with spoils from the enemy passed over this bridge; they were drawn by real horses ...'93 In 1679, the same journal also reported that 'chariots drawn by real horses, and cavaliers also on horseback' were used in Alessandro Magno in Sidone at SS. Giovanni and Paolo.94 And Andrea Perrucci, writing in 1699, confirms Ivanovich's report of flying horses, remarking on the 'flights not only of men, but of live horses' seen in Venetian theatres.95 A review of Lafortuna tra le disgratie, staged at S. Angelo in 1688, indicates that, at least on one occasion, even wild animals were brought into the opera house: One sees in the second [scene] a stag running followed by dogs and hunters, which causes extraordinary admiration and the necessity to confess that the Venetians even make wild animals adapt to the stage.96 91 Cristoforo Ivanovich, Memorie teatrali di Venezia (Venice, 1687; rpt. Lucca, 1993), 388-89. 92 This special attraction was reported in the manuscript mercuri, dated 25 December 1683, I-Vnm Cod. It. VI 460 (=12104). M3 ercure galant (August 1677), 80-83. Live horses are not mentioned in the libretto, which in the prefatory pages (p. 10) describes the scene at the end of the first act as an 'Abbatimento fra Vandali e Romani', whereas at the end of Act I the same scene is referred to as a 'Ballo de Cavallieri'. A propos of the second-act dances, the libretto lists a 'Ballo di Pastori con Fiere' in the prefatory pages, but a 'Ballo di Soldati inseguiti da due Orsi' at the end of Act II. 94 Mercure galant (April 1679), 129-30. 95 Perrucci (see n.1), 52. 96 Pallade Veneta (January 1688), 64-71, quoted in Selfridge-Field (see n.63), 202. A similar scene was described by Chassebras de Cramailles as having occurred in Berenice vedicativa (1680), presented in the private theatre of Marco Contarini at Piazzola sul Brenta: 'That which astonished the most was a real hunt of live deer, bears, and wild boars, that were killed by the hunters.' Mercure galant (February 1681), 245-46. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 246 Irene Alm Although this scene did not include a ballo, it raises the possibility that some of the animals mentioned for balli involving hunters may have been real. Comedy Choreographers at Venetian opera houses had to be adept at creating comic balli, since these were staples of the repertoire. Sometimes called a scherzo or a burla, the comic ballo could be tailored to many different settings. In some operas it was danced by pages or servants, as in Giacomo Dall'Angelo's 1/Demetrio (1666), where insolent pages tease Geliro and an old woman. In others the dancers are dressed as buffoons ('buffoni') or fools ('scemi'). Some comic balli depended on physical traits, such as hunchbacks ('gobbi'), dwarves ('nani'), or cripples ('zoppi').97 The adjectives 'cappricioso' (whimsical) and 'bizzaro' (bizarre) are used to describe the style of some balli. At the close of Act I in VeremondaAmazzone diAragona (1652/3), Vendetta commands a celebration 'con danza bizzara'. 'Cappriciose danze' are performed by prisoners with chains on their feet at the end of Act II in Diocletiano (1675), and in Act II, scene 19 ofAriberto e Flavio regi de Longobardi (1685), the Muses dance a 'capriccioso ballo'. The familiar commedia dell'arte characters, however, never crossed over into the theatrical dances in the Venetian opera houses, perhaps because they could be seen in other theatres in Venice where spoken comedies played. Pantomime was used in many comic situations. For example, the three-part ballo which closes Act I of La Bradamante (1650) begins with baboons (possibly danced by children) who are fleeing from hunters. When the hunters catch up with them, the baboons imitate their gestures, then climb up into the trees and escape. Pompeo Magno (1666) has two comic balli involving mime. Act II closes with a chorus of twelve shades, or spirits, who torment the comic character Delfo, surrounding him and tying him up while dancing (see below, Example 1). Even Santa Catterina d'Alessandria (1675), a rappresentatione sacra, uses pantomimed mockery. At the end of Act I, Labinia calls the children to dance in a ballo that mocks a group of fake doctors. Madness Madness, whether feigned or real, was one of the conventions in seventeenth- century opera and required specialized choreography. In La finta pa za (1641), the opera that established madness as a convention of Venetian opera, a ballo danced by madmen closes Act II.98 The first edition of the libretto includes a scenario, which describes the dance: 97 It is not known whether Venetian theatres hired hunchbacks and dwarves or if these roles were played by dancers in costume and by children. 98 On mad scenes in Venetian opera, with particular reference to La finta pazza, see Rosand, Opera in Seventeenth-Century Venice, 121-24, and Paolo Fabbri, 'On the Origins of an Operatic Topos: The Mad-scene', in Con che soavitai: Studies in Italian Opera, Song and Dance, 1580-1740, ed. Tim Carter and Iain Fenlon (Oxford, 1995), 157-95. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 247 In any case, the nurse having arrived unexpectedly to tie her up, Deidamia is helped by some court buffoons, crazy in the head, who with shovels drive off the nurse and the others, who were speaking with her; after which action Deidamia invites these madmen to perform a ballo for the happiness of having won a victory. And here ends the second Act.99 The Cannocchiale per la finta pazza gives a more detailed account: Then followed the court Eunuch, who gave Diomede a report on the fury of Deidamia, who then turned up with the captain of the armed chorus, and [Deidamia] pretended much frivolity speaking nonsense in such a way that the Eunuch, Diomede, and the Captain considered her completely mad; and because the nurse arrived and tried to tie her up with chains, the young lady gave cry, calling for help, and so the court buffoons, crazy in the head, who came out bizarrely dressed in various colours and sizes, made everyone withdraw and set her free. Deidamia then invited them with song to perform as a sign of happiness a ballo, as they did a very bizarre one, and as if madmen, except not so crazy that the art, the tempos, and the metres were very well marked, which not only gave delight to the eyes, but also to the intellect, seeing that even ridiculousness and discord are subjects of art and of ingenuity, and this was the end of the Epitasis, or second act.100 The libretto also emphasizes this choreographic madness. Deidamia first sings verses urging the madmen to dance. After they have danced a little, she interrupts them and urges them 'alle corde', which may have a double meaning of dancing on tightropes (she tells them not to be frightened) and of playing stringed instruments (she mentions chromatic and diatonic strings). The crazy buffoons then sing five strophes, with many references to crazy steps, crazy feet, crazy dancing, and so forth, all of which is suggestive both in terms of the style of dance and music. The second strophe begins 'Pazzo e il pie, che un pazzo segue' (Crazy is the foot that follows a madman), and the third and fourth strophes are: 3. Pazzo core hi pazzo piede Che leggiero Quinci e quindi errar si vede. Pur ch'io resti un pazzo vero, Voli il pi&, la gamba ondeggi, E di un pazzo brillar l'alma festeggi. 4. Pazzo suono, e questa accanto Pazza danza Accompagni il pazzo canto. Pazzo ballo hi pazza usanza, E noi pazzi, e saltellanti Per un pazzo desir siam pazzi amanti. [A crazy heart has a crazy foot, that is seen to wander lightly here and there. If only I remain a true madman, my foot flies, my leg sways, and grace celebrates in the sparkle of a madman. Crazy music, and nearby this crazy dance accompanies the crazy song. A crazy ballo has a crazy style, and we madmen and dancers are crazy lovers for a crazy desire.] 9 'Argomento e scenario', La finta paza (1641), 17. 100oo I1 cannocchiale per la finta pazza, 40-41. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 248 Irene Alm Madness of heart and mind thus leads to madness of the dance, all of which accompanies music no less driven by insanity. Martial subjects The ancient Greek use of military or pyrrhic dances was cited by many contempo- rary writers on dance, and many operas had at least one battle scene, staged as a ballo for soldiers, warriors, archers, gladiators, or fencers.101 The terms abbattimento and combattimento are in some cases used interchangeably with ballo; for instance the first dance in Noris's Astiage (1677) is labelled 'Mock battle by soldiers' ('Abbatimento finto de Soldati').102 A review of Astiage reports that 'The first scene was the camp of an entire army, where some soldiers performed a pyrrhic dance, accompanied by a marvellous symphony. This dance was interrupted by the arrival of a princess followed by officers of her army, all on horseback.'103 Some of these battle scenes were done on a huge scale complete with horses. Other choreographed fights are labelled 'rissa' (brawl), 'gioco di lotta' (game of wrestling), 'gioco di spada' (sword play), or 'gioco d'armi' (weapons play). The moresca, a battle dance dating back to the early fifteenth century, was seen in the theatres as well as on the streets and the bridges of the city during Carnival.104 Act III of Badoaro's Le nozze d'Enea con Lavinia (1641) ends with a 'combattimento in the style of a moresca which serves as the ballo'. Veremonda Amazzone di Aragona, by Strozzi, used a moresca in both the Neapolitan (1652) and the Venetian (1653) productions. In operas, however, morescas were not always serious battle dances. The Apparati sceniciper lo Teatro Novissimo describes the ballo at the end of Act II in Venere gelosa (1643), led by the 'buffoon' Trulla: The King laughed at these blunders, and commanded that [Trulla] sing, and he, having begun, had not yet finished the first strophe, when some other buffoons dressed as toys appeared, who after each strophe danced an amusing quasi-moresca.105 The moresca could also be used on joyful occasions; the final ballo in La Venere gelosa, which follows the wedding ceremony near the end of Act III, is a moresca. The Apparati scenici describes: 101 See 'arcieri', 'gladiatori', 'guerrieri', 'schermitori', and 'soldati' in the 'Index of Balli', Alm, Catalog of Venetian Librettos. 102 See also Minato's Xerse (1654), set to music by Cavalli; the libretto states that Acts I and II end with combattimenti, while the Venetian score labels them balli. I-Vnm Cod. It. IV 374 (=9898). 103 Mercure galant (August 1677), 87. 104 There is some debate over the origins of the moresca - whether it is actually a Moorish dance or if it has its roots in ancient fertility rites. Although historically Moors were white as well as black, the term 'Moorish' is often taken to mean dark-skinned, and is thought to come from the Greek word mauros, meaning dark. Renaissance morescas were often performed with blackened faces and portrayed battles between Muslims and Christians. The guerra de' pugni, or forte d'Ercole, was described by some as a form of the moresca fought by members of different sestieri (sections of the city). The Comte de Caylus, who travelled in Italy during 1714 and 1715, wrote that 'the Arsenalotti and the Nicolotti do dances in the style of the moresca and with turns, which one calls in Venice the forze d'Ercole'. Comte de Caylus, Voyage d'Italie 1714-1715, ed. A. Pons (Paris, 1914), 118. 105 Apparati scenici, 28. [Ed. note: see also Heller, 'Dancing Desire' below.] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 249 He was accompanied by a chorus of fauns and of satyrs, who were going around singing; at the end the buffoon Trulla arrived, who, so happy from the wedding concluded in court; was carrying a flask, and he began to jest with them, and these were their words: Chorus: Horns and pipes Flutes and cymbals And cymbals and kettledrums.106 In the libretto these verses are marked 'Chorus of Satyrs with a ballo', followed by the directions 'They fight a moresca with clubs'. Exoticism and foreign styles Seventeenth-century Venice was a city in which the 'four corners of the earth' met; its residents had first-hand contact every day in the piazzas and marketplaces with people from many nations brought there through commerce and trade. Furthermore, during Carnival Venetians indulged their fascination with other cultures by adopting foreign identities through costumes and masks. Thus, these 'foreign' dances not only reflected the commercial and political interests of the government, but also the daily life of the city and the fantasy life of Carnival. On the Venetian stage, dancers could be found garbed in African, Albanian, American, Armenian, Asian, Assyrian, Belgian, English, Egyptian, Ethiopian, French, German, Greek, Iberian, Indian, Macedonian, Moorish, Persian, Spanish, Slavic, and Turkish costumes, as well as portraying gypsies, savages, and pygmies. Foreign dance styles also are specified in libretti, which list balli 'alla francese', 'alla greca', 'alla spagnola', and 'popolare d'inghilterra', among others. A number of dances are simply described as being by foreigners ('genti straniere' or 'forastieri'). The four corners of the earth ('Le quattro parti del mondo') is a theme used in several balli, including one in Piccoli's L'incostanza trionfante overo II Theseo (1658), which has a battle among European, Africans, Asians, and Americans, who accompany the four parts of the world. Many of the dances reflect Venice's contact with and interest in the Muslim world to its east and south. Various groups appear, but perhaps the most popular were balli featuring the Moors: Muslim people of mixed Arab and Berber descent living chiefly in northern Africa. There are, for example, Ethiopian Moors in 1R Ciro (1654), the Egyptian Moors in Laodicea e Berenice (1695), or Indian Moors in II coloref/ la regina (1700). The two most common images of Moors in theatrical dances are as slaves, or as warriors or corsairs wielding the traditional curved sword, or scimitar. Another Eastern people, the Turks, also had a long history with Venice, one sharply defined by war. The seventeenth and early eighteenth centuries saw the last of Venice's many Turkish wars - the war of Crete (1645-69) and the two wars of Morea (1684-99 and 1714-18). Significantly, Turkish dances are not found in Venetian operas until the later part of the seventeenth century. Perhaps during the war of Crete, in which Venice suffered many losses, the Turks were too sensitive a 106 'Era egli accompagnato da un Choro di Fauni, e di Sattiri, che andavano cantando; alla fine vi giunse Trulla buffone, che tutto allegro delle nozze concluse in Corte, portava un fiasco, e si diede a buffoneggiare con coloro, e tali furono di questi le parole. Coro: Corni e Piffari / Flauti e cembali / E cembali e naccare'. Apparati scenici, 37. