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https://blog.csdn.net/qq_41933331/article/details/79797039
C语言：已知A大于B大于C，且A+B+C小于100，求满足1/A平方+1/B平方=1/C平方的共有多少组？_已知a>b>c,abc为整数,且a+b+c<100-CSDN博客 博客 下载 学习 社区 GitCode InsCodeAI 会议 搜索 AI 搜索 登录 登录后您可以： 复制代码和一键运行 与博主大V深度互动 解锁海量精选资源 获取前沿技术资讯 立即登录 会员·新人礼包 消息 历史 创作中心 创作 C语言：已知A大于B大于C，且A+B+C小于100，求满足1/A平方+1/B平方=1/C平方的共有多少组？ C语言求解勾股定理变式 Caleb_Sung于 2018-04-02 23:30:31 发布 阅读量8.9k收藏 11 点赞数 4 CC 4.0 BY-SA版权 分类专栏：大学C语言程序设计C语言入门基础 版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。 本文链接： 大学C语言程序设计 同时被 2 个专栏收录 81 篇文章 订阅专栏 C语言入门基础 80 篇文章 订阅专栏 题目来源：大工慕课 链接 作者：Caleb Sung 题目要求 C语言：已知A>B>C，且A+B+C<100，求满足1/A^2+1/B^2=1/C^2的共有多少组？ 注意事项 这道题的要求没有加入大于零的范围，因而从这一点考虑会导致无穷多解，所以我强行加了这一条件。 解答示例 ``` include int main(){ int a, b, c, num=0; for(c=0; c<100; c++){ for(b=c+1; b<100; b++){ for(a=b+1; a<100; a++){ if((bbcc+aacc==aabb) && (a+b+c<100)){ printf("A=%d, B=%d, C=%d\n", a, b, c); num++; } } } } printf("共有%d种组合。", num); return 0; } ``` AI运行代码 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 运行结果 确定要放弃本次机会？ 福利倒计时 : : 立减 ¥ 普通VIP年卡可用 立即使用 Caleb_Sung 关注关注 4点赞 踩 11 收藏 觉得还不错? 一键收藏 0评论 分享复制链接 分享到 QQ 分享到新浪微博 扫一扫 举报 举报 专栏目录 参与评论 您还未登录，请先 登录 后发表或查看评论 关于我们 招贤纳士 商务合作 寻求报道 400-660-0108 kefu@csdn.net 在线客服 工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号 京网文〔2020〕1039-165号 经营性网站备案信息 北京互联网违法和不良信息举报中心 家长监护 网络110报警服务 中国互联网举报中心 Chrome商店下载 账号管理规范 版权与免责声明 版权申诉 出版物许可证 营业执照 ©1999-2025北京创新乐知网络技术有限公司 Caleb_Sung 博客等级 码龄8年 143 原创560 点赞 1833 收藏 515 粉丝 关注 私信 分类专栏 C语言入门基础80篇 Linux入门基础11篇 大学C语言程序设计81篇 Java与面向对象编程18篇 Visual Basic程序设计19篇 Флаг Ленина 数据结构11篇 Python1篇 数据挖掘1篇 折腾3篇 Windows2篇 C++9篇 Coursera7篇 展开全部收起 上一篇： C语言解决猴子吃桃子问题 下一篇： C语言：求1+1/2+1/4+1/7+1/11+1/16+1/22+…的值，直到最后一项的值小于10的-5次幂 最新评论 让Onedrive云盘同步本地任意一个文件夹（适用于Windows） Steven Wu 666:感谢，地址路径必须加引号才能连接成功。 支持Win10：科学安装Visual Basic 6.0教程 baidu_40781980:分享的安装链接失效了，大佬可以再给分享一下吗 C语言：使用函数计算一个数的阶乘 天蓝小新:long fac(int x)这个语句执行的时候怎么报错那，换成int就行了是因为什么呐，溢出了吗 C语言：利用指针编写程序，统计输入的字符串中大写字母、小写字母和数字字符的数量 LiNaKRbCs:elseif那一行多了个； C语言：求4x4矩阵外围元素之和 2401_83938797:可是这个最后输出的不是矩形方阵啊 大家在看 为什么90%的C#开发者在SQLite SELECT上栽了跟头？5个关键技巧让你轻松驾驭 800 DeepSeek“出圈热”启示录：AI技术赋能下的商业模式创新与正能量实践 在 WPF 中，StackPanel 是一个布局容器控件，用于按水平或垂直方向按顺序排列子控件，适合快速构建简单、线性布局的界面 人工智能在航空领域的应用、挑战与未来发展路径探析 614 最新文章 批处理实现删除上一级文件夹保留本级文件夹及内容 利用按键精灵实现QQ群发和微信群发 按键精灵实现自动化点qq名片赞 2021年 1篇 2020年 1篇 2019年 13篇 2018年 130篇 目录 题目要求 注意事项 解答示例 运行结果 展开全部 收起 目录 题目要求 注意事项 解答示例 运行结果 展开全部 收起 上一篇： C语言解决猴子吃桃子问题 下一篇： C语言：求1+1/2+1/4+1/7+1/11+1/16+1/22+…的值，直到最后一项的值小于10的-5次幂 分类专栏 C语言入门基础80篇 Linux入门基础11篇 大学C语言程序设计81篇 Java与面向对象编程18篇 Visual Basic程序设计19篇 Флаг Ленина 数据结构11篇 Python1篇 数据挖掘1篇 折腾3篇 Windows2篇 C++9篇 Coursera7篇 展开全部收起 登录后您可以享受以下权益： 免费复制代码 和博主大V互动 下载海量资源 发动态/写文章/加入社区 ×立即登录 评论 被折叠的 条评论 为什么被折叠?到【灌水乐园】发言 查看更多评论 添加红包 祝福语 请填写红包祝福语或标题 红包数量 个 红包个数最小为10个 红包总金额 元 红包金额最低5元 余额支付 当前余额 3.43 元 前往充值 > 需支付：10.00 元 取消 确定 成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝 规则 hope_wisdom 发出的红包 实付 元 使用余额支付 点击重新获取 扫码支付 钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。 余额充值 确定 取消 举报 选择你想要举报的内容（必选） 内容涉黄 政治相关 内容抄袭 涉嫌广告 内容侵权 侮辱谩骂 样式问题 其他 原文链接（必填） 请选择具体原因（必选） 包含不实信息 涉及个人隐私 请选择具体原因（必选） 侮辱谩骂 诽谤 请选择具体原因（必选） 搬家样式 博文样式 补充说明（选填） 取消 确定 AI助手 AI 搜索 智能体 AI 编程 AI 作业助手 下载APP 程序员都在用的中文IT技术交流社区 公众号 专业的中文 IT 技术社区，与千万技术人共成长 视频号 关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
5001
https://economicsfromthetopdown.com/2020/10/19/working-with-google-ngrams-a-data-wrangling-tale/
Economics from the Top Down New ideas in economics and the social sciences Working With Google Ngrams: A Data-Wrangling Tale This post begins with a sigh. For the last month, I’ve been working on a project that analyzes word frequency in economics textbooks. I’d hoped to have the final write up done by now. But I don’t … for reasons explained here. I’m calling this post ‘Working with Google Ngrams’. But even if you’re not interested in ngrams (i.e. word frequency), there’s useful information here. Hence the subtitle ‘A Data-Wrangling Tale’. When you look at good empirical work, the results seem effortless — as if they leapt out of the data. But the reality is that doing empirical work can be frustratingly difficult. That’s because just getting the data and learning how to work with it can take a lot of time. With that in mind, I’m going to recount my journey into analyzing Google Ngrams. It follows a similar trajectory to most of my empirical work. The deeper you go into the analysis, the harder it gets. So strap on your boots and let’s wade into the empirical muck. Answering a question … with data Empirical analysis begins with a question. In this case my question was — how does word frequency in economics textbooks differ from word frequency in mainstream English? After you’ve asked the question, the next step is to look for data that can answer it. Finding the right data can take a lot of time. Here are two tips. First, check if anyone else has asked a similar question. If they have, read their work and see where they got their data. Second, try the Google Dataset Search. It’s surprisingly good, but also frustrating because it will show you all the paywalled data that you probably can’t get. Third, try a generic internet search. Whatever you do, don’t expect fast results. Wading through the empirical muck takes time. Back to my question. When it comes to word frequency in economics textbooks, obviously I need a sample of textbooks. Two decades ago, that would have meant spending thousands of dollars buying books. Considering economics textbooks are filled with garbage, that’s money poorly spent. Fortunately, we now have Library Genesis, a pirate database with millions of books. There you can get a PDF of nearly any economics textbook you want (or any other book, for that matter). What about a sample of mainstream English writing? For that, the best data comes from the Google Books corpus. According to Google, this the world’s most comprehensive index of full-text books. As part of this books corpus, Google has compiled an ‘ngram’ dataset that analyzes text frequency. What’s an ‘ngram’? It’s a combination of text separated by spaces. So the text ‘Bob’ would be a 1-gram. ‘Bob is’ would be a 2-gram. And ‘Bob is fun’ would be a 3-gram. In this example, our ngrams are combinations of words. But they need not be. “hjk” counts as a 1-gram. “hjk sdf” is a 2-gram. And “hjk sdf oiu” is a 3-gram. As long as text is separated by a space, it counts as an ngram. Google has created the Ngrams database, which analyzes text frequency in its books corpus. If you’re interested in quantitative analysis of language, the Ngrams data is a wonderland. (Side note: I used to think that Google created the Ngram database out of scientific curiosity. But now I realize that analyzing text frequency is an essential part of creating a search algorithm. So the Ngram database probably stems from Google’s search index for its books corpus.) Exploring the data After you’ve found the data, the next step is to start exploring it (visually, if it’s convenient). If you’re dealing with government statistics, many agencies have pre-built tools for data exploration. Use them. (For US government data, the Federal Reserve Economic Data portal is the most convenient.) Think of the data-exploration step as getting a feel for the data. You’re not doing any analysis yet. You’re just seeing what the data looks like. Back to the Ngram database. Google provides a portal for exploring its ngram data — the Ngram Viewer. It’s beautifully simple. You enter words (or phrases) and the Ngram Viewer will plot their relative frequency over time. Here’s what it looks like: Pretty cool, right? Downloading the data You’ve found some data that looks interesting. The next step is to download it onto your computer so you can work with it. Often, this is as simple as clicking Download. But in the case of the Ngram Viewer, this option isn’t available. (Look for yourself. The button in the top right corner looks promising. It has an option to Download raw data. But that doesn’t download the data on your graph. It takes you to the bulk repository for the Ngram database, which we’ll discuss shortly.) Although there’s no Download button, you can still download data from the Ngram Viewer. But you’ll have to do some hacking. Software developer Johannes Filter walks you through the steps here. Basically, you get the data you want on the screen. Then you open your web browser’s developer tools and look at the page’s source code. From there, you copy and paste the ngram data into a text file where you can work with it. (Note: the Ngram viewer automatically smooths the data. For empirical work, you want the unsmoothed data. Get this by setting Smoothing to 0.) This browser hack isn’t pretty, but it works. If you want data for one or two words, this hack is probably the fastest method. (It’s how I got the data for free-market speak analyzed here). That said, if you want data for more than a few words, the browser hack is too clunky. Using R packages I love R. If you’re not familiar, R is a programming language for statistical computing. Over the years, it’s evolved into a huge ecosystem of tools for doing analysis. If there’s empirical work you want to do, chances are there’s an R package for it. In the case of the Google Ngram dataset, there’s a convenient R package called ngramr. Before we dive into the ngramr package, let’s talk about when/why you should use software packages to get data. To use a package, you have to spend time learning its syntax. If you’re doing a one-off analysis, you probably don’t want to invest this time. But if you’re getting data repeatedly, it’s worth your time to learn how to use a dedicated package. Back to the R ngramr package. In this case, the syntax is simple. The package provides a function called ngram() that retrieves data from Google. To get data for the word ‘monopoly’, you’d enter: It will return a data frame with the relative frequency of the word ‘monopoly’ in different years: That’s easy! Now suppose you want to look up data for several words. Still easy. The ngram() function can take a ‘vector’ input (i.e. a list). To use this feature, first bind the words you want in a vector using R’s c() function. Then dump that vector into the ngram function. Suppose we want data for the words ‘monopoly’, ‘demand’ and ‘price’. First, bind the words into a vector: Then throw that vector into the ngram function: Perhaps you only want data starting in 2018. No problem. Just specify the start year: Here’s the result: As you can see, the ngramr package is easy to use. If you want Ngram data for a few hundred words (or even a few thousand), this package is your best bet. Scaling up A rule of thumb for empirical work is that your method will depend on how much data you’re analyzing. Let’s use the Ngram data to illustrate. If you want data for one or two words, the fastest approach is probably the browser hack. If you want data for hundreds of words, your best bet is the ngramr package. But what if you want data for hundreds of thousands of words? Then you need a different approach. The problem with the ngramr package is that it’s ‘slow’. I’ve used scare quotes here because you have to take the word ‘slow’ in context. The ngram() function limits you to retrieving data for 12 words at a time. So if you want data for hundreds of words, you’ll have to make a loop and use the ngram() function multiple times. That’s no big deal. However, the main limitation is that ngram() takes about 1 second to get data for 12 words. If you want data for a few hundred words, it will take a few minutes of computation. That’s not bad. But if you want data for 500,000 words, you’re looking at more than 11 hours of computation. That’s too slow for my taste. The reason ngram() is slow, it turns out, is because it’s actually an automated version of the browser hack I discussed above. Under the hood, ngram() goes to the Ngram Viewer website, searches for the words you want, and then scrapes the data from the html. That’s pretty clever. But it’s also slow. Back to empirical work in general. A rule of thumb is that tools that are convenient for small-scale analysis don’t work when you move to the large scale. When you want to analyze big data, you need to do a bulk download. Bulk downloads When you’re doing large-scale analysis, bulk download centers are your friend. They allow you to download an entire dataset in one go. That’s helpful when you want to analyze a lot of data. Here’s an example. A few years ago, I analyzed the price change of every commodity on the consumer price index (CPI). (See the results here). That would have been unthinkable if I had to download each price series individually. Instead, I bulk downloaded the entire CPI dataset here. Bulk downloads, however, come with a cost. They’re usually frighteningly opaque. Unlike with small-scale data, bulk downloads generally don’t describe the data. Instead, you’ll see a bunch of codes. Here, for example, is the bulk data for the consumer price index. There’s nothing but numbers and codes. The cost of using bulk downloads is that you have to spend time figuring out how the dataset is coded. You do this using a codebook. (Here’s the CPI codebook.) The problem is that codebooks are usually written by statisticians who already know the data well. For the novice analyst, codebooks can be as opaque as the raw data. I’ve spent many frustrating hours trying to decipher a database’s codebook. Back to the Ngram data. The bulk download center for the Google Ngram data lives here: Unlike with the Ngram viewer, here there are no graphics, no search bars, and no buttons. There’s just text and hyperlinks. It’s up to you to figure out what to do. After using the data for a few months, here’s what I’ve learned. First, there are three different versions of the Ngram dataset: Second, the entire datasets are unmanagably huge. So you’ll have to work with a subset of the data. For me, that means downloading only the ‘1-gram’ data (text frequency for single words). Third, each version of the database is structured differently. In short, when you start working with the raw Ngram data, you’re opening a can of worms. The 2012 Ngram corpus Let’s start with the 2012 corpus, because it’s the easiest to work with. The data comes in compressed text files. When you unpack them, each file will contain something like this: As is typical with bulk downloads, there’s no labels. It’s up to you to figure out what the data means. It turns out that the first column is the ngram. The second column is the year. The third column is the number of times the ngram occurs. And the fourth column is the number of ‘volumes’ (i.e. books) in which the ngram is observed. Alright, we’ve figured out what the data means. Now how do we work with it? To work with data, we usually read the whole dataset into the computer’s memory. But with the Ngram data, that’s not an option. The 2012 1-gram dataset is about 24 GB. Even if your computer has that much RAM, you’d need even more if you want to manipulate the data. So reading the whole dataset into memory is probably not going to work. So what should we do? The solution is to do our analysis in pieces. The 2012 1-gram dataset comes in 26 different files — one for each letter of the alphabet. Each file is about 1 GB, which can easily fit into your computer’s memory (unless you’re using something ancient). So here’s the approach. You read each file into memory. Then you look for the words that you want. You keep the data for these words and discard the rest. Then you move on to the next Ngram file. Repeat. Data formats Alright, we’re caught up to where I was two weeks ago. I’d first retrieved ngram data using the browser hack. Then I’d run some analysis with the ngramr package. After realizing that this was going to be too slow, I downloaded and analyzed the 2012 1-gram dataset. I was writing up my results when … I discovered that there was a 2020 Ngram dataset. No big problem, you’d think. Just download the new data and rerun your analysis. If only it were that simple. The problem, I discovered, was that the 2020 Ngram data comes in a completely different format — one that is not easy to use. To understand the problem, let’s talk data formats. Big datasets are usually stored as plain text. That means if you open up the file with a text browser, you’ll see a bunch of text. As I’ve already discussed, the 2012 Ngram data looks like this: This is called ‘tab separated data’. Within each line, the different entries are separated by a TAB. Another common approach is to separate data by commas, like this: The comma-separated approach is a bit harder for humans to look at. But for your computer, it’s easy to read. In principle, data can be separated (‘delimited’) by any character. But commas and tabs are the most common. What’s most important here is not the data-separating character, but the fact that the 2012 Ngram data is ‘rectangular’. Each line has the same number of entries. In other words, the data has a fixed number of columns. This is important because most data-reading tools work best when the data is rectangular. Let’s use R as an example. In R, there’s a whole ecosystem for reading rectangular data. In ‘base R’, there’s read.csv(), which parses comma-separated files. For tab-separated data, there’s read.table(). These functions will read the data and format it into an easily-manipulable data frame. For big datasets, these base functions are a bit slow. If you want speed, I recommend the fread() function from the data.table package. There’s also a family of functions available in the readr package. What all of these functions have in common is that they’re built for parsing rectangular data. If your data comes in some other format, you’re out of luck. That brings me to the 2020 Ngram corpus. The 2020 Ngram corpus Unlike the 2012 corpus, the 2020 Ngram corpus is not stored in a ‘rectangular’ format. Instead, the data looks like this: Here, each ngram gets its own line. Year, count and volume information then follow, separated first by a TAB, and then by commas. What’s important is that the number of entries in each line varies by ngram. (Some ngrams have data covering only a few years. Others have data covering hundreds of years.) In other words, the 2020 Ngram data is not rectangular. So standard data-reading functions won’t work. Why the odd formatting? My guess is that Google used it to save space. In the rectangular format, notice that each ngram gets repeated for each different year in which it occurs. In the new format, each ngram appears once. When you have billions of ngrams, that saves a lot of space. Still, the 2020 1-gram corpus is huge — about 44 GB. So we need a way to parse the data fast and efficiently. Let’s pause here and talk about a trade off in data analysis (or any software development, for that matter). There’s always a trade off between computation time and programmer time. If you want code that executes quickly, it will take more time to program. If you want code that’s written quickly, it will execute more slowly. A good rule of thumb is that your time is more important than your computer’s time. (Remember that you can do other things while your computer crunches numbers.) If you spend 10 hours writing code that saves the computer 10 seconds, you’re wasting your time. But if 10 minutes of coding will save 10 hours of computation, that’s worth your while. Back to the Ngram data. In R, there’s a base function called readLines() that does what it says — it takes lines from a file and reads them into memory. What’s useful about this function is that it’s form agnostic. It doesn’t care if the data is rectangular. The cost, however, is that the data gets returned in a way that’s not very usable. You’ll get a character vector, where each element contains the characters on a given line. When read with readLines(), the 2020 Ngram data would look like this (\t is computer code for TAB): This is ugly. Yes, the data is in your computer’s memory. But without more cleaning and processing, the data is useless. So the question is, how do you want to clean it? Answering this question depends on how much data you have. In a file with a few hundred lines, I’d use R to read and clean the data. The code would take me a half hour to write and would execute in milliseconds. The Google Ngram data, however, has billions of lines to read and clean. Doing that in R won’t take milliseconds. It could take hours. If I’m going to run the code once, that’s not so bad. I press ENTER, let the code run, and write some emails for an hour. The problem, though, is that empirical analysis almost never gets done once. You inevitably find mistakes, get new ideas, or expand your analysis. And so you rerun the code. That 1 hour run-time just to read and clean the data quickly becomes frustrating. If you’re going to run analysis many times, it’s probably worth it to write faster code. For me, that means turning to the programming language C++. Unlike R, which is an ‘interpreted language’, C++ is a ‘compiled’ language. That makes it much faster (to execute, not to code). One of the things I love about R is that it’s easy to integrate with C++. There’s a package called Rcpp that lets you write C++ code and import it seamlessly into R. Whenever I want speed, this is my go-to method. Back to the Ngram data. To crunch the 2020 data, I ended up writing two C++ functions. (The code lives here.) The first function is called get_ngrams(). As it suggests, this function parses an Ngram 2020 file and returns a list of all the ngrams in it. Run the function like this: The get_ngrams() function is useful because in most cases, we’re interested in only a tiny subset of the ngrams in the Ngram file. So we don’t want to waste time parsing all of the data. Instead, we run get_ngrams(), which will take a few seconds to execute. Then we have a list of the ngrams in the file. The next step is to select the ngrams we want, and get data for them. In my case, I’m interested in the ngrams that are actual English words. As an example, let’s get data for all of the ngrams that are in the grady_augmented word list (a list provide by the R lexicon package). The code would look like this Here’s what the code does. First we get a list of all the ngrams in the file. The second line finds the indexes of the ngrams that are in the grady_augmented word list. The third line gets data for these ngrams. This code allows me to extract data for hundreds of thousands of ngrams in about 5 seconds. That’s fast. With this code, I can parse and analyze the whole 2020 1-gram database in about 5 minutes. So when I inevitably rerun the analysis, I won’t have to wait long. (Just enough time to make a coffee.) That said, it took me a few days to write the C++ code. That’s because I took several wrong turns along the way. Did I waste my time? Perhaps. But I like to consider three factors besides raw time. First, I enjoy coding in C++ (I always learn something new). Second, I plan on using this code many times in the future. Third, I’m sharing my code (here). That means I’m not just saving my time. I’m potentially helping other researchers save time. How deep do you want to go? Alright, let’s wrap up this foray into Ngram analysis. If you’re a researcher who’s interested in Ngram data, I hope the above information is useful. That said, I know that the majority of readers aren’t Ngram researchers. Still, there are general principles that are useful. When you do empirical work, the deeper you go, the harder it gets. If you want to do exploratory analysis, it’s usually simple. Dump some data in a spreadsheet and plot it. But if you want to go deeper, it will take more time. You’ll probably have to use a new package. And if you want to go deeper still, you may have to write custom code. At each step, you need to think before you dive in. Do you want to pay the costs of going deeper? The reward is better analysis. The cost is time … and lot’s of it. And that brings me to the third reason for telling you about my foray into Ngram analysis. When I finally (hopefully next week) show you my results for relative word frequency in economics textbooks, I want you to know how much damn work it took. Support this blog Economics from the Top Down is where I share my ideas for how to create a better economics. If you liked this post, consider becoming a patron. You’ll help me continue my research, and continue to share it with readers like you. Stay updated Sign up to get email updates from this blog. Email Address Keep me up to date This work is licensed under a Creative Commons Attribution 4.0 License. You can use/share it anyway you want, provided you attribute it to me (Blair Fix) and link to Economics from the Top Down. Share this: Like this: Published by Blair Fix Political economist. Blogger. Muckraker. Foe of neoclassical economics. View all posts by Blair Fix Post navigation 4 comments […] you’re interested in working with Ngram data, I’ve written a post about my data-wrangling experience. I’ve also written some custom R code for importing Ngram 2020 data. It’s available at […] This is a really interesting post, thank you. Do you have a feeling of how to approach the following query: How many 2-grams contain a particular 1-gram? For example, how many 2-grams contain (not as a non-trivial substring) the 1-gram ‘went’? Instances: ‘I went’, ‘went to’, … I don’t think the the n-grams viewer allows you to do such a query, but if you have any suggestions I’d appreciate it. You are correct, the n-gram viewer will not answer such a query. To do that sort of analysis, you’d have to download the raw data yourself. I have not worked with the 2-gram data myself. Note that the datasets get progressively bigger as the n-gram length increases. Still, the technique would be straightforward. Get all the 2-gram data, find the strings that contain your 1 gram, and then analyze their frequency. Given the size of the dataset, though, this is going to be a computationally intensive task. […] the format of this ngram data is a bit unwieldy to work with. I wrote a post about my journey here. It includes some helpful coding […] Leave a ReplyCancel reply Support my research Follow by email Email Address Follow Mastodon: @blair_fix BlueSky: @blairfix Twitter: @blair_fix Discover more from Economics from the Top Down Subscribe now to keep reading and get access to the full archive. Type your email… Subscribe Continue reading
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https://study.com/academy/lesson/genetic-drift-activities-games.html
Genetic Drift Activities & Games | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Science Courses / Life Science Lesson Plans & Games Course Genetic Drift Activities & Games Instructor Nora JarvisShow bio Nora has a Master's degree in teaching, and has taught a variety of elementary grades. Cite this lesson Genetic drift is one of several forces that make up evolution. This lesson will provide activities and games for your class as you teach them about genetic drift. Create an account Table of Contents What Is Genetic Drift? Hypothetical Genetic Drift Simulating Genetic Drift Genetic Drift Speed Sort Show What Is Genetic Drift? ---------------------- Evolution is a long process, and there are many different factors that affect how organisms change over time. One aspect of evolution is called genetic drift, in which a population is altered by a chance event. This event results in a change in the genetic makeup of that population. As your students learn about genetic drift, it will be important to give them opportunities to identify examples and learn about what this form of adaptation entails through hands-on and engaging activities and games. Lesson Course 1.3K views Hypothetical Genetic Drift -------------------------- After your students have a basic understanding of genetic drift, challenge them by having them make up one way that an organism could experience genetic drift. Divide your students into pairs and have them decide on an organism to use for their project. They might do a plant or animal, but encourage them to choose one that they're already fairly familiar with. Students then decide on an event that could cause genetic drift in their population. These ideas can be funny or unrealistic, but they should be able to explain how the alleles change as the generations reproduce. Ask your students to create a genetic map that depicts how the alleles in their populations experience a genetic drift to alter the organism. Simulating Genetic Drift ------------------------ Use classroom materials to simulate a random event in a population and the genetic drift that could result. Divide your students into groups of 5 to 6. Give each group a bag of pom-poms of varying colors (choose 4-5 colors), and have each student pull out 10 pom-poms from the bag. They should divide their pom-poms into color groups and determine the percentage of each color. For instance, they might have 3 red pom-poms, which means the reds make up 30% of the population. Students combine all the pom-poms that each member picks out, along with the rest of the bag of pom-poms, and complete the same percentage process for the whole population. Then each student chooses 10 pom-poms at random, and re-calculates the percentage that the pom-poms make. In their groups, students answer the following questions: How are the percentages different from each individual pull versus the whole population? How do the pom-poms represent the alleles in a population? Genetic Drift Speed Sort ------------------------ Help your students better understand the difference between bottleneck and the founder effect as causes of genetic drift. With your students, develop a list of different events that can cause genetic drift. They might suggest things like: Earthquake that kills a lot of the population An animal is close to being extinct Disease wipes out significant part of the population Small group gets stranded on an island Insular community that keeps away from others Adventurers who start a small settlement where there aren't many other people (like the American colonies) Divide your students into pairs and have each group write each event on an index card. Once all the groups have the cards written, set a timer and have each group work together to sort the card into a pile of Bottleneck and Founder Effect. Whichever group finishes first, wins. Make sure to check their piles to correct for any misunderstandings. Next, have each pair compete against each other. One partner sets the timer while the other partner sorts the cards. Then, they switch. Whoever has the lowest time, wins the game. 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Evolution Lesson Plans How Darwin's Observations Showed the Process of Natural Selection 7:24 Evolution Lesson Plan Theories of Evolution: Lamarck vs. Darwin 7:30 Evolution Lesson Plan for Elementary School Natural Selection Lesson for Kids 3:29 Evolution Lesson for Kids: Definition & Theory Natural Selection Activities Natural Selection & Antibiotic Resistance Activity Artificial Selection Lesson Plan Artificial Selection | Overview, Processes & Examples 5:05 Genetic Variability, Random Mutation & Artificial Selection Selective Breeding Activities Genetic Drift Lesson Plan What is Genetic Drift? | Types, Examples & Importance 4:00 Genetic Drift Activities & Games Next Lesson Animal Adaptation Activities Animal Adaptation Games Charles Darwin Lesson Plan Charles Darwin | Theory of Evolution & Natural Selection 7:01 Charles Darwin: Voyages, Theories & Works 5:50 Evolution Activities & Games Natural Selection Classroom Games & Activities Social Darwinism Classroom Activities & Games Charles Darwin Lesson Plan for Elementary School Charles Darwin Lesson for Kids: Biography & Facts Speciation Lesson Plan Ecological Speciation Definition & Examples 4:55 Charles Darwin Activities Monohybrid & Dihybrid Cross Activities Lamarckism Activities Darwinian Revolution Activities Charles Darwin Activities for Kids Natural Selection Lesson Plan for Kids Convergent & Divergent Evolution Lesson Plan Natural Selection Lesson Plan Evolution Unit Plan for Middle School Natural Selection Project Ideas Your Inner Fish Discussion Questions Evolution Unit Plan for High School Hibernation and Migration Activities Ch 9. 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https://www.sciencedirect.com/science/article/pii/S1674205214603944
Phenylpropanoid Biosynthesis - ScienceDirect Skip to main contentSkip to article Journals & Books View PDF Download full issue Search ScienceDirect Outline ABSTRACT Key words INTRODUCTION THE CENTRAL PHENYLPROPANOID PATHWAY LIGNIN PHENYLPROPANOID ESTERS AND AMIDES (ACYLATED POLYAMINES) ANTHOCYANINS, AURONES, (ISO)FLAVONOIDS, AND STILBENES HYDROLYZABLE AND NON-HYDROLYZABLE PROANTHOCYANIDINS AND TANNINS C6–C1 METABOLITES PHENYLPROPENE VOLATILES COUMARINS EXOTIC PHENYLPROPANOIDS MIXED POLYMERS, SUBERIN, CUTIN, AND SPOROPOLLENIN CONCLUDING REMARKS FUNDING Acknowledgments REFERENCES Cited by (2335) Figures (5) Molecular Plant --------------- Volume 3, Issue 1, January 2010, Pages 2-20 Review Article Phenylpropanoid Biosynthesis Author links open overlay panelVogt Thomas a Show more Outline Add to Mendeley Share Cite rights and content Under an Elsevier user license Open archive ABSTRACT The general phenylpropanoid metabolism generates an enormous array of secondary metabolites based on the few intermediates of the shikimate pathway as the core unit. The resulting hydroxycinnamic acids and esters are amplified in several cascades by a combination of reductases, oxygenases, and transferases to result in an organ and developmentally specific pattern of metabolites, characteristic for each plant species. During the last decade, methodology driven targeted and non-targeted approaches in several plant species have enabled the identification of the participating enzymes of this complex biosynthetic machinery, and revealed numerous genes, enzymes, and metabolites essential for regulation and compartmentation. Considerable success in structural and computational biology, combined with the analytical sensitivity to detect even trace compounds and smallest changes in the metabolite, transcript, or enzyme pattern, has facilitated progress towards a comprehensive view of the plant response to its biotic and abiotic environment. Transgenic approaches have been used to reveal insights into an apparently redundant gene and enzyme pattern required for functional integrity and plasticity of the various phenylpropanoid biosynthetic pathways. Nevertheless, the function and impact of all members of a gene family remain to be completely established. This review aims to give an update on the various facets of the general phenylpropanoid pathway, which is not only restricted to common lignin or flavonoid biosynthesis, but feeds into a variety of other aromatic metabolites like coumarins, phenolic volatiles, or hydrolyzable tannins. Previous article in issue Next article in issue Key words Phenylpropanoid biosynthetic pathway flavonoid anthocyanin tannin coumarin volatiles lignin INTRODUCTION Phenylpropanoids contribute to all aspects of plant responses towards biotic and abiotic stimuli. They are not only indicators of plant stress responses upon variation of light or mineral treatment, but are also key mediators of the plants resistance towards pests (La Camera et al., 2004). They promote invasion of new habitats (Bais et al., 2003) and provide the biochemical resources for successful reproduction (Dudareva et al., 2004). Phenylpropanoid-based polymers, like lignin, suberin, or condensed tannins, contribute substantially to the stability and robustness of gymnosperms and angiosperms towards mechanical or environmental damage, like drought or wounding. The magnificent diversity of phenylpropanoids, in principle, is the result of efficient modification and amplification of a very limited set of core structures, derived from the shikimate pathway (Herrmann, 1995). Aromatic amino acid lyases link the pool of phenylalanine and tyrosine to biosynthetic pathways sometimes downgraded as secondary metabolism, which, however, turn out to be equivalently relevant to plant survival as photosynthesis or the citrate cycle. A clear differentiation between secondary and primary metabolites may seem obsolute (Hartmann, 2007). The diversity and plasticity of the resulting natural products, more specifically the phenylpropanoid profile, are then achieved by a set of enzymes organized usually in superfamilies, like oxygenases, ligases, oxidoreductases, and various superfamilies of transferases (Figure 1). Some of these enzymes might exhibit overlapping specificities in vitro, but their developmentally and spatially controlled expression specifically contributes to tissue and plant-specific chemical phenotypes. Download: Download full-size image Figure 1. Schematic View of the Diversity of the Secondary Metabolite Formation. In the last decade, knockout mutations and RNAi-mediated suppression in Arabidopsis and other model plants have facilitated the identification and functional characterization of genes and enzymes (Alonso et al., 2003; Ossowski et al., 2008). A plethora of mutants is available for candidate genes central to the various branches of the phenylpropanoid pathway. Identification of single loss-of-function mutants may not be immediately accessible based on a clear phenotype such as impaired growth or fertility (Briggs et al., 2006). In the absence of any phenotype, high throughput screening of phenylpropanoid patterns by GC– or UPLC–MS may enable the correlation of metabolite pattern with transcript profiles in an organ or tissue specificity manner (Matsuda et al., 2008). In some cases, this approach may reveal the impact of a single phenylpropanoid pathway gene on other metabolic pathways (Rohde et al., 2004). Metabolite profiling can be combined with immunofluorescence or imaging techniques based on intrinsically fluorescent proteins, like GFP, to follow also the expression and interaction of participating enzymes down to the cellular level (Dixit et al., 2006). In Arabidopsis and poplar, nearly all genes required for phenylpropanoid and lignin formation are encoded by small gene families (Hamberger et al., 2007). Clustering of these sequences in cladograms suggests a variety of functions other than the bona fide suggested roles of selected members of a gene family based on sequence information. If genes of a single pathway are compared, whole genome data indicate that in poplar, bean, or alfalfa, the pattern of genes involved in flavonoid hydroxylation appears more complex as compared to Arabidopsis (Hamberger et al., 2007). In recent years, various excellent reviews summarized the current knowledge on structural genes involved in phenylpropanoid, specifically lignin and flavonoid formation, regulatory transcription factors, hormonal control of the whole pathways by jasmonate or auxin and evolution of pathway genes from primary metabolism (Weisshaar and Jenkins, 1998; Pichersky and Gang, 2005; D'Auria and Gershenzon, 2005; Ober, 2005; Boudet, 2007). This report tries to emphasize that phenylpropanoid derived metabolites comprise and contribute to multiple biosynthetic branches other than lignin and flavonoid biosynthesis only (Figure 2). Some decisive steps in important pathways were resolved only recently, such as 2′-hydroxylation in coumarin biosynthesis (Kai et al., 2008). In a few cases, new organ-specific pathways could be established in the last couple of years based on functional transcriptomics (Alves-Ferreira et al., 2007; Ehlting et al., 2008) including the formation of tapetum-specific trisacyl-polyamine conjugates of Arabidopsis (Matsuno et al., 2009; Fellenberg et al., 2009). Other pathways, like sinapate ester formation in Arabidopsis, have been clarified in detail in the last decade and may have direct practical impacts on current breeding programs, resulting in consumer or farmer benefits (Bhinu et al., 2009). Several new ideas and insights into the complex organization of pathways are due to the advancement of analytical, chemical, and biomolecular tools, like whole genome tilling arrays and application of fluorescent biosensors (Herbst et al., 2009). These techniques will be a critical driving force towards understanding plant biology (Yonekura-Sakakibara et al., 2008; Zeller et al., 2009). They promote rapid annotation of genes, the structural and functional characterization of enzymes, realize quantification of even trace amounts of metabolites, and enable identification of signal transduction pathways in living cells towards a comprehensive view of the plant organism. Download: Download full-size image Figure 2. Diversification of Phenylpropanoids Based on the General Phenylpropanoid Pathway. The metabolites of the shikimate pathway and the central metabolite, 4-coumaroyl CoA, are shaded in gray. THE CENTRAL PHENYLPROPANOID PATHWAY The plant shikimate pathway is the entry to the biosynthesis of phenylpropanoids. Its intracellular, plastidial location and complex regulation have been investigated for more than a decade (Schmid and Amrhein, 1995). Regulation of the entire pathway in plants can be performed at multiple levels. Transcriptional control was shown for 3-deoxy-D-arabinose-heptulosonate (DHAP) synthase (Herrmann and Weaver, 1999), whereas arogenate and prephenate dehydratase are inhibited by phenylalanine, one end-product of the pathway (Yamada et al., 2008). The plastidal location of two elusive steps of the shikimate pathway, arogenate dehydratase and dehydrogenase, leading to phenylalanine and tyrosine biosynthesis has recently been proven (Rippert et al., 2009). No recent molecular evidence is available for the location of prephenate aminotransferase, although its activity within a plastidal fraction of Sorghum bicolor was shown (Siehl et al., 1986). Therefore, a location of the complete pathway within plastids appears plausible. It is noteworthy that genes of the shikimate pathway in plants apparently do not originate from a single prokaryotic ancestor of cyanobacterial origin, but are likely derived from at least three different sources. A complex, multi-step shuffling of loss and gain of function occurred during phylogeny of this pathway, which might explain the multiple factors contributing to the genomic organization and expression of the pathway genes in Arabidopsis and other plants (Richards et al., 2006). With respect to its important contribution to the carbon flow within the plant, the response of individual genes of the shikimate pathway to variations in light or nutrient content appears tightly regulated and more complex than the transcriptional responses of the entry genes of phenylpropanoid and flavonid biosynthesis, encoding phenylalanine ammonia lyase (PAL) or chalcone synthase (CHS) (Lillo et al., 2008). PAL and tyrosine ammonia lyase (TAL) catalyze the non-oxidative deamination of phenylalanine to trans-cinnamate and direct the carbon flow from the shikimate pathway to the various branches of the general phenypropanoid metabolism. A crystal structure became available for a codon-optimized, heterologously expressed enzyme from Petroselinum crispum (Ritter and Schulz, 2004). The structural data are consistent with the unusual deamination reaction, which is independent of any co-factor. The required electrophilic prosthetic group, 4-methylidene-imidazole-5-one, is auto-catalytically formed. These data also provide evidence for the requirements of antioxidative SH-protective reagent, like dithiothreitol (DTT), for full enzyme activity in vitro, since an oxidized DTT-derivative was detected in the electron density map of the crystal, when DTT was present into the crystallization buffer. Replacing the characteristic amino acid His89 of the active site of TAL in bacteria by phenylalanine, a characteristic feature of PALs, converts TAL to PAL and vice versa (Watts et al., 2006; Louie et al., 2006). This is one rare example that a single amino acid exchange leads to a drastic change in substrate specificity. From an evolutionary perspective, plant and microbial PALs and TALs are part of a superfamily of enzymes from plants, funghi, and bacteria, and are likely derived from a precursor of the widespread histidine ammonia lyase (HAL) family in the histidine degradation pathway (Röther et al., 2002). Both can be distinguished from HALs by an additional N-terminal domain and by a more effective shielding of their active sites from molecular oxygen. Several copies of the PAL-genes are found in all plant species, comprising four genes in Arabidopsis, to five in poplar and nine in rice (Hamberger et al., 2007). The individual genes may respond differentially to biotic and abiotic stressors and their expression is developmentally and spatially controlled (Bhuiyan et al., 2009; Lillo et al., 2008). In poplar, an organ-specific expression of three of the PAL-genes has been suggested—all display a different function (Kao et al., 2002). Two genes are expressed in lignifying tissues; only one is involved in lignin formation. The second PAL-gene is specifically targeted to condensed tannin formation. The third gene is associated with flowering, although its functional role remains to be established (Hamberger et al., 2007). In contrast to Arabidopsis and poplar, in the Solanaceae, the entry step of the phenylpropanoid pathway is represented by a remarkable set of an estimated 20 putative PAL-genes, as shown for Lycopersicon esculentum (Chang et al., 2008). However, only a single gene appears to be strongly expressed in all tissues, whereas the remaining genes appear effectively silenced. This underutilization is peculiar and the precise genetic mechanisms of this extensive gene silencing remain to be established. The initial three steps of the pathway catalyzed by PAL, cinnamate 4-hydroxylase, and 4-coumaroyl CoA-Ligase are mandatory and provide the basis for all subsequent branches and resulting metabolites. Subsequent steps downstream of PAL, leading to phenylpropanoid monomers (with exception of cinnamate 4-hydroxylase), are encoded by small gene families in all species investigated so far (Hamberger et al., 2007). With respect to its impact on the phenylpropanoid metabolism, it seems surprising that among nearly 300 functional genes encoding cytochrome P450s in Arabidopsis, 4-hydroxylation of trans-cinnamate to 4-coumarate is encoded by a single gene only, At2g30490, encoding CYP73A5. As expected, mutations in At2g30490 severely affect growth and development of the plants and result in the accumulation of the quite unusual cinnamoylmalate in leaves (Schilmiller et al., 2009). The subsequent step, encoded by the small gene family of 4-CL ligases, channels the resulting aromatic CoA-esters to different biosynthetic pathways. 4-coumaroyl CoA probably represents the most important branchpoint within the central phenylpropanoid biosynthesis in plants. It is either the direct precursor for flavonoid or H-lignin biosynthesis, or is fed into the production of increasingly methoxylated guaiacyl (G)- and syringyl (S)-monolignols. A decisive step is initiated by the enzyme hydroxycinnamoyl CoA:shikimate/quinate hydroxycinnamoyltransferase (HCT). HCT belongs to the large family of BAHD-like acyltransferases (D'Auria, 2006) and may require regulation at multiple levels. Its down-regulation results in an increase in H-units, a depletion in S-units, and a significant increase in caffeoylquinate esters in stems, with only minor effects in leaves (Hoffmann et al., 2004). Silencing of the SHT gene in tracheary elements forming Pinus radiata cell cultures demonstrated the importance of this gene for lignin quantity and quality, most notably a drastic shift towards the H-type of monolignols (Wagner et al., 2007). Only the shikimate and quinate esters, but not the free phenylpropanoid acid or its CoA and glucose ester, are the substrates of Cyp98A3, which catalyze the hydroxylation of the meta-position of the hydroxycinnamoyl shikimate or quinate esters, respectively (Schoch et al., 2001; Abdulrazzak et al., 2006). Supported by reports of a high in vitro KM of HCT for shikimate and quinate, two explanations for this fuzzy and apparently complicated meta-hydroxylation appear most likely: (1) effective protection and channeling of the reactive catechol moiety for subsequent methylation by CCoAOMT and (2) a versatile metabolic control for channeling of the shikimate pathway intermediates into the biosynthesis of secondary metabolites only when photosynthesis is running well. Under sub-optimal photosynthesis, when plants are stressed, low concentration of shikimate pathway intermediates may re-direct the entire phenylpropanoid pathway towards the production of phytoalexins, volatiles, flavonoids, and anthocyanins and de novo synthesis of (defence?) proteins (Schoch et al., 2006; Abdulrazzak et al., 2006). A sophisticated cross-talk and effective transport system between photosynthetic active parenchyma cells and the cambial cells or tracheary elements of the vascular bundles would, however, be required. Caffeoyl quinate (chlorogenic acid) and caffeoyl phenyl lactic acid (rosmarinic acid) are the predominant soluble phenylpropanoids in the Solanaceae and Lamiales, respectively. Possible roles for both esters as defence compounds or as potential antioxidants are discussed (Petersen et al., 2009). To explore the potential benefits as a nutraceutical antioxidant, overproduction of chlorogenic acid was achieved in tomato either by functional expression of a BAHD-like HCT (Niggeweg et al., 2004) or specifically in potato tubers by overexpressing the StMtf1-MYB-transcription factor (Rommens et al., 2009). Diversification of this small subset of phenylpropanoids like coumaroyl- and caffeoyl CoA esters is achieved by an array of subsequent oxygenases, reductases, and transferases and results in the observed complexity and heterogeneity of plant phenylpropanoid metabolite profiles. LIGNIN Besides cellulose, lignin is the most prominent polymer on Earth. Lignin is nearly exclusively based on phenylpropanoid units derived from the oxidative polymerization of hydroxycinnamoyl alcohol derivatives. As early as 1960, the contribution of the shikimate pathway, the reduction of ferulic acid to coniferyl alcohol, and the polymerization of monolignols had already been realized (Neish, 1960). The term lignocellulose implies that the phenolic polymer can be cross-linked to the cellulosic material by feruloyl-linkages, a characteristic feature of grasses (Burr and Fry, 2009). The various aspects of lignin and lignan formation have been summarized in a recent review by Davin et al. (2008), including a detailed biochemical evaluation of the requirements leading to controlled versus random coupling of 4-hydroxyphenyl (H)-, guaiacyl (G)-, and syringyl (S)-derived monolignols. This review also illustrates progress and experimental limitations in structural elucidation of the various forms of lignin in monocots and dicots. Due to its economical value for timber and biofuel formation, lignin biosynthesis and manipulation has been a central research focus (Rubin, 2008; Li et al., 2008). Numerous efforts to manipulate the lignin content have largely focused on structural and regulatory genes leading the accumulation of gymnosperm and angiosperm lignin (Sederoff et al., 1999; Boudet et al., 2003; Vanholme et al., 2008). The contribution of the corresponding enzymes leading to different monomer composition and resulting in various ratios of guaiacyl or syringyl units have been summarized and the effects of the knockout mutations of the corresponding genes have been analyzed in detail (Anterola and Lewis, 2002; Dixon and Reddy, 2003; Vanholme et al., 2008). The activation of phenylpropanoid precursors, by the initial formation of coenzyme A-thioesters and the subsequent reduction in two successive steps by cinnamoyl CoA reductase (CCR) and cinnamyol alcohol dehydrogenase (CAD), is thoroughly documented (reviewed in Davin et al., 2008). Both enzymes are considered the committed steps into monolignol, lignan, and lignin biosynthesis. Arabidopsis knockout mutants of the CCR1-gene show a reduced lignin formation, which can be partly compensated for by CCR-2, but they also display an enhanced pool of feruloyl malate at the expense of sinapoyl malate accumulation (Mir Derikvand et al., 2008). This again indicates that metabolic fluxes within a complex pathway can be, at least in part, redirected towards a pool of soluble metabolites by eliminating a single, central member of this pathway, without otherwise affecting the essential, vital functions of the plant. Partly due to the substrate specificity of most 4CL-ligases earlier in the pathway, sinapoyl CoA is usually considered not to be a direct precursor for lignin monomer production—this may, however, depend on the species investigated (Hamada et al., 2004). Comparison of annuals and perennials with respect to lignin biosynthesis has revealed an apparent functional redundancy of most participating genes and enzymes. Although CAD is considered a key enzymatic step in monolignol biosynthesis, a detailed analysis of 17 CAD encoding genes in Arabidopsis revealed that most insertion lines of individual CAD-genes only had a marginal effect on the overall lignin composition (Kim et al., 2004). This suggests at least some functional redundancy among these genes in case of Arabidopsis and single CADs are not rate-limiting. On the other hand, it is less likely that all of these genes encode bona fide CADs and actually contribute to monolignol formation. A complex grid of temporal and organ-specific regulatory mechanisms may exist to respond rapidly to various biotic and abiotic stressors and ensure the growth and development under adverse conditions. The development of functional genomics and genetic mapping of several species like poplar, pine, eucalyptus, and alfalfa has already enabled to decipher the numerous possibilities of various metabolic routes towards lignin monomer and oligomer formation, oxidative phenol coupling, and analysis of its polymeric structure (Boerjan et al., 2003; Bhalerao et al., 2003; Davin et al., 2008). Remarkably, information on enzymatic reactions from coniferyl alcohol leading to the various types of lignans and neolignans is limited. Data on the 8–8′-stereospecific coupling of two coniferyl alcohol moieties resulting in the formation of pinoresinol have been reported (Davin and Lewis, 2003). The lignans, in contrast, received considerable interest based on pharmacological properties, such as in the case of the cytostatic and anti-cancerous, but also severely toxic, podophyllotoxin. (Enzymatically) controlled, regioselective 8–8’ coupling of isoeugenol-derived lignans without an 9–9′-hydroxygroup was recently demonstrated in roots of the Aristolochiaceae Holystis reniformis upon feeding radiolabeled precursors (Messiano et al., 2009). The identification of two pinoresinol reductase genes, catalyzing specifically the formation of lariciresinol from pinoresinol in Arabidopsis, may facilitate a more thorough investigation of the enzymatic steps of this pathway on a molecular level in a model organism (Nakatsubo et al., 2008). In addition to the entry enzymes into lignin biosynthesis, hydroxylation and methylation steps by cytochrome P450s and S-adenosyl-L-methionine (AdoMet)-dependent O-methyltransferases essentially determine the contribution of guaiacyl and syringyl monomers to the gymnosperm and angiosperm pattern. Overexpression of ferulate-5 hydroxylase (F5H), also termed coniferylaldehyde hydroxylase, leads nearly exclusively to shift to syringyl-lignin in poplar, emphasizing the critical role of this gene for the structure and physical properties of angiosperm lignin (Stewart et al., 2009), whereas knockdown of the f5h gene reveals pleiotropic changes in the Arabidopsis metabolome (Huang et al., 2009). Decoration of the various hydroxylated monolignol precursors in Arabidopsis is performed either by cation-dependent CCoAOMT1, required for the 3-O-methylation of caffeoyl-CoA, or by a cation-independent caffeic acid OMT (COMT), which is involved in the formation of sinapoyl residues and S-lignin from the 5-hydroxyferuloyl derivatives (Davin et al., 2008). The role of CCoAOMT1, originally considered to be only associated with monolignol formation, has changed considerably in the last couple of years (Do et al., 2007; Kai et al., 2008; Fellenberg et al., 2009). Its involvement in various pathways, including sinapoylmalate formation, scopoletin biosynthesis, and methylation of acylspermidine precursors, has proven its central role in the modification of phenylpropanoids. Engineering substrate promiscuous OMTs to act as monolignol 4-O-methyltransferases and impaire oxidative radical coupling is a novel approach to modify lignin composition (Bhuiya and Liu, 2009). The key issue, if lignin formation proceeds via a metabolic grid or by precise channeling of individual precursors through metabolons as proposed for other pathways in plant secondary metabolism, remains to be resolved (Humphreys and Chapple, 2002; Jørgensen et al., 2005; Conrado et al., 2008). Based on current data, a clear decision favoring either one of both possibilities cannot be made. However, one might assume that precise channeling would be a superior mode of control. Additional work is also required on the transport of monolignols prior to assembly. Two possibilities are discussed, either golgi-mediated vesicle transport (Samuels et al., 2002) or specific translocation of monolignols by unknown (ATP binding cassette?) transporters suggested for xylem formation in angio- and gymnosperms (Ehlting et al., 2005; Kaneda et al., 2008). Since the basic principles in lignin and lignocellulose formation may be similar in dicots and monocots, molecular work on fast-growing crops with high biomass, severe drought resistance, or low fertilizer requirements, like Miscanthus spec., should be encouraged to grow lignin and lignocellulose-producing plants under extreme climatic conditions (Dohleman and Long, 2009) and enable developing countries to benefit from recent progress in biotechnology. PHENYLPROPANOID ESTERS AND AMIDES (ACYLATED POLYAMINES) Arabidopsis has been an excellent tool to investigate the biosynthesis of a specific class of phenylpropanoids characteristic for the Brassicaceae: the sinapic acid esters. Sinapoyl choline (sinapine) accumulates to high levels in the seeds of Brassica napus (rapeseed), and, besides tannins and glucosinolates, can negatively affect its use as a protein supplement (Bhinu et al., 2009). The reduction of sinapine by genetic engineering of various biosynthetic enzymes was one of the driving forces in the analysis of the pathway, but also led to important discoveries with respect to structural information and evolution of the participating enzymes. In contrast to lignin or flavonoid biosynthesis, where hydroxycinnamic acids are activated by CoA-esters, in sinapic ester metabolism, 1-O-acyl-glucose esters serve as the energy-rich metabolites during seed maturation and germination (Mock and Strack, 1992). The corresponding glucosyltransferases have been investigated in molecular detail (Milkowski et al., 2000; Sinlapadech et al., 2007; Meissner et al., 2008). Two transferases, sinapoylglucose:choline sinapoyltransferase (SCT) and sinapoylglucose:malate sinapoyltransferase (SMT), are involved in the accumulation of sinapine in the seed or sinapoylmalate in the seedlings, respectively. Compared with the presumed ancestors, the serine carboxypeptidases (SCPs), the SMT from Arabidopsis, has maintained the catalytic triad, Ser–His–Asp (Milkowski and Strack, 2004; Stehle et al., 2009). A single Asp/Glu substitution at the amino acid position close to the catalytic Ser enables binding of 1-O-glucose (Figure 3). In the absence of L-malate, a low esterase activity resulting in the formation of sinapic acid can still be observed in this SCP-like protein, which is reminiscent of the hydrolytic ancestor. The drastic consequences of only slight differences in the catalytic site can be perceived as one of the driving forces resulting in a total pool of 51 genes encoding SCPLs in Arabidopsis (Fraser et al., 2007). Even a small subset of this reservoir of novel catalytic activities may already be able to amplify existing metabolites and prove advantageous to the plant under changing environmental conditions. Download: Download full-size image Figure 3. Model Structure of the Active Site of the AtSMT Modified from Stehle et al. (2006) with the Docking Interaction between the Substrates Sinapoylglucose and L-malate (A) and the Mutated Peptidase Motif Gly–Asp–Ser–Tyr–Ser (B). The amino acids replaced by the residues of serine carboxypeptidases are marked in green. The bulky side chain of Glu at position 172 would restrict binding of the glucose moiety of sinapoylglucose sterically. Therefore, the Glu to Asp substitution enables the acyltrasferases to accommodate 1-O-glucose esters as substrates and is the hallmark of glucose ester-dependent acyltransferases. Residues forming the catalytic triad are colored in black. The substrates sinapoylglucose and L-malate are colored in orange. Dotted lines indicate hydrogen bonds. The existence of acylated polyamine conjugates has been described for several decades and their widespread occurrence, predominantly in flowers and pollen grains of wind-pollinated species, has been attributed to UV-protection, fertility, as well as defence (Meurer et al., 1988; Kakkar and Rai, 1993; Walters, 2003). The rationale of their existence has been disputed since then. The recent identification of phenylpropanoid polyamine conjugates in different organs of Arabidopsis wild-type and insertion lines by LC–MS analysis was accompanied by simultaneous identification of specific genes and enzymes in the acyl conjugate formation (Böttcher et al., 2008; Matsuda et al., 2008; Luo et al., 2009). The rapid annotation of several genes and enzymes involved in the biosynthesis of these conjugates illustrates the combined power of Arabidopsis knockout and knockdown insertion lines with sensitive UPLC–MS analyses (Fellenberg et al., 2008; Grienenberger et al., 2009). In the tapetum of Arabidopsis flower buds, a new pathway for the biosynthesis of acylated spermidine conjugates was established. Hydroxycinnamic acids are transferred to spermidine by a BAHD-like spermidine hydroxycinnamic acid transferase (SHT). Knockout mutation of this gene results in anthers and pollen grains, completely devoid of trisacyl conjugates (Grienenberger et al., 2009). Two other enzymes critical for this pathway, a cytochrome P450 and a methyltransferase (Figure 4), were recruited from enzymes in an established biosynthetic pathway to catalyze the precise modifications of acylspermidine derivatives. CYP98A3 hydroxylates quinate and shikimate esters of 4-coumaric acid in lignin biosynthesis (Schoch et al., 2001). The new, pathway-specific CYP98A8 and its paralog CYP98A9 belong to a monophyletic CYP98A clade and are duplications of an ancestor of CYP98A3, which evolved rapidly via retroposition and positive Darwinian selection (Matsuno et al., 2009). Knockout mutants of CYP98A8 display a shift in the major acylated spermidine conjugate pattern, from 5-hydroxyferuloyl to feruloyl conjugates. Likewise, AtTSM1, a cation-dependent OMT with significant similarity to CCoAOMT1 in lignin biosynthesis, catalyzes the final step in acyl spermidine conjugate formation. The enzyme methylates a single 5-hydroxyferuloyl moiety and yields N, N'-bis(5-hydroxyferuloyl)-N'’ sinapoyl spermidine (Fellenberg et al., 2008). CCoAOMT1 and potentially CYP98A3 play additional roles earlier in methylation and hydroxylation of the quinate/shikimate esters, respectively. Corresponding knockouts of CCoAOMT1 also resulted in a severe decrease in acyl-spermidine formation (Figure 4; Fellenberg et al., 2009). This pathway convincingly demonstrates that genetic mechanisms like retroposition and gene duplication are also active in plant secondary metabolism and may result in an accelerated establishment of organ-specific new pathways from ancestral genes (Matsuno et al., 2009). The reason for the exclusive appearance of the pathway in the tapetum is not yet understood, since neither the overall plant phenotype nor the plant fertility appears affected (Grienenberger et al., 2009). Regulatory elements of the pathway have up to now not been identified, although candidate MYB transcription factors are described that regulate phenylpropanoid–polyamine conjugate accumulation in other plants (Gális et al., 2006). Download: Download full-size image Figure 4. A Simplified Version of the Biosynthetic Pathway Resulting in Acylated Spermidine Conjugates in flower buds Is Based on Recent Data by Fellenberg et al. (2008, 2009), Grienenberger et al. (2009), and Matsuno et al. (2009). The exact sequence of the initial hydroxylation and Werner methylation steps by CYP98A3 and CCoAOMT1, the subsequent hydroxylation by Cyp98A9 and the transferase reaction by SHT is not resolved yet. Sequential steps catalyzed by CYP98A9 and AtTSM1 are established. AtTSM1, localized exclusively in the tapetum, may serve as a tapetal marker in young flower buds of Arabidopsis. ANTHOCYANINS, AURONES, (ISO)FLAVONOIDS, AND STILBENES Due to ease of visualization and control of mutants and genetic imbalances, anthocyanin biosynthesis was one of the first branches of the general phenylpropanoid metabolism, for which biosynthetic enzymes and corresponding transcription factors were identified and has been reviewed extensively (Allan et al., 2008). The key enzymatic step of the whole pathway, the formation of anthocyanidins via (2R-3R)-dihydroflavonols and oxidation of (2R, 3S, 4S)-leucoanthocyanidins by anthocyanidin synthase (ANS) has recently been questioned due the apparent flavonol synthase activity of Arabidopsis ANS and the identification of (+)-catechin as a substrate of Gerbera hybrida ANS (Wellmann et al., 2006). In a monomeric state, anthocyanins never display a blue color. Therefore, investigation of blue color formation of anthocyanins either by pH-shift, metal complex formation, modification of the core unit by additional hydroxylation, methylation, or acylation and subsequent molecular stacking has been a major focus of anthocyanin research in the last decade (Yoshida et al., 2009). This research was partly driven by the cut-flower industry to develop blue varieties of those flowers, which naturally do not exhibit this phenotype, like roses or carnations. During the process of developing a bluish tone, important contributions to anthocyanin biosynthesis have been made in terms of structural and regulatory genes. It was shown that roses, in contrast to most other plants, have a unique bifunctional glucosyltransferase that catalyses the glycosylation of both the 3- and the 5-OH group (Ogata et al., 2005). However, they are devoid of a 5′-hydroxylase activity, which partly explains the lack of any natural bluish color. Expression of the Viola spp. 3′, 5′-hydroxylase and functional replacement of the endogenous rose dihydroflavonol reductase (DFR) for an orthologous enzyme from Iris hollandica resulted in plants that nearly exclusively accumulated delphinidin in the petals and displayed a bluish, slightly purple tone (Katsumoto et al., 2007). Myricetin, in addition to quercetin, was detected as a co-pigment in the transgenic plants, indicating that the hydroxylase modifies both flavonols and anthocyanins. Further modification by co-expression of an additional acyltransferase did not result in the desired increase of acylated compounds and intensified color formation (Katsumoto et al., 2007). To further enhance color formation, additional methylation of these B-ring hydroxyl groups may be required. Methylation of vicinal dihydroxy groups of anthocyanins from a petunidin to malvidin type was recently demonstrated to proceed by a cation-dependent OMT in grapevine (Hugueney et al., 2009). It further extends the biosynthetic niches of this type of enzyme towards modification of chromogenic compounds with an aromatic cis-diol structure. Yellow flower colors are not necessarily due to carotinoid accumulation, as in sunflowers. In Anthirrhinum majus, the yellow aurones display a bright yellow color with an additional fluorescence and are considered to help attract pollinators. Ono and co-workers (2006) were able to produce ornamental Torrenia hybrida flowers with bright yellow petals. This was achieved by the combined down-regulation of the anthocyanin pathway and the co-expression of two genes involved in aurone formation, the vacuolar localized aureosidin synthase (AmSA1) and a cytoplasmatic chalcone 4′-O-glucosyltransferase. Co-epxression of the modifying UGT together with the AmSA1 was absolutely required, in order to enable transport of the 4′-O-glycosylated chalcones to the vacuoles, where enzymatic conversion to the aurone 6-O-glucosides was performed (Ono et al., 2006). The complex structures of flavonoids and anthocyanins require precise annotation of the participating modifying and decorating enzymes. Most acyl-, glucosyl-, and methyltransferases identified up to now exhibit overlapping or promiscuous substrate specificities in vitro (Bowles et al., 2006). If a potential gene function is not immediately apparent by sequence analysis of a multigene family, comparison of its gene expression with the expression profiles of known genes, complementation of the metabolite pattern combined with reverse genetics help to annotate a single member from large superfamilies, like CYPs or UGTs. CYP98A8/A9 was correlated to the phenylpropanoid polyamine conjugate biosynthesis in the Arabidopsis tapetum (Ehlting et al., 2008; Matsuno et al., 2009). UGT89C1 was linked to the late steps in flavonol biosynthesis based on transcript and metabolite pattern and therefore correctly identified as Arabidopsis 7-O-rhamnosyltransferase (7RhaT) (Yonekura-Sakakibara et al., 2007). Several other late steps in the formation of the flavonoid skeleton have only been poorly investigated. B-ring hydroxylation at positions 6 and 8 of the flavonoid skeleton is limited to a few plant families only, like the Asteraceae. For Chrysanthemum segetum, Halbwirth and Stich (2006) identified a new type of 8-monohydroxylase dependent on both NADPH and FAD, and was, therefore, clearly different from classical cytochrome P450s. Obviously, there is a lack of enzymatic and genetic information for this as well as other exotic plants. Scutellaria baicalensis is one of several prominent examples used in traditional folk medicine. Its roots harbor flavonoids with unusual hydroxylation, methylation, or glycosylation pattern in the flavonoid A- and B-ring (Li-Weber, 2009). Exponentially growing databases and sequence information may enable the identification and annotation of the corresponding genes and enzymes in due course, if we manage to prevent these valuable resources from eradication. Three other elusive modifications or decorations of the flavonoid molecules have been recently established at the molecular level. These modifications include the characterization and identification of the first soluble, rare plant glucuronosyl-transferase (GlurT) UGT94B1 from Bellis perennis (Sawada et al., 2005). The key amino acid distinguishing this type of enzyme from the common glycosyltransferases was identified as an arginine at position 25 outside of the conserved UDP–sugar binding-motif, the PSPG-box (Osmani et al., 2008). A second modification has been known to phytochemists for more than 20 years, but the corresponding enzymatic step has not been resolved: the C-glycosylation of the flavonoid A-ring. Unequivocal annotation of flavone 6- and 8 C-glycosylating enzyme was recently achieved for a C-glucosyltransferase (CGT) from rice (Brazier-Hicks et al., 2009). A biochemical and molecular approach was used in the case of the Oryza sativa CGT. This CGT could not be identified by a simple sequence similarity approach, since no CGT-specific amino acid sequence motif of OsCGT1 was detected and the most similar enzymes with 30% overall identity were described as UGTs involved in monolignol formation in Arabidopsis. The authors propose that C-glycosylation activity in rice and wheat may proceed via a mechanism involving a 2-hydroxyflavone intermediate, although this hydroxylation activity requires further confirmation. The 2-hydroxyflavone exists in equilibrium with its open chain form, which is proposed as the substrate C-glucosyltransferase. Last but not least, the cloning and characterization of enzymatically active naringenin 8-prenyltransferase pave the way for subsequent investigations of this important, yet poorly characterized class of membrane enzymes (Sasaki et al., 2008). Regulation of flavonoid and anthocyanin formation by basic TTG/basic helix-loop-helix bHLH and MybR2R3-type transcription factors has been the topic of many reports, such as Stracke et al. (2007), and several reviews (Broun, 2005; Allan et al., 2008). The complex pattern of factors regulating anthocyanin formation has recently been extended by Matsui et al. (2008) and Zhu et al. (2009). They discovered that single repeat R3 Myb transcription factors may also repress anthocyanin formation, probably by competing with R2R3-Mybs for the binding sites of bHLH proteins, thereby down-regulating structural genes and counteracting, such as the Myb-like PAP1/2 enhancer of anthocyanin formation in Arabidopsis (Gonzales et al., 2008). The bHLH-domain of the maize R-gene, participating in anthocyanin formation, links flavonoid formation to histone modification by interacting with an EMSY-like factor, which is involved in chromatin repair (Hernandez et al., 2007). Identification of metabolite pattern and elucidation of the structural genes and enzymes, deciphering the complex network of hundreds of transcription factors and their hormonal control (Gális et al., 2006) in Arabidopsis and other plant species, will provide a tremendous challenge to understand phenylpropanoid metabolism and its contribution to plant biology. Non-targeted approaches correlating co-expression data and metabolomics at the organ or tissue level can provide solutions to unravel parts of this intriguing puzzle (Tohge et al., 2005). Terpenophenolics, stilbenes, and isoflavonoids are limited to a few species only, but have attracted major interest due to potential health-promoting affects on mammals (Stevens and Page, 2004; Barnes and Prasain, 2005; Shen et al., 2009; Dixon, 1999; Veitch, 2009). The biosynthesis of prenylchalcone and other prenylated chalcones represent a deviation from the standard flavonoid biosynthesis. Xanthohumol is a major metabolite accumulating specifically in glandular trichomes of hops, Humulus lupulus. By an EST-approach, Nagel et al. (2008) could recently show that in the glandular trichomes, naringenin chalcone is first prenylated and subsequently methylated by a substrate-specific 6′-O-desmethylxanthohumol OMT, which is responsible for the stabilization of the resulting xanthohumol. This result provides the basis for further investigations of this important pathway, leading to the formation of a family of compounds with considerable phytoestrogenic properties. The biosynthesis of isoflavonoid and related pterocarpans in model legumes like Medicago sativa and M. truncatula or Lotus japonicus has been a major research focus for the last two decades. The oxidative migration of the aryl ring from 2S-liquiritigenin, the 5-deoxynaringenin formation catalyzed by the cytochrome P450 isoflavone synthase and the subsequent 4′-O-methylation are the decisive branch points leading to the formation of isoflavonoids. In contrast, stabilization of the isoflavanone by a 4′-OMT, a dehydratase and second cytochrome P450, isoflavone 2′-hydroxylase, are required for the pathway to proceed towards the biosynthesis of pterocarpan-type phytoalexins, like pisatin (Aoki et al., 2000). A promiscuous 4′-isoflavonoid OMT with identical catalytic properties towards the 4′- hydroxyl group of 2,7,4′-trihydroxyisoflavanone at an initial step of the pathway as well as towards the 3-hydroxy group of the pterocarpan 6a-hydroxymaackiain resulting in the formation of pisatin has been identified and structurally characterized (Liu et al., 2006). Careful analysis of both substrates revealed their close to identical three-dimensional structure. A single enzyme able to convert two metabolites of the same pathway suggests that substrate promiscuity does not only provide evolutionary advantages in response to novel, exogenous phytoalexins, but may also be maintained to serve those rare cases of metabolic serendipity. This situation is somewhat different from the methylation step in acyl-spermidine formation in Arabidopsis flowers, where not CCoAOMT1, but a similar cation-dependent OMT (AtTSM1) with distinct properties was recruited to catalyze the final decorating step of conjugate formation (Fellenberg et al., 2009). As in case of the isoflavonoids, also the stilbenes like resveratrol and its glycoside piceid are of limited distribution in the plant kingdom. Both have been shown to display numerous biological activities in vitro, including antitumor activity and induction of apoptosis (Pervaiz, 2004; Crozier et al., 2009). Trans-resveratrol apparently retards the ageing process and increases longevity in various organisms from yeast to insects and rodents (Orallo, 2008). Since dietary polyphenolics usually display low absorption and are chemically modified by the small intestine, the contribution to human health in vivo will require profound epidemiological studies (Crozier et al., 2009). The relative ease to produce transgenic edible crops by introducing only a single gene, in this case stilbene synthase under the control of a 35S-promoter, results in substantial levels of stilbenes in tomato fruits and may encourage other scientists to explore the proposed positive in vitro effects in controlled experimental set ups in vivo (Giovinazzo et al., 2005; Schijlen et al., 2006). HYDROLYZABLE AND NON-HYDROLYZABLE PROANTHOCYANIDINS AND TANNINS Proanthocyanidin biosynthesis branches off the flavonoid pathway from either 2,3-cis or trans flavanols and is initiated by anthocyanidin reductase or leucoanthocyanidin reductase, respectively (Dixon et al., 2005; Tian et al., 2008). The resulting epicatechin and catechin derivates can be oxidized to quinones, which are polymerized. However, it is not clear whether this polymerization of the soluble precursors proceeds enzymatically by a laccase type of enzyme (encoded by the TT10 gene in Arabidopsis) or non-enzymatically (Pourcel et al., 2005; Dixon et al., 2005). Transport of precursors across vacuoles membranes is facilitated by multidrug-extrusion MATE-type transporters in Arabidopsis (encoded by the TT12-locus) and Medicago truncatula (Marinova et al., 2007; Zhao and Dixon, 2009). Similar to anthocyanin formation, the deposition of proanthocyanidin oligomers in the seed coat of Arabidopsis is regulated by the interaction of a series of transcriptional regulators R2R3 MYB-type proteins (TT2), bHLH-factors (TT8), and WD40-repeat proteins (TTG1), which act synergistically. Overexpression of those transcription factors may create improved crop varieties with enhanced proanthocyanidin contents (Baudry et al., 2004; Mellway et al., 2009; Terrier et al., 2009). The reported health-promoting activities of 3-O-esterified proanthocyanidins like epigallocatechin-gallate (EGCG) (Rezai-Zadeh et al., 2005), a major metabolite in green tea leaves, can be hampered by a low bioavailability upon modification of the core structure by sulfatation, methylation, and glucuronylation (Henning et al., 2008). Similar to the proanthocyanidins (non-hydrolyzable tannins), the hydrolyzable tannins, like galloyl-glucose esters and ellagitannins, characterized by oxidative coupling of neighboring galloyl groups, display a strong tendency to scavenge radicals. Due to multiple aromatic hydroxyl groups, they cross-link and precipitate enzymes, making them interesting defence compounds specifically when present in high concentrations in oat or in sumac (Barbehenn et al., 2006; Gross, 2008). Remarkably little information is available on the biosynthesis of gallic acid, which respresents the core structure in gallo- and ellagitannin structures. Retrobiosynthetic 13C-NMR-studies, by feeding 13C-labeled glucose to young leaves of Rhus typhina and measurments of oxygen isotope abundance, performed by Werner and colleagues (1997, 2004), showed that gallic acid is derived from the shikimate pathway. The formation via phenylalanine and hydroxylated cinnamic acid derivatives could be ruled out by these experiments. Based on these studies, it appears most likely that gallic acid is formed via dehydrogenation of 5-dihydroshikimate, although a dehydratation step via protocatechuic acid and subsequent monooxygenation cannot be ruled out. Enzymatic evidence for the existence of a specific cytochrome P450 monooxygenase or a 2-oxoglutarate dependent dioxygenase as shown in coumarin formation would help to resolve the contribution of both possibilities. Clearly, a lot of research remains to be done specifically on the hydrolyzable tannins, although enzymatic studies appear difficult due to the protein precipitant properties of both, substrates and products. C6–C1 METABOLITES In contrast to the established biosynthesis of phenylpropanoids with a C6–C3 carbon skeleton, the formation of phenolics with a C6–C1 skeleton, like benzoic acid (BA), appears complex and is still fragmentary. Starting from trans-cinnamate, two biosynthetic scenarios are proposed. A β-oxidative pathway via CoA-activation and a 3-keto-acyl CoA-thiolase was described for Petunia hybrida (Van Moerkercke et al., 2009). An alternative non-oxidative pathway, via benzoyl CoA and benzaldehyde, with a final step catalyzed by a benzaldehyde dehydrogenase, was identified in Anthirrhinum majus (Long et al., 2009). Additional routes, specific for certain plants or specific organs, have also been described (Abd El-Mawla and Beerhues, 2002; Ibdah et al., 2009). From the sum of data, it is evident that several pathways for benzoic acid biosynthesis may co-exist in a single plant and relative fluxes from one or the other pathway may be controlled by hormones or abiotic stressors. Feeding deuterium-labeled phenylalanine showed that metabolic fluxes of benzaldehyde formation of the non β-oxidative pathway from cinnamate to benzaldehyde to benzoic acid was about twice that of the β-oxidative pathway involving benzoyl-CoA in P. hybrida petals (Boatright et al., 2004). The 2-hydroxyl derivative of benzoic acid, salicylic acid (SA), has long been known as a key component of plant innate immunity (Lee and Raskin, 1999; Nawrath and Metreux, 1999) and also plays a central role in systemic-acquired resistance (SAR) (Vlot et al., 2008). Biochemical evidence for a direct pathway from BA to SA was obtained first by the identification of benzoic acid 2-hydroxylase, a cytochrome P450 monooxygenase from tobacco (León et al., 1995), and was also based on an increase of PAL-transcripts and activity in tobacco plants upon tobacco mosaic virus (TMV) infection (Ogawa et al., 2006). However, in contrast to the prokaryotic BA 2-hydroxylase encoded by the NahG gene, there is no Arabidopsis insertion line indicating a similar gene and no CYP-candidate genes have been annotated yet. Transgenic tobacco plants overexpressing PAL also did not show any significant change in the SA levels (Shadle et al., 2003), suggesting that SA biosynthesis is not derived from trans-cinnamate and benzoic acid. In contrast, experimental evidence from the Arabidopsis sid2 mutant, encoding isochorismate synthase 1 (ICS1) (Wildermuth et al., 2001; Strawn et al., 2007), as well as virus-mediated RNA-silencing in N. benthamiana (Catinot et al., 2008) suggest that chorismate and ICS may be central to and required for salicylic acid biosynthesis. The exact contribution of the several pathways to SA bioynthesis may therefore be species-specific and research may require a comparative investigation of several mono- and dicot families. Modification of SA results in a significant increase in susceptibility to pathogens (Vlot et al., 2008). Modification of salicylic acid can be achieved either by methylation (Ross et al., 1999) or by glycosylation, at the carboxy- or at the 2-hydroxy group, respectively (Lee and Raskin, 1999; Lim et al., 2002). The methyltransferase activity towards the carboxygroup of benzoic acid and salicylic acid is part of the SABATH-family of enzymes that are generally implicated in hormone homoeostasis or phenylpropanoid volatile formation in plants (Zhao et al., 2008; Barkman et al., 2007; Kapteyn et al., 2007). Methyl salicylate has been identified as the critical mobile signal for SAR. Its perception may include additional regulatory steps, such as by a salicylic acid binding protein (SABP2), identified as a methylesterase (Forouhar et al., 2005; Zhao et al., 2009). Glycosylation of salicylic acid is much less investigated. A systematic screen within Arabidopsis plant natural product UGTs revealed only enzymes with activities towards the carboxyl group, but none towards the 2-OH group (Lim et al., 2002). This suggests that regiospecific SA-glycosylating enzymes share limited distribution in plants and that glycosylation is not the common way of inactivating this hormone. PHENYLPROPENE VOLATILES Benzenoid volatile formation in Petunia flowers has been shown to be under strict diurnal and hormonal control, and appears to follow the transcriptional control of central shikimate pathway genes, like 5-enol-pyruvylshikimate-3-phosphate synthase (EPSPS) (Schuurink et al., 2006). Methyl salicylate and methyl benzoate are not the only phenolic compounds emitted as volatiles by flowers. They are part of a complex species-specific bouquet of compounds, including terpenoids and phenylpropenes (Dudareva et al., 2004). These volatiles are also either emitted as floral attractants to pollinators, as defence compounds against microorganisms, insects, and mammalian predators, or can display allelopathic effects on other plants (Dudareva et al., 2004; Horiuchi et al., 2007). For phenylpropenes from the Lamiales and simple catechol derivatives, like orcinol from Chinese roses, several species and organ-specific biosynthetic pathways have been elucidated in the last decade (Scalliet et al., 2006; Koeduka et al., 2006; Kapteyn et al., 2007). The biosynthesis of antimicrobial phenylpropenes like eugenol, isoeugenol, or chavicol could be resolved due to the high production levels of these compounds either in the corolla of Petunia hybrida and the peltate glands of Ocymum basilikum. Both compounds are formed via the general phenylpropanoid pathway, including the conversion of coniferyl alcohol by a BAHD-like coniferyl alcohol acetyltransferase and a subsequent NADPH-dependent reduction of the coniferyl acetate (Dexter et al., 2007; Koeduka et al., 2006). The structure of the key enzyme, eugenol synthase (EGS), a member of the short chain dehydrogenases/reductases, catalyzing the reductive replacement of the propenyl side chain has been resolved by X-ray crystallography (Louie et al., 2007). Methylation of the 4-hydroxy group of phenylpropenes is achieved by a specific subfamily of methyltransferases. They may be either localized specifically in the glandular trichomes as observed for sweet basil or could be part of the phenylpropene pathway in the leaves, as in the case of Pimpinella anisum (anise) (Koeduka et al., 2009). As far as coniferylacetate-derived phenylpropenes are concerned, a CCoAOMT1-like protein is again responsible for the methylation of the 3-OH group, presumably at the caffeoyl CoA level. Several levels of metabolic control at the branch points of volatile emission are apparent when selected basil lines were tested by a systems biology approach combining transciptome, proteome, and volatile metabolite profiles (Xie et al., 2008). With the exception of cinnamate 4-hydroxylase, all other transcript and protein levels of phenylpropanoid pathway genes correlate well with metabolite formation. Nevertheless, additional control by post-translational phosphorylation or ubiquintinylation is described for PAL and chavicol OMT (CVOMT), respectively. Engineering of transcription factors has not only been used to control expression of flower color. Co-engineering scent and flower color by a single transcription factor PAP1, known to regulate anthocyanin formation, surprisingly also led to a drastic increase in phenylpropanoid volatile formation when expressed in Petunia flowers (Ben-Zvi et al., 2008). This fascinating aspect of simultaneously enhancing flower color and scent production illustrates only one part of the biotechnological prospects this kind of research may lead into for potential benefits to the consumer. Volatile emission from (genetically) enhanced scent producers might be used to ‘prime’ and thereby protect non-transgenic crops against pests (Dudareva and Pichersky, 2008) and pave the way to an alternative way of molecular farming approaches in an ongoing debate on the sense and non-sense of transgenic crops. COUMARINS High expression levels in all organs, from seeds to roots to flowers, suggested participation of CCoAOMT1 not only in lignin biosynthesis, but in several biosynthetic pathways leading to soluble phenylpropanoids (Do et al., 2007). Knockout of the gene encoding CCoAOMT1 also led to a reduction in the coumarin scopoletin in Arabidopsis roots (Kai et al., 2008), suggesting that feruloyl CoA is the phenylpropanoid-derived precursor of the UV-fluorescent coumarin scopoletin. The decisive step in coumarin biosynthesis in Arabidopsis roots was verified by T-DNA insertion mutants (Kai et al., 2008). The key ortho-oxygenation step is not performed by a CYP-like monooxygenase, but by a position specific 2-oxoglutarate dependent dioxygenase (F6′H1). This enzyme catalyzes the formation of the 6′-O-hydroxylated trans-isomer from feruloyl CoA, which undergoes isomerization of an enolate anion and lactonization to form scopoletin (Kai et al., 2008; Figure 5). The assumption that substrate specificities of orthologous F6′H1-dioxygenase-like enzymes affect the substitution pattern of coumarins justifies investigations of exotic plant species containing unusual highly methoxylated coumarins, such as in the roots of Pelargonium sidoites (Kolodziej, 2007). This might lead to the detection of orthologous enzymes earlier in the phenylpropanoid pathway, with novel substrate specificities, like unusual CoA-ligases. The resulting coumarin aglycones can be modified by a set of UDP-glucose-dependent glucosyltransferases to result, for example, in scopolin or esculin, the most prominent soluble coumarins in plants. Alternatively, in the Apiaceae, simple coumarins can be prenylated (Stanjek et al., 1999) and after a series of subsequent cytochrome P450-catalyzed reactions (Larbat et al., 2007), result in the formation of furanocoumarins, generally considered as phototoxic defense compounds. Download: Download full-size image Figure 5. Recently Established Decisive Steps of Scopoletin Biosynthesis, Slightly Modified According to Kai et al. (2008). Starting from feruloyl–CoA, the 2-oxoglutarate dependent dioxygenase F6′H1 catalyzes the formation of the 6′-O-hydroxylated trans-isomer from feruloyl CoA, which undergoes subsequent isomerization, rotation, and lactonization to form scopoletin and free coenzyme A. EXOTIC PHENYLPROPANOIDS Several phenylpropanoid-derived metabolites are restricted to a few species or families only. These include the curcumins, benzophenones, biphenyls, or phenylphenalenones. Although of exotic distribution, some possess pharmacological and health-promoting effects, which make them an interesting target for biosynthetic analyses (Bengmark et al., 2009). The biosynthesis of curcumins with linear diarylheptanoid structure is not sufficiently clarified. Specifically, the introduction of the oxygenation pattern, either before or after the formation of the curcumin skeleton remains to be established (Kita et al., 2008). That plant polyketide synthase type III (PKS III), which catalyzes the formation of bisdemethoxycurcumin, from p-coumaroyl and malonyl CoA, was described originally from rice (Katsuyama et al., 2007). Its presence in the actual curcumin producing Curcuma longa has most recently been demonstrated and the corresponding enzymes have been characterized (Katsuyama et al., 2009). Benzophenone-synthase (BPS) and biphenyl-synthase (BIS) are specific PKS III, use benzoyl CoA as a starter, and initiate the carbon skeleton formation of the benzophenones, xanthones, biphenyls, and dibenzofurans (Beerhues and Liu, 2009). The compounds are considered as phytoalexins and their occurrence is limited to a few known plant taxa like the Hypericaceae (benzophenones and xanthones) and the Maloideae (benzophenones and biphenyls) (Beerhues and Liu, 2009). A new kind of type III PKS was also suggested to be involved in the biosynthesis of cyclic diarylheptanoids and phenylphenalenons from Musaceae (Brand et al., 2006). The general phenylpropanoid pathway is not different from the pathway from established species. This was demonstrated for xanthone biosynthesis from the pharmaceutically relevant Hypericum perforatum (Franklin et al., 2009). Nevertheless, investigations of the genetic resources and metabolic diversity of plants with exotic phenylpropanoid-derived compounds are worth exploring to detect enzymes with unusual specificities for biotechnological or molecular farming approaches. MIXED POLYMERS, SUBERIN, CUTIN, AND SPOROPOLLENIN The contribution of the general phenylpropanoid metabolism towards the biosynthesis of mixed polymers is a challenging task for structural biochemists and physiologists. The complexity of the pathways and robustness of cutin, suberin, or sporopollenin towards degradation requires harsh oxidative chemical breakdown. Inherently, there is considerable chance of artefact formation, which hampers the precise determination of the contribution of the individual constituents of phenylpropanoid metabolism to the structural integrity of these polymers. These polymers essentially serve completely different functions, as a water barrier for the sessile plant in case of suberin and cutin, or as a highly protective and rigid scaffold for the male gametophyte. Nevertheless, their complex and variable, highly rigid structures are based on only two building blocks, aliphatic and aromatic. The contribution of the latter is apparently lower than in the case of lignin. The production of suberin is limited to a few plant tissues, like the endodermis with the casparian strip or the phellogen or cork cambium; its principle components have been established (Bernards, 2002). As summarized in recent reviews by Pollard et al. (2008) and Franke and Schreiber (2007), besides linear aliphatic, oxygenated long chain dicarboxylic acids and esters, in Arabidopsis, feruloyl esters comprise a significant part of the suberin structure. These are virtually absent in cutin. Poor substrate availability and problematic detection of intermediates render any biochemical investigations and identification of the participating enzymes of the phenylpropanoid pathway difficult. This includes the acyltransferase(s) connecting feruloyl CoA to the aliphatic, oxygenated building blocks either in or outside the symplast. A genomic approach towards suberin biosynthesis and cork differentiation in Quercus robur (cork oak) stresses the major contribution of two gene families in the assembly of this mixed polyester. Both the cytochrome P450 and the BAHD-like HC-tranferases (Soler et al., 2007). The exact mechanism, namely whether there is either coordinated or random enzymatic linkage of the phenylpropanoid-derived aromatic blocks and the aliphatic esters in the apoplast, either catalyzed by peroxidases or by laccases, remains to be established. Recent data demonstrate that a sterol glycosyltransferase determines efficient trafficking of cutin and suberin to the Arabidopsis seed coat (De Bolt et al., 2009). Much less is known about the regulatory networks affecting composition, transport, and assembly of both polymers. Candidates of R2R3s and WRKY type transcription factors have been identified, but await further structural and functional characterization (Soler et al., 2007). Sporopollenin, the major part of the pollen exine, apparently contains similar building blocks compared to cutin and suberin, namely phenylpropanoids and hydroxylated fatty acids. In contrast to cutin and suberin this polymer is even more resistant to chemical degradation, although advances in chemical analysis have enabled the complete identification and separation of its constituents (Dominguez et al., 1999). Based on labeling experiments, the presence of phenylpropanoid residues is generally accepted (Schulze-Osthoff and Wiermann, 1987). The mechanical strength and durability of the pollen exine are probably due to a substantial amount of ether-linkages between fatty acid and/or phenylpropanoids. Conclusive data on the assembly and the exact contribution of the phenylpropanoid conjugates are virtually non-existent. Whether the variable phenylpropanoid pattern deposited as soluble pollenkit serves as a fingerprint of individual units incorporated into the polymer is also unknown. Based on functional characterization of Arabidopsis knockout mutants, initial progress has been made recently at least on the aliphatic components of the polymer. Tissue-specific and developmentally controlled members of the cytochrome P450 family control the hydroxylation pattern of the aliphatic acids, with CYP703 and CYP704 unequivocally involved in hydroxylation of lauric acid and omega-hydroxylation of longer chain fatty acids, respectively (Morant et al., 2007; Dobritsaa et al., 2009). The coordinate expression of several genes of the general phenylpropanoid metabolism with CYP703 indicates the tight association of fatty acid biosynthesis and phenylpropanoid biosynthesis structures in the establishment of the exine barrier. The precise participation of phenylpropanoids in the proposed ether and ester linkages remains to be established, although it is already known that structures with vicinal dihydroxy groups are required to meet the current perceptions of polymer organization (Morant et al., 2007). The functional aspects of phenylpropanoids in sporopollenin and exine formation besides the contribution to UV-light protection may reside in protection against various pathogens. Biosynthesis and transport of aliphatic and aromatic building blocks are performed by the tapetum in specific organelles, termed tapetosoms (Hsieh and Huang, 2007), whereas the factors contributing to transport and polymerization are still unknown. CONCLUDING REMARKS The diversity within the structural and regulatory genes of several superfamilies of enzymes involved in secondary metabolism are the direct consequence of a complex plant–environment interaction mediated in part by soluble and insoluble secondary metabolites, including phenylpropanoids. The rapid progress in sequencing, structural elucidation, and analytical tools will ultimately result in the desired systems biology approach, which may start with a few model plants and crops, like Arabidopsis, alfalfa, poplar, and rice at the beginning. A thorough understanding of the biosynthetic pathway leading to the various branches of phenylpropanoid formation and function should also consider in more detail the following key aspects: (1) A more detailed investigation of transport and shuttling of metabolites across membranes and cells. (2) Elucidation and identification of enzyme complexes and metabolon formation to understand how several pathways may co-exist but are independently regulated. (3) A thorough understanding of the molecular mechanisms resulting in the formation of unusual phenylpropanoids in medicinal and exotic plants in a continuous search for new metabolites and new pathways. New analytical and biophysical methods, such as the implementation and application of nanotechnology to solve biological problems, are currently developed for new approaches towards resolving a most complex scenario of biological systems (Borch and Hamann, 2009; Zhong, 2009). Combining traditional methods with these new fascinating molecular and chemistry tools will not only contribute to a better understanding of plant natural product and phenylpropanoid biosynthesis, but encourage farmers and consumers to benefit from the diversity and plasticity of plant natural product formation and modification. FUNDING Financial support by the Deutsche Forschungsgemeinschaft (VO 719/8-1) is gratefully acknowledged. Acknowledgments I would like to thank Felix Stehle (IPB Halle, Germany) for the contribution of Figure 3. I am also grateful to Christin Fellenberg, Dieter Strack (both IPB Halle, Germany), Jörg Ziegler (University of Calgary, Canada), and two anonymous reviewers for critical reading of the manuscript and many helpful suggestions. No conflict of interest declared. Recommended articles REFERENCES Abd El Mawla and Beerhues, 2002 Abd El Mawla A.M., Beerhues L. Benzoic acid biosynthesis in cell cultures of Hypericum androsaemum Planta, 214 (2002), pp. 727-733 View in ScopusGoogle Scholar Abdulrazzak, 2006 Abdulrazzak N., et al. 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Many of these compounds are also beneficial to human health and welfare. V-myb myeloblastosis viral oncogene homolog (MYB) proteins belong to a large family of transcription factors and are key regulators of the synthesis of phenylpropanoid-derived compounds. This review summarizes the current understanding of MYB proteins and their roles in the regulation of phenylpropanoid metabolism in plants. New Insights into the shikimate and aromatic amino acids biosynthesis pathways in plants 2010, Molecular Plant Show abstract The aromatic amino acids phenylalanine, tyrosine, and tryptophan in plants are not only essential components of protein synthesis, but also serve as precursors for a wide range of secondary metabolites that are important for plant growth as well as for human nutrition and health. The aromatic amino acids are synthesized via the shikimate pathway followed by the branched aromatic amino acids biosynthesis pathway, with chorismate serving as a major intermediate branch point metabolite. Yet, the regulation and coordination of synthesis of these amino acids are still far from being understood. Recent studies on these pathways identified a number of alternative cross-regulated biosynthesis routes with unique evolutionary origins. Although the major route of Phe and Tyr biosynthesis in plants occurs via the intermediate metabolite arogenate, recent studies suggest that plants can also synthesize phenylalanine via the intermediate metabolite phenylpyruvate (PPY), similarly to many microorganisms. Recent studies also identified a number of transcription factors regulating the expression of genes encoding enzymes of the shikimate and aromatic amino acids pathways as well as of multiple secondary metabolites derived from them in Arabidopsis and in other plant species. Contribution of phenylpropanoid metabolism to plant development and plant–environment interactions 2021, Journal of Integrative Plant Biology Bright Side of Lignin Depolymerization: Toward New Platform Chemicals 2018, Chemical Reviews Bioavailability of dietary polyphenols and gut microbiota metabolism: Antimicrobial properties 2015, BioMed Research International The shikimate pathway and aromatic amino acid biosynthesis in plants 2012, Annual Review of Plant Biology View all citing articles on Scopus Published by the Molecular Plant Shanghai Editorial Office in association with Oxford University Press on behalf of CSPP and IPPE, SIBS, CAS. Copyright © 2010 The Author. All rights reserved. All rights reserved. Recommended articles Should I stay or should I go: are chlorogenic acids mobilized towards lignin biosynthesis? 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5004
https://mathspace.co/textbooks/syllabuses/Syllabus-500/topics/Topic-9567/subtopics/Subtopic-127201/
Book a Demo Log inSign upBook a Demo India Class X Determine a valid triangle from given conditions LessonPractice Lesson This applet starts with a side of length 12 units, and two other lengths of 7 and 4. Can you make this into a triangle? Then explore other lengths. What connection can you see between lengths b and c to length a? What kind of values do b and c take to make a triangle with base a. Created with Geogebra Adapted from This next applet shows the same relationship between the three sides in a different way. There are 2 circles whose raidii form 2 sides of the triangle. The base of the triangle is formed by a line segment joining the two centres together. Play with this applet and see if you can confirm the same relationship you discovered above. Created with Geogebra Adapted from Triangle Inequality Theorem The triangle inequality theorem states that the sum of two side lengths of a triangle is always greater than the third side. (if you hadn't already found it, just go back to the applet above and test it out) The triangle inequality theorem can be used to help us determine if a triangle can exist from the lengths of given sides. Example 1 Can a triangle be formed from $$3 segments of lengths $$2,5 and $$4. Think: Check the combinations of sides Do: $$2+5=7 which is greater than the third side of length $$4 $$2+4=6, which is greater than the third side of length $$5 and $$5+4=9 which is greater than the third side of length $$2 So a triangle can be made using the lengths $$2,4 and $$5. Example 2 Can a triangle be formed from $$3 segments of lengths $$6,2 and $$3. Think: Check the combinations of sides Do: $$6+2=8 which is greater than the third side of length $$3 $$2+3=5, which is NOT greater than the third side of length $$6 and $$3+6=9 which is greater than the third side of length $$2 So because the second combination did not satisfy the triangle inequality, then a triangle can NOT be made using the lengths $$6,2 and $$3. Triangle rules Any of the triangle rules we know can be used to determine a triangle. Sum of angles in a triangle is equal to $$180 degrees. The triangle inequality theorem states that the sum of two side lengths of a triangle is always greater than the third side. In right-angled triangles, we have the trigonometric ratios of tangent, sine and cosine $$tanθ=oppositeadjacent​,cosθ=adjacenthypotenuse​,sinθ=oppositehypotenuse​ Cosine rule, $$a2=b2+c2−2bccosa Sine rule, $$asin(A)​=bcos(B)​=ctan(C)​ Worked Examples question 1 Consider the adjacent figure: Reload the image Is this a valid triangle? Yes A No question 2 Given three side lengths $$9, $$4, and $$5: Complete the following statements, using $$=, $$>, or $$<: $$9+45 $$9+54 $$5+49 2. Do these lengths form a valid triangle? Yes A No B question 3 Outcomes 10.T.IT.1 Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios, whichever are defined at 0° and 90°. Values (with proofs) of the trigonometric ratios of 30°, 45° and 60°. Relationships between the ratios. What is Mathspace About Mathspace
5005
https://www.savemyexams.com/ap/chemistry/college-board/22/revision-notes/unit-6-thermochemistry/calorimetry/calorimetry-calculations/
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Tell us Study Guides 1. Unit 1: Atomic Structure & Properties 5 Topics · 18 Study Guides 1. Moles & Molar Mass ##### The Mole Concept ##### Masses & Particles ##### Molar Mass 2. Mass Spectra of Elements ##### Mass Spectrum of an Element ##### Average Atomic Mass 3. Elements & Mixtures ##### Elemental Composition ##### Composition of Mixtures ##### Elemental Analysis 4. Atoms & Electrons ##### Atomic Structure ##### Using Coulomb’s Law ##### Electron Configuration ##### The Aufbau Principle ##### Ionization Energy ##### Photoelectron Spectroscopy 5. Periodic Trends ##### Structure of the Periodic Table ##### Periodic Trends ##### Predicting Properties of Elements ##### Valence Electrons & Ionic Compounds 2. Unit 2: Compound Structure & Properties 4 Topics · 17 Study Guides 1. Chemical Bonding ##### Electronegativity Values ##### Covalent Bonds ##### Ionic vs Covalent ##### Valence Electrons in a Metallic Solid ##### Intramolecular Force & Potential Energy ##### Coulomb's Law & Attractive Forces 2. Ionic Crystals & Metals ##### Ionic Crystals ##### Representing Metallic Bonding ##### Interstitial Alloys ##### Substitutional Alloys 3. Lewis Diagrams ##### Lewis Diagrams ##### Resonance & Lewis Structures ##### Formal Charge ##### Limitations of Lewis Structures 4. VSEPR & Hybridization ##### VSEPR Theory ##### Hybridization ##### Sigma & Pi Bonds 3. Unit 3: Properties of Substances & Mixtures 5 Topics · 33 Study Guides 1. Intermolecular & Interparticle Forces ##### London Dispersion Forces ##### Dipole-Induced Dipole Interactions ##### Dipole-Dipole Interactions ##### Ion-Dipole Interactions ##### Molecular Dipole Moment ##### Hydrogen Bonding ##### Interactions in Large Biomolecules 2. Properties of Solids, Liquids & Gases ##### Explaining the Properties of Solids & Liquids ##### Properties of Ionic Solids ##### Covalent Network Solids ##### Molecular Solids ##### Metallic Solids ##### Noncovalent Interactions in Large Molecules ##### The Structure of Solids ##### The Liquid Phase ##### The Gas Phase 3. Gas Laws ##### The Ideal Gas Law ##### Partial Pressure & Total Pressure ##### Graphical Representations of the Gas Laws ##### Kinetic Molecular Theory ##### The Maxwell Boltzmann Distribution ##### Non-Ideal Behavior of Gases 4. Solutions & Mixtures ##### Calculations About Solutions ##### Homogeneous & Heterogeneous Mixtures ##### Molarity ##### Particulate Representations of Solutions ##### Separation By Chromatography ##### Separation by Distillation ##### Solubility of Ionic & Molecular Compounds 5. Spectroscopy ##### The Electromagnetic Spectrum ##### Transitions Associated with Radiation ##### Properties of Photons ##### Beer-Lambert Law 4. Unit 4: Chemical Reactions 4 Topics · 15 Study Guides 1. Chemical Changes ##### Physical & Chemical Changes ##### Representing Chemical Changes ##### Balancing Chemical Equations ##### Physical & Chemical Processes 2. Stoichiometry ##### Conservation of Mass ##### Mole Calculations ##### Ideal Gas Law & Solutions ##### Introduction to Titration 3. Types of Chemical Reactions ##### Identifying Acid-Base Reactions ##### Oxidation-Reduction Reactions ##### Assigning Oxidation Numbers ##### Oxidation-Reduction (Redox) Reactions ##### Precipitation Reactions 4. Brønsted-Lowry Acids & Bases ##### Brønsted-Lowry Acids & Bases ##### Conjugate Acids & Bases 5. Unit 5: Kinetics 3 Topics · 18 Study Guides 1. Reaction Rates ##### Measuring Reaction Rate ##### Stoichiometry & Reaction Rate ##### Factors Affecting Reaction Rate ##### Introduction to Rate Law ##### Concentration-Time Graphs ##### Manipulating Concentration-Time Graphs ##### Concentration-Time Graphs & Rate Constants ##### Concentration-Time Graphs & Half-Life ##### Catalysts & Reaction Rate ##### Catalysts in Action 2. Elementary Reactions & Collisions ##### Elementary Reactions ##### Collision Model ##### Reaction Energy Profiles ##### The Arrhenius Equation 3. Reaction Mechanisms ##### Reaction Mechanisms ##### Reaction Mechanism & Rate Law ##### Pre-Equilibrium Approximation ##### Multistep Reaction Energy Profile 6. Unit 6: Thermochemistry 3 Topics · 10 Study Guides 1. Chemical Energy ##### Energy of Phase Changes ##### Exothermic & Endothermic Reactions ##### Energy Diagrams ##### Thermal Energy & Molecular Collisions 2. Calorimetry ##### Energy Transfers ##### Calorimetry Calculations 3. Enthalpy Changes ##### Enthalpy of Reaction ##### Enthalpy & Bond Energies ##### Enthalpy of Formation ##### Hess's Law 7. Unit 7: Equilibrium 5 Topics · 14 Study Guides 1. Equilibrium ##### Reactions & Equilibrium ##### Reversible Reactions 2. Equilibrium Constants ##### The Reaction Quotient ##### Calculating the Equilibrium Constant ##### Magnitude of the Equilibrium Constant ##### Manipulating the Equilibrium Constant 3. Equilibrium Representation & Calculations ##### Calculating Equilibrium Concentrations ##### Representations of Equilibrium 4. Le Chtelier’s Principle ##### Le Chtelier’s Principle ##### Reaction Quotient & Le Chtelier’s Principle 5. Solubility Equilibria ##### Solubility-Product Constant ##### Common-Ion Effect ##### pH & Solubility ##### Free Energy of Dissolution 8. Unit 8: Acids & Bases 4 Topics · 15 Study Guides 1. pH & pOH ##### pH & pOH ##### The Ionic Product of Water ##### pH & pOH of Strong Acids & Bases 2. Acids & Bases ##### Ionization of Weak Acids & Weak Bases ##### Acid-Base Reactions ##### Relative Strength of Acids & Bases ##### pH & pKa 3. Acid-Base Titrations ##### Titration Curves & Equivalence Points ##### Titration Curves & Half-Equivalence Points ##### Titration Curves of Polyprotic Acids ##### Acid-Base Indicators 4. Buffers ##### Properties of Buffers ##### The Henderson-Hasselbalch Equation ##### Buffer Capacity ##### pH & Solubility 9. Unit 9: Thermodynamics & Electrochemistry 3 Topics · 14 Study Guides 1. Entropy ##### Introduction to Entropy ##### Entropy Calculations 2. Free Energy ##### Gibbs Free Energy Change ##### Thermodynamic Favorability ##### Gibbs Free Energy Calculations ##### Thermodynamic & Kinetic Control ##### Free Energy & Equilibrium ##### Coupled Reactions ##### Free Energy of Dissolution 3. Electrochemical Cells ##### Electrochemical Cells ##### Cell Potential & Thermodynamic Favorability ##### Cell Potential Under Nonstandard Conditions ##### Cell Potential & the Nernst Equation ##### Electrolysis & Faraday’s Law AP®ChemistryCollege BoardStudy GuidesUnit 6: Thermochemistry Calorimetry Calorimetry Calculations Calorimetry Calculations(College Board AP® Chemistry):Study Guide Download PDF Written by:Philippa Platt Reviewed by:Stewart Hird Updated on 24 August 2024 Calorimetry Calculations Measuring enthalpy changes Calorimetryis a technique used to measure changes in enthalpy of chemical reactions A calorimeter can be made up of a polystyrene drinking cup, a vacuum flask or metal can Diagram to show how to set up a simple calorimeter _A polystyrene cup can act as a calorimeter to find enthalpy changes in a chemical reaction_ The energy needed to raise the temperature of 1 g of a substance by 1 K is called the specific heat capacity(_c_) of the liquid The specific heat capacity of water is 4.18 J g-1 K-1 The energy transferred as heat can be calculated by: _Equation for calculating energy transferred in a calorimeter_ Worked Example The energy from 0.01 mol of propan-1-ol was used to heat 250 g of water. The temperature of the water rose from 298K to 310K (the specific heat capacity of water is 4.18 J g-1 K-1). Calculate the enthalpy of combustion. Answer: Step 1:_q_ = _m_ x _c_ x Δ _T_ _m_(of water) = 250 g _c_(of water) = 4.18 J g-1 K-1 Δ _T_(of water) = 310 - 298 K = 12 K Step 2:_q_ = 250 x 4.18 x 12 _q_= 12 540 J Step 3: This is the energy released by 0.01 mol of propan-1-ol Total energy Δ _H = q_÷ n = 12 540 J ÷ 0.01 mol = 1 254 000 J mol-1 Total energy = - 1254 kJ mol-1 Examiner Tips and Tricks There's no need to convert the temperature units in calorimetry as the change in temperature in o C is equal to the change in temperature in K Worked Example Excess iron powder was added to 100.0 cm 3 of 0.200 mol dm-3 copper(II) sulfate solution in a calorimeter. The reaction equation was as follows Fe (s) + CuSO 4(aq)→ FeSO 4(aq) + Cu (s) The maximum temperature rise was 7.5 o C. Determine the enthalpy of reaction, in kJ. Answer: Step 1: Calculate _q:_ _q_ = _m_ x _c_ x Δ _T_ _q_ = 100 g x 4.18 J g-1 K-1 x 7.5 K = - 3135 J Step 2: Calculate the amount of CuSO 4(aq) moles = volume in dm 3 x concentration = 0.1 x 0.2 = 0.02 mol Step 3: Calculate Δ _H_ Δ _H_ = _q_ ÷ _n_ = -3135 J ÷ 0.02 mol = - 156 750 J = -156.75 kJ Δ _H_= -160 kJ_(2 sig figs)_ Enthalpy of combustion experiments The principle here is to use the heat released by a combustion reaction to increase the heat content of water A typical simple calorimeter is used to measure the temperature changes in the water Diagram to show the set-up of a typical calorimeter _Not all the heat produced by the combustion reaction is transferred to the water. Some heat is lost to the surroundings and absorbed by the calorimeter_ To minimise the heat losses the copper calorimeter should not be placed too far above the flame and a lid placed over the calorimeter Shielding can be used to reduce draughts In this experiment the main sources of error are Heat losses Incomplete combustion Worked Example 1.023 g of propan-1-ol (_M_= 60.11 g mol-1) was burned in a spirit burner and used to heat 200 g of water in a copper calorimeter. The temperature of the water rose by 30 o C. Calculate the enthalpy of combustion of propan-1-ol using this data. Answer: Step 1: Calculate _q:_ _q_ = _m_ x _c_ x Δ _T_ _q_ = 200 g x 4.18 J g-1 K-1 x 30 K = - 25 080 J Step 2: Calculate the amount of propan-1-ol burned: moles = mass ÷ molar mass = 1.023 g ÷ 60.11 g mol-1= 0.01702 mol Step 3: Calculate Δ _H_: Δ _H_ = _q_ ÷ _n_ = -25 080 J ÷ 0.01702 mol = - 1 473 560 J = -1 474 kJ Δ _H_=-1.5 x 103kJ Unlock more, it's free! 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It's removed a lot of stress from the exams. Sameera Parent > Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years Mark Chemistry Teacher Excellent Read more reviews Join now for free Test yourself Did this page help you? Yes No Previous:Energy TransfersNext:Enthalpy of Reaction More Exam Questions you might like Calorimetry Enthalpy Changes Equilibrium Equilibrium Constants Equilibrium Representation & Calculations Le Chtelier’s Principle Solubility Equilibria pH & pOH Acids & Bases Acid-Base Titrations Author Reviewer Author:Philippa Platt Expertise:Chemistry Content Creator Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener Reviewer:Stewart Hird Expertise:Chemistry Content Creator Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies. 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https://pubmed.ncbi.nlm.nih.gov/4048460/
Superior vena caval obstruction: detection using CT - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Superior vena caval obstruction: detection using CT R E Bechtold,N T Wolfman,N Karstaedt,R H Choplin PMID: 4048460 DOI: 10.1148/radiology.157.2.4048460 Item in Clipboard Superior vena caval obstruction: detection using CT R E Bechtold et al. Radiology.1985 Nov. Show details Display options Display options Format Radiology Actions Search in PubMed Search in NLM Catalog Add to Search . 1985 Nov;157(2):485-7. doi: 10.1148/radiology.157.2.4048460. Authors R E Bechtold,N T Wolfman,N Karstaedt,R H Choplin PMID: 4048460 DOI: 10.1148/radiology.157.2.4048460 Item in Clipboard Full text links Cite Display options Display options Format Abstract A review of 210 chest computed tomographic (CT) examinations demonstrating upper chest masses revealed 16 cases of superior vena caval obstruction (SVCO); 11 of these were clinically occult. Two of the occult cases subsequently became clinically evident. Contrast-enhanced chest CT permits the diagnosis of clinically occult SVCO and should be considered in patients with upper chest masses, even in the absence of any physical signs of SVCO. PubMed Disclaimer Similar articles Clinical and radiological grading of superior vena cava obstruction.Plekker D, Ellis T, Irusen EM, Bolliger CT, Diacon AH.Plekker D, et al.Respiration. 2008;76(1):69-75. doi: 10.1159/000108921. Epub 2007 Sep 25.Respiration. 2008.PMID: 17895621 Systemic-to-pulmonary venous shunt in superior vena cava obstruction revealed on dynamic helical CT.Grayet D, Ghaye B, Szapiro D, Dondelinger RF.Grayet D, et al.AJR Am J Roentgenol. 2001 Jan;176(1):211-3. doi: 10.2214/ajr.176.1.1760211.AJR Am J Roentgenol. 2001.PMID: 11133568 No abstract available. [Endoprosthetic dilatation of the vena cava eliminates obstruction in superior vena cava syndrome].Lindström F, Lunderquist A, Jakobsson P.Lindström F, et al.Lakartidningen. 1991 Jan 9;88(1-2):52-4.Lakartidningen. 1991.PMID: 1848336 Swedish. Computed tomography of superior vena cava obstruction.Yedlicka JW Jr, Cormier MG, Gray R, Moncada R.Yedlicka JW Jr, et al.J Thorac Imaging. 1987 Jan;2(1):72-8. doi: 10.1097/00005382-198701000-00006.J Thorac Imaging. 1987.PMID: 3302288 Review. Superior vena cava obstruction evaluation with MDCT.Sheth S, Ebert MD, Fishman EK.Sheth S, et al.AJR Am J Roentgenol. 2010 Apr;194(4):W336-46. doi: 10.2214/AJR.09.2894.AJR Am J Roentgenol. 2010.PMID: 20308479 Review. See all similar articles Cited by CT for intracardiac thrombi and tumors.Tatli S, Lipton MJ.Tatli S, et al.Int J Cardiovasc Imaging. 2005 Feb;21(1):115-31. doi: 10.1007/s10554-004-5342-x.Int J Cardiovasc Imaging. 2005.PMID: 15915945 Review. A retrospective cohort study of the clinical safety of endobronchial ultrasound in patients with superior vena cava obstruction.Tu JSY, Witt A, Giudice C, Steinfort D, Heinze S, Rangamuwa K.Tu JSY, et al.Medicine (Baltimore). 2025 Jun 13;104(24):e42969. doi: 10.1097/MD.0000000000042969.Medicine (Baltimore). 2025.PMID: 40527767 Free PMC article. Endovascular management of chronic upper extremity deep vein thrombosis and superior vena cava syndrome.Warren P, Burke C.Warren P, et al.Semin Intervent Radiol. 2011 Mar;28(1):32-8. doi: 10.1055/s-0031-1273938.Semin Intervent Radiol. 2011.PMID: 22379274 Free PMC article. Endovascular stenting to treat obstruction of the superior vena cava.Watkinson AF, Yeow TN, Fraser C.Watkinson AF, et al.BMJ. 2008 Jun 21;336(7658):1434-7. doi: 10.1136/bmj.39562.512789.80.BMJ. 2008.PMID: 18566082 Free PMC article.Review. 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https://www.xometry.com/resources/materials/tensile-stress-vs-compressive-stress/
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From tensile to compressive to shear stress and many more in between, they show the strength of materials and their point of deformity. But it’s how they react to these forces that matters to designers, manufacturers, and engineers who are aiming to create products and structures that can handle these forces in real-life scenarios. Two such examples of stresses that are important for determining what materials to use are tensile stress and compressive stress, which we’ll explore the differences between. What Is Tensile Stress? Tensile stress is all about elongation, and it describes when an applied force stretches the length of a material along the axis of the force. If you’re working with a formula, you can represent tensile stress through this: stress = F/A Measurements for this type of stress, and most other kinds, are usually expressed through Pascals (Pa) or through pounds per square inch (psi). While tensile elongates, it’s helpful to see how it compares to compressive and shear forces, too, which you can see in the diagrams below. Examples of Tensile Stress There are tons of examples that exist in the real world of tensile stress and how it looks. We’ve listed a few to give you a good idea of how it appears: Pulling a rubber band apart Two people pulling opposite ends a rope away from each other A crane cable that’s holding something suspended in the air Cables on a suspension bridge while vehicles pass over it What Is Compressive Stress? This type of stress could be viewed as the reverse of tensile stress. Rather than elongating a material, compressive stress makes it become more compact, exactly as its name states. It’s also expressed in Pascals or pounds per square inch and it, too, deforms and dislocates the material as force is applied. Examples of Compressive Stress There are plenty of examples of this type of stress, too, many of which you’ll come across in everyday life: Concrete slabs as flooring in buildings or as sidewalks that deform as people, animals, and machines walk and roll over them. Structural columns in buildings that support the weight of every floor Furniture items like chairs and couches which compress under the weight of people on top The image below shows a visual comparison of before and after testing and how tensile stress impacts a material versus how compressive stress affects it. [IMAGE] Comparing These Stress Types The main difference between tensile stress and compressive stress is the type of force applied and how the material deforms as a response. With tensile stress, the force stretches and elongates the material, whereas compressive stress presses it together and makes the material more compact. Although they’re different and basically opposite in nature, they can still happen at the same time. It may sound impossible, but it’s a common occurrence in materials and objects that bend. For example, if you bend a pencil until it breaks, you can see both of these stresses in action. As the pencil bends into a U-shape, you can witness compressive stress in the upper portion of the U and tensile stress in the bottom area. How Xometry Can Help Whether you’re working with objects made of materials that are frequently exposed to compressive stress or tensile stress, our instant service quotes can help you find the right process, from 3D printing to CNC machining. Disclaimer The content appearing on this webpage is for informational purposes only. Xometry makes no representation or warranty of any kind, be it expressed or implied, as to the accuracy, completeness, or validity of the information. Any performance parameters, geometric tolerances, specific design features, quality and types of materials, or processes should not be inferred to represent what will be delivered by third-party suppliers or manufacturers through Xometry’s network. Buyers seeking quotes for parts are responsible for defining the specific requirements for those parts. Please refer to our terms and conditions for more information. Kat de Naoum Kat de Naoum is a writer, author, editor, and content specialist from the UK with 20+ years of writing experience. Kat has experience writing for a variety of manufacturing and technical organizations and loves the world of engineering. Alongside writing, Kat was a paralegal for almost 10 years, seven of which were in ship finance. She has written for many publications, both print and online. Kat has a BA in English literature and philosophy, and an MA in creative writing from Kingston University. Read more articles by Kat de Naoum
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https://www.medsci.cn/article/show_article.do?id=f5a882e500a7
肿瘤 MedSci主页 / 所有栏目 / 肿瘤 王孝深教授组稿｜周鑫：EB病毒DNA在鼻咽癌中的应用现状、机遇与挑战 2024-05-25 医悦汇 医悦汇 发表于上海 鼻咽癌EB病毒 肿瘤新前沿 企业/ 关注 互相关注 血浆游离EBV DNA定量检测方便、快捷、创伤小、费用低，是具有良好敏感度和特异度的新型鼻咽癌肿瘤标志物。其在鼻咽癌诊疗中临床价值已得到众多数据验证。 编者按：在华南和东南亚等地区，鼻咽癌的发生和流行与Epstein-Barr病毒（EBV）感染密切相关。一般认为，鼻咽癌大体肿瘤由EBV感染的单个肿瘤细胞增殖而成，EBV的复制和表达水平从一定程度上反应了体内肿瘤的负荷。因此，EBV相关指标一贯被视作重要的鼻咽癌生物标志物，广泛应用于鼻咽癌诊疗工作中。 本期「专家组稿」由复旦大学附属眼耳鼻喉科医院王孝深教授担任执行主编，他将与复旦大学附属肿瘤医院周鑫博士在本期内容中分享EB病毒DNA在鼻咽癌中的应用现状、机遇与挑战，为医者和患者提供更多参考。 传统的EBV特异性抗体应用时间较长，但诊断特异性较差，半衰期较长，因此不能实时反映体内肿瘤活动。而近年来，血浆/血清游离EBV DNA的临床价值逐渐得到验证，并被公认为鼻咽癌诊断、评估与生存预后中最有价值的生物标志物之一。 一、游离EBV DNA的来源和应用基础 迄今为止，鼻咽癌中外周血游离EBV DNA的来源尚未完全阐明，多数观点认为可能的产生机制包括：（1）鼻咽癌细胞的凋亡；（2）治疗相关的肿瘤坏死；（3）原发肿瘤脱落的鼻咽癌活细胞及其克隆；（4）潜在感染的B淋巴细胞；（5）游离的EBV颗粒等。Chan等的测序结果显示，外周血游离EBV DNA为82-181 Kb的小片段DNA，其来源很可能是肿瘤细胞崩解后基因组DNA碎裂并释放入血。而另一项研究通过肿瘤抑制基因启动子甲基化对肿瘤源性EBV DNA片段比例进行了定量计算，发现肿瘤细胞坏死和凋亡是鼻咽癌相关EBV DNA的主要来源。也有研究认为，游离EBV DNA部分来自巨噬细胞对鼻咽癌肿瘤坏死和凋亡DNA产物的吞噬、消化和释放。此外，肿瘤活细胞及克隆也可能主动释放游离EBV DNA。据结直肠癌中提出的理论模型，当实体肿瘤增殖至超过100g，相当于3×1010个肿瘤细胞时，每天即会有3.3%的肿瘤DNA入血，类似现象可能同样存在于鼻咽癌中。值得一提的是，外周血游离EBV DNA的水平还受到治疗模式、入血途径等多种因素影响，呈现显著的个体差异和时间依赖性动态变化。 二、EBV DNA与鼻咽癌的诊断/筛查/评估 早在1996年，Lo YM等就发现鼻咽癌患者中EBV DNA阳性率高达96%，而正常对照组仅为7%，提示该指标对鼻咽癌的潜在诊断价值。此后，多项临床研究相继将其用于鼻咽癌诊断试验。截止目前，各中心EBV DNA检测多靶向EBV 基因组中高度稳定和保守的BamH1-W片段，其诊断敏感度从53%-99%不等（平均94%），而特异度达87%-100%（平均94%），相较于应用已逾30年的EBV相关抗体（VCA/IgA，EA/IgA），EBV DNA的诊断性能具有显著优势，并逐渐呈替代之势，成为鼻咽癌中最具前景的液体活检指标之一。 基于其优秀的诊断效能，EBV DNA逐渐被用于鼻咽癌的健康人群筛查，以冀实现鼻咽癌的早期诊断。在一项香港地区20174人的大规模筛查研究中，309人（1.5%）血浆EBV DNA呈持续阳性；在接受鼻咽镜、鼻咽MRI等检查后，34人最终被证实为鼻咽癌，其中大多为早期患者（71%），远超既往自然患者人群中的早期比例（20%-25%）。而广东四会和中山市开展的一项人群研究显示，基于EBV IgA筛查阳性结果之上进一步加做EBV DNA检测，其阳性预测值和阴性预测值分别可达30%和99.3%，提示EBV DNA和血清IgA具有潜在的协同筛查价值。值得一提的是，正常健康人体也可出现一过性EBV DNA升高，但在随访中很快转阴。相比之下，EBV DNA呈现持续阳性、甚至进行性升高的患者，应当高度怀疑存在EBV相关肿瘤风险，建议进一步重点排查鼻咽等。 除定性诊断外，游离EBV DNA定量结果还可用于定量评估鼻咽癌进展程度。早期研究显示，初治鼻咽癌患者血浆/血清EBV DNA浓度与肿瘤分期呈明显正相关，进展期鼻咽癌患者EBV DNA的中位水平可达早期患者的8倍。影像学定量分析则发现，无远处转移鼻咽癌患者的EBV DNA水平、肿瘤体积、18F-FDG PET-CT的SUVmax三者之间均具有显著相关性，强烈提示外周EBV DNA与肿瘤负荷有关。一项来自复旦大学附属肿瘤医院的研究结果则进一步提示，无远转鼻咽癌患者的血浆EBV DNA水平更大程度上决定于阳性淋巴结的体积，而原发灶对之影响较弱，其解剖和生物学机制尚有待进一步挖掘。 三、EBV DNA与鼻咽癌近期疗效评估 基于其肿瘤源性，血浆/血清游离EBV DNA的变化情况可灵敏反映治疗中肿瘤退缩情况，包括对诱导化疗、放疗的敏感性等。监测EBV DNA在各个静态治疗节点的检出情况、乃至密切追踪其动态变化，将有助于准确评估疗效，以便及时调整治疗策略。 静态检测节点涉及放疗后、系统性治疗疗程后、复发术后、一线治疗中多个时间点。在根治性放疗中，Lo YM等结合影像学评估，认为根治性放疗后EBV DNA跌落至阈值以下预示临床完全缓解，而持续EBV DNA阳性则提示肿瘤残留可能性大，可能需要追加放疗剂量。接受诱导化疗的局晚期鼻咽癌中，诱导后EBV DNA状态可有效反映放化疗敏感度，诱导后转阴者放疗后肿瘤完全退缩概率较高；此外，诱导后EBV DNA转阴者，后续同步化疗的强度对预后的影响减弱，提示诱导后EBV DNA阴性有望指导后续治疗，为后续减强度放化疗提供支持。放疗后颈淋巴结残留或复发、接受颈清扫术的鼻咽癌患者中，术后持续EBV DNA阳性提示手术清扫不全、肿瘤残留以及复发和转移危险，需考虑辅助治疗和密切随访。而在POLARIS-02研究中，化疗抵抗的复发转移鼻咽癌接受PD-1单抗治疗4周时EBV DNA衰减>50%者，其后续影像学客观缓解率显著更高，提示EBV DNA可早期反映免疫治疗疗效，以便后续药物调整。 动态监测EBV DNA清除速度亦具有重要意义。血液动力学研究发现，对根治性放疗敏感者，放疗开始后1周内均观察到一过性EBV DNA升高，峰值甚至可高达治疗前2.9倍，其成因可能与放疗所致坏死或凋亡瘤细胞释放大量核酸片段有关，一般认为是应答良好的标志。在此之后，由于体内吞噬和清除作用，EBV DNA水平快速衰减至较低水平。目前认为，治疗过程中EBV DNA清除越快，肿瘤完全缓解率和总生存率越高。相反，延迟清除乃至持续高水平通常提示高复发风险，例如，放疗中期（第4周）EBV DNA仍未转阴者，其治疗失败率显著上升。上述现象提示，血浆/血清EBV DNA的清除速度有望成为早期评估疗效的敏感指标之一，将为后期自适应放疗、更改系统治疗方案等精准治疗手段提供重要依据。 四、EBV DNA与鼻咽癌长期生存预后 （一）治疗前EBV DNA 治疗前基线EBV DNA水平与可预示鼻咽癌长期生存，尤其是远处转移。一项经典的台湾研究显示，以1500 copies/L为截断值，可以显著区分局部区域晚期鼻咽癌患者的DFS和OS，其预后作用独立于TNM分期。另一项香港研究也发现，治疗前EBV DNA 大于4000 copies/L者远处转移风险显著升高。基于其预后价值，基线EBV DNA水平有望在TNM分期基础上进一步细化风险分层。Leung等发现， I/II期鼻咽癌合并治疗前高水平EBV DNA者的死亡率逼近III期患者，而低水平EBV DNA者则和I期相当。在局部区域晚期（III-IV期）患者中，EBV DNA也可进一步筛选出预后更差的患者亚组。 鉴于基线EBV DNA水平的预后价值，NCCN头颈部肿瘤临床实践指南从2010版开始正式将EBV DNA列为治疗前诊断和评估的推荐辅助手段，有望对UICC国际分期的不足作出补充。多项最新研究均显示，通过递归分割分析（Recursive partitioning analysis, RPA）将EBV DNA和TNM分期进行整合后，其衍生出的新分期系统的预测效能较单纯的TNM分期有了明显提高。但迄今为止，将EBV DNA与TNM分期系统整合仍然困难重重，其中最大难点就在于各中心EBV DNA检测结果差异极大，采用的预后界值也各不相同，很难得出统一结论。多中心之间EBV DNA检测方法的标准化、结果互认仍有待改进，以期未来更好地将EBV DNA应用于临床。 （二）治疗后EBV DNA浓度与预后 多项研究表示，治疗后DNA的预后效能要优于治疗前水平，其对预后的区分度明显大于治疗前DNA。越来越多证据表明，治疗后EBV DNA状态和治疗前EBV DNA在鼻咽癌中代表着不同的临床价值。治疗前EBV DNA水平体现了肿瘤的基线负荷，而治疗后EBV DNA持续阳性则可能提示肿瘤残留，高敏感性的检测手段更是使得“微小残留灶”（minimal residual disease）的发现成为可能。因此，一些学者提出，鼻咽癌接受根治性放疗后，即使影像学上完全缓解，EBV DNA阳性的患者仍具有较高复发和转移风险，可考虑酌情予以辅助化疗。 五、EBV DNA与鼻咽癌治疗后随访监测 以往的鼻咽癌常规随访监测往往依赖于影像学检查，血液学指标缺乏。EBV相关抗体如VCA/IgA等在反映肿瘤活动上往往存在滞后性，例如，肿瘤达根治后，VCA/IgA还将在相当长时间内持续阳性。相比之下，血浆/血清EBV DNA可以实时、动态反映体内肿瘤活动情况，用于随访监测也更为敏感。研究证实，治疗后随访过程中，EBV DNA升高可以比临床症状和影像学检查提早6个月发现复发转移。Ngan等则对鼻咽癌患者根治性放化疗后进行连续血清EBNA-1 EBV DNA监测，结果发现，62%患者在临床复发之前即出现DNA升高（平均提前时间17.4周），38%则与复发同时出现。多项研究提示，将血浆EBV DNA作为筛查手段，辅以18F-FDG PET-CT，有助于进一步提高复发、转移的准确度，且节省了大笔费用。 六、EBV DNA研究新动向 （一）放疗后EBV DNA阳性者是否需要辅助化疗 一项回顾性研究显示，根治性放疗后EBV DNA仍呈阳性的鼻咽癌患者约占13.4%，其中部分人接受口服和/或静脉辅助化疗后，远处转移率和死亡率较无辅助化疗组显著降低。然而，HKNPCSG-0502研究将治疗后EBV DNA阳性患者随机分组至辅助化疗组和观察随访组，却发现二组OS、局部复发率和远处转移率均无显著差异，提示并非所有治疗后EBV DNA阳性者均能从辅助化疗获益。Hui EP通过结合TNM分期和治疗后EBV DNA，进一步将0502辅助组分为3个预后亚组，并认为其中低风险组长期预后较好，可豁免辅助化疗。随着亚组分析和联合分析的细化，对治疗后EBV DNA的预后及治疗指导价值的认识还有待进一步深化。 （二）复查随访中EBV DNA能否精准预测复发转移 一项回顾性研究对1984例初治无转移鼻咽癌患者进行治疗后监测，其中767在随访中出现EBV DNA阳性，其复发率远大于阴性者（63.8% vs 8.6%）；血浆EBV DNA能有效预测局部（准确度78.2%）、区域（85.9%）和远处转移（92.8%），其中对于远处转移的预示尤为突出。值得一提的是，随访中EBV DNA阳性相较于临床影像学复发转移平均提前2.3月，为临床早期干预提供了宝贵时间窗。在此基础上，另一项研究通过倾向配比评分探索了随访中EBV DNA阳性者提前采用口服化疗干预的有效性，结果提示，相较于传统密切观察随访的处理，提前口服化疗可使EBV DNA阳性患者全组和各亚组的DFS显著提高，提示EBV DNA这一生物标志物对于随访复查中治疗选择的重要参考价值。 （三）诱导化疗期间EBV DNA早期清零能否指导后续减强度放疗 Lv JW等发现，鼻咽癌诱导化疗过程中，不同的EBV DNA缓解类型（早期、中期、晚期缓解、治疗抵抗）对应不同治疗敏感性和复发风险，并进而提出，诱导1周期后EBV DNA即清零的患者，后续可考虑省略同步化疗等减强度治疗；而诱导2周期后仍持续阳性者，可能具有原发化疗抵抗，后续可能需要考虑更换化疗方案或引入靶向药物等，必要时需排查治疗中远处转移可能性。相应的前瞻性临床研究正在进行中，其结果值得期待。 七、EBV DNA的局限和挑战 EBV DNA在鼻咽癌中的应用面很广，却并非万能。在鼻咽癌非流行区域，多数角化性鼻咽癌（WHO I型）以及部分非角化性癌与EBV并无关联，EBV DNA的价值较为局限。其次，受肿瘤负荷和入血途径的限制， I期鼻咽癌EBV DNA阳性率仍然较低，仅50%-86%。此外，手术、放疗等对肿瘤周围血管造成的不可逆损伤可能导致EBV DNA入血障碍，敏感度极大下降。如何提高这部分患者的DNA检出率，以及如何联用其他指标提高诊断准确性，仍是亟需解决的问题之一。 总结和展望 血浆游离EBV DNA定量检测方便、快捷、创伤小、费用低，是具有良好敏感度和特异度的新型鼻咽癌肿瘤标志物。其在鼻咽癌诊疗中临床价值已得到众多数据验证。然而，目前的检测方法学仍存在异质性，中心之间结果互认存在障碍。将来应进一步鼓励各中心的数据交流，鼓励研究检测标准化流程，以推广EBV DNA检测结果的普适性。 此外，外周游离EBV DNA的生物学角色仍未完全阐明，其在体内产生、分布、清除的过程尚需要基础研究论证。同时，EBV DNA体现出的某些特性，如其浓度受转移淋巴结、远处转移影响更明显，其背后的生物学机制及临床应用，将值得进一步研究解答。 阅读全文 微信扫一扫 分享 版权声明：本网站所有内容来源注明为“梅斯医学”或“MedSci原创”的文字、图片和音视频资料，版权均属于梅斯医学所有。非经授权，任何媒体、网站或个人不得转载，授权转载时须注明来源为“梅斯医学”。其它来源的文章系转载文章，或“梅斯号”自媒体发布的文章，仅系出于传递更多信息之目的，本站仅负责审核内容合规，其内容不代表本站立场，本站不负责内容的准确性和版权。如果存在侵权、或不希望被转载的媒体或个人可与我们联系，我们将立即进行删除处理。 在此留言 评论区 (1) #插入话题 [GetPortalCommentsPageByObjectIdResponse(id=2206639, encodeId=b62922066393d, content=#鼻咽癌# #EB病毒#, beContent=null, objectType=article, channel=null, level=null, likeNumber=142, replyNumber=0, topicName=null, topicId=null, topicList=[TopicDto(id=6468, encryptionId=16826468b2, topicName=EB病毒), TopicDto(id=103423, encryptionId=c3f910342343, topicName=鼻咽癌)], attachment=null, authenticateStatus=null, createdAvatar=null, createdBy=cade5395722, createdName=梅斯管理员, createdTime=Sun May 26 00:54:58 CST 2024, time=2024-05-26, status=1, ipAttribution=上海)] 2024-05-26 梅斯管理员 来自上海 #鼻咽癌# #EB病毒# 142 0 举报 相关资讯 探索人体的告急信号：了解鼻咽癌的症状与防治 中国是鼻咽癌患者数量最多的国家，了解危险因素和有效的防控措施对于尽早采取干预措施、降低患病风险以及提升预后至关重要。 European Radiology：双能CT对鼻咽癌诱导化疗早期反应和生存的预测！ 双能量CT (DECT) 是一种由两个不同能量数据集生成的功能成像技术，可以精确测量碘含量，并作为组织血管生成和细胞密度的定量生物标志物。使用DECT估计碘含量差异可预测NPC的疾病进展。 European Radiology:多模态放射组学分析在鼻咽癌预后预测中的应用 [18F] FDG PET/CT和多序列MRI在管理鼻咽癌患者方面发挥着重要作用，包括诊断、分期和复发监测。来自PET/CT和MRI图像的传统和新兴指标（如纹理特征）都显示出对鼻咽癌预后预测的潜力。 中山大学贾卫华教授团队《自然·通讯》：口腔微生物异位转移或促进鼻咽癌？ 该研究通过双中心人群的多组学数据揭示了口腔微生物，尤其是有害细菌，能够迁移到鼻咽处，重塑鼻咽微生态，进而渗透到肿瘤中，影响肿瘤微环境并于EB病毒感染相关。 Genes & Diseases：Jiang Zhu/Youquan Bu教授团队发现鼻咽癌相关关键分子网络和基因 该研究表明了细胞周期紊乱在鼻咽癌中的关键作用，揭示了Kinesin家族成员2C（KIF2C）这一有丝分裂中关键枢纽基因在鼻咽癌细胞中的重要调节作用，并提出KIF2C可能是鼻咽癌中治疗中的新型靶标。 姜新教授：同步放化疗后辅助顺铂-吉西他滨与顺铂-氟尿嘧啶治疗N2-3期鼻咽癌 本期文献内容由吉林大学第一医院姜新教授课题组边晨缤分享肿瘤治疗相关领域及病理生物学研究领域最新进展，以供交流！ 梅斯医学 App 前往APP阅读全文，体验更佳 取消 前往 扫描二维码下载梅斯医学APP,参与讨论！ 梅斯医学MedSci APP 医路相伴，成就大医 x 科室 订阅+ 更多科室 工具 服务
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https://medium.com/all-math-before-college/function-transformations-bad4493d7a04
Sign up Sign in Sign up Sign in All Math Before College Notes on important math concepts that are prerequisites for further study in college. Function Transformations Function Transformations -- Listen Share Refer to math is fun: Function Transformations Function Transformations Express different transformations in the function. Refer to the previous note Graph transformations. Let’s assume: f(x) = x^2 g(x) Translation (Shift) Translation C f(x) C f(x) C f(x) C f(x) Reflection (Flip/Mirror) Reflection x-axis y-axis Dilation (Scale/Stretch/Compress) Dilation Stretching the graph when C bigger than 1, compress it when C less than 1. C C y-axis x-axis Summary: -- -- Published in All Math Before College Notes on important math concepts that are prerequisites for further study in college. Written by Solomon Xie Jesus follower, Yankees fan, Casual Geek, Otaku, NFS Racer. No responses yet Help Status About Careers Press Blog Privacy Rules Terms Text to speech
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https://www.quora.com/How-do-you-convert-an-algebraic-expression-into-an-equation
How to convert an algebraic expression into an equation - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Conversion Process Simplifying Algebraic Exp... Algebra Class Mathematical Equations Mathematics and Algebra Equations and Expressions Algebra Conversions 5 How do you convert an algebraic expression into an equation? All related (31) Sort Recommended Aquiles Fischer Studied at School of Hard Knocks · Author has 781 answers and 137K answer views ·2y How about your example? In the mean time, let me give you some examples Twice a number is seven: twice means 2, a number could be n, is =, and 7. Another the high of a triangle is trice its base what is its area: A for area, b for base, trice is 3, so A = 1/2 b (3b). And another, the length of a rectangle is four times it width. P for perimeter, l for lenth, and w for width. P = 2 l + 2 w , so l is fourtimes w, then the formula changes to P = 2 (4 w) + 2 w = 8 w + 2 w = 10 w, Also remember that someting is “larger” or “older” or “more than” or “less than” and so on. Upvote · Sponsored by JetBrains DataGrip, a powerful GUI tool for SQL. Smart code completion, on-the-fly analysis, quick-fixes, refactorings that work in SQL files, and more. Download 999 740 Related questions More answers below Can you give an example of an equation that can be made into an algebraic expression? How do you find an equivalent algebraic expression? How do you find the final result from an algebraic equation or expression? What is the difference between an algebraic equation and an algebraic expression? What is the difference between an algebraic expression and a formula? Which algebraic expression is equivalent to the expression? Craig Kennedy Ruby on Rails Developer ·Jan 21 Related How can you simplify an algebraic expression? Method 1 of 4:Understanding Important Concepts Define "like terms" by their variables and powers. In algebra, "like terms" have the same configuration of variables, raised to the same powers. In other words, for two terms to be "like", they must have the same variable or variables, or none at all, and each variable must be raised to the same power, or no power at all. The order of variables within the term does not matter. For example, 3x and 4x are like terms because each contains the variable x raised to the second power. However, x and x are not like terms because each term has x raised to Continue Reading Method 1 of 4:Understanding Important Concepts Define "like terms" by their variables and powers. In algebra, "like terms" have the same configuration of variables, raised to the same powers. In other words, for two terms to be "like", they must have the same variable or variables, or none at all, and each variable must be raised to the same power, or no power at all. The order of variables within the term does not matter. For example, 3x and 4x are like terms because each contains the variable x raised to the second power. However, x and x are not like terms because each term has x raised to a different power. Similarly, -3yx and 5xz are not like terms because each term has a different set of variables. Factor by writing numbers as the product of two factors. Factoring is the concept of representing a given number as the product of two factors multiplied together. Numbers can have more than one set of factors - for instance, the number 12 can be formed by 1 × 12, 2 × 6, and 3 × 4, so we can say that 1, 2, 3, 4, 6, and 12 are all factors of 12. Another way of thinking of this is that a number's factors are the numbers by which it is evenly divisible. For example, if we wanted to factor 20, we might write it as . Note that variable terms can also be factored - 20x, for instance, can be written as . Prime numbers can't be factored because they are only evenly divisible by themselves and 1. Use the acronym PEMDAS to remember the order of operations. Sometimes, simplifying an expression means nothing more than performing the operations in the expression until no more can be done. In these cases, it's important to remember the order of operations so that no arithmetic errors are made. The acronym PEMDAS can help you remember the order of operations - the letters correspond to the types of operations you should perform, in order. If there is multiplication and division in the same problem, you must complete those operations from left to right when you get to that point. The same goes for addition and subtraction. The image above gives the incorrect answer. The last step did not work the addition and subtraction from left to right. It did the addition first. It should show 25-20 = 5 and then 5 + 6 = 11. arentheses xponents ultiplication ivision ddition ubtraction Method 2 of 4:Combining Like Terms Write your equation. The simplest algebraic equations, those involving just a few variable terms with whole number coefficients and no fractions, radicals, etc., can often be solved in just a few steps. As with most math problems, the first step to simplifying your equation is to write it out! As an example problem, for the next few steps, let's consider the expression . Identify like terms. Next, search your equation for like terms. Remember that like terms have both the same variable(s) and exponent(s). For example, let's identify like terms in our equation 1 + 2x - 3 + 4x. 2x and 4x both have the same variable raised to the same exponent (in this case, the x's aren't raised to any exponent at all). In addition, 1 and -3 are like terms, as neither has any variables. So, in our equation, and are like terms. Combine like terms. Now that you've identified like terms, you can combine them to simplify your equation. Add terms together (or subtract in the case of negative terms) to reduce each set of terms with the same variables and exponents to one singular term. Let's add the like terms in our example. 2x + 4x = 1 + -3 = Create a simplified expression from your simplified terms. After combining your like terms, construct an expression from your new, smaller set of terms. You should get a simpler expression that has one term for each different set of variables and exponents in the original expression. This new expression is equal to the first. In our example, our simplified terms are 6x and -2, so our new expression is . This simplified expression is equal to the original (1 + 2x - 3 + 4x), but is shorter and easier to manage. It's also easier to factor, which, as we'll see below, is another important simplifying skill. Obey the order of operation when combining like terms. In extremely simple expressions like the one dealt with in the example problems above, identifying like terms is simple. However, in more complex expressions, like ones that involve terms in parentheses, fractions, and radicals, like terms which can be combined may not be immediately apparent. In these cases, follow the order of operations, performing operations on the terms in your expression as necessary until only addition and subtraction operations remain. For example, let's consider the equation 5(3x-1) + x((2x)/(2)) + 8 - 3x. It would be incorrect to immediately identify 3x and 2x as like terms and combine them because the parentheses in the expression dictate that we're supposed to do other operations first. First, let's perform the arithmetic operations in the expression in accordance with the order of operations to obtain terms we can use. See below: 5(3x-1) + x((2x)/(2)) + 8 - 3x 15x - 5 + x(x) + 8 - 3x 15x - 5 + x + 8 - 3x. Now, since the only operations left are addition and subtraction, we can combine like terms. x + (15x - 3x) + (8 - 5) Method 3 of 4:Factoring Identify the greatest common factor in the expression. Factoring is a way to simplify expressions by removing factors that are common across all the terms in the expression. To start, find the greatest common factor that all of the terms in the expression share - in other words, the largest number by which all the terms in the expression are evenly divisible. Let's use the equation 9x + 27x - 3. Notice that every term in this equation is divisible by 3. Since the terms aren't all evenly divisible by any larger number, we can say that is our expression's greatest common factor. Divide the terms in the expression by the greatest common factor. Next, divide every term in your equation by the greatest common factor you've just found. The resulting terms will all have smaller coefficients than in the original expression. Let's factor our equation by its greatest common factor, 3. To do so, we'll divide each term by 3. 9x/3 = 3x 27x/3 = 9x -3/3 = -1 Thus, our new expression is . Represent your expression as the product of the greatest common factor and the remaining terms. Your new expression isn't equal to your old one, so it isn't accurate to say that it's simplified. To make our new expression equal to the old one, we'll need to account for the fact that it has been divided by the greatest common factor. Enclose your new expression in parentheses and set the greatest common factor of the original equation as a coefficient for the expression in parentheses. For our example expression, 3x + 9x - 1, we would enclose the expression in parentheses and multiply by the greatest common factor of the original equation to get . This equation is equal to the original, 9x + 27x - 3. Use factoring to simplify fractions. You may now be wondering why factoring is useful if, after removing the greatest common factor, the new expression must be multiplied by it again. In fact, factoring allows a mathematician to perform a variety of tricks to simplify an expression. One of the easiest of these involves taking advantage of the fact that multiplying a fraction's numerator and denominator by the same number gives an equivalent fraction. See below: Let's say our original example expression, 9x + 27x - 3, is the numerator of a larger fraction with 3 in the denominator. This fraction would look like this: (9x + 27x - 3)/3. We can use factoring to simplify this fraction. Let's substitute the factored form of our original expression for the expression in the numerator: (3(3x + 9x - 1))/3 Notice that now, both the numerator and the denominator share the coefficient 3. Dividing the numerator and denominator by 3, we get: (3x + 9x - 1)/1. Since any fraction with "1" in the denominator is equal to the terms in the numerator, we can say that our original fraction can be simplified to . Method 4 of 4:Applying Additional Simplification Skills Simplify fractions by dividing through by common factors. As noted above, if the numerator and denominator of an expression share factors, these factors can be removed from the fraction entirely. Sometimes this will require factoring the numerator, denominator, or both (as was the case in the example problem above) while other times the shared factors are immediately apparent. Note that it is also possible to divide the numerator terms by the expression in the denominator individually to obtain a simplified expression. Let's tackle an example that doesn't necessarily require drawn-out factoring. For the fraction (5x + 10x + 20)/10, we may want to divide every term in the numerator by the 10 in the denominator to simplify, even though the "5" coefficient in 5x isn't bigger than 10 and thus can't have 10 as a factor. Doing so gets us ((5x)/10) + x + 2. If we like, we may want to rewrite the first term as (1/2)x to get (1/2)x + x + 2. Use square factors to simplify radicals. Expressions under a square root sign are called radical expressions. These can be simplified by identifying square factors (factors that are themselves squares of an integer) and performing the square root operation on these separately to remove them from under the square root sign. Let's tackle a simple example - √(90). If we think of the number 90 as the product of two of its factors, 9 and 10, we can take the square root of 9 to give the whole number 3 and remove this from the radical. In other words: √(90) √(9 × 10) (√(9) × √(10)) 3 × √(10) Add exponents when multiplying two exponential terms; subtract when dividing. Some algebraic expressions require multiplying or dividing exponential terms. Rather than computing each exponential term and multiplying or dividing manually, simply add exponents when multiplying and subtract when dividing to save time. This concept can also be used to simplify variable expressions. For example, let's consider the expression 6x × 8x + (x/x). In each occasion where it's necessary to multiply or divide by exponents, we'll subtract or add the exponents, respectively, to quickly find a simplified term. See below: 6x × 8x + (x/x) (6 × 8)x + (x) For an explanation of why this works, see below: Multiplying exponential terms is essentially like multiplying long strings of non-exponential terms. For example, since x = x × x × x and x = x × x × x × x × x, x × x = (x × x × x) × (x × x × x × x × x), or x. Similarly, dividing exponential terms is like dividing long strings of non-exponential terms. x/x = (x × x × x × x × x)/(x × x × x). Since each term in the numerator can be canceled out by a matching term in the denominator, we're left with two x's in the numerator and none in the bottom, giving us an answer of x Upvote · 9 1 Allen Ries Math Major University of Alberta · Author has 25.1K answers and 9.7M answer views ·9mo Related How do you use the algebraic equation with an example? How do you use the algebraic equation with an example? You want to see an example how an algebraic equation is solved. 2a+3 = 6 2a+3–3 = 6–3 2a+(3–3) = 3 2a=3 a=1.5 Upvote · 9 2 Chris Studied Engineering (Graduated 2024) ·2y Related How do you find different forms and expressions of an algebraic equation? This is quite a vague question, but there are a few ways in which you can change the form of an algebraic equation. I don’t know what you mean by the ‘expression’ of an equation. You can perform operations or functions to both sides of the equation to change its form. One reason would be to move terms between the LHS and RHS to solve for different variables. For example, in the equation 6x + 3y = 3, you can subtract 3y from both sides and then divide by 6 to change its form and solve for x in terms of y. You can often factorise parts to simplify it. For example, the equation ‘6x + 3y = 3’ can be Continue Reading This is quite a vague question, but there are a few ways in which you can change the form of an algebraic equation. I don’t know what you mean by the ‘expression’ of an equation. You can perform operations or functions to both sides of the equation to change its form. One reason would be to move terms between the LHS and RHS to solve for different variables. For example, in the equation 6x + 3y = 3, you can subtract 3y from both sides and then divide by 6 to change its form and solve for x in terms of y. You can often factorise parts to simplify it. For example, the equation ‘6x + 3y = 3’ can be factorised to 3(2x+y) = 3, which simplifies to 2x + y = 1 when you divide both sides by 3. You can sometimes substitute identities into equivalent parts of the equation. For example, for the equation ‘y = sin^2(x) + cos^2(x)’, you could replace the RHS with ‘1’ since it’s equivalent to 1. This would give you the solution y = 1. There are probably other ways but that’s all I could think of. Upvote · 9 1 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 620 Related questions Can you give an example of an equation that can be made into an algebraic expression? How do you find an equivalent algebraic expression? How do you find the final result from an algebraic equation or expression? What is the difference between an algebraic equation and an algebraic expression? What is the difference between an algebraic expression and a formula? Which algebraic expression is equivalent to the expression? Can you explain the process of converting algebraic equations into expressions and vice versa? How do you determine if an algebraic expression is a solution to an equation? Why do we use letters to represent variables in an algebraic expression or equation? What is the definition of an algebraic expression? How do you solve algebraic expressions? Can all algebraic expressions be expressed as an equation in two ways? What are some examples of algebraic equations and expressions? Is it possible to solve an algebraic expression without using an algebraic formula or equations? If so, how would you do it? How can you simplify an algebraic expression? How do you translate word problems to algebraic expression/equation? What is the difference between an algebraic expression and an algebraic formula? How can they be used to solve equations? Related questions Can you give an example of an equation that can be made into an algebraic expression? How do you find an equivalent algebraic expression? How do you find the final result from an algebraic equation or expression? What is the difference between an algebraic equation and an algebraic expression? What is the difference between an algebraic expression and a formula? Which algebraic expression is equivalent to the expression? Can you explain the process of converting algebraic equations into expressions and vice versa? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-series-optional/v/infinite-geometric-series
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: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search DonateLog inSign up Search for courses, skills, and videos Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active 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 [x] 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 [x] 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 [x] Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices
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https://flexbooks.ck12.org/cbook/ck-12-cbse-maths-class-11/section/3.6/primary/lesson/trigonometric-functions-of-multiple-and-sub-multiple-angles/
Skip to content 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 3.6 Trigonometric Functions of Multiple and Sub-multiple of Angles Written by:Neha Khandelwal Fact-checked by:The CK-12 Editorial Team Last Modified: Jul 04, 2025 Understanding how trigonometric functions behave for multiple and sub-multiple angles is essential for solving complex mathematical problems in calculus, physics, and geometry. These functions are used in wave analysis, signal processing, and even astronomical calculations. The study of multiple and sub-multiple angles allows us to express trigonometric functions in simplified forms and derive important identities that are useful in higher-level mathematics. In the previous lesson, Trigonometric Functions of Sum and Difference of Two Angles, we learned about basic trigonometric functions and identities. Now, we extend these concepts to angles that are multiples or fractions of a given angle, which helps in transforming expressions and solving equations more efficiently. For example, if we know the sine or cosine of an angle, we can determine the sine or cosine of 2x,3x, or x2 using established formulas. Trigonometric Functions of Multiple Angles Multiple angle formulas express trigonometric functions of angles like 2x,3x, and so on in terms of functions of x. These formulas help in simplifying expressions and solving trigonometric equations efficiently. Double Angle Formulas The double-angle formulas can be derived using sum formulas of trigonometric functions. Sine Double Angle Formula We use the sum identity:sin(2x)=sin(x+x)Applying the sine sum formula:sin(2x)=sinxcosx+cosxsinx=2sinxcosx | | | Sine Double Angle Formula | | sin(2x)=2sinxcosx | Cosine Double Angle Formula Using the cosine sum formula:cos(2x)=cos(x+x)=cosxcosx−sinxsinx=cos2x−sin2xUsing the Pythagorean identity sin2x=1−cos2x and cos2x=1−sin2x, we derive two different forms:cos(2x)=2cos2x−1cos(2x)=1−2sin2x | | | Cosine Double Angle Formula | | cos(2x)=cos2x−sin2xcos(2x)=2cos2x−1cos(2x)=1−2sin2x | Tangent Double Angle Formula Using the identity tanx=sinxcosx, we derive:tan(2x)=sin(2x)cos(2x)Substituting the double-angle formulas:tan(2x)=2sinxcosxcos2x−sin2xDividing the numerator and denominator by cos2x, we get: tan(2x)=2tanx1−tan2x,(when tanx≠±1) | | | Tangent Double Angle Formula | | tan(2x)=2tanx1−tan2x,(when tanx≠±1) | Triple Angle Formulas The triple-angle formulas can be derived from double-angle identities and sum formulas. Sine Triple Angle Formula Using the sine sum identity:sin(3x)=sin(2x+x)Applying the sum formula:sin(3x)=sin2xcosx+cos2xsinxSubstituting the double-angle identities:sin(3x)=(2sinxcosx)cosx+(cos2x−sin2x)sinx=2sinxcos2x+sinxcos2x−sin3x=3sinxcos2−sin3x=3sinx(1−sin2)−sin3x=3sinx−3sin3x−sin3x=3sinx−4sin3x | | | Sine Triple Angle Formula | | sin(3x)=3sinx−4sin3x | Cosine Triple Angle Formula Using the cosine sum identity:cos(3x)=cos(2x+x)Applying the sum formula:cos(3x)=cos2xcosx−sin2xsinxSubstituting the double angle formulas:cos(3x)=(2cos2x−1)cosx−(2sinxcosx)sinx=2cos3x−cosx−2sin2xcosx=2cos3x−cosx−2(1−cos2x)cosx=2cos3x−cosx−2cosx+2cos3x=4cos3x−3cosx | | | Cosine Triple Angle Formula | | cos(3x)=4cos3x−3cosx | Tangent Triple Angle Formula Using the identity:tan(3x)=tan(2x+x)Applying the tangent sum formula:tan(3x)=tan2x+tanx1−tan2xtanxWe use the double angle formula for tangent:tan(2x)=2tanx1−tan2xSubstituting this into the above equation:tan(3x)=2tanx1−tan2x+tanx1−(2tanx1−tan2x)tanxOn simplifying the numerator, we get:2tanx1−tan2x+tanx=2tanx+tanx(1−tan2x)1−tan2x=2tanx+tanx−tan3x1−tan2x=3tanx−tan3x1−tan2xOn simplifying the denominator, we get:1−(2tanx1−tan2x)tanx=1−2tan2x1−tan2x=1−tan2x−2tan2x1−tan2x=1−3tan2x1−tan2x | | | Tangent Triple Angle Formula | | tan(3x)=3tanx−tan3x1−3tan2x | Trigonometric Functions of Sub-Multiple Angles Sub-multiple angles refer to fractions of an angle, such as x2. These formulas help simplify expressions in integration, equation-solving, and transformations. Half-Angle Formulas Half-angle formulas are derived from double angle identities and are useful in integration, trigonometric simplifications, and solving equations. Derivation of Half-Angle Identities We start with the double-angle identity for cosine:cos(2x)=2cos2x−1Rearranging for cos2x:cos2x=1+cos2x2Taking the square root of both sides, we get:cosx=±√1+cos2x2Similarly, using the double-angle identity for sine:cos(2x)=1−2sin2xRearranging for sin2x:sin2x=1−cos2x2Taking the square root:sinx=±√1−cos2x2Now, using x=2A, we replace 2Awith xin the above derivations: Sine Half-Angle Formulasinx2=±√1−cosx2 Cosine Half-Angle Formulacosx2=±√1+cosx2 Tangent Half-Angle Formula Using the relation tanA=sinAcosA, we get:tanx2=±√1−cosx1+cosxAlternatively, using sine and cosine ratios: tanx2=sinx1+cosx=1−cosxsinx | | | Half-Angle Formulas | | sinx2=±√1−cosx2,cosx2=±√1+cosx2,tanx2=±√1−cosx1+cosx | Sum to Product Identities The product-sum identities are useful trigonometric transformations that allow us to express the sum or difference of two sine or cosine functions as a product of sines and cosines. These formulas help simplify expressions, solve equations, and perform integrations. The Sum of Sines Formula We start with the sum and difference identities for sine:sinA=sin(A+B2+A−B2)sinB=sin(A+B2−A−B2)Using the sine sum and difference identities, we get:sin(X+Y)=sinXcosY+cosXsinYsin(X−Y)=sinXcosY−cosXsinYSubstituting X=A+B2 and Y=A−B2, we expand:sinA=sin(A+B2+A−B2)=sinA+B2cosA−B2+cosA+B2sinA−B2sinB=sin(A+B2−A−B2)=sinA+B2cosA−B2−cosA+B2sinA−B2Now, on adding these equations, we get:sinA+sinB=(sinA+B2cosA−B2+cosA+B2sinA−B2)+(sinA+B2cosA−B2−cosA+B2sinA−B2) The terms cosA+B2⋅sinA−B2 cancel, leaving:sinA+sinB=2sinA+B2⋅cosA−B2 | | | The Sum of Sines Formula | | sinA+sinB=2sinA+B2⋅cosA−B2 | The Difference of Sines Formula Using the previously derived sine expressions, we now subtract instead of adding: sinA−sinB=(sinA+B2cosA−B2+cosA+B2sinA−B2)−(sinA+B2cosA−B2−cosA+B2sinA−B2) The sinA+B2⋅cosA−B2terms cancel, leaving:sinA−sinB=2cosA+B2⋅sinA−B2 | | | The Difference of Sines Formula | | sinA−sinB=2cosA+B2⋅sinA−B2 | The Sum of Cosines Formula Using the cosine sum and difference identities:cosA=cos(A+B2+A−B2)cosB=cos(A+B2−A−B2)Expanding using cosine sum and difference formulas:cos(X+Y)=cosXcosY−sinXsinYcos(X−Y)=cosXcosY+sinXsinY⟹cosA=cosA+B2cosA−B2−sinA+B2sinA−B2⟹cosB=cosA+B2cosA−B2+sinA+B2sinA−B2Adding these equations, we get: cosA+cosB=(cosA+B2cosA−B2−sinA+B2sinA−B2)+(cosA+B2cosA−B2+sinA+B2sinA−B2) The sine terms cancel, giving:cosA+cosB=2cosA+B2cosA−B2 | | | The Sum of Cosines Formula | | cosA+cosB=2cosA+B2cosA−B2 | The Difference of Cosines Formula Subtracting the cosine expressions: cosA−cosB=(cosA+B2cosA−B2−sinA+B2sinA−B2)−(cosA+B2cosA−B2+sinA+B2sinA−B2) Now, the cosine terms cancel, leaving:cosA−cosB=−2sinA+B2sinA−B2 | | | The Difference of Cosines Formula | | cosA−cosB=−2sinA+B2sinA−B2 | Case Study - Biochemical Beats: Decoding Enzyme Oscillations with Trigonometry In a biochemistry experiment, a researcher is studying the oscillatory behaviour of certain metabolic processes in cells, which follow periodic patterns over time. The concentration of a specific enzyme was found to follow the equation: C(t)=Asin(3t)+Bcos(3t), where A and B are constants and t is time in seconds. To understand the behaviour of this function, the researcher uses trigonometric identities of multiple angles to simplify and analyse the changes in enzyme concentration over time. To accurately interpret the enzyme’s rhythmic changes, a mathematical understanding of trigonometric transformations, especially of multiple-angle identities, amplitude, and periodicity, is essential. Answer the following questions based on the given case. Examples of Trigonometric Functions of Multiple and Sub-multiple of Angles Example 1 If sinx=35, where 0<x<π2, find the values of sin2x,cos2x,tan2x, and sin4x. Given sinx=35,0<x<π2 We need to find the values of sin 2x, cos 2x, tan 2x, and sin 4x. First, let us find cosx. Using the Pythagorean identity:cos2x=1−sin2xcos2x=1−(35)2=1−925=1625cosx=45(∵ x is in the first quadrant, cosx>0)Now, we compute tanx.tanx=sinxcosx=3/54/5=34Using the double-angle formula:sin2x=2sinxcosx=2×35×45=2425Using the cosine double-angle formula:cos2x=1−2sin2x=1−2×(35)2=1−2×925=1−1825=725Using the tangent double-angle formula:tan2x=2tanx1−tan2x=2×341−(34)2=6/41−9/16=6/47/16=64×167=247Now, let us use the double-angle formula for computing sin4x:sin4x=2sin2xcos2x=2×2425×725=2×168625=336625 Example 2 Prove that: cos4x=1−8sin2xcos2x We start with the double-angle formula for cosine:cos4x=cos2(2x)Using cos2A=1−2sin2A, we get:cos4x=1−2sin2(2x)From earlier, we know:sin2x=2sinxcosxSo,sin22x=(2sinxcosx)2=4sin2xcos2x⟹cos4x=1−2(4sin2xcos2x)cos4x=1−8sin2xcos2xwhich is the same as the right-hand side (RHS). Hence proved. Example 3 Prove that: (cosx+cosy)2+(sinx−siny)2=4cos2x+y2 Let us first expand the left-hand side (LHS)2(cosx+cosy)2+(sinx−siny)2Expanding both squares, we get:(cos2x+2cosxcosy+cos2y)+(sin2x−2sinxsiny+sin2y)Rearrange the terms:(cos2x+sin2x)+(cos2y+sin2y)+2(cosxcosy−sinxsiny)Using the Pythagorean identity cos2x+sin2x=1 and cos2y+sin2y=1:1+1+2(cosxcosy−sinxsiny)Using the cosine sum formula:cosxcosy−sinxsiny=cos(x+y)Thus, we have: 2+2cos(x+y)Now, let us look at the right-hand side RHS. 4cos2x+y2Using the identity:cos2θ2=1+cosθ2We rewrite:2+2cos(x+y)=2(1+cos(x+y))=2(2cos2x+y2)=4cos2x+y2which matches the right-hand side (RHS). Example 4 Prove that (1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8)=18 Using known angle properties:cos7π8=−cosπ8,cos5π8=−cos3π8Rewriting the given expression:(1+cosπ8)(1−cosπ8)(1+cos3π8)(1−cos3π8)Using the identity 1−cos2A=sin2A:(1+cosπ8)(1−cosπ8)(1+cos3π8)(1−cos3π8)=(1−cos2π8)(1−cos23π8)=sin2π8sin23π8Using sin2x=1−cos2x2, we get:sin2π8=1−cosπ42,sin23π8=1−cos3π42Since cosπ4=1√2,cos3π4=−1√2: sin2π8sin23π8=14(1−1√2)(1+1√2)=14(1−12)=18which matches the RHS. | | | Summary of Trigonometric Functions of Multiple and Sub-multiple of Angles | | Double Angle Formulas: + Sine:sin(2x)=2sinxcosx + Cosine:cos(2x)=cos2x−sin2x, which can also be expressed as cos2x−1 or 1−2sin2x. + Tangent:tan(2x)=2tanx1−tan2x Triple Angle Formulas: + Sine:sin(3x)=3sinx−4sin3x + Cosine:cos(3x)=4cos3x−3cosx + Tangent:tan(3x)=3tanx−tan3x1−3tan2x Half-Angle Formulas: + Sine Half-Angle:sin(x2)=±√1−cosx2 + Cosine Half-Angle:cos(x2)=±√1+cosx2 + Tangent Half-Angle:tan(x2)=±√1−cosx1+cosx=sinx1+cosx=1−cosxsinx Sum and Difference of Sines and Cosines: These identities transform the sum or difference of two trigonometric functions into products, simplifying expressions and solving equations. + Sum of Sines: sinA+sinB=2sin(A+B2)cos(A−B2) + Difference of Sines: sinA−sinB=2cos(A+B2)sin(A−B2) + Sum of Cosines: cosA+cosB=2cos(A+B2)cos(A−B2) + Difference of Cosines: cosA−cosB=−2sin(A+B2)sin(A−B2) | Review Questions of Trigonometric Functions of Multiple and Sub-multiple of Angles Prove that: sin2x1+cos2x=tanx. Prove that: sin2x1−cos2x=cotx. Prove that: sin3x=3sinx−4sin3x. Prove that: cos3x=4cos3x−3cosx. If cosx=−35 and x lies in quadrant III, find cosx2,sinx2, and sin2x. If sinx=45 and x lies in quadrant II, find cosx2,sinx2, and tanx2. Prove that: tan(π4+x)+tan(π4−x)=2sec2x. Prove that: cot2x−tan2x=4cot2xcsc2x. Prove that: sin3x+sin2x−sinx=4sinxcosx2cos3x2. Prove that: tan82.5∘=(√3+√2)(√2+1)=√2+√3+√4+√6. Prove that: cotπ8=√2+1. Prove that: cos2π15cos4π15cos8π15=−116. Student Sign Up Are you a teacher? Having issues? Click here or By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? No Results Found Your search did not match anything in . Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)62/ 100 This lesson has been added to your library. |Searching in: | | | Looks like this FlexBook 2.0 has changed since you visited it last time. 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https://arxiv.org/pdf/1503.05930
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https://pmc.ncbi.nlm.nih.gov/articles/PMC3568203/