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 250 Irene Alm subject to be used in theatrical entertainment. However, during the First War of Morea (1684-99), when General Francesco Morosini regained Venice's lost territory and more, Turkish subjects would have been a reminder of the Republic's victories, and they became popular topics for dance. For the majority of the exotic dances, the libretti give no indications of style or steps, although several do make a point of stating that balli are danced according to the practice of a certain country - for example, the Armenian masquerade 'in the style of that nation' at the close of Act II in Sesto Tarquinio (1679), and the 'ballo for Moors, who dance according to their custom' in I1 prodigio dell'innocenza (1695). Although characterization undoubtedly relied on distinctive costumes and sets, descriptions and engravings suggest that the choreography itself also reflected national styles. Dances by Moors were apparently lively. For example, in La Dori, overo lo schiavo reggio (1663), Erindo sings an aria to introduce the dance by Moorish eunuchs of the seraglio, and he commands: Il Ballo movete Veloci col pi& Danzate Correte Venite con me [Begin the ballo with swift feet. Dance, run, come with me.] In La Semiramide (1671)107 and Oratio (1688), choreographers make use of Moorish soldiers with their swords to literally spell out messages, rather than do battle. The libretto for La Semiramide gives these stage directions: 'Ballo by Moorish Soldiers with Creonte who with their scimitars create in so many beats as many words, that form the verses given below'.108 In Orazio, Moors with weapons form verses that were also sung by a chorus, but Tosi's score does not survive.109 Francesco Coli reviewed Orazio in the Venetian journal, the Pallade veneta, writing, 'Bold and courageous battles, fights, and feats of great wonder are seen in this theatre, and in Act III one enjoys a ballo by Moors who, with weapons in hand, form various words, clear and easily understood'.110 Nine balli are listed as either in the French style or by French characters.111 Seven of these date from the 1680s and 1690s, the period during which French social 107 Libretto by G. A. Moniglia and M. Noris and music by Pietro Andrea Ziani. 108 'Ballo di Mori Soldati con Creonte i quail con scimitarre compongono in tante cadenze tante parole, che formano li sottoscritti versi: Dio di Gnido / Io rido di te / Se a volo / Ogni duolo / Ho fugge dal R&.' Unfortunately, the ballo music is not extant. 109 'Siegue intreccio giocoso di mori con armi, che formano le seguenti parole, che da un coro Vengono espresso. Goda Roma./ Alba e Doma./ Rida il Lazio./ Viva Tullo./ E viva Orazio.' 110 January 1688, 73-81; quoted in Selfridge-Field, Pallade Veneta, 205. 111 These are: a correntefrancese by six peasant girls in L'Eupatra (1655); a ballo by four French cavaliers in L'Adelaide regia principessa di Susa (1670); a ballo by two young French girls and boys, slaves in the seraglio in Pompeo Magno in Cilicia (1681); a ballo allafrancese in Pub. Elio Pertinace (1684); two balli alla francese in Il trionfo di Amore e di Marte (1689); a gran danza francese in Onorio in Roma (1692); a ballo by French ladies and cavaliers in Sigismondo primo al diadema (1696); and a ballo by French and Spanish cavaliers in L'innocenza giustfiycata (1699). This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 251 dance was beginning to arrive in Italy. Only two other foreign dances, one English and one Spanish, have verses that mention steps or style. The English ballo popolare in La barbarie del caso (Murano, 1664) seems to have been a lively dance. The verses exhort the dancers to 'invite your feet to leaps' and mention correnti and salti mortali (somersaults). Ballroom scenes The rich blend of cultures that made up Venetian society participated in what was to become a convention both in Venetian life and theatre: the festa di ballo. At least nineteen operas incorporate an extendedfesta di ballo modelled directly on the typical Venetian balls that took place during Carnival season.112 This type of scene was especially popular in the 1680s and 1690s, after the elegant Teatro S. Giovanni Grisostomo opened in 1678 and began an annual tradition of hosting a ball for the nobility on the last night of Carnival. During one season, 1683, four of the six opera houses (S. Cassiano, SS. Giovanni e Paolo, S. Salvatore, and Cannaregio) presented operas with dances of this sort, and the following season S. Angelo followed suit.113 Feste di ballo were given throughout the year by members of the Venetian nobility to entertain and impress prominent visitors. The number of balls peaked during Carnival season for two reasons: first, the arrival of throngs of visiting nobility from throughout Italy and Europe; and second, a relaxation of Venetian sumptuary laws (regulations which restricted expenditures on luxuries, affecting everything from clothing to food and lighting). During Carnival, the ballrooms could be transformed into scenes of lavish display for an audience of invited guests, as reported in the April 1679 issue of the Mercure galant. As the palaces are quite spacious, the ballroom is among eight or ten rooms, all of which are decorated with rich drapes, paintings, and very expensive furniture. ... The invited ladies are seated in the ballroom, where the noblemen come to take them to dance. Their dance is only a type of promenade, occasionally continued from room to room, where the occupants can have the pleasure of seeing the whole ball pass by.114 The nature of the Venetian festa di ballo made it well suited for adaptation to the dramatic stage. Unlike French courtly dances with complex patterns of steps, often performed by a single couple for a critically observing audience, the Venetian promenade was an ideal setting for amorous conversations and intrigues. Librettists, therefore, did not have to interrupt the development of the plot when inserting a festa di ballo, but could use the dancing to advantage in having characters express their thoughts or emotions 'privately' against the colourful backdrop of a crowded ballroom. Seleuco in 1666 was the first opera to portray a typical festa di ballo (see 112 See Irene Alm, 'Operatic Ballroom Scenes and the Arrival of French Social Dance in Venice', Studi musicali, 25 (1996), 345-71, which includes a list of all the operas in question (358-61); see also the descriptions of Venetian balls in Aim, Theatrical Dance, chapter 6, 194-202, and a discussion of the music in chapter 7, 269-72. 113 In 1683, S. Salvatore also hosted a ball in the theatre on the last night of Carnival. Chassebras de Cramailles described the event in the Mercure galant (April 1983), 78-81. 114 Mercuregalant (April 1679), 120-22. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 252 Irene Alm below, Ex. 7). Ten years passed, however, before another ballroom scene appeared, in Galieno (1676). Then within the space of a decade (1679-89) fourteen operas includedfeste di ballo. A French review of Nerone mentioned that for the festa di ballo there were 'many extraordinary instruments that were on the stage',115 and it seems likely that in most of these scenes one or more dance bands would have played on stage, visually as well as aurally re-creating the atmosphere of thefesta di ballo for the singers. That the geographical or historical setting of the opera might be at odds with a seventeenth-century Venetian ballroom seems to have mattered little to the librettist or the audience. A striking juxtaposition of locales occurs at the start of Circe abbandonata da Ulisse (autumn 1697 at SS. Giovanni e Paolo), in which a Venetian festa di ballo is immediately followed by a ballo alla greca. Whereas the opening dance would have transported the opera to a seventeenth-century Venetian ballroom, the ballo alla greca serves to relocate the opera in time and place so that the plot can begin to unfold. By the last decade of the century French social dances are mentioned more frequently in first-hand accounts of Venetian balls, although the descriptions were not always complimentary. On 18 February 1695, James Drummond, Fourth Earl of Perth, cynically observed in a letter to the Countess Marischall that the music seemed almost superfluous to the promenade and derided the Venetian attempts at French figured dancing: The grand dance is to walk about (with or without musick is all one) for three or four hours, every gallant with a lady, and this procession ends with an English country dance; then they dance, four together, a dance which I cannot remember how it is called, I think it is a Rigadoun, but it is to a Minuete; and then comes the French dances, as awkwardly as can be...116 During this same period French dances became part of the operatic festa di ballo and standard social dances appear more and more frequently in the scores for ballroom scenes from the 1690s. Minuets and other social dances also began to be used for a variety of balli on other subjects as well, whereas until the 1680s virtually none of the music for Venetian theatrical dances had been based on standard dance forms."' Thus, when the French style of dancing arrived in Venice, the ballroom scenes, reflecting this change, lost much of their dramatic function and became simply an element of spectacle. The convention of the dramatically integrated ballroom scene eventually would resurface in striking examples throughout the eighteenth and nineteenth centuries (from Mozart to Verdi), but never again with such frequency or such special power to bind audience and opera. 115 Mercuregalant (April 1679), 134-35. 1"6 Letters from James Earl of Perth (London, 1845), 52-53. 11v This is not true, however, of revivals of Venetian operas in other Italian cities. See the discussion above (n. 32) on the new dances that Stradella supplied for Giasone (as I1 novello Giasone) and Scipione africano at the Teatro Tordinona in Rome, 1671. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 253 Music for the dance Instrumental dance music Whereas the bulk of the instrumental pieces within Venetian operas were balli, some served other purposes, and the function of individual pieces is not always immediately apparent. There are nonetheless patterns in the ways scores were put together that help resolve the basic question, even if some individual cases remain problematic. Most operas open with a sinfonia in several short sections, and often each act is preceded by a shorter instrumental piece. It is not uncommon for some of these pieces to have dance titles such as 'Balletto', 'Corrente', or 'Giga'; they may also be marked 'Sinfonia' or 'Ritornello', or have no label at all. Pollarolo's IR colorefa la regina Ballo di Fantasme -' '" " " .. .. , _ , do _ aw I I I r -= F-p ': - TI" I I I I II I I I I .............MONO fdo As Jk IJ J J J-NI,- . . . . . . . I II V op # Al i -? 01&6ct 10/ I a, I I-I C . It , , vI ? r i I " ,- ,.I _ i d i Li i h i i i Ex. 1: Pompeo Magno (1666), Act II, scene 22 (I-Vnm, Cod. It IV 377 (-9901), fols. 105v-106'). The scoring here has been reduced from the original five parts. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 254 Irene Alm Ballo b A , I-.,, F w F Id "F II p 11 1 )[ " W w, ,o~ ~ I l u op IF b--d -opJ • : DE" 1w r. r I I r ' :-. - -, I Ex. 2: L'Argia (1669), Act II, scene 20 (I-VLevi, C.F.A. 8, fols. 47v-50v). (1700) opens with three movements titled 'Sinfonia Presto / Corrente staccato / Gigue'. In this and similar instances, the instrumental pieces appear to have functioned as 'symphonic' introductions, not as dances, a conclusion that finds support in cases where more than one score is available for comparison. The Venetian score for Pietro Andrea Ziani's L'Annibal in Capua (1661), for example, includes no music for the ballo danced by spirits at the end of Act I, but has a 'Corrente' at the beginning of Act II. That this piece should not be used for the ballo is made clear by the Roman score, which includes music for the ballo at the end of Act I, and then begins Act II with the same 'Corrente' as in the Venetian score. Similarly, the label 'balletto' does not necessarily mean that a piece was danced. In fact, in opera scores the term 'balletto' is more often associated with sinfonie or non-dance music than with choreographed balli. On the other hand, some untitled instrumental pieces were almost certainly dances, based on their style as well as on information from libretti. Most balli employ a homophonic texture, whereas sinfonie, This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 255 Ballo d'Eunuchi 6 6E [sic]7 Ex. 3: La Dori (1661) Act I, scene 12 (I-Vnm, Cod. It IV 410 (=9934), fol. 41r). ritornelli, and instrumental pieces for set changes often use imitative textures. These are general tendencies, however, not absolutes, as a few dances do have brief passages of imitation. And despite these guidelines, there are still scores in which it is difficult to decide whether a piece was used for a ballo or for some other function. The fifty-six instrumental pieces included in Table 3 (see Appendix) are those for which the available evidence suggests that they were actually danced. They range in length from four to ninety-six bars (counting repeated sections), with an average length of about twenty-six bars. In the earliest dances various metres and forms are employed. Duple metre and binary form are increasingly favoured for instrumental dance music in the 1670s, and overall more than half of these balli are in duple metre. Twelve dances use sections of contrasting metres, while eleven are in triple metre (eight in 3 and three simply marked 3), and four in compound metre (two in C~ and two in 12). The formal structure of most of these dances consists of two or more repeated sections; nearly half (twenty-five) are binary. Nine dances have three sections, and six are in four or more sections. The others lack repeat signs and are generally through-composed. Very few Venetian balli bear the titles of standard seventeenth-century dances. In fact, most of the instrumental pieces in these scores labelled with standard dance titles are not balli, but are movements of the opening sinfonia or are introductions to the second or third acts of the opera. Since so many Venetian theatrical dances were meant to convey a particular subject through movement and pantomime, the music was often specifically composed to suit their character and action. Interestingly, symmetrical four-bar phrasing is not used often in Venetian balli; more often the phrasing is asymmetrical and unpredictable. A four-bar phrase may be answered by five bars, or a series of two-bar phrases finish with a three-bar phrase. Irregular phrasing often occurs in conjunction with metre changes, and fermatas are used in a number of balli, suggesting that This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 256 Irene Aim Balletto de Pazzi ( ".., ..Ai " L 1b. "fJb F J4 AL.OF i i/ lop I ff • t' W O- , _ ,- od O O A2:do • - , F F • ' F i I ".. I ? "I - . ,,.. . r " " . J • " ?' d. P" r. ,OP ORa9"9 OP.. .. . IF wI I I I I 3:0o •.. 9 piano forte tJ piano A ? 1 I O! i P Ex. 4: Pompeo Magno (1666), Act I, scene 20 (I-Vnm, Cod. It IV 377 ( -9901), fols. 56r-57v). The scoring here has been reduced from the original five parts. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 257 1 - doo 4:0 m 1f, Ao 1 --, .li I0001 p , A. OP-M, f I F . 1 ., . - I I I MI , l -•:. r = wI .. .. F-ok I . . o " -- -- 4 -, - J I Ex. 4: continued. the choreography and music were carefully coordinated, and that the composer may well have written the music after the basic movements were set. For example, dances involving supernatural creatures often suggested the unpredict- able, the abnormal. It is not surprising, therefore, that composers and choreogra- phers used sudden changes, irregular phrasing, and rhythms interrupted by fermatas in balli danced by ghosts, phantoms, spirits, or other supernatural beings. Seven examples of supernatural balli appear in Venetian scores.118 Cavalli, for instance, called for a darker register by using soprano rather than treble clefs in two of his balli danced by phantoms, the first in the prologue of Gli amori d'Apollo e di Dafne (1640) and the second at the close of Act II in Pompeo Magno (1666). In the latter, the phrases are short and abrupt: mysterious dotted rhythms hesitate in the second bar on a whole note with a fermata, and the second section has a furious rush of semi-quavers, followed by an impish dance in compound metre reminiscent of a giga. (See Ex. 1.)119 In Act II, scene 20 of L'Argia (1669), the phantoms have mysterious music filled with suspensions and chromaticism. Somewhat surprisingly, only this ballo for phantoms (and the one in Pompeo Magno) emphasize the minor mode. (See Ex. 2.) Music survives for only about a dozen balli with foreign themes, although the music remained largely within the language of seventeenth-century Venetian opera, even when portraying Eastern cultures, as for example in this ballo for Moorish eunuchs from La Donr (1663). (See Ex. 3.) 