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Published in final edited form as: Biochim Biophys Acta. 2012 Oct 23;1829(3-4):318–330. doi: 10.1016/j.bbagrm.2012.10.006 Search in PMC Search in PubMed View in NLM Catalog Add to search Transcription termination by the eukaryotic RNA polymerase III Aneeshkumar G Arimbasseri Aneeshkumar G Arimbasseri 1 Intramural Research Program on Genomics of Differentiation, Eunice Kennedy Shriver National Institute of Child Health and Human Development, National Institutes of Health, Bethesda, MD USA Find articles by Aneeshkumar G Arimbasseri 1, Keshab Rijal Keshab Rijal 1 Intramural Research Program on Genomics of Differentiation, Eunice Kennedy Shriver National Institute of Child Health and Human Development, National Institutes of Health, Bethesda, MD USA Find articles by Keshab Rijal 1, Richard J Maraia Richard J Maraia 1 Commissioned Corps, U.S. Public Health Service Find articles by Richard J Maraia 1, Author information Article notes Copyright and License information 1 Intramural Research Program on Genomics of Differentiation, Eunice Kennedy Shriver National Institute of Child Health and Human Development, National Institutes of Health, Bethesda, MD USA 1 Commissioned Corps, U.S. Public Health Service To whom correspondence should be directed at: 31 Center Drive, Rm 2A25, Bethesda, MD 20892-2426, Phone: 301-402-3567, maraiar@mail.nih.gov Issue date 2013 Mar. PMC Copyright notice PMCID: PMC3568203 NIHMSID: NIHMS416900 PMID: 23099421 The publisher's version of this article is available at Biochim Biophys Acta Abstract RNA polymerase (pol) III transcribes a multitude of tRNA and 5S rRNA genes as well as other small RNA genes distributed through the genome. By being sequence-specific, precise and efficient, transcription termination by pol III not only defines the 3′ end of the nascent RNA which directs subsequent association with the stabilizing La protein, it also prevents transcription into downstream DNA and promotes efficient recycling. Each of the RNA polymerases appears to have evolved unique mechanisms to initiate the process of termination in response to different types of termination signals. However, in eukaryotes much less is known about the final stage of termination, destabilization of the elongation complex with release of the RNA and DNA from the polymerase active center. By comparison to pols I & II, pol III exhibits the most direct coupling of the initial and final stages of termination, both of which occur at a short oligo(dT) tract on the non-template strand (dA on the template) of the DNA. While pol III termination is autonomous involving the core subunits C2 and probably C1, it also involves subunits C11, C37 and C53, which act on the pol III catalytic center and exhibit homology to the pol II elongation factor TFIIS, and TFIIFα/β respectively. Here we compile knowledge of pol III termination and associate mutations that affect this process with structural elements of the polymerase that illustrate the importance of C53/37 both at its docking site on the pol III lobe and in the active center. The models suggest that some of these features may apply to the other eukaryotic pols. Keywords: RNA polymerase III, transcription termination, RPC11, RPC53, RPC37, RPC2, intrinsic transcript cleavage, oligo(U), RNA:DNA hybrid INTRODUCTION Transcription of a DNA template into a complementary RNA is a most fundamental process of cellular life. Bacterial and archaeal cells each use a single RNA polymerase (pol) for transcription that are evolutionarily related to the eukaryotic pols I, II and III (plants also have pols IV & V, variants of pol II) [1, 2]. Pol I transcribes a single gene type, the rRNA genes, while pol II transcribes the thousands to tens of thousands of protein coding and non-coding genes that vary over orders of magnitude in transcription output, controlled by a vast combinatorial set of promoters and enhancers. Pol III transcribes hundreds of tRNA genes which bear similar promoters, as well as 5S rRNA, U6 snRNA and several other noncoding RNA genes. Transcription involves three steps: 1) initiation, -recruitment of the polymerase to the gene promoter and formation of initial phosphodiester bonds; 2) elongation, -processive synthesis of the RNA chain, albeit with intermittent pausing in some cases, and 3) termination, -cessation of RNA synthesis and dissociation of the three components: nascent RNA, polymerase, and DNA. While each of the eukaryotic pols distinguish these steps, their specialization appears to include differences in how the steps are executed, the relative time spent at each step, and how the steps may be linked to each other. For example, termination and reinitiation by pol III have been shown to be mechanistically linked [3–5]. Also, while the three pols use highly similar mechanisms of transcription initiation, there is much more complexity in assembling pol II initiation complexes than pols I & III . Differences in elongation include the time spent in this mode, with pol III differing most since at 75–300 nt, its transcripts are relatively short . Another difference is that pol II uses pausing as a control point, regulated by the positive transcription elongation factor-b (P-TEFb) . Transcription termination must involve destabilization of the elongation complex followed by release of the nascent RNA and polymerase. This is a big transition because elongation complexes must be very stable in order to avoid the deleterious effects of premature termination . Failure to terminate can interfere with downstream genes, produce 3′ extended RNAs with potential adverse effects, and deplete the pool of polymerase that should be available for regulated initiation . TERMINATORS SIGNAL THE INITIATION OF A TWO-STAGE PROCESS Different RNA polymerases use different mechanisms to direct termination (Fig 1A–E). Each responds to a specific signal, a terminator element in the DNA or the elongating RNA that prompts the beginning of the termination process . The process initiates with recognition of a termination signal and this is followed by induced cessation of RNA synthesis and release. Some terminators work at a distance; recognition of the termination signal by the polymerase or a trans-acting factor is the first stage and this leads, albeit more directly for some than others, to the second stage, destabilization of a paused complex with release of the DNA and RNA from the polymerase active center. Figure 1. Schematic of termination mechanisms by multisubunit RNA polymerases. Open in a new tab The mechanisms reveal three themes common to more than one pol, i) 5′-3′ exonuclease digestion of a downstream fragment of the nascent transcript attached to the polymerase, ii) weakly base pairing oligo(rU:dA) hybrid and iii) involvement of a helicase, Rho or Sen1. A) Rho mediated termination by bacterial RNA polymerase. Rho helicase binds to the nascent transcript at a C rich region and travels along the RNA in a 5′ to 3′ direction toward the polymerase to induce destabilization and termination. B) Intrinsic termination by bacterial RNA polymerase. A hairpin formed by the nascent RNA followed by a T-rich stretch on non-template strand comprise the intrinsic termination signal. Transcript release occurs within the T stretch. C) Pol II termination mechanism for poly(A)-containing mRNA-coding genes. Transcript is cleaved downstream of the poly(A) addition site by cleavage factors associated with the pol II CTD. The polymerase-attached RNA fragment is digested by Rat1/Xrn2 in 5′-3′ direction. Sen1 helicase also binds to pol II and is required for termination. This figure is over simplified to emphasize similarities with other pols. Also, there are different pathways for termination at different classes of genes (see text). D) Eukaryotic pol I termination mechanism. Most of the transcripts are terminated upstream of the Nsi1 (Ydr026C) binding site within the T stretch. Cotranscriptional cleavage of the transcript followed by processive 5′-3′ cleavage by Rat1 is also involved in termination. E) Pol III termination mechanism. A stretch of Ts on the non-template strand is sufficient to direct termination. Bacterial RNA polymerase can use two kinds of terminators, factor-dependent and intrinsic . Rho factor is a helicase that recognizes the nascent RNA at a C-rich sequence after it emerges from the polymerase and propels along in a 5′-3′ direction to catch up with the elongation complex, inducing destabilization and termination (Fi. 1A) . Intrinsic terminators work more directly, i.e., within a relatively short, albeit bipartite terminator element, to coordinate pausing and destabilization. This involves formation of a hairpin structure in the transcribed RNA followed by transcription of oligo(dA)-rich sequence to produce an RNA with an oligo(U)-rich 3′ end [12, 14]. Termination occurs as the oligo(rU:dA) hybrid is melted in the active center of the polymerase (Fig. 1B). Pol II termination is more complex, involving post translational modifications of the polymerase as well as association of a number of trans-acting factors [10, 15]. Similar to bacteria different termination signals distinguish at least two types of pol II termination mechanisms, for poly(A)-containing mRNAs and for poly(A)-independent small nuclear RNAs . For the first type, the AAUAAA poly(A) addition site in the elongating RNA recruits a complex of factors that endonucleolytically cleave the transcript which is then further processed to become a mRNA. This is followed by 5′-3′ exonucleolytic digestion of the segment of RNA still attached to the polymerase (Fig. 1C). Thus for most pol II-transcribed mRNA genes the termination signal is the poly(A) addition site in the newly synthesized RNA which acts at a distance; its recognition by RNA-binding factors initiates a process that ends in termination at a downstream site. Pol I uses a termination factor, Ydr026C/Nsi1 bound to a specific element on the downstream DNA as well RNA endonucleolytic cleavage followed by 5′ exonucleolytic digestion [18–20]. Pol I transcript release occurs within an oligo(dA) tract on the template DNA (Fig. 1D). According to a current ‘torpedo’ models of termination by pols I & II, the 5′-3′ exonucleases catch up with the elongating polymerases to induce pausing and destabilization leading to the final stage, release, reminiscent of Rho-dependent termination [11, 20]. Pol III has a most direct acting termination signal, more similar to but distinct from intrinsic termination by bacterial RNA polymerase, than pols I or II. Oligo(dA) on the template DNA is sufficient to commence and complete all steps leading to termination by pol III (Fig. 1E) [22–24]. The pol III enzyme incorporates five stably associated subunits, the heterotrimer C31/34/82 and heterodimer C37/53 as two subcomplexes with homology to the pol II ancillary factors, TFIIE and TFIIF, as well as C11, a two-domain polypeptide with homology to Rpb9 and the elongation factor TFIIS, that collectively promote efficient initiation, termination and reinitiation (Fig. 2A, B) [25–30]. Unlike initiation by pol III which can be directed by gene types that differ in promoter structure and the transacting factors that recognize them, termination appears to occur by the same basic mechanism for all (with exceptions that likely reflect variations on the oligo(dT) theme; see below). These characteristics make pol III an attractive model for the study of the mechanism of termination by a eukaryotic RNA polymerase. Figure 2. RNA polymerase III subunits and structural elements that affect termination. Open in a new tab A) Comparison of subunit compositions of pols II and III. The termination subcomplex that shows homology to TFIIFα/β subunits, the initiation subcomplex that shows similarity to TFIIEα/β, and C11 which is homologous to RPB9 at its N terminus and to the C-terminal motif of TFIIS at its C terminus, are highlighted. B) Cartoon of surface features of pol III derived from the electron microscopic structure of pol III (EMD-1802) deposited by . Blue color reflects the proposed position of the C53/37 dimerization domains and pink shows the initiation subcomplex . The cartoon of the RNA:DNA template hybrid was obtained from pol II elongation structure (PDB ID: 3HOV). The non-template strand is not shown. Yellow color reflects localization of the N-terminal, Rpb9-homologous domain of C11 only. Other features discussed in the text and/or elsewhere in the figure are indicated. For other structural representations, refer to ; for additional models see figures 3 & 5 in . C) Cartoon depicting the structural elements that are involved in transcription elongation and termination. The catalytic site along with the hybrid binding region and other elements such as the for loops, the trigger loop and bridge helix are considered as part of a larger, extended active center. In summary, the eukaryotic RNA polymerases use different signals to initiate the termination process. It is much less clear as to how their active centers work to execute the final stage of termination, release from the RNA and DNA by their active centers, and to what extent they may share mechanisms involved in this stage of the process. TERMINATION BY POL III: THE SIGNAL IS OLIGO(dT) Comparison of the 3′ oligo(U) sequence at the ends of 5S rRNA with the 3′ regions of Xenopus 5S rRNA genes suggested that a stretch of 4 or more T residues in the DNA could act as a pol III terminator [31–34]. It was shown that pol III in Xenopus oocytes or in solution would terminate at oligo(dT) at the 3′ ends of 5S rRNA genes [22, 24]. While independent studies confirmed oligo(dT) as a universal terminator for pol III, further analysis revealed species-specificity in the minimum length of the oligo(dT) tract required for termination. For vertebrates, as few as 4 Ts can act as an efficient terminator while for yeasts variably longer T tracts are required [22, 24, 35]. In general, S. cerevisiae requires 6 or more Ts [36–38] while S. pombe requires 5 or more Ts, for efficient (~90%) termination , and this is supported by genome-wide analyses of tRNA genes [38, 39]. It was further suggested that fission yeast Schizosaccharomyces Japonicus requires only 4 Ts at a large number of tRNA gene terminators . It is expected that these differences are manifested in sequence disparities in the termination-relevant regions of the pol III subunits of these species. Noteworthy is that the minimal T length requirement of the different species correlates with the α-amanitin sensitivities of their pols III , the implications of which will be discussed in a later section. Oligo(dT) flanking sequence context effects Early studies revealed that the sequence surrounding the oligo(dT) signal can influence termination efficiency [22, 24]. This effect may be most pronounced for vertebrate pol III for oligo(dT) length of 4 . 4Ts flanked by AA is very inefficient as a terminator, and similar to Xenopus, most human pol III will read through, whereas replacing AA with GC increases termination efficiency dramatically . By contrast, a 5T tract is highly efficient for pol III termination [22, 24] and relatively insensitive to flanking sequence . Nonetheless, a 4T terminator flanked by GC can be as efficient as a 5T terminator . This context-dependency of 4T tracts is critical for some pol III transcribed genes with 4Ts in their coding region, such as all lysine tRNA genes and adenovirus VA RNA II . One may suspect that the necessity for four contiguous Ts in lysine tRNA genes (the invariant U33 in the mature tRNA followed by the UUU anticodon at positions 34–37) may have set an ancient limit of no fewer than 5Ts as the minimal pol III terminator which was later adjusted in multicellular eukaryotes by the ability to utilize sequence-context effects. In S. cerevisiae a 5T stretch is relatively inefficient and can act as a terminator only in the right context although the magnitude of the effect of flanking sequence may be less than on a 4T terminator for human pol III. However, there are no hard ‘universal’ rules with regard to flanking sequence. In S cerevisiae, CT following the T5 tract can weaken the terminator while A or G strengthens it . Also, termination efficiency of a Xenopus lysine tRNA gene with a 4T terminator was influenced by flanking sequence differently than for Xenopus 5S rRNA genes . Notably, these sequence context effects do not require other factors as pol III itself from yeast or Xenopus, exhibited context dependent termination [24, 38]. Flanking sequence effects might suggest the involvement of secondary structure. Although proximal hairpins appeared to promote pol III termination in one study , others indicate no such effects, confirming oligo(dT) alone as the major terminator element for pol III. Later in this review we note that for some genes, flanking sequence effects may extend to binding sites for extraneous factors that influence termination, e.g., the NF1 site adjacent to the terminator of the VA1 RNA gene . While oligo(dT) is clearly the most prevalent terminator for pol III, isolated reports cite non-canonical pol III terminators, and although most of these are interrupted oligo(dT) tracts, one was a run of A residues downstream of a mouse 5S rRNA gene . Human Pol III was reported to terminate 3′ of an Alu repeat which did not have 4 or more consecutive T residues and the potential for RNA hairpin formation was noted . However mapping pol III terminators using cellular extract can be complicated. Following in vitro transcription of a B1-Alu gene element using cellular extract, 3′-5′ digestion of the nascent RNA was so robust that mapping of the terminator that gave rise to the barely detectable primary transcript required rapid pulse-chase conditions . Remarkably, this robust 3′ digestion occurred following transcription of the B1-Alu gene that had a 4T terminator but not the same gene with a 5T terminator, and this was later attributed to the 3′-protective activity of the 3′ oligo(U) length-specific RNA binding protein, La . Another case is the avian adenovirus CELO VA RNA gene; TTATT caused inefficient termination . Termination within a T stretch shorter than 4 Ts or an interrupted T stretch has been noted for a human tRNAmeti gene . As noted above and discussed later in more detail, a large number of human tRNA genes that use non-canonical terminators have been catalogued but these are variations of oligo(dT), generally a T3, T2 or T1 stretches separated by another nucleotide (interrupted T5 or T4) . These studies together with the effects of flanking sequence on termination efficiencies suggest a complex mechanism(s) that control the efficiency of pol III termination on short oligo(dT) terminators. Primary and secondary oligo(dT) terminators Bacteria use a variety of anti-termination mechanisms to control the polar inclusion or exclusion of segments of polycistronic mRNAs . An intriguing concept for pol III is that of secondary terminators, i.e., an oligo(dT) stretch downstream of a weak primary terminator that can be used to produce a unique longer transcript. In one scenario, a fraction of pol III would terminate at the first oligo(dT) while the rest would read through to terminate at the downstream terminator [22, 33, 49]. Human adenoviral VA RNA I and avian adenovirus CELO VA RNA genes provide examples that can produce two different RNAs by this mechanism [41, 51]. Recent identification of biological functions for tRNA 3′ fragments derived from the 3′ trailer sequences of pre-tRNAs suggests that partial read through of the first terminator may be used to generate such tRNA 3′ fragments [52–54, reviewed in 55]. In accord with this, Orioli et al. identified a large number of tRNA genes in humans with weak terminators followed by a strong secondary terminator . A fraction of pol III terminates at the weaker terminators while the rest read through. These observations further suggest the exciting possibility that pol III termination may be modulated to produce a subclass of non-coding RNAs, although this remains to be determined. A genome wide chromatin immunoprecipitation (ChIP) followed by sequence analysis of human pol III showed that a significant fraction of pol III was unexpectedly found to be accumulated somewhat downstream of the tRNA gene terminators which may reflect among other things terminator read-through and/or a state of pausing . THE TERMINATION MECHANISM In simple accordance with a kinetic coupling model in which termination efficiency is inversely related to elongation rate, pausing provides a window of time during which alterations can occur in the polymerase that lead to transcript release [56, 57]. Indeed, RNA polymerase pausing at a terminator is prerequisite for termination [23, 58–60]. Kinetic coupling is supported by the fact that methods that slow elongation lead to increased termination efficiency . While oligo(dT)-rich tracts are recognized as pause sites, they are not sufficient to cause termination by other RNA polymerases which must traverse many kilobases (for pol II megabases) that often contain oligo(dT) by chance or necessity. The bacterial polymerase terminates when an oligo(dT)-rich tract closely follows a hairpin in the transcript. This arrangement increases specificity by enriching the signal with more complexity than oligo(dT) and allows both elements to function toward destabilization of the complex within the confines of the active center and adjacent elements . Multiple hypotheses may explain the mechanism(s) of action of the hairpin during intrinsic termination. A forward translocation model suggests that hairpin formation induces polymerase to translocate forward without nucleotide incorporation, leading to shortening and destabilization of the hybrid and transcript release . A second model suggests that the hairpin induces melting of the proximal end of the hybrid and destabilization . A third suggests an allosteric mechanism with direct interaction of the hybrid with the polymerase trigger loop inducing structural changes that loosen the grip on the hybrid, hybrid melting and destabilization of the complex . Termination must overcome the stability of the elongation complex One of the most important features of efficient transcription is the very high degree of processivity of RNA polymerases, without which there would be an overwhelming number of incomplete transcripts [9, 63]. Poor processivity would not only be wasteful but probably also cause havoc to RNA processing systems. Thus by design RNA polymerases must be processive and this is achieved via highly stable grips on the RNA and the DNA. Part of this stability is rooted in the complementary base pairing interactions that hold together the transcribed RNA and the DNA template within the active center of the polymerase [64, 65]. Elegant studies of bacterial RNA polymerase and pol II have shown that the nascent RNA forms an 8–9 nt hybrid with the template DNA [64, 65]. A strong RNA:DNA hybrid is one of the primary elements that is critical to the remarkable stability of the elongation complex and its processivity [65–67]. Thus the termination process must destabilize a very stable elongation complex, and it must do it precisely on cue and only on cue. As reviewed below it appears that many if not all RNA pols take advantage of a similar characteristic to do so, the outstanding weakness of rU:dA hybrid base pairs . The weak rU:dA hybrid is likely an underlying component of pol III termination Oligo(dT) (oligo(dA) in the template) constitutes a part of the termination complexes of bacterial, phage, archaeal, eukaryotic, and viral RNA polymerases [11, 12, 21, 69, 70]. The demonstration that a (rU:dA)5 hybrid is at least 200 times less stable than the corresponding hybrid containing (rA:dT)5 or other sequences, led to the proposal that a decrease in the stability of the hybrid as it acquires rU:dA richness may be a driving force for termination . A weak hybrid such as oligo(rU:dA) causes alterations of the elongation complex such as pausing followed by backtracking [64, 71, 72]. Shortening is an alternate means of hybrid weakening and this is also implicated in transcript release by bacterial RNA polymerase . As alluded to above, pol III termination appears to be somewhat similar to intrinsic termination by bacterial RNA polymerase. Intrinsic terminators are bipartite, with a 7-to-8 nt T-rich tract preceded by a G+C-rich dyad repeat that forms a stem-loop hairpin in the nascent RNA upstream of the 3′ U-rich tract [59, 61, 73–76]. While an encounter of either a T-rich tract or a G+C-rich dyad may in some contexts cause pausing, both are required for complete termination . Yet only an oligo(dT) tract (dA in the template) is required for pol III termination, but not a dyad-repeat, hairpin, or other cis- element. This suggests that pol III may be exquisitely sensitive to the destabilizing effects of an oligo(rU:dA) hybrid, a feature that could be accommodated by a polymerase whose substrate genes are short enough as to not require internal T tracts. The challenge is to understand the mechanisms that provide pol III the ability to respond so dramatically and efficiently to a simple signal that induces both pausing and destabilization. Although there is no direct evidence that rU:dA hybrid weakness provides a mechanism of pol III termination, some observations support this. Replacement of rU with Bromo-rU, which forms a more stable hybrid with dA, in in vitro transcription reactions promotes terminator read through . Lower reaction temperature which among other things increases hybrid stability, increases terminator read through . Also, the template strand oligo(dA) is required for termination while the non-template oligo(dT) can be mutated with little effect (A.G.A. & R.M., in preparation). The exact position at which pol III releases its RNA from within the oligo(dA) tract is heterogeneous as reflected by a variable number of 3′ Us on the RNAs released from a single gene [57, 63, 77]. A positive correlation between oligo(U) length and transcript release can be discerned , consistent with the idea that longer rU:dA hybrids destabilize the elongation complex leading to better termination. These observations support the idea that an extensive rU:dA hybrid destabilizes the elongation complex and promotes termination. However, none of them exclude the possibility that a contributing influence may be due to sequence-specific recognition of the rU:dA hybrid with allosteric alteration of pol III as an underlying mechanism. Is backtracking involved in pol III termination? A common response of RNA polymerases upon encountering oligo(dT) is pausing, followed by a more stable block to elongation caused by backtracking, a process in which the polymerase ‘backs-up’ i.e., slides upstream while maintaining a ~10 bp RNA:DNA hybrid in the active center [64, 71]. During backtracking the RNA 3′ end is displaced from the catalytic site and ejected out the secondary channel of the polymerase. RNA synthesis can be restarted from this state after the catalytic center is converted to a endonucleolytic transcript cleavage site by the cleavage factor TFIIS for pol II and the GreB and GreA factors for bacterial RNAP [78–80]. However, polymerase backtracking that is unresolved by transcript cleavage remains in a state of transcription arrest. It was proposed that backtracking may be involved in termination by pol III . Certainly pol III undergoes 3′ retraction, i.e. e., the cleavage of 3′ terminal residues from RNA in some cases allowing otherwise stalled complexes to move forward [25, 82–84]. Whether pol III undergoes backtracking as part of termination is an outstanding question. POL III SUBUNITS INVOLVED IN TERMINATION The 17 subunits of pol III are intricately connected via an extensive interaction network . Experimental data indicate that several pol III subunits can affect the termination process, C2, C11, C37 & C53, and possibly C1 and others [25, 26, 29, 86–89]. In addition, there have been multiple reports that extraneous factors can also affect termination by pol III (below). A challenge is to determine the mechanisms used by these subunits and extraneous factors to affect termination and to distinguish between direct and indirect (e.g., allosteric effects via other subunits) effects. The pol III active center is organized from parts of several subunits For the purposes of this review we shall refer to the active center as that which contains the catalytic site and its multiple structural elements such as bridge helix, trigger loop and fork loops that comprise the catalytic site responsible for nucleotide selection and phosphodiester bond formation, as well as the adjacent RNA:DNA hybrid of nearly ten base pairs (Fig. 2C). By comparison of electron micrographs of pol III with crystal structures of pol II it seems clear that the core structures of these are conserved (Fig. 2A). Indeed C1 and C2 are most homologous to their pol II homologs along the catalytic center including the invariant NADFDGD motif in the largest subunits of all multisubunit RNA polymerases, and surrounding regions [90, 91]. Thus we can be reasonably sure that RPC2, the second largest subunit of pol III together with the largest subunit C1, form an extensive active center similar to but with distinctive differences from pol II [6, 28, 91–93]. Certainly, the core pol II structure fits very well into the electron micrograph envelope of pol III (Fig. 2B) [6, 28, 91, 93]. Peripheral surfaces of pol III such as the jaws and lobe are in proximity to incoming DNA as polymerase moves along the template (Fig. 2B) . While the dimerization domains of the C37 and C53 polypeptides have been localized to a bulge comprising part of the upper jaw adjacent to the lobe, biochemical and physical evidence indicate that other parts of these proteins extend into the catalytic site. A region downstream of the dimerization domain of S. cerevisiae C37 was localized by physical proximity methods near the active site in vitro and the homologous region of S. pombe C37 is a hot spot for mutations that impair termination in vivo . In addition, a region of C11 with strong homology to elongation factor TFIIS likely inserts into the active site to mediate intrinsic transcript cleavage with effects on RNA 3′ end formation during termination [25, 26, 29, 88]. Part of C53 lies close to the RNA 3′ end in the catalytic center of the pol III elongation complex . These observations suggest that the pol III active center is a busy place during termination, comprised of parts of multiple subunits. As might be expected for an extended active center and peripheral nucleic acid-interacting motifs, more than one region of the C1 and C2 subunits would also be involved in termination. The pol III lobe is involved in termination probably via C53/37 and C11 A major advance toward understanding pol III termination came from a genome-wide screen in S. cerevisiae to identify genes that affect termination led to isolation of ret1-1, a transcription mutant with a mutated allele of RPC2 [86, 96]. ret1- 1 was selected as a mutant whose pol III could read through an otherwise functional oligo(dT) terminator placed within the intron of a suppressor tRNA gene . Three regions of RPC2 were then chosen for random mutagenesis and selection for either gain or loss of function, yielding alleles with mutations that either increased or decreased termination in vivo and in vitro . Further analyses revealed that many of these C2 termination-altering mutations also increased or decreased the intrinsic RNA 3′ cleavage activity of pol III suggesting a natural relationship between termination and RNA 3′ cleavage . The three regions of RPC2 were chosen based on termination mutants in the second largest subunit of E. coli RNA polymerase . Of particular interest is the E. coli polymerase β500–575 which overlap with the RPC2 455-524 region and E. coli β1230–1342 region that overlap with the RPC2 1061–1081. Termination altering mutations in similar regions of bacterial and eukaryotic RNA polymerases indicate crucial roles in termination in general [87, 97] and implies that all multisubunit RNA polymerases may undergo similar changes during transition from elongation to termination. Although the 3-dimensional locations of the three pol III regions were unknown at the time of their analyses, presently available structures of S. cerevisiae pol II and pol III allow their localization to three conserved elements referred to as the lobe, fork loops and anchor regions [87, 90] (Fig. 2B). Mutations in lobe residues 300–325 of S. cerevisiae RPC2 are generally associated with decreased termination as is the loss of function mutant, ret1-1: T311K . Curiously, a strong gain of function mutation in this region is K310T suggesting that threonines at either 310 or 311 promote termination. An unbiased genetic screen of zebra fish for digestive system disruption uncovered a deletion in C2 corresponding to aa 259–300 in S. cerevisiae C2 that caused a developmental malformation referred to as the sjm (slimJim) mutant . Recapitulating this deletion in S. pombe C2 led to dissociation of C11 from pol III confirming that this region is required for efficient C11-C2 interaction (Fig. 2) [also see Fig 5 in 29]. Most remarkably, overexpression of zebra fish C11 reversed the gross digestive anomaly, attributing the phenotype caused by the C2 mutation to decreased association of C11 and C2 . Since association of C11 is also required (in yeast) for stable association of the C53/37 heterodimer , the sjm mutation may further compromise the zebra fish pol III. This study, which used molecular modeling based on comparisons of yeast pols II and III strongly supports the idea that the RPC2-RPC11 interface is similar to the RPB2-RPB9 interface and functionally conserved from yeast to vertebrates . In addition, sjm was the first to reveal that mutations in a pol III subunit, a housekeeping enzyme, can have devastating tissue-specific phenotypic effects , including in humans [99–102]. Therefore the sjm deletion, corresponding to ScRPC2 259–300, together with the ScRPC2 300–325 mutations identified by Shaaban et al. form a contiguous region extending from a C11-interacting surface to what appears to be the upper ridge of the pol III lobe adjacent to the cleft from which incoming DNA moves into the active center (Fig. 2B, also see figure 5C–G in . This region likely affects the association of C11 and the C53/37 termination subcomplex with pol III. In remarkably good agreement with the structural localization of C53/37 dimerization domains on pol III are physical proximity data (photo-crosslinking and Fe-BABE mediated cleavage) . The lobe region of C2 that harbors termination mutations was extensively cross-linked to the dimerization domains of C53/37 including numerous interactions between C37 and C2 303–329, as well as between C53 and C2 276–280 . suggesting the lobe as docking site for C53/37 adjacent to the C2 259–300 region that interacts with the Rpb9-homologous domain of C11 (see . The fork loops of C2 in the pol III active center are involved in termination A large number of ret1 mutations occurred in a conserved region of C2 encompassing fork loop 2, the majority of which reduced termination (Fig. 2C). The fork loops are mobile elements that play critical roles in RNA strand separation and maintenance of the transcription bubble [103, 104]. Deletion of fork loop 2 from pol II is known to affect the rate of catalysis by impairing sequestration of substrate NTPs and to cause increased pausing and elongation arrest [105, 106]. The pol II fork loop 2 is a very highly conserved “IGRDGKLA” motif spanning aa 502–509 of S. cerevisiae RPB2. Structural studies of pol II had shown that the tip of the loop, containing RDGK, is mobile and was captured in two conformations, interacting with either the bridge helix or the non-template DNA [107, 108]. The positively charged side chains at the positions occupied by Arg (R) and Lys (K) are conserved by all multisubunit RNA polymerases and are suggested to be involved together with an invariant Arg 512, in DNA strand separation as well as NTP binding and sequestration . The corresponding region, “FEKT/SRKVS” spanning aa 477–484 of S. cerevisiae C2 is highly conserved in pols III but divergent from pol II. Mutations of E478 of S. cerevisiae C2 which is conserved by pols III affects termination [57, 87]. Substitution with Lys increased while Asp decreased termination [57, 87]. The region between fork loops 1 and 2 has an imperfect tripeptide repeat in which every third amino acid is hydrophobic. Most mutations to these hydrophobic residues led to increased termination suggesting that a hydrophobic network in this region contributes to maintaining stability of the elongation complex [57, 87]. Another advance in pol III termination was the physical mapping of a region of C37 to the fork loops and other elements near the catalytic site . Evidence that this was relevant to termination came from transcription analysis of a mutated form C37 that a 5-amino acid tract deletion (scC37 226–230) surrounding the fork loop and (βDloopII) interaction site . Pol III reconstituted with the mutated C37 read through the SUP4 tRNA gene terminator significantly more than did the nonmutated enzyme . The anchor region of C2 All mutations in the anchor region of C2 led to increased termination. The anchor region of pol II connects the RNA:DNA hybrid-binding domain to the clamp and lies between flexible switch domains 3 and 4 that control the mobility of the clamp (Fig 2B). The role of the anchor region in pol III termination is currently unknown although it is expected to control allosteric alterations. Results of in vitro transcription with the C2 mutants generally adhered to the kinetic coupling model; faster elongation rate was associated with reduced termination and slower elongation was associated with increased termination [57, 87] although some exceptions are noteworthy. Very interesting is the C2 double mutant T455I, E478K. As a single mutation, T455I shows increased elongation, decreased termination and efficient RNA release [57, 87]. Not surprisingly, the equivalent S. pombe mutant in rpc2-T455I also exhibits decreased termination [29, 30]. The single E478K mutant has slow elongation with increased terminator recognition (i.e., pausing in the terminator) but poor release of RNA . When these mutations were combined as in T455I, E478K, the net effect was decreased elongation and decreased termination, apparently uncoupling the kinetics of elongation rate and termination. In this case uncoupling occurred in association with decreased RNA release, reminiscent of the effects of nonreleased RNA on the uncoupling of pausing and termination . However, these two cases are different; for E478 mutants, the pause was more like an arrest. In any case, according to the model, E478 would be in fork loop 2 while T455 would flank fork loop 1, in close proximity to the RNA backbone of the RNA:DNA hybrid. Another mutant that uncouples elongation rate and termination is C2 K512N . The corresponding pol II residue shows extensive interactions with others in the region suggesting an intricate interaction network . So, a mutation here might affect the integrity of the region leading to the termination defect. In summary, it is remarkable that nearly 15 years after mapping these S. cerevisiae C2 mutants, the same regions were identified as in close proximity to the C53/37 termination subcomplex . The lobe region shows extensive crosslinks with the dimerization domains of C53/37 suggesting the lobe as a docking site of C53/37 adjacent to C11. The C terminal region of C37 was cross-linked to the fork loop region where another set of mutations were observed. Careful analysis of the C2 mutants revealed that most mutations in the lobe produced reduced termination which now are most likely due to decreased association of C53/37. Mutations in the region 455-524 that encompass fork loop 2 led to either increased or decreased termination reflecting a pivotal role in setting a balance between elongation and termination. Mutations in the anchor region produced only increased termination suggesting that mutations in this region could be destabilizing the elongation complex. C1 and potential effects of α-amanitin on pol III termination α-amanitin is a fungal cyclic octapeptide that inhibits transcription by eukaryotic RNA polymerases to varying extents [110, 111]. It was noted that α-amanitin termination by Xenopus pol III although this remains unsubstantiated . Co-crystallization with pol II has shown that α-amanitin is bound via interactions with the bridge helix and trigger loop, trapping the latter and limiting it’s mobility [112, 113]. While pols II are typically most sensitive to amanitin, pols III show species-specific sensitivities. S cerevisiae pol III is quite resistant, vertebrate pol III is more sensitive and S pombe pol III shows intermediate sensitivity . Intriguingly, these amanitin sensitivity patterns correlate with the oligo(dT) minimal length requirement for termination: human pol III is most sensitive and requires fewer Ts, S. cerevisiae pol III is least sensitive and requires the longest T tract, and S. pombe is intermediate in both amanitin sensitivity and T length . This correlation suggests that the sequence differences between different pol III species in the bridge helix and trigger loop might contribute to termination. These elements are involved in forward translocation of polymerase and thus constitute motifs that determine elongation rate [113–115]. Mutations in this region of C1 affect pausing, RNA cleavage and transcriptional transitions . Random mutagenesis followed by in vivo screening of this region of C1 for termination mutants might reveal further insight into mechanisms of termination by pol III. C53/C37: A dynamic duo Another major advance in pol III termination came from characterization of a C11 mutant that produced pol III enzyme devoid of C11 as well as C53 and C37, known as pol IIIΔ [25, 88]. It was shown that the C53 and C37 subunits are required for oligo(dT) terminator recognition (pausing) and form a stable heterodimer whose interaction domains are attached to the surface of pol III near the leading edge of incoming DNA close to the Rpb9-homologous domain of C11, [28, 88, 91, 92, 94]. The C53/37 dimerization domains are homologous to the dimerization domains of transcription initiation factor TFIIFβ/α which appear to occupy a similar surface on pol II [27, 94]. While these domains hold C53/37 to a surface location, biochemical evidence indicate that other parts of these polypeptides reach into the pol III active center. Association of C53/37 with pol III is dependent on C11. Purified pol III fails to terminate at some terminators and also lacks the intrinsic transcript 3′ cleavage activity . Add back experiments show that recombinant C11 restored cleavage activity but not termination whereas C53/37 restored terminator recognition [25, 88]. Kinetic analysis suggest that C53/37 reduces the elongation rate of pol III on the SUP4 tRNA gene such that pol IIIΔ appears resistant to several gene-internal pause sites . This suggested a mechanism by which C53/37 promotes termination, by reducing elongation rate thereby increasing the pause or residence time of pol III on the terminator, consistent with kinetic coupling. In accordance with this model, reducing the pol IIIΔ elongation rate by decreasing the NTP concentration corrects the termination defect [25, 88]. Although this model is nicely supported by experimental data, there are some notable points. First, the decreased elongation rate caused by C53/37 is manifested as pausing throughout the gene not only at the terminator . Thus, while reduced elongation rate is consequential to termination according to kinetic coupling, it is not specific to the terminator. Another indication that C53/37 action is not specific to termination is evidence that it can function in initiation by participating in open complex formation similar to certain transcription initiation factors including its pol II homolog, TFIIFα/β [94, 95]. Thus, C53/37 appears to be quite dynamic in the breadth of its activities. Moreover, related to both this activity and termination is the requirement for C53/37 along with C11 for facilitated reinitiation by pol III, which is mechanistically dependent on termination [3, 4, 88, 117]. Facilitated reinitiation dependent on termination has also been observed for mammalian pol III [40, 44, 48, 118–121]. Facilitated reinitiation, which has been observed in vitro, is a most appealing process because it can account for the extraordinary efficiency with which pol III and the TFIIIB/C stable transcription complexes can be recycled to produce the large amounts of tRNAs, 5S rRNA and other components required for cellular proliferation [122, 123]. Facilitated reinitiation. A recent study examined for the first time a potential link between termination deficiency and overall transcription output. The results showed that termination deficiency was not accompanied by a decrease in transcription output and question the degree to which if any, a link between termination and recycling observed in vitro is operational in vivo . More studies will be needed to address the mechanisms by which pol III is able to reinitiate with apparent high efficiency in vivo. Photo cross-linking and other physical methods revealed key interactions between C37 and C2 as well as other subunits . As noted above the dimerization domains of C53 and C37 reside near the lobe domain of C2 and the Rpb9-homologous region of C11 (Fig. x) [28, 92, 94]. Residues 226–230 in the C terminal region of C37 far downstream of its dimerization domain were found to react in the immediate vicinity of C2 fork loops 1 and 2, βDloopII and a hybrid binding region in the active center of pol III . Deletion of this tract from C37 led to deficiency of pol III termination in vitro . In support of function of the C-terminal region of C37 in termination, are point mutations in multiple residues in a homologous region of S. pombe C37 mutants isolated from a genetic termination screen . The mutants produced increased terminator read-through transcripts from various tRNA genes in vivo . The observations that C37 residues localize with C2 fork loops and that C2 fork loop mutations had been independently isolated as termination mutants argue that these regions participate in termination. Moreover, proximity of these regions of C2 and C37 to the pol III active site further suggests that termination involves alterations to catalytic activity that are more complex than passive cessation of phosphodiester bond formation (below). The cumulative data suggest that C37 contributes to and modulates the active center of pol III . As such, C37 may be considered as one of several polypeptides that can access the catalytic site of an RNA polymerase to affect its activity, the prototypical examples of which are TFIIS for pol II and its bacterial counterpart GreB [80, 124]. However, while TFIIS and presumably the homologous cleavage-domain of C11 access the catalytic center through the secondary channel [80, 107], the path of the C37 and C53 polypeptides to the catalytic center may be via a more frontal approach perhaps through the DNA cleft region (Fig. X) although this remains to be determined. Physical proximity data indicate that the association of the C37 C-region with the RNA:DNA hybrid binding region of C2 is more robust in the elongation complex than in the preinitiation complex . Proximity of the C37 C-region to the βDloopII of C2 was also increased in the elongation complex. The corresponding loop in the pol II model is part of ‘fraying site II’ which interacts with the frayed, i.e., non-annealed, 3′ terminal nucleotide of the RNA . Bioinformatics suggested a possible structural homology between C53 and MLE, an RNA helicase . Since C53 can interact with the 3′ end of the RNA in the elongation complex the data suggest that it may participate in RNA:DNA hybrid melting thereby contributing to destabilization and termination . Finally it should be noted that while lack of C37/53 leads to an increased elongation rate it does not render pol IIIΔ completely incapable of recognition of the SUP4 tRNA bipartite gene terminator . This suggests that there is a C37/53/11-independent mode of termination intrinsic to the core pol III enzyme that is enhanced by C53/37 (A.A. & R.M., in preparation). C11 involvement in intrinsic RNA 3′ cleavage, termination and facilitated recycling As first reported, C11 was believed to be directly involved in termination because it was not known at that time that C11 mediates association of C53/37 with pol III . C11 is a short polypeptide of 110 aa first identified as having strong homology to the pol II elongation factor TFIIS and responsible for the previously recognized robust intrinsic transcript cleavage activity of pol III [25, 82]. It has two Zn ribbon motifs separated by what may be a flexible linker. The N terminal Zn ribbon of C11 is homologous to the N terminal Zn ribbon of subunit RPB9 of pol II and the C terminal Zn ribbon is highly homologous to the Zn ribbon of the pol II RNA 3′ cleavage factor, TFIIS . Similar to TFIIS, the C11 Zn ribbon has the same two acidic residues at the tip of its loop that presumably position Mg 2+ within the pol III catalytic center similar to the homologous residues of TFIIS in pol II [80, 107]. The N terminal domain of RPB9 is attached to the lobe domain of second largest subunit RPB2 forming a structure called the jaw of the polymerase . As noted, the Rpb9-homologous domain of C11 (yellow in Fig. 2B) occupies a similar position on the C2 lobe of pol III and is required for association of C53/37 with pol III [28, 88, 91, 94]. Mutations to the C- and N-terminal motifs of C11 have different effects on termination; RNA 3′ oligo(U) nibbling and prevention of terminator read through [26, 29]. Although the latter may be due to allosteric effects on C53/37 the cumulative data nonetheless provide clear evidence that C11 is positioned to affect pol III termination [25, 29, 88]. As detailed below, the RNA 3′ cleavage activity of C11 is active during termination [26, 29] although the extent of its role in the termination process is unresolved. A challenge is to understand how the two domains of C11 work with each other and with C53/37 both at their surface locations and via their domains that approach the catalytic center of pol III, during termination. 3′ terminal U residues are preferred substrates of C11-mediated cleavage by pol III [82, 84]. An enticing model would be that shortening of the rU:dA hybrid by cleavage-mediated removal of 3′ terminal or frayed Us would lead to hybrid shortening, weakening and destabilization of the complex. However, this model was not supported by mutations in C11 that compromise RNA cleavage activity since although these mutations do lead to increased length of the 3′ oligo(U) tract of the nascent RNA by 1–2 nts both in vivo and in vitro, this was not accompanied by increased terminator read through [26, 29]. The data suggest that the cleavage activity of C11 mediates 3′ U cleavage during termination but not termination efficiency per se. In striking contrast to the C-terminal mutants, a cluster of N-terminal C11 mutants exhibited increased terminator read through but not 3′ oligo(U) lengthening . The N-terminal mutations of C11 may affect termination via effects on C53/37 . Far more C11 C-terminal domain mutants were isolated that impaired RNA cleavage than were N-terminal domain mutants that impaired terminator recognition, despite their isolation from the same library [26, 29]. Using the same screening approach applied to C37, the opposite was true; far more C37 mutants were isolated that impaired terminator recognition than 3′ oligo(U) lengthening . The same approach yielded even fewer C53 mutants using either both the terminator recognition and cleavage-sensitive screens . These results provide evidence that of the C11, C53 and C37 subunits, terminator recognition was by far most sensitive to C37 . It is noteworthy that C11 exhibits cleavage-dependent and cleavage-independent activities in pol III termination and associated processes. While C11 is required for facilitated recycling along with C53/37, a point mutant that disables its cleavage activity is nonetheless as active as the wild type C11 for facilitated recycling . EXTRANEOUS FACTORS MAY AFFECT POL III TERMINATION Several trans-acting factors were shown to augment pol III termination in vitro. Among them La protein ranks first by historical perspective and because of its appealing specificity, affinity for the 3′ oligo(U) tracts of pol III-terminated transcripts. La exhibits sequence- and length-specific binding to oligo(U), with the 3′-OH group contributing substantially . These features along with early findings that La is physically associated with all newly synthesized pol III transcripts in vivo made it a likely candidate for involvement in the termination process [127–131]. Experimental data obtained from in vitro assays supported that La is a transcription termination, transcript release and reinitiation factor for pol III [40, 48, 77, 118, 120, 132, 133], although such activities have been widely contested [134–136]. Only the non-phosphorylated fraction of human La was transcriptionally active in vitro while the phosphorylated fraction was inactive . Consistent with this, non-phosphorylated La is associated with pol III transcribed genes in vivo while phosphorylated La is not . Yet phosphorylated La is found associated with nascent pre-tRNAs and other pol III transcripts in HeLa cells whereas non-phosphorylated La was not . Although in vivo deletion of the La-homologous protein (Lhp1) from S. cerevisiae caused no decrease in pol III transcription levels , it led to increased accumulation of pol III on 5S rRNA genes in vivo, consistent with a role as a limiting factor for pol III termination . Thus, although by 3′ oligo(U) binding La protein provides a link between termination by pol III and processing of its transcripts, a role in the termination process per se and potential effects on recycling remain questionable. The mechanistic details of the link between pol III termination and RNA processing, and its implications, are described in a later section. Other factors reported to promote pol III termination These include transcription factor IIIC, topoisomerase-1 and PC4. Both topo-1 and PC4 copurify with TFIIIC and enhance the TFIIIC footprint on the downstream promoter . Yeast TFIIIC subunit TFC6 can be cross-linked to the terminators of tRNA and 5S rRNA genes suggesting a role in termination [141, 142]. In mammalian systems a TFIIIC-associated activity caused a specific footprint on the terminator [143, 144]. Genetic mutagenesis-based screens of S. pombe that yielded C11 and C37 mutants that impair terminator recognition and/or 3′ oligo(U) length [26, 29, 30], yielded no mutants for similarly-mutagenized sfc6+, the S. pombe homolog of TFC6 and mammalian TFIIICβ , (K. R. and R. J. M., unpublished observation). Thus, to date, there is no data linking TFC6 or its counterparts to a functional role in termination. The intriguing means by which NF1 promotes pol III termination of VA1 gene transcription may provide a clue into a mechanism by which some extraneous factors may operate . Effects of NF1 on VA1 appear to be gene specific because a NF1 sequence-specific binding site is found adjacent to the pol III terminator of the VA1 but not other pol III-transcribed genes . Consistent with this, ChIP analysis did not show preferred localization of NF1 to pol III transcribed genes . It is plausible that NF1 bound to terminator-adjacent DNA acts as a roadblock that induces pausing at the terminator thereby promoting termination. Such a roadblock may also alter pol III confirmation by allosteric transmission upon collision. Consistent with this, a roadblock caused by a peptide nucleic acid was shown to increase termination even from a suboptimal terminator . A role for chromatin in pol III termination Although several studies show active involvement of chromatin in the control of pol III transcription [147–151], little is known about a role in termination. tRNA genes are generally nucleosome free . Genome-wide ChIP that focused on the yeast histone variant H2A.Z revealed a large number of tRNA genes flanked by nucleosomes . A recent study revealed that the dynamics of a nucleosome that abuts the terminator of the SUP4 tRNA gene in S. cerevisiae can modulate the expression level of the tRNA under different conditions . Genome-wide pol III ChIP showed a significant fraction of pol III accumulation just beyond the terminators of tRNA and other class III genes, which may reflect pausing at a downstream nucleosome , similar to pol II accumulation downstream of its transcribed genes . Accordingly, it seems possible that nucleosomes abutting terminators could increase pausing and thereby modulate termination and/or pol III release and reinitiation. COUPLING OF POL III TERMINATION WITH RNA PROCESSING As noted above, data indicate that the C11-mediated RNA 3′ cleavage that occurs during termination is responsible, at least in part, for the variable lengths of the oligo(U) termini of several nascent pol III transcripts [Recently reviewed in 77]. The 1–2 nucleotide increase in 3′ oligo(U) length can have striking effects on pre-tRNA turnover and maturation in vivo . The subtle increase in 3′ oligo(U) length that occurs in C11 cleavage mutants significantly increases the affinity of the transcripts for La protein with consequent protection from transcript degradation, more efficient RNA processing and increased levels of mature functional tRNA . In the absence of La binding pre-tRNAs may succumb to nuclear surveillance-mediated decay by the exosome . Indeed La levels are limiting for tRNA maturation in S. pombe but increasing La by ectopic expression or increasing the 3′ oligo(U) length on pre-tRNAs in C11 cleavage mutants drives more nascent transcripts into the La-dependent productive pathway of tRNA maturation . These observations suggest a role for termination-associated C11-mediated transcript cleavage in affecting functional tRNA production and raises the possibility that termination-associated nibbling by C11 helps control tRNA levels and thus translation. Consistent with this possibility is that a genome-wide search for candidate factors regulated by upstream open reading frames (uORFs) similar to uORFs that regulate GCN4 under general translational stress, revealed C11 as one of just a few S. cerevisiae genes . While La has been found to transiently bind to all of the different types of nascent pol III transcripts examined, and its effects are most extensively characterized for tRNAs [reviewed in 77], it is also functionally involved in U6 snRNA maturation [158, 159]. As detailed below this may reflect a conserved mechanistic coupling between pol III transcription termination and U6 RNA 3′ processing which involves an intricate series of what appear to be 3′ uridylate-specific activities. Mature U6 snRNA has a 2′-3′ cyclic phospho UMP residue at the 3′ end, formed as part of a deuridylation-urydylation reaction . The sequential terminal Us on pre-U6 snRNA that are formed as a result of pol III termination may be required for this process. La is a functional component of U6 3′ end metabolism [158, 159]. Curiously, U6 snRNA is transcribed in all species by pol III while the other spliceosomal RNAs (U1, U2, U4, U5) are transcribed by pol II. However, unlike tRNA and 5S, the U6 snRNA sequence apparently can not accommodate an internal B box promoter element. Different species have developed varying mechanisms to circumvent the problematic need for a B box sequence in functional U6 snRNA genes. S. cerevisiae has the B box downstream of the terminator ; binding of TFIIIC to the A and B box elements is facilitated by a positioned nucleosome between them [147, 149]. An alternative solution was achieved by S. pombe, in which the B box resides in an intron in the U6 gene that is spliced out upon maturation of U6 snRNA . Vertebrates use yet another strategy; entirely upstream promoters for pol III-mediated U6 transcription . In vertebrates this additionally involved the emergence of a Brf1-homologous protein known as Brf2 . These observations suggest that U6 snRNA production can accommodate a variety of promoter types and positions which in some cases employ specially positioned nucleosomes, intron insertion, and factor-specific modes of initiation by pol III. These examples suggest that it is important that the U6 snRNA is a pol III transcript in these species and that the intricate uridylate-specific 3′ end processing accounts for its dependence on pol III. The coupling of a pol III termination mechanism that produces RNAs with 3′ oligo(U) ends that mediate specific association with the La protein maturation factor is undoubtedly a means to afford pol III transcripts a processing pathway separate from the pols I & II transcript maturation pathways. It was recently reported that pol III transcription and precursor tRNA processing are linked via a pathway that involves the pol III repressor Maf1 although this is likely due to saturation of processing and/or export machinery . A role for RNA’se P in coupling pol III transcription and tRNA 5′ processing has been proposed [164, 165], whereas all other links between the transcription and processing of class III genes appears to be via 3′ end processing as described above. CONCLUDING REMARKS Transcription termination involves pausing followed by destabilization of the elongation complex with dissociation of its components. Bacterial RNA polymerase as well as pol II can terminate transcription in response to more than one terminal signal. Pol II termination of poly(A)-containing mRNA synthesis uses a cis-acting terminator element, the poly(A) addition site, that acts at a distance to initiate the process, followed by dissociation of the complex somewhere downstream, somewhat similar to the Rho-dependent mechanism of E. coli transcription termination. Pol III can achieve the same outcome on a short oligo(dA) template that serves to both pause and dissociate the complex with efficiency and near nucleotide precision, somewhat similar to the intrinsic termination by E. coli RNA polymerase. Yet, despite the efficiency of what appears to be a simple termination signal for pol III, involvement of multiple subunits suggests a complex mechanism of termination. Recent reports of termination defective mutants of C37 and C11, and localization of these plus C53 near the catalytic site suggest active alteration of the pol III active center during termination, as opposed to polymerase jaw-mediated brakes clamping down on incoming DNA and passive cessation of RNA synthesis. Control of elongation rate as well as the degree to which the oligo(rU:dA) hybrid confers instability to the pol III complex are likely underlying components of the termination mechanism that are sensitive to the activities of the termination subunits. Because C37, C53 and C11 have homologs in the pols I & II systems it is suggested that these polymerases may use similar mechanisms to execute their final stages of termination, complex destabilization, dissociation and release. High-resolution crystal structures of pol III elongation complexes with and without C11/C53 and C11 will undoubtedly reveal more about these subunits and their involvement in the catalytic center. Another challenge will be to obtain crystals of a pol III termination complex. An aspect that we believe will provide unique insight into the mechanisms of pol III termination is to understand the species-specific differences in the minimal number of Ts required, as well as oligo(dT) context-dependent effects, and how terminator-adjacent binding sites for DNA-binding factors may promote termination. In addition, ‘secondary terminators’ and non-canonical pol III terminators are also a potential trove of important means to production of a newly emerging class of tRNA-downstream short RNAs. HIGHLIGHTS. The rU:dA(5–6) hybrid is a destabilizing component of transcription termination. Pols I, II & III initiate termination differently but may share features thereafter. Several pol III subunits, including C2, C11, C53 and C37 contribute to termination. C37/53 and C11 are homologous to TFIIFα/β, respectively. Extraneous factors may be recruited to affect pol III termination in some cases. Footnotes Publisher's Disclaimer: This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final citable form. 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[DOI] [PMC free article] [PubMed] [Google Scholar] ACTIONS View on publisher site PDF (959.3 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract INTRODUCTION TERMINATORS SIGNAL THE INITIATION OF A TWO-STAGE PROCESS TERMINATION BY POL III: THE SIGNAL IS OLIGO(dT) THE TERMINATION MECHANISM POL III SUBUNITS INVOLVED IN TERMINATION EXTRANEOUS FACTORS MAY AFFECT POL III TERMINATION COUPLING OF POL III TERMINATION WITH RNA PROCESSING CONCLUDING REMARKS 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
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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Principles of Microeconomics 3e Key Concepts and Summary Principles of Microeconomics 3eKey Concepts and Summary Contents Contents Highlights Table of contents Preface 1 Welcome to Economics! 2 Choice in a World of Scarcity 3 Demand and Supply 4 Labor and Financial Markets 5 Elasticity 6 Consumer Choices 7 Production, Costs, and Industry Structure Introduction to Production, Costs, and Industry Structure 7.1 Explicit and Implicit Costs, and Accounting and Economic Profit 7.2 Production in the Short Run 7.3 Costs in the Short Run 7.4 Production in the Long Run 7.5 Costs in the Long Run Key Terms Key Concepts and Summary Self-Check Questions Review Questions Critical Thinking Questions Problems 8 Perfect Competition 9 Monopoly 10 Monopolistic Competition and Oligopoly 11 Monopoly and Antitrust Policy 12 Environmental Protection and Negative Externalities 13 Positive Externalities and Public Goods 14 Labor Markets and Income 15 Poverty and Economic Inequality 16 Information, Risk, and Insurance 17 Financial Markets 18 Public Economy 19 International Trade 20 Globalization and Protectionism A | The Use of Mathematics in Principles of Economics B | Indifference Curves C | Present Discounted Value Answer Key References Index Search for key terms or text. Close 7.1 Explicit and Implicit Costs, and Accounting and Economic Profit ------------------------------------------------------------------- Privately owned firms are motivated to earn profits. Profit is the difference between revenues and costs. While accounting profit considers only explicit costs, economic profit considers both explicit and implicit costs. 7.2 Production in the Short Run ------------------------------- Production is the process a firm uses to transform inputs (e.g., labor, capital, raw materials, etc.) into outputs. It is not possible to vary fixed inputs (e.g., capital) in a short period of time. Thus, in the short run the only way to change output is to change the variable inputs (e.g., labor). Marginal product is the additional output a firm obtains by employing more labor in production. At some point, employing additional labor leads to diminishing marginal productivity, meaning the additional output obtained is less than for the previous increment to labor. Mathematically, marginal product is the slope of the total product curve. 7.3 Costs in the Short Run -------------------------- For every input (e.g., labor), there is an associated factor payment (e.g., wages and salaries). The cost of production for a given quantity of output is the sum of the amount of each input required to produce that quantity of output times the associated factor payment. In a short-run perspective, we can divide a firm’s total costs into fixed costs, which a firm must incur before producing any output, and variable costs, which the firm incurs in the act of producing. Fixed costs are sunk costs; that is, because they are in the past and the firm cannot alter them, they should play no role in economic decisions about future production or pricing. Variable costs typically show diminishing marginal returns, so that the marginal cost of producing higher levels of output rises. We calculate marginal cost by taking the change in total cost (or the change in variable cost, which will be the same thing) and dividing it by the change in output, for each possible change in output. Marginal costs are typically rising. A firm can compare marginal cost to the additional revenue it gains from selling another unit to find out whether its marginal unit is adding to profit. We calculate average total cost by taking total cost and dividing by total output at each different level of output. Average costs are typically U-shaped on a graph. If a firm’s average cost of production is lower than the market price, a firm will be earning profits. We calculate average variable cost by taking variable cost and dividing by the total output at each level of output. Average variable costs are typically U-shaped. If a firm’s average variable cost of production is lower than the market price, then the firm would be earning profits if fixed costs are left out of the picture. 7.4 Production in the Long Run ------------------------------ In the long run, all inputs are variable. Since diminishing marginal productivity is caused by fixed capital, there are no diminishing returns in the long run. Firms can choose the optimal capital stock to produce their desired level of output. 7.5 Costs in the Long Run ------------------------- A production technology refers to a specific combination of labor, physical capital, and technology that makes up a particular method of production. In the long run, firms can choose their production technology, and so all costs become variable costs. In making this choice, firms will try to substitute relatively inexpensive inputs for relatively expensive inputs where possible, so as to produce at the lowest possible long-run average cost. Economies of scale refers to a situation where as the level of output increases, the average cost decreases. Constant returns to scale refers to a situation where average cost does not change as output increases. Diseconomies of scale refers to a situation where as output increases, average costs also increase. The long-run average cost curve shows the lowest possible average cost of production, allowing all the inputs to production to vary so that the firm is choosing its production technology. A downward-sloping LRAC shows economies of scale; a flat LRAC shows constant returns to scale; an upward-sloping LRAC shows diseconomies of scale. If the long-run average cost curve has only one quantity produced that results in the lowest possible average cost, then all of the firms competing in an industry should be the same size. However, if the LRAC has a flat segment at the bottom, so that a firm can produce a range of different quantities at the lowest average cost, the firms competing in the industry will display a range of sizes. The market demand in conjunction with the long-run average cost curve determines how many firms will exist in a given industry. If the quantity demanded in the market of a certain product is much greater than the quantity found at the bottom of the long-run average cost curve, where the cost of production is lowest, the market will have many firms competing. If the quantity demanded in the market is less than the quantity at the bottom of the LRAC, there will likely be only one firm. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: David Shapiro, Daniel MacDonald, Steven A. Greenlaw Publisher/website: OpenStax Book title: Principles of Microeconomics 3e Publication date: Dec 14, 2022 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. Our mission is to improve educational access and learning for everyone. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. Give today and help us reach more students. Help Contact Us Support Center FAQ OpenStax Press Newsletter Careers Policies Accessibility Statement Terms of Use Licensing Privacy Policy Manage Cookies © 1999-2025, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License. Advanced Placement® and AP® are trademarks registered and/or owned by the College Board, which is not affiliated with, and does not endorse, this site.
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https://math.answers.com/math-and-arithmetic/How_many_3_digit_numbers_are_have_digits_in_increasing_order
How many 3 digit numbers are have digits in increasing order? - Answers Create 0 Log in Subjects>Math>Math & Arithmetic How many 3 digit numbers are have digits in increasing order? Anonymous ∙ 17 y ago Updated: 11/4/2022 123 234 345 456 567 678 789 I assume this is what you meant. Thanks, but I need all numbers such as 245, 246, 247, 345, 346, 347, 125,137,etc...any number with 3 increasing digits. There are 84 numbers that meet the criterion. 123, 124, 125, ..., 178, 179, 189 = 28 234, 235, 236, ..., 278, 279, 289 = 21 345, 346, 347, ..., 378, 379, 389 = 15 456, 457, 458, 459, 467, 468, 469, 478, 479, 489 = 10 567, 568, 569, 578, 579, 589 = 6 678, 679, 689 = 3 789 = 1 Wiki User ∙ 17 y ago Copy Add Your Answer What else can I help you with? Search Continue Learning about Math & Arithmetic ### What are the number combinations for 7498? There is only one 4-digit combination which is 7498, or 4789 if you prefer the digits in increasing order. ### How many decreasing numbers are between 100 and 500? If you mean 3-digit numbers whose digits are in decreasing order, then the answer is 20. ### What are the least and greatest digits in negative numbers? The positional place values of digits in negative numbers are in ascending order from least to greatest as for example in the number -987 the least value digit is 9 and the greatest value digit is 7 because -900 < -80 < -7 The positional place values of digits in positive numbers are in descending order from greatest to least as for example in the number 987 the least value digit is 7 and the greatest value digit is 9 because 900 > 80 > 7 ### The sum of the digits of a two digit number is 9 the number obtained by reversing the order of the digits of the gven numbers exceeds the given number by 27 find the given numbers? 3 and 6 ### What is the smallest 7 digit number with no two digits alike? The smallest 7-digit number with no two digits alike is 1023456. This number begins with the smallest non-zero digit (1) to ensure it remains a 7-digit number, followed by the smallest available digits (0, 2, 3, 4, 5, and 6) arranged in increasing order. Related Questions Trending Questions What were booths in biblical days?15 24 25 5 14 15 what is the pattern?How many ounces in 53.352 kg?How many times does 45 go into 68?What is 87 equals?How many minutes of fame are there?How do you use the word puzzle in a sentence?How many angles are on an isosceles triangle?What is the smallest number that has 235711 as its prime factors?Does 144 pieces equal a gross?Is 7 a square?Which plane of refernce would always divide a four-legged animal into two equal halves that are mirror images of each other?What is 0.029 as a percentage?How do you round 543 to the nearest tenth?Why do you cut notches at the end of sticks on a kite?Who are all the Marys in the Bible and what verses do we find them in?5 x 5 x 1 and one half?What are eight groups of 4?How many radians is in one second?What are the odds of rolling a seven with two dice fifty six times in succession? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
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https://ocw.mit.edu/courses/5-61-physical-chemistry-fall-2017/75f72c140e34cecc9463968404456a98_MIT5_61F17_lec3.pdf
5.61 Fall 2017 Lecture #3 page 1 revised 9/13/17 4:00 PM Lecture #3: Two-Slit Experiment. Quantum Weirdness Last time: 1. Wave character of e– X-ray and e– diffraction using atom-spacings in a crystal as ruler to measure λ(p). Find λ = h/p. de Broglie hypothesis: λ = h p for all particles Unasked questions: what happens when λ is comparable to the size of the container, or λ3 is large compared to the volume available to each atom: V/N? 2. Rutherford — postulated (based on Geiger-Marsden experiments) planetary atom in order to “solve” the space-filling requirement. But: no quantization radiative collapse 3. Bohr  =  r ×  p = r p ≡n n = 1, 2, 3… (quantization of l!) de Broglie: 2πr n circumference of Bohr orbit   = nλn Two different ad hoc hypotheses to “prevent” radiative collapse. Both lead to requirement of quantized energy levels. 4. Idea (Ritz, Balmer, Rydberg) that spectral lines are transitions between quantized energy levels. “Explains” spectra of 1e– atoms. nothing about radiative lifetimes or relative transition strengths effect of magnetic fields (transition line splits into too many components → e– spin) not a clue to explain the spectrum of a 2e– atom: Helium. Today: 2-slit experiment. interference taste of quantum measurement theory qualitative stuff about waves glimpse of uncertainty principle This stuff is weird! It should bother you. 5.61 Fall 2017 Lecture #3 page 2 revised 9/13/17 4:00 PM (needs to be monochromatic) why? d source two slits detector screen strip of screen 0 z s1 s2 L x 0 L distance to screen: L ≫d slit widths δs = δz = d = L Width of each slit = d. Choose the width of each slit to be ~λ and the distance between slits d > 10λ. We expect to see an interference pattern when both slits are open but no interference when one of the slits is covered. Constructive interference results when the paths from s1 and s2 to same point on the screen differ by an integer multiple of λ. Call the direction along the screen z and the direction along the ⊥ path from plane of slits to screen x. Here is a blow-up of the region near the 2 slits 5.61 Fall 2017 Lecture #3 page 3 revised 9/13/17 4:00 PM d 2 almost parallel paths to same spot on distant screen θ θ δ x z because L ? d, can treat the two rays as parallel yet intersecting at the same point on the screen. For constructive interference it is necessary that the two paths differ in length by δ = nλ δ = d sin θ (see diagram) so we get a set of θ-values at which constructive interference occurs nλ = d sin θ θn = sin −1 nλ d θ0 = 0 (central spot for n = 0) in the small θ limit, θn ≈nλ d (d λ) On the screen you see a series of equally spaced bright regions (constructive interference) separated by dark regions (destructive interference). On the screen, the bright regions are at z = 0,±Lsinθn ≈± L d nλ . OK. Now what happens if we cover one of the slits? Interference pattern disappears. [Width of central bright zone remains broadened by diffraction.] The plane of the screen is at a distance L from the slit plane. 5.61 Fall 2017 Lecture #3 page 4 revised 9/13/17 4:00 PM Does the pattern on the screen tell us which slit was covered? How? Are we allowed to know? Yes. Asymmetry! What does this mean? Suppose we reduce the intensity of light entering the 2-slit apparatus so much that, at any given time there is either 0 or 1 photon in the apparatus. It is rather straightforward to measure the intensity and know that the intensity is small enough to satisfy this requirement. What do you need to know to compute the < 1 photon at a time intensity? What will we see? Be ready for the weird beauty of Quantum Mechanics! no interference pattern? weak interference on top of a constant background, which suggests that only rare fluctuation events yield 2 photons simultaneously traversing the apparatus? the usual, full 100% modulated interference pattern? We expect the intensity distribution to exhibit interference, based on the wave nature of light. But we know that light also has particle characteristics. What do we see on a 2-D detector with single-event sensitivity and time resolution? The continuous distribution “collapses” into localized single events (dots). Each event is independent of all other events and one point cannot resemble a distribution. Reconcile? two slits open one slit open # of events 5.61 Fall 2017 Lecture #3 page 5 revised 9/13/17 4:00 PM See one-photon event as a dot on the screen. Initially the dots look randomly distributed. Eventually, once a sufficient number of dots has accumulated, the interference pattern emerges. You should be amazed by this! The interference pattern goes away when either slit is covered. (What happens if we use white light instead of monochromatic light?) Quantum Weirdness You are not allowed to know which one of 2 open slits each single photon went through. The interference is at the single-event level, not at the many-event level. The interference is of one photon with itself, not with another photon. Is this weird or what? We need to describe the two-slit experiment by some sort of probability amplitude distribution and to describe an experiment as the sum of interfering amplitudes followed by some sort of operation that expresses the action of the detector (i.e. collapse of each one-photon signal to a single dot). And now a note from our sponsor: Looking ahead: Light follows a wave equation. The probability amplitude will look like this: u(x,t) = A sin (kx – ωt) Figure removed due to copyright restrictions. See Fig. 1.18 in: McQuarrie, Donald A. Quantum Chemistry. University Science Books, 2007. ISBN: 9781891389504. 5.61 Fall 2017 Lecture #3 page 6 revised 9/13/17 4:00 PM If u(x,t) is a probability amplitude, what is the probability density? What is the difference between a probability amplitude and a probability density? Can either one be negative at any values of x and t? Wavelength: u(x + λ,t) = u(x,t) (λ defined as the spatial repeat distance) A sin[kx + kλ – ωt] = A sin[kx – ωt] if kλ = 2π k = 2π λ “wave number” λ= h p →k = 2π h p = p /  (in 3-D,  k points in the direction of wave motion. Large k implies small λ and large p.) k is 2π times the number of wavelengths per unit length. Velocity: Take a snapshot of a wave in time. How does the constant-phase point, xφ, move? phase: φ = kxφ −ωt , pick φ = 0 xφ = ωt k dxφ dt = ω k velocity of phase point For a wave of the form u(x,t) ∝ sin(kx – ωt) velocity is +ω k (moving in + x direction) not too surprising ω = 2πν k = 2π λ ω k = 2πν 2π / λ = νλ ⇒ for light ν = c / λ! ω/k = c (for monochromatic electromagnetic radiation propagating in vacuum). 5.61 Fall 2017 Lecture #3 page 7 revised 9/13/17 4:00 PM Intensity of an electromagnetic wave is ∝ ε 2 I(x,t) = aiu i ∑ kix +ω it ( ) 2 superposition of amplitudes Sum first, then square. A taste of the uncertainty principle Suppose we want to spatially localize a particle. Put it through a slit of width δs = δz 0 z L δs slit screen The wave nature of the particle λ = h p implies that there will be “diffraction” of the particle by the slit. This results in spreading of the image of the slit on the screen. Use the same algebra as for the 2-slit experiment. Paths from top and bottom edges of the slit to a point z on the screen must differ by λ/2 to get destructive interference. [Interference is less complete from points not at edges of slit.] amplitude of ith component 5.61 Fall 2017 Lecture #3 page 8 revised 9/13/17 4:00 PM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  FWHM -δz -0 z n0 ⇥n0,n = En0 −En n n 1 δz = 2 λ δs L ⎛ ⎝ ⎞ ⎠ This means that pz is uncertain because  p is a vector quantity. (  p is conserved, but the angular uncertainty results in a magnitude uncertainty of pz.) -0 L x 6 0 z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ✓ # # ✓ δpx # # # ✓ δpz |p| |p| Photon that hits at center has px = |p|, pz = 0. Photon that hits at edge has smaller px, larger pz. δ pz ≈p λ δs p = h λ ≈h λ λ δs = h δs This is the uncertainty in pz that results from slit of width δs. δzδpz ≈ h is an uncertainty principle. An attempt to restrict position (δz) results in uncertainty in pz (δpz). QM is based on what could, in principle, be measured. Every experiment must be analyzed in this way. 5.61 Fall 2017 Lecture #3 page 9 revised 9/13/17 4:00 PM Today: 2-slit experiment. Can’t know which slit. Photon interferes with itself. waves: c, λ, k amplitude and intensity a taste of uncertainty of joint measurement of x and px. Next Lecture: classical wave equation in preparation for Schrödinger Equation. Read Chapter 2 of McQuarrie before next lecture! Non-Lecture POSTULATES. The postulates of Quantum Mechanics are starting to emerge, and here they are. Think about them during the next few lectures. You will see them in action rather than in a mind-numbing review. POSTULATE 1 The state of a quantum--mechanical system is completely specified by a function Ψ(r,t) that depends on the coordinates of the particle and on time. This function, called the wave function or state function, has the important property that Ψ∗(r,t) Ψ(r,t) dxdydz is the probability that the particle lies in the volume element dxdydz located at r at time t. POSTULATE 2 To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics. POSTULATE 3 In any measurement of the observable associated with the operator A , the only values that will ever be observed are the eigenvalues an, which satisfy the eigenvalue equation A Ψa = aΨa POSTULATE 4 If a system is in a state described by a normalized wave function Ψ, then the average value of the observable corresponding to A  is given by a = Ψ −∞ ∞ ∫ A Ψdτ You might think about how to handle a non-normalized Ψ. POSTULATE 5 The wave function or state function of a system evolves in time according to the time-dependent Schrödinger equation 5.61 Fall 2017 Lecture #3 page 10 revised 9/13/17 4:00 PM H Ψ x,t ( ) = i∂Ψ ∂t MIT OpenCourseWare 5.61 Physical Chemistry Fall 2017 For information about citing these materials or our Terms of Use, visit:
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https://www.expii.com/t/base-binary-numbers-9192
Expii Base 2: Binary Numbers - Expii Base 2 uses powers of 2. The possible digits are 0 and 1. Most computers are coded in binary at the most basic level!. Explanations (4) Alex Federspiel Text 10 Binary Binary might be the second most known number system. Whereas our number system uses 10 digits, binary uses only 2 digits: 0,1 "Computers use binary to do things." -Alex on the topic of binary. From Decimal to Binary Convert from decimal to binary can be a little tedious. We want to take the number and continuously divide it by 2, rounding down for decimals, and listing the numbers out. Each even number we represent with a 0 and each odd number we represent with 1. That's a lot of words. Maybe an example will make things clearer. Let's convert 4210 (the little 10 means base-10) to base-2. Start by dividing 42 by 2 until we get down to a 1. Looking from the bottom number up, we have the numbers: 125102142 The final step is to put a 1 under the odd numbers and a 0 under the even. 125102142101010 So: 4210=1010102 From Binary to Decimal Luckily, going from binary to decimal isn't as difficult. Starting from the last number, we want to multiply each digit by 2 raised to increasing powers (starting from 0 and moving up). We then add all these up. Let's see it in action with our newly converted base-2 number, 101010. 101010=(1⋅25)+(0⋅24)+(1⋅23)+(0⋅22)+(1⋅21)+(0⋅20)=(1⋅32)+(0⋅16)+(1⋅8)+(0⋅4)+(1⋅2)+(0⋅1)=32+0+8+0+2+0=42 So: 1010102=4210 Practice Now that you're basically a glorified computer, try some out on your own. Report Share 10 Like Related Lessons Computer Science: Pseudocode Base 16: Hexadecimal Numbers Data Structures: Queues Base 3: Ternary Numbers View All Related Lessons Alex Federspiel Video 6 (Video) Convert Numbers into Binary by Mind Your Decisions Here's a video by MindYourDecisions explaining how to convert from base-10 to base-2. Summary To convert from 10 to 2, divide the number by 2 (ignoring the remainder) until you reach 1. Then any odd number gets a 1 below it and any even gets a 0. The following string of 1's and 0's is your new binary number. To go from base-2 to base-10 you use a similar method explained in the base-3 section but this time multiply by 2n instead of 3n Report Share 6 Like OC Science Text 5 Think about a decimal (base 10) number: 7. The binary number representation of 7 is 111. The base 2 representation (or equivalently, binary representation) of 21 is 10101. Let's analyze this: 21=2×10+1. In base 10, each digit represents a power of 10. Now going back to 7, it is 7=1×22+1×2+1×1. Here, the digits are just powers of the base. Report Share 5 Like Edwin Peng Text 2 There are many equivalent ways to represent a number. For example, twelve can be represented as 2×6,9+3, or just 12. Normally, we represent numbers by using the digits 0 through 9. This is known as a base-ten number. A base-ten number is equal to 1 times its first digit (from the right), plus 10 times its second digit, plus 100 times its third digit, and so on. For example, the base-ten number 12 is equal to 2×1+1×10. Another way to represent numbers is by using the digits 0 and 1. This is known as a binary number. A binary number is equal to 1 times its first digit (from the right), plus 2 times its second digit, plus 4 times its third digit, and so on. For example, the binary number 11002 is equal to 0×1+0×2+1×4+1×8. Although 12 and 11002 look nothing alike, they are actually equivalent numbers! When we write out binary numbers, we append a subscripted 2 to the end of the number. This is to differentiate a binary number from a base-ten number. We do not need to add a subscripted 10 to the end of base-ten numbers because they are the default. Now that you have some idea of how binary numbers work, try to do some of the examples! You've reached the end How can we improve? General Bug Feature Send Feedback
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https://pmc.ncbi.nlm.nih.gov/articles/PMC9924773/
The sperm mitochondria: clues and challenges - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Advanced Search Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice Anim Reprod . 2023 Feb 13;19(4):e20220131. doi: 10.1590/1984-3143-AR2022-0131 Search in PMC Search in PubMed View in NLM Catalog Add to search The sperm mitochondria: clues and challenges Diego Bucci Diego Bucci 1 Department of Veterinay Medical Sciences, University of Bologna, Bologna, Italy conceptualizationmanuscript writing Find articles by Diego Bucci 1,, Marcella Spinaci Marcella Spinaci 1 Department of Veterinay Medical Sciences, University of Bologna, Bologna, Italy revision of the manuscript Find articles by Marcella Spinaci 1, Ivan Cunha Bustamante-Filho Ivan Cunha Bustamante-Filho 2 Universidade do Vale do Taquari, Lajeado, RS, Brasil revision of the manuscript Find articles by Ivan Cunha Bustamante-Filho 2, Salvatore Nesci Salvatore Nesci 1 Department of Veterinay Medical Sciences, University of Bologna, Bologna, Italy conceptualizationmanuscript writing Find articles by Salvatore Nesci 1 Author information Article notes Copyright and License information 1 Department of Veterinay Medical Sciences, University of Bologna, Bologna, Italy 2 Universidade do Vale do Taquari, Lajeado, RS, Brasil Corresponding author: diego.bucci3@unibo.it Conflicts of interest: The authors have no conflict of interest to declare. Author contributions: DB and SN: Conceptualization and manuscript writing; MS and ICB: Revision of the manuscript. Roles Diego Bucci: conceptualizationmanuscript writing Marcella Spinaci: revision of the manuscript Ivan Cunha Bustamante-Filho: revision of the manuscript Salvatore Nesci: conceptualizationmanuscript writing Received 2022 Dec 20; Accepted 2023 Jan 24; Collection date 2022. Copyright © The Author(s). This is an Open Access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC9924773 PMID: 36819482 Abstract Sperm cells rely on different substrates to fulfil thei energy demand for different functions and diverse moments of their life. Species specific mechanism involve both energy substrate transport and their utilization: hexose transporters, a protein family of facilitative passive transporters of glucose and other hexose, have been identified in spermatozoa of different species and, within the species, their localization has been identified and, in some cases, linked to specific glycilitic enzyme presence. The catabolism of hexose sources for energy purposes has been studied in various species, and recent advances has been made in the knowledge of metabolic strategies of sperm cells. In particular, the importance of aerobic metabolism has been defined and described in horse, boar and even mouse spermatozoa; bull sperm cells demonstrate to have a good adaptability and capacity to switch between glycolysis and oxidative phosphorylation; finally, dog sperm cells have been demonstrated to have a great plasticity in energy metabolism management, being also able to activate the anabolic pathway of glycogen syntesis. In conclusion, the study of energy management and mitochondrial function in spermatozoa of different specie furnishes important base knowledge to define new media for preservation as well as newbases for reproductive biotechnologies. Keywords: sperm metabolism, bioenergetics, hexose uptake Introduction Spermatozoa are highly specialized cells that are aimed at delivering the male DNA into the new generation subjects. To do so, after being produced by the testicle, matured and stored in the epididymis, they are released in the female genital tract, where they begin the long journey to the female gamete in order to reach it and act for the fertilization process. In mammalian sperm, between ejaculation and fertilization, there can also be another important step in sperm life, that is not occurring in natural breeding animals, but usually occurs when artificial reproduction techniques are used: conservation. Irrespective of the imminent fate of sperm cells, it should be stated that they need some energy substrate to adapt to the environment, maintaining homeostasis and movement. This review aims at pointing out the most recent knowledge on sperm metabolism in terms of fuelling supply, utilization of substrates and metabolic strategies, and balance between anaerobic and aerobic pathways. Most of the knowledge on different mammalian species has been presented and reviewed by outstanding research groups and colleagues during the last years (Boguenet et al., 2021; Moraes and Meyers, 2018; Peña et al., 2022; Rodriguez-Gil, 2006; Rodríguez-Gil and Bonet, 2016; Varner et al., 2015) and we also focused on some specific aspects of sperm metabolism (Bucci et al., 2011). We therefore invite the reader to refer also to those review papers to widen the knowledge on the theme. In this review, we will also present some unpublished data from our labs regarding the balance and equilibrium between anaerobic and oxidative metabolism. Energy sources for sperm cells. Not only sugar An interesting paper by Storey (2008) focused on the regulatory and fuelling role of sugars in mammalian sperm life and activity (Storey, 2008). The author thoroughly revised a large number of studies dating back to the 1940s, in which it was first studied and reported how important and precious sugar fuelling was for sperm function. Of particular interest, among the first studies on sperm metabolism, were the researches reported in Storey’s review by Lardy and Philips and co-workers, as well as the impressive work published by Mann (for reviewing and references see (Storey, 2008)); those first experimental work were aimed at defining the role of fuelling sugars for maintaining the most evident sperm function, motility, as well as to maintain sperm fertilizing ability. From that point on, a large interest was given to sugars and their role in sperm preservation under liquid storage conditions. It is well known that different species may rely on hexoses for their metabolism (Fernández-Novell et al., 2004; Medrano et al., 2005; Peña et al., 2022). Nevetheless, other mechanisms of fuelling have been demonstrated over time (Brooks and Mann, 1972, 1973; Hutson et al., 1977; Medrano et al., 2006a) in different species, thus demonstrating that the direct impact of mitochondrial metabolization of energy sources could play a major role in energy obtainment from different species. The uptake of hexose monosaccharides is exerted by different members of a protein family of carriers and these comprises the so called GLUTs (glucose transporters) which have different specificities for the substrates hexoses (Bucci et al., 2011): CLASS I transporters, (GLUT 1, 2, 3, 4, and 14) are mainly glucose transporters (excepting for GLUT2, that transport also fructose); CLASS II transporters (Glut 5, 7, 9, 11), fructose or double affinity transporters; CLASS III transporters (GLUT 6, 8, 10, 12 and HMIT), with hight affinity for glucose and a different structure if compared with CLASS I and II ones. These proteins have been studied in sperm from different species (Angulo et al., 1998; Bucci et al., 2010a, b, 2011; Sung and Moley, 2007) such as human, rat, and bull sperm cells. GLUTs 1, 2, 3, 4, and 5 (Angulo et al., 1998) show species specific localization within sperm head and tail, and each GLUT shows a different distribution within the same species. The immunocytochemistry results were also validated with Western Blot analysis. Our group studied GLUT 1, 2, 3, 5 in horse, donkey, boar and dog sperm cells (Bucci et al., 2010a), defining, with the same experimental design as Angulo, the presence and abundance of glucose transporters (immunocytochemistry and western blotting). Finally, GLUT 8 and 9 were studied in mouse testis and mature spermatozoa (Sung and Moley, 2007) Interesting studies have regarded the relationship between GLUTs activity and the metabolism of sperm cells: in the early 2000s, Rigau and colleagues showed that metabolic plasticity of dog spermatozoa could be related to GLUTs localization (Fernández-Novell et al., 2004; Rigau et al., 2001, 2002); similarly, it was demonstrated that GLUT 3 co-localizes with Hexokinase I in pig spermatozoa(Medrano et al., 2006b), and this could strictly link the activity of the transporter and the enzymes responsible for the metabolization of the transported substance. Finally, we demonstrated that GLUT 3 and 5 in dog spermatozoa (Bucci et al., 2010a) undergo relocalization after incubation under capacitating conditions; again, a swift in metabolic rate of the sperm cells induces a modification of the localization of the suppliers of energy substrates. Monocarboxilate transporters (MCTs) have been recently described in spermatozoa, specifically MCT1 has been identified in the sperm head; these transporters are responsible for transport of pyruvate/lactate and their presence could play a promising role in the production of next-generation sperm preservation extenders (Peña et al., 2022). Different strategies for energy obtainment and mitochondria Since a long time, it was recognised that spermatozoa from different species have different metabolic strategies to obtain energy for their metabolic activity (Bucci et al., 2011; Gibb and Aitken, 2016; Peña et al., 2022; Rodriguez-Gil, 2006; Rodríguez-Gil and Bonet, 2016; Varner et al., 2015); this section will furnish a brief description of the different energy obtainment strategies exerted by different species to sustain sperm function. It is noteworthy to point out that sperm cells may use preferably the anaerobic pathway (glycolysis) or the aerobic one (oxidative phosphorylation) to obtain energy (see Figure 1). Figure 1. Representation of the anaerobic and aerobic energy obtainment pathways. Open in a new tab Boar spermatozoa Boar sperm cells are recognized as a typical phenotype of slow, short-living cells (Marin et al., 2003; Rodriguez-Gil, 2006); this fact is related to the physiology of reproduction in this species, in which the female, thus having a long lasting estrous (2-4 days) has a receptibility peak in the middle 24 hours of the estrous period; it is reported that sperm cells in the female genital tract could live no more than 18 hours (Johnson et al., 2000). Studies on metabolism of boar sperm cells defined these cells as primarily glycolytic, showing up to 95% of anaerobic metabolism, as revealed by mass spectrometry studies (Marin et al., 2003). In addition, the presence of hexokinase and the metabolizing rate of glucose have led the researcher to sustain this dogma (Fernández-Novell et al., 2004; Medrano et al., 2005, 2006b). Anyway, the intervention of mitochondria in boar sperm cells metabolism cannot be discarded, as different studies have shown the presence of an active metabolism of mitochondrial substrates (Brooks and Mann, 1973; Medrano et al., 2006a); in addition, a recent study from our laboratories, carried out using different specific inhibitors of the electron transfer chain, demonstrated that boar spermatozoa have an active mitochondrial metabolism and that mitochondria preferably rely on complex I instead of complex II to oxidize substrates (Nesci et al., 2020). This finding is opening a new interest on boar sperm mitochondria and their actual role in energy supply. Dog spermatozoa Dog spermatozoa are retained as the opposite phenotype of boar sperm: they are fast cells, with a great surviving capacity in the female genital tract (more than 10 days). Again, this situation mirrors the reproductive physiology demands of this species, in which the bitch has a long lasting estrous (till 9 days) and sexual receptivity is not always synchronous with ovulation (up to 11 days from ejaculation to fertilization) (Foutouhi and Meyers, 2022). Therefore, sperm cells from the dog must have the possibility to remain alive and functional for longer time, and evolute different metabolic strategies: high capacity to metabolize glucose, (Fernández-Novell et al., 2004), metabolic plasticity (Bucci et al., 2010a; Rigau et al., 2001, 2002), ability to activate anabolic glycogen synthesis pathways (Ballester et al., 2000) and to further use glycogen for highly demanding functions (Albarracín et al., 2004). Meyers and colleagues (Foutouhi and Meyers, 2022) report that canine spermatozoa demonstrated to have a high capacity to increase their oxidative metabolism when properly treated and that, in case of disruption of oxidative metabolism, they can in a certain way supply with hexose utilization. Horse spermatozoa Horse spermatozoa are mostly oxidative cells: it is reported that their metabolic strategy could be defined as “live fast, dye young”. Several researches from the group of Fernando Peña (Davila et al., 2015, 2016; González-Fernández et al., 2009; Ortiz-Rodriguez et al., 2021; Peña et al., 2022, 2015; Plaza Dávila et al., 2015) have clearly demonstrated that active and fit mitochondria are needed for horse sperm functioning in the proper way; in particular, the approach chosen in these studies, (recently applied also in our labs to better delineate the relationship between sperm motility, ROS production and mitochondrial function (Giaretta et al., 2022)) was selective inhibition of electron transfer chain (ETC) of respiratory complexes to determine their role in ETC function and ROS production. Other studies, by Meyers and colleagues (Darr et al., 2016a , b; Foutouhi and Meyers, 2022; Meyers et al., 2019; Moraes and Meyers, 2018) deepened the study of the role of mitochondria in stallion semen function, using different techniques to determine Oxygen Consumption Rate (OCR) and ATP production. Finally, the group of Zamira Gibb furnished some very interesting clues on mitochondria potentiality (Gibb et al., 2014, 2015; Gibb and Aitken, 2016; Swegen et al., 2016; Varner et al., 2015). These intense studies have demonstrated that an impairment of the mitochondrial function is strongly deleterious for horse sperm function and that the key point to support a good functionality and survival of horse sperm cells is the presence of intact mitochondria. Bull spermatozoa Bull spermatozoa have been the first ones to be studied, as reported in the milestone review by Storey (Storey, 2008). After a relatively long period in which sperm metabolism was not the center of the studies in bull semen, new interest was focused on these features. (Bulkeley et al., 2021; Chatterjee et al., 2001; Contri et al., 2010; Moraes et al., 2021; Thys et al., 2009). Bull sperm cells are probably the most widely used in AI techniques all over the world and, generally speaking, they are easily cryopreserved; as a consequence, we register a great advance in the application of AI techniques in spite of basic research on metabolic features. Bull spermatozoa can rely both on glycolysis and oxidative phosphorylation pathways: in normal conditions the two pathways play an integrated role as expected from somatic cells metabolism (Vishwanath and Shannon, 2000). In these conditions, mitochondria are “coupled” and their functionality guarantees a good function of the metabolic machinery (Bulkeley et al., 2021; Moraes et al., 2021) and sustains motility. Anyway, after cryopreservation, bull sperm mitochondria do not work properly, probably because of cryo-injuries, as we demonstrated in recent research from our labs (Algieri et al., 2022), in which we showed that bull frozen sperm mitochondria are uncoupled, as mitochondrial respiration does not support the ATP synthesis, in contrast with what was observed by other Authors in freshly ejaculated semen (Bulkeley et al., 2021). In another research (under review) we studied the action of different ETC inhibitors on bull frozen sperm cells. The results clearly showed that these cells are more resistant than horse ones to ETC inhibition, and that only inhibition of complex III is able to significantly decrease mitochondrial membrane potential and motility (as observed also in fresh semen) (Bulkeley et al., 2021). These findings together seem to contrast with the fact that frozen semen from bull have overall a good fertilizing ability; we believe that in case of oxidative phosphorylation breakdown, the glycolytic pathway is able to sustain motility and sperm cell homeostasis. Mouse spermatozoa Mouse sperm cells were studied under different aspects, one of these was sperm metabolism, and they were considered for a long time strictly glycolytic cells (Ford, 2006; Krisfalusi et al., 2006; Mukai and Okuno, 2004), as motility is strictly related to an active glycolytic pathway within the cell. After almost a decade from these studies, some insights were focused also in mouse sperm mitochondrial activity (Tourmente et al., 2015), thus demonstrating that mitochondria may have a role in energy production; a recent work by the same Authors (Tourmente et al., 2022), applying novel techniques already used in bull, boar and canine spermatozoa(Foutouhi and Meyers, 2022), demonstrated that mitochondrial ATP production plays a pivotal role in capacitation process in mouse spermatozoa, which shift their metabolism from a highly glycolytic one toward an oxidative one. As a technical note, perhaps some investigation on possible differences in sperm metabolism of the most used mouse strains could reveal the best model for translational reseach. Concluding remarks and future perspectives The study of sperm metabolism has undergone different moments of interest by the scientific community; anyway, this brief review underlines that knowledge on this topic is not only essential, but can represent an interesting research field. New methodologies and instruments have been developed in the last ten years, thus permitting to shift the approach toward more sensible analysis with respect to those available in the past decades. This led to new discoveries and to update some dogma that seemed to be, as per dogma definition, untouchable. Instead, the role of sperm cells mitochondria has grown in importance and the possibility to study more deeply these organelles showed that they have different roles in different species, and could show a really surprising metabolic plasticity that could be well exploited to develop new preservation strategies or to permit a better control of in vitro sperm activation (capacitation and acrosome reaction). Only in mouse sperm mitochondrial metabolism was studied under capacitating conditions (Tourmente et al., 2022), but this approach is becoming really precious also to control capacitation in species in which in vitro fertilization IVF works well, such as bovine and porcine, and also in species in which only recently IVF protocols have been described and actuated (Felix et al., 2022). The knowledge of basic sperm metabolism of each species, and the possibility to study at individual level the metabolic features will be of absolute interest in the future for the formulation of new extenders; current research is looking for new preservation strategies (Gibb et al., 2015; Rizkallah et al., 2022) possibly avoiding the need to cool semen to too low temperature. This technique could have a great impact on sperm preservation and business, but new extenders should be formulated in order to get the best results in terms of sperm survival, bacterial growth control and fertility. Acknowledgements This research was supported by the European Commission - NextGenerationEU - ALMArie CURIE 2021 SupER - University of Bologna funded by Italian Ministry of University and Research (D.M. 737/2021)(CUP: J45F21001470005) and by EDITAL FAPERGS/CAPES 06/2018. Funding Statement Financial support: DB was financially supported by the European Commission - NextGenerationEU - ALMArie CURIE 2021 SupER - University of Bologna funded by Italian Ministry of University and Research (D.M. 737/2021) (CUP: J45F21001470005) and by EDITAL FAPERGS/CAPES 06/2018. Footnotes Financial support: DB was financially supported by the European Commission - NextGenerationEU - ALMArie CURIE 2021 SupER - University of Bologna funded by Italian Ministry of University and Research (D.M. 737/2021) (CUP: J45F21001470005) and by EDITAL FAPERGS/CAPES 06/2018. How to cite: Bucci D, Spinaci M, Bustamante-Filho IC, Nesci S. The sperm mitochondria: clues and challenges. Anim Reprod. 2022;19(4):e20220131. References Albarracín JL, Fernández-Novell JM, Ballester J, Rauch MC, Quintero-Moreno A, Peña A, Mogas T, Rigau T, Yañez A, Guinovart JJ, Slebe JC, Concha II, Rodríguez-Gil JE. 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[DOI] [PMC free article] [PubMed] [Google Scholar] Varner DD, Gibb Z, Aitken RJ. Stallion fertility: a focus on the spermatozoon. Equine Vet J. 2015;47(1):16–24. doi: 10.1111/evj.12308. [DOI] [PubMed] [Google Scholar] Vishwanath R, Shannon P. Storage of bovine semen in liquid and frozen state. Anim Reprod Sci. 2000;62(1-3):23–53. doi: 10.1016/S0378-4320(00)00153-6. [DOI] [PubMed] [Google Scholar] Articles from Animal Reproduction are provided here courtesy of Brazilian College of Animal Reproduction ACTIONS View on publisher site PDF (369.1 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Energy sources for sperm cells. Not only sugar Different strategies for energy obtainment and mitochondria Concluding remarks and future perspectives Acknowledgements Funding Statement 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
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https://www.mathsisfun.com/algebra/standard-form.html
Algebra Index Show Ads Hide Ads | About Ads We may use Cookies Standard Form What is "Standard Form"? that depends on what you are dealing with! I have gathered some common "Standard Form"s here for you.. Note: Standard Form is not the "correct form", just a handy agreed-upon style. You may find some other form to be more useful. Standard Form of a Decimal Number In Britain this is another name for Scientific Notation, where you write down a number this way: In this example, 5326.6 is written as 5.3266 × 103, because 5326.6 = 5.3266 × 1000 = 5.3266 × 103 In other countries it means "not in expanded form" (see Composing and Decomposing Numbers): | | | --- | | 561 | 500 + 60 + 1 | | Standard Form | Expanded Form | Standard Form of an Equation The "Standard Form" of an equation is: (some expression) = 0 In other words, "= 0" is on the right, and everything else is on the left. Example: Put x2 = 7 into Standard Form Answer: x2 − 7 = 0 Standard Form of a Polynomial The "Standard Form" for writing down a polynomial is to put the terms with the highest degree first (like the "2" in x2 if there is one variable). Example: Put this in Standard Form: 3x2 − 7 + 4x3 + x6 The highest degree is 6, so that goes first, then 3, 2 and then the constant last: x6 + 4x3 + 3x2 − 7 Also, within each term, it is nice to have the variables in alphabetical order (if it does not make things more confusing): Example: Put this in Standard Form: yzx2 + 4yx3 The highest degree is 3, so that goes first, also put the variables in alphabetical order 4x3y + 3x2yz Standard Form of a Linear Equation The "Standard Form" for writing down a Linear Equation is Ax + By = C A shouldn't be negative, A and B shouldn't both be zero, and A, B and C should be integers. Example: Put this in Standard Form: y = 3x + 2 Bring 3x to the left: −3x + y = 2 Multiply all by −1: 3x − y = −2 Note: A = 3, B = −1, C = −2 This form: Ax + By + C = 0 is sometimes called "Standard Form", but is more properly called the "General Form". Standard Form of a Quadratic Equation The "Standard Form" for writing down a Quadratic Equation is (a not equal to zero) Example: Put this in Standard Form: x(x−1) = 3 Expand "x(x−1)": x2 − x = 3 Bring 3 to left: x2 − x − 3 = 0 Note: a = 1, b = −1, c = −3 Standard Form of a Circle Equation With a circle like this: The Standard Form is this: (x−a)2 + (y−b)2 = r2 See Circle Equations for more details. Algebra Index Copyright © 2024 Rod Pierce
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https://math.stackexchange.com/questions/4836391/finish-my-proof-that-there-are-infinitely-many-pairs-of-distinct-prime-numbers
contest math - Finish my proof that there are infinitely many pairs of distinct prime numbers $(p, q)$ such that $p \mid (2^{q - 1} - 1)$ and $q \mid (2^{p-1} - 1)$. - 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 Finish my proof that there are infinitely many pairs of distinct prime numbers (p,q)(p,q) such that p∣(2 q−1−1)p∣(2 q−1−1) and q∣(2 p−1−1)q∣(2 p−1−1). Ask Question Asked 1 year, 9 months ago Modified1 year, 9 months ago Viewed 416 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This is from the Romania IMO Team Selection Test, 2009, Day 7, Problem 3. Here is my work so far. Problem: Prove that there are infinitely many pairs of distinct prime numbers (p,q)(p,q) such that p∣(2 q−1−1)p∣(2 q−1−1) and q∣(2 p−1−1)q∣(2 p−1−1). Solution: Let p p be an arbitrary prime divisor of the Fermat number F n=2 2 n+1 F n=2 2 n+1, and let q q be an arbitrary prime divisor of F n+1=2 2 n+1+1 F n+1=2 2 n+1+1 not equal to p p. By Lucas, we know that 2 n+2∣(p−1)2 n+2∣(p−1), so (2 2 n+2−1)∣(2 p−1−1)(2 2 n+2−1)∣(2 p−1−1). Then, since 2 2 n+2−1=(2 2 n+1−1)(2 2 n+1−1)2 2 n+2−1=(2 2 n+1−1)(2 2 n+1−1), we know that F n+1∣(2 p−1−1)F n+1∣(2 p−1−1). Since q∣F n+1 q∣F n+1 by definition, the first condition is met. Similarly, by Lucas, we know that 2 n+3∣(q−1), so (2 2 n+3−1)∣(2 q−1−1). Note that (2 2 n+1)∣(2 2 n+3−1) since (x+1)∣(x 8−1), so F n∣(2 q−1−1). Since p∣F n by definition, the second condition is met. We will let n vary. To generate infinitely many pairs (p,q), what remains to be shown is that not all Fermat numbers are divisible by a finite set of primes. —————————— Sadly, this last step seems like such a minutiae that it might even be ignored, but I can’t seem to prove it. Zsigmondy’s Theorem doesn’t work because of the +1, not −1. Any help? elementary-number-theory contest-math Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Dec 31, 2023 at 9:25 Bill Dubuque 284k 42 42 gold badges 339 339 silver badges 1k 1k bronze badges asked Dec 31, 2023 at 4:07 solaskysolasky 220 1 1 silver badge 8 8 bronze badges 8 2 Welcome to Math SE. FYI, an Approach0 search shows Question on existence of infinite pairs of primes, but this uses a rather different method to solve the problem, which is why I'm not voting to close your question as a duplicate of that one. There's also the AoPS threads Nice NT - Infinitely many primes. and ...John Omielan –John Omielan 2023-12-31 05:11:43 +00:00 Commented Dec 31, 2023 at 5:11 1 (cont.) Number theory, divisibility with primes wanted, which basically use your method. In addition, the AoPS thread p divides 2^{q-1}-1 and q divides 2^{p-1}-1 uses several methods, including the Fermat numbers you're using. It also states the source is "Romania TST 7 2009, Problem 3", matching what you provided.John Omielan –John Omielan 2023-12-31 05:13:49 +00:00 Commented Dec 31, 2023 at 5:13 Mersenne Prime Variation? See Tower of Hanoi. 2 p−1−1=(2 p−1)−1 2.Hudjefa –Hudjefa 2023-12-31 06:22:33 +00:00 Commented Dec 31, 2023 at 6:22 Do all Olympiad problems have the same format? You have to know the known knowns ... thoroughly (RIP Donald Rumsfeld).Hudjefa –Hudjefa 2023-12-31 06:29:20 +00:00 Commented Dec 31, 2023 at 6:29 Duplicate of Infinitely many primes derived from coprimality of Fermat numbers. and many others.Bill Dubuque –Bill Dubuque 2023-12-31 09:26:25 +00:00 Commented Dec 31, 2023 at 9:26 |Show 3 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Never mind, I think figured out how to clear my small doubt. Note that the Fermat numbers are pairwise relatively prime. This is because F n+1=F n F n−1⋯F 0+2. To prove this, we will use induction. Note that 5=F 1=F 0+2, so the base case is clear. Then, assume that F k=F k−1 F k−2⋯F 0+2 for some integer k. Subtracting 1 from both sides and then multiplying both sides by F k, we get that (F k−1)2=F k F k−1⋯F 0+1. However, note that F k+1−1=(F k−1)2, so F k+1=F k F k−1⋯F 0+2 as desired. Then, assume for the sake of contradiction that p∣F m and p∣F n for m<n. The Fermat numbers are odd, so p is odd. Then, 0≡F n≡F n−1 F n−2⋯F 0+2≡2(mod p), since 0≤m≤n−1. This is a contradiction, so the Fermat numbers are relatively prime. Once we have shown that the Fermat numbers are relatively prime, then it is clear that infinitely many primes (p,q) can be generated, because each F k+1 contains a prime factor that none of the previous Fermat numbers had. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Dec 31, 2023 at 6:02 solaskysolasky 220 1 1 silver badge 8 8 bronze badges 1 1 FYI, there's what I believe is an easier way to show the Fermat numbers are pairwise relatively prime. Assume the prime p is a factor of F m=2 2 m+1 and F n=2 2 n+1, with m<n. Then from p∣F m we have 2 2 m≡−1(mod p)→2 2 m+1≡1(mod p). Squaring a total of n−m times leads to 2 2 n≡1(mod p), contradicting 2 2 n≡−1(mod p) from p∣F n.John Omielan –John Omielan 2023-12-31 06:34:55 +00:00 Commented Dec 31, 2023 at 6:34 Add a comment| This answer is useful 1 Save this answer. Show activity on this post. How I would solve: If there are infinitely many positive integers n such that both 2 2 n+1 and 2 2 n+1+1 are both prime: Then let n be any sufficiently large positive integer such that p=2 2 n+1 and q=2 2 n+1+1 are both prime. Then 2 p−1≡q 1 and 2 q−1≡p 1 and the result follows for Case 1. [Indeed, p−1=2 2 n [namely 2 n+2 multiplied by many more factors of 2 [as 2 n>>n+2 for n large enough]] and q−1=2 2 n+1. Furthermore, 2 2 n+2≡q 1 gives a desired 2 p−1≡q 1 [as p−1 is 2 n+2 times many more factors of 2, so 2 p−1 is 2 2 n+2 squared many times in particular an integral power of 2 2 n+2]. And likewise, 2 2 n+1≡p 1 gives 2 q−1≡p 1.] Otherwise there are infinitely many positive integers n such that 2 2 n+1 is composite. So for each such n, let p n and q n be distinct primes that both divide 2 2 n+1. Then both equations 2 p n−1≡q n 1 and 2 q n−1≡p n 1 hold. [Indeed, as p n and q n are primes dividing 2 2 n+1, Claim 0 [below] implies the following: 2 n+1 must divide both (p n−1) and (q n−1). This however gives 2 p n−1 an integral power of 2 2 n+1 [because 2 n+1 divides p n−1] and as the equation 2 2 n+1≡q n 1 holds, it follows that the equation 2 p n−1≡q n 1 also holds. Likewise, 2 q n−1 an integral power of 2 2 n+1 and so the equation 2 q n−1≡p n 1 also holds.] Finish this case 2. by noting that the Fermat numbers are relatively prime to each other, so the pairs (p n,q n); n a positive integer satisfying 2 2 n+1 composite; p n,q n distinct primes dividing 2 2 n+1; are distinct pairs and are in fact disjoint pairs. Claim 0:Let n be a positive integer, and let p n be a prime that divides 2 2 n+1. Then 2 n+1 divides p n−1. Indeed, let b n be the smallest positive integer such that the equation 2 b n≡p n 1. Then it follows that b n must divide (p n−1). However, we claim that b n must infact be 2 n+1. [Indeed, for any other positive integer y, the equation 2 y≡p n 1 holds iff b n divides y. Now from this b n does not divide y=2 n [as p n divides 2 y+1=2 2 n+1 so 2 y=2 2 n≡p n−1≠1], but b n does divide 2 n+1 [as 2 2 n+1=(2 y)2≡p n(−1)2=1]. So b n must be a power of 2 [as it divides another power of 2 namely 2 n+1] and must infact be 2 n+1.[Indeed, as 2 2 n≡p n−1≠1 it follows that b n does not divide 2 n. This together with b n being a power of 2 and b n dividing 2 n+1, leaves b n=2 n+1 as the only possibility.]] So from this it follows from the above that, as b n divides p n−1 and b n is 2 n+1, that indeed, b n=2 n+1 divides p n−1. And so Claim 0 follows. √ I am sorry I could not fix/follow what you had above in the OP, I was not seeing how your idea could be made to work. In particular, going by the notation in your OP, I'm still not seeing why 2 n+2 divides p−1. [Yes, 2 2 n+2≡p 1 but that does not imply 2 n+2 divides p−1.] Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jan 2, 2024 at 23:34 answered Dec 31, 2023 at 4:39 MikeMike 23.5k 4 4 gold badges 24 24 silver badges 40 40 bronze badges Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory contest-math See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 4Question on existence of infinite pairs of primes 3Infinitely many primes derived from coprimality of Fermat numbers. Related 14Prime divisors of Fermat numbers 2 2 n+1 have form k⋅2 n+2+1 1"If m divides two Fermat numbers, m divides 2." Why? 3Assume there exists infinitely many prime numbers l such that 2 l−1 is NOT a prime, show the following 3Number Theory: Prove there are infinitely many primes p satisfying n∣(p−1) 3Prove that A={4 n+3∣n∈N} contains infinitely many prime numbers 3Let p be a prime divisor of the Fermat number F n=2 2 n+1, and let a=2 2 n−2(2 2 n−1−1). Prove that o r d p(a)=2 n+2. 5Show that there are infinitely many composite integers n such that n∣(3 n−1−2 n−1) Hot Network Questions How do you create a no-attack area? Interpret G-code What is the meaning and import of this highlighted phrase in Selichos? Is encrypting the login keyring necessary if you have full disk encryption? Does the Mishna or Gemara ever explicitly mention the second day of Shavuot? Can you formalize the definition of infinitely divisible in FOL? What were "milk bars" in 1920s Japan? 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https://www.crns.ca/wp-content/uploads/2023/04/Urinary-Tract-Infection-Cystitis-and-Pyelonephritis-Adult-CDT.pdf
Urinary Tract Infection (Cystitis and Pyelonephritis): Adult Genitourinary Clinical Decision Tools for RNs with Additional Authorized Practice [RN(AAP)s] Effective Date: October 1, 2024 Background A urinary tract infection (UTI) can occur in the urethra, bladder, ureter, or kidney, can be classified as uncomplicated or complicated, and can be recurrent (Huether, 2019). Acute cystitis is an inflammation of the bladder epithelium and the most common site of a UTI, which is usually caused by bacteria from intestinal flora (Huether, 2019). Pyelonephritis is characterized by an infection of one or both of the renal parenchyma and renal pelvis, often producing localized flank or back pain combined with systemic symptoms such as fever, chills, and nausea (Huether, 2019). Bacteria may be introduced into the urinary tract from fecal contamination or unprotected intercourse, particularly anal (Conner, Thomas, & Porter, 2019). Fungi (especially Candida species) should be suspected in complicated UTI cases that fail to respond to antibiotic therapy (Conner et al., 2019). The most common bacterial pathogen in males and females is Escherichia coli (E. coli), the causative organism in 80% to 90% of cases of uncomplicated cystitis. Enterobacteriaceae (e.g., Klebsiella, Proteus mirabilis, Enterobacter, Pseudomonas, Serratia) are less common, but often the cause of complicated UTI cases including pyelonephritis (Conner et al., 2019). 2 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Immediate Consultation Requirements The RN(AAP) should seek immediate consultation from a physician/NP when any of the following circumstances exist: • altered level of consciousness, • sepsis (e.g., fever, tachycardia, hypotension, tachypnea, altered mental status), • pregnancy, • renal failure, • recurrent pyelonephritis (within one year) after a course of appropriate therapy, • immunocompromised client (e.g., diabetic, multiple sclerosis), • history of renal stones or other urologic concerns (e.g., urinary tract obstruction, neurogenic bladder), • prior urologic surgery, • gross hematuria, • acute abdominal pain, • immunocompromised client, • new onset edema, and/or • complicated UTIs (e.g., male client; client with an indwelling catheter) (Interprofessional Advisory Group [IPAG], personal communication, October 2, 2019). Classification Uncomplicated UTI • A UTI is considered uncomplicated when it occurs in clients who have a normal, unobstructed genitourinary tract; no history of recent instrumentation, and whose symptoms are confined to the lower urinary tract. • Uncomplicated UTIs are most common in young, sexually active women. Complicated UTI • A complicated UTI is an infection of the lower or upper urinary tract in the presence of an anatomic abnormality, a functional abnormality, or a urinary catheter. • All males and pregnant females are considered to present with complicated UTIs. Recurrent UTI • A recurrent UTI is a symptomatic UTI that follows resolution of an earlier episode, usually after appropriate treatment. No single definition of the frequency of recurrent UTI exists but two uncomplicated, culture positive UTIs in 6 months or three or more infections per year are commonly understood to be “recurrent”. • Recurrence may be related to an anatomical cause and warrant further investigation. 3 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Pyelonephritis • Pyelonephritis has a wide spectrum of presentation, from mild illness to septic shock. It is more common in women because it is frequently caused by fecal flora that colonize the vaginal introitus and subsequently ascend along the urinary tract to the kidney. • Most cases (75% to 95%) are caused by ascension of gram-negative organisms from the bladder including Escherichia coli (over 80% of cases), Proteus mirabilis, Klebsiella, and Pseudomonas aeruginosa. (Anti-infective Review Panel, 2019; Conner et al., 2019; Huether, 2019; RxFiles Academic Detailing Program, 2021) Predisposing and Risk Factors Predisposing and risk factors for acute cystitis include: • female anatomy (e.g., short urethra), • sexually active, • pregnancy, • recent antibiotic use, • spermicide use, • estrogen deficiency (e.g., postmenopausal females), • indwelling catheter, • diabetes mellitus, • immunosuppression, • inadequately treated prostatitis, • neurogenic bladder, and • urinary tract obstruction (e.g., benign prostatic hypertrophy, renal stones) (Conner et al., 2019; Huether, 2019). Predisposing and risk factors for pyelonephritis include: • female anatomy (e.g., short urethra), • recent untreated or undertreated UTI, • episode of acute pyelonephritis within the last 12 months, • older adults who are institutionalized or hospitalized, • immunocompromised clients, • anatomical abnormalities (e.g., ureterovesical reflux), • urinary obstruction, • stress incontinence, • multiple or recurrent UTIs, • renal disease, • renal trauma, • pregnancy, • metabolic disorders (e.g., diabetes mellitus), 4 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association • indwelling catheter, • neurogenic bladder, • female sexual trauma, or • urogenital instrumentation (Conner et al., 2019; Huether, 2019). Health History and Physical Exam Subjective Findings The circumstances of the presenting complaint should be determined. For acute cystitis these include: • sudden onset of symptoms; • dysuria (pain or burning during urination); • urgency (a sudden, almost uncontrollable need to urinate); • frequency; • sensation of incomplete bladder emptying; • blood in urine; • lower abdominal pain or cramping; • offensive odour of urine; • nocturia; • altered mental status or vague abdominal discomfort, especially in the elderly; • sudden onset of urinary incontinence; and/or • low back pain (Anti-infective Review Panel, 2019; Conner et al., 2019; Huether, 2019). For pyelonephritis these include: • fever (> 38°C oral), rigours, or chills; • nausea and vomiting; • abdominal, back, flank, or groin pain may be present (e.g., costovertebral angle tenderness); • headache; • malaise; • dysuria, frequency, and urgency may or may not be present; • fatigue; • diarrhea; • cloudy foul-smelling urine; or • mental status changes or general deterioration, especially in the elderly (Conner et al., 2019; Huether, 2019). 5 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association For male clients, enquire as to all of the above symptoms plus symptoms that may indicate prostatitis, including: • hesitancy, • slow urinary stream, • dribbling of urine, • nocturia, • suprapubic tenderness and urethral discomfort (Conner et al., 2019). Objective Findings Physical findings for acute cystitis may include suprapubic tenderness on palpation of the abdomen. The signs and symptoms of pyelonephritis may include: • moderately to acutely ill appearance; • fever (> 38°C oral) and chills; • tachycardia; • decreased blood pressure (e.g., septic shock); • mild, generalized abdominal discomfort may be present; • mild to severe pain with deep abdominal palpation of kidney may be present; and/or • costovertebral angle tenderness to percussion may be present (Conner et al., 2019; Huether, 2019). Urinary Tract Infection Symptoms The following chart provides guidance on symptom severity. Moderate to severe symptoms (or infections) require immediate consultation with a physician/NP. Local Symptoms General Symptoms Systemic Symptoms Systemic Symptoms These symptoms are typical of UTI (acute cystitis). These symptoms are suggestive of mild pyelonephritis. These symptoms are suggestive of moderate to severe pyelonephritis. These symptoms are suggestive of severe pyelonephritis. • dysuria, • frequency, • urgency, and/or • bladder tenderness. • fever, • flank pain, and/or • nausea. • fever, • rigours, • dehydration, and/or • vomiting. • circulatory failure, • organ dysfunction, • organ failure, and/or • altered level of consciousness. (Conner et al., 2019; IPAG, personal communication, October 2, 2019) 6 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Differential Diagnosis The following should be considered as part of the differential diagnosis for acute cystitis: • tumours, • pyelonephritis, • vaginitis, • sexually transmitted infections, • pelvic inflammatory disease, • interstitial cystitis, • prostate disorders (e.g., benign prostatic hypertrophy, prostatitis), or • infection in other sites of the genitourinary tract (e.g., epididymis) (Conner et al., 2019; Huether, 2019). The following should be considered as part of the differential diagnosis for pyelonephritis: • acute cholecystitis with fever, • lower lobe pneumonia, • perforated abdominal organ, • appendicitis, • acute pancreatitis, • pelvic inflammatory disease, • renal colic (kidney stone), • renal cancer, • bladder obstruction, • musculoskeletal pain, • shingles, • prostatitis, • ectopic pregnancy, or • diverticulitis (Johnson & Russo, 2018). Making the Diagnosis Presumptive diagnosis of UTI in female clients can be made if either dysuria and frequency are present or the dipstick urinalysis shows a positive result for nitrite or leukocyte esterase. Women with recurrent UTI may accurately self-diagnose. Urinary tract infections are less common in males and alternate diagnoses should be considered. Elderly clients with UTIs may present with atypical symptoms (e.g., delirium) (Conner et al., 2019). Clinical diagnosis of acute pyelonephritis is based on a combination of history and physical findings suggestive of pyelonephritis (fever, chills, flank pain, nausea, vomiting, costovertebral angle tenderness), recent or concurrent cystitis, and urine dipstick result positive for leukocytes and nitrites (Conner et al., 2019; Huether, 2019). The diagnosis is confirmed with a positive urine culture result. 7 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Investigations and Diagnostic Tests Diagnostic tests for UTIs in males and females include dipstick urinalysis and urine culture. In acute cystitis, a dipstick urinalysis typically indicates pyuria (leukocyte esterase) and the presence of red blood cells (Conner et al., 2019). The presence of nitrites in the urine are typically the result of the reduction of nitrates by gram-negative bacteria (e.g., E. coli). Gram-positive bacteria such as S. saprophyticus and Enterococcus do not produce nitrites (Huether, 2019). In acute cystitis, a urine culture is only indicated if the diagnosis is unclear or if the client has recurrent infections (greater than three infections per year) and antibiotic resistance is suspected. Other situations where urine culture is recommended include all complicated UTI cases, persistent or recurrence of symptoms within one month of empirical antibiotic therapy, or client presents with atypical symptoms (Anti-infective Review Panel, 2019; Conner et al., 2019). The diagnosis of acute pyelonephritis is made by urine culture, urinalysis, and clinical signs and symptoms (Huether, 2019). A midstream urine for urinalysis is often positive for bacteria, proteinuria, leukocyte esterase, urinary nitrites, hematuria, pyuria, and white blood cell casts (Conner et al., 2019). However, the urinalysis results may be altered by: current antibiotic therapy, esterase in the presence of kidney stones, tumours, urethritis, or contamination, creating a false positive result. Furthermore, the nitrites may provide a false negative result if the bladder has not been full for at least four hours (e.g., diuretic use), and if the uropathogen is a gram-positive bacteria (e.g., S. saprophyticus) (Huether, 2019). The gold standard confirmatory test of acute pyelonephritis is a urine culture (Johnson & Russo, 2018). The urine culture typically will show ≥ 10,000 CFU/mL of a single uropathogen. A urine culture must be collected in all pregnant females. Management and Interventions Goals of Treatment The primary goals of immediate treatment are to relieve symptoms, eradicate infection, prevent ascending infection, or in the case of pyelonephritis damage to the kidneys, prevent recurrent infection, and prevent complications (Huether, 2019). Early or mild pyelonephritis infections may be treated on an outpatient basis in those clients who: • are hemodynamically stable, • are young and otherwise healthy, • have mild general symptoms (low-grade fever, normal, or slightly elevated peripheral leukocyte count without vomiting), • are able to tolerate oral medications and fluids, • have stable coexisting medical conditions, and • for whom adequate follow-up has been arranged and is likely to occur (Anti-infective Review Panel, 2019; Johnson & Russo, 2018). 8 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Non-Pharmacological Interventions The RN(AAP) should recommend, as appropriate, the following non-pharmacological options: • Drink eight to 10 glasses of water and other fluids every day. • Empty bladder as soon as the urge is felt. • Females should wipe from front to back after using the toilet. • Females should void before and immediately after intercourse (Conner et al., 2019; Johnson & Russo, 2018). Pharmacological Interventions The pharmacological interventions recommended for the treatment of UTIs are in accordance with the Anti-infective Guidelines for Community-acquired Infections (Anti-infective Review Panel, 2019) and the RxFiles: Drug comparison charts (Rx Files Academic Detailing Program, 2021). Treatment of recurrent cases of acute cystitis should be guided by culture and sensitivity results. All cases of complicated UTIs should be referred to a physician/NP. Treatment options for acute cystitis in pregnancy are listed on page 9. Oral Antibiotics Drug Dose Route Frequency Duration Adult (uncomplicated acute cystitis) Nitrofurantoin (Macrobid) 100 mg p.o. b.i.d. 5 days OR Sulfamethoxazole- Trimethoprim 400/80 mg, 2 tabs p.o. b.i.d. 3 days OR Sulfamethoxazole- Trimethoprim 800/160 mg, 1 tab p.o. b.i.d. 3 days OR Trimethoprim 200 mg p.o. once daily 3 days OR Fosfomycin 3 g mixed in ½ cup water p.o. once n/a 9 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Drug Dose Route Frequency Duration Adult (pregnant client) Amoxicillin 500 mg p.o. q8h 7 days OR Cephalexin 250 to 500 mg p.o. q.i.d. 7 days Drug Dose Route Frequency Duration Adult (pregnant client) OR Nitrofurantoin (Macrobid) 100 mg p.o. b.i.d. 5- 7 days OR Fosfomycin 3 g mixed in ½ cup water p.o. once n/a Adult (First line for mild pyelonephritis) Sulfamethoxazole- Trimethoprim (SMX/TMP) 400/80 mg, 2 tabs p.o. b.i.d. 10-14 days Drug Dose Route Frequency Duration OR Sulfamethoxazole- Trimethoprim (SMX/TMP) 800/160 mg, 1 tab p.o. b.i.d. 10-14 days OR Trimethoprim 100 mg p.o. b.i.d. 10-14 days Alternatives to SMX/TMP and trimethoprim should be considered when local resistance is anticipated to be > 20% and if there is a history of recurrent UTI and previous treatment with SMX/TMP. Treat all pregnant women. Do not prescribe nitrofurantoin in pregnancy at term (36-42 weeks gestation) or during labour. 10 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Drug Dose Route Frequency Duration Adult (Second line for mild pyelonephritis) Ciprofloxacin 500 mg p.o. q12h 7-14 days OR Amoxicillin/ Clavulanate 500 mg p.o. q8h 7-14 days OR Amoxicillin/ Clavulanate 875 mg p.o. q12h 7-14 days Analgesics and Antipyretics Drug Dose Route Frequency Duration Adult Acetaminophen 500-1000 mg (maximum dose of 4 g/day) p.o. q4-6h prn 5-7 days AND/ OR Ibuprofen 400-600 mg (maximum dose of 3200 mg/day) p.o. q6-8h prn 5-7 days Client and Caregiver Education The RN(AAP) provides client and caregiver education as follows: • Counsel on appropriate use of medications (dose, frequency, side effects, the need to complete the entire course of treatment, etc.). • Instruct proper perineal hygiene (wiping from front to back) to prevent recurrence. • Wear cotton underclothes versus nylon to prevent moisture accumulation. • Inform that harsh soaps or feminine hygiene products can irritate the urethra and encourage showers instead of tub baths or bubble baths to prevent urethral irritation. • Encourage barrier protection (e.g., condom) during intercourse. • Advise to avoid the use of spermicide. • Advise to report recurrence of symptoms immediately. • Teach double void (void, wait three minutes, void again) in clients with post void residual. • Increase fluid intake to achieve light coloured urine. • Advise to establish a voiding routine every two to three hours (Conner et al., 2019; Huether, 2019). 11 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association Monitoring and Follow-Up Non-pregnant young or middle-aged clients diagnosed with acute cystitis require no follow-up if symptoms resolve with treatment. Pregnant clients should be seen following treatment to ensure infection has resolved, where typically a urine for culture and sensitivity is repeated (Connor et al., 2019). The RN(AAP) should reassess the client diagnosed with pyelonephritis in 24-48 hours to determine clinical response to therapy. If there is no improvement after 72 hours of therapy or if symptoms are worsening, consult a physician/NP. The client should also be reviewed after completion of therapy to assess for continuing symptoms. If the client is asymptomatic, except for pregnant clients, there is no need to repeat the urinalysis and culture. Complications The following complications may be associated with acute cystitis: • pyelonephritis or sepsis, • renal abscess, • acute urinary outlet obstruction, and • recurrent infection (Conner et al., 2019). The following complications may be associated with pyelonephritis: • acute or chronic renal failure (rare following acute pyelonephritis), • renal abscess, • sepsis, • focal renal scarring, • renal papillary necrosis, • emphysematous pyelonephritis and/or cystitis, • respiratory dysfunction, and • preterm labour and low birth weight infant in pregnancy (Conner et al., 2019; Huether, 2019). Referral Refer to a physician/NP if client presentation is consistent with those identified in the Immediate Consultation Requirements section, clients who do not respond to prescribed treatment, or clients who have recurrent UTIs (IPAG, personal communication, October 2, 2019). 12 UTI (Cystitis and Pyelonephritis - Adult ©2024 College of Registered Nurses of Saskatchewan GU | Urinary Tract Infection (Cystitis and Pyelonephritis) - ©2019 Saskatchewan Registered Nurses Association References Anti-infective Review Panel. (2019). Anti-infective guidelines for community-acquired infections. MUMS Guideline Clearinghouse. Canadian Pharmacists Association. (2021). CPS drug information. Conner, D., Thomas, D., & Porter, B. (2019). Urinary tract disorders. In L. Dunphy, J. Winland-Brown, B. Porter, & D. Thomas (Eds.), Primary care: The art and science of advanced practice nursing – an interprofessional approach (5th ed., pp. 628-649). F. A. Davis. Huether, S. (2019). Alterations of renal and urinary tract function. In K. McCance & S. Huether (Eds.), Pathophysiology: The biologic basis for disease in adults and children (8th ed., pp. 1246-1277). Elsevier. Johnson, J. R., & Russo, T. A. (2018). Acute pyelonephritis in adults. New England Journal of Medicine, 378(1), 48 -59. doi.org/10.1056/NEJMcp1702758 Rx Files Academic Detailing Program. (2021). RxFiles: Drug comparison charts (13th ed.). Saskatoon Health Region. NOTICE OF INTENDED USE OF THIS CLINICAL DECISION TOOL This CRNS Clinical Decision Tool (CDT) exists solely for use in Saskatchewan by an RN with additional authorized practice as granted by the CRNS. The CDT is current as of the date of its publication and updated every three years or as needed. A member must notify the CRNS if there has been a change in best practice regarding the CDT. This CDT does not relieve the RN with additional practice qualifications from exercising sound professional RN judgment and responsibility to deliver safe, competent, ethical and culturally appropriate RN services. The RN must consult a physician/NP when clients’ needs necessitate deviation from the CDT. While the CRNS has made every effort to ensure the CDT provides accurate and expert information and guidance, it is impossible to predict the circumstances in which it may be used. Accordingly, to the extent permitted by law, the CRNS shall not be held liable to any person or entity with respect to any loss or damage caused by what is contained or left out of this CDT. CRNS © This CDT is to be reproduced only with the authorization of the CRNS.
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https://en.wikipedia.org/wiki/Bowfin
Jump to content Bowfin العربية Asturianu Azərbaycanca Беларуская Български Català Cebuano Čeština Deutsch Español Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն Hrvatski Bahasa Indonesia Íslenska Italiano עברית ქართული Кыргызча Кырык мары Lietuvių Ligure Magyar مصرى Nederlands 日本語 Norsk bokmål Occitan Polski Русский Simple English Slovenčina Suomi Svenska ไทย Türkçe Українська Tiếng Việt Winaray 中文 Edit links From Wikipedia, the free encyclopedia Species of ray-finned fish For the Balao-class submarine, see USS Bowfin. | Bowfin | | --- | | | | | Bowfin in aquarium | | | Conservation status | | | Least Concern (IUCN 3.1) | | | Unranked (NatureServe) | | | Scientific classification | | | Kingdom: | Animalia | | Phylum: | Chordata | | Class: | Actinopterygii | | Clade: | Halecomorphi | | Order: | Amiiformes | | Family: | Amiidae | | Genus: | Amia | | Species: | A. calva | | Binomial name | | | Amia calva Linnaeus, 1766 | | | Synonyms | | | Species Amia occidentalis DeKay 1842 Amia marmorata Valenciennes 1847 Amia ornata Valenciennes 1847 Amia viridis Valenciennes 1847 Amia cinerea Valenciennes 1847 Amia reticulata Valenciennes 1847 Amia canina Valenciennes 1847 Amia lintiginosa Valenciennes 1847 Amia subcoerulea Valenciennes 1847 Amia thompsonii Duméril 1870 Amia piquotii Duméril 1870 Amiatus calvus (Linnaeus 1766) | | The ruddy bowfin (Amia calva) is a ray-finned fish native to North America. Common names include mudfish, mud pike, dogfish, grindle, grinnel, swamp trout, and choupique. It is regarded as a relict, being one of only two surviving species of the Halecomorphi, a group of fish that first appeared during the Early Triassic, around 250 million years ago. The bowfin is often considered a "living fossil" because they have retained some morphological characteristics of their early ancestors. It is one of two species in the genus Amia, along with Amia ocellicauda, the eyespot bowfin. The closest living relatives of bowfins are gars, with the two groups being united in the clade Holostei. Bowfins are demersal freshwater piscivores, commonly found throughout much of the eastern United States, and in southern Ontario and Quebec. Fossil deposits indicate Amiiformes were once widespread in both freshwater and marine environments across North and South America, Europe, Asia, and Africa. Now, their range is limited to much of the eastern United States and adjacent southern Canada, including the drainage basins of the Mississippi River, Great Lakes, and various rivers exiting in the Eastern Seaboard or Gulf of Mexico. Their preferred habitat includes vegetated sloughs, lowland rivers and lakes, swamps, and backwater areas; they are also occasionally found in brackish water. They are stalking, ambush predators known to move into the shallows at night to prey on fish and aquatic invertebrates such as crawfish, mollusks, and aquatic insects. Like gars, bowfin are bimodal breathers—they have the capacity to breathe both water and air. Their gills exchange gases in the water allowing them to breathe, but they also have a gas bladder that serves to maintain buoyancy, and also allows them to breathe air by means of a small pneumatic duct connected from the foregut to the gas bladder. They can break the surface to gulp air, which allows them to survive conditions of aquatic hypoxia that would be lethal to most other species. The bowfin is long-lived, with age up to 33 years reported. Morphology [edit] The typical length of a bowfin is 50 cm (20 in); females typically grow to 65–70 cm (26–28 in), males to 50–65 cm (20–26 in). They can reach 109 cm (43 in) in length, and weigh 9.75 kg (21.5 lb). Young of the year typically grow to 13–23 cm (5.1–9.1 in) by October. Females tend to grow larger than males. The body of the bowfin is elongated and cylindrical, with the sides and back olive to brown in color, often with vertical bars and dark reticulations or another camouflaged pattern. The dorsal fin has horizontal bars, and the caudal fin has irregular vertical bars. The underside is white or cream, and the paired fins and anal fin are bright green. During larval stage, hatchlings from about 7–10 mm (0.28–0.39 in) total length are black and tadpole-like in appearance. At approximately 25 mm (0.98 in) total length they have been described as looking like miniature placoderms. They grow quickly, and typically leave the nest within 4 to 6 weeks after hatching. Young males have a black eyespot on the base of the tail (caudal peduncle) that is commonly encircled by an orange-yellowish border, while the female's is black, if present at all. It is thought the purpose of the eyespot is to confuse predators, deflecting attacks away from the head of the fish to its tail, which affords the bowfin an opportunity to escape predation. The bowfin is so named for its long, undulating dorsal fin consisting of 145 to 250 rays that runs from the middle of the back to the base of the tail. The skull of the bowfin is made of two layers of skull, the dermatocranium and the chondrocranium. The chondrocranium layer cannot be seen because it is located below the dermal bones. The bowfin skull is made up of 28 fused bones, which compose the dermatocranium. The roof of the mouth is made up of three bones, the ectopterygoid, the palantine, and the vomer. They have two sets of teeth, including one set of larger sharp teeth coming out of the mandibular and premaxillary bones to grasp and control the prey. The other set of teeth, located posteriorly and connected to the hyomandibular bone, is made up of pharyngeal tooth patches, which are used for sorting out nutrients and grinding down larger pieces of food. Another three bones make up the lower jaw: the dentary, the angular, and the surangular. The cranial surface of the skull is made up of the nasals, the antorbital, the lacrimal, the parietal, the intertemporal, the post parietal, the supratemporal, the extra scapular, the post temporal, and the opercular. The entirety of the skull is attached to the girdle through another set of bones. Bowfin are often referred to as "living fossils" or "primitive fish" because they retained some of the primitive characters common to their ancestors, including a modified (rounded externally) heterocercal caudal fin, a highly vascularized gas bladder lung, vestiges of a spiral valve, and a bony gular plate. The bony gular plate is located underneath the head on the exterior of the lower jaw between the two sides of the lower jaw bone. Other distinguishing characteristics include long, sharp teeth, and two protruding tube-like nostrils. Unlike all of the most primitive actinopterygians, the scales of bowfin differ in that they are not ganoid scales, rather they are large, single-layered cycloid scales closer in similarity to more derived teleosts. Fish similar in appearance [edit] Northern snakeheads (Channa argus) are commonly mistaken for bowfin because of similarities in appearance, most noticeably their elongated, cylindrical shape, and long dorsal fin that runs along their backs. Northern snakeheads are piscivorous fish native to the rivers and estuaries of China, Russia, and Korea that have been introduced and become established in parts of North America. However, unlike bowfin which are native to North America, the northern snakehead is considered an invasive species and environmentally harmful there. Some contrasting differences in bowfin include a black eyespot on their caudal peduncle, a tan and olive coloration, a shorter anal fin, a more rounded head, pelvic fins at a greater distance from the pectoral fins than in the northern snakehead and the presence of the gular plate on the ventral side of the lower jaw. Another noticeable difference is that bowfin scales do not continue uniformly from their body to their head. Bowfin heads are smooth and free of scales, whereas the northern snakehead has scales that uniformly continue from their body through to their head. The burbot (Lota lota), a predatory fish native to streams and lakes of North America and Eurasia, is also commonly mistaken for bowfin. Burbots can be distinguished by their flat head and chin barbel, long anal fin, and pelvic fins situated beneath the pectoral fins. Bowfin body-shape evolution and development [edit] The first fish lacked jaws and used negative pressure to suck their food in through their mouths. The jaw in the bowfin is a result of their evolutionary need to be able to catch and eat bigger and more nutritious prey. As a result of being able to gather more nutrients, Bowfin are able to live a more active lifestyle. The jaw of a bowfin has several adaptations. The maxilla and premaxilla are fused and the posterior chondrocranium articulates with the vertebra which allows the jaw freedom to rotate. The suspensorium includes several bones and articulates with the snout, brain case, and the mandible. When the jaw opens epaxial muscles lift the chondrocranium, which is attached to the upper jaw, while adductor muscles act to close the lower jaw. This ability to open and close the jaw helps the bowfin to be an active predator that can catch bigger prey and digest them. The vertebral column in bowfin is ossified and in comparison to earlier fish, the centra are the major support for the body, whereas in earlier fish the notochord was the main form of support. In bowfin neural spines and ribs also increase in prominence, an evolutionary aspect that helps provide additional support and stabilize unpaired fins. The evolution of the vertebral column allows the bowfin to withstand lateral bending that puts the column under compression without breaking. This, in turn, allows the bowfin to have more controlled and powerful movements, in comparison to fish that had only a notochord. The bowfin has a rounded heterocercal tail that resembles a homocercal tail. This type of tail gives the body a streamlined shape which allows the bowfin to improve its swimming ability by reducing drag. These types of tails are common in fish with gas bladders, because the bladder supplies the fish with natural buoyancy. The bowfin is a member of actinopterygii which means that the pectoral girdle is partly endochondral but mostly dermal bone. In this group of fish the fins function to maneuver, brake, and for slight positional adjustments. The pectoral girdle of the bowfin has six parts. The post temporal, supracleithrum, postcleithrum, cleithrum, scapulacoracoid, and the clavicle make up the pectoral girdle. The pectoral girdle is attached to the skull. The paired pectoral and pelvic fins of fish are homologous with the limbs of tetrapods. Physiology [edit] Bowfin are physostomes, meaning they have a small "pneumatic duct" that connects their swim bladders to their digestive tract. This allows them, like lungfish, to "breathe" in two ways: they can extract oxygen from the water when breathing through their gills, but can also break the water's surface to breathe or gulp air through the pneumatic duct. When performing low-level physical activity, bowfin obtain more than half of their oxygen from breathing air. The fish have two distinct air-breathing mechanisms used to ventilate the gas bladder. Air breathing type I is consistent with the action of exhale / inhale exchange, stimulated by either air or water hypoxia, to regulate O2 gas exchange; type II air breaths are inhalation alone, which is believed to regulate gas bladder volume, to control buoyancy. Bimodal respiration helps bowfin survive and maintain their metabolic rate in hypoxic (low-oxygen) conditions. Bowfin air breathe more frequently when they are in darkness, and correspondingly more active. Bowfin blood can adapt to warm, acidic waters. The fish becomes inactive in waters below 10 °C (50 °F); at this temperature they breathe almost no air; however, with increasing temperature their air breathing increases. Their preferred temperature range is between 12–26 °C (54–79 °F), with 18 °C (64 °F) the temperature of maximum activity. Air breathing is at a maximum in the range 18.4–29.6 °C (65.1–85.3 °F). Bowfin do not use central chemoreceptor regulation for respiration control. Experiments manipulating the oxygen content, carbon dioxide content, and pH of bowfin extradural fluid did not affect breathing rate, heart rate, or blood pressure pointing to a lack of central chemoreceptor regulation. Instead, bowfin respiratory patterns respond to water oxygen content and water temperature, as water temperatures play a role in oxygen content. In the lab, bowfin showed an increase in the breathing rate when the temperatures were raised above 10°C. Bowfin also showed an increase in breathing rate when exposed to lower oxygen levels in the water. Herpetologist W. T. Neill reported in 1950 that he unearthed a bowfin aestivating (in a dormant state) in a chamber 4 inches (10 cm) below the ground surface, 8 inches (20 cm) in diameter, .25 miles (0.4 km) from a river. It was further noted that flood levels had previously reached the area, and receded. It is not unusual for riverine species like bowfin to move into backwaters with flood currents, and become trapped when water levels recede. While aestivation is anecdotally documented by multiple researchers, laboratory experiments have suggested instead that bowfin are physiologically incapable of surviving more than three to five days of air exposure. However, no field manipulation has been performed. Regardless of the lack of evidence confirming the bowfin's ability to aestivate, it has been noted that bowfin can survive prolonged conditions of exposure to air because they have the ability to breathe air. Their gill filaments and lamellae are rigid in structure which helps prevent the lamellae from collapsing and aids gas exchange even during air exposure. Evolution and phylogeny [edit] Competing hypotheses and debates continue over the evolution of Amia and relatives, including their relationship among basal extant teleosts, and organization of clades. Bowfin are the last remaining member of Halecomorphi, a group that includes many extinct species in several families. Halecomorphs were generally accepted as the sister group to Teleostei but not without question. While a halecostome pattern of neopterygian clades was produced in morphology-based analyses of extant actinopterygians, a different result was produced with fossil taxa which showed a monophyletic Holostei. Monophyletic Holostei were also recovered by at least two nuclear gene analyses, in an independent study of fossil and extant fish, and in an analysis of ultraconserved genomic elements. The extant ray-finned fish of the subclass Actinopterygii include 42 orders, 431 families and over 23,000 species. They are currently classified into two infraclasses, Chondrostei (holosteans) and Neopterygii (teleost fishes). Sturgeons, paddlefish, bichirs and reed fish compose the thirty-eight species of chondrosteans, and are considered relict species. Included in the over 23,000 species of neopterygians are eight relict species comprising gars and the bowfin. Infraclass Neopterygii [edit] Neopterygians are the second major occurrence in the evolution of ray-finned fish and today include the majority of modern bony fish. They are distinguished from their earlier ancestors by major changes to the jaws, shape of the skull, and tail. They are divided into three divisions: Division 1. Order Lepisosteiformes – the relict gars which include extant species of gars that first appeared in the Cretaceous. Division 2. Order Amiiformes – the relict bowfin, (halecomorphids), the only extant species in the order Amiiformes which date back to the Triassic period. Division 3. Division Teleostei – the stem group of Teleostei from which modern fish arose, including most of the bony fish we are familiar with today. Species [edit] The following is a species list[citation needed] †Amia depressa Marsh 1871 †Amia dictyocephala Cope 1875 †Amia elegans Leidy 1873 †Amia exilis Lambe 1908 †Amia fragosa (Jordan 1927) †Amia godai Yabumoto & Grande 2013 †Amia gracilis Leidy 1873 †Amia lewesiensis Mantell 1822 †Amia macrospondyla Cope 1891 †Amia media Leidy 1873 †Amia morini Priem 1911 †Amia newberriana Marsh 1871 †Amia selwyniana Ami 1891 †Amia uintaensis Leidy 1873 †Amia whiteavesiana Cope 1891 †Amia pattersoni †Amia scutata Cope 1875 Amia calva Linnaeus 1766 (Bowfin) Amia ocellicauda Todd 1837 (Eyetail bowfin) Genome evolution [edit] The bowfin genome contains an intact ParaHox gene cluster, similar to the bichir and to most other vertebrates. This is in contrast, however, with teleost fish, which have a fragmented ParaHox cluster, probably because of a whole genome duplication event in their lineage. The presence of an intact ParaHox gene cluster suggests that bowfin ancestors separated from other fish before the last common ancestor of all teleosts appeared. Bowfin are thus possibly a better model to study vertebrate genome organization than common teleost model organisms such as zebrafish. Feeding behavior [edit] Bowfin are stalking, ambush predators that customarily move into the shallows at night to prey on fish, amphibians, and aquatic invertebrates such as crawfish, other crustaceans, mollusks, and aquatic insects. Young bowfin feed mostly on small crustaceans, while adults are mostly piscivorous, but also known to be opportunistic. Some common examples of prey include frogs, bass, other bowfin, dragonflies, sunfish, crawfish, etc. Bowfin are remarkably agile, can move quickly through the water, and they have a voracious appetite. Their undulating dorsal fin propels them silently through the water while stalking their prey. The attack is straightforward and swift with a movement that lasts approximately 0.075 seconds. There were also some studies regarding the capacity of the bowfin to survive without food. In 1916, a female bowfin was starved for twenty months. It was the longest period that any vertebrate had been without food, as far the writer was aware during the observation. Some independent studies focus on the bowfin's ability to use organic material as a source of food and studied the structure of the gill raker. They concluded that it did not benefit from the organic material in the water because the gill rakers were short with blunt processes and a short space between them. Even bacteria could enter and exit through the gill easily. Its structure alone indicated that the Amia do not use microorganisms as a source of food.[citation needed] Distribution and habitat [edit] Fossil deposits indicate amiiforms included freshwater and marine species that were once widely distributed in North America, South America, Eurasia and Africa. Today, bowfin (Amia calva) are the only remaining species in the order Amiiformes; they are demersal freshwater piscivores, and their range is restricted to freshwater environments in North America, including much of the eastern United States and adjacent southern Canada from the St. Lawrence River and Lake Champlain drainage of southern Ontario and Quebec westward around the Great Lakes in southern Ontario into Minnesota. Historically, their distribution in North America included the drainage basins of the Mississippi River from Quebec to northern Minnesota, the St. Lawrence-Great Lakes, including Georgian Bay, Lake Nipissing and Simcoe, Ontario, south to the Gulf of Mexico; Atlantic and Gulf Coastal Plain from the Susquehanna River drainage in southeastern Pennsylvania to the Colorado River in Texas. Stocking [edit] Research from the late 1800s to the 1980s suggests a trend of intentional stockings of non-indigenous fish into ponds, lakes and rivers in the United States. At that time, little was known about environmental impacts, or long-term effects of new species establishment and spread as a result of "fish rescue and transfer" efforts, or the importance of nongame fish to the ecological balance of aquatic ecosystems. Introductions of bowfin to areas they were considered a non-indigenous species included various lakes, rivers and drainages in Connecticut, Delaware, Georgia, Illinois, Iowa, Kansas, Kentucky, Maryland, Massachusetts, Minnesota, Missouri, New Jersey, New York, North Carolina, Ohio, Oklahoma, Pennsylvania, Virginia, West Virginia, and Wisconsin. Many of the introductions were intentional stockings; however, there is no way to positively determine distribution resulting from flood transfers, or other inadvertent migrations. Bowfin are typically piscivorous, but as an introduced species are capable of being voracious predators that pose a threat to native fish and their prey. Preferred habitat [edit] Bowfin prefer vegetated sloughs, lowland rivers and lakes, swamps, backwater areas, and are occasionally found in brackish water. They are well camouflaged, and not easy to spot in slow water with abundant vegetation. They often seek shelter under roots, and submerged logs. Oxygen-poor environments can be tolerated because of their ability to breathe air. Life cycle [edit] Bowfin spawn in the spring or early summer, typically between April and June, more commonly at night in abundantly vegetated, clear shallow water in weed beds over sand bars, and also under stumps, logs, and bushes. Optimum temperatures for nesting and spawning range between 16–19 °C (61–66 °F). The males construct circular nests in fibrous root mats, clearing away leaves and stems. Depending on the density of surrounding vegetation there may be a tunnel-like entrance at one side. The diameter of the nests commonly range between 39–91 cm (15–36 in), at a water depth of 61–92 cm (24–36 in). During spawning season, the fins and underside of male bowfin often change in color to a bright lime green. The courtship/spawning sequence lasts one to three hours, and can repeat up to five times. Courtship begins when a female approaches the nest. The ritual consists of intermittent nose bites, nudges, and chasing behavior by the male until the female becomes receptive, at which time the pair lie side by side in the nest. She deposits her eggs while he shakes his fins in a vibratory movement, and releases his milt for fertilization to occur. A male often has eggs from more than one female in his nest, and a single female often spawns in several nests. Females vacate the nest after spawning, leaving the male behind to protect the eggs during the eight to ten days of incubation. A nest may contain 2,000 to 5,000 eggs, possibly more. Fecundity is usually related to size of the fish, so it is not unusual for the roe of a large gravid female to contain over 55,000 eggs. Bowfin eggs are adhesive, and will attach to aquatic vegetation, roots, gravel, and sand. After hatching, larval bowfin do not swim actively in search of food. During the seven to nine days required for yolk-sac absorption, they attach to vegetation by means of an adhesive organ on their snout, and remain protected by the parent male bowfin. Bowfin aggressively protect their spawn from the first day of incubation to a month or so after the eggs have hatched. When the fry are able to swim and forage on their own, they will form a school and leave the nest accompanied by the parent male bowfin who slowly circles them to prevent separation. Bowfin reach sexually maturity at two to three years of age. They can live up to 33 years in the wild, and 30 years in captivity. Bowfin may live decades at adult size. Diseases [edit] A common parasite of bowfin is the anchor worm (Lernaea). These small crustaceans infest the skin and bases of fins, with consequences ranging from slowed growth to death. The mollusk Megalonaias gigantea lays eggs in the bowfin gills, that are then externally fertilized by sperm passing in the water flow. The small glochidia larvae then hatch and develop in the gill tubes. Bowfin with liver cancer and with fatal leukemia have been reported. Utilization [edit] As a sport fish, bowfin are not considered desirable to many anglers. They were once considered a nuisance fish by anglers and early biologists who believed the bowfin's predatory nature was harmful to sport fish populations. As a result, efforts were taken to reduce their numbers. Research has since proven otherwise, and that knowledge together with a better understanding of maintaining overall balance of ecosystems, regulations were introduced to help protect and maintain viable populations of bowfin. Bowfin are strong fighters, a prized trait in game fish. However, they do have a jaw full of sharp teeth which requires careful handling. The current tackle record is 21.5 lb (9.8 kg) Bowfin were once considered to have little commercial value because of its poor-tasting meat which has been referred to as "soft, bland-tasting and of poor texture". However, it is considered quite palatable if cleaned properly and smoked, or prepared fried, blackened, used in courtbouillion, or in fishballs or fishcakes. Over the years, global efforts have imposed strict regulations on the international trade of caviar, particularly on the harvest of sturgeons from the Caspian Sea where the highly prized caviar from the beluga sturgeon originates. The bans imposed on Caspian sturgeons have created lucrative markets for affordable substitutes in the United States including paddlefish, bowfin, and various species of sturgeon. In Louisiana, bowfin are harvested in the wild, and cultured commercially in hatcheries for their meat and roe. The roe is processed into caviar, and sold as "Cajun caviar", or marketed under the trade name "Choupiquet Royale". Accumulation of toxic substances [edit] In some areas of the United States where aquatic environments have tested positive for elevated levels of toxins, such as mercury, arsenic, chromium, and copper, there are posted signs with warnings about the consumption of fish caught in those areas. Concentration of mercury biomagnifies as it passes up the food chain from organisms on lower trophic levels to apex predators. It bioaccumulates in the tissues of larger, long-lived predatory fish. When compared to smaller, short-lived fish, bowfin tend to concentrate mercury at higher levels thereby making them less safe for human consumption. See also [edit] Snakehead (fish) References [edit] ^ Jump up to: a b NatureServe (2013). "Amia calva". IUCN Red List of Threatened Species. 2013: e.T201942A2730796. doi:10.2305/IUCN.UK.2013-1.RLTS.T201942A2730796.en. 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"Control of gill ventilation and air-breathing in the bowfin Amia calva". J. Exp. Biol. 202 (1): 87–94. doi:10.1242/jeb.202.1.87. PMID 9841898. ^ Jackson, D.C.; Farmer, C.G. (March 5, 1998). "Air-breathing during activity in the fish Lepisosteus oculatus and Amia calva" (PDF). The Journal of Experimental Biology (201): 943–948. Retrieved 8 June 2014. ^ Boutilier, R.G. (1990). "Control and Co-Ordination of Gas Exchange in Bimodal Breathers". Vertebrate Gas Exchange. Advances in Comparative and Environmental Physiology. Vol. 6. pp. 279–345. doi:10.1007/978-3-642-75380-0_9. ISBN 978-3-642-75382-4. ^ McKenzie, David J.; Steffensen, John F.; Taylor, Edwin W. & Abe, Augusto S. (19 December 2011). "The contribution of air breathing to aerobic scope and exercise performance in the banded knifefish Gymnotus carapo L.". Journal of Experimental Biology. 215 (8): 1323–1330, Introduction. doi:10.1242/jeb.064543. PMID 22442370. ^ "Canon Envirothon Water Quality Study" (PDF). Water Quality Assessment Division. Louisiana Department of Environmental Quality. 2005. p. 18. Archived (PDF) from the original on April 13, 2013. Retrieved June 13, 2014. ^ Jump up to: a b c d e f g h i Ross, Stephen T. (2001). The Inland Fish of Mississippi. Univ. Press of Mississippi. p. 94. ISBN 978-1-57806-246-1. ^ HEDRICK, M. S.; BURLESON, M. L.; JONES, D. R.; MILSOM, W. K. (1991-01-01). "An Examination of Central Chemosensitivity in an Air-Breathing Fish (Amia Calva)". Journal of Experimental Biology. 155 (1): 165–174. doi:10.1242/jeb.155.1.165. ISSN 0022-0949. ^ Horn, M. H.; Riggs, C. (1973). "Effects of Temperature and Light on the Rate of Air Breathing of the Bowfin, Amia calva". Copeia. 1973 (4): 653–657. doi:10.2307/1443063. JSTOR 1443063. S2CID 86886787. ^ Glass, Mogens L.; Wood, Stephen C., eds. (2009). Cardio-Respiratory Control in Vertebrates. doi:10.1007/978-3-540-93985-6. ISBN 978-3-540-93984-9. ^ Neill, W.T. (1950). Copeia, 1950, An estivating bowfin. American Society of Ichthyologists and Herpetologists. p. 240. ISBN 9780521739672. {{cite book}}: ISBN / Date incompatibility (help) ^ Loftus, William F.; Kushlan, James A. (1987). "Freshwater fish of southern Florida". Florida State Museum Biological Sciences. p. 183. ^ Wolfgang J. Plunk; Peter B. Bayley; Richard E. Sparks (1989). The Flood Pulse Concept in River-Floodplain Systems (PDF) (Report). University of Florida. pp. 117, 8–pdf. Archived from the original (PDF) on 2010-06-21. Retrieved 2014-06-09. ^ McKenzie, D. J.; Randall, D. J. (1990). "Does Amia calva aestivate?". Fish Physiology and Biochemistry. 8 (2): 147–58. Bibcode:1990FPBio...8..147M. doi:10.1007/BF00004442. PMID 24221948. S2CID 20401237. ^ Gene Helfman; Bruce B. Collette; Douglas E. Facey; Brian W. Bowen (2009). The Diversity of Fish: Biology, Evolution, and Ecology. Wiley-Blackwell. p. 257, Chapter 13. ISBN 9781405124942. ^ Melvin L. Warren Jr.; Brooks M. Burr (July 2014). Freshwater Fish of North America. Vol. 1: Petromyzontidae to Catostomidae. JHU Press. pp. 288–289. ISBN 9781421412016. ^ Arratia, Gloria (2001). "The Sister-Group of Teleostei: Consensus And Disagreements". Journal of Vertebrate Paleontology. 21 (4): 767–773. doi:10.1671/0272-4634(2001)021[0767:TSGOTC]2.0.CO;2. S2CID 85895344. ^ Guang-Hui Xu; Li-Jun Zhao; Michael I. Coates (May 2014). "The oldest ionoscopiform from China sheds new light on the early evolution of halecomorph fish". Biology Letters. 10 (5): 20140204. doi:10.1098/rsbl.2014.0204. PMC 4046378. PMID 24872460. ^ Imogen A. Hurley; Rachel Lockridge Mueller; Katherine A. Dunn; Eric J. Schmidt; Matt Friedman; Robert K. Ho; Victoria E. Prince; Ziheng Yang; Mark G. Thomas & Michael I. Coates (February 2007). "A New Time-Scale for Ray-Finned Fish Evolution". Proceedings of the Royal Society B: Biological Sciences. 274 (1609): 489–498. doi:10.1098/rspb.2006.3749. JSTOR 25223804. PMC 1766393. PMID 17476768. ^ Broughton, Richard E. (2013). "Multi-locus phylogenetic analysis reveals the pattern and tempo of bony fish evolution". PLOS Currents. 5. doi:10.1371/currents.tol.2ca8041495ffafd0c92756e75247483e (inactive 5 July 2025). PMC 3682800. PMID 23788273.{{cite journal}}: CS1 maint: DOI inactive as of July 2025 (link) ^ Faircloth, Brant C. (2013). "A Phylogenomic Perspective on the Radiation of Ray-Finned Fish Based upon Targeted Sequencing of Ultraconserved Elements (UCEs)". PLOS ONE. 8 (6): e65923. Bibcode:2013PLoSO...865923F. doi:10.1371/journal.pone.0065923. PMC 3688804. PMID 23824177. ^ Jump up to: a b c "SUBCLASS ACTINOPTERYGII: RELICT SPECIES OF RAY-FINNED FISH & ORIGIN OF TELEOSTS". University of Edinburgh. January 2007. Retrieved September 29, 2014. ^ Kanae Kikugawa; Kazutaka Katoh; Shigehiro Kuraku; Hiroshi Sakurai; Osamu Ishida; Naoyuki Iwabe; Takashi Miyata (March 11, 2004). "Basal jawed vertebrate phylogeny inferred from multiple nuclear DNA-coded genes". BMC Biology. 2 (1): 3. doi:10.1186/1741-7007-2-3. PMC 387836. PMID 15070407. ^ Thom Holmes (June 28, 2008). The First Vertebrate. Chelsea House Publishers. p. 144. ISBN 9780816059584. ^ Mulley, John F.; Chiu, Chi-hua; Holland, Peter W. H. (2006). "Breakup of a homeobox cluster after genome duplication in teleosts". Proceedings of the National Academy of Sciences USA. 103 (27): 10369–10372. Bibcode:2006PNAS..10310369M. doi:10.1073/pnas.0600341103. PMC 1502464. PMID 16801555. ^ Indiana Department of Fish & Wildlfife. "Bowfin (Amia calva)" (PDF). Indiana Department of Natural Resources. Retrieved June 13, 2014. ^ Jump up to: a b c d e f Arndt, Rudolf G.; Foltz, Jeffrey W. (2009). Freshwater Fish of South Carolina. Univ of South Carolina Press. p. 80. ISBN 978-1-57003-680-4. ^ Smallwood, W.M. (1916-01-01). "Twenty months of starvation in Amia calva". Biological Bulletin. 31 (6): 453–464. doi:10.2307/1536322. JSTOR 1536322. ^ Hart, Paul J.B.; Reynolds, John D. (2002). Handbook of Fish Biology and Fisheries. Wiley. p. 27. ISBN 978-0-632-05412-1. ^ Jump up to: a b c d Fuller, Pam (31 March 2020) [Peer reviewed on 11 April 2006]. "Amia calva Linnaeus, 1766". USGS Nonindigenous Aquatic Species Database. Gainesville, Florida: US Geological Survey. Retrieved 24 March 2024. ^ Jump up to: a b c John Acorn (7 February 2007). Deep Alberta: Fossil Facts and Dinosaur Digs. University of Alberta. p. 10. ISBN 978-0-88864-481-7. ^ "Early Limnological and Fishery Research" (PDF). Wisconsin Fish and Fishery Management. University of Wisconsin. p. 29. Retrieved 9 August 2014. ^ J. Richard Arthur; Rohana P. Subasinghe. "Potential Adverse Socio-Economic And Biological Impacts Of Aquatic Animal Pathogens Due To Hatchery-Based Enhancement Of Inland Open-Water Systems, And Possibilities For Their Minimisation". Primary aquatic animal health care in rural, small-scale, aquaculture. Inland Water Resources and Aquaculture Service. Retrieved August 9, 2014. ^ Jump up to: a b c d Miller, Rudolph John; Robison, Henry W. (2004). Fish of Oklahoma. University of Oklahoma Press. p. 58. ISBN 978-0-8061-3610-3. ^ Joshua Laerm; B. J. Freeman (January 2008). Fish of the Okefenokee Swamp. University of Georgia Press. p. 37. ISBN 978-0-8203-3135-5. ^ Jump up to: a b c d e Thomas, Chad; Bonner, Timothy; Whiteside, Timothy H. (18 June 2007). Freshwater Fish of Texas. Texas A&M University Press. p. 23. ISBN 9781585445707. Retrieved 1 July 2017. ^ Jackson, Randy. "The Bowfin: New York's disrespected living fossil". New York Department of Environmental Conservation. Archived from the original on 27 April 2021. Retrieved 11 June 2014. ^ Berra, Tim M. (2001). Freshwater Fish Distribution. San Diego, CA: Academic Press. ISBN 0-12-093156-7. ^ Jump up to: a b Randy Jackson. "The Bowfin, New York's Disrespected Living Fossil". New York Department of Environmental Conservation. Archived from the original on April 27, 2021. Retrieved June 4, 2014. ^ IGFA. "Bowfin". International Game Fish Association. Archived from the original on October 12, 2018. Retrieved June 4, 2014. ^ Bourgeois, David A. (April 5, 2009). "Choupique may be a trash fish for some, treasure to others". Daily Comet. Archived from the original on March 14, 2016. Retrieved June 8, 2014. ^ Saulny, Susan (9 June 2012). "A roe, by any other name". The New York Times. ^ Scharpf, Christopher (30 December 2013). "Bowfin: North America's Freshwater Thug". USF&WS. ^ "Fish Consumption Advisories". Environmental Protection Agency. Retrieved 6 June 2014. ^ "What fish are safe to eat?" (PDF). North Carolina Division of Public Health. Archived from the original (PDF) on 20 September 2013. Retrieved 1 July 2017. Further reading [edit] McCormick, Catherine A. (1981). "Central Projects of the lateral line and eight nerves in the bowfin,Amia Calva" (PDF). The Journal of Comparative Neurology. 197 (1): 1–15. doi:10.1002/cne.901970102. hdl:2027.42/50011. PMID 6164698. S2CID 8836961. Conlon, J.M.; Youson, J.H.; Whittaker, J. (1991). "Structure and receptor-binding activity of insulin from a holostean fish, the bowfin:Amia Calva". Biochem. J. 276 (Pt 1): 261–264. doi:10.1042/bj2760261. PMC 1151174. PMID 2039477. Nguyen, T. M.; Mommsen, T. P.; Mims, S. M.; Conlon, J. M. (1994). "Characterization of insulins and proglucagon-derived peptides from a phylogenetically ancient fish, the paddlefish: Polyodon spathula". Biochem. J. 300 (2): 339–345. doi:10.1042/bj3000339. PMC 1138167. PMID 8002937. Conlon, J. M.; Youson, J. H.; Mommsen, T. P. (1993). "Structure and biological activity of glucagon and glucagon-like peptide from a primitive bony fish, the bowfin: Amia calva.". Biochem. J. 295 (3): 857–861. doi:10.1042/bj2950857. PMC 1134640. PMID 8240302. Sepkoski, Jack (2002). "A compendium of fossil marine animal genera". Bulletins of American Paleontology. 364: 560. Archived from the original on 2009-02-20. Retrieved 2011-05-17. External links [edit] Wikispecies has information related to Amia calva. Wikimedia Commons has media related to Amia calva. MN DNR Bowfin Anglers' Group FishBase info for Amiidae Bowfin info by Brent Courchene Archived 2011-09-27 at the Wayback Machine | v t e Extant orders of Actinopterygii (ray-finned fish) | | --- | | Kingdom: Animalia Phylum: Chordata Subphylum: Vertebrata Infraphylum: Gnathostomata Superclass: Osteichthyes | | | | | | | | --- --- | | | | | --- | | Cladistia | Polypteriformes | | | | Actinopteri | | | | | | --- --- | | | | | --- | | Chondrostei | Acipenseriformes | | | | Neopterygii | | | | | | --- --- | | | | | --- | | Holostei | Amiiformes Lepisosteiformes | | | | Teleostei | | | | | | | | | | | | | | --- --- --- --- --- --- | | | | | --- | | Elopomorpha | Elopiformes Albuliformes Anguilliformes Notacanthiformes | | Osteoglossomorpha | Hiodontiformes Osteoglossiformes | | Otocephala | | | | --- | | Clupeiformes Alepocephaliformes | | | Ostariophysi | Gonorynchiformes Cypriniformes Gymnotiformes Characiformes Siluriformes | | | | | Euteleostei | | | | --- | | Lepidogalaxiiformes Argentiniformes Salmoniformes Osmeriformes Stomiiformes Galaxiiformes Ateleopodiformes Aulopiformes Myctophiformes | | | Acanthomorpha | | | | --- | | Lampriformes Percopsiformes Polymixiiformes Gadiformes Zeiformes Trachichthyiformes Beryciformes | | | Percomorpha | | | | --- | | Ophidiiformes Batrachoidiformes Gobiiformes Scombriformes Syngnathiformes Carangiformes Anabantiformes Synbranchiformes | | | Ovalentaria | Cyprinodontiformes Atheriniformes Beloniformes Cichliformes Blenniiformes Mugiliformes | | Eupercaria | Perciformes Centrarchiformes Labriformes Acropomatiformes Acanthuriformes Lophiiformes Tetraodontiformes | | | | | | | | | | v t e | | --- | | | Kingdom: Animalia Phylum: Chordata Class: Actinopterygii Subclass: Neopterygii Infraclass: Holostei | | --- | | | | | | | | | | | | | --- --- --- --- --- | | Actinopterygii | | | | --- | | + see Actinopterygii | | | Neopterygii | | | | --- | | + see Neopterygii | | | Halecomorphi | + see below↓ | | | | Halecomorphi | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | Halecomorphi | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- | | | | | --- | | †Angolaichthys? †Cipactlichthys †Peia? †Promecosomina? †Prosantichthys | | | †Parasemionotidae | Albertonia Broughia Candelarialepis Devillersia Icarealcyon Jacobulus Jurongia Lehmanotus Lombardina? Ospia Parasemionotus Piveteaunotus Qingshania Stensionotus Suius Thomasinotus Watsonulus | | †Ionoscopiformes | | | | --- | | Ainia Allolepidotus? Altmuehlfuro Archaeosemionotus Asialepidotus? Brachyichthys Cerinichthys Elongofuro Eoeugnathus? Furo Gymnoichthys? Heterolepidotus Holzmadenfuro Italophiopsis Ohmdenfuro Oligopleurus? Ophiopsis Panxianichthys? Robustichthys? Sanctusichthys Schernfeldfuro Sinoeugnathus? Subortichthys Xingyia? Zandtfuro | | | Ionoscopidae | Ionoscopus Oshunia Quetzalichthys Spathiurus | | Ophiopsiellidae | Congophiopsis Macrepistius Ophiopsiella Petalopteryx Placidichthys Teoichthys | | | | | | | | | | | | | --- --- --- | | †Caturoidea | | | | --- | | Eurypoma Liodesmus Strobilodus | | | Caturidae | Amblysemius Caturus Catutoichthys | | | Amioidea | | | | --- | | †Tomognathus | | | †Sinamiidae | Ikechaoamia Siamamia Sinamia | | Amiidae | Amia †Amiopsis †Calamopleurus †Cratoamia †Cyclurus †Hispanamia †Maliamia †Melvius †Nipponamia †Pachyamia †Pseudamiatus †Solnhofenamia †Vidalamia | | | | | | | | Taxon identifiers | | --- | | Amia calva | Wikidata: Q491934 Wikispecies: Amia calva ADW: Amia BOLD: 30342 CoL: CRZ7 EoL: 223782 FishBase: 2600 GBIF: 5207891 iNaturalist: 58611 IRMNG: 10841152 ITIS: 161104 IUCN: 201942 NAS: 305 NatureServe: 2.104567 NCBI: 7924 ODNR: bowfin Open Tree of Life: 329623 Paleobiology Database: 149596 Plazi: FA4BDF92-0F59-0EE5-1F48-3D19C67379D2 WoRMS: 157877 ZooBank: 7895E785-6F85-42EA-B316-1FF0E262A954 | Retrieved from " Categories: IUCN Red List least concern species Amia (fish) Freshwater fish of North America Fish of the Great Lakes Freshwater fish of the Southeastern United States Fish described in 1766 Hidden categories: CS1 maint: multiple names: authors list CS1 errors: ISBN date CS1: long volume value CS1 maint: DOI inactive as of July 2025 Articles with short description Short description is different from Wikidata Good articles Articles with 'species' microformats All articles with unsourced statements Articles with unsourced statements from December 2021 Articles with unsourced statements from August 2019 Commons link from Wikidata Webarchive template wayback links Articles containing video clips
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https://www.cpd4dentalnurses.co.uk/PDFs/Medical%20Emergencies%20Annual%20CPR%20and%20AED%20Update%202023.pdf
Medical Emergencies: Annual CPR and AED Update 2023 Aims: To describe the steps involved in Cardiopulmonary Resuscitation and the use of an Automated External Defibrillator. To provide an update from the Resuscitation Council on providing CPR during the Coronavirus pandemic. Learning Outcomes: On completion of this verifiable CPD article, the participant will be able to demonstrate, through completion of a questionnaire, the ability to: • Identify statistics relating to cardiac arrest in the UK • Identify the changes in the 2021 Resuscitation Council (UK) Guidelines • Demonstrate knowledge of the Chain of Survival • Be able to identify the steps and methods involved in cardiopulmonary resuscitation • Know the importance of using an Automated External Defibrillator • Be able to identify which are the shockable heart rhythms ` • Know how to use an Automated External Defibrillator • Know how to carry out basic life support • Know the latest update from the Resuscitation Council regarding conducting basic life support during COVID 19 • Know the training requirements in relation to medical emergencies. Introduction Heart and circulatory disease (Cardiovascular disease), causes around a quarter of all deaths in the UK. Of these, most deaths from coronary heart disease are caused by a heart attack which can lead to cardiac arrest.1 Cardiac arrest is a serious medical emergency, where the heart stops pumping blood around the body. Unless treated immediately, it leads to death within minutes. The latest April 2023 factsheet from the British Heart Foundation estimates that there are around 7.6 million people in the UK that are living with heart and circulatory diseases. Heart and circulatory diseases cause around a quarter of all deaths in the UK. However, since the British Heart Foundation was established, the number of deaths from Cardiovascular disease has approximately halved. In the 1960s more than seven out of ten heart attacks in the UK were fatal. Today more than seven out of ten people survive1 In the UK, approximately 30,000 people sustain a cardiac arrest outside of the hospital environment and are treated by the emergency services each year. 2 The General Dental Council state that, “a patient could collapse at any time, whether they have received treatment or not. It is therefore essential that all registrants must be trained in dealing with Medical Emergencies, including resuscitation, and possess up to date evidence of capacity.”3 Therefore, although cardiac arrest is rare in Primary Dental Practice, it is important that dental staff are competent in dealing with such an eventuality. The new Resuscitation Council UK Adult Basic Life Support Guidelines were published in May 2021. These can be downloaded using the link at the end of this article. This article will describe Cardiopulmonary Resuscitation (CPR) and the use of Automated External Defibrillator. The heart, cardiac arrest and ventricular fibrillation The heart is a highly efficient muscular pump that pushes blood around the body at an average rate of 100,000 times each day. An electrical system initiates the power that enables the heart to do its work. If the electrical system and muscular pump work together in a co-ordinated manner, the result is our normal heartbeat.4 The term cardiac arrest means that the heart has stopped pumping blood around the body. The heart may have stopped beating altogether (asystole) or be twitching in an irregular and ineffective way (ventricular fibrillation). In either case, there is no circulation of blood. Within seconds, the casualty will lose consciousness and, if the heart is not restarted, will die within a few minutes. This means that the dental team will need to act quickly and efficiently.5 The Chain of Survival and Basic Life Support (Differences in basic life support guidance during COVID-19 are detailed in a separate section) Only 7-8% of people in whom resuscitation is attempted, survive hospital discharge. Survival chances are significantly increased by the early use of CPR. Currently, only 30-40% of victims of out of hospital cardiac arrest receive bystander CPR.2 The key stages in the life support sequence can be described as the Chain of Survival. The following steps together, maximise the chance of survival following cardiac arrest: 1) Early recognition and call for help- to prevent cardiac arrest 2) Early CPR- to buy time 3) Early Defibrillation- to restart the heart 4) Post-resuscitation care- to restore quality of life 2 The resuscitation council state that “dental practitioners and other dental care professionals must be trained in cardiopulmonary resuscitation (CPR) so that in the event of cardiorespiratory arrest occurring they can: • Recognise cardiorespiratory arrest; • Summon help immediately (dial 999); • Start CPR, using chest compressions and providing ventilation with a pocket mask or bag-mask device and supplemental oxygen. (Evidence suggests that chest compressions can be performed effectively in a fully reclined dental chair); • Attempt defibrillation (if appropriate) within 3 minutes of collapse, using an AED; and, • Provide other advanced life support skills if appropriate and if trained to do so.”6 Resuscitation Council (UK) Guidelines 2021 The Resuscitation Council (UK) Guidelines 2021 were published in May 2021. Although there were no major changes, there is a heightened emphasis on: • Cardiac arrest recognition remains a key priority as it is the first step in triggering the emergency response to cardiac arrest • Recognise cardiac arrest has occurred in any unresponsive person with absent or abnormal breathing • The ambulance call handler will assist with instructions for confirming cardiac arrest, starting compression-only CPR, and locating, retrieving, and using an AED • Provide chest compressions as soon as possible after cardiac arrest is confirmed • Send someone to fetch an AED and bring it to the scene of the cardiac arrest. The British Heart Foundation database, “The Circuit” serves as a national resource for the location of AEDs. • Use the recovery position, only if a person’s conscious level is reduced and they do not meet the criteria for starting CPR7 Adult Basic Life Support Recap Danger: Check for danger: sight, sound, and smell. Is there anything that may be dangerous to you, the person, or bystanders? Response: Speak loudly to the person and gently shake/tap the shoulders if there is no response to your voice. Shout for help, but do not leave the person yet. Airway: Open the airway using a head tilt and chin lift. Place two fingers under the point of the chin gently and lift, whilst placing the other hand on the person’s forehead and gently tilting the head back. Breathing: Look for signs of life such as movement, coughing, or normal breathing. Take no more than ten seconds to do this. If you do not think the person is breathing normally, or you do not think the person is breathing at all, treat this as a cardiac arrest. Ambulance: Begin chest compressions whilst asking someone else to call the ambulance service. If you are alone, you will need to leave now and make the call yourself before starting chest compressions. If an AED is available, send somebody to get it- do not delay starting chest compressions in order to get an AED. Circulation: Perform high quality, uninterrupted cardiopulmonary resuscitation (CPR). Deliver chest compressions on the lower half of the sternum (the centre of the chest). The compression rate is 100-120 per minute, with a compression depth of 5-6cm. Allow an equal amount of time for compression and release, ensuring the chest rises fully after each compression. Give 30 compressions then two gentle rescue breaths sufficient to make chest rise as in a normal breath. The breaths must not be forceful or rapid and should take one second each to deliver. If either or both rescue breaths do not result in chest movement, return immediately to chest compressions. Continue chest compressions and rescue breaths at a ratio of 30:2. If you are unable or unwilling to provide ventilations, give continuous chest compressions. 80% of the CPR cycle should comprise of chest compressions. Resuscitation of children and infants (babies) The adult BLS sequence for a child is far better than not doing anything at all. However, there are modifications that will make it more suitable for children and infants: • Start by giving 5 rescue breaths • Look for signs of life such as movement, coughing, or normal breathing. Take no more than ten seconds to do this. If there are no signs of life, start CPR • If you are on your own, perform CPR for one minute before leaving to get help • Compress the chest at least one-third of its depth. This will be approximately 5 cm for an older child and approximately 4 cm for the infant. Use one or two hands for a child over 1 year and two fingers for the infant under 1. Never press deeper than the adult 6 cm limit (approximately the length of an adult’s thumb) • To avoid compressing the abdomen, press down on the sternum one finger’s breadth above the xiphisternum • Use a compression: ventilation ratio of 15:2 if you are confident and competent in your resuscitation skills. This ratio will usually only be provided by healthcare professionals with a duty to respond to paediatric emergencies. In dental practice, it may be that staff are not used to dealing with cardiac arrest in children and infants and therefore may be more comfortable with the compression: ventilation ratio of 30:2. Time off the chest should be minimised as much as possible, therefore a 15:2 ratio should only be provided if the rescuer is confident in delivering rescue breaths. Within dental and orthodontic practices, one person alone may be able to provide compressions and ventilations. If two or more staff are available as rescuers, then one could deliver compressions while the other delivers rescue breaths. In such a case, a two-thumb encircling technique can be used. The team approach is key • A lone rescuer with a mobile phone should call for help after the five rescue breaths, then move on to the next step while waiting for the emergency services to answer (activate the speaker function on the phone) When performing rescue breaths: • Ensure head tilt and chin lift in a child over 1 year • Keep the head of an infant in a neutral position and lift the chin (when an infant is supine, the head is usually flexed and may require a degree of extension) • If the infant’s mouth and nose cannot be covered by the rescuer’s mouth, then the rescuer can try to seal only the mouth or nose of the infant whilst closing the other Age definitions: • An infant is under the age of one year • A child is between one year and 18 years of age The early use of a defibrillator significantly increases the patient’s likelihood of survival and does not increase the risk of infection. Resuscitation Council UK Statement on COVID-19 in relation to CPR and resuscitation in first aid and community settings The Resuscitation Council UK Statement was updated in August 2022 and is as follows8: “New evidence has emerged suggesting a low likelihood that airway management manoeuvres are aerosol generating, leading to the removal of airway management manoeuvres from the list of aerosol generating procedures. We await further evidence on whether chest compressions generate aerosol. Until such evidence emerges, we remain concerned that the provision of chest compressions and the proximity of the rescuer to the patient may constitute a risk of aerosol transmission. In light of this new information we recommend: • the curriculum for training members of the public and healthcare professionals reverts to the guidance set out in our quality standards • members of the public and healthcare professionals follow our 2021 guidelines for resuscitation • for those working in healthcare settings, the use of FFP3 masks or respirators as well as eye protection is still recommended when performing chest compressions for patients with suspected or confirmed COVID-19. AGP PPE, in particular FFP3 mask/respirator and eye protection, should be donned as swiftly as possible to avoid any delays in treatment. The above mentioned quality standards and 2021 guidelines can be accessed in the further reading section of this article. Automated External Defibrillator (AED) An AED is a simple to use device that recognises the life-threatening arrhythmias of ventricular fibrillation and ventricular tachycardia and is able to treat them through defibrillation - the application of electrical therapy which stops the arrhythmia, allowing the heart to re-establish an effective rhythm. AEDs are very simple to use – some are semi-automatic (the user pushes a button to deliver the shock) while others are fully-automatic (the shock is delivered without user intervention). Unlike any other piece of medical equipment or first aid kit, there is nothing that can substitute defibrillation. If a collapsed victim of cardiac arrest is in ventricular fibrillation, the only piece of medical equipment that can stop it is a defibrillator.4 Defibrillation within 3-5 minutes of collapse can produce survival rates as high as 50-70%. Each minute of delay to defibrillation recues the probability of survival to hospital discharge by 10%. Guidelines for use of an AED The Resuscitation Council state that: “All clinical dental areas should have immediate access (within the first minutes of a cardiorespiratory arrest) to oxygen, resuscitation equipment for airway management including suction, and an automated external defibrillator (AED). The standard AED sign should be used in order to reduce delay in a defibrillator in an emergency.”9 The standard AED signage10 reinforces the following key messages: • Anyone can use and AED- you do not need prior medical or first aid training • It is easy to use- you just follow the instructions • It is for use on an unconscious person who is not breathing normally Immediately the AED arrives, switch it on and follow the voice prompts, which include how to place the two electrode pads onto the patient’s bare chest. One pad should be placed on the front of the chest to the right of the breastbone, below the clavicle (collar bone) and the other pad should be placed on the left side of the chest, wrapped slightly towards the back. It is important that this pad is placed sufficiently laterally and that it is clear of any breast tissue. The pads can be placed front-to-back on a child. Dental practices should check the age range for their specific AED. If possible, do not interrupt chest compressions during placement of the pad. As soon as the pads are placed, the AED will determine whether or not the victim is in a shockable rhythm. If a shockable rhythm is identified, the AED will charge and prompt the user to stand clear. The patient must not be touched whilst the shock is delivered. A semi-automatic AED requires the user to push a button to deliver the shock, whereas an automatic AED will count down and deliver the shock without user intervention. If an AED does not detect a shockable rhythm, the machine will not charge. Two minutes of CPR is then required, after which time the machine will determine again whether or not the victim is in a shockable rhythm. Do not interrupt the process unless the victim begins to show signs of life. If you think that is the case; stop CPR, perform a quick ‘shout’ and ‘shake’ and reassess the victim’s breathing. If there is any doubt as to whether the victim is breathing normally, continue CPR. An unconscious person, who does not appear to be breathing normally, needs good quality, uninterrupted CPR. Unlike performing CPR, the use of an AED requires the victim to have a bare chest. Essential adjuncts that should be with every AED are: • Shears (to cut clothing if the clothing cannot be ripped) • Surgical Prep Razor (to shave the upper right chest if the chest is so hairy that skin is not visible beneath it) • Cloth (the area where the pads are placed, and in between, needs to be dry so that the pads are able to function properly) • Spare Defibrillation Pads Other useful items to consider keeping with an AED are: • Pocket Mask • Metronome Think about the six Ps • Perspiration – ensure the chest is dry • Pacemaker – do not place a pad over a pacemaker. Instead, position it to the side of, or below, the pacemaker • Pendants – do not have any metal resting between the pads during defibrillation • Piercings – there is nothing you can do about these • Patches/Plasters – avoid placing the pad over these • Playtex – when removing clothes, cut through a bra also Medical Emergencies Training The General Dental Council state that, “registrants must know their role in the event of a medical emergency, and ensure they are sufficiently trained and competent to carry out that role.” The GDC highly recommend that DCPs carry out 10 hours of medical emergency training in each five-year cycle. Although this article covers BLS and the use of an AED, it is important that dental professionals also carry out additional practical life support training. The Resuscitation Council have the following standards for Primary Dental Care practice in the UK: • “There is a public expectation that dental practitioners and all other dental care professionals should be competent in treating cardiorespiratory arrest • All primary care dental facilities should have a process for medical risk-assessment of their patients • Specific resuscitation equipment should be available immediately in all primary care dental premises. This equipment list should be standardised throughout the UK • All clinical areas should have immediate access to an automated external defibrillator (AED) • Primary dental care providers, general dental practitioners and all other dental healthcare professionals should undergo training in cardiopulmonary resuscitation (CPR) including basic airway management and the use of an AED • Each primary dental care facility should have a plan for summoning assistance in the event of a cardiorespiratory arrest. For most practices this will mean calling 999 immediately • There should be regular practice and teaching using simulation-based cardiorespiratory arrest scenarios • Dental staff’s knowledge and skills in resuscitation should be updated at least annually • A system must be in place for identifying which equipment requires special training, (such as AEDs, bag-mask devices and oropharyngeal airway insertion) and for ensuring that such training takes place • All new members of dental staff should have resuscitation training as part of their induction programme • Training can be undertaken locally within the dental practice or within local or regional training centres • For all staff, various methods to acquire, maintain and assess resuscitation skills and knowledge can be used for updates (e.g. life support courses, simulation training, mock-drills, ‘rolling refreshers’, e-learning, video-based training/self-instruction) • Hands-on’ simulation training and assessment is recommended for clinical staff • Dental practitioners and other dental care professionals who work with children should learn the differences in CPR (from CPR in adults) for use in children and practise these on paediatric manikins • Training in resuscitation must be a fundamental requirement for dental practitioners and other dental care professional qualifications • All primary dental care providers should recognise the need for and make provision for dental staff to have sufficient time to train in resuscitation skills as part of their employment • All training should be recorded in a database • Training and retraining should be a mandatory requirement for Continuing Professional Development and maintenance on professional healthcare registers. It may be appropriate for some retraining to be undertaken using ‘e-learning’”6 Conclusion Although cardiac arrest in Primary Dental Practice is rare, dental staff must be confident in dealing with a medical emergency should one arise. The chances of survival increase with good quality, uninterrupted CPR and early defibrillation. It is important that dental staff update their resuscitation skills at least annually, and that they carry out ‘hands on’ simulation training. ©2023 Jon Kyle Anderson and Nicky Gough BSc (Hons) RDH Personal Development Plan and Reflective Learning This CPD is linked to the following GDC Enhanced CPD Development Outcome: C. Maintenance and development of knowledge and skill within your field of practice. Reflective learning is now a requirement of the GDC Enhanced Professional Development Scheme. As such, you will now be given the opportunity to answer some reflective learning questions, before your certificate is generated. If you take a few moments to write your reflection on completion, you will have fulfilled the Enhanced CPD requirements. Further Reading The British Heart Foundation Heart Statistics 2023 Resuscitation Council (UK) (2021) Adult basic life support Guidelines Resuscitation Council UK Quality Standards Jon Andersen of ST4 Training Jon Andersen is the sole proprietor of ST4 Training and has personally delivered over 2000 courses to a range of organisations. The majority of Jon’s training is with GP and dental practices. Previously, Jon was a Paramedic, Operational Station Officer, Aircrew Paramedic (one of the first six in Sussex), Advanced Exercise Referral Instructor, and Phase IV Cardiac Rehabilitation Exercise Specialist. Jon can be contacted on 07883 703256 References 1 British Heart Foundation (2023) CVD Statistics-BHF UK Factsheet. Available at: 5A158 (accessed 26/04/2023) 2 Resuscitation to Recovery (2017) A National Framework to improve care of people with out-of-hospital cardiac arrest in England. Available at: (accessed 26/04/2023) 3. General Dental Council (2018)GDC Guidance. Available at: (accessed 26/04/2023) 4. Jon Andersen (2016) AED. www.cpd4dentalnurses.co.uk 5. Royal Life Saving Society UK (2012) Guarding Lives Saving Lives: Warwickshire 6. Resuscitation Council (UK) (2017) ‘Quality standards for cardiopulmonary resuscitation and training’. Available at: (accessed 26/04/2023) 7. Resuscitation Council (UK) (2021) Adult basic life support Guidelines. Available at: (accessed 26/04/2023) 8. Resuscitation Council UK (2023) Guidance: COVID-19 Update to Resuscitation Council UK (RCUK) guidance for practice. Available at: (accessed 27/04/2023) 9. Resuscitation Council (UK) (2017) Quality Standards: Primary dental care equipment list. Available at: (accessed 27/04/2023) 10. Resuscitation Council (UK) 2017 Standard sign for AED. Available at: Accessed (26/04/2023)
5025
https://brilliant.org/landing/basic-number-theory/divisibility/
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5026
https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Book%3A_Applications_of_Maxwells_Equations_(Cochran_and_Heinrich)/12%3A_Waveguides/12.01%3A_Simple_Transverse_Electric_Modes
Skip to main content 12.1: Simple Transverse Electric Modes Last updated : Jun 21, 2021 Save as PDF 12: Waveguides 12.2: Higher Order Modes Buy Print CopyView on Commons Donate Page ID : 22734 John F. Cochran and Bretislav Heinrich Simon Fraser University ( \newcommand{\kernel}{\mathrm{null}\,}) Consider two infinite plane waves of circular frequency ω oscillating in phase, and such that their propagation vectors lie in the x-z plane and make the angles θ with the z-axis: one wave has a positive x-component of wavevector, the other has a negative x-component of wave-vector, as illustrated in Figure . Explicit expressions for the electric and magnetic field components of these waves for the case in which the electric field in each wave has the same amplitude and is polarized along the y-direction are as follows: Wave Number (1) Wave Number (2) In writing these equations it has been assumed that the waves are propagating in a medium characterized by a real dielectric constant , and a magnetic permeability µ0. The wave-vector is , and the wave impedance is Ohms, where Z0 = 377 Ohms. The above fields satisfy Maxwell’s equations. One can now introduce two perfectly conducting infinite planes that lie parallel with the xz plane and which are separated by an arbitrary spacing, b. The plane waves of Figure still satisfy Maxwell’s equations between the conducting surfaces: they also satisfy the required boundary conditions on the electric and magnetic fields. In the first place, there is only one electric field component, Ey, and it is normal to the conducting planes, consequently the tangential component of is zero on the perfectly conducting surfaces as is required. In the second place, the magnetic field components lie parallel with the conducting planes so that the normal component of is zero at the perfectly conducting planes as is required by the considerations discussed in Chpt.(10). The total electric field at any point in the space between the two conducting planes is given by or The components of the magnetic field are given by and Notice that Ey and Hx are both zero, independent of z, on the planes defined by (xk sin θ) = ±/2, ±3/2, ±5/2, etc, i.e. on the planes This means that the wave defined by Equations (),(), and () can propagate along the hollow rectangular pipe bounded by perfectly conducting planes spaced b apart along the y-direction, and spaced a apart along the x-direction where a= m/(k sin θ), and where m is an odd integer, and yet satisfy the boundary conditions imposed by the presence of the perfectly conducting surfaces. The distribution of the electric and magnetic fields across the section of the wave-guide formed by the intersection of the four conducting planes is shown in Figure for the mode corresponding to k sin θ = /a. The width of the wave-guide along the x-direction, a, determines the propagation angle for waves that satisfy the boundary condition Ey = 0 on x=±a/2: The component of the propagation vector parallel with the wave-guide axis, along z, is given by The sum of the squares of these two components must be equal to the square of the wave-vector k, where : from which where is an odd integer. The most important wave-guide mode is that for which m=1, the mode illustrated in Figure (12.1.2). In most applications the wave-guide is filled with air for which = 1. For this m=1 mode, and assuming that (\epsilon_{r})=1.0, the fields are given by where . The mode of Equations (), Figure , is called a transverse electric mode, or a TE mode, because the electric field has no component along the guide axis, i.e. no component along the direction of propagation of the wave-guide mode. Notice that the ratio Ey/Hx = ZG is independent of position inside the wave-guide; in particular, it is independent of position across the wave-guide cross-section. The magnetic field Hx is equivalent to a surface current density Amps/m (from , and Ey has the units of Volts/m. The wave impedance ZG = Ey/Hx therefore has the units of Ohms: it plays a role for wave-guide problems that is similar to the role played by the characteristic impedance for transmission line problems. The analogy between transmission lines and wave-guides is discussed in a very clear manner in the article ” The Elements of Wave Propagation using the Impedance Concept” by H.G.Booker, Electrical Engineering Journal, volume 94, pages 171-202, 1947. The Poynting vector, , associated with the TE10 mode, Equation (), has two components: and The time averaged value of Sx is zero; this corresponds to the fact that no energy, on average, is transported across the guide from one side to the other. There is a non-zero time average for the z-component of the Poynting vector corresponding to energy flow along the guide: It is useful to integrate the time-averaged value of the Poynting vector over the cross-sectional area of the wave-guide in order to obtain the rate at which energy is carried past a particular section of the guide. A simple integration gives The time-averaged energy density associated with a wave-guide mode is given by or For the fundamental TE10 mode, Equations (), one obtains Integrate this energy density across a section of the guide in order to obtain the average energy per unit length of the wave-guide: [ where we have used , and for a waveguide filled with air The velocity with which energy is transported down the guide is called the group velocity, vg. The group velocity must have a value such that its product with the energy density per unit length of guide, Equation (), gives the rate at which energy is transported past a wave-guide section, Equation (): i.e. Thus It is easy to verify by direct differentiation of Equation () that this velocity is also given by the relation Equation () is valid for an arbitrary relative dielectric constant: from () The phase velocity, vphase, on the other hand is obtained from the condition That is z must increase at the rate in order to remain on a crest as the wave propagates along the guide. As the guide wave-vector, kg, approaches zero the phase velocity may become very large- much larger than the velocity of light in vacuum. This occurs because the phase velocity measures the rate of propagation down the guide of two intersecting wave fronts as these waves bounce back and forth across the guide ( see Figure (12.1.3)). This intersection velocity clearly becomes infinitely large in the limit as the wavefronts become parallel with the guide axis, i.e. in the limit as the guide wave-number, kg, goes to zero. The velocity of energy transport down the guide, the group velocity, goes to zero as θ approaches /2, the condition corresponding to waves that simply bounce forth and back along the x-direction between the perfectly conducting planes at x = ±a/2. The group velocity, the velocity with which information can be transmitted down the guide, is always less than the velocity of light in vacuum. The frequency at which the group velocity goes to zero can be calculated from Equation () by setting kg = 0, since the group velocity is proportional to kg from (): The wave-guide is a high pass filter that will transmit energy for frequencies larger than the cut-off frequency ωm. For = 1 and a=1 cm, the cut-off frequency is ωm = 9.42 × 1010 radians/sec. corresponding to a frequency of f=15 GHz. It should be clear from the above construction that Equations () represents the solution of Maxwell’s equations for the TE10 mode that carries energy in the positive z-direction. The TE10 mode that carries energy in the negative z-direction is described by where B is an arbitrary amplitude (NOT the magnetic field!). In order to answer the question of what happens if the frequency is less than the cut-off frequency, ωm, it is best to start from Maxwell’s equations. Consider the case for which there is only a y-component of electric field. From for a time dependence ∼ exp (−iωt), and for the permeability of free space, one obtains and From one obtains The above equations can be combined to give a single second order equation for Ey: For an electric field having the form it follows that or For a frequency less than the cut-off frequency corresponding to ωm = the square of the wave-vector kg becomes negative and therefore its square root becomes pure imaginary. A pure imaginary wave-vector where is a real number, corresponds to a disturbance that decays away exponentially along the guide either to the right or to the left. For example, kg = +i gives a disturbance of the form with magnetic field components (from Equation () and Equation ()) and Using these components, it is easy to show that the time-averaged z-component of the Poynting vector, Sz = −EyHx, is exactly equal to zero. The average energy density stored in the fields is not zero: and These expressions correspond to the energy density stored in the electric field, ( ), and to the energy density stored in the magnetic field, ( ). If a source of energy oscillating at a frequency less than the cut-off frequency is introduced into a wave-guide at some point, the resulting electromagnetic fields will remain localized around the source, and the effective load on the source will be purely reactive for a wave-guide whose walls are perfectly conducting. In the case of a real guide whose walls have some finite resistivity, the load on a source oscillating at a frequency which is less than the cut-off frequency will appear to be partly resistive but mainly reactive. 12: Waveguides 12.2: Higher Order Modes
5027
https://www.dictionary.com/browse/telo
Advertisement Skip to telo-1 telo-2 Advertisement telo- variant of tele-. telodynamic. telo- variant of tele-. telophase. telo- combining form complete; final; perfect telophase end; at the end telencephalon Word History and Origins Origin of telo-1 Example Sentences We were not convinced by a single DNA analysis done in Dolly showing slightly shorter telo­meres, the repetitive DNA sequences at the end of chromosomes that “count” how many times a cell divides. The 24-year-old Neymar is an extremely popular figure in Brazil and around the world, and several times he has appeared on stage with singers at concerts, including joining Michel Telo to sing his hit “Ai se eu te pego.” "And it's a very good party!" he said above speakers blasting hits by Brazilian star Michel Telo, Cuban-American rapper Pitbull, and American pop star Taylor Swift. But for a really authentic experience, no visit to the Argentine capital – if you are travelling as a couple – is complete without a cheeky visit to a “telo”. Discretion is one element of the telo experience that isn’t about to change soon though. Advertisement Advertisement Advertisement Advertisement Browse Follow us Get the Word of the Day every day! By clicking "Sign Up", you are accepting Dictionary.com Terms & Conditions and Privacy Policies.
5028
https://handwiki.org/wiki/Sylvester%27s_criterion
Sylvester's criterion - HandWiki Anonymous Not logged in Create account Log in Hand W iki Search Sylvester's criterion From HandWiki bookmark [x] Namespaces Page Discussion More More Page actions Read View source History ZWI Export Short description: Criterion of positive definiteness of a matrix In mathematics, Sylvester’s criterion is a necessary and sufficient criterion to determine whether a Hermitian matrix is positive-definite. It is named after James Joseph Sylvester. Sylvester's criterion states that a n × n Hermitian matrix M is positive-definite if and only if all the following matrices have a positive determinant: the upper left 1-by-1 corner of M, the upper left 2-by-2 corner of M, the upper left 3-by-3 corner of M, ⋮ M itself. In other words, all of the leading principal minors must be positive. By using appropriate permutations of rows and columns of M, it can also be shown that the positivity of any nested sequence of n principal minors of M is equivalent to M being positive-definite. An analogous theorem holds for characterizing positive-semidefinite Hermitian matrices, except that it is no longer sufficient to consider only the leading principal minors: a Hermitian matrix M is positive-semidefinite if and only if all principal minors of M are nonnegative. [x] Contents 1 Simple proof for special case 2 Proof for general case 3 Proof for the case of positive semidefinite matrices 4 Notes 5 References Simple proof for special case Suppose M n is n×n Hermitian matrix M n†=M n. Let M k,k=1,…n be the principal minor matrices, i.e. the k×k upper left corner matrices. It will be shown that if M n is positive definite, then the principal minors are positive; that is, det M k>0 for all k. M k is positive definite. Indeed, choosing x=(x 1⋮x k 0⋮0)=(x→0⋮0) we can notice that 00 since the determinant is the product of the eigenvalues. To prove the reverse implication, we use induction. The general form of an (n+1)×(n+1) Hermitian matrix is M n+1=(M n v→v→†d)(∗), where M n is an n×n Hermitian matrix, v→ is a vector and d is a real constant. Suppose the criterion holds for M n. Assuming that all the principal minors of M n+1 are positive implies that det M n+1>0, det M n>0, and that M n is positive definite by the inductive hypothesis. Denote x=(x→x n+1) then x†M n+1 x=x→†M n x→+x n+1 x→†v→+x¯n+1 v→†x→+d|x n+1|2 By completing the squares, this last expression is equal to (x→†+v→†M n−1 x¯n+1)M n(x→+x n+1 M n−1 v→)−|x n+1|2 v→†M n−1 v→+d|x n+1|2=(x→+c→)†M n(x→+c→)+|x n+1|2(d−v→†M n−1 v→) where c→=x n+1 M n−1 v→ (note that M n−1 exists because the eigenvalues of M n are all positive.) The first term is positive by the inductive hypothesis. We now examine the sign of the second term. By using the block matrix determinant formula det(A B C D)=det A det(D−C A−1 B) on (∗) we obtain det M n+1=det M n(d−v→†M n−1 v→)>0, which implies d−v→†M n−1 v→>0. Consequently, x†M n+1 x>0. Proof for general case The previous proof is only for nonsingular Hermitian matrix with coefficients in R, and therefore only for nonsingular real-symmetric matrices. Theorem I: A real-symmetric matrix A has nonnegative eigenvalues if and only if A can be factored as A=B T B, and all eigenvalues are positive if and only if B is nonsingular. Proof:Forward implication: If A ∈ Rn×n is symmetric, then, by the spectral theorem, there is an orthogonal matrix P such that A = PDP T , where D = diag(λ 1, λ 2, . . . , λ n) is real diagonal matrix with entries being eigenvalues of A and P is such that its columns are the eigenvectors of A. If λ i ≥ 0 for each i, then D 1/2 exists, so A = PDP T = PD 1/2 D 1/2 P T = B T B for B = D 1/2 P T, and λ i> 0 for each i if and only if B is nonsingular. Reverse implication: Conversely, if A can be factored as A = B T B, then all eigenvalues of A are nonnegative because for any eigenpair (λ,x): λ=x T A x x T x=x T B T B x x T x=‖B x‖2‖x‖2≥0. Theorem II (The Cholesky decomposition): The symmetric matrix A possesses positive pivots if and only if A can be uniquely factored as A = R T R, where R is an upper-triangular matrix with positive diagonal entries. This is known as the Cholesky decomposition of A, and R is called the Cholesky factor of A. Proof:Forward implication: If A possesses positive pivots (therefore A possesses an LU factorization: A=L·U'), then, it has an LDU factorization A=LDU=LDL T in which D=diag(u 11, u 22, . . . , u nn) is the diagonal matrix containing the pivots u ii>0. A=L U′=[1 0⋯0 ℓ 12 1⋯0⋮⋮⋮ℓ 1 n ℓ 2 n⋯1][u 11 u 12⋯u 1 n 0 u 22⋯u 2 n⋮⋮⋮0 0⋯u n n]=L D U=[1 0⋯0 ℓ 12 1⋯0⋮⋮⋮ℓ 1 n ℓ 2 n⋯1][u 11 0⋯0 0 u 22⋯0⋮⋮⋮0 0⋯u n n][1 u 12/u 11⋯u 1 n/u 11 0 1⋯u 2 n/u 22⋮⋮⋮0 0⋯1] By a uniqueness property of the LDU decomposition, the symmetry of A yields: U = L T, consequently A=LDU=LDL T. Setting R = D 1/2 L T where D 1/2 = diag(u 11,u 22,…,u 11) yields the desired factorization, because A=LD 1/2 D 1/2 L T =R T R, and R is upper triangular with positive diagonal entries. Reverse implication: Conversely, if A = RR T, where R is lower triangular with a positive diagonal, then factoring the diagonal entries out of R is as follows: R=L D=[1 0⋯0 r 12/r 11 1⋯0⋮⋮⋮r 1 n/r 11 r 2 n/r 22⋯1][r 11 0⋯0 0 r 22⋯0⋮⋮⋮0 0⋯r n n]. R = LD, where L is a lower triangular matrix with a unit diagonal and D is the diagonal matrix whose diagonal entries are the r ii ’s. Hence D has a positive diagonal and hence D is non-singular. Hence D 2 is a non-singular diagonal matrix. Also, L T is an upper triangular matrix with a unit diagonal. Consequently, A = LD 2 L T is the LDU factorization for A, and thus the pivots must be positive because they are the diagonal entries in D 2. Uniqueness of the Cholesky decomposition: If we have another Cholesky decomposition A = R 1 R 1 T of A, where R 1 is lower triangular with a positive diagonal, then similar to the above we may write R 1 = L 1 D 1, where L 1 is a lower triangular matrix with a unit diagonal and D 1 is a diagonal matrix whose diagonal entries are the same as the corresponding diagonal entries of R 1. Consequently, A = L 1 D 1 2 L 1 T is an LDU factorization for A. By the uniqueness of the LDU factorization of A, we have L 1 = L and D 1 2 = D 2. As both D 1 and D are diagonal matrices with positive diagonal entries, we have D 1 = D. Hence R 1 = L 1 D 1 = LD = R. Hence A has a unique Cholesky decomposition. Theorem III: Let A k be the k×k leading principal submatrix of A n×n. If A has an LU factorization A = LU, where L is a lower triangular matrix with a unit diagonal, then det(A k) = u 11 u 22 · · · u kk, and the k-th pivot is u kk = det(A 1) = a 11 for k = 1, u kk=det(A k)/det(A k−1) for k = 2, 3, . . . , n, where u kk is the (k, k)-th entry of U for all k = 1, 2, . . . , n. Combining Theorem II with Theorem III yields: Statement I: If the symmetric matrix A can be factored as A=R T R where R is an upper-triangular matrix with positive diagonal entries, then all the pivots of A are positive (by Theorem II), therefore all the leading principal minors of A are positive (by Theorem III). Statement II: If the nonsingular n × n symmetric matrix A can be factored as A=B T B, then the QR decomposition (closely related to Gram-Schmidt process) of B (B = QR) yields: A=B T B=R T Q T Q R=R T R, where Q is orthogonal matrix and R is upper triangular matrix. As A is non-singular and A=R T R, it follows that all the diagonal entries of R are non-zero. Let r jj be the (j, j)-th entry of R for all j = 1, 2, . . . , n. Then r jj ≠ 0 for all j = 1, 2, . . . , n. Let F be a diagonal matrix, and let f jj be the (j, j)-th entry of F for all j = 1, 2, . . . , n. For all j = 1, 2, . . . , n, we set f jj = 1 if r jj> 0, and we set f jj = -1 if r jj< 0. Then F T F=I n, the n × n identity matrix. Let S=FR. Then S is an upper-triangular matrix with all diagonal entries being positive. Hence we have A=R T R=R T F T F R=S T S, for some upper-triangular matrix S with all diagonal entries being positive. Namely Statement II requires the non-singularity of the symmetric matrix A. Combining Theorem I with Statement I and Statement II yields: Statement III: If the real-symmetric matrix A is positive definite then A possess factorization of the form A=B T B, where B is nonsingular (Theorem I), the expression A=B T B implies that A possess factorization of the form A=R T R where R is an upper-triangular matrix with positive diagonal entries (Statement II), therefore all the leading principal minors of A are positive (Statement I). In other words, Statement III proves the "only if" part of Sylvester's Criterion for non-singular real-symmetric matrices. Sylvester's Criterion: The real-symmetric matrix A is positive definite if and only if all the leading principal minors of A are positive. Proof for the case of positive semidefinite matrices Let M n be an n x n Hermitian matrix. Suppose M n is semidefinite. Essentially the same proof as for the case that M n is strictly positive definite shows that all principal minors (not necessarily the leading principal minors) are non-negative. For the reverse implication, it suffices to show that if M n then for all t>0, all leading principal minors of the Hermitian matrix M n+t I n are strictly positive, where I n is the n x n identity matrix. Indeed, from the positive definite case, we would know that the matrices M n+t I n are strictly positive definite. Since the limit of positive definite matrices is always positive semidefinite, we can take t→0 to conclude. To show this, let M k be the k th leading principal submatrix of M n. We know that q k(t)=det(M k+t I k) is a polynomial in t, related to the characteristic polynomial p M k via q k(t)=(−1)k p M k(−t). We use the identity in Characteristic polynomial to write q k(t)=∑j=0 k t k−j tr(⋀j M k), where tr(⋀j M k) is the trace of the j th exterior power of M k. From Minor (linear algebra), we know that the entries in the matrix expansion of ⋀j M k (for j > 0) are just the minors of M k. In particular, the diagonal entries are the principal minors of M k, which of course are also principal minors of M n, and are thus non-negative. Since the trace of a matrix is the sum of the diagonal entries, it follows that tr(⋀j M k)≥0. Thus the coefficient of t k−j in q k(t) is non-negative for all j > 0. For j = 0, it is clear that the coefficient is 1. In particular, q k(t)>0 for all t > 0, which is what was required to show. Notes ↑Horn, Roger A.; Johnson, Charles R. (1985), Matrix Analysis, Cambridge University Press, ISBN978-0-521-38632-6. See Theorem 7.2.5. ↑Carl D. Meyer, Matrix Analysis and Applied Linear Algebra. See section 7.6 Positive Definite Matrices, page 566 ↑Prussing, John E. (1986), "The Principal Minor Test for Semidefinite Matrices", Journal of Guidance, Control, and Dynamics9 (1): 121–122, doi:10.2514/3.20077, Bibcode:1986JGCD....9..121P, archived from the original on 2017-01-07, retrieved 2017-09-28 ↑Carl D. Meyer, Matrix Analysis and Applied Linear Algebra. See section 7.6 Positive Definite Matrices, page 558 ↑Carl D. Meyer, Matrix Analysis and Applied Linear Algebra. See section 3.10 The LU Factorization, Example 3.10.7, page 154 ↑Carl D. Meyer, Matrix Analysis and Applied Linear Algebra. See section 6.1 Determinants, Exercise 6.1.16, page 474 References Gilbert, George T. (1991), "Positive definite matrices and Sylvester's criterion", The American Mathematical Monthly (Mathematical Association of America) 98 (1): 44–46, doi:10.2307/2324036, ISSN0002-9890. Horn, Roger A.; Johnson, Charles R. (1985), Matrix Analysis, Cambridge University Press, ISBN978-0-521-38632-6. Theorem 7.2.5. Carl D. Meyer (June 2000), Matrix Analysis and Applied Linear Algebra, SIAM, ISBN0-89871-454-0. fr:Matrice définie positive#Critère de Sylvester 0.00 (0 votes) Original source: criterion. 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5029
https://my.clevelandclinic.org/health/diseases/25138-chorioretinitis
Chorioretinitis: Symptoms, Causes & Treatment Options Locations: Abu Dhabi|Canada|Florida|London|Nevada|Ohio| 800.223.2273 100 Years of Cleveland Clinic MyChart Need Help? Careers Donate now Search ClevelandClinic.org Find a Provider Locations & Directions Patients & Visitors Health Library Institutes & Departments Appointments Home/ Health Library/ Diseases & Conditions/ Chorioretinitis Advertisement Advertisement Chorioretinitis Chorioretinitis — a type of posterior uveitis — is inflammation of your retina and choroid. As your choroid supports your retina’s outer layers, inflammation can lead to complications that threaten your vision. Immediate treatment by a healthcare provider is critical. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services.Policy Care at Cleveland Clinic Eye Care Find a Doctor and Specialists Make an Appointment ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With Overview What is chorioretinitis? Chorioretinitis is a type of uveitis, an eye inflammation. In chorioretinitis, the inflammation affects the posterior (back parts) of your eye — your choroid and retina. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services.Policy Your choroid is a layer of eye tissue that has blood vessels located between your retina and sclera. Your sclera is the strong tissue that you might call “the white of your eye.” You can lose vision if you have issues with the choroid because it provides your retina with nutrients. Many things can cause your retina to swell, including infections. Some of these infections are present at birth (congenital). Others happen over time (acquired). When the inflammation is bad enough to harm your eye, the chorioretinitis becomes chorioretinopathy. The “-pathy” refers to the damage. Contact an eye care professional if you have symptoms like blurred vision, eye pain or sensitivity to light. Symptoms and Causes What are the symptoms of chorioretinitis? Chorioretinitis symptoms may include: Seeing black spots, floaters or flashes of light. Red eyes. Blurred vision. Eye pain. Sensitivity to light (photophobia). Excessive eye watering or tearing. Difficulty seeing colors. What causes chorioretinitis? There are infectious causes of chorioretinitis and non-infectious causes. Infectious causes of chorioretinitis These infections can be present when you’re born or happen over time. Infectious diseases that can cause chorioretinitis include: Toxoplasmosis. This is the most common cause of chorioretinitis. Tuberculosis. Cytomegalovirus. Syphilis. Advertisement Other infections are caused by viruses like Zika, rubella and varicella-zoster, as well as fungal infections. Non-infectious causes of chorioretinitis Non-infectious conditions that can cause chorioretinitis include: Eye injuries. Autoimmune and inflammatory illnesses, including rheumatoid arthritis, sarcoidosis and Behçet’s disease. Some types of cancer, like lymphoma or metastatic disease Some genetic conditions. What are the complications of chorioretinitis? Untreated, chorioretinitis can cause loss of vision that can become permanent. Diagnosis and Tests How is chorioretinitis diagnosed? An eye care provider will be able to see your retina if they do a dilated eye exam. They’ll give you eye drops to make your eyes wider (dilate) so they can see all the way to the back of your eye. Sometimes, providers can see chorioretinal scars, or pieces of tissue or patches of color at the back of your eye. What tests will be done to diagnose chorioretinitis? In addition to the eye exam, your provider may do the following tests: Lab tests on material swabbed from your eye. These tests can help identify infections. Optical coherence tomography. This is an imaging test using reflected light to take pictures of the back of your eye. Fluorescein angiography. This test lets your provider see the blood vessels at the back of your eye. Indocyanine green (ICG) angiography. This imaging test lets your provider see the blood vessels of the choroid, the layer under your retina. If you have another condition that is causing chorioretinitis, your provider might order additional tests. Management and Treatment How is chorioretinitis treated? Ophthalmologists (eye doctors) treat most cases of chorioretinitis with steroids to reduce inflammation. But you won’t be able to take steroids for an extended length of time. So, your eye care provider may work with a rheumatologist to prescribe steroid-sparing medications that suppress your immune system. In chorioretinitis caused by an infection, your provider will prescribe something to treat the infection, depending on the cause of the infection. These medicines include: Antibiotics. Antifungals. Antiparasitics. Antivirals. In very rare cases, a provider might suggest a vitrectomy. This surgery removes fluid and debris from your eye. You should always take your medications for as long as your prescription is written and in the way that your provider recommends. Care at Cleveland Clinic Eye Care Find a Doctor and Specialists Make an Appointment Outlook / Prognosis What can I expect if I have chorioretinitis? If you have chorioretinitis and you receive treatment for it and/or the underlying condition causing it, your outlook is good. When you have this condition, you’ll need frequent visits with your ophthalmologist. Many people don’t have lasting issues with their vision. But chorioretinitis can affect your vision if it’s not treated. One complication is choriogenin atrophy, which means that the affected parts of your eye — your retina and choroid — may become smaller and not work as well as they should. Advertisement Untreated chorioretinitis can lead to vision loss, even blindness. Prevention Can chorioretinitis be prevented? You may be able to reduce your risk of chorioretinitis by: Taking care to prevent infections. These actions can include washing your hands thoroughly, avoiding people who are sick and practicing safe sex. Making and keeping appointments with eye care providers according to the schedule they suggest. Contacting an eye care provider if you have any changes in your vision in between appointments. Taking care of your general health by eating well, getting exercise and managing chronic conditions. Living With When should I see my healthcare provider? If you have a chorioretinitis diagnosis, your provider will tell you what issues may come up with your eyes. If you have a treatment plan and you feel like it’s not working, you should discuss this with your provider. At any time, with or without an eye disease diagnosis, you should contact an eye care provider or go to an emergency room if you have any type of eye injury or a sudden loss of vision. What questions should I ask my doctor? You may want to ask your healthcare provider questions about chorioretinitis, including: What’s the cause of my illness? What treatment do you recommend? Are there side effects to this treatment? What complications should I look out for? Is this illness likely to affect anyone else in my family? Advertisement A note from Cleveland Clinic Most people have times when their eyes water or feel sore. But if you have blurred vision, eye pain, trouble seeing colors or a lot of floaters, contact a healthcare provider right away — it could be chorioretinitis, inflammation of the choroid and retina of your eye. The sooner you receive treatment, the better the outcome. Advertisement Care at Cleveland Clinic Your eyes let you see the world. That’s why it’s important to take care of them. Cleveland Clinic offers comprehensive ophthalmology services to help you do that. Eye Care Find a Doctor and Specialists Make an Appointment Medically Reviewed Last reviewed on 07/19/2023. Learn more about the Health Library and our editorial process. 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5030
https://brilliant.org/wiki/reciprocal-trigonometric-functions/
Reciprocal Trigonometric Functions Sign up with Facebook or Sign up manually Already have an account? Log in here. Mei Li and Jimin Khim contributed We first explore the reciprocal trigonometric functions by studying the relationships between side lengths in a right triangle. Contents Reciprocal Trigonometric Functions Behavior of Reciprocal Trigonometric Functions Examples Reciprocal Trigonometric Functions Recall that the trigonometric functions relate the angles in a right triangle to the ratios of the sides. Given the following triangle: with 0∘<θ<2π​, we have the basic trigonometric functions ​sinθ=cb​,​cosθ=ca​,​tanθ=ab​​ and the reciprocal trigonometric functions cscθsecθcotθ​=sinθ1​=bc​=cosθ1​=ac​=tanθ1​=ba​.​ We can also connect the reciprocal trigonometric functions to the unit circle, similar to the way we connected the basic trigonometric functions to the unit circle: Unit Circle We have cscθsecθcotθ​=sinθ1​=yradius​=y1​=cosθ1​=xradius​=x1​=tanθ1​=yx​.​ There are certain values for the reciprocal trigonometric functions which are useful to remember: θcscθsecθcotθ​0∘∞4​2​∞​6π​=30∘1​2​3​2​3​​4π​=45∘2​2​2​2​1​3π​=60∘3​2​1​2​3​1​​2π​=90∘4​2​∞0​​ By remembering the specific values of basic trigonometric functions and the relationships between the basic and reciprocal functions above, these specific values are readily recalled. Behavior of Reciprocal Trigonometric Functions We have seen that the basic trigonometric functions have the following behavior in the four quadrants of the plane: Since all of the basic trigonometric functions are positive in the first quadrant, the reciprocal trigonometric functions are also positive in the first quadrant. Similarly, we have the behavior for the reciprocal trigonometric functions in the remaining quadrants of the plane: Examples What are the values of θ in the range 0≤θ<2π such that cotθ=1? From the unit circle visualization above, we see that θ=4π​and θ=45π​ satisfy sin(4π​)sin(45π​)​=cos(4π​)=22​​=cos(45π​)=−22​​.​ Therefore, these values satisfy cotθ=sinθcosθ​=1. We also observe that the line y=x intersects the unit circles for only these two values of θ, so the values of θ satisfying the required conditions are θ=4π​ and θ=45π​. □​ For which values of θ in the range 0≤θ<2π are the functions cscθ,secθ, and cotθ not defined? From the above definition of cscθ,secθ, and cotθ, we see that cscθ=sinθ1​ is not defined when sinθ=0, which occurs for θ=0,π. Similarly, secθ=cosθ1​ is undefined for cosθ=0, which occurs when θ=2π​,23π​; cotθ=sinθcosθ​ is undefined for sinθ=0, which occurs when θ=0,π. □​ Cite as: Reciprocal Trigonometric Functions. Brilliant.org. Retrieved 12:16, September 28, 2025, from
5031
https://eventuallyalmosteverywhere.wordpress.com/tag/isogonal-conjugate/
Skip to primary content Skip to secondary content Eventually Almost Everywhere A blog about probability and olympiads by Dominic Yeo Tag Archives: isogonal conjugate Harmonic ranges and Balkan MO 2018 Q1 Posted on by dominicyeo A discussion of the non-geometry questions {Q2,Q3,Q4} on the Balkan MO 2018, held in Serbia, may be found here. A blog post about the UK team’s experience is here, and a more formal pdf report is here. Balkan MO 2018 Problem One A quadrilateral ABCD is inscribed in a circle , where AB>CD, and AB is not parallel to CD. Point M is the intersection of the diagonals AC and BD and the perpendicular from M to AB intersects the segment AB at the point E. If EM bisects the angle CED, prove that AB is a diameter of . I do not think that this was the hardest question on the paper, but I have the most to say about it, so it gets its own post. The section entitled ‘Step One’ contains (including the exercise at the end) a complete solution which only uses familiar material. The remaining sections have to quote some more obscure material, and may be of less interest to inexperienced readers, for whom many other Balkan and IMO geometry problems might be more appropriate. Although I’ve been working hard to improve my geometry over the past couple of years, my attitude to the subject remains recreational. I prefer problems with a puzzle-like quality rather than this sort of question, whose statement is, after a little thought, not so surprising, even if most proof methods are either complicated (but elementary) or exotic. I feel most approaches to this problem require three steps: it’s easy to read a solution and forget that the first step really is a step! I’m fairly vigorously opposed to software diagrams, as at least for me they discourage exactly the sort of insights one is generally hoping for. If you are reading this section carefully, almost certainly the most useful approach is to draw your own diagram, moderately accurate. There are only five points, though you might like to peek at Step Zero to inform drawing an accurate enough diagram without needing to apply the condition by eye. Step Zero: Introduce X, the intersection of AD and BC. To follow through any synthetic approach, it’s essential to have a good perspective on what the diagram means, and you will almost certainly need to introduce X to get such a perspective. Here are a couple of reasons why you might think to introduce X: If the conclusion is true, then , and so M lies on two altitudes, and thus is the orthocentre of some triangle. Which triangle? It’s triangle AXB. Alternatively, the corresponding altitude is an angle bisector of the pedal triangle, and so the given diagram might remind you very strongly of this. Which triangle has pedal triangle CED? It’s AXB again. If your diagram was accurate enough (and since part of the statement is a ‘given…’ this is not so easy) you might have noticed that AD, ME and BC were concurrent. Where? At X:= AD n BC, obviously. In a similar vein, if the conclusion is true, then ADME and BEMC are both cyclic, and we are given ABCD cyclic. The radical axes of these three circles are AD, ME, and BC, so it is reasonable to guess that X, the (hypothesised) point of concurrence is relevant. See later. You are given part of a complete quadrilateral (since M is one of the intersection points of quadrilateral ABCD$ – it might well be useful to complete it! Random luck. It’s not unreasonable to consider arbitrary intersections, though this can be a low-reward strategy in general. If you did introduce X for no reason, you then had to guess, observe or realise that X, M and E should be collinear. Step One: Proving X, M, and E are collinear. This is harder than Step Two I think, so is postponed. Step Two: showing the result, given X,M,E collinear The official solution proposes introducing the reflection of A in E, which is certainly a good way to get lots of equal angles into useful places rather than not-quite-useful places. However, probably one didn’t spot this. Whether or not this was your motivation in the first place, once X is present, it’s natural to look for an argument based on the radical axis configuration. Our conclusion is equivalent to showing that ADME or BEMC are cyclic, and obviously ABCD is given as cyclic. However, motivated by the radical axis configuration (Which you can look up – but I recommend not getting distracted by what radical axis means at this stage. It’s a theorem concerning when three pairs of points form three cyclic quadrilaterals, and it has a valid converse! I also recommend not drawing any circles when thinking about the diagram.) let E’ be second intersection of circles ADM and BMC. We know that E’ lies on line XM, and so it suffices to show that E’=E. But by chasing angles in the cyclic quadrilaterals involving E’, we find that if , then , and so , which after a bit of thought implies triangle AXB is isosceles, which contradicts the given assumptions. Step One: Proving X, M, and E are collinear By introducing enough extra notation and additional structure, one can prove this part by similar triangles. I think a natural approach in a question with significant symmetry is to use the sine rule repeatedly. This has pros and cons: Disadvantage: it’s easy to get into an endless sequence of mindless calculations, which don’t go anywhere and leads more towards frustration than towards insight. Advantage: one can plan out the calculation without actually doing it. Imagine, to give a completely hypothetical example, trying to plan such an approach in a lurching Serbian minibus with only one diagram. You establish which ratios can be calculated in terms of other ratios, and wait until you’re back in a quiet room actually to do it. You might try to show that directly by such a method, but I couldn’t make it work. I could plan out the following though: Start with some labelling. I write for , and a,b for . The goal is to prove that and are complementary by showing that . Will also refer to for when necessary. The first ratio of sines is the easier one. Using the equal length MX in triangle DXM, triangle CMX, and then the sine rule in triangle DXC, obtain . We can obtain , but this could get complicated. However, by exploiting the equal angles , we can derive . But of course, ABCD$is cyclic, and so there are relevant similar triangles, from which . So in fact we have shown , as we wanted since now we know: (1) We need to be careful as this doesn’t immediately imply and . (For example, we need to exclude ! It’s useful to exploit the fact that both a and b are obtuse here. For this type of thing, it’s more useful to focus on showing uniqueness (we definitely know one solution!) rather than finding all solutions. We are essentially asked to show uniqueness of a solution to an equation like (2) where . After suitable rearranging, (2) determines , and so certainly has at most one solution in any interval of width less than . This is a standard issue when using this type of argument and it’s important to know how roughly how to resolve such issues, as you wouldn’t want to waste significant competition time on such technicalities. As an exercise, you can try to prove Step Two using this method. A hint: suppose M is not the orthocentre of triangle AXB. Introduce points C’,D’ such that . Now AE bisects but also . Can you use this to find two congruent triangles which can’t possibly actually be congruent? An alternative synthetic approach UK student Alex started with the following observation. Simple angle-chasing in cyclic quadrilateral ABCD reveals that (3) But we are given that M lies on the angle bisector of $\angle CED$. So we make the following claim. Claim: the only point M which lies on the angle bisector and satisfies (3) is the incentre of triangle CED. Remark: This claim is false. However, it is true that such a point can only be the incentre or E-excentre of triangle CED. One could salvage the original by restricting M to lie inside the triangle. Remark: As was heavily discussed, this claim is certainly not well-known. It is very believable, but it is also not obvious either. An approach by ratios of sines, for example, as in the solution given above, seems rather tricky. Aron’s argument below is lovely, but again brief is not equal to easy’! Proof of claim (Aron): Write and . Consider the altitude MX in triangle MDC. This is isogonal in this triangle to line ME, because the angles and are interchanged at M. This means that the circumcentre of triangle MDC lies on ME. (Perhaps you are more familiar with the stronger statement that the orthocentre and circumcentre – eg of triangle MDC – are isogonal conjugates.) But the circumcircle of triangle MDC also lies on the perpendicular bisector of CD, and this meets the angle bisector on the circumcircle of triangle CED. Indeed, this intersection point is the arc midpoint of CD, and this really is well-known to be the circumcentre of , the circle which includes the incentre and the E-excentre, and so this characterises the two possibilities for M, as required. Harmonic ranges In the end, the most straightforward approach to this question was to use harmonic ranges. Personally, I would use this to complete what I referred to as Step One, namely showing X,M,E collinear. I feel the radical axis argument given above is a more natural way to handle the second step, though one can also deploy projective theory for this too in relatively few steps. This is not the place for an in-depth introduction to harmonic ranges. However, I think less experienced students are often confused about when they should consider looking for them, so I’ll try to focus on this. What is it? Study four points A,B,C,D on a line , grouped into two pairs (A,B),(C,D)$ Then define the cross-ratio to be (4) We say that (A,B;C,D) form a harmonic range (or harmonic bundle, harmonic system etc etc.) if their cross-ratio is -1. This certainly implies that one of (C,D) lies between A and B, and the other lies outside. Note that this is a property of two pairs of points, not of four points! (A,B;C,D) harmonic does not imply (A,C; B,D) harmonic and so on. Crucially, there is an analogous definition for two pairs of points lying on a given circle. What can you do with harmonic ranges? There are two reasons why they are useful in solving geometry problems: They often appear in standard configurations and given configurations! Given one harmonic range, there are natural ways to generate other harmonic ranges. We’ll discuss both of these in a second, but a rough outline of a typical proof using harmonic ranges is as follows. First, identify a harmonic range in the configuration, perhaps using a standard sub-configuration; then, project this first harmonic range around to find some new, perhaps less obvious, harmonic ranges; finally, use some converse result to recover a property about the diagram from your final harmonic range. We need to discuss the two useful reasons given above in more detail: Take a triangle ABC, and consider the intersection points D,E of the internal and external A-angle bisectors with the opposite side BC. Can you prove (for example using a theorem about lengths in the angle bisector configuration…) that (B,C; D,E) is harmonic? A related example occurs when you have both Ceva’s configuration and Menelaus’s transversal present in a given triangle, as you then have a harmonic range too. (See the suggested notes.) One of the points may be the point at infinity on . Without getting into philosophy, can you see how to choose C so that is harmonic? This is a very very useful example. There are plenty of good examples for cyclic ranges too, which you can explore yourself. Harmonic ranges live in the world known as projective geometry. What this means in general is not relevant here, but it’s a good mnemonic for remembering that one can project one harmonic range to acquire another. The most simple example is this. Given A,B,C,D on a line , let P be some point not on . The set of lines (PA,PB,PC,PD) is often referred to as a pencil. Now, consider intersecting this pencil with a different line (again not through P) to obtain a new set of points (A’,B’,C’,D’). The key fact is that if (A,B; C,D) is harmonic, then (A’,B’; C’,D’) is also harmonic! Not only does this give a new harmonic range, it establishes that the harmonic property really depends on the pencil of lines, rather than the choice of . Letting vary, we get an infinite collection of harmonic ranges. So if your diagram has a suggestive pencil of four lines, this is a promising sign that harmonic ranges may have value. One can also project between lines and circles and from circles to circles, and typically you will need to do this. How do you prove the results? If you proved the first example above using the angle bisector theorems, you might ask `how do you prove the angle bisector theorem’? Well, there are elegant synthetic methods, but the sine rule is a fail-safe mode of attack too. Essentially, almost all results about harmonic ranges can be proved using the sine rule, perhaps with a bit of help from other standard length-comparison results, in particular Menelaus, Ceva, and trigonometric Ceva. As we’ve seen in the first attempt at Step One, sine rule calculations can be arduous. Projecting harmonic ranges can be a shortcut through such calculations, provided you know enough examples. How do I know when to use them? This is really just a reiteration: If you are given a configuration and you recognise part of the diagram as a harmonic range, it might well be worth pursuing this. If you can’t project it into any useful other harmonic range (even after, for example, introducing one extra intersection point), this might lead nowhere, but you’ll probably find something. If you see that part of the diagram is well-suited for projecting harmonic ranges into other harmonic ranges, this is relevant. For example, if there are several lines through one point, particularly if that point also lies on a relevant circle. Similarly, if you require some sort of symmetric result like ‘points have some tangency condition iff points have the same tangency condition’, then consider whether the condition has a harmonic range interpretation, and whether can be projected onto . If it feels like the problem could be solved by a giant sine rule calculation comparing various ratios, it might be amenable to harmonic range analysis, so long as you find a first example! Where can I find actual details?Because this is a report on a contest, rather than a set of lecture notes, the level of detail given here is intentionally very low. Though I hope it gives a useful overview of why such approaches might be useful, perhaps especially for those students who have a passing familiarity with harmonic ranges, but are not yet fluent at successfully applying the methods in actual problems. The detail is important though, and I recommend these resources, among many articles on the internet: Alexander Remorov’s sheet on Projective Geometry, which also includes a discussion of polars. My own knowledge of the subject is particularly indebted to this source. I like Question 4. Sections 9.2–9.4 of Evan Chen’s recent book Euclidean Geometry in Mathematical Olympiads includes an ideally compact repository of useful statements. Problems, some of which veer into more challenging territory, are at the end of the section. Symmedians and Balkan MO 2017 Q2 Posted on by dominicyeo 1 While I was away, I wrote about my latest approach to teaching geometry at olympiad camps. This post will end up being about Q2 from the Balkan MO which took place yesterday in Macedonia, but first there is quite a long prelude. My solution, and probably many solutions, to this problem made use of a standard configuration in triangle geometry, namely the symmedian. I want to introduce the configuration, give some simpler examples in practice, and along the way talk about my slightly patched-together philosophy about the merits of practising Euclidean geometry. The symmedian Draw a triangle ABC, with A at the top of the page, and extend the rays AB and AC. The median is the line from A through M, the midpoint of BC. Now take points D and E on AB and AC respectively. The following properties are equivalent: DE is parallel to BC; triangle ADE is similar to triangle ABC; the median of ABC passes through the midpoint of DE, and thus is also the median of ADE. I think it’s a little awkward to prove either of the first two from the third – ratios of areas works – but the rest of the equivalences are straightforward. Later I’m going to talk about the difference between an exercise and a problem. These are all, at best, exercises. Now take B’ on the ray AC, and C’ on the ray AB such that triangle AB’C’ is similar to triangle ABC. One way to achieve this is to take B’ and C’ to be the reflections in the angle bisector of A of B and C respectively (so then AB’=AB and AC’=AC). We say the line B’C’ is antiparallel to BC, as is any other line DE parallel to B’C’. (Probably this should say ‘with respect to triangle ABC’ or similar, but the context here is very clear, and I want this to seem natural rather than opaque.) Note that DE is an antiparallel line iff BCED is a cyclic quadrilateral. We should remember that, as cyclic quadrilaterals are the signposts for progress in both exercises and problems. The median of triangle AB’C’ obeys the same equivalences as described above, and so bisects any antiparallel segment. We call the median of triangle AB’C’ the symmedian of triangle ABC. Using the first set of equivalences, the symmedian of triangle ABC bisects any line antiparallel to BC. Furthermore, by construction, the symmedian is the image of the median of ABC under reflection in the bisector of the angle at A. We sometimes say that the symmedian is the isogonal conjugate of the median. That’s my definition. Note that there was essentially one definition then a couple of easy equivalent definitions. At no point again will I discuss the equivalence of these definitions – we have to take that for granted if we want to get on to more interesting things. Intersection of tangents + concurrency Now, in triangle ABC, draw the tangents to the circumcircle at B and C. These meet at P. It turns out that AP is the symmedian. This could have been our definition of a symmedian, but it wasn’t, so let’s quickly prove this. Trigonometric arguments are very accessible, but I’ll give a Euclidean argument. Draw the antiparallel DE through P, as shown. Our task is to show that EP=PD. At this point, I would again say that this is an exercise. We colour the angle ABC in green. Two angles around point C share this measure by the alternate segment theorem. The angle at E shares this measure because DE is antiparallel. Therefore CPE is isosceles, and so EP=CP. But CP=BP, so by applying the same argument for the orange angles, we get EP=CP=BP=DP as required. Pause to regroup. Proving this wasn’t hard, but it was perhaps surprising. If this was all new to you, and I told you to consider the reflection of the median in the angle bisector, you probably wouldn’t instantly exclaim “it goes through the tangent intersection!” So this is a useful piece of knowledge to have gained, in case we ever have to work with the intersection of two tangents like this. Maybe it won’t be useful, but maybe it will. Maybe the statement itself won’t but some extra insights from the proof will be useful, like the fact that we actually showed P is the centre of the circle BCED, and thus angles ECD=EBD=90. A second property is that in a triangle ABC, the symmedian from A, the symmedian from B and the symmedian from C intersection at, naturally, the symmedian point, which is usually denoted K. This comes from the fact that each symmedian is the isogonal conjugate of the respective median, and the medians are known to concur at the centroid. I’m not going to get into this now. Configurations – an example Here’s a problem. Take an isosceles trapezium ABCD as shown (ie throughout I don’t want to worry about alternative diagrams). Let M be the midpoint of AD, and let E be the point on CM such that angle DBM = EBA. Prove that ABCDE is cyclic. Well, certainly ABCD is cyclic. So we just need to show E also lies on that circle. And we have two equal angles, but they aren’t in the right place to conclude this immediately. However, we have angle MCA = DBM = EBA, so ABCE is cyclic, and the result follows. Why is angle MCA = DBM? Well, the isosceles trapezium has an axis of (reflective) symmetry, and MCA is the is image of DBM under that reflection. Simple. If we wanted to do it with congruent triangles, this would all be a bit more laborious. First have to show BD=AC using one set of congruent triangles, then CM=BM using another, finally finishing using DM=MA. This is much less interesting. The symmetry of the configuration is a higher-level observation which could be proved from the axioms of geometry if necessary, but gives us more information more quickly. When we use a configuration like the symmedian configuration, we are really doing a higher-again-level version of this. Anyway, that problem is fine, but it’s not especially difficult. Consider instead the following problem. (I saw this online, possibly with slightly different notation, a few days ago and can no longer find the link. If anyone can help, I will add the link.) Let AB be a chord of a circle, with midpoint M, and let the tangents at A and B meet at P. Consider a line through P which meets the circle at C and D in that order. Extend CM to meet the circle again at E. Show DME is isosceles. Here’s a diagram, though it includes some clues. I thought this was a fun problem, and for a while I couldn’t do it because despite lots of equal angles and equal lengths, I couldn’t conjure any congruent triangles in the right places, and I didn’t care enough about solving it to get involved in trigonometry. Then came the moment of insight. We have a midpoint, and also the intersection of the tangents. So DP is the symmedian of triangle DAB, and DM is the median. This gives us the two equal orange angles. Cyclicity gives us an extra equal angle at E as well. Note now that the situation is very very similar to the previous question (after changing some of the labels), only this time we know ACBDE is cyclic, but don’t know that ABDE is an isosceles trapezium. If ABDE is an isosceles trapezium, we are clearly finished, as then by the same symmetry argument, EM=DM. This direction is probably harder to prove than the direction of the previous problem. Again there are a couple of ways to proceed, but one way is to consider the point E’ such that ABDE’ is an isosceles trapezium, and arguing that E’ lies on the given circle, and the circle through BME, and thus must coincide with E, in a reverse reconstruction argument. Anyway, this is all slightly a matter of taste, but I would say the second problem is much much more fun than the first problem, even though the second part of the solution is basically the first problem but in a more awkward direction. If you’re going to do Euclidean geometry at all (very much another question), I think you should do questions like the second question wherever possible. And the enjoyable ‘aha moment’ came from knowing about the symmedian configuration. Is it really plausible that you’d look at the original diagram (without the dashed orange lines) and think of the antiparallel to AB in triangle DAB through point P? Probably not. So knowing about the configuration gave access to the good bit of a nice problem. ‘Philosophy of this sort of thing’ If the goal was to solve the second problem in a competition, knowing about the symmedian configuration would be a big advantage. I’ve tried to justify a related alternative view that knowing about the configuration gave access to an enjoyable problem. The question is how many configurations to study, and how hard to study them? We tend not to think of cyclic quadrilaterals as a special configuration, but that is what they are. We derived circle theorems from the definition of a circle so that we don’t always have to mark on the centre, every single time we have a cyclic quadrilateral. So becoming familiar with a few more is not unreasonable. In particular, there are times when proofs are more important than statements. In research (certainly mine), understanding how various proofs work is the most important aspect, for when you try to extend them or specialise. And in lots of competition problems, the interesting bit is normally finding the argument rather than basking in wonder at the statement (though sometimes the latter is true too!). To digress briefly. In bridge, I don’t know enough non-obvious motifs in bidding or gameplay to play interesting hands well. I trust that if I thought about some of it very very carefully, I could come up with some of them, especially in gameplay, but not in real time. And it is supposed to be fun right?! Concentrating very very hard to achieve a basic level of competence is not so enjoyable, especially if it’s supposed to be a break from regular work. The end result of this is that I don’t play bridge, which is a shame, because I think the hurdles between where I am currently and a state where I enjoy playing bridge are quite low. If I knew I was going to play bridge regularly, a bit of time reading about conventions would be time well spent. And obviously this applies equally in pursuits which aren’t directly intellectual. Occasionally practising specific skills in isolation broadens overall enjoyment in sport, music, and probably everything. As anyone who’s played in an orchestra knows, there are standard patterns that come up all the time. If you practise these occasionally, you get to a stage where you don’t really need to concentrate that hard in the final movement of Beethoven 5, and instead can listen to the horns, make funny faces at the first violins, and save your mental energy for the handful of non-standard tricky bits. And of course, then move on to more demanding repertoire, where maybe the violas actually get a tune. This is highly subjective, but my view is that in all these examples are broadly similar to configurations in geometry, and in all of them a little goes a long way. How? In lots of the geometry configurations you might meet in, for example, a short session at a training camp, most of the conclusions about the configurations have proofs which, like in our symmedian case, are simple exercises. Once you’ve got over some low initial experience hurdles, you have to trust that you can normally solve any simple exercise if required. If you can’t, moving on and returning later, or asking for help is a good policy. The proof shown above that symmedians pass through tangent meet points (and especially a trigonometric alternative) really isn’t interesting enough to spend hours trying to find it. The statements themselves are more useful and interesting here. And it can often be summarised quite quickly: “symmedians are the isogonal conjugates of the medians, so they bisect antiparallels, meet at K, and pass through the alternate tangent meeting points.” Probably having a picture in your mind is even simpler. There’s a separate question of whether this is worthwhile. I think solving geometry problems occasionally is quite fun, so I guess yes I do think it is worthwhile, but I understand others might not. And if you want to win maths competitions, in the current framework you have to solve geometry problems under time pressure. But from an educational point of view, even though the statements themselves have no real modern research value, I think a) that’s quite a high bar to set, and there’s no a priori reason why they should – >99.9% of things anyone encounters before university have no value to modern research maths; b) in terms of knowledge acquisition, it’s similar in spirit to lots of things that are relevant to later study. I don’t have to solve PDEs very often, but when I do, I hope they are equivalent or similar to one of the small collection of PDEs I do know how to solve. If I worked more with PDEs, the size of this collection would grow naturally, after some initial struggles, and might eventually match my collection of techniques for showing scaling limits of random processes, which is something I need to use often, so the collection is much larger. Maybe that similarity isn’t enough justification by itself, but I think it does mean it can’t be written off as educationally valueless. Balkan MO 2017 Question Two An acute angled triangle ABC is given, with AB Ok, so why have I already written 1500 words about symmedians as a prelude to this problem? Because AL is a symmedian. This was my first observation. This observation is then a route into non-Euclidean solutions. It means, for example, that you can describe the point of concurrency fairly explicitly with reference to triangle ABC. If you wish, you can then proceed using areal coordinates. One member of the UK team, whom I know is perfectly capable of finding a synthetic solution, did this. And why not? It’s a competition, and if you can see a method that will definitely work, and definitely take 45 minutes (or whatever) then that’s good. I was taking a break from work in my office, and had no interest in spending the time evaluating determinants because that isn’t enjoyable at any level, so I focused on the geometry. I think there’s a good moral from the diagram above, which is the first moderately correct one I drew. I often emphasise that drawing an accurate diagram is important, as it increases the chance that you’ll spot key properties. In this case though, where you’re trying to examine a known configuration, I think it’s more important what you choose to include on your diagram, than how accurately you draw it. (In a moment, we’ll see why it definitely wasn’t very accurate.) In particular, what’s not on the diagram? E is not on the diagram, and S got added later (as did the equal length signs in TB and CS, which rather spoil what’s about to happen). My first diagram was wildly incorrect, but it also suggested to me that the line ST was hard to characterise, and that I should start by deducing as much as possible about S and T by themselves. So by symmetry, probably it was enough just to deduce as much as possible about T. Label the angles of triangle ABC as We haven’t yet made any use of the symmedian, but this is clearly where it’ll be useful. Note that if we didn’t know about everything in the prelude, we might well have deduced all of this, but we might not have thought to prove that AL bisects TB unless we’d drawn a very accurate diagram. At this point, we have to trust that we have enough information to delete most of the diagram, leaving just {A,B,C,S,T} and the line AL. There are a few ways to finish, including similar triangles if you try very hard or trigonometry if you do it right, but again knowledge of some standard configurations is useful. Probably the quickest way is to use Ceva’s theorem in triangle ACS. You can also use Menelaus’ theorem in ABC, so long as you know a little bit about where the symmedian meets the opposite side. An alternative is the following. We have a complete quadrilateral here, namely BTCS, and the intersection of all its diagonals. One is A, one is the proposed point of concurrency, and one is the point at infinity, since TB || CS. You can chase that, but I found it more clear to let P be the intersection of ST and BC (which we want to prove lies on AL), then look at the complete quadrilateral ATPB. Then AT and BP meet at C, and AB and TP meet at S. So if we look at where the diagonals of ATPB meet the line CS, we have a harmonic range. If I’d wanted, I could instead have written the prelude about harmonic ranges, but I had fewer ideas how to set these up in a slick Euclidean way. Also, it feels better to talk about the start, rather than the end of a proof, especially when there were alternative endings. Anyway, a harmonic range is a collection of two pairs of points on a line (A, B; C, D), satisfying the following ratio of directed lengths: A classic example is when D is the point at infinity, the RHS is -1, and so C is the midpoint of AB. Being happy about using the point at infinity is a property of projective geometry, of which this is a good first example. Anyway, returning to the problem, we are looking at where the diagonals of ATPB meet line CS, and this pair of points forms a harmonic range with (C,S). TB meets CS at the point at infinity, and so AP meets CS at the midpoint of CS. But from the symmedian configuration, AL bisects CS, so AP and AL are in fact the same line, and so P lies on AL as required. I think was a brilliant example of when knowing a bit of theory is enjoyable. It wasn’t at all obvious initially how to use the symmedian property, but then the observation that TB is antiparallel felt like a satisfying breakthrough, but didn’t immediately kill the problem. Subscribe Subscribed Eventually Almost Everywhere Already have a WordPress.com account? Log in now. Eventually Almost Everywhere Subscribe Subscribed Sign up Log in Report this content View site in Reader Manage subscriptions Collapse this bar Loading Comments... Design a site like this with WordPress.com Get started
5032
https://radiologyassistant.nl/chest/lung-cancer/tnm-classification-8th-edition-1
TNM classification 9ᵗʰ edition Onno Mets and Robin Smithuis Radiology department of the University Medical Center Amsterdam and the Alrijne Hospital, Leiden, the Netherlands Publicationdate This is a summary of the 9th Edition of TNM in Lung Cancer, which is the standard of lung cancer staging as per January 1st, 2025. It is issued by the IASLC (International Association for the Study of Lung Cancer) and replaces the TNM 8th edition. Introduction The purpose of TNM classification is to have a consistent nomenclature for the anatomic extent of a tumor, which facilitates communication on outcomes and applicability of data to an individual patient.The TNM classification system for lung cancer applies to both non-small cell lung carcinoma (NSCLC) as well as all neuroendocrine neoplasms ranging from small cell lung carcinoma (SCLC) to typical carcinoid.It does not apply to pulmonary sarcomas, lymphomas, and other rare tumors. Tables of the TNM 9ᵗʰ edition TNM 9ᵗʰ edition The 9th edition of lung cancer classification is shown in the table . There are only minor differences from the 8th edition.These are shown in red: No changes to the T component. Subdivision of N2 involvement into N2a (single N2 station) and N2b (multilevel stations). Subdivision of M1c status into M1c1 (multiple metastases in a single organ system) and M1c2 (metastases in multiple organ systems). These changes result in a rearrangement of some stage groups as shown in the next table. Several prefix can be used that specify the context of the TNM classification. Clinical stage (c) is determined by all information available pretreatment. Pathologic stage (p) is defined by the results after surgical resection only, and should not be used outside this context. Restaging (y) is used after part or all of the treatment has been given. It can be applied both in the absence of resection (ycTNM) or after resection (ypTNM). It is further suggested to add an evaluation categorization (‘E’) to indicate the type of testing involved: E1 means physical information E2 imaging information E3 tissue information (3a: cytology; 3b: histology) E4 resection. So for example, cT2aN2aM0 E3a means that cytologic proof of N status was used to stage the extent of disease pretreatment, while pT1cN0M0 E4 means that staging was based on definite resection. Subsolid lesions Subsolid lesions are lung cancer with ground glass morphology, which represents lepidic growth (ie. tumor cells proliferating along the surface of alveoli, without invasive growth).Subsolid lesions may or may not exhibit an internal solid component, which is thought to represent the invasive portion of the tumor.For subsolid lung cancers the radiology report should always mention both total lesion size and – if present - also solid component dimensions, as both are of importance.This should be obtained on contiguous thin slice (<1.5 mm) CT reconstructions in lung window. Subsolid lung cancers are classified as shown in the figures on the left. Pure Ground GlassWhen the total size is ≤30 mm, subsolid lesions with pure ground glass morphology are classified based on greatest total lesion dimension, and can at most be classified as ‘in situ’ cancer (cTis).When total lesion size is over the threshold of 30 mm, subsolid lesions are always at least cT1a, regardless of pure ground glass or part-solid morphology. Ground Glass with Solid componentIf an internal solid component is present, subsolid lesions are no longer classified based on total lesion size alone, but on the greatest dimension of the solid component present, which presumes to represent the invasive component as seen on histopathology.For example, a total lesion size of ≤ 30 mm with a small solid component of ≤ 5 mm is classified as cT1mi (minimally invasive). However, a total lesion size of > 30 mm with the same ≤ 5 mm solid component is classified as cT1a, as the total lesion size is over the 30 mm threshold for cT1a classification.Depending on invasive component size, part-solid lesions may be classified as cT1b (solid component of 11-20 mm) or even cT1c (solid component of 21-30 mm) as well. Subsets of T, N and M categories are grouped into certain stages, because these patients share similar prognosis.For example, cT1N0M0 disease (stage IA) has a 5-year survival of ~82%.On the other end of the spectrum, M1c disease (stage IVB) has a 5-year survival of ~7%. The changes in TNM9 result in a rearrangement of T and N categories included in the stage groups IIA, IIB, IIIA, and IIIB (see Table).Other categories are unchanged from the prior edition. T-staging T1 - T4 T1-tumor Tumor size is determined by the greatest dimensions of invasive tumor, based on either pathological information or clinical data (ie. the solid component on contiguous thin slice CT reconstructions, in lung window). Please note that the largest tumor dimension may be found on multiplanar reconstructions instead of the axial plane. T1-tumor ≤ 3 cm, surrounded by lung/visceral pleura, or in lobar or more peripheral bronchus. T1mi Minimally invasive adenocarcinoma. T1a ≤ 1 cm. T1b >1 cm but ≤ 2 cm. T1c >2 cm but ≤ 3 cm. T1 tumor A typical T1 tumor in the left lower lobe (23 mm, cT1c), completely surrounded by pulmonary parenchyma. T2a-tumoris a tumor with any of the following features: 3 cm but ≤ 4 cm. Invades visceral pleura or adjacent lobe. Involves main bronchus (up to but not including the carina) or Associated with atelectasis or obstructive pneumonitis extending to the hilar region, involving either part of or the entire lung. T2b-tumor 4 cm but ≤ 5 cm with or without other T2a tumor features. T2 tumor A typical T2 tumor with atelectasis/pneumonitis of the left lower lobe up to the hilum, due to involvement of the left lower lobe bronchus (31 mm, pT2a). T3-tumoris a tumor with any of the following features: 5 cm but ≤ 7 cm in greatest dimension. Invades parietal pleura or chest wall. Invades pericardium, phrenic nerve, or azygos vein. Invades thoracic nerve roots (ie. T1, T2) or stellate ganglion. Separate tumor nodule(s) in the same lobe. Invasion of other structures by the primary tumor (e.g. phrenic nerve, aorta, chest wall) counts to determine the T category.Direct invasion into lymph nodes is classified as lymph node involvement.Contrarily, direct invasion of an extrathoracic organ (e.g. liver) is not counted as M1 involvement. T3 tumor A typical T3 tumor in the right upper lobe with invasion of the chest wall (59 mm, cT3). T4-tumor 7 cm or invasion mediastinum, thymus, trachea, carina, recurrent laryngeal nerve, vagus nerve, esophagus, diaphragm or invasion heart, great vessels, intrapericardial pulmonary arteries/veins, supra-aortic arteries, brachiocephalic veins, subclavian vessels, vertebral body, lamina, spinal canal, cervical nerve roots, brachial plexus. or separate nodules in different ipsilateral lobe T4 tumor A typical T4 tumor in the right upper lobe with invasion of the mediastinum. Invasion of the right pulmonary artery. Invasion of the carina. Invasion of the left atrium. Staging multiple sites of malignancy Sometimes multiple sites of lung cancer are suspected in a single patient. If possible, try to establish whether the multifocality is based on: Two synchronous lung cancers Separate tumor nodules of the same malignancy Multifocal ground glass adenocarcinoma Pneumonic-type adenocarcinoma These patterns are distinguished because they exhibit different biological behaviour, and different rules of TNM classification apply to them. 1. Synchronous primary malignancies Separate TNM for each tumor. FiguresThis was a T4N0M0 in RLL.This was a T2bN1M0 in LUL 2. Separate tumor nodules of the same malignancy T3 if in same lobe T4 if same side, but different lobe M1a if in contralateral lobe Single N and M for all FiguresThis was a T3N2aM0 in RLL.T3 is based on the separate nodule in the same lobe. 3. Multifocal groundglass or lepidic lesions T-stage according to the most T-dominant lesion Single N and M for all (#/m) indicates multiplicity. When there are 3 lesions it is indicated as 3/m. FiguresThis was a T1aN0M0 (2/m).T1a is based on the dominant lesion in the RLL 4. Diffuse pneumonia-like malignancies T3 if in same lobe T4 if same side, different lobe M1a if in contralateral lobe Single N and M for all FiguresThis was a T4N1M0.T4 is based on the lesion in the RLL N-Staging Adapted from the American Thoracic Society mapping scheme Regional Lymph Node Classification System Regional lymph node involvement in lung cancer includes the intrathoracic, scalene, and supraclavicular nodes. Less common involvement of for example parasternal or axillary nodes is regarded metastatic. The IASLC lymph node classification is used: Low cervical, supraclavicular and sternal notch nodes Upper Paratracheal. Pre-vascular 3A: nodes not adjacent to the trachea like the nodes in station 2, but anterior to the vessels.Pre-vertebral 3P: nodes not adjacent to the trachea, but behind the esophagus, which is prevertebral (3P). Lower paratracheal nodes. Subaortic (A-P window) nodes Para-aortic (ascending aorta or phrenic) nodes. Subcarinal. Paraesophageal. Pulmonary Ligament: nodes lying within the pulmonary ligaments. Hilar nodes: Nodes located outside the mediastinum. CT is unreliable in staging lymph nodes in patients with NSCLC regardless of the threshold size that is chosen. PET-CT is much more reliable in determining the N-status, although false-positives do occur in patients with for example sarcoid, TB and other infections. PET-CT has a high negative predictive value. Level 1: There is an important separation to be made between level 1 (N3) and level 2 and 3 nodes (N2). The lower border of level 1 is the clavicles bilaterally and, in the midline, the upper border of the manubrium. Level 2R: The upper border is the apex of lung and in the midline the upper border of the manubrium. The lower border is intersection of the caudal margin of the innominate vein with the trachea. The medial border is along the left lateral border of the trachea. Level 2L: From the apex of the lung and the upper border of the manubrium to the superior border of the aortic arch. Level 4R: Includes right paratracheal and pretracheal nodes extending to the left lateral border of trachea. From the intersection of caudal margin of innominate vein with the trachea to the lower border of the azygos vein. Level 4L: Nodes to the left of the left lateral border of the trachea, but medial to the ligamentum arteriosum. From the upper margin of the aortic arch to the upper rim of the left main pulmonary artery. Level 5: Subaortic nodes lateral to ligamentum arteriosum. These nodes are not located between the aorta and the pulmonary trunk, but lateral to these vessels. Level 10R: Hilar nodes up to the inferior border of the azygos vein, including those in a precarinal position. Level 10L: Hilar nodes extend all the way up to the upper border of the left pulmonary artery. N1 - N3 nodes N1Metastasis in ipsilateral intrapulmonary, peribronchial or hilar lymph nodes, including nodal involvement by direct extension. N2aMetastasis to a single ipsilateral mediastinal or subcarinal lymph node station. N2bMetastasis to multiple ipsilateral mediastinal and/or subcarinal lymph node stations. N3Metastasis in contralateral hilar or mediastinal lymph nodes or in scalene or supraclavicular lymph nodes . N1 nodesN1-nodes are ipsilateral nodes within the lung and hilar nodes.N1 alters the prognosis but not the management. ImageT2a tumor (33 mm) in the right lower lobe with ipsilateral hilar node metastasis (N1). N2 NodesN2-nodes represent ipsilateral mediastinal or subcarinal lymphadenopathy. N2a - Single N2 station involvement N2b - Multiple N2 station involvement ImageRight sided tumor with lymph node metastases in multiple (4R and 2R) ipsilateral mediastinal stations (N2b). N3-stage disease. N3 NodesN3-nodes represent contralateral mediastinal or contralateral hilar lymphadenopathy or scalene or supraclavicular nodes.This is considered irresectable disease. ImageRight sided tumor with N3 lymph node metastases, including the contralateral mediastinal 4L and 5 station. Two patients with N3-disease. These images are of two different patients with lung cancer in the right lung. ImagesThere are lymph nodes on the contralateral side.If these lymph nodes contain tumor cells, it means N3 disease. Images of a patient with a right sided tumor.There are N3-nodes on the contralateral side and in the right supraclavicular region. Scroll through the images. For a tumor in the right lung the N-stages are: N1 -Ipsilateral peribronchial and/or hilar lymph nodes 10R-14R N2 -Ipsilateral mediastinal and/or midline lymph nodes 2R, 3a/p, 4R, 7, 8R, 9R N3 -Contralateral mediastinal and/or hilar, as well as any supraclavicular lymph nodes 1, 2L, 3aL, 4L, 5, 6, 8L, 9L, 10L-14L For a tumor in the left lung the N-stages are: N1 -Ipsilateral peribronchial and/or hilar lymph nodes 10L-14L N2 -Ipsilateral mediastinal and/or midline lymph nodes 2L, 3a/p, 4L, 7, 8L, 9L N3 -Contralateral mediastinal and/or hilar, as well as any supraclavicular lymph nodes 1, 2R, 3aR, 3pR, 4R, 8R, 9R, 10R-14R M-Staging M-staging is based on the presence of metastases, their location and multiplicity.M0 is no metastases and M1 means distant metastases.A distinction is made between regional or intrathoracic metastatic disease (M1a) and distant metastatic disease, either solitary (M1b) or multiple (M1c), in a single organ system (M1c1) or in multiple organs (M1c2). FigureM1a: Regional (or intrathoracic) metastatic disease defined as malignant pleural or pericardial effusion or nodules, as well as contralateral separate pulmonary nodules. M1b is a single extrathoracic metastasis in a single organ system. It is important to emphasize, that an organ system denotes all sites of an organ system distributed throughout the body, or both sides in case of a paired organ. This means that multiple lesions in different sites of the skeletal system or the skin is M1c1 disease.However, multiple lesions in both the skeletal system and the adrenals is M1c2 disease. FigureM1b: Single extrathoracic metastasis in a single organ system. This can be in a non-regional lymph node (without a liver metastasis) or a single metastasis in the liver (without a non-regional lymph node). FigureM1c1: Multiple extrathoracic metastases in a single organ. Almost every organ may be involved in metastatic disease.Metastases are commonly seen in the adrenals, lymph nodes, the brain, bones and liver. FigureM1c2: Multiple extrathoracic metastases in multiple organs systems. Privacy and cookie notification This website uses cookies.With this we ensure that the website works. Select one of the two options and click 'Accept'. Read the privacy and cookie statement. Optimal: With this choice the website works optimally. We use cookies that are necessary to visit our website. By accepting cookies from external parties you also see videos. Minimal: We use cookies that are necessary to visit our website. Your visit and data will not be tracked.
5033
https://precalculus.flippedmath.com/39-inverse-trig-functions.html
| | | | | --- --- | | Pre-Calculus | | | | --- | | | | | | Home AP PreCalc + Unit 1A Polynomial and Rational Functions > - 1.1 Change in Tandem - 1.2 Rates of Change - 1.3 Rates of Change in Linear and Quadratic Functions - 1.4 Polynomial Functions and Rates of Change - 1.5A Polynomial Functions and Complex Zeros - 1.5B Even and Odd Polynomials - 1.6 Polynomial Functions and End Behavior - Unit 1A Review + Unit 1B Polynomial and Rational Functions > - 1.7A Rational Functions and End Behavior - 1.7B Rational Functions and End Behavior - 1.8 Rational Functions and Zeros - 1.9 Rational Functions and Vertical Asymptotes - 1.10 Rational Functions and Holes - 1.11A Equivalent Expressions and Binomial Thm - 1.11B Polynomial Long Division and Slant Asymptotes - 1.12A Translations of Functions - 1.12B Dilations of Functions - 1.13 Function Model Selection - 1.14 Function Model Construction - Unit 1B Review + Unit 2A Exponential and Logarithmic Functions > - 2.1 Change in Arithmetic and Geometric Sequences - 2.2 Change in Linear and Exponential Functions - 2.3 Exponential Functions - 2.4 Exponential Function Manipulation - 2.5.A Exponential Function Context and Data Modeling - 2.5.B Exponential Function Context and Data Modeling - 2.6 Competing Function Model Validation - 2.7A Composition of Functions - 2.7B Composition of Functions - 2.8 Inverse Functions - Unit 2A Review + Unit 2B Exponential and Logarithmic Functions > - 2.9 Logarithmic Expressions - 2.10 Inverses of Exponential Functions - 2.11 Logarithmic Functions - 2.12 Logarithmic Function Manipulation - 2.13A Exponential and Logarithmic Equations and Inequalities - 2.13B Exponential and Logarithmic Equations and Inequalities - 2.14 Logarithmic Function Context and Data Modeling - 2.15 Semi-Log Plots - Unit 2B Review + Unit 3A Trigonometric and Polar Functions > - 3.1 Periodic Phenomena - 3.2A Radians - 3.2B Sine, Cosine, and Tangent - 3.3A Sine and Cosine Function Values - 3.3B Sine and Cosine Function Values - 3.4 Sine and Cosine Function Graphs - 3.5 Sinusoidal Functions - 3.6A Sinusoidal Function Transformations - 3.6B Sinusoidal Function Transformations - 3.7 Sinusoidal Function Context and Data Modeling - Unit 3A Review + Unit 3B Trigonometric and Polar Functions > - 3.8 The Tangent Function - 3.9 Inverse Trig Functions - 3.10 Trig Equations and Inequalities - 3.11 Cosecant, Secant, Cotangent - 3.12A Equivalent Representations of Trig Functions - 3.12B Equivalent Representations of Trig Functions - 3.13 Trig and Polar Coordinates - 3.14A Polar Function Graphs - 3.14B Polar Function Graphs - 3.15 Rates of Change in Polar Functions - Unit 3B Review + Unit 4A Parametrics, Vectors, and Matrices > - 4.1 Parametric Functions - 4.2 Parametric Functions Modeling Planar Motion - 4.3 Parametric Functions and Rates of Change - 4.4 Parametrically Defined Circles and Lines - 4.5 Implicitly Defined Functions - 4.6A Conic Sections - Parabolas - 4.6B Conic Sections - Ellipses - 4.6C Conic Sections - Hyperbolas - 4.7 Parametrization of Implicitly Defined Functions - Unit 4A Review + Unit 4B Parametrics, Vectors, and Matrices > - 4.8A Vectors - 4.8B Vectors - 4.9 Vector-Valued Functions - 4.10 Matrices - 4.11 The Inverse and Determinant of a Matrix - 4.12 Linear Transformations and Matrices - 4.13A Matrices as Functions - 4.13B Matrices as Functions - 4.14 Matrices Modeling Contexts - Unit 4B Review PreCalc (traditional) + Semester 1 > - Unit 1 Intro Skills > 1.0 Marking the Text 1.1 Multiple Representations 1.2 Linear Functions 1.3 Basic Factoring Unit 1 Review Unit 1 Skillz Review - Unit 2 Functions and Limits > Unit 2 Skillz Review 2.1 Function Introduction 2.2 Domain & Range Graphically 2.3 Limits Graphically 2.4 Limits to Infinity Unit 2 Review - Unit 3 Function Analysis > Unit 3 Skillz Review 3.1 Discontinuity and Domain Analytically 3.2 Extrema 3.3 Piecewise FunctIons Unit 3 Review - Unit 4 Function Algebra > 4.1 Transformations 4.2 Even and Odd Functions 4.3 Operations with Functions 4.4 Inverse Functions Unit 4 Review Unit 4 Skillz Review - Unit 5 Polynomials > Unit 5 Skillz Review 5.1 Operations on Polynomials 5.2 Polynomial Divison and Factor Theorem 5.3 Polynomial Graphs Unit 5 Review - Unit 6 Rational Functions > 6.1 Solving Rational Functions 6.2 Complex Fractions and Rationalizing 6.3 Graphing Rational Functions 6.4 Variation and Modeling Unit 6 Review Unit 6 Skillz Review - Unit 7 Exponential and Logarithmic Functions > 7.1 Exponential Functions 7.2 Logarithmic Functions 7.3 Exp and Log Problem Solving Unit 7 Review Unit 7 Skillz Review - SEMESTER EXAM + Semester 2 > - Unit 8 Intro to Trigonometry > 8.1 Angles and Degrees 8.2 Radians 8.3 Velocity 8.4 Basic Trig Functions Unit 8 Review Unit 8 Skillz Review - Unit 9 Unit Circle > 9.1 Reference Triangles and Reciprocal Trig Functions 9.2 Reference and Special Triangles 9.3 Evaluating Trig Functions Unit 9 Review Unit 9 Skillz Review - Unit 10 Graphing Trig Functions > Unit 10 Skillz Review 10.1 Graphing Sine and Cosine 10.2 Phase Shift and Tangent 10.3 Reciprocal Trig Graphs 10.4 Inverse Trig Functions Unit 10 Review - Unit 11 Trig Identities > 11.1 Basic Identities and Trig Algebra 11.2 Negative and Pythagorean Identities 11.3 Sum and Difference Identities 11.4 Double and Half Angle Identities 11.5 Trigonometric Equations Unit 11 Review - Unit 12 Solve Triangles > 12.1 Law of Sines 12.2 Law of Cosines 12.3 Area of Triangles Unit 12 Review Unit 12 Skillz Review - Unit 13 Coordinate Systems > 13.1 Three Dimensional Graphs 13.2 Polar Coordinates Unit 13 Review - Unit 14 Series and Sequences > 14.1 Sequences, Series and Summation 14.2 Arithmetic and Geometric Sequences 14.3 The Binomial Theorem Unit 14 Skillz Review Review Unit 14 - Unit 15 Calculus Intro > Unit 15 Skillz Review 15.1 Limits Analytically 15.2 Definition of the Derivative 15.3 Power Rule 15 Review Teacher Resources FlippedMath.com | | | | | --- | Previous Lesson | 3.9 Inverse Trigonometric Functions | Next Lesson | If you find errors in our work, please let us know at [email protected] so we can fix it. | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- | Packet | | | appc_3.9_packet.pdf | | File Size: | 1108 kb | | File Type: | pdf | Download File ​Want to save money on printing? Support us and buy theAP Pre-Calculus workbookwith all the packets shipped to you as a physical copy. | Practice Solutions | | | appc_3.9_solutions.pdf | | File Size: | 2133 kb | | File Type: | pdf | Download File | Corrective Assignments | | | appc_3.9_ca1.pdf | | File Size: | 221 kb | | File Type: | pdf | Download File | | | appc_3.9_ca2.pdf | | File Size: | 401 kb | | File Type: | pdf | Download File | AP Learning Objectives:3.9A Construct analytical and graphical representations of the inverse of the sine, cosine, and tangent functions over a restricted domain. AP®is a trademark registered and owned by the CollegeBoard, which was not involved in the production of, and does not endorse, this site.
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https://samplecontents.library.ph/wikipedia/wp/m/Modular_arithmetic.htm
Modular arithmetic Modular arithmetic Related subjects: Mathematics About this schools Wikipedia selection SOS Children have produced a selection of wikipedia articles for schools since 2005. SOS Children is the world's largest charity giving orphaned and abandoned children the chance of family life. Modular arithmetic (sometimes called modulo arithmetic, or clock arithmetic) is a system of arithmetic for integers, where numbers "wrap around" after they reach a certain value — the modulus. Modular arithmetic was introduced by Carl Friedrich Gauss in his book Disquisitiones Arithmeticae, published in 1801. A familiar use of modular arithmetic is its use in the 24-hour clock: the arithmetic of time-keeping in which the day runs from midnight to midnight and is divided into 24 hours, numbered from 0 to 23. If the time is 19:00 now — 7 o'clock in the evening — then 8 hours later it will be 3:00. Usual addition would suggest that the later time should be 19 + 8 = 27, but this is not the answer because clock time "wraps around" at the end of the day. Likewise, if the clock starts at 12:00 (noon) and 21 hours elapse, then the time will be 09:00 the next day, rather than 33:00. Since the hour number starts over when it reaches 24, this is arithmetic modulo 24. It should be noted that in this system 24:00 is not a valid time because this is equal to the 0:00 of the next day, in much the same way 2:60 is not a valid time because it is equal to 3:00. Time-keeping on a clock gives an example of modular arithmetic. The congruence relation Modular arithmetic can be handled mathematically by introducing a congruence relation on the integers that is compatible with the operations of the ring of integers: addition, subtraction, and multiplication. For a fixed modulus n, it is defined as follows. Two integers a and b are said to be congruentmodulon, if their difference a−b is an integer multiple of n. If this is the case, it is expressed as: The above mathematical statement is read: "a is congruent to bmodulon". For example, because 38 − 14 = 24, which is a multiple of 12. For positive n and non-negative a and b, congruence of a and b can also be thought of as asserting that these two numbers have the same remainder after dividing by the modulus n. So, because, when divided by 12, both numbers give 2 as remainder. The same rule holds for negative values of a: A remark on the notation: Because it is common to consider several congruence relations for different moduli at the same time, the modulus is incorporated in the notation. In spite of the ternary notation, the congruence relation for a given modulus is binary. This would have been clearer if the notation a≡n b had been used, instead of the common traditional notation. The properties that make this relation a congruence relation (respecting addition, subtraction, and multiplication) are the following. If and , then: The ring of congruence classes Like any congruence relation, congruence modulo n is an equivalence relation, and the equivalence class of the integer a, denoted by , is the set . This set, consisting of the integers congruent to a modulo n, is called the congruence class or residue class of a modulo n. Another notation for this congruence class, which requires that in the context the modulus is known, is . The set of congruence classes modulo n is denoted as and defined by: When n ≠ 0, has n elements, and can be written as: When n = 0, does not have zero elements; rather, it is isomorphic to , since . We can define addition, subtraction, and multiplication on by the following rules: The verification that this is a proper definition uses the properties given before. In this way, becomes a commutative ring. For example, in the ring , we have as in the arithmetic for the 24-hour clock. The notation is used, because it is the factor ring of by the ideal containing all integers divisible by n, where is the singleton set . In terms of groups, the residue class is the coset of a in the quotient group , a cyclic group. The set has a number of important mathematical properties that are foundational to various branches of mathematics. Rather than excluding the special case n = 0, it is more useful to include (which, as mentioned before, is isomorphic to the ring of integers), for example when discussing the characteristic of a ring. Remainders The notion of modular arithmetic is related to that of the remainder in division. The operation of finding the remainder is sometimes referred to as the modulo operation and we may see "2 = 14 (mod 12)". The difference is in the use of congruency, indicated by ≡, and equality indicate by =. Equality implies specifically the "common residue", the least non-negative member of an equivalence class. When working with modular arithmetic, each equivalence class is usually represented by its common residue, for example "38 ≡ 2 (mod 12)" which can be found using long division. It follows that, while it is correct to say "38 ≡ 14 (mod 12)", "2 ≡ 14 (mod 12)" and "2 ≡ 14 (mod 12)", it is incorrect to say "38 = 14 (mod 12)" (with "=" rather than "≡"). Parentheses are sometimes dropped from the expression, e.g. "38 ≡ 14 mod 12" or "2 = 14 mod 12", or placed around the divisor e.g. "38 ≡ 14 mod (12)". Notation such as "38(mod 12)" has also been observed, but is ambiguous without contextual clarification. The congruence relation is sometimes expressed by using modulo instead of mod, like "38 ≡ 14 (modulo 12)" in computer science. The modulo function in various computer languages typically yield the common residue, for example the statement "y = MOD(38,12);" gives y = 2. Applications Modular arithmetic is referenced in number theory, group theory, ring theory, knot theory, abstract algebra, cryptography, computer science, chemistry and the visual and musical arts. It is one of the foundations of number theory, touching on almost every aspect of its study, and provides key examples for group theory, ring theory and abstract algebra. In cryptography, modular arithmetic directly underpins public key systems such as RSA and Diffie-Hellman, as well as providing finite fields which underlie elliptic curves, and is used in a variety of symmetric key algorithms including AES, IDEA, and RC4. In computer science, modular arithmetic is often applied in bitwise operations and other operations involving fixed-width, cyclic data structures. The modulo operation, as implemented in many programming languages and calculators, is an application of modular arithmetic that is often used in this context. In chemistry, the last digit of the CAS registry number (a number which is unique for each chemical compound) is a check digit, which is calculated by taking the last digit of the first two parts of the CAS registry number times 1, the next digit times 2, the next digit times 3 etc., adding all these up and computing the sum modulo 10. In the visual arts, modular arithmetic can be used to create artistic patterns based on the multiplication and addition tables modulo n (see external link, below). In music, arithmetic modulo 12 is used in the consideration of the system of twelve-tone equal temperament, where octave and enharmonic equivalency occurs (that is, pitches in a 1∶2 or 2∶1 ratio are equivalent, and C- sharp is considered the same as D- flat). The method of casting out nines offers a quick check of decimal arithmetic computations performed by hand. It is based on modular arithmetic modulo 9, and specifically on the crucial property that 10 ≡ 1 (mod 9). More generally, modular arithmetic also has application in disciplines such as law (see e.g., apportionment), economics, (see e.g., game theory) and other areas of the social sciences, where proportional division and allocation of resources plays a central part of the analysis. Some neurologists (see e.g., Oliver Sacks) theorize that so-called autistic savants utilize an "innate" modular arithmetic to compute such complex problems as what day of the week a distant date will fall on. Computational complexity Since modular arithmetic has such a wide range of applications, it is important to know how hard it is to solve a system of congruences. A linear system of congruences can be solved in polynomial time with a form of Gaussian elimination, for details see the linear congruence theorem. Solving a system of non-linear modular arithmetic equations is NP-complete. For details, see for example M. R. Garey, D. S. Johnson: Computers and Intractability, a Guide to the Theory of NP-Completeness, W. H. Freeman 1979. Retrieved from " Subjects Art Business Studies Citizenship Countries Design and Technology Everyday life Geography History Information Technology Language and literature Mathematics Music People Portals Religion Science Title Word Index Wikipedia for Schools is a selection taken from the original English-language Wikipedia by the child sponsorship charity SOS Children. It was created as a checked and child-friendly teaching resource for use in schools in the developing world and beyond.Sources and authors can be found at www.wikipedia.org. See also our Disclaimer. These articles are available under the Creative Commons Attribution Share-Alike Version 3.0 Unported Licence. This article was sourced from .
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https://study.com/academy/lesson/isolated-systems-in-physics-definition-and-examples.html
Isolated Systems in Physics | Overview, Types & Examples - Lesson | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Copyright Test Prep Courses / AP Physics 1 Study Guide and Exam Prep Course Isolated Systems in Physics | Overview, Types & Examples Lesson Transcript Katherine Kaylegian-Starkey, Elizabeth Friedl Author Katherine Kaylegian-Starkey Katherine has a bachelor's degree in physics, and she is pursuing a master's degree in applied physics. She currently teaches struggling STEM students at Lane Community College. View bio Instructor Elizabeth Friedl Elizabeth, a Licensed Massage Therapist, has a Master's in Zoology from North Carolina State, one in GIS from Florida State University, and a Bachelor's in Biology from Eastern Michigan University. She has taught college level Physical Science and Biology. View bio This lesson explores isolated systems in physics. Examples of isolated and other types of systems in physics are provided. Updated: 11/21/2023 Table of Contents What is an Isolated System Isolated System vs Open System vs Closed System Isolated System Examples Perfect Isolated Systems Lesson Summary Show Frequently Asked Questions What are examples of open, closed, and isolated systems? An open system is a system that exchanges matter and energy with its surroundings. A melting ice cube is an example of this. A closed system is a system that only exchanges energy with its surroundings. A tea kettle before the whistle blows is an example of a closed system. An isolated system exchanges neither energy or matter with its external environment. A sealed vacuum chamber is an example of an isolated system. What is meant by an isolated system? An isolated system in physics is a system that has no net external force. This means that energy within an isolated is conserved and momentum is constant. What is the difference between isolated and closed systems? An isolated system is a system that does not have a net external force and does not exchange matter or energy with its surroundings. A closed system, on the other hand, is a system that exchanges energy with its surroundings. Because of this, energy exchanged in a closed system has a net external force acting on it. Is the universe an example of an isolated system? The universe is an isolated system because, as far as we know, there is nothing else other than our universe; so there is no external environment that our universe can exchange matter or energy with. Create an account Table of Contents What is an Isolated System Isolated System vs Open System vs Closed System Isolated System Examples Perfect Isolated Systems Lesson Summary Show What is an Isolated System -------------------------- In physics, a system is a collection of objects. For example, a system could be a sled and the person on the sled; or a system could be a person, a sled, and a dog pulling the sled. There is no limit to how many objects can be part of a system; and there are three different types of systems in physics which will be described in the next section. The main focus of this lesson is an isolated system in physics. An isolated system is a system that has no net external force acting on it. Since there is no net external force acting on the system, the momentum of the system remains constant. One implication of an isolated system having no net external force is the fact that the total energy of the system is conserved. The person and sled system is an isolated system if the sled never loses momentum due to friction between the sled and the earth. In this case, the energy of the system would change between kinetic energy and gravitational potential energy; but the total energy of the person and sled system would remain constant. Figure 1: A free body diagram representing a block losing energy to an unbalanced friction force. If an isolated system cannot have a net external force acting on it, what exactly is an external force? In the person/sled system example, a lack of friction would cause the system to maintain its momentum and be considered an isolated system. In this case, friction is the external force. Think about a sled coasting down a snowy hill: eventually the sled stops because friction between the sled and the earth causes the system to lose energy. Therefore, the system will lose its momentum. This stopping force is one sided—there is no equal opposite force opposing it because, if there was, energy would enter the system at the same rate it was lost; and the sled would never stop. The key idea here is that of unbalanced forces. An unbalanced force will change the momentum of a system, and, therefore, an unbalanced force is an external force. To unlock this lesson you must be a Study.com Member. Create your account Click for sound 5:55 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Try it now. Already registered? Log in here for access 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 Coming up next: Center of Mass | Definition, Equation & Calculation You're on a roll. Keep up the good work! Take QuizWatch Next Lesson Replay Just checking in. Are you still watching? Yes! Keep playing. Your next lesson will play in 10 seconds 0:01 What Is a System? 1:52 Examples of Isolated Systems 4:04 The Reality of… 5:06 Lesson Summary View Video Only Save Timeline 93K views Video Quiz Course Video Only 93K views Isolated System vs Open System vs Closed System ----------------------------------------------- As stated above, there are three different types of systems in physics. Other than isolated systems, there are also open systems and closed systems. An open system is a system that can freely exchange energy and matter with external objects. Consider an ice cube as a system. If the ice cube is held for long enough, the ice cube will melt and the hand holding the ice cube will get cold. Energy is transferred from the hand to the ice cube and, in turn, this energy transfer causes the ice cube to melt and exchange matter with its surroundings. A closed system is a system that only exchanges energy with its surroundings. This time, consider an ice cube in a glass as a system. Holding the glass for long enough will cause the hand to get cold and the ice within the glass to melt, but the ice never melts out of the glass. In this example, energy is transferred between the hand, glass, and ice; but no matter is transferred. An isolated system differs from an open system and a closed system because an isolated system exchanges neither energy nor matter with its external environment. Go back to the ice cube and glass example. Imagine holding a glass full of ice; only this time, imagine that, no matter how long the glass is held, the hand never gets cold and the ice does not melt. This fictitious example is an example of a system that does not exchange energy or matter. It is an isolated system. Figure 2: A chart showing how open systems, closed systems, and isolated systems exchange matter. To unlock this lesson you must be a Study.com Member. Create your account Isolated System Examples ------------------------ The previous examples of isolated systems were fictitious examples. What are some real life examples of isolated systems that do not require stretching the laws of reality? The inside of a vacuum thermos is an isolated system. The purpose of a vacuum thermos is to retain the temperature of the liquid inside. This means that the thermos is designed so that there is no energy transfer between the inside and the outside of the thermos. There is also no transfer of matter between the inside and the outside of the thermos; because no one would buy a leaky thermos! In this example, there is no unbalanced external force to change the system. Our universe is an isolated system because, as far as we know, there is nothing other than our universe; so there is nothing external that can exchange energy or matter with our universe. The calorimeters of the Large Hadron Collider are an isolated system because they are designed to be sealed vacuum chambers. No matter is exchanged between the inside and the outside of the calorimeter. The LHC calorimeters are also designed to capture all of the energy produced during a proton-proton collision, so all of the energy inside of the system stays within the system. Because there is no net external force in this system, this system is considered isolated. To unlock this lesson you must be a Study.com Member. Create your account Perfect Isolated Systems ------------------------ A perfect isolated system is a system that is truly isolated. This means that, to be a perfect isolated system, there can be no possibility of the system ever exchanging matter or energy with its external environment. This stringent condition makes a perfect isolated system almost impossible to find. Think back to the vacuum thermos example: a vacuum thermos can get old or break, and it can begin exchanging energy or matter with its surroundings. In this case, the vacuum thermos is no longer an isolated system. Similarly, the calorimeters of the LHC, while designed to capture 100% of the energy from a collision, have the potential to malfunction. In the event of a calorimeter malfunction, there is the possibility that the calorimeter system could exchange matter or energy with its surroundings. If this happened, the calorimeter system would no longer be a closed system. The only system that has no probability of ever interacting with its external environment is our universe because, as far as we know, there is nothing outside of our universe that it can interact with. To unlock this lesson you must be a Study.com Member. Create your account Lesson Summary -------------- A system in physics is a collection of objects. There are three types of systems: there are open systems, closed systems, and isolated systems. An isolated system is a system that has no net external force acting on it. The momentum of the system is constant and the energy of the system is conserved. An external force is an unbalanced force acting on the system. Because the force is unbalanced, the momentum of the system will change, and the energy of the system may not be conserved. A defining characteristic of an isolated system is that an isolated system does not exchange matter or energy with its external environment. This is contrary to a closed system, which exchanges energy with its environment. It is also contrary to an open system, which exchanges energy and matter with its enviroment. To unlock this lesson you must be a Study.com Member. Create your account Video Transcript What Is a System? We talk a lot about systems in physics, but this tends to be a tough concept for many students. The idea of a system is simple, but the application of it can be somewhat tricky. Simply put, a system is two or more objects. But this is a bit vague, isn't it? To make it even trickier, what we are really interested in understanding is an isolated system. Now, instead of just any old set of objects, we have the added stipulation that the total amount of energy in the system doesn't change. Another way of looking at it is to say that there is no external net force on the objects in the system, or that the objects in the system are isolated from external forces. An external net force on a system comes from something outside of the system. In other words, it comes from something other than the objects within the system itself. If you need to get a stalled car moving, you probably won't get very far if you sit in the driver's seat and push from the inside. Instead, you get a bunch of people to apply an external force from the outside. In fact, you could get 5 of the strongest men in the world to push from the inside, but in that position, they're part of the system, so the force doesn't do much good there. In order for the force to be external, the force has to be unbalanced, meaning that there is no other force that opposes it with the same magnitude. If you and your friend push on opposite ends of the stalled car, it's going to stay right where it is because the two forces are balanced - they're acting with the same magnitude but in opposing directions. They are working against each other so they essentially cancel out, making both of you part of the isolated system! It would be much wiser to have your friend help you push from the same end so that the force is unbalanced, taking you out of the isolated system with the car. Examples of Isolated Systems The car is a pretty straightforward example of an isolated system because it's easy to see how you wouldn't move it by pushing from the inside or by pushing from opposite sides. But objects in an isolated system aren't always clearly contained like this. For example, a head-on collision between two cars could either be an isolated system or not, depending on the forces involved. If the head-on collision is between two vehicles traveling on a road that provides a large amount of friction, the cars are not part of an isolated system because there is an external force of friction at work there. However, take friction out of the equation and the crashing cars become an isolated system. This is because the forces they exert on each other cancel out as they collide, and there are no external forces involved. Billiard balls are part of an isolated system because as they collide, they too exert balanced forces on each other. When the cue ball hits another ball, the momentum from the first ball is transferred to the second ball, sending it rolling across the table. In the absence of friction from the table, the two balls are part of an isolated system because there are no external forces at work. It might help to think about isolated systems in terms of your morning coffee. You may notice that your coffee stays hotter longer if it is in a covered mug than one without a lid. In the covered mug, the heat from the coffee is 'isolated' from the external environment. However, the heat (a form of energy) is free to escape into the air in the uncovered mug because the lid 'barrier' doesn't exist. Of course, some heat is also conducted through the mug itself (which is why you pick it up by the handle), but if under ideal conditions there was no heat loss through the mug, the covered one would resemble an isolated system since there would be no loss of heat from this container. The people in the car, the billiard balls on the table, and the cars in the frictionless collision are all like coffee inside a closed mug. Sometimes the mug is easy to see, like with the stalled car. Other times though, the 'mug' is invisible, as is the case with the head-on collision and the billiard balls. The Reality of Isolated Systems Like many things in physics, we describe ideal situations such as cars driving on frictionless roads. But here on Earth, the absence of such forces is unrealistic, so there really aren't any perfectly isolated systems. External net forces are almost always present, and like our coffee cup, some heat is lost. The universe is considered the only real isolated system, because it acts like a big, humongous stalled car that contains everything inside of it. Try as you might, you can't escape from the universe. All of the energy and heat inside of it simply gets transferred, transformed, and recycled as it moves through different forms and objects. However, we only take up a tiny, tiny fraction of a fraction of the universe, so it's easier to understand how an isolated system COULD exist on a more local scale, with some minor assumptions of course. And understanding these small-scale examples helps us understand how the everyday world around us works in a way that is relevant and meaningful to us on Earth. Lesson Summary Physics can be a confusing subject because it challenges us to look at the world around us differently. And sometimes, the things we see are not always what they appear to be because there are other things going on that we don't see. An isolated system is one of these tricky concepts because this is a system where the total amount of energy doesn't change. We can't necessarily see energy, so we have to use other knowledge to help us understand what's going on. Like hot coffee in a sealed mug, objects in isolated systems are isolated from external forces. So as long as the total energy of the system doesn't change and there are no external forces acting on the objects in the system, we can say that the system is isolated. Learning Outcomes After reviewing this lesson, you'll have the ability to: Describe what an isolated system is Identify examples of isolated systems Explain why the universe is the only true isolated system 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 Recommended Lessons and Courses for You Related Lessons Related Courses Recommended Lessons for You An Introduction to Thermal Physics | Overview & Key Areas Closed & Open Systems | Definition, Differences & Thermodynamics Brownian Motion | Definition, Causes & Examples Specific Volume Definition, Formulas & Units Forms of Energy | Overview & Examples Thermodynamics Practice Problems & Solutions Heat Energy | Definition, Examples & Types Specific Heat of Water & Metals | Overview, Factors & Calculation Latent Heat of Fusion | Definition, Equation & Calculation Heat Transfer Examples: Problems & Solutions Conduction | Definition & Examples Heat Energy Lesson for Kids: Definition & Examples Mechanical Equivalent of Heat Closed & Open Systems | Definition, Differences & Thermodynamics Related Courses AP Physics C - Mechanics Study Guide and Exam Prep NY Regents - Physics Study Guide and Exam Prep UExcel Physics: Study Guide & Test Prep GED Science: Life, Physical and Chemical AP Chemistry Study Guide and Exam Prep AP Biology Study Guide and Exam Prep FTCE Physics 6-12 (032) Study Guide and Test Prep SAT Subject Test Chemistry: Practice and Study Guide SAT Subject Test Biology: Practice and Study Guide NY Regents Exam - Living Environment: Test Prep & Practice AP Environmental Science Study Guide and Exam Prep NY Regents - Chemistry Study Guide and Exam Prep NY Regents - Earth Science Study Guide and Exam Prep Physics: High School DSST Environmental Science Study Guide and Test Prep DSST Health and Human Development Study Guide and Test Prep DSST Principles of Physical Science: Study Guide & Test Prep DSST Astronomy Study Guide and Test Prep DSST Foundations of Education Study Guide and Test Prep AP Physics 2 Study Guide and Exam Prep Start today. Try it now AP Physics 1 Study Guide and Exam Prep 16 chapters | 152 lessons Ch 1. AP Physics 1: Systems Ch 2. AP Physics 1: Vectors Ch 3. AP Physics 1: Kinematics Ch 4. AP Physics 1: Graphing... Ch 5. AP Physics 1: Newton's First Law of... Ch 6. AP Physics 1: Newton's Second Law of... Ch 7. AP Physics 1: Newton's Third Law of... Ch 8. AP Physics 1: Work, Energy, &... Ch 9. AP Physics 1: Linear Momentum Linear Momentum | Definition, Formula & Examples 4:58 Impulse-Momentum Theorem | Definition, Formula & Examples 6:11 Conservation of Linear Momentum | Definition & Formula 5:32 Inelastic vs. Elastic Collision | Definition & Differences 4:52 Elastic Collision Definition, Characteristics & Examples 9:11 Isolated Systems in Physics | Overview, Types & Examples 5:54 7:39 Next Lesson Center of Mass | Definition, Equation & Calculation Center of Mass | Formula, Velocity & Acceleration 6:32 Ch 10. AP Physics 1: Motion Ch 11. AP Physics 1: Oscillations Ch 12. AP Physics 1: Rotational Motion Ch 13. AP Physics 1: Electrical Forces and... Ch 14. AP Physics 1: Direct Current... Ch 15. AP Physics 1: Mechanical Waves Ch 16. AP Physics 1 Flashcards Isolated Systems in Physics | Overview, Types & Examples Related Study Materials Related Lessons An Introduction to Thermal Physics | Overview & Key Areas Closed & Open Systems | Definition, Differences & Thermodynamics Brownian Motion | Definition, Causes & Examples Specific Volume Definition, Formulas & Units Forms of Energy | Overview & Examples Thermodynamics Practice Problems & Solutions Heat Energy | Definition, Examples & Types Specific Heat of Water & Metals | Overview, Factors & Calculation Latent Heat of Fusion | Definition, Equation & Calculation Heat Transfer Examples: Problems & Solutions Conduction | Definition & Examples Heat Energy Lesson for Kids: Definition & Examples Mechanical Equivalent of Heat Closed & Open Systems | Definition, Differences & Thermodynamics Related Courses AP Physics C - Mechanics Study Guide and Exam Prep NY Regents - Physics Study Guide and Exam Prep UExcel Physics: Study Guide & Test Prep GED Science: Life, Physical and Chemical AP Chemistry Study Guide and Exam Prep AP Biology Study Guide and Exam Prep FTCE Physics 6-12 (032) Study Guide and Test Prep SAT Subject Test Chemistry: Practice and Study Guide SAT Subject Test Biology: Practice and Study Guide NY Regents Exam - Living Environment: Test Prep & Practice AP Environmental Science Study Guide and Exam Prep NY Regents - Chemistry Study Guide and Exam Prep NY Regents - Earth Science Study Guide and Exam Prep Physics: High School DSST Environmental Science Study Guide and Test Prep DSST Health and Human Development Study Guide and Test Prep DSST Principles of Physical Science: Study Guide & Test Prep DSST Astronomy Study Guide and Test Prep DSST Foundations of Education Study Guide and Test Prep AP Physics 2 Study Guide and Exam Prep Related Topics Browse by Courses Chemistry: High School Physics: High School Geology 101: Physical Geology Science 102: Principles of Physical Science Praxis 5551 Study Guide - Health Education Exam Prep FTCE Physics 6-12 (032) Study Guide and Test Prep SAT Subject Test Chemistry: Practice and Study Guide ILTS Science - Chemistry (240) Study Guide and Test Prep Praxis Environmental Education (0831) Study Guide and Test Prep UExcel Microbiology: Study Guide & Test Prep Chemistry 101: General Chemistry Biology 102: Basic Genetics Science 101: Intro to Life Sciences Biology 105: Anatomy & Physiology GED Science: Life, Physical and Chemical Browse by Lessons Types of Energy | Overview & Examples Internal Energy of a System: Object Model vs. System Model Using Thermodynamic Properties to Solve Problems Thermal Energy Activities for Middle School Internal Energy of a System: Qualitative & Quantitative Calculations Predicting the Direction of Energy Transfer Heat Transfer Project Ideas Heat Transfer Activities & Games Heat Transfer Lesson Plan for Elementary School Joule | Definition, Conversion & Examples Create an account to start this course today Used by over 30 million students worldwide Create an account 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 AP Physics 1 Study Guide and Exam Prep Enrolling in a course lets you earn progress by passing quizzes and exams. 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https://www.livescience.com/planet-earth/earthquakes/why-do-earthquakes-happen-far-away-from-plate-boundaries
Skip to main content Why do earthquakes happen far away from plate boundaries? News By Alice Sun published It's well known that earthquakes can rock fault-filled places like the U.S. West Coast. But why do earthquakes happen in the middle of tectonic plates? Volunteers restock the shelves at Millers Market, which was damaged by the magnitude 5.8 earthquake that struck the Northeast U.S. on Aug. 24, 2011. This earthquake, with an epicenter in Mineral, Virginia, happened far away from tectonic plate boundaries. (Image credit: Scott Olson / Staff via Getty Images) It's commonly assumed that earthquakes occur only near the boundaries of tectonic plates, and roughly 90% of earthquakes do happen in these areas. These boundaries include, for example, the San Andreas Fault, which runs roughly along the west coast of California, where the North American and Pacific plates meet. But not all earthquakes occur along plate boundaries. For example, an earthquake near New Madrid, Missouri in the winter of 1811 was thousands of miles from the nearest fault, yet the magnitude 7.2 to 8.2 quake violently shook the region, triggering a series of powerful aftershocks collectively called the 1811-1812 New Madrid earthquakes. So how was this possible? How do earthquakes happen far away from plate boundaries? First, as a point of comparison, it's important to understand the way conventional earthquakes form along boundary lines. These areas experience more earthquakes because Earth's interior — namely, the mantle — move the planet's tectonic plates, causing them to split apart and collide. The cracks in between these plates, called faults, are fragile. So, when stress starts accumulating at these weak points, plates can break, sending a shudder through the planet. This is what we feel as earthquakes, explained Attreyee Ghosh, a geophysicist at the Centre for Earth Sciences at the Indian Institute of Science in Bangalore. Related: How many tectonic plates does Earth have? But every once in a while, a shudder can happen in the middle of a tectonic plate. Scientists call this an intraplate earthquake. Exactly why it happens remains much of a mystery, said Christine Powell, a geophysicist at the University of Memphis. She and other scientists have studied places with a high concentration of intraplate earthquakes, called intraplate seismic zones. These zones exist, for example, in parts of the central and eastern United States. After researching these areas, experts have some theories as to why temblors may occur in unexpected places. One possible explanation is that intraplate earthquakes may be caused by old glaciers, a 2001 study proposed. Around 20,000 years ago, much of North America was covered under a giant ice sheet, and the ground was weighed down considerably. As the ice sheet melted, the ground slowly rose, so the earthquakes may be the result of this adjustment. Evidence for this theory, however, is sparse. "The orientation of the earthquake axis and the glacial isostatic adjustment doesn't match," Ghosh said. Sign up for the Live Science daily newsletter now Get the world’s most fascinating discoveries delivered straight to your inbox. Another idea is that intraplate earthquakes are occurring around old faults on the insides of tectonic plates. For billions of years, Earth's crust has split apart and come back together, and old wounds leave scars. When forces propagate through to the plates' interior and put too much stress on these old faults, they may get reactivated, Ghosh said. The complicated composition of Earth's crust and interior could also be a factor. Sometimes, remains of an ancient slab of rock gets stuck in the middle of a plate, causing instability, as posited by a 2007 study in Geophysical Research Letters. Pipes of hot fluids could add pressure, resulting in movement on the planet's surface, Powell said, who co-authored a study on this upwelling in 2016. Hydraulic factoring, or fracking — the act of injecting water, sand and chemicals into underground rocks to extract oil or gas — can trigger earthquakes, too. Wastewater fluid from these operations are injected into deep wells, which can seep into cracks, lubricate old faults and cause seismic activity, according to a 2013 review in Science. For instance, fracking was tied to a number of earthquakes in Ohio in 2015. Scientists are trying to get a better understanding of these complexities with data from projects such as EarthScope, which use sensors to capture the dynamics underneath Earth's crust. Powell recalls that, when the project first started, some scientists didn't think the sensors would find anything that could lead to the generation of earthquakes except for within the West Coast, where the plate boundary was. But the project "really opened our eyes to what is going on inside our Earth here," said Powell, who is based in Tennessee. "It was a remarkable experiment." RELATED MYSTERIES —How big is the largest possible earthquake? —Why are rare earth elements so rare? —Why don't we feel Earth spinning? It's important to understand intraplate earthquakes because they pose a considerable risk for people who live in these seismic zones. The three earthquakes in New Madrid, Missouri in 1811-1812 caused considerable destruction, even altering the course of the Mississippi River and causing it to temporarily run backward. A magnitude 5.8 quake in Virginia shook Washington, D.C. in 2011, damaging monuments and cathedrals. "Nobody thinks about earthquakes in the central and eastern U.S.," Powell said. "We must be prepared. You have to be aware that earthquakes can happen in these places." TOPICS Alice Sun Live Science Contributor Alice Sun is a science journalist based in Brooklyn. She covers a wide range of topics, including ecology, neuroscience, social science and technology. Her work has appeared in Audubon, Sierra, Inverse and more. For her bachelor's degree, she studied environmental biology at McGill University in Canada. She also has a master's degree in science, health and environmental reporting from NYU. LATEST ARTICLES 1 Dozens of mysterious blobs discovered inside Mars may be the remnants of 'failed planets' 2. 2 1,000-year-old 'king' game piece with a distinctive hairstyle is 'as close as we will ever get to a portrait of a Viking' 3. 3 1.8 million-year-old human jawbone discovered in Republic of Georgia — and it may be earliest evidence yet of Homo erectus 4. 4 'Extremely alarming': ChatGPT and Gemini respond to high-risk questions about suicide — including details around methods 5. See what would happen to Tokyo if Mount Fuji erupted 'without any warning' in new AI-generated video Please wait...
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https://www.sigmaaldrich.com/US/en/product/sial/denwat?srsltid=AfmBOorMdRx5N2iS6XCSL6pyOQGdb1N_OAht6plpk1Cr1Hf2Ld1ht7AR
Pure Water Density Standard UKAS ISO/IEC17025 and ISO Guide 34 certified, 0.9982g/mL 20°C, 0.9970g/mL 25°C 7732-18-5 Skip to Content Products Cart 0 IL EN Products Products Applications Services Documents Support Analytical ChemistryCell Culture & AnalysisChemistry & BiochemicalsClinical & DiagnosticsFiltration Greener Alternative Products Industrial MicrobiologyLab AutomationLabwareMaterials ScienceMolecular Biology & Functional GenomicsmAbs Development & ManufacturingmRNA Development & ManufacturingPharma & Biopharma ManufacturingProtein BiologyWater Purification Analytical ChemistryCell Culture & AnalysisChemistry & SynthesisClinical & DiagnosticsEnvironmental & Cannabis TestingFood & Beverage Testing & ManufacturingGenomicsMaterials Science & EngineeringMicrobiological TestingmAbs Development & ManufacturingmRNA Development & ManufacturingPharma & Biopharma ManufacturingProtein BiologyResearch & Disease AreasWater Purification Contract ManufacturingContract TestingCustom ProductsDigital Solutions for Life ScienceIVD Development & ManufacturingProduct ServicesSupport Safety Data Sheets (SDS) Certificates of Analysis (COA) Certificates of Origin (COO) Certificates of Quality (COQ) Customer Support Contact Us Get Site Smart FAQ Quality & Regulatory Calculators & Apps Webinars Login Order Lookup Quick Order Cart 0 Products Analytical Chemistry DENWAT Key Documents SDS COA COO View All Documentation Skip To Compare Similar Items Safety Information Documentation Protocols and Articles Peer Reviewed Papers Questions & Answers Reviews DENWAT Share Pure Water Density Standard UKAS ISO/IEC17025 and ISO Guide 34 certified, density: 0.9982 g/mL at 20°C, density: 0.9970 g/mL at 25°C No rating value Same page link. (0) Write a review Ask a question Synonym(s): Water Sign Into View Organizational & Contract Pricing Select a Size Change View 60 ML ₪1,385.00 3 X 60 ML ₪2,348.00 About This Item Linear Formula: H 2 O CAS Number: 7732-18-5 Molecular Weight: 18.02 Beilstein: 2050024 EC Number: 231-791-2 MDL number: MFCD00011332 UNSPSC Code: 41116107 PubChem Substance ID: 329798917 NACRES: NA.24 DENWAT-60ML ₪1,385.00 Estimated to ship on 15 October 2025 Details Add to Cart Request a Bulk Order Technical Service Need help? Our team of experienced scientists is here for you. Let Us Assist Recommended Products Slide 1 of 10 1 of 3 Sigma-Aldrich 306975 tert-Butyl methyl ether Quick View DENWAT3 Pure Water Density Standard Quick View Sigma-Aldrich L092004 Ethyl acetate Quick View Sigma-Aldrich L092003 Ethyl acetate Quick View Sigma-Aldrich 270989 Ethyl acetate Quick View Sigma-Aldrich W4502 Water Quick View Sigma-Aldrich L090000 Dichloromethane Quick View Sigma-Aldrich 322415 Methanol Quick View Supelco 1.88051 Water standard 0.1% Quick View Supelco 1.15333 Water Quick View Properties grade certified reference material Quality Level 100 Agency according to ASTM® D1480 shelf life limited shelf life, expiry date on the label manufacturer/tradename Paragon Scientific Ltd refractive index n 20/D 1.34 (lit.) bp 100°C (lit.) mp 0°C (lit.) density 0.9982 g/mL at 20°C 0.9970 g/mL at 25°C 1.000 g/mL at 3.98°C (lit.) application(s) cleaning products cosmetics environmental flavors and fragrances food and beverages industrial qc personal care petroleum Show More Looking for similar products? Visit Product Comparison Guide Related Categories Analytical Chemistry Certified Reference Materials Petrochemical Standards Physical & Chemical Property Standards Reference Materials Compare Similar Items View Full Comparison Show Differences [x] 1 of 1 | ###### This Item | EM3234 | 900682 | DENWAT3 | --- --- | | DENWAT Pure Water Density Standard Quick View | Supelco EM3234 Water, Deionized Distilled, ASTM Type II Quick View | Sigma-Aldrich 900682 Water Quick View | DENWAT3 Pure Water Density Standard Quick View | | format mixture | format - | format - | format mixture | | bp 100°C (lit.) | bp 100°C (lit.) | bp 100°C (lit.) | bp 100°C (lit.) | | mp 0°C (lit.) | mp 0°C (lit.) | mp 0°C (lit.) | mp 0°C (lit.) | | grade certified reference material | grade - | grade - | grade certified reference material | | application(s) cleaning products cosmetics environmental flavors and fragrances food and beverages industrial qc personal care petroleum | application(s) - | application(s) food and beverages | application(s) cleaning products cosmetics environmental flavors and fragrances food and beverages industrial qc personal care petroleum | | manufacturer/tradename Paragon Scientific Ltd | manufacturer/tradename - | manufacturer/tradename - | manufacturer/tradename Paragon Scientific Ltd | Description General description Please note that the value stated on this page is nominal. All certified values for the current lots of all Paragon Density Standards can be found on the Paragon Scientific webpageThis product is a certified reference material (CRM) certified to International Standards BS EN ISO / IEC 17025 Application This pure water density CRM is designed for the calibration of instruments used to measure density. And although typically used, it is not limited to the verification of digital density meters. Features and Benefits Manufactured and verified in an ISO 17025 and ISO 17034 accredited laboratory, as per UKAS Carries an associated uncertainty of +/- 0.01% Certified density value following ASTM D1480 Provided in a tamper-evident packaging to assure safe delivery Meets the definition of Type II in the specification ASTM D1193 Accompanied with a certificate of analysis (CoA) having information on expiry date and storage conditions for an effective audit Legal Information ASTM is a registered trademark of American Society for Testing and Materials Safety Information Storage Class Code 10 - Combustible liquids WGK nwg Flash Point(F) Not applicable Flash Point(C) Not applicable Documentation SDS Certificate of Analysis Certificate of Origin More Documents Certificates of Analysis (COA) Search for Certificates of Analysis (COA) by entering the products Lot/Batch Number. Lot and Batch Numbers can be found on a product’s label following the words ‘Lot’ or ‘Batch’. Lot/Batch Number Search Need A Sample COA? This is a sample Certificate of Analysis (COA) and may not represent a recently manufactured lot of this specific product. View Sample COA Already Own This Product? Find documentation for the products that you have recently purchased in the Document Library. Visit the Document Library Customers Also Viewed Slide 1 of 7 1 of 2 DEN2009 Density Standard (20 °C) Quick View DEN2004 Density Standard (20 °C) Quick View Supelco 76081 Density Standard 749 kg/m 3 Quick View DEN2008 Density Standard (20 °C) Quick View Sigma-Aldrich L092003 Ethyl acetate Quick View CRMUDEKR Density Standard Quick View CRMDEGA Density Standard Quick View Peer Reviewed Papers Effects of 1,25-dihydroxyvitamin D3 and vitamin D3 on the expression of the vitamin d receptor in human skeletal muscle cells. Rachele M Pojednic et al. Calcified tissue international, 96(3), 256-263 (2014-12-07) Vitamin D receptor (VDR) expression and action in non-human skeletal muscle have recently been reported in several studies, yet data on the activity and expression of VDR in human muscle cells are scarce. We conducted a series of studies to Gene expression profiling of giant cell tumor of bone reveals downregulation of extracellular matrix components decorin and lumican associated with lung metastasis. M Lieveld et al. Virchows Archiv : an international journal of pathology, 465(6), 703-713 (2014-10-12) Giant cell tumor of bone (GCTB) displays worrisome clinical features such as local recurrence and occasionally metastatic disease which are unpredictable by morphology. Additional routinely usable biomarkers do not exist. Gene expression profiles of six clinically defined groups of GCTB Widespread decline of Congo rainforest greenness in the past decade. Liming Zhou et al. Nature, 509(7498), 86-90 (2014-04-25) Tropical forests are global epicentres of biodiversity and important modulators of climate change, and are mainly constrained by rainfall patterns. The severe short-term droughts that occurred recently in Amazonia have drawn attention to the vulnerability of tropical forests to climatic Water in Photosystem II: structural, functional and mechanistic considerations. Katrin Linke et al. Biochimica et biophysica acta, 1837(1), 14-32 (2013-08-28) Water is clearly important for the functioning of Photosystem II (PSII). Apart from being the very substrate that needs to be transported in this water oxidation enzyme, water is also vital for the transport of protons to and from the Epistatic interactions between neuraminidase mutations facilitated the emergence of the oseltamivir-resistant H1N1 influenza viruses. Susu Duan et al. Nature communications, 5, 5029-5029 (2014-10-10) Oseltamivir-resistant H1N1 influenza viruses carrying the H275Y neuraminidase mutation predominated worldwide during the 2007-2009 seasons. Although several neuraminidase substitutions were found to be necessary to counteract the adverse effects of H275Y, the order and impact of evolutionary events involved remain View All Related Papers Protocols and Articles Articles GC Analysis of Permanent Gases and Light Hydrocarbons on Carboxen-1006 PLOT and a Packed Column (Carboxen® 1004) Separation of Methane; Acetylene; Carbon monoxide; Water; Nitrogen; Carbon dioxide; Ethane; Ethylene ★★★★★ ★★★★★ No rating value for Pure Water Density Standard Search questions and answers ϙ Search questions and answers ✘ Clear Search field ϙ Search 0 Reviews 2 Questions 2 Answers Questions Ask a question 1–2 of 2 Questions Sort by: ▼ Menu Anonymous ·3 months ago ### How can I determine the shelf life / expiration / retest date of this product? 1 answer Technical Support ·3 months ago If this product has an expiration or retest date, it will be shown on the Certificate of Analysis (COA, CofA). If there is no retest or expiration date listed on the product's COA, we do not have suitable stability data to determine a shelf life. For these products, the only date on the COA will be the release date; a retest, expiration, or use-by-date will not be displayed. For all products, we recommend handling per defined conditions as printed in our product literature and website product descriptions. We recommend that products should be routinely inspected by customers to ensure they perform as expected. For products without retest or expiration dates, our standard warranty of 1 year from the date of shipment is applicable. For more information, please refer to the Product Dating Information document: Helpful? Yes · 0 No · 0 Report Anonymous ·11 months ago ### How is shipping temperature determined? And how is it related to the product storage temperature? 1 answer Technical Support ·11 months ago Products may be shipped at a different temperature than the recommended long-term storage temperature. If the product quality is sensitive to short-term exposure to conditions other than the recommended long-term storage, it will be shipped on wet or dry-ice. If the product quality is NOT affected by short-term exposure to conditions other than the recommended long-term storage, it will be shipped at ambient temperature. As shipping routes are configured for minimum transit times, shipping at ambient temperature helps control shipping costs for our customers. For more information, please refer to the Storage and Transport Conditions document: Helpful? 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https://unacademy.com/content/upsc/study-material/chemistry/units-of-van-der-waals-constants/
© 2023 Sorting Hat Technologies Pvt Ltd Units of Van der Waals Constants The Van der Waals equation (or Van der Waals equation of state) is a state equation in science and thermodynamics that stretches out the best gas regulation to represent the impacts of interaction between atoms in a gas, as well as the finite size of the particles. The Van der Waals equation describes the connection between pressure, volume, temperature, and number of genuine gasses. The equation for a real gas with ‘n’ moles is as per the following: The tension, volume, temperature, and moles of the gas are addressed as P, V, T, n. Constants ‘a’ and ‘b’ that are interesting to each gas. The Units of Vander Waals equation: In the van der Waals equation, there are two rectification factors. In the best gas equation, the first an²⁄ V², adjusts the strain. The intermolecular alluring interactions between gas atoms are represented. The strength of the intermolecular alluring power is indicated by the size of a. and is composed of units . (L² atm) ⁄ mol² The volume involved by the gas atoms is addressed by the component – nb. The units of b are L ⁄ mol. Because b addresses the general volume per mole filled by gas atoms, it is basically the same as the volume per mole of the fluid state, which has firmly stacked particles. The extent of b is generally altogether more modest than that of a. With the size and intricacy of the particle, the upsides of an and b will quite often rise. For the calculation of units and aspects of constants a and b, the Van der Waals equation is used. an² ⁄ V² = is the Van der Waals equation for n-mole real gasses, which is the unit of internal tension. Subsequently, the Van der Waals constant is estimated in, a= atm lit² mol-² . Again, nb = volume unit, therefore unit of Van der Waals constant, and b = lit mol-¹ . Vendor was constant: The van der Waals equation is a state equation that records for two parts of real gases: gas molecule prohibited volume and gas atom appealing powers. The van der Waals equation is usually expressed as follows: (P+an² ⁄ V²)(V-nb) = nrt The constants a and b are gas-explicit and describe the greatness of intermolecular fascination and rejected volume, individually. Van der Waals constant significance: 1.The worth of the tension amendment factor The intermolecular fascination causes the tension rectification factor (a) = an² ⁄ V², and therefore, the strain remedy component can be utilized to determine the gas’ internal tension. Nonetheless, the more prominent the worth of the strain revision factor, the more grounded the intermolecular fascination and the simpler the gas will melt. Subsequently, the Van der Waals constant (a) for carbon dioxide gas is 3.95, but hydrogen gas is 0.2. 2.The worth of the volume amendment factor The particle size as well as frightful powers are estimated by another constant volume rectification factor (b). The bigger the gas atom, the higher the worth of b. Therefore, the Van der Waals constant b for carbon dioxide is 0.04 while it is 0.02 for hydrogen. 3.Van der Waals gases At 0°C, appealing powers dominated the behavior of Van der Waals gasses (carbon dioxide, nitrogen, and hydrocarbons like methane, ethane, and acetylene), but the sub-atomic size impact dominated the behavior of hydrogen gas. Because these gasses are challenging to condense, the unit worth of Van der Waals constant hydrogen, helium, nitrogen, and other gasses utilized in science and physical science is tiny. Unit of b in real gas equation: A gas that doesn’t carry on like an ideal gas is alluded to as a “real gas.” The interactions between the gaseous atoms assist with explaining their behavior. Real gasses don’t keep the ideal gas regulation because of these intermolecular interactions between gas particles. Therefore, Real gasses are non-ideal gasses with particles that take up a particular measure of room and can interact with each other. A few components should be examined in request to appreciate how real gasses behave. The different contemplations that should be made while dealing with genuine gasses are outlined below. Ø Impacts of compressibility on the real gas Ø Different real gasses have different explicit hotness limits. Ø Van der Waals powers affect the interactions between atoms in a real gas. Ø The framework’s true capacity for non-equilibrium thermodynamic impacts. Ø The gas’ different organization and changes in structure because of atomic separation, as well as any rudimentary cycles that might happen. Positive qualities are allotted to the constants a and b, which are special to each gas. Intermolecular fascination is revised by the term involving the constant a. Alluring powers between atoms bring down the strain of a real gas, causing particles to dial back and decrease impacts with the dividers. The more prominent the fascination between atoms and the more straightforward the gas packs, the higher the worth of a. The prohibited volume of the gas, or the volume involved by the gas particles, is addressed by the b term. The units of b are L ⁄ mol . Because b addresses the general volume per mole filled by gas atoms CONCLUSION: For a real gas, the van der Waals equation of state is: (P+an² ⁄ V²)(V-nb) = nrt P addresses pressure, V addresses volume, T addresses temperature, n addresses how much substance (in moles), and R addresses the gas constant. The van der Waals constants a and b are properties of the substance and are temperature independent. Frequently asked questions Get answers to the most common queries related to the UPSC Examination Preparation. In the Van der Waals equation, what are the units of the constant A? What do the letters A and B in the real gas equation mean? What are the flaws in the van der Waals equation in the case of a real gas? Ans. : (L² atm) ⁄ mol² is the Van der Waals equation for n-mole real gasses, which is the unit of internal pressure. As a result, the Van der Waals constant is measured in atm . Ans. : (L² atm) ⁄ mol² is the Van der Waals equation for n-mole real gasses, which is the unit of internal pressure. As a result, the Van der Waals constant is measured in atm . Ans. In the van der Waals equation, the constant ‘a’ represents the magnitude of attractive forces between gas molecules, whereas ‘b’ represents the effective volume occupied by the gas molecules. It’s also known as excluded volume or co-volume. Ans. In the van der Waals equation, the constant ‘a’ represents the magnitude of attractive forces between gas molecules, whereas ‘b’ represents the effective volume occupied by the gas molecules. It’s also known as excluded volume or co-volume. Ans. Van der Waals equation has some limitations: (P +an²⁄ V² )(V – nb) = nRT For any gas, the values of a and b are not perfect constants. They differ slightly depending on the temperature. The values of a and b vary depending on the gas. In the equation, three constants (a, b, and R) are employed. Ans. Van der Waals equation has some limitations: (P +an²⁄ V² )(V – nb) = nRT For any gas, the values of a and b are not perfect constants. They differ slightly depending on the temperature. The values of a and b vary depending on the gas. In the equation, three constants (a, b, and R) are employed. Unacademy is India’s largest online learning platform. Download our apps to start learning Starting your preparation? Call us and we will answer all your questions about learning on Unacademy Call +91 8585858585 Company Help & support Products Popular goals Trending exams Study material © 2025 Sorting Hat Technologies Pvt Ltd Share via COPY
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https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/06%3A_Applications_of_Integration/6.03%3A_Volumes_of_Revolution_-_Cylindrical_Shells
Published Time: 2016-07-11T03:03:34Z 6.3: Volumes of Revolution - Cylindrical Shells - Mathematics LibreTexts Skip to main content Table of Contents menu search Searchbuild_circle Toolbarfact_check Homeworkcancel 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 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 6: Applications of Integration Calculus (OpenStax) { "6.3E:_Exercises_for_Section_6.3" : "property get Map 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MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.<PageSubPageProperty>b__1" } Fri, 17 Jan 2025 19:35:44 GMT 6.3: Volumes of Revolution - Cylindrical Shells 2521 2521 Delmar Larsen { } Anonymous Anonymous 2 false false [ "article:topic", "method of cylindrical shells", "solid of revolution", "Volume by Shells", "authorname:openstax", "license:ccbyncsa", "showtoc:no", "program:openstax", "licenseversion:40", "source@ "author@Gilbert Strang", "author@Edwin \u201cJed\u201d Herman" ] [ "article:topic", "method of cylindrical shells", "solid of revolution", "Volume by Shells", "authorname:openstax", "license:ccbyncsa", "showtoc:no", "program:openstax", "licenseversion:40", "source@ "author@Gilbert Strang", "author@Edwin \u201cJed\u201d Herman" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents Home Bookshelves Calculus Calculus (OpenStax) 6: Applications of Integration 6.3: Volumes of Revolution - Cylindrical Shells Expand/collapse global location Calculus (OpenStax) Front Matter 1: Functions and Graphs 2: Limits 3: Derivatives 4: Applications of Derivatives 5: Integration 6: Applications of Integration 7: Techniques of Integration 8: Introduction to Differential Equations 9: Sequences and Series 10: Power Series 11: Parametric Equations and Polar Coordinates 12: Vectors in Space 13: Vector-Valued Functions 14: Differentiation of Functions of Several Variables 15: Multiple Integration 16: Vector Calculus 17: Second-Order Differential Equations Appendices Back Matter 6.3: Volumes of Revolution - Cylindrical Shells Last updated Jan 17, 2025 Save as PDF 6.2E: Exercises for Section 6.2 6.3E: Exercises for Section 6.3 picture_as_pdf Full BookPage Downloads Full PDFImport into LMSIndividual ZIPBuy Print CopyPrint Book Files Buy Print CopySubmit Adoption ReportPeer Review Donate Page ID 2521 Gilbert Strang & Edwin “Jed” Herman OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}\) Table of contents Learning Objectives The Method of Cylindrical Shells Rule: The Method of Cylindrical Shells Example 6.3.16.3.1: The Method of Cylindrical Shells I Solution Exercise 6.3.16.3.1 Example 6.3.26.3.2: The Method of Cylindrical Shells II Solution Exercise 6.3.26.3.2 Rule: The Method of Cylindrical Shells for Solids of Revolution around the xx-axis Example 6.3.36.3.3: The Method of Cylindrical Shells for a Solid Revolved around the xx-axis Solution Exercise 6.3.36.3.3 Example 6.3.46.3.4: A Region of Revolution Revolved around a Line Solution Exercise 6.3.46.3.4 Example 6.3.56.3.5: A Region of Revolution Bounded by the Graphs of Two Functions Solution Exercise 6.3.56.3.5 Which Method Should We Use? Example 6.3.66.3.6: Selecting the Best Method Solution Exercise 6.3.66.3.6 Key Concepts Key Equations Glossary Learning Objectives Calculate the volume of a solid of revolution by using the method of cylindrical shells . Compare the different methods for calculating a volume of revolution. In this section, we examine the method of cylindrical shells , the final method for finding the volume of a solid of revolution . We can use this method on the same kinds of solids as the disk method or the washer method ; however, with the disk and washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical shells , we integrate along the coordinate axis perpendicular to the axis of revolution. The ability to choose which variable of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method . In the last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to help you determine which method to use in a given situation. The Method of Cylindrical Shells Again, we are working with a solid of revolution . As before, we define a region RR, bounded above by the graph of a function y\=f(x)y\=f(x), below by the xx-axis, and on the left and right by the lines x\=ax\=a and x\=bx\=b, respectively, as shown in Figure 6.3.1a6.3.1a. We then revolve this region around the yy-axis, as shown in Figure 6.3.1b6.3.1b. Note that this is different from what we have done before. Previously, regions defined in terms of functions of xx were revolved around the xx-axis or a line parallel to it. Figure 6.3.16.3.1: (a) A region bounded by the graph of a function of xx. (b) The solid of revolution formed when the region is revolved around the yy-axis. As we have done many times before, partition the interval [a,b][a,b] using a regular partition , P\=x0,x1,…,xnP\=x0,x1,…,xn and, for i\=1,2,…,ni\=1,2,…,n, choose a point x∗i∈[xi−1,xi]xi∗∈[xi−1,xi]. Then, construct a rectangle over the interval [xi−1,xi][xi−1,xi] of height f(x∗i)f(xi∗) and width ΔxΔx. A representative rectangle is shown in Figure 6.3.2a6.3.2a. When that rectangle is revolved around the yy-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in Figure 6.3.26.3.2. Figure 6.3.26.3.2: (a) A representative rectangle. (b) When this rectangle is revolved around the yy-axis, the result is a cylindrical shell. (c) When we put all the shells together, we get an approximation of the original solid. To calculate the volume of this shell, consider Figure 6.3.36.3.3. Figure 6.3.36.3.3: Calculating the volume of the shell. The shell is a cylinder , so its volume is the cross-sectional area multiplied by the height of the cylinder . The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius xixi and inner radius xi−1xi−1. Thus, the cross-sectional area is πx2i−πx2i−1πxi2−πxi−12. The height of the cylinder is f(x∗i).f(xi∗). Then the volume of the shell is Vshell\=f(x∗i)(πx2i−πx2i−1)\=πf(x∗i)(x2i−x2i−1)\=πf(x∗i)(xi+xi−1)(xi−xi−1)\=2πf(x∗i)(xi+xi−12)(xi−xi−1).Vshell\=f(xi∗)(πxi2−πxi−12)\=πf(xi∗)(xi2−xi−12)\=πf(xi∗)(xi+xi−1)(xi−xi−1)\=2πf(xi∗)(xi+xi−12)(xi−xi−1). Note that xi−xi−1\=Δx,xi−xi−1\=Δx, so we have Vshell\=2πf(x∗i)(xi+xi−12)Δx.Vshell\=2πf(xi∗)(xi+xi−12)Δx. Furthermore, xi+xi−12xi+xi−12 is both the midpoint of the interval [xi−1,xi][xi−1,xi] and the average radius of the shell, and we can approximate this by x∗ixi∗. We then have Vshell≈2πf(x∗i)x∗iΔx.Vshell≈2πf(xi∗)xi∗Δx. Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate (Figure 6.3.46.3.4). Figure 6.3.46.3.4: (a) Make a vertical cut in a representative shell. (b) Open the shell up to form a flat plate. In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height f(x∗i)f(xi∗), width 2πx∗i2πxi∗, and thickness ΔxΔx (Figure). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get Vshell≈f(x∗i)(2πx∗i)Δx,Vshell≈f(xi∗)(2πxi∗)Δx, which is the same formula we had before. To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain V≈∑i\=1n(2πx∗if(x∗i)Δx).V≈∑i\=1n(2πxi∗f(xi∗)Δx). Here we have another Riemann sum , this time for the function 2πxf(x).2πxf(x). Taking the limit as n→∞n→∞ gives us V\=limn→∞∑i\=1n(2πx∗if(x∗i)Δx)\=∫ba(2πxf(x))dx.V\=limn→∞∑i\=1n(2πxi∗f(xi∗)Δx)\=∫ab(2πxf(x))dx. This leads to the following rule for the_ method of cylindrical shells _. Rule: The Method of Cylindrical Shells Let f(x)f(x) be continuous and nonnegative. Define RR as the region bounded above by the graph of f(x)f(x), below by the xx-axis, on the left by the line x\=ax\=a, and on the right by the line x\=bx\=b. Then the volume of the solid of revolution formed by revolving RR around the yy-axis is given by V\=∫ba(2πxf(x))dx.V\=∫ab(2πxf(x))dx. Now let’s consider an example. Example 6.3.16.3.1: The Method of Cylindrical Shells I Define RR as the region bounded above by the graph of f(x)\=1/xf(x)\=1/x and below by the xx-axis over the interval [1,3][1,3]. Find the volume of the solid of revolution formed by revolving RR around the yy-axis. Solution First we must graph the region RR and the associated solid of revolution , as shown in Figure 6.3.56.3.5. Figure 6.3.56.3.5: (a) The region RR under the graph of f(x)\=1/xf(x)\=1/x over the interval [1,3][1,3]. (b) The solid of revolution generated by revolving RR about the yy-axis. Figure 6.3.56.3.5 (c) Visualizing the solid of revolution with CalcPlot3D. Then the volume of the solid is given by V\=∫ba(2πxf(x))dx\=∫31(2πx(1x))dx\=∫312πdx\=2πx∣∣∣31\=4πunits3.V\=∫ab(2πxf(x))dx\=∫13(2πx(1x))dx\=∫132πdx\=2πx|13\=4πunits3. Exercise 6.3.16.3.1 Define R as the region bounded above by the graph of f(x)\=x2f(x)\=x2 and below by the xx-axis over the interval [1,2][1,2]. Find the volume of the solid of revolution formed by revolving RR around the yy-axis. Hint Use the procedure from Example 6.3.16.3.1. Answer 15π2units315π2units3 Example 6.3.26.3.2: The Method of Cylindrical Shells II Define RR as the region bounded above by the graph of f(x)\=2x−x2f(x)\=2x−x2 and below by the xx-axis over the interval [0,2][0,2]. Find the volume of the solid of revolution formed by revolving RR around the yy-axis. Solution First graph the region RR and the associated solid of revolution , as shown in Figure 6.3.66.3.6. Figure 6.3.66.3.6: (a) The region RR under the graph of f(x)\=2x−x2f(x)\=2x−x2 over the interval [0,2].[0,2]. (b) The volume of revolution obtained by revolving RR about the yy_-axis._ Then the volume of the solid is given by V\=∫ba(2πxf(x))dx\=∫20(2πx(2x−x2))dx\=2π∫20(2x2−x3)dx\=2π[2x33−x44]∣∣∣20\=8π3units3V\=∫ab(2πxf(x))dx\=∫02(2πx(2x−x2))dx\=2π∫02(2x2−x3)dx\=2π[2x33−x44]|02\=8π3units3 Exercise 6.3.26.3.2 Define RR as the region bounded above by the graph of f(x)\=3x−x2f(x)\=3x−x2 and below by the xx-axis over the interval [0,2][0,2]. Find the volume of the solid of revolution formed by revolving RR around the yy-axis. Hint Use the process from Example 6.3.26.3.2. Answer 8πunits38πunits3 As with the disk method and the washer method , we can use the method of cylindrical shells with solids of revolution, revolved around the xx-axis, when we want to integrate with respect to yy. The analogous rule for this type of solid is given here. Rule: The Method of Cylindrical Shells for Solids of Revolution around the xx-axis Let g(y)g(y) be continuous and nonnegative. Define QQ as the region bounded on the right by the graph of g(y)g(y), on the left by the yy-axis, below by the line y\=cy\=c, and above by the line y\=dy\=d. Then, the volume of the solid of revolution formed by revolving QQ around the xx-axis is given by V\=∫dc(2πyg(y))dy.V\=∫cd(2πyg(y))dy. Example 6.3.36.3.3: The Method of Cylindrical Shells for a Solid Revolved around the xx-axis Define QQ as the region bounded on the right by the graph of g(y)\=2y√g(y)\=2y and on the left by the yy-axis for y∈[0,4]y∈[0,4]. Find the volume of the solid of revolution formed by revolving QQ around the xx-axis. Solution First, we need to graph the region QQ and the associated solid of revolution , as shown in Figure 6.3.76.3.7. Figure 6.3.76.3.7: (a) The region QQ to the left of the function g(y)g(y) over the interval [0,4][0,4]. (b) The solid of revolution generated by revolving QQ around the xx-axis. Label the shaded region QQ. Then the volume of the solid is given by V\=∫dc(2πyg(y))dy\=∫40(2πy(2y√))dy\=4π∫40y3/2dy\=4π[2y5/25]∣∣∣40\=256π5units3V\=∫cd(2πyg(y))dy\=∫04(2πy(2y))dy\=4π∫04y3/2dy\=4π[2y5/25]∣04\=256π5units3 Exercise 6.3.36.3.3 Define QQ as the region bounded on the right by the graph of g(y)\=3/yg(y)\=3/y and on the left by the yy-axis for y∈[1,3]y∈[1,3]. Find the volume of the solid of revolution formed by revolving QQ around the xx-axis. Hint Use the process from Example 6.3.36.3.3. Answer 12π12π units3 For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells . Recall that we found the volume of one of the shells to be given by Vshell\=f(x∗i)(πx2i−πx2i−1)\=πf(x∗i)(x2i−x2i−1)\=πf(x∗i)(xi+xi−1)(xi−xi−1)\=2πf(x∗i)(xi+xi−12)(xi−xi−1).Vshell\=f(xi∗)(πxi2−πxi−12)\=πf(xi∗)(xi2−xi−12)\=πf(xi∗)(xi+xi−1)(xi−xi−1)\=2πf(xi∗)(xi+xi−12)(xi−xi−1). This was based on a shell with an outer radius of xixi and an inner radius of xi−1xi−1. If, however, we rotate the region around a line other than the yy-axis, we have a different outer and inner radius. Suppose, for example, that we rotate the region around the line x\=−k,x\=−k, where kk is some positive constant. Then, the outer radius of the shell is xi+kxi+k and the inner radius of the shell is xi−1+kxi−1+k. Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line x\=−k,x\=−k, the volume of a shell is given by Vshell\=2πf(x∗i)((xi+k)+(xi−1+k)2)((xi+k)−(xi−1+k))\=2πf(x∗i)((xi+xi−22)+k)Δx.Vshell\=2πf(xi∗)((xi+k)+(xi−1+k)2)((xi+k)−(xi−1+k))\=2πf(xi∗)((xi+xi−22)+k)Δx. As before, we notice that xi+xi−12xi+xi−12 is the midpoint of the interval [xi−1,xi][xi−1,xi] and can be approximated by x∗ixi∗. Then, the approximate volume of the shell is Vshell≈2π(x∗i+k)f(x∗i)Δx.Vshell≈2π(xi∗+k)f(xi∗)Δx. The remainder of the development proceeds as before, and we see that V\=∫ba(2π(x+k)f(x))dx.V\=∫ab(2π(x+k)f(x))dx. We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the xx- term in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example. Example 6.3.46.3.4: A Region of Revolution Revolved around a Line Define RR as the region bounded above by the graph of f(x)\=xf(x)\=x and below by the xx-axis over the interval [1,2][1,2]. Find the volume of the solid of revolution formed by revolving RR around the line x\=−1.x\=−1. Solution First, graph the region RR and the associated solid of revolution , as shown in Figure 6.3.86.3.8. Figure 6.3.86.3.8: (a) The region RR between the graph of f(x)f(x) and the xx-axis over the interval [1,2][1,2]. (b) The solid of revolution generated by revolving RR around the line x\=−1.x\=−1. Note that the radius of a shell is given by x+1x+1. Then the volume of the solid is given by V\=∫212π(x+1)f(x)dx\=∫212π(x+1)xdx\=2π∫21x2+xdx\=2π[x33+x22]∣∣∣21\=23π3units3V\=∫122π(x+1)f(x)dx\=∫122π(x+1)xdx\=2π∫12x2+xdx\=2π[x33+x22]|12\=23π3units3 Exercise 6.3.46.3.4 Define RR as the region bounded above by the graph of f(x)\=x2f(x)\=x2 and below by the xx-axis over the interval [0,1][0,1]. Find the volume of the solid of revolution formed by revolving RR around the line x\=−2x\=−2. Hint Use the process from Example 6.3.46.3.4. Answer 11π611π6 units3 For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions. Example 6.3.56.3.5: A Region of Revolution Bounded by the Graphs of Two Functions Define RR as the region bounded above by the graph of the function f(x)\=x−−√f(x)\=x and below by the graph of the function g(x)\=1/xg(x)\=1/x over the interval [1,4][1,4]. Find the volume of the solid of revolution generated by revolving RR around the yy-axis. Solution First, graph the region RR and the associated solid of revolution , as shown in Figure 6.3.96.3.9. Figure 6.3.96.3.9: (a) The region RR between the graph of f(x)f(x) and the graph of g(x)g(x) over the interval [1,4][1,4]. (b) The solid of revolution generated by revolving RR around the yy-axis. Note that the axis of revolution is the yy-axis, so the radius of a shell is given simply by xx. We don’t need to make any adjustments to the x- term of our integrand . The height of a shell, though, is given by f(x)−g(x)f(x)−g(x), so in this case we need to adjust the f(x)f(x) term of the integrand . Then the volume of the solid is given by V\=∫41(2πx(f(x)−g(x)))dx\=∫41(2πx(x−−√−1x))dx\=2π∫41(x3/2−1)dx\=2π[2x5/25−x]∣∣∣41\=94π5units3.V\=∫14(2πx(f(x)−g(x)))dx\=∫14(2πx(x−1x))dx\=2π∫14(x3/2−1)dx\=2π[2x5/25−x]|14\=94π5units3. Exercise 6.3.56.3.5 Define RR as the region bounded above by the graph of f(x)\=xf(x)\=x and below by the graph of g(x)\=x2g(x)\=x2 over the interval [0,1][0,1]. Find the volume of the solid of revolution formed by revolving RR around the yy-axis. Hint Hint: Use the process from Example 6.3.56.3.5. Answer π6π6 units3 Which Method Should We Use? We have studied several methods for finding the volume of a solid of revolution , but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. Figure 6.3.106.3.10 describes the different approaches for solids of revolution around the xx-axis. It’s up to you to develop the analogous table for solids of revolution around the yy-axis. Figure 6.3.106.3.10 Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them. Example 6.3.66.3.6: Selecting the Best Method For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the xx-axis, and set up the integral to find the volume (do not evaluate the integral). The region bounded by the graphs of y\=x,y\=2−x,y\=x,y\=2−x, and the xx-axis. 2. The region bounded by the graphs of y\=4x−x2y\=4x−x2 and the xx-axis. Solution a. First, sketch the region and the solid of revolution as shown. Figure 6.3.116.3.11: (a) The region RR bounded by two lines and the xx-axis. (b) The solid of revolution generated by revolving RR about the xx-axis. Looking at the region , if we want to integrate with respect to xx, we would have to break the integral into two pieces, because we have different functions bounding the region over [0,1][0,1] and [1,2][1,2]. In this case, using the disk method , we would have V\=∫10πx2dx+∫21π(2−x)2dx.V\=∫01πx2dx+∫12π(2−x)2dx. If we used the shell method instead, we would use functions of y to represent the curves, producing V\=∫102πy[(2−y)−y]dy\=∫102πy[2−2y]dy.V\=∫012πy[(2−y)−y]dy\=∫012πy[2−2y]dy. Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case. b. First, sketch the region and the solid of revolution as shown. Figure 6.3.126.3.12: (a) The region RR between the curve and the xx-axis. (b) The solid of revolution generated by revolving RR about the xx-axis. Looking at the region , it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function . Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then V\=∫40π(4x−x2)2dxV\=∫04π(4x−x2)2dx Exercise 6.3.66.3.6 Select the best method to find the volume of a solid of revolution generated by revolving the given region around the xx-axis, and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of y\=2−x2y\=2−x2 and y\=x2y\=x2. Hint Sketch the region and use Figure 6.3.126.3.12 to decide which integral is easiest to evaluate. Answer Use the method of washers; V\=∫1−1π[(2−x2)2−(x2)2]dxV\=∫−11π[(2−x2)2−(x2)2]dx Key Concepts The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution . This method is sometimes preferable to either the method of disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other. The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use. Key Equations Method of Cylindrical Shells V\=∫ba(2πxf(x))dxV\=∫ab(2πxf(x))dx Glossary method of cylindrical shells a method of calculating the volume of a solid of revolution by dividing the solid into nested cylindrical shells; this method is different from the methods of disks or washers in that we integrate with respect to the opposite variable This page titled 6.3: Volumes of Revolution - Cylindrical Shells is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 6.2E: Exercises for Section 6.2 6.3E: Exercises for Section 6.3 Was this article helpful? Yes No Recommended articles 6.3: Volumes of Revolution - Cylindrical ShellsIn this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method... 2.3: Volumes of Revolution - Cylindrical ShellsIn this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method... 7.3: Volumes of Revolution - Cylindrical ShellsIn this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method... 7.3: Volumes of Revolution - Cylindrical ShellsIn this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method... 8.2: Volumes of Revolution - Cylindrical ShellsIn this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method... Article type Section or Page Author OpenStax License CC BY-NC-SA License Version 4.0 OER program or Publisher OpenStax Show Page TOC no Tags author@Edwin “Jed” Herman author@Gilbert Strang method of cylindrical shells solid of revolution source@ Volume by Shells © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries are Powered by NICE CXone Expert and 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 at info@libretexts.org. Support Center How can we help? Contact SupportSearch the Insight Knowledge BaseCheck System Status× contentsreadabilityresourcestools ☰ 6.2E: Exercises for Section 6.2 6.3E: Exercises for Section 6.3 Complete your gift to make an impact
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https://www.geeksforgeeks.org/dsa/square-root-of-a-perfect-square/
Babylonian method for square root Last Updated : 23 Jul, 2025 Suggest changes 18 Likes Algorithm: This method can be derived from (but predates) Newton–Raphson method. 1 Start with an arbitrary positive start value x (the closer to the root, the better). 2 Initialize y = 1. 3. Do following until desired approximation is achieved. a) Get the next approximation for root using average of x and y b) Set y = n/x Implementation: C++ ```` include using namespace std; class gfg { /Returns the square root of n. Note that the function / public: float squareRoot(float n) { /We are using n itself as initial approximation This can definitely be improved / float x = n; float y = 1; float e = 0.000001; / e decides the accuracy level/ while (x - y > e) { x = (x + y) / 2; y = n / x; } return x; } }; / Driver program to test above function/ int main() { gfg g; int n = 50; cout << "Square root of " << n << " is " << g.squareRoot(n); getchar(); } ```` ``` include #include ``` using namespace std; using namespace std class gfg {class gfg /Returns the square root of n. Note that the function / /Returns the square root of n. Note that the function / public: public float squareRoot(float n) float squareRoot float n { /We are using n itself as initial approximation /We are using n itself as initial approximation This can definitely be improved / This can definitely be improved / float x = n; float x = n float y = 1; float y = 1 float e = 0.000001; / e decides the accuracy level/ float e =0.000001 / e decides the accuracy level/ while (x - y > e) {while x - y> e x = (x + y) / 2; x = x + y/ 2 y = n / x; y = n/ x } return x; return x } }; ​ ​ / Driver program to test above function/ / Driver program to test above function/ int main() int main { gfg g; gfg g int n = 50; int n = 50 cout << "Square root of " << n << " is " << g.squareRoot(n); cout<< "Square root of "<< n<< " is "<< g squareRoot n getchar(); getchar } C ```` include /Returns the square root of n. Note that the function / float squareRoot(float n) { /We are using n itself as initial approximation This can definitely be improved / float x = n; float y = 1; float e = 0.000001; / e decides the accuracy level/ while (x - y > e) { x = (x + y) / 2; y = n / x; } return x; } / Driver program to test above function/ int main() { int n = 50; printf("Square root of %d is %f", n, squareRoot(n)); getchar(); } ```` Java ```` class GFG { /Returns the square root of n. Note that the function / static float squareRoot(float n) { /We are using n itself as initial approximation This can definitely be improved / float x = n; float y = 1; // e decides the accuracy level double e = 0.000001; while (x - y > e) { x = (x + y) / 2; y = n / x; } return x; } / Driver program to test above function/ public static void main(String[] args) { int n = 50; System.out.printf("Square root of " + n + " is " + squareRoot(n)); } } // This code is contributed by // Smitha DInesh Semwal ```` Python 3 ```` Returns the square root of n. Note that the function def squareRoot(n): # We are using n itself as # initial approximation This # can definitely be improved x = n y = 1 # e decides the accuracy level e = 0.000001 while(x - y > e): x = (x + y)/2 y = n / x return x Driver program to test above function n = 50 print("Square root of", n, "is", round(squareRoot(n), 6)) This code is contributed by Smitha Dinesh Semwal ```` C# ```` // C# Program for Babylonian // method of square root using System; class GFG { // Returns the square root of n. // Note that the function static float squareRoot(float n) { // We are using n itself as // initial approximation This // can definitely be improved float x = n; float y = 1; // e decides the // accuracy level double e = 0.000001; while (x - y > e) { x = (x + y) / 2; y = n / x; } return x; } // Driver Code public static void Main() { int n = 50; Console.Write("Square root of " + n + " is " + squareRoot(n)); } } // This code is contributed by nitin mittal. ```` PHP ```` php // Returns the square root of n. // Note that the function function squareRoot($n) { // We are using n itself // as initial approximation // This can definitely be // improved $x = $n; $y = 1; / e decides the accuracy level / $e = 0.000001; while($x - $y $e) { $x = ($x + $y)/2; $y = $n / $x; } return $x; } // Driver Code { $n = 50; echo "Square root of $n is ", squareRoot($n); } // This code is contributed by nitin mittal. ?> ```` JavaScript ```` // javascript Program to find the area // of triangle /Returns the square root of n. Note that the function / function squareRoot( n) { /We are using n itself as initial approximation This can definitely be improved / let x = n; let y = 1; let e = 0.000001; / e decides the accuracy level/ while (x - y > e) { x = (x + y) / 2; y = n / x; } return x; } / Driver program to test above function/ let n = 50; document.write( "Square root of "+n+" is " + squareRoot(n).toFixed(6)); // This code is contributed by todaysgaurav ```` Output : Square root of 50 is 7.071068 Time Complexity: O(n1/2) Auxiliary Space: O(1) Example: n = 4 /n itself is used for initial approximation/ Initialize x = 4, y = 1 Next Approximation x = (x + y)/2 (= 2.500000), y = n/x (=1.600000) Next Approximation x = 2.050000, y = 1.951220 Next Approximation x = 2.000610, y = 1.999390 Next Approximation x = 2.000000, y = 2.000000 Terminate as (x - y) > e now. If we are sure that n is a perfect square, then we can use following method. The method can go in infinite loop for non-perfect-square numbers. For example, for 3 the below while loop will never terminate. C++ ```` // C++ program for Babylonian // method for square root include using namespace std; class gfg { / Returns the square root of n. Note that the function will not work for numbers which are not perfect squares/ public: float squareRoot(float n) { / We are using n itself as an initial approximation. This can definitely be improved / float x = n; float y = 1; while (x > y) { x = (x + y) / 2; y = n / x; } return x; } }; / Driver code/ int main() { gfg g; int n = 49; cout << "Square root of " << n << " is " << g.squareRoot(n); getchar(); } // This code is edited by Dark_Dante_ ```` // C++ program for Babylonian // C++ program for Babylonian // method for square root // method for square root ``` include #include ``` using namespace std; using namespace std ​ ​ class gfg {class gfg ​ ​ / Returns the square root of / Returns the square root of n. Note that the function n. Note that the function will not work for numbers will not work for numbers which are not perfect which are not perfect squares/ squares/ public: public float squareRoot(float n) float squareRoot float n { / We are using n itself as an initial / We are using n itself as an initial approximation. This can definitely be approximation. This can definitely be improved / improved / float x = n; float x = n float y = 1; float y = 1 ​ ​ while (x > y) {while x> y x = (x + y) / 2; x = x + y/ 2 y = n / x; y = n/ x } return x; return x } }; ​ ​ / Driver code/ / Driver code/ int main() int main { gfg g; gfg g int n = 49; int n = 49 cout << "Square root of " << n << " is " << g.squareRoot(n); cout<< "Square root of "<< n<< " is "<< g squareRoot n getchar(); getchar } ​ ​ // This code is edited by Dark_Dante_ // This code is edited by Dark_Dante_ C ```` // C program for Babylonian // method for square root include / Returns the square root of n. Note that the function will not work for numbers which are not perfect squares/ unsigned int squareRoot(int n) { int x = n; int y = 1; while (x > y) { x = (x + y) / 2; y = n / x; } return x; } // Driver Code int main() { int n = 49; printf("root of %d is %d", n, squareRoot(n)); getchar(); } ```` Java ```` // Java program for Babylonian // method for square root import java.io.; public class GFG { / Returns the square root of n. Note that the function will not work for numbers which are not perfect squares/ static long squareRoot(int n) { int x = n; int y = 1; while (x > y) { x = (x + y) / 2; y = n / x; } return (long)x; } // Driver Code static public void main(String[] args) { int n = 49; System.out.println("root of " + n + " is " + squareRoot(n)); } } // This code is contributed by anuj_67. ```` Python3 ```` python3 program for Babylonian method for square root Returns the square root of n. Note that the function will not work for numbers which are not perfect squares def squareRoot(n): x = n; y = 1; while(x > y): x = (x + y) / 2; y = n / x; return x; Driver Code n = 49; print("root of", n, "is", squareRoot(n)); This code is contributed by mits. ```` C# ```` // C# program for Babylonian // method for square root using System; public class GFG { / Returns the square root of n. Note that the function will not work for numbers which are not perfect squares/ static uint squareRoot(int n) { int x = n; int y = 1; while (x > y) { x = (x + y) / 2; y = n / x; } return (uint)x; } // Driver Code static public void Main() { int n = 49; Console.WriteLine("root of " + n + " is " + squareRoot(n)); } } // This code is contributed by anuj_67. ```` PHP ```` php // PHP program for Babylonian // method for square root / Returns the square root of n. Note that the function will not work for numbers which are not perfect squares / function squareRoot( $n) { $x = $n; $y = 1; while($x $y) { $x = ($x + $y) / 2; $y =$n / $x; } return $x; } // Driver Code $n = 49; echo " root of ", $n, " is ", squareRoot($n); // This code is contributed by anuj_67. ?> ```` JavaScript ```` // javascript program for Babylonian // method for square root / Returns the square root of n. Note that the function will not work for numbers which are not perfect squares / function squareRoot(n) { var x = n; var y = 1; while (x > y) { x = (x + y) / 2; y = n / x; } return x; } // Driver Code var n = 49; document.write("root of " + n + " is " + squareRoot(n)); // This code contributed by shikhasingrajput ```` Output : root of 49 is 7 Time Complexity: O(n1/2) Auxiliary Space: O(1) References; Asked by Snehal Please write comments if you find any bug in the above program/algorithm, or if you want to share more information about Babylonian method. 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http://backreaction.blogspot.com/2008/04/elegant-proofs.html
Sabine Hossenfelder: Backreaction: Elegant proofs Pages Home About Friday, April 11, 2008 Elegant proofs Here is a little riddle: Take a checkerboard, and remove two squares at opposite corners. Is it then possible to find an exact and complete cover of the remaining board using dominoes (two are shown in orange), without overlap and overhang? There is a surprisingly simple and elegant proof for the negative answer to this question. I've just learned it this afternoon, in a great public talk by mathematician Günter Ziegler, coauthor of "Proofs from the book", current president of the German Association of Mathematicians, and main organiser of the "Year of Mathematics 2008" in Germany. Starting with the "Year of Physics" in 2000, the German Federal Ministry of Science and Education has dedicated each year since to one particular discipline, and following the humanities in 2007, this year is all about math. As Ziegler writes in the March 2008 issue of the Notices of the American Mathematical Society, "The entire year 2008 has been officially declared Mathematics Year in Germany. This has created an unprecedented opportunity to work on the public's view of the subject." And he used the opportunity, in a talk this afternoon on the occasion of the opening of the Mathematics Year for Frankfurt. He discussed the role of proof in mathematics, and then gave examples of actual elegant proofs of geometrical problems using colorations of the plane. The checkerboard riddle was just the first of them - he ended explaining the steps of a quite surprising proof that a square can not be decomposed into an odd number of triangles of equal area. I was amazed to see how I was guided by him through the steps and the idea of the proof - it's exciting to follow a talk like this! And my impression was that the 300 or so people in the audience have felt the same. Most of them, however, were faces I knew from the math department, or students and teachers. It would be great if such an event will attract even more people from the interested public. Have a nice weekend - and if you want to solve the checkerboard riddle by yourself, don't read the comments - I'm convinced the answer will be there pretty soon! Unfortunately, the slides of the talk are not online. Here are, roughly, the steps of the proof of the impossibility to decompose the square into an odd number of triangles of equal area: Start by colouring the rational points of the unit square in three different colours, using a scheme depending on the enumerator and denominator of the coordinates of the point. Then, convince yourself that each decomposition of the square into triangles (with corners in the rational points) contains at least one triangle with three different colours for the three corners. It comes out that this triangle, because of the rules chosen for colouring, has an area with an even denominator. Hence, in a decomposition into triangles with equal area, it cannot be part of a decomposition into an odd number of triangles. Use some heavy machinery to promote this proof from rational corner points to any corner points of the triangles, and you're done. If you find an error in this description, it's probably my fault - you may consult the original papers, "A Dissection Problem" by John Thomas, Mathematics Magazine 41 No. 4 (Sep. 1968), 187-190 (via JSTOR; subscription required), and "On Dividing a Square Into Triangles" by Paul Monsky, The American Mathematical Monthly 77 No. 2 (Feb. 1970) 161-164 (via JSTOR, subscription required). Tag: Mathematics Year Posted by stefan at 7:40 PM Labels: Germany, Science Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest 9 comments: CarlBrannen7:56 PM, April 11, 2008 Hah! An invitation to point out that counting white and black squares proves the impossibility. Which reminds me, at the local chess club, a game was discussed where white ended up with two white square bishops. The discussers concluded that there had been a mistake in the play as no pawns had been promoted. So this raises the question. Is it possible to change a white square bishop into a black square bishop by using an other than finite number of moves? Perhaps if one makes enough moves, the board will suffer a quantum transformation and the color of some of its squares will change... ReplyDelete Replies Reply Anonymous9:02 PM, April 11, 2008 Parity check! Parity might falsify General Relativity and string theory - parity Eötvös experiment opposing single crystal alpha-quartz enantiomorphs, parity calorimetry experiment opposing single crystal benzil enantiomorphs. Physics routinely chokes on parity - right hand rule, gyroscope precession, spacetime torsion... and that little Christmas 1956 experimental embarrassment. ReplyDelete Replies Reply Phil Warnell7:29 AM, April 12, 2008 Hi Uncle Al, “Physics routinely chokes on parity - right hand rule, gyroscope precession, spacetime torsion... and that little Christmas 1956 experimental embarrassment.” Yes, not only parity yet more importantly symmetry. That is our world is not simply that of the other side of the mirror and therefore just simply a different perspective of reality. For me it has always served to indicate that reality is thus not a special case of the possible, yet what is the possible is rather a special case of what we call reality. The other way to put it is, if Alice had stepped through the looking glass, she wouldn’t have found that all was simply strange, yet rather impossible and therefore non existent. The implications of breaking symmetry are one thing, while the disregard of it is another. I feel this to must form to be part of the consideration and explanation. Best, Phil ReplyDelete Replies Reply oleg8:02 AM, April 12, 2008 Stefan, Maybe I misunderstand something, but isn't there a much simpler proof by noting that the two removed squares are both black? Since every domino covers exactly one black and one white square, they can't cover 32 white and 30 black squares. ReplyDelete Replies Reply Phil Warnell8:22 AM, April 12, 2008 Hi Stefan, Nice post and again I find yours’ and Bees access at times both to be wondrous and yet at the same time a little envious. Yet, when I think about it I am grateful that you bring in part some of this experience to life within this blog. The puzzle you present I also find interesting, as to the fact that a proof can be discovered and presented. What I find interesting to considerer is that this proof is obviously discoverable since the spaces presented do have finite limit. The question that this then has one wonder about is if the answers that are sought within physics have the same and if not does this then form to be an obstacle that can or cannot be overcome? This I would understand to be something that first must be dealt with within the confines of mathematics and thus physics must in this sense wait for the judgment, unless it can demonstrate that the first limit is all that is to be considered. Best, Phil ReplyDelete Replies Reply Sabine Hossenfelder9:46 AM, April 12, 2008 Dear Stefan, Interesting post. I vaguely remember I came across this triangle proof elsewhere previously but can't recall where. Regarding the dominoes, what Oleg says above sounds good to me, what do you think? Best, B. ReplyDelete Replies Reply stefan4:40 AM, April 13, 2008 Hi Carl, Oleg, dear Bee, yes, the answer is indeed as simple as that - there are 32 white squares and 30 black squares, and each domino covers exactly one white and one black square - hence no complete covering is possible. I didn't know the riddle before, neither this answer. Günter Ziegler in his talk had presented the problem first for a board without the black and white coloration of the fields. The key step is to introduce this coloration, which then makes the counting argument quite obvious. When he explained this reasoning, I had a very surprised and pleasant "Aha" effect, which is, in my eyes, hallmark of an elegant proof. And Ziegler then went on applying similar surprising coloration arguments to more complicated and less obvious problems, like that of the decomposition of the square into triangles of equal area. Best, Stefan ReplyDelete Replies Reply Anonymous3:25 PM, April 13, 2008 Parity check! Every domino is 01 or 10. Remove two squares of the same color to give 00 or 11 residual. No domino can cover it. Order of rotations of a sphere; current, magnetic field, force; Weak Interaction; Cartesian coordinates... Physics assumes isotropic, commutative, abelian conditions then adds exceptions. Perhaps symmetry "breakings" were never unbroken. Gravitation is defined immune to symmetry breakings (General Relativity and spacetime curvature) or not (teleparallel gravitation and spacetime torsion transforming like EM Lorentz force). Parity Eötvös and parity calorimetry experiments answer the question. Somebody should look. ReplyDelete Replies Reply Arun8:07 AM, April 14, 2008 Sylvester-Gallai also has, IMO, an elegant proof. "The Sylvester–Gallai theorem asserts that given a finite number of points in the plane, either All the points are collinear; or There is a line which contains exactly two of the points." It remained open from 1893 to 1944. You might want to try your hand at it before looking at the answer (no, I didn't get it). Hint: proof by contradiction. ReplyDelete Replies Reply Add comment Load more... COMMENTS ON THIS BLOG ARE PERMANENTLY CLOSED. You can join the discussion on Patreon. 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https://www.cs.unm.edu/~saia/classes/561-f18/lec/lec-divideAndConquer.pdf