118 In addition to those discussed below, these include the end of Act II of Legrenzi's Germanico sul Reno (1676) and the ultimo intramezzo of Pollarolo's Ilpastore d'Anfriso (1695). The Roman revival of Giasone, Stradella's Il novello Giasone (Rome, 1671), also included a supernatural ballo associated with Medea's incantation scene. 119 Ed. note: Alm's transcriptions in all of the following music examples adhere to her sources, which, in some cases, may transmit problematic harmonies, rhythms, or text underlay. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 258 Irene Alm Ba~e ,-•I , f m r .• - dr op Ir OR " •.• • ! I 1ir' I I I W. L4 I p . v .. . .P I IF .... t • 'I Ir :--- , f lip- •-,•~~~~~~ .. . .L= - ,I1 t op 7 -'• ? r .. . " r ' " : ; "OR .. . . kf 6wo I O-P• " ....r r' I" r I r -- ? .. OP •:•,• r•I r r OP J • rof I Ex. 5: ]Eliogabalo (1668), Act III, scene 5 (I-Vnm, Cod. It IV 413 (=9937), fol. 78r). This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 259 Ballo [follows Sibilla's aria 'a la Francese'] A-4 Ex. 6: Amulio e Numitore (1689), end of Act I (A-Wn, Mus. Hs. 17689 [no folio numbers]), bars 1-4. Cymbals or percussion may have been added to create a Moorish flavour, although they are not indicated in the score. Turkish instruments are mentioned in the directions for some of these dances - see IC Gran Tamerlano (1689)120 and Irene (1695)121 - but unfortunately the score for the first is lost and the score for Irene does not include music for the dances. Music associated with comedy or madness could also be highly idiosyncratic, inspiring what might have been a parody of dance steps. Cavalli's Pompeo Magno (1666) includes an entire suite danced by eight madmen: two for music, two for painting, two for alchemy, and two for poetry - each pair being characterized with its own short binary piece in a contrasting style. (See Ex. 4.) In the 'Burla tri Giardinieri e Buffoni di Corte' in Act III of Eliogabalo (1668) the stop-and-start rhythms offset the regular four-bar phrases, and the dotted rhythms seem to add a touch of mock courtliness.122 (See Ex. 5.) Among the dances with social dance titles that functioned as balli, some seem to have been chosen for their comic effect. The brief corrente in Cavalli's second opera, Gli amori d'Apollo e di Dafne (Act I, scene 2), has a conventional profile - compound metre (ci), binary form, and cadences marked by hemiolas - but the intent may well have been comic, in that it immediately follows a short aria in which the old woman, Cirella, sings of moving slowly with shaky steps.123 Comedy may also have been the inspiration for gighe, such as one found in Pietro Andrea Ziani's L'Antigona delusa d'Alceste.124 It occurs in the midst of the closing recitative of Act II, sung by the stuttering comic character Lesbo. The libretto describes a 'ballo for cavaliers of various nations with Lesbo amidst them', and the stage directions state, 'Here 120 Text by G. C. Corradi and music by Marc'Antonio Ziani. 121 Text by G. Frigimelica Roberti and music by C. F. Pollarolo. 122 Ed. note: Except for the regular four-bar phrases, this piece has the rhythmic profile of a French-style courante, and perhaps was intended as a reference to courtly French ballrooms. 123 See the facsimile score to Francesco Cavalli, Gli amori d'Apollo e di Dafne, ed. Howard Mayer Brown (New York, 1978), 17v. 124 This opera was revived at S. Salvatore in 1670, having first been performed in 1660 at SS. Giovanni e Paolo under somewhat hasty circumstances, with borrowed balli. A number of changes were made for the revival in 1670, including new balli. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 260 Irene Alm [Ersistrato] [Eurindo] Mi- raquan- e bel- le - Lo [libretto: Antioco sta sedendo] ....-..- K s WN - --m • . . . sguar- do si con - fon - de nel Con - i 1 nuo pas sag gio di Splen dor in sp-e-- dor di Ex. 7: Seleuco (1666), Act II, scene 18 (I-Vnm, Cod. It IV 454 (= 9978), fols. 68'71r), bars 1-17. (Ersistrato: Look, so many beautiful women! Eurindo: The eye gets confused by the continuous passing of splendour after splendour, of gleam after gleam ...) This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 261 rag- gi in rag- gio di splen dor in splen dor oooI r op4 rag- . , 1. - gio di rag- rag - go m rag - gio di splen dor in splen dor di: .' " r If r1 rag gio di rag - gio in rag- gio op =PIE Ex. 7: continued. people of various nations come out, who introduce the ballo, beginning to disagree among themselves on account of some tokens of love".125 Lesbo interrupts them with much stuttering, and urges them to dance with him; this ballo may have been labelled giga because it was intended as a parody of a popular dance, as appropriate to the comic situation. The use of standard court dances later in the century is particularly evident in works that consciously invoke the French style. Tosi's Amulio e Numitore (1689), for example, includes three dances in triple metre (1) with the two-bar phrasing characteristic of the minuet (despite the time signature, they are barred every six beats). The French connection is even explicit, as the first of these directly borrows the music from the aria ' la Francese' that precedes it. (See Ex. 6.) The majority of balli using social dance titles appear in ballroom scenes - the staged versions of the Venetian feste di hallo that were held both on and off the operatic stage during Carnival. In the first libretto to contain a staged Venetian-style festa di ballo, Seleuco (1666), Sartorio re-created the sound of the dance band by composing an eight-and-a-half-bar instrumental ground. The extra half bar causes 125 See the prefatory pages and Act II, scene 22. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 262 Irene Aim Rigadon [Minuet] Da si - nI ni - stra i1 Ciel ba - le- - - na piD O ri - den - te ap - pa - re ii di 40• " " " " L ,va.k ,U . -slylyJ I J Ex. 8: Neronefatto Cesare (1693), Act III, scene 16 (D-SWI, Mus. 4189, 294-5). First strains of the rigaudon and the sung minuet. the pattern to shift within the bar, and there are nearly seven full repetitions of this instrumental music. The rhythmic pattern of the bass line establishes the leisurely pace of the promenade. Against this music, Sartorio composed a variety of vocal lines, suited to the different characters attending this ball. Overall the music provides two levels of awareness: the large picture of the ongoing dance and the close-ups of various characters on stage - a cinematic effect of focusing on the whole as well as parts of the scene, cutting back and forth between the two. (See Ex. 7.) Towards the end of the century, the scenes with continuous duple-metre instrumental music supporting sung conversations disappear and are supplanted by French figured dances, with more rigid structures and generic celebratory texts sung by soloists or the entire ensemble. Pollarolo's Onorio in Roma (1692) embodies this This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 263 moment of change in Venetian social dance via its two lengthy social dances. The first, in Act II, scene 9, is a passeggio, or promenade, in duple metre. The voices and instruments are used in a concertato style, however, and the duet text is a simple expression of rejoicing, not an intimate conversation. The finale calls for a 'gran Danza Francese' and the score contains a lengthy ciaccona, or French chaconne, in triple metre. Pollarolo skilfully alternates the different vocal and instrumental ensembles; the text is celebratory and neither the libretto nor the musical setting suggest conversation.126 Other examples of French social dances appear more and more frequently in the scores for ballroom scenes during the 1690s. For example, the ballofigurato that ends Furio Camillo (1692) is clearly a minuet. A 'Borea' (bourree) is paired with a balletto in Act III, scene 1 of Neronefatto Cesare (1693) and the finale of that opera employs a 'Rigadon' (rigaudon), which alternates with two strophes of an aria based on the minuet. (See Ex. 8.) Vocal dance music The same question raised with instrumental music - that is, 'which pieces were danced?' - can be asked about many vocal pieces, both choral and solo. Did a vocal piece introduce the dance, was the dance performed to the ritornello, or was the piece sung and danced simultaneously? The only instances in which it is certain that vocal music accompanied dance are those in which either the score or the libretto specifically states that a chorus is danced and sung, or that an aria accompanies the dance. In addition, the special circumstances of the festa di ballo, or ballroom, scenes often involved sung conversation while the instrumental dance music continues in the background. Seventeenth-century writers disagreed as to whether choruses should (or could) sing and dance simultaneously. Based on his interpretation of Greek practice, Cavalieri recommended that a final, 'formal' ballo should be sung and played by the dancers, but Doni strongly disagreed with this notion, and stated that Greeks never sang and danced simultaneously, thus expressing a more practical approach to performance.127 A letter from Monteverdi (6 January 1617), concerning his Le nozze di Tetide, specifically refers to separate groups of singers and dancers.128 Indeed, relatively few choruses are specified as dances in Venetian libretti or scores, and the 126 Ed. note: These two pieces may be seen as Examples 79 and 80 in Alm, 'Theatrical Dance', 506-20. 127 Doni, Trattato della musica secnica, in Lyra Barberina II, 115-17. See Alm, 'Theatrical Dance', 25-27, and 'Humanism' (n.46). 128 'And if at the same time you accommodate to a dance measure the lines which the Nereids have to sing (to the tempo of which you could make expert dancers dance gracefully), it seems to me that it would be a much more suitable thing.' Trans. Denis Stevens, The Letters of Claudio Monteverdi (Cambridge, 1980), 126. Monteverdi discusses his approach to composing for dance in a number of other letters as well. See ibid., 46-47, 106-09, 115-18, 140-41, and 167. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 264 Irene Alm stage directions often imply that the ballo is danced to verses sung by someone other than the dancers.129 Glover has proposed that the singers hired for the chorus would have also danced the balli at the close of the acts with instrumental accompaniment: With the increasingly spectacular historical librettos in the 1660s, choruses were once more adopted. Yet it would seem that their chief contribution was visual rather than musical. Choral music was generally short and simple, whether in six parts or two, and the chorus was probably intended also to dance the balli at the end of the first two acts and to execute any formal movement during the opera.130 Account books from the 1660s, however, record payments for dancers, and it seems almost certain that in Venice the dancers and singers were separate groups. Whereas the use of the chorus waxes and wanes over the course of several decades, the balli persist, and the number of dancers seems to have remained constant. It thus seems unwarranted to assume that chorus members would have doubled as dancers. Table 4 (see Appendix) provides a detailed list of the thirty-nine balli with vocal accompaniments, solo, ensemble, and choral. On average the vocal pieces appear to be twice as long as the instrumental dance pieces; this may simply be a result of the fact that there was less need to write out the instrumental dance music in the score of the opera, or to indicate repeated sections. Whereas dances with instrumental accompaniments could be kept in separate scores, with brief cues copied into the full score, those with vocal accompaniments needed to be rehearsed with the singers, and thus were necessarily included in the full score. The vocal pieces range in length from nine to 174 bars. In some of these, of course, dancing alternated with singing, and then combined for a grand finale. For the most part vocal dance pieces favour triple metre or contrasting sections of both duple and triple metre. The notable exceptions to this are the ballroom scenes, which as we have seen above, often employ duple metre for the promenades typical of the Venetian festa di ballo, since the dancing in these scenes is more closely tied to the instrumental accompaniment than to the conversation sung by the soloists. The earlier choruses do not have instrumental accompaniment (some even lack continuo parts), although instruments may have doubled the vocal lines. Only in the lavish ensembles by Pollarolo in the 1690s do full complements of instruments ('tutti gl'istromenti') join the chorus. Danced choral music is most often celebra- tory, and characteristically relies on simple homophonic textures and short repeated phrases of text. This is equally true in the early choruses by Cavalli and the large-scale scenes for chorus and instruments by Pollarolo from the 1690s. Many choruses not specified as dances also use this style, such as a chorus of hunters singing 'alla caccia' or a crowd singing 'viva, viva'. These may have been accompanied by some simple gestures and stage movement, but unless a ballo is 129 See, for instance, Act II, scene 11 of Cavalli's Le nozze di Teti e di Peleo (1639). The stage directions in the score state 'Ballo sung by Bacco and by Sileno and answered by the chorus of gods; Dance by fauns and bacchantes'. 130 Jane Glover, 'The Peak Period of Venetian Public Opera: The 1650s', Proceedings of the Royal MusicalAssociation, 102 (1975-76), 73. See also her dissertation, 'The Teatro Sant'Apollinare' (n.75), 118. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 265 indicated immediately after, or in the vicinity of one of these choruses (as in the final scene of Act I in Cavalli's Elena), it probably was not accompanied by dancing. Cavalli's Le nozze di Teti e di Peleo (1639) contains two long dance scenes that involve both solo and choral singing. In Act I, Mercurio summons four groups of spirits to celebrate. The scenario states: [Act I] Scene Seven Mercurio descends from heaven, followed by Momo, who reveals to Peleo the will of Giove. Peleo, not content with exalting the heavenly graces, asks the plants, the stones, the breezes and the waves themselves to express the praises of Giove; Mercurio joins these spirits, who accompany the joy of Peleo with dances in Scene Eight The trees peel back their bark, the rocks open up, and two Dryads and two Oreads come out to dance; then rising from the sea, and flying from the heavens, come two Nereids and two Aure, who together form a ballo cantato; at the end each god takes with him a pair of nymphs, and the dance ends ...131 Cavalli's score emphasizes the central role of the dances in this scene, which unites music and dance in an unusually complex and intricate fashion. Mercurio invokes each pair of spirits, exhorting them to dance, and they answer with two stanzas of song. Each of his invitations to dance is set to different music; the first and third are in triple metre, while the second and fourth are in recitative. The stanzas for the spirits are each sung to the same music in a graceful triple metre and are marked as balli.132 For the chorus of all the spirits, Cavalli set the four stanzas of text as a lively duet in duple metre (with two soprano parts left blank) and added the stage direction 'Qui ballano tutte insieme' (Here they all dance together). At the end of the scene, the verses sung by Momo, Meleagro, Mercurio, and Peleo alternate with an instrumental ciaccona, suggesting that this entire section would have been danced for the exit of the spirits. (See Ex. 9.) The other extended dance scene in Le noZZe di Teti e di Peleo is a bacchanalian celebration in Act II, scene 11. The scenario states: 'Bacco and Sileno join with a chorus of fauns and another of Bacchantes; here Bacco and Sileno praise the virtues of wine, and the choruses dance to their melody'.133 Directions in the score suggest that even the end of the scene was danced: 'Ballo sung by Bacco and by Sileno, and answered by the chorus of gods / Danced by fauns and Bacchantes'.134 The length of the scene and variety of music offer the opportunity for a sophisticated choreography to match the changing textures. The music includes several solos, a duet for Sileno and Bacco, two different ritornelli, and a six-voice chorus of gods. 131 Breve espositione della festa teatrale, 11. 