CS 561, Divide and Conquer: Induction, Recurrences, Master Method, Recursion Trees, Annihilators Jared Saia University of New Mexico Recurrence Relations “Oh how should I not lust after eternity and after the nuptial ring of rings, the ring of recurrence” - Friedrich Nietzsche, Thus Spoke Zarathustra • Getting the run times of recursive algorithms can be chal-lenging • Consider an algorithm for binary search (next slide) • Let T(n) be the run time of this algorithm on an array of size n • Then we can write T(1) = 1, T(n) = T(n/2) + 1 1 Alg: Binary Search bool BinarySearch (int arr[], int s, int e, int key){ if (e-s<=0) return false; int mid = (e+s)/2; if (key==arr[mid]){ return true; }else if (key < arr[mid]){ return BinarySearch (arr,s,mid,key);} else{ return BinarySearch (arr,mid,e,key)} } 2 Recurrence Relations • T(n) = T(n/2) + 1 is an example of a recurrence relation • A Recurrence Relation is any equation for a function T, where T appears on both the left and right sides of the equation. • We always want to “solve” these recurrence relation by get-ting an equation for T, where T appears on just the left side of the equation 3 Recurrence Relations • Whenever we analyze the run time of a recursive algorithm, we will first get a recurrence relation • To get the actual run time, we need to solve the recurrence relation 4 Substitution Method • One way to solve recurrences is the substitution method aka “guess and check” • What we do is make a good guess for the solution to T(n), and then try to prove this is the solution by induction 5 Example • Let’s guess that the solution to T(n) = T(n/2)+1, T(1) = 1 is T(n) = O(log n) • In other words, T(n) ≤c log n for all n ≥n0, for some positive constants c, n0 • We can prove that T(n) ≤c log n is true by plugging back into the recurrence 6 Proof We prove this by induction: • B.C.: T(2) = 2 ≤c log 2 provided that c ≥2 • I.H.: For all j < n, T(j) ≤c log(j) • I.S.: T(n) = T(n/2) + 1 (1) ≤ (c log(n/2)) + 1 (2) = c(log n −log 2) + 1 (3) = c log n −c + 1 (4) ≤ c log n (5) Second step holds by IH. Last step holds for all n > 0 if c ≥1. Thus, entire proof holds if n ≥2 and c ≥2. 7 Recurrences and Induction Recurrences and Induction are closely related: • To find a solution to f(n), solve a recurrence • To prove that a solution for f(n) is correct, use induction For both recurrences and induction, we always solve a big prob-lem by reducing it to smaller problems! 8 Some Examples • The next several problems can be attacked by induction/recurrences • For each problem, we’ll need to reduce it to smaller problems • Question: How can we reduce each problem to a smaller subproblem? 9 Sum Problem • f(n) is the sum of the integers 1, . . . , n 10 Tree Problem • f(n) is the maximum number of leaf nodes in a binary tree of height n Recall: • In a binary tree, each node has at most two children • A leaf node is a node with no children • The height of a tree is the length of the longest path from the root to a leaf node. 11 Binary Search Problem • f(n) is the maximum number of queries that need to be made for binary search on a sorted array of size n. 12 Dominoes Problem • f(n) is the number of ways to tile a 2 by n rectangle with dominoes (a domino is a 2 by 1 rectangle) 13 Simpler Subproblems • Sum Problem: What is the sum of all numbers between 1 and n −1 (i.e. f(n −1))? • Tree Problem: What is the maximum number of leaf nodes in a binary tree of height n −1? (i.e. f(n −1)) • Binary Search Problem: What is the maximum number of queries that need to be made for binary search on a sorted array of size n/2? (i.e. f(n/2)) • Dominoes problem: What is the number of ways to tile a 2 by n −1 rectangle with dominoes? What is the number of ways to tile a 2 by n −2 rectangle with dominoes? (i.e. f(n −1), f(n −2)) 14 Recurrences • Sum Problem: f(n) = f(n −1) + n, f(1) = 1 • Tree Problem: f(n) = 2 ∗f(n −1), f(0) = 1 • Binary Search Problem: f(n) = f(n/2) + 1, f(2) = 1 • Dominoes problem: f(n) = f(n −1) + f(n −2), f(1) = 1, f(2) = 2 15 Guesses • Sum Problem: f(n) = (n + 1)n/2 • Tree Problem: f(n) = 2n • Binary Search Problem: f(n) = log n • Dominoes problem: f(n) = 1 √ 5 󰀕 1+ √ 5 2 󰀖n −1 √ 5 󰀕 1− √ 5 2 󰀖n 16 Inductive Proofs “Trying is the first step to failure” - Homer Simpson • Now that we’ve made these guesses, we can try using induc-tion to prove they’re correct (the substitution method) • We’ll give inductive proofs that these guesses are correct for the first three problems 17 Sum Problem • Want to show that f(n) = (n + 1)n/2. • Prove by induction on n • Base case: f(1) = 2 ∗1/2 = 1 • Inductive hypothesis: for all j < n, f(j) = (j + 1)j/2 • Inductive step: f(n) = f(n −1) + n (6) = n(n −1)/2 + n (7) = (n + 1)n/2 (8) Where the second step holds by the IH. 18 Tree Problem • Want to show that f(n) = 2n. • Prove by induction on n • Base case: f(0) = 20 = 1 • Inductive hypothesis: for all j < n, f(j) = 2j • Inductive step: f(n) = 2 ∗f(n −1) (9) = 2 ∗(2n−1) (10) = 2n (11) (second step holds by IH) 19 Binary Search Problem • Want to show that f(n) = log n. (assume n is a power of 2) • Prove by induction on n • Base case: f(2) = log 2 = 1 • Inductive hypothesis: for all j < n, f(j) = log j • Inductive step: f(n) = f(n/2) + 1 (12) = log n/2 + 1 (13) = log n −log 2 + 1 (14) = log n (15) (second step holds by IH) 20 In Class Exercise • Consider the recurrence f(n) = 2f(n/2) + 1, f(1) = 1 • Guess that f(n) ≤cn −1: • Q1: Show the base case - for what values of c does it hold? • Q2: What is the inductive hypothesis? • Q3: Show the inductive step. 21 Recurrences and Inequalities • Often easier to prove that a recurrence is no more than some quantity than to prove that it equals something • Consider: f(n) = f(n −1) + f(n −2), f(1) = f(2) = 1 • “Guess” that f(n) ≤2n 22 Inequalities (II) Goal: Prove by induction that for f(n) = f(n −1) + f(n −2), f(1) = f(2) = 1, f(n) ≤2n • Base case: f(1) = 1 ≤21, f(2) = 1 ≤22 • Inductive hypothesis: For all j < n, f(j) ≤2j • Inductive step: f(n) = f(n −1) + f(n −2) (16) ≤ 2n−1 + 2n−2 (17) < 2 ∗2n−1 (18) = 2n (19) (second step holds by IH) 23 Recursion-tree method • Each node represents the cost of a single subproblem in a recursive call • First, we sum the costs of the nodes in each level of the tree • Then, we sum the costs of all of the levels 24 Recursion-tree method • Can use to get a good guess which is then refined and verified using substitution method • Best method (usually) for recurrences where a term like T(n/c) appears on the right hand side of the equality 25 Example 1 • Consider the recurrence for the running time of Mergesort: T(n) = 2T(n/2) + n, T(1) = O(1) n n/2 n/2 n/4 n/4 n/4 n/4 n/8 n/8 n/8 n/8 n/8 n/8 n/8 n/8 n n n n 26 Example 1 • We can see that each level of the tree sums to n • Further the depth of the tree is log n (n/2d = 1 implies that d = log n). • Thus there are log n + 1 levels each of which sums to n • Hence T(n) = Θ(n log n) 27 Example 2 • Let’s solve the recurrence T(n) = 3T(n/4) + n2 • Note: For simplicity, from now on, we’ll assume that T(i) = Θ(1) for all small constants i. This will save us from writing the base cases each time. (n/16)^2 (n/16)^2 n^2 (n/4)^2 (n/4)^2 (n/16)^2 (n/4)^2 (n/16)^2 (n/16)^2 (n/16)^2 (n/16)^2 (n/16)^2 (n/16)^2 n^2 (3/16)n^2 (3/16)^2n^2 ... 28 Example 2 • We can see that the i-th level of the tree sums to (3/16)in2. • Further the depth of the tree is log4 n (n/4d = 1 implies that d = log4 n) • So we can see that T(n) = 󰁓log4 n i=0 (3/16)in2 29 Solution T(n) = log4 n 󰁛 i=0 (3/16)in2 (20) < n2 ∞ 󰁛 i=0 (3/16)i (21) = 1 1 −(3/16)n2 (22) = O(n2) (23) 30 Master Theorem • Divide and conquer algorithms often give us running-time recurrences of the form T(n) = a T(n/b) + f(n) (24) • Where a and b are constants and f(n) is some other function. • The so-called “Master Method” gives us a general method for solving such recurrences 31 Master Theorem • Unfortunately, the Master Theorem doesn’t work for all func-tions f(n) • Further many useful recurrences don’t look like T(n) • However, the theorem allows for very fast solution of recur-rences when it applies 32 Master Theorem • Master Theorem is just a special case of the use of recursion trees • Consider equation T(n) = a T(n/b) + f(n) • We start by drawing a recursion tree 33 The Recursion Tree • The root contains the value f(n) • It has a children, each of which contains the value f(n/b) • Each of these nodes has a children, containing the value f(n/b2) • In general, level i contains ai nodes with values f(n/bi) • Hence the sum of the nodes at the i-th level is aif(n/bi) 34 Details • The tree stops when we get to the base case for the recur-rence • We’ll assume T(1) = f(1) = Θ(1) is the base case • Thus the depth of the tree is logb n and there are logb n + 1 levels 35 Recursion Tree • Let T(n) be the sum of all values stored in all levels of the tree: T(n) = f(n)+a f(n/b)+a2 f(n/b2)+· · ·+ai f(n/bi)+· · ·+aL f(n/bL) • Where L = logb n is the depth of the tree • Since f(1) = Θ(1), the last term of this summation is Θ(aL) = Θ(alogb n) = Θ(nlogb a) 36 A “Log Fact” Aside • It’s not hard to see that alogb n = nlogb a alogb n = nlogb a (25) alogb n = aloga n∗logb a (26) logb n = loga n ∗logb a (27) • We get to the last eqn by taking loga of both sides • The last eqn is true by our third basic log fact 37 Master Theorem • We can now state the Master Theorem • We will state it in a way slightly different from the book • Note: The Master Method is just a “short cut” for the re-cursion tree method. It is less powerful than recursion trees. 38 Master Method The recurrence T(n) = aT(n/b) + f(n) can be solved as follows: • If a f(n/b) ≤Kf(n) for some constant K < 1, then T(n) = Θ(f(n)). • If a f(n/b) ≥K f(n) for some constant K > 1, then T(n) = Θ(nlogb a). • If a f(n/b) = f(n), then T(n) = Θ(f(n) logb n). 39 Proof • If f(n) is a constant factor larger than a f(n/b), then the sum is a descending geometric series. The sum of any geometric series is a constant times its largest term. In this case, the largest term is the first term f(n). • If f(n) is a constant factor smaller than a f(n/b), then the sum is an ascending geometric series. The sum of any ge-ometric series is a constant times its largest term. In this case, this is the last term, which by our earlier argument is Θ(nlogb a). • Finally, if a f(n/b) = f(n), then each of the L + 1 terms in the summation is equal to f(n). 40 Example • T(n) = T(3n/4) + n • If we write this as T(n) = aT(n/b) + f(n), then a = 1,b = 4/3,f(n) = n • Here a f(n/b) = 3n/4 is smaller than f(n) = n by a factor of 4/3, so T(n) = Θ(n) 41 Example • Karatsuba’s multiplication algorithm: T (n) = 3T (n/2) + n • If we write this as T(n) = aT(n/b) + f(n), then a = 3,b = 2,f(n) = n • Here a f(n/b) = 3n/2 is bigger than f(n) = n by a factor of 3/2, so T(n) = Θ(nlog2 3) 42 Example • Mergesort: T (n) = 2T (n/2) + n • If we write this as T(n) = aT(n/b) + f(n), then a = 2,b = 2,f(n) = n • Here a f(n/b) = f(n), so T(n) = Θ(n log n) 43 Example • T(n) = T(n/2) + n log n • If we write this as T(n) = aT(n/b) + f(n), then a = 1,b = 2,f(n) = n log n • Here a f(n/b) = n/2 log n/2 is smaller than f(n) = n log n by a constant factor, so T(n) = Θ(n log n) 44 In-Class Exercise • Consider the recurrence: T (n) = 4T (n/2) + n lg n • Q: What is f(n) and a f(n/b)? • Q: Which of the three cases does the recurrence fall under (when n is large)? • Q: What is the solution to this recurrence? 45 In-Class Exercise • Consider the recurrence: T (n) = 2T (n/4) + n lg n • Q: What is f(n) and a f(n/b)? • Q: Which of the three cases does the recurrence fall under (when n is large)? • Q: What is the solution to this recurrence? 46 Take Away • Recursion tree and Master method are good tools for solving many recurrences • However these methods are limited (they can’t help us get guesses for recurrences like f(n) = f(n −1) + f(n −2)) • For info on how to solve these other more difficult recur-rences, review the notes on annihilators on the class web page. 47 Intro to Annihilators “Listen and Understand! That terminator is out there. It can’t be bargained with, it can’t be reasoned with! It doesn’t feel pity, remorse, or fear. And it absolutely will not stop, ever, until you are dead!” - The Terminator • Suppose we are given a sequence of numbers A = 〈a0, a1, a2, · · · 〉 • This might be a sequence like the Fibonacci numbers • I.e. A = 〈a0, a1, a2, . . . ) = (T(1), T(2), T(3), · · · 〉 48 Annihilator Operators We define three basic operations we can perform on this se-quence: 1. Multiply the sequence by a constant: cA = 〈ca0, ca1, ca2, · · · 〉 2. Shift the sequence to the left: LA = 〈a1, a2, a3, · · · 〉 3. Add two sequences: if A = 〈a0, a1, a2, · · · 〉and B = 〈b0, b1, b2, · · · 〉, then A + B = 〈a0 + b0, a1 + b1, a2 + b2, · · · 〉 49 Annihilator Description • We first express our recurrence as a sequence T • We use these three operators to “annihilate” T, i.e. make it all 0’s • Key rule: can’t multiply by the constant 0 • We can then determine the solution to the recurrence from the sequence of operations performed to annihilate T 50 Example • Consider the recurrence T(n) = 2T(n −1), T(0) = 1 • If we solve for the first few terms of this sequence, we can see they are 〈20, 21, 22, 23, · · · 〉 • Thus this recurrence becomes the sequence: T = 〈20, 21, 22, 23, · · · 〉 51 Example (II) Let’s annihilate T = 〈20, 21, 22, 23, · · · 〉 • Multiplying by a constant c = 2 gets: 2T = 〈2 ∗20, 2 ∗21, 2 ∗22, 2 ∗23, · · · 〉= 〈21, 22, 23, 24, · · · 〉 • Shifting one place to the left gets LT = 〈21, 22, 23, 24, · · · 〉 • Adding the sequence LT and −2T gives: LT −2T = 〈21 −21, 22 −22, 23 −23, · · · 〉= 〈0, 0, 0, · · · 〉 • The annihilator of T is thus L −2 52 Distributive Property • The distributive property holds for these three operators • Thus can rewrite LT −2T as (L −2)T • The operator (L −2) annihilates T (makes it the sequence of all 0’s) • Thus (L −2) is called the annihilator of T 53 0, the “Forbidden Annihilator” • Multiplication by 0 will annihilate any sequence • Thus we disallow multiplication by 0 as an operation • In particular, we disallow (c−c) = 0 for any c as an annihilator • Must always have at least one L operator in any annihilator! 54 Uniqueness • An annihilator annihilates exactly one type of sequence • In general, the annihilator L −c annihilates any sequence of the form 〈a0cn〉 • If we find the annihilator, we can find the type of sequence, and thus solve the recurrence • We will need to use the base case for the recurrence to solve for the constant a0 55 Example If we apply operator (L −3) to sequence T above, it fails to annihilate T (L −3)T = LT + (−3)T = 〈21, 22, 23, · · · 〉+ 〈−3 × 20, −3 × 21, −3 × 22, · · · 〉 = 〈(2 −3) × 20, (2 −3) × 21, (2 −3) × 22, · · · 〉 = (2 −3)T = −T 56 Example (II) What does (L−c) do to other sequences A = 〈a0dn〉when d ∕= c?: (L −c)A = (L −c)〈a0, a0d, a0d2, a0d3, · · · 〉 = L〈a0, a0d, a0d2, a0d3, · · · 〉−c〈a0, a0d, a0d2, a0d3, · · · 〉 = 〈a0d, a0d2, a0d3, · · · 〉−〈ca0, ca0d, ca0d2, ca0d3, · · · 〉 = 〈a0d −ca0, a0d2 −ca0d, a0d3 −ca0d2, · · · 〉 = 〈(d −c)a0, (d −c)a0d, (d −c)a0d2, · · · 〉 = (d −c)〈a0, a0d, a0d2, · · · 〉 = (d −c)A 57 Uniqueness • The last example implies that an annihilator annihilates one type of sequence, but does not annihilate other types of sequences • Thus Annihilators can help us classify sequences, and thereby solve recurrences 58 Lookup Table • The annihilator L −a annihilates any sequence of the form 〈c1an〉 59 Example First calculate the annihilator: • Recurrence: T(n) = 4 ∗T(n −1), T(0) = 2 • Sequence: T = 〈2, 2 ∗4, 2 ∗42, 2 ∗43, · · · 〉 • Calulate the annihilator: – LT = 〈2 ∗4, 2 ∗42, 2 ∗43, 2 ∗44, · · · 〉 – 4T = 〈2 ∗4, 2 ∗42, 2 ∗43, 2 ∗44, · · · 〉 – Thus LT −4T = 〈0, 0, 0, · · · 〉 – And so L −4 is the annihilator 60 Example (II) Now use the annihilator to solve the recurrence • Look up the annihilator in the “Lookup Table” • It says: “The annihilator L −4 annihilates any sequence of the form 〈c14n〉” • Thus T(n) = c14n, but what is c1? • We know T(0) = 2, so T(0) = c140 = 2 and so c1 = 2 • Thus T(n) = 2 ∗4n 61 In Class Exercise Consider the recurrence T(n) = 3 ∗T(n −1), T(0) = 3, • Q1: Calculate T(0),T(1),T(2) and T(3) and write out the sequence T • Q2: Calculate LT, and use it to compute the annihilator of T • Q3: Look up this annihilator in the lookup table to get the general solution of the recurrence for T(n) • Q4: Now use the base case T(0) = 3 to solve for the con-stants in the general solution 62 Remaining Outline • Annihilators with Multiple Operators • Annihilators for recurrences with non-homogeneous terms • Transformations 63 Multiple Operators • We can apply multiple operators to a sequence • For example, we can multiply by the constant c and then by the constant d to get the operator cd • We can also multiply by c and then shift left to get cLT which is the same as LcT • We can also shift the sequence twice to the left to get LLT which we’ll write in shorthand as L2T 64 Multiple Operators • We can string operators together to annihilate more compli-cated sequences • Consider: T = 〈20 + 30, 21 + 31, 22 + 32, · · · 〉 • We know that (L−2) annihilates the powers of 2 while leaving the powers of 3 essentially untouched • Similarly, (L −3) annihilates the powers of 3 while leaving the powers of 2 essentially untouched • Thus if we apply both operators, we’ll see that (L−2)(L−3) annihilates the sequence T 65 The Details • Consider: T = 〈a0 + b0, a1 + b1, a2 + b2, · · · 〉 • LT = 〈a1 + b1, a2 + b2, a3 + b3, · · · 〉 • aT = 〈a1 + a ∗b0, a2 + a ∗b1, a3 + a ∗b2, · · · 〉 • LT −aT = 〈(b −a)b0, (b −a)b1, (b −a)b2, · · · 〉 • We know that (L−a)T annihilates the a terms and multiplies the b terms by b −a (a constant) • Thus (L −a)T = 〈(b −a)b0, (b −a)b1, (b −a)b2, · · · 〉 • And so the sequence (L −a)T is annihilated by (L −b) • Thus the annihilator of T is (L −b)(L −a) 66 Key Point • In general, the annihilator (L −a)(L −b) (where a ∕= b) will anihilate only all sequences of the form 〈c1an + c2bn〉 • We will often multiply out (L−a)(L−b) to L2 −(a+b)L+ab • Left as an exercise to show that (L −a)(L −b)T is the same as (L2 −(a + b)L + ab)T 67 Lookup Table • The annihilator L−a annihilates sequences of the form 〈c1an〉 • The annihilator (L −a)(L −b) (where a ∕= b) anihilates se-quences of the form 〈c1an + c2bn〉 68 Fibonnaci Sequence • We now know enough to solve the Fibonnaci sequence • Recall the Fibonnaci recurrence is T(0) = 0, T(1) = 1, and T(n) = T(n −1) + T(n −2) • Let Tn be the n-th element in the sequence • Then we’ve got: T = 〈T0, T1, T2, T3, · · · 〉 (28) LT = 〈T1, T2, T3, T4, · · · 〉 (29) L2T = 〈T2, T3, T4, T5, · · · 〉 (30) • Thus L2T −LT −T = 〈0, 0, 0, · · · 〉 • In other words, L2 −L −1 is an annihilator for T 69 Factoring • L2 −L −1 is an annihilator that is not in our lookup table • However, we can factor this annihilator (using the quadratic formula) to get something similar to what’s in the lookup table • L2 −L −1 = (L −φ)(L −ˆ φ), where φ = 1+ √ 5 2 and ˆ φ = 1− √ 5 2 . 70 Quadratic Formula “Me fail English? That’s Unpossible!” - Ralph, the Simpsons High School Algebra Review: • To factor something of the form ax2 + bx + c, we use the Quadratic Formula: • ax2 + bx + c factors into (x −φ)(x −ˆ φ), where: φ = −b + 󰁴 b2 −4ac 2a (31) ˆ φ = −b − 󰁴 b2 −4ac 2a (32) 71 Example • To factor: L2 −L −1 • Rewrite: 1 ∗L2 −1 ∗L −1, a = 1, b = −1, c = −1 • From Quadratic Formula: φ = 1+ √ 5 2 and ˆ φ = 1− √ 5 2 • So L2 −L −1 factors to (L −φ)(L −ˆ φ) 72 Back to Fibonnaci • Recall the Fibonnaci recurrence is T(0) = 0, T(1) = 1, and T(n) = T(n −1) + T(n −2) • We’ve shown the annihilator for T is (L −φ)(L −ˆ φ), where φ = 1+ √ 5 2 and ˆ φ = 1− √ 5 2 • If we look this up in the “Lookup Table”, we see that the sequence T must be of the form 〈c1φn + c2ˆ φn〉 • All we have left to do is solve for the constants c1 and c2 • Can use the base cases to solve for these 73 Finding the Constants • We know T = 〈c1φn + c2ˆ φn〉, where φ = 1+ √ 5 2 and ˆ φ = 1− √ 5 2 • We know T(0) = c1 + c2 = 0 (33) T(1) = c1φ + c2ˆ φ = 1 (34) • We’ve got two equations and two unknowns • Can solve to get c1 = 1 √ 5 and c2 = −1 √ 5, 74 The Punchline • Recall Fibonnaci recurrence: T(0) = 0, T(1) = 1, and T(n) = T(n −1) + T(n −2) • The final explicit formula for T(n) is thus: T(n) = 1 √ 5 󰀣 1 + √ 5 2 󰀤n −1 √ 5 󰀣 1 − √ 5 2 󰀤n (Amazingly, T(n) is always an integer, in spite of all of the square roots in its formula.) 75 A Problem • Our lookup table has a big gap: What does (L −a)(L −a) annihilate? • It turns out it annihilates sequences such as 〈nan〉 76 Example (L −a)〈nan〉 = 〈(n + 1)an+1 −(a)nan〉 = 〈(n + 1)an+1 −nan+1〉 = 〈(n + 1 −n)an+1〉 = 〈an+1〉 (L −a)2〈nan〉 = (L −a)〈an+1〉 = 〈0〉 77 Generalization • It turns out that (L −a)d annihilates sequences of the form 〈p(n)an〉where p(n) is any polynomial of degree d −1 • Example: (L −1)3 annihilates the sequence 〈n2 ∗1n〉= 〈1, 4, 9, 16, 25〉since p(n) = n2 is a polynomial of degree d −1 = 2 78 Lookup Table • (L −a) annihilates only all sequences of the form 〈c0an〉 • (L−a)(L−b) annihilates only all sequences of the form 〈c0an+ c1bn〉 • (L −a0)(L −a1) . . . (L −ak) annihilates only sequences of the form 〈c0an 0 + c1an 1 + . . . ckan k〉, here ai ∕= aj, when i ∕= j • (L−a)2 annihilates only sequences of the form 〈(c0n+c1)an〉 • (L −a)k annihilates only sequences of the form 〈p(n)an〉, degree(p(n)) = k −1 79 Lookup Table (Final!) (L −a0)b0(L −a1)b1 . . . (L −ak)bk annihilates only sequences of the form: 〈p1(n)an 0 + p2(n)an 1 + . . . pk(n)an k〉 where pi(n) is a polynomial of degree bi −1 (and ai ∕= aj, when i ∕= j) 80 Examples • Q: What does (L −3)(L −2)(L −1) annihilate? • A: c01n + c12n + c23n • Q: What does (L −3)2(L −2)(L −1) annihilate? • A: c01n + c12n + (c2n + c3)3n • Q: What does (L −1)4 annihilate? • A: (c0n3 + c1n2 + c2n + c3)1n • Q: What does (L −1)3(L −2)2 annihilate? • A: (c0n2 + c1n + c2)1n + (c3n + c4)2n 81 Annihilator Method • Write down the annihilator for the recurrence • Factor the annihilator • Look up the factored annihilator in the “Lookup Table” to get general solution • Solve for constants of the general solution by using initial conditions 82 Annihilator Method • Write down the annihilator for the recurrence • Factor the annihilator • Look up the factored annihilator in the “Lookup Table” to get general solution • Solve for constants of the general solution by using initial conditions 83 Lookup Table (L −a0)b0(L −a1)b1 . . . (L −ak)bk annihilates only sequences of the form: 〈p0(n)an 0 + p1(n)an 1 + . . . pk(n)an k〉 where pi(n) is a polynomial of degree bi −1 (and ai ∕= aj, when i ∕= j) 84 Examples • Q: What does (L −3)(L −2)(L −1) annihilate? • A: c01n + c12n + c23n • Q: What does (L −3)2(L −2)(L −1) annihilate? • A: c01n + c12n + (c2n + c3)3n • Q: What does (L −1)4 annihilate? • A: (c0n3 + c1n2 + c2n + c3)1n • Q: What does (L −1)3(L −2)2 annihilate? • A: (c0n2 + c1n + c2)1n + (c3n + c4)2n 85 Example Consider the recurrence T(n) = 7T(n−1)−16T(n−2)+12T(n− 3), T(0) = 1, T(1) = 5, T(2) = 17 • Write down the annihilator: From the definition of the sequence, we can see that L3T −7L2T + 16LT −12T = 0, so the annihilator is L3 −7L2 + 16L −12 • Factor the annihilator: We can factor by hand or using a computer program to get L3−7L2+16L−12 = (L−2)2(L−3) • Look up to get general solution: The annihilator (L − 2)2(L −3) annihilates sequences of the form 〈(c0n + c1)2n + c23n〉 • Solve for constants: T(0) = 1 = c1 + c2, T(1) = 5 = 2c0 + 2c1 + 3c2, T(2) = 17 = 8c0 + 4c1 + 9c2. We’ve got three equations and three unknowns. Solving by hand, we get that c0 = 1,c1 = 0,c2 = 1. Thus: T(n) = n2n + 3n 86 Example (II) Consider the recurrence T(n) = 2T(n −1) −T(n −2), T(0) = 0, T(1) = 1 • Write down the annihilator: From the definition of the se-quence, we can see that L2T −2LT +T = 0, so the annihilator is L2 −2L + 1 • Factor the annihilator: We can factor by hand or using the quadratic formula to get L2 −2L + 1 = (L −1)2 • Look up to get general solution: The annihilator (L −1)2 annihilates sequences of the form (c0n + c1)1n • Solve for constants: T(0) = 0 = c1, T(1) = 1 = c0 + c1, We’ve got two equations and two unknowns. Solving by hand, we get that c0 = 0,c1 = 1. Thus: T(n) = n 87 At Home Exercise Consider the recurrence T(n) = 6T(n −1) −9T(n −2), T(0) = 1, T(1) = 6 • Q1: What is the annihilator of this sequence? • Q2: What is the factored version of the annihilator? • Q3: What is the general solution for the recurrence? • Q4: What are the constants in this general solution? (Note: You can check that your general solution works for T(2)) 88 Non-homogeneous terms • Consider a recurrence of the form T(n) = T(n −1) + T(n − 2) + k where k is some constant • The terms in the equation involving T (i.e. T(n −1) and T(n −2)) are called the homogeneous terms • The other terms (i.e.k) are called the non-homogeneous terms 89 Example • In a height-balanced tree, the height of two subtrees of any node differ by at most one • Let T(n) be the smallest number of nodes needed to obtain a height balanced binary tree of height n • Q: What is a recurrence for T(n)? • A: Divide this into smaller subproblems – To get a height-balanced tree of height n with the smallest number of nodes, need one subtree of height n −1, and one of height n −2, plus a root node – Thus T(n) = T(n −1) + T(n −2) + 1 90 Example • Let’s solve this recurrence: T(n) = T(n −1) + T(n −2) + 1 (Let Tn = T(n), and T = 〈Tn〉) • We know that (L2−L−1) annihilates the homogeneous terms • Let’s apply it to the entire equation: (L2 −L −1)〈Tn〉 = L2〈Tn〉−L〈Tn〉−1〈Tn〉 = 〈Tn+2〉−〈Tn+1〉−〈Tn〉 = 〈Tn+2 −Tn+1 −Tn〉 = 〈1, 1, 1, · · · 〉 91 Example • This is close to what we want but we still need to annihilate 〈1, 1, 1, · · · 〉 • It’s easy to see that L −1 annihilates 〈1, 1, 1, · · · 〉 • Thus (L2 −L−1)(L−1) annihilates T(n) = T(n−1)+T(n− 2) + 1 • When we factor, we get (L−φ)(L−ˆ φ)(L−1), where φ = 1+ √ 5 2 and ˆ φ = 1− √ 5 2 . 92 Lookup • Looking up (L −φ)(L −ˆ φ)(L −1) in the table • We get T(n) = c1φn + c2ˆ φn + c31n • If we plug in the appropriate initial conditions, we can solve for these three constants • We’ll need to get equations for T(2) in addition to T(0) and T(1) 93 General Rule To find the annihilator for recurrences with non-homogeneous terms, do the following: • Find the annihilator a1 for the homogeneous part • Find the annihilator a2 for the non-homogeneous part • The annihilator for the whole recurrence is then a1a2 94 Another Example • Consider T(n) = T(n −1) + T(n −2) + 2. • The residue is 〈2, 2, 2, · · · 〉and • The annihilator is still (L2 −L −1)(L −1), but the equation for T(2) changes! 95 Another Example • Consider T(n) = T(n −1) + T(n −2) + 2n. • The residue is 〈1, 2, 4, 8, · · · 〉and • The annihilator is now (L2 −L −1)(L −2). 96 Another Example • Consider T(n) = T(n −1) + T(n −2) + n. • The residue is 〈1, 2, 3, 4, · · · 〉 • The annihilator is now (L2 −L −1)(L −1)2. 97 Another Example • Consider T(n) = T(n −1) + T(n −2) + n2. • The residue is 〈1, 4, 9, 16, · · · 〉and • The annihilator is (L2 −L −1)(L −1)3. 98 Another Example • Consider T(n) = T(n −1) + T(n −2) + n2 −2n. • The residue is 〈1 −1, 4 −4, 9 −8, 16 −16, · · · 〉and the • The annihilator is (L2 −L −1)(L −1)3(L −2). 99 In Class Exercise • Consider T(n) = 3 ∗T(n −1) + 3n • Q1: What is the homogeneous part, and what annihilates it? • Q2: What is the non-homogeneous part, and what annihi-lates it? • Q3: What is the annihilator of T(n), and what is the general form of the recurrence? 100 Limitations • Our method does not work on T(n) = T(n−1)+ 1 n or T(n) = T(n −1) + lg n • The problem is that 1 n and lg n do not have annihilators. • Our tool, as it stands, is limited. • Key idea for strengthening it is transformations 101 Transformations Idea • Consider the recurrence giving the run time of mergesort T(n) = 2T(n/2) + kn (for some constant k), T(1) = 1 • How do we solve this? • We have no technique for annihilating terms like T(n/2) • However, we can transform the recurrence into one with which we can work 102 Transformation • Let n = 2i and rewrite T(n): • T(20) = 1 and T(2i) = 2T(2i 2 ) + k2i = 2T(2i−1) + k2i • Now define a new sequence t as follows: t(i) = T(2i) • Then t(0) = 1, t(i) = 2t(i −1) + k2i 103 Now Solve • We’ve got a new recurrence: t(0) = 1, t(i) = 2t(i −1) + k2i • We can easily find the annihilator for this recurrence • (L−2) annihilates the homogeneous part, (L−2) annihilates the non-homogeneous part, So (L−2)(L−2) annihilates t(i) • Thus t(i) = (c1i + c2)2i 104 Reverse Transformation • We’ve got a solution for t(i) and we want to transform this into a solution for T(n) • Recall that t(i) = T(2i) and 2i = n t(i) = (c1i + c2)2i (35) T(2i) = (c1i + c2)2i (36) T(n) = (c1 lg n + c2)n (37) = c1n lg n + c2n (38) = O(n lg n) (39) 105 Success! Let’s recap what just happened: • We could not find the annihilator of T(n) so: • We did a transformation to a recurrence we could solve, t(i) (we let n = 2i and t(i) = T(2i)) • We found the annihilator for t(i), and solved the recurrence for t(i) • We reverse transformed the solution for t(i) back to a solu-tion for T(n) 106 Another Example • Consider the recurrence T(n) = 9T(n 3) + kn, where T(1) = 1 and k is some constant • Let n = 3i and rewrite T(n): • T(30) = 1 and T(3i) = 9T(3i−1) + k3i • Now define a sequence t as follows t(i) = T(3i) • Then t(0) = 1, t(i) = 9t(i −1) + k3i 107 Now Solve • t(0) = 1, t(i) = 9t(i −1) + k3i • This is annihilated by (L −9)(L −3) • So t(i) is of the form t(i) = c19i + c23i 108 Reverse Transformation • t(i) = c19i + c23i • Recall: t(i) = T(3i) and 3i = n t(i) = c19i + c23i T(3i) = c19i + c23i T(n) = c1(3i)2 + c23i = c1n2 + c2n = O(n2) 109 In Class Exercise Consider the recurrence T(n) = 2T(n/4) + kn, where T(1) = 1, and k is some constant • Q1: What is the transformed recurrence t(i)? How do we rewrite n and T(n) to get this sequence? • Q2: What is the annihilator of t(i)? What is the solution for the recurrence t(i)? • Q3: What is the solution for T(n)? (i.e. do the reverse transformation) 110 A Final Example Not always obvious what sort of transformation to do: • Consider T(n) = 2T(√n) + log n • Let n = 2i and rewrite T(n): • T(2i) = 2T(2i/2) + i • Define t(i) = T(2i): • t(i) = 2t(i/2) + i 111 A Final Example • This final recurrence is something we know how to solve! • t(i) = O(i log i) • The reverse transform gives: t(i) = O(i log i) (40) T(2i) = O(i log i) (41) T(n) = O(log n log log n) (42) 112
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Relationship Between Sides and Angles in a Triangle - Effortless Math: We Help Students Learn to LOVE Mathematics File failed to load: Effortless Math X +eBooks +ACCUPLACER Mathematics +ACT Mathematics +AFOQT Mathematics +ALEKS Tests +ASVAB Mathematics +ATI TEAS Math Tests +Common Core Math +CLEP +DAT Math Tests +FSA Tests +FTCE Math +GED Mathematics +Georgia Milestones Assessment +GRE Quantitative Reasoning +HiSET Math Exam +HSPT Math +ISEE Mathematics +PARCC Tests +Praxis Math +PSAT Math Tests +PSSA Tests +SAT Math Tests +SBAC Tests +SIFT Math +SSAT Math Tests +STAAR Tests +TABE Tests +TASC Math +TSI Mathematics +Worksheets +ACT Math Worksheets +Accuplacer Math Worksheets +AFOQT Math Worksheets +ALEKS Math Worksheets +ASVAB Math Worksheets +ATI TEAS 6 Math Worksheets +FTCE General Math Worksheets +GED Math Worksheets +3rd Grade Mathematics Worksheets +4th Grade Mathematics Worksheets +5th Grade Mathematics Worksheets +6th Grade Math Worksheets +7th Grade Mathematics Worksheets +8th Grade Mathematics Worksheets +9th Grade Math Worksheets +HiSET Math Worksheets +HSPT Math Worksheets +ISEE Middle-Level Math Worksheets +PERT Math Worksheets +Praxis Math Worksheets +PSAT Math Worksheets +SAT Math Worksheets +SIFT Math Worksheets +SSAT Middle Level Math Worksheets +7th Grade STAAR Math Worksheets +8th Grade STAAR Math Worksheets +THEA Math Worksheets +TABE Math Worksheets +TASC Math Worksheets +TSI Math Worksheets +Courses +AFOQT Math Course +ALEKS Math Course +ASVAB Math Course +ATI TEAS 6 Math Course +CHSPE Math Course +FTCE General Knowledge Course +GED Math Course +HiSET Math Course +HSPT Math Course +ISEE Upper Level Math Course +SHSAT Math Course +SSAT Upper-Level Math Course +PERT Math Course +Praxis Core Math Course +SIFT Math Course +8th Grade STAAR Math Course +TABE Math Course +TASC Math Course +TSI Math Course +Puzzles +Number Properties Puzzles +Algebra Puzzles +Geometry Puzzles +Intelligent Math Puzzles +Ratio, Proportion & Percentages Puzzles +Other Math Puzzles +Math Tips +Articles +Blog Relationship Between Sides and Angles in a Triangle This article provides a comprehensive guide on understanding the relationship between the sides and angles in a triangle. By the end of this article, you will have a clear understanding of how to find and apply this relationship in various problem-solving situations. A step-by-step guide to finding the relationship between sides and angles in a triangle In a triangle, the sum of the measures of the three angles is always 180 degrees. This means that if you know the measures of two of the angles, you can find the measure of the third angle by subtracting the sum of the other two angles from 180 degrees. Additionally, there is a relationship between the lengths of the sides of a triangle and the measures of the angles opposite those sides. Specifically, the longest side of a triangle is always opposite the largest angle, and the smallest side is always opposite the smallest angle. For example, if you have a triangle with sides of lengths 5, 8, and 10, you can determine which angle is the largest and which angle is the smallest. The largest angle will be opposite the longest side (10), and the smallest angle will be opposite the shortest side (5). You can use this relationship to help you find missing angles or side lengths in a triangle if you know the measures of some of the other angles or side lengths. Angle-Side Relationships in a triangle: The biggest side is opposite the biggest angle and vice versa The smallest side is opposite the smallest angle and vice versa. An isosceles triangle is one that has 2 equal side lengths. The opposite angles of these two identical sides are similarly equal. In a triangle, the sum of any of the two sides must be higher than the 3rd side. Relationship Between Sides and Angles in a Triangle – Example 1 Find the smallest side of the triangle. Solution: Since, 30∘<40∘<110∘,∡𝐺<∡𝐻<∡𝐼 30°<40°<110°,∡G<∡H<∡I, their opposite sides are in the same order, from smallest to largest: (IH) ̅<<(IG) ̅<<(HG) ̅ So, the smallest side is (IH) ̅. Relationship Between Sides and Angles in a Triangle – Example 2 Find the smallest angle of the triangle. Solution: Since 14<14.5<15 14<14.5<15, (MO) ̅<<(ON) ̅<<(MN) ̅, their opposite angles are in the same order, from smallest to largest: ∡𝑂<∡𝑁<∡𝑀∡O<∡N<∡M So, ∡𝑂∡O is the smallest angle by: Effortless Math Team about 3 years ago (category: Articles) Effortless Math Team 3 years ago Effortless Math Team Related to This Article Triangle More math articles Powerful Decimals: How to Uncover the Missing Number in Division by Powers of 10 Top 10 Pre-Algebra Practice Questions Proportional Predictions: How to Use Proportions to Estimate Population Size PERT Math Formulas 10 Most Common ISEE Middle-Level Math Questions How to Use Partial Products to Multiply Two-Digit Numbers By Two-digit Numbers 10 Most Common 3rd Grade NYSE Math Questions Compare the Temperatures Above and Below Zero Best Coronavirus Resources for Students ASVAB Math Practice Test Questions What people say about "Relationship Between Sides and Angles in a Triangle - Effortless Math: We Help Students Learn to LOVE Mathematics"? 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Skip to content Home » Math Theory » Algebra » Algebraic Identities Algebraic Identities Introduction A fundamental group of mathematical formulas is algebraic identities. They assist you in carrying out calculations in quick, uncomplicated steps and are the building blocks of algebra. Many mathematical operations must be performed to solve certain algebraic problems. For all possible values of the variables, an algebraic identity is a relationship in which the left side of the equation is precisely equivalent to the right side. In this article, we’ll aim to become familiar with all the algebraic identities, their proofs, and how to apply them to mathematical calculations. Algebraic Identities Definition Algebraic identities are equations in which the right-hand side of the equation’s value is exactly equal to the left-hand side of the equation’s value. Any value for the variables satisfies them. Identity vs. Conditional Number relationships can be expressed using identity and conditional equations. The term “identity equation” refers to an equation that holds true for every value of the variable. A conditional equation is one that is false for at least one value. The table below shows some examples of the identity and conditional equations. | | | --- | | Examples of Conditional Equations | Examples of Identity Equations | | 3x = 12 This equation is false when the value of x is not equal to 4. 10x – 3 = 17 This equation is true when x is equal to 2. | ( x + 1 )2 = x2 + 2x + 1 x2 – 6x + 9 = ( x – 3 )2 Both equations are satisfied for any value of x. | List of Algebraic Identities Algebraic identity states that the left and right sides of the equation are equal for all values of the variables. To determine the values of unknown variables, algebraic identities are used. These algebraic identities are some of the most common ones: | | | ( a + b )2 = a2 + 2ab + b2 | | ( a – b )2 = a2 – 2ab + b2 | | ( a + b ) ( a – b ) = a2 – b2 | | ( a + b )3 = a3 + 3a2 b + 3ab2 + b3 | | ( a – b )3 = a3 – 3a2 b + 3ab2 – b3 | Algebraic Identities Proof We will be able to visualize each of the identities and understand them more fully with the help of the algebra identities proofs that are provided below. Proof: ( a + b )2 = a2 + 2ab + b2 LHS = ( a + b )2 Starting with the left-hand side of the equation LHS = ( a + b ) ( a + b ) LHS = a2 + ab + ab + b2 Multiplying each term LHS = a2 + 2ab + b2 LHS = RHS Visual Proof of ( a + b )2 = a2 + 2ab + b2 Proof: ( a – b )2 = a2 – 2ab + b2 LHS = ( a – b )2 Starting with the left-hand side of the equation LHS = ( a – b ) ( a – b ) LHS = a2 – ab – ab + b2 Multiplying each term LHS = a2 – 2ab + b2 LHS = RHS Visual Proof of ( a – b )2 = a2 – 2ab + b2 Proof: ( a + b ) ( a – b ) = a2 – b2 LHS = ( a + b ) ( a – b )Starting with the left-hand side of the equation LHS = a2 – ab + ab + b2 Multiplying each term LHS = a2 – b2 LHS = RHS Visual Proof of ( a + b ) ( a – b ) = a2 – b2 Proof: (a + b + c)2= a2 + b2 + c2 + 2ab + 2bc + 2ac LHS = (a + b + c)2 Starting with the left-hand side of the equation LHS = ( a + b + c ) ( a + b + c ) LHS = a2 + ab + ac + ab + b2 + bc + ac + bc + c2 Multiplying each term LHS = a2 + b2 + c2 + 2ab + 2bc + 2ac LHS = RHS Visual Proof of (a + b + c)2= a2+ b2+ c2+ 2ab + 2bc + 2ac Proof: ( x + a ) ( x + b ) = x2+ x ( a + b ) + ab LHS = ( x + a ) ( x + b )Starting with the left-hand side of the equation LHS = x2 + ax + bx + ab Multiplying each term LHS = x2 + x ( a + b ) + ab Common Factor LHS = RHS Visual Proof of ( x + a ) ( x + b ) = x2+ x ( a + b ) + ab Algebraic Identities for Two Variables a and b The identities in algebra with two variables are as follows. These identities can be quickly confirmed by multiplying polynomials and extending the square or cube. A. ( a + b )2 = a2 + 2ab + b2 B. ( a – b )2 = a2 – 2ab + b2 C. ( a + b ) ( a – b ) = a2 – b2 D. ( a + b )3 = a3 + 3a2 b + 3ab2 + b3 E. ( a – b )3 = a3 – 3a2 b + 3ab2 – b3 Examples Example 1: Expand ( x + 7 )2. Solution Let us substitute a = x and b = 7 into the formula ( a + b )2 = a2 + 2ab + b2. ( x + 7 )2 = x2 + 2 ( x ) ( 7 ) + 72 ( x + 7 )2 = x2 + 14x + 49 Example 2: Expand ( 3x + 2y )2. Solution Substituting a = 3x and b = 2y into the formula ( a + b )2 = a2 + 2ab + b2, we have, ( 3x + 2y )2 = ( 3x )2 + 2 ( 3x ) ( 2y ) + ( 2y )2 ( 3x + 2y )2 = 9x2 + 12xy + 4y2 Example 3: Expand ( x – 3y )2. Solution Let us substitute a = x and b = 3y into the formula ( a – b )2 = a2 – 2ab + b2. ( x – 3y )2 = ( x )2 – 2 ( x ) ( 3y ) + ( 3y )2 ( x – 3y )2 = x2 – 6xy + 9y2 Example 4: Expand ( 2x + 2y )3. Solution Substituting a = 2x and b = 2y into the formula ( a + b )3 = a3 + 3a2 b + 3ab2 + b3, we have, ( 2x + 2y )3 = ( 2x )3 + 3 ( 2x )2 ( 2y ) + 3 ( 2x ) ( 2y )2 +( 2y )3 ( 2x + 2y )3 = 8x3 + 3 ( 4x2) ( 2y ) + 3 ( 2x ) ( 4y2) + 8y3 ( 2x + 2y )3 = 8x3 + 24x2y + 24xy2 + 8y3 Example 5: Expand ( 3x – 4y )3. Solution Let us substitute a = 3x and b = 4y into the formula ( a – b )3 = a3 – 3a2 b + 3ab2 – b3, we have, ( 3x – 4y )3= ( 3x )3 – 3 ( 3x )2 ( 4y ) + 3 ( 3x ) ( 4y )2 – ( 4y )3 ( 3x – 4y )3= 27x3 – 3 ( 9x2 ) ( 4y ) + 3 ( 3x ) ( 16y2) – ( 64y3 ) ( 3x – 4y )3= 27x3 – 108x2y + 144xy2 – 64y3 Algebraic Identities for Three Variables a, b, and c Just as the identities for the two variables, the identities for three variables in algebra have also been determined. Furthermore, using the fewest possible steps, these identities make it simple to work with algebraic expressions. A. ( a + b + c )2 = a2 + b2 + c2 + 2ab + 2bc + 2ac B. a2 + b2 + c2 = ( a + b + c )2 – 2 ( ab + bc + ac ) C. a3 + b3 + c3 – 3abc = ( a + b + c ) ( a2 + b2 + c2 – ab – ac – bc ) D. ( a + b ) ( b + c ) ( c + a ) = ( a + b + c) ( ab + ac + bc ) – 2abc Examples Example 1: Expand ( x + 2y + 3z )2. Solution Let us substitute a = x, b = 2y, and c = 3z into the formula ( a + b + c )2 = a2 + b2 + c2 + 2ab + 2bc + 2ac. ( x + 2y + 3z )2 = ( x )2 + ( 2y )2 + ( 3z )2 + 2 ( x ) ( 2y ) + 2 ( 2y ) ( 3z ) + 2 ( x ) ( 3z ) ( x + 2y + 3z )2 = x2 + 4y2 + 9z2 + 4xy + 12yz + 6xz Example 2: Expand ( 3x + 4y + 2z )2. Solution Let us substitute a = 3x, b = 4y, and c = 2z into the formula ( a + b + c )2 = a2 + b2 + c2 + 2ab + 2bc + 2ac. ( 3x + 4y + 2z )2 = ( 3x )2 + ( 4y )2 + ( 2z )2 + 2 ( 3 x ) ( 4y ) + 2 ( 4y ) ( 2z ) + 2 ( 3x ) ( 2z ) ( 3x + 4y + 2z )2 = 9x2 + 16y2 + 4z2 + 24xy + 16yz + 12xz Factoring Identities For quickly factoring algebraic expressions, algebraic identities are quite helpful. Some more complex algebraic expressions can easily be factored in using basic identities. A. a2 – b2 = ( a – b ) (a + b ) B. a3 – b3 = ( a – b ) ( a2 + ab + b2 ) C. a3 + b3 = ( a + b ) ( a2 – ab + b2 ) D. x2 + x ( a + b ) + ab = ( x + a ) ( x + b ) Examples Example 1: Factorize m4 – 81. Solution m4 – 81 = ( m2 )2 – ( 9 )2 m4 – 81 = ( m2 + 9 ) ( m2 – 9 ) m4 – 81 = ( m2 + 9 ) ( m – 3 ) ( m + 3 ) Example 2: Factorize p4 – ( q + r )4. Solution p4 – ( q + r )4 = ( p2 )2 – ( ( q + r )2 )2 p4 – ( q + r )4 = [ ( p2 + ( q + r )2 ] [ ( p2 – ( q + r )2 ] p4 – ( q + r )4 = [ ( p2 + ( q + r )2 ] ( p + q + r ) ( p – q – r ) Example 3: Factorize n3 + 216. Solution Let us use the identity a3 + b3 = ( a + b ) ( a2 – ab + b2 ) where a = n and b = 6. n3 + 216 = n3 + 63 n3 + 216 = ( n + 6 ) ( n2 – 6n + 36 ) Example 4: Factorize r3 – 64. Solution Let us use the identity a3 – b3 = ( a – b ) ( a2 + ab + b2 ) where a = r and b = 4. r3 – 64 = r3 – 43 r3 – 64 = ( r – 4 ) ( r2 + 4r + 16 ) Summary Algebraic identities are equations in which the right-hand side of the equation’s value is exactly equal to the left-hand side of the equation’s value. Any value for the variables satisfies them. Algebraic Identities for Two Variables a and b A. ( a + b )2 = a2 + 2ab + b2 B. ( a – b )2 = a2 – 2ab + b2 C. ( a + b ) ( a – b ) = a2 – b2 D. ( a + b )3 = a3 + 3a2 b + 3ab2 + b3 E. ( a – b )3 = a3 – 3a2 b + 3ab2 – b3 Algebraic Identities for Three Variables a, b and c A. ( a + b + c )2 = a2 + b2 + c2 + 2ab + 2bc + 2ac B. a2 + b2 + c2 = ( a + b + c )2 – 2 ( ab + bc + ac ) C. a3 + b3 + c3 – 3abc = ( a + b + c ) ( a2 + b2 + c2 – ab – ac – bc ) D. ( a + b ) ( b + c ) ( c + a ) = ( a + b + c) ( ab + ac + bc ) – 2abc Factoring Identities A. a2 – b2 = ( a – b ) (a + b ) B. a3 – b3 = ( a – b ) ( a2 + ab + b2 ) C. a3 + b3 = ( a + b ) ( a2 – ab + b2 ) D. x2 + x ( a + b ) + ab = ( x + a ) ( x + b ) Frequently Asked Questions on Algebraic Identities ( FAQs ) What is meant by algebraic identities? Algebraic identities are equations in which the right-hand side of the equation’s value is exactly equal to the left-hand side of the equation’s value. Any value for the variables satisfies them. What are examples of standard algebraic identities? The following are examples of some standard algebraic identities for variables a, b, and c. | | | Algebraic Identities for Two Variables a and b | | A. ( a + b )2 = a2 + 2ab + b2 B. ( a – b )2 = a2 – 2ab + b2 C. ( a + b ) ( a – b ) = a2 – b2 D. ( a + b )3 = a3 + 3a2 b + 3ab2 + b3 E. ( a – b )3 = a3 – 3a2 b + 3ab2 – b3 | | Algebraic Identities for Three Variables a, b, and c | | A. ( a + b + c )2 = a2 + b2 + c2 + 2ab + 2bc + 2ac B. a2 + b2 + c2 = ( a + b + c )2 – 2 ( ab + bc + ac ) C. a3 + b3 + c3 – 3abc = ( a + b + c ) ( a2 + b2 + c2 – ab – ac – bc ) D. ( a + b ) ( b + c ) ( c + a ) = ( a + b + c) ( ab + ac + bc ) – 2abc | | Factoring Identities | | A. a2 – b2 = ( a – b ) (a + b ) B. a3 – b3 = ( a – b ) ( a2 + ab + b2 ) C. a3 + b3 = ( a + b ) ( a2 – ab + b2 ) D. x2 + x ( a + b ) + ab = ( x + a ) ( x + b ) | How do we verify algebraic identities? The substitution approach is used to confirm the algebraic identities. Use the values in place of the variables in this method, then carry out the arithmetic operation. How do we prove that an equation is an identity? Applying known identities to show that one side of an equation may be converted into the other is necessary to establish an equation’s identity. To demonstrate that an equation is not an identity, we must identify one input where the values of the two sides of the equation disagree. The following are examples of identity equations. ( x + 5 )2 = x2 + 10x + 25 ( a – 4 )2 = a2 – 8a + 16 Notice that if we try to substitute values to the variables, the left-hand side is equal to the right-hand side of the equation. On the other hand, these are examples of equations that are conditional. 4x = 32 x + 2 = 10 Notice that for both equations, only x = 8 will make the equations true. If we try to use other values, the left-hand side will not be equal to the right-hand side of the equation. What is the importance of algebraic identities? A fundamental group of mathematical formulas is algebraic identities. They assist you in carrying out calculations in quick, uncomplicated steps and are the building blocks of algebra. Many mathematical operations must be performed to solve certain algebraic problems. What is meant by algebraic proofs? The logical justifications for an algebraic solution are displayed in an algebraic proof. When a problem is presented, the task is to demonstrate that the solution is correct. What are some examples of algebraic identity proofs? The following are examples of standard algebraic proofs. Proof: ( a + b )2 = a2 + 2ab + b2 LHS = ( a + b )2 Starting with the left-hand side of the equation LHS = ( a + b ) ( a + b ) LHS = a2 + ab + ab + b2 Multiplying each term LHS = a2 + 2ab + b2 LHS = RHS Proof: ( a – b )2 = a2 – 2ab + b2 LHS = ( a – b )2 Starting with the left-hand side of the equation LHS = ( a – b ) ( a – b ) LHS = a2 – ab – ab + b2 Multiplying each term LHS = a2 – 2ab + b2 LHS = RHS Proof: ( a + b ) ( a – b ) = a2 – b2 LHS = ( a + b ) ( a – b )Starting with the left-hand side of the equation LHS = a2 – ab + ab + b2 Multiplying each term LHS = a2 – b2 LHS = RHS Proof: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac LHS = (a + b + c)2 Starting with the left-hand side of the equation LHS = ( a + b + c ) ( a + b + c ) LHS = a2 + ab + ac + ab + b2 + bc + ac + bc + c2 Multiplying each term LHS = a2 + b2 + c2 + 2ab + 2bc + 2ac LHS = RHS Recommended Worksheets Algebraic Identities (World Environment Day Themed) Math Worksheets Algebra (Rosh Hashanah Themed) Math Worksheets Formulas (National Aviation Day Themed) Math Worksheets Browse All Worksheets Link/Reference Us We spend a lot of time researching and compiling the information on this site. If you find this useful in your research, please use the tool below to properly link to or reference Helping with Math as the source. We appreciate your support! Algebraic Identities "Algebraic Identities". Helping with Math. Accessed on August 23, 2025. "Algebraic Identities". Helping with Math, Accessed 23 August, 2025. Algebraic Identities. Helping with Math. Retrieved from Additional Algebra Theory: Standard Form Slope of a Line Arithmetic Progression Properties of Addition Identifying Arithmetic Patterns of Numbers Division of Polynomials Ratio and Proportions Symbols in Algebra Rate of Change & Slope of a Line Function Latest Worksheets The worksheets below are the mostly recently added to the site. 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https://www.geeksforgeeks.org/maths/uniform-distribution-formula/
Uniform Distribution | Formula, Definition and Examples Last Updated : 24 Jul, 2025 Suggest changes 5 Likes A Uniform Distribution is a type of probability distribution in which every outcome in a given range is equally likely to occur. That means there is no bias—no outcome is more likely than another within the specified set. It is also known as rectangular distribution (continuous uniform distribution). It has two parameters a and b: a = minimum and b = maximum. The distribution is written as U (a, b). For example: Rolling a fair six-sided die. Each number (1, 2, 3, 4, 5, or 6) has an equal chance of being rolled, with a probability of 61​ for each outcome. Graph of Uniform Distribution Rolling a Fair Die ( Discrete Uniform distribution) When rolling a fair six-sided die, each face (1, 2, 3, 4, 5, 6) has an equal probability of 1/6 of landing face up. This is a classic example of a discrete uniform distribution. Graph: Probability Mass Function (PMF) of Rolling a Fair Die The graph for this example will show each possible outcome (1 through 6) on the x-axis and the probability (1/6) on the y-axis. Random Number Generation (Continuous Uniform Distribution) Suppose a random number generator is programmed to produce a real number between 0 and 1, with each number in this range being equally likely. This is an example of a continuous uniform distribution. Graph: Probability Density Function (PDF) of Random Number Generation The graph will show the range [0, 1] on the x-axis, with a constant probability density of 1 across this interval. Uniform Distribution Formula A random variable X is said to be uniformly distributed over the interval -∞ < a < b < ∞. Formulae for uniform distribution: | | | --- | | Probability density function(pdf) | f(x) = (b−a)1​ , a ≤ x ≤ b | | Mean(μ) | ∫ab​x.f(x)dx=b−a1​[2x2​]ab​ = 2(a+b)​ | | Variance (σ2 ) | ∫ab​x.f(x)dx=b−a1​[2x2​]ab​ = μ2' - μ2 = ∫ab​x2.b−a1​dx−(2a+b​)2 = 12(b−a)2​ | | Standard Deviation (σ) | =12(b−a)2​​ | | Cumulative Distribution function (CDF) | = (b−a)(x−a)​ for x ∈ [a , b] | | Median | = 2(a+b)​ | | For the conditional probability = P( c < x < d ) | = (d - c ) × f(x) = (b−a)(d−c)​ | Types of Uniform Distribution Types of uniform distribution are: Discrete Uniform Distribution Applies to finite and countable outcomes. Each possible value of the random variable has an equal probability. Example: Rolling a fair die (each number from 1 to 6 has an equal chance: 1/6). Continuous Uniform Distribution This applies when outcomes can take on any value within a continuous interval [a, b]. All values within this range are equally likely, but since there are infinitely many values, we use a probability density function (PDF) instead of direct probabilities. Example: Choosing a random number between 0 and 1. Let's discuss these types in detail as follows. Discrete Uniform Distribution Discrete uniform distribution is a probability distribution that describes the likelihood of outcomes when each outcome in a finite set is equally likely. It's characterized by a constant probability mass function (PMF) over a finite range of values. Its properties and applications extend across various disciplines, making it a versatile tool in data analysis and decision-making processes. P(X=xi​)=n1​,fori=1,2,…,n Example: Rolling fair dice or flipping fair coins, where each outcome has an equal probability. Estimation of Maximum In statistics, the estimation of the maximum refers to methods used to estimate the largest value or the maximum observation in a dataset. Techniques such as order statistics and maximum likelihood estimation are commonly employed for this purpose. Random Permutation A random permutation is a random arrangement of a set of items or elements. It's often used in various fields such as cryptography, statistics, and computer science. Generating random permutations is essential in algorithms, simulations, and experimental designs. Properties of Discrete Uniform Distribution Each outcome in the sample space has an equal probability of occurrence. The probability mass function (PMF) is constant over the range of possible outcomes. The mean of a discrete uniform distribution is the average of the minimum and maximum values. The variance of a discrete uniform distribution is[12(n2−1)​], where n is the number of possible outcomes. Applications of Discrete Uniform Distribution Modeling scenarios where there is no preference or bias towards any particular outcome. Sampling without replacement, such as selecting random samples from a finite population. Generating random numbers for simulations, Monte Carlo methods, and randomized algorithms. Creating random permutations for shuffling decks of cards, designing experiments, and cryptographic applications. Continuous Uniform Distributions or Rectangular Distributions Continuous uniform distributions, also known as rectangular distributions, are probability distributions where the probability density function (PDF) is constant within a certain interval and zero elsewhere. This means that all outcomes within the interval are equally likely. Continuous uniform distributions provide a simple yet powerful framework for understanding and modeling randomness within defined intervals, making them essential tools in probability theory and applied statistics. Example: Generating a random number between 0 and 1. Note: The density is constant and non-zero over an interval. The probability at a particular point is always zero. P(a ≤ X ≤ b) = 1 (entire interval). P(X = x) = 0 for any exact value of x. Probability Density Function (PDF) The probability density function (PDF) of a continuous uniform distribution defines the probability of a random variable falling within a particular interval. For a continuous uniform distribution over the interval [a, b], the PDF is given by: f(x) = (b−a)1​ for a ≤ x ≤ b and f(x) = 0 otherwise. The height of the PDF is constant between a and b. The total area under the curve (which represents the total probability) is 1. Every value within the interval is equally likely. Cumulative Distribution Function (CDF) The cumulative distribution function (CDF) of a continuous uniform distribution gives the probability that a random variable is less than or equal to a certain value. For the continuous uniform distribution over [a, b], the CDF is defined as: F(x) = (x - a) / (b - a) for a ≤ x ≤ b and F(x) = 0 for x < a, F(x) = 1 for x > b. For values less than a: the probability is 0. For values greater than b, the probability is 1. For values within the interval [a,b]: probability increases linearly from 0 to 1. This function is a rising straight line between a and b, showing how probability accumulates across the interval. Generating Functions Generating functions provide a way to represent sequences of numbers as power series. In probability theory, generating functions are often used to manipulate sequences of random variables. They can simplify calculations and help derive important properties of random variables and distributions. Standard Uniform Distribution The standard uniform distribution is a special case of the continuous uniform distribution where the interval is [0, 1]. It is widely used in simulations, random number generation, and various statistical applications. Properties of Continuous Uniform Distributions Equal probability density within the interval. Cumulative distribution function increases linearly within the interval. The mean of a continuous uniform distribution is the midpoint of the interval. The variance of a continuous uniform distribution is 12[(b−a)2]​. Applications of Continuous Uniform Distributions Modeling uncertainty in various fields such as engineering, finance, and physics. Random number generation for simulations and games. Used in statistical quality control to model uniformity in manufacturing processes. In cryptography generate keys and create random permutations. As a baseline distribution for comparison with other distributions in statistical analysis. Difference between Discrete and Continuous Uniform Distribution The following table shows the difference between Continuous and Discrete Uniform Distribution: | Discrete Uniform Distribution | Continuous Uniform Distribution | --- | | Finite or countable number of possible outcomes | An infinite number of possible outcomes within a continuous interval | | Each outcome has a specific, equal probability of occurring. P(X = xi​) = 1/n ​,for i=1,2,…,n | The probability density function (PDF) is constant over the given interval [a,b]. P(a≤X≤b) = 1 (entire interval) But zero at a specific point, i.e. P(X=x)=0 for any exact value of x. | | - Rolling a fair die (6 possible outcomes) - Drawing a card from a deck (52 possible outcomes) | - Generating a random number between 0 and 1 - Time of arrival at a bus stop (between 0 and 10 minutes) | | Probability Mass Function (PMF) with individual bars representing each outcome with equal height | Probability Density Function (PDF) with a flat, continuous line over the interval indicating equal likelihood across all points | Related Articles: Poisson Distribution Binomial Distribution Normal Distribution Sample Questions on Uniform Distribution Question 1: A random variable X has a uniform distribution over(-2, 2), (i) Find k for which P(X>k) = 1/2 (ii) Evaluate P(X<1) (iii) P[|X-1|<1] Solution: (i) X =f(x) = 1/(b-a) =1/(2-(-2)) = 1/4 We want the value of k such that the probability to the right of k is 1/2. In a uniform distribution, probability is proportional to length, so we find k such that: 2-k/2-(-2) = 1/2 2-k/4 = 1/2 2-k = 2 k = 0 By solving we get k = 0 (ii) Evaluate P(X < 1) We find the proportion of the interval [−2,2] that is less than 1: P(X < 1) = 1 - (-2)/4 = 3/4 (iii) Evaluate P( ∣X − 1∣ < 1 ) This is equivalent to: P (−1 < X − 1 <1 )⇒P (0 < X < 2) Now compute: P( 0 < X < 2 ) = 2 − 0/4 = 2/4 = 1/2 Question 2: If X is uniformly distributed in (-1,, 4) then (i) Its mean is ______________. (ii) Its variance is ______________. (iii) Spade's standard deviation is ___________. (iv) Its median is ______________. Solution: Here, a = -1 and b = 4 (i) Mean (μ) = (4-1)/2 = 1.5 (ii) Variance(σ2) = (4+1)2 /12 = 2.08 (iii) Standard deviation(σ) =√2.08 = 1.443 (iv) Median = (4-1)/2 = 1.5 Question 3: If there are 52 cards in the traditional deck of cards with four suits: hearts, a clubs, and diamonds. Each suite contains 13 cards of which 3 cards are face cards. The new deck is formed by excluding a . Then what is the probability of getting a heart card from the modified deck? Solution: In the question, the given number of cards is finite so it is a discrete uniform distribution. Given: Original deck: 52 cards 4 suits: Hearts, Spades, Clubs, Diamonds Each suit has 13 cards Each suit contains 3 face cards (Jack, Queen, King) Modified Deck Composition: Cards per suit after removing number cards: 4 cards Total suits: 4 Total cards in modified deck: 4 cards/suit × 4 suits=16 cards Formula for the probability in discrete uniform distribution is P(X) = 1/n Probability of getting heart in the modified deck P(Heart) = 4/16 ​= 1/4 = 0.25 Question 4: Using the uniform distribution probability density function for random variable X. in (0, 20), find P(3< X < 16). Solution: Here, a = 0, b =20 f(x) = 1/(20 - 0) = 1/20 P(3< X < 16) = (16 - 3) × (1/20) = 13/20 Question 5: A random variable X has a uniform distribution over (-5 , 6), find cumulative distribution function for x = 3. Solution: Here, a = -5, b = 6, x = 3 CDF = (3 - (-5))/(6 - (-5)) = 8/11 Practice Problems on Uniform Distribution Question 1. A random variable X follows a uniform distribution over the interval [2,10]. Find the probability that X lies between 4 and 8. Question 2. Suppose a continuous random variable Y is uniformly distributed over the interval [0,5] (a) Calculate the expected value E(Y) (b) Calculate the variance Var(Y). Question 3. A random variable Z is uniformly distributed over [3,15]. Spade's Derive the CDF of Z and use it to find the probability that Z is less than or equal to 9. 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https://artofproblemsolving.com/wiki/index.php/Arbitrarily_close?srsltid=AfmBOor_5wS5bZbbQ9P_RPa93g-eB4aaUayjfJLCTHl8hES_GFkm2AEd
Art of Problem Solving Arbitrarily close - 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 Arbitrarily close Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Arbitrarily close Informally, a set contains points arbitrarily close to some point if, for any positive distance , there are members of which are less than a distance from . More formally, in a metric space (such as the Euclidean space or just the real numbers) with distance function , a set is said to contain members arbitrarily close to some point if for all there exists some such that . Examples In the particular case of the real numbers with the usual distance , the set contains points arbitrarily close to 0. The set of real numbers contains points arbitarily close to any given real number. This is also true of the rational numbers, but it is not true of the integers. We can see this last fact is true because (for example) there is no integer at distance or less from . (A set which contains points arbitrarily close to every point is said to be dense.) See also This article is a stub. Help us out by expanding it. Retrieved from " Categories: Analysis Stubs 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.
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https://www.acs.psu.edu/drussell/Demos/Pendulum/Pendula.html
| Acoustics and Vibration AnimationsDaniel A. Russell, Ph.D.Graduate Program in AcousticsThe Pennsylvania State UniversityAll text and images on this page are ©2004-2011 by Daniel A. Russelland may not used in other web pages or reports withoutpermission. | The Simple Pendulum | A simple pendulum consists of a massmhanging from a string of lengthLand fixed at a pivot point P. When displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. By applying Newton's secont law for rotational systems, the equation of motion for the pendulum may be obtained,and rearranged as.If the amplitude of angular displacement is small enough that the small angle approximation () holds true, then the equation of motion reduces to the equation of simple harmonic motion.The simple harmonic solution iswithbeing the natural frequency of the motion. Small Angle Approximation and Simple Harmonic Motion | With the assumption of small angles, the frequency and period of the pendulum are independent of the initial angular displacement amplitude. All simple pendulums should have the same period regardless of their initial angle (and regardless of their masses). This simple approximation is illustrated in the (48 kB) mpeg movie at left. All three pendulums cycle through one complete oscillation in the same amount of time. | The period for a simple pendulum does not depend on the mass or the initial anglular displacement, but depends only on the lengthLof the string and the value of the gravitational field strengthg, according toThe mpeg movie at left (39.5 kB) shows two pendula, with different lengths.How many complete oscillations does the shorter (blue) pendulum make in the time for one complete oscillation of the longer (black) pendulum?From this information and the definition of the period for a simple pendulum, what is the ratio of lengths for the two pendula? The Real (Nonlinear) Pendulum Small Initial Amplitude Large Initial Amplitude
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https://bestpractice.bmj.com/topics/en-gb/415?locale=en_GB
When viewing this topic in a different language, you may notice some differences in the way the content is structured, but it still reflects the latest evidence-based guidance. Subarachnoid hemorrhage Epidemiology Etiology Case history Approach History and exam Tests Differentials Criteria Screening Approach Treatment algorithm Prevention Patient discussions Monitoring Complications Prognosis Guidelines Images and videos References Patient information Evidence Summary Subarachnoid hemorrhage (SAH) presents as a sudden severe headache, often described as "the worst headache of life," with nausea, vomiting, and photophobia. This topic focuses on the diagnosis and management of subarachnoid hemorrhage caused by aneurysm (i.e., aneurysmal SAH). Examination can be normal or may reveal altered consciousness, meningismus, intraocular hemorrhages, or focal findings. Computed tomography (CT) indicated if subarachnoid hemorrhage is clinically suspected. Lumbar puncture is indicated if CT is unrevealing. Cerebral angiography confirms the presence of aneurysms. Initial stabilization followed by surgical clipping or endovascular coil embolization is standard therapy. Enteral nimodipine should be started early to prevent vasospasm, delayed cerebral ischemia and improve functional outcomes. Complications are common and include rebleeding, acute hydrocephalus, and delayed cerebral ischemia. Definition SAH is bleeding into the subarachnoid space and is an emergency. The most common cause of nontraumatic SAH is rupture of an intracranial aneurysm, which is the focus of this topic.Suarez JI, Tarr RW, Selman WR. Aneurysmal subarachnoid hemorrhage. N Engl J Med. 2006 Jan 26;354(4):387-96. Aneurysmal SAH causes substantial morbidity and mortality. When a cerebral aneurysm ruptures, blood flows into the subarachnoid space, sometimes seeping into brain parenchyma and/or ventricles. The sudden increase in intracranial pressure, as well as the destructive and toxic effects of blood on brain parenchyma and cerebral vessels, accounts for most complications. History and exam Key diagnostic factors Other diagnostic factors Risk factors Diagnostic tests 1st tests to order Tests to consider Treatment algorithm all patients Contributors Authors Salah Keyrouz, MD, FAHA, FCCM Professor Neurology and Neurosurgery Washington University School of Medicine St. Louis MO Disclosures SK is an author of a reference cited in this topic. Acknowledgements Dr Salah Keyrouz would like to gratefully acknowledge Dr Michael N. Diringer, a previous contributor to this topic. Disclosures MND is an author of a number of references cited in this topic. Peer reviewers Venkatesh Aiyagari, MD Associate Professor Department of Neurology and Rehabilitation University of Illinois at Chicago Chicago IL Disclosures VA declares that he has no competing interests. Peter Martin, MA, BM BCh, MD, FRCP Consultant Neurologist Addenbrookes Hospital Cambridge UK Disclosures PM declares that he has no competing interests. Giovanni Grasso, M.D., PhD Aggregate Professor of Neurosurgery Neurosurgical Clinic Department of Clinical Neuroscience University of Palermo Palermo Italy Disclosures GG declares that he has no competing interests. References Key articles Thompson BG, Brown RD Jr, Amin-Hanjani S, et al. Guidelines for the management of patients with unruptured intracranial aneurysms: a guideline for healthcare professionals from the American Heart Association/American Stroke Association. Stroke. 2015 Aug;46(8):2368-400.Full textAbstract Hoh BL, Ko NU, Amin-Hanjani S, et al. 2023 Guideline for the management of patients with aneurysmal subarachnoid hemorrhage: a guideline from the American Heart Association/American Stroke Association. Stroke. 2023 Jul;54(7):e314-70.Full textAbstract Steiner T, Juvela S, Unterberg A, et al. European Stroke Organization guidelines for the management of intracranial aneurysms and subarachnoid haemorrhage. Cerebrovasc Dis. 2013 Feb 7;35(2):93-112.Full textAbstract Reference articles A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice. Differentials Guidelines Patient information Subarachnoid hemorrhage Videos Diagnostic lumbar puncture in adults: animated demonstration Tracheal intubation: animated demonstration Use of this content is subject to our disclaimer Log in or subscribe to access all of BMJ Best Practice Log in or subscribe to access all of BMJ Best Practice Log in to access all of BMJ Best Practice Browse Services Legal External link opens in a new window External link opens in a new window External link opens in a new window External link opens in a new window External link opens in a new window © BMJ Publishing Group 2025 ISSN 2515-9615 Help us improve BMJ Best Practice Please complete all fields. 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https://math.stackexchange.com/questions/2896638/how-do-i-expand-z-1-z-22
complex numbers - How do I expand $|z_{1} - z_{2}|^2$? - 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 How do I expand |z 1−z 2|2|z 1−z 2|2? Ask Question Asked 7 years, 1 month ago Modified7 years, 1 month ago Viewed 493 times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. I'm trying to expand |z 1−z 2|2|z 1−z 2|2 where z 1 z 1 and z 2 z 2 are complex numbers. Is it exactly the same as with real numbers z 1 z 1 and z 2 z 2? complex-numbers absolute-value Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Aug 27, 2018 at 21:38 User42User42 95 5 5 bronze badges 2 1 Not for reals, as the user below noted, i thought you had said complex, my mistake. Recall that |z|2=z z∗|z|2=z z∗, where z∗z∗ is the complex conjugate. Set z 3=z 1−z 2 z 3=z 1−z 2 and just use the definition.Shinaolord –Shinaolord 2018-08-27 21:42:06 +00:00 Commented Aug 27, 2018 at 21:42 4 not quite the same as reals, (z 1−z 2)(z 1¯−z 2¯).(z 1−z 2)(z 1¯−z 2¯). In the reals, the absolute value signs become irrelevant because of squaring. Here, we need to keep track of complex conjugates.Will Jagy –Will Jagy 2018-08-27 21:43:41 +00:00 Commented Aug 27, 2018 at 21:43 Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Also take a look at user57404 and Will Jagy's comments. To wrap this up in an answer, you have z z¯=|z|2 z z¯=|z|2 Thus, |z 1−z 2|2=(z 1−z 2)z 1−z 2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=(z 1−z 2)(z 1¯−z 2¯)|z 1−z 2|2=(z 1−z 2)z 1−z 2¯=(z 1−z 2)(z 1¯−z 2¯) as complex conjugation distributes over sums and products. Now, if you want to go further, you may multiply this out: (z 1−z 2)(z 1¯−z 2¯)=z 1 z 1¯−z 1 z 2¯−z 2 z 1¯+z 2 z 2¯=|z 1|2+|z 2|2−(z 1 z 2¯+z 2 z 1¯)=|z 1|2+|z 2|2−2 R e(z 1 z 2¯)(z 1−z 2)(z 1¯−z 2¯)=z 1 z 1¯−z 1 z 2¯−z 2 z 1¯+z 2 z 2¯=|z 1|2+|z 2|2−(z 1 z 2¯+z 2 z 1¯)=|z 1|2+|z 2|2−2 R e(z 1 z 2¯) So structurally, there is some similarity to the real case. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 28, 2018 at 12:03 answered Aug 27, 2018 at 22:00 user369816 user369816 4 Maybe adding that z 1 z 2¯+z 1¯z 2=2 R(z 1 z 2¯)z 1 z 2¯+z 1¯z 2=2 ℜ(z 1 z 2¯) it would be clearer that all is real Joe –Joe 2018-08-27 22:04:36 +00:00 Commented Aug 27, 2018 at 22:04 Ah yes, I was going to. Thank you.user369816 –user369816 2018-08-27 22:05:07 +00:00 Commented Aug 27, 2018 at 22:05 It should be |z 1|2+|z 2|2−2 Re(z 1 z¯2)|z 1|2+|z 2|2−2 Re⁡(z 1 z¯2).Bernard –Bernard 2018-08-27 22:08:58 +00:00 Commented Aug 27, 2018 at 22:08 Oh damn, yes, missed that. Thank you as well. Feel free to edit my post as well if you find something else missed in the heat of the battle.user369816 –user369816 2018-08-27 22:10:32 +00:00 Commented Aug 27, 2018 at 22:10 Add a comment| This answer is useful 2 Save this answer. Show activity on this post. Using polar coordinates, write z 1=ρ e i ϕ z 1=ρ e i ϕ and z 2=r e i θ z 2=r e i θ where ρ=|z 1|ρ=|z 1|, r=|z 2|r=|z 2|, ϕ=arg(z 1)ϕ=arg⁡(z 1), and θ=arg(z 2)θ=arg⁡(z 2). Then we see that |ρ e i ϕ−r e i θ|2=|(ρ cos(ϕ)−r cos(θ))+i(ρ sin(ϕ)−r sin(θ))|2=(ρ cos(ϕ)−r cos(θ))2+(ρ sin(ϕ)−r sin(θ))2=ρ 2+r 2−2 ρ r cos(θ−ϕ)=|z 1|2+|z 2|2−2|z 1||z 2|cos(arg(z 1)−arg(z 2))|ρ e i ϕ−r e i θ|2=|(ρ cos⁡(ϕ)−r cos⁡(θ))+i(ρ sin⁡(ϕ)−r sin⁡(θ))|2=(ρ cos⁡(ϕ)−r cos⁡(θ))2+(ρ sin⁡(ϕ)−r sin⁡(θ))2=ρ 2+r 2−2 ρ r cos⁡(θ−ϕ)=|z 1|2+|z 2|2−2|z 1||z 2|cos⁡(arg⁡(z 1)−arg⁡(z 2)) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 27, 2018 at 23:02 Mark ViolaMark Viola 185k 12 12 gold badges 154 154 silver badges 264 264 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 complex-numbers absolute-value See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 0Diagonals of a parallelogram bisect each other 2If z z and w w are complex numbers can we use the proof in R R to demonstrate that |z w|=|z||w||z w|=|z||w|? 3Finding a substitution that eliminates the squared term from a cubic equation 3Expand (2 i−2)38(2 i−2)38 using de Moivre's formula 3Redundancy of numbers 1Can one expand (a+b)i(a+b)i? 1Which field should I read to understand extending the natural numbers to integers, up to complex numbers? Hot Network Questions Is there a way to defend from Spot kick? How different is Roman Latin? Where is the first repetition in the cumulative hierarchy up to elementary equivalence? 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I have a lot of PTO to take, which will make the deadline impossible Languages in the former Yugoslavia Discussing strategy reduces winning chances of everyone! Direct train from Rotterdam to Lille Europe в ответе meaning in context The rule of necessitation seems utterly unreasonable Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. 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https://thestarman.pcministry.com/math/rec/TLR.htm
Sponsoring website: Emergency Boot Kit Truthteller, Liar and Randomizer A Recreational Logic Problem There are three people who will answer any question you ask with "yes" or "no." One will always tell the truth, one will always lie and the last one will answer either way at random (meaning that it's quite possible for this third person to give the same answer two or even three times in a row... so asking the same person the same question might not be very productive). We'll label these three people as X, Y and Z since we haven't a clue (yet) as to who they are... but, X, Y and Z do know who each of the other two people are (hint!). Your assignment is to determine who is the truthteller (T), the liar (L) and the randomizer (R) by asking one or more of these people only three questions. [ To be clear: That's only three questions in total that you are allowed to ask! ]. This may require you to come up with (devise) more than just three questions, since your second and third questions can be based upon the answers you receive from the first and second questions. Therefore, in order to solve this problem, you must show how any group of three answers, to whatever set of questions you ask, allows you to prove the true identity of X, Y and Z. Please do spend some time thinking about this, if you don't, then what was the point of reading the page? If you really have tried to figure out how to at least start working on the problem, you may click here for more hints on how to solve it... But I will never give you the complete solution (unless you show me some work on your part! — Or am I being a Liar or a Randomizer? Just teasing!). ____________ Although Martin Gardner presented this puzzle in Scientific American some years before it was included in his work, the Sixth Book of Mathematical Games from Scientific American (W.H. Freeman, 1971), chapter 20, section 4, "Truthers, Liars and Randomizers," page 197, he said there, "Logic problems involving truth-tellers and liars are legion, but the following unusual variation — first called to my attention by Howard De Long of West Hartford, Connecticut — had not to my knowledge been printed before it appeared in Scientific American." So, if you really want to know when it first came into existence and by whom, you'll have to find out from someone else. Oh, this may be the same Howard DeLong who wrote, A Profile of Mathematical Logic (Dover Books on Mathematics). The Starman's Math Index
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https://chemistry.stackexchange.com/questions/107489/oxidation-of-sodium-thiosulfate-by-iodine
inorganic chemistry - Oxidation of sodium thiosulfate by iodine - Chemistry Stack Exchange Join Chemistry 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 Chemistry helpchat Chemistry 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 Oxidation of sodium thiosulfate by iodine Ask Question Asked 6 years, 8 months ago Modified6 years, 8 months ago Viewed 10k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. When N a X 2 S X 2 O X 3 N a X 2 S X 2 O X 3 reacts with I X 2 I X 2, it undergoes a redox reaction as 2 N a X 2 S X 2 O X 3+I X 2⟶N a X 2 S X 4 O X 6+2 N a I 2 N a X 2 S X 2 O X 3+I X 2⟶N a X 2 S X 4 O X 6+2 N a I Sulfur is being oxidized from a +2+2 state to a +2.5+2.5 state. Why isn't it getting oxidized any further, say to 3+3+, to form a compound like N a X 2 H S O X 3 N a X 2 H S O X 3? I do notice that there's no hydrogen involved in the equation, but maybe the reaction occurs in an acidic medium? Could someone please tell me whether the above scenario is possible or not and why? inorganic-chemistry redox titration Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jan 5, 2019 at 9:16 andselisk♦ 38.7k 14 14 gold badges 134 134 silver badges 225 225 bronze badges asked Jan 5, 2019 at 8:32 QuasarChaserQuasarChaser 113 1 1 gold badge 1 1 silver badge 7 7 bronze badges Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Whether compound X X is going to be reduced/oxidized and to what degree is mostly dictated by the corresponding redox potential for the given medium. The thiosulfate anion S X 2 O X 3 X 2−S X 2 O X 3 X 2− is used in quantitative analysis and in iodometric titration in particular since it is a reasonably strong reducing agent [1 pp. 714-717]: S X 4 O X 6 X 2−+2 e X−↽−−⇀2 S X 2 O X 3 X 2−E∘=0.169 V S X 4 O X 6 X 2−+2 e X−↽−−⇀2 S X 2 O X 3 X 2−E∘=0.169 V 2 S X 2 O X 3 X 2−+I X 2⟶S X 4 O X 6 X 2−+2 I X−2 S X 2 O X 3 X 2−+I X 2⟶S X 4 O X 6 X 2−+2 I X− However, stronger oxidants such as chlorine are capable of oxidizing thiosulfate further down to sulfur(VI), e.g. forming a sulfate instead: S X 2 O X 3 X 2−+4 C l X 2+5 H X 2 O⟶2 H S O X 4 X−+8 H X++8 C l X−S X 2 O X 3 X 2−+4 C l X 2+5 H X 2 O⟶2 H S O X 4 X−+8 H X++8 C l X− Bromine, being intermediate between iodine and chlorine, can cause S X 2 O X 3 X 2−S X 2 O X 3 X 2− to act either as a 1-electron or an 8-electron reducer according to conditions. For example, in an amusing and instructive experiment, if concentrated aqueous solutions of S X 2 O X 3 X 2−S X 2 O X 3 X 2− and B r X 2 B r X 2 are titrated, and the titration is then repeated after having diluted both the S X 2 O X 3 X 2−S X 2 O X 3 X 2− and B r X 2 B r X 2 solutions 100-fold, then the titre will be found to have increased by a factor of exactly 8. Note, however, that it is not entirely correct to state that sulfur in thiosulfate has oxidation state +2+2. In fact, there are two nonequivalent S S atoms with oxidation states -1 and +5. Same goes for tetrathionate anion S X 4 O X 6 X 2−S X 4 O X 6 X 2− where two central S S atoms have oxidation state 0 0 and the two terminal ones +5+5. References Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 10, 2020 at 14:04 CommunityBot 1 answered Jan 5, 2019 at 9:16 andselisk♦andselisk 38.7k 14 14 gold badges 134 134 silver badges 225 225 bronze badges Add a comment| Your Answer Reminder: Answers generated by AI tools are not allowed due to Chemistry Stack Exchange's artificial intelligence policy Thanks for contributing an answer to Chemistry Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions inorganic-chemistry redox titration See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 22Oxidation state of the sulfur atoms in the thiosulfate Ion 2What are the oxidation states of sulfur in the tetrathionate ion? Related 7Prediction of ionic reduction? 5How much lactose is there in milk (mechanism)? 11What is the correct structure of sodium thiosulfate? 0Standardization of sodium thiosulfate using potassium dichromate 1Difficulty understanding redox in terms of hydrogen and oxygen transfer? 1Redox reaction between Potassium thiocyanate and iodine Hot Network Questions Storing a session token in localstorage Exchange a file in a zip file quickly Interpret G-code How to start explorer with C: drive selected and shown in folder list? What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left? How do you emphasize the verb "to be" with do/does? How long would it take for me to get all the items in Bongo Cat? Checking model assumptions at cluster level vs global level? Cannot build the font table of Miama via nfssfont.tex What "real mistakes" exist in the Messier catalog? How to locate a leak in an irrigation system? 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https://emedicine.medscape.com/article/852734-overview
No Results No Results Medscape Editions Medscape Editions No Results No Results processing.... Mandibular Cysts and Odontogenic Tumors Overview Multiple radiolucent or mixed radiolucent/radiopaque lesions of the mandible may present as incidental findings on radiographs or as the chief symptom of a patient. This article is not intended to be an all-inclusive discussion of such lesions; instead, it confines itself to an overview of the major odontogenic cysts and tumors with a brief discussion of other mandibular lesions that are often called cysts but are not true cystic lesions. Although often similar in radiographic presentation, malignant tumors (both primary and metastatic), benign salivary tumors, and vascular lesions are not addressed herein. However, such lesions should be included in the differential diagnoses of a patient presenting with mandibular radiolucency or swelling. As a corollary, before the biopsy of any such lesions, the area should be aspirated to exclude a vascular lesion. Odontogenic Mandibular Cysts Odontogenic cysts are defined as epithelial-lined structures derived from odontogenic epithelium. Most odontogenic cysts are defined more by their location than by any histologic characteristics. Accordingly, the surgeon must provide the pathologist with appropriate history and radiographs when submitting such specimens for examination. Periapical cyst A periapical (radicular) cyst is the most common odontogenic cyst. The usual etiology is a tooth that becomes infected, leading to necrosis of the pulp. Toxins exit the apex of the tooth, leading to periapical inflammation. This inflammation stimulates the Malassez epithelial rests, which are found in the periodontal ligament, resulting in the formation of a periapical granuloma that may be infected or sterile. Eventually, this epithelium undergoes necrosis caused by a lack of blood supply, and the granuloma becomes a cyst. The lesions are not usually clinically detectable when small but most often are discovered as incidental findings on radiographic survey. Radiographically, distinguishing between a granuloma and a cyst is impossible, although some say that if the lesion is quite large it is more likely to be a cyst. They both present as radiolucent lesions in association with the apex of a nonvital tooth. Occasionally, these lesions can become quite large because they grow in response to pressure. However, granulomas and cysts are not neoplastic. Microscopically, the epithelium is a nondescript stratified squamous epithelium without keratin formation. Inflammatory changes may be observed in the cyst wall, and these changes, in turn, may lead to epithelial changes (eg, ulceration, atrophy, hyperplasia). In particularly inflamed lesions, cholesterol slits and/or foamy macrophages may be apparent. Several treatment options exist for such cysts. Many cysts resolve with endodontic therapy of the involved tooth. Those lesions should be monitored radiographically to ensure such resolution. Lesions that fail to resolve with such therapy should be surgically removed and histopathologically examined. Although these cysts arise from a mature resting epithelium and thus have a relatively low growth potential, a squamous cell carcinoma occasionally may arise de novo in a radicular cyst, thus the recommendation for histopathologic examination of all tissues removed. Dentigerous cyst The second most common odontogenic cyst is the dentigerous cyst, which develops within the normal dental follicle that surrounds an unerupted tooth. The dentigerous cyst is not thought to be neoplastic. It most frequently is found in areas where unerupted teeth are found: mandibular third molars, maxillary third molars, and maxillary canines, in decreasing order of frequency. Dentigerous cysts can grow very large and can move teeth, but, more commonly, they are relatively small. Most dentigerous cysts are asymptomatic, and their discovery is usually an incidental finding on radiography. The usual radiographic appearance is that of a well-demarcated radiolucent lesion attached at an acute angle to the cervical area of an unerupted tooth. The border of the lesion may be radiopaque. The radiographic differentiation between a dentigerous cyst and a normal dental follicle is based merely on size. However, histologically, a distinction other than size is found. The dental follicle is normally lined by the reduced enamel epithelium (see Odontogenesis), while the dentigerous cyst is lined with a stratified squamous nonkeratinizing epithelium. Dystrophic calcification and clusters of mucous cells may be found within the cysts. Dentigerous cysts develop from follicular epithelium, and follicular epithelium has greater potential for growth, differentiation, and degeneration than the epithelium from which radicular cysts arise. Occasionally, other more ominous lesions arise within the walls of the dentigerous cyst, including mucoepidermoid carcinoma arising from mucous cells within the cyst walls, ameloblastoma (see Odontogenic tumors; 17% of ameloblastomas arise within a dentigerous cyst), and squamous cell carcinoma. As previously mentioned, dentigerous cysts also can become quite large and can place the patient at risk for pathologic jaw fracture. These findings comprise most of the medical rationale for removal of impacted third molars with pericoronal radiolucencies; however, impacted teeth with small pericoronal radiolucencies (suggesting the presence of normal dental follicle rather than dentigerous cyst) may also be monitored with serial radiographic examination. Any increase in the size of the lesion should prompt removal and histopathologic examination. Any lesion that appears larger than a normal dental follicle indicates removal and histopathologic examination. Primordial cyst By definition, the primordial cyst develops instead of a tooth. Presumably, the dental follicle forms and subsequently undergoes cystic degeneration without ever completing odontogenesis. This is the rarest odontogenic cyst, and lesions designated as primordial cysts may represent residual cysts. The histology of these lesions is a nondescript stratified squamous epithelium. A complete dental history is important to establish a diagnosis of primordial (versus residual) cyst, although such a diagnosis often has little clinical significance in terms of treatment planning and decision making. Residual cyst Residual cyst is a term of convenience because no teeth are left by which to identify the lesion. Most commonly, these are actually retained periapical cysts from teeth that have been removed. The histology is a nondescript stratified squamous epithelium. Lateral periodontal cyst The name lateral periodontal cyst is a misnomer. These cysts are not inflammatory, they do not arise from periodontitis, and they are not a phenomenon associated with lateral canals within the tooth structure. These cysts are always well demarcated, relatively small, and radiolucent (sometimes with a radiopaque roof). They are most commonly associated with the mandibular premolar area and are occasionally found in the maxillary anterior. They are usually not clinically apparent but, rather, are detected through radiographic examination. These cysts have a distinctive histology consisting of a thick fibrous noninflamed cyst wall, and the lining epithelium is made of thin cuboidal cells. This lining is incomplete and easily sloughs away with mural thickenings of clear cells at periodic intervals. These cysts develop from the postfunctional dental lamina, and no good explanation is known for the localization that is shown. Gingival cyst of the newborn Gingival cysts of newborns generally occur in multiples but occasionally occur as solitary nodules. They are located on the alveolar ridges of newborns or young infants. These structures originate from remnants of the dental lamina and are located in the corium below the surface epithelium. Occasionally, they may become large enough to be clinically noticeable as discrete white swellings on the ridges. They are generally asymptomatic and do not produce any discomfort for the infant. Bohn nodules and Epstein pearls are 2 similar lesions with which gingival cysts sometimes may be confused; however, the location and etiology of these lesions are somewhat different. Epstein pearls are cystic keratin-filled nodules found along the midpalatine raphe and are thought to be derived from entrapped epithelial remnants along the line of fusion. Bohn nodules are keratin-filled cysts scattered all over the palate, but they are most apparent at the junction of the hard and soft palate. These are thought to be derived from palatal salivary gland structures. Histologically, the gingival cyst of the newborn is a true cyst with a thin epithelial lining. The lumen is usually filled with keratin but may contain some inflammatory cells, dystrophic calcifications, and hyaline bodies, such as those often found in dentigerous cysts. No treatment is required for these lesions, which usually disappear either by opening onto the surface mucosa or through disruption by erupting teeth. These cysts are most likely what older literature describes as predeciduous dentition. Gingival cyst of the adult Gingival cysts of the adult are found only in soft tissue in the lower premolar areas. These cysts present as tense, fluctuant, vesicular, or bullous lesions. Histologically, they look like lateral periodontal cysts, and they probably represent the same lesion when found in soft tissue. Odontogenic keratocyst This lesion is relatively common. Originally described in 1956, by a Danish surgeon, it was first called an odontogenic keratocyst. In 2006, the parakeratinized (and more common) variant of the lesion was renamed the keratocystic odontogenic tumor (KCOT), to reflect the fact that this lesion represents a benign odontogenic tumor rather than a simple cyst. Of note, the much less common orthokeratinized version, which has a far lower incidence of recurrence, is still considered a cyst. In 2017, with the publication of the fourth edition of the WHO Classification of Head and Neck Tumours, from the World Health Organization (WHO), the official name of this entity reverted to odontogenic keratocyst (OKC). Clinically and radiographically, the OKC may have any appearance; it is a great mimic, and the diagnosis is a histologic one. These lesions can be both aggressive and difficult to remove, growing quite rapidly, and recurrences are frequent. They are the third most common odontogenic cyst/tumor and belong in the differential diagnoses of any radiolucency of the jaws. Although 40% of OKCs appear in a dentigerous relationship, 9% of dentigerous cysts are OKCs when the histology is examined. These lesions are additionally found as part of the basal cell nevus syndrome, also known as Gorlin syndrome (see the next section, Basal cell nevus syndrome). Histologically, OKCs are formed with a stratified squamous epithelium that produces orthokeratin (10%), parakeratin (83%), or both types of keratin (7%). The epithelial lining appears corrugated when viewed under a microscope. A well-polarized, hyperchromatic basal layer is observed, and the cells remain basaloid almost to the surface. No rete ridges are present; therefore, the epithelium often sloughs from the connective tissue (94% of the time). The epithelium is thin, and mitotic activity is frequent; therefore, OKCs grow in a neoplastic fashion and not in response to internal pressure. The lumen frequently is filled with a foul-smelling, cheeselike material that is not pus but rather collected degenerating keratin. The lesions grow in a multilocular, bosselated fashion, with daughter cysts that extend into the surrounding bone. Because of this relationship, the tendency for recurrence is high, particularly if the original surgical treatment does not result in complete removal of the lesion. Enucleations with peripheral ostectomy and/or cryosurgery are the most common forms of treatment. Preliminary marsupialization and decompression prior to definitive treatment is particularly helpful when treating lesions that are large or in close proximity to important anatomic structures such as the inferior alveolar nerve or maxillary sinus. Long-term (lifetime) clinical and radiographic follow-up is imperative. If these lesions are left untreated, they can become quite large and locally destructive. As mentioned, the variant that produces only orthokeratin acts somewhat differently than the lesions that produce parakeratin. The orthokeratin variation is almost always found in a dentigerous association, usually around the mandibular third molar, and is much less aggressive. It does not exhibit a hyperchromatic basal layer; in fact, the basal layer is flattened. This variant is not associated with basal cell nevus syndrome. Basal cell nevus syndrome This symptom complex includes hypertelorism, midface hypoplasia, relative frontal bossing and prognathism, intellectual disability, schizophrenia, multiple basal cell carcinomas, calcification of the falx cerebri, bifid ribs, palmar pitting (the pits later develop into basal cell carcinoma), and multiple OKCs. Multiple OKCs are diagnostic for basal cell nevus syndrome until proven otherwise. This is a hereditary disease with autosomal dominant inheritance and high penetration. Of patients with OKC, 5% have basal cell nevus syndrome. Early identification of these patients and their lesions is key to improving long-term survival and quality of life. Glandular odontogenic cyst (GOC) The glandular odontogenic cyst (GOC) is a very rare and relatively new entity. First reported in 1987, by Padaychee and Van Wyk, it was initially called a sialodontogenic cyst. By 1992, however, it had become obvious that this cyst is not salivary but odontogenic in origin and as such the name was changed to glandular odontogenic cyst. This lesion represents approximately 0.2% of all odontogenic cysts, making reliable statistics difficult to obtain. One of the largest literature reviews of the cyst was performed by Kaplan et al and included 111 published case reports. In this cohort, the age ranged from 14-75 years, with a mean age of 45.7; there was a slight male predilection (1.3:1) and a significant mandibular over maxillary predilection (70:30), with no significant difference in the number of cases found between anterior and posterior tooth-bearing structures. GOCs range in size from very small to rather large and may present as a unilocular or multilocular radiolucency. They can be very aggressive, not only expanding but also causing thinning (24.4%) and perforation (61%) of cortical bone. Recurrence rates after treatment vary and appear to be proportional to a number of factors, including the size of the initial lesion at diagnosis and the extent of the treatment offered. Published recurrence rates thus range from as low as 1.44% in very small lesions to 85.6% in large lesions. Kaplan et al defined a large lesion as one that involves more than two teeth and/or extends beyond alveolar bone or into adjacent structures. GOCs, though quite rare, are significant because they are easily confused with multiple types of lesions, most of which are benign (eg, mucous metaplasia in an odontogenic cyst, surgical ciliated cysts, botryoid cysts). Importantly, they are also easily confused with low-grade mucoepidermoid carcinoma, a slow-growing but relentless malignancy with excellent short-term survival rates but a very poor long-term prognosis. Thus, it is essential to identify and appropriately treat GOCs not only to address local destruction but also to ensure that a proper diagnosis is made. Histologically, GOCs are characterized by a nonkeratinized, stratified squamous epithelium of varying thickness with focal areas of proliferation. There is a flat interface with the underlying connective tissues, and there is no palisading of the basal layer. Hobnail (cuboidal eosinophilic) cells are common in the lining epithelium, and periodic acid-Schiff–positive material and/or clefts lined by mucous cells within the lining, along with intraepithelial ducts, are common. Less commonly, papillary conformations, cilia, vacuolated cells, and daughter cysts have been described. An excellent flow chart for the histologic differentiation of these lesions is provided in the following study: Kaplan I, Anavi Y, Manor R, Sulkes J, Calderon S. The use of molecular markers as an aid in the diagnosis of glandular odontogenic cyst. Oral Oncol. 2005 Oct. 41 (9):895-902. Treatment options vary depending on the size and location of the GOC at discovery. Minor treatment for a very small lesion confined to the apex of a single tooth may consist of endodontic therapy, apicoectomy, curettage, and submission for histopathologic diagnosis with peripheral osteotomy. Less aggressive surgical treatment of larger lesions may include enucleation/curettage with subsequent treatment with Carnoy’s solution (now difficult to obtain in United States, as the chloroform in this preparation has been deemed a carcinogen), cryosurgery, or peripheral osteotomy. For lesions that are very large and/or in close proximity to a significant anatomic structure, marsupialization and decompression prior to definitive treatment may improve success, decrease morbidity, and increase patient acceptance of treatment. Finally, for very large lesions, marginal or segmental resection may be indicated. Nonodontogenic Mandibular Cysts Stafne bone cyst A Stafne bone cyst is an unusual form of slightly aberrant salivary gland tissue wherein a developmental inclusion of glandular tissue is found within or, more commonly, adjacent to the lingual surface of the body of the mandible within a deep and well-circumscribed depression. The oldest described occurrence of this phenomenon is in a skull dated to the sixth to fourth centuries BC. The phenomenon was first recognized by Stafne in 1942, hence the eponym. However, this cyst has been referred to by many other names, including static bone cavity, defect of the mandible, lingual mandibular bone cavity, static bone cyst, latent bone cyst, and Stafne bone defect. The incidence of occurrence has been reported as 0.1-1.3% in various studies. The general consensus is that this is a congenital defect, although it rarely has been observed in children. These lesions may generally be regarded as developmental rather than pathologic defects. A predilection for males over females seems to exist. Radiographically, the lesion usually appears as an ovoid radiolucency located between the inferior alveolar canal and the inferior border of the mandible in the region of the second or third molars. It can be differentiated from the traumatic or hemorrhagic bone cyst, which by location almost invariably lies superior to the inferior alveolar canal. Although the classic Stafne cyst is described in the posterior mandible, an anterior variant presenting as a round or ovoid radiolucency in the area between the central incisors and first premolars exists; however, it is far less common. These lesions generally represent benign developmental anomalies that normally do not require any treatment. A complication occasionally reported in the literature is the development of a true salivary gland neoplasm in the tissue associated with one of the cortical defects. Therefore, recording the finding of these lesions and periodically observing them radiographically seem prudent. Clinical or radiographic changes may indicate the need for further investigation. Traumatic bone cyst The traumatic bone cyst also is known as solitary bone cyst, hemorrhagic cyst, extravasation cyst, unicameral bone cyst, simple bone cyst, and idiopathic bone cavity. The traumatic bone cyst is a relatively frequent lesion both in the jaws and elsewhere in the skeleton. The specific etiology of the lesion is unknown, although several mechanisms have been proposed. The most widely accepted is that these lesions originate from intramedullary hemorrhage caused by trauma. In these cases, failure of organization of the blood clot occurs followed by subsequent degeneration of the clot, eventually leading to an empty bone cavity. Restricted venous drainage leads to increasing edema, which in turn causes continued resorption of trabeculae and expansion of the lesion. Expansion of the lesion tends to stop when cortical bone is reached, thus these lesions are not characterized by any cortical expansion. Instead, they are usually incidental findings on radiographs taken for other purposes. However, it is not unusual for a patient to be unable to recall any trauma to the involved jaw. The lesion is most commonly found in young persons (median age, 18 y); the male-to-female incidence ratio is 3:2. The lesions occasionally have been reported in the maxilla but are far more common in the mandible. When the cavities are opened surgically, they are generally either empty or filled with a small amount of straw-colored fluid. Shreds of necrotic clot and fragments of fibrous connective tissue have been reported less commonly. Histologically, these cysts may have a thin connective tissue membrane lining or no lining at all. Radiographically, these lesions tend to appear as smoothly outlined radiolucencies that scallop around the roots of the teeth. They do not displace teeth or resorb roots, and the lamina dura is left intact. They may range from very small (< 1 cm) to very large (involving most of the mandible). They tend to occur above the inferior alveolar canal. These lesions are usually surgically explored to establish a diagnosis, which is made upon finding an empty cavity. No further treatment is generally necessary because surgical manipulation causes the cavity to fill with blood. Soft tissues are closed, and the lesion tends to heal without further intervention. The extreme rarity of such lesions in older patients suggests that the lesions may be self-limiting and/or subject to resolution over time. Focal osteoporotic bone marrow defect Bone marrow may be stimulated in response to unusual demands for increased blood cell production. This hyperplastic marrow may present as focal radiolucencies in the jaws. Of jaw lesions, 75% are reported to occur in female patients, and 85% of jaw lesions are found in the mandible. The lesions are almost always asymptomatic and are discovered as incidental findings on radiographs taken for other indications. Radiographically, these lesions present as ill-defined radiolucencies of variable size, more commonly in edentulous areas. This suggests that in some cases, the lesions represent failure of normal bone regeneration postextraction. Histologically, the tissue in these areas is composed primarily of red marrow, yellow marrow, or a combination of both with long thin irregular trabeculae that are missing the osteoblastic layer. The radiographic appearance of these lesions is not pathognomonic; therefore, these lesions are usually diagnosed surgically. Once diagnosed, they require no further specific treatment; however, if the etiology for the increased hematopoietic demands is unknown, consider an investigation. Aneurysmal bone cyst In 1942, Jaffe and Lichtenstein first classified the aneurysmal bone cyst as a distinct lesion; it is neither a cyst nor an aneurysm. It was not reported in the jaws until 1958, and although theories abound, the etiology and pathogenesis of this lesion are still unknown. Proposed mechanisms for the formation of the aneurysmal bone cyst include alterations in local hemodynamics leading to venous engorgement, resorption, and replacement with connective tissue and osteoid; futile attempts at repair of a hematoma (eg, what occurs in the giant cell granuloma); and microcyst formation secondary to cellular edema associated with other lesions. Often, but not always, these lesions appear in association with other lesions of bone, such as the unicameral cyst, dentigerous cyst, osteoclastoma, central giant cell tumor, fibrous dysplasia, and osteosarcoma. Aneurysmal bone cysts have been observed in every part of the skeleton, although more than 50% of lesions occur in the long bones and vertebral column. They occur in jaws of people of all ages, but a predilection for younger patients and female patients exists. Aneurysmal bone cysts are more likely to occur in the mandible than in the maxilla. They may displace but usually do not resorb the dentition, and sensory disturbances generally are not present. The radiographic appearance often is described as cystic, honeycomb, or soap bubble with eccentric expansion. Cortical bone may be thinned or destroyed, and a periosteal reaction may be present. Histologically, the aneurysmal bone cyst reveals a fibrous connective tissue stroma with many cavernous or sinusoidal blood-filled spaces. Young fibroblasts are present throughout the stroma, and multinucleated giant cells are scattered throughout the lesion. Without the cavernous spaces, this lesion would appear almost identical to the central giant cell granuloma. Treatment of the aneurysmal bone cyst requires complete removal, and complete removal of the lesion with aggressive curettage is the most common treatment modality. Excellent exposure to facilitate this treatment is necessary because these lesions may bleed copiously, and timely but thorough removal of the lesion helps to decrease blood loss. Failure to completely remove all traces of the lesion carries a significant risk (21-59%) of recurrence. Recommendations for bone grafting of the resultant defect vary according to the remaining clinical situation after removal of the lesion. Some authors recommend excision with cryosurgery for recurrent lesions, while other authors advocate block excision or resection with reconstruction. In the past, radiation was proposed for the treatment of these lesions, but radiation may fail to arrest the lesion and, more importantly, can result in sarcomatous changes. Odontogenesis Odontogenic tumors represent the results of interruptions in or reactivation of tissues involved in the normal sequence of odontogenesis. The nature of the neoplasm is determined by the stage of development at which the arrest occurs. A brief review of odontogenesis is helpful in understanding the pathogenesis and behaviors of odontogenic tumors. At the sixth week of gestation, odontogenesis begins with proliferation of certain areas of oral ectoderm to form the dental lamina. At each of the locations where a tooth will be formed, a downgrowth from the dental lamina forms the beginning of the enamel organ. Together, the enamel organ, dental papilla, and dental sac are the formative structures for the entire tooth and supporting structures. The dental lamina, which originally connected the enamel organ to the oral epithelium, breaks up, thus separating the developing tooth bud from the epithelium of the oral cavity. Each stage in the development of the tooth is associated with certain events that are summarized below. Interruptions in this sequence may lead to the formation of odontogenic tumors. Bud stage: Initiation and formation of the enamel organ occurs. Cap stage: Proliferation occurs. Unequal growth leads to the characteristic shape. The peripheral cells are cuboidal and are termed the outer enamel epithelium (OEE), and the cells in the concavity are tall columnar cells termed the inner dental epithelium. At the same time, polygonal cells between outer and inner enamel epithelia begin to separate and form the delicate cellular network known as the stellate reticulum (SR), the spaces of which are filled with a mucoid fluid. Histologically, this material resembles Wharton jelly. The proliferation of the epithelial components just discussed leads to condensation of the enclosed ectomesenchyme and formation of the dental papilla. The cells of the dental papilla eventually form tooth pulp and dentin. Similarly, condensation of the ectomesenchyme surrounding the enamel organ leads to creation of the dental sac. The cells of the dental sac eventually form cementum and the periodontal ligament. Bell stage: Histodifferentiation (early) and morphodifferentiation (late) occur. The enamel organ now assumes an elongated bell shape and has 4 distinct types of epithelial cells: inner enamel epithelium (IEE), stratum intermedium, SR, and OEE. The IEE organizes and induces the adjacent cells of the dental papilla to differentiate into odontoblasts, which form dentin. The dentin, in turn, induces the IEE to differentiate into ameloblasts, which lay down enamel matrix opposite the dentin. This reciprocal induction is essential for the formation of the tooth. The stratum intermedium consists of a few layers of squamous cells between the IEE and the SR. This layer is apparently essential to enamel formation because it is absent in the part of the tooth germ that outlines areas of the tooth without enamel. The SR expands by increasing the amount of intracellular fluid. It collapses before the formation of enamel, leaving the ameloblasts closer to the nutrient capillaries adjacent to the OEE. The OEE, which was formerly smooth, becomes folded. In the folds, the adjacent mesenchyme of the dental sac forms papillae with capillary loops to provide a supply of nutrients for the metabolic activity of the avascular enamel organ. The enamel organ also forms Hertwig epithelial root sheath, which determines the shape of the roots and initiates the formation of dentin in the roots. Apposition: The deposition of the matrix of the hard dental structures occurs next. These structures then begin calcification, eruption, and, with time, attrition. Odontogenic Tumors Ameloblastoma Ameloblastoma (see the images below) is an entirely epithelial tumor arising from the dental lamina, Hertwig sheath, the enamel organ, or the lining of dental follicles/dentigerous cysts. Ameloblastoma is the most common epithelial odontogenic tumor. Ameloblastomas usually occur in individuals aged 20-40 years; however, the unicystic variant (see Surgical considerations, below) most often occurs in adolescents. This lesion occurs in both the maxilla and mandible, but the posterior mandible is the most common location; only 20% of lesions are found in the maxilla. The lesion is distributed equally between males and females. Although ameloblastoma generally is not classified as a malignant lesion (a rare malignant variant exists), it is extremely aggressive and infiltrative. Many have suggested that this lesion should be considered a low-grade or indolent malignancy, similar to basal cell carcinoma. Many histologic and behavioral similarities are found between the 2 lesions. It generally does not metastasize but is slow growing, persistent, and hard to eradicate. If ameloblastoma is not noticed as an incidental finding on radiographs taken for other purposes, the first symptom is usually painless bony expansion. Based on a literature review and meta-analysis, Chae et al estimated that ameloblastomas have a mean growth rate of 87.84% per year, with solid, multicystic ameloblastomas having the highest growth rate, and the peripheral form of the tumor having the lowest such rate. Radiographic findings Ameloblastomas typically appear as an expansile multilocular radiolucency in the area of the lower third molar, but they may be found anywhere in the jaws (see the image below). These lesions may be unilocular when small, and they often resorb the teeth they contact. These lesions are never radiopaque. Histologic characteristics Ameloblastoma does not have a capsule. The neoplastic component is purely epithelial and resembles the cap stage of odontogenesis (ie, polarized tall columnar cells on the outer aspect of the lesion with SR on the inner aspect, which may form a cyst). The lesion may have a reactive connective tissue component that is not neoplastic. This is a nonfunctional tumor, ie, it does not induce the surrounding connective tissue, which in turn is unable to induce enamel formation. In effect, these tumors represent arrested odontogenesis. Multiple histologic varieties exist, eg, the acanthomatous type in which the SR is replaced by squamous cells and pearls, the granular cell type in which the SR is replaced by granular cells, and the plexiform type in which the SR is reduced or absent. Treatment The treatment of ameloblastoma is surgical excision with wide free margins (see Surgical considerations, below). Appropriate reconstruction may be performed at the same time. All patients with ameloblastoma, regardless of surgical treatment method or histologic type, must be monitored radiographically throughout their lifetime. If excision is inadequate, recurrence is common. Surgical considerations See the list below: The maxillary ameloblastoma is not confined by the strong cortical plate found in the mandible. In addition, the posterior maxilla lies in close relationship to many vital structures. These factors make strong arguments for aggressive and definitive surgical treatment of the maxillary ameloblastoma. In the mandible, 1-cm clear margins are considered the standard. This may be accomplished with block or segmental resection, depending on the relationship of the lesion to the inferior cortical border. The single exception to this may be the unicystic ameloblastoma. This variant most commonly appears in late adolescence and, as the name suggests, is characterized by a unicystic radiolucency that most commonly is found in the area of the mandibular third molars. Unlike other types of ameloblastomas, it is believed that this lesion is encapsulated and can be removed with enucleation/curettage procedures alone. These lesions may recur, and recurrences may require more aggressive treatment. Most authors believe that if left untreated, this lesion becomes an ameloblastoma of one of the classic varieties, leading to the corollary conclusion that these lesions simply represent an early stage in the development of ameloblastoma. For peripheral ameloblastoma, a more conservative excision with close clinical follow-up is the standard of care. Relationship to other lesions See the list below: Basal cell carcinoma: Basal cell carcinoma is another infiltrative, essentially nonmetastasizing adnexal neoplasm. Basal cell carcinomas and ameloblastomas are slow growing but persistent, and they may cause death via local extension into vital structures. If one considers that the tooth is an oral adnexal structure, then it is easy to understand why ameloblastoma may be seen as an analogue to basal cell carcinoma. Tibial adamantinoma: This lesion is histologically similar to the plexiform variety of ameloblastoma. It is considered a low-grade malignancy and, as the name suggests, is found in the tibia. Craniopharyngioma: This pituitary tumor arises from Rathke pouch, part of the oral stomadeum that histologically appears somewhat like ameloblastoma. However, it is actually more like the Gorlin cyst. Peripheral ameloblastoma: This lesion is histologically identical to the central ameloblastoma, but it does not involve bone and is confined entirely to the gingiva. It has a lower potential for growth and invasion than the central ameloblastoma, and, quite possibly, it is responsible for reported cases of basal cell carcinoma in the gingiva. Malignant ameloblastoma: Approximately 2% of ameloblastomas metastasize, usually to the lungs. These lesions may actually be the result of aspiration of material from fungating lesions in the oral cavity and, therefore, do not represent true metastases. Ameloblastic carcinoma: These are cytologically malignant lesions with hyperchromatism, pleomorphism, and high mitotic activity. Real metastases occur with ameloblastic carcinoma. Adenomatoid odontogenic tumor The adenomatoid odontogenic tumor (AOT) is a fairly uncommon tumor, but it usually can be easily identified from its clinical and radiographic appearance. It often is remembered as the "two-thirds tumor." It most commonly occurs in the second and third decades of life (12-20 y). Two thirds of the cases occur in the anterior maxilla, one third occur in the anterior mandible, and it is never found posterior to the premolars. Two thirds of the cases occur in females, and two thirds of the cases are associated with an impacted tooth (usually the cuspid). This tumor originates from the reduced enamel epithelium of the dental follicle and histologically reproduces the IEE. It is generally asymptomatic but may present with mild swelling or in association with a clinically missing tooth. Radiographic findings This lesion generally appears as a well-demarcated radiolucency. In 75% of cases, it is associated with an unerupted tooth, usually the canine. It may contain radiopaque flecks, which represent calcified material. If associated with a tooth, it generally attaches to the tooth further apical on the root than the typical dentigerous cyst. Histologic characteristics Technically, this is a hamartoma rather than a true neoplasm because it has a limited growth potential. It has a thick fibrous capsule filled with a proliferation of epithelial elements that form nodules and ductlike structures (eg, organoid nodules of cuboidal or low columnar cells separated by spindly epithelium). In the absence of connective tissue to induce the formation of enamel, the product of these cells, a pre-enamel matrix, is thought to degenerate and ultimately leave areas of dystrophic calcification and amyloid. Treatment The recommended treatment of these lesions is simple removal. If left alone, these structures probably involute. However, they can become quite large. Most are removed at biopsy. If AOT is incompletely removed at biopsy, the literature suggests that the remainder of the lesion degenerates. They are not known to recur. Calcifying epithelial odontogenic tumor The calcifying epithelial odontogenic tumor (CEOT), or Pindborg tumor, is a benign infiltrative odontogenic tumor that is one of the rarest. It is named after Jens Pindborg, the Danish oral pathologist. It is most often found in the mandibular molar/premolar region, but 33% of cases are found in the maxilla. It is associated with an unerupted or impacted tooth in 50% of cases. CEOT is an infiltrative neoplasm and causes destruction with local expansion. It is derived from the stratum intermedium and has a lower growth potential than ameloblastoma. Not surprisingly, it is less aggressive than ameloblastoma. In a review of 20 cases of CEOT, by Ruddocks et al, the condition was found equally in males and females, with more of the tumors arising in the mandible than in the maxilla (60% and 40%, respectively). With regard to CEOT variants, two patients had incipient CEOT, and two had peripheral CEOT. Radiographic findings These lesions can be radiolucent, but they more characteristically are mixed lucent and opaque masses, exhibiting a snow-driven appearance. Histologic characteristics The histologic view of this lesion is worrisome because it appears as invasive infiltrative islands in bone. These islands look like pure squamous cells with a high degree of nuclear pleomorphism; however, Liesegang rings (ovoid dystrophic calcifications), a normal mature cytoplasm (large polyhedral cells with good intercellular bridges and filled with mature keratin granules), and the lack of mitotic figures help to distinguish this lesion from squamous cell carcinoma. The pleomorphism noted is secondary to degeneration of the nuclei and necrobiosis, and dystrophic calcification and amyloid conversion are characteristic in dying epithelial cells. Treatment The treatment of this lesion is complete surgical excision. The recurrence rate for CEOT is 4%. The lesion is slow growing and requires long-term follow-up monitoring for recurrence (at least 5-10 y). No cases of malignant transformation are reported. Keratinizing and calcifying odontogenic cyst/calcifying odontogenic cyst The keratinizing and calcifying odontogenic cyst (KCOC), or Gorlin cyst, is not actually a cyst but rather a neoplasm with cystic tendencies. Some KCOC lesions are actually solid. This is a very rare lesion with no age, sex, or location predilections. KCOC may be found anywhere in the jaws, and one fourth of lesions are found in peripheral soft tissue (eg, gingiva). If KCOC is not discovered as an incidental finding on radiographic examination, the earliest clinical presentation usually is a localized swelling. These lesions arise from a more mature enamel epithelium than ameloblastoma, and, accordingly, they have less growth potential. Radiographic findings These are nondescript radiolucencies that may contain flecks of opacity. They may become quite large if not discovered serendipitously. Histologic characteristics These lesions are lined by an epithelium that is similar in appearance to ameloblastoma (eg, polarized basal layer and SR are present). KCOC appears somewhat similar to the unicystic ameloblastoma, with masses of keratinized squamous epithelial cells within the SR. However, these cells have no nuclei and are called ghost cells. This ghost epithelium eventually herniates into the connective tissue, causing a connective tissue foreign body response that results in dentinoid dystrophic calcification and the formation of granulation tissue. More simply, this lesion represents enamel epithelium that has a tendency to mature but is unable to form enamel. The result is the formation of ghost keratin that may, in turn, induce dentinoid. The similarities between this lesion and the craniopharyngioma have previously been mentioned. Treatment These lesions are surgically removed and rarely recur after excision. A case report by Emam et al found that in KCOC surgery, use of a two-stage approach, in which decompression was followed by enucleation and curettage, led to a significant reduction in deformity. However, long-term follow-up was necessitated by the possibility of recurrence. A note on terminology: Often there is discussion, particularly amongst pathologists, as to the correct name for a specific entity and, from time to time, terminology changes. A recent discussion on the Bulletin Board of Oral Pathology noted that at one institution the term calcifying odontogenic cyst (Gorlin cyst) is used for the cystic version of this lesion, while dentinogenic ghost cell tumor is used for the “solid, non-cystic lesions” composed of dentinoid and ghost cells. A similar terminology discussion has simultaneously been entertained on the subject of the odontogenic keratocyst. The important take-home point is that good communication with the pathologist examining the lesion is a necessity to make sure that the treating surgeon and the pathologist are indeed talking about the same lesion and to ensure that the treatment options to be offered are appropriate to the disease entity sampled. Odontogenic myxoma This is a benign infiltrative lesion that is clinically indistinguishable from ameloblastoma. It is found in tooth-bearing areas, and a slight predilection for the mandible exists. It generally appears in the early third to fourth decades of life as a slow-growing expansile lesion. If odontogenic myxoma is left untreated, it is invasive and destructive. It is derived from dental mesenchyme (papilla) or follicle. Radiographic findings The radiographic appearance of this lesion is not distinctive. It appears quite similar to ameloblastoma (eg, multilocular radiolucency), though some authors believe that the individual loculations are somewhat smaller in odontogenic myxoma (myxofibroma). Histologic characteristics A few stellate fibroblasts with copious amounts of hyaluronic acid, scant collagen fibrils, and no capsule describe the histologic appearance of this lesion. The hyaluronic acid component stains with Alcian blue, reminding one of the appearance of Wharton jelly. This lesion looks like developing pulp and may be confused with a developing third molar. Treatment As with ameloblastoma, this lesion is treated with block excision. Recurrences occur although with somewhat less frequency than with ameloblastoma. Ameloblastic fibroma Ameloblastic fibroma is a true mixed tumor arising from a combination of 2 embryonic tissues. The epithelial component is able to induce mesenchyme but not to the extent of developing dental hard tissues. It is a relatively uncommon tumor of young people (aged 5-20 y); 75% of ameloblastic fibromas are found in the posterior mandible in the area of a developing tooth. It is benign and expansile, growing as a pushing front rather than invading surrounding tissues. Radiographic findings This lesion appears as a uniocular or bilocular radiolucency, most often in the posterior mandible. The radiographic appearance is identical to that of unicystic ameloblastoma, and both lesions should be differential diagnoses because they affect similar age groups and have similar clinical and radiographic appearances. Histologic examination differentiates the two. Histologic characteristics The epithelial component of this lesion is almost identical to that of ameloblastoma; however, the connective tissue component looks like dental pulp. It is a young, cellular, homogenous connective tissue without much dense collagen. The epithelial and connective tissue components grow together inside a capsule. Treatment The treatment of this lesion is block excision with a border of normal bone. With simple enucleation, recurrence rates of 20-40% have been reported. Sarcomatous change (ameloblastic fibrosarcoma) has also been reported with recurrence or inadequate excision. Ameloblastic fibro-odontoma Ameloblastic fibro-odontoma is an extremely rare odontogenic lesion that develops dental hard tissues (eg, dentin, enamel, cementum). It appears most often before age 20 years and has a slight predilection for the premolar area in either jaw. Although these lesions can become quite large, they generally do not have much growth potential. Radiographic findings Ameloblastic fibro-odontoma appears as a well-demarcated radiolucency with a large central opacity. These lesions almost always are found in association with a tooth. Some similarity exists in radiographic appearance to that of the Pindborg tumor and Gorlin cyst because this is a mixed radiolucent/radiopaque lesion. If located in the anterior mandible, AOT also may be included in the radiographic differential diagnoses. Histologic characteristics This lesion has the histologic characteristics of ameloblastic fibroma. However, epithelial induction of connective tissue occurs to the point of histodifferentiation, leading to presence of dentin, enamel, and/or cementum in the microscopic presentation. Treatment Surgical excision of ameloblastic fibro-odontoma is the usual proposed treatment. These lesions do not often recur. The literature contains some support for sarcomatous change with recurrence. Complex odontoma Complex odontoma lesions represent further histologic differentiation of the odontogenic pluripotential epithelium. Findings from this lesion resemble those of ameloblastic fibro-odontoma but extended one step further. The epithelium in this lesion has involuted, leaving disorganized dental hard tissues in place. These are common lesions, and they persist throughout life. They are usually detected in adolescence and have a predilection for the mandibular molar regions; however, they can be found in other areas of the jaws. Radiographic findings These lesions are generally described as sunburst radiopacities surrounded by a thin, uniform, radiolucent rim. Although this description may have some superficial resemblance to the radiographic description of osteosarcoma, the association with a tooth, the clear demarcation of the lesions' borders, and the lack of pain and/or swelling serve to delineate this very benign lesion from osteosarcoma. Histologic characteristics The histodifferentiation of this lesion is extended one step further than that of ameloblastic fibro-odontoma. In complex odontoma, a physiologic reduction of the ameloblastic epithelium is found. A mixed honeycomb presentation of enamel, cementum, dentin, and pulpal tissue is present. Histodifferentiation but no morphodifferentiation is observed in this lesion. Treatment Simple removal or radiographic observation is the method of treatment. These lesions do not recur. Compound odontoma This is the most common odontogenic tumor. It represents the product of both histodifferentiation and morphodifferentiation of odontogenic tissues, resulting in what appears as a cluster of multiple abortive teeth. It is most commonly found in the maxillary anterior alveolar bone but may be located anywhere within the tooth-bearing segments of the jaws. It is often responsible for preventing normal tooth eruption, thus it is usually discovered during adolescence. Radiographic findings Multiple tiny toothlike structures are contained within a fine radiolucent rim. Histologic characteristics Histology of compound odontomas approaches normal tooth structure. Gross clinical examination is usually sufficient for diagnosis. Treatment Simple removal is the method of treatment. These lesions do not recur. Cementoblastoma Cementoblastoma, as distinguished from cementoma, is a true neoplasm of cementum. This benign neoplasm is rare and is usually observed in patients younger than 25 years. It is most often found in association with the apex of the mandibular first molars (50% of lesions), and it is never found in association with the anterior dentition. The lesion is usually asymptomatic, although occasionally the associated tooth may be slightly sensitive to percussion. Radiographic findings A round opaque sunburst mass attached to the apex of a tooth that is well-demarcated and surrounded by a thin radiolucent rim is observed. The lesion obscures the lamina dura. Students sometimes confuse it with condensing osteitis, a common lesion resulting from low-grade periapical irritation that stimulates bone growth. Although the most usual location for the 2 lesions is the same, condensing osteitis does not obscure the periodontal ligament (PDL) space and tends to be more irregular in outline. The mature cementoma, also known as periapical cemental dysplasia, is another common lesion that students may confuse with cementoblastoma. However, cementoma is usually located in the mandibular anterior region and does not obscure the PDL space. Cementomas actually have 3 developmental stages: osteolytic (at which point the lesion appears as a radiolucency), cementoblastic (mixed radiolucent/radiopaque), and mature (radiopaque). Histologic characteristics Plump cementoblasts separated by cemental partitions form the histology of this encapsulated lesion. Treatment Removal of attached tooth and tumor is the traditional method of treatment. No recurrences are reported. A case report by Borges et al described a conservative, tooth sparing methodology involving endodontic treatment of the affected tooth, followed 30 days later by excision of the lesion and the apical third of the tooth’s root. In the first 7 years of follow-up, no recurrence was observed, and the tooth remained functional. References Siar CH, Ong ST, Ng KH. Adenomatoid odontogenic tumour: a case study with radiographic differential diagnostic considerations. Singapore Dent J. 1996 Jul. 21(1):19-22. [QxMD MEDLINE Link]. El-Naggar AK, Chan JKC, Grandis JR, Takata T, Slootweg PJ, eds. WHO Classification of Head and Neck Tumours. 4th ed. Lyon, France: IARC; 2017. Pogrel MA. The keratocystic odontogenic tumour (KCOT)--an odyssey. Int J Oral Maxillofac Surg. 2015 Dec. 44 (12):1565-8. [QxMD MEDLINE Link]. Kaplan I, Anavi Y, Hirshberg A. Glandular odontogenic cyst: a challenge in diagnosis and treatment. Oral Dis. 2008 Oct. 14 (7):575-81. [QxMD MEDLINE Link]. Kaplan I, Anavi Y, Manor R, Sulkes J, Calderon S. The use of molecular markers as an aid in the diagnosis of glandular odontogenic cyst. Oral Oncol. 2005 Oct. 41 (9):895-902. [QxMD MEDLINE Link]. Chae MP, Smoll NR, Hunter-Smith DJ, et al. Establishing the natural history and growth rate of ameloblastoma with implications for management: systematic review and meta-analysis. PLoS One. 2015. 10(2):e0117241. [QxMD MEDLINE Link].[Full Text]. Ruddocks LA, Fitzpatrick SG, Bhattacharyya I, Cohen DM, Islam MN. Calcifying epithelial odontogenic tumor: a case series spanning 25 years and review of the literature. Oral Surg Oral Med Oral Pathol Oral Radiol. 2021 Jun. 131 (6):684-93. [QxMD MEDLINE Link]. Emam HA, Smith J, Briody A, Jatana CA. Tube Decompression for Staged Treatment of a Calcifying Odontogenic Cyst-A Case Report. J Oral Maxillofac Surg. 2017 Sep. 75 (9):1915-20. [QxMD MEDLINE Link]. Borges DC, Rogerio de Faria P, Junior HM, Pereira LB. Conservative Treatment of a Periapical Cementoblastoma: A Case Report. J Oral Maxillofac Surg. 2019 Feb. 77 (2):272.e1-e7. [QxMD MEDLINE Link]. Tables Contributor Information and Disclosures Kim E Goldman, DMD Assistant Clinical Professor, Department of Oral and Maxillofacial Surgery, New York University College of Dentistry Kim E Goldman, DMD is a member of the following medical societies: American Academy of Oral and Maxillofacial Pathology, American Association of Oral and Maxillofacial Surgeons, American Trauma Society, International Association of Oral and Maxillofacial Surgeons, Kentucky Medical Association Disclosure: Nothing to disclose. Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Received salary from Medscape for employment. Karen H Calhoun, MD, FACS, FAAOA Professor, Department of Otolaryngology-Head and Neck Surgery, Ohio State University College of Medicine Karen H Calhoun, MD, FACS, FAAOA is a member of the following medical societies: American Academy of Facial Plastic and Reconstructive Surgery, American Head and Neck Society, Association for Research in Otolaryngology, Southern Medical Association, American Academy of Otolaryngic Allergy, American Academy of Otolaryngology-Head and Neck Surgery, American College of Surgeons, American Medical Association, American Rhinologic Society, Society of University Otolaryngologists-Head and Neck Surgeons, Texas Medical Association Disclosure: Nothing to disclose. Arlen D Meyers, MD, MBA Emeritus Professor of Otolaryngology, Dentistry, and Engineering, University of Colorado School of Medicine Arlen D Meyers, MD, MBA is a member of the following medical societies: American Academy of Facial Plastic and Reconstructive Surgery, American Academy of Otolaryngology-Head and Neck Surgery, American Head and Neck Society Disclosure: Serve(d) as a director, officer, partner, employee, advisor, consultant or trustee for: Cerescan; Neosoma; MI10;Invitrocaptal,MedtechsyndicatesReceived income in an amount equal to or greater than $250 from: Neosoma; Cyberionix (CYBX);MI10;Invitrocaptal;MTSReceived ownership interest from Cerescan for consulting for: Neosoma, MI10 advisor. Ted L Tewfik, MD Professor of Otolaryngology-Head and Neck Surgery, Professor of Pediatric Surgery, McGill University Faculty of Medicine; Senior Staff, Montreal Children's Hospital, Montreal General Hospital, and Royal Victoria Hospital Ted L Tewfik, MD is a member of the following medical societies: American Society of Pediatric Otolaryngology, Canadian Society of Otolaryngology-Head & Neck Surgery Disclosure: Nothing to disclose. What would you like to print? Policies Medscape About For Advertisers
5053
https://cdn.kutasoftware.com/Worksheets/Alg1/Factoring%201.pdf
©7 42e061n2U UKXu0tgak zSPo0fNtPwCalroe6 RLhL4Cw.j b yAoldlr XrBiEgoh5t7sa RrmePs3ecr4v8eqdg.z H sMeaDdeet EwMiWtGhK 8Iyntf8iinzi4tge4 PADlqgceFbtrsaX W1W.m Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Period_ Date___ Factoring Trinomials (a = 1) Factor each completely. 1) b2 + 8 b + 7 2) n2 − 11 n + 10 3) m2 + m − 90 4) n2 + 4 n − 12 5) n2 − 10 n + 9 6) b2 + 16 b + 64 7) m2 + 2 m − 24 8) x2 − 4 x + 24 9) k2 − 13 k + 40 10) a2 + 11 a + 18 11) n2 − n − 56 12) n2 − 5 n + 6 -1-©L 12H0b1K2T zKZudtqas bSSomfDtdwea4rqeG PLTLUC1.Q t XAdlQlZ zrfi6gRhatQst tr5eWs0eOrMv7ewdO.5 9 qMNaQdVeJ TwliAt8hT FIwnkfBiRnsi9tveN 7ALlfglecbGryaq s1v.9 Worksheet by Kuta Software LLC 13) b2 − 6 b + 8 14) n2 + 6 n + 8 15) 2 n2 + 6 n − 108 16) 5 n2 + 10 n + 20 17) 2 k2 + 22 k + 60 18) a2 − a − 90 19) p2 + 11 p + 10 20) 5 v2 − 30 v + 40 21) 2 p2 + 2 p − 4 22) 4 v2 − 4 v − 8 23) x2 − 15 x + 50 24) v2 − 7 v + 10 25) p2 + 3 p − 18 26) 6 v2 + 66 v + 60 -2-©1 t2t0w1v2Y PKOuct4aN ISpo9fbtywGaZr2eh 3LDLNCR.v Y gAhlcll XrBiugGhWtdsd FrleZsvepr7vQexdC.p v dMnaMdfev lwTiSt1ht HIbnZfdifngiktleO sAOl1gfeGb8r6ae Q1Y.M Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 1 Name_____ Period Date__ Factoring Trinomials (a = 1) Factor each completely. 1) b2 + 8 b + 7 ( b + 7)( b + 1) 2) n2 − 11 n + 10 ( n − 10)( n − 1) 3) m2 + m − 90 ( m − 9)( m + 10) 4) n2 + 4 n − 12 ( n − 2)( n + 6) 5) n2 − 10 n + 9 ( n − 1)( n − 9) 6) b2 + 16 b + 64 ( b + 8)2 7) m2 + 2 m − 24 ( m + 6)( m − 4) 8) x2 − 4 x + 24 Not factorable 9) k2 − 13 k + 40 ( k − 5)( k − 8) 10) a2 + 11 a + 18 ( a + 2)( a + 9) 11) n2 − n − 56 ( n + 7)( n − 8) 12) n2 − 5 n + 6 ( n − 2)( n − 3) -1-©K W2v091S23 QKVu8tLaH sSpoFfjtLwUaArfet zLZLfCS.e w lAFlllX MrbixgPhptosK treejsNeir9vaegdi.G W cMragdFeD bw0iktDhx CIznZfXiZnaiqtseA FAPlZgteebOrfaj m1z.a Worksheet by Kuta Software LLC 13) b2 − 6 b + 8 ( b − 4)( b − 2) 14) n2 + 6 n + 8 ( n + 2)( n + 4) 15) 2 n2 + 6 n − 108 2( n + 9)( n − 6) 16) 5 n2 + 10 n + 20 5( n2 + 2 n + 4) 17) 2 k2 + 22 k + 60 2( k + 5)( k + 6) 18) a2 − a − 90 ( a − 10)( a + 9) 19) p2 + 11 p + 10 ( p + 10)( p + 1) 20) 5 v2 − 30 v + 40 5( v − 2)( v − 4) 21) 2 p2 + 2 p − 4 2( p − 1)( p + 2) 22) 4 v2 − 4 v − 8 4( v + 1)( v − 2) 23) x2 − 15 x + 50 ( x − 10)( x − 5) 24) v2 − 7 v + 10 ( v − 5)( v − 2) 25) p2 + 3 p − 18 ( p − 3)( p + 6) 26) 6 v2 + 66 v + 60 6( v + 10)( v + 1) -2-Create your own worksheets like this one with Infinite Algebra 1. Free trial available at KutaSoftware.com
5054
https://math.dartmouth.edu/~carlp/CarmAPs6.pdf
On Carmichael numbers in arithmetic progressions William D. Banks Department of Mathematics University of Missouri Columbia, MO 65211 USA bbanks@math.missouri.edu Carl Pomerance Department of Mathematics Dartmouth College Hanover, NH 03755-3551 USA carl.pomerance@dartmouth.edu January 19, 2010 We dedicate this paper to our friend Alf van der Poorten Abstract Assuming a conjecture intermediate in strength between one of Chowla and one of Heath-Brown on the least prime in a residue class, we show that for any coprime integers a and m ⩾1, there are infinitely many Carmichael numbers in the arithmetic progression a mod m. 1 1 Introduction For every prime number n, Fermat’s little theorem states that bn ≡b mod n for all b ∈Z. (1) Around 1910, Carmichael began an in-depth study of composite numbers n with this property, which are now known as Carmichael numbers. In 1994 the existence of infinitely many Carmichael numbers was established by Alford, Granville and Pomerance . Since prime numbers and Carmichael numbers are linked by the common property (1), it is natural to ask whether certain known results about primes can also be established for Carmichael numbers. In the present note, we focus on the question of whether an analogue of Dirichlet’s theorem on primes in an arithmetic progression holds for the set of Carmichael numbers. Below, we give a conditional proof in support of the following: Conjecture. There are infinitely many Carmichael numbers in any arith-metic progression a mod m with gcd(a, m) = 1. In fact, we believe a stronger assertion holds. A necessary condition that there is at least one Carmichael number in the residue class a mod m is that gcd(g, 2 ϕ(g)) = 1, where g = gcd(a, m) and ϕ is Euler’s function. It is reasonable to conjecture that this condition is also sufficient for the existence of infinitely many Carmichael numbers in the residue class a mod m. However, our conditional argument does not appear to extend to this more general case. The idea behind our argument is to construct Carmichael numbers with the special form n·p, where n is a Carmichael number congruent to 1 mod m, and p is a prime congruent to a mod m. To produce Carmichael numbers in the arithmetic progression 1 mod m, we use a straightforward variant of the Alford–Granville–Pomerance construction. The real difficulty in our approach lies in finding primes p ≡a mod m such that n and np are both Carmichael numbers. To do this, we must assume (a weak version of) a con-jecture of Heath-Brown on the smallest prime in an arithmetic progression, hence our principal result is conditional. More precisely, for any integers a, d with 1 ⩽a ⩽d−1 and gcd(a, d) = 1, let ϱ(d, a) be the least prime p in the arithmetic progression a mod d, and put ϱ(d) = max{ϱ(d, a) : 1 ⩽a ⩽d −1, gcd(a, d) = 1}. 2 In 1934, Chowla showed that the bound ϱ(d) ≪ε d2+ε with any ε > 0 follows from the generalized Riemann hypothesis, and he conjectured that the stronger bound ϱ(d) ≪ε d1+ε holds. In 1978, Heath-Brown , following thoughts of Cram´ er on gaps between consecutive primes, conjectured that the bound ϱ(d) ≪d(log d)2 holds uniformly for all d ⩾2. In the present note we shall assume that the bound ϱ(d) ≪d1+ξ/ log log d (d ⩾2) (2) holds with a specific real number ξ > 0 identified in the proof of our principal result. Note that this hypothesis is stronger than the conjecture of Chowla but weaker than that of Heath-Brown. Theorem 1. There is a value of ξ > 0 such that if (2) holds for ξ, then there are infinitely many Carmichael numbers in any arithmetic progression a mod m with gcd(a, m) = 1. We remark that Rotkiewicz [10, 11] proved this result unconditionally for pseudoprimes (that is, composite integers n such that 2n ≡2 mod n), and later, van der Poorten and Rotkiewicz established the same result for strong pseudoprimes relative to an arbitrary (fixed) base b ⩾2. In Section 3, we give a quantitative version of Theorem 1 under the same hypothesis (2); see Theorem 2. We also give a quantitative version of Theorem 1 under the slightly stronger hypothesis that the bound ϱ(d) ≪d exp((log d)κ) (d ⩾2) (3) holds with some fixed real number κ < 1; see Theorem 3. In Section 4 we conclude with a few additional remarks. Acknowledgements. The authors would like to thank Igor Shparlinski for numerous helpful conversations. The work on this paper began during a visit by the first author to Macquarie University; the hospitality and sup-port of this institution are gratefully acknowledged. The second author was supported in part by NSF grant DMS-0703850. 2 Preliminaries In what follows, the letters p and q always denote prime numbers. We denote by π(x) the prime counting function, and by λ(n) the Carmichael function, 3 i.e., the order of the largest cyclic subgroup of (Z/nZ)∗. For an integer n > 1, we denote by P(n) the largest prime that divides n. The following result, which is [2, Proposition 1.5], is crucial for our con-struction of Carmichael numbers in the next section. Lemma 1. There exists a constant c0 > 0 such that for any fixed coprime integers a and m ⩾1, if x is sufficiently large (depending on m) and if L is a squarefree integer that is coprime to m, then there is an integer k ⩽x3/5 such that {d | L : p = dk + 1 is prime, p ⩽x, and p ≡a mod m} > c0 ϕ(m) log x {d | L : d ⩽x2/5} . For a finite abelian group G, the Davenport constant D(G) is the least positive integer D with the property that for any sequence of D elements from G, there is a nonempty subsequence whose product is the identity. Clearly D(G) ⩾λ(G), where λ(G) denotes the maximal order of an element in G. We shall use the following result, which is a weakened form of [1, Theorem 1.1]. Lemma 2. If G is a finite abelian group, then D(G) < λ(G)(1 + log |G|). Finally, we need the following lemma, which is [1, Proposition 1.2]. Lemma 3. Let G be a finite abelian group and let r > t > n = D(G) be integers. Then any sequence of r elements of G contains at least r t  / r n  distinct subsequences of length at most t and at least t −n, whose product is the identity. 3 Quantitative results Theorem 2. There is a value of ξ > 0 such that if (2) holds for ξ, then there is a constant c > 0 such that for any fixed coprime integers a and m ⩾1, one has {n ⩽X : n is Carmichael and n ≡a mod m} ⩾Xc/ log log log X for all sufficiently large X (depending on the choice of m). 4 Proof. Let the coprime integers a and m ⩾1 be fixed. Let y be a real parameter which we shall choose to be large and put Q = {q prime : y3/ log y < q ⩽y3, P(q −1) ⩽y}. Note that if y is large enough, then no prime factor of m lies in Q. Let L denote the product of the primes in Q. We know from Friedlander that |Q| ⩾c1π(y3) for some absolute constant c1 > 0; therefore, L ⩾exp((c1 + o(1))y3) (y →∞). On the other hand, since L divides (hence does not exceed) the product of all primes up to y3, we have L ⩽exp((1 + o(1))y3) (y →∞). (4) We apply Lemma 1 with x = L5/2. Since all of the 2|Q| divisors d of L satisfy d ⩽x2/5, we see that there is an integer k ⩽x3/5 for which the set P = {p prime : p ⩽x, p = dk + 1 for some d | L, and p ≡a mod m} has cardinality |P| ⩾ c0 ϕ(m) log x 2|Q| ⩾exp 1 5c1y3/ log y  (5) for all large y, since 1/5 < (log 2)/3. It is not so important that members of P are congruent to a mod m; what is important is that the progressions 1 mod kL and a mod m are compatible (since P is nonempty and gcd(L, m) = 1) and thus may be glued to a single progression a′ mod lcm[kL, m], where gcd(a′, lcm[kL, m]) = 1. Let p0 be the least prime in this progression, so that assuming (2) with ξ = 1 10c1 we have p0 ≪kLm exp 1 10c1 log(kLm)/ log log(kLm)  . Since m is fixed and k ⩽x3/5 = L3/2, using (4) we derive the bound p0 ⩽kL exp ( 1 12c1 + o(1))y3/ log y  (y →∞). (6) Write p0 = 1 + ukL, so that u ⩽exp 1 11c1y3/ log y  (7) for large y. 5 We now remove from P any prime which happens to divide uLp0, denoting the remaining set by P′. It is easy to see that there are ≪y3/ log y distinct primes dividing uLp0, hence from (5) it follows that |P′| ⩾exp (1 5c1 + o(1))y3/ log y  (y →∞). (8) Let N be the set of integers n such that gcd(n, uLm) = 1 and n ≡1 mod k; note that P′ ⊂N. Let G be the subgroup of (Z/ukLZ)∗×(Z/mZ)∗consisting of pairs (α, β) with α ≡1 mod k, and let Ψ : N →G be the natural map that takes each integer n ∈N to the pair Ψ(n) = (n mod ukL, n mod m). We claim that if S is any subset of P′ with more than one element, nS is the element of N given by nS = Q p∈S p, and Ψ(nS) is the identity in G, then NS = nSp0 is a Carmichael number in the arithmetic progression a mod m. To show this, we shall apply Korselt’s criterion. We have bn ≡b mod n for all integers b if and only if n is squarefree and p −1 divides n −1 for every prime p dividing n. The proof is elementary and in fact was found by Korselt before he knew of the existence of any composite examples. Now consider NS = nSp0. Since p0 ̸∈P′, NS is squarefree. Since Ψ(nS) is the identity element in G, we have nS ≡1 mod m, hence NS ≡p0 ≡a mod m. Further, p0 −1 = ukL | nS −1, and it follows that p0 −1 | NS −1. Similarly, for each prime p ∈S, we have p −1 | kL | nS −1, which implies p −1 | NS −1 since p0 ≡1 mod kL. Thus, NS is a Carmichael number by Korselt’s criterion, and the claim is established. To estimate the number of Carmichael numbers produced in this manner, we first need to bound the Davenport constant for the group G. From the definition of G and using [1, Equation (4.3)], we see that for large y, λ(G) ⩽um λ(L) ⩽um e6y. In view of (7) and the fact that m is fixed, it follows that λ(G) ⩽exp ( 1 11c1 + o(1))y3/ log y  (y →∞). Further, using (4), (7), and the fact that m is fixed, we have |G| = ϕ(ukL)ϕ(m) ϕ(k) ⩽uLm ⩽exp((1 + o(1))y3) (y →∞). 6 Hence, applying Lemma 2 we derive that D(G) ⩽exp ( 1 11c1 + o(1))y3/ log y  (y →∞). (9) Put t = exp 1 10c1y3/ log y  and X = exp(3ty3). Note that log log log X = (3 + o(1)) log y (y →∞). (10) If NS is a Carmichael number of the type constructed above, and |S| ⩽t, then by (6) we have, as y →∞, NS = p0 Y p∈S p ⩽xt 1 + kL exp 1 11c1y3/ log y  = x(1+o(1))t, where we have used the inequality kL ⩽x in the last step. Taking into account (4) we have log x = 5 2 log L ⩽(5 2 + o(1))y3, hence NS ⩽exp(3ty3) = X if y is sufficiently large. Finally, one sees that the number T of Carmichael numbers NS produced in this manner is equal to the number of distinct nonempty subsets S ⊂P′ such that Ψ(nS) is the identity in G, and |S| ⩽t. Applying Lemma 3 with r = |P′| and n = D(G), we obtain the lower bound T ⩾ |P′| ⌊t⌋  |P′| D(G)  ⩾ |P′| ⌊t⌋ ⌊t⌋ |P′|−D(G) = |P′|⌊t⌋−D(G) ⌊t⌋−⌊t⌋. (11) By (8), (9), and our definition of t, it follows that T ⩾exp (1 5c1 −1 10c1 + o(1))ty3/ log y  (y →∞). From (10) it follows that ty3/ log y ∼log X/ log log log X as y →∞, so it follows that with c = 1 11c1 log T ⩾ c log X log log log X for all sufficiently large y. Since the value of y can be uniquely determined from X, the result follows. 7 Theorem 3. Suppose that (3) holds with some real number κ < 1. Then there is a constant c > 0, depending only on κ, such that for any coprime integers a and m ⩾1, one has {n ⩽X : n is Carmichael and n ≡a mod m} ⩾Xc for all sufficiently large X (depending on the choice of m). Proof. Our proof follows closely that of Theorem 2, so we focus mainly on the modifications that are needed. Let 0 < κ < 1 be arbitrary and suppose that (3) holds with κ. Let Q and L be defined as before. Let c2 be a fixed real number larger than 5/(2 −2κ). Let y be a real parameter which is assumed large. Applying Lemma 1 with x = exp(c2y3κ), we see that there is an integer k ⩽x3/5 for which |P| ⩾ c0 ϕ(m) log x {d | L : d ⩽x2/5} , where P is defined as before. The product of any s = log(x2/5) log(y3)  =  2c2y3κ 15 log y  distinct primes in Q is a divisor d of L with d ⩽x2/5. Since |Q| ≫π(y3), it follows that as y →∞, {d | L : d ⩽x2/5} ⩾ |Q| s  ⩾ |Q| s s ⩾exp((2−2κ 5 c2 + o(1))y3κ). Consequently, |P| ⩾exp(c3y3κ) for all large y, where c3 is any fixed number such that 1 < c3 < 2−2κ 5 c2. As before, we glue the progressions 1 mod kL and a mod m to a single progression a′ mod lcm[kL, m], where gcd(a′, lcm[kL, m]) = 1. By (3) the least prime p0 in this progression satisfies the bound p0 ≪kLm exp((log(kLm))κ). Since m is fixed and k ⩽x3/5 = Lo(1), using (4) we derive the bound p0 ⩽kL exp(c4y3κ) (12) 8 for all large y, where c4 is any fixed number such that 1 < c4 < c3. Write p0 = 1 + ukL, so that u ⩽exp(c4y3κ). We now proceed as in the proof of Theorem 2 to form the sets P′ and N, the group G, and the map Ψ : N →G. Arguing as before, we derive the bounds |P′| ⩾exp((c3 + o(1))y3κ) (y →∞) (13) and D(G) ⩽exp((c4 + o(1))y3κ) (y →∞). (14) Let c5, c6 be fixed real numbers such that c4 < c5 < c3 and c6 > c2, and put t = exp(c5y3κ) and X = exp(c6ty3κ). If NS = nSp0 is a Carmichael number of the type constructed in Theorem 2, with |S| ⩽t, then by (12) and the fact that log x = c2y3κ, we have NS = p0 Y p∈S p ⩽xt 1 + kL exp(c4y3κ)  = x(1+o(1))t = exp((c2 + o(1))ty3κ) as y →∞. Hence, NS ⩽X for large y. Using the lower bound (11) together with (13), (14), and our definition of t, we see the number T of Carmichael numbers NS produced in this manner satisfies T ⩾exp((c3 −c5 + o(1))ty3κ) = X(c3−c5)/c6+o(1) (y →∞). Thus, if 0 < c < (c3 −c5)/c6 is fixed, then T ⩾Xc for all sufficiently large y. Since the value of y can be uniquely determined from X, the result follows. 4 Remarks We first remark that for the residue class 1 mod m we do not need the prime p0 as in the previous section, and without a need for bounding p0, the rest of the proof is completely rigorous. In fact, one can easily amend the existing proofs of the infinitude of Carmichael numbers in or to prove the following result, where the exponent 1/3 is from . Theorem 4. Let m be an arbitrary fixed positive integer. The number of Carmichael numbers n ⩽X with n ≡1 mod m exceeds X1/3 once X is sufficiently large (depending on the choice of m). 9 In this result it would be interesting to let the modulus m vary more dynamically with X; we leave this as a project for the interested reader. Could it be that for each m, all but finitely many Carmichael numbers are congruent to 1 mod m? Certainly not if our Conjecture holds, and so certainly not if the weaker form of Heath-Brown’s conjecture described above holds. Of course, there may be a cheaper way of disproving such a possibility. We have not found such a path, but we remark that it is easy to see the following: For each number B there is some pair a, m with a ̸≡1 mod m such that there are at least B Carmichael numbers n ≡a mod m. Indeed, let C(x) denote the number of Carmichael numbers in [1, x], and let x be large. Since no positive integer n ⩽x is congruent to 1 mod m for every modulus m ⩽2 log x, there is a pair a, m with a ̸≡1 mod m and m ⩽2 log x such that at least C(x)/(2 log2 x) Carmichael numbers lie in the residue class a mod m. From it follows that for all large x, C(x)/(2 log2 x) > x1/3, which thus proves the assertion in a stronger form. Finally, let Ca,m(x) denote the number of Carmichael numbers n ⩽x with n ≡a mod m. It may be that C1,m(x) ∼C(x) as x →∞for each fixed m. For example, using computations from at x = 2.5 × 1010, we have Ca,3(x)/C(x) ≈0.0116, 0.9792, 0.0092 for a = 0, 1, 2, respectively. Also, Ca,4(x)/C(x) ≈0.9783, 0.0217 for a = 1, 3, respectively. From statistics in at x = 1015, we have Ca,5(x)/C(x) ≈0.0553, 0.8570, 0.0290, 0.0291, 0.0297 for a = 0, 1, 2, 3, 4, respectively. References W. Alford, A. Granville, and C. Pomerance, ‘There are infinitely many Carmichael numbers’, Ann. of Math. (2) 139 (1994), 703–722. W. Alford, A. Granville, and C. Pomerance, ‘On the difficulty of finding reliable witnesses’, Algorithmic number theory (Ithaca, NY, 1994), 1–16, Lecture Notes in Comput. Sci., 877, Springer, Berlin, 1994. S. Chowla, ‘On the least prime in an arithmetical progression’, J. Indian Math. Soc. (N.S.) 1 (1934), 1-3. J. B. Friedlander, ‘Shifted primes without large prime factors’, in Num-ber theory and applications (ed. R. A. Mollin), (Kluwer, NATO ASI, 1989), 393–401. 10 G. Harman, ‘Watt’s mean value theorem and Carmichael numbers’, Int. J. Number Theory 4 (2008), 241–248. D. R. Heath-Brown, ‘Almost-primes in arithmetic progressions and short intervals’, Math. Proc. Cambridge Philos. Soc. 83 (1978), 357–375. R. G. E. Pinch, ‘The Carmichael numbers up to 1015’, Math. Comp. 61 (1993), 381–391. C. Pomerance, J. L. Selfridge, and S. S. Wagstaff, Jr., ‘The pseudoprimes to 25 × 109’, Math. Comp. 35 (1980), 1003–1026. A. J. van der Poorten and A. Rotkiewicz, ‘On strong pseudoprimes in arithmetic progressions’, J. Austral. Math. Soc. Ser. A 29 (1980), 316– 321. A. Rotkiewicz, ‘Sur les nombres pseudopremiers de la forme ax + b’, C. R. Acad. Sci. Paris 257 (1963), 2601–2604. A. Rotkiewicz, ‘On the pseudoprimes of the form ax + b’, Proc. Cam-bridge Philos. Soc. 63 (1967), 389–392. 11
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Art of Problem Solving Set - 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 Set Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Set The notion of a set is one of the fundamental notions in mathematics that is difficult to precisely define. Of course, we have plenty of synonyms for the word "set," like collection, ensemble, group, etc., but those names really do not define the meaning of the word set; all they can do is replace it in various sentences. So, instead of defining what sets are, one has to define what can be done with them or, in other words, what axioms the sets satisfy. These axioms are chosen to agree with our intuitive concept of a set, on one hand, and to allow various, sometimes quite sophisticated, mathematical constructions on the other hand. For the full collection of these axioms, see Zermelo-Fraenkel Axioms. In this article we shall present just a brief discussion of the most common properties of sets and operations related to them. Contents [hide] 1 Rough Definition 2 Relation of Belonging 3 Specifying a Set 3.1 Listing its Elements 3.2 Stating the Common Property of its Elements 3.3 Set Notation 4 Subsets 5 Power Sets 6 Operations 6.1 Union and Intersection 6.2 Cartesian Product 7 Empty Set 8 Infinite Sets 9 Cardinality 10 Problems 10.1 Introductory 10.2 Intermediate 10.3 Olympiad 11 See Also 12 External Links Rough Definition A set is a collection of objects. The objects can be anything: numbers, letters, libraries that have at least 20 male staff, or absolutely nothing. Order does not matter. What does matter is what is in the set. There might be a finite number of objects in the set, in which case it is called a finite set. Otherwise we call it an infinite set. The objects in a set are called the elements of the set. A common misconception is that a set can have multiple indistinct elements, such as the following: Such an entity is actually called a multiset. Relation of Belonging The most important property of sets is that, for every object and a set , we can say whether belongs to (written as ), or not (written as ). Two sets and are equal if they include the same objects, i.e., if for every object , we have if and only if . Specifying a Set There are many ways to specify a set, using different notation. Listing its Elements This means that in order to identify a particular set, it suffices to tell which objects belong to this set. If the set contains just several such objects, all you need to do is list them. So, you can specify the set consisting of the numbers , and , for example. (The standard notation for this set is . Note that the order in which the terms are listed is completely unimportant: we have to follow some order when writing things in one line, but you should actually imagine those numbers flowing freely inside those curly braces with no preference given to any of them. What matters is that these four numbers are in the set and everything else is out). But how do you specify sets that have very many (maybe infinitely many) elements? You cannot list them all even if you spend your entire life writing! Stating the Common Property of its Elements Another way to specify a set is to use some property to tell when an object belongs to this set. For instance, we may try to think (alas, only try!) of the set of all objects with green hair. In this case, we do not even try to list all such objects. We just decide that something belongs to this set if it has green hair and doesn't belong to it otherwise. This is a wonderful way to describe a set. Unfortunately, this method has several potential pitfalls. It turns out, counter-intuitively, that not every collection of objects with a certain property is a set. The most famous example of this problem is Russell's Paradox: consider the property, "is a set and does not contain itself." (Remember that, given a set, we should be able to tell about every object whether it belongs to this set or not; in particular, we can ask this question about the set itself). The set specified by this property can neither belong, nor not belong to itself. There are a variety of ways to resolve this paradox, but the problem is clear: this way to describe sets should be used with extreme caution. One way to avoid this problem is as follows: given a property , choose a known set . Then the collection of elements of which have property will always be a set. (In particular, for our previous example to lead to a paradox, we would need to choose . However, it turns out that it can be proven that the set of all sets does not exist -- the collection of all sets is too big to be a set.) Set Notation There is a notation used just for sets: That symbolizes the set of all reals not equal to 0. This is probably the fastest way of describing a large set. Also, the empty set can be specified using set notation: Since there are no reals such that the square of it is less than 0, that set is the empty set. Subsets We say that a set is a subset of a set if every object that belongs to also belongs to . This is denoted or . For example, the sets and are subsets of the set , but the set is not. Thus we can say that two sets are equal if and only if each is a subset of the other. A special kind of subset is the empty set. Power Sets Main article: power set The power set of a set , denoted is defined as the set of all its subsets. For example, the power set of is . If a is a finite set of size then has size . Operations Union and Intersection The union of two or more sets is the set of all objects that belong to one or more of the sets. The union of A and B is denoted . For example, the union of and is . Unions can also be represented just as sums and products can be. would be the union of all sets that satisfy the statement. So, for example, would be the set of all natural numbers . The intersection of two or more sets is the set of all objects that belong to all of the sets. The intersection of A and B is denoted . For example, the intersection of and is . Just like unions, intersections can be represented as such: . For example, , or the empty set defined next. Cartesian Product The Cartesian Product of two sets and is defined as the set of Ordered Pairs such that and Empty Set Main article: empty set An empty set, denoted is a set with no elements. An empty set has some special properties: It is a subset of every other set. The union of any other set and an empty set is the original set. The intersection of any other set and an empty set is an empty set. Infinite Sets An infinite set can be defined as a set that has the same cardinality as one of its proper subsets. Alternatively, infinite sets are those which cannot be put into correspondence with any set of the form . Cardinality The cardinality of a set , denoted , is (informally) the size of the set. For a finite set, the cardinality is simply the number of elements. The empty set has cardinality 0. iff there is a bijective function meaning that there is a function that maps all elements of to all the elements of with one-to-one correspondence. iff there exists an injective function meaning there is a function that maps all elements of to (not necessarily all) elements of . can be defined similarily by expressing it as . iff there exists an injective function and there is no bijective function meaning but . is defined similarly. Although showing that and implies that is easy to prove when using finite sets, it is more complicated when using infinite sets. This theorem is called the Cantor-Bernstein-Schröder theorem and was proven by Georg Cantor, Felix Bernstein, and Ernst Schröder. Problems Introductory The regular 5-point star is drawn and in each vertex, there is a number. Each and are chosen such that all 5 of them came from set . Each letter is a different number (so one possible way is ). Let be the sum of the numbers on and , and so forth. If and form an arithmetic sequence (not necessarily in increasing order), find the value of . (Source) Intermediate Let set be a 90-elementsubset of and let be the sum of the elements of Find the number of possible values of (Source) Olympiad Let be a fixed positive integer, and let be an infinite family of sets, each of size , no two of which are disjoint. Prove that there exists a set of size that meets each set in . (Source) See Also Set theory Function External Links Naive Set Theory by Paul R. Halmos. Set Theory and Logic by Robert Roth Stoll. 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https://teaching.martahidegkuti.com/shared/lnotes/6_calculus/integral/trigsubstitutions/trigsubstitutions.pdf