132 The same opening stanza of verse is used by each pair, with slight changes to reflect their domains - the woods for the Driads, the rocks for the Oreads, the sea for the Nereids, and the heavens for the Aure. Although only a bass line is given, there are blank staves with clefs for three upper voices, two soprano and one alto. 133 'Giungono Bacco e Sileno con un Coro di Fauni & un'altro di Baccanti; qui Bacco, e Sileno commendano la virtii del vino, & i Cori danzano alla lor melodia.' 134 'Ballo cantato da Bacco e da Sileno e risposto dal Coro degli Dei / Ballato da fauni e da Baccanti.' This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 266 Irene Alm Mercurio Sii dunque in lie - te e ful - gi- de sem - bian - ze sor - ge - te, 6 L I Di - ve di fron zu - - te scor - ze, e lo- dan - do d'A - mor l'armi, e le L J L I for - ze, ac - comrn - pa- gnate al suon con - cen - ti e dan- ze. Lf Ballo di Driadi A -I II Si6 fes- to - so lie - to spo - so, go- da teco il no - stro bos - co, non sia fe - ra si se - ve - ra, che non -N.." I I I,,I-- •-•, , las - si l'ira e'l tos - co. . [libretto has a second strophe] Mercurio Figli or di Nin - fe o- gn'in- sen- sa- to sas- so, al sel- vag- gio drap pel Co- ro con cor- de ch'al- la dol- cez- za di temrn pra- te cor - de, unis-ca il la - bro 6 non dis-codi il pas so. Ballo di Oreadi [=repeat of Ballo di Driadi] Ex. 9: Le nozze di Teti e di Peleo (1639), Act I, scene 6 (I-Vnm, Cod. It IV 365 (= 9989), fols. 35'39V). Excerpts. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 267 Mercurio Sti fuor de salsi e li - qui- di cris- tal - Ii ven - ganl'on - do se L J r • ver - gi- ni stil - lan - ti ch'ar- ti - fi - ciosi ar- ti-co lando i Can - ti for - min fes- tivi e re-go -la - ti bal - i. Ballo di Ninfe maritime [=repeat of Ballo di Driadi] Mercurio E voi fig- lie del Ciel suo-re de Ven- ti Ven- ti la- te leg- gie- re il pie vo- lan- te mi- su- ran- do al suon le snel- le pian- te cre sce-te al gran Pe leo gio - - - ie e con- ten - ti. o , I " Ballo di Aure [=repeat of Ballo di Driadi] 'Qui ballano tutte insieme' [The Driadi, Oreadi, Ninfe, and Aure dance together] San pig- liar sensi et ef - fet - ti per go- der a tuoi con-ten- ti fino i sas - si in- a - ni - Son fes- tan- ti son ri den- ti per gio- ir a tuoi di - let- ti, fi-no gl'ar- bo - ri in- sen- sa - ti ma- ti, Son fes - tan - ti son ri- den- ti per gio - ir a tuoi di - let- ti, fi - no gl'ar- bo - ri in - sen - sa - ti Ex. 9: continued. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 268 Irene Alm Meleagro - - (I) Con le ser - - - ve di fa - vo - nio dun - que an - di - nan - L J e, A ban - dir ii ma - tri - rmo - nio, dal mar indo all' on- de ma re. Sas - si, pian l - - te, et on - - de,et au - - - re L J Ciaccona I.m :,,,a .• F F F.I / .I ! .1 ,, Ex. 9: continued. The predominant metre is triple, with typical hemiola patterns, and is interrupted by a few sections in duple metre, creating the shifting rhythmic patterns common in late Renaissance balli. In Cavalli's Gli amori d'Apollo e di Dafne (1640), Act I, scene 4 contains choral dance music for nymphs and shepherds that alternates with a solo aria by Dafne; only the choral music accompanies the ballo, as annotations in the score make clear.135 In another instance, the alternation between chorus and solo might have been used to comic effect: In Alessandro vincitor di se stesso, there is a ballo for hunchbacks introduced by a solo sung by the stuttering Bleso. After beginning in duple metre, Bleso shifts to triple, calling to his companions to join him - which might well have been the cue for the beginning of the dance. The chorus then joins Bleso with music using nearly the same bass line and the same pattern of hemiolas. In the midst of the dancing, Bleso's comic stuttering is heard, an element that might well have been emphasized in the choreography. (See Ex. 10.) 135 See the facsimile score of Cavalli, Gli amori d'Apollo. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 269 Momo (2) A lodar belti, che splende, Dunque andianne, Dove il Sole i Campi ascende; Dov'il mar l'acque diffonde, Piante e sassi, et aure et onde. Ciaccona Mercurio (3) A bandir le regie feste Dunque andianne, Dal paese almo, e celeste Fino ai Regni oscuri, e bassi, Aure, et onde, e piante, e sassi. Ciaccona Pelio (4) A portar le Gioie mie Dunque andianne, Da i confini ond'esce il die Fino a termini d'Atlante, Onde et aure, e sassi, e pianti. Ciaccona Ex. 9: continued. In addition, solo arias with instruments or ritornelli also accompanied some balli. The score for L'Argia (1669), for example, indicates that the aria at the end of Act I is for Alceo and a chorus of slaves who dance. The instruments and voices alternate in concertato style, and the choreography may have also reflected this alternation. Another such example is Eudemo's aria with instruments 'Compagni correte' in Act I, scene 16 of La caduta di Elio Seiano (1667). Here, the running eighth notes used in both the vocal and instrumental parts, might have accompanied dance, or the instrumental part may have been extracted and repeated for the ballet. (See Ex. 11.) If a ritornello is to be danced, there will often be an indication in the score. In Cesti's L'Orontea (1666) the ritornello of Gelone's aria in Act I, scene 13 is marked 'balla' ('he dances'), and the ritornello for Euridice's aria 'Ninfe danzate' in L'Orfeo (1673), is titled 'Balletto per la Danza' in two of the three scores, although the libretto does not specify a ballo here. In L'amazaone corsare, overo L'Alvilda regina de Goti (1686) an aria and ritornello in Act I serves for three dances. In scene 6, the aria introduces a fencing lesson which is 'danced' to the ritornello; scene 7 indicates a return to fencing when the ritornello is repeated. The same aria returns later in scene 7 with a new text for Gilda's dancing lesson, and she dances to the vitornello. However, the style seems more suited to the fencing lesson, with repeated quavers and pauses evoking the swordplay. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 270 Irene Alm Si suo - ni si can - ti si bal - s si si suo - m si Si so - s ca - ca- ca- Si suo - ni si can - ti si bal - ii si si si suo- ni si can - ti si bal - li si si, al - Ile - grie fe - stan - ti pas sia - mo - ci ii si can - ti bal - li si si al - le - gri e fe - stan - ti pas - sia -mo-ciil al - le - grie fe - stan - ti pas - sia - mo - ciil di Ca - C a- C a- Si di al - le - grie fe - stan - ti pas - sia - mo - ciil di. si suo - ni si suo - ni i ca - ca- ca - i su - ni si si si suo - ni si Ex. 10: Cesti, Alessandro vincitor di se stesso (1651), Act I, scene 12 (I-Rvat, Chigi Q.V.61, fols. 68v-72v). Arias labelled as introductions to balli did not necessarily provide music for the dance; in fact, many of them seem unsuitable for balli, but must have been followed by instrumental music that was not copied into the score. The final arias of Acts I and II of L'Orfeo (1673) and those in L'Almerico in Cpro (1675) are also not characteristic of dance music, although all of these are clearly meant to introduce the balli. The inclusion in scores of arias that introduce balli is yet another indication that the dances were indeed performed and not an optional part of the production. The lengthy dance scenes by Pollarolo from the end of the century extend this principle of alternation between chorus, solo voice, and instruments. Many of these scenes consist of linked sections of accompanied arias, choruses, and instrumental music, and the larger structure is built of the alternation of two or three small This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 271 Com- pag- ni " cor- re - - - te IF - ri cor-re - te la vec - chia ve - de- te che fmi- ge co- Ex. 11: La caduta di Elio Seiano (1667), Act I, scene 16 (I-Vnm, Cod. It IV 397 (= 9921), fol. 31". The continuo part, doubled by the bass throughout much of the example, is omitted. sections. The few independent instrumental pieces for balli are often thematically connected to the vocal music that precedes or follows, and in many scenes the dance clearly continues through both instrumental and vocal sections. Often the entire ensemble of soloists, chorus, and instruments combines for the final section of the dance. This type of choral accompaniment for balli may be seen as evidence of neo-classical trends, French influence, or both.136 136 Ed. note: This type of integrated structure is typical of the divertissements Jean-Baptiste Lully composed into his operas; see Rebecca Harris-Warrick, 'Recovering the Lullian Divertissement', in Dance and Music in French Baroque Theatre: Sources and Interpretations, ed. Sarah McCleave (London, 1998), 55-80. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 272 Irene Alm Conclusion The surviving dance music, both vocal and instrumental, represents only a very small portion of the repertoire, but it is clear even from this small sample that Venetian theatrical dance was a thriving art during the seventeenth century. The dazzling array of subjects found in the libretti resulted in equally diverse musical styles, and what would seem to have been a highly athletic and varied style of dance. Unlike the French stage, where social and theatrical dance almost seamlessly merged, the Venetian opera houses presented a style of dance that was entirely the domain of professional dancers and choreographers. The imagination of these artists seems limitless; the dances and choreographers of Venice created a kinaesthetic world that matched the eccentric and highly idiosyncratic nature of the operas themselves. Indeed, this is certainly the sentiment expressed in 1688 by Cristoforo Ivanovich, who saw the developing opera industry as a reflection of the Republic's own increasing perfection: From here it arose that Carnival became rather more amazing than it was in the past, each year all types and a considerable quantity of strangers converging to enjoy such a delightful entertainment, and seeing the most sublime talents embodied in a virtuoso, the same in poetry as in music, the most exquisite voices of men and women chosen, and the most extraordinary creations found in costumes, sets, machines, 'flights', and balli.137 While we may not be able to re-create Ivanovich's experience in the Venetian opera theatres, it is indeed time for historians and opera producers to acknowledge that the balli were an integral element in the spectacle that was Venetian opera. Tragedy and comedy, pathos and satire - expressed through a dazzling array of subjects in the hundreds of operas produced in this period - not only found expression through the singing voice, but also through the mute eloquence of the many dancers who graced the Venetian opera stage. Appendix Theatre abbreviations used in tables SC S. Cassiano Nov Novissimo SSGP SS. Giovanni e Paolo SSAp SS. Apostoli SM S. Moise SApol S. Apollinare SSalv S. Salvatore (S. Luca) aiSal ai Saloni SAng S. Angelo Cr Cannaregio SGG S. Giovanni Grisostomo SF S. Fantino (which took over the 1699 production from Cr). 137 Cristoforo Ivanovich, Minerva al tavolino, 392. This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Winged feet and mute eloquence 273 Table 1. Venetian theatres: number of balli per opera Year SC SSGP SM SSalv SAng SGG Nov SSAp SApol aiSal Cr 1637 3 1638 3 1639 4 2-0 1640 3 2-1 0-0 1641 1 4-1 2 2 1642 1 1-0 3-3 0-1 1643 0 0-2 3 1644 0 0-4 2 1645 0-0 0-2 0 1646 - 1647 - 2 (May) - 1648 0 2 0 1649 2 0 2 0 1650 3 2 0 1651 0 2 0 0-2 1652 - 4 -0 2-2 1653 - 0-2 - - 0 1654 - 2-2 2 0 1655 - 1 -- 2 1656 - 2 2 1657 - - - 2 1658 2 2 - 2 1659 3 2 1660 2 2-2 - 2 1661 - 2-2 1-2 1662 - 1-2 2 1663 - 2-2 2 1664 - 2-3 2 1665 3-2 2 1666 2-2 2-2 2-2 2-3 1667 - 2-2 2 2-2 1668 - 2-2 - 2-2 1669 3-2 - 2 1670 - 2-2 2-0 2 1671 - 2-2-2 2-2 2 1672 - 2-2 - 2-2 1673 - 2 3 2-2 1674 2-2 2 1675 - 2-2 2 2-2 2 1676 - 3 2-2 2-3 1677 - 3-2 2-2 2-2 2 1678 - 3-2 2-2 2-2-2 3 1679 - 2 3-2 2-2 2 0 1680 2-2 2 2P 3 2 3-3 1 1681 - 2-? 1p 2 2-5 3-2 1682 - 0-? OP-2p 0-1 3-2 ?-3 - 1683 0-1 4-2-2 - 1-2 2-1 3 2-1 1684 - 1 2-2 1-0 3-2 1685 - 2 0-2 2-0 2-0-3 0 1686 - 2-3 1-2 0-0 1-0 6 - 1687 - 2 2-2 0 2-2 3-3 - - - 1688 - 0-3 2-1 1 2 3-2 0 1689 - 4 2-3 2-2 2 3 1690 - 3 - 2 2 3 2 1691 2-0 2-0 1-2 2-2 2-2 1692 - 2-0-2 - 0-3 - 1-3 1693 - 3-3 2 3 0-2 3 1694 - 2 - 3-6 3-2 5 - 1695 - 2 - 4 2-2 2-5 This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions 274 Irene Alm Table 1. continued Year SC SSGP SM SSalv SAng SGG Nov SSAp SApol aiSal Cr 1696 2-4 2-2-3 5-1 3-3 5-4 1697 2-2 3-0-2 4-3-2 2-2 1-1 1698 4-3 3 - 2-2 3-1 1699 2-2-0 3-2 2-3-3 2 3-2 1700 2-3-3 1-? 2-2 2 Cr-SF This table indicates all operas performed in Venice, by season and theatre, according to how many balli they contain. Thus a season that offered 3 operas, the first containing 2 balli, the second none, and the third 2, would be listed as 2-0-2. A season with a single opera containing 3 balli is listed as 3; if the single opera performed had no balli, it is listed as 0. -Indicates that a theatre did not present operas that season. ? Means balli were indicated, but their number and location was not specified. Indicates five-act operas. "Indicates puppet operas. Years indicate the carnival season; autumn productions are counted with the following carnival. Thus an opera that opened in November 1678 is listed under 1679. Table 2. Frequency of balli: percentages by decade Years Number of operas Operas with balli Percentage 1637-1640 10 7 70% 1641-1650 36 21 58.3% 1651-1660 30 23 76.7% 1661-1670 41 40 97.6% 1671-1680 59 58 98.3% 1681-1690 85 70 82.3% 1691-1700 85 78 91.8% Total 1637-1700 346 297 85.8% This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Table 3. Instrumental dance music Year Theatre Opera Composer Ballo location Ballo title Tex 1639 SC Le nozze di Teti e di Peleo Cavalli end of Act I Corrente Grave Tr; Corrente Grave 1639 SC Le nozge di Teti e di Peleo Cavalli Act II scene 7 Ballo [di Coro di Centauri] 2Tr 1639 SC Le nozze di Teti e di Peleo Cavalli end of Act II Corrente B [2 1640 SC Gli amori d'Apollo e di Cavalli Prologue Ballo de fantasmi 2S; Dafne 1640 SC Gli amori d'Apollo e di Cavalli Act I scene 2 Corrente 2Tr Dafne 1642 SC La virtd de' strali d'amore Cavalli end of Act I Ballo delle Maghe [Ballo di Tr; varii spiriti] 1654 SSGP Xerse Cavalli end of Act I Ballo [Combattimento] B 1654 SSGP Xerse Cavalli end of Act II Entrata 2" / Balletto B [Combattimento] 1654 SSGP Xerse Cavalli end of Act II Ballo 2d [Combattimento] B 1661 SSGP L'Annibal in Capua P.A. Ziani end of Act II Balletto 2Tr 1663 SSalv La Dosi Cesti end of Act I Ballo d'Eunuchi 2Tr 1664 SSGP Scpione africano Cavalli Act I scene 2 Sinfonia [Gioco de' 2Tr Gladiatori] 1666 SM II Demetrio Pallavicino end of Act I Giga [Restano li paggi 2Tr insolenti, e formano il ballo] 1666 SSalv Pompeo Magno Cavalli Act I scene 1 Balletto de Cavalli [Ballo di 2Tr Quattro cavalli naturali vivi] 1666 SSalv Pompeo Magno Cavalli end of Act I Segue il Ballo di 8 impazziti: 2T 2 per la musica; 2 per la pittura; 2 per alchimia; 2 per la poesia / Balletto de Pazzi 1666 SSalv Pompeo Magno Cavalli end of Act II Ballo di Fantasme [Ballo di 2S 12 Ombre] 1668 SSGP Eliogabalo Boretti Act II scene 5 Ballo [Burla tri Giardinieri e 2T Buffoni di Corte] 1669 SSGP II Genserico Partenio end of Act II [Ballo] B 1669 SSalv L'Argia Cesti Act II scene 20 [Ballo di Fantasmi] 2T 1670 SSGP L'Antgona delusa d'Alceste P.A. Ziani end of Act I Ballo [di Pastorelle e