Lecture Notes Trigonometric Substitutions page 1 Sample Problems Compute each of the following integrals. 1. Z 1 p x2 + 4 dx 2. Z p 1 x2 dx 3. Z 1 p x2 9 dx 4. Z x2 p 16 x2 dx 5. Z p x2 + 4 dx 6. Z x2 p x2 + 9 dx 7. Z x6 p 1 x14 dx 8. 1 Z 0 tan1 x x2 + 1 dx Practice Problems Compute each of the following integrals. Please note that some of the integrals can also be solved using other, previously seen methods. 1. Z 1 p x2 25 dx 2. Z x p x2 25 dx 3. Z x2 p x2 25 dx 4. Z p x2 25 dx 5. Z p 16 x2 dx 6. Z 1 p 16 x2 dx 7. Z x p 16 x2 dx 8. Z x2 p 16 x2 dx 9. Z p x2 + 1 dx 10. Z 1 p x2 + 1 dx 11. Z x p x2 + 1 dx 12. Z x2 p x2 + 1 dx c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 2 Sample Problems - Answers 1.) ln x + p x2 + 4 + C 2.) 1 2 sin1 x + 1 2x p 1 x2 + C 3.) ln x + p x2 9 + C 4.) 8 sin1 x 4  1 2x p 16 x2 + C 5.) 1 2x p x2 + 4 + 2 ln x + p x2 + 4 + C 6.) 1 2x p x2 + 9 9 2 ln x + p x2 + 9 + C 7.) 1 7 sin1 x7 + C 8.) 2 32 Practice Problems - Answers 1.) ln x + p x2 25 + C 2.) p x2 25 + C 3.) 1 2x p x2 25 + 25 2 ln x + p x2 25 + C 4.) 1 2x p x2 25 25 2 ln x + p x2 25 + C 5.) 8 sin1 x 4  + 1 2x p 16 x2 + C 6.) sin1 x 4  + C 7.) p 16 x2 + C 8.) 8 sin1 x 4  1 2x p 16 x2 + C 9.) 1 2x p x2 + 1 + 1 2 ln x + p x2 + 1 + C 10.) ln x + p x2 + 1 + C 11.) p x2 + 1 + C 12.) 1 2x p x2 + 1 1 2 ln x + p x2 + 1 + C c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 3 Sample Problems - Solutions Trigonometric substitution is a technique of integration. It is especially useful in handling expressions under a square root sign. Case 1. The substitution x = a tan . This is useful in handling an integral involving p x2 + a2. Let x = a tan  where  2 <  <  2 . (That is the same thing as stating that  = tan1 x a. The interval between  2 and  2 is the domain of the inverse function tan1 x.) The picture below shows the reference triangle we use for this substitution. Using this triangle, we do not have to do heavy duty algebra because we can read (up to sign) the trigonometric functions of  in terms of x and a. Example 1: Compute the integral Z 1 p x2 + 4 dx. Solution: We will use a trigonometric substitution. We start with a reference triangle where the hypotenuse is the denominator. Using the substitution x = 2 tan , (where  2 <  <  2 ) we will transform the integral into one in . From the triangle, x = 2 tan . Then dx = 2 sec2  d. The expression p x2 + 4 becomes 2 sec  - using the picture, or using algebra. Recall the identity tan2 x + 1 = sec2 x p x2 + 4 = q (2 tan )2 + 4 = p 4 tan2  + 4 = p 4 p tan2  + 1 = 2 p sec2  = 2 jsec j Because  is in the interval   2 ;  2  , sec x is positive and so jsec xj = sec x. Z 1 p x2 + 4 dx = Z 1 2 sec  2 sec2  d  = Z sec  d This is an integral we have already seen: we can either use substitution (see in that handout) or partial fraction(see in that handout). Either way, Z sec  d = ln jsec  + tan j + C Now we need to reverse the substitution and write the result as an expression of x: This is where the reference triangle comes handy. sec  = p x2 + 4 2 and tan  = x 2 Thus the answer is Z 1 p x2 + 4 dx = ln p x2 + 4 2 + x 2 + C: This expression can be further simpli…ed: ln p x2 + 4 2 + x 2 + C = ln p x2 + 4 + x 2 + C = ln p x2 + 4 + x ln 2 + C = ln p x2 + 4 + x + C c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 4 and so the …nal answer is ln p x2 + 4 + x + C . Case 2. The substitution x = a sin  where  2 <  <  2 . This is useful in handling an integral involving p a2 x2. The picture below shows the reference triangle we use for this substitution. Using this triangle, we can read (up to sign) the trigonometric functions of  in terms of x and a. Example 2: Compute the integral Z p 1 x2 dx. Solution: This is a very famous integral because it leads to the area formula of the unit circle. We will use a trigonometric substitution. We start with a reference triangle where the p 1 x2 is one of the legs. Using the substitution x = sin , ( 2 <  <  2 ) we will transform the integral into one in . From the triangle, x = sin . Then dx = cos  d. The expression p 1 x2 becomes p 1 x2 = p 1 sin2  = p cos2  = jcos j Because  is in the interval   2 ;  2  , cos x is positive and so jcos xj = cos x. Z p 1 x2 dx = Z cos  (cos  d) = Z cos2  d This is an integral we have already seen; we can simplify it using the double angle formula for cosine. cos 2 = 2 cos2  1 = ) cos2  = 1 2 (cos 2 + 1) Z cos2  d = Z 1 2 (cos 2 + 1) d = 1 2 Z cos 2 + 1 d = 1 2 1 2 sin 2 +   + C = 1 2 1 2 (2 sin  cos ) +   + C = 1 2 sin  cos  + 1 2 + C Now we need to reverse the substitution and write the result as an expression of x: This is where the reference triangle comes handy. sin  = x, cos  = p 1 x2 and  = sin1 x Thus the answer is Z p 1 x2 dx = 1 2 sin  cos  + 1 2 + C = 1 2x p 1 x2 + 1 2 sin1 x + C Note that if we now compute 1 Z 1 p 1 x2 dx the result is the area of the unit semi-circle,  2 . c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 5 Case 3. The substitution x = a sec  where 0 <  <  2 . This is useful in handling an integral involving p x2 a2. The picture below shows the reference triangle we use for this substitution. Using this triangle, we can read (up to sign) the trigonometric functions of  in terms of x and a. Example 3: Compute the integral Z 1 p x2 9 dx. Solution: We will use a trigonometric substitution. We start with a reference triangle where the hypotenuse is x and one shorter side is 3. Using the substitution x = 3 sec , we will transform the integral into one in . From the triangle, x = 3 sec . Then dx = 3 sec  tan  d. The expression p x2 9 becomes 3 tan  - either from the picture or using algebra. Recall the identity sec2 x = tan2 x + 1 p x2 9 = q (3 sec )2 9 = p 9 sec2  9 = p 9 p sec2  1 = 3 p tan2  = 3 jtan j Because 0 <  <  2 , tan  is positive and so jtan j = tan . Z 1 p x2 9 dx = Z 1 3 tan  (3 sec  tan  d) = Z sec  d Again, Z sec  d = ln jsec  + tan j + C Now we need to reverse the substitution and write the result as an expression of x: This is where the reference triangle comes handy. sec  = x 3 and tan  = p x2 9 3 Thus the answer is Z 1 p x2 9 dx = ln jsec  + tan j + C = ln x 3 + p x2 9 3 + C. We can still simplify this result a bit: ln x 3 + p x2 9 3 + C = ln x + p x2 9 3 + C = ln x + p x2 9 ln 3 + C = ln x + p x2 9 + C2 Thus the …nal answer is Z 1 p x2 9 dx = ln x + p x2 9 + C . c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 6 Example 4: Compute the integral Z x2 p 16 x2 dx Solution: Let x = 4 sin  where  2     2 . Then dx = 4 cos  d and p 16 x2 = p 16 16 sin2  = p 16 p 1 sin2  = 4 p cos2  = 4 jcos j Because  2     2 ; cos  is non-negative, and jcos j = cos . So the integral is Z x2 p 16 x2 dx = Z 16 sin2  4 cos  (4 cos  d) = Z 16 sin2  d = 16 Z sin2  d By the double angle formula for cosine, cos 2 = 1 2 sin2  = ) sin2  = 1 2 (1 cos 2) Z x2 p 16 x2 dx = 16 Z sin2  d = 16 Z 1 2 (1 cos 2) d = 8 Z 1 cos 2 d = 8   1 2 sin 2 + C  = 8 4 sin 2 + C Now we need to reverse the substitution and write the result as an expression of x: This is where the reference triangle comes handy. Recall that x = 4 sin  and so  = sin1 x 4  and sin 2 = 2 sin  cos  = 2 sin  p 1 sin2  = 2 x 4  r 1 x 4 2 = x 2 r 1 16 (16 x2) = x 2 1 4  p 16 x2 = 1 8x p 16 x2 And so the …nal answer is Z x2 p 16 x2 dx = 8 4 sin 2 + C = 8 sin1 x 4  1 2x p 16 x2 + C Example 5: Compute the integral Z p x2 + 4 dx Solution: Let x = 2 tan where  2    2 . Then dx = 2 sec2 d and Z p x2 + 4 dx = Z p 4 tan2 + 4 2 sec2 d  = Z 2 p tan2 + 1 2 sec2 d  = 4 Z jsec j sec2 d  = = 4 Z sec sec2 d  = 4 Z sec3 d We will compute Z sec3 d by parts. Let u = sec and dv = sec2 d . Then du = sec tan d and v = Z dv = Z sec2 d In short, u = sec v = tan du = sec tan d dv = sec2 d c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 7 Z u dv = uv Z v du becomes Z sec sec2 d = sec tan Z tan sec tan d Z sec3 d = sec tan Z tan2 sec d recall tan2 + 1 = sec2 Z sec3 d = sec tan Z sec2 1  sec d Z sec3 d = sec tan Z sec3 sec d Z sec3 d = sec tan Z sec3 d + Z sec d 2 Z sec3 d = sec tan + Z sec d 2 Z sec3 d = sec tan + ln jsec + tan j + C Z sec3 d = 1 2 sec tan + 1 2 ln jsec + tan j + C Now the original integral is Z p x2 + 4 dx = 4 Z sec3 d = 4 1 2 sec tan + 1 2 ln jsec + tan j  + C = 2 sec tan + 2 ln jsec + tan j + C Now we need to reverse the substitution and write the result as an expression of x: Recall that x = 2 tan . Then tan = x 2 and sec = p tan2 + 1 = rx 2 2 + 1 = r 1 4x2 + 1 = r 1 4 (x2 + 4) = 1 2 p x2 + 4 and so Z p x2 + 4 dx = 2 sec tan + 2 ln jsec + tan j + C = 2 1 2 p x2 + 4  x 2  + 2 ln 1 2 p x2 + 4 + x 2 + C = 1 2x p x2 + 4 + 2 ln x + p x2 + 4 2 + C = 1 2x p x2 + 4 + 2  ln x + p x2 + 4 ln 2  + C = 1 2x p x2 + 4 + 2 ln x + p x2 + 4 2 ln 2 + C = 1 2x p x2 + 4 + 2 ln x + p x2 + 4 + C c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 8 Example 6: Compute the integral Z x2 p x2 + 9 dx Let x = 3 tan  where  2 <  <  2 . Then dx = 3 sec2 d and Z x2 p x2 + 9 dx = Z 9 tan2  p 9 tan2  + 9 3 sec2  d  = Z 9 tan2  3 p tan2  + 1 3 sec2  d  = Z 9 tan2  3 jsec j 3 sec2  d  = Z 9 tan2  3 sec  3 sec2  d  = 9 Z tan2  sec  d = 9 Z sec2  1  sec  d = 9 Z sec3  sec  d = 9 Z sec3  d 9 Z sec  d We know that Z sec  d = ln jsec  + tan j + C and from the previous computation we have that Z sec3  d = 1 2 sec  tan  + 1 2 ln jsec  + tan j + C. So that the integral is Z x2 p x2 + 9 dx = 9 Z sec3  d 9 Z sec  d = 9 1 2 sec  tan  + 1 2 ln jsec  + tan j  9 ln jsec  + tan j + C = 9 2 sec  tan  + 9 2 ln jsec  + tan j 9 ln jsec  + tan j + C = 9 2 sec  tan  9 2 ln jsec  + tan j + C Now we need to reverse the substitution and write the result as an expression of x: Recall that x = 3 tan . Then tan  = x 3 and sec  = p tan2  + 1 = rx 3 2 + 1 = r 1 9x2 + 1 = r 1 9 (x2 + 9) = 1 3 p x2 + 9 and so Z x2 p x2 + 9 dx = 9 2 sec  tan  9 2 ln jsec  + tan j + C = 9 2 1 3 p x2 + 9  x 3  9 2 ln 1 3 p x2 + 9  + x 3 + C = 1 2x p x2 + 9 9 2 ln x + p x2 + 9 3 + C = 1 2x p x2 + 9 9 2  ln x + p x2 + 9 ln 3  + C = 1 2x p x2 + 9 9 2 ln x + p x2 + 9 + 9 2 ln 3 + C = 1 2x p x2 + 9 9 2 ln x + p x2 + 9 + C Example 7: Z x6 p 1 x14 dx Solution: Let u = x7. Then du = 7x6dx and so dx = du 7x6 . Then the integral becomes Z x6 p 1 u2 du 7x6 = 1 7 Z 1 p 1 u2 du We can either recognize that this is the derivative of sin1 u: 1 7 Z 1 p 1 u2 du = 1 7 sin1 u + C = 1 7 sin1 x7 + C c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016 Lecture Notes Trigonometric Substitutions page 9 If we do not recognize the derivative, then we can use trigonometric substitution  = sin1 u. Then u = sin  and so du = cos d and  is in h  2 ;  2 i . 1 7 Z 1 p 1 u2 du = 1 7 Z 1 p 1 sin2  cos d = 1 7 Z 1 p cos2  cos d = 1 7 Z 1 jcos j cos d Since  is in h  2 ;  2 i , cos  is non-negative and so jcos j = cos  and so 1 7 Z 1 jcos j cos d = 1 7 Z 1 cos  cos d = 1 7 Z 1d = 1 7 + C = 1 7 sin1 x7 + C So the answer is 1 7 sin1 x7 + C . Example 8: 1 Z 0 tan1 x x2 + 1 dx Solution: Let u = tan1 x. Then du = 1 1 + x2 dx. For the limits of the integral, when x = 0; then u = tan1 (0) = 0 and when x = 1; u = tan1 (1) =  4 : So our integral becomes 1 Z 0 tan1 x x2 + 1 dx = =4 Z 0 u du = u2 2 =4 0 = 1 2  4 2 02  = 1 2  2 16 = 2 32 For more documents like this, visit our page at and click on Lecture Notes. E-mail questions or comments to mhidegkuti@ccc.edu. c copyright Hidegkuti, Powell, 2012 Last revised: February 6, 2016
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Math Worksheets Name: ____ Date: _____ … So Much More Online! Please visit: EffortlessMath.com Writing Polynomials in Standard Form  Write each polynomial in standard form. 1) 9𝑥 − 7𝑥 = 2) − 3 + 16 𝑥 − 16 𝑥 = 3) 3𝑥 2 − 5𝑥 3 = 4) 3 + 4𝑥 3 − 3 = 5) 2𝑥 2 + 1𝑥 − 6𝑥 3 = 6) −𝑥 2 + 2𝑥 3 = 7) 2𝑥 + 4𝑥 3 − 2𝑥 2 = 8) −2𝑥 2 + 4𝑥 − 6𝑥 3 = 9) 2𝑥 2 + 2 − 5𝑥 = 10) 12 − 7𝑥 + 9𝑥 4 = 11) 5𝑥 2 + 13 𝑥 − 2𝑥 3 = 12) 10 + 6𝑥 2 − 𝑥 3 = 13) 12 𝑥 2 − 7𝑥 + 9𝑥 3 = 14) 5𝑥 4 − 3𝑥 2 − 2𝑥 3 = 15) −12 + 3𝑥 2 − 6𝑥 4 = 16) 5𝑥 2 − 9𝑥 5 + 8𝑥 3 − 11 = 17) 4𝑥 2 − 2𝑥 5 + 14 − 7𝑥 4 = 18) −𝑥 2 + 2𝑥 − 5𝑥 3 − 4𝑥 = 19) 8𝑥 5 + 11 𝑥 3 − 6𝑥 5 − 8𝑥 2 = 20) 5𝑥 2 − 12 𝑥 4 + 4𝑥 2 + 5𝑥 3 = 21) 7𝑥 3 − 6𝑥 4 − 3𝑥 2 + 22 𝑥 3 = 22) 9𝑥 2 + 𝑥 4 + 12 𝑥 3 − 5𝑥 4 = 23) 3𝑥 (2𝑥 + 5 − 2𝑥 2) = 24) 11 𝑥 (𝑥 5 + 2𝑥 3) = 25) 5𝑥 (3𝑥 2 + 2𝑥 + 1) = 26) 7𝑥 (3 − 𝑥 + 6𝑥 3) = 27) 2𝑥 (3𝑥 2 − 4𝑥 4 + 3) = 28) 6𝑥 (4𝑥 5 + 7𝑥 3 − 2) = 29) 5𝑥 (3𝑥 2 + 2𝑥 3 + 𝑥 ) = 30) 7𝑥 (3𝑥 − 𝑥 2 + 6𝑥 4) = KV Find more at Math Worksheets Name: ____ Date: _____ … So Much More Online! Please visit: EffortlessMath.com Answers Writing Polynomials in Standard Form 1) 2𝑥 2) −3 3) −5𝑥 3 + 3𝑥 3 4) 4𝑥 3 5) −6𝑥 3 + 2𝑥 3 + 𝑥 6) 2𝑥 3 − 𝑥 2 7) 4𝑥 3 − 2𝑥 2 + 2𝑥 8) −6𝑥 3 − 2𝑥 2 + 4𝑥 9) 2𝑥 2 − 5𝑥 + 2 10) 9𝑥 4 − 7𝑥 + 12 11) −2𝑥 3 + 5𝑥 2 + 13 𝑥 12) −𝑥 3 + 6𝑥 2 + 10 13) 9𝑥 3 + 12 𝑥 2 − 7𝑥 14) 5𝑥 4 − 2𝑥 3 − 3𝑥 2 15) −6𝑥 4 + 3𝑥 2 − 12 16) −9𝑥 5 + 8𝑥 3 + 5𝑥 2 − 11 17) −2𝑥 5 − 7𝑥 4 + 4𝑥 2 + 14 18) −5𝑥 3 − 𝑥 2 − 2𝑥 19) 2𝑥 5 + 11 𝑥 3 − 8𝑥 2 20) −12 𝑥 4 + 5𝑥 3 + 9𝑥 2 21) −6𝑥 4 + 29 𝑥 3 − 3𝑥 2 22) −4𝑥 4 + 12 𝑥 3 + 9𝑥 2 23) −6𝑥 3 + 6𝑥 2 + 15 𝑥 24) 11 𝑥 6 + 22 𝑥 4 25) 15 𝑥 3 + 10 𝑥 2 + 5𝑥 26) 42 𝑥 4 − 7𝑥 2 + 21 𝑥 27) −8𝑥 5 + 6𝑥 3 + 6𝑥 28) 24 𝑥 6 + 42 𝑥 4 − 12 𝑥 29) 10 𝑥 4 + 15 𝑥 3 + 5𝑥 2 30) 42 𝑥 5 − 7𝑥 3 + 21 𝑥 2
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Terms and Conditions of Use: this document downloaded from vulcanhammer.info the website about Vulcan Iron Works Inc. and the pile driving equipment it manufactured All of the information, data and computer software (“information”) presented on this web site is for general information only. While every effort will be made to insure its accuracy, this information should not be used or relied on for any specific application without independent, competent professional examination and verification of its accuracy, suit­ ability and applicability by a licensed professional. Anyone making use of this information does so at his or her own risk and assumes any and all liability resulting from such use. The entire risk as to quality or usability of the information contained within is with the reader. In no event will this web page or webmaster be held liable, nor does this web page or its webmaster provide insurance against liability, for any damages including lost profits, lost savings or any other incidental or consequential damages arising from the use or inability to use the information contained within. This site is not an official site of Prentice-Hall, Pile Buck, or Vulcan Foundation Equipment. All references to sources of software, equipment, parts, service or repairs do not constitute an endorsement. Visit our companion site Final Report ESTIMATING DRIVEN PILE CAPACITIES DURING CONSTRUCTION FDOT No.: 99700-3600-119 UF No.: 4910450460312 WPI No.: 0510852 Contract No.: BB-349 Principal Investigator: Michael C. Mc Vay Researchers: Victor Alvarez Limin Zhang Ariel Perez Andrew Gibsen University of Florida Civil and Coastal Engineering PO Box 116580 Gainesville, FL 32611-6580 Submitted to: FDOT August 2002 i TABLE OF CONTENTS Page CHAPTER 1 ................................................................................................................................1 INTRODUCTION ............................................................................................................1 CHAPTER 2 REVIEW OF FLORIDA PILE DRIVING PRACTICES.....................................................5 CURRENT FLORIDA PRACTICE .................................................................................5 Bearing Requirements.................................................................................................5 Blow count criteria................................................................................................5 Practical refusal.....................................................................................................6 Set-checks and pile redrive...................................................................................6 Pile redrive............................................................................................................6 Pile heave..............................................................................................................6 Pile with insufficient bearing................................................................................7 Methods to Determine Pile Capacity..........................................................................7 Wave equation ......................................................................................................7 Bearing formulas...................................................................................................9 Dynamic load tests................................................................................................9 Static load tests .....................................................................................................9 EVALUATION OF FLORIDA PRACTICE CHANGES..............................................10 Bearing Requirements...............................................................................................10 Methods to Determine Pile Capacity........................................................................11 CHAPTER 3 PILE CAPACITY ASSESSMENT USING STATIC AND DYNAMIC METHODS .............13 DAVISSON’S CAPACITY............................................................................................13 DYNAMIC METHODS REVIEW.................................................................................14 Momentum Conservation ENR................................................................................15 Modified Engineering News Record Formula..........................................................15 FDOT........................................................................................................................16 Gates method ......................................................................................................16 Combined Wave Mechanics and Energy Conservation............................................17 Sakai et al. Japanese energy method...................................................................17 Paikowsky’s method ...........................................................................................17 Wave Mechanics.......................................................................................................18 PDA method........................................................................................................18 CAPWAP program .............................................................................................21 ii CHAPTER 4 FLORIDA DEPARTMENT OF TRANSPORTATION PILE DATABASE..........................23 GENERAL INFORMATION AND HISTORY.............................................................23 PILE INFORMATION...................................................................................................25 General......................................................................................................................25 Soil Classification.....................................................................................................25 Driving Information..................................................................................................26 Dynamic Data (CAPWAP and PDA) .......................................................................26 Load Test Results......................................................................................................27 SPT97 Capacity ........................................................................................................27 GATHERING NEW INFORMATION..........................................................................28 Additional Information .............................................................................................28 Criteria for New Entries in Database........................................................................28 CHAPTER 5 ASD AND LRFD CONCEPTS...............................................................................29 ALLOWABLE STRESS DESIGN (ASD) METHOD...................................................29 LOAD RESISTANCE FACTOR DESIGN (LRFD) METHOD....................................30 Advantages of LRFD Over ASD..............................................................................30 Limitation of LRFD..................................................................................................31 CALIBRATION OF LRFD............................................................................................31 Engineering Judgement.............................................................................................31 Fitting ASD to LRFD................................................................................................31 Reliability Calibration...............................................................................................33 Statistical data.....................................................................................................33 Probability density function................................................................................34 LRFD Approach........................................................................................................35 Probability of failure...........................................................................................35 Reliability index..................................................................................................37 Resistance factor, φ.............................................................................................39 Definition φ/λR and its importance.....................................................................41 CHAPTER 6 LRFD ASSESSMENT ...........................................................................................43 DATA REDUCTION .....................................................................................................43 LRFD ANALYSIS OF RESULTS.................................................................................48 Effect of Bridge Span Length and Probability of Failure.........................................49 Level of Conservatism and Accuracy Indicators......................................................52 φ/λR Ratio.................................................................................................................55 Comparison of φ/λR for Each Method......................................................................55 iii EOD versus BOR................................................................................................56 Evaluation of smaller piles..................................................................................57 Recommended Safety Factors...................................................................................57 Review of Typical ASD Safety Factors....................................................................59 CHAPTER 7 DETERMINATION OF CASE DAMPING, Jc , AND SKIN AND TIP RESISTANCE FROM PDA TRACES.........................................................................61 METHOD I FOR ESTIMATING SKIN AND TIP RESISTANCE...............................61 METHOD II FOR ESTIMATING SKIN AND TIP RESISTANCE .............................63 Description and Main Assumptions..........................................................................63 Case Damping Coefficient, Jc , versus Tip to Skin Ratio..........................................66 Automation of Proposed Method II..........................................................................68 SENSITIVITY ANALYSIS OF CASE DAMPING COEFFICIENT, Jc ......................69 SELECTION OF STATIC AND DYNAMIC LOAD TEST DATA.............................70 COMPARISON OF SKIN, TIP AND TOTAL PILE CAPACITIES ............................70 Results.......................................................................................................................70 Total Capacity Predictions........................................................................................72 Skin and Tip Capacity Predictions............................................................................77 Sensitivity of Jc and Pile Resistance .........................................................................82 CHAPTER 8 DEVELOPMENT OF NEW FIELD INSTRUMENTATION..............................................85 CHAPTER 9 INSTRUMENTATION REQUIREMENTS FOR DYNAMIC TESTING EQUIPMENT...........89 TRANSDUCERS............................................................................................................89 SIGNAL CONDITIONING............................................................................................90 SIGNAL TRANSMISSION ...........................................................................................90 APPARATUS FOR RECORDING, REDUCING AND DISPLAYING DATA...........90 DATA QUALITY CHECK............................................................................................91 PILE DRIVING ANALYZER (PDA) EQUIPMENT....................................................91 SUMMARY OF REQUIREMENTS AND SUGGESTIONS........................................94 iv CHAPTER 10 SIGNAL CHARACTERISTICS................................................................................97 FORCE AND VELOCITY TRACES.............................................................................97 STRAIN AND ACCELERATION TRACES ................................................................99 ACCURACY OF STRAIN AND ACCELERATION MEASUREMENTS................102 FREQUENCY OF STRAIN AND ACCELERATION TRACES ...............................105 ACCELEROMETER AND STRAIN GAGE OUTPUT TRACES IN TERMS OF VOLTAGE .........................................................................................106 CHAPTER 11 THE NEW WIRELESS ANALYSIS PACKAGE (WAP) ..........................................110 NON-RECOVERABLE UNIT.....................................................................................111 Strain Gages............................................................................................................111 Accelerometers .......................................................................................................113 Signal Conditioning and Sending Unit ...................................................................115 Batteries ............................................................................................................117 Power control board..........................................................................................118 Strain gage conditioners....................................................................................118 Accelerometer conditioners ..............................................................................119 Digitalization and data encoding ......................................................................120 Transmitter........................................................................................................121 Transmitter Antenna ...............................................................................................122 RECEIVER AND DATA PROCESSING UNIT .........................................................122 Receiver Antenna....................................................................................................123 Signal Receiver and Conditioning Unit..................................................................124 Digital receiver..................................................................................................125 Signal conditioning at the receiver ...................................................................126 Data Acquisition Card.............................................................................................126 Laptop Computer ....................................................................................................128 CALIBRATION ...........................................................................................................129 MOUNTING OF THE NON-RECOVERABLE UNIT ...............................................135 CHAPTER 12 ACQUISITION AND DATA PROCESSING SOFTWARE ..........................................137 LAB OVERVIEW .........................................................................................................137 FRONT PANEL DESCRIPTION AND OPERATION ...............................................138 BLOCK DIAGRAMS...................................................................................................145 v CHAPTER 13 LABORATORY AND SMALL SCALE FIELD TESTS ..............................................155 MINIATURE PILE.......................................................................................................155 STRAIN GAGE AND CONDITIONERS....................................................................157 ACCELEROMETER, CONDITIONERS AND FILTERS..........................................160 SMALL SCALE DRIVING TEST...............................................................................165 CHAPTER 14 FULL SCALE TESTS AT CYPRESS CREEK ..........................................................172 SR-54 SITE DESCRIPTION........................................................................................172 PILE INSTRUMENTATION.......................................................................................175 DRIVING TEST...........................................................................................................180 Pile #3 at End Bent #1 ............................................................................................183 Pile #1 at End Bent #1 ............................................................................................188 CHAPTER 15 CONCLUSIONS AND RECOMMENDATIONS ........................................................195 LRFD CALIBRATION FOR EIGHT DYNAMIC METHODS..................................195 Recommendations...................................................................................................196 Determination of case damping coefficient, Jc , and skin & tip resistance from measured stress wave propagation in a pile .............................................197 WIRELESS ACQUISITION OF A PILE’S (WAP) DYNAMIC RESPONSE............198 Recommendations...................................................................................................200 REFERENCES .........................................................................................................................201 APPENDICES A - J vi LIST OF FIGURES FIGURES Page 3-1 Construction of Davisson’s Pile Capacity..................................................................14 3-2 Typical Force and Velocity Traces from PDA...........................................................21 5-1 Lognormal Probabililty Density Function..................................................................34 5-2 Probability Density Functions for Normally Distributed Load and Resistance.........36 5-3 Definition of Reliability Index, β for Lognormal Distributions of R and Q..............37 5-4 Reliability Definition Based on Standard Normal Probability Density Function......39 5-5 Comparison of Esteva and Withiam Methods to Obtain Reliability Index, β ...........40 6-1 Davisson Capacity vs. PDA BOR Capacity...............................................................44 6-2 Measured Over Predicted Capacity for PDA at BOR ................................................44 6-3 Lognormal Probability Distribution at EOD..............................................................46 6-4 Lognormal Probability Distribution at BOR..............................................................47 7-1 GRL Procedure to Determine the Total Skin Capacity of a Driven Pile ...................62 7-2 Typical Force, Velocity and Double Wave Up Traces from PDA.............................64 7-3 Procedure to Determine Tip to Skin Ratio for the Suggested Method.......................66 7-4 Suggested Tip to Skin Ratio versus Case Damping Coefficient................................67 7-5 Davisson Capacity vs. Suggested Method II Capacity...............................................75 7-6 Davisson Capacity vs. CAPWAP Capacity ...............................................................75 7-7 Davisson Capacity vs. PDA Rmax Capacity..............................................................76 7-8 Davisson Capacity vs. Suggested Method Capacity ..................................................78 7-9 Davisson Capacity vs. CAPWAP Capacity ...............................................................79 vii 7-10 Davisson Capacity vs. Method I Skin Capacity.........................................................79 7-11 Davisson Capacity vs. Suggested Method II..............................................................80 7-12 Davisson Capacity vs. CAPWAP Capacity ...............................................................80 7-13 Davisson Capacity vs. Method I Capacity .................................................................81 7-14 Case Damping Coefficient Jc Sensitivity Analysis Results.......................................84 8-1 System Concept..........................................................................................................86 8-2 System Operation .......................................................................................................87 8-3 Stress Wave Acquisition Device ................................................................................88 9-1 PDA Strain Gages and Accelerometers......................................................................89 9-2 CAPWAP Format.......................................................................................................94 10-1 PDA Force Traces of 22 Piles in PILEUF .................................................................98 10-2 PDA Velocity (ZVel) Traces for 22 Piles of PILEUF .............................................98 10-3 General Form of Strain and Acceleration Traces.....................................................100 10-4 PDA Strain Traces for 22 Piles of PILEUF .............................................................101 10-5 PDA Acceleration Traces for 22 Piles of PILEUF...................................................102 10-6 Strain Increase for 22 Piles in Database...................................................................103 10-7 Acceleration Increase for 22 Piles in Database........................................................104 10-8 FFT Strain for 22 Piles in Database .........................................................................105 10-9 FFT Acceleration for 22 Piles in the Database.........................................................106 10-10 PDA Strain Gage Output for 22 Piles in Database...................................................108 10-11 PDA Piezoelectric Accelerometer Output for 22 Piles in Database ........................108 10-12 PDA Piezoresistive Accelerometer Output for 22 Piles in Database.......................109 11-1 New Instrumentation System ...................................................................................110 viii 11-2 Special Purpose Embedment Strain Gage EGP-5-120.............................................112 11-3 Endevco Accelerometer ...........................................................................................114 11-4 (a) Accelerometer Mounting Technique; (b) Protective Enclosure .........................115 11-5 Signal Conditioning and Sending Unit.....................................................................116 11-6 Strain Gage Conditioner...........................................................................................119 11-7 Accelerometer Conditioner ......................................................................................120 11-8 Digital Data String....................................................................................................121 11-9 Digital Transmitter T900V, from Applied Wireless ................................................122 11-10 Receiver Antennas....................................................................................................123 11-11 Signal Receiver and Conditioning Unit....................................................................124 11-12 DAQCard 6533 Data Acquisition Card (Digital).....................................................127 11-13 DAQP-12 Data Acquisition Card (Analog) .............................................................127 11-14 Calibration Equipment .............................................................................................129 11-15 Calibrating the Strain Gauge Equipment..................................................................131 11-16 Calibrating the Accelerometer Channel ...................................................................133 11-17 Instrumentation Placement.......................................................................................135 11-18 Signal Conditioning and Sending Unit Placement in the Concrete Forms ..............136 12-1 Pile-Monitoring.vi Front Panel.................................................................................139 12-2 Input-Data Front Panel .............................................................................................139 12-3 Pile-Monitoring.vi Front Panel.................................................................................141 12-4 View Raw Data Window..........................................................................................143 12-5 Code Used to Select Directory and File Paths .........................................................145 12-6 Code Used to Input Data ..........................................................................................146 ix 12-7 Input Data Sub VI Code ...........................................................................................146 12-8 Code Used for Run Mode.........................................................................................147 12-9 Code Used for Standby Mode ..................................................................................147 12-10 Strain and Acceleration Sub VI Code ......................................................................148 12-11 Integration Sub VI Code...........................................................................................148 12-12 Capacity Sub VI Code..............................................................................................149 12-13 Capacity Graph Sub VI Code...................................................................................150 12-14 FFT Filter Sub VI Code............................................................................................150 12-15 Save in CAPWAP Format Sub VI Code..................................................................151 13-1 Miniature Pile...........................................................................................................155 13-2 Installation of Non-Recoverable Unit and Wired Sensors .......................................156 13-3 Output Linearity Test of Strain Gage and Conditioner ............................................158 13-4 Comparison Between PDA and Transducers Embedded in the Miniature Pile.......159 13-5 Force Traces Obtained with the Proposed Equipment and PDA, and FFT..............160 13-6 Accelerometer Output Comparison Between Proposed Equipment and PDA in Terms of Acceleration, Frequency Spectrum and Velocity Trace ...........................162 13-7 Accelerometer Output Comparison Between Proposed Equipment and PDA in Terms of Acceleration, Frequency Spectrum and Velocity Trace ...........................162 13-8 Accelerometer Installation at the Pile Tip................................................................163 13-9 Accelerometer Output Comparison Between Proposed Equipment and DC Accelerometer in Terms of Acceleration, Frequency Spectrum and Velocity Trace.........................................................................................................................164 13-10 Cone Penetration Test in Green Cove Springs. 30 m (100 ft) Away from the Driving Location ......................................................................................................165 13-11 Shows the Pile in the Leads and the Receiver and Data Processing Unit Ready to Start Driving.........................................................................................................166 x 13-12 PDA Transducers Attached to the Pile.....................................................................167 13-13 Pile-Monitoring.vi Front Panel for Blow #7 During the Driving Test in Green Cove Springs ............................................................................................................168 13-14 Typical Force and Velocity Traces from the PDA During the Driving Test in Green Cove Springs..................................................................................................169 14-1 Job Site .....................................................................................................................173 14-2 Bridge Elevation.......................................................................................................173 14-3 Pile Distribution in Two Abutments and One Center Pier.......................................174 14-4 SPT Boring on the West Side of the Creek .............................................................174 14-5 Instrumentation Mounting Technique .....................................................................175 14-6 Instrumentation of Pile #3 (Transmitter Unit #4).....................................................176 14-7 Instrumentation of Pile #1 (Transmitter Unit #5).....................................................177 14-8 Access Points............................................................................................................178 14-9 Testing the Instrumentation Package on the Casting Yard by Striking the Piles with a Hand Held Hammer.......................................................................................179 14-10 Driving Equipment and Setup..................................................................................180 14-11 Pile Setup Preparation ..............................................................................................181 14-12 Set Up.......................................................................................................................182 14-13 WAP Records for the Last 500 Blows on Pile #3 ....................................................184 14-14 Traces versus PDA Traces for Blow 1345 (last Blow Pile #3)................................185 14-15 Traces versus PDA Traces for Blow 1340 (Pile #3) ................................................185 14-16 Traces versus PDA Traces for Blow 1335 (Pile #3) ................................................186 14-17 Comparison of Maximum Compression Stress between WAP and PDA Records at the Pile Top ............................................................................................187 14-18 Determination of Tension Stresses in the Pile..........................................................188 xi 14-19 Comparison of Maximum Tension Stress Between WAP and PDA .......................189 14-20 WAP Records for the Last 500 Blows on Pile #1 ....................................................189 14-21 WAP Traces versus PDA Traces for Blow 909 (last Blow Pile #1) ........................190 14-22 WAP Traces versus PDA Traces for Blow 908 (Pile #1) ........................................191 14-23 WAP Traces versus PDA Traces for Blow 905 (Pile #1) ........................................191 14-24 Comparison of Maximum Compression Stress Between WAP and PDA at the Pile Top and WAP at Pile Bottom............................................................................192 14-25 Comparison of Maximum Tension Stress Between WAP and PDA at the Pile Top and WAP at Pile Bottom............................................................................193 14-26 Comparison of Dynamic Capacity Assessment WAP and PDA..............................193 xii LIST OF TABLES TABLES Page 4-1 Classification of Florida Driven Piles.......................................................................24 4-2 Classification of Non-Florida Driven Piles...............................................................24 6-1 φ Values Evaluated for PDA (EOD).........................................................................50 6-2 ASD Design Safety Factor Values Evaluated for PDA (EOD) ................................50 6-3 φ Factors for Safety Factor = 2.50 – PDA (EOD) ....................................................50 6-4 φ Values Evaluated for PDA (BOR).........................................................................51 6-5 ASD Design Safety Factor Values Evaluated for PDA (BOR) ................................51 6-6 φ Factors for Safety Factor = 2.50 – PDA (BOR) ....................................................51 6-7 Load and Resistance Factor Design (LRFD) of Driven Piles Using Dynamic Methods at EOD .......................................................................................................53 6-8 Load Resistance Factor Design (LRFD) of Driven Piles Using Dynamic Methods at BOR .......................................................................................................54 6-9 Recommended Safety Factors for Dynamic Methods ..............................................58 6-10 Summary of Results for Fitting the ASD to the LRFD ............................................60 7-1 Geometric and Driving Information of Load Test Piles...........................................71 7-2 Suggested Method II Results and Static Load Test Results .....................................73 7-3 CAPWAP and Method I Results ..............................................................................74 7-4 Comparison of Mean, Standard Deviation, and COVR.............................................76 7-5 Statistical Analysis for Skin and Tip Predictions .....................................................78 7-6 Sensitivity Analysis Results for Variation of Case Damping Coefficient and Pile Resistance ..........................................................................................................83 9-1 PDA Sensors Specifications .....................................................................................92 xiii 9-2 Summary of Requirements and Suggestions ............................................................95 11-1 Strain Gage Specifications, Requirements, and Recommendations .......................112 11-2 Accelerometer Specifications, Requirements, and Recommendations...................114 11-3 Signal Conditioning and Sending Unit Specifications, Requirements, and Recommendations...................................................................................................117 13-1 Driving Records......................................................................................................167 1 CHAPTER 1 INTRODUCTION Dynamic testing has been used for estimating pile capacities and hammer suitability since 1888 when the first driving formula, i.e., the Engineering News formula, was published. Up to the early seventies, most if not all-driving formulas adopted into codes were derived from the principles of impulse-momentum conservation. In the late sixties, research focused on predicting both stresses and pile capacities based on wave mechanics. The results were the creation of programs such as WEAP (GRL, 1993), PDA (Pile Dynamics Inc., 1992), and CAPWAP (GRL, 1996). More recently, energy approaches based on both wave mechanics and energy conserva-tion (Paikowsky, 1992) have been developed to determine the pile capacity. However, until recently the accuracy of the older versus the newer methods was unknown, especially for Florida soils conditions. Recently, the American Association of State Highway and Transportation Officials (AASHTO) and FDOT has changed from the Allowable Stress Design (ASD) to the newer Load Resistance Factor Design (LRFD). The latter method allows the resistance factor, φ, to be com-puted in terms of the probability of failure (i.e., reliability index). In order to determine accurate resistance factors (φ), a database to assess mean and standard deviation of measured versus predicted is required. Once the resistance factor, φ, and the bias (ratio of measured to predicted) for a given method is known, then its percentage of Davisson capacity available for design may be determined. The latter may be subsequently used to compare all of the current and past prediction methods based on a predetermined risk (i.e., failure probability). Based on the FDOT (at University of Florida) pile database (242 piles), AASHTO’s recommended reliability index, and live to dead load ratios, the resistance factors (LRFD) and 2 equivalent safety factors (ASD) were developed for many of the current and past dynamic pile prediction methods. The latter included four stress wave and energy approaches (CAPWAP, PDA, Paikowsky Energy, and Sakai Energy) as well as four driving formulas (ENR, modified ENR, FDOT, and Gates). In the case of the older driving formulas, the database was broken into both small (i.e., Davisson capacity less than 1779 kN [200 tons]) and large (Davisson capacity larger than 1779 kN) capacity piles. It was found that on the whole, the newer methods were more accurate and higher percentage of Davisson’s capacity is available for design; however, for smaller piles and lower capacities (i.e., capacities less than 1779 kN [200 tons]), a few of the older methods (i.e., Gates) did quite well. Since 1994, the Florida Department of Transportation (FDOT) construction specification (A455) recommends the use of a Wave Equation approach to assess the suitability of driving systems, as well as estimate pile capacity. In the case of pile capacities, a two step approach is generally employed. First the PDA (Pile Driving Analyzer) collects data (strain and accelera-tion) and performs a field estimate of pile capacity based on a user supplied Case Damping Coefficient, Jc . The latter is usually selected on the site based on soil type. Subsequently, CAPWAP (Case Pile Wave Analysis Program) is performed in the office on the recovered data to assess distribution of skin and tip resistance, as well as determine an improved capacity pre-diction (i.e., over PDA). CAPWAP will also estimate a Jc value, which may be used in PDA pile capacity assessment. However, a review of the FDOT database, revealed that over seventy percent of the PDA pile capacity predictions was computed from Jc values estimated from soil type, not CAPWAP. It was also determined that a thirty percent change in Jc resulted in only a ten percent change in estimated pile capacity. 3 Since the CAPWAP program requires very experienced users and may not result in a unique solution (total capacity, skin and tip resistance), this research focused on estimating Jc , and skin and tip resistance real time (in the field). It was found that the static tip to skin resistance ratio was approximately equal to the measured PDA dynamic value, and that the Jc value was proportional as well. The latter allowed for a field assessment of total pile capacity and corresponding breakdown of skin and tip resistance. The latter was compared with measured (FDOT database) field values, as well as CAPWAP predictions. The proposed method was found as accurate as CAPWAP and allows for immediate assessment of pile capacity and its distribution automatically (no user input required) in the field. The focus of the research mainly consisted of improving the field instrumentation. A number of different technologies were investigated: laser, optical, and radio. Given the economical constraints, location of the needed information (i.e., pile tip), the radio option (wireless) was pursued. The effort started from initially transmitting an analogue signal from embedded strain gauges and accelerometers cast in the pile. The latter had significant noise interference, resulting in very poor signal recovery. Next, a frequency approach was tried. However, due to limited bandwidth of the transmitters, the approach resulted in a limited the number of channels, which could be broadcasted. Finally, multiple analogue (i.e., multiple gauges) signals were converted to a single digital signal which was transmitted through one transmitter (wireless) which was picked up by a receiver and decoded (recover multiple chan-nels). Also, due to cost constraints (gauges, transmitters, etc. were not reusable, i.e., lost with pile), a new accelerometer was required. Using new technology, a piezoelectric accelerometer was developed for this application with an estimated mass production cost of thirty dollars. 4 For each of the three generations of instrumentation development, model piles were con-structed to test the instrumentation. In the case of the final system, i.e., digital, two full-scale twenty-four inch piles were tested. These piles were production piles in SR-54 (north of Tampa) new bridge over Cypress Creek. One of the piles was instrumented on the top and the other was instrumented on the top and bottom in the casting yard (SCG). During the driving, the new wireless system and PDA system were employed. The wireless analysis of a pile (WAP) employed a standard laptop computer, labview software running a vi (virtual interface--i.e., program), and a receiver/decoder, along with the cast insitu transmitter and gauges. The results (accelerations, strains) at the top of the piles compared very favorably between the PDA and WAP systems. Also, the WAP system gave a real-time assessment of pile capacity (skin, tip and total). The report is arranged as follows. First, a discussion of LRFD, phi factors, and ASD safety factors along with pile capacity assessment methods is presented. Next, a comparison of the accuracy of pile capacity methods is presented. Then the development of tip, skin and total capacity from PDA dynamic measurements is given. Finally, the new wireless instrumentation package is discussed, along with field trials. The report recommends the further development of the new wireless technology for the installation of pile foundations. 5 CHAPTER 2 REVIEW OF FLORIDA PILE DRIVING PRACTICES In this chapter, a review of pile driving practices in the state of Florida will be presented. Because the Florida Department of Transportation (FDOT) uses a large percentage of driven piles compared to the private industry, the information presented herein is based on the former recommendations. Both the current driving practice together with the most relevant changes through the years are presented. This discussion is focussed on such aspects as bearing require-ments, and methods to determine pile capacity. CURRENT FLORIDA PRACTICE The information presented is in relation to the current Florida practice, which was obtained from the FDOT Standard Specifications for Road and Bridge Construction of 1999. For more details, the reader is referred to the latest FDOT specifications. Bearing Requirements As a general criterion the engineer in charge of the driving process may accept a driven pile if it has achieved the minimum penetration, the blow count has a tendency to increase and the minimum bearing capacity is obtained for 2 ft (600 mm) of consecutive driving. The engi-neer may also accept a driven pile if the minimum penetration was reached and the driving has achieved practical refusal in firm strata. Aspects such as practical refusal and others driving criteria will be discussed in detail in the following sections. Blow count criteria. Using the Wave Equation Analysis for Piles (WEAP) the engineer can determine the number of blows per specific penetration to reach a design pile capacity. The blow count has to be averaged for every 12 inches (250 mm) of pile penetration or through the last 10 to 20 blows 6 of the hammer. It should be noted that the driving equipment must be selected in order to pro-vide the required resistance at a blow count ranging from 36 blows per foot (36 blows for 300 mm) to 120 blows per foot (120 blows per 300 mm). Practical refusal. Practical refusal is defined as a blow count of 20 blows per inch (20 blows per 25 mm) for 2 inches (50 mm) of driving. The FDOT specifications recommend that driving cease after driving to practical refusal conditions for 12 inches (300 mm). If the required penetration can not be achieved by driving without exceeding practical refusal, other alternatives should be con-sidered such as jetting or Preformed Pile Holes. Set-checks and pile redrive. Set-checks - Set checks are performed in the event that the Contractor has driven the pile up to the point that the pile top elevation is within 300 mm of the cut-off elevation and the pile has not reached the required resistance. Prior to a set check, the driving process is interrupted for 15 minutes. Then, the engineer is provided with a level or other suitable equipment to determine ele-vation in such a way that the pile penetration during the set-checks could be determine in a very accurate manner. If the initial set-check results are not satisfactory, additional set-checks could be performed. The pile is then accepted if the pile has achieved the minimum required pile bearing. Pile redrive. Pile redrive consists of redriving the pile after 72 hours from original driving. The pile redrive is considered when time effect is important in the pile capacity. Other considerations include the pile heave. Pile heave. Pile heave is defined as the upward movement of a pile from its originally driven eleva-tion. In occasions, driving a pile can cause excessive heave and/or lateral displacement of the 7 ground. The previously driven pile should be monitored, and in the event of pile heave (6 mm or more), all piles must be redriven unless the engineer has determined that the heave is not detri-mental to the pile capacity. Piles with insufficient bearing. In the event that the pile top has reached the cut-off elevation without achieving the required bearing resistance, the FDOT specifications recommends: 1. Splice the pile and continue driving. 2. Extract the pile and drive a pile of greater length. 3. Drive additional piles until reducing the adjusted required bearing per pile to the bearing capacity of the piles already driven. Methods to Determine Pile Capacity The FDOT Specifications recommend the use of Wave Equation to determine pile capacity for all structures or projects. The use of static load tests or dynamic load tests, or both, is recommended to verify the capacity estimated from Wave Equation predictions. Nevertheless, the prediction by the Wave Equation (blow count criteria) could be adjusted to match the resistance determined from the static or dynamic load tests, or both. Wave equation. The FDOT Specifications recommends to use the WEAP program to predict the pile capacity. This program allows the engineer to evaluate other aspects of the driving process. In the following paragraphs, a description of these aspects will be presented. Evaluation of driving system. Evaluate the suitability of the driving system (including hammer, follower, capblock and pile cushions. The driving system must be capable of driving the pile to a resistance of 3.0 times the design load, plus the scour and down drag resistance or the ultimate resistance, whichever is higher. 8 Determine pile driving resistance. The pile driving resistance, in blows per 300 mm or blows per 25 mm are to be determined. The required driving resistance is defined as the design load multiplied by the appropriate factor of safety plus the scour and down drag resistance or the ultimate bearing capacity, whichever is higher. Evaluate pile driving stresses. The engineer must evaluate the driving system to avoid overstressing the pile at any moment during the driving. If the Wave Equation analyses show that the hammer will overstress the pile, the driving system has to be rejected. The FDOT Specifications 455-5.11.2 presents the allowable stresses for piles made out of concrete, steel and timber. Equation 2-1, 2-2, and 2-3 give the maximum allowable tensile and compression stresses for prestressed concrete piles. The allowable compressive stress is, 0.7 0.75 = − 1 apc c pc s f f (2-1) For piles length less than 15 meters the allowable tensile stress is given by ( ) pc c apt f f s 05 . 1 54 . 0 5 . 0 1 + = (2-2a) For piles length less than 50 m, ( ) 0.5 6.5 1.05 1 apt c pc s = f + f (2-2b) And for piles length greater than 15 meters ( ) pc c apt f f s 05 . 1 27 . 0 5 . 0 1 + = (2-3a) For piles length greater than 50 feet, ( ) 0.5 3.25 1.05 = + 1 apt c pc s f f (2-3b) where Sapc Maximum Allowable Pile Compressive Stress, MPa/psi Sapt Maximum Allowable Pile Tensile Stress, MPa/psi 9 fc 1 Specified Minimum Compressive Strength of Concrete, MPa/psi fpc Effective Prestresses at Time of Driving. For steel piles the maximum allowable compression and tensile stresses are equal to ninety percent (90 %) of the yield strength (0.9 fy) of the steel. Dynamic load tests. Dynamic load testing consists of predicting pile capacity from blows of the hammers during drive and/or redrive of an instrumented pile. Chapter 3 includes more details of how the dynamic load test is performed (see PDA and CAPWAP sections). Static load tests. Static load testing consists of applying a static load to the pile to determine its capacity. The FDOT recommends the Modified Quick Test. For more details about the static load test, the reader is referred to the FDOT specification 455-2.2.1. Some general information about this test, and the procedure to obtain the pile capacity are explained in Chapter 3. EVALUATION OF FLORIDA PRACTICE CHANGES In the following sections the most relevant changes in the Florida practice (i.e., bearing requirements and proposed methods to determine pile capacity) for the last 10 years approxi-mately will be discussed. For this purpose, the actual practice will serve as a reference for any comparison. To facilitate the comparison process, only the changed criteria will be discussed. The latter does not mean that the aspects not mentioned within this document did not vary (i.e., only the topics related to this report will be investigated). Because the largest change in FDOT specifications in regards to pile foundation were found in the 1994 version versus 1991 version specifications, the discussion will be based on these two references. To simplify the comparison, 10 the FDOT specifications of 1991 and prior to 1991 will be called “old specifications” and any other specification after 1991 will be called the “new specifications”. Bearing Requirements In general, there was a great change in the FDOT specifications of 1994 in comparison to the older FDOT specifications. In the old specifications the piles were allowed to be driven to grade. Even if the practical resistance had not been reached at that point, the engineer was able to drive the pile below grade and build up. After driving 12 inches (0.305 m) below grade, a set-check could be performed after 12 hour of initial driving. The latter criterion differs from the new practice in the elevation at which the set-check is recommended. The new practice recom-mends the set-check to be performed at approximately 10 inches above the cut-off elevation. Another important difference is related to the bearing formulas. In the old specifications, the FDOT recommend the use of bearing formulas to determine the pile bearing capacity for piles made out of timber, concrete, composite concrete-steel and steel. Then, from 1994 to date the specifications limited the use of bearing formulas to timber piles driven with power hammers only. Methods to Determine Pile Capacity It was noted that in the older specifications no requirement for the use of Wave Equation programs to determine the pile capacity was given. The same observation applies to the use of dynamic testing as a method to determine the pile capacity. Prior to 1994, the FDOT recom-mended the use of static load test to determine the pile capacity of any pile that did not reach the required resistance at the end of drive or as directed by the engineer. The new specifications recommend the use of Pile Driving Analyzer (PDA), the Wave Equation Analysis for Pile 11 (WEAP), and the static load test separately or in a combination of each, as recommended by the engineer (the safety factor for design depends upon type of test performed). Other difference were noted in the criterion for determining pile capacity from the static load test. In the old specifications, the failure criteria is given by either or both conditions shown below: 1. One and one-half times the yield load settlement develops. The yield load is defined as that load beyond which the total additional settlement exceeds 0.03 inch per ton, for the last increment applied. 2. The total permanent settlement of the top of the pile is greater than ¼ of an inch. The new specifications present two criteria to determine the static pile capacity. Those criteria are: 1. Davisson – for shafts with diameter up to 24″ (600 mm), the load that causes a shaft top deflection equal to the calculated elastic compression, plus 4 mm, plus 1/120 of the shaft diameter in millimeters. 2. FHWA – for shafts with diameter larger than 24″ (600 mm), the load that causes a shaft top deflection equal to the calculated elastic compression, plus 1/30 of the shaft diameter. The changes in criteria for selecting the failure load reflect, first, an increase in the use of larger piles in the construction field, and second, the FDOT recognizes that for larger piles (diameter larger than 24″ (600 mm)) the capacity according to Davisson’s criterion is conservative. As a general observation, the FDOT has abandoned the old methods to determine the bearing capacity of piles (i.e., bearing formulas, based on momentum conservation). At the same 12 time, the FDOT has adopted other prediction methods such as Wave Equation, PDA, and CAPWAP, which are based on wave propagation through the pile to estimate static pile capacity. Other old methods such as Gates, ENR, and Modified ENR are not considered as alternates in estimating the pile capacity, neither are the relatively new methods such as Paikowsky’s method and Sakai et. al. method. It was proposed by the FDOT to investigate the new FDOT specifications in relation to the older methods based on momentum conservation (i.e., FDOT, Gates, ENR, Modified ENR). Another important consideration was to evaluate the older methods for large capacity piles, which are prevalent today, separately from small capacity piles (i.e., piles with capacity up to 2000 kN approximately). The latter reflects the magnitude of design loads for which piles were designed in the past in comparison to the present practice. 13 CHAPTER 3 PILE CAPACITY ASSESSMENT USING STATIC AND DYNAMIC METHODS The Florida Department of Transportation (FDOT) under contract No. BB-349 required UF to evaluate the older empirical methods for determining pile capacity and compare them to the modern instrumented methods. In order to perform the latter, the Davisson’s capacity served as the measured capacity for each case. In the following sections, a brief description of the Davisson criterion together with the description of the dynamic methods investigated will be presented. DAVISSON’S CAPACITY The Davisson method (Davisson, 1972) is one of many methods developed to determine the pile capacity based on a static load test results. Davisson defined the pile capacity as the load corresponding to the movement which exceeds the elastic compression of the pile by a value of 4-mm (0.15 inches) plus a factor equal to the diameter of the pile in millimeter divided by 120. Figure 3-1 presents the load-displacement curve resulting from a static load test. From this curve, the Davisson’s pile capacity can be obtained. The steps to obtain the Davisson’s capacity are as follow: Plot a line with slope representing the elastic deformation of the pile (Slope m): L AE m = (3-1) where: A Cross-sectional Area of the Pile E Elastic Modulus of the Pile Material L Pile Length 14 Figure 3-1. Construction of Davisson’s Pile Capacity Draw a line parallel to the elastic deformation line with an intercept, x, on the settlement (movement) axis given as 120 0 . 4 D x + = (3-2) where: D Diameter of Pile in millimeters x horizontal displacement of elastic deformation line in millimeters The Davisson’s capacity (point D on Figure 3-1) is defined as the intersection point between the load-settlement curve and the elastic deformation line. DYNAMIC METHODS REVIEW Eight methods were considered in this study, which are subdivided in three categories: momentum conservation, combined wave mechanics with energy conservation, and wave mechanics alone. The methods are the Engineering News Record (ENR), Modified ENR, 15 FDOT, Gates, Paikowsky, Sakai (Japanese), Pile Driving Analyzer (PDA) and the Case Pile Wave Analysis Program (CAPWAP). In the following sections, a brief description of each method is presented. Momentum Conservation ENR One of the older formulas developed to estimate the driven pile capacity was the formula published in the Engineering News Record (ENR) (Coduto after Wellington, 1994). It has since become known as the Engineering News Record formula: ) 1 . 0 ( + = s F h W P r a (3-3) where: Pa Allowable Pile Load Wr Hammer Ram Weight h Hammer Stroke (the distance the hammer falls) F Factor of Safety s Pile Set (penetration) per Blow in Inches Wellington (1888) recommended using a Safety Factor of 6.0. Modified Engineering News Record Formula In 1961, the Michigan Highway Department (Housel, 1966) performed a series of pile driving tests with the objective of evaluating the accuracy of the ENR formula. After evaluating 88 piles, the investigators found that the ENR formula overpredicted the pile capacities by a factor of 2 to 6. The findings mean that piles designed with a SF of 6 will have a real factor of safety of 1 and 3. Based on their results, the Michigan Highway Department developed the Modified Engineering News Formula: 16 ) )( 1 . 0 ( ) ( 0025 . 0 2 P r P r a W W s W e W E P + + + = (3-4) where: Pa Allowable Pile Load (kips) E Rated Hammer Energy Per Blow (ft-lb) WP Weight of Pile plus Driving Appurtenances (lb) Wr Weight of Hammer Ram (lb) s Pile Set (in/blow) e Coefficient of Restitution FDOT The Florida Department of Transportation under specification 455-3.3 (1991) recom-mends the following bearing formula (FDOT, 1991). P S E R 01 . 0 1 . 0 2 + + = (3-5) where: R Safe Bearing Value in Tons P Weight of Pile as Driven, in Tons S Average Penetration per Blow, in Inches E Energy per Blow of Hammer, in Foot-Tons The last formula was used for concrete piles, composite concrete-steel piles and steel piles. The bearing capacity obtained using the latter FDOT approach either coincided or exceed the design capacity (suggested FS = 1.0). Gates method. The method was the results of a research performed by Marvin Gates, J.M (1957). The basic assumption is that the resistance is directly proportional to the squared root of the net ham-mer energy. This relationship is presented by: 17 ( ) s b E e a P h h u log − = (3-6) where: Pu Static Pile Resistance eh Hammer Efficiency (0.85 used for all Cases) Eh Hammer Energy a, b 27 and 1.0 Respectively (English units) s Point Permanent Penetration per Blow - Set A suggested safety factor equal 3.0 is recommended. Combined Wave Mechanics and Energy Conservation Sakai et al. Japanese energy method. Sakai’s pile driving formula was developed based on stress-wave theory. According to Sakai, this consideration introduced two advantages, it is theoretically accurate as well as easy to use (Sakai et al., 1996). For a blow by an elastic hammer Sakai recommends ( ) s D M M L AE R H P P u −         = max 2 (3-7) where: A Pile Cross Sectional Area E Young’s Modulus of Pile Material LP Length of the Pile MP Mass of the Pile Dmax Maximum Penetration of Pile per Blow s Permanent Set Paikowsky’s method. The Paikowsky method or “Energy Approach” is a simplified energy approach formula-tion for the prediction of pile resistance based on the dynamic measurements recorded during 18 driving. The basic assumption of the method is an elasto-plastic load displacement pile-soil reaction. The Paikowsky method uses as input parameters the maximum calculated transferred energy and maximum pile displacement from the measured data together with the field blow count. Equation 3-8 presents the solution for the dynamic pile capacity Ru (Paikowsky, 1994). ( ) 2 max Set D Set E R m u − + = (3-8) where: Ru Dynamic Pile Capacity Em Maximum Energy Entering the Pile Dmax Maximum Pile Top Pile Movement Set Point permanent penetration per blow The static pile resistance Pu can be obtained by u sp u R K P = (3-9) where: Ksp ‘Static Pile’ Correlation Factor Accounting for all Dynamic Energy Looses. For easy driving of piles with small area ratios, Paikowsky recommends a value of Ksp smaller than 1.0, while for hard driving cases with large area ratios, the recommended Ksp value must be larger than 1.0. A value of Ksp equals 1 was used in our calculations. Wave Mechanics PDA method. In the 1960’s a new method to determine the pile capacity was developed at the Case Institute of Technology in Cleveland, Ohio. This new method called Pile Driving Analyzer (PDA) is based on electronic measurements of the stress waves occurring in the pile while driving. Some advantages of dynamic pile testing are (GRL, 1996): 19 Bearing Capacity – The bearing capacity can be found at the time of testing. For the prediction of a pile’s long term bearing capacity, measurements can be taken during restriking (Beginning of Restrike – BOR) Dynamic Pile Stresses – While the pile is driving the stresses within the pile can be monitored. This avoids any possibility of pile damage due to compression or tension stresses. Bending stresses caused by asymmetry of the hammer impact can be also monitored. Pile Integrity – To detect any existing damage within the pile. Hammer Performance – The performance of the hammer is monitored for productivity and construction control purpose. The PDA is considered as field equipment for measuring the forces and accelerations in a pile during driving. The methodology is standardized and is described in ASTM standard D4945. The equipment includes three components (Coduto, 1994): 1. A pair of strain transducers mounted near the top of the pile on each side. 2. A pair of accelerometers mounted near the top of the pile. 3. A pile driving analyzer (PDA). The main purpose of the PDA is to compute the static resistance of the pile using the Case method as it is driven. To perform the latter, the dynamic capacity has to be separated from the static capacity by mean of a damping value Jc , or Case damping value. In the following paragraph a summary of the basic equations used by PDA is presented. The pile wave speed, c, can be determined prior to pile installation while the pile is still on the ground. The accelerometers are installed and the pile is hit with a hammer. Knowing the pile length and the wave travel time, the wave speed can be calculated using Equation 3-10. t L c 2 = (3-10) 20 where: L Length of the Pile t Time Required for the Pulse to Travel Twice the Pile Length The dynamic modulus of the pile material, E, is presented in Equation 3-11. The mass density of the pile material is represented by ρ and the wave speed c. 2 c E ρ = (3-11) Equation 3-12 presents the impedance, Z, of a pile as a function of the dynamic modulus, E, the wave speed, c, and the pile cross-sectional area, A. c EA Z = (3-12) The force within the pile can be obtained from the strain transducers and knowing the elastic modulus of the pile material and cross-sectional area, according to Equation 3-13. EA P ε = (3-13) The velocity is obtained from the integration of the acceleration signal acquired by the accelerometers. It should be noted that the force and velocity used in the PDA calculation is the result from the average of the two strain transducers and the two accelerometers. The velocity is then converted to force units by multiplying by the pile’s impedance (Z). The result of the force and velocity are then plotted in a graph versus time for a given blow. Figure 3-2 shows an example of the force and velocity traces for a given hammer blow. The main equation used by PDA to determine static pile capacity was derived assuming that the pile is linearly elastic with constant cross sectional properties along its length. This equation is based on the force and velocity records at time T1, time T2 equal to T1 plus 2L/c, and the Case damping constant, Jc . Equation 3-14 presents the PDA equation for determining the static pile capacity. The reader is referred to the PDA manual for detailed information and more thorough derivation. 21 -1000 -500 0 500 1000 1500 2000 2500 0 0.01 0.02 0.03 0.04 0.05 0.06 Time (s) Force Velocity Figure 3-2. Typical Force and Velocity Traces from PDA ( )[ ] ( )[ ] 2 1 2 1 2 2 1 1 ZV P J ZV P J RSP c c − + + − − = (3-14) where: RSP Total Static Capacity Jc Case Damping Constant P1,P2 Force at Time T1 and T2 Respectively V1,V2 Velocities at Time T1 and T2 Respectively Z Impedance CAPWAP program. The Case Pile Wave Analysis Program (CAPWAP) is a computer program that combines the wave equation’s pile and soil model with the Case method of forces and velocities from PDA. The CAPWAP solution includes the static total resistance, skin friction and toe bearing of 22 the pile, in addition to the soil resistance distribution, damping factors, and soil stiffness. The program calculates acceleration, velocities, displacements, waves up, waves down and forces at all points along the pile. The procedure used by CAPWAP includes inputting the force trace obtained from PDA and adjust the soils parameters until the velocity trace obtained from PDA can be recreated. It should be noticed that the opposite procedure (i.e., input velocity trace and generate the force trace) can also be performed. When the match obtained is unsatisfactory, it is necessary to modi-fy the soil parameters, until satisfactory match results. Consequently, the process of running CAPWAP is considered an iterative one. 23 CHAPTER 4 FLORIDA DEPARTMENT OF TRANSPORTATION PILE DATABASE GENERAL INFORMATION AND HISTORY The University of Florida in conjunction with the Florida Department of Transportation (FDOT) maintains a database on driven piles inside and outside the state of Florida. The data-base, originally called PILEUF, began with the collection of driven pile load tests to study pile failure (Davisson, Fuller-Hoy, Debeer, etc.) in mid 1980s. Since then, the database has expanded to H piles, pipe piles and drilled shafts with conventional load tests, Osterberg tests and Statnamic tests. Recently, efforts are under way to collect data on cylinder piles (soil properties and conventional load test data). Originally, the database was in a Lotus 123 format (spreadsheet) and recently it was transferred to Microsoft’s Access (Geotech.mdb) in 1998. For this research, the database infor-mation was transferred to a Microsoft Excel format for data analysis. New pile information (driving data) was obtained from the original Geotechnical reports for this research. For this study, a total of 242 piles were in the database. Out of these, 198 were concrete piles (both square and round), 21 were steel pipe piles, and 9 were H-Piles. Table 4-1 sum-marizes the number of piles, classification and diameter for the Florida cases while Table 4-2 summarizes the same information for the Non Florida cases. The total in state piles (Table 4-1) includes 175 piles obtained from 60 sites and repre-sents 218 cases. The difference between the number of piles and cases is due to the multiple attempts to determine the same pile’s capacity (i.e., multiple piles on same site). The Non-Florida total from Table 4-2 represents 22 sites. For the latter, the number of cases is equal to the number of piles. 24 Table 4-1. Classification of Florida Driven Piles Pile Number of Piles Diameter Squared Pipe Sub-Total (inches) Concrete Piles 10 4 3 7 12 14 2 16 14 26 8 34 18 34 0 34 20 12 0 12 24 44 0 44 30 26 0 26 36 2 0 2 Sub-Total 162 13 175 Table 4-2. Classification of Non-Florida Driven Piles Pile Number of Piles Diameter Squared Round Pipe H-Piles Sub-Total (inches) Concrete Concrete Piles 10 3 0 1 6 10 10.75 2 0 0 0 2 12 15 2 0 1 18 12.75 3 0 0 0 3 14 2 0 3 2 7 16 5 0 0 0 5 18 2 0 0 0 2 24 2 0 0 0 2 59.06 0 0 4 0 4 Sub-Total 34 2 8 9 53 It should be noted that for the Non-Florida information in Table 4-2, the total number of piles does not include 14 piles, which had unspecified diameter. The following sections describe briefly the information gathered in the database for this study. 25 PILE INFORMATION General Each pile record contains general description of the site location, together with pier or bent number (if available). Four pile types were selected to describe the piles within the data-base based on shape and materials. The four categories were: 1. Square concrete pile 2. Round concrete pile 3. Pipe pile 4. H-pile The geometry of the piles was described by the pile width, total length, embedded length, and cross sectional area. In addition, the dates when the piles were driven and tested were recorded. Soil Classification The soil information includes the soil stratigraphy (description and properties) with in-situ tests such as SPT, CPT, etc. Nine soil types were used to allow consideration of several combinations of cohesive and non-cohesive soil. The nine soil categories were presented according to the following numbering: 1. Plastic clay 2. Silt-sand-clay, silts and marls 3. Clean sand 4. Limestone, very shelly sands 5. Clayey sand 6. Sandy Clay 7. Silty clay 8. Rocks 9. Sandy gravel, tills 26 The original database combined the side and tip soil number to form a two-digit code, in which the first digit is the side soil type and the second digit is the tip soil type. Driving Information The driving information includes the driving system type, hammer and pile weight and manufacturer’s rated energy together with the efficiency of the hammer. Additional information includes the dynamic modulus, wave speed and the pile impedance. If the impedance was not available from CAPWAP or other results, it was calculated as EA/c. The average set for EOD (End of Drive) and BOR (Beginning of Restrike) was taken as the inverse of the blow counts as near as possible to the blow used in PDA or CAPWAP analysis. The latter may represent an average of the last foot of driving in some cases, if inch-by-inch information was not available. A record of the depth of penetration and blows per foot (calculated for penetration intervals less than one foot) facilitated the determination of set, knowing the tip depth at the time of the blow. Dynamic Data (CAPWAP and PDA) The CAPWAP and PDA results were sometimes available only for EOD or BOR. Fur-thermore, not all CAPWAP analyses have complete PDA results available or vice versa. Having both results was not a requirement during the construction of the database. The PDA results include date, RMX (maximum Case Static Resistance calculated during the blow analysis) or other PDA calculated capacity as listed in the source report. The database also presents the PDA Case damping used for calculating the Total Static Resistance. CAPWAP results include date, tip and friction capacities, total capacity, and Case and Smith damping factors for side and tip, where the Case damping factors were calculated from the Smith damping factors. The latter was performed by dividing the Smith damping value by the impedance and multiplying the result by the side or tip resistance. 27 Load Test Results The database contains load test information, measured at the top of the piles. It includes the load in tons and settlements in inches at failure for a given criterion. The failure criteria pre-sented in the database are: 1. Davisson 2. Fuller-Hoy 3. DeBeer 4. FDOT The database also includes the maximum load in tons from the static load test, in addition to the date at which the load test was performed. SPT97 Capacity SPT97 is a pile capacity prediction program based on the original Research Bulletin 121 (RB-121), “Guidelines for use in the Soils Investigation and Design of Foundations for Bridge Structures in the State of Florida”. The latter has under a number of changes (1989) and recently with the inclusion of “Design of Steel Pipe and H-piles” by McVay et al in 1994. The method calculates pile capacity based on N values obtained from the Standard Penetration Test. SPT97 is capable of evaluating round and square concrete piles, H-piles, and steel pipe piles (open or close end). It calculates an Estimated Davisson capacity by summing the Ultimate Side Friction and 1/3 of the Ultimate End Bearing (Mobilized End Bearing) capacity of the pile. SPT97 predictions presented in the database include the Ultimate Side Friction, Ultimate Tip Capacity, Mobilized Tip Capacity, Ultimate Total Capacity, and Davisson’s Capacity. Other related information presented in the database is the input data for SPT97 program. It includes the layering and the soil properties (i.e., unit weight and SPT blow count). 28 GATHERING NEW INFORMATION During the course of evaluating the eight dynamic methods (Chapter 3), some extra infor-mation was necessary. The latter required that the original Geotechnical reports on many of these sites be found and catalogued. A discussion on the new information follows. Additional Information Two parameters that were not found in the original database were the maximum displace-ment and the maximum energy transfer to the pile. They were essential to obtain the Paikowsky and Sakai capacities (Chapter 3). Both, the maximum energy transfer to the pile and the maxi-mum displacement were obtained from the CAPWAP output printout in the geotechnical reports. Criteria for New Entries in Database As a general criterion for this research, new entries should be from within the State of Florida. Also, because the evaluation of the dynamic methods was performed in correlation to the Davisson’s capacity, any new entries should have the Load Tests carried to the point for which Davisson’s capacity could be determined. Other information needed depends on the method, which is to be evaluated. The more information obtained for a particular record will assure a larger number of dynamic methods which may be evaluated, as well accurate statistical analysis (bias, standard deviation, etc.). 29 CHAPTER 5 ASD AND LRFD CONCEPTS Over the years, multiple design procedures have been developed which provide satisfac-tory margins of safety. Safety in design is obtained when the material properties exceed the de-mand put on them by any load or loads combination. Another way to describe the same principle is that the resistance of the structure must exceed the effect of the loads, i.e.: Loads of Effect Resistance ≥ (5-1) When a specific loading condition reaches its limit, failure occurs. Two general states of interest to engineers are Strength and Service Limit. Strength Limit State involves the total or partial collapse of the structure (i.e., bearing capacity failure, sliding, and overall instability). On the other hand, Service Limit State only affects the function of the structure under regular service loading conditions (i.e., excessive settlement and/or lateral deflection, structural deterioration, etc). ALLOWABLE STRESS DESIGN (ASD) METHOD In Geotechnical engineering, the ASD has been the primary method used in U.S.A. ASD procedures are different for Service Limit and Strength Limit States. For the Strength Limit State, safety is obtained in the foundation elements by restricting the ultimate loads to values less than the ultimate resistance divided by a factor of safety, (FS): ∑ ≥ i n Q FS R (5-2) where: Rn Nominal Resistance ΣQi Load Effect (Dead, Live and Environmental Loads) FS Factor of Safety 30 For the Service Limit State, the deformations (i.e., settlements) are calculated using the unfactored loads, and the values obtained are compared to the allowable deformation for that structure. LOAD RESISTANCE FACTOR DESIGN (LRFD) METHOD The LRFD specifications as approved by AASHTO in 1998 recommend the use of load(s) factors to account for uncertainty in the load(s) and a resistance(s) factor to account for the uncertainty in the material resistance(s). This safety criterion can be written as: ∑ = i i n Q R γ η φ (5-3) where: φ Statistically Based Resistance Factor Rn Nominal Resistance 0 Load Modifier to Account for Effects of Ductility, Redundancy and Operational Importance γi Statistically Based Load Factor Qi Load effect Even though the LRFD method differs from the accustomed ASD procedure, it has been widely approved by the Geotechnical engineers. Some of the advantages and disadvantages of the LRFD method over the ASD method are as follows (Withiam et al., 1997). Advantages of LRFD Over ASD • Account for variability in both resistance and load. • Achieves relatively uniform levels of safety based on the strength of soil and rock for different limit states, foundation types, and design methods. • Provide more consistent levels of safety in the superstructure and substructure when the same probabilities of failure are employed. 31 • Using load and resistance factors provided in the code, no complex probability and statistical analysis is required. Limitation of LRFD • Implementation requires a change in design procedures for engineers accustomed to ASD. • Resistance factors vary with design methods and are not constants. • The most rigorous method for developing and adjusting resistance factors to meet individual situations requires availability of statistical data and probabilistic design algorithms. CALIBRATION OF LRFD Calibration is defined, as the process of assigning values to resistance factors and load factors, required for the LRFD approach. This process can be performed by use of engineering judgement, fitting to other codes (e.g. ASD method), use of reliability theory, or a combination of them. In the following sections these approaches will be discussed. Engineering Judgement The calibration of a code using engineering judgement requires experience. Such ex-perience is usually obtained through years of engineering practice. Sometimes, using such an approach results in certain level of conservatism with little validation. Also under varying con-ditions where no experience exist both excessive conservatism or ever unconservatism may develop. Fitting ASD to LRFD Fitting ASD to LRFD is the selection of LRFD resistance factor, φ, that will result in equivalent physical dimensions of a substructure or superstructure as given by ASD. It does not 32 provide a better or more uniform margin of safety. In order to calibrate the ASD method, the first step is to rewrite equations 5-2 and 5-3 as: L D n Q Q FS R + ≥ (5-4) D D L L n Q Q R γ γ φ + ≥ (5-5) It should be noted that the loads only include dead and live loads. Environmental loads (i.e., wind, earthquake, etc) were not taken into consideration for the derivation of the ASD fitting equation herein. Solving both equations for Rn gives: ( ) L D n Q Q FS R + ≥ (5-6) ( ) φ γ γ D D L L n Q Q R + ≥ (5-7) Setting Equation 5-6 equal to Equation 5-7 and solving for φ results in: ( ) D L D D L L Q Q FS Q Q + + = γ γ φ (5-8) Dividing both the numerator and the denominator of Equation 5-8 by QL:         + + = 1 L D L D L D Q Q FS Q Q γ γ φ (5-9) Equation 5-9 is the resulting calibration equation for LRFD based on ASD factor of safety, FS, dead load/live load ratio, (QD/QL), and load factors (γD and γL). For deep foundation design, the values of γD and γL recommended by LRFD Highway Bridges Design Specifications (AASHTO, 1994) are 1.25 and 1.75 respectively. The QD/QL definition and values will be pre-sented in more detail in latter sections. Calibration by fitting is recommended when there is insufficient measured and predicted data to perform a more sophisticated calibration by statistical 33 analysis. When measured and predicted data is available, reliability theory is strongly recom-mended. Reliability Calibration Statistical data. In order to perform a reliability calibration for deep foundations (obtain resistance factor, φ), such as piles and drilled shafts, the designer must have available statistical data for the method of interest. This statistical data is based on the real or measured capacities and the estimated or nominal capacities of individual piles. First, the bias is defined as: n m Ri R R = λ (5-10) where: λRi Bias Factor for an individual pile Rm Measured Resistance for an individual pile Rn Predicted (nominal) Resistance for an individual pile After the biases for all cases are determined for the database for a given method, its mean, stan-dard deviation and coefficient of variance are found. Equations 5-11, 5-12 and 5-13 (Withiam et al., 1997) identify the process. N Ri R ∑ = λ λ (5-11) ( ) 1 2 − − = ∑ N R Ri R λ λ σ (5-12) R R R COV λ σ = (5-13) where: λR Average Resistance Bias Factor N Total Number of Χασεσ in Database 34 σR Resistance Standard Deviation COVR Resistance Coefficient of Variance It should be noted that the measured resistance, Rm (Eq. 5.10) was obtained from the computed Davisson’s capacity from the individual field load test. The nominal resistances (Rn) were obtained from the various dynamic equations under study (Chapter 3). Probability density function. For the LRFD design, the probability of failure to occur for a given load and resistance distribution is of fundamental importance. The latter is calculated with the “probability density function” which is defined as the probability that X occurs in the interval x to x + dx as fx(x)dx (See Figure 5-1). The total area under the curve fx(x) must be equal to unity because a proba-bility of 1 includes all possible outcomes. Figure 5-1. Lognormal Probability Density Function 35 Based on the distribution of the resistance data, a lognormal probability distribution was recommended for the resistance data by the AASHTO Specification. A normal function was used to represent the observed distribution of load data. Equation 5-14 presents the lognormal probability density equation. ( )                 − − = 2 ln 2 1 exp 2 1 ξ θ ξ π x x x f x (5-14) In Equation 5-14 the values of θ and ζ are the lognormal mean and lognormal standard deviation respectively,        + = 2 2 2 1 ln R R λ σ ξ (5-15) 2 2 1 ln ξ λ θ − = R (5-16) Where σR and λR are the standard deviation and the mean of the resistance as defined earlier. LRFD Approach Probability of failure. The LRFD approach defines the probability of failure of a structure based on the load and resistance distribution curves. Figures 5-2 shows the probability density functions for normally distributed load and resistance. The shaded area represents the region of failure where the resis-tance is smaller than the loads. For the load and resistance curves, the margin of safety can be defined in terms of the probability of survival as: ( ) Q R P ps > = (5-17) And the probability of failure, pf may be represented as ) ( 1 Q R P p p s f < = − = (5-18) 36 Figure 5-2. Probability Density Functions for Normally Distributed Load and Resistance Where the right hand of Equation 5-18 represents the probability, P, that R is less than Q. It should be noted that the probability of failure can not be calculated directly from the shaded area in Figure 5-2. That area represents a mixture of areas from the load and resistance distribution curves that have different ratios of standard deviation to mean values. To evaluate the probability of failure, a single combined probability density curve function of the resistance and load may be developed based on a normal distribution, i.e., ( ) Q R Q R g − = , (5-19) If a lognormal distribution is used, the limit state function g(R,Q) can be written as: ( ) Q R Q R Q R g ln ) ln( ) ln( ) , ( = − = (5-20) For both Equation 5-19 and 5-20, the limit state is reached when R=Q and failure will occurs when g(R,Q)<0. 37 Reliability index. The reliability index is a simple method of expressing the probability of failure using function g(R, Q) (Eq. 5-20). The frequency distribution of g(R,Q) looks similar to the curve shown in Figure 5-3. Figure 5-3. Definition of Reliability Index, β for Lognormal Distributions of R and Q Evident from the curve is that if the standard deviation is small or the mean value is located further to the right, the probability of failure will be smaller. The reliability index β, is defined as the number of standard deviations, ξg, between the mean value, g (average), and the origin, or: g g ξ β = (5-21) If the resistance, R, and load, Q, are both lognormally distributed random variables and are statistically independent, it can be shown that the mean values of g(R, Q) is:         + + = 2 2 1 1 ln R Q COV COV Q R g (5-22) 38 and its standard deviation is: ( )( ) [ ] 2 2 1 1 ln Q R g COV COV + + = ξ (5-23) Substituting Equations 5-22 and 5-23 into Equation 5-21, the relationship for the reliabil-ity index, β, can be expressed as: ( ) ( ) ( ) [ ] ( )( ) [ ] 2 2 2 2 1 1 ln 1 / 1 ln Q R R Q COV COV COV COV Q R + + + + = β (5-24) Equation 5-24 is very convenient because it depends only on statistical data and not on the distribution of the combined function g(R, Q). A very precise definition of probability of failure, pf, is in terms of reliability index, Fu(β) (Withiam et al. 1997). ( ) β u f F p − =1 (5-25) A graphical representation of Equation 5-26 is presented in Figure 5-4. The shaded area in Figure 5-4 represents the probability of failure, pf, to achieve a target reliability index, βT. In the latter equation, Fu(x) is the standard normal cumulative distribution function. ( ) dx x Fu      − − = ∫ ∞ 2 2 1 exp 2 1 1 β π β (5-26) Another commonly accepted relationship between the reliability index, β, and the prob-ability of failure, pf, has been developed by Rosenblueth and Esteva (1972) for values between 2 and 6 as: ( ) β 3 . 4 exp 460 − = f p (5-27) 39 Figure 5-4. Reliability Definition Based on Standard Normal Probability Density Function Figure 5-5 presents a comparison of the results for both, the Rosenblueth and Esteva method and the Withiam method, to determine the reliability index, β. It can be observed that Rosenblueth and Esteva approximation method will yield good values of probability of failure for values of reliability index between 2.0 and 6.0 as recommended by the authors of the method. Resistance factor, φ φ φ φ. Once the reliability index, β, is selected then a resistance factor, φ, may be calculated. Assuming lognormal distributions of load and resistance in Eq. 5-5 substituted into Eq. 5-24 gives the follow resistance, φ, equation: 40 Reliability Index, β β β β, vs. Probability of Failure, pf (Comparison of Methods) 1.0E-10 1.0E-09 1.0E-08 1.0E-07 1.0E-06 1.0E-05 1.0E-04 1.0E-03 1.0E-02 1.0E-01 1.0E+00 0 1 2 3 4 5 6 7 Reliability index,β β β β pf = 460 exp (-4.3β) Esteva et al. (1972) True value Withiam et al. (1997) Figure 5-5. Comparison of Esteva and Withiam Methods to Obtain Reliability Index, β ( ) )] 1 )( 1 ln[( exp ) ( 1 1 ) ( 2 2 2 2 2 2 QL QD R T QL L D QD R QL QD L L D D R COV COV COV Q Q COV COV COV Q Q + + + + + + + + = β λ λ γ γ λ φ (5-28) where: φ Resistance Factor λR Resistance Bias Factor COVR Resistance Coefficient of Variance βT Target Reliability Index λQD, λQL Bias (Dead and Live Load) QD/QL Dead to Live Load Ratio 41 The dead to live load ratio (QD/QL) in Eq. 5-28 varies with the bridge span. For any bridge the live load is obtain by a standard procedure, while the dead load is based upon the size of the structure. The dead to live load ratio has been correlated to the span of the bridges by the following representation (Hansell and Viest, 1971): ( )( ) L IM Q Q L D 0132 . 0 1 / + = (5-29) where: QD/QL Dead to Live Load Ratio IM Dynamic Load Allowance Factor (equal 0.33) L Span Length (feet) Table 5-1 presents typical values of bridges span and the corresponding dead to live load ratios. Definition φ φ φ φ/λ λ λ λR and its importance. The axial design capacity of a pile may be represented as n R P φ = Design (5-30) Table 5-1. Values of QD/QL Based on Bridge Span Length Span Length, L (m) Dynamic Load Allowance, IM QD /QL 9 18 27 36 45 60 0.33 0.33 0.33 0.33 0.33 0.33 0.52 1.06 1.58 2.12 2.64 3.53 However, from Eq. 5-10, we know the nominal resistance may be expressed in terms of measured value as 42 Ri m n R R λ = (5-31) Substituting Equation 5-31 into Equation 5-30 the PDesign becomes m Ri Design R P λ φ = (5-32) Equation 5-32 gives the design capacity of a shaft for a specific site. If the bias λRi is replaced by the bias factor λR, (which represents the average of the bias for the method), then the design capacity of the method is directly related to measured resistance (i.e., Davisson's capacity) through the LRFD fitting parameters φ, and λR . And the φ/λR ratio represents the percentage of Davisson capacity, which is allowed for design for a given probability of failure. Obviously, the higher the ratio, the fewer the number of piles which will be required and the better the method. 43 CHAPTER 6 LRFD ASSESSMENT DATA REDUCTION The number of cases for each method was determined based on the availability of param-eters needed to obtain the estimated capacity for the corresponding method. For all dynamic methods, plots of measured (Davisson Capacity) vs. predicted capacity at End of Drive (EOD) and Beginning of Restrike (BOR) was constructed. Subsequently, the bias, λ, standard devia-tion, σ, and LRFD phi factors, φ were computed. The statistics, as well as some of the LRFD results (tables & graphs) for each method are contained within separate Appendices (i.e., Appendix A for all CAPWAP Procedure results, Appendix B for all PDA results, Appendix C for Paikowsky’s Energy Method results, etc.). The specifics for some of the methods will be presented. Figure 6-1 shows PDA’s predicted BOR capacity versus the measured Davisson Capac-ity. A line with slope equal 45 degrees (prefect correlation) has been drawn to show the com-parison between the predicted and measured pile capacity. Also shown in the plot is a regression line (fit between measured and predicted) with the corresponding slope and R2 (data fit). This latter graph is ideal to visually determine how scattered the predictions are for each method. The second graph (Figure 6-2, PDA at BOR) presents the ratio of measured to predicted capacity on the vertical axis and the measured Davisson capacity on the horizontal axis. On each graph, the number of cases for each dynamic method, the mean (or bias factor, λR), and standard deviation, σR, and the ratio of measured to predicted capacity is also presented. Appendices A throughout H present the resulting plots for all the methods studied. 44 y = 0.9966x R 2 = 0.4851 0 1000 2000 3000 4000 5000 6000 7000 8000 0 2000 4000 6000 8000 Davisson Capacity (kN) Figure 6-1. Davisson Capacity vs. PDA BOR Capacity 0.0 0.5 1.0 1.5 2.0 2.5 0 1000 2000 3000 4000 5000 6000 7000 Davisson Capacity (kN) Cases=42 Mean=1.036 Standard Dev.=0.32 Figure 6-2. Measured Over Predicted Capacity for PDA at BOR 45 In addition to the mean and standard deviation, a lognormal probability density function was computed for each method. The latter may be used to represent the resistance distribution of a single pile. However, if the ratio of measured to predicted capacity is plotted, the subsequent distribution is a good indicator of a method's accuracy. Using Equations 5-14 to 5-16, the log normal probability density functions of measured to predicted capacities were determined for all of the methods. Figures 6-3 and 6-4 present a summary of the lognormal probability density functions for the eight dynamic methods at EOD and BOR studied, respectively. By inspection it can be observed that dynamic methods such as ENR and Modified ENR have a very small mean and standard deviation in comparison to the rest of the methods. On the other hand, the Energy (Paikowsky) method has an excellent mean (close to one) but a larger standard devia-tion. Figure 6-4 presents the lognormal probability density function of measured to predicted capacity for the BOR cases. Evident, the Energy (Paikowsky) has a mean close to one for EOD but a values less than one (over predicts) at BOR. Similarly CAPWAP underpredicts capacity at EOD, and does a better job at BOR. At an early stage of this research, it was noted that the old methods (i.e., Gates, FDOT, ENR, and Modified ENR) gave good estimates of the Davisson capacity for piles with Davisson capacity less than 200 tons (1779 kN). This finding is clearly justified because the range of pile capacity in the past did not exceed 1779 kN for driven piles (limitation of construction equip-ment). Consequently, it was decided to consider the older methods under three load ranges: 1) piles with Davisson capacity less than 1779 kN; 2) piles with capacity larger than 1779 kN; and 3) no load limitations at all (i.e., combined). 46 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 0 1 2 3 4 5 6 Davisson Capacity / EOD Capacity 1 ENR Cases=77 mean=0.30 std. dev=0.16 2 Modified ENR Cases=61 mean=0.45 std. dev=0.27 3 FDOT Cases=72 mean=2.38 std. dev=1.34 4 Gates Cases=74 mean=1.74 std. dev=0.79 5 Energy Cases=27 mean=1.11 std. dev=0.37 6 CAPWAP Cases=44 mean=1.60 std. dev=0.56 7 PDA Cases=48 mean=1.34 std. dev=0.44 8 Japanese Energy Cases=21 mean=1.50 std. dev=1.26 6 7 2 1 3 4 5 8 Figure 6-3. Log Normal Probability Distribution at EOD 47 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 0 1 2 3 4 5 6 Davisson Capacity / BOR Capacity 1 ENR cases=71 mean=0.24 std dev=0.16 2 Modified ENR cases=63 mean=0.36 std dev=0.25 3 FDOT cases=63 mean=2.57 std dev=1.29 4 Gates cases=71 mean=1.89 std dev=0.72 5 Energy cases=72 mean=0.84 std dev=0.30 6 CAPWAP cases=79 mean=1.26 std dev.=0.44 7 PDA cases=42 mean=1.04 std dev=0.32 8 Japanese Energy cases=52 mean=1.46 std dev=0.78 5 2 1 6 3 4 7 8 Figure 6-4. Log Normal Probability Distribution at BOR 48 Based solely on the statistical analysis and the lognormal probability distribution, a com-parison of the eight dynamic methods is very difficult. For instance a number of the methods have means close to one, but large standard deviation (Figures 6-1 and 6-2) or vice versa. Con-sequently, it was decided to evaluate each approach by the percentage of measured Davisson capacity that was available for design given the same reliability (risk) for each method. The LRFD provides the framework to evaluate the latter. LRFD ANALYSIS OF RESULTS In order to obtain the LRFD resistance factor, φ, two approaches were considered. The first consists of fitting ASD to LRFD, by use of Equation 5-9. For this approach, the φ factor depends on the safety factor (ASD) and the ratio of dead load to live load for a given bridge. The second approach used was Reliability Calibration (See Chapter 5) in which the φ factor was determined based on the covariance of the measured to predicted (λ, bias), the reliability index, β, and the ratio of dead load to live load. In order to perform the LRFD analysis, the στατιστιχαλ results calculated earlier were employed. For instance, the bias factor (λR) and standard deviation (σR) of the bias were used to calculate the covariance of the resistance (COVR). These parameters were subsequently used with AASHTO’s load statistics (See Chapter 5) and the failure probabilities also recommended by AASHTO to obtain the resistance factor, φ, from Eq. 5-28. Additional attention was paid to the target reliability index, βT. For this study, the relia-bility index was calculated using the Withiam method instead of the Rosenblueth and Esteva method, recommended by AASHTO (See Chapter 5). The values recommended with the Withiam method yield an exact reliability index, while the values obtained using the Rosenblueth 49 and Esteva result in an approximate quantity. Consequently, the probability of failure, pf, sug-gested by AASHTO had to be back calculated from the reliability indexes. Tables 6-1 through Table 6-6 show typical output results for each LRFD analysis per-formed. Tables 6-1 to 6-3 present the results for PDA at EOD, while Tables 6-4 to 6-6 present the commensurate for PDA at BOR. Tables 6-1 and 6-4 summarize the resistance factors, φ, evaluated for four different failure probabilities (i.e., reliability indexes) and eight bridge span lengths based on LRFD calibration of the method. Tables 6-2 and 6-5 show the equivalent ASD safety factors based on the resistance factors calculated using the LRFD calibration method for the same bridge span lengths and failure probabilities. Finally, Tables 6-3 and 6-6 present the resistance factor, reliability index and probability of failure that correspond to the actual safety factor that have been used in current ASD procedures. Appendices A throughout H present in detail the LRFD analysis results for each dynamic method studied. Effect of Bridge Span Length and Probability of Failure The bridge span length evaluated in this study range from 9.0 to 60.0 meters, each length corresponding to a recommended (AASHTO) dead to live load ratio. After observing the results for the PDA, EOD and/or BOR it is evident that the bridge span length has negligible effect on the resistance factor, φ, or the corresponding safety factor. At a target reliability index of 2.5 for both EOD and BOR, the φ factor decreases only 6.8% and the safety factor decreases only 1.6% when the span length increased from 15.0 meters to 50.0 meters. The latter phenomenon was observed throughout all the methods evaluated. Thus, for practical purposes the rest of LRFD analysis results are summarized for a 27 meters bridge span length. 50 Table 6-1. φ Values Evaluated for PDA (EOD) Failure Probability pf /Reliability Index βT 2.50 E-02 6.22 E-03 1.22 E-03 1.79 E-04 Span length (m) QD /QL 1.96 2.50 3.03 3.57 9 15 18 27 36 45 50 60 0.52 1.00 1.06 1.58 2.12 2.64 3.00 3.53 0.856 0.821 0.818 0.796 0.781 0.770 0.765 0.758 0.694 0.666 0.663 0.645 0.633 0.625 0.620 0.615 0.565 0.542 0.540 0.525 0.515 0.508 0.505 0.500 0.458 0.439 0.437 0.426 0.418 0.412 0.409 0.405 Table 6-2. ASD Design Safety Factor Values Evaluated for PDA (EOD) Failure Probability pf /Reliability Index βT 2.50 E-02 6.22 E-03 1.22 E-03 1.79 E-04 Span length (m) QD /QL 1.96 2.50 3.03 3.57 9 15 18 27 36 45 50 60 0.52 1.00 1.06 1.58 2.12 2.64 3.00 3.53 1.845 1.827 1.825 1.814 1.806 1.801 1.798 1.795 2.276 2.253 2.251 2.237 2.228 2.221 2.218 2.214 2.796 2.769 2.766 2.749 2.738 2.730 2.725 2.720 3.449 3.415 3.412 3.391 3.377 3.367 3.362 3.356 Table 6-3. φ Factors for Safety Factor = 2.50 – PDA (EOD) Span length (m) QD/QL φ βT pf 9 0.52 0.632 2.742 0.0035 15 1.00 0.600 2.767 0.0031 18 1.06 0.597 2.770 0.0031 27 1.58 0.578 2.786 0.0029 36 2.12 0.564 2.797 0.0028 45 2.64 0.555 2.804 0.0027 50 3.00 0.550 2.808 0.0026 60 3.53 0.544 2.813 0.0026 51 Table 6-4. φ Values Evaluated for PDA (BOR) Failure Probability pf /Reliability Index βT 2.50 E-02 6.22 E-03 1.22 E-03 1.79 E-04 Span length (m) QD /QL 1.96 2.50 3.03 3.57 9 15 18 27 36 45 50 60 0.52 1.00 1.06 1.58 2.12 2.64 3.00 3.53 0.682 0.654 0.652 0.634 0.622 0.614 0.609 0.604 0.557 0.534 0.532 0.518 0.508 0.501 0.498 0.493 0.457 0.438 0.436 0.425 0.417 0.411 0.408 0.404 0.373 0.358 0.356 0.347 0.340 0.336 0.333 0.330 Table 6-5. ASD Design Safety Factor Values Evaluated for PDA (BOR) Failure Probability pf /Reliability Index βT 2.50 E-02 6.22 E-03 1.22 E-03 1.79 E-04 Span length (m) QD /QL 1.96 2.50 3.03 3.57 9 15 18 27 36 45 50 60 0.52 1.00 1.06 1.58 2.12 2.64 3.00 3.53 2.316 2.293 2.291 2.277 2.267 2.260 2.257 2.253 2.835 2.807 2.805 2.787 2.776 2.768 2.763 2.758 3.458 3.424 3.421 3.400 3.386 3.376 3.370 3.364 4.234 4.192 4.188 4.163 4.145 4.133 4.127 4.119 Table 6-6. φ Factors for Safety Factor = 2.50 – PDA (BOR) Span length (m) QD/QL φ βT pf 9 0.52 0.632 2.164 0.0418 15 1.00 0.600 2.191 0.0373 18 1.06 0.597 2.193 0.0369 27 1.58 0.578 2.210 0.0344 36 2.12 0.564 2.221 0.0328 45 2.64 0.555 2.229 0.0317 50 3.00 0.550 2.233 0.0311 60 3.53 0.544 2.238 0.0304 52 In terms of the probability of failure, pf, a decrease in the probability of failure is directly related to an increase in reliability index, β, and a diminishment in the resistance factor, φ. This effect could be easily understood, because lowering the probability of failure for a given struc-ture should be accompanied by a reduction of the resistance factor, which means, penalizing the structure resistance and producing a more conservative design. Different to the effect of the bridge span length, the reliability index has a big influence in the resistance factor φ, and the equivalent safety factor. The relation of φ factor and reliability index, β, is inversely propor-tional and linear. For piles AASHTO recommends a reliability index, β, of 2.0 to 2.5. Level of Conservatism and Accuracy Indicators A similar analyses as the PDA (Tables 6-1 to 6-6) was performed for each of the eight dynamic methods, as well as the older methods which were separated into capacities less than 1779 kN and greater than 1779 kN. A summary of these results is presented in Tables 6-7 and 6-8 for EOD and BOR respectively. The results show that the higher the bias factor λR, for a given method the higher the resulting resistance factor. For example, for CAPWAP at BOR the mean is equal to 1.260 and φ is equal to 0.58 (for β = 2.50) while FDOT method shows a mean and resistance factor of 2.574 and 0.97 respectively for the same reliability index. This effect may be explained by the bias factors for each method. As defined previously, the bias factor is the ratio of measured to pre-dicted capacity. The larger the bias factor, the more conservative the method. Consequently, LRFD will raise the resistance factor to generate the same probability of failure as another method with a much lower bias. Although the resistance factor is a good qualitative measure of the degree of conserva-tism or unconservatism, it does not indicate the accuracy of the method. The combination of 53 Table 6-7. Load and Resistance Factor Design (LRFD) of Driven Piles Using Dynamic Methods at EOD Prediction Method Number Mean, λR Standard COVR Pf=0.62% (βT=2.50) Pf=2.50% (βT=1.96) Of cases (Dav/Pred) Deviation φ FS φ/λR φ FS φ/λR CAPWAP 44 1.597 0.559 0.350 0.733 1.970 0.459 0.912 1.584 0.571 PDA 48 1.344 0.443 0.329 0.645 2.237 0.480 0.796 1.814 0.592 Paikowsky Energy 27 1.110 0.372 0.335 0.527 2.740 0.475 0.651 2.216 0.587 Sakai et al Energy 21 1.504 1.256 0.835 0.231 6.254 0.153 0.348 4.150 0.231 FDOT (overall) 72 2.381 1.341 0.563 0.669 2.160 0.281 0.909 1.588 0.382 FDOT (<1779 kN) 34 1.490 0.782 0.525 0.457 3.161 0.307 0.611 2.362 0.410 FDOT (>1779 kN) 38 3.158 1.248 0.395 1.307 1.104 0.414 1.658 0.871 0.525 ENR (overall) 77 0.299 0.159 0.532 0.090 16.024 0.301 0.121 11.935 0.405 ENR (<1779 kN) 34 0.250 0.129 0.515 0.078 18.395 0.314 0.105 13.801 0.419 ENR (>1779 kN) 43 0.338 0.171 0.507 0.108 13.388 0.319 0.143 10.074 0.424 Modified ENR (overall) 61 0.446 0.267 0.599 0.115 12.533 0.258 0.159 9.086 0.357 Modified ENR (<1779 kN) 25 0.325 0.222 0.683 0.069 20.818 0.214 0.099 14.604 0.305 Modified ENR (>1779 kN) 36 0.530 0.321 0.606 0.135 10.720 0.254 0.186 7.749 0.352 Gates (overall) 74 1.742 0.787 0.452 0.633 2.280 0.363 0.822 1.756 0.472 Gates (<1779 kN) 32 1.071 0.351 0.328 0.515 2.802 0.481 0.635 2.272 0.593 Gates (>1779 kN) 42 2.254 0.717 0.318 1.109 1.302 0.492 1.361 1.061 0.604 54 Table 6-8. Load Resistance Factor Design (LRFD) of Driven Piles Using Dynamic Methods at BOR Prediction Method Number Mean, λR Standard COVR Pf=0.62% (βT=2.50) Pf=2.50% (βT=1.96) Of cases (Dav/Pred) Deviation φ FS φ/λR φ FS φ/λR CAPWAP 79 1.260 0.438 0.347 0.581 2.485 0.461 0.722 1.999 0.573 PDA 42 1.036 0.322 0.311 0.518 2.787 0.500 0.634 2.277 0.612 Paikowsky Energy 72 0.836 0.301 0.360 0.374 3.857 0.448 0.468 3.086 0.560 Sakai et al Energy 52 1.457 0.784 0.538 0.433 3.334 0.297 0.583 2.477 0.400 FDOT (overall) 63 2.574 1.293 0.502 0.832 1.735 0.323 1.103 1.309 0.429 FDOT (<1779 kN) 8 1.355 0.380 0.280 0.724 1.995 0.534 0.875 1.650 0.646 FDOT (>1779 kN) 55 2.751 1.284 0.467 0.966 1.495 0.351 1.262 1.144 0.459 ENR (overall) 71 0.235 0.160 0.681 0.050 28.597 0.215 0.072 20.080 0.306 ENR (<1779 kN) 9 0.186 0.057 0.306 0.094 15.348 0.505 0.115 12.562 0.617 ENR (>1779 kN) 62 0.242 0.169 0.698 0.050 28.841 0.207 0.072 20.120 0.296 Modified ENR (overall) 63 0.363 0.246 0.676 0.079 18.314 0.217 0.112 12.881 0.308 Modified ENR (<1779 kN) 8 0.277 0.062 0.224 0.166 8.704 0.598 0.196 7.356 0.708 Modified ENR (>1779 kN) 55 0.376 0.260 0.692 0.079 18.321 0.210 0.113 12.810 0.300 Gates (overall) 71 1.886 0.715 0.379 0.810 1.783 0.429 1.020 1.416 0.541 Gates (<1779 kN) 9 1.067 0.201 0.189 0.681 2.121 0.638 0.796 1.815 0.746 Gates (>1779 kN) 62 2.005 0.684 0.341 0.938 1.540 0.468 1.162 1.242 0.580 55 both, mean and standard deviation through the COVR (i.e., ratio of standard deviation to the mean or bias) is a better indicator of the accuracy of a method. Under the latter criterion, PDA, Paikowsky, and CAPWAP are better, while ENR and modified ENR are not as accurate. A simpler and more direct comparison of all the methods is through Eq. 5-32, i.e., φ/λR, which follows. φ/λ φ/λ φ/λ φ/λR Ratio The efficiency or performance of every dynamic method can be evaluated by its φ/λR ratio (Eq. 5-22). The latter indicates the percentage of the measured Davisson capacity that can be utilized for design for pre-defined structure reliability. It should be noted that the average bias factor, λR, is an average of all the pile biases, therefore, the φ/λR ratio is an “average” percentage of the measured Davisson capacity. The latter consideration makes the φ/λR ratio remarkably valuable from an economic point of view. The higher the φ/λR ratio, the more cost effective the method (shorter piles or fewer required). Comparison of φ/λ φ/λ φ/λ φ/λR for Each Method Table 6-7 (LRFD Results for Dynamic Methods at EOD) presents the φ/λR ratio for all the methods at end of drive (EOD) at the reliability index, β = 2.50. Evident, is that all the newer methods: CAPWAP (φ/λR=0.459), PDA (φ/λR=0.480), and Paikowsky (φ/λR=0.475) give the highest φ/λR ratio, resulting in more economical design. The lower φ/λR values are given by the older driving formulas: FDOT (φ/λR=0.281), ENR (φ/λR=0.301), Modified ENR (φ/λR=0.258), and Sakai et al. (φ/λR=0.153) methods. The same trend was observed for a reliability index of 1.96. In the case of Beginning of Restrike (BOR), the newer methods again resulted in higher φ/λR ratios (Table 6-8) as compared to the older methods. However, there was one exception, 56 the Gates method, which had an φ/λR ratio of 0.541 at a reliability index of 1.96. The latter com-pares favorably with the newer driving methods. EOD versus BOR From Tables 6-7 (EOD) and 6-8 (BOR), the bias (λR: ratio of measured to predicted), is clearly higher for EOD than BOR. The latter suggest that the predicted pile capacities are increasing from EOD to BOR on a whole for the database (i.e., independent of soil type). This effect may be attributed to pile freeze (increase in skin friction due to dissipation of pore water pressure, or increase in lateral total stress, or even soil strength). However, LRFD phi factors are influenced by both the bias, λR , and coefficient of variation, COVR. In the case of the latter, the standard deviation, σ, is diminishing from EOD to BOR. However, the combined effect, COVR (ratio of σ/λ) which controls accuracy is changed little. Consequently, since the COVR is changing little, but the bias, λR, is diminishing (i.e., predicted capacity agree more closely to measured), the LRFD phi, φ, is diminishing. For instance, PDA, which has a bias, λ, of 1.344, standard deviation, σ, of 0.443, coefficient of variation, COVR of .329, and a resulting LRFD φ of 0.645 at a reliability of 2.5. Whereas, at BOR, the PDA has a bias, λ, of 1.036, standard deviation, σ, of 0.322, coefficient of variation, COVR of .311, and a resulting LRFD φ of 0.518 at a reliability of 2.5. However, the φ/λR ratio, indicating the percentage of the measured Davisson capacity that can be utilized for design, is 0.48 vs. 0.50 at EOD vs. BOR respectively. The latter suggests that if F.S. for ASD, and φ for LRFD at EOD vs. BOR were assigned different values (based on failure probabilities) then similar φ/λR ratios would be obtained. If so, similar pile lengths for design would occur and the resulting accuracy of EOD versus BOR would be com-parable at least for the PDA, CAPWAP, and Paikowsky methods. The latter analysis is based on the present database (45 to 75 cases). 57 Evaluation of smaller piles (i.e., capacities less than 1779 kN). In the case of the older methods (ENR, Gates, etc.) the cases were separated into capaci-ties (i.e., bins) larger than 1779 kN and capacities smaller than 1779 kN. Subsequently, the statistics (mean, standard deviation, etc.) for each bin was performed and LRFD phi factors were determined for each method. The results are summarized in Tables 6-7 and 6-8. The following general conclusions were obtained. For the Gates and FDOT methods, the bias for the capacities smaller than 1779 kN tended closer to unity in comparison to the overall (i.e., all cases) bias which tended to be a higher value. However, in the cases of the ENR and Modified ENR the biases are significantly less than one which makes them very unconservative. Moreover, the bias for cases with capacity smaller than 1779 kN tends to be even more unconservative (i.e., smaller). In terms of φ/λR, as a general observation, the piles with capacity larger than 1779 kN have larger values than the overall cases, and the piles with capacity smaller than 1779 kN have smaller values than the overall cases. Therefore, the equivalent ASD factor of safety decreases for cases with capacity larger than 1779 kN. An example of this pattern is shown with the Modi-fied ENR at EOD (β = 1.96), which safety factors decrease from 14.6 for pile with capacity smaller than 1779 kN to 7.75 for piles with capacity larger than 1779 kN. For a reliability index, β, equals 2.5 the decrease in safety factor is also by a half. Recommended Safety Factors The Factors of Safety shown in Tables 6-7 and 6-8 were calculated using Equation 5-9 with φ computed from Equation 5-28 (given in tables). All of the factors of safety that were calculated from Eq. 5-9 used a bridge span length of 27 meters to determine QD/QL. Table 6-9 summarizes the FS obtained for EOD and BOR in Tables 6-7 and 6-8 for the failure probabili- 58 Table 6-9. Recommended Safety Factors for Dynamic Methods FS at FS at FS at FS at Prediction Method pf =0.62% pf =2.50% pf =0.62% pf =2.50% Recommended FS BOR BOR EOD EOD EOD BOR CAPWAP 2.48 2.00 1.97 1.58 1.80 2.25 PDA 2.79 2.28 2.24 1.81 2.00 2.50 Paikowsky Energy 3.86 3.09 2.74 2.22 2.50 3.50 FDOT (overall) 1.74 1.31 2.16 1.59 1.90 1.50 FDOT (<1779 kN) 2.00 1.65 3.16 2.36 2.75 1.80 FDOT (>1779 kN) 1.49 1.14 1.10 0.87 1.00 1.30 ENR (overall) 28.60 20.08 16.02 11.93 14.00 24.00 ENR (<1779 kN) 15.35 12.56 18.40 13.80 16.00 14.00 ENR (>1779 kN) 28.84 20.12 13.39 10.07 12.00 24.00 Modified ENR (overall) 18.31 12.88 12.53 9.09 11.00 16.50 Modified ENR (<1779 kN) 8.70 7.36 20.82 14.60 17.70 8.00 Modified ENR (>1779 kN) 18.32 12.81 10.72 7.75 9.20 16.50 Gates (overall) 1.78 1.42 2.28 1.76 2.00 1.60 Gates (<1779 kN) 2.12 1.81 2.80 2.27 2.50 2.00 Gates (>1779 kN) 1.54 1.24 1.30 1.06 1.20 1.40 ties, pf, of 0.62 % and 2.50 %, respectively. The latter represent AASHTO suggested range of failure probabilities. The recommended Factors of Safeties in the last two columns of Table 6-9 for each method at EOD and BOR is the average from the two failure probabilities. It is evident, that the recommended FS in Table 6-9 are much higher than typically used for ENR and Modified ENR safety factors (FS=6.0). For such cases, Factor of Safety as high as 28.60 are recommended for a failure probability of 0.62 % at BOR. The latter suggests that ENR and Modified ENR are very unconservative (i.e., overestimate the pile capacity) methods. The old FDOT method recommends a safety factor equal to 1, but according to the LRFD approach, a higher safety factor should be used for design. It is recommended that a FS of 1.50 59 for BOR and 1.90 for EOD be used for the old FDOT method. Other methods such as CAPWAP, PDA, and Gates should use different values for FS at EOD and BOR (freeze, etc.). For the Paikowsky’s Energy method, it is recommended to modify only the safety factor at BOR from 2.50 to 3.50, while at EOD the actual safety factor of 2.50 is appropriate. Review of Typical ASD Safety Factors Based on the database, the current ASD Factor of Safeties may be evaluated for the various dynamic methods. Shown in Table 6-10 is the typical ASD Safety Factors presented in the literature. Based on the ASD Safety Factors, and a typical 27 m bridge span length (i.e., fixed QD/QL ), Equation 5-9 was used to compute the resistance factors, φ. Next based on the COVR and bias, λR , from the database, the reliability index, β, for a given method was computed from Equation 5-28. Due to the range of the reliability index, β, the probability of failure was subsequently computed from Figure 5-4. Both the reliability index, β, and the probability of failure, pf, for each method based on the assumed ASD Safety Factors are given in Table 6-10. Based on the results of Table 6-10 and AASHTO’s recommended probability of failure for single piles [i.e., should be between 0.62 % for (reliability index) β = 2.50 and 2.50% for β = 1.96] a number of conclusions are evident: The older ENR and Modified ENR procedures show extremely high probability of fail-ures, which reflect the level of unconservatism as discussed in preceding sections. On the other hand, Gates method show a probability of failure as between 0.010 % to 0.11% which is con-servative. Of the newer methods, CAPWAP is generally conservative (pf of .1% at EOD and .596 % at BOR), and PDA is less conservative (pf of 0.267 % at EOD and 1.36% at BOR). Table 6-9 presents the recommended Safety Factors for these methods based on AASHTO's recommended probabilities of failure. 60 Table 6-10. Summary of Results for Fitting the ASD to the LRFD ASD LRFD Equiv. ASD Design (EOD) ASD Design (BOR) Prediction Method FS φ φ/λR β Pf (%) φ/λR β Pf (%) CAPWAP 2.5 0.578 0.362 3.089 0.100 0.458 2.515 0.596 PDA 2.5 0.578 0.430 2.786 0.267 0.557 2.210 1.357 Paikowsky Energy 2.5 0.578 0.520 2.267 1.171 0.691 1.449 7.374 Sakai et al Energy 2.5 0.578 0.384 1.293 9.810 0.396 1.976 2.411 FDOT (overall) 1.0 1.444 0.606 1.146 12.600 0.561 1.445 7.430 FDOT (<1779 kN) 1.0 1.444 0.969 0.369 35.630 1.065 0.536 29.620 FDOT (>1779 kN) 1.0 1.444 0.457 2.274 1.150 0.525 1.688 4.575 ENR (overall) 6.0 0.241 0.805 0.700 24.210 1.024 0.115 45.440 ENR (<1779 kN) 6.0 0.241 0.963 0.395 34.660 1.293 -0.032 51.300 ENR (>1779 kN) 6.0 0.241 0.713 0.976 16.470 0.994 0.146 44.220 Modified ENR (overall) 6.0 0.241 0.540 1.263 10.340 0.662 0.788 21.550 Modified ENR (<1779 kN) 6.0 0.241 0.741 0.605 27.280 0.868 1.306 9.586 Modified ENR (>1779 kN) 6.0 0.241 0.454 1.534 6.258 0.640 0.815 20.770 Gates (overall) 3.0 0.481 0.276 3.067 0.108 0.255 3.718 0.010 Gates (<1779 kN) 3.0 0.481 0.450 2.676 0.363 0.451 3.703 0.010 Gates (>1779 kN) 3.0 0.481 0.214 4.694 0.0001 0.240 4.177 0.0015 61 CHAPTER 7 DETERMINATION OF CASE DAMPING, JC , AND SKIN AND TIP RESISTANCE FROM PDA TRACES As discussed in the previous chapter, the PDA Case method was found to be as accurate as other dynamic methods, Paikowsky, CAPWAP, etc. to estimate static pile capacity. This trend was observed for driven piles at both, EOD and BOR. However, to determine the static pile capacity from the PDA Case method, the user is required to estimate the damping coeffi-cient, Jc . The open literature suggests that the Jc values vary according to soil type, hammer size, pile size, etc. Typically from the Florida experience, engineers either employ GRL recom-mended Jc values based on soil type or use CAPWAP’s estimate of Jc . CAPWAP selects the Coefficient of Damping, skin resistance, toe resistance and quake along the pile in such a way that the predicted and measured force traces match, from a given velocity trace. Unfortunately, due to CAPWAP's multivariable input, there is some question to its uniqueness, requiring sig-nificant expertise by the user. Consequently, it would be very beneficial if the PDA's Case damping, Jc could be determined in the field, along with both skin and tip resistance for the pile. The latter would combine the features of the PDA with CAPWAP, as well as provide real time results without delays. In the next sections, two methods to determine the skin and the tip static capacity from the PDA traces are presented. The results will be compared to the Davisson’s capacity deter-mined by the static load test. METHOD I FOR ESTIMATING SKIN AND TIP RESISTANCE In this method, the GRL procedure to determine the total dynamic skin capacity is used as starting point. The GRL procedure (Likins et. al., 1988) makes use of the pile top measure- 62 ments of force and velocity histories during a hammer blow as recorded by the PDA. Figure 7-1 illustrates this procedure. Figure 7-1. GRL Procedure to Determine the Total Skin Capacity of a Driven Pile To obtain the total dynamic skin friction, the rise time, tr , must be determined. The rise time is defined as the time between the beginning of blow (force pulse rise) and its peak. This rise time, tr , is used to determine point “b” which is one rise time earlier than time “a.” At time “a,” the separation between force and velocity (Fa –ZVa) represents the total pile skin friction above that point on the pile. The point “a” was moved back one rise time from 2 L/c to exclude any reflections from the pile tip of the initial input wave. The assumption of this method is that the skin friction at the bottom one rise length is the same as the one additional rise 63 time length above. Based on this assumption, the skin resistance is the separation of force and velocity at point “a” added to the increase in resistance from point “b” to point “a” as stated below. ( ) ( ) ( ) [ ] b b a a a a ZV F ZV F ZV F SFT − − − + − = (7-1) Once the estimate of dynamic skin friction is determined using the GRL procedure, the static tip and skin capacity were calculated according to the following methodology (Method I). First, the total dynamic capacity, RTL, is determined using the Case solution for dynamic capac-ity (Equation 3-14 for Jc equals 0). Then, knowing the total dynamic capacity and the GRL esti-mate of skin dynamic capacity, the dynamic tip capacity, TT, is computed as: SFT RTL TT − = (7-2) where: TT Dynamic Tip Capacity RTL Total Dynamic Capacity SFT GRL Estimate of Dynamic Skin Capacity Next, the total tip to skin capacity ratio, T/S, is determined: SFT TT S T = / (7-3) The latter was developed to negate the damping in each (i.e., skin and tip), i.e., it is assumed that the tip to skin ratio for the dynamic capacity was similar to the tip to skin ratio for the static capacity. Finally, the total static capacity of the pile must be found. For this purpose, the PDA Rmax obtained from the driving records is used. METHOD II FOR ESTIMATING SKIN AND TIP RESISTANCE (PROPOSED) Description and Main Assumptions For this method, the tip/skin ratio is obtained directly from the force and velocity traces of the PDA. Figure 7-2 presents an example of a PDA force and velocity traces. This method 64 . -2000 -1000 0 1000 2000 3000 4000 5000 6000 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 Time (s) Force Velocity P - ZV 2L/c Figure 7-2. Typical Force, Velocity and Double Wave Up Traces from PDA differs from Method I in the procedure to obtain the tip to skin ratio. Instead of determine the dynamic skin resistance (GRL procedure) by extrapolating the skin resistance from the rise time before the tip reflection, the method considers the skin capacity prior the tip reflection and the tip reflection itself. For driven piles, when the hammer suddenly hits the pile, a compression wave is pro-duced. The wave travels at a constant speed through the pile, eventually reaching the pile tip, and, depending on the soil tip resistance, reflects back as a tension or a compression wave. Before the compression wave reaches the pile tip, a series of compression waves are reflected back to the top due to the skin resistance of the pile. By measuring the Force and Velocity (and velocity multiplied by impedance) at the pile top, the skin resistance can be calculated as the force minus the velocity (twice the wave up). This principle only applies before any reflection from the tip arrives. 65 In order to determine the time at which the pile tip reflection arrives to the pile top, the rise time must be considered. The rise time accounts for how fast the tip reflection is developed and it will depend on how fast the hammer transmits the energy to the pile top. The larger the rise time, the more difficult to have an estimate of the skin capacity of the pile. Knowing the rise time allow us to determine the skin resistance at a distance "x" from the top of the pile as: c RT L x 2 − = (7-4) where: L Length of Pile Below Gages c Wave Speed RT Rise Time Under this consideration all the skin capacity near the tip of the pile (length equals RT/2c) can not be calculated. Figure 7-3 shows the plot of force minus velocity for the same PDA signal presented in Figure 7-2 (i.e., double wave up). After that point in time (point "b") the reflection from the tip begins to arrive. From point "b" to a time equals 2L/c from the force peak (point “c”), both skin and tip reflections interact creating an abrupt change in the force up trace. The decrease in the force up trace shown in Figure 7-3 corresponds to the increase in velocity and decrease in the force at time 2L/c (see Figure 7-2). This condition is attributed to piles with small tip capacity, and piles that, although they possess large tip capacity, the energy imparted by the hammer is too large to mobilize the tip capacity. The proposed method makes use of the two concepts mentioned to determine the tip to skin ratio. The basic assumption of this method is that the tip/skin ratio is equal to the decrease in the wave up trace from point "b" (time 2L/c minus rise time) to point "c" (time 2L/c from 66 Figure 7-3. Procedure to Determine Tip to Skin Ratio for the Suggested Method maximum force) divided by the net increase in the wave up trace from point "a" to point "b." The mathematical expression for the proposed method is ( ) ( ) [ ] ( ) ( ) [ ] a a b b c c b b ZV F ZV F ZV F ZV F Skin Tip − − − − − − = (7-5) The use of wave up at point "a" is to account for any error in the instrumentation readings. Because a decrease in the wave up is needed to obtain a tip to skin ratio, the hammer must be able to mobilize the whole pile capacity. Case Damping Coefficient, Jc , versus Tip to Skin Ratio In the process of finding a method to obtain the Case damping coefficient, Jc , directly from the PDA signal, the relation between the tip to skin ratio and the Jc value was studied. For the latter, the measured Davisson’s capacity was substituted as RSP in Equation 3-14, and the Jc 67 coefficient was back calculated. Subsequently, the computed damping value, Jc was plotted versus the suggested tip to skin ratio. The results are presented in Figure 7-4. 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.0 2.0 4.0 6.0 8.0 10.0 Suggested Tip / Skin Ratio Suggested Regression Curve Jc = -0.09744 ln (Tip/Skin) + 0.2686 Figure 7-4. Suggested Tip to Skin Ratio versus Case Damping Coefficient Two observations can be made from Figure 7-4. First, for tip to skin ratios smaller than 1.0, the Jc is on the order of 0.4 to 0.6, and there is a tendency for Jc to decrease as the tip to skin ratio is increased. Second, the values of Case damping are scattered. It is believed that there are two reasons for the latter: 1) the hammer may or may not impart enough energy to mobilize the static pile capacity (results in very low Jc values in Figure 7-4). And 2) PDA capacity predic-tions assume that all pile damping occurs at pile toe (may not always be true). However, as part of this study a sensitivity analysis was performed on the influence of Jc on the static pile capacity 68 (see next section). An important finding of this study is that a 30% variation in Jc results in only a 10 % variation in static pile capacity. Figure 7-4 shows the proposed regression curve for the Jc versus Tip/Skin Ratio. This curve is significant, because it allows us to obtain a Case damping coefficient, Jc, directly from the PDA traces, without taking into consideration the soil type in which the pile is embedded. The suggested equation to correlate tip to skin ratio to Jc is: 2686 . 0 ln 09744 . 0 +       − = Skin Tip Jc (7-6) Automation of Proposed Method II A Fortran program was developed in order to simplify the calculation process. This pro-gram is able to read directly the PDA recorded force and velocity signal and perform the data reduction. The input parameters for the program are the length of the pile below the gages, the wave speed, the material modulus, the cross-sectional area, and the data-sampling interval. The output file begins with a general description of the pile itself and the driving infor-mation entered. Next, it presents in tabular form the time increment, velocity, force, velocity times impedance, displacement, and the double wave up (P – ZV) results (Figure 7-3). Finally, the program computes the Tip to skin ratio, and the case damping, Jc value. Appendix I contains the PDA traces (force, velocity, double force up, and displacement) for the signals studied. At the end of the output file, the results from the Method I and II (sug-gested) are presented. In addition, the significant times such as maximum force, T1, time T2 (T1 + 2L/c), rise time, etc, are also presented. Appendix J presents an example of the Fortran output. Only the first five pages and the last page of the output are included. 69 SENSITIVITY ANALYSIS OF CASE DAMPING COEFFICIENT, JC In order to measure how sensitive the Case resistance (RSP) solution is to the variation of Jc , a sensitivity study was performed. The following steps were followed to obtain a relation between the coefficient of variance for the resistance and the coefficient of variance for Jc . Using the Case resistance approach for estimating pile capacity, R, the average (bars for average) resistance based on the average value of Jc is given as ( )( ) ( )( ) [ ] 2 2 1 1 1 1 2 1 ZV F J ZV F J R c c − + + + − = (7-7) If the standard deviation of the resistance is defined in terms of the standard deviation of the Case damping, we obtain: ( ) ( ) [ ] 2 2 1 1 2 1 ZV F ZV F J R − + + − = σ σ (7-8) Then the covariance of the resistance can be written as: ( ) ( ) [ ] ( )( ) ( )( ) 2 2 1 1 2 2 1 1 1 1 ZV F J ZV F J ZV F ZV F R COV c c J R R − + + + − − + + − = = σ σ (7-9) If we define the constant η in terms of the force and velocity at time T1 and time T2 (T1 plus 2L/c), then ( ) ( ) 1 1 2 2 ZV F ZV F + − = η (7-10) Next divide the numerator and denominator of Equation 7-10 by the average Jc and making the appropriate substitutions for η we obtain:               − + + = 1 2 1 1 1 η c R J J COV COV (7-11) Equation 7-11 allows us to determine the variance in the Jc for a given percent of vari-ance in the resistance, or vice-versa. The next section provides the results of a study of 18 piles. 70 SELECTION OF STATIC AND DYNAMIC LOAD TEST DATA The criteria for selecting a load test for inclusion in this study involved: 1) a static load test was performed and Davisson’s failure was obtained and 2)PDA force and velocity traces were available for the pile. For the dynamic load test, the selected blow (PDA) was chosen as close to the static load test date as possible (i.e., minimize freeze). For all the cases reported, the closest dynamic load test date occurred after the static load test was performed (i.e., the static capacity was compared to the Beginning of Restrike). Other useful information includes having a measured skin and tip resistance. The latter occurred through either pull out static load tests or Osterberg load cell test. For each pile, it was required to know both geometric and driving parameters, such as, wave speed, pile length below the gages, cross-sectional area, and material modulus. Based on the latter requirements, 18-test piles were considered. Table 7-1 summarizes general information of each load test pile. The pile population represents 9 sites and 3 diameters, while the range of depths of the piles below the gages varies from 9.15 to 34.45 meters. Only one test pile is from a location outside the State of Florida. COMPARISON OF SKIN, TIP AND TOTAL PILE CAPACITIES Results In the following sections, the results from the proposed Method II will be presented in addition to the results of other methodologies: Method I, PDA (total capacity only), and CAPWAP (skin, tip and total). The reader is referred to Appendix I for the plots of force, velocity, double wave up, and displacement traces for each case studied. The first results pre-sented are predicted versus measured total pile capacity. For Methods I and II, the Jc used in Equation 3-14 was obtained from Figure 7-4 with the Tip to skin ratio obtained from the force up traces and Equations 7-3 (Method I) and 7-5 (Method II). Next, the predicted and measured skin 71 Table 7-1. Geometric and Driving Information of Load Test Piles Shaft Shaft Wave Material Cross-Sec. Data Sampling Length Diameter Speed Modulus Area Increment File Name Site Location Pile Location (m) (m) (m/s) (kN/m2) (m2) (s) TS4I Pascagoula, Mississippi Test Site 4 34.45 0.61 4325.0 44905380 0.37 0.0001 TS1B102 Vilano Bridge Test Site 1 29.88 0.61 3929.9 37684870 0.37 0.0001 B5RS2 Escambria Bridge Bent 5 / Pile 27 26.83 0.61 4425.0 47005420 0.37 0.0001 TS13A1-A Buckman Bridge Test Site 13 33.84 0.76 3935.7 37184330 0.48 0.0001 TS19FRCB Buckman Bridge Test Site 19 32.74 0.76 3993.9 38292260 0.48 0.0001 TS24ALTA Buckman Bridge Test Site 24 30.98 0.76 3937.1 37211220 0.48 0.0001 TS29RC Buckman Bridge Test Site 29 29.27 0.76 3984.2 39470520 0.48 0.0001 F6-58A Acosta Bridge Pier F6 / Pile 44 19.21 0.61 3872.0 35989530 0.37 0.0001 G13-37 Acosta Bridge Pier G13 / Pile 95 17.68 0.61 4325.0 39470519 0.37 0.0002 H2-27B Acosta Bridge Pier H2 / Pile 26 10.64 0.61 4116.2 40656360 0.37 0.0001 VLWA-61D Vilano West Bridge STA. 142+74 19.05 0.46 4186.6 42075920 0.21 0.0001 VLE-32C Vilano East Bridge STA. 183+47 10.98 0.46 3730.2 33402750 0.21 0.0001 BKM30J Buckman Bridge STA. 362+90 9.15 0.46 3900.6 36523850 0.21 0.0001 BZ83N Seebreze Bridge STA. 353+15 25.91 0.46 3926.6 37011970 0.21 0.0001 AUC63K Aucilla Bridge STA. 494+47 20.43 0.46 3993.8 38291580 0.21 0.0001 B1-76F Choctawhatche Bridge Pier FSB3 / Pile 2 24.36 0.61 4172.7 41796700 0.37 0.0001 PR5R2 Choctawhatche Bridge Pier 5 / Pile x 18.63 0.76 4329.3 43612000 0.42 0.0001 B8-97R2 Choctawhatche Bridge Pier 11 / Pile 38 29.76 0.76 4643.0 51750160 0.42 0.0001 B14-89R2 Choctawhatche Bridge Pier 23 / Pile 13 29.27 0.76 4520.8 49062030 0.42 0.0001 B17-94R2 Choctawhatche Bridge Pier 29 / Pile 7 29.00 0.76 4208.8 42524060 0.42 0.0001 PR35FIN Choctawhatche Bridge Pier 35 / Pile 7 27.16 0.76 4292.1 44224220 0.42 0.0001 B23-76F2 Choctawhatche Bridge Pier 41 / Pile x 24.09 0.76 4412.0 46728270 0.42 0.0001 B27-63F Choctawhatche Bridge FSB 26 / Pile 3 19.82 0.61 4283.2 44041520 0.37 0.0001 72 and tip resistance for Methods I, II and CAPWAP are presented. For the predicted skin capacity of the piles, the latter was evaluated by either pullout tests or Osterberg tests in which skin fric-tion failure occurred. Table 7-2 summarizes the results for the suggested Method II and the load tests. The tip to skin ratio is presented together with the recommended Case damping coefficient, Jc . Based on the latter parameter, the PDA Case solution for total capacity was found. Also presented in Table 7-2 are the measured Davisson’s capacity for all compression tests, and the skin capacity for piles with a performed tensile load test or an Osterberg test with skin failure. Table 7-3 presents the estimated capacities using Method I and CAPWAP. For Method I, the total dynamic capacity was obtained using the Case solution for time T1 located at the maximum force entering the pile. Then, the tip and the tip to skin ratio were calculated using Equations 7-2 and 7-3. With this tip to skin ratio and the PDA Rmax the static skin and tip capacity for the Method I were calculated. Total Capacity Predictions A series of figures show the results from the three methods studied, i.e., CAPWAP, Method I (using PDA Rmax), and the suggested Method II in comparison with the measured capacity. Figure 7-5 shows the Davisson’s capacity versus Method II (suggested) estimated static pile capacity. The CAPWAP estimated capacity versus the Davisson’s capacity is presented in Figure 7-6, and Method I's estimated capacity vs. Davisson's capacity is presented in Figure 7-7. For the total capacity analysis, only 18 cases out of 23 cases were used. The bias factor λR, the standard deviation, σ, and the coefficient of variance, COVR, were calculated (see bias factor, standard deviation, and coefficient of variance definition on Chapter 5). Table 7-4 presents the findings for the three methods studied. 73 Table 7-2. Suggested Method II Results and Static Load Test Results Suggested Method Static Capacities Load Test Results (kN) Location Tip/Skin Jc Total Skin Tip Type of Davisson File Name Ratio (kN) (kN) (kN) Test Total Skin Tip TS4I Pascagoula, Mississippi 0.57 0.33 5069.7 3261.9 1807.8 Osterb.(T&T) 3282.6 1316.6 1966.0 TS1B102 Vilano Bridge 0.8 0.29 4140.9 2270.8 1870.1 Static (C&T) 4919.5 2775.6 2143.9 B5RS2 Escambria Bridge 1.34 0.24 3256.3 1389.9 1866.4 Static (C) 3780.8 n/a n/a TS13A1-A Buckman Bridge 0.97 0.27 5456.4 2772.2 2684.2 Static (C&T) 4092.2 1601.3 2490.9 TS19FRCB Buckman Bridge 2.16 0.19 5279.6 1671.2 3608.3 Static (C&T) 4376.8 1396.7 2980.2 TS24ALTA Buckman Bridge 0.48 0.34 6599.9 4455.8 2144.1 Static (C&T) 4892.8 2490.9 2401.9 TS29RC Buckman Bridge 2.98 0.16 4797.8 1207.0 3590.8 Static (C&T) 4519.2 1734.7 2784.4 F6-58A Acosta Bridge 0.58 0.32 4096.2 2589.1 1507.1 Static (C) 3451.6 n/a n/a G13-37 Acosta Bridge 1.05 0.26 3990.6 1948.9 2041.7 Static (C) 4964.0 n/a n/a H2-27B Acosta Bridge 5.50 0.10 3530.5 543.1 2987.3 Static (C) 2570.9 n/a n/a VLWA-61D Vilano West Bridge 4.70 0.12 1403.8 246.5 1157.4 Osterberg n/a 382.5 n/a VLE-32C Vilano East Bridge 1.06 0.26 2722.7 1325.2 1397.5 Osterberg n/a 1150.3 n/a BKM30J Buckman Bridge 9.18 0.05 1561.5 153.4 1408.1 Osterberg n/a 120.1 n/a BZ83N Seebreze Bridge 0.99 0.27 3244.8 1629.5 1615.3 Osterberg n/a 1470.5 n/a AUC63K Aucilla Bridge 3.33 0.15 2419.9 558.4 1861.4 Osterberg n/a 938.5 n/a B1-76F Choctawhatche Bridge 0.84 0.29 3349.8 1822.7 1527.1 Static (C) 2215.1 n/a n/a PR5R2 Choctawhatche Bridge 5.53 0.10 3722.9 570.2 3152.7 Static (C) 5444.4 n/a n/a B8-97R2 Choctawhatche Bridge 0.33 0.38 6618.2 4986.3 1631.9 Static (C) 6360.6 n/a n/a B14-89R2 Choctawhatche Bridge 1.37 0.24 4734.1 1997.2 2736.9 Static (C) 2846.7 n/a n/a B17-94R2 Choctawhatche Bridge 0.16 0.45 3196.9 2752.9 444.0 Static (C) 4074.4 n/a n/a PR35FIN Choctawhatche Bridge 0.46 0.35 5013.0 3440.3 1572.7 Static (C) 6485.2 n/a n/a B23-76F2 Choctawhatche Bridge 3.62 0.14 4355.2 942.9 3412.3 Static (C) 6253.9 n/a n/a B27-63F Choctawhatche Bridge 1.17 0.25 3547.6 1633.8 1913.8 Static (C) 4270.1 n/a n/a 74 Table 7-3. CAPWAP and Method I Results Method I CAPWAP Static Results Location Tip/Skin Total (PDA) Skin Tip Tip/Skin Total Skin Tip File Name Ratio (kN) (kN) (kN) Ratio (kN) (kN) (kN) TS4I Pascagoula, Mississippi 1.6 4581.4 1785.4 2796.0 2.6 4518.3 1265.9 3252.4 TS1B102 Vilano Bridge 0.1 3433.9 3116.0 317.8 0.3 3915.1 2930.8 984.3 B5RS2 Escambria Bridge 1.1 2526.5 1217.6 1308.9 0.6 2610.1 1623.1 987.0 TS13A1-A Buckman Bridge 0.9 4803.8 2468.6 2335.3 1.1 4007.6 1950.0 2057.6 TS19FRCB Buckman Bridge 1.9 5640.1 1947.5 3692.5 2.1 6510.1 2087.9 4422.2 TS24ALTA Buckman Bridge 9.0 6636.4 661.8 5974.6 0.5 6240.5 4123.7 2116.8 TS29RC Buckman Bridge 3.1 5239.7 1276.1 3963.6 3.6 5115.2 1123.6 3991.6 F6-58A Acosta Bridge 0.3 3771.9 2926.2 845.7 0.1 3474.8 3052.7 422.1 G13-37 Acosta Bridge 1.1 5177.5 2457.7 2719.8 1.2 4826.1 2224.0 2602.1 H2-27B Acosta Bridge 3.2 3967.6 948.1 3019.6 5.2 4091.3 658.3 3433.0 VLWA-61D Vilano West Bridge 1.5 1080.9 430.1 650.8 0.9 916.3 494.6 421.7 VLE-32C Vilano East Bridge 0.8 2570.9 1395.0 1176.0 0.9 2691.0 1400.7 1290.4 BKM30J Buckman Bridge 1.5 1961.6 772.6 1189.0 1.2 1939.3 881.6 1057.7 BZ83N Seebreze Bridge 0.2 2628.8 2183.4 445.4 0.3 2361.9 1774.8 587.1 AUC63K Aucilla Bridge 4.6 1757.0 313.4 1443.6 0.7 1823.7 1045.3 778.4 B1-76F Choctawhatche Bridge 0.0 3220.4 3233.3 -12.9 0.2 2231.6 1907.3 324.3 PR5R2 Choctawhatche Bridge 8.7 3255.9 334.7 2921.2 6.0 2596.3 371.0 2225.3 B8-97R2 Choctawhatche Bridge 1.4 4554.8 1863.6 2691.1 0.2 3621.6 3033.1 588.5 B14-89R2 Choctawhatche Bridge 0.5 3131.4 2029.4 1102.0 0.1 1975.4 1802.3 173.0 B17-94R2 Choctawhatche Bridge 0.0 3700.7 3780.1 -79.4 0.0 3452.5 3288.9 163.7 PR35FIN Choctawhatche Bridge 0.3 4661.5 3650.4 1011.1 0.3 4045.0 3231.9 813.1 B23-76F2 Choctawhatche Bridge -7.9 4448.0 -645.7 5093.7 3.8 2155.9 453.7 1702.2 B27-63F Choctawhatche Bridge 0.8 2673.2 1452.1 1221.2 3.6 2503.3 542.2 1961.1 75 0 1000 2000 3000 4000 5000 6000 7000 8000 0 1000 2000 3000 4000 5000 6000 7000 8000 Davisson Capacity (kN) Number of Cases = 18 Mean = 1.00 Std. Dev. = 0.29 COV = 0.29 Figure 7-5. Davisson Capacity vs. Suggested Method II Capacity 0 1000 2000 3000 4000 5000 6000 7000 8000 0 1000 2000 3000 4000 5000 6000 7000 8000 Davisson Capacity (kN) Number of Cases = 18 Mean = 1.28 Std. Dev. = 0.58 COV = 0.45 Figure 7-6. Davisson Capacity vs. CAPWAP Capacity 76 0 1000 2000 3000 4000 5000 6000 7000 8000 0 1000 2000 3000 4000 5000 6000 7000 8000 Davisson Capacity (kN) Number of Cases = 18 Mean = 1.09 Std. Dev. = 0.35 COV = 0.32 Figure 7-7. Davisson Capacity vs. PDA Rmax Capacity (used in Method I) Table 7-4. Comparison of Mean, Standard Deviation, and COVR. Method Mean, λR Std. Dev. σ COVR Suggested Method II 1.00 0.29 0.29 CAPWAP 1.28 0.58 0.45 Method I 1.09 0.35 0.32 It should be noted that although the number of cases studied were only 18, the mean, standard deviation, and coefficient of variance for CAPWAP and PDA show good agreement with the database. According to the database, for PDA capacity at BOR, the mean, standard deviation, and COV are 1.04, 0.32 and 0.31, respectively. For CAPWAP at BOR the mean is 1.26, the standard deviation is 0.44, and the COVR is 0.35. 77 From Table 7-4, it can be observed that the best method to estimate the Davisson’s capacity is the suggested Method II with a mean value of 1.0 (it is important to remember that the suggested method was calibrated to have a mean value of 1.00). The PDA underestimated the Davisson’s capacity by 9 percent, while the CAPWAP procedure underestimated the Davisson’s capacity by 26 percent at BOR. In Chapter 6, it was explained how the coefficient of variance could be a useful tool to measure the accuracy of a dynamic method. Making use of this criterion, the suggested method (COVR equals 0.29) can be considered the most accurate of the three methods followed by the PDA (COVR equals 0.32). The CAPWAP procedure was the less accurate of the three methods (COVR equals 0.45). Skin and Tip Capacity Predictions The number of cases to evaluate the skin and tip predictions is much smaller than the number of cases used for total capacity prediction. For the skin prediction, the total number of cases is 10, while for the tip prediction the number of cases drops to only six. The reason for this decrease in the number of cases is the limited number of tensile test and Osterberg tests. From the six Osterberg tests, one test failed at the tip and five failed at the skin). The tensile tests are very helpful to determine the skin capacity. With the skin capacity calculated, and knowing the total static capacity the tip capacity can be calculated. Table 7-5 summarizes the statistical results for both, tip and skin predictions. Figures 7-8 to 7-10 present the predicted skin capacity versus the estimated skin capacity for the suggested Method II, CAPWAP, and Method I, respectively. For the same methods, the tip predictions versus the measured tip capacity are shown in Figures 7-11 to 7-13. 78 Table 7-5. Statistical Analysis for Skin and Tip Predictions Method Skin Capacity Prediction (10 Cases) Tip Capacity Prediction (6 Cases) Mean, λR Std. Dev. σ COVR Mean, λR Std. Dev. σ COVR Suggested Method 1.04 0.40 0.39 0.96 0.16 0.17 CAPWAP 0.80 0.35 0.43 1.08 0.59 0.55 Method I 1.29 1.15 0.89 1.94 2.47 1.27 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Davisson Skin Capacity (kN) Number of Cases = 10 Mean = 1.04 Std. Dev. = 0.40 COV = 0.39 Figure 7-8. Davisson Capacity vs. Suggested Method Capacity (Skin Capacity) 79 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Davisson Skin Capacity (kN) Number of Cases = 10 Mean = 0.80 Std. Dev. = 0.35 COV = 0.43 Figure 7-9. Davisson Capacity vs. CAPWAP Capacity (Skin Capacity) 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Davisson Skin Capacity (kN) Number of Cases = 10 Mean = 1.29 Std. Dev. = 1.15 COV = 0.89 Figure 7-10. Davisson Capacity vs. Method I Skin Capacity 80 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 Davisson Tip Capacity (kN) Number of Cases = 6 Mean = 0.96 Std. Dev. = 0.16 COV = 0.17 Figure 7-11. Davisson Capacity vs. Suggested Method II (Tip Capacity) 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 Davisson Tip Capacity (kN) Number of Cases = 6 Mean = 1.08 Std. Dev. = 0.59 COV = 0.55 Figure 7-12. Davisson Capacity vs. CAPWAP Capacity (Tip Capacity) 81 0 1000 2000 3000 4000 5000 6000 0 1000 2000 3000 4000 5000 6000 Davisson Tip Capacity (kN) Number of Cases = 6 Mean = 1.94 Std. Dev. = 2.47 COV = 1.27 Figure 7-13. Davisson Capacity vs. Method I Capacity (Tip Capacity) In terms of skin prediction, the suggested Method II, on average, slightly underestimate the measured skin capacity by 4 percent, while the CAPWAP procedure overestimate the mea-sured skin capacity by 20 percent. On the other hand, Method I underestimates the skin capacity by 29 percent. The accuracy of the methods (govern by the coefficient of variance) is not as good as for determining the total capacity. Both, the suggested Method II and CAPWAP have very similar COVR values (0.39 and 0.43, respectively), while Method I is the less accurate method with a COVR of 0.89. The tip predictions show excellent results for the suggested Method II with a mean of 0.96 and COVR of 0.17. The CAPWAP procedure tends to underestimate the tip capacity by 8 percent and Method I by 94 percent. The accuracy is quite good for the suggested method 82 (COVR = 0.17), followed by CAPWAP (COVR = 0.55), and Method I with the worst accuracy of the three methods (COVR = 1.27). Sensitivity of Jc and Pile Resistance The 23 cases studied were analyzed using Equation 7-8 to determine the sensitivity of the Jc . Table 7-6 presents the results for a change in covariance of the resistance of +/- 10 %, and for a change in covariance of Jc for the same amount. An average value for each condition has been calculated at the bottom of the table. According to Table 7-6, if the coefficient of variance of Jc is modified by 10 %, the coefficient of variance of the resistance will be altered only 3.15 %. Another way to look at this finding is by changing the coefficient of variance of the resistance. If the coefficient of variance of the resistance is altered by 10 %, the resulting variation in Jc is on average 33.67 %. The latter proves that the pile’s static resistance is not very sensitive to the Case-damping coefficient. The scattered pattern in the back calculated Jc on Figure 7-4 reflects the latter fact. Figure 7-14 shows the sensitivity analysis results for variation on the pile resistance capacity by 10 and 20 %. The solid line represents the suggested curve (Equation 7-6) relating the tip to skin ratio to Jc . The dashed lines show how much the suggested line must be altered to obtain a change in the resistance of 10 or 20 percent. 83 Table 7-6. Sensitivity Analysis Results for Variation of Case Damping Coefficient and Pile Resistance Change COV (Rs) Change COV (Jc ) η Constant (+10%) (-10%) (+10%) (-10%) File Name Site Location P1 (kN) ZV1 (kN) P2 (kN) ZV2 (kN) Jc (Jc ) (Jc ) (Rs) (Rs) TS4I Pascagoula, Mississippi 6137.9 5713.3 772.1 -849.3 0.326 0.14 -3.040 -30.40 30.40 -3.29 3.29 TS1B102 Vilano Bridge 5662.7 5526.2 606.1 35.7 0.287 0.05 -2.859 -28.59 28.59 -3.50 3.50 B5RS2 Escambria Bridge 4936.9 3949.5 158 355.5 0.240 -0.02 -2.985 -29.85 29.85 -3.35 3.35 TS13A1-A Buckman Bridge 7051.8 7688.8 -545.9 -682.4 0.271 0.01 -2.759 -27.59 27.59 -3.62 3.62 TS19FRCB Buckman Bridge 7941 4895 -46.2 2216.1 0.193 -0.18 -2.629 -26.29 26.29 -3.80 3.80 TS24ALTA Buckman Bridge 6690.3 6280.7 227.6 -3231.4 0.340 0.27 -4.080 -40.80 40.80 -2.45 2.45 TS29RC Buckman Bridge 8014.4 8730 -238.5 3577.9 0.162 -0.23 -2.881 -28.81 28.81 -3.47 3.47 F6-58A Acosta Bridge 5252.8 4527.1 276.5 -898.5 0.321 0.12 -2.966 -29.66 29.66 -3.37 3.37 G13-37 Acosta Bridge 5395 5157.5 576.8 407.2 0.264 0.02 -2.912 -29.12 29.12 -3.43 3.43 H2-27B Acosta Bridge 5214.7 4994.4 807.9 2717.5 0.102 -0.19 -5.714 -57.14 57.14 -1.75 1.75 VLWA-61D Vilano West Bridge 4074.9 3549.8 -231.1 3276.7 0.120 -0.46 -2.082 -20.82 20.82 -4.80 4.80 VLE-32C Vilano East Bridge 4061.2 2975.7 1029.3 923.5 0.260 0.02 -2.964 -29.64 29.64 -3.37 3.37 BKM30J Buckman Bridge 4383.7 3757.4 371.8 4735.9 0.050 -0.54 -5.041 -50.41 50.41 -1.98 1.98 BZ83N Seebreze Bridge 4688.7 3506.6 965.3 571.3 0.270 0.05 -3.078 -30.78 30.78 -3.25 3.25 AUC63K Aucilla Bridge 4328.3 3626.9 -160.3 1502.9 0.151 -0.21 -3.332 -33.32 33.32 -3.00 3.00 B1-76F Choctawhatche Bridge 5251.1 3389 111.7 -297.9 0.286 0.05 -2.845 -28.45 28.45 -3.52 3.52 PR5R2 Choctawhatche Bridge 6337.9 5708.3 713.5 3777.6 0.102 -0.25 -4.828 -48.28 48.28 -2.07 2.07 B8-97R2 Choctawhatche Bridge 8312.8 7291.1 1625.4 -928.8 0.377 0.16 -2.691 -26.91 26.91 -3.72 3.72 B14-89R2 Choctawhatche Bridge 7144.5 6375.7 859.1 1537.4 0.238 -0.05 -2.800 -28.00 28.00 -3.57 3.57 B17-94R2 Choctawhatche Bridge 2904.7 2483.8 842 -1515.5 0.446 0.44 -4.730 -47.30 47.30 -2.11 2.11 PR35FIN Choctawhatche Bridge 5709.8 5624 1159.1 -772.8 0.345 0.17 -3.090 -30.90 30.90 -3.24 3.24 B23-76F2 Choctawhatche Bridge 6795.9 6487 -485.4 1853.4 0.143 -0.18 -3.899 -38.99 38.99 -2.56 2.56 B27-63F Choctawhatche Bridge 5008.1 3975.9 420.5 114.7 0.253 0.03 -3.231 -32.31 32.31 -3.09 3.09 Averages -33.67 33.67 -3.15 3.15 84 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.0 2.0 4.0 6.0 8.0 10.0 Suggested Tip / Skin Ratio Back Calculated Jc Suggested Curve + 10 % Resistance Change - 10 % Resistance Change + 20 % Resistance Change - 20 % Resistance Change Figure 7-14. Case Damping Coefficient Jc Sensitivity Analysis Results 85 CHAPTER 8 DEVELOPMENT OF NEW FIELD INSTRUMENTATION Currently, the Florida Department of Transportation (FDOT) recommends the use of the Pile Driving Analyzer (PDA instrumentation system) for dynamic testing of piles. Approxi-mately 10% of the piles (Test Piles) in a foundation system are dynamically tested with the PDA (90% of the production piles in the foundation system are untested). The information from the Test Piles are used together with GRLWEAP and CAPWAP software to produce a driving criteria and production pile length for the rest of the production piles. Without discrediting the immeasurable benefit of the PDA, several shortcomings in its application should be pointed out. The PDA is a very expensive (software, hardware, instrumen-tation) package to license and operate (> $70,000). Operation of PDA requires a well-trained engineer/technician. Moreover, the test is intrusive to the construction process, as it requires a technician to climb the leads to attach the sensors to the pile (close to the pile head). This is dangerous for the technician and delays the driving procedure, sometimes by about two hours. Finally, among several capacity assessment theories available, the PDA field unit only uses the CASE method for total capacity with no estimate of skin and tip resistances. This eliminates the possibility of using other dynamic methods for capacity assessment (Total, Skin, and Tip). Recent new technology developed for LAN computer systems has made it possible to design and build new equipment for dynamic load testing and pile monitoring. Using state of the art in wireless communication technology, the equipment (Figure 8-1) transmits acceleration and strain information from the pile to a laptop computer in real time (10,000 readings/sec) for stor-age and signal processing (capacity assessment: total, skin and tip). Because of its small size, 86 Figure 8-1. System Concept low cost (instrumentation and transmitter), the equipment may be permanently cast in the pile. The latter does away with the consuming process of sensor installation, as well as climbing the pile leads to attach cables to the pile head which now occurs with standard PDA site work. Due to the low cost of the new system, it would be possible to monitor every pile in the foundation, removing all uncertainty and eliminating the need for a driving criterion for produc-tion piles. Furthermore, the transducers could be installed in different locations along the pile axis. Example of the latter is at the pile tip (Figure 8-2), which would develop a clear differen-tiation between tip and skin resistances. Another use would be determining pile damping for improved static pile capacity assessment. The equipment is separated into two major components (Figure 8-1): a Non-Recoverable Unit embedded in the concrete pile, and a Receiver and Data Processing Unit used to recover the information from the first unit and process it. Figure 8-1 depicts the system and Figure 8-2 schematically shows its operation. 87 Figure 8-2. System Operation The non-recoverable unit is composed of two accelerometers, two strain transducers, a signal conditioning unit and a sending unit with antenna (Figure 8-3 (a)). Every component is embedded in the pile including the antenna, which is mounted flush to the pile face. The receiver and data processing module consists of three independent apparatus (Figure 8-3(b)), a receiver and conditioning unit, a data acquisition card, and a laptop computer with built in acquisition and data processing software. This entire unit is portable and can be placed up to 150 m (500 ft) away from the pile head (from transmitter antenna). A more in depth description of each component is presented in Chapter 10. The acquisition and data processing software, called Pile Monitoring, controls the data acquisition card, retrieves the sensors signals, processes and presents the information, records it, and provides pertinent information about pile capacity and pile damage according to methods that will be described in Chapter 7. A detailed description of this software is given in Chapter 9. 88 (b) (a) Figure 8-3. Stress Wave Acquisition Device (a) Non-recoverable Unit (b) Receiver and Data Processing Unit In addition to the CASE method for capacity determination, used by the PDA, other capacity methods are implemented in the software. These methods, encouraged by the Federal Highway Administration (FHWA) and the State Transportation Department, are the Paikowsky Energy method (FHWA) and the recently proposed University of Florida Method (UF Method). Next a brief discussion of the instrumentation (strain gauges and accelerometers) requirements will be presented. 89 CHAPTER 9 INSTRUMENTATION REQUIREMENTS FOR DYNAMIC TESTING EQUIPMENT The new Instrumentation system should be composed of transducers, Signal Condi-tioning, and an apparatus for recording, reducing and displaying data. This chapter covers the ASTM D4945 requirements and the currently used apparatus, PDA, characteristics. A summary of all the requirements is presented later in this chapter. TRANSDUCERS According to ASTM specifications, the transducers should be capable of independently measuring strain and acceleration versus time at a specific location along the pile axis. For this purpose, at least two strain gages and two accelerometers should be used, as shown in Figure 9-1 (One accelerometer and one strain gage on one side and another accelerometer and strain gage on the opposite side). The latter is to eliminate bending. Also, the instrumentation should be securely attached so they do not slip. (a) (b) Figure 9-1. PDA Strain Gages and Accelerometers (a) Installation Top View (b) Installation Side View 90 Force or strain transducers shall have a linear output over the entire range of possible strains, and their natural frequency higher than 2000 Hz. Acceleration, velocity or displacement transducers can be used to determine velocity, through the integration of acceleration records or derivation of displacements measurements with respect to time. These calculations should be automatically performed by an apparatus for reducing data. Accelerometers should have a resonant frequency above 2500 Hz and their output shall be linear to at least 1000g (g = 9.81 m/sec2 = 32.2 ft/sec2) and 1000 Hz when used in concrete piles. If DC accelerometers are used, they should be damped with low pass filters having a minimum frequency of 1500 Hz. The transducers should be calibrated to an accuracy of 3% through the applicable mea-surement range. SIGNAL CONDITIONING ASTM requires that the signal conditioning for force and velocity have equal frequency response curves to avoid relative phase shifts and relative amplitude differences. Signal condi-tionings are electrical circuits that provide power to the sensors, transforming the sensor’s signals to an analog format, and increasing the amplitude of signal (gain) if required. SIGNAL TRANSMISSION ASTM specifies that the signals from transducers shall be transmitted to the apparatus for recording, reducing, and displaying data by means of cable or equivalent in such a way that electronic or other interferences are not limited. APPARATUS FOR RECORDING, REDUCING AND DISPLAYING DATA Signals from the transducers shall be recorded electronically in either analog or digital format with frequency components below 1500 Hz. The digitalization sample frequency shall be 91 at least 5000 Hz for each transducer channel. The apparatus for reducing data might be an ana-log or digital computer capable of providing signal conditioning, amplification, and calibration factors for the transducers. If strain gages are used the apparatus shall be capable of computing force from them. In the case of accelerometers, the apparatus shall integrate acceleration to obtain velocity. If required, the apparatus shall zero the Force and Velocity between impact events to account for the changing of zero offset (zero drift). Signals from the transducers might be displayed on a LCD graphics screen in terms of Force (P) and Velocity (ZVel) for each hammer blow. The apparatus shall be capable of holding and displaying the signal from each selected blow for a minimum period of 30 seconds. DATA QUALITY CHECK The confirmation of data quality is a two step procedure. First, the force (P) and Velocity (ZVel) should be compared at the moment of impact for proportionality agreement. Ideally, these values should be identical but differences of 10% or less are usually acceptable. The second step is a consistency check, for this, Force and Velocity Traces of consecutive impact events are compared for repeatability (ASTM D 4945). PILE DRIVING ANALYZER (PDA) EQUIPMENT The Pile Driving Analyzer equipment, PDA, complies with ASTM D4945 (Standards for High-strain Dynamic Testing). It has been used worldwide for more than 30 years and it is recommended by the FDOT for Dynamic Load testing and Dynamic Pile Monitoring. The PDA is composed of accelerometer and strain transducers (Figure 9-1b), which are connected by cables to a PC based apparatus, Model PAK that processes the incoming signals. The sensors, which are specially manufactured for this application, are attached to the pile by means of bolt connections, and are recovered after each test. The strain and acceleration 92 within the pile are transferred to the sensors by friction between the pile face and friction humps on the transducers. The force provided by the bolts guarantee enough frictional resistance for this purpose. Table 9-1 presents the properties of these transducers. Table 9-1. PDA Sensors Specifications (PDA Manual) Strain Gage Effective Length (mm) 50 Size (mm) 115 x 35 x 11 Material Aluminum Circuit Full Bridge Sensitivity (µε/mV/V) 380 Strain Range (µε) 2000 Shock Range (g) 5000 Operating Temperature Range oC -50 to 120 Piezo Electric Accelerometer Circuit Integral Impedance Converting Electronics. Sensitivity (mV/g) 1 Acceleration Range (g) 5000 Frequency Range (Hz) 0.25 to 7000 Operating Temperature Range oC -50 to 120 Piezoresistive Accelerometer Circuit Full Bridge Sensitivity (mV/g) 0.07 Acceleration Range (g) 10000 Frequency Range (Hz) 3 Operating Temperature Range oC -50 to 90 The PAK is a PC based Pentium computer with special electrical components built into it. These elements include: signal conditioners for 4 accelerometers and 4 strain gages, analog 93 integrators to obtain velocity from accelerometers, amplifiers, low-pass filters, and eight 12-bit analog to digital converters that sample 8 channels simultaneously. A built in software is used to calculate and display Force and Velocity Traces, pile capacity, pile stress during driving, and damage detection. It also performs data quality checks and internal calibration. The amount of data recorded in the PAK is limited to 860 blows (14.09 Mbytes). The PDA manual states that no accelerometer is perfect at all frequencies, therefore the velocity curve may not always return to zero at the end of the blow, in this instance, the PDA would perform a velocity adjustment. In this adjustment the entire Velocity Trace is rotated with reference to a pivot point (20 ms on the trace) until the last recorded Velocity (at 1024 ms on the trace) becomes zero. The PAK is suited with various modes to record the Force and Velocity Traces; among them the most important is the CAPWAP format. This format is intended to reduce hard drive storage memory and is used for the final step of dynamic testing (matching the recorded traces with the modeling software CAPWAP). A brief description of the CAPWAP format is shown in Figure 9-2. On this Figure, Force (P) and Velocity (VVel) are transformed to FOR and VEL using Equations 9-1 and 9-2 respectively. Then FOR and VEL are transformed to integers and stored in that form. Column 1, in Figure 9-2, contains the values of FOR and VEL obtained from such equations, while columns 2 to 11, contain the relative increase (FOR or VEL) with respect to the previous column. 100 / 100 ) ( : × = × × = Vel Z Vel Z VEL (9-1) Z P FOR / 100 ) ( : × = (9-2) 94 Figure 9-2. CAPWAP Format (incomplete file) SUMMARY OF REQUIREMENTS AND SUGGESTIONS Table 9-2 presents a summary of the identified requirements plus some suggestions. The requirements are the provision by ASTM, and the suggestions include some of the PDA charac-teristics. It should be noted that the PDA is not only used for concrete piles, but also for steel piles, which are out of the scope of this report. Some assumptions have also been taken in this table: the maximum pile length is 33 m (100 ft), the maximum driving time is 4 hours, and the prestress level might be zero if the sensors are located close to the pile ends (transfer length). 