di B Cacciatori] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Table 3. continued Year Theatre Opera Composer Ballo location Ba//llo title Te 1670 SSGP L'Antigona de/usa d'l/ceste P.A. Ziani end of Act 11 Gighe [Ballo d'Artegiani] B 1671 SSalv L'Ercole in 7Tbe Boretti Act II scene 19 Qui si fa il ballo / Ritornello B [Ballo di Eroi] 1673 SSalv Orfeo Sartorio Act I scene 1 Ritornello [A-Wn: Balletto 2T per la Danza] 1674 SM La schiava.f/rtunata M.A. Ziani end of Act I Ballo [di Guerrieri] B 1674 SM La schiavafoirtunata M.A. Ziani end of Act II Ballo [di Naiadi, ch'escono B dal Tigri] 1675 SSGP Diocletiano Pallavicino end of Act I Ballo de Paggi e de Pazzi Tr 1675 SSGP Diocletiano Pallavicino end of Act 11 Ballo de Persiani [Stuolo di Tr persiani prigionieri con le catene al piede] 1676 SSalv Germanico sul Reno Legrenzi end of Act I Baletto [Ballo di Soldati 2T Belgici] 1676 SSalv Germanico sul Reno Legrenzi Act II scene 8 Baletto di combatim. [gioco] 2T 1676 SSalv Germanico sul Reno Legrenzi end of Act II Balletto [Ballo di Fantasmi e 2T Spiriti] 1678 SGG II tVespesiano Pallavicino finale [Ballo di Muse in terra, di 2T Ninfe in acqua, e d'Amorini in aria] 1680 SGG II Vespasiano Pallavicino finale Ballo [di Personaggi che 2T figurano la Terra, Acqua, Aria ed il Foco] 1680 SC Candaule P.A. Ziani Act II scene 18 Balletto [Ballo di Ninfe] 2T (I-Vnm) 1680 SC Candaule P.A. Ziani Act II scene 18 Ballo per il Cigno [Ballo di B (D-AN) Ninfe] 1680 SSGP L'Alcibiade M.A. Ziani end of Act II Ballo [di Scultori Discepoli di 2T Prassitele] 1680 SGG Il ratto delle Sabine Augustini end of Act II Ballo [di dodeci Romani B Rattori] 1681 SA Pompeo Magno in Cilicia Freschi end of Act I Ballo [di quattro Mori e B quattro Nani Spagnoli] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Table 3. continued Year Theatre Opera Composer Ballo location Ballo title Tex 1681 SA Pompeo Magno in Cilicia Freschi end of Act II Ballo [di quattro Paggi al B levar d'una Mensa Reale] 1685 SM Rodoaldo re d'Italia Gabrielli end of Act I Ballo B 1685 SM Rodoaldo re d'Italia Gabrielli end of Act II B1llo d'Armi B 1685 SM Clearco in Negroponte Gabrielli end of Act I Ballo [di Pittor in forma B d'Academia] 1685 SM Clearco in Negroponte Gabrielli end of Act II Ballo [di Soldati coronati B d'Ulivo] 1686 SSGP L'amazzone corsara, overo Pallavicino Act I scene 12 and Sonata con tromba [Ballo di 2T L'Alvilda regina de Goti scene 15 Popolo Festante; varij giochi a suono di Trombe] 1686 SSalv Le generose gare tra Cesare e Gabrielli end of Act II Ballo B Pompeo 1689 SGG Amulio e Numitore Tosi end of Act I Ballo [d'Amorini in aria, di 2T Ninfe in terra, e di Mostri nel mare] 1689 SGG Amulio e Numitore Tosi Act III scene 3 Ballo di Paggi 2T 1689 SGG Amulio e Numitore Tosi Act III scene 3 Ballo d'altri Paggi e Damigelle 2T 1689 SGG Amulio e Numitore Tosi Act III scene 3 Danza di Dame e Cavalieri 2T 1689 SGG Amulio e Numitore Tosi finale 'Ballo' [Grandanza di Dame e 2T Cavaglieri] 1692 SSalv Furio Camillo Perti Act II scene 1 Ballo [di Alfieri] B 1693 SSalv Neronefatto Cesare Perti Act III scene 1 Boria / segue il Balletto 2T doppo il quale si replica la Boria 1693 SSalv Neronefatto Cesare Perti Act III scene 16 Rigadon [Ballo] Tr 1693 SGG Laforza della virt~i Pollarolo Act II scene 5 Introduttione al Ballo/Ballo Tr Sinfonia Tr [Ballo di Greci e d'Amazoni] A; 1693 SGG Laforga della virtsi Pollarolo Act II scene 14 Ballo / Ciaccona [Ballo di 2T Dame e Cavalieri Spagnoli] un 1694 SSalv Alfonso primo Pollarolo Act I scene 2 Sinfonia [Ballo di Amorini] 2T 1694 SSalv Alfonso primo Pollarolo before Act II [Danza] [Ballo di Araldi 2T dell'Alba] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Table 4. Vocal dance music Year Theatre Opera Composer Ballo location Ba//llo title Te 1639 SC Le nozze di Teti e di Peleo Cavalli Act I scene 6 Ballo di Driadi B j Ballo di ()readi Ballo di Ninfe Maritime Ballo di Aure Qui Ballano tutte insieme A; Ciaccona A;B 1639 SC Le nosze di Teti e di Peleo Cavalli Act II scene 11 Ballo cantato da Bacco e da B;B Silenot, e risposto dal Coro 2T degli Dei; Ballato da fauni e 2T da Baccanti B;B 2S; 1639 Le nozze di Teti e di Peleo Cavalli Act III scene 9 Coro d'Amorini 2S; Venere [Ballo di Coro S;B d'Amoretti] 1640 SC Gli amori d'Apollo e di Cavalli Act I scene 4 Choro [Ballo di Ninfe e S; A Dafne Pastorelli] 1649 SC Giasone Cavalli Act I scene 14 Choro di spirit A 4 [Ballo di A; spiriti] 1651 SSGP Alessandro vincitor di se stesso Cesti Act I scene 12 Bleso; [Choro] A 4 [Ballo de' T; B Gobbi Discepoli d'Apelle] S; A 1659 SC Elena Cavalli Act I scene 16 Cacciatori A; 2 Ballo con gl'Orsi [Li A; Cacciatori prendono gl'Orsi e ballano] 1662 SSGP Lefatiche d'Ercole per P.A. Ziani Act II scene 22 Allegrezza; Armonia; Diletto 2S; Deianira [Ballo di Spiriti Beati] 1665 SSGP Ciro Cavalli Act I scene 17 Fatama; Choro [Ballo di Mori A; B Etioppi] 1666 SSGP Orontea Cesti Act I scene 13 Gelone solo 2Tr 1666 SSalv Seleuco Sartorio Act II scene 18 Festa di ballo [Ballo di Dame solo e Cavallieri] role 1667 SSalv La caduta di Elio Seiano Sartorio Act I scene 16 Eudemo [Giardinieri e Paggi solo fanno un Ballo] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Table 4. continued Year Theatre Opera Composer Ballo location Ballo title Tex 1669 SSalv L'Argia Cesti Act I scene 16 Alceo; Choro di Schiavi che sol ballano [Ballo di Schiavi] 2B 1671 aiSal Iphide greca Partenio Act I scene 12 Choro di Popolo; Ligdo; S; A Teletusia; Iphide [Giuochi sol d'Armi in forma di Ballo] 1673 SSalv Massenzio Sartorio Act II scene 26 Choro de Marinari che fanno sol il ballo [Ballo di Marinari 2T Africani] 1676 SSGP Galieno Pallavicino Act I scene 2-6 [Danza di Galieno e Fulvia, sol Cavalieri e Dame] 1679 SSGP Alessandro Magno in Sidone M.A. Ziani Act II scene 4 Alessandro; Eusonia [Danza] T; 2T 1680 SGG 1P ratto delle Sabine Augustini Act 1 scene 21 ritornello [Ballo] 2T aria sol 1681 SA Pompeo Magno in Cilicia Freschi Act I scene 1 Ballo B Alimene; Gemmira sol [Ballo ... Danza] 1682 SA Olimpia vendicata Freschi Act III scene 15 [Danza] voi 1683 SSalv Giustino Legrenzi Act I scene 15 Allegrezza [in machina guida'l sol Ballo de Cavalieri e Dame] + 2 1685 SM Massimo Puppieno Pallavicino Act III scene 2 [Ballo di seguaci S; B dell'Allegrezza] 2T 1686 SSGP L'amazzone corsara, overo Pallavicino Act I scene 6 Olmiro 'Con la scherma' S; B L'Alvilda regina de Goti Act I scene 7 Giocano di spada 2T Act I scene 7 Novo giocano di spada / B ritornello ut supra S; B Gilde 'Con la danza' (aria to same music) 1692 SSalv Furio Camillo Perti finale [Ballo figurato] sol 2 T 1692 SGG Onorio in Roma Pollarolo Act II scene 9 Tutto il concerto [Ballo di voi Dame e Cavalieri] 2B 1692 SGG Onorio in Roma Pollarolo finale Segue il Ballo subito voi intrecciato dal Canto di tutte T; le parte [Ballo di Soggetti di Corte per la gran Danza Francese] This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions Table 4. continued Year Theatre Opera Composer Ballo location Ba//o title Tex 1693 SGG Laforga della virti Pollarolo finale duct and Ballo [di Ninfe e voi Fiumi seguaci del Tago] et o 1695 SGG Ilpastore d'An/iso Pollarolo Primo intramezzo [Ninfe che suonano, cantono, voi e ballanol 1689 SGG Ilpastore d'An/riso Pollarolo Secondo intramezzo [Satiri c Cacciatori che voi suonano, cantono, e ballano] 1695 SGG Ilpastore d'An/isos Pollarolo Terzo intramezzo [Aure c Zeffiri che suonano, voi cantono,, e ballano] (fla 1695 SGG Ilpastore d'Anriso Pollarolo Quatro intramezzo [Ninfe c Pastori che suonano, voi cantono, e ballano] B 1695 SGG Ipastore d'Anjiso Pollarolo Ultimo intramczzo [Sogni e Fantasme che voi suonano, cantono, e ballano] 1696 SGG Rosimonda Pollarolo Act I scene 1 [Gran Ballo] voic A; 1696 SGG Rosimonda Pollarolo Act I scene 8 Coro di Damigelle, Coro di voic Cavalieri d'Alsuinda T; 1696 SGG Rosimonda Pollarolo Act II scene 8 Coro di Giardinieri Uomini e voic Donne Cantano e Ballano T; 1696 SGG Rosimonda Pollarolo Act III scene 7 Coro di Cavalieri e di voic donzelle che esprimono T; l'afflizione per la morte del R& 1696 SGG Rosimonda Pollarolo Act IV scene 8 Coro di Uomini e Donne che voic fanno voti per la salvezza del Regno 1699 SSalv Faramondo Pollarolo finale Coro [I seguaci della Virtu voic accompagnano il canto del B Coro col Ballo] 1700 SGG I1 colore fi la regina Pollarolo Act II scene 7 Ballo Tutti Gl'Istromenti [di voic Baccanti] (ob This content downloaded from 128.112.200.107 on Fri, 05 Jun 2015 15:05:41 UTC All use subject to JSTOR Terms and Conditions
15597	https://www.quora.com/What-are-some-examples-of-pigeonhole-principles-in-problem-solving	Something went wrong. Wait a moment and try again. Mathematical Sciences Problem Solving Pigeonhole Principle Combinatorics of Words Mathematical Concepts Algorithms, Combinatorics Combinatorics C Math Solving Problems Combinatorial Math 5 What are some examples of pigeonhole principles in problem solving? · The pigeonhole principle is a fundamental concept in combinatorics that states if items are put into containers, with , then at least one container must contain more than one item. Here are some examples of how this principle can be applied in problem-solving: Birthday Paradox In a group of 23 people, the probability that at least two people share the same birthday is surprisingly high (over 50%). This can be explained using the pigeonhole principle: with 23 people and only 365 days in a year, at least one day must be the birthday of at least two people. Socks in a Drawer If you hav The pigeonhole principle is a fundamental concept in combinatorics that states if items are put into containers, with , then at least one container must contain more than one item. Here are some examples of how this principle can be applied in problem-solving: Birthday Paradox In a group of 23 people, the probability that at least two people share the same birthday is surprisingly high (over 50%). This can be explained using the pigeonhole principle: with 23 people and only 365 days in a year, at least one day must be the birthday of at least two people. Socks in a Drawer If you have 10 pairs of socks in a drawer and you randomly pick 11 socks, you are guaranteed to have at least one matching pair. Here, the 10 pairs represent the pigeonholes, and the 11 picked socks represent the items. Handshakes In a group of people, if each person shakes hands with at least others, then at least two people must have shaken hands with the same number of people. This is because the possible number of handshakes ranges from 0 to , creating a limited number of outcomes. Coloring Problems In a scenario where you have 5 colors and need to color 6 points on a plane, by the pigeonhole principle, at least two points must be the same color. This principle is often used in graph theory and coloring problems. Divisibility If you take any group of 13 integers, at least two of them must have the same remainder when divided by 12. This is because there are only 12 possible remainders (0 to 11), and if you have 13 integers, at least one remainder must repeat. Seating Arrangements In a situation where you are seating 10 people at a round table, at least two people must sit next to each other if there are only 9 distinct pairs of adjacent seats. This is a simple application of the principle in combinatorial arrangements. Distinct Elements If you have a set of integers where the maximum integer is , and , then at least two integers must be equal when considering their values modulo . This is useful in modular arithmetic problems. Conclusion The pigeonhole principle is a powerful and often intuitive tool in problem-solving across various fields, including mathematics, computer science, and probability theory. By recognizing situations where the principle applies, you can derive conclusions that might not be immediately obvious. Justin Conlon Studied Mathematics · 9y Related What are some interesting applications of pigeon hole principle? My father used to ask me and my sibling riddles when we were kids and one that sticks out in my mind is as follows A man wakes up very early for work every morning, so early in fact that it is still dark outside. This man is considerate of his wife sleeping so takes his clothes from the dresser in the dark and gets dressed in another room. His sock drawer is unorganized and is full of unpaired socks that are either white, black or brown. What is the minimum number of socks the man must take in order to guarantee he has a matching pair? If the man has 4 socks then by the pigeon hole principle, 4 My father used to ask me and my sibling riddles when we were kids and one that sticks out in my mind is as follows A man wakes up very early for work every morning, so early in fact that it is still dark outside. This man is considerate of his wife sleeping so takes his clothes from the dresser in the dark and gets dressed in another room. His sock drawer is unorganized and is full of unpaired socks that are either white, black or brown. What is the minimum number of socks the man must take in order to guarantee he has a matching pair? If the man has 4 socks then by the pigeon hole principle, 4 pigeons (socks) and 3 holes (colors), he will have at least one pair and it is easy to show that a smaller number than 4 could result in a set of socks with no pairs. Sponsored by NoBuzzZone This 2-Minute Ritual Will Keep Fruit Flies Out of Your Kitchen. Tired of sprays and vinegar traps? This chemical-free fix works 24/7—without the mess. Scott Brickner Programming almost every day for the last 40 years. · Author has 18.9K answers and 6.5M answer views · 7y Originally Answered: What is an example of the pigeonhole principle in problem solving? · You have a drawer full of socks, which are a mixture of black, blue, and brown. If you’re drawing socks from the drawer without looking, how many socks must you draw to guarantee you have a matched pair? There are three colors. If you’ve drawn three socks, then you either have a pair or you have three different colors, in which case your next sock guarantees a pair. So, four socks are required. This is the pigeonhole principle. There are three “holes” (black, blue, and brown), and items get assigned to holes. At some point, you’ve filled all the holes and the next item is guaranteed to be assign You have a drawer full of socks, which are a mixture of black, blue, and brown. If you’re drawing socks from the drawer without looking, how many socks must you draw to guarantee you have a matched pair? There are three colors. If you’ve drawn three socks, then you either have a pair or you have three different colors, in which case your next sock guarantees a pair. So, four socks are required. This is the pigeonhole principle. There are three “holes” (black, blue, and brown), and items get assigned to holes. At some point, you’ve filled all the holes and the next item is guaranteed to be assigned to a hole that’s already got something in it. Related questions How can I solve the pigeonhole principle problem? What are some interesting applications of pigeon hole principle? What are some of the toughest problems you have solved using the pigeonhole principle? What are the most awesome applications of the Pigeonhole principle that you are aware of? Why do we need to prove the pigeonhole principle? Gia Mukherjee imaginative and intuitive · 8y Related What are some interesting applications of pigeon hole principle? Thanks for the A2A. I can think of two pigeon-hole principle applications. If you pick five numbers from the integers 1 to 8, then two of them must add up to nine. Every number can be paired with another to sum to nine. In all, there are four such pairs: the numbers 1 and 8, 2 and 7, 3 and 6, and lastly 4 and 5. Each of the five numbers belongs to one of those four pairs. By the pigeonhole principle, two of the numbers must be from the same pair–which by construction sums to 9. If you draw five points on the surface of an orange in permanent marker, then there is a way to cut the orange in half Thanks for the A2A. I can think of two pigeon-hole principle applications. If you pick five numbers from the integers 1 to 8, then two of them must add up to nine. Every number can be paired with another to sum to nine. In all, there are four such pairs: the numbers 1 and 8, 2 and 7, 3 and 6, and lastly 4 and 5. Each of the five numbers belongs to one of those four pairs. By the pigeonhole principle, two of the numbers must be from the same pair–which by construction sums to 9. If you draw five points on the surface of an orange in permanent marker, then there is a way to cut the orange in half so that four of the points will lie on the same hemisphere (suppose a point exactly on the cut belongs to both hemispheres). Two points determine a great circle on a sphere, so for any two points, cut the orange into half. The remaining three points can be on either one of the two resulting hemispheres. By the pigeonhole principle, at least two of them belong to the same hemisphere, bringing the total to 4 points. Cheers! :D Aryak Sen knows high school Maths · Author has 70 answers and 493.3K answer views · 7y Related What are some interesting applications of pigeon hole principle? Below are some applications. I will start with the easy ones and as I proceed the complexity will increase. 