95 This is important because the proposed strain gages will be embedded in the concrete (they will measure the prestress in the concrete and wave stresses, summed together). Table 9-2. Summary of Requirements and Suggestions Requirements and Suggestions Source Non-Intrusive: no leads, no setup, etc. Monitor Force and Velocity at the tip of the pile Assess pile Stresses (for damage) at top and bottom of pile Determine both Skin and Tip Resistance (Scour) Use PDA Case method, Paikowsky and UF methods Minimum expertise required (automatic) Improvements Cheap: Monitor every pile – Elimination of Driving Criteria Non-recoverable should be securely mounted before concrete is poured Non-recoverable unit temperature resistance up to 160oF (Concrete curing) Non-Recoverable unit should withstand stresses due to the concrete pouring Mounting Instruments should be properly aligned and in the proper direction Concrete pouring procedure 96 Table 9-2. Summary of Requirements and Suggestions (Cont.) Requirements and Suggestions Source Real time capacity assessment and pile monitoring ASTM D4945 Non-Recoverable unit should withstand stresses up to 30.9 MPa and -11.5 MPa FDOT Current Practice Sensor’s operating temperature form -50 to 120 oC PDA System Strain gage range at least 1014 µε το -377.9 µε FDOT Current Practice Strain gage should have a linear output ASTM D4945 Strain gage natural frequency above 2000 Hz ASTM D4945 Strain accuracy of 2.14 µε Signal Characteristics Minimum strain gage effective length of 50 mm PDA System Accelerometer resonant frequency above 2500 Hz ASTM D4945 Accelerometer should have a linear output up to at least 1000g and 1000 Hz ASTM D4945 Acceleration accuracy of 5.1 g Signal Characteristics Accelerometer sensitivity of 1 mV/g or smaller value PDA System Low Pass filter minimum frequency of 1500 Hz ASTM D4945 Sensor calibration to 3% accuracy ASTM D4946 Signal conditioning of both sensors should have same frequency response ASTM D4947 Minimum sampling frequency of 5000 Hz for each channel ASTM D4948 Testing duration of 4 hours and reuse at 72 hours after driving FDOT Current Practice Minimum data transmission distance of 34 m (110 ft) FDOT Current Practice Sensors have to be attached 2 to 3 diameters below the pile top FDOT Current Practice Use at least two accelerometer and two strain gages ASTM D4948 Signal displayed in an LCD screen at least for 30 seconds ASTM D4949 Perform real time data quality tests ASTM D4950 Use simultaneous 12 bit A/D converters for signal digitalization PDA System Record traces in digital format ASTM D4950 Minimum storage capacity of 860 blows PDA System Store Data in CAPWAP Format PDA System General Present recovered data over a period of 0.1024 seconds and the acceleration signal sent after 0.02 sec. Signal Characteristics Non-recoverable unit is the instrumentation package to be permanently embedded in the pile. 97 CHAPTER 10 SIGNAL CHARACTERISTICS The final product with any new instruments, conditioners, and data transfer is a signal. Knowing the ranges of signal for both the strain and acceleration are required to capture the whole signal. Consequently, this chapter focuses on past signals recovered from earlier FDOT projects. The twenty-two PDA traces studied were the signals used in Chapter 7 (Table 7-1 and 7-2) for the development of skin and tip resistance capacities. The latter signals were used to find signal characteristics such as frequency response, amplitude, and signal accuracy. FORCE AND VELOCITY TRACES The Force and Velocity Traces are not exactly the same with each other, as shown in Figure 10-1 and 10-2. However, the trends in Force Trace from one pile to another are similar, as well as the trends in Velocity Trace for the different piles. This similarity as well as ampli-tudes is studied for the twenty-two piles. From all the Force Traces, Figure 10-1, their average (thick line) was found. Similarly, the average Velocity Trace (thick line) is shown in Figure 10-2. Maximum values: Maximum Force = 8312.968 KN (1868.83 Kips) Maximum Velocity (ZVel) = 8730.061 KN (1962.60 Kips) Minimum Force = -1240.337 KN (-278.84 Kips) Minimum Velocity (ZVel) = -4278.126 KN (961.76 Kips) General observations were that the maximum Force and maximum Velocity values are very similar and are located at the beginning of the trace. Also, the Velocity shows a high negative value after the peak, indicative of a tension wave coming back up the pile. 98 Figure 10-1. PDA Force Traces of 22 Piles in PILEUF Figure 10-2. PDA Velocity (ZVel) Traces for 22 Piles of PILEUF -2000 0 2000 4000 6000 8000 10000 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 Relative Time (sec) Force (KN) -6000 -4000 -2000 0 2000 4000 6000 8000 10000 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 Relative Time (sec) Force (KN) 99 Other observations were both of the Force and Velocity Traces start at a zero value and increase to a maximum value, after which their magnitude decreases almost to a null value. From this behavior it can be inferred that the stress wave is still traveling through the pile but with very small amplitude. The difference between Force and Velocity Traces is that the Velocity Trace decreases to a value smaller than zero before the stress wave vanishes. Every trace was composed of 1024 discrete data points distributed uniformly in a time period of 0.1024 seconds. It is interesting to note that 210 = 1024, making the data very suitable for performing a Fast Fourier analysis (FFT) (discussed later). Also, this short time period con-tains all the necessary information for obtaining pile capacities (from a single blow), hammer efficiency, and pile damage. Figure 10-2 shows a more consistent initiation of the stress wave than Figure 10-2. The Velocity Traces start at 0.016 seconds, while the Force Traces start somewhere between 0.01 and 0.018 seconds. The information presented was originally obtained from strain gage and accelerometer output signals. The raw signal may be obtained using the reverse procedure. The raw signals (voltage) will serve as a guide for instrumentation adequacy, data acquisition, and electrical circuitry requirements. STRAIN AND ACCELERATION TRACES Strain and acceleration traces can be back calculated from Force and Velocity Traces by using simple structural and dynamic equations as described in Chapters 5-7. PDA traces are discrete points; therefore, the following equations (discrete format) are used: A E P i i × = ε (10-1a) 100 ( ) ( ) [ ] te SamplingRa Vel Z Vel Z Vel Z Vel Z Z a i i i i i × × − × + × − × × × = − + 1 1 2 1 (10-1b) where: Pi Force (from Force Trace) E Modulus of Elasticity of Pile A Cross Sectional Area of Pile Z Pile Impedance EA/c c Wave Speed Veli Velocity (from ZVel/Z) Sampling rate = usually 10,000 samples/second Figure 10-3 shows the strain and acceleration traces obtained by this procedure using the PDA traces that were shown in Figures 10-1 and 10-2. Values of E, A, Z and c, have been ob-tained from Table 7-1. Figure 10-3. General Form of Strain and Acceleration Traces -500 -400 -300 -200 -100 0 100 200 300 400 500 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 Time (s) Accelertion (m/sec^2) -0.0005 -0.0004 -0.0003 -0.0002 -0.0001 0.0000 0.0001 0.0002 0.0003 0.0004 0.0005 Strain (ε ε ε ε) Acceleration Strain 101 The thick line in Figures 10-4 and 10-5 again represents the average among the 22 traces in each Figure. It will subsequently be referred to as average-strain and average-acceleration. The average-strain curve has a similar shape to the 22 individual strain traces. The average-accelerometer curve does not closely follow the same trend as the individual ones, nevertheless, its shape is very similar over the fist 0.025 seconds. Over this period, the wave return is observed even for the longest pile under study (Table 7.1: 2L/c = 234.45/4325 = 0.016sec). Consequently, both average traces will be used as the representative curves for future reference. From these figures some additional and valuable information can be recovered as: Maximum Strain = 0.0006036 = 0.60364 mε Minimum Strain = -0.0000654 = -0.06547 mε Maximum Acceleration = 1400 m/s2 = 142.71g Minimum Acceleration = -1200 m/s2 = -122.32g Figure 10-4. PDA Strain Traces for 22 Piles of PILEUF -1.0E-04 0.0E+00 1.0E-04 2.0E-04 3.0E-04 4.0E-04 5.0E-04 6.0E-04 7.0E-04 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 Relative Time (sec) Strain ε ε ε ε 102 Figure 10-5. PDA Acceleration Traces for 22 Piles of PILEUF ACCURACY OF STRAIN AND ACCELERATION MEASUREMENTS The accuracy of the acceleration and strain measurements is the main factors that control the accuracy of Force and Velocity Traces and further calculations (including pile capacities). These values are calculated from the traces obtained in the last section. The strain and acceleration plots shown in Figures 10-4 and 10-5 are discrete series of points with their accuracy defined as the minimum increase/decrease between two consecutive data points (zero increase not considered). The following equation is used for this intention: 1 − − = ∆ i i i ε ε ε and 1 − − = ∆ i i i a a a (10-2) where: εi Strain (from Figure 10-4) ai Acceleration (from Figure 10-5) -1500 -1000 -500 0 500 1000 1500 2000 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 Relative Time (sec) Acceleration (m/s^2) 103 Using this equation with the information shown in Figures 10-4 and 10-5, the relative change is obtained for strain and acceleration. The results are presented as a cloud of points in Figures 10-6 and 10-7 for strain and acceleration respectively. These clouds of points are orga-nized in horizontal groups. The same distance separates the groups in the vertical direction. This distance is the minimum relative increase and may be considered as the accuracy (very repetitive), and was found to be: Strain accuracy = 2.137µε Acceleration accuracy = 5.097g It is important to point out that the above analyzed accuracy is applicable only to the 22 piles in the database and not to the PDA equipment. As described in the PDA section, the PAK (PDA) is available with various modes to record the Force and Velocity Traces; among them the CAPWAP format (Figure 9-2) that was used for the 22 piles in the database. This type of format alters the accuracy of the measured signal (Equations 9-1 and 9-2). Figure 10-6. Strain Increase for 22 Piles in Database -2.0E-05 -1.5E-05 -1.0E-05 -5.0E-06 0.0E+00 5.0E-06 1.0E-05 1.5E-05 2.0E-05 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 Relative Time (sec) Strain Increase (ε ε ε ε) 104 Figure 10-7. Acceleration Increase for 22 Piles in Database It should be noted that the PDA PAK does not measure acceleration. The PAK is sup-plied with analog integrators to transform the accelerometer output into velocity, which is subse-quently digitized for further analysis. Therefore, the above mentioned acceleration-accuracy is questionable, and should be used carefully. Nevertheless, that accuracy will be used as a refer-ence point for the proposed instrumentation. The same can be said about the strain accuracy. Even though the PDA strain-measure-ments are altered by storing this information in the CAPWAP format, this distorted accuracy is further used to assess the pile capacities in CAPWAP software. Since the output of this software is accepted in practice, the calculated strain-accuracy will also be used as a reference point. A more in depth analysis of the strain-accuracy reveals that this value equals the inverse of the wave speed times 100 (Equation 9-2). -60 -40 -20 0 20 40 60 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 Relative Time (sec) Acceleration Increase (g) 105 FREQUENCY OF STRAIN AND ACCELERATION TRACES The frequency of the acceleration and strain traces, Figures 10-4 and 10-5, was calculated using Fast Fourier analysis (FFT). This analysis was performed in LabVIEW for the 22 traces, and is presented in Figures 10-8 and 10-9 for strain and acceleration respectively. Disregarding small amplitudes, the useful frequency content was found to be: Strain: 0 to 300 Hz Acceleration: 10 to 3,000 Hz Within these ranges, the significant values are those with the highest amplitude, and are referred as the predominant periods. They are the major contributor to the signal, which make up the strain and acceleration traces (Figures 10-4 and 10-5). It is found, from Figures 5-8 and 5-9, that these frequencies are concentrated within narrow ranges: Major Strain frequency content: 5 Hz and 30 Hz (highest at 10 Hz) Major Acceleration frequency content: 10Hz -500 Hz, 1,100Hz - 1,700Hz. Figure 10-8. FFT Strain for 22 Piles in Database 0.0E+00 5.0E-10 1.0E-09 1.5E-09 2.0E-09 2.5E-09 3.0E-09 0 100 200 300 400 500 Frequency (Hz) Amplitude 106 Figure 10-9. FFT Acceleration for 22 Piles in the Database ACCELEROMETER AND STRAIN GAGE OUTPUT TRACES IN TERMS OF VOLTAGE The raw product from most types of instruments is voltage, which is directly correlated to the intended measurement. When it’s linearly proportional to the physical action being mea-sured, the constant factor is called sensitivity. The sensitivity of the sensors used in each one of the 22 piles in the database was re-quired in order to obtain strain and acceleration traces in terms of voltage. Unfortunately, the sensitivity values were not recorded in the database. Nevertheless, the sensors used by PDA are of a standard type, with sensitivity values close to a nominal value provided by the PDA bro-chure, shown in Table 9-1. From the table, the nominal strain gage sensitivity is 380 µε/mV/V, 0 10 20 30 40 50 60 0 500 1000 1500 2000 2500 3000 3500 Frequency (Hz) Amplitude 107 and the accelerometer nominal sensitivity is 1.0 mV/g or 0.07 mV/g, (for a piezoelectric accelerometer and piezoresistive accelerometer, respectively). It will be assumed that the excitation voltage for the strain gage is 10 Volts. This is a common value and provides a stable signal in terms of noise immunity (i.e., signal to noise ratio). Furthermore, a perfectly linear transducer response will be assumed for the analysis; but it be should remembered that the latter is particularly not true for piezoelectric accelerometers. The analog traces for the 22 files were calculated, and are plotted in Figures 10-10, 10-11, and 10-12. They correspond to the standard PDA strain gage, piezoelectric accelerometer, and piezoresistive accelerometer. Using the sensitivity values given above, the traces were calculated as follows: A E P i i × = ε (10.3a) ( ) ( ) [ ] Rate Samplin Vel Z Vel Z Vel Z Vel Z Z a i i i i i × × − × + × − × × × = − + 1 1 2 1 (10.3b) It is clear, that the traces in Figures 10-10, 10-11, and 10-12 have the same shape as the strain and acceleration traces in Figures 10-4 and 10-5 (but the magnitudes are different). There-fore, the shape and frequency characteristics, pointed out earlier, are applicable to the analog traces. Additional information from Figures 5-10, 5-11, and 5-12 were: Maximum Strain gage Output = 0.01588 Volts Minimum Strain gage Output = -0.001723 Volts Maximum Acceleration Output = 0.1427 Volts Minimum Acceleration Output = -0.1223 Volts The general trend of the traces shown in Figures 10-4 and 10-10 shows that up to 0.0045 seconds the strain in the pile is almost zero. The peak strain value is recorded between 0.02 seconds and 0.025 seconds. Finally, after 0.09 seconds the strain goes back to zero, or close to it. 108 Figure 10-10. PDA Strain Gage Output for 22 Piles in Database Figure 10-11. PDA Piezoelectric Accelerometer Output for 22 Piles in Database -4.0E-03 -2.0E-03 0.0E+00 2.0E-03 4.0E-03 6.0E-03 8.0E-03 1.0E-02 1.2E-02 1.4E-02 1.6E-02 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 Relative Time (sec) Strain gage output (Volts) -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 Relative Time (sec) Accelerometer Output (Volts) 109 Figure 10-12. PDA Piezoresistive Accelerometer Output for 22 Piles in Database In Figures 10-5, 10-11, and 10-12 the traces show that the peak acceleration is located between 0.19 seconds to 0.21 seconds. Most of the acceleration traces present two peak values within this time frame. -0.01 -0.008 -0.006 -0.004 -0.002 0 0.002 0.004 0.006 0.008 0.01 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 Relative Time (sec) Acceleration (Volts) 110 CHAPTER 11 THE NEW WIRELESS ANALYSIS PACKAGE (WAP) A new wireless analysis package for driven pile has been developed and is shown in Fig-ure 11-1. The package has been designed to satisfy the requirements stated in Chapter 4, with applicability to the current practice (Chapter 9). (a) (b) Figure 11-1. New Instrumentation System (a) Non- Recoverable Unit; (b) Signal Receiver and Data Processing Unit To achieve the improvements stated in Table 9.2 the system is physically divided into two modules, which communicate, by a wireless link. One of the modules is called Non-Recoverable Unit (Figure 11-1a). This unit is capable of being embedded in concrete piles and broadcasting stress wave information to the second module (for data processing) during installa-tion. Since the Non-Recoverable Unit is pre-installed in the pile, no installation setup is required at the time of driving. The second module receives the information from the Non-Recoverable Unit to calculate Force Traces, Velocity Traces, and pile capacities. This module is called Receiver and Data Processing Unit (Figure 11-1b). The cost of the Non-Recoverable unit is 111 below $300. The cost of the Receiver and Data Processing Unit is $6,400 and it may be used in other tests. In this chapter, the two units are presented in more detail. Some aspects, such as the electrical design, will be avoided since the system is protected by patent law and disclosure agreements. Nevertheless, a detailed explanation of their operation is given as well as a description of their installation. NON-RECOVERABLE UNIT The Non-Recoverable Unit is shown in Figure 11-1a. This unit gathers six components that will be described independently: - Two strain gages - Two accelerometers - A Signal Conditioning and Sending Unit - Antenna Since the unit is designed for installation in concrete piles, it must withstand harsh envi-ronments. Examples of the latter are hydration temperatures of 70oC (160oF); damage in the casting process by portable concrete vibrators; and possibility of high shock and vibration (200g and 3,000Hz) during the driving process. A brief discussion on system preparation (signal con-ditioning and sending unit) is presented later. Strain Gages Micro-Measurements Group Inc manufactures the selected strain gages, Model EGP-5-120. These transducers (Figure 11-2) are specifically designed for measuring mechanical strains inside concrete structures. They have a foil-sensing grid that is protected by a rugged outer con-crete resin. This body is used to resist mechanical damage during concrete pouring and provides protection from moisture as well as corrosive attack. The strain gages are provided with a three- 112 Figure 11-2. Special Purpose Embedment Strain Gage EGP-5-120 conductor cable, intended to eliminate thermal effects on the cable. See specifications in Table 11-1. Table 11-1. Strain Gage Specifications, Requirements, and Recommendations Strain Gage EGP-5-120 Requirements Recommendations Resistance (Ohms) 120 + /- 0.8 ------- ------- Gage Factor 2.05 + /- 1 ------- ------- Effective Length (mm) 100 ------- > 50 Size (mm) 17 x 130 x 10 ------- 115 x 35 x 11 Material Concrete Resin no aluminum ------- Natural Frequency (Hz) not available > 2000 ------- Shock Limit (g) > 200 ------- 5000 Circuit Quarter Bridge ------- Full Bridge Excitation Voltage 2.5 ------ 10 System Sensitivity (µε/V) 313.5 ------- 380 Strain Range (µε) 1561 ------- 2000 Strain Accuracy (µε) 3.13 ------ < 2.14 Calibration Accuracy (%) 4.74 < 3 ------- Linear Output (µε) 1,100 to –461 1,014 to –377.9 ------- Operating Temperature Range oC -45 to 55 ------ -50 to 120 Maximum Temperature Range oC > 70 > 70 ------ 113 The outer protective concrete resin has a modulus of elasticity of 11,031 Mpa (1.9106-psi) which is approximately one third of the minimum specified for the FDOT piles. The latter ensures that the protective body will deform properly while providing protection to the mea-suring grid. Furthermore, this protective body is dimpled to provide a good bonding/mechanical interlocking with the surrounding concrete. The measuring grid, 100 mm (4 in) length, offers a good span to measure strain in aggre-gate materials such as concrete. The grid is made of nickel-chromium alloy, similar to Karma. A good fatigue life and excellent stability characterize this material, which is required for our dynamic measurements. It is also self-temperature compensated to minimize thermal effects on the output signal. A quarter bridge configuration, with 5.0 volt DC excitation, is used with the strain gage. Further details on the electrical conditioner for this transducer and placement in the pile, are pre-sented later. Accelerometers A piezoelectric accelerometer, manufactured by Endevco was selected for this applica-tion (Figure 11-3). This is a rugged, low-cost, high-performance, transducer that requires no electrical power for operation. It meets the ASTM requirements for dynamic testing (Chapter9) and its specifications are presented in Table 11-2. The accelerometer is composed of piezo electric crystals inside an alumina case with two thick film gold connectors. Its output is characterized by a linear charge that is transformed to DC voltage by a charge converter circuit in the Signal Conditioning and Sending Unit. 114 (a) (b) Figure 11-3. Endevco Accelerometer (a) Characteristics (b) Size Compared with a Mechanical Pencil Table 11-2. Accelerometer Specifications, Requirements, and Recommendations Accelerometer Endevco Requirements Recommendations Material Alumina ------- ------- Charge Sensitivity (pC/g) 2 ------- ------- Transverse Sensitivity (%) 5 ------ ------- Resonant Frequency (Hz) 10000 > 2500 40000 Frequency Range (Hz) 10 to 2000 ------- 0.25 to 7000 Acceleration Range (g) 1000 1000 5000 Circuit Charge Converter ------- Integral Impedance Converting Electronics System Sensitivity (g/V) 93.22 ------- 1000 Acceleration Accuracy (g) 2.7 >5.1 ------ Calibration Accuracy (%) 4.43 < 3 ------- Linear Output (g), (Hz) +/-1000, 1000 1000, 1000 ------- Operating Temperature Range oC -65 to 150 ------- -50 to 120 Maximum Temperature Range oC > 150 > 70 ------- The accelerometer is connected to the Signal Conditioning and Sending Unit with a thin, but high-resistant, coaxial low-noise cable (manufactured by Endevco). Its major task is to elim-inate electrical noise and spurious signals generated by stressing the cable (as the stress wave 115 passes through). Unacceptable signal generation was observed on regular coaxial cables during slight bending. The size of the Endevco accelerometer makes it ideal for pile driving application because it displaces a very small volume of concrete. However, its size also makes it difficult to properly mount and align it inside the pile. Therefore, an enclosure (3.813.81cm1 cm) was fabricated with the transducer installed in it (Figure 11-4b). Besides providing the means for alignment, this enclosure protects the transducer (harsh and humid environment), as well as waterproofing it. Moreover, the enclosure acts as a mechanical filter to eliminate frequencies above 3,000 Hz. The latter guarantees that the accelerometer will not be subjected to frequencies close to and above 10,000 Hz (natural frequency of accelerometer) and associated zero shifts. (a) (b) Figure 11-4. (a) Accelerometer Mounting Technique; (b) Protective Enclosure Signal Conditioning and Sending Unit The Signal Conditioning and Sending Unit is shown in Figure 11-5. This unit is used to provide power source to the sensors, condition the sensors, filter background noise, digitize the signal, and send the information to the Receiver and Data Processing Unit. The character-istics of the Signal Conditioning and Sending Unit are presented Table 11-3. 116 (a) (b) Figure 11-5. Signal Conditioning and Sending Unit (a) To be used at the pile top; (b) Alternative tip sensor’s conditioners. The unit encloses a number of electrical systems, which are contained in a single plastic box: - Three 9-Volt batteries - Power control board - Two Strain Gage Conditioners - 2 Accelerometer Conditioners - Digitalization and Data Encoding circuit - Transmitter These elements can be gathered, according to their task, into four components that will be described next. The plastic box is filled with a liquid epoxy, which hardens after 24 hours. This compound protects, as well as supports all the electrical elements during the shock and vibration of pile driving. Also, the compound is temperature retardant which eliminates dynamic thermal effects on the electronic part from rapid changes in temperature during the driving process. 117 Table 11-3. Signal Conditioning and Sending Unit Specifications, Requirements, and Recommendations Signal Recovery and Transmission Specifications Requirements Source Transmission Real time Real time ASTM D4945 Type of transmission Digital Carrier Frequency (MHz) 908/922 Unlicensed Unlicensed FFCC Selectable Carrier Frequencies 8 Channels Antenna Type Whip Style Antenna Size (cm) 23 Transmission distance (m) 152 > 34 FDOT Current Practice Stress Limits inside the pile (MPa) Eliminated with Foam -11.5 to 30.9 FDOT Current Practice Low Pass filter for both conditioners (Hz) 5,000 >1,500 ASTM D4945 Sampling frequency per channel (Hz) 10,000 5,000 ASTM D4948 Operation time (Hours) 6 >6 FDOT Current Practice Power Supply Three 9-Volt Batteries Power Saving Automatic turn on-turn off Accelerometer Channels 2 > 2 ASTM D4947 Strain Gage Channels 2 > 2 ASTM D4948 Digitalization Accuracy (bits) 15 > 12 PDA System Signal digitalization Simultaneously Simultaneously PDA System Max. Temperature (Concrete Curing) > 70 o C (160 o F) > 70 o C (160 o F) FDOT Current Practice Concrete resistant (%) 100 100 FDOT Current Practice Size (mm) 1468925.4 Cost ($) <200 Batteries. Three 9 Volt batteries are used to provide DC power for a minimum period of 12 hours. The selected batteries were Lithium for two reasons. First, the electrical circuits had to have a 118 minimum voltage, as well as a significant active and dormant life. Second, the batteries had to survive the casting process and the associated thermal rises due to hydration and possible steam curing. To help with the active life cycle, three 9-Volt batteries, connected in parallel were selected which provided at least 7.1 Volts for 12 hours. For the second issue, temperature, of the 15 companies approached only one would guarantee battery survival for temperatures of 70°C (160°F) expected during the concrete hydration. Power control board. Since the concrete aging process may take 28 days, and the dynamic testing may last for 72 hours (EOD vs. BOR), the recording process (strain and acceleration) could not be continu-ous. Since the pile driving process as well as restrike will be less than a few hours, it was decided to shut down the system in order to preserve the batteries when in inactivity (i.e. not driving). Consequently, a Power Control Board (electrical circuit) was built into the unit, which activated the system with the first strike and turned the system off if another strike was not received within 3.5 minutes. A waiting period of 3.5 minute was selected due to the possibility of set checks and/or forms of pile inspection at End of Driving. It should be noted that the first blow, which turns the system on, has incomplete strain and acceleration data. Strain gage conditioners. The strain gage conditioners use a quarter bridge configuration, analog amplifier, and low pass filter. Its size is 76.228.63.2 mm3 (31.1250.125 in3), and shown in Figure 11-6. The entire conditioner works with a 5-Volt DC input, and outputs the signal in analog format from 0 to 5 Volt. The conditioner was developed with economy and accuracy in mind. In particular, the excitation (5-volts DC) was selected to reduce power consumption. However the 0-5 volts 119 (a) (b) Figure 11-6. Strain Gage Conditioner (a) Entire Circuit Board; (b) Quarter Bridge output signal was significant enough to eliminate or minimize background noise, and improve the signal quality. The bridge is composed of four resistors: two 5 KΩ, one 120Ω and the 120Ω strain gage. The thermal coefficients of these resistors match each other so that the bridge maintains balance. The common practice of using potentiometers on the bridge was found unacceptable due to their instability under thermal variation and vibration conditions. In addition, they may gen-erate a dynamic signal on the strain gage channel, as well as result in drift of the signal beyond the sampling limits (0 to 5 volts). Since the output signal from the bridge by itself is very low, an analog amplifier was used to introduce a gain factor of 1,000. This value is intended to accommodate the expected strain in the 0 to 5 Volts range. The signal is subsequently filtered with a 5,000 Hz low pass filter. Accelerometer conditioners. The accelerometer conditioning uses a charge converter, an analog signal amplifier, and low pass filters of the same type as the strain gage conditioner. Its size is 73 × 28.6 × 3.2 mm3 120 (2.88 × 1.125 × 0.125 in.3) and is shown in Figure 11-7. The entire conditioner works with a 5 Volts DC input voltage, and outputs the signal in analog format from 0 to 5 Volts. Figure 11-7. Accelerometer Conditioner The charge converter is a standard circuit recommended by Endevco to convert the accelerometer output from Pico-Coulombs to DC voltage. Background noise on this signal is minimized using a 5,000 Hz low pass analog filter. The signal is amplified by a factor of 100, using an analog amplifier, to bring the expected acceleration signals within 0 to 5 Volts. Digitalization and data encoding. The analog signal provided by the two-accelerometer conditioners and two strain gage conditioners, is transformed to a digital format. The four analog signals are sampled simultane-ously using four sample-and-hold analog circuits. They deliver the information sequentially (Channel 1, 2, 3, 0) to a single Analog to Digital converter (A/D converter) of 16 bit resolution. The A/D converter works with a 40,000 Hz clock in order to provide a sampling rate of 10,000 Hz in each one of the four channels. The input and output range is from 0 to 5 Volts. The digital data complies with TTL format and uses Manchester encoding technology. TTL is a digital format that defines voltages from 0 to 1.5 Volts with 0 logic level, and 3.5 to 5 Volts as 1 logic level. Manchester guarantees logic levels change for every bit time, and has a 121 self-clock generation (Figure 11-8) at the receiver end, that will be used to properly decode the data. Figure 11-8. Digital Data String (a) TTL Format; (b) Numeric Format The digital signal is further improved by introducing a Sink Pulse and two additional digital levels. The Sink Pulse introduced on the signal, as shown in Figure 11-8, defines the start of the 16–bit word (one sample in one channel). Two digital levels are then added right after the sink pulse to carry channel information. This is necessary in order to separate the four signals at the receiver. Transmitter. A digital transmitter, T900V (Figure 11-9), manufactured by Applied Wireless is used for this application. This transmitter was selected on the basis of affordability and its high band width (5 MHz). Transmitting in the 908/922 MHz unlicensed band, this transmitter is capable of broadcasting up to 5 Mbps of information. The actual speed used is 1.04 Mbps (described in the previous section) for which the signal is both stable and immune to noise interference. The transmission distance can reach up to 150 m (500 ft) but it was tested only up to 45 m (150 ft). 122 The transmission frequency can be pre-selected among 8 different values, ranging from 908 to 922 MHz. With the latter, multiple piles and systems may be working in close proximity with another without interfering with each other’s signal. Figure 11-9. Digital Transmitter T900V, from Applied Wireless In the development stages, the manufacturer was concerned that the crystal component might be damaged under high shock vibrations. Nevertheless, they were confident that a foam enclosure would provide sufficient energy absorption to ensure the continuous function of their device. As will be discussed in the next chapter, the latter will be proven true. Transmitter Antenna The transmitter antenna is a flexible 23 cm (9 in) quarter whip style antenna, shown in Figure 11-5a. A rubber membrane protects it, which is resistant to the concrete environment, corrosion, UV rays, water, and pile handling. It is also detachable so that it can be replaced. This antenna is connected to the transmitter with a thick coaxial cable. RECEIVER AND DATA PROCESSING UNIT The Receiver and Data Processing Unit is shown in Figure 11-1b. This unit can be located up to 150 m (500 ft) from the transmitter antenna (Non-Recoverable Unit in pile) and still receive and decode a signal. 123 The Receiver and Data Processing Unit is made of several components and will be described separately: - Receiver Antenna - Receiver and Conditioning Unit - Data Acquisition Card - Laptop Computer - Data Acquisition and Processing Software Receiver Antenna Different types of antennas can be used for the receiver. In terms of this application which has a directional signal (pile antenna), a directional 1-meter antenna as shown in Figure 11-10a (similar to a TV antenna) was employed. The receiver antenna is mounted separately from the data-conditioning unit and is connected to it through a rugged coaxial 50-Ohm cable. The antenna provides the longest transmission distance, 150 m (500 ft). A smaller antenna may be used (Figure 11-10b) for closer transmission distances, up to 45 m (150 ft). However, it is important to place this antenna in a vertical position in order for it to work properly. (a) (b) Figure 11-10. Receiver Antennas (a) Directional 1 m Antenna; (b) 23 cm Whip Style Antenna 124 It should be noted that the antenna might also receive unwanted information from other nearby transmitters, etc. However, if the antenna size is reduced, the undesirable noise will also be diminished or eliminated because the broadcast distance is also reduced. Signal Receiver and Conditioning Unit The Signal Receiver and Conditioning Unit (Figure 11-11) receives the information from the antenna, decodes the signal, and parses the signal for the Data Acquisition Card. As ex-plained in Figure 11-8, the transmitted signal is arranged in a serial string, with Manchester coding, which involves sink pulses, and channel ID information. Also, the signal needs to be separated into four independent (separate) channels (strain, and acceleration) along with time for the Data Acquisition Card. All the components of the Signal Receiver and Conditioning Unit are contained in a portable enclosure (Figure 11-11). The components can be grouped into two categories: the digital receiver and the signal conditioning. Figure 11-11. Signal Receiver and Conditioning Unit Six LED indicators specify the proper function of the unit. The first (bottom box –Figure 11-11), located at the top middle of box, indicates the unit is powered up. The second (top box – Figure 11-11), shown on the left of the top box indicates the digital receiver is powered up. The remaining four LEDs (upper right box Figure 11-11) indicate the strength of the transmis- 125 sion signal. If all four LEDs are lit the wireless link is working at optimum antenna location, orientation, etc. However, if none of them are lit then there is a poor signal or no signal at all. The carrier frequency for signal transmission can be selected with a knob control at the back of the digital receiver. Any of eight available frequencies (from 908 to 922 MHz) may be used. The selection of a different carrier frequency will allow multiple transmitters to be in proximity to one another without interference. The unit is provided with a 12-Volt commercial Battery and AC plug-in socket. The batteries will work for up to 10 hours, and they can be replaced any time the unit is not oper-ating, or recharged using the AC adapter. Some precautions when using the receiver with antennae are: 1) The antenna needs to be attached firmly with proper orientation; 2) The antenna should not be touched during the dy-namic testing; and 3) The receiver needs to be placed in a stable position. Testing experience with this unit has noted that excessive vibration induces undesired interference (noise) in the dynamic signal. Digital receiver. A R900V commercial receiver (Figure 11-11) is used to capture the broadcasted signal. This receiver is manufactured by Applied Wireless to work with the T900V transmitter described in the previous section. This receiver has the same transmission characteristics as the transmitter. As described before, the information coming out of this receiver includes Manchester encoding technology, sink pulse, and channel ID information in a serial string (Figure 11-8). A clock is also generated to indicate the location of proper data bits. This clock is generated using Manchester technology. 126 Signal conditioning at the receiver. Using the digital information, and clock from the receiver, the Signal Conditioning iden-tifies the sink pulses, determines the 16 bit word, and finally separates the bit words in the proper format required for the Data Acquisition Card (Analog and Digital). For Digital Data Acquisition Cards the signal is arranged in 18-bit parallel strings. The first 2 bits carry the channel information, and the remaining 16 bits carry the digital value of one measurement in one channel. TTL (Figure 11-8) is also used for this information. Two extra lines (request and acknowledgement) are also provided to work with the handshaking mode. Handshaking mode involves an interaction between the Data Acquisition Card and Receiver Conditioning Unit. A low-bit on the request line, generated by the receiver conditioner, informs the Data Acquisition Card that the data is ready to be read. When the Data Acquisition Card has properly read the information, it sends a low-bit to the Signal Condition Unit (using the acknowl-edge line) reporting that it is ready for the next reading. At this time, the request line is reset to a high-bit, waiting for the next transmitted measurement. For the Analog card, the digital signal is transformed back into a voltage reference signal and divided into four channels, each one containing the information for one sensor. The output range is from 0 to 5 volts, and it may be changed to cover the sampling range for any specific Data Acquisition Card. To change the signal from digital to analog format, four independent digital to analog converters are used with a 5,000 Hz low pass filter. Data Acquisition Card Any analog or digital Data Acquisition Card may be used for this application, provided they have the required speed and resolution. PCMCIA cards are preferred since they are portable 127 and easily inserted into a Type II slot of a standard laptop computer. Two models are currently used, DAQCard 6533 (digital, Figure 11-12) and DAQP-12 (analog, Figure 11-13). Figure 11-12. DAQCard 6533 Data Acquisition Card (Digital) Figure 11-13. DAQP-12 Data Acquisition Card (Analog) The DAQCard 6533 is a high-speed 32-bit digital I/O Card. The maximum sampling rate is 740 Kbytes/second for 32 parallel lines (digital TLL) of incoming information at 100 Hz (ex-ceeds the required speed and resolution of dynamic pile testing). It is also provided with a 68-pin connector cable, Figure 11-12, which is connected to the Signal Receiver and Conditioning Unit. The Analog DAQP-12 card, from Quatech, is shown in Figure 11-13. This card allows up to 16 single-ended (8 differential) input channels. The maximum sample rate is 100 KHz, and the resolution is 12 bits. This is more than adequate for this application since the current require- 128 ment is 20,000 Hz (5,000 Hz in each cannel) with less than 12 bit resolution. The input range is selectable, from 0 to 5 Volts. Laptop Computer The laptop computer may be any commercial system, which is fast enough (CPU) to pro-cess the incoming data and have sufficient memory (RAM) to hold the data (buffers) for real time analysis and display of all the signals. Also, the machine must have a PCMCIA Type II socket or USB connector for the data acquisition card. Currently a DELL model Latitude C600 computer is used (Figure 11-1b). Its specifications are as follows: • P-III 800 MHz processor • 14 inch screen • 256 MB RAM • 10 GB HD • 100 MHz buss speed • 32 bit resolution • 24x CD-ROM • 1.44 MB floppy • 2 type II PCMCIA slots • 2nd Lithium ion battery (backup) • AC adapter The laptop’s batteries will work for 6 hours, slightly less than the Non-Recoverable Unit operation. If further piles were to be tested, extra batteries would have to be used. Another option would be to recharge the batteries. The latter would take 30 minutes to recharge the batteries to their full capacity. 129 CALIBRATION The sensitivity of the accelerometers and strain gages is provided by the manufactures. However, as identified from the previous discussion, the signal from the instrumentation is trans-formed (analog to digital and back) many times before reaching the Data Acquisition Card. Therefore the entire system needs to be calibrated. For this purpose two calibration devices are used (Figure 11-14). These devices simulate strain and acceleration signals of the same charac-teristics as the proposed transducers. These signals are pre-selected (known) and sent to the Non-Recoverable Unit, to be broadcasted to the Signal Receiver and Data Processing Unit. By knowing the input signal characteristics, the output signal can be calibrated. (a) (b) Figure 11-14. Calibration Equipment (a) Portable Simulator 4830A (b) Precision Calibrator 1550 A Portable Simulator (Figure 11-14a) manufactured by ENDEVCO, the same brand as the accelerometers, was used to calibrate the accelerometer channels. A sinusoidal signal of the same characteristics and sensitivity as the accelerometer chip is generated at any selectable acceleration and frequency. Generating and transmitting a known signal allows its comparison on the laptop computer. 130 For the strain gage calibration a Precision Calibrator (Figure 11-14b), model 1550 form Measurements Group, was used. It embodies a true whetstone bridge, which provides a known repeatable resistance. For instance, the latter allows a known strain to be sent from the trans-mitter unit and be compared at the receiver end; this can be done for at least 20 selectable strain values. The calibration is performed for each strain gauge channel independently from one another. The results are presented in numerical and graphical format in the following pages. This calibration also provides the means to verify the proper function of the proposed system, like output linearity, output range, and accuracy. The strain gage indicator has selectable strains, but they represent half the strain if a quarter bridge is used (as in our application). The gage factor for this indicator is 2 and the gage factor for the EGP-5-120 is 2.05; therefore, the selected strain on the indicator has to be modified according to the following equation: εEGP-5-120 = (εIndicator/2)2/2.05 (11-1) To test the system, the following steps were performed. First, the indicator was substi-tuted in the place of one of the strain gages in the Non-Recoverable Unit. Next, its signal was sent wireless to the Receiver and Data Processing Unit, where the output was stored in the laptop computer as output (Volts). The recorded signal had a duration of one second, for which the maximum, minimum, and average values were found. The average value was then plotted versus the calculated strain for the EGP-5-120 and the slope of the trend line is the calibration factor. The following plot (Figure 11-15) shows the process. 131 STRAIN GAGE #1 CALIBRATION USING STRAIN INDICATOR Output Relative Output (respect to 0 µε) Strain of Indicator Strain of EGP-5-120 Average Maximum Minimum Average Maximum Minimum Max Output Error µε µε Volts Volts Volts Volts Volts Volts Volts µε 0 0.000 0.30086 0.32119 0.28174 0 0.020328 -0.019119 0.0203 6.37 200 97.561 0.60454 0.6281 0.58027 0.303682 0.327239 0.279407 0.0243 7.61 400 195.122 0.88267 0.92794 0.79403 0.581811 0.627074 0.493171 0.0886 27.79 600 292.683 1.10059 1.20239 0.91712 0.79973 0.901527 0.616258 0.1835 57.52 800 390.244 1.36881 1.50336 1.20289 1.067944 1.202499 0.902031 0.1659 52.01 1000 487.805 1.74217 1.82878 1.54037 1.441311 1.527914 1.239506 0.2018 63.26 1200 585.366 2.12769 2.15877 1.96982 1.82683 1.857912 1.668954 0.1579 49.49 1400 682.927 2.4506 2.47567 2.42514 2.149738 2.174813 2.124276 0.0255 7.98 1600 780.488 2.76449 2.78972 2.73855 2.463623 2.488861 2.437693 0.0259 8.13 1800 878.049 3.10065 3.12443 3.07785 2.799789 2.823567 2.776992 0.0238 7.45 2000 975.610 3.42376 3.44839 3.39815 3.122901 3.147532 3.097292 0.0256 8.03 2200 1073.171 3.74242 3.76858 3.71903 3.441559 3.467717 3.418164 0.0262 8.20 2400 1170.732 4.05886 4.08612 4.03414 3.757997 3.785258 3.733276 0.0273 8.55 2600 1268.293 4.37535 4.39814 4.35237 4.074493 4.097276 4.051504 0.0230 7.21 2800 1365.854 4.69008 4.71438 4.6659 4.389217 4.413519 4.365037 0.0243 7.62 3000 1463.415 4.99766 5.02369 4.97475 4.696795 4.722828 4.673884 0.0260 8.16 3200 1560.976 5.30235 5.32637 5.28184 5.001491 5.02551 4.980974 0.0240 7.53 3400 1658.537 5.42516 5.42642 5.42415 5.124297 5.125559 5.12329 0.0013 0.40 3600 1756.098 5.41818 5.42049 5.41749 5.117316 5.119632 5.116626 0.0023 0.73 3800 1853.659 5.41596 5.41711 5.41559 5.1151 5.116247 5.114732 0.0011 0.36 4000 1951.220 5.41497 5.41662 5.41293 5.114103 5.115759 5.112067 0.0020 0.64 4200 2048.780 5.41193 5.4131 5.41152 5.111073 5.112243 5.110654 0.0012 0.37 4400 2146.341 5.41182 5.41311 5.41166 5.110958 5.112249 5.110796 0.0013 0.40 4600 2243.902 5.41191 5.41305 5.41139 5.111049 5.11219 5.110528 0.0011 0.36 Figure 11-15a. Calibrating the Strain Gauge Equipment 132 NOTE: The offset is different for the strain gage EGP-5-120 Sensitivity = 313.49 µε/ µε/ µε/ µε/Volt STRAIN LIMITS BASED ON TANDARD FDOT PILES AND GIDELINES Maximum expected ε due to driving = 603.64 µε = ∆V = 1.93 Volts Maximum ε due to prestress = 251.23 µε = ∆V = 0.80 Volts Maximumε due to prestress and drivability = 1014.07 µε = ∆V = 3.23 Volts Minimum ε due to prestress and drivability = -320.84 µε = ∆V = -1.02 Volts Overall Output Range = 4.26 Volts Max Error (+/-)= 63.26 µε = µε = µε = µε = 4.7392 % Strain Indicator Used: 1550-A STRAIN INDICATOR CALIBRATOR (From Micro-Measurements) Fig. 11-15b. Calibrating the Strain Gauge Equipment (continued) 133 To calibrate the accelerometer equipment, the portable simulator (Figure 11-16) was employed. The frequency and amplitude are selectable, and they are increased from 0 to 1,500 Hz and 0 to +/-250g. The calibration process and data process is the same as the strain indicator. ACCELEREMOTER #1 CALIBRATION USING ACCELEROMETER SIMULATOR Induced Average Output Average Among Maximum Acceleration Induced Acceleration Frequency Offset Peak Values (Absolute Value V) Error (+/-) g Hz Volts ∆V ∆V Volts 0 0 -0.001258907 0.001258907 0.045127811 0.043869 50 30 -0.001555631 1.003324447 1.062897494 0.059573 50 60 -0.001386824 1.025655778 1.063371477 0.037716 50 100 -0.00116358 1.043060605 1.079769037 0.036708 50 200 -0.001467657 1.048808691 1.08739953 0.038591 50 500 -0.001425273 1.050460419 1.091104024 0.040644 50 1000 -0.00143403 1.041308545 1.093231497 0.051923 100 30 -0.001147773 1.997966151 2.08991668 0.091951 100 60 -0.001169683 2.048700051 2.104543208 0.055843 100 100 -0.001680954 2.081011782 2.1385168 0.057505 100 200 -0.001101422 2.095838708 2.139096332 0.043258 100 500 -0.001070968 2.100534039 2.150723465 0.050189 100 1000 -0.000975458 2.081674582 2.153260382 0.071586 150 30 -0.000960448 3.038234501 3.107254884 0.06902 150 60 -0.000812205 3.05425425 3.129375783 0.075122 150 100 -0.001413189 3.09677124 3.163034283 0.066263 150 200 -0.002037587 3.128117588 3.177588878 0.049471 150 500 -0.001343433 3.127686278 3.183471021 0.055785 150 1000 -0.001424479 3.099656386 3.191732096 0.092076 200 30 -0.001054307 3.937560051 4.098954698 0.161395 200 60 -0.001128641 4.077471123 4.17703298 0.099562 200 100 -0.001704511 4.10906499 4.19019002 0.081125 200 200 -0.002160764 4.138022655 4.200463987 0.062441 200 500 -0.002088994 4.153284008 4.220228642 0.066945 200 1000 -0.001664201 4.115014624 4.236006697 0.120992 250 30 -0.00242682 4.906060003 5.10008307 0.194023 250 60 0.001994552 5.070773265 5.199116288 0.128343 250 100 -0.000384666 5.114732497 5.213848732 0.099116 250 200 -0.000754616 5.165591347 5.237587669 0.071996 250 500 -0.000763443 5.164991672 5.26229808 0.097306 250 1000 -0.001797519 5.11715747 5.270419278 0.153262 Figure 11-16a. Calibrating the Accelerometer Channel 134 Sensitivity = 48.878 g/Volt ACCELEROMETER LIMITS BASED ON PILE-UF Maximum expected acceleration (+/-) = 250g Overall Output Range (+/-) = 5.11 Volts Max Error due to noise (+/-) = 0.19 Volts = 9.4834 g Calibration Accuracy % = 6.87 Signal Generator used: Portable Generator Model 4830A ( from ENDEVCO) No filter used in the in the VI, Therefore the error is big. Accelerometer Sensitivity = 1.74 Pc/g # 3 P.Order 10039584; Box No. 17 Figure 11-16b. Calibrating the Accelerometer Channel (continued) 135 MOUNTING OF THE NON-RECOVERABLE UNIT It is very important that the strain gages and accelerometers be positioned in the pile cor-rectly. The latter ensure that the signals include only the intended measurements. To accom-plish the latter the transducers must be aligned with the axial direction of the pile. The proper alignment can be achieved using the strands as reference lines as well as support for the trans-ducers. The strain gages and accelerometers can be firmly supported using plastic ties as shown in Figure 11-17. If only one strain gage is used, it should be placed at the center of gravity of the of the pile cross section, as show in Figure 11-17 (a). This will guaranteed that the measurement is the average strain in the pile area (bending effects eliminated). If two strain gages are used they should be equally spaced from the pile center of gravity, as shown in Figure 11-17 (b). The latter assures that bending effects are eliminated if the average of the two gages is used. This arrangement will also provide information about bending in the pile. The accelerometers should be placed close to the strain gages, and as far as possible from the strands to avoid measuring the wave in the strands (Figure 11-17 a). (a) (b) Figure 11-17. Instrumentation Placement (a) When One Strain Gage is Used; (b) When Two Strain Gages Are Used 136 The Signal Conditioner and Sending Unit should be placed between the sensors and the pile top. The unit should be separated from the transducers by at least 2 times the diameter of the pile. The latter will minimize the interference with the stress wave propagation in the pile. Also, any wave reflection from the unit will not unduly influence the instrumentation signal. Finally, the Unit should be placed at the pile’s center of gravity to avoid induced bending stresses (Figure 11-18). Figure 11-18. Signal Conditioning and Sending Unit Placement in the Concrete Forms 137 CHAPTER 12 ACQUISITION AND DATA PROCESSING SOFTWARE (Pile-Monitoring.vi) As mentioned in Chapter 8, an acquisition and data processing software was written for pile capacity determination, as well as pile stress monitoring during driving. This program is called Pile-Monitoring.vi. Its principal task is to obtain pile capacity from the accelerometer and strain gage signals. For this purpose, the program uses three theoretical methods of capacity assessment: PDA Case Method, Paikowsky Method (FHWA Method), and UF Method (Chapter 7). The software also performs pile integrity and stress analysis. In order to apply the methods, the software calculates three types of traces for each blow event: the Force (P) Trace, the Velocity (ZVel) Trace, and displacement (z) trace. The traces are obtained using the pile properties, and signals acquired by the Data Acquisition Card (raw signal). The raw signal is a discrete series of digital strain and acceleration voltages during driving. These values are transformed to real strain and acceleration, and then to Force, and Velocity. The displacement is obtained from the double integral of acceleration. This chapter is divided into three sections which 1) provides a brief explanation of the programming language used, LabVIEW, 2) describes the User Interface (Front Panel of the software), and 3) describes the written code (Block diagram of the software). LabVIEW OVERVIEW LabVIEW was the programming language used to create Pile-Monitoring.vi. Developed by National Instruments Corporation, its main purpose is data acquisition and instrumentation control. Its tight integration with hardware allows for rapid development of data acquisition, analysis, and visual presentation of results. 138 LabVIEW is a high-end language written to simplify data acquisition. With the latter in mind, LabVIEW uses a graphical programming language, G, to create programs (Virtual Instru-ments, VIs) in block diagram form similar to a electrical circuit. LabVIEW provides libraries of functions and tools for specific application, which are available in other and programs. Each VI has three basic components: 1. The user interface, or front panel, controls the operation of the VI. It resembles the front of an actual instrument with buttons and displays. The user can input parameters using the keyboard or mouse. 2. The block diagram, is the source code for the data acquisition evolution. It uses the input parameters from the user interface to direct the process. 3. Icon connections that allow lower level VI’s to be called for operation by a higher-level VI. These are basically subroutines called SubVI. The input and output information is transferred between the front panel and the block dia-gram using special connections. There is a great variety of connectors available in LabVIEW. They are divided into two major groups: controls and indicators. Controls are used to input data into the program at any time. Indicators are use to present the output of the program. FRONT PANEL DESCRIPTION AND OPERATION The Front Panel, shown in Figure 12-1, is the user interface of the software. It is used to start and stop the program and also to input and display data in numerical and graphical format. The upper left buttons (arrow and stop sign) are used to start and stop the program at any time. As soon as the arrow button is pressed (left mouse click), the program starts operating and brings up a dialog-box asking for a directory to save data. Once the directory has been selected, a new window shows up (Figure 12-2) asking for input parameters (to be introduced by the user). These parameters are shown in Figure 12-2 and 139 Figure 12-1. Pile-Monitoring.vi Front Panel Figure 12-2. Input-Data Front Panel 140 they have been gathered into two groups: Pile Properties and Transducer Information. They are required for capacity determination and incoming data reduction, respectively. Figure 12-2 shows two square options (check or no check) named SI Units and English Units. They are used to select the type of Units to input the pile properties. The pile properties are introduced in the third column and their specific units, either in SI or English, is shown in the right column. On the bottom half of the screen, four columns show the transducer input information (Figure 12-2). The first column is used to select the transducers to be used. The second column is used to specify the transducer location (pile tip or pile top). The third column specifies the transducer name. Finally, the fourth column is used to input the sensor sensitivity. The sensi-tivity is obtained from the calibration of the entire wireless system, as described in Chapter 11. Once the required data is inputted, the Continue button (Figure 12-2) should be pressed, and the Front Panel (Figure 12-1) reappears. At this time the data acquisition process starts. After each pile strike, the Front Panel will output new or updated information, as shown in Figure 12-3. Four graphs show the program output (numbered from 1 to 4 in Figure 12-3). Graphs 1 and 2 show two Force Traces and two Velocity Traces respectively, in terms of force versus time. These graphs are internally obtained from the output signal of two strain gages and two accelerometers respectively. Graph 3 shows two plots, one is the average of the two Force Traces (Graph 1) and the other is the average of the two Velocity Traces (Graph 2). In the event that one set of transducers is placed at the pile tip, Graph 3 disregards it and will display only the Force and Velocity from the transducers at the pile top. The last graph, Graph #4, shows the pile 141 Figure 12-3. Pile-Monitoring.vi Front Panel (after 22 pile strikes) capacities (Skin, Tip and Total) versus the pile penetration depth. Graphs 1 to 3 are specific for a single blow while Graph 4 represents the entire driving history. Every time the hammer strikes the pile head, a new set of accelerations and strains are delivered to the software. Traces on Graphs 1 to 3 are replaced by the new incoming information while Graph 4 is updated with the new capacity and penetration (from the new blow). The new capacity and pile displacements are calculated from the new traces on Graph 3. Graphs 1 and 2 are used to provide information about the transducers performance. Force Traces and Velocity Traces should look similar to the ones presented in Chapter 5, other-wise there might be a problem with the instrumentation package. If one of the Force Traces or 142 one of the Velocity Traces looks incorrect, they can be turned off with the ON/OFF controls on the Front Panel (upper left in Figure 12-1). The signal will not be used, but it will be stored for further reference. If all transducers are at the same location along the pile axis, both Force Traces and both Velocity Traces should look very similar. If there is a great difference between Velocity Traces, one accelerometer might not be working properly and should be turned off. The same could happen with the Force Traces, one strain gage might be damaged, or more likely indicate bending stresses in the pile. Bending stresses should be kept within safe limits (FDOT specs.). For the purpose of capacity determination, taking the average of the two Force Traces (Graph 3) eliminates the bending effect. Graph 3 is used for data quality checks (Chapter 10), pile stress monitoring, and pile damage detection. The total pile capacity, in Graph 4, is calculated using the CASE method or the Paikowsky Method (Chapter 7). The method is selected in the Total Capacity Method selector on the left side of the front panel. The Skin and Tip Capacities are obtained using the UF Method described in Chapter 7. The pile penetration, in Graph 4, is calculated from the cumulative penetration obtained after each blow. However, since penetration during each blow is calculated with the double integration of the acceleration signal, errors in the cumulative penetration can arise. To account for this error, a Depth Correction selector has been incorporated in the Front Panel. Conse-quently, operator has the option of periodically updating the total penetration depth. The Velocity Fix control knob, shown on the Front Panel, is used to correct the Velocity Trace in Graph 4. The Velocity Trace may be forced to come back to 0 at the end of the trace 143 using the method described in Chapter 6. Using the start of the trace as a pivot point, the entire trace is rotated until the velocity at time 0.1024 seconds becomes 0 (similar to PDA). The Trigger Channel selector, shown on the Front Panel, is used to pick up one of the four incoming signals for triggering purposes. During the acquisition process a few of the incoming signals carrying the dynamic activity of the pile are selected for further analysis. Some typical incoming signals were presented in Figures 10-10 and 10-11 for the strain gage and the accelerometer respectively. The program will use the selected channel to look for these types of specific signals for further analysis. The View Raw Data control knob is used to display a new window where the incoming data from the sensors can be continually observed (Figure 12-4). This graph is used to analyze the operation of the instrumentation system, including excessive offset, background noise, and Figure 12-4. View Raw Data Window 144 signal stability. The input signal shown in this graph should be between 0 and 5 volts, with the offset close to 3.75 Volts for the strain gage channels and close to 2.5 Volts for the accelerometer channels. The noise level should be below 10 mV for each channel. If the signal shows unex-pected sharp spikes (digitalization error), then the receiver antenna is not well oriented, or it is too far from the pile. The Run Mode control knob, shown on the Front Panel, is used to select the operation mode of the software. The program has two modes of operation: Run and Standby. The Run Mode is used during the driving procedure to pick up the transferred signals from the pile and perform the activities described above. The Standby Mode is used to replay past blow traces that have been stored in the hard drive of the computer. In Standby Mode the software performs the same activities as in Run Mode, except data acquisition. The same controls and graphs described above are available. Additional features of the Front Panel include the Blow No. Indicator and the Current Blow Data save control. The Blow No. indicates for which blow the traces are plotted in Graphs 1 to 3. It is also used to replay previous blows when the program is in Standby Mode. The Current Blow Data save control is used to store the blow information as raw data, in the previously selected directory. They are individually stored in text format under the name of BLOW #, where # indicates the actual blow number. In Standby Mode, this knob will resave the stored information in CAPWAP format (Figure 9-2). The pile capacity and depth are saved after each pile strike by using the excel format (in the previously selected directory). The input data (Figure 12-1) is also stored in this directory under the parameter name. 145 BLOCK DIAGRAMS As explained in the introduction, the Block Diagram, shown in Figures 12-5 to 12-15, is the actual program code. The details of the language are not discussed herein, but a brief layout of the VI is given below. The Block Diagram of the software is divided into two main sequences. The first group (sequence # 0) is in charge of the data input, and the second group (sequence # 1) is in charge of the data acquisition, storage and display. These sequences will be referred as Input Sequence (sequence #0) and Data-Processing Sequence (sequence # 1). The Input Sequence is composed of two sections shown in Figures 12-5 and 12-6. The first section is in charge of asking the user for the input directory. Using this directory, file paths are created to store further information (parameters, raw data, capacity, and a temporary direc-tory to keep input parameters). The second section asks the user for input parameters (Figure 12-2), stores the latter information and transfers it to the Data-Processing Sequence. Figure 12-5. Code Used to Select Directory and File Paths 146 Figure 12-6. Code Used to Input Data (a) (b) (c) Figure 12-7. Input Data Sub VI Code (a) Read Parameters from Previous File; (b) Check Units Selected by the User; (c) Transfer Parameters to the Input Sequence 147 Figure 12-8. Code Used for Run Mode Figure 12-9. Code Used for Standby Mode 148 Figure 12-10. Strain and Acceleration Sub VI Code Figure 12-11. Integration Sub VI Code 149 (a) (b) Figure 12-12. Capacity Sub VI Code 150 Figure 12-13. Capacity Graph Sub VI Code Figure 12-14. FFT Filter Sub VI Code 151 Figure 12-15. Save in CAPWAP Format Sub VI Code Right in the middle of Figure 12-6 is a small square called Input. This SubVI (subrou-tine) is in charge of displaying the input window shown in Figure 12-2. The code for this SubVI is shown in Figure 12-7. It is composed of three groups that operate in sequence. Figure 12-7a is in charge of recovering previously stored parameters in the event that the user requires a replay of previous driving records. Figure 12.7b analyzes the units selected by the user. And Figure 12.7c gathers the input parameters and sends them to the higher level code. Figures 12-8 and 12-9 shows the Data–Processing Sequence (mentioned at the begin-ning of this section). Figure 12-8 presents the code used when the program is in Run Mode and Figure 12-9 presents the code used when the program is in Standby Mode. Both areas of the pro-gram are similar, with the major difference that one recovers the strain and acceleration signals 152 from the transducers and the other recovers that information from a file. Since both codes are similar, only the Run Mode will be described. The first step (Figure 12-8) involves querying the Data Acquisition Card as to its readi-ness. The Data Acquisition Card is configured to recover four signals (two accelerometers and two strain gages) at 10,000 samples per second each. A buffer size capable of holding 4,000 samples is set up to transfer the signals from the Data Acquisition Card to the computer. This buffer size has proven (hundreds of hours of testing) to be stable during the transfer process. The incoming signals have the following channel allocations: Channel 0: accelerometer #1 Channel 1: accelerometer #2 Channel 2: strain gage #1 Channel 3: strain gage #2 Note, the specified sampling rate (10,000 Hz) is higher than the minimum specified by ASTM D4945 (Chapter 9) but results in good signal resolution. The second step of the code in Run Mode is to recover the information in the while loop shown in Figure 12-8. A number of other activities also take place within this while loop. A trigger determines if there is dynamic activity in one of the four channels (selected with the Trigger Channel knob on the front panel). If the amplitude between maximum and minimum spikes overpass a predefined value (0.1 Volts = 27g), the data processing will be trig-gered. The incoming data will then be processed and plotted in the front panel. For this purpose four SubVIs (Figure 12-8) are used. Special icons as shown identify them. Next to each icon is the name of each SubVIs followed by a brief description of its activity in the software. Strain and Acceleration.vi. This sub VI transforms the raw data into strain and acceleration, as well as select which channels are in use (ON/OFF control in the front 153 panel). The process starts by filtering the signal from 0 to 5 Hz and 3,000 to 10,000 Hz using an FFT filter (Figure 12-14). Next it changes the signal offset to 0, and rounds out to 0 any values that are smaller than the noise level (<10 mV). Finally, it multiplies each signal by its calibration factor to obtain strain and acceleration records. The code is shown in Figure 12-10. Integration.vi. Uses the strain and acceleration from the last SubVI to obtain Force, Velocity (ZVel), and displacement traces. Integrates the acceleration to obtain velocity, performs the PDA velocity fix (if selected on the front panel), obtains Force and Velocity Traces to be plotted in the Front Panel (Graph 1 and 2). Calculates the average Force Trace and average Velocity Trace (from sensors at the pile top) to be plotted on the Front Panel (Graph # 3), and to be used in further analysis. Obtains displacement trace from the average velocity and finds the start of that trace, the rise time, 2L/c, set, and displace-ment. The code is presented in Figure 12-11. Capacity.vi. Uses the Force average, Velocity average, set, and displacement from the last SubVI to estimate pile capacity (Tip, Skin, Total). The methods described in Chapter 7 are used for this purpose. This code is presented in Figures 12-12. Capacity Graph.vi. Uses the capacity assessments from the last SubVI to update the previous assessments (prior blow) for plotting on the Front Panel. The code is presented in Figure 12-13. Store CAPWAP.vi. Stores the average Force and Average Velocity (from the Inte-gration.vi) in CAPWAP Format. The code is presented in Figure 12-15. 154 Evident from the prior discussion, the processing of the information involves many activ-ities. Also, the trigger plays a key role in this program. The amount of information recovered at 10,000 Hz is enormous, and much of it is not used. As described in Chapter 9, only 0.1024 seconds of sampled data is required to analyze a single blow. To show the magnitude of the data, consider a hammer with blow rate of 60 blows per minute. In one minute (60 blows), 6.14 seconds of data is of interest, and the data recorded over the remaining 53.86 seconds (60 sec – 6.14 sec) is redundant. The latter represents 2.15 million values of useless data (53.86 sec 4 channel 10,000 samples/sec/channel). Consequently, minimizing operations (removal of unwanted data) was a major concern in this software. As can be seen, in Run Mode, more than 4,000 values need to be managed to obtain the pile capacities at a given blow, another 12,300 values are analyzed for redundancy, and 4,000 new incoming data points are sampled and saved. Therefore, the redundant informa-tion must be detected and eliminated as soon as possible with minimum amount of operations. 155 CHAPTER 13 LABORATORY AND SMALL SCALE FIELD TESTS In developing the wireless instrumentation equipment, a significant amount of laboratory and small scale field-testing was performed. In the case of laboratory work, each component was tested for calibration purposes, as well as minimizing signal noise. The small-scale field-testing was to ensure that the instrumentation package held up to the rigors of pile driving, as well as the identification of any other problems (transmission distance, noise, etc.). The field tests involved embedding the non-recoverable units in a miniature prestressed concrete pile and driving it into to the ground (more than 60 blows) with a small air/steam hammer. The small pile was also used in laboratory for calibrating and validating equipment response. The laboratory testing and field testing are described as follows. MINIATURE PILE A miniature prestressed concrete pile, shown in Figure 13-1, was built in the University of Florida’s Structures Lab. The purpose of this pile was to create a proper environment, similar to FDOT concrete piles to test the instrumentation package in the laboratory and in the field. The dimensions and properties of the pile are presented in Figure 13-1. (a) (b) Figure 13-1. Miniature Pile (a) In the structures laboratory; (b) Properties 156 The pile was instrumented in such way that it could be monitored by two independent systems: I) the proposed wireless system and II) a separate wired system. The wired system was a simplified version of the wireless system with the same instrumentation package but hard wired directly to the data acquisition system. Consequently, there were eight sensors (Figure 13-2) embedded in the pile, four of which form part of the wireless system (Non-Recoverable Unit), and the remaining (four) which were connected to external conditioners outside the pile using cables. The difference between the two systems was intended to localize problems in the wire-less link, as well as identify shock and vibration issues with the conditioners and transmitter. Furthermore, the additional sensors served as backup instrumentation for the Non-Recoverable Unit (in the event that they were damaged during casting of the pile). Figure 13-2. Installation of Non-Recoverable Unit and Wired Sensors 157 Figure 13-2 shows the instrumentation arrangement prior to pouring the concrete. The four transducers of the Non-Recoverable Unit were installed two pile diameters below the pile top and two diameters above the pile tip; one set of transducers (1 accelerometer and 1 strain gage) at each location. The four wired sensors were installed in the same way and in the same cross section as the other four transducers. The strain gages were aligned using four plastic ties each; one pair to support the strain gages from the top strand, and the other pair to keep the strain gage alignment during the con-crete pouring (attached to bottom strands). The strands and plastic ties provided the proper alignment, in one direction, and uniform separation between strain gages and strands achieved the alignment in the other direction. It is important to note that the accelerometers (Endevco) were mounted in this pile using an older technique than the one presented in Chapter 11. The accelerometers were first mounted in a 10-cm by 5-cm (4 in. by 2 in.) concrete cylinder using adhesive mounting. A low noise cable was soldered directly to the top of the accelerometer. Then, the top and lateral sides of the accelerometers were covered with a layer of silicone to avoid compression stresses and to protect the accelerometer from moisture. Finally, the side of the concrete cylinder was covered with more concrete to protect the accelerometer during the casting procedure. This cylinder was aligned in the pile using four plastic ties attached to two strands. STRAIN GAGE AND CONDITIONERS Two different types of tests were performed to verify the suitability of the proposed strain gages and conditioners. The first test was to verify the output linearity of the gages and condi-tioners, and the second was to ensure proper dynamic response time. 158 For the output linearity, one strain gage was embedded in a concrete cylinder and loaded as shown in Figure 13-3. The strain gage was connected to conditioners outside the cylinder (Black Box) and then to a voltmeter. A special dial gage connected to the cylinder provided information about the induced strain and the voltmeter indicated the output voltage from the strain gage and conditioners. The plot of induced strain versus output voltage, Figure 13-3, reveals the linearity with a maximum error of 2.16 % (30 µε 100/1391.9 µε) over the intended measuring range (Table 9.2). This test was repeated many times to verify repeatability and the calibration procedure presented in Chapter 11. Figure 13-3. Output Linearity Test of Strain Gage and Conditioner It is important to note that the voltmeter output conceals the channel noise because it averages the voltage over a time interval. Nevertheless, during the strain gage calibration pro-cess (Chapter 11), it was observed that the noise influence was low. The measured amplitude of noise was 10 mV (2.78 µε), which represents an additional 0.2% error. The total maximum error is 2.36% (below the required maximum 3% in Table_9.2). Next, the dynamic response of the strain gages and conditioners was checked against the PDA strain gages with a Fast Fourier analysis (FFT). The PDA strain gages were attached to the 159 miniature pile (Figure 13-4a) at the same location of the embedded strain gages. The wired strain gage present at this location was connected to external conditioners and then to a laptop computer (with the same software as the wireless system). The pile was struck with a hand held hammer to create a force wave to be recorded by both apparatus independently (Figure 13-4b). (a) (b) Figure 13-4. Comparison Between PDA and Transducers Embedded in the Miniature Pile (a) PDA sensors on miniature pile; (b) Simultaneous monitoring using PDA and the proposed new system Shown in Figure 13-5 is the Force Trace recorded in the laptop computer versus the one recorded by PDA. The bottom graph in Figure 13-5 shows the frequency content of both traces (FTT). Both graphs show good frequency and fair amplitude agreement, considering that the transducers were not in the same exact position (PDA vs. Proposed). Moreover, the frequency response shows frequencies above the minimum expected (300 Hz). This test also served to verify that the strain gage calibration procedure was correct. In Figure 13-5 the maximum error in terms of amplitude is 13% with respect to the maxi-mum recorded force, and 0.22 % with reference to the maximum expected force (EA1391.9). Note this error is in the same range as the noise (10 mV). 160 Figure 13-5. Force Traces Obtained with the Proposed Equipment and PDA, and FFT ACCELEROMETER, CONDITIONERS AND FILTERS Many tests were conducted to verify the suitability of the proposed accelerometers, con-ditioners and filters. Many of the early lab tests revealed problems, that were fixed and tested again. Other problems were discovered in driving the miniature pile, which were later fixed. A brief discussion of both the laboratory testing and miniature pile testing of accelerometers, condi-tioners, and filters are described herein. From early in the work, a major concern was the development of a viable, cheap accelerometer, which was accurate for the pile dynamic testing. After identifying the Endevco accelerometer, as a likely candidate, an appropriate housing and conditioner circuit had to be designed and subsequently tested. In the case of the latter, the accelerometer, conditioners, and 161 filters had to be able sustain high shock and vibration for concrete piles (+/-200g and 10 to 3000 Hz). To test the latter, the accelerometer and conditioner were cast in a miniature pile and struck with a hand held hammer to induce accelerations, which were monitored. In addition, a high quality accelerometer (DC), as well as the PDA accelerometers were attached to the outside of the miniature pile for comparison purposes. Both amplitude and frequency content were of interest. For the first tests, the Endevco accelerometer was mounted on the back of the PDA accelerometer, and the PDA accelerometer was mounted on the miniature pile. In this test the Endevco accelerometer was installed on a flat plate, and the plate was glued to the PDA accelerometer. Figure 13-6 shows the signal comparison for one blow event in three forms: acceleration, frequency content and Velocity Trace. Figure 13-7 shows the same graphs as in Figure 13-6 but in a smaller time frame. In these graphs a 5 Hz low pass filter was used on the acceleration signal to eliminate any zero shift effect (Charter 9). These figures show a good match between both accelerometers, in terms of acceleration and frequency response. The Velocity Traces also show agreement up to 0.05 seconds, which is more than the time of interest (0.03 seconds to capture the wave return). It should be noted that the Velocity Traces in Figures 13-6 and 13-7 were forced to return back to zero at the end of the trace (0.1024 seconds), as explained in Chapter 12. From the latter tests it can be seen that the accelerometer and its installation on the flat plate are adequate for the intended application (similar to PDA). Also, it is shown that the cali-bration, in accordance with Chapter 11, is reliable. The maximum acceleration achieved during these tests was 81g. The maximum-recorded error was 20% with respect to the maximum accel-eration measured, but only 4% with respect to the expected range of accelerations (+/-500g). 162 Figure 13-6. Accelerometer Output Comparison Between Proposed Equipment and PDA in Terms of Acceleration, Frequency Spectrum and Velocity Trace (0-0.1 sec.) Figure 13-7. Accelerometer Output Comparison Between Proposed Equipment and PDA in Terms of Acceleration, Frequency Spectrum and Velocity Trace (0-0.055 sec.) 163 The maximum error found in the Velocity Trace was 18% with respect to the maximum Velocity recorded, but only 1.1% with respect to the expected range of Force (ZVel). A second series of tests were performed on the Endevco accelerometer after placed in its new sealed enclosure. The accelerometer was subsequently glued to the end of the miniature pile along with a DC (piezoresistive) accelerometer. The DC accelerometer, model 7270A (from Endevco), is considered a top quality (price: $5000) accelerometer used for calibration and com-parison purposes. Figure 13-8 shows how the accelerometers attached to the miniature pile. Figure 13-9 shows the signal comparison for one blow in terms of acceleration, frequency response and Velocity Trace. The velocity adjustment and the 5 Hz low pass filter were used, as in the last tests, to obtain the Velocity Trace. Evident from the figure is the good agreement between the two accelerometers. It should be noted that the DC accelerometer is a piezoresistive accelerometer, which is not likely to present zero shifts. Also, the signal from DC accelerometer showed good similarity with the PDA accelerometer. This was observed during a test where the DC accelerometer was Figure 13-8. Accelerometer Installation at the Pile Tip 164 Figure 13-9. Accelerometer Output Comparison Between Proposed Equipment and DC Accelerometer in Terms of Acceleration, Frequency Spectrum and Velocity Trace mounted on top of the PDA accelerometer, and the PDA accelerometer on the pile. In these tests the maximum acceleration recorded was close to 150g, and the frequency was below 3000 Hz. These are typical values as described in Chapter 5. The maximum error recorded was 3% with respect to the maximum measurement, but only 1% with respect to the expected range of acceler-ations (+/-500g). The maximum error found in the Velocity Trace was 22.7% with respect to the maximum velocity recorded, but only 0.1% with respect to the expected range of forces (ZVel). 165 SMALL SCALE DRIVING TEST The miniature pile described earlier was driven in Green Cove Springs, Florida. An 8.9 KN (2 kips) air/steam hammer, with 0.30 m (1 ft) stroke, was used to drive the pile 0.91 m (3ft) into a compacted layer of limestone overlaying a layer of silty sand to sandy silt (Figure 13-10). Three independent apparatus were used to monitor the driving: the wireless system, an additional wired system, and the PDA. It is important to remember that the accelerometers, embedded in this pile, were mounted using an older technique than the technique explained previously. Also, the wireless system used for this pile is an older version, which tend to show more noise than the latest version (65 mV versus 10 mV). Figure 13-10. Cone Penetration Test in Green Cove Springs. 30 m (100 ft) Away from the Driving Location 166 The pile properties were obtained in the laboratory (one-day prior driving). These properties are shown in Figure 13-1. The wave speed was checked in the field and showed no significant variation with respect to the laboratory measurements. The pile was positioned on the leads and the verticality checked with a hand level. The Receiver and Data Processing Unit was located 10 m (30-ft) away from the pile with the 1-meter directional antenna in direct alignment with the transmitter antenna (inside the pile). Figure 13-11 (a) also shows the cables running out of the pile from the wired transducers. These cables are connected to a laptop computer shown in Figure 13-11 (b) (upper laptop). The lower laptop computer in this figure belongs to the wireless system. (a) (b) Figure 13-11. Shows the Pile in the Leads and the Receiver and Data Processing Unit Ready to Start Driving Figure 13-12 shows the PDA sensors being attached to the pile. The PDA transducers were installed 21.3 cm (0.7 ft) below the embedded sensors. They were not installed exactly in the same section as the embedded sensors because the pile cracked at the anchor bolt locations were the PDA sensors were to be attached initially. Consequently, only one accelerometer and one strain gage were used because of the cracking, and space constrains (small pile). 167 Figure 13-12. PDA Transducers Attached to the Pile The driving consisted of 62 blows (Table 13-1). During the first 22 blows only the wire-less system was used. Over the remaining 40 blows, the PDA package and the wired package were also used for comparison with the proposed instrumentation package. Table 13-1. Driving Records Penetration Number of Blows 0.1 m (1 ft) 26 0.2 m (2 ft) 29 0.3 m (3 ft) 7 For the fist 22 blows, the driving was stopped several times to evaluate the operation of the instrumentation package components. Figure 13-13 shows the software output for one of these blows. The system seemed to work properly at this point in time. The following issues were checked: - The Non-Recoverable Unit woke (sent signals) after the first strike - The wireless link showed a full strength signal (4 LEDs were lighted) 168 - The software displayed Force and Velocity Traces for each blow. - The capacity graph in the software was updated for each blow Figure 13-13. Pile-Monitoring.vi Front Panel for Blow # 7 During the Driving Test in Green Cove Springs Over the last 40 blows the wireless system, the PDA, and wired System were used. Typi-cal Force and Velocity Traces are shown in Figures 13-14 for the PDA System. Some problems were identified as follows: - The PDA Force Traces showed high negative values not recorded by the wireless and wired packages - The Velocity Traces from the PDA and the wireless system equipment did not match after the first peak value 169 Figure 13-14. Typical Force and Velocity Traces from the PDA During the Driving Test in Green Cove Springs. Blow # 35 - The data quality check failed for both the PDA and the wireless equipment. - The wave return was not observed at 2L/c but at a later time - The top accelerometer for the wireless system did not work The recorded difference in the Force Traces between the equipment is attributed to two known and one possible problem. The fist known was that only one strain gage was used in each instrumentation system; therefore, bending stresses strongly influence the Force Traces, espe-cially for the PDA strain gage (far from the pile’s center of gravity). The second known was that the measuring point for the PDA transducers was below the measuring point of the embedded strain gages. Finally, a possible problem is that the PDA strain gage might have been wrongly 170 attached. This last issue is addressed by the fact that laboratory tests conducted in the laboratory showed fairly good agreement among embedded gages and the PDA strain gages using the same pile. The difference between Velocity Traces is attributed to two possible problems. Fist the measuring point as mentioned in the last paragraph. Secondly, zero shift effect on the accelera-tion signal. The observed zero shift effect was the reason to improve the mounting technique as explained in Chapter 11. The data quality failure and the late wave return are attributed to the small size of the pile and the placement of the instrumentation in the top of the pile (Non-Recoverable components). For further testing, i.e. full-scale piles, the components would be placed between the top sensors and the pile top, or between the tip sensors and the pile tip. This would result in no interruption of the stress wave propagation. Even though this pile did not performed 100% as expected, some valuable information was obtained: - The instrumentation Package survived the shock and vibration during driving - The instrumentation Package survived the stresses under driving - The Non-recoverable Unit transmits the signal without being influenced by the pile movement, shock, vibration, and compression. - The transmitter antenna works properly when it is partially embedded in the pile, and show no interference due to vibration - The software performed properly, triggering for every blow, and displaying Force and Velocity Traces, plus pile capacities Other addition problems identified and fixed after the field tests were: 171 • The noise was reduced from 66 mV to 10 mV. • The accelerometer mounting technology was changed and tested to avoid zero shift effect. • The Signal Receiver and Conditioning Unit output range (0 to 2.5 Volts) was increased to 0 to 5 Volts. • The gain factor on the strain gage channels has increased by a factor of 2. • Finally, a protective circuit was added to the conditioners to avoid damage to the transmitter in the event of a shorted or open strain gage. 172 CHAPTER 14 FULL SCALE TESTS AT CYPRESS CREEK Two full-scale piles were driven in Tampa, Florida (February 2002) with the proposed wireless equipment. The piles were 24 inch prestressed concrete, sixteen meters (52.5 ft) long, and part of a bridge pier bent. The bridge is located on State Road 54 (SR 54) over Cypress Creek, North of Tampa. Each pile was monitored during driving using 2 accelerometers and 2 strain gages. The first pile had all of instrumentation at the pile top, whereas the second pile had one pair of sensors near the pile top and the other pair close to the pile tip. Additionally, both piles were monitored with the PDA for comparison purposes. The difference in layout of the instrumentation for the two piles was to check repeata-bility, consistency (pile 1), as well as measure stresses at the pile tip (pile 2, proposed future studies). For pile #1, the output from the two accelerometers should be identical, but depending on bending behavior, the two top strain gages may show slight differences. For pile #2, the transducers at the pile tip were intended to measure the stresses, rather than estimate them based on the top measurements. It is also proposed that the force and velocity measurements at the pile tip be used for future studies to better assess damping, tip resistance, residual stresses, and skin friction on piles. SR-54 SITE DESCRIPTION A forty-two meter long multiple span concrete bridge is under construction in Pasco County, North of Tampa. Located on SR 54, the bridge spans Cypress Creek (Figure 14-1), and it replaces a smaller single span bridge. 173 (a) (b) Figure 14-1. Job Site (a) Location; (b) Picture of the Creek and Piles on the East Side The new bridge foundation was designed with two abutments and a center pier. Both abutments (ends) and center pier are supported on standard 24-inch FDOT prestressed concrete piles. The abutments had 8 piles each and the center pier had 6 piles. The bridge elevation is shown in Figure 14-2 and the pile distribution in Figure 14-3. Two of the eight piles on the west side of the creek were instrumented with the wireless instrumentation packages. The piles are labeled as pile number 1 and 3, and are located in the left bent shown in Figure 14-3. There were three SPT bring logs for the whole site, with only Figure 14-2. Bridge Elevation 174 Figure 14-3. Pile Distribution in Two Abutments and One Center Pier one of them, BB-2, located on the west side of the creek. The latter boring, BB-2 was west of pile # 7, and is to a depth of 17 meters. Boring log BB-2 is presented in Figure 14-4 b. (a) (b) Figure 14-4. SPT Boring on the West Side of the Creek (a) Location; (b) Boring Log There was approximately 10 ft of mixed soils overlying limestone. Since, a portion of the upper soil was compacted fill; it was predrilled in accordance with FDOT specifications for pile driving. 175 It is important to note again that piles #1 and #3 in end bent #1 were part of the produc-tion piles. Their length was set at 16 meters based on earlier test pile monitoring in later Decem-ber 2001 with the PDA. Three test piles were performed at that time, pile #4 in end bent #1, pile #3 in bent #2, and pile #7 in end bent #3. PILE INSTRUMENTATION Each pile was instrumented with four sensors: two accelerometers and two strain gages, plus backup sensors and conditioners to be used in case of transducer breakage, cable damage, misalignment or unexpected problems. Each transducer was pre-installed on a 31-mm (1/8-inch) Plexiglas plate, as shown in Figure 14-5, to allow ease of installation and alignment during pile construction in the casting yard. Each plate held one accelerometer and one strain gage as shown in Figure 14-5 (a). The plate was installed in the pile’s steel form using two steel rebars attached to the pile’s longitudi-nal pretension strands. The latter both aligned the instrumentation packages as well as anchored them during the concrete placement. (a) (b) Figure 14-5. Instrumentation Mounting Technique (a) Schematics; (b) Picture 176 The Plexiglas plates were connected to the steel bars using plastic ties and the bars were connected to the strands using wire ties. The transducers were connected to the plate using 10 cm (4 inch) plastic ties. The accelerometer was glued to the Plexiglas plate due to provide resis-tance to concrete flow during casting. In order to ensure good bonding between the strain gage and the concrete, the Plexiglas plate was cut out underneath the strain gage’s active length (i.e., sensing area). Pile #3 was instrumented with two accelerometers and two strain gages at the same cross sectional area (6 ft from the pile top). Both strain gages were equally spaced from the pile cen-terline line to ensure average stresses (account for bending moments), consistency, and similar-ity. Additionally, a backup accelerometer and strain gage was installed between the active trans-ducers. Figure 14-6 shows the instrumentation set up. The alignment of the transducers attached to the Plexiglas plate was with a steel square placed against the pile’s formwork. The distance (a) (b) (c) Figure 14-6. Instrumentation of Pile #3 (Transmitter Unit #4) (a) Instrumentation Placement in Plan View; (b) Picture of Sensors; (c) Picture of Transmitter Plus Electronics 177 of the sensors to the Transmitter (Signal conditioning and Sending Unit) was 2B (48”) in order to avoid unwanted wave return or end effects. The transmitter used for this pile was defined as Unit #4. The antenna was located between the sensors and the transmitter to avoid signal inter-ference by the pile cap. Pile #2 was instrumented with one accelerometer and one strain gage near the pile’s top (6 ft from the top) and with one strain gage and one accelerometer close to the pile’s bottom (6 ft from the pile tip). Additionally a backup accelerometer and strain gage was used near the pile top, as well as the pile tip (5 ft from the pile top and 5 ft from the pile tip). Figure 14-7 shows (a) (b) (c) Figure 14-7. Instrumentation of Pile #1 (Transmitter Unit #5) (a) Instrumentation Placement in Plan View; (b) Picture of Sensors at the Top; (c) Picture of Transmitter Plus Electronics 178 the instrumentation set up. For pile #2, the strain gages were aligned at the pile’s central axis using a square and a ruler. As in pile #3, the distance from the sensors to the Transmitter (Signal conditioning and Sending Unit) was 2 B to avoid unwanted wave returns. The antenna was located between the top sensors and the transmitter. The transmitter for this pile was identified as Unit #5. Since this was the first full-scale test, all of the sensors had a section of their cables coming to the pile surface (flush with the concrete) for access, testing and possible switching in case of problems. Pile #3 had one access point close to the pile top, and pile #1 had two access points, one at the pile tip and the other at the pile top. A typical access point is shown in Figure 14-8. The backup transducers and conditioners were not needed for pile #3 but they were used for the top sensors of pile #1. (a) (b) Figure 14-8. Access Points: (a) Covered; (b) Open to Test Transducers and Conditioners For pile #1 the strain gage conditioners at the pile top failed, so the backup conditioner and transducer replaced it. Also, the accelerometer at this location was replaced (backup used), not because it was defective but to recover both strain and velocity measurements at the same pile cross section. 179 It should be noted that UF researchers placed the instrumentation in both piles and subse-quently left the casting yard. The personnel at Standard Concrete poured the concrete for the two test piles along with the rest of the piles in the bed later that night. Consequently, it is be-lieved that the instrumentation package placement, and concrete pouring may be handled in the casting yard by casting personnel. Next, striking the pile with a hand held hammer tested the activity on each sensor. First striking the pile on its side, close to the transmitter turned on the system. It was noted that striking the pile with a hand held hammer close to its end was very difficult to wake system up (Figure 14-9). After waking the pile, the accelerometer response was verified with and external accelerometer. The wave speed measured with the external accelerometer showed good agree-ment with the embedded accelerometer, giving indication of proper alignment (accelerometer not moved by the abrasive activity during the casing). Figure 14-9. Testing the Instrumentation Package on the Casting Yard by Striking the Piles with a Hand Held Hammer 180 DRIVING TEST Both piles were driven with an ICE 80-S single acing diesel Hammer. The hammer, Fig-ure 14-10, has a total weight of 15.4 tons, a ram weight of 8 ton, maximum stroke of 10 ft (Energy 80,000 - 32,000 lb-ft), and a maximum blow rate of 55 blows per minute. (a) (b) Figure 14-10. Driving Equipment and Setup (a) Diesel Hammer ICE 80-S and Leads; (b) 150 Ton Crane, Leads and Hammer, Template, Pile #3 Driven, Pile #1 to be Driven Each pile was driven with a new pile cushion made of plywood. Several sheets of ply-wood were used to make each cushion 190 mm thick. Before driving, pile on the ground, the WAP instrumentation was turned on, the wave speed measured, and the PDA transducer holes were installed (Figure 14-11b). The WAP equip-ment was turned on with a strike of a hand held hammer. Using the wireless link and the em-bedded accelerometer the wave speed was measured several times following the procedure described in Chapter 8. For Pile # 1, the wave speed was also checked using the time of wave travel from the top to the tip transducer and the distance separating them. Later the PDA trans- 181 (a) (b) Figure 14-11. Pile Setup Preparation (a) Pre Drilling; (b) PDA Transducer Holes ducer holes and anchor bolts were installed in the same location as the WAP transducers (6 ft below the pile top for pile #3 and 5 ft below the pile top for pile #1). The PDA transducers were installed on the lateral sides of the pile to avoid interference from our antenna (front face the pile - Figure 14-11b). Before driving each pile, a template was placed on top of the ground to locate and sup-port the pile. The template was made of steel I-beams and kept in place with concrete weights (Figure 14-11a). Prior to placement of the pile, a hole was predrilled through the template. The hole, predrilled to a depth of six meters with a continuous auger (Figure 14-11a) passed through the existing compacted fill to natural soil. Next, the pile was inserted and then the leads plus hammer was installed around and on top of the pile. The orientation of the pile was checked with a hand level on the leads. Prior to driving, the directional 1-m antenna was mounted on a pole and installed as shown in Figure 14-12c. The laptop computer was installed at the back of a Van and connected to the antenna using a long coaxial cable (Figure 14-12a). 182 (a) (b) (c) Figure 14-12. Set Up (a) WAP Laptop; (b) PDA PAK; (c) Receiver Antenna and Pile #1 on the Leads The pile was placed such the embedded antenna was facing the directional antenna, as shown in Figure 14-12c. It was discovered due to lead construction that some of its frame ele-ments were periodically in front of the embedded antenna. However no problems were recorded other than a minimum decrease in signal strength whenever those elements (Figure 14-12c) were in front of the embedded antenna. During the actual driving the transmission distance was keep within 50 ft (horizontal dis-tance) from the pile. For the fist driven pile, pile #3, the distance was 20 ft and for the second pile, pile #1, the distance was 50 ft. 183 During the early stages of driving, the WAP system was not able to wake up from the hammer strike accelerations. The latter was attributed to low g levels recorded with the PDA (acceleration less than 60g). To partially solve the problem and have some records for early stages of driving, the pile was struck with a hand held hammer close to the embedded Signal Conditioning and Sending Unit. This turned the system on and allowed the recording of 2.5 minutes approximately, after which the pile had to be struck by hand again to turn the system on again. It was not until in later stages of driving (after 12m) that the wakeup system worked as expected. This is attributed to higher accelerations (a > 70 g) recorded by the PDA as well as the WAP equipment. Consequently, only the last WAP records (last 175 blows for pile #3 and last 235 blows for pile #1) were correlated with the PDA values. They will be shown in the following two sections. Once the pile was driven (met the driving criteria), the transmission distance was further tested by moving the receiver antenna away from the pile in multiple directions and striking the pile with a hand held hammer. The maximum distance achieved was 200 ft with the maximum possible signal strength. It is important to note that the directional 1 m antenna was used for the latter tests. Regardless of the small problems mentioned, the WAP output signals were strong with little noise (less than 10 milivolts) and compare very satisfactorily with the PDA outputs. The results as well as other special during details are described in the following sections. Pile # 3 at End Bent #1 Pile # 3 was driven on February 14 of 2002. The actual driving occurred over a one-hour time frame with 7.25 m of penetration. The PDA recorded 1345 blows and the WAP recorded 184 827 blows. The latter was due to the previously described problem with the wake up system. The contractor personnel recorded 1297 blows. Figure 14-13 shows the WAP output at the very last blow. It is important to note that the capacity graph shows the driving history for the last 500 blows. The total pile capacity shown is based on the CASE method. The UF method (Chapters 7 & 8) was used to assess the case damping constant, Jc , as well as the tip and skin capacities. Figure 14-13. WAP Records for the Last 500 Blows on Pile #3 Force and velocity traces recorded in this pile were very consistent as shown in Figure 14-14 through 4-16. The top two graphs in each figure are the forces based on the strains and accelerometers. In each window there are two graphs, which report the response from two 185 Figure 14-14. Traces versus PDA Traces for Blow 1345 (last Blow Pile #3) Figure 14-15. Traces versus PDA Traces for Blow 1340 (Pile #3) 186 Figure 14-16. Traces versus PDA Traces for Blow 1335 (Pile #3) gauges (strains or accelerometers), at similar elevation but at different offsets. Note the very similar response of each. Also shown in the larger plot below the smaller graphs is the com-bined force plots from both strain and accelerometer. Included in this plot are the response recorded with the PDA equipment. The comparison is very good up to 2L/c (green vertical lines), after which the computed force from the PDA accelerometer deviates slightly from the WAP value. It is not known if the latter is due to the zeroing process that the PDA software employs (rotates acceleration trace from maximum peak to zero at 0.1 sec). A further and more in depth comparison between the WAP output and PDA is presented in terms of compression and tension stresses in the pile. Shown in Figure 14-17 are maximum compression stresses predicted both by PDA and WAP. Agreement is very good, especially for the last 175 blows were both instruments were recording simultaneously. 187 Figure 14-17. Comparison of Maximum Compression Stress Between WAP and PDA Records at the Pile Top Tension stresses are very important in concrete piles. Using the Wave Up one can easily investigate whether tension stresses are present. Generally, if the soil resistance is small, the Wave Up will show tensile stresses, which will arrive at time 2L/c. The maximum tension stress occurs at the minimum point of the downward compression stress (Wave Down at time t3 in Fig-ure 14-18). The maximum tension is found using Equation 14-1a and 14-1b for easy and hard driving, respectively. [ ] [ ] 2 2 3 3 2 2 ) 3 , ( ) 2 , ( Vel Z P Vel Z P F F T t DOWN t UP MAX × + + × − = + = (14-1a) [ ] [ ] 2 2 2 2 3 3 ) 2 , ( ) 3 , ( Vel Z P Vel Z P F F T t DOWN t UP MAX × + + × − = + = (14-1b) 188 Figure 14-18. Determination of Tension Stresses in the Pile Using Eqs. 14, the maximum tension stresses in the pile were computed for all the blows using WAP and PDA data (Figure 14-19). Agreement is very good, especially for the last 175 blows were both instruments were recording simultaneously. Pile # 1 at End Bent #1 Pile # 1 was driven on February 15 of 2002. The driving was from 11:15 am to 11:50 am with a driving penetration of 6.0 meters. The PDA recorded 909 blows and the WAP recorded 770 blows. The difference was that the early driving did not exceed 50 to 60 gs which is the threshold to wake up the WAP system. The contractor personnel recorded only 891 blows. Figure 14-20 shows the WAP output recorded for the very last blow. The capacity graph (right picture) in the figure shows the driving history for the last 500 blows (latter figure is up-dated for each blow). The total capacity prediction was based on the CASE method and the UF method was used to assess the case-damping constant, as well as the tip and skin capacities. 189 Figure 14-19. Comparison of Maximum Tension Stress Between WAP and PDA Figure 14-20. WAP Records for the Last 500 Blows on Pile #1 190 Evident from the capacity graph is that the total pile capacity is gradually increasing with depth and that the pile’s resistance is approximately 30% skin and 70% end bearing. A comparison of WAP and PDA response at the top of the pile is given in Figures 14-21 to 14-23. The comparison is very good up to 2L/c (green vertical lines), after which the com-puted force from the PDA accelerometer deviates slightly from the WAP value. It is not known if the latter is due to the zeroing process that the PDA software employs (rotates acceleration trace from maximum peak to zero at 0.1 sec). Also shown in upper to graphs of Figures 14-21 to 14-23 are the forces at the pile tip (yellow lines). This data is not available with the PDA equipment, but only through the WAP system. The data in the left graph is based on strain and the values in the right are based on the velocities (integration of acceleration). The latter shows that for most blows that the velocity is very small, which is typical for piles with considerable tip capacity (see capacity graph, Fig 14-20). From the soil borings (Fig 14-4b), it was noted that the piles were tipped in limestone. Figure 14-21. WAP Traces versus PDA Traces for Blow 909 (last Blow Pile #1) 191 Figure 14-22. WAP Traces versus PDA Traces for Blow 908 (Pile #1) Figure 14-23. WAP Traces versus PDA Traces for Blow 905 (Pile #1) 192 Also, the force plot (from the strain gauge) shows considerable offset at the end of trace. However, the latter offset generally returns back to the vicinity of zero after 1.5 to 2 seconds. The latter is not shown in the figures, but it is recorded with the WAP equipment (continuously moni-toring the pile between blows). Consequently, it is believed that the early offset is tip stresses, which diminish due to creep, and/or are redistributed through the pile as residual stresses. Similar to pile #3, a more in depth comparison between the WAP and PDA outputs was performed and is presented in Figures 14-24 to 14-26. Figure 14-24 shows the maximum com-pression stress recorded with the strain gages at both the top and bottom of the pile. The PDA recorded values at the top of the pile are also shown for comparison. Agreement between PDA and WAP at the top of the pile are very good. Figure 14-24. Comparison of Maximum Compression Stress Between WAP and PDA at the Pile Top and WAP at Pile Bottom 193 Figure 14-25. Comparison of Maximum Tension Stress Between WAP and PDA at the Pile Top and WAP at Pile Bottom Figure 14-26. Comparison of Dynamic Capacity Assessment WAP and PDA 194 Shown in Figure 14-25 is the tension stresses in pile #3 computed from Equations 14-1 for both the PDA and WAP records. WAP data was available for both the top and bottom of the pile, whereas the PDA was available for top of pile only. Again similarity between PDA and WAP at the top of the pile is excellent. As expected for predominately tip resistance pile (founded in limestone), WAP shows very little if any tension stresses at the pile’s tip. Figure 14-26 plots the total pile capacity with the Jc = 0 as calculated with equation 3-14 for both PDA and WAP. It should be noted that some values for the WAP are copied from the previous ones due to the loss of data from wake up problems described earlier. Consequently, the blow number recorded by WAP might not be exactly the same as the PDA for earlier values but are the same for the last 235 blows. Again, the figure shows very good agreement between PDA and WAP especially for the last blows. Finally, URS which performed the PDA and CAPWAP analysis on pile # 1 indepen-dently of this research, reported the following CAPWAP results: Total capacity - 1600 kN, skin resistance - 700 kN, and tip resistance - 900 kN at the end of drive. Comparing the latter with the WAP values given in Figure 14-20 (Total resistance, 1550-1600 kN, skin resistance, 500-600 kN, and tip resistance, 1000-1100 kN), agreement is very good for total capacity assessment. The slight difference in end bearing may be attributed to the difficulty CAPWAP has in differ-entiating skin and tip resistance near the end of the pile. 195 CHAPTER 15 CONCLUSIONS AND RECOMMENDATIONS The conclusions and recommendations presented herein are subdivided into three sec-tions. The first deals with the assessment of dynamic methods to estimate static pile capacity from Load and Resistance Factored Design (LRFD). The second concerns the determination of PDA Case damping coefficient, Jc , as well as predicting the skin and tip resistance from mea-sured stress wave propagation in a pile. Finally, a new cast insitu instrumentation package and wireless transmission system is presented. In the case of the latter, comparisons with current practice (PDA), as well as future applications are discussed. LRFD CALIBRATION FOR EIGHT DYNAMIC METHODS Eight dynamic methods were evaluated based on a Florida database and LRFD design procedures. The following conclusions can be drawn from this study: 1. The bridge span length has negligible effect on the resistance factor, φ, or the cor-responding safety factor. 2. The COVR or φ/λR-values, rather than the absolute values of the φ factor, should be used to evaluate the accuracy and efficiency of a dynamic method. 3. The accuracy (COVR & φ/λR) of dynamic methods from testing at BOR show negligible increase over the same testing at EOD. 4. The evaluation of φ/λR shows that the newer dynamic methods (PDA, Paikowsky, CAPWAP) are generally more cost effective to meet a reliability index in comparison with the older methods. 196 5. The CAPWAP procedure tends to underestimate the Davisson capacity by 28 percent, while the PDA underestimate the Davisson capacity by only 9 percent at EOD. 6. The older methods (i.e., Gates, FDOT, ENR, and Modified ENR) result in better esti-mates of Davisson capacity for pile capacity less than 200 tons (1779 kN). 7. The Gates formula, when used separately for Davisson capacity larger than 1779 kN and less than 1779 kN, may have comparable accuracy with the modern methods. 8. For Allowable Stress Design (ASD) design with dynamic methods, the following safety factors are recommended to meet a probability of failure of 0.62 ~ 2.50 %: 1.8, 2.0 and 2.5 for CAPWAP, PDA, and Paikowsky’s energy method at EOD, and 2.25, 2.5 and 3.5 at BOR, respectively. Currently, a safety factor of 2.5 is used for CAPWAP, PDA, and Paikowsky’s method, for both, BOR, and EOD. The driving formulas are usually applied at EOD only, and safety factors of 1.9, 14.0, 11.0, and 2.0 are recommended for FDOT, ENR, modified ENR, and Gates formulas, respec-tively. Recommendations The results presented in this study reflect the pile driving history of Florida. The number of piles considered is based on the availability of dynamic and static load test data. In addition, sometimes the available information is not sufficient to evaluate a given dynamic method. Based on the latter facts, the following recommendations are presented. 1. In order to have a more representative LRFD calibration for the Florida practice, any pile driving and static load test information not included in this study should be added to the existing Florida database. 197 2. The LRFD resistance factors should be monitored on a semi annual event as more information (cases or pile types: cylinder, etc.) is obtained from the field or other sources (National). Determination of Case damping coefficient, JC , and skin & tip resistance from measured stress wave propagation in a pile. An important finding of this research was that the coefficient of variance of the static capacity changed by only 10 percent when the Case damping, Jc , was modified by 30 percent. The latter makes the PDA static pile estimate Equation very attractive. However, two short-comings of the PDA equipment are 1) determining Jc ; and 2) estimating the skin and tip resis-tance of the pile in the field real time. Based on the PDA traces of force and velocity, two meth-ods were developed to determine the ratio of static Tip/skin resistance for a pile. In addition, the Tip/skin ratio was correlated to the Case Damping, Jc value. Subsequently, the developed meth-ods, as well as CAPWAP were compared to a database of 23 piles, which had static results. The following was concluded: 1. For the total static capacity prediction, proposed Method II (COVR = 0.29) was the most accurate of the three methods evaluated, followed by the original PDA with user supplied Jc (COVR = 0.32) and then the CAPWAP procedure (COVR = 0.45). 2. In terms of the static skin resistance, the suggested Method II (λR = 1.04, COVR = 0.39) proved to be the best method followed by CAPWAP (λR = 0.80, COVR = 0.43) and then Method I (λR = 1.29, COVR = 0.89). 3. The tip static prediction from the suggested Method II (λR = 0.96, COVR = 0.17) was superior to the predictions of CAPWAP (λR = 1.08, COVR = 0.55) and Method 1 (λR = 1.94, COVR = 1.27). 198 4. The suggested method II is simple to perform, and its calculations may be performed automatically. 5. A programming code has been developed to obtain the total, skin, and tip static capacities from Method II. The latter permits the rapid (computer) assessment of skin, tip and total capacities for every blow in the field real time. Recommendations The proposed Method II has been proven to be an accurate tool in determining the driven pile static capacity. However, there have been some limitations in this study. The following suggestions take into consideration those limitations and the feasibility for further research in the evaluation of this new method. 1. The number of cases used to evaluate the total, skin, and tip static resistances were 18, 10 and 6, respectively. It is recommended that more driving information for a pile, which includes static load testing, which monitors skin and tip resistance be undertaken. 2. Method II does not take into consideration the soil properties. It is recommended to investigate the distribution of soil for the piles studied and for any other new pile to be evaluated. WIRELESS ACQUISITION OF A PILE’S (WAP) DYNAMIC RESPONSE Chapters nine through fourteen describe the design, construction and testing of a new wireless instrumentation package for high strain dynamic testing of piles. The package has proven to perform properly under both laboratory and field conditions. The latter involved installing the WAP system at SR-54 at Cypress Springs, Florida. In the latter case top of one pile and the top and bottom of another pile was monitored. More than one thousand blows were 199 analyzed for two the 24-inch prestressed concrete piles. The signals, stresses, etc. compared very favorably with externally mounted PDA instrumentation. In the case of the internally cast WAP system, information at the pile tip (stresses, displacements, etc.) was also available. Many problems arose in the development of the equipment, which had to be overcome. Three of the biggest problems had to do with the wireless transmission, the accelerometer (price, mounting, etc.) and overall cost of the system. Two transmitter modules were developed initially before the final one presented herein. The first was an analog transmitter, which failed because it was incapable of transmitting static strain signals. The second, a radio transmitter (FM), failed because it was incapable of transmitting four signals at once (signal overlap was recorded) and noise. The major problem with the accelerometer was its cost and mounting technique (to avoid the zero shift effect). The former was overcome by using an accelerometer developed for the automotive industry. The mounting was overcome through a significant amount of testing of materials, adhesives and dimensions, with the accelerometer in the laboratory. All options were benchmarked against both high quality accelerometers and existing PDA accelerometers. In this research, the cost was a major limitation. The system was required to comply with a maximum cost of $300 in parts, for the non-recoverable equipment (transmitters, conditioners, and instruments: accelerometers, strain gauges, etc.). The latter could not of been feasible ten years ago due to unavailability of cheap transmitters, accelerometers, etc. However, with rapid technology growth in wireless LAN systems, automotive industry, cell phone etc., a digital 5 Mbs system (cast insitu) was developed for $250 per pile. The latter included four 10,000 words per second signals (two strain and two acceleration), which are transmitted digitally. A receiver, PCMCA card, laptop computer, and LABVIEW software (costs less than $3000) is required to 200 process the signals. Developed in the LABVIEW software (VIs) are routines to display forces, stresses, total capacities, as well as skin and tip resistance based on earlier work. Comparison of the results (Chapter 14) for SR-54 from PDA, and CAPWAP are very favorable. Recommendations Given the cost of the new WAP system, it is now viable to cast the system into every pre-cast concrete pile. The latter would allow the elimination of the current driving criterion based on blow count, which does not handle changing driving conditions (soil, rock hammer, etc.). For instance, consider the two SR-54 test piles (chapter 14) which had estimated capacities of 1600 kN, and 2200 kN (WAP and CAPWAP), but had similar blow counts at the end of drive. Also, if every pile was monitored, current FDOT Factors of Safety could possibly be lowered from over two to less than two, resulting in less pile lengths and total costs. It is expected with WAP’s successful installation of instrumentation at the pile tip (Chapter 14), improved analysis of total capacity, as well as skin and tip resistance will continue to occur. Finally, it is recommended that the FDOT convene a panel of construction, geotechnical and structural engineers to identify how the WAP system may be implemented into FDOT’s “Standard Specifications for Road and Bridge Construction Manual.” 201 REFERENCES AASHTO (1994). LRFD Highway Bridges Design Specifications, SI Units, First Edition. American Association of State Highway and Transportation Officials, Washington, D.C. ACI (1999), “Building Code Requirements for reinforced Concrete (ACI 318-99)”, American Concrete Institute, Detroit, MI. American Society of Civil Engineers (2000), “Annual Book of ASTM Standards” Volume 04.08- ASTM D4945-96 Standard Test Method for High-Strain Dynamic Testing of Piles, West Conshohocken, PA. Chu, Anthony (1987), Technical Paper, “Zero Shift of Piezoelectric Accelerometers in Phyro-shock Measurements”, The Shock and Vibration Information Center, Naval Research Laboratory, Washington, DC. Coduto, Donald P. (1994). Foundation Design Principles and Practices. Prentice Hall Inc., Englewood Cliffs, N.J., 403-416. Davisson, M.T. (1972). “High Capacity Piles, Proceedings, Soil Mechanics Lecture Series on Innovations in Foundation Construction.” American Society of Civil Engineers, ASCE, Illinois Section, Chicago, 81-112. Florida Department of Transportation (1991). “Standard Specifications for Road and Bridge Construction.” Tallahassee, Florida, 419-443. Florida Department of Transportation (1994). “Supplemental Specifications to the 1991 Standard Specifications for Road and Bridge Construction.” Tallahassee, Florida, 109-161. Florida Department of Transportation (1999). “Standard Specifications for Road and Bridge Construction.” Tallahassee, Florida, 398-451. Gates, M. (1957). “Empirical Formula for Predicting Pile Bearing Capacity.” Engineering New Record, 26-28, 33-34. Goble Rausche Likins and Associates Inc. (2000), “Dynamic Load Testing”, Cleveland, Ohio. GRL (1996). Case Pile Wave Analysis Program – CAPWAP Manual. Cleveland, Ohio. GRL (1993). GRLWEAP – Wave Equation Analysis of Pile Driving Manual. Cleveland, Ohio. Hansel, W.C. and I.M. Viest (1971). “Load Factor Design for Steel Highway Bridges.” AISC Engineering Journal, Vol. 8, No. 4, 113-123. 202 Housel, William S. (1966). “Pile Load Capacity: Estimates and Test Results.” ASCE Journal of the Soil Mechanics and Foundations Division, Vol. 92, No. SM4, 1-30. Likins, Garland E. and Hussein, Mohamad (1988). “A Summary of the Pile Driving Analyzer Capacity Methods – Past and Present.” Cleveland, Ohio. Measurements Group Inc. (1999), EGP-5-120 Embedment Strain Gage Data Sheet, Raleigh, North Carolina. Measurements Group, (1993), “Catalog 5000 Precision Gages”, Measurements Group Inc., Raleigh, North Carolina. National Instruments (1998), “LabVIEW User Manual”, Austin, Texas National Instruments (1997), “DIO 6533 User Manual”, National Instruments Corporation, Austin, Texas. Paikowsky, S.G., Regan, J.E., and McDonnell, J.J. (1994). “A simplified field Method for Capacity Evaluation of Driven Piles.” Publication No. FHWA-RD-94-042. Pile Buck (1992), “Foundations”, Pile Buck Production, Jupiter, Florida Pile Dynamics, Inc. (1990), “Model GCX Pile Driving Analyzer Manual”, Cleveland, Ohio. Pile Dynamics, Inc. (1992). Pile Driving Analyzer (PDA) Manual. Cleveland, Ohio. Pile Dynamics, Inc. (1995), “Pile Driving Analyzer Manual PAK”, Cleveland, Ohio. Pile Dynamics, Inc. (1996), “Strain and Motion Sensors from Pile Dynamics” Data Sheet, Cleveland, Ohio. Rosenblueth, E. and Esteva, L. (1972). “Reliability Basis for Some Mexican Codes.” ACI Publi-cation SP-31, American Concrete Institute, Detroit, Mi. Sakai, T., Sawai, H., and Shioi, Y. (1996). “Theoretical Analysis of the Pile Driving Formula.” Proceeding, 5th International Conference on the Application of Stress Wave Theory to Piles. September 11-13, 1996. Orlando, Florida, 81-88. Shearman, Sam (2001), Signal Processing Easies Vibration Transducer Selection, National Instruments, Austin Texas. Shear Robert (1999), “Piezoelectric Sensors for OEM Applications”, The journal of Applied Sensing Technology, Endevco, San Juan Capistrano, California. Willington, A.M. (1893). Piles and Pile-Driving. Engineering News Publishing Co., New York. 203 Withiam, J.L., Voytko, E.P., Barker, R.M., Duncan, J.M., Kelly, B.C., Musser, S.C., and Elias, V. (1997). Load and Resistance Factor Design (LRFD) for Highway Bridge Structures. FHWA Report DTFH61-94-C-00098. Federal Highway Administration, Washington, D.C. APPENDIX A LRFD ANALYSIS RESULTS - CAPWAP PROCEDURE A-1 y = 0.6745x R 2 = 0.6264 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Figure A-1. Davisson Capacity vs. CAPWAP EOD Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Cases=44 Mean=1.597 Standard Dev.=0.56 Figure A-2. Measured Over Predicted Capacity for CAPWAP at EOD A-2 Table A-1. φ Values Evaluated for CAPWAP (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.980 0.788 0.636 0.511 15 1.00 0.940 0.756 0.610 0.490 18 1.06 0.937 0.753 0.608 0.489 27 1.58 0.912 0.733 0.591 0.475 36 2.12 0.894 0.719 0.580 0.466 45 2.64 0.882 0.709 0.572 0.460 50 3.00 0.876 0.704 0.568 0.457 60 3.53 0.868 0.698 0.563 0.453 Table A-2. ASD Design Safety Factor Values Evaluated for CAPWAP (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 1.611 2.004 2.483 3.089 15 1.00 1.595 1.984 2.459 3.059 18 1.06 1.594 1.982 2.456 3.056 27 1.58 1.584 1.970 2.441 3.037 36 2.12 1.577 1.962 2.431 3.024 45 2.64 1.572 1.956 2.424 3.015 50 3.00 1.570 1.953 2.420 3.011 60 3.53 1.567 1.949 2.415 3.005 Table A-3. φ Factors for Safety Factor = 2.50 – CAPWAP (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.632 3.047 0.00116 15 1.00 0.600 3.071 0.00107 18 1.06 0.597 3.074 0.00106 27 1.58 0.578 3.089 0.00101 36 2.12 0.564 3.099 0.00097 45 2.64 0.555 3.107 0.00095 50 3.00 0.550 3.111 0.00093 60 3.53 0.544 3.115 0.00092 A-3 y = 0.7743x R 2 = 0.2553 0 1000 2000 3000 4000 5000 6000 7000 0 1000 2000 3000 4000 5000 6000 7000 Davisson Capacity (kN) Figure A-3. Davisson Capacity vs. CAPWAP BOR 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 0 1000 2000 3000 4000 5000 6000 7000 Davisson Capacity (kN) Cases=79 Mean=1.26 Standard Dev.=0.44 Figure A-4. Measured Over Predicted Capacity for CAPWAP at BOR A-4 Table A-4. φ Values Evaluated for CAPWAP (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.776 0.625 0.505 0.406 15 1.00 0.745 0.599 0.484 0.390 18 1.06 0.742 0.597 0.482 0.388 27 1.58 0.722 0.581 0.469 0.378 36 2.12 0.708 0.570 0.460 0.371 45 2.64 0.699 0.562 0.454 0.366 50 3.00 0.694 0.558 0.451 0.363 60 3.53 0.688 0.553 0.447 0.360 Table A-5. ASD Design Safety Factor Values Evaluated for CAPWAP (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 2.033 2.527 3.128 3.888 15 1.00 2.013 2.502 3.098 3.850 18 1.06 2.012 2.500 3.095 3.846 27 1.58 1.999 2.485 3.076 3.823 36 2.12 1.991 2.474 3.063 3.806 45 2.64 1.985 2.467 3.054 3.795 50 3.00 1.982 2.463 3.049 3.789 60 3.53 1.978 2.459 3.043 3.782 Table A-6. φ Factors for Safety Factor = 2.50 – CAPWAP (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.632 2.473 0.00671 15 1.00 0.600 2.498 0.00625 18 1.06 0.597 2.500 0.00622 27 1.58 0.578 2.515 0.00596 36 2.12 0.564 2.526 0.00578 45 2.64 0.555 2.533 0.00566 50 3.00 0.550 2.537 0.00560 60 3.53 0.544 2.542 0.00552 APPENDIX B LRFD ANALYSIS RESULTS – PDA METHOD B-1 y = 0.7822x R 2 = 0.6042 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Figure B-1. Davisson Capacity vs. PDA EOD Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Cases=48 Mean=1.344 Standard Dev.=0.44 Figure B-2. Measured Over Predicted Capacity for PDA at EOD B-2 Table B-1. φ Values Evaluated for PDA (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.856 0.694 0.565 0.458 15 1.00 0.821 0.666 0.542 0.439 18 1.06 0.818 0.663 0.540 0.437 27 1.58 0.796 0.645 0.525 0.426 36 2.12 0.781 0.633 0.515 0.418 45 2.64 0.770 0.625 0.508 0.412 50 3.00 0.765 0.620 0.505 0.409 60 3.53 0.758 0.615 0.500 0.405 Table B-2. ASD Design Safety Factor Values Evaluated for PDA (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 1.845 2.276 2.796 3.449 15 1.00 1.827 2.253 2.769 3.415 18 1.06 1.825 2.251 2.766 3.412 27 1.58 1.814 2.237 2.749 3.391 36 2.12 1.806 2.228 2.738 3.377 45 2.64 1.801 2.221 2.730 3.367 50 3.00 1.798 2.218 2.725 3.362 60 3.53 1.795 2.214 2.720 3.356 Table B-3. φ Factors for Safety Factor = 2.50 – PDA (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.632 2.742 0.003058 15 1.00 0.600 2.767 0.002833 18 1.06 0.597 2.770 0.002807 27 1.58 0.578 2.786 0.002672 36 2.12 0.564 2.797 0.002583 45 2.64 0.555 2.804 0.002528 50 3.00 0.550 2.808 0.002496 60 3.53 0.544 2.813 0.002458 B-3 y = 0.9966x R 2 = 0.4851 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Figure B-3. Davisson Capacity vs. PDA BOR 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Cases=42 Mean=1.036 Standard Dev.=0.32 Figure B-4. Measured Over Predicted Capacity for PDA at BOR B-4 Table B-4. φ Values Evaluated for PDA (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.682 0.557 0.457 0.373 15 1.00 0.654 0.534 0.438 0.358 18 1.06 0.652 0.532 0.436 0.356 27 1.58 0.634 0.518 0.425 0.347 36 2.12 0.622 0.508 0.417 0.340 45 2.64 0.614 0.501 0.411 0.336 50 3.00 0.609 0.498 0.408 0.333 60 3.53 0.604 0.493 0.404 0.330 Table B-5. ASD Design Safety Factor Values Evaluated for PDA (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 2.316 2.835 3.458 4.234 15 1.00 2.293 2.807 3.424 4.192 18 1.06 2.291 2.805 3.421 4.188 27 1.58 2.277 2.787 3.400 4.163 36 2.12 2.267 2.776 3.386 4.145 45 2.64 2.260 2.768 3.376 4.133 50 3.00 2.257 2.763 3.370 4.127 60 3.53 2.253 2.758 3.364 4.119 Table B-6. φ Factors for Safety Factor = 2.50 – PDA (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.632 2.164 0.01525 15 1.00 0.600 2.191 0.01424 18 1.06 0.597 2.193 0.01417 27 1.58 0.578 2.210 0.01357 36 2.12 0.564 2.221 0.01319 45 2.64 0.555 2.229 0.01292 50 3.00 0.550 2.233 0.01279 60 3.53 0.544 2.238 0.01263 APPENDIX C LRFD ANALYSIS RESULTS – PAIKOWSKY’S ENERGY METHOD C-1 y = 1.0236x R 2 = 0.7241 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 Davisson Capacity (kN) Figure C-1. Davisson Capacity vs. Paikowsky’s Method EOD Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 Davisson Capacity (kN) Cases=27 Mean=1.11 Standard Dev.=0.37 Figure C-2. Measured Over Predicted Capacity for Paikowsky’s Method at EOD C-2 Table C-1. φ Values Evaluated for Paikowsky’s Method (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.700 0.567 0.460 0.372 15 1.00 0.672 0.544 0.441 0.357 18 1.06 0.669 0.541 0.440 0.356 27 1.58 0.651 0.527 0.428 0.346 36 2.12 0.639 0.517 0.420 0.340 45 2.64 0.630 0.510 0.414 0.335 50 3.00 0.626 0.506 0.411 0.333 60 3.53 0.620 0.502 0.408 0.330 Table C-2. ASD Design Safety Factor Values Evaluated for Paikowsky’s Method (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 2.254 2.787 3.432 4.242 15 1.00 2.232 2.759 3.398 4.201 18 1.06 2.230 2.757 3.395 4.197 27 1.58 2.216 2.740 3.374 4.171 36 2.12 2.207 2.728 3.360 4.153 45 2.64 2.200 2.720 3.350 4.141 50 3.00 2.197 2.716 3.345 4.135 60 3.53 2.193 2.711 3.338 4.127 Table C-3. φ Factors for Safety Factor = 2.50 – Paikowsky’s Method (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.632 2.224 0.01309 15 1.00 0.600 2.249 0.01227 18 1.06 0.597 2.251 0.01221 27 1.58 0.578 2.267 0.01171 36 2.12 0.564 2.278 0.01138 45 2.64 0.555 2.285 0.01117 50 3.00 0.550 2.289 0.01105 60 3.53 0.544 2.294 0.01091 C-3 y = 1.2096x R 2 = 0.4088 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 Davisson Capacity (kN) Figure C-3. Davisson Capacity vs. Paikowsky’s Method BOR 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 Davisson Capacity (kN) Cases=72 Mean=0.84 Standard Dev.=0.30 Figure C-4. Measured Over Predicted Capacity for Paikowsky’s Method at BOR C-4 Table C-4. φ Values Evaluated for Paikowsky’s Method (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.503 0.403 0.323 0.259 15 1.00 0.483 0.386 0.310 0.248 18 1.06 0.481 0.385 0.309 0.247 27 1.58 0.468 0.374 0.301 0.241 36 2.12 0.459 0.367 0.295 0.236 45 2.64 0.453 0.362 0.291 0.233 50 3.00 0.449 0.360 0.289 0.231 60 3.53 0.445 0.356 0.286 0.229 Table C-5. ASD Design Safety Factor Values Evaluated for Paikowsky’s Method (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 3.139 3.923 4.881 6.100 15 1.00 3.108 3.884 4.834 6.040 18 1.06 3.105 3.881 4.829 6.034 27 1.58 3.086 3.857 4.799 5.997 36 2.12 3.073 3.840 4.779 5.972 45 2.64 3.064 3.829 4.765 5.955 50 3.00 3.059 3.823 4.758 5.945 60 3.53 3.054 3.816 4.749 5.934 Table C-6. φ Factors for Safety Factor = 2.50 – Paikowsky’s Method (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.632 1.408 0.07964 15 1.00 0.600 1.432 0.07614 18 1.06 0.597 1.434 0.07586 27 1.58 0.578 1.449 0.07374 36 2.12 0.564 1.460 0.07221 45 2.64 0.555 1.467 0.07126 50 3.00 0.550 1.471 0.07071 60 3.53 0.544 1.475 0.07017 APPENDIX D LRFD ANALYSIS RESULTS – SAKAI ET AL (JAPANESE) METHOD D-1 y = 1.0282x R 2 = 0.4452 0 2000 4000 6000 8000 10000 12000 0 2000 4000 6000 8000 10000 12000 Davisson Capacity (kN) Figure D-1. Davisson Capacity vs. Sakai Method EOD Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 2000 4000 6000 8000 10000 12000 Davisson Capacity (kN) Cases=21 Mean=1.504 Standard Dev.=1.26 Figure D-2. Measured Over Predicted Capacity for Sakai Method at EOD D-2 Table D-1. φ Values Evaluated for Sakai Method (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.374 0.248 0.166 0.110 15 1.00 0.359 0.238 0.159 0.106 18 1.06 0.358 0.237 0.159 0.105 27 1.58 0.348 0.231 0.154 0.102 36 2.12 0.341 0.226 0.151 0.100 45 2.64 0.337 0.223 0.149 0.099 50 3.00 0.334 0.222 0.148 0.098 60 3.53 0.331 0.220 0.147 0.098 Table D-2. ASD Design Safety Factor Values Evaluated for Sakai Method (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 4.221 6.361 9.515 14.341 15 1.00 4.179 6.299 9.422 14.201 18 1.06 4.175 6.293 9.413 14.188 27 1.58 4.150 6.254 9.355 14.100 36 2.12 4.132 6.228 9.316 14.041 45 2.64 4.120 6.210 9.289 14.000 50 3.00 4.114 6.200 9.274 13.978 60 3.53 4.106 6.188 9.257 13.952 Table D-3. φ Factors for Safety Factor = 2.50 – Sakai Method (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.632 1.271 0.10195 15 1.00 0.600 1.284 0.09966 18 1.06 0.597 1.285 0.09948 27 1.58 0.578 1.293 0.09809 36 2.12 0.564 1.299 0.09706 45 2.64 0.555 1.302 0.09654 50 3.00 0.550 1.305 0.09603 60 3.53 0.544 1.307 0.09569 D-3 y = 0.7446x R 2 = 0.3119 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson's Capacity (kN) Figure D-3. Davisson Capacity vs. Sakai Method BOR 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Cases=52 Mean=1.46 Standard Dev.=0.78 Figure D-4. Measured Over Predicted Capacity for Sakai Method at BOR D-4 Table D-4. φ Values Evaluated for Sakai Method (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.627 0.466 0.348 0.258 15 1.00 0.601 0.447 0.334 0.248 18 1.06 0.599 0.445 0.332 0.247 27 1.58 0.583 0.433 0.323 0.240 36 2.12 0.572 0.425 0.317 0.236 45 2.64 0.564 0.419 0.313 0.233 50 3.00 0.560 0.416 0.311 0.231 60 3.53 0.555 0.412 0.308 0.229 Table D-5. ASD Design Safety Factor Values Evaluated for Sakai Method (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 2.520 3.391 4.540 6.110 15 1.00 2.495 3.358 4.495 6.050 18 1.06 2.493 3.355 4.491 6.044 27 1.58 2.477 3.334 4.463 6.007 36 2.12 2.467 3.320 4.444 5.982 45 2.64 2.460 3.311 4.431 5.964 50 3.00 2.456 3.305 4.424 5.955 60 3.53 2.451 3.299 4.416 5.944 Table D-6. φ Factors for Safety Factor = 2.50 – Sakai Method (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.632 1.946 0.02586 15 1.00 0.600 1.964 0.02479 18 1.06 0.597 1.965 0.02474 27 1.58 0.578 1.976 0.02411 36 2.12 0.564 1.984 0.02366 45 2.64 0.555 1.990 0.02332 50 3.00 0.550 1.992 0.02321 60 3.53 0.544 1.996 0.02299 APPENDIX E LRFD ANALYSIS RESULTS – FDOT METHOD E-1 y = 0.3457x R 2 = -0.0051 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Figure E-1. Davisson Capacity vs. FDOT Method EOD Capacity 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Cases=72 Mean=2.38 Standard Dev.=1.34 Figure E-2. Measured Over Predicted Capacity for FDOT Method at EOD E-2 Table E-1. φ Values Evaluated for FDOT – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.977 0.719 0.532 0.391 15 1.00 0.938 0.690 0.510 0.375 18 1.06 0.934 0.687 0.508 0.374 27 1.58 0.909 0.669 0.495 0.364 36 2.12 0.892 0.656 0.485 0.357 45 2.64 0.880 0.647 0.479 0.352 50 3.00 0.873 0.642 0.475 0.349 60 3.53 0.866 0.637 0.471 0.346 Table E-2. ASD Design Safety Factor Values Evaluated for FDOT – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 1.616 2.196 2.969 4.037 15 1.00 1.600 2.175 2.940 3.997 18 1.06 1.598 2.173 2.937 3.994 27 1.58 1.588 2.160 2.919 3.969 36 2.12 1.582 2.150 2.907 3.952 45 2.64 1.577 2.144 2.899 3.941 50 3.00 1.575 2.141 2.894 3.934 60 3.53 1.572 2.137 2.889 3.927 Table E-3. φ Factors for Safety Factor = 1.00 – FDOT – Overall (EOD) QD/QL φ βΤ Pf Span length (m) 9 0.52 1.579 1.117 0.13210 15 1.00 1.500 1.134 0.12850 18 1.06 1.493 1.136 0.12808 27 1.58 1.444 1.146 0.12600 36 2.12 1.410 1.154 0.12435 45 2.64 1.387 1.159 0.12333 50 3.00 1.375 1.162 0.12272 60 3.53 1.360 1.165 0.12211 E-3 Table E-4. φ Values Evaluated for FDOT < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.657 0.491 0.369 0.276 15 1.00 0.631 0.471 0.354 0.264 18 1.06 0.628 0.469 0.352 0.263 27 1.58 0.611 0.457 0.343 0.256 36 2.12 0.600 0.448 0.336 0.251 45 2.64 0.592 0.442 0.332 0.248 50 3.00 0.587 0.439 0.330 0.246 60 3.53 0.582 0.435 0.327 0.244 Table E-5. ASD Design Safety Factor Values Evaluated for FDOT < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 2.402 3.215 4.281 5.730 15 1.00 2.378 3.184 4.239 5.674 18 1.06 2.376 3.181 4.235 5.669 27 1.58 2.362 3.161 4.209 5.634 36 2.12 2.352 3.148 4.191 5.610 45 2.64 2.345 3.139 4.179 5.594 50 3.00 2.341 3.134 4.172 5.585 60 3.53 2.337 3.128 4.164 5.575 Table E-6. φ Factors for Safety Factor = 1.00 – FDOT <1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 1.579 0.338 0.36787 15 1.00 1.500 0.356 0.36111 18 1.06 1.493 0.358 0.36036 27 1.58 1.444 0.369 0.35625 36 2.12 1.410 0.377 0.35327 45 2.64 1.387 0.382 0.35142 50 3.00 1.375 0.385 0.35030 60 3.53 1.360 0.389 0.34882 E-4 Table E-7. φ Values Evaluated for FDOT > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 1.783 1.406 1.114 0.878 15 1.00 1.710 1.349 1.068 0.843 18 1.06 1.703 1.343 1.064 0.839 27 1.58 1.658 1.307 1.036 0.817 36 2.12 1.626 1.283 1.016 0.801 45 2.64 1.604 1.265 1.002 0.791 50 3.00 1.593 1.256 0.995 0.785 60 3.53 1.579 1.245 0.986 0.778 Table E-8. ASD Design Safety Factor Values Evaluated for FDOT > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.886 1.123 1.418 1.798 15 1.00 0.877 1.112 1.404 1.780 18 1.06 0.876 1.111 1.403 1.778 27 1.58 0.871 1.104 1.394 1.767 36 2.12 0.867 1.100 1.388 1.760 45 2.64 0.865 1.096 1.384 1.755 50 3.00 0.863 1.095 1.382 1.752 60 3.53 0.862 1.093 1.379 1.749 Table E-9. φ Factors for Safety Factor = 1.00 – FDOT >1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 1.579 2.236 0.01269 15 1.00 1.500 2.258 0.01199 18 1.06 1.493 2.260 0.01193 27 1.58 1.444 2.274 0.01150 36 2.12 1.410 2.284 0.01120 45 2.64 1.387 2.291 0.01100 50 3.00 1.375 2.294 0.01091 60 3.53 1.360 2.299 0.01077 E-5 y = 0.3733x R 2 = -0.6585 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Figure E-3. Davisson Capacity vs. FDOT Method BOR 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 0 1000 2000 3000 4000 5000 6000 7000 8000 Davisson Capacity (kN) Cases=63 Mean=2.57 Standard Dev.=1.29 Figure E-4. Measured Over Predicted Capacity for FDOT Method at BOR E-6 Table E-10. φ Values Evaluated for FDOT – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.186 0.895 0.678 0.511 15 1.00 1.138 0.858 0.651 0.491 18 1.06 1.134 0.855 0.648 0.489 27 1.58 1.103 0.832 0.631 0.476 36 2.12 1.082 0.816 0.619 0.467 45 2.64 1.068 0.805 0.610 0.460 50 3.00 1.060 0.799 0.606 0.457 60 3.53 1.051 0.792 0.601 0.453 Table E-11. ASD Design Safety Factor Values Evaluated for FDOT – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.331 1.765 2.328 3.087 15 1.00 1.318 1.748 2.305 3.057 18 1.06 1.317 1.746 2.303 3.054 27 1.58 1.309 1.735 2.289 3.035 36 2.12 1.303 1.728 2.279 3.023 45 2.64 1.299 1.723 2.273 3.014 50 3.00 1.297 1.720 2.269 3.009 60 3.53 1.295 1.717 2.265 3.003 Table E-12. φ Factors for Safety Factor = 1.00 – FDOT – Overall (BOR) QD/QL φ β pf Span length (m) 9 0.52 1.579 1.413 0.07890 15 1.00 1.500 1.432 0.07614 18 1.06 1.493 1.433 0.07600 27 1.58 1.444 1.445 0.07430 36 2.12 1.410 1.453 0.07318 45 2.64 1.387 1.459 0.07235 50 3.00 1.375 1.462 0.07194 60 3.53 1.360 1.466 0.07139 E-7 Table E-13. φ Values Evaluated for FDOT < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.941 0.778 0.646 0.534 15 1.00 0.903 0.747 0.620 0.512 18 1.06 0.899 0.744 0.617 0.510 27 1.58 0.875 0.724 0.601 0.497 36 2.12 0.858 0.710 0.589 0.487 45 2.64 0.847 0.700 0.581 0.481 50 3.00 0.841 0.695 0.577 0.477 60 3.53 0.833 0.689 0.572 0.473 Table E-14. ASD Design Safety Factor Values Evaluated for FDOT < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.678 2.029 2.445 2.956 15 1.00 1.662 2.009 2.421 2.927 18 1.06 1.660 2.007 2.419 2.925 27 1.58 1.650 1.995 2.404 2.907 36 2.12 1.643 1.987 2.394 2.894 45 2.64 1.638 1.981 2.387 2.886 50 3.00 1.636 1.978 2.383 2.881 60 3.53 1.633 1.974 2.378 2.876 Table E-15. φ Factors for Safety Factor = 1.00 – FDOT <1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 1.579 0.488 0.3130 15 1.00 1.500 0.516 0.3031 18 1.06 1.493 0.519 0.3021 27 1.58 1.444 0.536 0.2962 36 2.12 1.410 0.548 0.2920 45 2.64 1.387 0.557 0.2889 50 3.00 1.375 0.561 0.2876 60 3.53 1.360 0.566 0.2859 E-8 Table E-16. φ Values Evaluated for FDOT > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.357 1.039 0.799 0.611 15 1.00 1.302 0.996 0.766 0.587 18 1.06 1.297 0.992 0.763 0.584 27 1.58 1.262 0.966 0.743 0.569 36 2.12 1.238 0.947 0.729 0.558 45 2.64 1.221 0.935 0.719 0.550 50 3.00 1.212 0.928 0.714 0.546 60 3.53 1.202 0.920 0.708 0.542 Table E-17. ASD Design Safety Factor Values Evaluated for FDOT > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.164 1.520 1.976 2.582 15 1.00 1.152 1.505 1.957 2.557 18 1.06 1.151 1.504 1.955 2.555 27 1.58 1.144 1.495 1.943 2.539 36 2.12 1.139 1.488 1.935 2.528 45 2.64 1.136 1.484 1.929 2.521 50 3.00 1.134 1.482 1.926 2.517 60 3.53 1.132 1.479 1.923 2.512 Table E-18. φ Factors for Safety Factor = 1.00 – FDOT >1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 1.579 1.654 0.0491 15 1.00 1.500 1.674 0.0471 18 1.06 1.493 1.676 0.0469 27 1.58 1.444 1.688 0.0458 36 2.12 1.410 1.697 0.0449 45 2.64 1.387 1.703 0.0443 50 3.00 1.375 1.706 0.0441 60 3.53 1.360 1.710 0.0437 APPENDIX F LRFD ANALYSIS RESULTS – ENGINEERING NEWS RECORD (ENR) F-1 y = 3.6966x R 2 = 0.6617 0 5000 10000 15000 20000 25000 30000 35000 0 5000 10000 15000 20000 25000 30000 35000 Davisson Capacity (kN) Figure F-1. Davisson Capacity vs. ENR EOD Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 Davisson Capacity (kN) Cases=77 Mean=0.30 Standard Dev.=0.16 Figure F-2. Measured Over Predicted Capacity for ENR Method at EOD F-2 Table F-1. φ Values Evaluated for ENR – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.130 0.097 0.073 0.054 15 1.00 0.125 0.093 0.070 0.052 18 1.06 0.124 0.093 0.069 0.052 27 1.58 0.121 0.090 0.067 0.050 36 2.12 0.119 0.088 0.066 0.049 45 2.64 0.117 0.087 0.065 0.049 50 3.00 0.116 0.087 0.065 0.048 60 3.53 0.115 0.086 0.064 0.048 Table F-2. ASD Design Safety Factor Values Evaluated for ENR – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 12.139 16.298 21.763 29.220 15 1.00 12.020 16.138 21.549 28.933 18 1.06 12.009 16.123 21.530 28.906 27 1.58 11.935 16.024 21.397 28.729 36 2.12 11.884 15.956 21.307 28.607 45 2.64 11.850 15.910 21.245 28.524 50 3.00 11.831 15.885 21.211 28.479 60 3.53 11.809 15.855 21.172 28.425 Table F-3. φ Factors for Safety Factor = 6.00 – ENR – Overall (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.263 0.669 0.2519 15 1.00 0.250 0.687 0.2462 18 1.06 0.249 0.688 0.2459 27 1.58 0.241 0.700 0.2421 36 2.12 0.235 0.707 0.2399 45 2.64 0.231 0.713 0.2381 50 3.00 0.229 0.716 0.2372 60 3.53 0.227 0.719 0.2362 F-3 Table F-4. φ Values Evaluated for ENR < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.112 0.084 0.064 0.048 15 1.00 0.108 0.081 0.061 0.046 18 1.06 0.107 0.081 0.061 0.046 27 1.58 0.105 0.078 0.059 0.044 36 2.12 0.103 0.077 0.058 0.044 45 2.64 0.101 0.076 0.057 0.043 50 3.00 0.101 0.075 0.057 0.043 60 3.53 0.100 0.075 0.056 0.042 Table F-5. ASD Design Safety Factor Values Evaluated for ENR < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 14.037 18.709 24.806 33.063 15 1.00 13.899 18.526 24.562 32.739 18 1.06 13.886 18.509 24.540 32.709 27 1.58 13.801 18.395 24.389 32.508 36 2.12 13.742 18.317 24.285 32.370 45 2.64 13.702 18.264 24.215 32.276 50 3.00 13.681 18.235 24.177 32.225 60 3.53 13.655 18.201 24.131 32.165 Table F-6. φ Factors for Safety Factor = 6.00 – ENR <1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.263 0.363 0.3585 15 1.00 0.250 0.381 0.3518 18 1.06 0.249 0.383 0.3510 27 1.58 0.241 0.395 0.3466 36 2.12 0.235 0.403 0.3437 45 2.64 0.231 0.408 0.3418 50 3.00 0.229 0.411 0.3407 60 3.53 0.227 0.415 0.3393 F-4 Table F-7. φ Values Evaluated for ENR > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.154 0.116 0.088 0.066 15 1.00 0.148 0.111 0.084 0.063 18 1.06 0.147 0.111 0.084 0.063 27 1.58 0.143 0.108 0.082 0.061 36 2.12 0.141 0.106 0.080 0.060 45 2.64 0.139 0.104 0.079 0.059 50 3.00 0.138 0.104 0.078 0.059 60 3.53 0.136 0.103 0.078 0.058 Table F-8. ASD Design Safety Factor Values Evaluated for ENR > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 10.247 13.616 17.999 23.918 15 1.00 10.146 13.483 17.822 23.683 18 1.06 10.137 13.470 17.806 23.662 27 1.58 10.074 13.388 17.697 23.516 36 2.12 10.032 13.331 17.622 23.417 45 2.64 10.003 13.292 17.570 23.348 50 3.00 9.987 13.271 17.543 23.312 60 3.53 9.968 13.246 17.510 23.268 Table F-9. φ Factors for Safety Factor = 6.00 – ENR >1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.263 0.944 0.1727 15 1.00 0.250 0.962 0.1682 18 1.06 0.249 0.964 0.1676 27 1.58 0.241 0.976 0.1647 36 2.12 0.235 0.984 0.1627 45 2.64 0.231 0.989 0.1615 50 3.00 0.229 0.992 0.1607 60 3.53 0.227 0.996 0.1597 F-5 y = 4.8143x R 2 = 0.2881 0 5000 10000 15000 20000 25000 30000 35000 40000 45000 0 5000 10000 15000 20000 25000 30000 35000 40000 45000 Davisson Capacity (kN) Figure F-3. Davisson Capacity vs. ENR BOR Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.0 1000.0 2000.0 3000.0 4000.0 5000.0 6000.0 7000.0 8000.0 Davisson Capacity (kN) Cases=71 Mean=0.24 Standard Dev.=0.16 Figure F-4. Measured Over Predicted Capacity for ENR Method at BOR F-6 Table F-10. φ Values Evaluated for ENR – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.077 0.054 0.038 0.027 15 1.00 0.074 0.052 0.037 0.026 18 1.06 0.074 0.052 0.037 0.026 27 1.58 0.072 0.050 0.036 0.025 36 2.12 0.071 0.050 0.035 0.025 45 2.64 0.070 0.049 0.035 0.024 50 3.00 0.069 0.049 0.034 0.024 60 3.53 0.068 0.048 0.034 0.024 Table F-11. ASD Design Safety Factor Values Evaluated for ENR – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 20.423 29.086 41.153 58.608 15 1.00 20.223 28.800 40.749 58.033 18 1.06 20.204 28.774 40.712 57.980 27 1.58 20.080 28.597 40.461 57.623 36 2.12 19.995 28.476 40.290 57.379 45 2.64 19.937 28.393 40.172 57.212 50 3.00 19.905 28.348 40.109 57.122 60 3.53 19.868 28.296 40.034 57.015 Table F-12. φ Factors for Safety Factor = 6.00 – ENR – Overall (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.263 0.089 0.46474 15 1.00 0.250 0.104 0.45878 18 1.06 0.249 0.106 0.45799 27 1.58 0.241 0.115 0.45442 36 2.12 0.235 0.122 0.45165 45 2.64 0.231 0.126 0.45006 50 3.00 0.229 0.129 0.44888 60 3.53 0.227 0.131 0.44809 F-7 Table F-13. φ Values Evaluated for ENR < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.124 0.101 0.083 0.068 15 1.00 0.119 0.097 0.080 0.065 18 1.06 0.118 0.097 0.079 0.065 27 1.58 0.115 0.094 0.077 0.063 36 2.12 0.113 0.092 0.076 0.062 45 2.64 0.111 0.091 0.075 0.061 50 3.00 0.110 0.090 0.074 0.061 60 3.53 0.109 0.090 0.074 0.060 Table F-14. ASD Design Safety Factor Values Evaluated for ENR < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 12.777 15.610 19.002 23.217 15 1.00 12.651 15.457 18.816 22.989 18 1.06 12.640 15.443 18.798 22.968 27 1.58 12.562 15.348 18.683 22.827 36 2.12 12.509 15.283 18.604 22.730 45 2.64 12.472 15.239 18.549 22.664 50 3.00 12.453 15.215 18.520 22.628 60 3.53 12.429 15.186 18.486 22.586 Table F-15. φ Factors for Safety Factor = 6.00 – ENR <1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.263 -0.078 0.5313 15 1.00 0.250 -0.051 0.5205 18 1.06 0.249 -0.048 0.5193 27 1.58 0.241 -0.032 0.5130 36 2.12 0.235 -0.020 0.5082 45 2.64 0.231 -0.013 0.5054 50 3.00 0.229 -0.008 0.5034 60 3.53 0.227 -0.003 0.5014 F-8 Table F-16. φ Values Evaluated for ENR > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.077 0.054 0.038 0.026 15 1.00 0.074 0.052 0.036 0.025 18 1.06 0.074 0.051 0.036 0.025 27 1.58 0.072 0.050 0.035 0.025 36 2.12 0.070 0.049 0.034 0.024 45 2.64 0.069 0.048 0.034 0.024 50 3.00 0.069 0.048 0.034 0.024 60 3.53 0.068 0.048 0.033 0.023 Table F-17. ASD Design Safety Factor Values Evaluated for ENR > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 20.463 29.334 41.770 59.877 15 1.00 20.263 29.046 41.360 59.289 18 1.06 20.244 29.019 41.323 59.235 27 1.58 20.120 28.841 41.068 58.871 36 2.12 20.034 28.719 40.894 58.621 45 2.64 19.976 28.635 40.775 58.451 50 3.00 19.945 28.590 40.711 58.359 60 3.53 19.907 28.537 40.635 58.250 Table F-18. φ Factors for Safety Factor = 6.00 – ENR >1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.263 0.120 0.4524 15 1.00 0.250 0.135 0.4465 18 1.06 0.249 0.136 0.4461 27 1.58 0.241 0.146 0.4422 36 2.12 0.235 0.152 0.4398 45 2.64 0.231 0.156 0.4382 50 3.00 0.229 0.159 0.4370 60 3.53 0.227 0.162 0.4358 APPENDIX G LRFD ANALYSIS RESULTS – MODIFIED ENR G-1 y = 2.4493x R 2 = 0.5281 0 5000 10000 15000 20000 25000 0 5000 10000 15000 20000 25000 Davisson Capacity (kN) Figure G-1. Davisson Capacity vs. Modified ENR EOD Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Cases=61 Mean=0.45 Standard Dev.=0.27 Figure G-2. Measured Over Predicted Capacity for Modified ENR Method at EOD G-2 Table G-1. φ Values Evaluated for Modified ENR – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.171 0.124 0.090 0.065 15 1.00 0.164 0.119 0.087 0.063 18 1.06 0.163 0.118 0.086 0.063 27 1.58 0.159 0.115 0.084 0.061 36 2.12 0.156 0.113 0.082 0.060 45 2.64 0.154 0.111 0.081 0.059 50 3.00 0.153 0.111 0.081 0.059 60 3.53 0.151 0.110 0.080 0.058 Table G-2. ASD Design Safety Factor for Modified ENR – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 9.241 12.747 17.480 24.111 15 1.00 9.151 12.622 17.308 23.875 18 1.06 9.142 12.611 17.292 23.853 27 1.58 9.086 12.533 17.186 23.706 36 2.12 9.047 12.480 17.113 23.606 45 2.64 9.021 12.444 17.063 23.537 50 3.00 9.007 12.424 17.036 23.500 60 3.53 8.990 12.401 17.005 23.456 Table G-3. φ Factors for Safety Factor = 6.00 – Modified ENR – Overall (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.263 1.235 0.1085 15 1.00 0.250 1.251 0.1056 18 1.06 0.249 1.253 0.1052 27 1.58 0.241 1.263 0.1034 36 2.12 0.235 1.270 0.1021 45 2.64 0.231 1.275 0.1012 50 3.00 0.229 1.278 0.1007 60 3.53 0.227 1.281 0.1002 G-3 Table G-4. φ Values Evaluated for Modified ENR < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.106 0.075 0.053 0.037 15 1.00 0.102 0.072 0.051 0.035 18 1.06 0.102 0.071 0.050 0.035 27 1.58 0.099 0.069 0.049 0.034 36 2.12 0.097 0.068 0.048 0.034 45 2.64 0.096 0.067 0.047 0.033 50 3.00 0.095 0.067 0.047 0.033 60 3.53 0.094 0.066 0.047 0.033 Table G-5. ASD Design Safety Factor for Modified ENR < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 14.853 21.174 29.986 42.746 15 1.00 14.708 20.966 29.692 42.327 18 1.06 14.694 20.947 29.665 42.288 27 1.58 14.604 20.818 29.482 42.028 36 2.12 14.542 20.730 29.357 41.850 45 2.64 14.500 20.670 29.272 41.728 50 3.00 14.477 20.637 29.226 41.662 60 3.53 14.450 20.599 29.171 41.584 Table G-6. φ Factors for Safety Factor = 6.00 – Modified ENR <1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.263 0.579 0.2815 15 1.00 0.250 0.594 0.2764 18 1.06 0.249 0.596 0.2758 27 1.58 0.241 0.605 0.2728 36 2.12 0.235 0.612 0.2704 45 2.64 0.231 0.616 0.2691 50 3.00 0.229 0.618 0.2685 60 3.53 0.227 0.621 0.2675 G-4 Table G-7. φ Values Evaluated for Modified ENR > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.200 0.145 0.105 0.076 15 1.00 0.192 0.139 0.101 0.073 18 1.06 0.191 0.138 0.101 0.073 27 1.58 0.186 0.135 0.098 0.071 36 2.12 0.183 0.132 0.096 0.069 45 2.64 0.180 0.130 0.095 0.069 50 3.00 0.179 0.129 0.094 0.068 60 3.53 0.177 0.128 0.093 0.067 Table G-8. ASD Design Safety Factor for Modified ENR > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 7.881 10.903 14.993 20.742 15 1.00 7.804 10.796 14.846 20.538 18 1.06 7.797 10.786 14.833 20.519 27 1.58 7.749 10.720 14.741 20.393 36 2.12 7.716 10.675 14.679 20.307 45 2.64 7.694 10.644 14.636 20.248 50 3.00 7.682 10.627 14.613 20.216 60 3.53 7.667 10.607 14.586 20.178 Table G-9. φ Factors for Safety Factor = 6.00 – Modified ENR >1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.263 1.506 0.0661 15 1.00 0.250 1.523 0.0639 18 1.06 0.249 1.524 0.0638 27 1.58 0.241 1.534 0.0626 36 2.12 0.235 1.541 0.0617 45 2.64 0.231 1.546 0.0611 50 3.00 0.229 1.549 0.0608 60 3.53 0.227 1.552 0.0604 G-5 y = 3.0408x R2 = 0.1197 0 5000 10000 15000 20000 25000 30000 0 5000 10000 15000 20000 25000 30000 Davisson Capacity (kN) Modified ENR BOR Capacity (kN) Figure G-3. Davisson Capacity vs. Modified ENR BOR Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 Davisson Capacity (kN) Davisson Capacity / Modified ENR BOR Capacity Cases=63 Mean=0.36 Standard Dev.=0.25 Figure G-4. Measured Over Predicted Capacity for Modified ENR Method at BOR G-6 Table G-10. φ Values Evaluated for Modified ENR – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.121 0.085 0.060 0.042 15 1.00 0.116 0.081 0.058 0.040 18 1.06 0.115 0.081 0.057 0.040 27 1.58 0.112 0.079 0.056 0.039 36 2.12 0.110 0.077 0.055 0.039 45 2.64 0.108 0.076 0.054 0.038 50 3.00 0.108 0.076 0.054 0.038 60 3.53 0.107 0.075 0.053 0.037 Table G-11. ASD Design Safety Factor for Modified ENR – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 13.101 18.627 26.313 37.413 15 1.00 12.972 18.445 26.055 37.046 18 1.06 12.960 18.428 26.031 37.012 27 1.58 12.881 18.314 25.871 36.784 36 2.12 12.826 18.237 25.761 36.628 45 2.64 12.789 18.184 25.686 36.522 50 3.00 12.769 18.155 25.646 36.464 60 3.53 12.745 18.121 25.598 36.396 Table G-12. φ Factors for Safety Factor = 6.00 – Modified ENR – Overall (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.263 0.762 0.2232 15 1.00 0.250 0.777 0.2187 18 1.06 0.249 0.778 0.2184 27 1.58 0.241 0.788 0.2155 36 2.12 0.235 0.794 0.2137 45 2.64 0.231 0.799 0.2123 50 3.00 0.229 0.801 0.2117 60 3.53 0.227 0.804 0.2108 G-7 Table G-13. φ Values Evaluated for Modified ENR < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.211 0.178 0.151 0.128 15 1.00 0.202 0.171 0.145 0.123 18 1.06 0.202 0.170 0.144 0.122 27 1.58 0.196 0.166 0.141 0.119 36 2.12 0.193 0.163 0.138 0.117 45 2.64 0.190 0.161 0.136 0.115 50 3.00 0.189 0.159 0.135 0.114 60 3.53 0.187 0.158 0.134 0.113 Table G-14. ASD Design Safety Factor for Modified ENR < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 7.482 8.853 10.442 12.356 15 1.00 7.408 8.766 10.340 12.235 18 1.06 7.401 8.758 10.330 12.223 27 1.58 7.356 8.704 10.267 12.148 36 2.12 7.325 8.667 10.223 12.097 45 2.64 7.303 8.642 10.194 12.062 50 3.00 7.292 8.628 10.178 12.043 60 3.53 7.278 8.612 10.159 12.020 Table G-15. φ Factors for Safety Factor = 6.00 – Modified ENR <1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.263 1.252 0.1054 15 1.00 0.250 1.283 0.0998 18 1.06 0.249 1.286 0.0993 27 1.58 0.241 1.306 0.0959 36 2.12 0.235 1.320 0.0935 45 2.64 0.231 1.329 0.0920 50 3.00 0.229 1.334 0.0912 60 3.53 0.227 1.340 0.0902 G-8 Table G-16. φ Values Evaluated for Modified ENR > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.121 0.085 0.060 0.042 15 1.00 0.116 0.081 0.057 0.040 18 1.06 0.116 0.081 0.057 0.040 27 1.58 0.113 0.079 0.055 0.039 36 2.12 0.111 0.077 0.054 0.038 45 2.64 0.109 0.076 0.054 0.038 50 3.00 0.108 0.076 0.053 0.037 60 3.53 0.107 0.075 0.053 0.037 Table G-17. ASD Design Safety Factor for Modified ENR > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 13.029 18.635 26.475 37.864 15 1.00 12.901 18.452 26.215 37.493 18 1.06 12.890 18.435 26.191 37.458 27 1.58 12.810 18.321 26.030 37.228 36 2.12 12.756 18.244 25.920 37.070 45 2.64 12.719 18.191 25.844 36.962 50 3.00 12.699 18.162 25.804 36.904 60 3.53 12.675 18.128 25.756 36.835 Table G-18. φ Factors for Safety Factor = 6.00 – Modified. ENR >1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.263 0.790 0.2149 15 1.00 0.250 0.805 0.2106 18 1.06 0.249 0.806 0.2103 27 1.58 0.241 0.815 0.2077 36 2.12 0.235 0.822 0.2057 45 2.64 0.231 0.826 0.2045 50 3.00 0.229 0.828 0.2040 60 3.53 0.227 0.831 0.2031 APPENDIX H LRFD ANALYSIS RESULTS – GATES FORMULA H-1 y = 0.4487x R2 = -0.0733 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Gates EOD Capacity (kN) Figure H-1. Davisson Capacity vs. Gates EOD Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 Davisson Capacity (kN) Davisson Capacity / Gates EOD Capacity Cases=74 Mean=1.74 Standard Dev.=0.79 Figure H-2. Measured Over Predicted Capacity for Gates Method at EOD H-2 Table H-1. φ Values Evaluated for Gates – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.884 0.681 0.527 0.406 15 1.00 0.848 0.653 0.505 0.389 18 1.06 0.845 0.651 0.503 0.388 27 1.58 0.822 0.633 0.490 0.377 36 2.12 0.806 0.621 0.481 0.370 45 2.64 0.796 0.613 0.474 0.365 50 3.00 0.790 0.608 0.471 0.363 60 3.53 0.783 0.603 0.467 0.359 Table H-2. ASD Design Safety Factor for Gates – Overall (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 1.786 2.319 2.997 3.891 15 1.00 1.769 2.297 2.967 3.853 18 1.06 1.767 2.294 2.965 3.849 27 1.58 1.756 2.280 2.946 3.826 36 2.12 1.749 2.271 2.934 3.809 45 2.64 1.744 2.264 2.925 3.798 50 3.00 1.741 2.260 2.921 3.792 60 3.53 1.738 2.256 2.915 3.785 Table H-3. φ Factors for Safety Factor = 3.00 – Gates – Overall (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.526 3.032 0.00122 15 1.00 0.500 3.053 0.00113 18 1.06 0.498 3.054 0.00113 27 1.58 0.481 3.067 0.00108 36 2.12 0.470 3.076 0.00105 45 2.64 0.462 3.082 0.00103 50 3.00 0.458 3.085 0.00102 60 3.53 0.453 3.089 0.00101 H-3 Table H-4. φ Values Evaluated for Gates < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 0.683 0.554 0.451 0.366 15 1.00 0.655 0.532 0.433 0.351 18 1.06 0.653 0.529 0.431 0.350 27 1.58 0.635 0.515 0.420 0.340 36 2.12 0.623 0.505 0.412 0.334 45 2.64 0.615 0.499 0.406 0.329 50 3.00 0.610 0.495 0.403 0.327 60 3.53 0.605 0.491 0.400 0.324 Table H-5. ASD Design Safety Factor for Gates < 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 2.311 2.850 3.500 4.316 15 1.00 2.289 2.822 3.466 4.273 18 1.06 2.286 2.819 3.463 4.269 27 1.58 2.272 2.802 3.441 4.243 36 2.12 2.263 2.790 3.427 4.225 45 2.64 2.256 2.782 3.417 4.213 50 3.00 2.253 2.777 3.411 4.206 60 3.53 2.248 2.772 3.405 4.198 Table H-6. φ Factors for Safety Factor = 3.00 – Gates <1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.526 2.632 0.00425 15 1.00 0.500 2.658 0.00394 18 1.06 0.498 2.660 0.00391 27 1.58 0.481 2.676 0.00373 36 2.12 0.470 2.687 0.00361 45 2.64 0.462 2.695 0.00352 50 3.00 0.458 2.699 0.00348 60 3.53 0.453 2.704 0.00343 H-4 Table H-7. φ Values Evaluated for Gates > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 1.464 1.192 0.974 0.793 15 1.00 1.404 1.144 0.935 0.761 18 1.06 1.399 1.139 0.931 0.758 27 1.58 1.361 1.109 0.906 0.738 36 2.12 1.335 1.087 0.889 0.724 45 2.64 1.318 1.073 0.877 0.714 50 3.00 1.308 1.065 0.871 0.709 60 3.53 1.296 1.056 0.863 0.703 Table H-8. ASD Design Safety Factor for Gates > 1779 kN (EOD) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.5 3.03 3.57 9 0.52 1.079 1.325 1.620 1.990 15 1.00 1.068 1.312 1.605 1.970 18 1.06 1.067 1.310 1.603 1.969 27 1.58 1.061 1.302 1.593 1.956 36 2.12 1.056 1.297 1.586 1.948 45 2.64 1.053 1.293 1.582 1.943 50 3.00 1.051 1.291 1.579 1.939 60 3.53 1.049 1.289 1.576 1.936 Table H-9. φ Factors for Safety Factor = 3.00 – Gates >1779 kN (EOD) QD/QL φ βΤ pf Span length (m) 9 0.52 0.526 4.649 0.00000167 15 1.00 0.500 4.675 0.00000147 18 1.06 0.498 4.678 0.00000145 27 1.58 0.481 4.694 0.00000134 36 2.12 0.470 4.705 0.00000127 45 2.64 0.462 4.713 0.00000122 50 3.00 0.458 4.717 0.00000120 60 3.53 0.453 4.722 0.00000117 H-5 y = 0.4806x R 2 = -0.6104 0 1000 2000 3000 4000 5000 6000 7000 8000 0 1000 2000 3000 4000 5000 6000 7000 8000 Davisson Capacity (kN) Figure H-3. Davisson Capacity vs. Gates BOR Capacity 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 1000 2000 3000 4000 5000 6000 7000 8000 Davisson Capacity (kN) Davisson Capacity / Gates BOR Capacity Cases=71 Mean=1.89 Standard Dev.=0.72 Figure H-4. Measured Over Predicted Capacity for Gates Method at BOR H-6 Table H-10. φ Values Evaluated for Gates – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.097 0.871 0.694 0.551 15 1.00 1.052 0.835 0.666 0.529 18 1.06 1.048 0.832 0.664 0.527 27 1.58 1.020 0.810 0.646 0.513 36 2.12 1.000 0.794 0.633 0.503 45 2.64 0.987 0.784 0.625 0.496 50 3.00 0.980 0.778 0.620 0.493 60 3.53 0.971 0.771 0.615 0.488 Table H-11. ASD Design Safety Factor for Gates – Overall (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.440 1.813 2.274 2.864 15 1.00 1.426 1.796 2.252 2.836 18 1.06 1.424 1.794 2.250 2.833 27 1.58 1.416 1.783 2.236 2.816 36 2.12 1.410 1.775 2.226 2.804 45 2.64 1.406 1.770 2.220 2.796 50 3.00 1.403 1.767 2.216 2.791 60 3.53 1.401 1.764 2.212 2.786 Table H-12. φ Factors for Safety Factor = 3.00 – Gates – Overall (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.526 3.679 0.0001173 15 1.00 0.500 3.702 0.0001072 18 1.06 0.498 3.704 0.0001063 27 1.58 0.481 3.718 0.0001006 36 2.12 0.470 3.728 0.0000967 45 2.64 0.462 3.735 0.0000940 50 3.00 0.458 3.739 0.0000926 60 3.53 0.453 3.743 0.0000911 H-7 Table H-13. φ Values Evaluated for Gates < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 0.855 0.732 0.628 0.538 15 1.00 0.821 0.702 0.603 0.516 18 1.06 0.817 0.700 0.600 0.514 27 1.58 0.796 0.681 0.584 0.500 36 2.12 0.780 0.668 0.573 0.491 45 2.64 0.770 0.659 0.566 0.484 50 3.00 0.764 0.654 0.561 0.480 60 3.53 0.758 0.648 0.556 0.476 Table H-14. ASD Design Safety Factor for Gates < 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.846 2.157 2.513 2.936 15 1.00 1.828 2.136 2.488 2.908 18 1.06 1.826 2.134 2.486 2.905 27 1.58 1.815 2.121 2.471 2.887 36 2.12 1.807 2.112 2.460 2.875 45 2.64 1.802 2.105 2.453 2.866 50 3.00 1.799 2.102 2.449 2.862 60 3.53 1.796 2.098 2.445 2.857 Table H-15. φ Factors for Safety Factor = 3.00 – Gates <1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.526 3.644 0.0001345 15 1.00 0.500 3.679 0.0001173 18 1.06 0.498 3.682 0.0001159 27 1.58 0.481 3.703 0.0001067 36 2.12 0.470 3.718 0.0001006 45 2.64 0.462 3.728 0.0000967 50 3.00 0.458 3.733 0.0000948 60 3.53 0.453 3.740 0.0000922 H-8 Table H-16. φ Values Evaluated for Gates > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.250 1.008 0.817 0.659 15 1.00 1.199 0.967 0.783 0.632 18 1.06 1.194 0.963 0.780 0.629 27 1.58 1.162 0.938 0.759 0.613 36 2.12 1.140 0.920 0.745 0.601 45 2.64 1.125 0.907 0.735 0.593 50 3.00 1.117 0.901 0.730 0.589 60 3.53 1.107 0.893 0.723 0.583 Table H-17. ASD Design Safety Factor for Gates > 1779 kN (BOR) Failure Probability pf / Reliability Index βΤ QD/QL 2.50E-02 6.22E-03 1.22E-03 1.79E-04 Span length (m) 1.96 2.50 3.03 3.57 9 0.52 1.263 1.566 1.934 2.397 15 1.00 1.251 1.551 1.915 2.374 18 1.06 1.250 1.549 1.913 2.371 27 1.58 1.242 1.540 1.901 2.357 36 2.12 1.237 1.533 1.893 2.347 45 2.64 1.233 1.529 1.888 2.340 50 3.00 1.231 1.526 1.885 2.336 60 3.53 1.229 1.524 1.881 2.332 Table H-18. φ Factors for Safety Factor = 3.00 – Gates >1779 kN (BOR) QD/QL φ β pf Span length (m) 9 0.52 0.526 4.134 0.0000179 15 1.00 0.500 4.159 0.0000160 18 1.06 0.498 4.161 0.0000159 27 1.58 0.481 4.177 0.0000148 36 2.12 0.470 4.187 0.0000142 45 2.64 0.462 4.195 0.0000137 50 3.00 0.458 4.199 0.0000134 60 3.53 0.453 4.203 0.0000132 APPENDIX I FORCE AND VELOCITY TRACES FROM PDA SIGNAL I-1 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 6000 7000 0 0.02 0.04 0.06 0.08 0.1 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-1. Force, Velocity, Double Wave Up, and Displacement Traces for TS4I (Pascagoula Bridge) I-2 -2000 -1000 0 1000 2000 3000 4000 5000 6000 0 0.02 0.04 0.06 0.08 0.1 0.12 Time (s) Force (kN) -0.001 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-2. Force, Velocity, Double Wave Up, and Displacement Traces for TS1B102 (Vilano Bridge) I-3 -2000 -1000 0 1000 2000 3000 4000 5000 6000 0 0.02 0.04 0.06 0.08 0.1 0.12 Time (s) -0.001 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 Force Velocity 2 Wave Up Displacement Figure I-3. Force, Velocity, Double Wave Up, and Displacement Traces for B5RS2 (Escambria Bridge) I-4 -4000 -2000 0 2000 4000 6000 8000 10000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-4. Force, Velocity, Double Wave Up, and Displacement Traces for TS13A1-A (Buckman Bridge) I-5 -6000 -4000 -2000 0 2000 4000 6000 8000 10000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.004 -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-5. Force, Velocity, Double Wave Up, and Displacement Traces for TS19FRCB (Buckman Bridge) I-6 -6000 -4000 -2000 0 2000 4000 6000 8000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.006 -0.004 -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-6. Force, Velocity, Double Wave Up, and Displacement Traces for TS24ALTA (Buckman Bridge) I-7 -6000 -4000 -2000 0 2000 4000 6000 8000 10000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 Force Velocity 2 Wave Up Displacement Figure I-7. Force, Velocity, Double Wave Up, and Displacement Traces for TS29RC (Buckman Bridge) I-8 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 6000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.001 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-8. Force, Velocity, Double Wave Up, and Displacement Traces for F6-58A (Acosta Bridge) I-9 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 6000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.001 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-9. Force, Velocity, Double Wave Up, and Displacement Traces for G13-37 (Acosta Bridge) I-10 -5000 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 6000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-10. Force, Velocity, Double Wave Up, and Displacement Traces for H2-27B (Acosta Bridge) I-11 -2000 -1000 0 1000 2000 3000 4000 5000 6000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.001 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-11. Force, Velocity, Double Wave Up, and Displacement Traces for B1-76F (Choctawhatche Bridge) I-12 -4000 -2000 0 2000 4000 6000 8000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-12. Force, Velocity, Double Wave Up, and Displacement Traces for PR5R2 (Choctawhatche Bridge) I-13 -4000 -2000 0 2000 4000 6000 8000 10000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.001 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-13. Force, Velocity, Double Wave Up, and Displacement Traces for B8-97R2 (Choctawhatche Bridge) I-14 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 6000 7000 8000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-14. Force, Velocity, Double Wave Up, and Displacement Traces for B14-89R2 (Choctawhatche Bridge) I-15 -2000 -1000 0 1000 2000 3000 4000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 -0.001 0.000 0.001 0.002 0.003 0.004 0.005 0.006 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-15. Force, Velocity, Double Wave Up, and Displacement Traces for B17-94R2 (Choctawhatche Bridge) I-16 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 6000 7000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.004 -0.002 0.000 0.002 0.004 0.006 0.008 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-16. Force, Velocity, Double Wave Up, and Displacement Traces for PR35FIN (Choctawhatche Bridge) I-17 -4000 -2000 0 2000 4000 6000 8000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) -0.001 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010 Force Velocity 2 Wave Up Displacement Figure I-17. Force, Velocity, Double Wave Up, and Displacement Traces for B23-76F2 (Choctawhatche Bridge) I-18 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 6000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-18. Force, Velocity, Double Wave Up, and Displacement Traces for B27-63F (Choctawhatche Bridge) I-19 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-19. Force, Velocity, Double Wave Up, and Displacement Traces for AUC63K (Aucilla Bridge) I-20 -6000 -4000 -2000 0 2000 4000 6000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.005 0.000 0.005 0.010 0.015 0.020 0.025 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-20. Force, Velocity, Double Wave Up, and Displacement Traces for BKM30J (Buckman Bridge) I-21 -2000 -1000 0 1000 2000 3000 4000 5000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-21. Force, Velocity, Double Wave Up, and Displacement Traces for SBZ83N (Seabreeze Bridge) I-22 -2000 -1000 0 1000 2000 3000 4000 5000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-22. Force, Velocity, Double Wave Up, and Displacement Traces for VLE-32C (Vilano East Bridge) I-23 -5000 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Time (s) Force (kN) -0.002 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 Displacement (m) Force Velocity 2 Wave Up Displacement Figure I-23. Force, Velocity, Double Wave Up, and Displacement Traces for VLWA-61D (Vilano West Bridge) APPENDIX J OUTPUT FILE FOR SUGGESTED METHOD & GRL PROCEDURE (FORTRAN) J-1 J-2 J-3 UNIVERSITY OF FLORIDA J-4 J-5 FLORIDA DEPARTMENT OF TRANSPORTATION SUGGESTED METHOD FOR DETERMINE SKIN & TIP CAPACITIES / METHOD 1 <<<<< PILE INFORMATION >>>>> LENGTH = 29.76 m WAVESPEED = 4643.00 m/s MODU = 51750160.00 kN/m2 AREA = .42 m2 DT TIME = .00010 s TIME VEL FOR ZVEL DISP 2 WAVE UP (s) (m/s) (kN) (kN) (m) (kN) .0001 .000 .000 .000 .00000 .000 .0002 .000 .000 .000 .00000 .000 .0003 .000 .000 .000 .00000 .000 .0004 .000 .000 .000 .00000 .000 .0005 .000 .000 .000 .00000 .000 .0006 .000 .000 .000 .00000 .000 .0007 .000 .000 .000 .00000 .000 .0008 .000 .000 .000 .00000 .000 .0009 .000 .000 .000 .00000 .000 .0010 .000 .000 .000 .00000 .000 .0011 .000 .000 .000 .00000 .000 .0012 .000 .000 .000 .00000 .000 .0013 .000 .000 .000 .00000 .000 .0014 .000 .000 .000 .00000 .000 .0015 .000 .000 .000 .00000 .000 .0016 .000 .000 .000 .00000 .000 .0017 .000 .000 .000 .00000 .000 .0018 .000 .000 .000 .00000 .000 .0019 .000 .000 .000 .00000 .000 .0020 .000 .000 .000 .00000 .000 .0021 .000 .000 .000 .00000 .000 .0022 .000 .000 .000 .00000 .000 .0023 .000 .000 .000 .00000 .000 .0024 .000 .000 .000 .00000 .000 .0025 .000 .000 .000 .00000 .000 .0026 .000 .000 .000 .00000 .000 .0027 .000 .000 .000 .00000 .000 .0028 .000 .000 .000 .00000 .000 .0029 .000 .000 .000 .00000 .000 .0030 .000 .000 .000 .00000 .000 .0031 .000 .000 .000 .00000 .000 .0032 .000 .000 .000 .00000 .000 .0033 .000 .000 .000 .00000 .000 .0034 .000 .000 .000 .00000 .000 .0035 .000 .000 .000 .00000 .000 .0036 .000 .000 .000 .00000 .000 .0037 .000 .000 .000 .00000 .000 .0038 .000 .000 .000 .00000 .000 .0039 .000 .000 .000 .00000 .000 J-6 .0040 .000 .000 .000 .00000 .000 .0041 .000 .000 .000 .00000 .000 .0042 .000 .000 .000 .00000 .000 .0043 .000 .000 .000 .00000 .000 .0044 .000 .000 .000 .00000 .000 .0045 .000 .000 .000 .00000 .000 .0046 .000 .000 .000 .00000 .000 .0047 .000 .000 .000 .00000 .000 .0048 .000 .000 .000 .00000 .000 .0049 .000 .000 .000 .00000 .000 .0050 .000 .000 .000 .00000 .000 .0051 .000 .000 .000 .00000 .000 .0052 .000 .000 .000 .00000 .000 .0053 .000 .000 .000 .00000 .000 .0054 .000 .000 .000 .00000 .000 .0055 .000 .000 .000 .00000 .000 .0056 .000 .000 .000 .00000 .000 .0057 .000 .000 .000 .00000 .000 .0058 .000 .000 .000 .00000 .000 .0059 .000 .000 .000 .00000 .000 .0060 .000 .000 .000 .00000 .000 .0061 .000 .000 .000 .00000 .000 .0062 .000 .000 .000 .00000 .000 .0063 .000 .000 .000 .00000 .000 .0064 .000 .000 .000 .00000 .000 .0065 .000 .000 .000 .00000 .000 .0066 .000 .000 .000 .00000 .000 .0067 .000 .000 .000 .00000 .000 .0068 .000 .000 .000 .00000 .000 .0069 .000 .000 .000 .00000 .000 .0070 .000 .000 .000 .00000 .000 .0071 .000 .000 .000 .00000 .000 .0072 .000 .000 .000 .00000 .000 .0073 .000 .000 .000 .00000 .000 .0074 .000 .000 .000 .00000 .000 .0075 .000 .000 .000 .00000 .000 .0076 .000 .000 .000 .00000 .000 .0077 .000 .000 .000 .00000 .000 .0078 .000 .000 .000 .00000 .000 .0079 .000 .000 .000 .00000 .000 .0080 .000 .000 .000 .00000 .000 .0081 .000 .000 .000 .00000 .000 .0082 .000 .000 .000 .00000 .000 .0083 .000 .000 .000 .00000 .000 .0084 .000 .000 .000 .00000 .000 .0085 .000 46.440 .000 .00000 46.440 .0086 .000 46.440 .000 .00000 46.440 .0087 .000 46.440 .000 .00000 46.440 .0088 .000 46.440 .000 .00000 46.440 .0089 .000 46.440 .000 .00000 46.440 .0090 .000 46.440 .000 .00000 46.440 .0091 .000 46.440 .000 .00000 46.440 .0092 .000 46.440 .000 .00000 46.440 .0093 .000 46.440 .000 .00000 46.440 .0094 .000 46.440 .000 .00000 46.440 .0095 .000 46.440 .000 .00000 46.440 .0096 .000 46.440 .000 .00000 46.440 J-7 .0097 .000 46.440 .000 .00000 46.440 .0098 .000 46.440 .000 .00000 46.440 .0099 .000 46.440 .000 .00000 46.440 .0100 .000 46.440 .000 .00000 46.440 .0101 .000 46.440 .000 .00000 46.440 .0102 .000 46.440 .000 .00000 46.440 .0103 .000 46.440 .000 .00000 46.440 .0104 .000 46.440 .000 .00000 46.440 .0105 .000 92.881 .000 .00000 92.881 .0106 .000 92.881 .000 .00000 92.881 .0107 .000 92.881 .000 .00000 92.881 .0108 .000 92.881 .000 .00000 92.881 .0109 .000 92.881 .000 .00000 92.881 .0110 .000 92.881 .000 .00000 92.881 .0111 .000 92.881 .000 .00000 92.881 .0112 .000 92.881 .000 .00000 92.881 .0113 .000 92.881 .000 .00000 92.881 .0114 .000 92.881 .000 .00000 92.881 .0115 .010 92.881 46.440 .00000 46.440 .0116 .010 92.881 46.440 .00000 46.440 .0117 .010 92.881 46.440 .00000 46.440 .0118 .010 92.881 46.440 .00000 46.440 .0119 .010 92.881 46.440 .00000 46.440 .0120 .010 92.881 46.440 .00001 46.440 .0121 .010 92.881 46.440 .00001 46.440 .0122 .010 92.881 46.440 .00001 46.440 .0123 .010 92.881 46.440 .00001 46.440 .0124 .010 92.881 46.440 .00001 46.440 .0125 .010 139.321 46.440 .00001 92.881 .0126 .010 139.321 46.440 .00001 92.881 .0127 .010 139.321 46.440 .00001 92.881 .0128 .010 139.321 46.440 .00001 92.881 .0129 .010 139.321 46.440 .00001 92.881 .0130 .010 139.321 46.440 .00002 92.881 .0131 .010 139.321 46.440 .00002 92.881 .0132 .010 139.321 46.440 .00002 92.881 .0133 .010 139.321 46.440 .00002 92.881 .0134 .010 139.321 46.440 .00002 92.881 .0135 .010 139.321 46.440 .00002 92.881 .0136 .010 139.321 46.440 .00002 92.881 .0137 .010 139.321 46.440 .00002 92.881 .0138 .010 139.321 46.440 .00002 92.881 .0139 .010 139.321 46.440 .00002 92.881 .0140 .010 139.321 46.440 .00003 92.881 .0141 .010 139.321 46.440 .00003 92.881 .0142 .010 139.321 46.440 .00003 92.881 .0143 .010 139.321 46.440 .00003 92.881 .0144 .010 139.321 46.440 .00003 92.881 .0145 .020 185.761 92.881 .00003 92.881 .0146 .020 185.761 92.881 .00003 92.881 .0147 .020 185.761 92.881 .00003 92.881 .0148 .020 185.761 92.881 .00004 92.881 .0149 .020 185.761 92.881 .00004 92.881 .0150 .020 185.761 92.881 .00004 92.881 .0151 .020 185.761 92.881 .00004 92.881 .0152 .020 185.761 92.881 .00004 92.881 .0153 .020 185.761 92.881 .00005 92.881 J-8 .0154 .020 185.761 92.881 .00005 92.881 .0155 .020 232.201 92.881 .00005 139.321 .0156 .020 232.201 92.881 .00005 139.321 .0157 .020 232.201 92.881 .00005 139.321 .0158 .020 232.201 92.881 .00006 139.321 .0159 .020 232.201 92.881 .00006 139.321 .0160 .020 232.201 92.881 .00006 139.321 .0161 .040 232.201 185.761 .00006 46.440 .0162 .030 232.201 139.321 .00007 92.881 .0163 .030 232.201 139.321 .00007 92.881 .0164 .030 232.201 139.321 .00007 92.881 .0165 .030 278.642 139.321 .00008 139.321 .0166 .030 278.642 139.321 .00008 139.321 .0167 .030 278.642 139.321 .00008 139.321 .0168 .030 278.642 139.321 .00009 139.321 .0169 .030 278.642 139.321 .00009 139.321 .0170 .030 278.642 139.321 .00009 139.321 .0171 .030 325.082 139.321 .00009 185.761 .0172 .030 325.082 139.321 .00010 185.761 .0173 .030 325.082 139.321 .00010 185.761 .0174 .040 325.082 185.761 .00010 139.321 .0175 .040 325.082 185.761 .00011 139.321 .0176 .040 325.082 185.761 .00011 139.321 .0177 .040 371.522 185.761 .00012 185.761 .0178 .040 371.522 185.761 .00012 185.761 .0179 .040 371.522 185.761 .00012 185.761 .0180 .040 371.522 185.761 .00013 185.761 .0181 .050 417.963 232.201 .00013 185.761 .0182 .040 417.963 185.761 .00014 232.201 .0183 .050 417.963 232.201 .00014 185.761 .0184 .050 464.403 232.201 .00015 232.201 .0185 .060 510.843 278.642 .00015 232.201 .0186 .070 557.283 325.082 .00016 232.201 .0187 .080 650.164 371.522 .00017 278.642 .0188 .100 743.044 464.403 .00017 278.642 .0189 .140 928.806 650.164 .00019 278.642 .0190 .150 1114.567 696.604 .00020 417.963 .0191 .200 1300.328 928.806 .00022 371.522 .0192 .230 1439.649 1068.126 .00024 371.522 .0193 .280 1671.850 1300.328 .00027 371.522 .0194 .310 1950.492 1439.649 .00030 510.843 .0195 .360 2182.693 1671.850 .00033 510.843 .0196 .420 2461.335 1950.492 .00037 510.843 .0197 .480 2832.857 2229.133 .00041 603.724 .0198 .530 3111.499 2461.335 .00046 650.164 .0199 .600 3436.581 2786.417 .00052 650.164 .0200 .640 3808.103 2972.178 .00058 835.925 .0201 .740 4179.625 3436.581 .00065 743.044 .0202 .800 4551.147 3715.222 .00073 835.925 .0203 .880 4829.789 4086.745 .00081 743.044 .0204 .920 5108.431 4272.506 .00090 835.925 .0205 .980 5340.632 4551.147 .00100 789.485 .0206 1.020 5526.393 4736.909 .00110 789.485 .0207 1.060 5712.154 4922.670 .00120 789.485 .1006 -.010 -278.642 -46.440 -.00013 -232.201 .1007 -.010 -278.642 -46.440 -.00013 -232.201 .1008 -.010 -278.642 -46.440 -.00013 -232.201 J-9 .1009 -.010 -278.642 -46.440 -.00013 -232.201 .1010 -.010 -325.082 -46.440 -.00013 -278.642 .1011 -.010 -278.642 -46.440 -.00013 -232.201 .1012 -.010 -325.082 -46.440 -.00013 -278.642 .1013 -.010 -278.642 -46.440 -.00013 -232.201 .1014 -.010 -325.082 -46.440 -.00013 -278.642 .1015 -.010 -278.642 -46.440 -.00013 -232.201 .1016 -.010 -325.082 -46.440 -.00014 -278.642 .1017 .000 -278.642 .000 -.00014 -278.642 .1018 .000 -325.082 .000 -.00014 -325.082 .1019 .000 -278.642 .000 -.00014 -278.642 .1020 -.010 -325.082 -46.440 -.00014 -278.642 .1021 -.010 -325.082 -46.440 -.00014 -278.642 .1022 -.010 -325.082 -46.440 -.00014 -278.642 .1023 -.010 -325.082 -46.440 -.00014 -278.642 .1024 -.010 -325.082 -46.440 -.00014 -278.642 MAX FORCE TIME (T1) = .02230 s TIME AT 2L/c (T2) = .03512 s SKIN TIME (T3) = .03320 s FORCE STARTING TIME = .01870 s RISE TIME = .00360 s MAXIMUM FORCE = 8312.81 kN MAXIMUM VELOCITY = 1.57 m/s <<<<< SUGGESTED METHOD >>>>> T/S RATIO = .327 Jc = .377 RST = 6618.246 kN = 1487.915 kips SKINCAP = 4986.349 kN = 1121.032 kips TIPCAP = 1631.896 kN = 366.883 kips EQUIV DAMP= 1.391 METHOD 1 METHOD 1 RATIO = 1.444 TOTAL CAPACITY = 9079.075 kN = 2041.159 kips TOTAL TIP CAP = 5363.852 kN = 1205.902 kips TOTAL SKIN CAP = 3715.222 kN = 835.257 kips
5059
https://artofproblemsolving.com/wiki/index.php/Triangle_Inequality?srsltid=AfmBOorTBv5IbxoVJCFKXOeu9Afl1ZFGPeZ8b8eu7H4xwYOkrIyVAAmp
Art of Problem Solving Triangle Inequality - 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 Triangle Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Triangle Inequality The Triangle Inequality says that in a nondegeneratetriangle: That is, the sum of the lengths of any two sides is larger than the length of the third side. In degenerate triangles, the strict inequality must be replaced by "greater than or equal to." The Triangle Inequality can also be extended to other polygons. The lengths can only be the sides of a nondegenerate -gon if for . Expressing the inequality in this form leads to , where is the sum of the , or . Stated in another way, it says that in every polygon, each side must be smaller than the semiperimeter. Contents [hide] 1 Problems 1.1 Introductory Problems 1.2 Intermediate Problems 1.3 Olympiad Problems 2 See Also Problems Introductory Problems 2003 AMC 12A Problems/Problem 7 2006 AMC 10B Problem 10 2006 AIME II Problem 2 Intermediate Problems 2010 AMC 12A Problem 25 Olympiad Problems Belarus 2002 Aops Topic Given , prove: See Also Algebra Inequality Geometric inequalities Triangle This article is a stub. Help us out by expanding it. Retrieved from " Categories: Stubs Geometry Theorems 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.
5060
https://en.wikipedia.org/wiki/Gas_exchange
Jump to content Search Contents (Top) 1 Physical principles of gas-exchange 1.1 Diffusion and surface area 1.2 Interaction with circulatory systems 2 Mammals 3 Exchange membrane 3.1 Alveolar air 3.2 Pulmonary circulation 4 Other vertebrates 4.1 Fish 4.2 Amphibians 4.3 Reptiles 4.4 Birds 5 Plants 6 Invertebrates 7 Summary of main gas exchange systems 8 See also 9 References Gas exchange العربية Արեւմտահայերէն Беларуская Català Чӑвашла Čeština Deutsch Eesti Emiliàn e rumagnòl Español Esperanto Français Gaeilge Galego 한국어 Հայերեն Bahasa Indonesia Italiano Қазақша Kurdî Bahasa Melayu Nederlands 日本語 Norsk bokmål Norsk nynorsk Polski Português Русский Simple English தமிழ் Українська اردو Tiếng Việt 粵語 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Process by which gases diffuse through a biological membrane | | | Part of a series on | | Continuum mechanics | | Fick's laws of diffusion | | Laws | Conservations | | Mass Momentum Energy | | Inequalities | | Clausius–Duhem (entropy) | | | Solid mechanics Deformation Elasticity + linear Plasticity Hooke's law Stress Strain + Finite strain + Infinitesimal strain Compatibility Bending Contact mechanics + frictional Material failure theory Fracture mechanics | | Fluid mechanics | Fluids | | Statics · Dynamics Archimedes' principle · Bernoulli's principle Navier–Stokes equations Poiseuille equation · Pascal's law Viscosity + (Newtonian · non-Newtonian) Buoyancy · Mixing · Pressure | | Liquids | | Adhesion Capillary action Chromatography Cohesion (chemistry) Surface tension | | Gases | | Atmosphere Boyle's law Charles's law Combined gas law Fick's law Gay-Lussac's law Graham's law | | Plasma | | | Rheology | | | Viscoelasticity Rheometry Rheometer | | Smart fluids | | Electrorheological Magnetorheological Ferrofluids | | | Scientists Bernoulli Boyle Cauchy Charles Euler Fick Gay-Lussac Graham Hooke Newton Navier Noll Pascal Stokes Truesdell | | v t e | Gas exchange is the physiological process by which gases move passively by diffusion across a surface. For example, this surface might be the air/water interface of a water body, the surface of a gas bubble in a liquid, a gas-permeable membrane, or a biological membrane that forms the boundary between an organism and its extracellular environment. Gases are constantly consumed and produced by cellular and metabolic reactions in most living things, so an efficient system for gas exchange between, ultimately, the interior of the cell(s) and the external environment is required. Small, particularly unicellular organisms, such as bacteria and protozoa, have a high surface-area to volume ratio. In these creatures the gas exchange membrane is typically the cell membrane. Some small multicellular organisms, such as flatworms, are also able to perform sufficient gas exchange across the skin or cuticle that surrounds their bodies. However, in most larger organisms, which have small surface-area to volume ratios, specialised structures with convoluted surfaces such as gills, pulmonary alveoli and spongy mesophylls provide the large area needed for effective gas exchange. These convoluted surfaces may sometimes be internalised into the body of the organism. This is the case with the alveoli, which form the inner surface of the mammalian lung, the spongy mesophyll, which is found inside the leaves of some kinds of plant, or the gills of those molluscs that have them, which are found in the mantle cavity. In aerobic organisms, gas exchange is particularly important for respiration, which involves the uptake of oxygen (O2) and release of carbon dioxide (CO2). Conversely, in oxygenic photosynthetic organisms such as most land plants, uptake of carbon dioxide and release of both oxygen and water vapour are the main gas-exchange processes occurring during the day. Other gas-exchange processes are important in less familiar organisms: e.g. carbon dioxide, methane and hydrogen are exchanged across the cell membrane of methanogenic archaea. In nitrogen fixation by diazotrophic bacteria, and denitrification by heterotrophic bacteria (such as Paracoccus denitrificans and various pseudomonads), nitrogen gas is exchanged with the environment, being taken up by the former and released into it by the latter, while giant tube worms rely on bacteria to oxidize hydrogen sulfide extracted from their deep sea environment, using dissolved oxygen in the water as an electron acceptor. Diffusion only takes place with a concentration gradient. Gases will flow from a high concentration to a low concentration. A high oxygen concentration in the alveoli and low oxygen concentration in the capillaries causes oxygen to move into the capillaries. A high carbon dioxide concentration in the capillaries and low carbon dioxide concentration in the alveoli causes carbon dioxide to move into the alveoli. Physical principles of gas-exchange [edit] Diffusion and surface area [edit] The exchange of gases occurs as a result of diffusion down a concentration gradient. Gas molecules move from a region in which they are at high concentration to one in which they are at low concentration. Diffusion is a passive process, meaning that no energy is required to power the transport, and it follows Fick's law: [citation needed] In relation to a typical biological system, where two compartments ('inside' and 'outside'), are separated by a membrane barrier, and where a gas is allowed to spontaneously diffuse down its concentration gradient:[citation needed] J is the flux, the amount of gas diffusing per unit area of membrane per unit time. Note that this is already scaled for the area of the membrane. D is the diffusion coefficient, which will differ from gas to gas, and from membrane to membrane, according to the size of the gas molecule in question, and the nature of the membrane itself (particularly its viscosity, temperature and hydrophobicity). φ is the concentration of the gas. x is the position across the thickness of the membrane. dφ/dx is therefore the concentration gradient across the membrane. If the two compartments are individually well-mixed, then this is simplifies to the difference in concentration of the gas between the inside and outside compartments divided by the thickness of the membrane. The negative sign indicates that the diffusion is always in the direction that - over time - will destroy the concentration gradient, i.e. the gas moves from high concentration to low concentration until eventually the inside and outside compartments reach equilibrium. Gases must first dissolve in a liquid in order to diffuse across a membrane, so all biological gas exchange systems require a moist environment. In general, the higher the concentration gradient across the gas-exchanging surface, the faster the rate of diffusion across it. Conversely, the thinner the gas-exchanging surface (for the same concentration difference), the faster the gases will diffuse across it. In the equation above, J is the flux expressed per unit area, so increasing the area will make no difference to its value. However, an increase in the available surface area, will increase the amount of gas that can diffuse in a given time. This is because the amount of gas diffusing per unit time (dq/dt) is the product of J and the area of the gas-exchanging surface, A: Single-celled organisms such as bacteria and amoebae do not have specialised gas exchange surfaces, because they can take advantage of the high surface area they have relative to their volume. The amount of gas an organism produces (or requires) in a given time will be in rough proportion to the volume of its cytoplasm. The volume of a unicellular organism is very small; thus, it produces (and requires) a relatively small amount of gas in a given time. In comparison to this small volume, the surface area of its cell membrane is very large, and adequate for its gas-exchange needs without further modification. However, as an organism increases in size, its surface area and volume do not scale in the same way. Consider an imaginary organism that is a cube of side-length, L. Its volume increases with the cube (L3) of its length, but its external surface area increases only with the square (L2) of its length. This means the external surface rapidly becomes inadequate for the rapidly increasing gas-exchange needs of a larger volume of cytoplasm. Additionally, the thickness of the surface that gases must cross (dx in Fick's law) can also be larger in larger organisms: in the case of a single-celled organism, a typical cell membrane is only 10 nm thick; but in larger organisms such as roundworms (Nematoda) the equivalent exchange surface - the cuticle - is substantially thicker at 0.5 μm. Interaction with circulatory systems [edit] In multicellular organisms therefore, specialised respiratory organs such as gills or lungs are often used to provide the additional surface area for the required rate of gas exchange with the external environment. However the distances between the gas exchanger and the deeper tissues are often too great for diffusion to meet gaseous requirements of these tissues. The gas exchangers are therefore frequently coupled to gas-distributing circulatory systems, which transport the gases evenly to all the body tissues regardless of their distance from the gas exchanger. Some multicellular organisms such as flatworms (Platyhelminthes) are relatively large but very thin, allowing their outer body surface to act as a gas exchange surface without the need for a specialised gas exchange organ. Flatworms therefore lack gills or lungs, and also lack a circulatory system. Other multicellular organisms such as sponges (Porifera) have an inherently high surface area, because they are very porous and/or branched. Sponges do not require a circulatory system or specialised gas exchange organs, because their feeding strategy involves one-way pumping of water through their porous bodies using flagellated collar cells. Each cell of the sponge's body is therefore exposed to a constant flow of fresh oxygenated water. They can therefore rely on diffusion across their cell membranes to carry out the gas exchange needed for respiration. In organisms that have circulatory systems associated with their specialized gas-exchange surfaces, a great variety of systems are used for the interaction between the two. In a countercurrent flow system, air (or, more usually, the water containing dissolved air) is drawn in the opposite direction to the flow of blood in the gas exchanger. A countercurrent system such as this maintains a steep concentration gradient along the length of the gas-exchange surface (see lower diagram in Fig. 2). This is the situation seen in the gills of fish and many other aquatic creatures. The gas-containing environmental water is drawn unidirectionally across the gas-exchange surface, with the blood-flow in the gill capillaries beneath flowing in the opposite direction. Although this theoretically allows almost complete transfer of a respiratory gas from one side of the exchanger to the other, in fish less than 80% of the oxygen in the water flowing over the gills is generally transferred to the blood. Alternative arrangements are cross current systems found in birds. and dead-end air-filled sac systems found in the lungs of mammals. In a cocurrent flow system, the blood and gas (or the fluid containing the gas) move in the same direction through the gas exchanger. This means the magnitude of the gradient is variable along the length of the gas-exchange surface, and the exchange will eventually stop when an equilibrium has been reached (see upper diagram in Fig. 2). Cocurrent flow gas exchange systems are not known to be used in nature. Mammals [edit] The gas exchanger in mammals is internalized to form lungs, as it is in most of the larger land animals.[citation needed] Gas exchange occurs in microscopic dead-end air-filled sacs called alveoli, where a very thin membrane (called the blood-air barrier) separates the blood in the alveolar capillaries (in the walls of the alveoli) from the alveolar air in the sacs. Exchange membrane [edit] The membrane across which gas exchange takes place in the alveoli (i.e. the blood-air barrier) is extremely thin (in humans, on average, 2.2 μm thick). It consists of the alveolar epithelial cells, their basement membranes and the endothelial cells of the pulmonary capillaries (Fig. 4). The large surface area of the membrane comes from the folding of the membrane into about 300 million alveoli, with diameters of approximately 75–300 μm each. This provides an extremely large surface area (approximately 145 m2) across which gas exchange can occur. Alveolar air [edit] Air is brought to the alveoli in small doses (called the tidal volume), by breathing in (inhalation) and out (exhalation) through the respiratory airways, a set of relatively narrow and moderately long tubes which start at the nose or mouth and end in the alveoli of the lungs in the chest. Air moves in and out through the same set of tubes, in which the flow is in one direction during inhalation, and in the opposite direction during exhalation. During each inhalation, at rest, approximately 500 ml of fresh air flows in through the nose. It is warmed and moistened as it flows through the nose and pharynx. By the time it reaches the trachea the inhaled air's temperature is 37 °C and it is saturated with water vapor. On arrival in the alveoli it is diluted and thoroughly mixed with the approximately 2.5–3.0 liters of air that remained in the alveoli after the last exhalation. This relatively large volume of air that is semi-permanently present in the alveoli throughout the breathing cycle is known as the functional residual capacity (FRC). At the beginning of inhalation the airways are filled with unchanged alveolar air, left over from the last exhalation. This is the dead space volume, which is usually about 150 ml. It is the first air to re-enter the alveoli during inhalation. Only after the dead space air has returned to the alveoli does the remainder of the tidal volume (500 ml - 150 ml = 350 ml) enter the alveoli. The entry of such a small volume of fresh air with each inhalation, ensures that the composition of the FRC hardly changes during the breathing cycle (Fig. 5). The alveolar partial pressure of oxygen remains very close to 13–14 kPa (100 mmHg), and the partial pressure of carbon dioxide varies minimally around 5.3 kPa (40 mmHg) throughout the breathing cycle (of inhalation and exhalation). The corresponding partial pressures of oxygen and carbon dioxide in the ambient (dry) air at sea level are 21 kPa (160 mmHg) and 0.04 kPa (0.3 mmHg) respectively. This alveolar air, which constitutes the FRC, completely surrounds the blood in the alveolar capillaries (Fig. 6). Gas exchange in mammals occurs between this alveolar air (which differs significantly from fresh air) and the blood in the alveolar capillaries. The gases on either side of the gas exchange membrane equilibrate by simple diffusion. This ensures that the partial pressures of oxygen and carbon dioxide in the blood leaving the alveolar capillaries, and ultimately circulates throughout the body, are the same as those in the FRC. The marked difference between the composition of the alveolar air and that of the ambient air can be maintained because the functional residual capacity is contained in dead-end sacs connected to the outside air by long, narrow, tubes (the airways: nose, pharynx, larynx, trachea, bronchi and their branches and sub-branches down to the bronchioles). This anatomy, and the fact that the lungs are not emptied and re-inflated with each breath, provides mammals with a "portable atmosphere", whose composition differs significantly from the present-day ambient air. The composition of the air in the FRC is carefully monitored, by measuring the partial pressures of oxygen and carbon dioxide in the arterial blood. If either gas pressure deviates from normal, reflexes are elicited that change the rate and depth of breathing in such a way that normality is restored within seconds or minutes. Pulmonary circulation [edit] Main article: Pulmonary circulation All the blood returning from the body tissues to the right side of the heart flows through the alveolar capillaries before being pumped around the body again. On its passage through the lungs the blood comes into close contact with the alveolar air, separated from it by a very thin diffusion membrane which is only, on average, about 2 μm thick. The gas pressures in the blood will therefore rapidly equilibrate with those in the alveoli, ensuring that the arterial blood that circulates to all the tissues throughout the body has an oxygen tension of 13−14 kPa (100 mmHg), and a carbon dioxide tension of 5.3 kPa (40 mmHg). These arterial partial pressures of oxygen and carbon dioxide are homeostatically controlled. A rise in the arterial , and, to a lesser extent, a fall in the arterial , will reflexly cause deeper and faster breathing until the blood gas tensions return to normal. The converse happens when the carbon dioxide tension falls, or, again to a lesser extent, the oxygen tension rises: the rate and depth of breathing are reduced until blood gas normality is restored. Since the blood arriving in the alveolar capillaries has a of, on average, 6 kPa (45 mmHg), while the pressure in the alveolar air is 13 kPa (100 mmHg), there will be a net diffusion of oxygen into the capillary blood, changing the composition of the 3 liters of alveolar air slightly. Similarly, since the blood arriving in the alveolar capillaries has a of also about 6 kPa (45 mmHg), whereas that of the alveolar air is 5.3 kPa (40 mmHg), there is a net movement of carbon dioxide out of the capillaries into the alveoli. The changes brought about by these net flows of individual gases into and out of the functional residual capacity necessitate the replacement of about 15% of the alveolar air with ambient air every 5 seconds or so. This is very tightly controlled by the continuous monitoring of the arterial blood gas tensions (which accurately reflect partial pressures of the respiratory gases in the alveolar air) by the aortic bodies, the carotid bodies, and the blood gas and pH sensor on the anterior surface of the medulla oblongata in the brain. There are also oxygen and carbon dioxide sensors in the lungs, but they primarily determine the diameters of the bronchioles and pulmonary capillaries, and are therefore responsible for directing the flow of air and blood to different parts of the lungs. It is only as a result of accurately maintaining the composition of the 3 liters alveolar air that with each breath some carbon dioxide is discharged into the atmosphere and some oxygen is taken up from the outside air. If more carbon dioxide than usual has been lost by a short period of hyperventilation, respiration will be slowed down or halted until the alveolar has returned to 5.3 kPa (40 mmHg). It is therefore strictly speaking untrue that the primary function of the respiratory system is to rid the body of carbon dioxide "waste". In fact the total concentration of carbon dioxide in arterial blood is about 26 mM (or 58 ml per 100 ml), compared to the concentration of oxygen in saturated arterial blood of about 9 mM (or 20 ml per 100 ml blood). This large concentration of carbon dioxide plays a pivotal role in the determination and maintenance of the pH of the extracellular fluids. The carbon dioxide that is breathed out with each breath could probably be more correctly be seen as a byproduct of the body's extracellular fluid carbon dioxide and pH homeostats If these homeostats are compromised, then a respiratory acidosis, or a respiratory alkalosis will occur. In the long run these can be compensated by renal adjustments to the H+ and HCO3− concentrations in the plasma; but since this takes time, the hyperventilation syndrome can, for instance, occur when agitation or anxiety cause a person to breathe fast and deeply thus blowing off too much CO2 from the blood into the outside air, precipitating a set of distressing symptoms which result from an excessively high pH of the extracellular fluids. Oxygen has a very low solubility in water, and is therefore carried in the blood loosely combined with hemoglobin. The oxygen is held on the hemoglobin by four ferrous iron-containing heme groups per hemoglobin molecule. When all the heme groups carry one O2 molecule each the blood is said to be "saturated" with oxygen, and no further increase in the partial pressure of oxygen will meaningfully increase the oxygen concentration of the blood. Most of the carbon dioxide in the blood is carried as HCO3− ions in the plasma. However the conversion of dissolved CO2 into HCO3− (through the addition of water) is too slow for the rate at which the blood circulates through the tissues on the one hand, and alveolar capillaries on the other. The reaction is therefore catalyzed by carbonic anhydrase, an enzyme inside the red blood cells. The reaction can go in either direction depending on the prevailing partial pressure of carbon dioxide. A small amount of carbon dioxide is carried on the protein portion of the hemoglobin molecules as carbamino groups. The total concentration of carbon dioxide (in the form of bicarbonate ions, dissolved CO2, and carbamino groups) in arterial blood (i.e. after it has equilibrated with the alveolar air) is about 26 mM (or 58 ml/100 ml), compared to the concentration of oxygen in saturated arterial blood of about 9 mM (or 20 ml/100 ml blood). Other vertebrates [edit] Fish [edit] The dissolved oxygen content in fresh water is approximately 8–10 milliliters per liter compared to that of air which is 210 milliliters per liter. Water is 800 times more dense than air and 100 times more viscous. Therefore, oxygen has a diffusion rate in air 10,000 times greater than in water. The use of sac-like lungs to remove oxygen from water would therefore not be efficient enough to sustain life. Rather than using lungs, gaseous exchange takes place across the surface of highly vascularized gills. Gills are specialised organs containing filaments, which further divide into lamellae. The lamellae contain capillaries that provide a large surface area and short diffusion distances, as their walls are extremely thin. Gill rakers are found within the exchange system in order to filter out food, and keep the gills clean. Gills use a countercurrent flow system that increases the efficiency of oxygen-uptake (and waste gas loss). Oxygenated water is drawn in through the mouth and passes over the gills in one direction while blood flows through the lamellae in the opposite direction. This countercurrent maintains steep concentration gradients along the entire length of each capillary (see the diagram in the "Interaction with circulatory systems" section above). Oxygen is able to continually diffuse down its gradient into the blood, and the carbon dioxide down its gradient into the water. The deoxygenated water will eventually pass out through the operculum (gill cover). Although countercurrent exchange systems theoretically allow an almost complete transfer of a respiratory gas from one side of the exchanger to the other, in fish less than 80% of the oxygen in the water flowing over the gills is generally transferred to the blood. Amphibians [edit] Amphibians have three main organs involved in gas exchange: the lungs, the skin, and the gills, which can be used singly or in a variety of different combinations. The relative importance of these structures differs according to the age, the environment and species of the amphibian. The skin of amphibians and their larvae are highly vascularised, leading to relatively efficient gas exchange when the skin is moist. The larvae of amphibians, such as the pre-metamorphosis tadpole stage of frogs, also have external gills. The gills are absorbed into the body during metamorphosis, after which the lungs will then take over. The lungs are usually simpler than in the other land vertebrates, with few internal septa and larger alveoli; however, toads, which spend more time on land, have a larger alveolar surface with more developed lungs. To increase the rate of gas exchange by diffusion, amphibians maintain the concentration gradient across the respiratory surface using a process called buccal pumping. The lower floor of the mouth is moved in a "pumping" manner, which can be observed by the naked eye. Reptiles [edit] All reptiles breathe using lungs. In squamates (the lizards and snakes) ventilation is driven by the axial musculature, but this musculature is also used during movement, so some squamates rely on buccal pumping to maintain gas exchange efficiency. Due to the rigidity of turtle and tortoise shells, significant expansion and contraction of the chest is difficult. Turtles and tortoises depend on muscle layers attached to their shells, which wrap around their lungs to fill and empty them. Some aquatic turtles can also pump water into a highly vascularised mouth or cloaca to achieve gas-exchange. Crocodiles have a structure similar to the mammalian diaphragm - the diaphragmaticus - but this muscle helps create a unidirectional flow of air through the lungs rather than a tidal flow: this is more similar to the air-flow seen in birds than that seen in mammals. During inhalation, the diaphragmaticus pulls the liver back, inflating the lungs into the space this creates. Air flows into the lungs from the bronchus during inhalation, but during exhalation, air flows out of the lungs into the bronchus by a different route: this one-way movement of gas is achieved by aerodynamic valves in the airways. Birds [edit] Main article: Bird anatomy § Respiratory system Birds have lungs but no diaphragm. They rely mostly on air sacs for ventilation. These air sacs do not play a direct role in gas exchange, but help to move air unidirectionally across the gas exchange surfaces in the lungs. During inhalation, fresh air is taken from the trachea down into the posterior air sacs and into the parabronchi which lead from the posterior air sacs into the lung. The air that enters the lungs joins the air which is already in the lungs, and is drawn forward across the gas exchanger into anterior air sacs. During exhalation, the posterior air sacs force air into the same parabronchi of the lungs, flowing in the same direction as during inhalation, allowing continuous gas exchange irrespective of the breathing cycle. Air exiting the lungs during exhalation joins the air being expelled from the anterior air sacs (both consisting of "spent air" that has passed through the gas exchanger) entering the trachea to be exhaled (Fig. 10). Selective bronchoconstriction at the various bronchial branch points ensures that the air does not ebb and flow through the bronchi during inhalation and exhalation, as it does in mammals, but follows the paths described above. The unidirectional airflow through the parabronchi exchanges respiratory gases with a crosscurrent blood flow (Fig. 9). The partial pressure of O2 () in the parabronchioles declines along their length as O2 diffuses into the blood. The capillaries leaving the exchanger near the entrance of airflow take up more O2 than capillaries leaving near the exit end of the parabronchi. When the contents of all capillaries mix, the final of the mixed pulmonary venous blood is higher than that of the exhaled air, but lower than that of the inhaled air. Plants [edit] Gas exchange in plants is dominated by the roles of carbon dioxide, oxygen and water vapor. CO2 is the only carbon source for autotrophic growth by photosynthesis, and when a plant is actively photosynthesising in the light, it will be taking up carbon dioxide, and losing water vapor and oxygen. At night, plants respire, and gas exchange partly reverses: water vapor is still lost (but to a smaller extent), but oxygen is now taken up and carbon dioxide released. Plant gas exchange occurs mostly through the leaves. Gas exchange between a leaf and the atmosphere occurs simultaneously through two pathways: 1) epidermal cells and cuticular waxes (usually referred as 'cuticle') which are always present at each leaf surface, and 2) stomata, which typically control the majority of the exchange. Gases enter into the photosynthetic tissue of the leaf through dissolution onto the moist surface of the palisade and spongy mesophyll cells. The spongy mesophyll cells are loosely packed, allowing for an increased surface area, and consequently an increased rate of gas-exchange. Uptake of carbon dioxide necessarily results in some loss of water vapor, because both molecules enter and leave by the same stomata, so plants experience a gas exchange dilemma: gaining enough CO2 without losing too much water. Therefore, water loss from other parts of the leaf is minimised by the waxy cuticle on the leaf's epidermis. The size of a stoma is regulated by the opening and closing of its two guard cells: the turgidity of these cells determines the state of the stomatal opening, and this itself is regulated by water stress. Plants showing crassulacean acid metabolism are drought-tolerant xerophytes and perform almost all their gas-exchange at night, because it is only during the night that these plants open their stomata. By opening the stomata only at night, the water vapor loss associated with carbon dioxide uptake is minimised. However, this comes at the cost of slow growth: the plant has to store the carbon dioxide in the form of malic acid for use during the day, and it cannot store unlimited amounts. Gas exchange measurements are important tools in plant science: this typically involves sealing the plant (or part of a plant) in a chamber and measuring changes in the concentration of carbon dioxide and water vapour with an infrared gas analyzer. If the environmental conditions (humidity, CO2 concentration, light and temperature) are fully controlled, the measurements of CO2 uptake and water release reveal important information about the CO2 assimilation and transpiration rates. The intercellular CO2 concentration reveals important information about the photosynthetic condition of the plants. Simpler methods can be used in specific circumstances: hydrogencarbonate indicator can be used to monitor the consumption of CO2 in a solution containing a single plant leaf at different levels of light intensity, and oxygen generation by the pondweed Elodea can be measured by simply collecting the gas in a submerged test-tube containing a small piece of the plant. Invertebrates [edit] The mechanism of gas exchange in invertebrates depends their size, feeding strategy, and habitat (aquatic or terrestrial). The sponges (Porifera) are sessile creatures, meaning they are unable to move on their own and normally remain attached to their substrate. They obtain nutrients through the flow of water across their cells, and they exchange gases by simple diffusion across their cell membranes. Pores called ostia draw water into the sponge and the water is subsequently circulated through the sponge by cells called choanocytes which have hair-like structures that move the water through the sponge. The cnidarians include corals, sea anemones, jellyfish and hydras. These animals are always found in aquatic environments, ranging from fresh water to salt water. They do not have any dedicated respiratory organs; instead, every cell in their body can absorb oxygen from the surrounding water, and release waste gases to it. One key disadvantage of this feature is that cnidarians can die in environments where water is stagnant, as they deplete the water of its oxygen supply. Corals often form symbiosis with other organisms, particularly photosynthetic dinoflagellates. In this symbiosis, the coral provides shelter and the other organism provides nutrients to the coral, including oxygen.[citation needed] The roundworms (Nematoda), flatworms (Platyhelminthes), and many other small invertebrate animals living in aquatic or otherwise wet habitats do not have a dedicated gas-exchange surface or circulatory system. They instead rely on diffusion of CO2 and O2 directly across their cuticle. The cuticle is the semi-permeable outermost layer of their bodies.[citation needed] Other aquatic invertebrates such as most molluscs (Mollusca) and larger crustaceans (Crustacea) such as lobsters, have gills analogous to those of fish, which operate in a similar way. Unlike the invertebrates groups mentioned so far, insects are usually terrestrial, and exchange gases across a moist surface in direct contact with the atmosphere, rather than in contact with surrounding water. The insect's exoskeleton is impermeable to gases, including water vapor, so they have a more specialised gas exchange system, requiring gases to be directly transported to the tissues via a complex network of tubes. This respiratory system is separated from their circulatory system. Gases enter and leave the body through openings called spiracles, located laterally along the thorax and abdomen. Similar to plants, insects are able to control the opening and closing of these spiracles, but instead of relying on turgor pressure, they rely on muscle contractions. These contractions result in an insect's abdomen being pumped in and out. The spiracles are connected to tubes called tracheae, which branch repeatedly and ramify into the insect's body. These branches terminate in specialised tracheole cells which provides a thin, moist surface for efficient gas exchange, directly with cells. The other main group of terrestrial arthropod, the arachnids (spiders, scorpion, mites, and their relatives) typically perform gas exchange with a book lung. Summary of main gas exchange systems [edit] | Surface area | Diffusion distance | Maintaining concentration gradient | Respiratory organs | --- --- | | Human | Total alveoli = 70–100 m2 | Alveolus and capillary (two cells) | Constant blood flow in capillaries; breathing | Lungs | | Fish | Many lamellae and filaments per gill | Usually one cell | Countercurrent flow | Gills | | Insects | Specialised tracheole cell | One cell | Buccal pumping | Spiracles | | Sponges | Ostia pores | One cell | Water movement | None | | Flatworms | Flat body shape | Usually one cell | Countercurrent flow | None | | Cnidarians | Oral arms | Usually one cell | Water movement | None | | Reptiles | Many lamellae and filaments per gill[clarification needed] | Alveolus and capillary (two cells) | Countercurrent flow | Lungs | | Amphibians | Many lamellae and filaments per gill | Alveolus and capillary (two cells) or one cell | Countercurrent flow | Lungs, skin and gills | | Plants | High density of stomata; air spaces within leaf | One cell | Constant air flow | Stomata | See also [edit] Respiratory system – Biological system in animals and plants for gas exchange References [edit] ^ Carlson, C. 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The trigeminal nerve. Part II: the ophthalmic division - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Report format: Send at most: [x] Send even when there aren't any new results Optional text in email: Save Cancel Create a file for external citation management software Create file Cancel Your RSS Feed Name of RSS Feed: Number of items displayed: Create RSS Cancel RSS Link Copy Full text links Atypon Full text links Actions Cite Collections Add to Collections Create a new collection Add to an existing collection Name your collection: Name must be less than 100 characters Choose a collection: Unable to load your collection due to an error Please try again Add Cancel Permalink Permalink Copy Display options Display options Format Page navigation Title & authors Abstract Similar articles Cited by MeSH terms LinkOut - more resources Cranio Actions Search in PubMed Search in NLM Catalog Add to Search . 2001 Jan;19(1):8-12. doi: 10.1080/08869634.2001.11746145. The trigeminal nerve. Part II: the ophthalmic division W E Shankland1 Affiliations Expand Affiliation 1 American Academy of Head, Neck, and Facial Pain, USA. drwes@drshankland.com PMID: 11842844 DOI: 10.1080/08869634.2001.11746145 Item in Clipboard The trigeminal nerve. Part II: the ophthalmic division W E Shankland. Cranio.2001 Jan. Show details Display options Display options Format Cranio Actions Search in PubMed Search in NLM Catalog Add to Search . 2001 Jan;19(1):8-12. doi: 10.1080/08869634.2001.11746145. Author W E Shankland1 Affiliation 1 American Academy of Head, Neck, and Facial Pain, USA. drwes@drshankland.com PMID: 11842844 DOI: 10.1080/08869634.2001.11746145 Item in Clipboard Full text links Cite Display options Display options Format Abstract The ophthalmic, or first division (V1) of the trigeminal nerve, is the smallest of the three divisions and is purely sensory or afferent in function. It supplies sensory branches to the ciliary body, the cornea, and the iris; to the lacrimal gland and conjunctiva; to portions of the mucous membrane of the nasal cavity, sphenoidal sinus, and frontal sinus; to the skin of the eyebrow, eyelids, forehead, and nose; and to the tentorium cerebelli, dura mater, and the posterior area of the falx cerebri. At first glance, one might not expect one interested in the diagnosis and treatment of orofacial pain and temporomandibular joint disorders to have a need to be concerned with the ophthalmic division. Although much of this division's influence is dedicated to structures within the orbit, nose, and cranium, still, the ophthalmic division may be afflicted with a lesion or structural disorder which can cause all sorts of orofacial pain. Ignorance of this or any portion of the trigeminal nerve will lead to diagnostic and therapeutic failures. In this, the second of four (4) articles concerning the trigeminal nerve, the first division of this vast cranial nerve will be described in detail. PubMed Disclaimer Similar articles Trigeminal Neuropathy.Veerapaneni KD, Kapoor N, Veerapaneni P, Lui F, Nalleballe K.Veerapaneni KD, et al.2024 Mar 1. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan–.2024 Mar 1. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan–.PMID: 32310586 Free Books & Documents. The trigeminal nerve. Part III: The maxillary division.Shankland WE 2nd.Shankland WE 2nd.Cranio. 2001 Apr;19(2):78-83. doi: 10.1080/08869634.2001.11746155.Cranio. 2001.PMID: 11842868 Human orbital sympathetic nerve pathways.Thakker MM, Huang J, Possin DE, Ahmadi AJ, Mudumbai R, Orcutt JC, Tarbet KJ, Sires BS.Thakker MM, et al.Ophthalmic Plast Reconstr Surg. 2008 Sep-Oct;24(5):360-6. doi: 10.1097/IOP.0b013e3181837a11.Ophthalmic Plast Reconstr Surg. 2008.PMID: 18806655 The trigeminal nerve. Part IV: the mandibular division.Shankland WE 2nd.Shankland WE 2nd.Cranio. 2001 Jul;19(3):153-61. doi: 10.1080/08869634.2001.11746164.Cranio. 2001.PMID: 11482826 Review. The trigeminal nerve. Part I: An over-view.Shankland WE 2nd.Shankland WE 2nd.Cranio. 2000 Oct;18(4):238-48. doi: 10.1080/08869634.2000.11746137.Cranio. 2000.PMID: 11202843 Review. See all similar articles Cited by Lacrimal Neuralgia: A Case Report and Comprehensive Review of the Literature.Lin W, Wu JW, Stern JI, Robertson CE, Chiang CC.Lin W, et al.Curr Pain Headache Rep. 2024 Sep;28(9):929-939. doi: 10.1007/s11916-024-01250-6. Epub 2024 Apr 27.Curr Pain Headache Rep. 2024.PMID: 38676822 Review. Murine nasal-associated lymphoid tissue (NALT) harbors human alphaherpesvirus 1 (HSV-1) DNA during latency, and dexamethasone triggers viral replication.Harrison KS, Cowan SR, Jones C.Harrison KS, et al.J Virol. 2025 Apr 15;99(4):e0225124. doi: 10.1128/jvi.02251-24. Epub 2025 Mar 26.J Virol. 2025.PMID: 40135894 Free PMC article. Effect of carbon dioxide on calcitonin gene-related peptide secretion from trigeminal neurons.Vause C, Bowen E, Spierings E, Durham P.Vause C, et al.Headache. 2007 Nov-Dec;47(10):1385-97. doi: 10.1111/j.1526-4610.2007.00850.x.Headache. 2007.PMID: 18052948 Free PMC article. Connections between the lacrimal gland and sensory trigeminal neurons: a WGA/HRP study in the cynomolgous monkey.Baljet B, VanderWerf F.Baljet B, et al.J Anat. 2005 Mar;206(3):257-63. doi: 10.1111/j.1469-7580.2005.00374.x.J Anat. 2005.PMID: 15733298 Free PMC article. The neuro-pathophysiology of temporomandibular disorders-related pain: a systematic review of structural and functional MRI studies.Yin Y, He S, Xu J, You W, Li Q, Long J, Luo L, Kemp GJ, Sweeney JA, Li F, Chen S, Gong Q.Yin Y, et al.J Headache Pain. 2020 Jun 19;21(1):78. doi: 10.1186/s10194-020-01131-4.J Headache Pain. 2020.PMID: 32560622 Free PMC article. See all "Cited by" articles MeSH terms Ciliary Body / innervation Actions Search in PubMed Search in MeSH Add to Search Conjunctiva / innervation Actions Search in PubMed Search in MeSH Add to Search Cornea / innervation Actions Search in PubMed Search in MeSH Add to Search Cranial Nerve Diseases / complications Actions Search in PubMed Search in MeSH Add to Search Dura Mater / anatomy & histology Actions Search in PubMed Search in MeSH Add to Search Eyebrows / innervation Actions Search in PubMed Search in MeSH Add to Search Eyelids / innervation Actions Search in PubMed Search in MeSH Add to Search Facial Pain / etiology Actions Search in PubMed Search in MeSH Add to Search Forehead / innervation Actions Search in PubMed Search in MeSH Add to Search Frontal Sinus / innervation Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Iris / innervation Actions Search in PubMed Search in MeSH Add to Search Lacrimal Apparatus / innervation Actions Search in PubMed Search in MeSH Add to Search Mucous Membrane / innervation Actions Search in PubMed Search in MeSH Add to Search Nasal Cavity / innervation Actions Search in PubMed Search in MeSH Add to Search Neurons, Afferent / physiology Actions Search in PubMed Search in MeSH Add to Search Nose / innervation Actions Search in PubMed Search in MeSH Add to Search Ophthalmic Nerve / anatomy & histology Actions Search in PubMed Search in MeSH Add to Search Ophthalmic Nerve / physiology Actions Search in PubMed Search in MeSH Add to Search Sphenoid Sinus / innervation Actions Search in PubMed Search in MeSH Add to Search Trigeminal Nerve / anatomy & histology Actions Search in PubMed Search in MeSH Add to Search LinkOut - more resources Full Text Sources Atypon Full text links[x] Atypon [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov
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https://openstax.org/books/contemporary-mathematics/pages/6-key-concepts
Skip to Content Go to accessibility page Keyboard shortcuts menu Log in Contemporary Mathematics Key Concepts Contemporary Mathematics Key Concepts Search for key terms or text. Key Concepts ## 6.1 Understanding Percent Know what a percent is as a fraction, a decimal, and as a part of the whole. Use the percent equation to find any of the three values that are related by the equation. Apply the percent equation in applications. ## 6.2 Discounts, Markups, and Sales Tax Discounts are markdowns from an original price. Mark-ups are increases to the price paid by a retailer to cover their costs. be able to calculate the markup based on a percentage of the cost Sales taxes vary from state to state and often county to county. Retail prices, sales prices and percent discounts can be calculated if the other two values are known. Original costs, retail prices, and percent markup can be calculated if the other two values are known. In calculations, sales tax acts like a markup. ## 6.3 Simple Interest Interest is money that is paid by a borrower for the privilege of borrowing the money. Simple interest is computed by substituting the principal, interest rate, and number of years into the formula I=P×r×t The payoff for a loan is the amount of principal remaining on a loan plus the interest that accumulated on the loan since the last payment. The future value of an investment yielding simple interest is the original principal plus the interest earned on the investment. When making a partial payment, some of the payment pays off all the accumulated interest, while the remainder of the payment is applied to the principal of the loan. Finding the present value of an investment is used to determine how much should be invested now in order to achieve a specific goal. ## 6.4 Compound Interest Compound interest means that the interest earned during one period will earn interest in later periods. Essentially, the amount of the principal grows from period to period. The important values in computing compound interest are the interest rate, the principal, the length of time the investment, and the number of times the investment is compounded. Compound interest has minimal impact early, but later has a very large impact. You can determine how much to invest today in order to reach a goal for some time later. Compound interest can be translated into an effective annual yield, which allows for comparison between investment options. ## 6.5 Making a Personal Budget A budget is a set of guidelines for how to allocate your income. Budgeting helps to plan for many of life’s expenses Budgets are used to compare income to expenses. When expenses exceed income, changes have to be made. Budgets can help evaluate the affordability of life changes. One guideline for setting a budget is the 50-30-20 budget philosophy. The guidelines suggest that 50% of income is allocated to necessary expenses, 30% to expenses that wants, and 20% to savings and other debt reduction. ## 6.6 Methods of Savings There are three main types of savings accounts, saving accounts, certificates of deposit (CD), and money market accounts. Savings account are very risk free, and so yield low interest rates. The differences in the three types of savings accounts relate to their convenience. Savings account typically have a lower interest rate that money market accounts, which typically have lower interest rates than CDs. Ordinary annuities more accurately reflect how we save, in that money is deposited repeatedly over time. Spreadsheet software, such as Google Sheets, have built in functions that can be used to quickly calculate both the future value of an ordinary annuity account, but also the payment necessary to reach a goal using an ordinary annuity. ## 6.7 Investments There are many different investments with different returns and risks. Bonds are loans form the purchaser to the entity selling the bond. Bonds have some tax benefits, low to no risk, and a low return. Stocks represent part ownership in a company. As such, stock holders share in the profits, and losses, of the company. Information, including price, P/E, yearly highs and lows, and dividend amount can be found in online stock tables available on many websites. Mutual funds represent collections of professionally administered investment vehicles. Have shares in a mutual fund has lower risk than ownership of stocks. Retirement accounts employ some of the same strategies as mutual funds, in that they spread the risk and are professionally managed. IRAs and Roth IRAs differ on when taxes are paid on the money, and who can use them. Roth IRAs have income limits while traditional IRAs do not. ## 6.8 The Basics of Loans There are many reasons for a loan, but primarily it is taken out for a large expense when cash is not available. Each payment for an installment loan consists of an interest portion and a principal portion. There is a formula to calculate the payment necessary to pay off a loan in installments. Amortization schedules, or tables, show how each payment is applied to principal and interest. It also includes other details such as remaining balance and total interest paid. Loans often have other fees associated with them such as origination fees or application fees. The total of the interest paid and the fees is the cost of finance. ## 6.9 Understanding Student Loans The FAFSA must be filled out each year that a student wishes to borrow for. A student’s funding gap determines how much they need in loans to pay for college. Federal subsidized student loans defer payments until after graduation and interest does not accrue on these loans. Unsubsidized student loans defer payment until after graduation but interest begins accruing as soon as the loan finds are disbursed. There are both yearly and aggregate limits for student loans to prevent over-borrowing, among other reasons. Federal direct loans have a low interest rate set by the government, but other student loans have varying rates of interest set by the banks. The standard repayment plan lasts 10 years and is made up of monthly payments. Consolidating or refinancing student loans merges many student loans into one loan. If only federal loans are consolidated, the interest rate is the same as the individual loans, currently set at 4.99%. If other loans are refinanced together, the interest rate may be lower with the new loan. Other repayment plans are available. Such a plan may have payment that start small and grow as the loan is paid off, or it may have a longer term, or may be based on the discretionary income of the student. Being delinquent on a student loan is a precursor to being in default. Making payments in a timely fashion allows the student to avoid this situation. ## 6.10 Credit Cards Credit cards can be a flexible way to pay for almost anything, but can become a financial hazard if used unwisely. When deciding which credit card to apply for, evaluate the interest rate, fees (annual and late), reward programs and credit limit. Be sure they meet your criteria. Paying off the balance of your credit card every month will control your spending and will never result in paying interest. Credit card statements hold all important information about your credit card, including payment, balances, charges and billing cycle dates. Although the minimum payment is attractive precisely because it is so small, paying only the minimum results is a long payoff term and higher interest costs. ## 6.11 Buying or Leasing a Car There are many factors to consider when choosing to buy or lease a car. The cost of the car is increased by a number of fees and sales tax. There are advantages to buying a car and advantages to leasing a car. The decision between the two depends on the preference of the buyer. Insurance covers costs associate with accidents. It is made up of various components. The costs of owning a car, including insurance and maintenance, should be a part of the budgeting process. Budgeting for unexpected repairs can ease the stress of encountering large repair bill. ## 6.12 Renting and Homeownership There are many points of comparison between renting and buying a house. Before deciding to buy a house, you should carefully consider all the responsibilities that come with home ownership. Renting comes with more restrictions on the renter, but with fewer costs and is easier to move from. Owning a house has more costs but has more freedom, plus the owner creates equity. Mortgages are loans, and payments are calculated in the same way as any other loan. Amortization tables help a homeowner understand the mortgage and how the payments are applied to the principal and interest. In addition to paying the amount financed for a mortgage, the monthly payment will include an escrow payment, which covers insurance and taxes. ## 6.13 Income Tax Federal income tax is based on income after certain adjustments. Gross income is income from all sources, including gifts and winnings. Before taxes are calculated, the taxable income is found by subtracting deductions and exemptions from gross income. Income tax is progressive, increasing in rate as income increases. Being in the 32% tax bracket means some of your income is taxed at 10%, some at 12%, some at 22%, some at 24%, and the rest at 32%. Income in each tax bracket is taxed at that bracket’s rate, which means in 2022 the first $10,275 earned is taxed at 10% only. Tax credits are subtracted from the taxes that are owed. Some tax credits are refundable, which means they can make the amount you owe negative, which results in a refund. Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Donna Kirk Publisher/website: OpenStax Book title: Contemporary Mathematics Publication date: Mar 22, 2023 Location: Houston, Texas Book URL: Section URL: © Oct 8, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
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https://www.oracle.com/cn/erp/cfo/revenue-forecasting/
收入预测详解 | Oracle 中国 跳到内容 单击查看我们的辅助功能政策 产品 行业 资源 客户 合作伙伴 开发人员 关于公司 关闭搜索 搜索 Oracle.com 快速链接 Oracle Cloud Infrastructure Oracle Fusion Cloud Applications Oracle Database 下载 Java Oracle 职业机会 搜索 Country Close Would you like to visit an Oracle country site closer to you? 您想访问附近国家/地区的 Oracle 网站吗？ Visit Oracle Israel 不，我要留在这里 See this page for a different country/region 查看帐户 返回 云帐户登录云注册云免费套餐 Oracle 帐户 登录 创建帐户 帮助 注销 联系销售 菜单 甲骨文中国 应用 ERP 首席财务官 收入预测详解 Alex Chan | 内容策略师 | 2024 年 5 月 10 日 本文目录 什么是收入预测？ 收入预测的定义 收入预测的重要性 预测收入的方法 八个预测方法 您应避免的六个收入预测错误 Oracle 解决方案助您预测并增加收入 收入预测的常见问题解答 企业可通过收入预测，为整个企业的财务和运营计划奠定基础。比如，今年需要哪些方面的生产力，是否需要新的资本投资？要花多少钱在广告和需求挖掘上？需要多少个销售人员，安排他们负责多少定额？收入预测将如何影响企业的股票价格和筹集资本的能力？企业想要构建准确的收入预测，就需要了解所拥有的产品和执行能力，以及材料或劳动力、竞争市场、宏观经济等限制。收入预测远不止是设定一个目标数字。预测流程应该帮助您评估业务状态、机会和风险，并制定成功的策略。 什么是收入预测？ 收入预测是用于预估企业在特定期间（例如每月、每季度或每年）销售产品或服务的收入所得的过程。您可以通过了解您的业务状况、之前的表现以及企业以外的因素，做出有根据性的假设来预测未来的总销售额。 收入预测评估的是整个业务，而不仅仅是销售目标、营销工作或收入运营团队执行的其它活动。收入预测还必须考虑企业的竞争格局、生产力和人员配置以及经济趋势。收入预测是制定企业预算时重要的初始预测。通过将收入预测与费用和投资预估相结合，企业可以创建利润和现金流预测。收入预测是一种定量分析，帮助企业做出其它各种数据驱动的决策，包括投多少广告和雇用多少人。 关键要点 收入预测是预测您的企业可从销售产品和服务中获得多少收入。收入预测以历史趋势、市场状况和您的业务战略为基础。 有效的收入预测能够为业务带来巨大的影响力，因为预测将会影响企业做出有关投资和支出的决策、投资者对企业的看法甚至是企业吸引人才的能力。 为了预测收入，分析师需要收集与业绩和财务相关的企业数据，充分考虑竞争和经济因素，然后根据业务选择合适的财务建模和预测技术。 收入预测的定义 收入预测尝试预测您的企业在某个时间段（通常为一个月、一个季度或一年）内的销售额。收入预测将会考虑到历史绩效、经济和竞争状况以及企业的业务计划，包括在某个预测期间内的具体产品、销售额和营销策略。自上而下的收入预测方法会从这些战略要素和历史数据开始，然后交由负责执行预测的业务实体进行预测。 此外，企业还可以使用自下而上的预测方法，也就是先预测各个部门的收入目标。例如，首席收入官的组织可以通过提出增加销售或营销活动，抓住新的市场机会，进而改变收入预测。或者，产品开发可以通过将新功能或产品推向市场来改变预测。在构建收入预测后，整个企业的财务和运营团队将使用收入预测进行预算和预测性计划。 收入预测的重要性 收入预测塑造了企业看待未来及其作出的业务决策的方式。这些假设有助于推动收入预测，制定企业的短期和长期目标，在帮助企业为未来做好准备方面发挥了重要作用。未来几个月甚至是一年内的收入前景，决定了新员工、营销活动、设施和设备以及研究和产品开发的预算。财务团队将该收入预测结果用于实现该销售目标所需的损益表中成本方面的所有要素，从而预估利润和现金流。 例如，如果您预测收入增长率为 5%，根据当前竞争市场的历史数据和预测策略来估计，您需要将广告收入增加 8%，并将销售人员数量增加 5%，才能实现预测的收入。您可能计算出，更高的收入会实现更大规模的经济，将销货成本从收入的 30% 减少到 28.5%。现有的财务人员能否在更高的销售级别处理应付账款和应收账款？您是否有足够的营运资本来支付库存和付款浮动？其中的重点是，收入预测可在企业中产生涟漪效应。 对于公开交易的企业，收入预测也经常与华尔街分享。分析师会密切关注企业增长率，将其作为投资决策的其中一个考量点。分析师会将这些收入预测纳入自己的模型中，如果该企业无法实现预期收入目标，特别是在多次失败后，分析师可能会调低该企业的股票级别，建议投资者出售股票。私营公司在寻求借贷、吸引私募股权或风险投资或出售部分或全部业务的时候，也需要提供收入预测。收入预测是衡量一个企业实力的核心问题的一个关键部分。那么，这个机会值得我们投入更多资金吗？ 预测收入的方法 通过了解业务的当前状态、历史绩效和外部因素，您可以对未来的收入进行有根据性的预估。此预测可以帮助您制定数据驱动的策略，并制定改善业务的决策。预测收入涉及一定的准备工作和流程规范，具体收入预测步骤如下： 一、收集准确的财务数据 您可以通过数据，了解企业历史。预测是基于历史数据和更当前的企业数据进行的，因此需要清楚地了解企业过去的结果和当前的财务状况。损益表、资产负债表和现金流量表可为预测奠定基础。如果要收集此类信息，在理想情况下，您可以使用自动跟踪交易、对费用进行分类以及生成财务报表的软件。 二、选择期限 典型的做法是进行年度收入预测；如果再加上一些较小的增量，通常会每季度预测一次。虽然预测未来几年的收入很有帮助，但长期预测的确定性自然就会更低。 三、考虑可能影响增长的内部因素 这应该从您销售的产品和服务开始，包括任何新产品或地理扩展。再纳入生产能力、人员配置、物流等方面的因素。策略也起着重要作用，例如大型营销活动或收购。 四、考虑外部因素 这些因素（也称为“驱动因素”）可能会推动或减缓您的业务增长，其中包括消费者需求、季节性、监管或法律变化、经济状况或重大全球、国家/地区或地方事件。 五、研究限制和风险因素 您的预测对消费者支出或企业投资等因素有多敏感？是否存在供应限制，例如可能限制产能的物料输入、熟练劳动力或运输问题？这些因素会影响预测的概率和可能取得的结果。 六、选择支持预测的软件 这可以是简单的电子表格或复杂的财务预测软件。专用软件有助于整合预测流程，自动执行部分数据收集和分析，并提供对预构建预测模型和方法的访问。了解哪些系统提供了所需的基础财务数据及其具备的功能，对于预测和持续监控和更新非常重要。 七、选择预测方法 根据您的数据、假设和工具，选择适合您的业务模型和假设的预测方法。市面上有多种可选择的预测方法，其中包括时间序列分析、回归分析或财务建模技术。有的方法更适合季节性企业，有的则是为可预测规模的企业而设计的。 八、跟踪预测 通过设置，使仪表盘生成预算差异报告，并按需根据实际收入和不断变化的经济、竞争和其它条件调整预测。 八个预测方法 任何预测方法都有优点和缺点，具体取决于企业的各种变量。您需要根据当前的收入增长模式，为您的企业选择合适的方法。您可以考虑两大主流预测方法：定性法和定量法。 定性预测依赖于专家意见，这可能包括来自您的销售人员、渠道合作伙伴、外部分析和高管的意见。定量预测则是使用数据并从中推断出未来的价值。您很可能会选择采用双管齐下的方法。 以下是一些常用的定量预测方法。 直线预测：这种方法是假设企业继续实现和过去一样的增长率，因此您需要将企业最近一年的收入乘以当前增长率。这意味着，如果您想了解未来一年的预计收入，您可以参考过去两年的收入。假设您在两年前赚了 1000 万美元，在去年赚了 1050 万美元，那么增长率就是 5%。如果要预测明年的收入，您只需将 1050 万美元乘以 1.05，得出结果为 1102.5 万美元。这种方法太简单了，无法完全借鉴。比较适合给过去一直拥有稳定增长率的企业用来进行粗略估算，或者是作为对增长范围和驱动因素的初步考量。 时间序列分析：这种技术使用定期的历史数据点来预测未来的结果。市面上存在许多时间序列预测方法，因此财务团队必须根据季节性和趋势波动等因素，确定哪种方法更适合其行业和使用场景。以下是时间序列预测的一个例子，即加权移动平均值。 加权移动平均预测：此时间序列方法使用数据点的加权平均值，按顺序预测下一个值。这是一种实际的方法，可以用于监视每月收入结果和调整近期预测。例如，您可以根据一月、二月、三月和四月的收入，预测五月的收入。加权移动平均值的公式如下： （一月收入 x 10%）+（二月收入 x 15%）+（三月收入 x 25%）+（四月收入 x 50%）= 五月收入 此方法适合进行短期、近期的预测，也适用于非季节性的企业。 4. 线性回归：此方法涉及使用收入和独立变量之间的关系来做出预测。这是一个数学模型，使用特定的因素来预测未来的收入。或者，该方法可以反向评估有多少给定因素可以推动您的收入。例如，如果您认为广告支出可以推动收入增长，您可以收集有关企业收入和广告支出的过去数据，运行简单的线性回归模型，以确定其中的关系。如果要将该结果转化为收入预测，您只需在模型中输入未来一年的预期广告支出，就可以估算收入。当然，其它因素也会推动收入增长，所以您可能需要建立一个更复杂的回归模型，将更多因素纳入考虑范围。 以下是一些常见的定性预测方法。 咨询您的高管：这种方法需要您组建一个高管小组；对于较小型的企业，您可能只需要创始人一个对象。当然，这听起来很简单，但您的高管可能会带来强有力的专业知识。如果您正在推出一款全新的产品或成立一家初创企业，那么您可能没有太多相关的数据。您无需为了追逐一个未知数字而在预测方面超支。 咨询您的销售人员：通过自下而上的预测方法，向销售人员了解收入潜力。同时，您也可以向经销商或其它渠道合作伙伴获取数据。他们的意见可以很好地帮助您检验自上而下的方法（如询问高管小组）是否符合实际情况。 咨询外部专家：在这种方法中，您需要依赖行业分析师和顾问、行业协会、学术研究人员和其他高度关注市场的人。例如，某家汽车制造商可以利用许多外部观点以及内部的专家观点，预测特定市场可以购买多少辆汽车，从而将结果纳入自己收入预算。 咨询您的客户：在 B2B 工业价值链等领域中，您的潜在买方有限，因此了解买方意图是很有价值的。对于消费品行业，您可能会通过更广泛的消费者信心调研来衡量整体支出意图，然后将结果应用于您的领域，包括食物、旅行、耐用品、家居装修等。 您应避免的六个收入预测错误 企业预测收入的准确程度取决于许多因素，包括输入数据质量和广度、内部信息（例如产品发布和营销工作）的可见性、对外部驱动因素的敏感性以及相关因素的波动性等等。以下是人们在预测收入时会犯的一些常见错误及其预防措施。 认为未来会重现过去：直线预测，或过于依赖历史数据来预测增长趋势的预测方法，可能会削弱外部力量（例如更强大的竞争对手或不断变化的经济情况）或内部限制（例如对产品或人员的投资不足）。 依赖有限或冲突的数据：您需要完整且可信的内部数据来生成精细的预测结果。除了总营业收入数字之外，您还可以深入了解按地域、产品和销售渠道划分的收入来源数据。在深挖的过程中，如果遇到数据点冲突或缺失，则预测将变得更加困难。此外，数据可能会受到季节性或异常情况的影响，例如零售商在短期预测中包含了年度数据，但没有列明假期季节，或制造商的某个需求期间发生了法规变化。 低估了外部条件的变化：人们相信自己的战略和计划，因此他们有理由专注于上升空间。但预测者必须充分权衡外部因素和风险，如竞争、监管、颠覆性技术和经济，以勘破任何看似美好的观点。根据理想情况、糟糕情况和可能发生的情况创建预测，效果更佳。 低估了变量的影响力：收入预测在企业中产生涟漪效应，影响着许多重大决策，例如招聘、采购和扩展。财务预测人员需要了解这些数据会推动哪些决策，并提供适当的背景，帮助员工为变量提前制定计划。执行敏感性分析，了解不同的变量如何影响总收入，并运行场景建模来预估收入波动的影响。较短期（例如一个月）的预测可能非常准确，而较长期（例如一年）的预测可能需要考虑更多的背景信息，并谨慎考虑可能出现的一系列结果。 模型过于复杂：在预测模型中添加太多变量，可能会导致一系列的问题。例如，使用高度相关的变量可能会过分强调某个因素。因此，如果您正在考虑气温、月份、降雪和水上乐园的客流量等变量如何影响冰淇淋的销量，您可能只是在用不同的方式来得出同一个结论：冰淇淋在天气炎热时销量更高。同样地，您构建了一个如此复杂的模型，不仅难以维护，更难以说服利益相关者（如高管、销售人员和投资者）。 过拟合的机器学习模型：当您建立一个模型，过于精确地解释某一组历史数据，以致于无法预测未来时，就会发生过拟合。过拟合的模型可能是由于没有足够的样本数据而导致的，因此您的模型会试图解释每个异常情况。 Oracle 解决方案助您预测并增加收入 企业在获取并利用财务和历史数据来预测未来收入时，这些由数据支持的预测结果将有助于加强和扩展业务。Oracle Fusion Cloud Enterprise Resource Planning (ERP) 和 Oracle Fusion Cloud Enterprise Performance Management (EPM) 提供数据和分析工具，可帮助您创建完善的收入预测流程。该平台为您的财务团队提供数据，以了解财务状况以及收入驱动因素，从而帮助您提高预测准确性并改善风险管理。企业高管可以通过仪表盘访问整个企业的数据，获取可信赖的财务和运营视图以及强大的预测基础。 收入预测的常见问题解答 收入预测如何影响业务决策？ 收入预测是预算流程的核心，为企业对业务的乐观程度奠定了基调。例如，强劲的收入预测将告诉首席收入官他们可以更激进地投资于营销和销售，而首席财务官也可以更有信心地对人力资源或生产力进行投资。 更准确的收入预测可带来哪些优势？ 准确的收入预测可以帮助您的企业做出更明智的业务支出决策，制定正确的长期目标，并赢得潜在投资者的信心。 如何预测收入？ 您可以通过收集有关企业历史绩效和财务的准确信息来预测收入，确定哪些外部和内部因素可能会改变增长率的方向，然后选择更相关的预测方法来预测未来的收入。 改善企业规划的五种方式 阅读真实案例，了解企业如何围绕财务、营销和人员制定规划策略。 获取电子书 注：为免疑义，本网页所用以下术语专指以下含义： 除Oracle隐私政策外，本网站中提及的“Oracle”专指Oracle境外公司而非甲骨文中国。 相关Cloud或云术语均指代Oracle境外公司提供的云技术或其解决方案。 按角色查看 招贤纳士 开发人员 投资者 合作伙伴 研究员 学生和教育工作者 为什么选择 Oracle 分析报告 基于云的优秀 ERP 云经济学 社会影响 文化与包容性 安全实践 学习 什么是主权云？ 什么是零信任安全? AI 如何推进财务转型 什么是向量数据库？ 什么是多云? 什么是 AI agent？ 新闻与活动 新闻 Oracle AI World Oracle Health Summit Oracle Dev Tour 活动 软件产品登记证书 完整使用程序使用通知申请流程 联系我们 IL Sales+972-3-9273700 销售： 400-699-8888 您需要什么帮助？ 订阅电子邮件 举报热线 辅助功能 © 2025 Oracle 使用条款和隐私政策 京ICP备10049020号-1 Cookie Preferences 广告选择 招贤纳士 咨询销售人员联系或致电 销售咨询考虑采购？致电销售人员咨询销售人员Cloud Support 聊天帐户/订阅，促销问题开始聊天Oracle 支持服务选项需要技术支持或其他支持？查看支持选项
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https://math.stackexchange.com/questions/3046182/how-to-prove-periodicity-modulo-b
elementary number theory - How to prove periodicity Modulo 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. 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 How to prove periodicity Modulo b Ask Question Asked 6 years, 9 months ago Modified6 years, 9 months ago Viewed 157 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. We have the sequence x n=a n mod b,x n=a n mod b, where a a and b b are positive integers. How to show that it's periodic? It is intuitively clear but I have no clue how to prove it rigorously from first principles... Thanks in advance elementary-number-theory discrete-mathematics ring-theory discrete-calculus Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Dec 19, 2018 at 8:54 plus1plus1 429 4 4 silver badges 19 19 bronze badges Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. Note first that periodic here means there are K K, P>0 P>0 such that a i=a i+P a i=a i+P for i>K i>K. One begins with showing that there are m>n m>n such that a m=a n a m=a n. Then for all i>n i>n we have a i=a i−n+n=a i−n a n=a i−n a m=a i+(m−n).a i=a i−n+n=a i−n a n=a i−n a m=a i+(m−n). So you can take K=n K=n and P=m−n P=m−n. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Dec 19, 2018 at 9:59 Andreas CarantiAndreas Caranti 71.2k 4 4 gold badges 78 78 silver badges 141 141 bronze badges 0 Add a comment| This answer is useful -1 Save this answer. Show activity on this post. Using the fact that a φ(b)≡1 mod b a φ(b)≡1 mod b and the fact that for any n n n=φ(b)k+r n=φ(b)k+r for some 0≤r<φ(b)0≤r<φ(b) we have the desired result. Here φ(b)φ(b) is the Euler Totient function. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Dec 19, 2018 at 9:03 Jacky ChongJacky Chong 26.3k 2 2 gold badges 23 23 silver badges 42 42 bronze badges 7 1 @plus1 Even simpler, note that a n mod b a n mod b can only take a finite range of values i.e. {0,1,2,…,b−1}{0,1,2,…,b−1} so eventually we have a n=a m mod b a n=a m mod b and so a n−m=1 mod b a n−m=1 mod b and the sequence is periodic with a period that is a factor of n−m n−m.gandalf61 –gandalf61 2018-12-19 09:33:13 +00:00 Commented Dec 19, 2018 at 9:33 Thanks! But isn't there a simpler proof? I am not familiar with the Euler Totient function and it's properties..plus1 –plus1 2018-12-19 09:33:31 +00:00 Commented Dec 19, 2018 at 9:33 @gandalf61 That only shows that a n=a m mod b a n=a m mod b for some m>n. Can you deduce from that that a n+1=a m+1 mod b a n+1=a m+1 mod b and get some kind of induction going ?plus1 –plus1 2018-12-19 09:38:54 +00:00 Commented Dec 19, 2018 at 9:38 1 @gandalf61, take a=2 a=2 and b=6 b=6. How do you go from a 3≡a 1(mod b)a 3≡a 1(mod b) to a 4≡1(mod b)a 4≡1(mod b)?Andreas Caranti –Andreas Caranti 2018-12-19 09:47:55 +00:00 Commented Dec 19, 2018 at 9:47 1 @AndreasCaranti yes, I was too hasty. If a a does not have an inverse mod b mod b then all we can conclude from the simpler argument is that the sequence is eventually periodic i.e. a n=a m,a n+1=a m+1,a n+2=a m+2…a n=a m,a n+1=a m+1,a n+2=a m+2…gandalf61 –gandalf61 2018-12-19 09:56:39 +00:00 Commented Dec 19, 2018 at 9:56 |Show 2 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-number-theory discrete-mathematics ring-theory discrete-calculus 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 1Modulo Theorem proof 2How can we be sure of periodicity by testing some terms? 6Prove existence of multiplicative inverse modulo p when p is prime 1Trees: how can I prove that if two vertices share the same descendant, then one of them must be the descendant of the other? Hot Network Questions Is encrypting the login keyring necessary if you have full disk encryption? Cannot build the font table of Miama via nfssfont.tex Vampires defend Earth from Aliens Calculating the node voltage Are there any legal strategies to modify the nature and rules of the UN Security Council in the face of one, or more SC vetos? 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https://en.wiktionary.org/wiki/controverted
controverted - Wiktionary, the free dictionary Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main Page Community portal Requested entries Recent changes Random entry Help Glossary Contact us Special pages Feedback If you have time, leave us a note. Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donations Preferences Create account Log in [x] Personal tools Donations 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 EnglishToggle English subsection 1.1 Verb 1.1.1 Derived terms controverted [x] 7 languages Deutsch فارسی မြန်မာဘာသာ Suomi தமிழ் తెలుగు Tiếng Việt Entry Discussion Citations [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 Create a book Download as PDF Printable version In other projects From Wiktionary, the free dictionary English [edit] Verb [edit] controverted simple past and pastparticiple of controvert Derived terms [edit] uncontroverted Retrieved from " Categories: English non-lemma forms English verb forms Hidden categories: Pages with entries Pages with 1 entry This page was last edited on 2 January 2025, at 02:04. Definitions and other text are available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Privacy policy About Wiktionary Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Search Search [x] Toggle the table of contents controverted 7 languagesAdd topic
5066
https://math.stackexchange.com/questions/4215736/products-of-sums-of-2n-squares
Skip to main content Products of sums of 2n squares Ask Question Asked Modified 4 years ago Viewed 288 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. It is well-known that the product of two squares is a square, and the product of sums of two squares is a sum of two squares (Brahmagupta-Fibonacci identity): (a2+b2)(c2+d2)=(ac−bd)2+(ad+bc)2 and an analogous result holds for sums of four squares (Euler's four square identity), eight squares (Degen's eight square identity) and 16 squares (Pfister's 16 square identity). According to the Wikipedia article, analogous results hold for all sums of 2n squares. But what about sums of 2n squares? Is it known whether there is a formula for the product of sums of 6 squares as a sum of 6 squares, i.e. whether, given ai,bj∈Z, (∑i=16a2i)(∑j=16b2j)=∑k=16c2k where ck=fk(a1,…,a6,b1,…,b6) for some polynomials fk? number-theory square-numbers sums-of-squares Share CC BY-SA 4.0 Follow this question to receive notifications edited Aug 3, 2021 at 12:25 a1402 asked Aug 3, 2021 at 11:48 a1402a1402 39911 silver badge88 bronze badges 6 This is trivially true, since every non-negative integer is the sum of six squares. But there is no such formula as for powers of two. – Dietrich Burde Commented Aug 3, 2021 at 11:51 @DietrichBurde Thank you, so doesn't this mean it's trivial for every sum of ≥4 squares? What is the significance of the 8 and 16 square identities? – a1402 Commented Aug 3, 2021 at 11:54 1 There are no nice formulas, but do such formulas exist? – a1402 Commented Aug 3, 2021 at 11:54 No, I don't think so. Wikipedia certainly would have mentioned those, right? K. Conrad's article is cited for m=16, there should be an argument given there, why we should have m=2n. – Dietrich Burde Commented Aug 3, 2021 at 11:56 1 I found it in Conrad's article. It is called "A Converse to Pfister's Theorem." – Dietrich Burde Commented Aug 3, 2021 at 12:02 | Show 1 more comment 1 Answer 1 Reset to default This answer is useful 2 Save this answer. Show activity on this post. There is no such formula. Pfister’s theorem tells us that in any field, when n is a power of 2, the set of sums of n squares is closed under multiplication. It follows that when n is a power of 2, there is a formula for the ck in terms of ai,bi, where (a21+…+a2n)(b21+…+b2n)=c21+…+c2n . Another theorem of Pfister (converse of Pfister’s above theorem) tells us that for every positive integer n that is not a power of 2, the nonzero sums of n squares are not closed under multiplication in the field R(x1,x2,…). This means that there are two sums of n squares in this field such that their product is not a sum of n squares in the field. Hence no formula exists when n is not a power of 2. Source: Share CC BY-SA 4.0 Follow this answer to receive notifications answered Aug 3, 2021 at 12:25 a1402a1402 39911 silver badge88 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 number-theory square-numbers sums-of-squares See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Related 6 Is it known whether any positive integer can be written as the sum of n different squares? 12 Integers of the form x2+2y2. 3 Combining sums and/or differences of squares 4 For which polynomials f is the subset {f(x):x∈Z} of Z closed under multiplication? 0 Is the difference between consecutive sums of N consecutive squares trivial or already a theorem? Hot Network Questions Function for displaying a value checked against another set of values A property of the rectangular (isosceles) hyperbola What happens when the acoustic pressure equals the ambient pressure during rarefaction? How long must a jet plane be for a passenger in the back row to survive a nose-dive crash? What would a "big dumb satellite" be like? How could the US promise, without a vote in Congress, not to expand NATO? 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5067
https://www.youtube.com/watch?v=TdgEH7-DqjQ
The Distributive Law Virtual High School 64700 subscribers 1084 likes Description 96940 views Posted: 24 Jul 2014 The distributive law allows us take different approaches to simplifying and/or evaluating algebraic expressions, and still arrive at the same answer. This video explains two methods, and shows algebraically and visually how they both reach the same final answer. All images are created and owned by Virtual High School or are in the Public Domain. For ownership and copyright information, please contact Virtual High School. 2 comments Transcript: the distributive law gives us an alternate way to evaluate expressions and still arrive at the same answer think of it this way you can drive to the store or walk to the store regardless of your choice you will arrive at the same destination you can think of the distributive law in the same way different methods same outcome here let's look at an example given this question one method we could use to to solve is to apply our knowledge of order of operations to start simplify within the brackets 3 + 4 is 7 now perform the multiplication 5 7 is 35 another way to solve this is to Use the distributive law this method tells us to distribute the leading term into each term within the set of brackets so 5 is multiplied by 3 and and 5 is Multiplied 4 simplify within the brackets 5 3 is 15 and 5 4 is 20 adding these values gives us 35 notice that this is the same outcome as the first method let's look at this visually using order of operations 3 + 4 can be shown like this and when multiplied by five it becomes five rows of seven using the distributive law 5 3 can be shown like this five rows of three 5 4 can be shown like this five rows of four adding them together looks like this five rows of seven from this we can see that there are different methods that provide us with the same outcome
5068
https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas?srsltid=AfmBOooXE1IUzbS6MmsKByM7pWA6J3pPYOo97iq9VWgoYkkmUDxpJhWs
Art of Problem Solving Vieta's Formulas - 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 Vieta's Formulas Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Vieta's Formulas In algebra, Vieta's formulas are a set of results that relate the coefficients of a polynomial to its roots. In particular, it states that the elementary symmetric polynomials of its roots can be easily expressed as a ratio between two of the polynomial's coefficients. It is among the most ubiquitous results to circumvent finding a polynomial's roots in competition math and sees widespread usage in many math contests/tournaments. Contents [hide] 1 Statement 2 Proof 3 Problems 3.1 Introductory 3.2 Intermediate 4 Advanced 5 See also Statement Let be any polynomial with complex coefficients with roots , and let be the elementary symmetric polynomial of the roots. Vieta’s formulas then state that This can be compactly summarized as for some such that . Proof Let all terms be defined as above. By the factor theorem, . We will then prove Vieta’s formulas by expanding this polynomial and comparing the resulting coefficients with the original polynomial’s coefficients. When expanding the factorization of , each term is generated by a series of choices of whether to include or the negative root from every factor . Consider all the expanded terms of the polynomial with degree ; they are formed by multiplying a choice of negative roots, making the remaining choices in the product , and finally multiplying by the constant . Note that adding together every multiplied choice of negative roots yields . Thus, when we expand , the coefficient of is equal to . However, we defined the coefficient of to be . Thus, , or , which completes the proof. Problems Here are some problems with solutions that utilize Vieta's quadratic formulas: Introductory 2005 AMC 12B Problem 12 2007 AMC 12A Problem 21 2010 AMC 10A Problem 21 2003 AMC 10A Problem 18 2021 AMC 12A Problem 12 Intermediate 2017 AMC 12A Problem 23 2003 AIME II Problem 9 2008 AIME II Problem 7 2021 Fall AMC 12A Problem 23 2019 AIME I Problem 10 Advanced 2020 AIME I Problem 14 See also Polynomial Retrieved from " Categories: Algebra Polynomials Theorems 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.
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https://juancardenas21.wordpress.com/segundo-corte-12m/ecuaciones-de-moviemiento-parabolico/alcance-maximo-y-altura-maxima/
Alcance maximo y Altura máxima | FISICA MECANICA FISICA MECANICA El amor es física, el matrimonio química. Alejandro Dumas Ir directamente al contenido Inicio 3° Tercer Corte 12M Biografias Galileo Galilei Isaac Newton Energia y Trabajo Equilibrio Rotacional Las Fuerzas Composición y Descomposición de Fuerzas Definición y Características Equilibrio: fuerza equilibrante Los principios de la Dinámica REFERENCIAS Pregunta Ecaes Proyecto Final Segundo Corte 12M Ecuaciones de moviemiento parabolico Alcance maximo y Altura máxima Tiempo de Vuelo Movimiento Browiniano Pregunta Ecaes Proyecto 1° Primer corte 12M ¿Que Es Ser Ing.Mecatronico? ¿Que Es Ser Ingeniero? Cifras Significativas ECAES HERRAMIENTAS BASICAS Kilogramo (Masa) Metro (Longitud) Otros Patrones Basicos del SI Segundo (Tiempo) Tabla de Factores de Conversion Producto Cruz o Vectorial Producto Punto o Escalar Proyecto avance Alcance maximo y Altura máxima Alcance maximo y altura máxima En el applet se trazan las trayectorias de proyectiles disparados con la misma velocidad inicial v 0 pero con los siguientes ángulos de tiro θ : 10º, 20º, 30º, 40º, 45º, 50º, 60º, 70º, 80º, 90º. Las ecuaciones del movimiento de los proyectiles son x=v 0 cos θ⋅t y=v 0 sin θ⋅t−1/2 g t 2 La parábola de seguridad El alcance horizontal de cada uno de los proyectiles se obtiene para y=0. x max=[v 0 2 sin(2 θ)]/g Su valor máximo se obtiene para θ =45º, teniendo el mismo valor para θ =45+α , que para θ =45-α . Por ejemplo, tienen el mismo alcance los proyectiles disparados con ángulos de tiro de 30º y 60º, ya que sin(2·30)=sin(2·60). La altura máxima que alcanza un proyectil se obtiene con v y=0. y max=[v 0 2 sin 2 θ]/2 g Su valor máximo se obtiene para el ángulo de disparo θ =90º. La envolvente de todas las trayectorias descritas por los proyectiles cuyo ángulo de disparo está comprendido entre 0 y 180º se denomina parábola de seguridad. Esta denominación hace referencia al hecho de que fuera de esta parábola estamos a salvo de los proyectiles disparados con velocidad v 0. Se trata de la parábola simétrica respecto del eje Y de ecuación y=-ax 2+b que pasa por los puntos (x=v 0 2 /g,y=0), y (x=0, y=v 0 2 /(2 g)) tal como se ve en la figura. La ecuación de dicha parábola es y=−1 g/2 v 0 2 x 2+1 v 0 2/2g Deducción alternativa de la ecuación de la parábola de seguridad Las ecuaciones paramétricas de la trayectoria son x=v 0 2 cos θ⋅t y=v 0 sin θ⋅t−1/2 g t 2 Eliminado el tiempo t, obtenemos la ecuación de la trayectoria y=x tan θ−g x 2/2 v 0 2 cos 2 θ Esta ecuación se puede escribir de forma alternativa tan 2 θ−2 v 0 2/g xtan θ+(1+2 v 0 2 y/g x 2)=0 Consideremos un punto arbitrario P del plano. Sustituimos las coordenadas (x, y) del punto en la ecuación de la trayectoria y puede ocurrir Que la ecuación de segundo grado en tan θ no tenga raíces reales. El punto P 1 no podría ser un punto de impacto para un proyectil disparado con velocidad inicial v 0. Que la ecuación de segundo grado tenga dos raíces reales, lo que implicará que el punto P 2 es accesible, y que hay dos ángulos de tiro θ 1 y θ 2 que dan en el blanco P 2. En la figura, vemos que cualquier punto en el interior de la envolvente es alcanzado por dos trayectorias. Cuando la raíz de la ecuación de segundo grado es doble θ 1=θ 2. Como vemos en la figura, solo hay una trayectoria que pasa por un punto P 3 dado de la envolvente. Para que las raíces sean iguales, se tiene que cumplir que el discriminante b 2-4 ac de la ecuación de segundo grado ax 2+bx+c=0 sea nulo. (2 v 0 2/g x)2−4(1+2 v 0 2 y/g x 2)=0 y=−g/2 v 0 2 x 2+v 0 2/2 g Esta es la ecuación de la envolvente que hemos obtenido anteriormente. Otra deducción alternativa es la siguiente: La ecuación de la trayectoria depende del ángulo θ con el que se dispara el proyectil. f(x, y, θ)=0 La ecuación de la envolvente de las trayectorias se obtiene derivando con respecto a θ o bien, respecto a tan θ e igualando a cero. ∂f/∂θ=0 2 tan θ−2 v 0 2/g x=0 tan θ=v 0 2/g x Combinamos esta ecuación con la de la trayectoria para eliminar el ángulo θ. F(x,y)=0 Es decir, introducimos la expresión de tan θ en la ecuación de la trayectoria, para obtener la ecuación de la envolvente y=−g 2 v 0 2 x 2+v 0 2 /2 g La elipse que une las posiciones de altura máxima La altura máxima se alcanza cuando v y=0, en el intante t=v 0·sin θ/g. La posición (x h, y h) del proyectil en este instante es x h=v 0 2/2 g sin(2 θ)y h=v 0 2/2 g sin 2 θ Teniendo en cuenta, la relación trigonométrica 1-cos(2 θ)=2sin 2 θ x h=v 0 2 2 g sin(2 θ)y h=v 0 2/4 g(1−cos(2 θ)) Despejando sin(2 θ) en la primera ecuación, cos(2 θ), en la segunda, elevando al cuadrado y sumando, eliminamos el ángulo 2 θ. x 2 h/4 b 2+(y h−b)2/b 2=1 b=v 0 2/4 g REFERENCIAS Compártelo: Haz clic para compartir en X (Se abre en una ventana nueva)X Haz clic para compartir en Facebook (Se abre en una ventana nueva)Facebook Me gusta Cargando... 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5070
https://www2.isye.gatech.edu/~hsharp/math2420/sets
Published Time: 2005-07-06T16:23:00.000Z MATH 2420 Discrete Mathematics Lecture notes Sets and Set Theory Objectives: Determine whether one set is a subset of another 2. Determine whether two sets are equal 3. Determine whether an element is in a set or not 4. Determine the union, intersection, difference, and complement of sets 5. Illustrate sets using Venn diagrams 6. Determine Cartesian product of two or more sets 7. Prove set identities 8. Use set identities to derive new set properties from old set prop-erties 9. Use Venn diagrams to prove set identities 10. Determine whether sets form a partition of a given set 11. Determine the power set of a set General definitions: set A collection of discrete items, whether numbers, letters, people, animals, cars, atoms, planets, etc. A set is identified when the items are grouped together between { and }. element One item in a set, whether a discrete item or a set itself. cardinality The number of elements in a finite set. Sets are defined by listing the elements between braces ( { } ). Any items listed between the braces are considered elements of the set, even if those items are sets themselves. An item by itself is just an element and not a set. {a} is a set a is not a set If the elements of a set are too numerous to list individually, the set may be defined by a property. We write the set defintion by listing first the membership and then the condition which must be met for membership. As an example {x ∈ Z | − 2 < x < 5} where Z is the set of all integers defines a set which has elements x which are integers between -2 and 5. 1Special Sets Z+ set of all positive integer numbers N set of all natural numbers (0 , 1, 2, 3, 4, . . . )Z set of all integer numbers Q set of all rational numbers I set of all irrational numbers R set of all real numbers U the universal set Subsets If A and B are subsets, A is called a subset of B if, and only if, every element of A is also an element of B. We write this as A ⊆ B. Written symbolically this is: A ⊆ B ⇔ ∀ x, if x ∈ A then x ∈ B. The definition of subset is rigid and inflexible. If any element in A does not appear in B then A cannot be a subset of B. That is: A 6 ⊆ B ⇔ ∃ x such that x ∈ A and x 6 ∈ B. Looking at the special sets above we have N ⊆ Z ⊆ Q ⊆ R A set can be a subset of itself, strange as this may seem. To differen-tiate subsets by type, we call a subset with at least one item missing from the larger set as a proper subset . That is, a set may be a proper subset if, and only if, every element of A is in B but there is at least one element of B that is not in A. Set Equality Two sets, A and B, can be said to be equal if every element in set A is also in set B, and vice versa. Stated another way A = B if ( x ∈ A ⇒ x ∈ B) and ( x ∈ B ⇒ x ∈ A) Operations on Sets We can operate on sets in a manner similar to other mathematical operations. Our operators are a bit different, though. union ∪ A ∪ B all elements in both sets intersection ∩ A ∩ B all elements common to both sets difference − A − B all elements in A but not in B complement various Ac all elements not in A 2Procedural definitions: union A ∪ B = {x ∈ U | x ∈ A or x ∈ B} intersection A ∩ B = {x ∈ U | x ∈ A and x ∈ B} difference A − B = {x ∈ U | x ∈ A and x 6 ∈ B} complement Ac = {x ∈ U | x 6 ∈ A} Examples: union {1, 2, 3} ∪ { 2, 4, 6} = {1, 2, 3, 4, 6} intersection {1, 2, 3} ∩ { 2, 4, 6} = {2} difference {1, 2, 3} − { 2, 4, 6} = {1, 3} Empty Set A set with no members, however weird, is part of set theory. It is called the empty set (or null set ) and is denoted by either empty brackets () or by the special symbol ∅. Partitions and Disjoint Sets Two sets are called disjoint if they have no elements in common. That is A ∩ B = ∅ If we increase the number of sets to more than two we must consider the property known as mutually disjoint . Also known as pairwise disjoint . This refers to the relationship between any two sets at a time. The definition is Sets A1, A 2, A − 3, . . . , A n are mutually disjoint if, and only if, no two sets Ai and Aj with distinct subscripts have any elements in common. More precisely, for all i, j =1, 2, . . . , n , Ai ∩ Aj = ∅ whenever i 6 = j Example: Let A1 = 1, 2, 3, A2 = 4, 5 and A3 = 6, 7, 8. Then A1 A2 A3 A1 na ∅ ∅ A2 ∅ na ∅ A3 ∅ ∅ na So what purpose does this serve? It provides the basis for constructing partitions of sets. A partition is a collection of mutually disjoint sets that when unioned together form the whole larger set. Each set must be nonempty, that is, must have at least one member and all sets must be pairwise, or mutually, disjoint. We use the following definition: A collection of nonempty sets A1,A2,. . . ,An is a partition of a set A if, and only if, 1. A = A1 ∪ A2 ∪ . . . ∪ An;2. A1,A2,. . . ,An are mutually disjoint. 3Power Sets So far we have seen that any given set can be broken into several different subsets. A subset is simply a set of elements from a (usually) larger set, though the set can be a subset of itself A ⊆ A, strange as that may seem. We have also seen that the set of no elements, the empty set, is always a subset of every set. This collection is referred to as the power set of a set and is denoted P(§). But how many sets are there in the power set? The number of sets is equal to 2 n where n is the number of elements in a set. For example, if we have a set A = {2, 4, 6} then the power set P(A)consists of ∅, {2}, {4}, {6}, {2, 4}, {2, 6}, {4, 6}, {2, 4, 6} of which all but {2, 4, 6} are proper subsets. This works even for the empty set ∅ thusly: There are no elements in the empty set since it is “empty”. Hence the cardinality is zero. If we compute 2 0 we get an answer of 1, which is the only subset of ∅ which is itself. Cartesian Products In sets, order and repetition usually don’t matter. But what if they do? Then we have an item known as an ordered n-tuple . Often seen with only two elements, these are called ordered pairs . Ordered n-tuples are unique based on elements and order of elements. Two n-tuples can be equal only if all elements are the same and in the same order. (1 , 2) 6 = (2 , 1) (3 , (−2) 2, 12 ) = ( √9, 4, 36 )Now, if X and Y are sets, we let X × Y denote the set of all ordered pairs ( x, y ) where x ∈ X and y ∈ Y . This is called the Cartesian product of X and Y . Symbolically, A × B = {(a, b ) | a ∈ A and b ∈ B} Example: Let A = 1, 2 and B = 1, 4, 7. Then A × B = {(1 , 1) , (1 , 4) , (1 , 7) , (2 , 1) , (2 , 4) , (2 , 7) } B × A = {(1 , 1) , (1 , 2) , (4 , 1) , (4 , 2) , (7 , 1) , (7 , 2) } This can be applied to more than two sets at a time. 4Properties of Sets Subset Relations Inclusion of Intersection: For all sets A and B, A ∩ B ⊆ A and A ∩ B ⊆ B Inclusion in Union: For all sets A and B, A ⊆ A ∪ B and B ⊆ A ∪ B Transitive Property of Subsets: For all sets A, B, and C if A ⊆ B and B ⊆ C, then A ⊆ C Procedural Versions of Set Definitions Let X and Y be subsets of a universal set U and suppose x and y are elements of U .1. x ∈ X ∪ Y ⇔ x ∈ X or x ∈ Y x ∈ X ∩ Y ⇔ x ∈ X and x ∈ Y x ∈ X − Y ⇔ x ∈ X and x 6 ∈ Y x ∈ Xc ⇔ x 6 ∈ X (x, y ) ∈ X × Y ⇔ x ∈ X and y ∈ Y Element Arguments Using these subset relations and set definitions, we can prove that sets are subsets of each other. To prove that X ⊆ Y requires two steps: 1. Suppose that x is a particular yet arbitrarily chosen element of X.2. Show that x is an element of Y .This is an element argument or also known as a “chasing x argu-ment” (in that we “chase” x from one side of the statement to the other). Using this method of proving subsets we can show that two sets are equal to each other. We know that two sets are equal if they are subsets of each other, so if we can show the subset relation then we have shown the equality. So first we show that X ⊆ Y and then we show that Y ⊆ X.This can best be explained with an example. Show that R ∩ (S ∪ T ) ⊆ S ∪ (R ∩ T ). We must show that each x in R ∩ (S ∪ T ) is also in S ∪ (R ∩ T ). Let x ∈ R ∩ (S ∪ T ). 52. Then x is in both R and S ∪ T by definition. 3. Since x is in S ∪ T , x is in S or x is in T . This gives us two cases. 4. (a) In the case that x is in S, we get that x ∈ S ∪(R∩T )by the definition of union. 5. (b) In the case that x is in T , we get that x ∈ (R∩T ) by the definition of intersection and the observation above that x ∈ R.6. Then, once again, we get x ∈ S∪(R∩T ) by the definition of union. 7. Thus in either case x ∈ S ∪ (R ∩ T ). 8. Thus each x in R ∩ (S ∪ T ) is in S ∪ (R ∩ T ) as well, so R ∩ (S ∪ T ) ⊆ S ∪ (R ∩ T ) From Bogart, p. 21-22 6
5071
https://math.stackexchange.com/questions/4829177/help-with-finding-probability-of-finding-same-colored-balls
Help with Finding Probability of Finding Same colored balls - 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 Help with Finding Probability of Finding Same colored balls Ask Question Asked 1 year, 9 months ago Modified1 year, 9 months ago Viewed 81 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I have a bag that contains 6 blue, 3 red and 1 yellow balls and I pick 3 balls at random. What is the probability that my 3 chosen balls contains only blue and red color? I tried to solve this using the combinations intuition, but I get 35% as result and the correct answer is 72%. I'm not sure how. Assuming that as soon as I pick, I put the balls back into the bag, how do I even approach this problem? I know that I should be using permutations and combinations, but I lack the intuition behind this. Could someone please explain me this? probability Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Dec 17, 2023 at 13:45 joesanjoesan 111 5 5 bronze badges 5 There are (9 3)(9 3) ways to pick three from the blue red balls. There are (10 3)(10 3) ways to do it with no restrictions. The ratio is 84 120=7 10 84 120=7 10. I don't see how you get .35.35 , That's clearly far too low. Why should there be a 65%65% chance of getting the yellow? Nor is 72%72% correct, but that's close.lulu –lulu 2023-12-17 13:53:25 +00:00 Commented Dec 17, 2023 at 13:53 Can you write that with the combinators formula that I can use to understrand the intuition behind problems like these?joesan –joesan 2023-12-17 13:56:50 +00:00 Commented Dec 17, 2023 at 13:56 I don't know what formulas you prefer. I wrote this using a standard method for denoting the number of combinations... As a quick variant, the probability of drawing th eyellow is clearly 3 10 3 10 so we subtract from 1 1 to get 7 10 7 10 as before.lulu –lulu 2023-12-17 13:57:38 +00:00 Commented Dec 17, 2023 at 13:57 Please edit your post to explain where the 35%35% comes from. And the 72%72% for that matter (as I say, that's closer, but it's still wrong).lulu –lulu 2023-12-17 13:58:24 +00:00 Commented Dec 17, 2023 at 13:58 1 As another variant, suppose you are drawing sequentially. The probability that the first ball is non-yellow is 9 10 9 10. Given that, the probability that the second is also non-yellow is 8 9 8 9, and given both of those the probability for the third is 7 8 7 8 making the total probability 9 10×8 9×7 8=7 10 9 10×8 9×7 8=7 10 as before.lulu –lulu 2023-12-17 14:02:13 +00:00 Commented Dec 17, 2023 at 14:02 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Before we start we have to be clear on how the ball selection work. No Reinsertion As is written I think there is no reinsertion, so our event A={picking 3 balls only blue and red}A={picking 3 balls only blue and red} could be written as the disjoint union of {picking 3 B}∪{picking 3 R}∪{picking 2 B,1 R}∪{picking 1 B, 2R}=A 1∪A 2∪A 3∪A 4{picking 3 B}∪{picking 3 R}∪{picking 2 B,1 R}∪{picking 1 B, 2R}=A 1∪A 2∪A 3∪A 4 How do we evaluate this P(A 1)=(6 3)(10 3)=from 6 possible choices we pick 3 from 10 possible choices we pick 3 P(A 1)=(6 3)(10 3)=from 6 possible choices we pick 3 from 10 possible choices we pick 3 I will not explain the remaining formulas and the conclusion is P(A)=(6 3)+(3 3)+(6 2)⋅(3 1)+(6 1)⋅(3 2)(10 3)=7 10 P(A)=(6 3)+(3 3)+(6 2)⋅(3 1)+(6 1)⋅(3 2)(10 3)=7 10 And I don't know how you get 35%35%. Reinsertion In this case at every pick the ball get reinserted in the bag so at each pick we have 9/10 9/10 chance of not picking a yellow ball then P(A)=9 3 10 3=0,729≈73%P(A)=9 3 10 3=0,729≈73% The fact that A A could be written as disjoint union is independent of the method of picking balls. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 17, 2023 at 14:33 answered Dec 17, 2023 at 14:04 Turquoise TiltTurquoise Tilt 2,571 1 1 gold badge 10 10 silver badges 28 28 bronze badges 2 I'm not sure if the first 2 cases are correct. I mean the ask is blue and red, so out of the 3 picked balls you should have a mix of blue and red and not just all blue and all red.joesan –joesan 2023-12-17 14:11:31 +00:00 Commented Dec 17, 2023 at 14:11 @joesan Then you should remove the first two and in this case you got P(A)=21 40=52.5%P(A)=21 40=52.5%, still not 72%72% nor 35%35%.Turquoise Tilt –Turquoise Tilt 2023-12-17 14:16:43 +00:00 Commented Dec 17, 2023 at 14:16 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability See similar questions with these tags. 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5072
https://www.statisticshowto.com/probability-and-statistics/probability-main-index/odds-ratio/
Skip to content Odds Ratio Calculation and Interpretation Contents: What are “Odds”? What is the Odds Ratio? How to Calculate the Odds Ratio What do the Results mean? Population Averaged vs. Subject Specific Odds Ratio Watch the video for an overview of the odds ratio and a couple examples of calculations, or read on below: Can’t see the video? Click here to watch it on YouTube. What are “Odds”? Statistics Definitions > Odds Definition Odds is usually defined in statistics as the probability an event will occur divided by the probability that it will not occur . In other words, it’s a ratio of successes (or wins) to losses (or failures). As an example, if a racehorse runs 100 races and wins 20 times, the odds of the horse winning a race is 20/80 = 1/4. The above odds definition is the odds in favor of an event. But there is another way to define odds in statistics— odds against an event happening. This is defined as the number of ways an event does not happen against the number of ways an event does happen . For our racehorse example, that would be 80 to 20, or 4 to 1. Odds vs. Probability The odds definition is very similar to the definition of probability, which is the fraction of times an event occurs in a certain number of trials. In the horse example, the probability of a win is 20/100. When there are a very small number of events, odds and probability are very similar. For example, let’s say the horse runs 4 races and wins 1. The odds of the horse winning a race is 1/3 (1 win to 3 losses) and the probability is 1/4 (1 win out of 4 races). One major differences between odds and probability is that while odds range from 0 and infinity, probability ranges from 0 to 1. Dictionary Definition of Odds The statistical definition of odds is quite precise compared to dictionary definitions, which are inconsistent and have different meanings, even in a plain language sense instead of a statistical one : “the chances of something happening” (Macmillan British Dictionary, 2012) “the probability that one thing is so or will happen rather than another; the ratio of the probability of one event to that of an alternative event” (Merriam-Webster Dictionary, 2012) “the ratio of the probability of an event’s occurring to the probability of its not occurring; the likelihood of the occurrence of one thing rather than the occurrence of another thing, as in a contest” (American Heritage Dictionary, 2012) What is the Odds Ratio? An odds ratio (OR) is a measure of association between a certain property A and a second property B in a population. Specifically, it tells you how the presence or absence of property A has an effect on the presence or absence of property B. The OR is also used to figure out if a particular exposure (like eating processed meat) is a risk factor for a particular outcome (such as colon cancer), and to compare the various risk factors for that outcome. You could use the OR to find out how much alcohol use leads to liver disease. Or you might want to find out if cell phone use has some link to brain cancer. As long as you have two properties you think are linked, you can calculate the odds. Back to Top How to Calculate the Odds Ratio You have two choices for the formula:(a/c) / (b/d) or, equivalently: (ad) / (bc) General Steps: Step 1: Calculate the odds that a member of the population has property “A”. Assume the person already has “B.” Step 2: Calculate the odds that a member of the population has property “A”. Assume the person does not have “B.” Step 3: Divide step 1 by step 2 to get the odds ratio (OR). Back to Top Odds Ratio Example The above image shows two levels of exposure to ice cream: those who ate it, and those who didn’t. The 2×2 table also shows two outcome levels: people who were ill (“cases”) and people who were not (“controls”). The odds ratio is calculated as follows: Ill people: people who ate ice cream / people who did not = 13/17 People who are not ill: people who ate ice cream / people who did not = 32/23 Dividing the two results, we get (13/17) / (32/23) = 0.55 The resulting odds ratio of .55 means that ill people were about half as likely to eat ice cream as well people. Back to Top Odds Ratio Interpretation; What do the Results mean? An odds ratio of exactly 1 means that exposure to property A does not affect the odds of property B. An odds ratio of more than 1 means that there is a higher odds of property B happening with exposure to property A. An odds ratio is less than 1 is associated with lower odds. However, it’s not quite as simple as that. You could think of the odds ratio as being a bit overly simplistic at describing real world situations. If, for example, you have a positive OR, it doesn’t mean that you have a statistically significant result. In order to figure that out, you need to consider the confidence interval and p-values (if you know it). The other issue is that even if you determine your results are statistically significant, that significance might not apply to all members of a population — there are nearly always a multitude of factors associated with risk. For example, this article points out that while overall, depression is strongly linked to suicide, “…in a particular sample, with a particular size and composition, and iBn the presence of other variables, the association may not be significant.” Back to top Population Averaged vs. Subject Specific Odds Ratio Population averaged models compare marginal distributions and give an overview of the effect on a whole population. The margins of a contingency table contain the totals, so it makes sense for them to be used to calculate the marginal odds ratio for a whole population. On the other hand, subject-specific models look at joint distributions: specific conditions or experiences within the model. The joint distributions are used to calculate conditional odds ratios. Marginal Odds Ratio Example (for Population Averaged Models) Michael Radelet studied death sentence data from Florida from 1976-77. Calculate the marginal odds ratio for the race of defendant and whether or not that made a difference about if they got the death penalty: | | | | | --- --- | | Defendant Race | Victim Race | Death Penalty Y | Death Penalty N | | W | W | 19 | 132 | | W | B | 0 | 9 | | B | W | 11 | 52 | | B | B | 6 | 97 | Solution: Sum (marginalize) the values in the table. We’re interested in only the race of the defendant and whether or not they got the death penalty. Therefore, we can marginalize (sum up) values for the race of the victim. This creates a new 2×2 table: | | | | --- | Defendant Race | Death Y | Death N | | W | 19 | 141 | | B | 17 | 149 | 2. Use the information in the marginal table to find the OR (using the OR formula from above):OR = (a/c) / (b/d) = (19/17)/(141/149) = 1.12/0.95 = 1.18.The odds are 1.18 times higher that a white defendant will get the death penalty compared to a black defendant. If you’re interested in his findings, he concluded that there isn’t any clear evidence to support the hypothesis that the defendant’s race is strongly associated with imposition of the death penalty. However, defendants were more likely to be sentenced to death if they had a white victim. Subject specific models calculate the odds ratio using the same formula as all of the examples above. The only difference is that instead of summing all the variables together, you’ll hold one variable constant (i.e. you’ll use joint distributions). Next: Adjusted Odds Ratio References The Difference Between “Probability” and “Odds”. Retrieved November 26, 2021 from: Fulton, L. et al. Confusion Between Odds and Probability, a Pandemic?Journal of Statistics Education, Volume 20, Number 3 (2012) 3www.amstat.org/publications/jse/v20n3/fulton.pdfAgresti A. (1990) Categorical Data Analysis. John Wiley and Sons, New York. Radelet, M. L. Racial Characteristics and the Imposition of the Death Penalty. American Sociological Review, v46 n6 p918-27 Dec 1981 Levine, D. (2014). Even You Can Learn Statistics and Analytics: An Easy to Understand Guide to Statistics and Analytics 3rd Edition. Pearson FT Press Comments? Need to post a correction? Please Contact Us.
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Published Time: 2005-04-05 (PDF) On the structure ofp-zero-sum free sequences and its application to a variant of Erdös-Ginzburg-Ziv theorem Home Ethics Philosophy Free Will Article PDF Available On the structure ofp-zero-sum free sequences and its application to a variant of Erdös-Ginzburg-Ziv theorem April 2005 Proceedings Mathematical Sciences 115(1):67-77 DOI:10.1007/BF02829840 Authors: Weidong Gao Nankai University A. Panigrahi A. Panigrahi This person is not on ResearchGate, or hasn't claimed this research yet. R. Thangadurai Download full-text PDFRead full-text Download full-text PDF Read full-text Download citation Copy link Link copied Read full-textDownload citation Copy link Link copied Citations (17)References (28) Abstract Letp be any odd prime number. Letk be any positive integer such that 2⩽k⩽[p+1 3]+1 2 \leqslant k \leqslant \left[ {\frac{{p + 1}}{3}} \right] + 1. LetS = (a 1,a 2,...,a 2p−k ) be any sequence in ℤp such that there is no subsequence of lengthp of S whose sum is zero in ℤp. Then we prove that we can arrange the sequence S as follows: S=(a,a,...,a,⏟u t i m e s b,b,...,b,⏟v t i m e s a 1′,a 2′,...,a 2 p−k−u−v′) S = (\underbrace {a,a,...,a,}{u times}\underbrace {b,b,...,b,}{v times}a'1 ,a'_2 ,...,a'{2p - k - u - v} ) (1) whereu ≥v,u +v ≥ 2p - 2k + 2 anda -b generates ℤp. This extends a result in to all primesp andk satisfying (p + 1)/4 + 3 ≤k ≤ (p + 1)/3 + 1. Also, we prove that ifg denotes the number of distinct residue classes modulop appearing in the sequenceS in ℤp of length 2p -k (2≤k ≤ [(p + 1)/4]+1), and g⩾2 2 k−2 g \geqslant 2\sqrt 2 \sqrt {k - 2} , then there exists a subsequence of S of lengthp whose sum is zero in ℤp. Discover the world's research 25+ million members 160+ million publication pages 2.3+ billion citations Join for free Public Full-text 1 Content uploaded by Weidong Gao Author content All content in this area was uploaded by Weidong Gao on Jan 15, 2014 Content may be subject to copyright. arXiv:math/0503095v1 [math.CO] 5 Mar 2005 Proc. Indian Acad. Sci. (Math. Sci.)V ol. 115, No. 1,February 2003,pp. 67–77. Printed in India On the structure of p p p-zero-sum free sequences and its application to a variant of Erd¨ os–Ginzburg–Ziv theor em W D GAO, A P ANIGRAHI∗and R THANGADURAI∗ Department of Computer Science and Techno logy, University of Petroleum, Changping Shuiku Road, Beijing 10220 0, China ∗School of Mathematics, Harish-Chandra Research Institute, Chhatnag Road, Jhusi, Allahabad 211 019, India E-mail: wdgao 1963@yahoo.com.cn; anupama@mri.ernet.in; thanga@mri.ernet.in MS received 8 September 2004 Abstract. Let p be any odd prime number. Let k be any positive integer such that 2≤k≤h p+1 3 i+1. Let S= (a 1,a 2,...,a 2 p−k)be any sequence in Z p such that there is no subsequence of length p of S whose sum is zero in Z p. Then we prove that we can arrange the sequence S as follows: S= (a,a,...,a |{z } u times ,b,b,...,b | {z } v times ,a′ 1,a′ 2,...,a′ 2 p−k−u−v) where u≥v,u+v≥2 p−2 k+2 and a−b generates Z p. This extends a result in to all primes p and k satisfying (p+1)/4+3≤k≤(p+1)/3+1. Also, we prove that if g denotes the number of distinct residue classes modulo p appearing i n the sequence S in Z p of length 2 p−k(2≤k≤[(p+1)/4] + 1), and g≥2√2√k−2, then there exists a subsequence of S of length p whose sum is zero in Z p. Keywords. Sequences; zero-sum problems; zero-free; Erd¨os–Ginzburg–Ziv theorem. Introduction Let n be any positive integer. Let S= (a 1,a 2,...,a ℓ)be a sequence (possibly with r epe- tition) in the cyclic group of ord er n(denoted by Z n) of length ℓ. We call a subsequence T= (b 1,b 2,...,b r)of S to be zero-sum subsequence if b 1+b 2+···+b r=0 in Z n. In 1961, Erd¨os–Ginzburg–Ziv proved the following theorem (which we call the EGZ theorem). EGZ Theorem..Given a sequen ce S in Z n of length 2 n−1, one can extract a zero- sum subsequence of length n in Z n. The EGZ theorem is tight in the following sense. If S= (0,0,...,0 |{z } n−1 times ,1,1,...,1 |{z } n−1 times ) is a sequence in Z n of leng th 2 n−2, then S does not have a zero-su m subsequence of length n. 67 68 W D Gao, A P anigrahi and R Th angadurai Many authors studied the chara cterization of the above extremal example. In p articular, Y uster and Peterson and independently Bialostocki and Dierker proved that any sequence S in Z n of length 2 n−2 having no zero-sum subsequence of length n will be of the form S= (a,a,...,a |{z } n−1 times ,b,b,...,b |{z } n−1 times ), where a 6=b∈Z n. Also, Flores and Ordaz proved the following result of this na ture. Suppose S is any sequence in Z n of length 2 n−3 such th at S has no zero-sum subsequen ce of length n. Then there exists a,b∈Z n such that Z n is generated by b−a and a appearing n−1 times in S and one of the following conditions h old: (i) b appearing exactly n−2 times; (ii) b appearing exactly n−3 times in S and also, 2 b−a appearing exactly once in S. In 1996, Gao proved the generalization of the above two results as follows. Theorem.. Let n be any positive integer. Let k be any positive integer such that 2≤k≤n+1 4+2. Let S = (a 1,a 2,...,a 2 n−k)be any sequence in Z n such that th er e is no subsequence of length n of S wh ose sum is zero in Z n. Then we can re-arrange the sequence S as fo llows: S= (a,a,...,a |{z } u times ,b,b,...,b |{z } v times ,a′ 1,a′ 2,...,a′ 2 n−k−u−v) where u ≥v,u+v≥2 n−2 k+2 and a −b generates Z n. One of our main theorems in this article is to extend the above result to all primes p and integer k for the range p+1 4+3≤k≤p+1 3+1. This extension is meaningful for all large primes p. Also, we shall study the problem of how many distinct residue classes modulo p occur in those sequences of length 2 p−k in Z p having a zero-sum subse- quence of length p in it. Befo re we state our main theorems, we shall fix up notations as follows. For every integer 1 ≤k≤ℓ, define ∑ k(S) = a i 1+a i 2+···+a i k|1≤i 1<i 2<···<i k≤ℓ and ∑(S) = ∪ℓ k=1∑k(S). For any subsequence T= (b 1,b 2,...,b r)o f S, we let σ (T) = ∑r i=1 b i. We denote ST−1 by the deleted sequen ce R which is obtained from S by deleting the elements of T. Also, if S= (a,a,...,a |{z } r times ,b 1,b 2,...), then we write S= (a r,b 1,b 2,...). For any b∈Z n, we denote by b+S the sequence (b+a 1,b+a 2,...,b+a ℓ). For ev ery x∈Z n, define x to be the least positive inverse image under the natural homomorphism from the additive group of integers Z onto Z n.For example, 0 =n. If A⊂Z n, th en we denote the cardinality of A by |A|. If A is a sequence in Z n, we denote the length of A by |A|(same notation as the cardinality). For any g∈Z n, we d efine v g(S)by the number of times g appears in S. Also, we define h=h(S) =ma x g∈Z n v g(S). Gao introduced the following definition. On the structure of p-zer o-sum free sequences 69 DEFINITION 1.1. Let S= (a 1,a 2,...,a ℓ)and T= (b 1,b 2,...,b ℓ)be two sequences in Z n of leng th ℓ. W e say that S is equivalent to T(written as S∼T)if there exist an in teger c coprime to n, an element x∈Z n, and a permutation π of {1,2,...,ℓ}such that a i=c(b π (i)−x)for every i=1,2,...,n. Clearly, ∼is an equivalence relation; and if S∼T,then 0 ∈∑n(S)if and only if 0 ∈∑n(T). In this article,we shall prove theorems 3.1 and 3.2. Theorem 3.1.Let p be any odd prime number.Let k b e any positive integer such that 2≤k≤h p+1 3 i+1. Let S = (a 1,a 2,...,a 2 p−k)be any sequen ce in Z p such that 0 6∈∑p(S). Then S∼(0 u,1 v,a′ 1,a′ 2,...,a′ 2 p−k−u−v), where u ≥v and u+v≥2 p−2 k+2. Using the information in Theorem 3.1, we consider the following problem of v ariant of EGZ theorem as follows. Before we state our theorem, we recall the following definition which was introduc ed in and state the known results. DEFINITION 1.2. Let n,k be positive integers, 1 ≤k≤n. Deno te by f(n,k)the least positive integer g for which the following holds: If S= (a 1,a 2,...,a g)is a sequence of elemen ts of Z n, the cyclic group of or der n, of length g such that the number of distinct a i’s is equal to k, then there are n indices i 1,i 2,...,i n belongin g to {1,2,...,g}such that a i 1+a i 2+···+a i n =0. Theorem. W e have f(n,k)≤2 n−1 for all n and for all 1≤k≤n(By EGZ theorem). f(n,n) = n,if n is odd n+1,if n is even. f(n,k) = n+2, for all n ≥5 an d 1+n/2<k≤n−1 [5,10]. fn,n 2+1=n+3 for all n ∈2 N. f(n,k) = 2 n−((k−1)/2)2−1 for all n ≥(k−1)2−4 for an odd k ≥5 . f(n,k) = 2 n−k(k−2)/4−1 for all n ≥k(k−2)−4 for an even k ≥6 . f(n,2) = 2 n−1,f(n,3) = 2 n−2 and f(n,4) =2 n−3 for all n . f(n,k)≤2 n−k+1 for all 2≤k≤n. f(p,k)≤2 p−3 k+11 for all 5≤k≤(p+15)/3 . Other than these results m any authors (for instance , an d ) consider some lower bounds for f(n,k)for various k. In this article, we shall prove the following result. Theorem 3.2.Let p be any odd prime number.Let k b e any positive integer such that 2≤k≤h p+1 3 i+1. Then f (p,ℓ)≤2 p−k for all ℓ≥2√2√k−2. 70 W D Gao, A P anigrahi and R Th angadurai Preliminaries W e shall start this section w ith a well-known fundamental inequality of subsets as follows. Cauchy–Davenport inequality [6,7].Let p be any prime numb er. Let A 1,A 2,...,A t be non-empty subsets of Z p. Then |A 1+A 2+···+A t| ≥ min(p, t ∑ i=1|A i|−t+1). Theorem 2.1..Let n and k b e any positive integer s such that n −2 k≥1.If S = (a 1,a 2,...,a n−k)is a sequence in Z n such that 0 6∈ ∑(S),then there exists a 6=0∈Z n which appear at least n−2 k+1 times in S. The following Theorem is crucial for the p roof of Theorem 3.1. Theorem 2.2.Let p be any prime number and 1≤k≤p−2. Let S be a sequence in Z p of length p +k. If 0 6∈ ∑p(S),then h(S)≥k+1. Proof. When k=1, the result follows from the Pigeon hole principle. So, we can assume that k≥2. If possible, we assume that h(S)≤k. Th en, we can distribute the elements of S into a union A 1⊔A 2⊔···⊔A k, so that in each A i, an element occurs only once. By the Cauchy–Davenport theorem,we see that  k ∑ i=1 A i≥min(p, k ∑ i=1|A i|−k+1) =min{p,p+k−k+1=p+1}=p. Therefore,A 1+A 2+···+A k=Z p. In particular, σ (S)∈∑k(S). Without loss of generality we shall assume that σ (S) = a 1+a 2+···+a k. Then we have a k+1+a k+2+···+a p+k=0 which implies 0 ∈∑p(S)as |S|=p+k.This contradicts the assumption that 0 6∈ ∑p(S). Therefore, h(S)≥k+1. 2 Theorem 2.3..Let n be any positive integer. Let 1≤k≤n+1 3,and let S be a sequence in Z n of length n −k such that 0 6∈ ∑(S). Then S∼(1 n−2 k+1,x 1,x 2,...,x k−1) with ∑k−1 i=1 x i≤2 k−2. Lemma 2.4. Let p be any odd prime and 1≤k≤h p+1 3 i.Let S = (1 p−2 k+1,x 1,x 2,..., x k−1)be a sequence in Z p{0}of length p −k such that ∑k−1 i=1 x i≤2 k−2. Then,for any x∈Z p satisfying p−2 k+1≤x≤p−2 k+1+∑k−1 i=1 x i,there exists a subsequence T of S such that |T| ≥p−2 k+1 with σ (T) = x. Proof. Let x∈Z p such that p−2 k+1≤x≤p−2 k+1+∑k−1 i=1 x i. If x=p−2 k+1, then x=∑p−2 k+1 i=1 1 and we are done; otherwise, we have p−2 k+2≤x≤p−2 k+1+ k−1 ∑ i=1 x i≤p−1. Therefore, we have 1 ≤x−(p−2 k+1)≤∑k−1 i=1 x i. On the structure of p-zer o-sum free sequences 71 Claim. For any positive integer k, if S′= (x 1,x 2,...,x k)be a sequen ce in Z p{0}such that |S′|=k and ∑k i=1 x i≤2 k, then, for every x∈Z p satisfying 1 ≤x≤∑k i=1 x i, either x∈∑(S′)or x+1∈∑(S′). If the claim is proven, then, we get, either x−(p−2 k+1)or x−(p−2 k+1) + 1 in ∑((x 1,x 2,...,x k−1)). That is, either x=1+1+···+1 |{z } p−2 k+1 +y or x=1+1+···+1 |{z } p−2 k +y where y∈∑((x 1,x 2,...,x k−1)). So, to end the proof of this lem ma, it is enou gh to prove this claim. When k=1,2, the claim is trivially true. So, we let k≥3. Assume the result is true for k−1 and we sh all prove for k. If necessary by renaming the indices, without loss of generality, we can assume that S′= (x 1,x 2,...,x k)with x 1≤x 2≤ ··· ≤ x k. Suppose x k−1=1. Then, we have x 1=x 2=···=x k−1=1. As ∑k i=1 x i≤2 k, we see that x k≤ 2 k−(k−1) = k+1. There fore, we see that ∑(S′) = ({1,2,...,x k+k−1},if x k≤k, {1,2,...,x k+k−1}{k},if x k=k+1 which clearly implies the claim.Th us, now, we can assume that 2 ≤x k−1≤x k. I f x≤x k+∑k−2 i=1 x i, then by induction, either x or x+1 in ∑((x 1,x 2,...,x k−2,x k)), an d we are through; otherwise, we have, x k+∑k−2 i=1 x i≤x<∑k i=1 x i. Therefor e, we have k−2≤x k−x k−1+ k−2 ∑ i=1 x i≤x−x k−1≤x k+ k−2 ∑ i=1 x i. Therefore, by the induction hypothesis, we see that either x−x k−1 or x−x k−1+1 in ∑((x 1,x 2,...,x k−2,x k)) and hence, we have either x or x+1 in ∑(S′).2 3.P roof of Theore ms 3.1 and 3.2 Proof of Theorem 3.1. Let S be a sequenc e in Z p of length 2 p−k wh ere 2 ≤k≤[p+1 3]+1. Given that 0 6∈ ∑p(S). Without loss of generality we can assume that 0 (if necessary, by translating by an elem ent) appears maximum number of, say u, times in S. By The- orem 2.2, it is clear that u≥p−k+1. Therefore, S= (0 u,a 1,a 2,...,a 2 p−k−u)where a i∈Z p{0}. Let S 1= (a 1,a 2,...,a 2 p−k−u)be a su bsequence of S. Since u≥p−k+1, we have 2 p−k−u≤2 p−k−p+k−1=p−1. T hat is, |S 1| ≤ p−1. Let |S 1|= p−m for some positive integer m. Note that p−m+u=2 p−k which implies u+k− p=m.As 0 6∈ ∑p(S), we have u≤p−1. Therefore, m=u+k−p≤p−1+k−p =k−1. If 0 6∈ ∑(S 1), then b y Theorem 2.1, we know that there exists an element a∈ Z p{0}such that v a(S 1)≥p−2 m+1. Th erefore, S= (0 u,a v,b 1,b 2,...,b 2 p−k−u−v) and 2 p−k−u−v≤m−1≤k−2 which implies 2 p−2 k+2≤u+v and we are done. Thus, we can assume that 0 ∈∑(S 1). Let W be the maximal zero-sum subsequence of S 1 of length w. Mo reover, s ince 0 6∈∑p(S)and S 1 is a sequ ence in Z p{0}and u≥p−k+1, we have 2≤w≤p−u−1=⇒2≤w≤k−2.(1) 72 W D Gao, A P anigrahi and R Th angadurai Also note that k+u+w≥k+p−k+1+w≥p+1. Put ℓ=k+u+w−p. Therefore, 2 p−k−u−w=p−ℓ. By the definition of W, we have 0 6∈ ∑(S 1 W−1)and |S 1 W−1|= p−ℓ. Let T=S 1 W−1. Also, by the inequality (1), we see that ℓ=k+u+w−p≤ k+u+p−u−1−p=k−1≤h p+1 3 i. Therefore, by Theorem 2.3, we see that T∼(1 p−2 ℓ+1,x 1,x 2,...,x ℓ−1)and ℓ−1 ∑ i=1 x i≤2 ℓ−2. Thus, the given sequence S=0 u S 1=0 u T W is equivalent to the following sequence: S∼(0 u,1 p−2 ℓ+1,x 1,x 2,...,x ℓ−1,z 1,z 2,...,z w) where all the x i 6=0 satisfying ∑ℓ−1 i=1 x i≤2 ℓ−2 and W∼(z 1,z 2,...,z w)is the maxim al zero-sum subsequence of S 1. Without loss of generality, we shall replace ‘∼’ by ‘=’above. Also, we denote the number of 1’s appearing in th e sequences (x 1,x 2,...,x ℓ−1)and (z 1,z 2,...,z w)by r and t respectively. P ut v=p−2 ℓ+1+r+t. T o end the proof of this theorem, it is enough to prove that u+v≥2 p−2 k+2. If 2 ≤z i≤p−2 ℓ+1 for some i satisfying 1 ≤i≤w, then as there are p−2 ℓ+1 number of 1’s in T, we can write z i= σ (L 1)where L 1= (1 z i)with |L 1|≥2. If p−2 ℓ+2≤ z i≤p−2 ℓ+1+∑ℓ−1 j=1 x i holds for some 1 ≤i≤w, then by Lemma 2.4, there exists a subsequence L 1 of T such that z i= σ (L 1)and |L 1| ≥ 2. By letting W 1=L 1 Wz−1 i, we see that σ (W 1) = 0 and |W 1| ≥ w+1 wh ich contradicts the maximality of W. Hence p−1≥z i≥p−2 ℓ+2+ ℓ−1 ∑ i=1 x i for each z i 6=1.(2) Since ∑ℓ−1 i=1 x i≤2 ℓ−2, we have 2 ℓ−2≤ ℓ−1 ∑ i=1 x i+r.(3) Therefore, by the inequalities (2) and(3), we get p−1≥z i≥p−r for each z i 6=1.(4) By rearranging the indices and renaming them, if necessary, we can assume that for 0 ≤ q≤w, we have z i 6=1 for 1 ≤i≤q and z i=1 for q+1≤i≤w.(5) On the structure of p-zer o-sum free sequences 73 Case i.(w=2) In this case, by the d efinition of W, we have z 1+z 2=0. Therefore, there are two cases, namely, z 1=1 and z 2=−1 or z 1 6=1 and hence z 2 6=−1. When z 1 6=1, by the inequ ality (4), we have p−2≥z 1≥p−r and in particular, we have r≥2.Since 2 ≤r≤p− 2 ℓ+1, we have a zero-sum subsequence Z= (z 1,1 p−z 1)which has length >2 which is a contradiction to the maximality of W. Thus, z 1=1 and z 2=−1. In this case, v≥ 2 p−k−u−ℓ+r. Therefor e, u+v=u+2 p−k−u−ℓ+r≥2 p−k−(k−1) + r= 2 p−2 k+1+r≥2 p−2 k+3. W e are done in this case. Case ii.(w≥3) Since W is a zero-su m sequence, q 6=0. So,we have 1 ≤q≤w.When q=1, fro m the inequality (4), we get p−1≥z 1≥p−r.(6) When q=2, we have 2 p−2≥z 1+z 2≥2 p−2 r≥2 p−2(ℓ−1)≥2 p−2(k−2) =2 p−2 k+4. Since k≤h p+1 3 i+1, it is clear that p≥3 k−4 and hen ce 2 p−2≥z 1+z 2>p=⇒p−2≥z 1+z 2−p>p−2 ℓ+2. Therefore, it follows th at p−2≥z 1+z 2>p−2 ℓ+2. If z 1+z 2≤p−2 ℓ+1+∑ℓ−1 i=1 x i, then z 1+z 2= σ (L 2)for some subsequence L 2 of T with |L 2| ≥ p−2 ℓ+1 (by Lemma 2.4). If we let W 2=L 2 W z−1 1 z−1 2, then σ (W 2) = 0 and |W 2|= |L 2|+w−2≥p−2 ℓ+1+w−2=w+p−(2 ℓ+1)>w(as ℓ <k−1≤(p+1)/3) which contradicts the maximality of W. Theref ore, we have z 1+z 2≥∑ℓ−1 i=1 x i+p−2 ℓ+2. Thus, by the inequality (3), we have p−2≥z 1+z 2≥p−r.(7) Now, we shall assume that q≥3. Let a=min{q,w−2}. Then we claim the following. Claim 1.For q≥3 an d for every s=1,2,...,a, we have p−s≥s ∑ i=1 z i!−(s−1)p= s ∑ i=1 z i≥p−r. By inequalities (4) and(7), the Claim 1 is true for s=1 and 2. Now, we shall assume that claim 1 is true for s−1 and we prove for s. By the inequality (4)and induction hypothesis, we have p−s≥s ∑ i=1 z i!−(s−1)p=s−1 ∑ i=1 z i!−(s−2)p+z s−p ≥p−r−r=p−2 r≥p−2 ℓ+2≥p−2 k+4>0. 74 W D Gao, A P anigrahi and R Th angadurai Hence, p−2 ℓ+2≤ s ∑ i=1 z i. If ∑s i=1 z i≤p−2 ℓ+1+∑ℓ−1 i=1 x i, then b y Lemma 2.4, there exists a subsequence L 3 of T with |L 3| ≥ p−2 ℓ+1 su ch that ∑s i=1 z i= σ (L 3). If we let W 3=L 3 W z−1 1 z−1 2...z−1 s, then we get σ (W 3) = 0. Since w≤k−2, ℓ≤k−1 and p≥3 k−1, we have |W 3|=w+|L 3|−s≥w+p−2 ℓ+1−(w−2) ≥w+p−2 k+4+1−k+4=w+p−(3 k−9)>w. This contrad icts the fact tha t W is the maximal zero-sum subsequ ence of S 1. Therefore, we have p−s≥ s ∑ i=1 z i≥p−2 ℓ+2+ ℓ−1 ∑ i=1 x i and by the inequ ality (3), we get Claim 1. Claim 2. q≤w−2. Assume, on the contrary that q≥w−1. Then q=w−1 or q=w. If q=w−1, then we have p−(w−2)≥z 1+z 2+···+z w−2≥p−r,p−1≥z w−1≥p−r and z w=1. Therefore, 2 p>2 p−w+2≥z 1+z 2+···+z w−2+z w−1+z w ≥2 p−2 r+1>p which is a contradiction to σ (W) = 0. Hence q 6=w−1. If q=w, then p−(w−2)≥z 1+z 2+···+z w−2≥p−r,p−1≥z w−1,z w≥p−r. Therefore, 3 p>3 p−w≥z 1+z 2+···+z w−2+z w−1+z w ≥3 p−3 r≥3 p−3(k−2)≥3 p−3 k+6=2 p+p−3 k+6>2 p, (as r≤ℓ−1≤k−2 and p≥3 k−4) which is also a contra diction to σ (W) = 0. Hence q 6=w. Thus Claim 2 is true. From Claims 1 and 2, we see that s varies fro m 1 to q. Since we have p−s≥p−r which implies r≥s. In particular, when s=q, we get q≤r.(8) But by the definition of q,we have q=w−t which implies that w=q+t. Therefore, b y the inequality (8), we have r+t≥q+t=w. Thus u+v=u+p−2 ℓ+1+r+t≥u+p−2 ℓ+1+w =2 p−k−(ℓ−1)≥2 p−2 k+2, as desired.2 On the structure of p-zer o-sum free sequences 75 Proof of Theorem 3.2. Let S be a given sequence in Z p of length 2 p−k. Suppose the num- ber of d istinct residue classes appearing in S is g≥2√2√k−2. If possible, we assume that 0 6∈ ∑p(S). The n by Theorem 3.1, S=0 u T W (notations as in th e proof of Theo- rem 3.1). Now, we shall count the nu mber of distinct residue classes mo dulo p appearing in T and in W separately. W e recall that T= (1 p−2 ℓ+1,x 1,x 2,...,x ℓ−1)with ∑ℓ−1 i=1 x i≤2 ℓ−2 and r=v 1((x 1,x 2,...,x ℓ−1)). Also, W= (z 1,z 2,...,z q,1,1,...,1 |{z } w−q times )where z i 6=1. Note th at by Claim 2 of Theorem 3.1, we have 1 ≤q≤w−2 and by (6) and (8) we have q≤r and r≥2. Let g 1(respectively, g 2) denote the number of distinct residue classes modu lo p appear- ing in T(respectively,in W). Thus, including 0, the total number of distinct residue classes modulo p appearing in S is g=g 1+g 2+1−1=g 1+g 2 because the residue 1 is cal- culated twice in g 1 and g 2. So, to end the proof of this theorem, it is enou gh to estimate g=g 1+g 2. Since ∑ℓ−1 i=1 x i≤2 ℓ−2 and r number of 1’s appearing in (x i)s, we have 1+2+···+g 1≤2 ℓ−2−(r−1) g 2 1+g 1≤4 ℓ−4−2(r−1)≤4 ℓ−4−2=2(ℓ−3). Therefore, since ℓ≤k−1, we have g 2 1+g 1≤2(k−4) =⇒g 1≤√2√k−4<√2√k−2.(9) Now, note that−z i=p−z i. The refore by Claim 1 of Theorem 3.1,we get ∑q i=1−z i≤r. Thus, 1+2+···+g 2≤r=⇒g 2≤√2 r. Since r≤ℓ−1≤k−2, we have g 2≤√2 k−4=√2√k−2.(10) Thus, from the inequalities (9) and (10) and cou nting 0, we have g 1+g 2<√2√k−2+√2√k−2=2√2√k−2, a contradictio n. Hence the theorem.2 W e shall end this section with the following open problem s. Open Problem.Let n and k be two po sitive inte gers such that k≤n−2. Determine th e constant defined by h(n,k) = min{h(S)| |S|=n+k}, where S runs over all sequences in Z n of length n+k such that 0 6∈ ∑n(S). 76 W D Gao, A P anigrahi and R Th angadurai It is proved in and that h(n,n−2) = n−1 and proved in that h(n,n−3) = n−1. Theorem 2.2 shows that h(p,k)≥k+1 for all 1 ≤k≤p−2. The main result in implies that h(n,k)≥k+1 when ever n−[(n+1)/4]−1≤k≤n−2. It is natural to ask if h(n,k)≥k+1 for every positive integer n and every k such that 1 ≤k≤n−2. However, the answer is ‘no’ in general. Recently,in [1 5] we provided a co unter example for k satisfying p≤k≤n/p−2. We conjectured the following. Conjecture.Let n>1 b e any positive integer and let p b e the smallest prime divisor of n. Let k be an integer such that k≥(n/p)−1. Then h(n,k)≥k+1. In ,it is proved that Conjecture 1 is true for n=p ℓ for any prime p. Also, it is not known whether Conjecture 1 is tru e for k<p/3. Acknowledgements The first author is supp orted by NSFC with grant No s 19971058 and 10271080. References Bialostocki A and Dierker P, On Erd¨os–Ginzb urg–Ziv theorem and the Ramsey numbers for stars and matchings, Disc. Math. 110 (1992) 1–8 Bialostocki A, Dierker P, Grynkiewicz D and Lo tspeich M, On some developments of the Erd¨os–Ginzbur g–Ziv theorem II, Acta Arith. 110(2) (2003) 173–184 Bialostocki A and Lotspeich M, Some develo pments on the Erd¨os–Ginzburg–Zi v the- orem I, Sets, graphs and numbers (Budapest, 1991) Colloq. Math. Soc. 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Publ. 20 (2000) 91–98 Gallardo L, Grekos G, Habsieger L, Hennecart F, Landreau B and Plagne A, Restricted addition in Z/n Z and an application to the Erd¨os–Ginzburg–Ziv problem, J. London Math. Soc. (2) 65(3) (2002) 513–523 Gao W D, An addition theorem for finite cyclic groups, Disc. Math. 163 (1997) 257–265 Gao W D, Zero sums in finite cyclic groups, Integers 0(2000) #A 12, pp. 1–7 Gao W D, Thangadurai R and Zhuang J, Addition theorems on the cyclic groups Z p ℓ (2004) preprint On the structure of p-zer o-sum free sequences 77 Hamidoune Y O, Ordaz O and Ortu˜no A, On a combinatorial theorem of Erd¨os, Ginzburg and Ziv, Combin. Prob. Comput. 7(4) (1998) 403–412 Panigrahi A and Thangadurai R, A variant of Erd¨os–Ginzburg–Ziv Theorem (2004) preprint Peterson B and Y uster T, A generalization of an addition theorem for solv able groups, Can. J. Math. XXXVI(3) (1984) 529–536 W ang C, Note on a variant of the Erd¨os–Ginzburg–Zi v problem, Acta Arith. 108(1) (2003) 53–59 Citations (17) References (28) ... Our theorem already yields a non-trivial conclusion when A has slightly more than p elements. A similar statement was proved in (see also ), but under the stronger assumption that |A| ≥ 3 2 p. ... Classification theorems for sumsets modulo a prime Preprint Nov 2008 Hoi Nguyen Van Vu Let Z/p Z\Z/pZ be the finite field of prime order p and A be a subsequence of Z/p Z\Z/pZ. We prove several classification results about the following questions: (1) When can one represent zero as a sum of some elements of A ? (2) When can one represent every element of Z/p Z\Z/pZ as a sum of some elements of A ? (3) When can one represent every element of Z/p Z\Z/pZ as a sum of l elements of A ? View Show abstract ... Gao proved a statement to this effect for n-zero-free sequences of length roughly greater than 7n/4. A recent work of Gao, Panigrahi and Thangadurai considered the same question in the case of a prime n, for length roughly greater than 5n/3. ... Long n-zero-free sequences in finite cyclic groups Preprint Apr 2006 Svetoslav Savchev Fang Chen A sequence in the additive group Z n{\mathbb Z}_n of integers modulo n is called n-zero-free if it does not contain subsequences with length n and sum zero. The article characterizes the n-zero-free sequences in Z n{\mathbb Z}_n of length greater than 3 n/2−1 3n/2-1. The structure of these sequences is completely determined, which generalizes a number of previously known facts. The characterization cannot be extended in the same form to shorter sequence lengths. Consequences of the main result are best possible lower bounds for the maximum multiplicity of a term in an n-zero-free sequence of any given length greater than 3 n/2−1 3n/2-1 in Z n{\mathbb Z}_n, and also for the combined multiplicity of the two most repeated terms. Yet another application is finding the values in a certain range of a function related to the classic theorem of Erd\H{o}s, Ginzburg and Ziv. View Show abstract ... Ordaz and Flores determined the n-zero-sum free sequences over G of length 2n − 3. Gao characterized the n-zero-sum free sequences of length roughly greater than 7n 4 . Gao, Panigrahi and Thangadurai considered the same question for sequences of length roughly greater than 5n 3 providing that n is prime. The best result so far is achieved by Savchev and Chen in 2008. ... On the structure of n-zero-sum free sequences over cyclic groups of order n Article Full-text available May 2014 INT J NUMBER THEORY Jiangtao Peng Guoyou Qian Fang Sun Linlin Wang Let G be a finite cyclic group of order n. The Erdős–Ginzburg–Ziv theorem states that each sequence of length 2n - 1 over G has a zero-sum subsequence of length n. A sequence without a zero-sum subsequence of length n is called n-zero-sum free. Savchev and Chen characterized all the n-zero-sum free sequences of length n + k - 1 over G, where ⌊n 2⌋+1≤k<n\lfloor \frac{n}{2}\rfloor + 1 \le k < n. In the present paper, we determine all the n-zero-sum free sequences of length n+⌊n 2⌋−1 n + \lfloor \frac{n}{2}\rfloor -1 over G. View Show abstract ... It is easy to see that elementary 2-groups have Property D and H. Harborth [22, Beweis von Hilfssatz 3] showed (not using this terminology) that elementary 3-groups have Property D as well. Furthermore, for cyclic groups the inverse problems are well investigated (see, e.g., [12,11,29,17]), and in particular it is known that cyclic groups have Property D. Yet, for groups of rank 2 only Conjecture 1.1 is confirmed in general (cf. Theorem 2.2); the more general claims are confirmed only in special cases: for instance it is known that C 2 n has Property D if n is not divisible by a prime greater than 7 (see and the references there). ... On short zero-sum subsequences. II Article Full-text available Jan 2007 Weidong Gao Qing-Hu Hou Wolfgang A. Schmid R. Thangadurai Let G be a finite abelian group of exponent n. In this paper we investigate the structure of the maximal (in length) sequences over G that contain no zero-sum subsequence of length [at most] n. Among others, we obtain a result on the multiplicities of elements in these sequences, which support well-known conjectures on the structure of these sequences. Moreover, we investigate the related invariants s(G) and η(G), which are defined as the smallest integer l such that every sequence over G of length at least l has a zero-sum subsequence of length n (at most n, respectively). In particular, we obtain the precise value of s(G) for certain groups of rank 3. Part I, see Ars Comb. 74, 231–238 (2004; Zbl 1201.11031). View Show abstract ... For cyclic groups these questions are completely settled. Indeed, the answer for the invariants D(G) and η(G) is straightforward (see Theorem 2.1), and the inverse problem for the invariant s(G), with G cyclic, gave rise to a great variety of investigations (see [5,8,18,9,19,4,38,26,31,29]). In this paper we study the problems for groups G of the form G = C r n , with n, r ≥ 2, where the emphasis is laid on groups of rank two. ... Inverse zero-sum problems Article Full-text available Jan 2007 Weidong Gao Alfred Geroldinger Wolfgang A. Schmid View ... When k = m − 2, the inverse problem was solved by Yuster and Peterson and Bialostocki and Dierker ; k = m − 3 was solved by Flores and Ordaz ; and when m − [(m + 1)/4] − 1 k m − 2, the inverse problem was tackled by Gao . Also, for m = p, a prime number, Gao et al. solved this inverse problem when p − [(p + 1)/3] − 1 k p − 2. But it becomes difficult to describe the structure of S completely, when k is much smaller than m. ... Addition theorems on the cyclic groups of order Article May 2008 DISCRETE MATH Weidong Gao R. Thangadurai Jujuan Zhuang Let p be a prime number and ℓ be any positive integer. Let G be the cyclic group of order pℓ and let S be any sequence in G of length pℓ+k for some positive integer k⩾pℓ-1-1 such that S do not admit a subsequence of length pℓ whose sum is zero in G. Then we prove that there exists an element of G which appears in S at least k+1 times. View Show abstract Analytic number theory in India during 2001–2010 Article Sep 2019 Ritabrata Munshi In this article we summarize the contribution of Indian mathematicians to analytic number theory during 2001–2010. View Show abstract Inverse zero-sum problems and arithmetical consequences Chapter Jan 2009 Alfred Geroldinger Imre Ruzsa The investigation of inverse problems has a long tradition in combinatorial number theory (see [107, 37]), and more recently it has been promoted by applications in the theory of non-unique factorizations. In this chapter we discuss the inverse problems associated with the invariants D(G), η(G) and s(G). More precisely, we investigate the structure of sequences of length D(G)−1 (η(G)−1 or s(G)−1, respectively) that do not have a zero-sum subsequence (of the required length). Recent results on the structure of Σ(S) for (long) zero-sum free sequences may be found in [9, 58, 127, 132, 60]. View Show abstract Two conjectures on an addition theorem Article Jan 2011 Xiangneng Zeng Pingzhi Yuan View On Erdős–Ginzburg–Ziv inverse theorems Article Jan 2007 David J. Grynkiewicz Oscar Ordaz Maria Teresa Varela Felicia Villarroel View Show more On a variant of the Erdős-Ginzburg-Ziv problem Article Full-text available Jan 1999 Luis H. Gallardo Georges Grekos Jukka Pihko View Show abstract Note on a variant of the Erdo"s-Ginzburg-Ziv problem Article Full-text available Jan 2003 Chao Wang In (?), A. Bialostocki and M. Lotspeich introduced a function f(n,k) to study the relation between the number of residue classes modulo n present in a sequence A = (a1,...,ag) and the possibility to have a relation like ai1+ai2+···+ain 0(mod n). In this paper, the author obtained a formula for f(n,k) when n is big enough relative to k. View Show abstract Some developments of the Erdős-Ginzburg-Ziv theorem. I Article Arie Bialostocki M. Lotspeich View Conditions for a Zero Sum Modulo n Article Mar 1975 J.D. Bovey Paul Erdős Ivan Niven In this paper the following result is proved. Let n > 0 and k ≥ 0 be integers with n — 2k ≥ 1. Given any n — k integers there is a non-empty subset of indices I ⊂ {1, 2,…, n — k} such that the sum Σ i∊I ≡ 0(mod n) if at most n — 2k of the integers (1) lie in the same residue class modulo n. The result is best possible if n ≥ 3k — 2 in the sense that if "at most n — 2k" is replaced by "at most n — 2k + 1" the result becomes false. This can be seen by taking a j = 1 for 1 ≤ j ≤ n — 2k + 1 and a j = 2 for n —2k + 2 ≤ j ≤ n — k, noting that the number of 2's here is n — k — (n — 2k + 1)= k — 1 ≤ n — 2k + 1. View Show abstract A Generalization of an Addition Theorem for Solvable Groups Article Jun 1984 CAN J MATH Thomas Yuster Bruce Peterson The “sets” in this paper are actually multi-sets. That is, we allow an element to occur several times in a set and distinguish between the number of elements in a set and the number of distinct elements in the set. On the few occasions when we need to avoid repetition we will use the term “ordinary set.“ Definition. Let G be a group and let S a set of elements of G. An r-sum in S is an ordered subset of S of cardinality r; the result of that r - sum is the product of its elements in the designated order. Definition. If S is a set, r ( x , S ) denotes the number of times x appears in S and [ x, S ] is a set consisting of r ( x, S ) copies of x. An n-set or n-subset is a set consisting of n elements. Hence [ x , S ] is an r ( x, S )-subset of S. View Show abstract On sequences with zero sum in abelian groups Article Oscar Ordaz Carlos Flores View On Brakemeier’s variant of the Erdős-Ginzburg-Ziv problem Article Jan 2000 Luis H. Gallardo Georges Grekos View On a Combinatorial Theorem of Erdös, Ginzburg and Ziv Article Dec 1998 COMB PROBAB COMPUT Yahya Ould Hamidoune Oscar Ordaz ASDRUBAL ORTUÑO Let G be an abelian group of order n and let k+1 taking k distinct values. Assuming that no value occurs n must include a non-null subgroup. As a corollary we show that if G is cyclic then Ginzburg–Ziv theorem for k=2. View Show abstract On some developments of the Erdo"s-Ginzburg-Ziv Theorem II Article Jan 2003 ACTA ARITH Arie Bialostocki P. Dierker David J. Grynkiewicz M. Lotspeich Let S be a sequence of elements from the cyclic group ℤ m . We say S is zsf (zero-sum free) if there does not exist an m-term subsequence of S whose sum is zero. Denote by g(m,k) the least integer such that every sequence S with at least k distinct elements and length g(m,k) must contain an m-term subsequence whose sum is zero. Furthermore, denote by E(m,s) the set of all equivalence classes of zsf sequences with length s, up to order and affine transformation, that are not a proper subsequence of another zsf sequence. In this paper, we first find for a sequence S of sufficient length, |S|≥2m-⌊m/4⌋-2, necessary and sufficient conditions in terms of a system of inequalities over the integers for S to be zsf. Among the consequences, we determine g(m,k) for large m, namely g(m,k)=2m-⌊(k 2 -2k+5)/4⌋ provided m≥k 2 -2k-3, which in turn resolves two conjectures of the first and fourth authors. Next, using independent methods, we evaluate g(m,5) for every m≥5. We conclude with the list of E(m,s) for every m and s satisfying 2m-2≥s≥max{2m-8,2m-⌊m/4⌋-2}. Part I, cf. Colloq. Math. Soc. János Bolyai 60, 97–117 (1992; Zbl 1042.11510). View Show abstract On the addition of residue classes mod p Article Mar 1996 Ösytein J. Rödseth Given distinct residue classesa1,...,ak modulo a primep, we consider the setSj of all sums , wherel≦i1ij≤k. We give a sufficient condition for the inequality |Sj|≧j(k−j)+1 to hold. View Show abstract Show more Recommended publications Discover more about:Free Will Chapter Littlewood’s theorem and Weyl’s method January 1970 Prof. Dr. K. Chandrasekharan We have proved in Chapter II, Theorem 4, that all the non-real zeros of Riemann’s zeta-function ζ(s) lie in the critical strip 0≤σ≤1, and are symmetrically situated about the lines σ =1/2 and t = 0, where σ = Re s, t = Im s. We shall now prove a theorem of J. E. Littlewood that there exists a positive constant A such that the region \sigma>1-\frac{{A\log\log(\left|t\right|+ ... [Show full abstract] 3)}}{{\log(\left|t\right|+3)}} (1) is free from any zero of ζ. Read more Article Full-text available New extremal Type II Z4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usep... December 2024 · Applicable Algebra in Engineering Communication and Computing Sara Ban Martinović Sanja Rukavina Type II Z 4{\mathbb {Z}}_4-codes are a class of self-dual Z 4{\mathbb {Z}}_4-codes with Euclidean weights divisible by eight. A Type II Z 4{\mathbb {Z}}_4-code of length n is extremal if its minimum Euclidean weight is equal to 8⌊n 24⌋+8.8\left\lfloor \frac{n}{24}\right\rfloor +8. A small number of such codes is known for lengths greater than or equal to 48. Based on the doubling method, in this ... [Show full abstract] paper we develop a method to construct new extremal Type II Z 4{\mathbb {Z}}_4-codes starting from a free extremal Type II Z 4{\mathbb {Z}}_4-code of length 48, 56 or 64. Using this method, we construct extremal Type II Z 4{\mathbb {Z}}_4-codes of length 64 and type 4 31 2 2 4^{31}2^2. Extremal Type II Z 4{\mathbb {Z}}_4-codes of length 64 of this type were not known before. Moreover, the residue codes of the constructed extremal Z 4{\mathbb {Z}}_4-codes are best known [64, 31] binary codes and the supports of the minimum weight codewords of the residue code and the torsion code of one of these codes yields self-orthogonal 1-design. View full-text Article Addition theorems on the cyclic groups of order May 2008 · Discrete Mathematics Weidong Gao R. Thangadurai Jujuan Zhuang Let p be a prime number and ℓ be any positive integer. Let G be the cyclic group of order pℓ and let S be any sequence in G of length pℓ+k for some positive integer k⩾pℓ-1-1 such that S do not admit a subsequence of length pℓ whose sum is zero in G. Then we prove that there exists an element of G which appears in S at least k+1 times. Read more Article Full-text available On zero-sum sequences of prescribed length December 2006 · Aequationes mathematicae Weidong Gao R. Thangadurai Let k ≥ 1 be any integer. Let G be a finite abelian group of exponent n. Let s k (G) be the smallest positive integer t such that every sequence S in G of length at least t has a zero-sum subsequence of length kn. We study this constant for groups G \cong {\user2{{\mathbb{Z}}}}^{d}{n} when d = 3 or 4. In particular, we prove, as a main result, that (s{k} ({\user2{{\mathbb{Z}}}}^{3}{p} ) = ... [Show full abstract] kp + 3p - 3) for every k ≥ 4, 5p + \frac{{p - 1}}{2} - 3 \leq s{2} ({\user2{{\mathbb{Z}}}}^{3}{p} ) \leq 6p - 3 and 6p - 3 \leq s{3} ({\user2{{\mathbb{Z}}}}^{3}_{p} ) \leq 8p - 7 for every prime p ≥ 5. View full-text Article A variant of Kemnitz Conjecture July 2004 · Journal of Combinatorial Theory Series A Weidong Gao R. Thangadurai For any integer n greater than or equal to 3, by g(Z(n) circle plus Z(n)) we denote the smallest positive integer t such that every subset of cardinality t of the group Z(n) circle plus Z(n) contains a subset of cardinality n whose sum is zero. Kemnitz (Extremalprobleme fur Gitterpunkte, Ph.D. Thesis, Technische Universitat Braunschweig, 1982) proved that g(Z(p) circle plus Z(p)) = 2p - 1 for p = ... [Show full abstract] 3, 5, 7. In this paper, as our main result, we prove that g(Z(p) circle plus Z(p)) = 2p - 1 for all primes p greater than or equal to 67. Read more Article Olson’s constant for the group ℤ p ⊕ℤ p July 2004 · Journal of Combinatorial Theory Series A Weidong Gao R. Thangadurai Imre Z. Ruzsa Let G be a finite abelian group. By Ol(G), we mean the smallest integer t such that every subset A⊂G of cardinality t contains a non-empty subset whose sum is zero. In this article, we shall prove that for all primes p>4.67×1034, we have Ol(Zp⊕Zp)=p+Ol(Zp)−1 and hence we have Ol(Zp⊕Zp)⩽p−1+⌈2p+5logp⌉. This, in particular, proves that a conjecture of Erdős (stated below) is true for the group ... [Show full abstract] Zp⊕Zp for all primes p>4.67×1034. Read more Last Updated: 07 Aug 2025 Interested in research on Free Will? Join ResearchGate to discover and stay up-to-date with the latest research from leading experts in Free Will and many other scientific topics. 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https://byjus.com/jee/coherence-and-coherent-sources/
Coherence describes the properties of the interrelation between physical quantities of a single wave or between several waves. Two waves are coherent when they have a constant relative phase or when they have zero or constant phase difference and the same frequency. Coherence is also the property which enables waves to have stationary interference. Besides, the amount of coherence is usually measured by interference visibility. For instance, two parallel slits lighted by a laser beam can be classified as two coherent point sources. We will learn about this concept in detail, along with types, characteristics, coherent sources and more here. Types of Coherence There are two types of coherence, namely temporal coherence and spatial coherence. These two types are explained below. Temporal Coherence Temporal coherence is a measure of the average relationship between the value of a wave and itself, making a delay of τ at any considerable pair of times. Temporal coherence gives the measurement of how monochromatic a source is. It characterises how well a wave can interfere with itself at different times. The delay above which the phase or amplitude should be by a significant amount (and hence the correlation decreases by a significant amount) is defined as the coherence time. Spatial Coherence If you take systems such as optics or water waves, you will find the dimension of waves extends from one or two spaces. Spatial coherence can, therefore, be described as the ability of two points in the space of a wave (x1 and x2) that will interfere. Spatial coherence, in simple words, can be said to be the cross-relation between two points in a wave at all times. When a wave has a single amplitude value over an infinite length, it is said to have perfect spatial coherence. The significant interference that is present between the range of separation and the two points can be used to define the diameter of the coherence area, Ac. Coherent Sources Coherent sources of light are those sources which emit a light wave having the same frequency, wavelength and in the same phase, or they have a constant phase difference. A coherent source forms sustained interference patterns when the superimposition of waves occurs, and the positions of maxima and minima are fixed. Two independent sources are never coherent, or they cannot be considered coherent sources, as all the above-mentioned factors cannot be present at the same time. Examples Some of the common examples of coherent sources are given below: Sound waves produced by speakers are driven by electrical signals that have the same frequency and a definite phase. A laser is also a type of coherent source. Lasers are commonly used as coherent sources as it uses a phenomenon called stimulated emission to generate highly coherent light. Small sources of light are at least partially coherent. Because of this, we can observe interference patterns on soap bubbles and appreciate the iridescence of butterfly wings. While sunlight is incoherent while small portions on small areas are generally partially coherent. Characteristics of Coherent Sources Coherent sources have the following characteristics: The waves generated have a constant phase difference (they are in phase with each other). The waves are of a single frequency. The waves should have the same amplitude. Methods of Producing Coherent Source Finding a coherent source of light is sometimes difficult. Therefore, we can use different methods to produce such sources, and they are as follows: By Dividing the Wavefront Wavefront is divided into several parts. We can use either different lenses, mirrors, and even prisms. Some of the techniques that can be followed are Young’s double-slit experiment, Lloyd’s mirror arrangement, Fresnel’s biprism method, etc. By Dividing the Amplitude A coherent source can be created by taking the incoming beam and dividing its amplitude into different parts through a process of partial reflection or refraction. Later, these parts that travel through new paths meet with each other creating an interference. We can take a reference of Newton’s ring phenomenon or make use of Michelson’s interferometer arrangement or configuration. Incoherent Sources Incoherent sources are the exact opposite of coherent sources. These types of sources emit light with frequency, and there are random changes in phase between the photons. Examples of incoherent sources include all conventional light sources. The transitions between energy levels in an atom are a completely random process. So, we have no control over when an atom is going to lose energy in the form of radiation. The below figure shows the diagram of an incoherent wave. Interference Interference is a phenomenon in which two waves superimpose to form a resultant wave which has greater, lower or the same amplitude. Interference is of types – constructive and destructive. They result from the combination of waves that are interrelated or coherent with each other, either because they come from the same source or because they have the same or nearly the same frequency. Read more about it here: Interference of Waves Why Are Coherent Sources Essential for Observable Interference? We need coherent sources of light mainly to observe the effects of interference. Coherent sources have the same phase, so the phase difference between the two sources remains constant. This is a necessary condition for observable and distinct interference. They help us identify the interference patterns that occur as a result of the change in phase at a given point (at source, at the screen ). This also gives us the minima and maxima, i.e. constructive and destructive interference, which can be seen at different points on the screen simultaneously. Sometimes the phases interfere at such a high rate that our eyes are not able to detect the simultaneous formation of fringes. Applications of Coherence Radiography is an old technique. Very recently, it took a new way that the coherence of a next-generation facility beam makes it possible in an easy way to break the usual barrier to absorption and to visualise phase features. The X-ray beam has, A high spatial coherence, which means that the size and the divergence of the beam are very small. A good temporal coherence after monochromatization. These characteristics of the beam, which are due to its super brilliance, allow new techniques to be developed in the X-ray field: Phase-contrast imaging and tomography In-line holography Photon correlation spectroscopy Additionally, the property of coherence is the basis for many other applications. Some of the popular ones are as follows: Holography Holography makes use of coherent superpositions of optical wave fields, and its uses are commonly found in credit cards. Non-optical Wavefield Opposite to holography, the superposition of non-optical wavefields is also considered. For example, in quantum mechanics, there is a consideration for the probability field, which is related to the wave function. The applications of coherent waves can be seen among the future technologies of field quantum computing, etc. Modal Analysis The coherence signal is used in modal analysis to assess the quality of transfer functions. Young’s Double Slit Experiment 59,650 Frequently Asked Questions on Coherence Q1 What are coherent sources? Coherent sources are two sources which emit light waves with the same frequency with a constant phase difference. Q2 Why should the source be coherent to obtain a sustained interference pattern? The intensity of the bright fringe or the dark fringe does not fluctuate only when the source is coherent. Q3 Why can’t we get an interference pattern with two different sources? Two independent sources will not be coherent, so they cannot be used to produce an interference pattern. Q4 The light waves from two coherent sources have intensities in the ratio of 36:16. What is the ratio of their amplitude? I1 ∝ A12 I2∝ A12 (I1/I2) = A12/A22 = 36/16 ⇒A1/A2 = 6/4 A1:A2 = 6:4 = 3:2 Test your Knowledge on Coherence And Coherent Sources Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Jee related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs
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https://www.quora.com/Why-is-Leonardos-Last-Supper-composition-more-famous-than-others-from-the-Renaissance
Why is Leonardo's 'Last Supper' composition more famous than others from the Renaissance? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Art History The Last Supper Leonardo da Vinci The Renaissance Famous Paintings and Pain... Art Appreciation Painting Composition Classic Works of Art Renaissance Artists 5 Why is Leonardo's "Last Supper" composition more famous than others from the Renaissance? All related (32) Sort Recommended Ellen B Cutler A love of art led to a career in museums and teaching. · Author has 5K answers and 3.9M answer views ·Updated 4y Hi, Sandy Ray. Leonardo’s Last Supper composition is more famous than others of that era because it was exceptional in its conception and execution, except for the actual painting because that was a disaster, technically speaking, from the get-go. Leonardo, Last Supper (1595–98), after most recent restoration You have requested a book not an answer. Leonardo did two things. The first was to create a remarkable composition that summarizes that extraordinary moment when Jesus tells his disciples that one of them will betray him. Like colliding waves, the four groups of three disciples rise, recoil, Continue Reading Hi, Sandy Ray. Leonardo’s Last Supper composition is more famous than others of that era because it was exceptional in its conception and execution, except for the actual painting because that was a disaster, technically speaking, from the get-go. Leonardo, Last Supper (1595–98), after most recent restoration You have requested a book not an answer. Leonardo did two things. The first was to create a remarkable composition that summarizes that extraordinary moment when Jesus tells his disciples that one of them will betray him. Like colliding waves, the four groups of three disciples rise, recoil, move forward, look to one another for some kind of reassurance, as Jesus is a still, calm figure at the center of the storm. The magical four (compass points, seasons, humors, etc.) appear in the wall hangings on the sides. The three of the Trinity are the windows behind Jesus. The center one also functions as a sort of mandorla or halo; together the Palladian structure with the arched lintel at the center remind us of the classical triumphal arch (think of the arch of Constantine) and the triumph to which it alludes is of course salvation. The perspective was seamless with the refectory where it was painted and so the monks must have felt every meal was that communion of the last supper. But magically, the perspective comes together only when the viewers eyes are on the same level as Jesus’ head. This was also a reminder to all there that they were in the sinful here and now, in no way on “the same level” as the disciples, let alone Christ himself. diagram of one point perspective in Last Supper Imaginative reconstruction of room depicted It was famous in its own time—and its quick degradation was noted and mourned—because it was simply visually, theologically, imaginatively, philosophically and aesthetically the greatest visualization of the event. Ever. In many ways this is the greatest example of Leonardo’s genius and how he was victim of his own hubris. Upvote · 9 4 9 1 Promoted by Betterbuck Anthony Madden Writer for Betterbuck ·Updated Aug 15 What are the weirdest mistakes people make on the internet right now? Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your pr Continue Reading Here are a couple of the worst mistakes I’ve seen people make: Not using an ad blocker If you aren’t using an ad blocker yet, you definitely should be. A good ad blocking app will eliminate virtually all of the ads you’d see on the internet before they load. No more YouTube ads, no more banner ads, no more pop-up ads, etc. Most people I know use Total Adblock (link here) - it’s about £2/month, but there are plenty of solid options. Ads also typically take a while to load, so using an ad blocker reduces loading times (typically by 50% or more). They also block ad tracking pixels to protect your privacy, which is nice. More often than not, it saves even more than 50% on load times - here’s a test I ran: Using an ad blocker saved a whopping 6.5+ seconds of load time. Here’s a link to Total Adblock, if you’re interested. Not getting paid for your screentime Apps like Freecash will pay you to test new games on your phone. Some testers get paid as much as £270/game. Here are a few examples right now (from Freecash's website): You don't need any kind of prior experience or degree or anything: all you need is a smartphone (Android or IOS). If you're scrolling on your phone anyway, why not get paid for it? I've used Freecash in the past - it’s solid. (They also gave me a £3 bonus instantly when I installed my first game, which was cool). Upvote · 999 557 99 60 9 4 Related questions More answers below Are there other depictions of The Last Supper before Leonardo da Vinci’s? Why is the Last Supper painting deteriorating? How has been Leonardo da Vinci's painting "The Last Supper" conserved? What does the Last Supper painting signify? What are the secrets inside Leonardo Da Vinci's "Last Supper"? Mark McKee Artist of a certain age. MFA, BFA Professor of Art · Author has 19.4K answers and 7.7M answer views ·4y The awe-inspiring fame of Leonardo da Vinci as the original Renaissance Man colors the polymath’s enter oeuvre. Just about anything and everything touch by him has turned to historic Gold. However, as seemingly wonderful as this painting may be, da Vinci’s “The Last Supper” is probably as well known among art and art history scholars as a major technical failure. Executed in and experimental secco technique on a surface of dubious quality, the mural is literally disintegrating as conservators desperately attempt to stabilize and preserve it a square inch at a time. That said, it is a pretty stra Continue Reading The awe-inspiring fame of Leonardo da Vinci as the original Renaissance Man colors the polymath’s enter oeuvre. Just about anything and everything touch by him has turned to historic Gold. However, as seemingly wonderful as this painting may be, da Vinci’s “The Last Supper” is probably as well known among art and art history scholars as a major technical failure. Executed in and experimental secco technique on a surface of dubious quality, the mural is literally disintegrating as conservators desperately attempt to stabilize and preserve it a square inch at a time. That said, it is a pretty straight-forward composition of a much covered subject. Save that it was executed by one who is considered by many thee supreme being in the canon of Western Art. (Mostly because for many this is the ONLY person they know of or recall regarding artists.) Upvote · 9 1 K Grace 30 years an artist, art consultant, instructor, 10 years certified framer · Author has 22.1K answers and 46.7M answer views ·4y Related Why is Da Vinci's 'Mona Lisa' painting considered better than his 'Last Supper' painting? The Last Supper is an oil and tempera painting on a wall, significantly deteriorated condition, from time and environment. It is by far a masterpiece, but they can’t compare, since it is a mural. Because of the experimental nature of the work, it hasn’t endured well. What makes the Mona Lisa more popular is the nature of the painting, it is a portrait, and with the enigmatic question of the smile, that in itself makes the painting intriguing. The Last Supper - which has its own meanings and interpretations - is a religious work, it is a celebration of Christianity, and that may offend some. That Continue Reading The Last Supper is an oil and tempera painting on a wall, significantly deteriorated condition, from time and environment. It is by far a masterpiece, but they can’t compare, since it is a mural. Because of the experimental nature of the work, it hasn’t endured well. What makes the Mona Lisa more popular is the nature of the painting, it is a portrait, and with the enigmatic question of the smile, that in itself makes the painting intriguing. The Last Supper - which has its own meanings and interpretations - is a religious work, it is a celebration of Christianity, and that may offend some. That separates it from the Mona Lisa, who conveys a great a mystery about the woman, what is behind that smile, and why. That mystery is what draws people in, it makes the Mona Lisa an enduring question. But better, the Mona Lisa has remained in far better condition considering its age though it is now at risk and is regularly evaluated to ensure it’s condition doesn’t deteriorate too severely. It was painted well, has survived well, and from that simple context, it is better than The Last Supper, Upvote · 9 7 9 1 Martin Fox Art Historian · Author has 3.6K answers and 8.9M answer views ·4y Related Why is Da Vinci's 'Mona Lisa' painting considered better than his 'Last Supper' painting? It’s not. The Last Supper is a superior painting in terms of its ambition and its advancement of Renaissance ideals. It’s a complex, multi-figure painting that conveys a moment that is central to Christianity in a innovative way, uses perspective in a fascinating fashion, and represents a high point in Leonardo da Vinci’s career. Unfortunately, it’s not been well preserved, as a tempera painting that has significantly deteriorated over time. The Mona Lisa is perhaps better known because it’s in one of the most-visited museums in the world and because it’s in much better condition. It’s not a be Continue Reading It’s not. The Last Supper is a superior painting in terms of its ambition and its advancement of Renaissance ideals. It’s a complex, multi-figure painting that conveys a moment that is central to Christianity in a innovative way, uses perspective in a fascinating fashion, and represents a high point in Leonardo da Vinci’s career. Unfortunately, it’s not been well preserved, as a tempera painting that has significantly deteriorated over time. The Mona Lisa is perhaps better known because it’s in one of the most-visited museums in the world and because it’s in much better condition. It’s not a better painting. Upvote · 9 6 9 1 Sponsored by Charles Stanley Direct Open an account in minutes. Get up to £1,500 cashback when you transfer an account. Terms apply. Investment involves risk. Sign Up 99 68 Related questions More answers below What is the significance of Leonardo da Vinci's "The Last Supper" painting? How much is the last supper worth? Why is The Last Supper (Leonardo da Vinci) such a source for conspiracies? Why is the last supper so expensive? What happened to Leonardo da Vinci's last supper? Archie bulsara . ·9y Related What are the secrets inside Leonardo Da Vinci's "Last Supper"? ​ ​ The Last Supper is a late 15th-century muralpainting by Leonardo da Vinci in the refectory of the Convent of Santa Maria delle Grazie, Milan. The painting represents the scene of The Last Supper of Jesus with his disciples, as it is told in the Gospel of John, 13:21. Leonardo has depicted the consternation that occurred among theTwelve Disciples when Jesus announced that one of them would betray him. Well, THE LAST SUPPER contain many secrets 1:- The geometry used by Sir Leonardo, as per all artist use a geometry before printing (sketch or something) but is not shown in the fina Continue Reading ​ ​ The Last Supper is a late 15th-century muralpainting by Leonardo da Vinci in the refectory of the Convent of Santa Maria delle Grazie, Milan. The painting represents the scene of The Last Supper of Jesus with his disciples, as it is told in the Gospel of John, 13:21. Leonardo has depicted the consternation that occurred among theTwelve Disciples when Jesus announced that one of them would betray him. Well, THE LAST SUPPER contain many secrets 1:- The geometry used by Sir Leonardo, as per all artist use a geometry before printing (sketch or something) but is not shown in the final painted piece. But,in the case of THE LAST SUPPER , Sir Leonardo showed the geometry, a triangle, in the final painted piece. ​ ​ 2:-There is a message or a text written at the down side of the painting (no one is able to read that). 3:-In the mirrored image there was seen a cup( known as holy grail )on the chest of Jesus. ​ ​ And By mirroring the image of Jesus (just Jesus)in THE LAST SUPPER we will find a image of goat or several goats [Weird! ,I know, but personally I think this mirror thing have not a great excestense ,else I don't know. It can be true and may be not.] ​ ​ 4:-The bread pieces represent something, well a new theory has discovered about it. 5:-There were 13 people in the painting- Left to right Bartholomew James minor Andrew Peter Judas John( or marry) Jesus Thomas James major Philip Mathew Thaddeus Simon And the Peter was holding a knife pointing towards Andrew Why he was doing so and why Sir Leonardo painted that? ​ ​ After Dan Brown's research in new directions 1:- The person who is setting after Jesus,named as John ,is marry . Jesus's wife There was a book found by a former, people says it's written by Jesus himself and the book cantains a note (sentence) with some missing letters, says 'a kiss to marry Magdalene '(something like that, if we are considering the missing litters correct) Mario Taddei Leonardo da Vinci expert disagreed this theory as he said John was more look like a women. He said it's John not marry. ​ ​ 2:- There is a image of a baby on Judas ,by seeing it closely and Andrew's eyes are on the baby ​ ​ 3:-marry and Jesus togetherly make a M which represent Marry ​ ​ 4:-By changing the position of the image if marry ,we find marry's head on Jesus's shoulder . ​ ​ 5:-marry and Jesus were wearing same type of clothes in same position.. ​ ​ 6:-Jesus and marry were representing male and female symbols as Jesus a triangle pointing towards up and marry a triangle pointing downwards ​ ​ ​ ​ [May be many things or secrets are left ,If you now some you can comment below, I will be thank full] Upvote · 99 29 99 11 9 1 Brian Overland Longtime student of history, foreign and US. · Author has 31.3K answers and 103.7M answer views ·3y Related Why is the "Last Supper" painting considered part of the Renaissance? Biblical subjects were often subjects of Renaissance art. This includes not only Leonardo’s Last Supper, but also Michelangelo’s greatest works: All the paintings on the ceiling on the Sistine Chapel, which show many scenes from the Bible. The statue of young David, who is destined to become King of Israel according to the Bible. How do you reconcile this? Well, remember that even during the Scientific Enlightenment, the new thinking did not reject Jesus. Instead, modern thinkers reinterpreted Jesus by finding in the New Testament an emphasis on Jesus’ humane and compassionate teachings. The is n Continue Reading Biblical subjects were often subjects of Renaissance art. This includes not only Leonardo’s Last Supper, but also Michelangelo’s greatest works: All the paintings on the ceiling on the Sistine Chapel, which show many scenes from the Bible. The statue of young David, who is destined to become King of Israel according to the Bible. How do you reconcile this? Well, remember that even during the Scientific Enlightenment, the new thinking did not reject Jesus. Instead, modern thinkers reinterpreted Jesus by finding in the New Testament an emphasis on Jesus’ humane and compassionate teachings. The is not a new religion, because this side of Jesus was always there, and even in the Scientific Enlightenment, he was not rejected, merely re-interpreted. But consider the Last Supper itself. It is a Renaissance masterpiece, because of the thirteen people shown, it is their humanness and individuality that are emphasized. This supper, as Leonardo shows it, is not a gathering of thirteen saints: far from it! Of those present, only Jesus appears particularly saintly — and this is not a great and powerful Jesus who is kicking ass and taking names: no, this is a Jesus of such humanity and modesty that his image is moving. And the others? This is a work of genius, because the other twelve show various degrees of mistrust, envy, suspicion, fear, pride, apprehension, wonder, and doubt… so much of the range of emotions of human beings. Upvote · 9 3 9 1 9 2 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 620 Richa Nagare Relishing the simple pleasures in Life! · Author has 127 answers and 650.7K answer views ·11y Related What makes the painting 'The Last Supper', painted by da Vinci, remarkable? In terms of technicality, if you observe closely you shall realize directly or indirectly every person has his final attention midpoint of the composition i.e. Christ. ( i remember this because my art teacher had mentioned this as the best example of art with a focal point where there is a centre of interest e.e Christ in the last supper painting). Here is another one for your refernce. I hope I answered your question well Regards, Richa Continue Reading In terms of technicality, if you observe closely you shall realize directly or indirectly every person has his final attention midpoint of the composition i.e. Christ. ( i remember this because my art teacher had mentioned this as the best example of art with a focal point where there is a centre of interest e.e Christ in the last supper painting). Here is another one for your refernce. I hope I answered your question well Regards, Richa Upvote · 9 1 9 1 Barbara Fracchia Lives in Kensington, California (2000–present) · Author has 2.1K answers and 489.5K answer views ·1y Related Why is Leonardo da Vinci Lord’s supper picture so famous? WHAT??????? Are you telling me he did ANOTHER painting besides Mona Lisa? I am shocked and amazed. Why was it so famous???? He captured the essence of the Religious Right with their administration of life under God (well something like that I am not too religious) in his Last Supper. It is a magnificent painting (huge) My husband and I viewed it when we visited Italy many years ago. It was not a little dinky portrait of Mona Lisa that everyone raves about and tries to understand the meaning. Upvote · 9 1 Sponsored by Giftster Never get the wrong gift again. Tired of guessing what your family wants? Giftster lets everyone make and share wish lists. Sign Up 999 624 Mark McKee Artist of a certain age. MFA, BFA Professor of Art · Author has 19.4K answers and 7.7M answer views ·4y Related Why is Da Vinci's 'Mona Lisa' painting considered better than his 'Last Supper' painting? Generally the material quality of the former is superior to the latter. La Giaconda (Mona Lisa) though actually a rather common subject and composition is the product of meticulous craftsmanship, executed over an extended period. da Vinci’s Last Supper on the other hand is an example of how NOT to execute a mural. Unlike the usual, approved and time tested “fresco” medium and technical, the original Renaissance Man, executed his famous failure in an experimental ‘secco’ or dry medium and technique. And this his did on a relatively porous surface which it turned out “bloomed” with moisture and va Continue Reading Generally the material quality of the former is superior to the latter. La Giaconda (Mona Lisa) though actually a rather common subject and composition is the product of meticulous craftsmanship, executed over an extended period. da Vinci’s Last Supper on the other hand is an example of how NOT to execute a mural. Unlike the usual, approved and time tested “fresco” medium and technical, the original Renaissance Man, executed his famous failure in an experimental ‘secco’ or dry medium and technique. And this his did on a relatively porous surface which it turned out “bloomed” with moisture and various mineral salts from behind, under the already dubiously applied preparation and subsequent paint layers. The result is a craftsman’s disaster which was disintegrating nearly as soon as it was completed. Bits and pieces of the da Vinci’s work have rained off continuously over the centuries (Note that Michelangelo’s Sistine Chapel works, though old indeed, have remained durably inplace, due to his proper preparation and execution of the traditional fresco technique.) For the past few decades heroic effort has been undertaken to painstakingly go over the Last Supper, inch by inch, desperately attempting to stabilized and preserve da Vinci’s biggest mistake. It is a constant battle that conservators may never truly win, only keep on fighting. All the while, La Giaconda serenely smiles, only requiring the occasional visit to the conservation lab for a freshening up. Upvote · Michael Bross Former Psychology Professor,Fellow Liberal Arts College, at Concordia University, Montreal (1974–2005) · Author has 1.7K answers and 12.5M answer views ·Updated 5y Related What makes Leonardo da Vinci's Last Supper beautiful? This answer will concern itself with an unusual projective technique Leonardo used in painting the last supper which results in what can be considered a somewhat perceptually enforced attention on part of the viewer. Notice that all perspective lines come together (the so called vanishing point) at the head of Jesus, but this is not how a visitor sees it, to take a picture of the Last Supper with the correct perspective (as above) it has to be done like this: But when you see the picture as a visitor this is how you see it: Now, in normal life and in paintings the vanishing point coincides with t Continue Reading This answer will concern itself with an unusual projective technique Leonardo used in painting the last supper which results in what can be considered a somewhat perceptually enforced attention on part of the viewer. Notice that all perspective lines come together (the so called vanishing point) at the head of Jesus, but this is not how a visitor sees it, to take a picture of the Last Supper with the correct perspective (as above) it has to be done like this: But when you see the picture as a visitor this is how you see it: Now, in normal life and in paintings the vanishing point coincides with the horizon line, i.e is the most distal part in the scene: So, looking at the picture from the dining hall of the monastery of Santa Maria delle Grazie means that to get the ‘correct’ perspective you would have to be on a platform ca 5 meters high. But, given how Leonardo used the perspective construction of it, as it is seen from below there opens a virtual, ethereal space behind the vanishing point enticing the viewer to project to a higher plane. We don’t know all of da Vinci’s intentions, but the result is that we must always look up to the picture, elevating the viewer to a high centre of projection producing thereby also a spiritual elevation. (Da Vinci actually constructed the Last Supper’s perspective lines contrary to his own advice on perspective: “If you want to represent an object near you which is to have the effect of nature, it is impossible that your perspective should not look wrong,....” Leonardo a Vinci: The literary works of Leonardo da Vinci, Ed. J.P. Richter, London: Phaidon, 1970, p325.) Upvote · 9 7 9 1 9 1 MIke Brenneman Minister & Teacher of the Bible for 43 Years (1979–present) ·2y Related What are some other versions of The Last Supper? How do they compare to Leonardo DaVinci's version? Leonardo was an incredible artist and impressive person. Having said that he got several things wrong in his “Last Supper”. All four of the gospels mention that Jesus was reclining at the table (Matthew 26:20). Leonardo and most artists paint Jesus and His disciples sitting at a table. The Jews at that time reclined on the floor to eat their formal meals. The table was 18 inches high. Typically they would lay on their sides, propping themselves up on the left arm and eating with their right hand. Photo from Bible Studies Net by permission in 2006 Remember the woman who washed Jesus’ feet as he w Continue Reading Leonardo was an incredible artist and impressive person. Having said that he got several things wrong in his “Last Supper”. All four of the gospels mention that Jesus was reclining at the table (Matthew 26:20). Leonardo and most artists paint Jesus and His disciples sitting at a table. The Jews at that time reclined on the floor to eat their formal meals. The table was 18 inches high. Typically they would lay on their sides, propping themselves up on the left arm and eating with their right hand. Photo from Bible Studies Net by permission in 2006 Remember the woman who washed Jesus’ feet as he was reclining at a table in Luke 7:37–38? It says she was standing behind Jesus washing his feet while He was reclining at a table. That would be impossible to do if Jesus was sitting in a chair. Leonardo’s mistake is understandable, because at that time, the Catholic church didn’t allow people to read the Bible in their native tongue. The New Testament was only available in Latin or Greek. A few priests could read those languages, not the common folk. This restriction was necessary for the Catholic authorities, because anyone who read and understood the Bible, would see that the Catholic church didn’t follow the scriptures in many critical doctrines. image from Shutterstock with permission Leonardo painted the disciple on Jesus’ right hand to look very effeminate. Leonardo probably intended this to be the apostle John who was a rugged fisherman along with several other apostles. Jesus called John and his brother James “sons of thunder” (Mark 2:17) They were as masculine as a man can be. John 13:23 uses a phrase that sounds strange and awkward to us. “There was reclining on Jesus’ bosom one of His disciples” This is an idiomatic phase that means ‘‘to dine in the place of honor.” See two of the best Lexicons (Greek-English dictionaries: Lowe & Nida, and BDAG). To recline next to Jesus was definitely the place of honor. Lastly most artists in the middle ages and now, depict Jesus as a handsome man. The scriptures say otherwise. Isaiah 53:2 “…He has no stately form or majesty, that we should look upon Him, nor appearance that we should be attracted to Him.” God wanted people to be attracted to the teachings of Jesus and His righteousness—not His physical appearance. Upvote · 9 3 Jeff Pellarin Works at Accounting · Author has 5.6K answers and 4.9M answer views ·9y Related Why is Mona Lisa much more revered than other portraits painted in the same era? I beg to differ with those who respond with the story of the Louvre guard who stole it. I think most today do not even know that story. I think the real reason is, that it was the first application of a technique of painting that would not be understood for centuries. Layers of paint so thin that even x-rays cannot detect the brushstrokes. Aside from the dedication to craft (to painstakingly paint so many layers) is the scientific achievement--it is difficult to paint that many layers, without the earlier layers rejecting the subsequent ones, the way water beads up on a freshly waxed car. Continue Reading I beg to differ with those who respond with the story of the Louvre guard who stole it. I think most today do not even know that story. I think the real reason is, that it was the first application of a technique of painting that would not be understood for centuries. Layers of paint so thin that even x-rays cannot detect the brushstrokes. Aside from the dedication to craft (to painstakingly paint so many layers) is the scientific achievement--it is difficult to paint that many layers, without the earlier layers rejecting the subsequent ones, the way water beads up on a freshly waxed car. And ultimately, it amounted to the groundbreaking sfumato technique, which adds to the allure. All of that is why art historians hold it in high regard. Their high regard then leads to an air of mystery that we common folk love to gobble up--the special room at the Louvre constructed to house her, the long lineups (she is, in a sense, famous for being famous, long before the Kardashians coined the phrase). I dare say, it is difficult to truly appreciate Mona today; she is under glass, and surrounded at any time by dozens of jostling tourists taking selfies. One cannot get more than a glimpse of this relatively small painting before being elbowed out. So we don't go see her to appreciate her (unfortunately), we go see her only so we can say we did. Upvote · 99 12 9 2 Ron Richards Studied at California State University, Los Angeles (Graduated 1968) · Author has 20.1K answers and 88.8M answer views ·2y Related What is the story behind The Last Supper? Who painted it and why is it so famous? Original question: “What is the story behind The Last Supper? Who painted it and why is it so famous?” Leonardo Da Vinci painted it. And the reason it is so famous is because it is considered one of Da Vinci’s greatest works of art. Some background. The Duke of Milan, Ludovico Sforza, commissioned Da Vinci to paint it as a mural for the Monastery of Santa Maria delle Grazie in Milan. The Last Supper by Leonardo Da Vinci Continue Reading Original question: “What is the story behind The Last Supper? Who painted it and why is it so famous?” Leonardo Da Vinci painted it. And the reason it is so famous is because it is considered one of Da Vinci’s greatest works of art. Some background. The Duke of Milan, Ludovico Sforza, commissioned Da Vinci to paint it as a mural for the Monastery of Santa Maria delle Grazie in Milan. The Last Supper by Leonardo Da Vinci Upvote · 9 4 Related questions Are there other depictions of The Last Supper before Leonardo da Vinci’s? Why is the Last Supper painting deteriorating? How has been Leonardo da Vinci's painting "The Last Supper" conserved? What does the Last Supper painting signify? What are the secrets inside Leonardo Da Vinci's "Last Supper"? What is the significance of Leonardo da Vinci's "The Last Supper" painting? How much is the last supper worth? Why is The Last Supper (Leonardo da Vinci) such a source for conspiracies? Why is the last supper so expensive? What happened to Leonardo da Vinci's last supper? Why is the painting “The Last Supper” so much more famous than other biblical works? How many times did Leonardo DaVinci paint “The Last Supper”? The Last Supper has been memorably depicted in Leonardo da Vinci's famous painting of the same name. In this painting, which disciple is shown as having his elbows on the table? How did Leonardo's "The Last Supper" and the "Mona Lisa" look the day the last brush stroke was applied? What was wrong with Leonardo da Vinci's painting "The Last Supper"? What is missing from the painting that would make it more realistic? Why didn't anyone notice it for so long? Related questions Are there other depictions of The Last Supper before Leonardo da Vinci’s? Why is the Last Supper painting deteriorating? How has been Leonardo da Vinci's painting "The Last Supper" conserved? What does the Last Supper painting signify? What are the secrets inside Leonardo Da Vinci's "Last Supper"? What is the significance of Leonardo da Vinci's "The Last Supper" painting? 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https://fiveable.me/key-terms/principles-physics-i/orbital-radius
new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintables upgrade All Key Terms Principles of Physics I Orbital radius 🍏principles of physics i review key term - Orbital radius Citation: MLA Definition Orbital radius refers to the distance from the center of an orbiting body, such as a planet or satellite, to the center of the body it is orbiting around. This measurement is crucial for understanding the dynamics of orbital motion, as it influences the gravitational forces at play and determines the shape and stability of the orbit according to Kepler's Laws. 5 Must Know Facts For Your Next Test The orbital radius is a critical factor in determining the period of an orbiting object, as described by Kepler's Third Law, which states that the square of the orbital period is proportional to the cube of the semi-major axis (orbital radius) of its orbit. In a circular orbit, the orbital radius remains constant, while in an elliptical orbit, it changes as the object moves closer to or farther from the central body. The gravitational force experienced by an object in orbit is inversely proportional to the square of its orbital radius; as the radius increases, this force decreases. Different celestial bodies can have varying orbital radii based on their mass and distance from other bodies; for instance, moons typically have smaller orbital radii than planets around their stars. Understanding orbital radius is essential for satellite placement and space missions, as it affects speed, trajectory, and energy requirements for launching and maintaining orbits. Review Questions How does orbital radius relate to Kepler's Third Law, and why is this relationship important in understanding celestial mechanics? Orbital radius is directly related to Kepler's Third Law, which states that the square of a planet's orbital period is proportional to the cube of its average distance from the sun. This relationship allows scientists to predict how long it takes for planets to complete their orbits based on their distances. Understanding this connection is crucial for calculating orbital dynamics and planning space missions effectively. Analyze how changes in orbital radius affect gravitational force and speed of an orbiting object. As an object's orbital radius increases, the gravitational force acting upon it decreases because gravitational force is inversely proportional to the square of the distance between two masses. Consequently, this reduced force leads to a slower orbital speed. Conversely, if an object's orbital radius decreases, it experiences stronger gravitational pull and must move faster to maintain its orbit without falling into the central body. Evaluate the significance of understanding orbital radius in real-world applications such as satellite deployment and space exploration. Understanding orbital radius is vital for satellite deployment and space exploration because it directly impacts a satellite's stability, coverage area, and communication capabilities. When planning missions, engineers must consider how changes in orbital radius can affect fuel consumption and mission success. For example, satellites in low Earth orbit require different parameters compared to those in geostationary orbit, making knowledge of these distances crucial for effective mission planning and execution. Related terms Kepler's Laws: A set of three scientific laws that describe the motion of planets around the sun, including how their orbital radii affect their speeds and periods. Gravitational Force:The attractive force between two masses, which depends on their masses and the distance between them, playing a key role in determining an object's orbital radius. Elliptical Orbit:An oval-shaped path that an object follows when it revolves around another object, characterized by varying distances from the center at different points in its orbit. 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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website. every AP exam is fiveable Study Content & Tools Study GuidesPractice QuestionsGlossaryScore Calculators Company Get $$ for referralsPricingTestimonialsFAQsEmail us Resources AP ClassesAP Classroom history 🌎 ap world history🇺🇸 ap us history🇪🇺 ap european history social science ✊🏿 ap african american studies🗳️ ap comparative government🚜 ap human geography💶 ap macroeconomics🤑 ap microeconomics🧠 ap psychology👩🏾‍⚖️ ap us government english & capstone ✍🏽 ap english language📚 ap english literature🔍 ap research💬 ap seminar arts 🎨 ap art & design🖼️ ap art history🎵 ap music theory science 🧬 ap biology🧪 ap chemistry♻️ ap environmental science🎡 ap physics 1🧲 ap physics 2💡 ap physics c: e&m⚙️ ap physics c: mechanics math & computer science 🧮 ap calculus ab♾️ ap calculus bc📊 ap statistics💻 ap computer science a⌨️ ap computer science p world languages 🇨🇳 ap chinese🇫🇷 ap french🇩🇪 ap german🇮🇹 ap italian🇯🇵 ap japanese🏛️ ap latin🇪🇸 ap spanish language💃🏽 ap spanish literature go beyond AP high school exams ✏️ PSAT🎓 Digital SAT🎒 ACT honors classes 🍬 honors algebra II🐇 honors biology👩🏽‍🔬 honors chemistry💲 honors economics⚾️ honors physics📏 honors pre-calculus📊 honors statistics🗳️ honors us government🇺🇸 honors us history🌎 honors world history college classes 👩🏽‍🎤 arts👔 business🎤 communications🏗️ engineering📓 humanities➗ math🧑🏽‍🔬 science💶 social science RefundsTermsPrivacyCCPA © 2025 Fiveable Inc. All rights reserved. AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website. Orbital radius - (Principles of Physics I) - Vocab, Definition, Explanations | Fiveable | Fiveable 0
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https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2010/c53bca93ceaede29a7983409ec7e233c_5RSMLgy06Ew.pdf
MITOCW | Lec 7 | MIT 6.042J Mathematics for Computer Science, Fall 2010 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, on Tuesday we talked about sex, so today we're going to talk about marriage. Now, in terms of graph theory, marriage is expressed as a matching problem, and today we're going to talk about a matching algorithm that is used in all sorts of applications. It's used by online dating agencies to match compatible people together. It's used for assignment problems, for example, matching interns to hospitals on match day. It's used for resource allocation problems, for example, load balancing traffic on the internet. And we'll talk about the applications at the end of class. In its simplest form of a matching problem, you have a graph where the edges represent compatibility. Two nodes can be paired together, or married, and the goal is to create the maximum number of compatible pairs. So let's define a matching, given a graph, G, with nodes, V, and edges, E. In matching, you can think of it as a collection of edges, or a subgraph of G where every node has degree 1. So everybody can be married just to one person. So let's draw an example, maybe not put that edge in. And let's label these nodes x1, x2, x3, x4, x5, x6, x7, and x8. Now x1, x6 and x2, x5 is a matching, so x1, x6 and x2, x5 is a matching with two edges, so we say it has size two. All right, so I can pair these guys up and pair these guys up. Is there any bigger matching in this graph? So I found one with two marriages, here and here, two edges. Yeah? AUDIENCE: x1, x7. PROFESSOR: x1, x7. AUDIENCE: x2, x6. PROFESSOR: x2, x6. AUDIENCE: [INAUDIBLE]. PROFESSOR: Good. All right, that's a matching of size three. So I got three couples together. Good. Can I make a bigger matching, one with size four-- four marriages here? No. Why not? Can anybody give me a reason why it can't be done? Yeah? AUDIENCE: x8 and x7 would have to be matched with someone. PROFESSOR: Yeah, so if-- Yeah? AUDIENCE: They could only be paired with x1, but x1 can't be paired with both. PROFESSOR: Good. If I were to have a matching with four edges, well, there's only eight nodes, so I'd have to have all eight nodes involved in the matching. And that means x7 and x8 would have to be in the matching, but they could only be paired with x1, and so it's not possible to do that. All right, so there is no matching of size four in this graph. Three is the best I can do. Now, when you get every node in a matching, then it's called a perfect matching. And so in this case, it doesn't exist, but sometimes it does. So a matching is perfect if it has size half the number of nodes. In other words, if the number of edges is v over 2, then every node is in the matching. All right, so that one doesn't have a perfect matching. What about this graph? So I got b1, b2, b3, b4, g1, g2, g3, g4, and I'll put in the compatibility edges here. OK, Does that graph have a perfect matching? Can you pair up every boy with a girl here so that everybody is compatible with their mate, and you have just one spouse? Can you do that? AUDIENCE: Yes. PROFESSOR: Yeah? All right, who do I start pairing up? AUDIENCE: 1, 1. AUDIENCE: 1, 1. AUDIENCE: b2, g3. PROFESSOR: b2, g3. AUDIENCE: b3, g2. PROFESSOR: b3, g2 AUDIENCE: b4, g4. PROFESSOR: There we go. All right, so there is a perfect matching in this graph. Very good. Now, in some cases, some pairings are more desirable than others, and this can be represented with a weight. And so you might have a weighted graph where every edge has a weight on it. For example, we might weight b1, g2 with 5, and b1, g1 gets a 10. And usually when you see weighted graphs in the matching context, a lower weight means it's more desirable, so that b1 and g2 would get along better than b1 and g1. And then the goal is to find a matching with minimum weight. Now, the weight of a matching, call it M, is the sum of the weights on the edges of M. Now, usually when you're looking at weighted matchings, you require yourself to have a perfect matching so that everybody gets paired up. And often in that case, you'll see all the edges present, some of them with very big weights, maybe even infinity if they just can't be put together. Because otherwise you just say, don't match anybody together, and you have weight zero. So for when you look at minimum weight matchings, you're looking for the perfect matching with minimum weight. So we say a min-weight matching for a graph, G, is a perfect matching for G with the minimum weight, overall perfect matchings. Let's try an example. Say I've got this graph, and call this node Brad, here's Billy Bob, here's Jennifer, and then Angelina. And the weights are as follows-- I put a 10 here, a 10 down here, a 5 here, because Brad really likes Angelina and vice versa, and a 16 between Jennifer and Billy Bob. What is the weight of the min weight matching in that graph? 20. And who gets paired with who there? Who does Brad get hooked up with? AUDIENCE: Jan. PROFESSOR: Jan, and that leaves Billy Bob with Angelina, and the weight is 20. So even though Brad really likes Angelina, if I go that route, my weight is 21, which is not as good. So the min-weight matching would be this one. OK. Now, it turns out that finding the maximum matching-- the maximum number of edges you OK. Now, it turns out that finding the maximum matching-- the maximum number of edges you can put together-- or finding the minimum weight perfect matching, those are both solvable, tractable problems. You don't get a million dollar prize for solving that. The algorithm is run in quadratic to cubic time, so not terrible, but they're not NP-complete, so people know how to do it. Now, they are pretty complicated, so we're not going to cover them in 6042. What we're going to do is look at a slightly different version of the problem that actually turns out to be more useful in practice, because there's a very nice algorithm for it. Now, in the version of the problem that we're going to look at, everybody has preferences-- a preference list. It's not weighted, but it's a priority order of who they want to get mated to, or matched up with. So it would look not quite like that, but it would look like this. So Brad, Billy Bob, Jennifer, and Angelina-- same players, but what we do is we know that Brad really likes Angelina, and Angelina really likes Brad, so they are first choices, at least was. Number two for Brad is Jennifer, but Jennifer really likes Brad first, and Billy Bob second. Billy Bob likes Angelina first, then Jennifer. Angelina thinks Billy Bob is number two. So it's not necessarily symmetric. Jennifer has Brad as the first choice. Brad has Angelina as the first choice. All right? So it's an asymmetric situation. Now, what would happen if we set up our marriages so that Brad is married to Jen, and Billy Bob is married to Angelina? What might happen if we made those as our pairings, and we put them on a desert island, all four of them? What's likely to happen there? AUDIENCE: Brad and Angelina are going to cheat on each other. PROFESSOR: Yeah, we're going to have a problem. Because Brad and Angelina have the hots for each other here, and they like each other better than their spouse. So before you know it, they're going to be doing their 6042 homework together late at night. All right? Now, when this happens, we have what's called a rogue couple. Given a matching, x and y, a boy and a girl, say, form a rogue couple if they prefer each other over their mates in M. All right. So here, if we married Brad to Jennifer, and Billy Bob to Angelina, Brad and Angelina form a rogue couple, because they like each other better than who they were hooked up with. And that's sort of a bad thing. It creates instability if you were to make the matchings that way. In fact, we say that a matching is stable if there aren't any rogue couples. And now, one thing to make clear is that your preferences can't change over time. So it's not a situation where you get bored with your spouse, and you change your mind, and then you go off and create a rogue couple. You're fixed in your preferences over all time, no playing the field, none of that stuff here, OK? So it's fixed once and for all, and your goal, of course, is to create or find a perfect matching that's stable. That's the goal. So get everybody married up and make it stable. All right, is it doable in that example? If I put those four people on a desert island, could I make a stable matching? Yeah, who would I match Brad to? AUDIENCE: Angelina. PROFESSOR: All right. Good, and then Billy Bob gets Jennifer. Now, I'm not saying that you make everybody happy, because Billy Bob and Jennifer are probably not happy there. They each got their number two choice. But it's stable, because Brad and Angelina aren't going anywhere. They are going to stay together because they like each other best, and so there's no chance that Jennifer is going to-- that Angelina is going to sneak off, sorry, with Billy Bob, or that Brad's going to sneak off with Jennifer. All right? So it's stable. Not everybody's happy, but it's a stable set of marriages. OK. Any questions about what we're trying to do here now? Yeah? AUDIENCE: [INAUDIBLE] the edges? PROFESSOR: That's a great question. You'll see it referred to both ways. Technically, it's a subgraph, so it has nodes and edges, but you'll see me, and you'll see everybody say, oh, it's a bunch of edges that don't share any nodes, and you'll see them refer to it as the edges. But really, underlying that, it's a subgraph, technically. Any other questions about what we're trying to do here to find stable, perfect matching? All right. Well, in this example, there was a stable, perfect matching. But what about in general? If I have a lot more people, and they have arbitrary preferences, how many people think you can always find a stable perfect matching? There's one optimistic person. How many people think there's some cases where you're just not going to be able to do it? Wow. OK, it's a pessimistic view. Well, in some sense you're both right. If you allow boys to prefer boys and girls to prefer girls, then it is not possible, and I'll give you an example. You can find examples where there's always a rogue couple, but if you require boys to only get matched to girls and vice versa, then it is possible to always find stable marriages, stable matchings. And we're going to talk about an algorithm for that. But before I show you the algorithm, let me show you the bad case when boys can prefer boys, or what's sort of a unisex scenario. So here's a bad example with four people, and the idea is to create a love triangle. So we have Alex, who prefers Bobby Joe, Bobby Joe prefers Robin, and Robin prefers Alex. And their second choices go in the opposite order there, all right? So Alex wants to be with Bobby Joe. Bobby Joe wants be with Robin. Robin wants to be with Alex. And then there's Mergatoid--[LAUGHTER] --and nobody likes Mergatoid. So that's choice three for all of them here. And Mergatoid's choices, preferences don't really matter in this case. [LAUGHTER] I hope nobody is named Mergatoid in the class. I would get complaints here. All right, so I want to claim and prove a theorem that says there is no stable matching for this group of preferences. So we'll state that as the theorem. There does not exist a stable matching for this graph. The proof is by contradiction-- assume there is one. All right, so assume there exists a stable matching. We're going to find a rogue couple in it. So call the stable matching M. Well, if there's a stable matching, Mergatoid's got to get married to somebody. Mergatoid will be matched with someone. All right. Now, here I'm going to do something that you can do in your proofs, but you've got to be careful when you do it. I'm going to say, without loss of generality, assume Mergatoid is matched to Alex, all right? And I can do that. So this is the abbreviation, without loss of generality by symmetry. And really, I should explain what I mean here in the proof. Well, that love triangle is symmetric. Each one has a preference for the next person around the triangle. In terms of graph isomorphism, with the weights on that, every node looks the same. So I can use symmetry. So we're going to say, without loss of generality, we're going to assume Mergatoid is matched to Alex. And I'm implying the argument is going to be the same. What I say next is the same no matter who Mergatoid is matched to, because it's symmetric. All right. If Mergatoid is matched to Alex, do you see a rogue couple up there? So you have Robin matched to Bobby Joe. Mergatoid is matched to Alex. AUDIENCE: Alex to Bobby Joe? PROFESSOR: Alex and Bobby Joe, no. AUDIENCE: Between Alex and Robin. PROFESSOR: Alex and Robin, yeah. Alex and Bobby Joe aren't rogue, because Bobby Joe likes Robin. He likes the person he or she is married to. All right, so they're not going to go off with Alex, but Alex and Robin are a rogue couple, because Robin likes Alex the best, and Alex for sure likes Robin better than Mergatoid. All right? So they both prefer each other to their mates, and so they form a rogue couple. All right, so Alex and Robin form a rogue couple, and that means that M was not stable. The matching was not stable. And that's a contradiction, because we assumed it was. So we have a contradiction, and the proof is done. Any questions about that? And I'm sort of implying here that you could match, you could have had Mergatoid matched to Bobby Joe, and then Bobby Joe and Alex would have been a rogue couple, or Mergatoid matched to Robin, and then Robin and Bobby Joe would be a rogue couple by the without loss of generality. So it's OK to use that, but you want to be careful that you're doing it OK. Questions? Yeah? AUDIENCE: So the without loss of generality only works if it's, like, perfectly symmetric? PROFESSOR: Yeah. Basically, the argument you're going to make is just going to be the same argument done in all three cases, and to save yourself some effort, you're saying, don't do the three cases, do one, the other two look the same. So technically, you could add case one, case two, case three, and they would've looked symmetric. OK? All right. Now, this is not very surprising, as you all voted. Almost all of you said it's hard to find stable matchings. You might not be able to do it always. And in fact, you can't in the unisex world. The surprising thing is, you can always find a stable matching in the world where boys could only be paired with girls and vice versa. Now, this statement, this result is pretty famous. The problem itself is known as the stable marriage problem. So let me just define it here, and then we'll talk about an algorithm to find the matching. So we have N boys and N girls. And it's important we have the same number of each. Now, actually, tomorrow, in recitation, you're going to look at the scenario where there's more girls than boys, or vice versa. And you'll be using a similar algorithm, but it'll be a different context, and that's the context that comes up in matching interns to hospitals, and stuff like that. But for our version it's an equal number of boys and girls. Each boy has his own ranked preference list of all the girls. So every boy sort of has his dance card of the girls that he likes in order. All right? The orders can be different for different boys. And each girl has the same thing. She has her own list, ranked 1 to n of all the boys. The lists are complete, and there's no ties, so all the ties have to be broken here. And the goal is to find a perfect matching without rogue couples. OK, so let's try an example. Before we do the algorithm, let me just do a bigger example with five boys and five girls, and we'll get some feel for, this is not completely obvious how to find it. OK, so let's put the boys over here. And here's boy 1, and his preference list is going to be--the girls will be C, B, E, A, D, and boy 2 is going to have the preference list A, B, E, C, D, and boy 3 is going to have D, C, B, A, E. Boy 4 is going to be A, C, D, B, E, and boy 5, I have to write his across like this, A, B, D, E, C. And the girls also have their lists, so let's put those up. So girl A likes the boys in order 3, 5, 2, 1, 4. Girl B likes them in order 5, 2, 1, 4, 3. Girl C has the order 4, 3, 5, 1, 2. Girl D, 1, 2, 3, 4, 5, and then the last girl, E, has 2, 3, 4, 1, 5. All right, so say that's our matching problem, and those are the preference lists. Any ideas for how we might try to make an algorithm to do this? Any thoughts about what you'd do? What are some approaches we could take? Yeah? AUDIENCE: Use the mating algorithm. PROFESSOR: Use the greedy algorithm? AUDIENCE: The mating algorithm. PROFESSOR: Oh, the mating algorithm. Well, yeah, the mating algorithm is going to do very well, but I haven't told you what that is, and it's a little complicated. So I sort of want to explore the things you might know from what we've done so far. What approaches have we seen so far for solving problems? Yeah? AUDIENCE: The greedy algorithm? PROFESSOR: The greedy algorithm-- yep. And let's try that. So let's choose the first boy, and we'll just go down and give each boy the best choice available and see what happens. First thing to try usually is the greedy algorithm, and about half the time it'll work in life, and half the time it won't. So let's try greedy. So boy 1 is going to get his first choice, girl C. Who does boy 2 get? A. Who does boy 3 get? D. Huh, this is going well. Who does boy 4 get? Boy 4, oh yeah, boy 4 has got to go all the way down to B, because A, C, and D got taken, so boy 4 gets B. And boy 5-- who's left? E, fourth choice again. All right, well, maybe that's a stable matching. All right, well let's see. To see if it's stable, we've got to see are there any rogue couples. Well, is there any rogue couple involving boy 1? No, boy 1 got his first choice. You're not going anywhere. Boy 2? No-- first choice, same with boy 3. These boys are quite happy. They're not doing anything. Boy 4 may not be so happy though, because boy 4 got paired with B, so there's possible rogue couples. Let's see, is 4, A a rogue couple? No. A hates boy 4, all right? No chance she's running off with boy 4. AUDIENCE: 4, C? PROFESSOR: 4, C? All right, let's see that. Where's C-- whoa, yeah. C has the hots for boy 4, so she likes him better than whoever her mate was, and boy 4 likes C better than who he got matched with, boy B. That is now a rogue couple, so the greedy algorithm did not work. That's too bad. Well, it would be short lecture, I guess, if it worked. [LAUGHTER] Well, what would you do next, sort of, if you were-- I mean, you could try to match 4 with C, and then try to patch things up, and you could start doing that, but you might create other rogue couples. In fact, I don't know an approach-- yeah? AUDIENCE: You could let it evolve and see if it gets it more stable. PROFESSOR: Say it again. AUDIENCE: You could let it evolve, let all the rogue couples evolve and see if it ends up any more stable. PROFESSOR: Yes, you could start swapping around to get rid of rogue couples. In doing that, you might create other rogue couples. In fact, I don't know of an algorithm that works like this, that works, that's known to work, where you start patching things up, because as you're patching things up, you might make other things much worse by doing that. So I don't know of an approach that way. What's another approach? Yeah? AUDIENCE: [INAUDIBLE] pertaining to what order they have each other. PROFESSOR: And then do? AUDIENCE: And then do the highest ordered [INAUDIBLE] PROFESSOR: Min-weight matching kind of thing? AUDIENCE: Maybe. PROFESSOR: Maybe? I don't of an approach like that that works. Also, min-weight matching, that algorithm is going to be more complicated than the one I'm going to show you, in the end. It takes more time to run. And I don't even know if it works. Like, I don't know if you can take these numbers and make weights on the edges get a min-weight matching. AUDIENCE: Could you do like a merge sort [INAUDIBLE] PROFESSOR: Oh, a merge sort, so you take the minimum weight edge, put that in, and recurse on that kind of thing. I don't know. It's possible. In fact, you know what, a recursive approach is a good idea. I don't know of a nice recursive algorithm for this. It's true that if you found a boy and a girl who liked each other best, you could then match them safely and recurse, because you know they're not going to be in a rogue couple, because they like each other best. Then you could recurse. But that might not exist here. You might not have a boy and girl that like each other best, in which case it's hard to know. I don't know now if you pick the minimum weight in some sense, like add the preference list or something to make a min-weight and recurse on that, if that works. I don't know of an approach like that. But those are the kinds of things you try. And as far as I know, all the simple things fail for this problem. But there is something that's a little more involved, but does work, and that's the mating algorithm. And does everybody have the handout? There, it's back up there. I got some copies down here if you need it, but pull that out. So we're going to read this and talk about what the algorithm is, and then prove that it performs well. So the initial condition is you have each of the N boys has an ordered list of the N girls and vice versa, and the ritual, we're going to view this is a mating ritual, and really, the program is doing it. The code is doing it. But think of it as real life. It takes place over several days. Now, the day is broken up into three parts-- the morning, the afternoon, and the evening. In the morning, each girl comes out to her balcony and stands on the balcony. Each boy goes to the balcony of his favorite girl who is still on his list that hasn't been crossed off. Now, initially, every girl is on his list. So he goes to his favorite girl, goes under her balcony, and serenades her. Now, if, over the course of the algorithm, the boy has nobody left on his list, he's out of luck. He just stays home and does homework. All right? There's no serenade anymore for him. Now, in the afternoon, the girls who have at least one suitor-- a boy down there serenading her-- looks at all the suitors, picks her favorite, and to the favorite she says, maybe I'll marry you. Come back tomorrow. Girls don't want to make it too easy here for the boys. Now, to all the other boys who are lower priority she says, I will never marry you. Go away. So she writes them off for good. Now, that night, any boy who heard a no-- like the girl said no, I'll never marry you-- crosses that girl off his list. Because, you know, it's the only practical thing to do at that point. Now, if the boy heard the maybe I'll marry you, well, he's going to go back tomorrow, because that's still his favorite girl that's not crossed off. So he goes back and serenades her again the next day in the hopes that eventually, she'll say yes. Now, we keep doing this every day, OK? And if we ever encounter a day where every girl has, at most, one suitor, the algorithm stops, and then every girl who has a suitor says, yes, I will marry you. Now, if a girl doesn't have a suitor, there's no one to marry her. We're going to prove that doesn't happen, OK? But determination condition is, you no longer have a situation with two or more boys under one balcony. OK? All right, so let's run that algorithm on this example, just so we make sure we understand it, because I'm going to try to prove theorems about it. So here are the serenades that are going on, and here's the girls, and the days. It's going to work over four days in this case. And then we're also going to keep track of the boy's lists-- who's gotten crossed off. All right, so these will be the cross outs down here. Actually, maybe I'll fit it up here if I can. Girls, let's see, A, B-- no, I'm going to have to space it out. And the boys have their lists here, and we have boys 1, 2, 3, 4, 5. And here I'm going to record the cross outs. All right, so let's look at day one. Who is under girl A's balcony on day one? AUDIENCE: 2, 4, and 5? PROFESSOR: 2, 4, and 5-- each like girl A the best, so she's got a lot of activity. These three boys show up. Anybody under girl B's balcony? No, nope, no, nothing there. C, does C have anybody? Boy 1, yeah. D? 3, and E, I don't think there's any action, right? Nope. All right, so that's the status on day one. So the action is all up here. All right, so what does girl A do? Who does she tell to hang around? AUDIENCE: Number 5. PROFESSOR: Number 5, yes. She's got 5, 4, and 2, and she likes 5 the best. She tells 5 to come back, and she says to these guys, we're not going to marry him. That means that boys 2 and 4 cross girl A off their list. So they say A is no longer possible for them. All right, now we go to day two, and 5 goes back to girl A, and, of course, boys 1 and 3 stay there. Where does boy 2 go on day two? B. All right, so boy 2 shows up here on day two, and, let's see, boy 5 is already there, and then we've got to get boy 4. Where does he go now? C. All right, boy 4 shows up here. All right, so now the action is with girl C. And what does she do? AUDIENCE: [INAUDIBLE] PROFESSOR: Yeah, she keeps 4, and she boots poor 1. You know, led him along for a day, and then he gets the boot because boy 4 showed up, right? Because girl C likes 4 better than 1, so bad luck for 1 there. So now one goes home that night and crosses off girl C. OK. All right, and where does 1 go on day three? All right, so he goes to-- he crossed off C--he goes to B. All right, so this is left over. We have 5, 2, 4, 3-- boy 1 now goes to B, right? OK. And then what does girl B do? Who does she keep around? Keeps 2, boots poor 1. All right, so 1 says, all right, I'm crossing B off my list, got the message. And now where does boy 1 go on day four? AUDIENCE: Girl E? PROFESSOR: E-- third choice. OK. So 1 shows up down here, and these guys keep returning. Wow, so on day four there's no more fighting. Every girl has at most one. The termination condition is invoked, and these are the marriages that take place. The girls say yes. All right, now is everybody--[LAUGHTER] --does everybody understand the algorithm we used? Any questions on the algorithm? Yeah? AUDIENCE: [INAUDIBLE] the first case [INAUDIBLE] or is that just in case the [INAUDIBLE] PROFESSOR: That's in case the algorithm doesn't work. Because we haven't proved it works yet, and I gotta have a possibility for, a boy crosses every girl off his list, he gets rejected everywhere. That is a possibility, so I've got to say what would happen in that possibility. Now, we will prove in a few minutes that condition never arises, OK? But I'm giving you what might happen, in which so it never does, so no boy ends up having to stay home and do homework here. All right, so that's the algorithm. Now let's see if we can see if there's any rogue couples. All right, so, for example, let's look at boy 1. Boy 1 got paired to his third choice, E, so 1, C might be a rogue couple. Is that possible? No, because C got 4 here, which is her first choice. She's not going with boy 1. That's not working. What about 1, B? Could that be rogue? No, B got 2, and B likes 2 better than 1, so 1, C and 1, B are not rogue. And 1 got E, so there's no other possibility for boy 1, so boy 1 is not in a rogue couple. What about boy 2? 2 got B, and that's 2's second choice. What about 2, A? Let's see, A got 5, and A likes 5 better than 2, so that's not rogue. Boy 2 is not rogue. All right, boy 3-- boy 3 got his first choice, right? 3 got D, so 3 is not going anywhere. Boy 4 got-- who did boy 4 get-- got C. 4 got his second choice. What about 4, A? A hates 4. A is not gonna get caught dead with 4, so that's not rogue. So 4 is OK. 4 is not in a rogue couple. And finally, boy 5-- boy 5 is paired with A, and that is boy 5's first choice, so he's not wandering off here. All right? So in fact, we've just argued that this is a stable set of marriages, a stable matching in this case. Any questions now about what we're trying to do? OK, because we're going to try to show now it always produces a stable matching. And to do that, we need to do a few things. All right, so what are the things we need to show to prove everything is going to be good here? What are some facts we to prove? Yeah? AUDIENCE: Don't we need to show that the algorithm, it does come to an end? PROFESSOR: Yes. [LAUGHTER] The algorithm terminates, so we need to show that the marriage algorithm, TMA, terminates. Otherwise, the boys are serenading forever, and that's not too good. All right, what else do we want to show? Yeah? AUDIENCE: If it does terminate, then one is left empty. PROFESSOR: Then? AUDIENCE: Everyone gets someone. PROFESSOR: Everyone gets married. Yep. Stability is easy if nobody gets married. All right, what else do we want to show here? Yeah? AUDIENCE: There are no rogue couples. PROFESSOR: No rogue couples. All right, what else might we like to show? These are the three main ones. There's a couple of other things you might like. Yeah? AUDIENCE: It runs quickly. PROFESSOR: It runs quickly. Well, you can only serenade for so long. In fact, it does run quickly, and that's why it's useful in practice. Anything else you might want to show about this algorithm? AUDIENCE: How many people you crossed out? PROFESSOR: How many people you crossed out-- yeah, you could, and that'll tie into how long it takes. Yeah. Anything else? Yeah? AUDIENCE: The average likeness between the couples. PROFESSOR: The average likeness, oh. So yeah, we haven't had a notion here of how happy people are at the end, because we don't have a weighting on the edges, but you might want to think about that. That's actually a good point. Yeah? AUDIENCE: If they [INAUDIBLE] if it matters who serenades [INAUDIBLE] girls or boys to get her. PROFESSOR: That's a great point. Is this algorithm good for girls, or good for boys? Yeah, that's a good point-- fairness. All right, so we'll take a look at fairness also. Is it better to be a serenader, or be on the balcony making your choices? OK, so this is what we've got to do. So let's start by showing TMA terminates, and that it terminates pretty quickly. Now, in fact, I'm going to prove a fairly crude bound on the time, but it actually does fairly well. So our first theorem is going to be that TMA terminates in, at most, N squared plus 1 days. N is the number of boys and girls. The proof is going to be by contradiction. This is probably the only day where we'll do a bunch of proofs and none of them use induction. They're pretty much all by contradiction. Suppose TMA does not terminate in N squared plus 1 days, because we're going to show that leads to a contradiction. We need to show that some kind of progress is made each day to show that it terminates. Yeah? AUDIENCE: [INAUDIBLE] the number of causes that have [INAUDIBLE], so if it has not terminated, then at least one girl is seeing a group of at least one guy--PROFESSOR: Yes. AUDIENCE: --so the number of crosses cannot be defined [INAUDIBLE] PROFESSOR: Very good. OK, let's state that as a claim, and you've given the proof of the claim, which is great. If we don't terminate, on a day, that must be because a girl had two boys there, or more, therefore she rejected some, at least one. And that night, the rejected boy crosses a girl off his list. So if we don't terminate, then at least one boy crosses at least one girl off his list. And so we're going to measure progress by the cross outs. So every day we don't terminate, a boy crossed a girl off his list, so if we didn't terminate at N squared plus 1 days, we must have crossed off N squared plus 1 girls across all the lists. Well, is that possible? To have done N squared plus 1 cross outs? What do you think? How many names are on each list? AUDIENCE: N. PROFESSOR: N, and how many lists are there? AUDIENCE: N. PROFESSOR: N, so there's N lists with N names implies there's, at most, N squared cross outs ever. But we just said we had N squared plus 1 cross outs. But we have also N squared plus 1 cross outs, and that's a contradiction. All right? So we're done. It has to terminate within N squared plus 1 days. Any questions? This is a very common proof technique in computer science. You're analyzing some system, and every step or every day or every time period, you want to argue progress got made, and then after you've made enough progress you have to be done, and therefore, the algorithm is completed. All right, so we know that TMA terminates. Now, we've still got to get everyone married and have them all be happy, or at least stable. Now, to do this, we're going to use an invariant. Yeah? AUDIENCE: If a girl has some guy in front of [INAUDIBLE] on something, then she will always have [INAUDIBLE] PROFESSOR: That is true. Something even stronger is true. AUDIENCE: All the [INAUDIBLE] PROFESSOR: Have a--PROFESSOR: If she has a preference for someone who [INAUDIBLE] only better guys. PROFESSOR: Yeah, that's a great invariant. As the girl is sitting there on her balcony, things only get better, because she always keeps the best one around that's there. And when new ones come in, they've got to be better than the last one for her to keep them. So as she rejects boys, she only does that because she's got better ones there, and whoever she says maybe to always comes back. So an invariant of this algorithm is that when a girl has a suitor, going forward, she only has suitors she likes at least as well. And if she ever rejects a boy, then she's got somebody better there forever. All right, so let's state that as an invariant and prove that. OK, so we're going to let P be our invariant here. P is the statement that if a girl, G, ever rejected a boy, B, then the girl, G, has a suitor, or if the algorithm is terminated, a husband who she prefers to B. All right, that's going to be our invariant, and now we've got to prove it's an invariant. So we'll do that with a lemma called lemma one. P is an invariant for TMA. All right, let's prove that. Now, what's the first thing you've got to prove when you're proving something as an invariant? There's two things you've got to do to establish a variant. What's the first one? AUDIENCE: Base case? PROFESSOR: Base case. Show it holds true at the beginning. Proof is going to be by induction, and so we've got to show that P holds true at the beginning. But what are we going to induct on here? What's the parameter we're inducting on? AUDIENCE: Time? PROFESSOR: Time, the number of days. Good. The base case is day zero, the beginning. All right. Well, is that statement true on day zero? Yeah, it's true for sort of a weird reason. Nobody's been rejected yet, so it's what's called vacuously true. Because no girl, G, has rejected any boy, so it's true. So no one is rejected yet. So it's vacuously true. All right, next we have the induction step. So we'll assume P holds at the end of day d, and we need to argue that it holds now at the end of day d plus 1. So say it's true up to now, up to day d. Why Is it true at the end of the next day? Well, there's two cases to look at here, depending on when G rejected B. So if she rejects B on this day, day d plus 1, well, why would she reject B on day d plus 1? There's only one scenario. Yeah? AUDIENCE: There's a better boy. PROFESSOR: There's a better boy. And so she says maybe to him, and he becomes her suitor. So, in fact, P is true. So then there was someone better, and that implies P is true on day d plus 1. Case two is very similar. It's a very simple proof. G rejected B before day d, before d plus 1. Well, now we use the fact that P was true at the end of day d, which means P now implies that G had at least as good a suitor. Actually, it's better, because it was rejected. A better suitor on day d, that's what the hypothesis says, the invariant says. And now we just have to look at what happens on d plus 1. Well, either she has the same suitor on d plus 1, or somebody better came along. And so we're going to be done. She has the same or better suitor on day d plus 1, and that implies P is true on d plus 1, and we're done. All right, so I went through this proof. It was sort of obvious, but this is the careful way you'd write it down to show that the invariant holds. Any questions about the invariant dilemma? All right, so things only get better for the girls. If she ever rejected somebody, she's got somebody better. So now we can prove the main result, that everyone is married. And again the proof will be by contradiction. So we assume not everyone was married. Assume, for the purpose of contradiction, that some boy, B, we'll call him, is not married. Because if everyone is not married then some boy is not. If not everyone is married then some boy is not married. So when it terminates, B is not married. Well, what do you know about B if he was not married at the end? PROFESSOR: He was rejected by everyone. PROFESSOR: He was rejected by everyone, because if at the end, he's still under a balcony, if he still had somebody on his list, he'd be there. And he'd be getting married, because it's the end. So this means that if B is not married, he's rejected by everybody. Yeah? AUDIENCE: [INAUDIBLE] list? PROFESSOR: Oh, he's on everybody's list. AUDIENCE: Yeah. PROFESSOR: Everybody has everybody of the opposite sex on their list, but he had to cross everybody off. That's true. Yeah? AUDIENCE: That would mean that everyone had somebody better on the same day that [INAUDIBLE] telling him she didn't want to be with him. PROFESSOR: That's true. That means that B crossed every girl off. B is rejected by every girl, which means every girl has somebody better than B, which is not possible, because that would mean every girl was married. And therefore, the equal number of boys and girls, that means B would have been married. Good. All right, let's write that down. This means that B was rejected by every girl. OK, that means that every girl, by lemma 1, has a better suitor, and that's where we use lemma 1. And that means that every girl is married, and that means that every boy is married, including B, and that's a contradiction, because we said B wasn't married. OK? Everybody buy that proof? Any questions about that? Yeah, proof by contradiction is a pretty powerful technique. Once you assume something is not going to be true, it gives you a lot of power to find a contradiction. All right, so now we know that the algorithm ends, and everybody gets married. All's we got to do is show that there's no rogue couples, so let's do that. TMA produces a stable matching. Now, how do you suppose we're going to prove this? What's going to be the approach to prove this? Yeah? AUDIENCE: Assume that there's a rogue couple. PROFESSOR: Assume that there's a rogue couple, that namely this is not true, so there must be rogue couple. So let Bob and Gail be any pair that are not married. I need to prove the Bob and Gail are not rogue, and then we'll be fine. Because if it says everybody who is not married is not rogue, then we know we have a stable matching. Now, there's a couple of cases here. Bob and Gail weren't married, so there's two ways that could have happened. What's one of them? What's one reason they might not be married, something that happened that made that impossible? Yeah? AUDIENCE: Gail rejected Bob. PROFESSOR: Good. Case one-- Gail rejected Bob. Well, what do we know in that case, if Gail rejected Bob? AUDIENCE: Gail had better suitors. PROFESSOR: Gail has another suitor that she likes better. And, in fact, what do we know about who Gail married? AUDIENCE: Better than Bob. PROFESSOR: Better than Bob-- things only get better for the girls. That's the lemma one. All right, so this means that Gail marries someone that she thinks is better than Bob. And that's by lemma one. Well, can Gail and Bob be a rogue couple here? No, because Gail likes her spouse better than Bob, so she's not going to be in a rogue affair with Bob. So that means that Gail and Bob are not rogue. All right, case two is Gail did not reject Bob. She never did. That's the other case, she didn't reject Bob. Could Bob have ever serenaded Gail in this case? No, because if he did and he was never rejected, they would have ended up married. All right? So that means the Bob never serenaded Gail. What does that mean about how Bob feels about Gail? Yeah? AUDIENCE: Bob must have never serenaded for Gail. PROFESSOR: Yeah, Bob never got far enough down on his list to serenade Gail. He got married before he got down there. All right, so that means that Gail is lower on Bob's list than Bob's wife. And that means that they're not rogue, because Bob likes his wife better than Gail. All right, so in each case, the cases clearly cover everything. Gail rejected Bob, or she didn't. Either way, Gail and Bob are not rogue. So that means there's no rogue couples, and that implies M is stable, TMA is stable. OK? All right, so TMA terminates, everyone gets married, no rogue couples, nice outlook. So we're done. We actually proved it works. One issue left to think about here. Any questions on that before we launch off into the last issue? AUDIENCE: Is it unique? PROFESSOR: Is it unique? TMA gives you a unique answer because it's an algorithm. It's deterministic. But there may be other stable matchings. OK, so there's not just one stable matching, necessarily. You could make examples with multiple stable matchings. That's a great question. Any other questions? All right. Oh, yeah? AUDIENCE: Is there generally any other way to assess optimality besides the fact that it's stable, or--PROFESSOR: Yeah, you can make up lots of them. You could put weights on the edges and get a min-weight matching. You could try to get the perfect matching with the least unfavorable marriage kind of thing. There's a lot of criteria you can make. This one turns out to be useful in practice in a variety ways that we'll talk about, and also have a nice, fast, simple algorithm that can actually run in a distributed environment, which is really nice. So it is probably the most practical approach to matching out there. Any other questions? OK, the last issue-- I don't know if we still have the issues up there-- is fairness. So who thinks the TMA is favorable to the boys? Just a couple. Who thinks it's favorable to the girls? More, that's the common response. Who thinks you can't even define it one way or the other? It's unclear and hopeless to decide. OK, well, it seems like maybe the girls, because they get the best of their suitors. They sit back, and they just take the best as they come along. On the other hand, the boys do try to go out and get their first choice. The girls have to wait. And Mr. Right may never come along. The boys are out there. I'm going to my first choice, and they get denied, OK, they just move right on to the next choice. So this is actually one of these questions of study in sociology. In the animal species, which is better, proposers or acceptors? What's the more powerful result? Who has the better power in courtship? It turns out that we can answer this question in a very clear way, and prove it here. And the answer is the boys have all the power here. This is very favorable to the boys, and we'll see why. Now, to prove that, we need some definitions. OK, so I've got to define a couple things here to be able to prove this. The first is, for any collection of preference lists, we're going to let S be the set of all stable matchings. Now, we know that S is not empty, right? How do we know that S is not empty here? AUDIENCE: [INAUDIBLE] stable. PROFESSOR: Yeah, because TMA produces a stable matching, so we know that S is not empty. There's at least one. And, in fact, there could be many. Now, for each person, P, we define the realm of possibility for P to be the set of mates that you might have in a stable matching. So it's a set queue of people for which there exists a matching that's stable such that you're mated to that person. So I've written this in some--that's sort of complicated mathematics, but somebody is in your realm of possibility if there is a stable matching where you married them. If it exists, there's a stable matching where you could marry them. And this is vaguely like you see sometimes, you're going out with somebody and your parents say, oh, they're not in your league, or something. This is a mathematical formulation for that, that there is some way in the world to marry everybody up, so it's all stable, and you could be married to that person. OK? Let's do an example with four people. All right, so for example, say we have Brad, Jen, and Angelina again. And of course Brad likes Angelina and vice versa. Gen likes Brad, and then there's Billy Bob here. All right. Now, Brad is not realistic for Jen. Brad Is not in Jen's realm of possibility, because in any perfect matching where Brad marries Jen, you're going to have a rogue couple. Brad's going to go for Angelina. So Brad is not within Jen's realm of possibility. And similarly, vice versa-- Jen is not in Brad's realm of possibility, because there's no stable matching where they're married, in this scenario, for this problem. All right? Now that we have that notion, we can define who your optimal mate is, and who your pessimal mate is, the worst case. All right, so we define a person's optimal mate is his or her favorite in the realm of possibility. So your optimal mate is not your favorite overall, it's just your favorite among those who are possible, that there is a stable matching that you could be matched to that person. And similarly, you have a person's pessimal mate is your least favorite from the realm of possibility. OK? Because you have all the people you might be married to in stable matchings. Your favorite is the optimal, your least favorite is the pessimal. Any questions about that? Does that make sense? And you don't count the ones you can't be married to, because if you were, it would be unstable. OK, now we can state the blockbuster theorems here. Theorem four says TMA marries every boy with his optimal mate. Theorem five, probably can guess, TMA marries every girl with her pessimal mate. All right, so it is optimal for every boy. They get the best possible they could have in stable marriages, and every girl gets the worst possible mate. Now, you wouldn't necessarily think that when you first look at that algorithm, where the boys are working down their list, and the girls are just getting better and better. But that's the case. Now, let's see, I'm not going to have time to prove them both, but I will prove theorem five. And I'm going to assume theorem four is true to prove theorem five. Maybe we'll do theorem four tomorrow in recitation. I don't know. So let's do theorem five so you can see how this works. The proof is by contradiction, so we'll assume theorem four is true, and we'll prove the proof of theorem five by contradiction. So assume there was a stable matching where a girl got worse off than in TMA. All right? Assume three and five is not true, then there's some girl in some stable matching. So suppose, for the purpose of contradiction, that there exists a stable matching, we'll call it M, where some girl, and she fares worse than in TMA. I want to show that results in a contradiction. And the proof is pretty simple, just a simple picture. So G did worse in M than in TMA. All right, so let's let B prime be the mate of G in M. So girl G did worse in M than in TMA, so here's her mate in M, and let B be her mate in TMA. Who does she like better, B or B prime? B, because we're saying that she did worse in M than she did in TMA, so she likes B better. And let G prime be the mate of B in M. Now, who does B like better, G or G prime? [INTERPOSING VOICES] PROFESSOR: Not G prime. B got married to G prime in M, but what does theorem four say? Yeah, G. Theorem four says that across all the stable matchings, B gets his favorite mate in TMA. So G prime is in the realm of possibility, and so is G, and we know from theorem four B likes G best, better than G prime, because of theorem four. So what happened here? There's a rogue couple in M. B and G are a rogue couple in M. And that means that M is not stable, and that is a contradiction. All right? And so theorem five is true, because we assumed we had a stable matching M where there's a girl who did worse off. So it really pays off to be aggressive party in courtship. Everybody lives happily ever after, especially the boys in TMA. Now, TMA arises in lots of applications. The most famous is the matching program that's used to match MDs to residency programs. So how many pre-meds are here? There's one, at least. So you're going to go to medical school some day, and at the end of medical school, there's a big day called match day, where you get assigned to a hospital for your internship, and the way that works is using this algorithm. And who do you suppose is the boy in this algorithm? AUDIENCE: The hospital. PROFESSOR: The hospital. And they want it so that when they make the matchings, that there's not some pair of an intern or doctor in a hospital where they'd like each other better than what they got. This is used in online dating for the obvious reasons, and we actually use this algorithm lot at Akamai to load balance traffic on the web. And here you have the boys are web servers, and the girls are requests for service. And the goals are to balance performance, getting a server that's nearby you that's fast. And on the other side, our cost, and by who we make a boy and a girl, we can trade off cost for performance in a very nice, distributed way. All right, so that's it for today.