1.Among 3 persons at least 2 are of the same sex. 2.Among 13 persons there are at least 2 persons born in the same month. 3.Nobody has more than 300000 hairs on his head.The capital of India has 300001 inhabitants.Then it can be asserted that at least 2 person have the same number of hairs on their heads. 4.A chessmaster has 77 days to prepare for a tournament. He wants to play at least 1 game per day, but not more than 132 games (in 77 days). Prove that there is a sequence of successive day Below are some applications. I will start with the easy ones and as I proceed the complexity will increase. 1.Among 3 persons at least 2 are of the same sex. 2.Among 13 persons there are at least 2 persons born in the same month. 3.Nobody has more than 300000 hairs on his head.The capital of India has 300001 inhabitants.Then it can be asserted that at least 2 person have the same number of hairs on their heads. 4.A chessmaster has 77 days to prepare for a tournament. He wants to play at least 1 game per day, but not more than 132 games (in 77 days). Prove that there is a sequence of successive days on which he plays exactly 21 games. 5.There are 12 computers and 8 laser printers in an office.Find the minimum number of connections to be made which will guarantee that if 8 or fewer computers want to print at the same time each of them will be able to use a different printer. Sponsored by Grammarly Does your paper sound smart or scripted? Grammarly’s new AI agents are with you at every stage—helping you shape your writing without replacing it Alessandro Felisi Graduate Student in Mathematics · 6y Related What are some of the toughest problems you have solved using the pigeonhole principle? From Miklos Bona’s A Walk Through Combinatorics: (Warming up) The set M consists of nine positive integers, none of which has a prime divisor larger than six. Prove that M has two elements whose product is the square of an integer. Solution : The possible prime factors for the numbers in consideration are 2, 3 and 5; for each one of them, let us denote the parity of the exponent in the prime factorization with 0 or 1. Then there are 2^3 = 8 possible triples of parities; because the set M contains 9 numbers, for the Pigeonhole Principle, there must exist two numbers in M with the same parity tripl From Miklos Bona’s A Walk Through Combinatorics: (Warming up) The set M consists of nine positive integers, none of which has a prime divisor larger than six. Prove that M has two elements whose product is the square of an integer. Solution: The possible prime factors for the numbers in consideration are 2, 3 and 5; for each one of them, let us denote the parity of the exponent in the prime factorization with 0 or 1. Then there are 2^3 = 8 possible triples of parities; because the set M contains 9 numbers, for the Pigeonhole Principle, there must exist two numbers in M with the same parity triple. Multiplying them together gives a square (note that multiplying numbers together corresponds to the component-wise addition mod 2 of the triples of parities; moreover, a number with parities (0,0,0) is a square). The set A consists of n+1 positive integers, none of which have a prime divisor that is larger than the nth smallest prime number. Prove that there exists a non-empty subset B in A so that the product of the elements of B is a perfect square. Solution: Let us represent the parity of each exponent in the prime factorization of a number in A with a vector of 0s and 1s; in other words, because there are in general n primes in the factorization, this vector can be thought as an element of the n-dimensional vector space over the fields of two elements. Because there are n+1 positive integers, and because the vector space in consideration has dimension n, the vectors in A are linearly dependent. Therefore, there exists a non-trivial linear combination of the vectors which is equal to 0. The coefficients in this linear combination are the scalars 0 and 1, which correspond respectively to discarding or joining numbers to the subset B. Because the linear combination is non-trivial, the corresponding set is non-empty. The fact that the product of the numbers in B is a square is a consequence of the fact that the linear combination in consideration is equal to the 0 vector. You might wonder where I used the Pigeonhole principle in the solution of the second problem. I like to think of the Dimension theorem for vector spaces as the linear algebra analogue of the Pigeonhole principle. Indeed, for vector spaces over finite fields (which is the case of the second problem), the theorem can be proved using the Pigeonhole principle! Theorem: Let K be the field of q elements; consider n+1 vectors in the n-dimensional vector space over K. Then these vectors are linearly dependent. Proof: There are q^(n+1) possible linear combinations of the vectors, while there are only q^n elements of the vector field. Therefore, for the Pigeonhole principle, there must exist two distinct linear combinations representing the same element. Considering their difference, we obtain a non-trivial linear combination which is equal to the 0 vector. Related questions What is the pigeonhole principle? Explain with a suitable example. What are some scenarios in which the pigeonhole principle is really useful? How do I solve pigeon hole principle related problems? What is the best question that you have ever faced on the Pigeonhole Principle? What is the pigeonhole principle? What are some applications of the pigeonhole principle in problem solving? Mingzhi Zhang Studying Combinatorics · 9y Related What are some interesting applications of pigeon hole principle? In the first chapter of , Miklos Bona elaborated the Pigeon-Hole Principle. Here I quote two examples in his book. There is an element in the sequence 7,77,777,7777,..., that is divisible by 2003. During the last 1000 years, the reader had an ancestor A such that there was a person P who was an ancestor of both the father and mother of A. Sponsored by Mutual of Omaha Comparing Medicare plans this fall? Answer a few questions to find a Medicare Plan that meets your needs. Anonymous 11y Related What is the pigeonhole principle? The pigeonhole principle is a fairly simple idea to grasp. Say that you have 7 pigeons and 6 pigeonholes. So, now you decide to start putting the pigeons one by one into each pigeonhole ___ ___ ___ ___ ___ ___ \| p \| \| p \| \| p \| \| p \| \| p \| \| p \| So, now, you have one pigeon left, and you can put it into any of the pigeonholes. ___ ___ ___ ___ ___ ___ \|pp \| \| p \| \| p \| \| p \| \| p \| \| p \| I'll put the pigeon into the first pigeonhole without over thinking the placement. The point is that when the number of pigeons > no. of pigeonholes, there wil The pigeonhole principle is a fairly simple idea to grasp. Say that you have 7 pigeons and 6 pigeonholes. So, now you decide to start putting the pigeons one by one into each pigeonhole ___ ___ ___ ___ ___ ___ \| p \| \| p \| \| p \| \| p \| \| p \| \| p \| So, now, you have one pigeon left, and you can put it into any of the pigeonholes. ___ ___ ___ ___ ___ ___ \|pp \| \| p \| \| p \| \| p \| \| p \| \| p \| I'll put the pigeon into the first pigeonhole without over thinking the placement. The point is that when the number of pigeons > no. of pigeonholes, there will be at least one pigeonhole with two or more pigeons. The applications of this principle are immense. One of my favourite applications of the pigeonhole principle is the hair counting problem. If the amount of hair is expressed in terms of the number of hair strands, the average human head has about 150,000 hair strands. It is safe to assume, then, that no human head has more than 1,000,000 strands of hair. If the population of a country is more than 1,000,000 at least two people in that country have the same amount of hair. A better explanation: Assume that you have lined 5,000,000 people in a huge ground to prove the hair counting problem. You have a special apparatus that can count the number of strands in someone's head by simply placing the apparatus over said head. The first person steps up and is found to be bald, you make them stand in queue number 0. The second person has 5 strands of hair, you make them stand in queue 5 and so on. In the worst case, after measuring a million people, you'd have a million queues with each queue consisting of one person who has the same amount of hair as the queue number, so queue 1,00,000 will have a person with 1,00,000 strands of hair. In that case, out of the remaining 4,00,000, there will be at least one person who goes into the existing queues. Are you able to identify the pigeons and the pigeonholes here? The pigeonholes(Or the queues) are the number of hair strands on someone's head while the pigeons are the people that are matched to a queue(pigeonhole) based on the amount of hair they have. Charles S. former mathematician, current patent lawyer · Upvoted by Yousef Daneshbod , PhD Mathematics & Engineering, Claremont Graduate University (2006) and Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) · Author has 7.6K answers and 60M answer views · 4y Related What is the pigeonhole principle? Explain with the help of an example. Let’s first start with an example: Suppose you want to put 10 pigeons into 9 pigeonholes. There are lots of ways to do that. But regardless of how you do it, you can be sure that there’s one hole that gets two pigeons. And that’s the pigeonhole principle. Stated only a teensy weensy bit more generally, if you have pigeons and you want to put them in holes (with ), then at least one hole will have two pigeons. This is a simple idea, bordering on self-evident. But its application are many, varied, and sometimes surprisingly subtle. For example, here’s a problem: I have decided to work medi Let’s first start with an example: Suppose you want to put 10 pigeons into 9 pigeonholes. There are lots of ways to do that. But regardless of how you do it, you can be sure that there’s one hole that gets two pigeons. And that’s the pigeonhole principle. Stated only a teensy weensy bit more generally, if you have pigeons and you want to put them in holes (with ), then at least one hole will have two pigeons. This is a simple idea, bordering on self-evident. But its application are many, varied, and sometimes surprisingly subtle. For example, here’s a problem: I have decided to work meditation into my daily routine. In fact, I have decided to meditate exactly 45 times in the next 30 days, while meditating at least once a day for the next 30 days. Prove there is a sequence of consecutive days during which I meditate a total of 14 times. You already have the hint that the solution involves the pigeonhole principle. But I suspect it’s a tough problem for the uninitiated (and maybe even for the initiated). I’ll put the solution in the comments, in case anyone would like to work on it. Sponsored by OrderlyMeds Is Your GLP-1 Personalized? Find GLP-1 plans tailored to your unique body needs. Sharad Srivastava The Inquisitor · Author has 1.9K answers and 2.4M answer views · 10y Related What are some interesting applications of pigeon hole principle? If my post is read by more than 366 people, atleast 2 of them will share their bithdays. How about this? Raziman T.V. IOI silver medallist (2007), ACM ICPC world finalist (2011, 2012) · Upvoted by Deepak Bakal , M.Sc. Mathematics, University of Hyderabad and Fabio García , MSc Mathematics, CIMAT (2018) · Author has 2.6K answers and 28M answer views · Updated 6y Related What are some of the toughest problems you have solved using the pigeonhole principle? The toughest pigeonhole problems are the ones where not only are the pigeons and holes not obvious, that there are pigeons and holes at all is not clear in the first place. An interesting problem of this kind relates to Dirichlet's approximation theorem. Suppose you are asked to prove that, given an irrational number , there exist infinitely many integer pairs such that If you haven’t seen the proof already, the immediate feeling might be: How exactly does this have anything to do with the pigeonhole princi Footnotes Dirichlet's approximation theorem - Wikipedia The toughest pigeonhole problems are the ones where not only are the pigeons and holes not obvious, that there are pigeons and holes at all is not clear in the first place. An interesting problem of this kind relates to Dirichlet's approximation theorem. Suppose you are asked to prove that, given an irrational number , there exist infinitely many integer pairs such that If you haven’t seen the proof already, the immediate feeling might be: How exactly does this have anything to do with the pigeonhole principle? The solution lies in proving a slightly easier-looking version of the theorem: given an irrational number and a positive integer , prove that there exist integers such that and It might not be still clear how this can be turned into the pigeonhole principle. The idea is to look for , the fractional parts of for . There are numbers and all of them lie in . The pigeons and holes should now be clear. We first split the range into parts: Since we have ranges and numbers which should fall in one of the ranges, some two should fall in the same range. That is, we have for some . This means that for . Writing , we have the result Going from here to the original statement is straightforward. Dividing by gives By repeating the process with sufficiently large values of , it is possible to find infinite values of . Footnotes Dirichlet's approximation theorem - Wikipedia Gaurav Pachpute Related What are the most awesome applications of the Pigeonhole principle that you are aware of? Here are some theorems that can be proven using the pigeonhole principle. They are simple applications of the principle and also surprising at first glance. Given any 5 points on a sphere there exists a closed hemisphere which contains 4 of them. The pumping lemma which is a useful test to prove that some languages are not regular. Every rational number has an eventually repeating decimal expansion. Every element of a finite group has an order. In a group of 6 friends, there will always be three people that are mutual friends or mutual strangers. In a group of two or more people, t Here are some theorems that can be proven using the pigeonhole principle. They are simple applications of the principle and also surprising at first glance. Given any 5 points on a sphere there exists a closed hemisphere which contains 4 of them. The pumping lemma which is a useful test to prove that some languages are not regular. Every rational number has an eventually repeating decimal expansion. Every element of a finite group has an order. In a group of 6 friends, there will always be three people that are mutual friends or mutual strangers. In a group of two or more people, there are at least two people with the same number of friends. The first one is my personal favourite. Steve Baker Senior Software Engineer (2013–present) · Author has 39.6K answers and 292.9M answer views · 6y Related What is a real life example that uses the pigeon hole theorem? It’s not really a “theorem” - it’s usually called “The Pigeon Hole Principle”. What is says is if you have N holes and M objects that must be placed in the holes - then if M>N then at least one hole must contain more than one object. It’s one of those “bloody obvious” things that is so easy to accept that nobody much thinks about the truth or otherwise of it. If you work at a company that has 100 emp It’s not really a “theorem” - it’s usually called “The Pigeon Hole Principle”. What is says is if you have N holes and M objects that must be placed in the holes - then if M>N then at least one hole must contain more than one object. It’s one of those “bloody obvious” things that is so easy to accept that nobody much thinks about the truth or otherwise of it. If you work at a company that has 100 employees - and the parking lot only has 99 spaces - then any idiot can tell you that there is a risk that someone will have nowhere to park if everyone drives in their own cars and everyone comes to work that day…Why? Because you can’t get two cars into one parking place…DUH! There are some ways to make interesting conclusions from this - if you have a drawer full of socks - some black, some white. If it’s too dark to see properly - how many socks do you have to grab in order to guarantee that you get a matching pair? Well, if you imagine having two pigeon holes - one where you put black socks that you pulled from the drawer and another where you put white ones - then you have two pigeon holes - and you only need to... Related questions How can I solve the pigeonhole principle problem? What are some interesting applications of pigeon hole principle? What are some of the toughest problems you have solved using the pigeonhole principle? What are the most awesome applications of the Pigeonhole principle that you are aware of? Why do we need to prove the pigeonhole principle? What is the pigeonhole principle? Explain with a suitable example. What are some scenarios in which the pigeonhole principle is really useful? How do I solve pigeon hole principle related problems? What is the best question that you have ever faced on the Pigeonhole Principle? What is the pigeonhole principle? What are some applications of the pigeonhole principle in problem solving? What are some great proofs that use the pigeonhole principle? What hints/indications in a problem statement suggest that the pigeonhole principle might be applicable? How do I apply pigeon-hole principle? Can you give a mathematical example of the Pigeonhole Principle? What is the pigeonhole principle? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15598	https://brainly.com/question/32985412	[FREE] How long (in years) will it take your money to triple at an annual percentage rate of 6% compounded - brainly.com 6 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 +52,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +24,7k Ace exams faster, with practice that adapts to you Practice Worksheets +6,9k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified How long (in years) will it take your money to triple at an annual percentage rate of 6% compounded annually? Use logarithms to solve. Round your answer to two decimal places. 1 See answer Explain with Learning Companion NEW Asked by hannahkelly5286 • 06/02/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 8548810 people 8M 1.0 0 Upload your school material for a more relevant answer It will take approximately 11.9 years to triple the amount of the initial investment at an annual percentage rate of 6% compounded annually. This is calculated using the compound interest formula. To calculate the duration required for your money to triple at an annual percentage rate of 6% compounded annually, you can use the formula for compound interest given as; A = P(1 + r/n)^(nt), Where; P is the principal amount, R is the annual interest rate, N is the number of times the interest is compounded per year, T is the time in years. For this question,P = $1 (since we don't know the amount) R = 6% = 0.06N = 1 (since the interest is compounded annually) We want to know the time required (T) to get triple the amount of the initial amount. This means that; A = 3P Therefore, 3P = P(1 + r/n)^(nt) Dividing both sides of the equation by P, 3 = (1 + r/n)^(nt) Since we want to calculate for time (T), we can use logarithms to solve. Therefore, taking the natural logarithm of both sides of the equation, ln(3) = ln(1 + r/n)^(nt) Applying logarithmic laws, ln(3) = nt ln(1 + r/n) Solving for time (T), T = ln(3)/(n ln(1 + r/n)) Substituting the values we have, we get; T = ln(3)/(1 ln(1 + 0.06/1)) = ln(3)/ln(1.06) ≈ 11.9 years It will take approximately 11.9 years to triple the amount of the initial investment at an annual percentage rate of 6% compounded annually. This is calculated using the compound interest formula. Learn more about compound interest visit: brainly.com/question/14295570 SPJ11 Answered by AndrewCarol •17K answers•8.5M people helped Thanks 0 1.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 8548810 people 8M 1.0 0 Principles of Economics 3e - Steven A. Greenlaw, David Shapiro, Daniel Macdonald Principles of Microeconomics 3e - David Shapiro, Daniel Macdonald, Steven A. Greenlaw Money and Banking - Robert E. Wright Upload your school material for a more relevant answer It will take approximately 18.85 years for your money to triple at an annual percentage rate of 6% compounded annually. This was calculated using the compound interest formula and natural logarithms. The final formula is t=l n(1.06)l n(3). Explanation To determine how long it will take for your money to triple at an annual percentage rate of 6% compounded annually, we can use the compound interest formula. The formula for compound interest is given by: A=P(1+r/n)n t Where: A is the amount of money accumulated after n years, including interest. P is the principal amount (the initial amount of money). r is the annual interest rate (decimal). n is the number of times that interest is compounded per year. t is the time the money is invested or borrowed for, in years. In this case, since we want the money to triple, we can set A to 3 P. Thus, we can rewrite the equation as: 3 P=P(1+0.06/1)1 t Next, we can divide both sides by P: 3=(1.06)t To find t, we will take the natural logarithm of both sides: ln(3)=ln((1.06)t) This simplifies using the logarithmic identity ln(a b)=b⋅ln(a): ln(3)=t⋅ln(1.06) Now, we can solve for t: t=ln(1.06)ln(3) Using a calculator, we find: ln(3)≈1.0986 ln(1.06)≈0.0583 Thus, substituting these values gives: t≈0.0583 1.0986≈18.85 Therefore, it will take approximately 18.85 years for your money to triple at an annual percentage rate of 6% compounded annually. Rounding to two decimal places, the answer is approximately 18.85 years. Examples & Evidence For example, if you initially invest $1, at an interest rate of 6%, after approximately 18.85 years, your investment will grow to $3. Similarly, an investment of $100 will grow to $300 in the same timeframe. The calculations were based on the mathematical principles of compound interest and logarithms, which are widely taught in high school mathematics courses. Thanks 0 1.0 (1 vote) Advertisement hannahkelly5286 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer At 5% annual interest compounded monthly, how long (in years) will it take to triple your money? Round-off your answer to two decimal places. Community Answer 12 Solve this application using logarithms. How long will money in savings take to double at 5% interest compounded annually? Round the answer to the nearest hundredth of a year. 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 New questions in Mathematics \| \| Day 2 \| Day 3 \| Day 4 \| :---: :---: \| \| Spent \| 7 \| 12 \| 9 \| \| Earned \| 16 \| 22 \| 18 \| Juan ran the lemonade stand for 3 more days after his first day profit of $12. Each day, he used the money from sales to purchase more lemons, cups, and sugar to make more lemonade. The table shows how much he spent and earned each day.What is the expression needed to find his total profit?What was his total profit? Using the discriminant, how many solutions will this quadratic have? Explain and show all work. −3 x 2−12 x+11=0 What is $\cos \left(6^{\circ}\right) ? A. 0.10 B. 0.60 C. 0.99 D. 0.06 The power g 2 is equivalent to 81. What is the value of g−2? A. −81 B. −9 C. 81 1 D. 9 1 What is the value of (−4 3)−4 ? A. −81 256 B. −256 81 C. 256 81 D. 81 256 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
15599	https://pressbooks.atlanticoer-relatlantique.ca/algebratrigonometryopenstax/chapter/the-hyperbola/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 81 The Hyperbola Learning Objectives In this section, you will: Locate a hyperbola’s vertices and foci. Write equations of hyperbolas in standard form. Graph hyperbolas centered at the origin. Graph hyperbolas not centered at the origin. Solve applied problems involving hyperbolas. What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom. See (Figure). Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom. Locating the Vertices and Foci of a Hyperbola In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See (Figure). Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all pointsin a plane such that the difference of the distances betweenand the foci is a positive constant. Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances. As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See (Figure). In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x– and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin. Deriving the Equation of a Hyperbola Centered at the Origin Letandbe the foci of a hyperbola centered at the origin. The hyperbola is the set of all pointssuch that the difference of the distances fromto the foci is constant. See (Figure). Ifis a vertex of the hyperbola, the distance fromtoisThe distance fromtoisThe sum of the distances from the foci to the vertex is Ifis a point on the hyperbola, we can define the following variables: By definition of a hyperbola,is constant for any pointon the hyperbola. We know that the difference of these distances isfor the vertexIt follows thatfor any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses. This equation defines a hyperbola centered at the origin with verticesand co-vertices Standard Forms of the Equation of a Hyperbola with Center (0,0) The standard form of the equation of a hyperbola with centerand transverse axis on the x-axis is where the length of the transverse axis is the coordinates of the vertices are the length of the conjugate axis is the coordinates of the co-vertices are the distance between the foci is where the coordinates of the foci are the equations of the asymptotes are See (Figure)a. The standard form of the equation of a hyperbola with centerand transverse axis on the y-axis is where the length of the transverse axis is the coordinates of the vertices are the length of the conjugate axis is the coordinates of the co-vertices are the distance between the foci is where the coordinates of the foci are the equations of the asymptotes are See (Figure)b. Note that the vertices, co-vertices, and foci are related by the equationWhen we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci. How To Given the equation of a hyperbola in standard form, locate its vertices and foci. Determine whether the transverse axis lies on the x– or y-axis. Notice thatis always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices. If the equation has the form then the transverse axis lies on the x-axis. The vertices are located at and the foci are located at If the equation has the form then the transverse axis lies on the y-axis. The vertices are located at and the foci are located at Solve forusing the equation Solve forusing the equation Locating a Hyperbola’s Vertices and Foci Identify the vertices and foci of the hyperbola with equation [reveal-answer q=”fs-id1934192″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1934192″] The equation has the form so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set and solve for The foci are located atSolving for Therefore, the vertices are located at and the foci are located at[/hidden-answer] Try It Identify the vertices and foci of the hyperbola with equation [reveal-answer q=”fs-id1369547″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1369547″]Vertices:Foci:[/hidden-answer] Writing Equations of Hyperbolas in Standard Form Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin. Hyperbolas Centered at the Origin Reviewing the standard forms given for hyperbolas centered atwe see that the vertices, co-vertices, and foci are related by the equationNote that this equation can also be rewritten asThis relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices. How To Given the vertices and foci of a hyperbola centered at write its equation in standard form. Determine whether the transverse axis lies on the x– or y-axis. If the given coordinates of the vertices and foci have the formandrespectively, then the transverse axis is the x-axis. Use the standard form If the given coordinates of the vertices and foci have the formandrespectively, then the transverse axis is the y-axis. Use the standard form Findusing the equation Substitute the values forandinto the standard form of the equation determined in Step 1. Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices What is the standard form equation of the hyperbola that has verticesand foci [reveal-answer q=”fs-id2157034″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2157034″] The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form The vertices aresoand The foci aresoand Solving for we have Finally, we substituteandinto the standard form of the equation,The equation of the hyperbola is as shown in (Figure). [/hidden-answer] Try It What is the standard form equation of the hyperbola that has verticesand foci [reveal-answer q=”fs-id2587698″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2587698″] [/hidden-answer] Hyperbolas Not Centered at the Origin Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translatedunits horizontally andunits vertically, the center of the hyperbola will beThis translation results in the standard form of the equation we saw previously, withreplaced byandreplaced by Standard Forms of the Equation of a Hyperbola with Center (h, k) The standard form of the equation of a hyperbola with centerand transverse axis parallel to the x-axis is where the length of the transverse axis is the coordinates of the vertices are the length of the conjugate axis is the coordinates of the co-vertices are the distance between the foci iswhere the coordinates of the foci are The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle isand its width isThe slopes of the diagonals areand each diagonal passes through the centerUsing the point-slope formula, it is simple to show that the equations of the asymptotes areSee (Figure)a The standard form of the equation of a hyperbola with centerand transverse axis parallel to the y-axis is where the length of the transverse axis is the coordinates of the vertices are the length of the conjugate axis is the coordinates of the co-vertices are the distance between the foci iswhere the coordinates of the foci are Using the reasoning above, the equations of the asymptotes areSee (Figure)b. Like hyperbolas centered at the origin, hyperbolas centered at a pointhave vertices, co-vertices, and foci that are related by the equationWe can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given. How To Given the vertices and foci of a hyperbola centered atwrite its equation in standard form. Determine whether the transverse axis is parallel to the x– or y-axis. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form Identify the center of the hyperbola,using the midpoint formula and the given coordinates for the vertices. Findby solving for the length of the transverse axis,, which is the distance between the given vertices. Findusingandfound in Step 2 along with the given coordinates for the foci. Solve forusing the equation Substitute the values for andinto the standard form of the equation determined in Step 1. Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices What is the standard form equation of the hyperbola that has vertices atandand foci atand [reveal-answer q=”fs-id1388532″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1388532″] The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form First, we identify the center,The center is halfway between the verticesandApplying the midpoint formula, we have Next, we findThe length of the transverse axis,is bounded by the vertices. So, we can findby finding the distance between the x-coordinates of the vertices. Now we need to findThe coordinates of the foci areSoandWe can use the x-coordinate from either of these points to solve forUsing the pointand substituting Next, solve forusing the equation Finally, substitute the values found forandinto the standard form of the equation. [/hidden-answer] Try It What is the standard form equation of the hyperbola that has verticesandand fociand [reveal-answer q=”fs-id1909176″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1909176″] [/hidden-answer] Graphing Hyperbolas Centered at the Origin When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard formfor horizontal hyperbolas and the standard formfor vertical hyperbolas. How To Given a standard form equation for a hyperbola centered atsketch the graph. Determine which of the standard forms applies to the given equation. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes. If the equation is in the formthen the transverse axis is on the x-axis the coordinates of the vertices are the coordinates of the co-vertices are the coordinates of the foci are the equations of the asymptotes are If the equation is in the form then the transverse axis is on the y-axis the coordinates of the vertices are the coordinates of the co-vertices are the coordinates of the foci are the equations of the asymptotes are Solve for the coordinates of the foci using the equation Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola. Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form Graph the hyperbola given by the equationIdentify and label the vertices, co-vertices, foci, and asymptotes. [reveal-answer q=”fs-id2013263″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2013263″] The standard form that applies to the given equation isThus, the transverse axis is on the y-axis The coordinates of the vertices are The coordinates of the co-vertices are The coordinates of the foci arewhereSolving forwe have Therefore, the coordinates of the foci are The equations of the asymptotes are Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in (Figure). [/hidden-answer] Try It Graph the hyperbola given by the equationIdentify and label the vertices, co-vertices, foci, and asymptotes. [reveal-answer q=”fs-id1829664″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1829664″] vertices:co-vertices:foci:asymptotes:[/hidden-answer] Graphing Hyperbolas Not Centered at the Origin Graphing hyperbolas centered at a pointother than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard formsfor horizontal hyperbolas, andfor vertical hyperbolas. From these standard form equations we can easily calculate and plot key features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes. How To Given a general form for a hyperbola centered at sketch the graph. Convert the general form to that standard form. Determine which of the standard forms applies to the given equation. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes. If the equation is in the formthen the transverse axis is parallel to the x-axis the center is the coordinates of the vertices are the coordinates of the co-vertices are the coordinates of the foci are the equations of the asymptotes are If the equation is in the formthen the transverse axis is parallel to the y-axis the center is the coordinates of the vertices are the coordinates of the co-vertices are the coordinates of the foci are the equations of the asymptotes are Solve for the coordinates of the foci using the equation Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola. Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form Graph the hyperbola given by the equationIdentify and label the center, vertices, co-vertices, foci, and asymptotes. [reveal-answer q=”fs-id2084849″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2084849″] Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation. Factor the leading coefficient of each expression. Complete the square twice. Remember to balance the equation by adding the same constants to each side. Rewrite as perfect squares. Divide both sides by the constant term to place the equation in standard form. The standard form that applies to the given equation is whereandorandThus, the transverse axis is parallel to the x-axis. It follows that: the center of the ellipse is the coordinates of the vertices areorand the coordinates of the co-vertices areorand the coordinates of the foci arewhereSolving forwe have Therefore, the coordinates of the foci areand The equations of the asymptotes are Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in (Figure). [/hidden-answer] Try It Graph the hyperbola given by the standard form of an equationIdentify and label the center, vertices, co-vertices, foci, and asymptotes. [reveal-answer q=”fs-id1837480″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1837480″] center:vertices:andco-vertices:andfoci:andasymptotes:[/hidden-answer] Solving Applied Problems Involving Hyperbolas As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See (Figure). For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide! The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In (Figure) we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides. Solving Applied Problems Involving Hyperbolas The design layout of a cooling tower is shown in (Figure). The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places. [reveal-answer q=”fs-id1673246″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1673246″] We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin:where the branches of the hyperbola form the sides of the cooling tower. We must find the values ofandto complete the model. First, we findRecall that the length of the transverse axis of a hyperbola isThis length is represented by the distance where the sides are closest, which is given asmeters. So,Therefore,and To solve forwe need to substitute forandin our equation using a known point. To do this, we can use the dimensions of the tower to find some pointthat lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore, The sides of the tower can be modeled by the hyperbolic equation [/hidden-answer] Try It A design for a cooling tower project is shown in (Figure). Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places. [reveal-answer q=”fs-id1669711″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1669711″] The sides of the tower can be modeled by the hyperbolic equation.[/hidden-answer] Access these online resources for additional instruction and practice with hyperbolas. Conic Sections: The Hyperbola Part 1 of 2 Conic Sections: The Hyperbola Part 2 of 2 Graph a Hyperbola with Center at Origin Graph a Hyperbola with Center not at Origin Key Equations \| \| \| --- \| \| Hyperbola, center at origin, transverse axis on x-axis \| \| \| Hyperbola, center at origin, transverse axis on y-axis \| \| \| Hyperbola, center attransverse axis parallel to x-axis \| \| \| Hyperbola, center attransverse axis parallel to y-axis \| \| Key Concepts A hyperbola is the set of all pointsin a plane such that the difference of the distances betweenand the foci is a positive constant. The standard form of a hyperbola can be used to locate its vertices and foci. See (Figure). When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. See (Figure) and (Figure). When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. See (Figure) and (Figure). Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. See (Figure). Section Exercises Verbal Define a hyperbola in terms of its foci. [reveal-answer q=”fs-id2227183″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2227183″] A hyperbola is the set of points in a plane the difference of whose distances from two fixed points (foci) is a positive constant. [/hidden-answer] What can we conclude about a hyperbola if its asymptotes intersect at the origin? What must be true of the foci of a hyperbola? [reveal-answer q=”fs-id2227199″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2227199″] The foci must lie on the transverse axis and be in the interior of the hyperbola. [/hidden-answer] If the transverse axis of a hyperbola is vertical, what do we know about the graph? Where must the center of hyperbola be relative to its foci? [reveal-answer q=”fs-id2227214″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2227214″] The center must be the midpoint of the line segment joining the foci. [/hidden-answer] Algebraic For the following exercises, determine whether the following equations represent hyperbolas. If so, write in standard form. [reveal-answer q=”fs-id2218627″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2218627″] yes[/hidden-answer] [reveal-answer q=”fs-id1639560″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1639560″] yes[/hidden-answer] For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. [reveal-answer q=”fs-id2129385″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2129385″] vertices:foci:asymptotes: [/hidden-answer] [reveal-answer q=”fs-id1688364″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1688364″] vertices:foci:asymptotes: [/hidden-answer] [reveal-answer q=”fs-id2609360″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2609360″] vertices:foci:asymptotes: [/hidden-answer] [reveal-answer q=”fs-id1681059″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1681059″] vertices:foci:asymptotes: [/hidden-answer] [reveal-answer q=”fs-id1407002″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1407002″] vertices:foci:asymptotes: [/hidden-answer] [reveal-answer q=”fs-id2272156″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2272156″] vertices:foci:asymptotes: [/hidden-answer] [reveal-answer q=”fs-id1673399″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1673399″] vertices:foci:asymptotes: [/hidden-answer] [reveal-answer q=”fs-id1914290″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1914290″] vertices:foci:asymptotes: [/hidden-answer] For the following exercises, find the equations of the asymptotes for each hyperbola. [reveal-answer q=”fs-id1673755″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1673755″] [/hidden-answer] [reveal-answer q=”fs-id1837820″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1837820″] [/hidden-answer] Graphical For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. [reveal-answer q=”1741219″]Show Solution[/reveal-answer][hidden-answer a=”1741219″][/hidden-answer] [reveal-answer q=”1741352″]Show Solution[/reveal-answer][hidden-answer a=”1741352″][/hidden-answer] [reveal-answer q=”2465308″]Show Solution[/reveal-answer][hidden-answer a=”2465308″][/hidden-answer] [reveal-answer q=”2342989″]Show Solution[/reveal-answer][hidden-answer a=”2342989″][/hidden-answer] [reveal-answer q=”2343121″]Show Solution[/reveal-answer][hidden-answer a=”2343121″][/hidden-answer] [reveal-answer q=”1743280″]Show Solution[/reveal-answer][hidden-answer a=”1743280″][/hidden-answer] [reveal-answer q=”1743413″]Show Solution[/reveal-answer][hidden-answer a=”1743413″][/hidden-answer] For the following exercises, given information about the graph of the hyperbola, find its equation. Vertices atandand one focus at [reveal-answer q=”fs-id1388398″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1388398″] [/hidden-answer] Vertices atandand one focus at Vertices atandand one focus at [reveal-answer q=”fs-id2303260″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2303260″] [/hidden-answer] Center:vertex:one focus: Center:vertex:one focus: [reveal-answer q=”fs-id1864081″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1864081″] [/hidden-answer] Center:vertex:one focus: For the following exercises, given the graph of the hyperbola, find its equation. [reveal-answer q=”fs-id1926837″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1926837″] [/hidden-answer] [reveal-answer q=”fs-id1926930″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1926930″] [/hidden-answer] [reveal-answer q=”fs-id1928090″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1928090″] [/hidden-answer] Extensions For the following exercises, express the equation for the hyperbola as two functions, withas a function ofExpress as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. [reveal-answer q=”fs-id1949870″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1949870″] [/hidden-answer] [reveal-answer q=”fs-id1944604″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1944604″] [/hidden-answer] Real-World Applications For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes and its closest distance to the center fountain is 5 yards. [reveal-answer q=”fs-id1925647″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1925647″] [/hidden-answer] The hedge will follow the asymptotes and its closest distance to the center fountain is 6 yards. The hedge will follow the asymptotesand and its closest distance to the center fountain is 10 yards. [reveal-answer q=”fs-id1925839″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1925839″] [/hidden-answer] The hedge will follow the asymptotesand and its closest distance to the center fountain is 12 yards. The hedge will follow the asymptotes and its closest distance to the center fountain is 20 yards. [reveal-answer q=”fs-id1274757″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1274757″] [/hidden-answer] For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object’s path. Give the equation of the flight path of each object using the given information. The object enters along a path approximated by the lineand passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line The object enters along a path approximated by the lineand passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line [reveal-answer q=”fs-id1920341″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id1920341″] [/hidden-answer] The object enters along a path approximated by the lineand passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line The object enters along a path approximated by the lineand passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line [reveal-answer q=”fs-id2590330″]Show Solution[/reveal-answer] [hidden-answer a=”fs-id2590330″] [/hidden-answer] The object enters along a path approximated by the lineand passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line Glossary center of a hyperbola : the midpoint of both the transverse and conjugate axes of a hyperbola conjugate axis : the axis of a hyperbola that is perpendicular to the transverse axis and has the co-vertices as its endpoints hyperbola : the set of all pointsin a plane such that the difference of the distances betweenand the foci is a positive constant transverse axis : the axis of a hyperbola that includes the foci and has the vertices as its endpoints

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