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https://backoffice.biblio.ugent.be/download/1092888/6745132
Philosophica 82 (2008), pp. 87-114 REGIOMONTANUS AND CHINESE MATHEMATICS Albrecht Heeffer ABSTRACT This paper critically assesses the claim by Gavin Menzies that Regiomontanus knew about the Chinese Remainder Theorem ( CRT ) through the Shù sh ū Ji ǔ zh āng (SSJZ ) written in 1247. Menzies uses this among many others arguments for his controversial theory that a large fleet of Chinese vessels visited Italy in the first half of the 15th century. We first refute that Regiomontanus used the method from the SSJZ . CRT problems appear in earlier European arithmetic and can be solved by the method of the Sun Zi , as did Fibonacci. Secondly, we pro-vide evidence that remainder problems were treated within the European abbaco tradition independently of the CRT method. Finally, we discuss the role of recre-ational mathematics for the oral dissemination of sub-scientific knowledge. Post-doctoral fellow of the Research Foundation Flanders (FWO). An early draft of this paper was presented at the Symposium in Honor of Chikara Sasaki, Mathematical Sciences & Philosophy in the Mediterranean & the East , 4 – 7 August, 2009, Kamena Vourla, Greece, and benefited from constructive com-ments by Roshdi Rashed and Li Wenlin. 88 A. H EEFFER 1 Introduction The title of this paper is inspired by the section heading of Gavin Men-zies’s latest book 1434 (Menzies 2008, p. 147) titled Regiomontanus’s Knowledge of Chinese Mathematics . Menzies already stirred considerable controversy with his first work titled 1421 (Menzies 2002) in which he claims that the Chinese fleet circumnavigated the world during the Ming and travelled to parts of the world yet undiscovered by the Europeans, such as the Americas. Menzies is a retired naval commander with no command of the Chinese language. His methods were criticized and his conclusions dismissed by several historians and sinologists but his hypo-thesis also attracted many followers, gathered around a website. They call themselves “friends of the 1421 website” and their aim is to support Menzies’s theories with “additional evidence”. In a critical assessment of Menzies’ “evidence” that has escaped “distinguished academics in the field”, Bill Richardson concluded that “‘[i]maginography’ and unin-formed, wildly speculative ‘translations’ of toponyms are not conducive to a credible rewriting of history” (Richardson 2004, p. 10). His latest book is no less controversial. The new hypothesis in 1434 becomes apparent from the subtitle “The Year a Magnificent Chinese Fleet Sailed to Italy and Sparked the Renaissance”. According to Menzies a large fleet of Chinese vessels visited Italy in the first half of the 15th century. As a consequence, many great men from the Renaissance such as Paolo Toscanelli, Leone Battista Alberti, Nicolas of Cusa, Regiomonta-nus, Giovanni di Fontana and Mariano Taccola found direct inspiration for their knowledge from the Chinese. He claims for example, that many of the inventions that Leonardo da Vinci is credited for actually depended on knowledge of Chinese contrivances through Taccola and Francesco di Giorgio. The evidence is presented sophistically by showing the similari-ty of Renaissance and Chinese illustrations side by side (see figure 1). Alberti’s knowledge of perspective would depend on the Shù sh ū Ji ǔ zh āng (數書九章 , Mathematical Treatise in Nine Sections , hence SSJZ ,REGIOMONTANUS & CHINESE MATHEMATICS 89 Libbrecht 1973) by Qín Ji ǔshào (秦九韶 ) written in 1247. 1 And so it goes on. From the Nung shu , 1313 (Need-ham, 1965) Late 15 th C. copy of Tac-cola (Ms. Palatino 767, f. 32r, BNC Florence) Figure 1: A typical example of Menzies’s “evidence” for Chinese influences. It is all too easy to dismiss the claims by Menzies by reasons of a lack of historical scrutiny and his limited knowledge of the Chinese language and discard all of his observations. However, his claims are challenging, the parallels are surprising, and some evidence cries out for an explana-tion. As it is our belief that scholars should not shy away from the claims presented in the book we will take up the challenge. We will focus here on one of Menzies’s arguments dealing with mathematics: because of his knowledge of the Chinese Remainder Theorem ( CRT ) Regiomontanus knew the relevant Chinese mathematical works. We will demonstrate that 1We will use the pinyin transcription for the names of Chinese books and au-thors. 90 A. H EEFFER the argument is false and that the criticism applies to many similar claims in his book. However, there has been a transfer of mathematical knowledge from East to West. The CRT method or Da-yan shu (大衍術 Great Extension Mathematics) is a Chinese invention. Therefore we will also address the question of how then to account for dissemination of knowledge through distant cultures without the availability of written evidence. 2 The da yan rule 2 Let us first recall that the da yan rule describes that for a set of congru-ences (mod )i ix a m≡ in which the im are pair wise relative primes, the solution is given by 11 niii M x a b m= = ∑ in which M is the product of all the im and the ib are derived by the congruence relation 1(mod )i i i M b m m ≡ .The Chinese version of the rule depends on a specific procedure which we will illustrate by a numerical example. See Needham ( SCC , III, pp. 119-120), Libbrecht 1973, pp. 333-354), Katz 1992, p. 188, or Martzloff 2006, pp. 310-323 for other examples. We will abbreviate con-gruence sets by the compact notation 1 1 2 2( ), ( ), , ( )n nx a m a m a m≡ … . Our example is x ≡ 2 (3), 3 (5), 4 (7). This would translate in natural lan-guage as: x is a number which divided by three leaves 2, and divided by five leaves 3, and divided by seven leaves 4. 2The name is in Western publications better known under its Wade-Giles transli-teration ta-yen. REGIOMONTANUS & CHINESE MATHEMATICS 91 1) Reduce the moduli ( ding mu or fixed denominators) to a multip-lication or a power of prime numbers unless they are prime or a power of a prime already (which is the case in our example). The relative prime moduli are called ding shu . This procedure is called da yan qiu yi shu (great expansion procedure for finding the unity) from which the da yan name is derived. 2) Find the least common multiple of the moduli, called the yan mu (multiple denominators). In our example the product of 3, 5 and 7 is 105. 3) Divide the yen mu by all the ding shu . The result is called yan shu (multiple numbers or operation numbers), in our example 35, 21 and 15 respectively. 4) Subtract from the yen shu the corresponding ding shu as many times as is possible. The remainders are called qi shu (surplus). Thus 35 – 3(11) = 2, 21 – 5(4) = 1, 15 – 7(2) = 1. 5) Calculate the chêng lü (multiplying terms) as 1(mod )i i i M b m m ≡ ,being 1 12 1(mod3), 2b b= = , 2 21 1(mod 5), 1b b= = and 33 1 1(mod7), 1b b= = .6) Multiply the chêng lü with the corresponding yan mu and the re-mainders. These are called yong shu (useful numbers). Thus 2.35 = 70, 3.21 = 63 and 4.15 = 60. 7) Multiply the yong shu with the remainders. Thus 70.2 = 140, 63.1 = 63 and 60.1 = 60. 8) Add these products together and you will get the zong shu (sum) thus 140 + 63 + 60 = 263. 9) Subtract the yan mu from this sum as many times as possible to get the solution, x = 263 – 105 – 105 = 53. The terminology is taken from Qín Ji ǔshào . An English translation of the complete text of the procedure is given by Libbrecht 1973, pp. 328-332 and Dauben 2007, pp. 314-15. 92 A. H EEFFER 3 An epistemic claim In this section we provide a fair representation of the epistemic argumen-tation developed by Menzies. The first premise consists of the fact that Regiomontanus had knowledge of the da yan rule. The evidence pre-sented stems from his correspondence with the astronomer Francesco Bianchini in 1463. The Latin correspondence was published by Curtze 1902. In a letter to Bianchini (not dated but late 1463 or early 1464) Re-giomontanus poses eight questions, the last one being “Quero numerum, qui si dividatur per 17, manent in residuo 15, eo autem diviso per 13, manent 11 residua, At ipso diviso per 10 manent tria residua: quero, quis sit numerus ille” (Curtze 1902, p. 219). Within the context of the quatro-cento such questions should be understood as challenges rather than ge-nuine enquiries. Regiomontanus probably had the solution before posing the question. But Bianchini met the challenge and produced a correct answer in his letter of 5 Feb 1464 (Curtze 1902, p. 237). Bianchini an-swers the above problem with the solutions 1103 and 3313 and adds that there are many more solutions, but that he does not wish to spend the labor for finding more (“Sed in hoc non curo laborem expendere, in aliis numeris invenire”). This comment leads Curtze to conclude that Bianchi-ni did not know the general method. In his answering letter (not dated, Curtze 1902, p. 254) Regiomontanus reveals that there is an infinite num-ber of solutions and that the solutions are easily generated, annotated by a figure in the margin (“Huic si addiderimus numerum numeratum ab ipsis tribus divisoribus, scilicet 17, 13 et 10, habebitur secundus, item eodem addito resultat tertius etc.”). Every time one adds the least common mul-tiple of the three divisors, additional solutions are generated, not much labor required at all. But concluding from this, as Curtze does, that Regi-omontanus knew the general solution for the remainder problem, is one step too far. As we shall demonstrate below, the solution can be obtained by tables or trial and error and the rule for generating additional solutions was known within the abbaco tradition. It even had its own name. It is impossible from the correspondence to establish the specific procedure he REGIOMONTANUS & CHINESE MATHEMATICS 93 used for finding the first solution. The second premise, that a general procedure for applying the CRT is explained in the Shù sh ū Ji ǔ zh āng , is of course justified. From this Menzies concludes that “it follows that Regiomontanus must have been aware of this Chinese book of 1247 un-less he had quite independently thought up the Da-yan rule, which he never claimed to have done” (Menzies 2008, p. 149). So, Menzies tacitly assumes that Regiomontanus did not reconstruct the CRT by himself. We also believe this to be true on basis of what we know of his writings and the historical context, discussed below. However, remark that Menzies’s argument thus can be used to pose the opposite: “it follows that Regi-omontanus must have independently thought up the Ta-Yen rule unless he was aware of this Chinese book of 1247, which he never claimed to have known”. He further adds: “The implications of Regiomontanus knowing of this massive book, which was the fruit of the work of thirty Chinese schools of mathematics, could be of great importance. (..) It may lead to a major revision of Ernst Zinner’s majestic work on Regiomontanus” (ib-id.). 3 In summary, in we keep tacit the premise that “Regiomontanus did not discover the CRT independently”, the argument runs as follows: Premisses: 1) Regiomontanus had knowledge of the CRT in 1463. 2) The CRT is explained in the SSJZ of 1247. Conclusion: Regiomontanus had knowledge of the SSJZ. In the sections below we will show that this argumentation is wrong. First we start with a historical reconstruction of the premises and conclusion and show that the first premise cannot be justified. Secondly, we will provide new historical evidence that before 1434 there was European tradition of dealing with remainder problems which was sufficient to explain Regiomontanus’s solutions without the CRT . Thirdly we will 3Ernst Zinner 1939 is the biographer of Regiomontanus. 94 A. H EEFFER demonstrate that even if one accepts Menzies’s premises, alternative ex-planations are still possible. 4 Other sources for the da yan rule 4.1 Earlier European sources 4 Already in 1202 Fibonacci discusses seven remainder problems in his Liber abbaci . His last problem is actually the one which we used as an example (Latin in (Boncompagni 1857, p. 304), English by (Sigler 2002, pp. 428-29), also discussed and translated by (Libbrecht 1973, pp. 236-238)). Fibonacci predates the Shù sh ū Ji ǔ zh āng and while he gives a general method for the moduli 3, 5 and 7 and another one for the moduli 5, 7 and 9, his procedure is different from the one we listed above (Sigler 2002, pp. 428-9): He divides the chosen number by 3, and by 5, and by 7, and always you ask what are the remainders from each division. You truly for each unit of the remainder upon division by 3 keep 70, and for each unit of the remainder upon division by five you keep 21, and for each unit of the remainder upon division by seven you keep 15. And whenever the total exceeds 105, you throw away 105, and that which remains for you after all the 105 are thrown away will be the chosen number. For exam-ple, it is put that after division by 3 there remains 2 for which you twice seventy, that is 140; from this you take away the 105 leaving 35 for you. 4 We will limit the discussion of European sources to the period before 1464. Shortly after the correspondence of Regiomontanus and Bianchini the problem appears in Maestro Benedetto da Firenze’s Trattato (c.1465, pp.68-69), in the Pseudo-dell’Abbaco of 1478 (ff. 69r-69v) and in Piero della Francesca’s Trattato d’abbaco of c. 1480 (f.122). For its later history and influence on Gauss see Bullynck 2008. REGIOMONTANUS & CHINESE MATHEMATICS 95 And after division by the five there remains 3 for which you keep three times 21, that is 63, which you add to the aforesaid 35; there will be 98. After the division by the seven there remains 4 for which you retain quadruple 15 that is 60 which you add to the aforesaid 98; there will be 158 from which you throwaway 105; there will remain for you 53 which is the chosen number. In Fibonacci’s procedure the zong shu are not calculated but given and the yan mu are subtracted from zong shu before they are summed togeth-er. In fact, Fibonacci’s method corresponds closely with the text of the Sun Zi Suan Ching for the similar problem x ≡ 2 (3), 3 (5), 2 (7). 5 (Mi-kami 1913, p. 32; further discussed below): In general, take 70, when the remainder of the repeated divisions by 3 is 1; take 21, when the remainder of the repeated divisions by 5 is 1; and take 15, when the remainder of the repeated divisions by 7 is 1. When the sum of these numbers is above 106, subtract 105, before we get the an-swer. (…) The remainder divided by 3 is 2, and so take 140. The re-mainder divided by 5 is 3, and so take 63. The remainder divided by 7 is 2, and so take 30. Adding these together we get 293. There from subtract 210, and we obtain the answer. Libbrecht (1973, p. 240) also discusses Fibonacci’s solution and con-cludes that he “does not give the slightest theoretical or general explana-tion of his method of the remainder problem, and for this reason his whole treatment is on a level no higher than that of Sun Zi”. Needham (SCC, I, p. 4, p. 34) and Libbrecht (1973, pp. 241-2) mention a four-teenth-century Byzantine manuscript of the Isagoge Arithmetica by Ni-chomachus of Gerase which contains an appendix dealing with a problem 5 Martzloff 2006, p. 322 comes to the same conclusion: “In 1202, Fibonacci (..) proposed a solution to the remainder problem similar to that of Sun Zi, in other words incomparably less powerful than that due to Qin Jiushao”. 96 A. H EEFFER finding a contrived number. The solution method as well as the recrea-tional context is very similar to Fibonacci. Two additional occurrences of the problem are documented in extant writings before Regiomontanus in Europe. The astronomer Giovanni Marliani was the teacher of Giorgio Valla and wrote a vernacular treatise on arithmetic. His Arte giamata arismeticha dates from c.1417 (codex A. II. 39, Biblioteca Universitaria de Genova) and is described and partly transcribed by Gino Arrighi 1965. The remainder problem is basically the same as the one by Fibonacci and the limited explanation provides no clues concerning the knowledge of the CRT. Two other remainder prob-lems appear in a pseudo-Paolo dell’abbaco of c.1440, with moduli 2 to 10 and the congruences one less than the moduli and 1: Truova uno numero che partjto per 2 ne rjmanghi uno, e partjto per 3 ne rjmanghi 2, e partjto per 4 ne rjmanghj 3, e partjto per 5 ne rimanghj 4, e choxj per insino in 10 (Arrighi 1964, p. 95). Truovamj uno numero che partjto per 2 ne rjmanghj uno, e partjto per 3 ne rjmanghi uno, e partjto per 4 ne rimanghj uno, e partjto per insino in 10 (Arrighi 1964, p. 96). Now, these two problems are quite interesting as the author seems to have heard of these or similar remainder problems but he does not know the answer or the method of the CRT . For the first problems he gives the answer 3628799 and for the second 75601. Both solutions are valid, but of course not the smallest integral solutions as provided by the CRT me-thod, 2521 and 2519 respectively. The author solves the first problem by multiplying all the moduli and subtracting one: Fa’ choxj e di’: in che si truova il 2 e ’l 3 e ’l 4 e ’l 5 e ’16 e ’l 7 e ’l 8 e ’l 9 e ’l 10? E però multjpricha 2 via 3, fa 6, e 4 via 6, fa 24, e 5 via 24, fa 120, e 6 via 120, fa 720, e 7 via 720, fa 5040, e 8 via 5040, 40320, e 9 via 40320, fa 362880, e 10 via 362880, fa 3628800. E ora puoj dire: jo òe trovato uno numero che partjto per 2 no’ rrimane alchuna choxa, e partjto per 3 no’ rimane nulla e choxj per 4, per 5, per 6, per insino in 10. E ora REGIOMONTANUS & CHINESE MATHEMATICS 97 di’: traendo uno di questo numero, si nne verrà da xxezzo 1 meno che nel numero nel quale io ò partjto, uno di questo numero cioè di 3628800 che rresta 3628799 e questo numero è desso. For the solution to the second problem he finds 7560 as a common mul-tiple, though not the least common multiple, and adds one, and 7561 in-deed is a solution. He then claims that 75601 is “ a more secure solution”. While we can relate some remainder problems to the Arabic tradition and Fibonacci, it seems that such kind of problems were known, discussed and solved by methods which do not reflect any knowledge of the da yan rule. The occurrences of the problems as discussed here exhaust all our published sources of abbaco texts. Convinced that there had to be found something more we looked at a number of previously unpublished manu-scripts. We will come back to the abbaco tradition in a further section. First we look at the problem in German cossic texts. Its appearance is unnoticed by Menzies but could be used as an argument in favour of his first premise. 4.2 The cossic tradition In a classic overview of fifteenth-century algebra in Germany Maximilian Curtze points out that by the middle of the fifteenth century the CRT was a well known subject. 6 As evidence he refers to the Deutsche Algebra, the manuscript 14908 of the State library in Munich written in a mixture of German and Latin. Curtze names the author Gerhard, but it is since then established that it is the monk Fredericus Amann who authored the text in 1461 (Folkerts 1996, Gerl 1999, Gerl 2002 and Vogel 1981). While this 6Curtze 1895, p. 65 note: “Ich will jetzt hier den Beweis führen, dass um die Mitte des 15. Jahrhunderts sie mit ihrem Beweise und ohne Benutzung des chine-sischen Beispiels eine ganz bekannte Sache war”. 98 A. H EEFFER treatise was written two years before the correspondence between Regi-omontanus and Bianchini it still is difficult to establish who got it from whom. According to Folkerts (1996, 26) Fredericus based his text on the still unpublished Regiomontanus’s manuscript on algebra written in 1456, Columbia University, Plimpton 188, (ff. 82v-96r). However, the algebra problems by Fredericus run from ff. 133v-157r and the method for solv-ing the CRT is discussed on the preceding pages (124v-125r, Curtze 1895, p. 65). Though we do not have a transcription of Regiomontanus’s problems we know that it includes one remainder problem (problem 60, f. 93r). According to Folkerts 2002, p. 421, it is “equivalent” to the problem in his correspondence. If at all Regiomontanus had a general method for solving remainder problems, which cannot be established from the cor-respondence, it is safe to assume that he and Amann used the same me-thod. As we have a transcription available of Amann’s solution let us therefore take this as a reference. In a curious mixture of old German and Latin, Amann provides the following solution to the problems with mod-uli (3, 5, 7) (Curtze 1895, p. 65; Vogel 1954, pp. 120-1): Item ich wil wissen, wie vil pfenning in dem peutel odor im synn hast. Machs also. Hays yn dy dn, dy er hat, zelen mit 3, darnach cum 15, postea cum 7, vnd alz oft eins vber pleibt mit 3, so merck 70, vnd alz oft 1 vber pleibt mit 5, merk 21, vnd mit 7, merk 15. Postea adde illos numeros in simul, et ab ista summa subtrahe radicem, hoc est multiplica 3per 5 et 7 erit 105, als oft du magst, vnd wz do pleibt, alz vil hat er ym sinn oder in peutel. Amann further lists a table with the moduli (2, 3, 5), (3, 4, 5), (3, 4, 7), (2, 3, 7), (2, 7, 9), (5, 6, 7), (5, 8, 9) and (9, 11, 13) including the least com-mon multiple of the moduli and but not what the Chinese called yen shu .So for the moduli (3, 5, 7), Amann gives 105 and then the numbers 70, 21, 15 instead of the yen shu 35, 21 and 15. We can thus conclude that Amann – and possibly Regiomontanus – followed the method of the Sun Zi Suan Ching also employed by Fibonacci instead of the one described REGIOMONTANUS & CHINESE MATHEMATICS 99 in the Shù sh ū Ji ǔ zh āng . As the method for solving remainder problems in fifteenth-century Europe is different from the superior one discussed in the Shù sh ū Ji ǔ zh āng it can actually be interpreted as evidence against the hypothesis posed by Menzies: Premisses: 1) If Regiomontanus knew the general solution to the CRT as explained in the SSJZ he would have used it. 2) Regiomontanus (and Amann) did not use the method of the SSJZ. Conclusion: Therefore Regiomontanus had no knowledge of the SSJZ . 4.3 The rule of numbers with no end From a study of a number of unpublished manuscripts we discovered that remainder problems were treated already in the Italian abbaco tradition since the fourteenth century. We identified a family of at least six manu-scripts which contain a table with remainders for moduli 3, 5 and 7. The table is listed in relation to one in a series of remainder problems. The purpose of that problem is to find numbers that satisfy x ≡ 2 (3), 3 (5), 1 (7). The unpublished manuscripts containing the table are the following (including sigla and folio location): • Z (c. 1395) Florence, BNC, Conv.Soppr.G7.1137, Inc. “Queste sonno le fegurre nostre dello aboccho ceh le guagli tu pot inscri-vere qualunque numero tu vogli..” ( corsiva mercantesca formata )ff. 241r-241v • α (c.1417), a lost archetype of which the following are derived copies; the table of contents in a later copy points to ff. 141 • A (c.1433) Florence, BNC, Magl. Cl. XI. 119, Inc: “Concio sia cosa che sono nove figure nell’abacho per le quali chi conoscie quelle agievolmente conosciera poi l’altre ..” (neat corsiva can-cellaresca formata ) ff. 141(?) 100 A. H EEFFER • B (c1440) Florence, Biblioteca Mediceo-Laurenziana, Ash. 608, Inc: “Concio sia chosa che sono nove fighure nel abacho per le quali chi chonoscie quele agievolmente chonosciera poi l’altre ..” (rapid corsiva mercantesca ) ff. 101v-106r • C (c1440) London, BL, Add. 10363, Inc: “Concio sie cosa che son 9 fighure nell’abbaco per le quali chi chonoscie quelle age-volmente conosciera poi l’altre …”, (very neat humanistic boo-khand) ff. 149r-152r • E (1442) London, BL, Add. 8784, Inc: “E choncio sia chosa che sono 9 fighure nell’abacho per le quale chi conosce quelle agie-volmente conoscera poi l’altre...”, (fairly neat Italian Gothic bo-okhand, by Agostino di Bartolo) ff. 138v-140v The manuscripts α and A to E are part of a related family of copies of an abbaco treatise which is described in Heeffer 2008. We will here use the same sigla. The table (see figure 2) in its earliest form is contained in a zibaldone (an author’s notebook, hence Z) from the end of the fourteenth century. All the cited manuscripts contain the table and the main problem to find 11 numbers which satisfy the congruence relation. The solution is listed (8, 113, 218, 323, 428, 533, 638, 743, 848, 953 and 1058) and it is verified that they have the same remainders for the three moduli 3, 5 and 7 but apart from the table no method is presented. REGIOMONTANUS & CHINESE MATHEMATICS 101 102 A. H EEFFER Figure 2: Table of remainders for moduli 3, 5 and 7 (used with permission © The British Library Board, ms. Add.10363 ff. 152rv) These texts therefore show no evidence of any knowledge of the da yan rule, not even the limited version from the Liber abbaci . The table starts from 8 and runs to 113, the first two numbers in the congruence set. The accompanying text explains how to generate additional solutions. First it is observed that if you add one to the solutions 8 and 113, the numbers 9 and 114 will have the same remainders for moduli 3, 5 and 7. When you add 2, 3 and more, the resulting numbers will also have the same rea-minders. Thus when one adds the difference between 8 and 113 (which is 105) to 8 and to 113, the remainders will also be the same. This allows us to generate infinitely many numbers with the same remainders for a given set of moduli. In fact, as far as we know, the reasoning is the earliest in-stance of mathematical induction in European writings (Heeffer 2010). Because of the demonstrated infinity of possible solutions the rule is REGIOMONTANUS & CHINESE MATHEMATICS 103 called “la reghola del numero sanza fini” or the rule of numbers with no end. We remark that the limited knowledge about remainder problems in these manuscripts is sufficient to explain the reasoning of both Regi-omontanus and Bianchini in their correspondence. There are infinitely many solutions and these solutions are in an arithmetical progression. This undermines Menzies’s argument since all what Regiomontanus needed to know was available already in fourteenth-century Italy. 4.4 Earlier sources in Asia A typical fallacy of Menzies’s argumentation is that he was looking for an explanation which necessarily involves Chinese knowledge, sources or artifacts which supposedly have been disseminated through a hypotheti-cal visit of the Chinese in Italy in 1434. He therefore looks for Renais-sance sources after 1434 and tries to match them with Chinese ones be-fore that date ignoring all other contextual explanations. In our case he matches a very specific problem from the correspondence of Regiomon-tanus with one specific Chinese book. We will now show that the CRT was known long before Shù sh ū Ji ǔ zh āng .It is generally acknowledged by historians of Chinese mathematics that remainder problems grew out of calendar calculations in China dur-ing the third century. The problem of finding a common cycle for the lunar months and the tropical year can be formulated as finding a period of N days in which )(mod )(mod )(mod 321 YrLrDrN ≡≡≡ with D as the length of a day, L the duration of a lunar month and Y the duration of a tropical year and the ri as the remainders. The zh āng cycle ( 章) of 19 tropical years corresponding with 235 lunar months was determined in China already in the sixth century BC. Such a cycle was 104 A. H EEFFER also known by the Babylonians around 500 BC and by the ancient Greeks (Meton at 432 BC). Remainder problems appear in different forms but usually within a practical setting. The earliest extant source dealing with such problems is the Sun Zi Suan Jing (孙子算 经, Master Sun’s Arithmetical Manual , c. 400). An English translation and discussion is given by Lam Lay Yong 2004. Sun Zi describes the problem x ≡ 2 (3), 3 (5), 2 (7) (Chap. 3, prob. 26, p. 10b, also translated and discussed by Needham ( SCC , III, p. 119), Mikami 1913, p. 32, Li and Du 1987, p. 93, Libbrecht 1973, p. 269). Remainder problems are also known in India, and first appear in the Āryabha ṭīya of Āryabha ṭa (of 499) where they are solved by the ku ṭṭ aka or pulverizer method (Clark 1930, pp. 42-50, Datta and Sing 1934, pp. 87-99, pp. 131-133, Libbrecht 1973, p. 229). In the Br āhmasphu ṭasiddh ānta by Br āhmagupta of 628 the method is applied within the context of astronomy (Colebrooke 1817, pp. 326-7). In the Bījaga ṇita of 1150 Bh āskara II uses algebra to calculate the solution (Chap. VI, stanzas 160 & 162, Colebrooke 1817, pp. 235-239). Although remainder problems are rare in Arabic treatises they were known by Ibn al-Hait am (c.1000), in particular the problem x ≡ 1(2, 3, 4, 5, 6), 0(7), as discussed by Wiedemann 1892, Libbrecht 1973, pp. 234-5, and Rashed 1994. Libbrecht points our attention to the similarity of the Arabic and Indian solutions. 5 Transmission from China to Europe We concluded that Regiomontanus did not get his knowledge of the CRT from the Shù sh ū Ji ǔ zh āng . Without conceding to Menzies’s claim that Chinese knowledge about mathematics was passed on to Europe through a visit during the fifteenth century, some questions remain. How did he and other Europeans like Fibonacci learn about the remainder problems and possible ways to solve them? The close correspondence between the REGIOMONTANUS & CHINESE MATHEMATICS 105 remainder problems in the Sun Zi Suan Ching and Fibonacci’s Liber Ab-baci remains puzzling. As with regard to transmission between cultures the Western historio-graphy is deeply influenced by the humanist prejudice that all higher intellectual culture, in particular all science, had risen from Greek soil. A most typical example is the monumental and influential four-volume work on the history of mathematics by Moritz Cantor (1880-1908). Like many men of his era Cantor was entrenched with humanist ideas about the cultural superiority of Western knowledge. For example, when deal-ing with Hindu algebra Cantor takes every opportunity to point out the Greek influences on India. Some examples: the Indians learned algebra through traces of algebra within Greek geometry; Br āhmagupta’s solution to quadratic equations has Greek origins; or the Indian method for solving linear problems in several unknowns depended on the Greek method of Epanthema and so on. 7 However, when he comes to the CRT , Cantor writes the following: 8 It is not impossible that the Chinese problem and its solution could have become known somewhere by a Byzantine who took note of it, possibly through Arabic mediation. The reverse course, which is so often the case, where Western knowledge penetrates China, is here hardly possible, be-cause it is only in Chinese texts that the explanation of the procedure is provided, quite difficult to understand, but comprehensible nevertheless, as experience has shown us. 7 See Heeffer 2009 for a discussion on the reception of the first translations of Hindu classics on mathematics. 8 Cantor (I, p. 644) “Es ist nicht unmöglich, dass die chinesische Aufgabe und ihre Auflösung etwa durch arabische Vermittlung irgend einem Byzantiner hekannt geworden sein kann, der sie sich aufnotirte. Ein umgelrehrter Gang, dass also hier wie so vielfach im Westen Bekanntes nach China drang, ist kaum anzunehmell, weil nur im chinesischen Texte die Begründung des Verfahrens angedeutet ist, freilich schwer zu verstehen, aber doch zu verstehen, wie die Erfahrung gezeigt hat.” (my translation). 106 A. H EEFFER This is one of the few instances were Cantor admits an influence from the East to the West although he is puzzled about how such a transmission could have taken place. The whole of his Vorlesungen solely depends on written sources. He knew about the Byzantine manuscript we cited before and did not know of any Arabic sources dealing with remainder prob-lems. The quotation reflects perfectly the current knowledge about writ-ten resources at that time. So does Menzies also rely on written sources is his latest book. However, remainder problems do not solely belong to the scholarly domain of mathematics, the kind of mathematics that supported the state and bureaucracy of ancient China. They also have a clear recreational value. These problems and their solutions belong to what Jens Høyrup has coined as sub-scientific mathematics. The narration of stories, riddles and recreational puzzles is the most important factor in the transition of arithmetical problems and their solution methods between generations, cultures and continents. They are passed on through oral traditions from master to apprentice and are mostly situated in the practices of merchants, lay surveyors and craftsmen. To get a grip on the oral tradition I proposed elsewhere a concrete implementation for sub-scientific knowledge in the form of proto-algebraic rules (Heeffer 2007). A proto-algebraic rule is a procedure or algorithm for solving one specific type of problem. Our main hypothesis is that many recipes or precepts for arithmetical problem solving, in abbacus texts and arithmetic books before the second half of the sixteenth century, are based on proto-algebraic rules. We call these rules proto-algebraic because they are or could be based originally on algebraic derivations. Yet their explanation, communication and applica-tion do not involve algebra at all. Proto-algebraic rules are disseminated together with the problems to which they can be applied. The problem functions as a vehicle for the transmission of this sub-scientific structure. Little attention has yet been given to sub-scientific mathematics or proto-algebraic rules. We provided a framework for identifying proto-algebraic rules which are common to Renaissance and Indian arithmetic and alge-REGIOMONTANUS & CHINESE MATHEMATICS 107 bra. Some Chinese rules such as of Yíng bù zú ‘excess and defi-cit’( 盈不足 ) definitely fall under this category. 9 The recreational aspect of remainder problems is prominent from its earliest occurrence. Libbrecht 1973 provides a comprehensive overview of all Chinese sources and we note that several formulations of the prob-lem are in verse and belong to the recreational domain. General Hán Xìn ’s method of counting soldiers ( 韩信乱点兵 ) is a cryptical formula-tion of a remainder problem but we can recognize the multipliers 70, 21 and 15 and the least common multiple 105 of moduli 3, 5 and 7, which occurs in many of the problems already cited. Hán Xìn was a general of Emperor Liu Bang of the Han Dynasty and lived around 200 AD: Not in every three persons is there one aged three score and ten, On five plum trees only twenty-one boughs remain, Every fifteen days rendezvous the seven learned men, We get our answer by subtracting one hundred and five over and again. The folk rhyme is still known in the oral tradition today by many Chinese and also Japanese. Li and Du 1987, pp. 93-94 discuss several other ver-sions over the next centuries. Libbrecht 1973, p. 286 quotes a stanza from the Zhiyatang zachao , written in 1290 by Zhou Mi: A child of three years old [when the father] is seventy, it is rare. 9 The problem appears already in the earliest extant Chinese treatise, the Suàn shù sh ū (筭數書 ) or ‘Writings on reckoning’: “In sharing cash, if [each] person [gets] 2 then the surplus is 3. If [each] person [gets] 3 then the deficit is 2. Ques-tion: how many persons, and how many cash? Result: 5 persons and 13 cash”. Christopher Cullen (2004, 81-88), who published the transcription of this trea-tise, identifies this rule with the rule of false position. However, the rule of false position, also known in Chinese arithmetic, is commonly considered as a general method for solving linear problems of the type ax = b. The Yíng bù zú however corresponds with the Indian rule gulik ā-antara or the regula augmenti in Euro-pean arithmetic, for problems of the form ax + b = cx – d (Heeffer 2007, 15-21). 108 A. H EEFFER At five to leave behind the things of 21, it is even more rare. At seven one celebrates the Lantern festival, again they meet together. The Cold meal holiday [on the 105 th day] then you will get it. Again we recognize the numbers of Master Sun’s original problem in a form which facilitates the recollection of the base numbers to solve prob-lems for moduli 3, 5 and 7. Alexei Volkov 2002, p. 402 describes a Viet-namese mathematical work of the fifteenth century Toan phap dai thanh in which a similar verse appears. So, with these examples we find evi-dence that the problem travelled in South-East Asia in the form of folk rhymes. As to its influence on Hindu mathematics opinions vary, with Indian scholars favoring the precedence of India over China. As men-tioned before, remainder problems in India are known as ku ṭṭ aka or pul-veriser problems and are very prominent in scholarly texts on mathemat-ics and astronomy. However, any sensible analysis, as Libbrecht 1973 does in his chapter 18, must come to the conclusion that the methods differ too widely to assume any influence from the one side to the other. However, the difference between the ku ṭṭ aka and the da-yen method do not exclude that the recreational version of the Chinese problem circu-lated in India. The early classic period of mathematics in India is charac-terized by the merging of two traditions, the scholarly astronomical tradi-tion to which the Āryabha ṭīya belongs, with the sub-scientific tradition of merchant and surveying mathematics, such as the Bakhsh ālī Manuscript Hayashi 1995. The ku ṭṭ aka method belongs to the first, the recreational version would fit in the second tradition. We finally come to the question of how the problem came to Europe. As we remarked already the treatment by Fibonacci very well reflects the Chinese method of Master Sun and bears little relation with methods from Hindu arithmetic. On the other hand, problems such as that of men finding a purse, where Fibonacci deals with many variants in the Liber Abbaci and the Flos seem to have originated in India (Heeffer 2007). If we look at the context in which the two remainder problems are discussed in the Liber Abbaci it comes as no coincidence that Fibonacci treats them REGIOMONTANUS & CHINESE MATHEMATICS 109 within a series of recreational problems. The section called de quibusdam divinationibus concerns divination problems in which a number or object has to be guessed (Boncompagni 1857, I, p. 303; Sigler 2002, p. 427). Such number divination or combinatorial divination problems were quite popular in Medieval Europe. Menso Folkerts (1968) found many in-stances in a study of 32 fourteenth and fifteenth century Latin manu-scripts. Also the German text by Fredericus Amann, cited earlier, sets the remainder problem as a divination problem: “I wish also to know how many coins he has in his purse or in his mind” (translation by Libbrecht 1972, pp. 244-5). After Regiomontanus we find remainder problems treated as part of divination problems in Luca Pacioli’s Viribus Quantita-tis (c. 1500, problems 22 to 25, Peirani 1997), as well as in the seven-teenth-century works Problèmes Plaisantes by Bachet 1612 (problem V) and the Recréations Mathématiques ([Leurechon] 1624, problems 51, 52). The problem continues to appear in books on recreational mathemat-ics during the following centuries. Our suspicion that the Chinese re-mainder problem travelled to Europe as part of an oral tradition is par-ticularly difficult to prove. However, the recreational version of the prob-lem, being part of this oral tradition of challenging others with riddles and puzzles, would not be out of place on the trade routes from East to West. We do not have to assume a single visit of a Chinese delegation in Italy in 1434 to explain the fact that mathematical knowledge travels between cultures and continents. Mathematics, being a product of culture, has been exchanged between cultures for centuries, together with other arti-facts and products of cultures. 6 Conclusion We have demonstrated that the argumentation for an epistemic claim by Menzies is fallacious. Menzies claims that Regiomontanus had know-ledge of a Chinese book written in 1247 because he could solve remaind-er problems in 1463. The structure of the argument and the reason for its 110 A. H EEFFER failure applies to many similar claims in the book. We have shown that some of Menzies premises cannot be proved from historical sources, in our case the claim that Regiomontanus knew the CRT . But even if we would accept his premises then his conclusion – throughout his book being the dependence of Renaissance scholars on Chinese sources – does not necessarily follows. For all of his claims there are contextual histori-cal sources which provide a more acceptable explanation than his wild speculation that the Chinese visited Renaissance Italy in 1434. We have provided two lines of explanations: 1) new historical evidence for solving remainder problems within the abbaco tradition which is sufficient to account for Regiomontanus’s solutions, and 2) the existence of an oral tradition of telling stories and riddles travelling along the silk route which could have included mathematical problem solving methods such as the CRT .Ghent University Center for Logic and Philosophy of Science Email: Albrecht.Heeffer@UGent.be REFERENCES Bachet, Claude-Gaspard. (1612). 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(1913). “The development of mathematics in China and Japan”, Abhandlungen zur geschichte der mathematischen wissenschaften mit einschluss ihrer anwendungen , 30, B. G. Teubner, Leipzig, pp. 1-347. Menzies, Gavin (2002). 1421, The Year China Discovered the World .London: Bantam Press. Menzies, Gavin (2008). 1434 The Year a Magnificent Chinese Fleet Sailed to Italy and Ignited the Renaissance , Harper Collins, Lon-don. Needham, Joseph. (1959). Science and civilisation in China. Vol. 3: Mathematics and the sciences of the heavens and the earth , Cam-bridge University Press, Cambridge. Needham, Joseph. (1965). Science and civilisation in China. Vol. 5: Part II: Mechanical engineering , Cambridge University Press, Cambridge. Peirani, Maria Garlaschi. (1997). Pacioli. De Viribus Quantitatis , Ente raccolta vinciana, Milano. Rashed, Roshdi. (1994). “Ibn al-Haytham and Wilson’s Theorem”, in The Development of Arabic Mathematics: Between Algebra and Arithmetic , Kluwer, Dordrecht, pp. 238-260. (translated from the French: Entre Arithmétique et Algèbre. Recherches sur l’Histoire des Mathématiques Arabes , Les Belles Letres, Paris, 1984). 114 A. H EEFFER Richardson, W.A.R. (2004). “Gavin Menzies‘ cartographic fiction: The case of the Chinese discovery‘ of Australia”, The Globe , 56, pp. 1-12. Sigler, Laurence (2002). Fibonacci’s Liber Abaci. A Translation into Modern English of Leonardo Pisano’s Book of Calculation .Springer, Heidelberg. Vogel, Kurt. (1954). Die Practica des Algorismus Ratisbonensis. Ein Rechenbuch des Benediktinerklosters St. Emmeram aus der Mitte des 15. Jahrhunderts nach den Handschriften der Münchner Staatsbibliothek und der Stiftsbibliothek St. Florian , C. H. Beck, Munich. Vogel, Kurt. (1981). Die erste Deutsche Algebra aus dem Jahre 1481. Nach einer Handschrift aus C80 Dresensis , Bayerische Akade-mie der Wissenschaften, Munich. Volkov, Alexei (2002). “On the origins of the Toan phap dai thanh (Great Compendium of Mathematical Methods)” In Y. 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[FREE] What is the solution set for the exponential inequality (\frac{1}{2})^{2x+1} \leq 4? Options: A. (0, - brainly.com 2 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 +51,1k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +21,9k Ace exams faster, with practice that adapts to you Practice Worksheets +8,9k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified What is the solution set for the exponential inequality (2 1​)2 x+1≤4? Options: A. (0, 4) B. (−3/2,4) C. (−3/2,∞) 1 See answer Explain with Learning Companion NEW Asked by anasandoval50371 • 09/15/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 6539985 people 6M 5.0 0 Upload your school material for a more relevant answer The solution set for the exponential inequality (1/2)^(2x+1) ≤ 4 is all x such that x ≤ -3/2. This can be written in interval notation as (-∞, -3/2]. So, Option 3: (-3/2, [infinity]) is the correct answer. Explanation The student is asking for the solution set for the exponential inequality (1/2)^(2x+1) ≤ 4. To solve this exponential inequality, we can first simplify it. The expression (1/2)^(2x+1) can be rewritten as 2^-(2x+1). We know that 4 = 2^2, so we can rewrite the inequality as 2^-(2x+1) ≤ 2^2. Since the base on both sides of this inequality is the same, we can set their exponents equal to each other to get - (2x+1) ≤ 2. Solving this inequality gives us x ≤ -3/2. Thus, the solution to this inequality is all x such that x ≤ -3/2. In interval notation, the solution set is (-∞, -3/2]. Therefore, among the given options, Option 3: (-3/2, [infinity]) is the correct answer. This option indicates that any value greater than or equal to -3/2 will satisfy the inequality. Learn more about exponential inequality here: brainly.com/question/33735279 SPJ11 Answered by shobhnahs •25.7K answers•6.5M people helped Thanks 0 5.0 (1 vote) Expert-Verified⬈(opens in a new tab) This answer helped 6539985 people 6M 5.0 0 Upload your school material for a more relevant answer The solution set for the exponential inequality (2 1​)2 x+1≤4 is all values of x such that x≥−2 3​, represented in interval notation as [-\frac{3}{2}, \infty)]. Therefore, the correct option is C: ((-\frac{3}{2}, \infty). Explanation To solve the exponential inequality (2 1​)2 x+1≤4, we first rewrite the expressions in terms of a common base. We know that 4=2 2 and that (\frac{1}{2})^{2x+1}") can be rewritten as (2^{-(2x+1)}. Therefore, the inequality becomes: 2−(2 x+1)≤2 2 Since the base is the same (2), we can compare the exponents. This gives us the inequality: −(2 x+1)≤2 Next, we simplify it: −2 x−1≤2 −2 x≤2+1 −2 x≤3 Dividing both sides by −2, we must flip the inequality sign: x≥−2 3​ Thus, the solution set indicates that x can be any value greater than or equal to −2 3​. In interval notation, this can be represented as ([-\frac{3}{2}, \infty)]. Since none of the options provided lists this notation directly, the solution set is best summarized in terms of the option that captures this greater condition. Therefore, since the values of x range from −2 3​ to infinity, we identify the most suitable option as C: (−2 3​,∞). Examples & Evidence For example, if we substitute x=−1 into the original inequality: (2 1​)2(−1)+1=(2 1​)−1=2, which satisfies the inequality 2≤4. Similarly, substituting x=−2 gives us (2 1​)−3=8, which does not satisfy the inequality since 8≤4. This demonstrates that values of x greater than or equal to −2 3​ satisfy the condition. The method used follows the rules of inequalities and exponentials, where both sides can be equated based on their exponents when the bases are the same. Additionally, flipping the inequality sign when multiplying/dividing by a negative number aligns with foundational mathematics principles. Thanks 0 5.0 (1 vote) Advertisement anasandoval50371 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.2 8 What is the solution set for the exponential inequality (1/2)^2x+1<4 A.(0, 4) B.(0, ∞) C.{-3/2,4} xxxxD.{-3/2,infinity) It's D Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics If Q(0,1) is equidistant from P(5,−3) and R(x,6), find the value(s) of x. Also, find distances QR and PR. What is the simplified form of 12 x 900 x 2 y 4 z 6​? A. 30 y 2​x z 3​ B. 30 x 2 y 2 z 4 C. 360 x y 2​x z 3​ D. 360 x y 2​z 3​ Example: Round 24.53 to the nearest one. 1. Round 68.23 to the nearest tenth. 2. Round 64.28 to the nearest one. 3. Round 92.02 to the nearest ten. 4. Round 0.45 to the nearest hundredth. 5. Round 412.218 to the nearest hundredth. Solve the simultaneous equations: [ \begin{array}{r} 5x+y=21 \ x-3y=9 \end{array} ] The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers. 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最大公约数 (四种解法) 和最小公倍数 (两种解法)_199因数分解-CSDN博客 博客 下载 学习 社区 GitCode InsCodeAI 会议 搜索 AI 搜索 登录 登录后您可以： 复制代码和一键运行 与博主大V深度互动 解锁海量精选资源 获取前沿技术资讯 立即登录 会员·新人礼包 消息 历史 创作中心 创作 最大公约数 (四种解法) 和最小公倍数 (两种解法) lllsure已于 2024-05-27 16:07:56 修改 阅读量1k收藏 10 点赞数 24 CC 4.0 BY-SA版权 分类专栏：C语言文章标签：c语言 于 2024-01-26 22:00:07 首次发布 版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。 本文链接： C语言 专栏收录该内容 4 篇文章 订阅专栏 本文介绍了求解最大公约数和最小公倍数的不同方法，包括穷举法、辗转相除法、递归等，并强调了巧用辗转相除法和公式法可以大大提高计算效率。 求两个数的原理： 最大公约数： x%gcd==0 && y%gcd==0 最小公倍数： lcm%x==0 && lcm%y==0 一.最大公约数 1.穷举法 采用循环暴力解决，比较简单 cobol int gcd(int x,int y){ if (x == 0) return y; else if (y == 0) return x; else if (x == y) return x; int min = x > y ? y : x; //找出两个的最小值 while (min) { if (x % min == 0 && y % min == 0) return min; min--; }} AI写代码 2.相减法 对上面穷举法的优化，提高效率 cobol int gcd(int x, int y){ if (x == 0) return y; else if (y == 0) return x; else if (x == y) return x; int n; //存差值的变量 while (x != y) { n= x > y ? (x -= y) : (y -= x); } return n;} AI写代码 3.辗转相除法 效率更高的一种方法，使用一些数学知识 cobol int gcd(int x, int y){ if (x == 0) return y; else if (y == 0) return x; else if (x == y) return x; int mod=x%y; while (mod) { x = y; y = mod; mod = x % y; } return y;} AI写代码 4.递归做法 这是最好的方法，是辗转相除法的进阶，用的是辗转相除法的原理，但更加简洁高效 csharp int gcd(int x, int y){ if (y == 0) return x; else return gcd(y, x % y);} AI写代码 看，是不是简洁了很大，但其实还可以更简洁 csharp int gcd(int x, int y){ return y == 0 ? x : gcd(y, x % y);} AI写代码 以后写最大公约数的代码采用上面这个代码就行了，太妙了^-^ 二.最小公倍数 1.穷举法 根据原理暴力求解，不断循环 cobol int lcm(int x, int y){ if (x y == 0) return 0; int max = x > y ? x : y; //找到两个值的最大值 while (max) { if (max % x == 0 && max % y == 0) return max; max++; }} AI写代码 2.公式法 两个数的乘积等于最大公约数乘以最小公倍数。所以说公式法要用到上面求的最大公约数，函数也相对简单些 cobol int gcd(int x, int y){ return y == 0 ? x : gcd(y, x % y);}int lcm(int x, int y){ if (x y == 0) return 0; return (x y) / gcd(x, y);} AI写代码 ---------------------------------------------------------------------------------------------------------------------------------总结：两个数都可以采用循环暴力直接解决，但效率较低，当数字比较特殊时，处理的时间较长。 巧解效率高，代码相对简洁，应该着力去记住。 谢谢大家！^-^ 确定要放弃本次机会？ 福利倒计时 : : 立减 ¥ 普通VIP年卡可用 立即使用 lllsure 关注关注 24点赞 踩 10 收藏 觉得还不错? 一键收藏 1评论 分享复制链接 分享到 QQ 分享到新浪微博 扫一扫 举报 举报 专栏目录 【牛客刷题】求四个数的最小公约数：两种 高效 解法 详解（枚举法 和 最大公约数 法） 字节卷动 08-17 112 本文探讨了寻找多个数的最小公共因子的 两种 高效 解法 及优化策略。问题要求：给定四个正整数，求大于1的最小公约数，若无则返回-1。解法 一采用直接枚举法，从2到最小值逐个检查能否整除所有数；解法 二基于 最大公约数 ( GCD ) 性质，先计算所有数的GCD，再寻找其最小质因子。后者效率更高，时间复杂度为O ( n×log ( max_val ) )。文章详细介绍了算法流程、数学证明、Java实现及复杂度分析，并扩展了质因子分解优化、并行计算等技巧，以及 最小公倍数、公约数列表等相关应用场景。两种 解法 适用于不同数据规模，GCD法在数值较大时 1 条评论 您还未登录，请先 登录 后发表或查看评论 分解质因数 ( C++代码实现 ) _c++分解质因数 9-16 输入可以是多个正整数,例如90、256 和 199,输出为它们的质 因数分解 形式。 分解质因数 题目描述:将一个正整数分解质因数。输入数据包含多行,每行是一个正整数n ( 1<n<10000 )。输出每个整数的质因数。 输入样例:90,256,199; 输出样例:2×3×3×5 = 90;2×2×2×2×2×2×2×2 = 256;199 = 199; 注: 每个合数都可以 分解因数的算法实现 8-25 n行,每行输出对应一个输入。输出应是一个正整数,指明满足要求的分解的种数。 【输入样例】 2 2 20 【输出样例】 1 4 题目分析 此题目为递归题目。 递归的因数必须满足每一次比上一次大。我们只需要递归到最后一个数不可再有因数即可。 例子,20. 最大公约数 和 最小公倍数（多种）的 解法 有时间可以指导毕业设计，可以私聊我。一个热爱编程的smallCuteMonkey，可以私信我 03-07 259 import java.util.Scanner; import java.util.; public class 两个数的 最小公倍数 { public static void main ( String[] args ) { Scanner sc = new Scanner ( System.in ); while ( true ){ System.out.println ( "请输入两个整数：" ); int x = sc.n 求 最大公约数 和 最小公倍数 的常用 解法 Gabriel_tzc的博客 10-20 1338 对于m，n两数求 最大公约数 的算法： （1）辗转相除法 1.c=m%n。 2.若c=0，则n就是两数的 最大公约数。 3.若c！=0，则m=n，n=c然后重新执行1。 代码实现过程如下： #include int sove ( int x,int y ) { return x%y==0?y:sove ( y,x%y ); } int main ( ) { int m,n,c; scanf ("%d % 提高Shor因式分解算法的成功率资源 9-7 资源浏览查阅178次。Shor的因式分解算法 ( SFA ) 是量子计算领域中的一项重要算法,它由数学家彼得·绍尔 ( Peter Shor ) 于 199 4年提出,用于在多项式时间内分解一个合数整数N为其素因数。这个算法在理论上证明了量子计算机对于特定的数学问题具有巨大的潜力,并且直接威胁,更多 【NOIP 199 7】【Luogu2626】斐波那契数列 ( 枚举,质 因数分解 ) 9-2 本文探讨了Fibonacci数列的计算及其第n项对2^31取模后的质 因数分解 问题。通过递推算法高效求解Fibonacci数,并采用直接枚举法进行质 因数分解,适用于n<=48的情况。 摘要生成于C知道,由 DeepSeek-R1 满血版支持,前往体验 > problem 输入n,求f ( n )%231f ( n )%231,并将其分解质因数 ... 求2个整数的 最大公约数 和 最小公倍数，Python 解法 05-14 4540 #coding:utf-8 #gdc.py ''' 求2个整数的 最大公约数 和 最小公倍数，Python 解法 ''' def gdc ( p,q ): '''求整数 p，q的 最大公约数 ''' if q==0: return p if p<q: p,q =q,p ... 最大公约数 和 最小公倍数，你知道有几种求法吗？ 平凡的指针的博客 06-27 3万+ 文章目录 最大公约数 求法一：暴力求解求法二：更相减损法求法三：辗转相除法求法四：递归写法 最小公倍数 求法一：暴力求解求法二：公式法总结 最大公约数 什么是 最大公约数 呢？定义如下： 如果数a能被数b整除，a就叫做b的倍数，b就叫做a的约数。 几个整数中公有的约数，叫做这几个数的公约数；其中最大的一个，叫做这几个数的 最大公约数。 比如1：12、16的公约数有1、2、4，其中最大的一个是4，则4是12与16的 最大公约数。 比如2：12、18的公约数有1、2、3、6，其中最大的一个是6，则6是12与18的 最大公约数。 洛谷P1128 [HNOI2001]求正整数_p1128 [hnoi2001] 求正整数 9-20 本文介绍洛谷P1128[HNOI2001]求正整数题目的解决思路。针对给定的正整数n,通过编程找出具有n个不同因子的最小正整数m。采用贪心算法,通过分解n为多个素数相乘的形式,并分配到特定位置来优化答案。 洛谷P1128 [HNOI2001]求正整数 题目 题目描述 对于任意输入的正整数n,请编程求出具有n个不同因子的最小正整数m。 整数因子分解问题的递归算法_分解因数递归算法,1210因子分解资源... 9-17 将一个正整数分解质因数.docx 浏览:122 5星 · 资源好评率100% 将一个正整数分解质因数。例如:输入90,打印出90=2335。 对n进行分解质因数,应先找到一个最小的质数k,然后按下述步骤完成: ( 1 ) 如果这个质数恰等于n,则说明分解质因数的过程已经结束,打印出即可。 ( 2 ) 如果n>k,但n能被k整除,则应打印... 【c 语言】—求 最大公约数 和 最小公倍数 多种方法 热门推荐 m0_74044018的博客 12-23 3万+ 求 最大公约数 最小公倍数 常见的几种 解法！ 求两个数的 最大公约数 和 最小公倍数 k154235的博客 10-29 277 质 因数分解 法 公共静态 int gcd（int a， int b） { int 结果 = 1;i++） { 而 （a % i == 0 && b % i == 0） { 结果 = i;更相减损法 公共静态 int gcd（int a， int b） { 而 （a！= b） { 如果 （a > b） { a = a - b;b = a % b;最小公倍数 公共静态 int lcm（int a， int b） { 返回 a b / gcd（a， b）; 【免费】1.1.4算数基本定理1_素 因数分解 唯一性资源 9-23 τ ( n ) 等于n的素 因数分解 中每个因数指数加1后的乘积。在例子中,20的素 因数分解 为2^2 5^1,所以τ ( 20 ) = ( 2+1 )( 1+1 ) = 6,因为20有1, 2, 4, 5, 10, 和 20这六个正因子。 2. σ ( n ):n的正因子之 和。σ ( n ) 可以通过将τ ( n ) 中每个因数的指数增加1,然后用这个指数乘以相应的素因数来... python 中 两种 求 最大公约数 和 最小公倍数 的方法 python学习者的博客 09-27 9718 def hu ( n1,n2 ): if n1 < n2: n1,n2 = n2,n1 while n2: n1,n2 = n2,n1%n2 return n1 def min ( n1,n2 ): if n1 < n2: n1,n2 = n2,n1 x = n1 n2 while n2: n1,n2 = n2,n1%n2 return x//n1 a = 5 b = 10 pr PTA-C 语言-最大公约数 和 最小公倍数（多 解法，有扩展，附解析） MoBai0619的博客 01-26 3096 本题要求两个给定正整数的 最大公约数 和 最小公倍数。 Python求 最大公约数 和 最小公倍数 qq_46857088的博客 04-07 2万+ 大家好，我是Python领域的博主，如果觉得我的写的不错的话， 请点赞关注支持我一下。如果你是初学者，可以 和 小编一起学习，在这里我每天都会发Python的基础知识，以及相关的代码 解题思路： 最大公因数，也称 最大公约数、最大公因子，指两个或多个整数共有约数中最大的一个；最小公倍数 是指两个或多个整数公有的倍数叫做它们的公倍数，其中除0以外最小的一个公倍数就叫做这几个整数的 最小公倍数。最小公倍数=两整数的乘积÷最大公约数 ， 所以怎么求 最大公约数 是关键。 求 最大公约数： 辗转相除法是求最大公. C 语言 求 最大公约数 及 最小公倍数.pdf 10-18 在C 语言 中，求 最大公约数（Greatest Common Divisor, GCD）和 最小公倍数（Least Common Multiple, LCM）是常见的算法问题，主要应用于数论 和 计算机科学领域。本文将详细介绍几种不同的方法来解决这个问题。 1. ... Codeforces Round 1052 ( Div. 2 ) C. Wrong Binary Searchong Binary Search xyk050216的博客 09-28 331 给一个二进制字符串，字符串下标从1开始，如果s[i]为1则i为稳定点，反正为不稳定点，需要我们依据伪代码构造一个长度为n的排序，使得排序中的每个数都满足对应s中的稳定点与不稳定点。如果非稳定点个数>=2，如非稳定点个数=2，举例：s=1001,1为稳定点，2为不稳定点，3为不稳定点 4为稳定点。如果非稳定点个数>=2，如非稳定点个数=2，举例：s=0011,1为不稳定点，2为不稳定点，3为稳定点 4为稳定点。如果非稳定点个数=1，举例：s=101的情况下，1为稳定点，2为不稳定点，3为稳定点。 一个简易版算法集合库主要用于测试目的性能表现一般但覆盖国密算法SM1SM2SM3SM4以及通用算法AESDESMD5SHA1SHA224SHA256SHA384SHA512RSA1.zip 09-28 一个简易版算法集合库主要用于测试目的性能表现一般但覆盖国密算法SM1SM2SM3SM4以及通用算法AESDESMD5SHA1SHA224SHA256SHA384SHA512RSA1.zip 项目极简说明这是一个目前处于开发初期阶段尚未完全成熟的软件项目其可执行文件位于dist文件夹中用户需自行探索许多功能若遇到进程未响应情况请耐心等待因为程序未采用多进程技术.zip 09-28 项目极简说明这是一个目前处于开发初期阶段尚未完全成熟的软件项目其可执行文件位于dist文件夹中用户需自行探索许多功能若遇到进程未响应情况请耐心等待因为程序未采用多进程技术.zip 健康追踪系统-JAVA-基于springBoot健康追踪系统软件设计与实现（毕业论文+PPT） 09-28 健康追踪设计 采用java 语言， springboot技术，mysql数据库开发 开发平台：idea，mysql5.7以上版本 是一款后台管理系统，管理员 和 普通用户 1.用户管理:该应用程序需要提供用户注册、登录、密码找回等功能，并对用户信息进行加密处理，保证用户信息的安全性。 2.健康情况:该应用程序需要实时监测用户的身体状况，包括心率、血压、血糖等等指标，并对异常数据进行提醒 和 预警。 3.健康管理:该应用程序需要根据用户的身体状况 和 健康，提供个性化健康管理方案，包括饮食、运动、用药等方面的建议。 4.运动跟踪:集成运动跟踪功能，记录用户的运动轨迹、步数 5.医生共享:用户可以选择与医生共享健康数据，实现更有效的医患协作，医生可以根据实时数据调整治疗方案。 6.健康资料:该应用程序需要提供健康小知识，让用户可以快速了解健康小知识。 7.健康反馈:用户可以对该应用程序、医生等进行评论留言。 8.日常签到:该应用程序需要提供每日签到功能，鼓励用户保持规律性的健康生活活。 基于LlamaIndex框架构建的剧本杀AI助手专用记忆管理系统_采用检索增强生成RAG架构实现对话历史与线索的持久化存储及智能检索_通过双层记忆结构分别保存原始对话片段 和 AI生成.zip 09-28 基于LlamaIndex框架构建的剧本杀AI助手专用记忆管理系统_采用检索增强生成RAG架构实现对话历史与线索的持久化存储及智能检索_通过双层记忆结构分别保存原始对话片段 和 AI生成.zip STM32灯控系统仿真源程序-C 语言 源码.7z 最新发布 09-28 源码纯C编写，寄存器+库双版本： 时钟：RCC->APB2ENR 使能GPIOA/GPIOC & TIM2时钟 GPIO：PA0-PA7推挽输出驱动8路灯，PC13下拉输入接按键，消抖用定时器10ms中断计数 PWM：TIM2_CH1-CH4，重映射到PA8-PA11，占空比0-100%可调，中断内双缓冲影子寄存器防闪烁 通信：USART1 115200，接收中断环形缓冲区，协议帧头0xAA+长度+校验，支持串口调光 低功耗：SLEEP模式下RTC每秒唤醒，电流降至2mA 代码遵循MISRA-C 2012，Keil5 AC5编译零警告，Proteus8.13仿真验证通过，可直接移植到F103C8T6最小系统板。 基于Matlab平台开发的火灾检测系统项目_该系统分为烟雾检测 和 火焰检测两大核心模块_烟雾检测采用边缘检测算法对图像中的烟雾区域进行识别_火焰检测则结合颜色空间分析与形态学处理方法.zip 09-28 基于Matlab平台开发的火灾检测系统项目_该系统分为烟雾检测 和 火焰检测两大核心模块_烟雾检测采用边缘检测算法对图像中的烟雾区域进行识别_火焰检测则结合颜色空间分析与形态学处理方法 永久免费开源的导师评价数据平台_包含导师评价数据数据爬虫工具无需编程基础的展示网页以及新信息补充功能_旨在为研究生 和 学术人员提供全面可靠的导师信息检索 和 分享服务_通过收集 和 整.zip 09-28 永久免费开源的导师评价数据平台_包含导师评价数据数据爬虫工具无需编程基础的展示网页以及新信息补充功能_旨在为研究生 和 学术人员提供全面可靠的导师信息检索 和 分享服务_通过收集 和 整.zip 关于我们 招贤纳士 商务合作 寻求报道 400-660-0108 kefu@csdn.net 在线客服 工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号 京网文〔2020〕1039-165号 经营性网站备案信息 北京互联网违法和不良信息举报中心 家长监护 网络110报警服务 中国互联网举报中心 Chrome商店下载 账号管理规范 版权与免责声明 版权申诉 出版物许可证 营业执照 ©1999-2025北京创新乐知网络技术有限公司 lllsure 博客等级 码龄2年 86 原创2923 点赞 2042 收藏 1641 粉丝 关注 私信 亚马逊云服务全新“免费计划”确保零花费，现在注册还可获取随身小挎包或健身提包👜广告 热门文章 详解C语言函数递归经典问题：汉诺塔（图+代码） 21493 详解：HTTP/HTTPS协议 7867 详解：TCP/IP五层（四层）协议模型 7361 Java 栈(Stack)与队列(Queue) 3237 Java 实现 红黑树（通俗易懂） 2887 分类专栏 Python1篇 MySQL10篇 Redis4篇 docker4篇 RabbitMQ3篇 Java9篇 数据结构13篇 Spring6篇 SpringCloud5篇 Linux4篇 C语言4篇 前端语言2篇 展开全部收起 上一篇： 随机数生成+猜数字游戏 下一篇： 详解C语言函数递归经典问题：汉诺塔（图+代码） 最新评论 【MySQL】SQL优化 CSDN-Ada助手:恭喜你这篇博客进入【CSDN每天值得看】榜单，全部的排名请看 Git使用与管理 CSDN-Ada助手:恭喜你这篇博客进入【CSDN每天值得看】榜单，全部的排名请看 详解C语言函数递归经典问题：汉诺塔（图+代码） 凤年徐:看了这么多人的，就这个讲的最明白 Spring 事务 CSDN-Ada助手:恭喜你这篇博客进入【CSDN每天值得看】榜单，全部的排名请看 Java 实现 红黑树（通俗易懂） 生鱼片片片:通俗易懂，感谢博主！ 大家在看 从PS导出到前端加载，聊聊被设计师忽略的“图片性能” Python序列操作详解 NLP基础 848 [超表面领域大牛-17] 南京大学-王振林-国家杰青|长江|南京大学副校长 编程的根本是数学吗？ 172 最新文章 Redis 主从复制 【快速入门】Nginx 【Docker】 网络 2025年 39篇 2024年 47篇 亚马逊云服务全新“免费计划”确保零花费，现在注册还可获取随身小挎包或健身提包👜广告 上一篇： 随机数生成+猜数字游戏 下一篇： 详解C语言函数递归经典问题：汉诺塔（图+代码） 分类专栏 Python1篇 MySQL10篇 Redis4篇 docker4篇 RabbitMQ3篇 Java9篇 数据结构13篇 Spring6篇 SpringCloud5篇 Linux4篇 C语言4篇 前端语言2篇 展开全部收起 登录后您可以享受以下权益： 免费复制代码 和博主大V互动 下载海量资源 发动态/写文章/加入社区 ×立即登录 评论 1 被折叠的 条评论 为什么被折叠?到【灌水乐园】发言 查看更多评论 添加红包 祝福语 请填写红包祝福语或标题 红包数量 个 红包个数最小为10个 红包总金额 元 红包金额最低5元 余额支付 当前余额 3.43 元 前往充值 > 需支付：10.00 元 取消 确定 成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝 规则 hope_wisdom 发出的红包 实付 元 使用余额支付 点击重新获取 扫码支付 钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。 余额充值 确定 取消 举报 选择你想要举报的内容（必选） 内容涉黄 政治相关 内容抄袭 涉嫌广告 内容侵权 侮辱谩骂 样式问题 其他 原文链接（必填） 请选择具体原因（必选） 包含不实信息 涉及个人隐私 请选择具体原因（必选） 侮辱谩骂 诽谤 请选择具体原因（必选） 搬家样式 博文样式 补充说明（选填） 取消 确定 AI助手 AI 搜索 智能体 AI 编程 AI 作业助手 下载APP 程序员都在用的中文IT技术交流社区 公众号 专业的中文 IT 技术社区，与千万技术人共成长 视频号 关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
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https://en.wikipedia.org/wiki/Zeller%27s_congruence
Jump to content Zeller's congruence Deutsch Español Italiano 日本語 Português Slovenščina Suomi 中文 Edit links From Wikipedia, the free encyclopedia Algorithm to calculate the day of the week Zeller's congruence is an algorithm devised by Christian Zeller in the 19th century to calculate the day of the week for any Julian or Gregorian calendar date. It can be considered to be based on the conversion between Julian day and the calendar date. Formula [edit] For the Gregorian calendar, Zeller's congruence is for the Julian calendar it is where h is the day of the week (0 = Saturday, 1 = Sunday, 2 = Monday, ..., 6 = Friday) q is the day of the month m is the month (3 = March, 4 = April, 5 = May, ..., 14 = February) K the year of the century . (The adjYear (adjusted year) is explained in the Note below.) J is the zero-based century (actually ) For example, the zero-based centuries for 1995 and 2000 are 19 and 20 respectively (not to be confused with the common ordinal century enumeration which indicates 20th for both cases). is the floor function or integer part mod is the modulo operation or remainder after division Note: In this algorithm January and February are counted as months 13 and 14 of the previous year. E.g. if it is 2 February 2010 (02/02/2010 in DD/MM/YYYY), the algorithm counts the date as the second day of the fourteenth month of 2009 (02/14/2009 in DD/MM/YYYY format) So the adjusted year above is: : = the actual year, for months from March to December. : = the previous year, for January and February. For an ISO week date Day-of-Week d (1 = Monday to 7 = Sunday), use Analysis [edit] These formulas are based on the observation that the day of the week progresses in a predictable manner based upon each subpart of that date. Each term within the formula is used to calculate the offset needed to obtain the correct day of the week. For the Gregorian calendar, the various parts of this formula can therefore be understood as follows: represents the progression of the day of the week based on the day of the month, since each successive day results in an additional offset of 1 in the day of the week. represents the progression of the day of the week based on the year. Assuming that each year is 365 days long, the same date on each succeeding year will be offset by a value of . Since there are 366 days in each leap year, this needs to be accounted for by adding another day to the day of the week offset value. This is accomplished by adding to the offset. This term is calculated as an integer result. Any remainder is discarded. Using similar logic, the progression of the day of the week for each century may be calculated by observing that there are 36,524 days in a normal century and 36,525 days in each century divisible by 400. Since and , the term accounts for this. The term adjusts for the variation in the days of the month. Starting from January, the days in a month are {31, 28/29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}. February's 28 or 29 days is a problem, so the formula rolls January and February around to the end so February's short count will not cause a problem. The formula is interested in days of the week, so the numbers in the sequence can be taken modulo 7. Then the number of days in a month modulo 7 (still starting with January) would be {3, 0/1, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3}. Starting in March, the sequence basically alternates 3, 2, 3, 2, 3, but every five months there are two 31-day months in a row (July–August and December–January). The fraction 13/5 = 2.6 and the floor function have that effect; the denominator of 5 sets a period of 5 months. The overall function, , normalizes the result to reside in the range of 0 to 6, which yields the index of the correct day of the week for the date being analyzed. The reason that the formula differs between calendars is that the Julian calendar does not have a separate rule for leap centuries and is offset from the Gregorian calendar by a fixed number of days each century. Since the Gregorian calendar was adopted at different times in different regions of the world, the location of an event is significant in determining the correct day of the week for a date that occurred during this transition period. This is only required through 1929, as this was the last year that the Julian calendar was still in use by any country on earth, and thus is not required for 1930 or later. The formulae can be used proleptically, but "Year 0" is in fact year 1 BC (see astronomical year numbering). The Julian calendar is in fact proleptic right up to 1 March AD 4 owing to mismanagement in Rome (but not Egypt) in the period since the calendar was put into effect on 1 January 45 BC (which was not a leap year). In addition, the modulo operator might truncate integers to the wrong direction (ceiling instead of floor). To accommodate this, one can add a sufficient multiple of 400 Gregorian or 700 Julian years. Examples [edit] For 1 January 2000, the date would be treated as the 13th month of 1999, so the values would be: So the formula evaluates as . (The 36 comes from , truncated to an integer.) However, for 1 March 2000, the date is treated as the 3rd month of 2000, so the values become so the formula evaluates as . Implementations in software [edit] Basic modification [edit] Further information: Modulo operation § Variants of the definition The formulas rely on the mathematician's definition of modulo division, which means that −2 mod 7 is equal to positive 5. Unfortunately, in the truncating way most computer languages implement the remainder function, −2 mod 7 returns a result of −2. So, to implement Zeller's congruence on a computer, the formulas should be altered slightly to ensure a positive numerator. The simplest way to do this is to replace − 2J with + 5J and − J with + 6J. For the Gregorian calendar, Zeller's congruence becomes For the Julian calendar, Zeller's congruence becomes One can readily see that, in a given year, the last day of February and March 1 are a good test dates. As an aside note, if we have a three-digit number abc, where a, b, and c are the digits, each nonpositive if abc is nonpositive; we have (abc) mod 7 = 9a + 3b + c. Repeat the formula down to a single digit. If the result is 7, 8, or 9, then subtract 7. If, instead, the result is negative, then add 7. If the result is still negative, then add 7 one more time. Utilizing this approach, we can avoid the worries of language specific differences in mod 7 evaluations. This also may enhance a mental math technique. Common simplification [edit] Zeller used decimal arithmetic, and found it convenient to use J and K values as two-digit numbers representing the year and century. But when using a computer, it is simpler to handle the year as a single 4-digit number. For the Gregorian calendar, Zeller's congruence becomes where is , defined in the section above. In this case there is no possibility of underflow due to the single negative term because . For the Julian calendar, Zeller's congruence becomes The algorithm above is mentioned for the Gregorian case in RFC 3339, Appendix B, albeit in an abridged form that returns 0 for Sunday. Other variations [edit] At least three other algorithms share the overall structure of Zeller's congruence in its "common simplification" type, also using an m ∈ [3, 14] ∩ Z and the "modified year" construct. Michael Keith published a piece of very short C code in 1990 for Gregorian dates. The month-length component () is replaced by . J R Stockton provides a Sunday-is-0 version with , calling it a variation of Zeller. Claus Tøndering describes as a Sunday-is-0 replacement. Both expressions can be shown to progress in a way that is off by one compared to the original month-length component over the required range of m, resulting in a starting value of 0 for Sunday. See also [edit] Determination of the day of the week Doomsday rule ISO week date Julian day References [edit] ^ The every five months rule only applies to the twelve months of a year commencing on 1 March and ending on the last day of the following February. ^ Jump up to: a b Stockton, J R. "Material Related to Zeller's Congruence". "Merlyn", archived at NCTU Taiwan. Archived from the original on 2023-09-18. Retrieved 2021-03-10. ^ Tøndering, Claus. "Week-related questions". www.tondering.dk. Bibliography [edit] Each of these four similar imaged papers deals firstly with the day of the week and secondly with the date of Easter Sunday, for the Julian and Gregorian calendars. The pages link to translations into English. Zeller, Christian (1882). "Die Grundaufgaben der Kalenderrechnung auf neue und vereinfachte Weise gelöst". Württembergische Vierteljahrshefte für Landesgeschichte (in German). V: 313–314. Archived from the original on January 11, 2015. Zeller, Christian (1883). "Problema duplex Calendarii fundamentale". Bulletin de la Société Mathématique de France (in Latin). 11: 59–61. Archived from the original on January 11, 2015. Zeller, Christian (1885). "Kalender-Formeln". Mathematisch-naturwissenschaftliche Mitteilungen des mathematisch-naturwissenschaftlichen Vereins in Württemberg (in German). 1 (1): 54–58. Archived from the original on January 11, 2015. Zeller, Christian (1886). "Kalender-Formeln". Acta Mathematica (in German). 9: 131–136. doi:10.1007/BF02406733. External links [edit] The Calendrical Works of Rektor Chr. Zeller: The Day-of-Week and Easter Formulae by J R Stockton, near London, UK. The site includes images and translations of the above four papers, and of Zeller's reference card "Das Ganze der Kalender-Rechnung". This article incorporates public domain material from Paul E. Black. "Zeller's congruence". Dictionary of Algorithms and Data Structures. NIST. Retrieved from " Categories: Gregorian calendar Julian calendar Calendar algorithms Modular arithmetic Eponymous algorithms of mathematics Hidden categories: Articles with short description Short description is different from Wikidata CS1 German-language sources (de) CS1 Latin-language sources (la)
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https://www.pearson.com/channels/general-chemistry/learn/jules/ch-5-gases/gas-stoichiometry
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Gases 3h 49m Pressure Units 6m + The Ideal Gas Law 18m + The Ideal Gas Law Derivations 13m + The Ideal Gas Law Applications 6m + Chemistry Gas Laws 13m + Chemistry Gas Laws: Combined Gas Law 12m + Mole Fraction of Gases 6m + Partial Pressure 19m + The Ideal Gas Law: Molar Mass 13m + The Ideal Gas Law: Density 14m + Gas Stoichiometry 18m + Standard Temperature and Pressure 14m + Effusion 13m + Root Mean Square Speed 9m + Kinetic Energy of Gases 10m + Maxwell-Boltzmann Distribution 8m + Velocity Distributions 4m + Kinetic Molecular Theory 14m + Van der Waals Equation 9m 8. Thermochemistry 2h 37m Nature of Energy 6m + Kinetic & Potential Energy 7m + First Law of Thermodynamics 7m + Internal Energy 8m + Endothermic & Exothermic Reactions 7m + Heat Capacity 19m + Constant-Pressure Calorimetry 24m + Constant-Volume Calorimetry 10m + Thermal Equilibrium 8m + Thermochemical Equations 12m + Formation Equations 9m + Enthalpy of Formation 12m + Hess's Law 23m 9. Quantum Mechanics 2h 58m Wavelength and Frequency 6m + Speed of Light 8m + The Energy of Light 13m + Electromagnetic Spectrum 10m + Photoelectric Effect 17m + De Broglie Wavelength 9m + Heisenberg Uncertainty Principle 17m + Bohr Model 14m + Emission Spectrum 5m + Bohr Equation 13m + Introduction to Quantum Mechanics 5m + Quantum Numbers: Principal Quantum Number 5m + Quantum Numbers: Angular Momentum Quantum Number 10m + Quantum Numbers: Magnetic Quantum Number 11m + Quantum Numbers: Spin Quantum Number 9m + Quantum Numbers: Number of Electrons 11m + Quantum Numbers: Nodes 6m 10. Periodic Properties of the Elements 3h 9m The Electron Configuration 22m + The Electron Configuration: Condensed 4m + The Electron Configurations: Exceptions 13m + The Electron Configuration: Ions 12m + Paramagnetism and Diamagnetism 8m + The Electron Configuration: Quantum Numbers 16m + Valence Electrons of Elements 12m + Periodic Trend: Metallic Character 3m + Periodic Trend: Atomic Radius 8m + Periodic Trend: Ionic Radius 13m + Periodic Trend: Ionization Energy 12m + Periodic Trend: Successive Ionization Energies 11m + Periodic Trend: Electron Affinity 10m + Periodic Trend: Electronegativity 5m + Periodic Trend: Effective Nuclear Charge 21m + Periodic Trend: Cumulative 12m 11. Bonding & Molecular Structure 3h 29m Lewis Dot Symbols 10m + Chemical Bonds 13m + Dipole Moment 11m + Octet Rule 10m + Formal Charge 9m + Lewis Dot Structures: Neutral Compounds 20m + Lewis Dot Structures: Sigma & Pi Bonds 14m + Lewis Dot Structures: Ions 15m + Lewis Dot Structures: Exceptions 14m + Lewis Dot Structures: Acids 15m + Resonance Structures 21m + Average Bond Order 4m + Bond Energy 15m + Coulomb's Law 6m + Lattice Energy 12m + Born Haber Cycle 14m 12. Molecular Shapes & Valence Bond Theory 1h 57m Valence Shell Electron Pair Repulsion Theory 5m + Equatorial and Axial Positions 10m + Electron Geometry 11m + Molecular Geometry 18m + Bond Angles 14m + Hybridization 12m + Molecular Orbital Theory 12m + MO Theory: Homonuclear Diatomic Molecules 10m + MO Theory: Heteronuclear Diatomic Molecules 7m + MO Theory: Bond Order 14m 13. Liquids, Solids & Intermolecular Forces 2h 23m Molecular Polarity 10m + Intermolecular Forces 20m + Intermolecular Forces and Physical Properties 11m + Clausius-Clapeyron Equation 18m + Phase Diagrams 13m + Heating and Cooling Curves 27m + Atomic, Ionic, and Molecular Solids 11m + Crystalline Solids 4m + Simple Cubic Unit Cell 7m + Body Centered Cubic Unit Cell 12m + Face Centered Cubic Unit Cell 6m 14. Solutions 3h 1m Solutions: Solubility and Intermolecular Forces 17m + Molality 15m + Parts per Million (ppm) 13m + Mole Fraction of Solutions 8m + Solutions: Mass Percent 12m + Types of Aqueous Solutions 8m + Intro to Henry's Law 4m + Henry's Law Calculations 12m + The Colligative Properties 14m + Boiling Point Elevation 16m + Freezing Point Depression 10m + Osmosis 19m + Osmotic Pressure 10m + Vapor Pressure Lowering (Raoult's Law) 16m 15. Chemical Kinetics 2h 53m Intro to Chemical Kinetics 4m + Energy Diagrams 9m + Catalyst 9m + Factors Influencing Rates 10m + Average Rate of Reaction 7m + Stoichiometric Rate Calculations 5m + Instantaneous Rate 5m + Collision Theory 7m + Arrhenius Equation 25m + Rate Law 24m + Reaction Mechanism 17m + Integrated Rate Law 23m + Half-Life 23m 16. Chemical Equilibrium 2h 29m Intro to Chemical Equilibrium 7m + Equilibrium Constant K 13m + Equilibrium Constant Calculations 9m + Kp and Kc 22m + Using Hess's Law To Determine K 9m + Calculating K For Overall Reaction 15m + Le Chatelier's Principle 20m + ICE Charts 34m + Reaction Quotient 15m 17. Acid and Base Equilibrium 5h 1m Acids Introduction 9m + Bases Introduction 7m + Binary Acids 15m + Oxyacids 10m + Bases 14m + Amphoteric Species 5m + Arrhenius Acids and Bases 5m + Bronsted-Lowry Acids and Bases 21m + Lewis Acids and Bases 12m + The pH Scale 16m + Auto-Ionization 9m + Ka and Kb 16m + pH of Strong Acids and Bases 9m + Ionic Salts 17m + pH of Weak Acids 31m + pH of Weak Bases 32m + Diprotic Acids and Bases 8m + Diprotic Acids and Bases Calculations 30m + Triprotic Acids and Bases 9m + Triprotic Acids and Bases Calculations 17m 18. Aqueous Equilibrium 4h 47m Intro to Buffers 20m + Henderson-Hasselbalch Equation 19m + Intro to Acid-Base Titration Curves 13m + Strong Titrate-Strong Titrant Curves 9m + Weak Titrate-Strong Titrant Curves 15m + Acid-Base Indicators 8m + Titrations: Weak Acid-Strong Base 38m + Titrations: Weak Base-Strong Acid 41m + Titrations: Strong Acid-Strong Base 11m + Titrations: Diprotic & Polyprotic Buffers 32m + Solubility Product Constant: Ksp 17m + Ksp: Common Ion Effect 18m + Precipitation: Ksp vs Q 12m + Selective Precipitation 9m + Complex Ions: Formation Constant 18m 19. Chemical Thermodynamics 1h 50m Spontaneous vs Nonspontaneous Reactions 7m + Entropy 23m + Entropy Calculations 13m + Entropy Calculations: Phase Changes 6m + Third Law of Thermodynamics 7m + Gibbs Free Energy 13m + Gibbs Free Energy Calculations 22m + Gibbs Free Energy And Equilibrium 14m 20. Electrochemistry 2h 42m Standard Reduction Potentials 9m + Intro to Electrochemical Cells 6m + Galvanic Cell 25m + Electrolytic Cell 10m + Cell Potential: Standard 13m + Cell Potential: The Nernst Equation 20m + Cell Potential and Gibbs Free Energy 16m + Cell Potential and Equilibrium 8m + Cell Potential: G and K 16m + Cell Notation 20m + Electroplating 16m 21. Nuclear Chemistry 2h 36m Intro to Radioactivity 10m + Alpha Decay 9m + Beta Decay 7m + Gamma Emission 7m + Electron Capture & Positron Emission 9m + Neutron to Proton Ratio 7m + Band of Stability: Alpha Decay & Nuclear Fission 10m + Band of Stability: Beta Decay 3m + Band of Stability: Electron Capture & Positron Emission 4m + Band of Stability: Overview 14m + Measuring Radioactivity 7m + Rate of Radioactive Decay 12m + Radioactive Half-Life 16m + Mass Defect 18m + Nuclear Binding Energy 14m 22. Organic Chemistry 5h 7m Introduction to Organic Chemistry 8m + Structural Formula 8m + Condensed Formula 10m + Skeletal Formula 6m + Spatial Orientation of Bonds 3m + Intro to Hydrocarbons 16m + Isomers 11m + Chirality 15m + Functional Groups in Chemistry 11m + Naming Alkanes 4m + The Alkyl Groups 9m + Naming Alkanes with Substituents 13m + Naming Cyclic Alkanes 6m + Naming Other Substituents 8m + Naming Alcohols 11m + Naming Alkenes 11m + Naming Alkynes 9m + Naming Ketones 5m + Naming Aldehydes 5m + Naming Carboxylic Acids 4m + Naming Esters 8m + Naming Ethers 5m + Naming Amines 5m + Naming Benzene 7m + Alkane Reactions 7m + Intro to Addition Reactions 4m + Halogenation Reactions 4m + Hydrogenation Reactions 3m + Hydrohalogenation Reactions 7m + Alcohol Reactions: Substitution Reactions 4m + Alcohol Reactions: Dehydration Reactions 9m + Intro to Redox Reactions 8m + Alcohol Reactions: Oxidation Reactions 7m + Aldehydes and Ketones Reactions 6m + Ester Reactions: Esterification 4m + Ester Reactions: Saponification 3m + Carboxylic Acid Reactions 4m + Amine Reactions 3m + Amide Formation 4m + Benzene Reactions 10m 23. Chemistry of the Nonmetals 2h 39m Main Group Elements: Bonding Types 4m + Main Group Elements: Boiling & Melting Points 7m + Main Group Elements: Density 11m + Main Group Elements: Periodic Trends 7m + The Electron Configuration Review 16m + Periodic Table Charges Review 20m + Hydrogen Isotopes 4m + Hydrogen Compounds 11m + Production of Hydrogen 8m + Group 1A and 2A Reactions 7m + Boron Family Reactions 7m + Boron Family: Borane 7m + Borane Reactions 7m + Nitrogen Family Reactions 12m + Oxides, Peroxides, and Superoxides 12m + Oxide Reactions 4m + Peroxide and Superoxide Reactions 6m + Noble Gas Compounds 3m 24. Transition Metals and Coordination Compounds 3h 16m Atomic Radius & Density of Transition Metals 11m + Electron Configurations of Transition Metals 7m + Electron Configurations of Transition Metals: Exceptions 11m + Paramagnetism and Diamagnetism 10m + Ligands 10m + Complex Ions 5m + Coordination Complexes 7m + Classification of Ligands 11m + Coordination Numbers & Geometry 9m + Naming Coordination Compounds 22m + Writing Formulas of Coordination Compounds 8m + Isomerism in Coordination Complexes 14m + Orientations of D Orbitals 4m + Intro to Crystal Field Theory 10m + Crystal Field Theory: Octahedral Complexes 5m + Crystal Field Theory: Tetrahedral Complexes 4m + Crystal Field Theory: Square Planar Complexes 4m + Crystal Field Theory Summary 8m + Magnetic Properties of Complex Ions 9m + Strong-Field vs Weak-Field Ligands 6m + Magnetic Properties of Complex Ions: Octahedral Complexes 11m Gases Gas Stoichiometry Gases Gas Stoichiometry: Videos & Practice Problems Video Lessons Practice Worksheet Topic summary Stoichiometry is the study of the quantitative relationships in a balanced chemical equation, and gas stoichiometry specifically deals with reactions that produce gases. To perform gas stoichiometric calculations, one must understand the ideal gas law: which allows us to determine the moles of a gas given its pressure (P), volume (V), and temperature (T). Using a stoichiometric chart, we can convert the known quantity of a reactant or product, such as the moles of a gas, to find the unknown quantity of another substance in the reaction. This process involves a "jump" from known to unknown quantities, requiring a mole-to-mole comparison based on the coefficients of the balanced equation. Mastery of stoichiometry is essential for predicting the amounts of reactants and products in chemical reactions involving gases. Gas Stoichiometry involves chemical reactions that contain gases. Gas Stoichiometry 1 concept Gas Stoichiometry Concepts Video duration: 47s Play a video: Gas Stoichiometry Concepts Video Summary Gas stoichiometry focuses on the numerical relationships between reactants and products in chemical reactions that produce gases. It builds upon the principles of stoichiometry, which involves using balanced chemical equations to relate the quantities of different substances involved in a reaction. To perform gas stoichiometry calculations, a stoichiometric chart is utilized. This chart allows you to take a known quantity of one compound, often a gas, and use it to calculate the unknown quantity of another compound in the balanced equation. The balanced chemical equation provides the necessary ratios of the reactants and products, which are essential for these calculations. Understanding gas stoichiometry is crucial for accurately predicting the outcomes of reactions involving gases, as it helps in determining how much of each substance will be produced or consumed. This knowledge is particularly important in fields such as chemistry, environmental science, and engineering, where gas reactions are common. 2 concept Gas Stoichiometry Chart Video duration: 2m Play a video: Gas Stoichiometry Chart Video Summary In stoichiometry, understanding the relationships between reactants and products in a chemical reaction is crucial. For instance, consider the reaction where 4 moles of solid silver react with 1 mole of oxygen gas to produce 2 moles of silver oxide solid. This relationship can be represented using a stoichiometric chart, which helps visualize the conversion between different substances involved in the reaction. To determine the grams of silver oxide produced, we can utilize the ideal gas law, expressed as: PV = nRT Here, P represents pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. By rearranging this equation, we can isolate the number of moles of the gas: n = \frac{PV}{RT} With the provided values for pressure, volume, and temperature, we can calculate the moles of oxygen gas. This step is essential as it allows us to transition from the known quantities to the unknown quantities in the reaction. Once we have the moles of oxygen, we can make a stoichiometric leap to find the moles of silver oxide produced. This leap involves using the coefficients from the balanced equation, which provides the necessary mole-to-mole ratios. For example, if we know the moles of oxygen, we can determine the moles of silver oxide using the ratio: 2 \text{ moles Ag}_2\text{O} : 1 \text{ mole O}_2 After calculating the moles of silver oxide, converting this quantity into grams is straightforward using the molar mass of silver oxide. This process illustrates the importance of stoichiometry in predicting the outcomes of chemical reactions and converting between different units of measurement. For those unfamiliar with stoichiometric charts, revisiting foundational concepts in stoichiometry can enhance understanding and application in gas stoichiometry problems. 3 example Gas Stoichiometry Calculation Example Video duration: 3m Play a video: Gas Stoichiometry Calculation Example Video Summary To determine the mass of silver oxide produced from the reaction of oxygen gas with excess solid silver, we start by using the ideal gas law, which relates pressure, volume, and temperature to the number of moles of a gas. The formula for calculating moles is given by: In this scenario, we are provided with the volume of oxygen gas (384 milliliters), the pressure (736 millimeters of mercury), and the temperature (25 degrees Celsius). First, we need to convert the pressure from millimeters of mercury to atmospheres, knowing that 1 atmosphere equals 760 millimeters of mercury: Next, we convert the volume from milliliters to liters: We also convert the temperature to Kelvin by adding 273.15: Now, substituting these values into the ideal gas law equation, we can calculate the moles of oxygen gas: With the moles of oxygen calculated, we can now perform a stoichiometric conversion to find the moles of silver oxide produced. The balanced chemical equation for the reaction is: This indicates that 1 mole of oxygen gas produces 2 moles of silver oxide. Therefore, we set up the mole ratio: Next, we convert moles of silver oxide to grams. The molar mass of silver oxide () is calculated as follows: Now, we can find the mass of silver oxide produced: In this calculation, we maintain two significant figures, as dictated by the initial temperature measurement. This process illustrates the application of stoichiometry in gas reactions, emphasizing the importance of understanding the relationships between moles, mass, and the ideal gas law. The metabolic breakdown of glucose (C6H12O6) (MW:180.156 g/mol) is given by the following equation: C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) Calculate the volume (in mL) of CO2 produced at 34°C and 1728.9 torr when 231.88 g glucose is used up in the reaction. 8.6×105 mL 6.8×104 mL 8.6×104 mL 4.3×104 mL Problem The oxidation of phosphorus can be represented by the following equation: P4 (s) + 5 O2 (g) → 2 P2O5 (g) If 1.85 L of diphosphorus pentoxide form at a temperature of 50.0 ºC and 1.12 atm, what is the mass (in g) of phosphorus that reacted? A 4.84 g B 7.02 g C 10.9 g D 12.4 g 17.7 g Problem Determine the mass (in grams) of water formed when 15.3 L NH3 (at 298 K and 1.50 atm) is reacted with 21.7 L of O2 (at 323 K and 1.1 atm). 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) A 12 g B 19 g C 25 g D 38 g E 51 g Do you want more practice? More sets Gas Stoichiometry Gases 7 problems 7. Gases - Part 1 of 4 5 topics 14 problems Jules 7. Gases - Part 2 of 4 5 topics 13 problems Chapter Jules 7. Gases - Part 3 of 4 5 topics 13 problems Chapter Jules 7. Gases - Part 4 of 4 5 topics 13 problems Chapter Jules Go over this topic definitions with flashcards More sets Gas Stoichiometry definitions Gases 15 Terms Gas Stoichiometry quiz Gases 10 Terms Take your learning anywhere! Prep for your exams on the go with video lessons and practice problems in our mobile app. Here’s what students ask on this topic: Gas stoichiometry involves calculating the volumes, pressures, or temperatures of gases involved in a chemical reaction using the ideal gas law and the stoichiometric coefficients from the balanced chemical equation. Here’s a step-by-step guide: Write and Balance the Equation: Ensure the chemical equation for the reaction is balanced. The coefficients will give you the ratio of moles of reactants to products. Convert to Moles: If you're given volumes at STP (Standard Temperature and Pressure, which is 0°C and 1 atm), you can use the fact that one mole of an ideal gas occupies 22.4 liters at STP to convert volumes to moles. If conditions are not at STP, use the Ideal Gas Law () to find moles. Stoichiometric Calculations: Use the stoichiometric coefficients from the balanced equation to calculate the moles of the desired substance based on the moles of the given substance. Convert Back to Desired Units: If you need to find the volume, and you’re at STP, multiply the moles by 22.4 L/mol. If not at STP, rearrange the Ideal Gas Law to solve for the desired variable (, , or ). Account for Non-Ideal Conditions: If Gas stoichiometry is the quantitative relationship between reactants and products in chemical reactions involving gases. It is based on the principles of the ideal gas law, which relates the volume, temperature, pressure, and amount of a gas. In gas stoichiometry, you use the molar volume of a gas at standard temperature and pressure (STP) to predict how much gas is produced or consumed in a reaction. For example, at STP (0°C and 1 atm), one mole of any ideal gas occupies 22.4 liters. If you're given a chemical equation and the amount of one reactant in moles, you can calculate the volume of a gaseous product by using the stoichiometric coefficients from the balanced equation. This allows you to predict the volume of gas produced from a given amount of reactant or vice versa. It's a crucial concept for understanding how chemical reactions proceed and for practical applications like synthesizing chemicals in the lab or industrial processes. Your General Chemistry tutor Jules Bruno General Chemistry, Analytical Chemistry and GOB lead instructor
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How can I get the complement of vector y in vector x Ask Question Asked 15 years, 6 months ago Modified11 years, 9 months ago Viewed 21k times Part of R Language Collective This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. That's x \ y using mathematical notation. Suppose r x <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,2,1,1,1,3) y <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1) How can I get a vector with ALL the values in x that are not in y. i.e the result should be: r 2,1,1,3 There is a similar question here. However, none of the answers returns the result that I want. R Language Collective r vector Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited May 23, 2017 at 11:44 CommunityBot 1 1 1 silver badge asked Mar 21, 2010 at 17:11 YorgosYorgos 30.6k 20 20 gold badges 113 113 silver badges 152 152 bronze badges 3 You need to define your problem more precisely. Why is 1 in your output even though 1 occurs in y? And then why only two 1s and not three? It would be helpful if you could some kind of pseudocode specifying what you want to compute.Jyotirmoy Bhattacharya –Jyotirmoy Bhattacharya 2010-03-21 18:19:36 +00:00 Commented Mar 21, 2010 at 18:19 y is a proper subset of x. And I am looking for its complement in x. I don't care about positions. Had y been c(0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,2,1,1,1,3) I should get the rest 4 zeros.Yorgos –Yorgos 2010-03-21 18:31:30 +00:00 Commented Mar 21, 2010 at 18:31 Y can not be a proper subset of X, because neither of them are sets!hadley –hadley 2010-03-21 19:06:44 +00:00 Commented Mar 21, 2010 at 19:06 Add a comment| 3 Answers 3 Sorted by: Reset to default This answer is useful 13 Save this answer. Show activity on this post. Here a solution using pmatch (this gives the "complement" as you require): r x <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,2,1,1,1,3) y <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1) res <- x[is.na(pmatch(x,y))] From pmatch documentation: "If duplicates.ok is FALSE, values of table once matched are excluded from the search for subsequent matches." Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Mar 22, 2010 at 10:18 teucerteucer 6,268 2 2 gold badges 28 28 silver badges 36 36 bronze badges Comments Add a comment This answer is useful 7 Save this answer. Show activity on this post. How about this: r R> x[x!=y] 2 1 1 1 3 Warning message: In x != y : longer object length is not a multiple of shorter object length R> This is difficult problem, I think, as you are mixing values and positions. The easier solution relies on one of the 'set' functions in R: r R> setdiff(x,y) 2 3 but that uses only values and not position. The problem with the answer I gave you is the implicit use of recycling and the warning it triggered: as your x is longer than your y, the first few values of y get reused. But recycling is considered "clean" on when the longer vector has an integer-multiple length of the length of the shorter vector. But that is not the case here, and hence I am not sure we can solve your problem all that cleanly. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Mar 21, 2010 at 18:16 Dirk is no longer hereDirk is no longer here 370k 60 60 gold badges 668 668 silver badges 742 742 bronze badges 1 Comment Add a comment William Doane William DoaneOver a year ago and setdiff(x, y) is, indeed, the standard definition for x \ y ... since it's a set operation, it first finds unique values and then compares the two vectors. 2010-03-21T18:19:51.04Z+00:00 2 Reply Copy link This answer is useful 3 Save this answer. Show activity on this post. If I understand the problem, you can use table to compute the difference in the number of elements in each set and then create a vector based on the difference of those counts (note that this won't necessarily give you the order you gave in your question). ```r diffs <- table(x) - table(factor(y, levels=levels(factor(x)))) rep(as.numeric(names(diffs)), ifelse(diffs < 0, 0, diffs)) 1 1 2 3 ``` Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Mar 21, 2010 at 18:53 Jonathan ChangJonathan Chang 25.5k 5 5 gold badges 36 36 silver badges 33 33 bronze badges 3 Comments Add a comment Yorgos YorgosOver a year ago Thanks, this one is working indeed!. Order does not matter. I am just curious to see how can this be achieved using library sets (using a function like set_complement perhaps) or another "one liner". I can hardly believe there's no way to get this directly. 2010-03-21T19:30:55.193Z+00:00 0 Reply Copy link William Doane William DoaneOver a year ago There would be if you were working with true sets. Sets don't have duplicates and order doesn't matter. All the set functions in R are going to follow from that definition... what you actually have is the set X whose elements are {0, 1, 2, 3} and the set Y whose elements are {0, 1}. Thus X \ Y is {2, 3}. While what you're looking for for output is well defined, it's NOT a set operation, so you're going to need to do a little work to get it. You can always wrap Jonathan's code in a function, if you must have a one-line solution. 2010-03-21T22:54:44.597Z+00:00 1 Reply Copy link Yorgos YorgosOver a year ago If sets don't accept duplicates, there are also multisets that do. 2010-03-22T13:06:00.88Z+00:00 0 Reply Copy link Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. 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Something went wrong. Wait a moment and try again. Number of Roots Polynomial Graph Graphing (functions) Degrees (polynomials) Roots (mathematics) Polynomial Theory Roots of Polynomials 5 What is the difference between odd and even multiplicity of a root of a polynomial and how that affects the graph? Michael Livshits Officially retired at USA Social Security (2011–present) · Author has 1.7K answers and 747.9K answer views · 5y Hopefully you remember the polynomial Factor theorem - Wikipedia. which says that x−b divides a polynomial f(x) if and only if b is a root of f. Now, b is a root of multiplicity k if (x−b)k divides f(x), but (x−b)k+1 does not. In this case f factors as f(x)=g(x)(x−b)k, where g(b)≠0. The graph of f(x) near b will look like g(b)(x−b)k, for k even it will have a local extremum at b, while for k>1 and odd b will be a horizontal inflection point for f(x), for k=1 b will be a simple root, with g(b)=f’(b). All this extends to functions that have sufficiently many derivatives. In fact differ Hopefully you remember the polynomial Factor theorem - Wikipedia. which says that x−b divides a polynomial f(x) if and only if b is a root of f. Now, b is a root of multiplicity k if (x−b)k divides f(x), but (x−b)k+1 does not. In this case f factors as f(x)=g(x)(x−b)k, where g(b)≠0. The graph of f(x) near b will look like g(b)(x−b)k, for k even it will have a local extremum at b, while for k>1 and odd b will be a horizontal inflection point for f(x), for k=1 b will be a simple root, with g(b)=f’(b). All this extends to functions that have sufficiently many derivatives. In fact differentiability of f can be viewed as divisibility of f(x)-f(a) by x-a. Considering divisibility in different classes of functions gives us differentiation of different flavors. Alex Moon BS in Pure Mathematics, Michigan State University · Author has 3.4K answers and 2.1M answer views · 5y If 𝒂 is a root of 𝒑(𝒙) then there exists some 𝒒(𝒙) such that 𝒒(𝒙)(𝒙-𝒂)ⁿ for 𝒏∈︎ℕ, and 𝒏 dictates the multiplicity of 𝒂 as a root, we choose the largest 𝒏 that still leaves 𝒒(𝒙) as a polynomial function. we can actually ignore 𝒒(𝒙) since we extracted all 0’s at 𝒂, 𝒒(𝒙) is strictly positive or negative about 𝒙=𝒂 as long as we get sufficiently close. If we take limits then we are definitely sufficiently close: 𝒍𝒊𝒎 𝒙→𝒂⁺ (𝒙-𝒂)ⁿ = 0, but if we take a finite distance where 𝒙>𝒂 but 𝒙≈︎𝒂 then we see we have a positive result. 𝒍𝒊𝒎 𝒙→𝒂⁻ (𝒙-𝒂)ⁿ = 0, but if we take a finite di If 𝒂 is a root of 𝒑(𝒙) then there exists some 𝒒(𝒙) such that 𝒒(𝒙)(𝒙-𝒂)ⁿ for 𝒏∈︎ℕ, and 𝒏 dictates the multiplicity of 𝒂 as a root, we choose the largest 𝒏 that still leaves 𝒒(𝒙) as a polynomial function. we can actually ignore 𝒒(𝒙) since we extracted all 0’s at 𝒂, 𝒒(𝒙) is strictly positive or negative about 𝒙=𝒂 as long as we get sufficiently close. If we take limits then we are definitely sufficiently close: 𝒍𝒊𝒎 𝒙→𝒂⁺ (𝒙-𝒂)ⁿ = 0, but if we take a finite distance where 𝒙>𝒂 but 𝒙≈︎𝒂 then we see we have a positive result. 𝒍𝒊𝒎 𝒙→𝒂⁻ (𝒙-𝒂)ⁿ = 0, but if we take a finite distance where 𝒙<𝒂 but 𝒙≈︎𝒂 then we see even 𝒏 implies positive result and odd 𝒏 implies negative result. We need to choose a small enough ε = |𝒙-𝒂| such that 𝒒(𝒙) does not change sign, and we can clearly do this as the roots of 𝒒(𝒙) are a finite distance from 𝒂 (if 𝒒 is nonconstant, and if it is constant life is easier) since none of them are 𝒂! If the closest root of 𝒒(𝒙) was 𝒃 then we choose ε = |𝒃-𝒂|/2 or something. In conclusion: 𝒏 is odd implies the graph of 𝒑(𝒙) turns negative to the immediate left of 𝒂 and positive to the immediate right of 𝒂. (𝒙-𝒂)ⁿ is positive for 𝒙>𝒂 and negative for 𝒙<𝒂, 𝒑(𝒙) as a whole has other contraints from 𝒒(𝒙). 𝒏 is even implies the graph of 𝒑(𝒙) turns positive to the immediate left of 𝒂 and to the immediate right of 𝒂. (𝒙-𝒂)ⁿ is positive for 𝒙≠𝒂, but again, 𝒑(𝒙) as a whole has other constraints from 𝒒(𝒙). Jack Smith Lived in Philadelphia, PA · Author has 4.2K answers and 11.3M answer views · 4y When the amount of times the factor appears is odd, the root is a single root and the graph crosses the x-axis at the root. Assuming k is a natural number, (x−a)2k−1 has a single root at a: (x−a)2k−1=0 2k−1√(x−a)2k−1=2k−1√0 x−a=0 x=a When the amount of times the factor appears is even, the root is a double root and the graph touches and bounces off the x-axis at the root. That’s because squares are always positive and the factor doesn’t change the sign of the function. Assuming k is a natural number, (x−a)2k has a double root at a: (x−a)2k=0 \sqrt[2k]{(x-a)^{2 When the amount of times the factor appears is odd, the root is a single root and the graph crosses the x-axis at the root. Assuming k is a natural number, (x−a)2k−1 has a single root at a: (x−a)2k−1=0 2k−1√(x−a)2k−1=2k−1√0 x−a=0 x=a When the amount of times the factor appears is even, the root is a double root and the graph touches and bounces off the x-axis at the root. That’s because squares are always positive and the factor doesn’t change the sign of the function. Assuming k is a natural number, (x−a)2k has a double root at a: (x−a)2k=0 2k√(x−a)2k=±2k√0 x−a=±0 x=a±0 x1=a+0, x2=a−0 This gives the roots x=a and x=a Related questions What exactly is the difference between a root of a polynomial, and a zero? Is it possible to have a real root with odd multiplicity in any polynomial? What is the minimum number of x-intercepts for an even degree polynomial? What is the difference between the multiplicity of a root and the degree of a polynomial? Is infinity odd or even? Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.5M answer views · 4y Related Is it possible to have a real root with odd multiplicity in any polynomial? ———————————————————————————————————— ———————————————————————————————————— Similarly the equation x^5 = 0 goes through x = 0 exactly 5 times and we say it has a root of multiplicity 5. 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Assistant Bot · 1y The multiplicity of a root of a polynomial refers to how many times that root appears as a solution. This characteristic significantly influences the behavior of the polynomial's graph at the root. Here’s a breakdown of odd and even multiplicities: Odd Multiplicity Definition: A root r of a polynomial has odd multiplicity if it appears an odd number of times (e.g., 1, 3, 5, etc.). Graph Behavior: The graph will cross the x-axis at the root r. Near the root, the graph will change direction (i.e., it will go from positive to negative or vice versa). Example: For the polynomial f(x)=(x−2)3, the ro The multiplicity of a root of a polynomial refers to how many times that root appears as a solution. This characteristic significantly influences the behavior of the polynomial's graph at the root. Here’s a breakdown of odd and even multiplicities: Odd Multiplicity Definition: A root r of a polynomial has odd multiplicity if it appears an odd number of times (e.g., 1, 3, 5, etc.). Graph Behavior: The graph will cross the x-axis at the root r. Near the root, the graph will change direction (i.e., it will go from positive to negative or vice versa). Example: For the polynomial f(x)=(x−2)3, the root x=2 has odd multiplicity (3), and the graph crosses the x-axis at this point. Even Multiplicity Definition: A root r has even multiplicity if it appears an even number of times (e.g., 2, 4, 6, etc.). Graph Behavior: The graph will touch the x-axis at the root r but will not cross it. Near the root, the graph will not change direction; it will either remain above or below the x-axis. Example: For the polynomial f(x)=(x−2)2, the root x=2 has even multiplicity (2), and the graph touches the x-axis at this point but does not cross it. Summary Odd Multiplicity: Crosses the x-axis, changes direction. Even Multiplicity: Touches the x-axis, does not change direction. Understanding the multiplicity of roots helps in sketching the graph of a polynomial and predicting its behavior at various points. Richard Johnston B.Sc in Physics, University of Western Ontario (Graduated 2019) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · 6y Related Why does any polynomial with odd degree have at least one real root? The easiest way to explain why any odd degree polynomial has at least one real root is to look at the end behaviors. That is, we look at what happens as x approaches infinity and negative infinity. For a polynomial, the highest power of x determines what happens as x becomes very large and tends towards +/- infinity. So for an odd polynomial, the highest power is odd and the function behaves like it was just the highest power for really large x. For example, the function f(x)=x^5+4x^4-2x^2+1 behaves as though it was just f(x)=x^5 when x is really large. Thus plugging in for x=infinity gives tha The easiest way to explain why any odd degree polynomial has at least one real root is to look at the end behaviors. That is, we look at what happens as x approaches infinity and negative infinity. For a polynomial, the highest power of x determines what happens as x becomes very large and tends towards +/- infinity. So for an odd polynomial, the highest power is odd and the function behaves like it was just the highest power for really large x. For example, the function f(x)=x^5+4x^4-2x^2+1 behaves as though it was just f(x)=x^5 when x is really large. Thus plugging in for x=infinity gives that f(x) approaches infinity and for x=-infinity we find that f(x) approaches negative infinity. So we see that at the two ends of the real axis that f(x) is unbounded above and below, and since any polynomial function is also continuous we sre forced to conclude that f(x) also takes all values between its highest and lowest value. It is clear to see that zero is in this range, and thus there must exist at least one x such that f(x)=0 in order for the function to satisfy these end behaviors and be continuous. Therefore, for any odd degree polynomial this holds as well. Related questions What is an intuitive explanation of zero/root multiplicity in a polynomial? What are the differences between scientific calculators and graphing calculators? What is the difference between 'an expression' and 'a polynomial? What is the difference between a root and a zero of a polynomial? Is Root 3 X-1 a polynomial? Gary Russell Former Professor at University of Iowa (1996–2025) · Upvoted by Clyde Monma , B.S. Computer Science & Mathematics, Washington State University (1974) · Author has 6K answers and 3M answer views · Updated 4y Related What do you notice about a graph when the polynomial function has a factor that occurs twice, three times? Let’s examine some different types of polynomials. QUADRATIC FUNCTION The function “bounces back” from the x-axis. Suppose that we have the quadratic equation f(x) = (x-q)^2 where q is a real number. It is clear that the f(x) is never negative. Thus, f(x) just touches the x axis at q and then “bounces back” to positive values of y. Here is an example of the quadratic f(x) for q = 5: CUBIC FUNCTION Consider the cubic equation g(x) = (x-q)^3 = h(x)f(x) h(x) = (x-q) f(x) = (x-q)^2 Notice, as before, that f(q) = 0. Otherwise, f(x) > 0. However, h(x) < 0 for x < q and h(x) > 0 for x > q. Of course, h(q) = 0. Let’s examine some different types of polynomials. QUADRATIC FUNCTION The function “bounces back” from the x-axis. Suppose that we have the quadratic equation f(x) = (x-q)^2 where q is a real number. It is clear that the f(x) is never negative. Thus, f(x) just touches the x axis at q and then “bounces back” to positive values of y. Here is an example of the quadratic f(x) for q = 5: CUBIC FUNCTION Consider the cubic equation g(x) = (x-q)^3 = h(x)f(x) h(x) = (x-q) f(x) = (x-q)^2 Notice, as before, that f(q) = 0. Otherwise, f(x) > 0. However, h(x) < 0 for x < q and h(x) > 0 for x > q. Of course, h(q) = 0. Thus, when we multiply the two functions, we can see that the left hand side of the quadratic will move into negative numbers, while the right hand side will stay within positive numbers. The function g(x) will, of course, cross the x axis at q. Here’s a graph of the cubic g(x) for q = 5: HIGHER ORDER POLYNOMIALS All higher degree polynomials with a repeated real root q are variants on these two themes. When the order of the polynomial is an even number, you will see something like the quadratic pattern. When the order of the polynomial is odd, you will see something like the cubic pattern. FINAL COMMENTS All these statements assume that q is a real number. The reason is the following: There does not exist any polynomial (having real coefficients) with a single repeated root q, such that q is a complex number (with an non-zero imaginary part). This is due to the fact that complex roots come in pairs. If a + bi is a root, then so is a - bi where i is the square root of minus one. Sponsored by Amazon Business Drive decisions with analytics. Make better data-driven decisions with real-time reporting and visibility into purchasing patterns. Daniel Claydon Learning mathematics · Author has 779 answers and 4.3M answer views · 6y Related Can all numbers be expressed as the root of a polynomial with integer coefficients? No. First, let's dig in to some terminology. We call any such number that is the root of a finite, integer polynomial an algebraic number, sometimes denoted by or [math]\overline{\mathbb{Q}}[/math]. Any number that is not algebraic is said to be transcendental. How many integer polynomials are there? This sounds like a dumb question — obviously there are infinitely many. But are there countably or uncountably many of them? To answer this, we start be recalling the following theorems: The set [math]\mathbb{Z}[/math] is countable. If [math]A[/math] is any countable set, and [math]B[/math] is the set of all [math]n[/math]-tuples math[/math] of No. First, let's dig in to some terminology. We call any such number that is the root of a finite, integer polynomial an algebraic number, sometimes denoted by [math]\mathbb{A}[/math] or [math]\overline{\mathbb{Q}}[/math]. Any number that is not algebraic is said to be transcendental. How many integer polynomials are there? This sounds like a dumb question — obviously there are infinitely many. But are there countably or uncountably many of them? To answer this, we start be recalling the following theorems: The set [math]\mathbb{Z}[/math] is countable. If [math]A[/math] is any countable set, and [math]B[/math] is the set of all [math]n[/math]-tuples math[/math] of elements in [math]A[/math], then [math]B[/math] is also countable. The union of countably many countable sets is countable. How does this help? Well, any integer polynomial of degree [math]n[/math] can be represented as an math[/math]-tuple of integers, corresponding to its coefficients. Thus, by theorems 1 and 2, the number of these polynomials is countable. Since each polynomial contributes a finite number of roots, the number of algebraic numbers generated by degree [math]n[/math] polynomials is countable. Then the total number of algebraic numbers is at most the union of the number generated by degree [math]1,2,3 \ldots[/math] polynomials, which by theorem 3 is countable. This immediately answers your question. The set [math]\mathbb{C}[/math] of complex numbers is uncountable, therefore not all its elements can be algebraic, and the same applies for [math]\mathbb{R}[/math]. In fact, almost no numbers are algebraic. Ok great, you might think, we've shown the existence of transcendental numbers, but do we have any actual examples? Yes, we do. But there are no known simple proofs for any of them, so I'll just state some of what we know. The first class of numbers whose transcendence was proven are the Liouville Numbers, including Liouville's constant: [math]L = \displaystyle \sum_{k=1}^{\infty} \frac{1}{10^{k!}} \tag{}[/math] Later on, it emerged that [math]e \approx 2.72[/math] is also transcendental. I just want to stress: this result is not at all surprising — in fact it would have been astonishing if it wasn't true — it's just that it's hard to prove. Moreover, we found that [math]e^a[/math] is transcendental whenever [math]a[/math] is a nonzero algebraic number. Here's a cool piece of logic: since [math]e^{ i \pi} = -1[/math] is algebraic, it follows that [math]i\pi[/math] is transcendental. If [math]\pi[/math] was the root of some integer polynomial [math]P(x)[/math], then [math]\pi[/math] must be the root of some equation [math]a_1x^{b_1} + \cdots + a_nx^{b_n} = c_1x^{d_1} + \cdots c_mx^{d_m} \tag{}[/math] where the exponents on the left side are all even and on the right side all odd. Squaring both sides, rearranging to sort the exponents that are [math]0[/math] and [math]2[/math] modulo [math]4[/math], and squaring again shows [math]\pi[/math] also satisfies some integer polynomial whose exponents are all multiples of [math]4[/math]. But then [math]i\pi[/math] satisfies this polynomial as well, so [math]i\pi[/math] is algebraic; a contradiction. Thus we conclude [math]\pi[/math] must be transcendental. We know [math]\sin(a), \cos(a),[/math] and [math]\tan(a)[/math] are transcendental for any nonzero algebraic number [math]a[/math], as well as their reciprocals. Also [math]W_n(a)[/math] is transcendental for any branch of the Lambert W Function and if we further require [math]a\neq 1[/math] then [math]\ln(a)[/math] is too. Also [math]\Gamma(1/3), \Gamma(1/4),[/math] and [math]\Gamma(1/6)[/math] are all known to be transcendental, as well as the Champernowne constant, [math]0.1234567891011 \ldots[/math]. Perhaps the “biggest” class of transcendental numbers comes from the Gelfond-Schneider Theorem, that tells us [math]a^b[/math] is transcendental whenever [math]a \neq 0,1[/math] is algebraic and [math]b[/math] is an irrational algebraic number. This includes the Gelfond-Schneider constant, [math]\sqrt{2}^{\sqrt{2}}[/math]. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.5M answer views · Apr 22 Related Can you explain the differences between even degree polynomials, odd degree polynomials, constant functions, and linear functions? The BEST way to answer this question is with examples! See if you agree… NOTE WHICH WAY THE ARROWS POINT These are EVEN degree polynomial graphs… (the highest power is even) ———————————————————————————————————- The BEST way to answer this question is with examples! See if you agree… NOTE WHICH WAY THE ARROWS POINT These are EVEN degree polynomial graphs… (the highest power is even) ———————————————————————————————————- Sponsored by ExpertsInSeniorHealth.com Watch Now: This 1 Thing Helps Mute Tinnitus & Fight Ringing Ears. Am I The Only One Who Didn't Know This Before? Gt Jacobs Ph.D. in Mathematics, University of North Texas (Graduated 2015) · 1y Related How does a complex root affect a polynomial? The question isn’t very precise, but we can say something. First of all, it depends what kind of polynomial we’re talking about, and then it also depends on what you mean by “affect”. For a polynomial with real coefficients, a complex root is always accompanied by a second complex root, its complex conjugate. For example, if the polynomial has real coefficients, and one root is 4+i, then another root is guaranteed to be 4–i. For a polynomial with complex coefficients, the above is not necessarily true, and the roots can be distributed however you like on the complex plane, with the only limitati The question isn’t very precise, but we can say something. First of all, it depends what kind of polynomial we’re talking about, and then it also depends on what you mean by “affect”. For a polynomial with real coefficients, a complex root is always accompanied by a second complex root, its complex conjugate. For example, if the polynomial has real coefficients, and one root is 4+i, then another root is guaranteed to be 4–i. For a polynomial with complex coefficients, the above is not necessarily true, and the roots can be distributed however you like on the complex plane, with the only limitation being that the number of them is bounded by the degree of the polynomial. It’s clear from what we’ve said so far that polynomials with real coefficients are more constrained by complex roots, so let’s talk about those; they’re usually the polynomials that we study, anyway. If we’re trying to factor our polynomial into irreducible polynomials with real coefficients, then having a complex root (really a pair of complex roots) means that one of our irreducible factors will be quadratic. For example, if we have the roots 4+i and 4–i, then one of the factors will have to be (x^2 - 8x + 17), which cannot be factored any further using real coefficients. There is also an influence on the polynomial’s graph, but this is a little more subtle. When a polynomial has a real root, the graph has an x-intercept corresponding to it. A pair of complex roots does not produce an x-intercept on an ordinary graph, where x- and y-coordinates are real numbers. However, it does produce a dip towards the x-axis, around the area corresponding to the real part of the complex roots. An example will help illustrate this. Consider again a pair of complex roots: 4+i and 4–i. Their real part is 4. Now consider a simple polynomial with one real root, for example, x-1. The graph of y = x-1 has an x-intercept at (1,0), and is otherwise just a straight line. If we multiply x-1 by the factor (x^2 - 8x + 17), we get a cubic polynomial: x^3 - 9x^2 + 25x - 17. If you look at its graph, you can see that it’s still an ascending path through the point (1,0), but it dips back down towards the point (4,0), as if it’s considering another x-intercept there, before changing its mind and heading off to infinity after all. That dip in the graph is the influence of the pair of complex roots. Wes Browning Studied Mathematics at Cornell University · Author has 8.8K answers and 6M answer views · 2y Related What is the number of roots for any polynomial equation? Why does this happen and what are its implications on mathematics? If the polynomial is defined on just any field and has degree n greater than zero, it can have any number of roots in the field between 0 and n inclusive, except for n-1. (If you have n-1 roots you can find an nth by polynomial long division.) If the underlying fied is algebraically closed the polynomial will completely factor into linear, degree one, polynomials each of which will contribute a root. Two or more linear factors can contribute the same root, and we say that root appears to some multiplicity. That lets us say the polynomial has exacly n roots “counting multiplicities.” It was prove If the polynomial is defined on just any field and has degree n greater than zero, it can have any number of roots in the field between 0 and n inclusive, except for n-1. (If you have n-1 roots you can find an nth by polynomial long division.) If the underlying fied is algebraically closed the polynomial will completely factor into linear, degree one, polynomials each of which will contribute a root. Two or more linear factors can contribute the same root, and we say that root appears to some multiplicity. That lets us say the polynomial has exacly n roots “counting multiplicities.” It was proved in the 1800s by Gauss and others that the complex field is algebraically closed. The theorem is called the Fundamental Theorem of Algebra, even though its proof is really of a theorem in complex analysis. The Fundamental Theorem of Algebra is extremely important in mathematics. It’s why we know that all nxn rational, real or complex matrices have n complex eigenvalues “up to mutiplicity.” There are applications of that to physics, and engineering. A special case is the study of dynamic systems given by a linear differial equation of the form a0(D^n)y + … +aj(D^j)y + … +an(D^0)y. The general solution is obtained by factoring the polynomial in the differential operator D as if it were a polynomial over C, Multiple linear factors with the same complex roots require special care, but that’s not a very hard problem. These systems become easy to understand, conceptually at least. Eric Hawk Author has 3K answers and 3.9M answer views · 1y Related Is it true that every odd degree polynomial has at least one odd root and every even degree polynomial has at least one even root? No, this isn’t true. I’m not sure what you mean by an odd or even root. Do you mean an odd or even numeric value for the root, as in 3 is odd and 2 is even? The roots of polynomials do not have to be integers. Even more basically, an odd degree polynomial with real coefficients has at least one real root, from the intermediate value theorem, since it is an odd function and goes to +/- infinity as x goes to +/- infinity (must cross the x-axis somewhere). An even degree polynomial might not have any real roots. Consider y(x) = 1 (0-degree polynomial, 0 is even) y(x) = x^2 + 1 (y(x) > 0 for all real No, this isn’t true. I’m not sure what you mean by an odd or even root. Do you mean an odd or even numeric value for the root, as in 3 is odd and 2 is even? The roots of polynomials do not have to be integers. Even more basically, an odd degree polynomial with real coefficients has at least one real root, from the intermediate value theorem, since it is an odd function and goes to +/- infinity as x goes to +/- infinity (must cross the x-axis somewhere). An even degree polynomial might not have any real roots. Consider y(x) = 1 (0-degree polynomial, 0 is even) y(x) = x^2 + 1 (y(x) > 0 for all real values of x) Hope this helps and does not mislead or confuse you. Alon Amit CS degree and many years of coding. · Upvoted by Aditya Garg , M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) and Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) · Author has 8.7K answers and 172.9M answer views · 6y Related Can all numbers be expressed as the root of a polynomial with integer coefficients? No. The numbers which are roots of single-variable polynomials with integer coefficients are called algebraic numbers. As a whole, they are denoted [math]\overline{\Q}[/math]. Some such numbers are [math]\sqrt{17-\sqrt{5}}[/math], [math]\cos(10^\circ)[/math] and googolplex. Real or complex numbers that are not algebraic are called transcendental. They are far more numerous than algebraic numbers: there are uncountably many of them, but only countably many algebraic numbers. Some famous transcendental numbers are [math]e[/math], [math]\pi[/math], [math]\ln(2)[/math] and [math]\sqrt{2}^{\sqrt{2}}[/math]. Related questions What exactly is the difference between a root of a polynomial, and a zero? Is it possible to have a real root with odd multiplicity in any polynomial? What is the minimum number of x-intercepts for an even degree polynomial? What is the difference between the multiplicity of a root and the degree of a polynomial? Is infinity odd or even? What is an intuitive explanation of zero/root multiplicity in a polynomial? What are the differences between scientific calculators and graphing calculators? What is the difference between 'an expression' and 'a polynomial? What is the difference between a root and a zero of a polynomial? Is Root 3 X-1 a polynomial? A polynomial function has a root of –5 with multiplicity 3, a root of 1 with multiplicity 2, and a root of 3 with multiplicity 7. If the function has a negative leading coefficient and is of even degree, which statement about the graph is true? What is the difference between an even and odd polynomial? Why do both the odd and even parts of a Hurwitz polynomial p(s) have roots on imaginary axis only? Is the set of all polynomials with odd constant terms an ideal of Z[x]? What is the process for finding the eigenspace of an eigenvalue with a multiplicity greater than one? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://www.teachoo.com/1653/518/Ex-6.2--1---In-figure-(i)-DE----BC.-Find-EC-(ii)-Find-AD/category/Ex-6.2/
Ex 6.2, 1 - In figure (i) DE || BC. Find EC (ii) Find AD Everything Class 6 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Class 11 Maths Class 11 English Class 11 Computer Science (Python) Class 12 Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education Courses GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Interesting Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance Popular GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Maths Updated 2025-26 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 9 Maths Class 10 Maths Class 11 Maths Class 12 Maths Sample Paper Class 10 Maths Sample Paper Class 12 Maths Accounts & Finance Learn Excel free Learn Tally free Learn GST (Goods and Services Tax) Learn Accounting and Finance GST and Income Tax Return Filing Course Tally Ledger List Popular Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 9 Science Class 10 Science Remove Ads Login Maths Remove ads Science GST Accounts Tax Englishtan Excel Social Science Class 6 Class 6 Maths (Ganita Prakash) Class 6 Maths (old NCERT) Class 6 Science (Curiosity) Class 6 Science (old NCERT) Class 6 Social Science Class 6 English Class 7 Class 7 Maths (Ganita Prakash) Class 7 Maths (old NCERT) Class 7 Science (Curiosity) Class 7 Science (old NCERT) Class 7 Social Science Class 7 English Class 8 Class 8 Maths (Ganita Prakash) Class 8 Maths (old NCERT) Class 8 Science (Curiosity) Class 8 Science (old NCERT) Class 8 Social Science Class 8 English Class 9 Class 9 Maths Class 9 Science Class 9 Social Science Class 9 English Class 10 Class 10 Maths Class 10 Science Class 10 Social Science Class 10 English Class 11 Class 11 Maths Class 11 Computer Science (Python) Class 11 English Class 12 Class 12 Maths Class 12 English Class 12 Economics Class 12 Accountancy Class 12 Physics Class 12 Chemistry Class 12 Biology Class 12 Computer Science (Python) Class 12 Physical Education Courses GST and Accounting Course Excel Course Tally Course Finance and CMA Data Course Payroll Course Interesting Learn English Learn Excel Learn Tally Learn GST (Goods and Services Tax) Learn Accounting and Finance Popular GST Demo GST Tax Invoice Format Accounts Tax Practical Tally Ledger List Remove Ads Login Chapter 6 Class 10 Triangles Serial order wise Ex 6.2 Ex 6.2 Ex 6.2, 1 You are here Ex 6.2, 2 (i) Ex 6.2, 2 (ii) Ex 6.2, 2 (iii) Important Ex 6.2, 3 Ex 6.2, 4 Important Ex 6.2, 5 Important Ex 6.2, 6 Important Ex 6.2, 7 Ex 6.2, 8 Ex 6.2, 9 Important Ex 6.2, 10 Ex 6.3→ Chapter 6 Class 10 Triangles Serial order wise Ex 6.1 Ex 6.2 Ex 6.3 Examples Theorems Case Based Questions (MCQ) NCERT Exemplar - MCQ Ratio of Area of Similar Triangles Pythagoras Theorem and it's important questions Important questions of Triangle (in Geometry) Ex 6.2, 1 - Chapter 6 Class 10 Triangles Last updated at December 13, 2024 by Teachoo ADVERTISEMENT Powered by VidCrunch Next Stay Playback speed 1x Normal Quality Auto Back 720p 360p 240p 144p Auto Back 0.25x 0.5x 1x Normal 1.5x 2x / Skip Ads by Next: Ex 6.2, 2 (i) →Remove AdsShare on WhatsApp Transcript Ex 6.2, 1 In figure(i) ,DE || BC. Find EC in (i) It is given that DE II BC Hence , / = / Putting the values 1.5/3 = 1/ 1.5 EC = 1 3 EC = (1 3)/1.5 EC = 2 cm Ex 6.2, 1 In figure (ii), DE || BC. Find AD. It is given that DE II BC Hence , 𝐴𝐷/𝐷𝐵 = 𝐴𝐸/𝐸𝐶 Putting the values 𝐴𝐷/7.2 = 1.8/5.4 AD × 5.4 = 1.8 ×7.2 AD = (1.8 × 7.2)/5.4 AD = (18 × 72 × 10)/(54 × 100) AD = 72/30 AD = 2.4 cm Show More Chapter 6 Class 10 Triangles Serial order wise Ex 6.2 Ex 6.2, 1 You are here Ex 6.2, 2 (i) Ex 6.2, 2 (ii) Ex 6.2, 2 (iii) Important Ex 6.2, 3 Ex 6.2, 4 Important Ex 6.2, 5 Important Ex 6.2, 6 Important Ex 6.2, 7 Ex 6.2, 8 Ex 6.2, 9 Important Ex 6.2, 10 Ex 6.3→ Made by Davneet Singh Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo Hi, it looks like you're using AdBlock :( Displaying ads are our only source of revenue. 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https://www.quora.com/How-do-you-find-a-spring-constant-without-force
Something went wrong. Wait a moment and try again. Spring Constant Hooke's Law Springs (device) Scientific Concepts Mechanics (general) Science Physics Physical Constant Mechanics and Energy 5 How do you find a spring constant without force? Ron Brown Decades of teaching physics to undergrads · Author has 13.5K answers and 83.4M answer views · Updated 1y It depends on what else you know. Clearly, just by definition of a spring constant, if you know the applied force as well as the extension or compression of that spring due to the applied force, you can calculate the spring constant. But if you don’t have that information, you have to look at an indirect way of determining the spring constant. So you look at something else that depends on the spring constant. For example, if you suspend a mass from the spring, and set that mass in oscillation, the period of oscillation will depend on the spring constant and the mass. So to address your homework p It depends on what else you know. Clearly, just by definition of a spring constant, if you know the applied force as well as the extension or compression of that spring due to the applied force, you can calculate the spring constant. But if you don’t have that information, you have to look at an indirect way of determining the spring constant. So you look at something else that depends on the spring constant. For example, if you suspend a mass from the spring, and set that mass in oscillation, the period of oscillation will depend on the spring constant and the mass. So to address your homework problem, look at how a mass on a spring oscillates, learn what it depends on and how, and if you have that information, you can solve for k. Assistant Bot · 1y Finding the spring constant k without directly measuring the force can be done through several indirect methods. Here are a few approaches you can consider: Using Hooke's Law with Displacement: If you can measure the displacement x of the spring from its equilibrium position, you can relate this to the spring constant using the following relationship: k=Fx If you don't know the force but can apply a known mass m to the spring and measure the displacement, you can use the weight of the mass as the force due to gravity: F=mg So, k=mgx Oscillation Method: Another method is to Finding the spring constant k without directly measuring the force can be done through several indirect methods. Here are a few approaches you can consider: Using Hooke's Law with Displacement: If you can measure the displacement x of the spring from its equilibrium position, you can relate this to the spring constant using the following relationship: k=Fx If you don't know the force but can apply a known mass m to the spring and measure the displacement, you can use the weight of the mass as the force due to gravity: F=mg So, k=mgx Oscillation Method: Another method is to use the oscillation characteristics of the spring. If you can attach a mass to the spring and allow it to oscillate, you can determine the spring constant from the period of oscillation. The formula for the period T of a mass-spring system is: T=2π√mk Rearranging this gives: k=4π2mT2 You can measure the period T of the oscillation and use the mass m to calculate the spring constant k. Energy Method: If you can compress or extend the spring and measure the potential energy stored in it, you could also find the spring constant. The potential energy U stored in a spring is given by: U=12kx2 If you can measure the displacement x and the potential energy U (for example, by measuring the height of a weight that was dropped onto the spring), you can rearrange this to find k: k=2Ux2 Summary Displacement and Mass: Measure how far the spring stretches with a known mass. Oscillation Period: Measure the oscillation period of a mass on the spring. Potential Energy: Measure the potential energy associated with a known displacement. Choose the method that best fits your tools and situation! Jake Forsyth Hughes Mphys in Physics, Northumbria University (Graduated 2024) · Author has 68 answers and 78.9K answer views · 5y Related How can you determine the value of a spring constant? There are different ways you can calculate the value of the spring constant. One of the ways of doing this is to first measure a test mass. Use a standard mass like 100g if available. Get a stand and attach it to a table with a clamp. Attach the spring to the stand so that it hags down and then attach the standard mass to it. Extend the mass with the spring and time for at least ten oscillations. Divide by ten to get the time period after measuring. Then Repeat this experiment with different masses attached to the spring. Using the equation T=2pi sqrt(m/k) Rearrange to obtain: T^2=4pi^2 m/k which There are different ways you can calculate the value of the spring constant. One of the ways of doing this is to first measure a test mass. Use a standard mass like 100g if available. Get a stand and attach it to a table with a clamp. Attach the spring to the stand so that it hags down and then attach the standard mass to it. Extend the mass with the spring and time for at least ten oscillations. Divide by ten to get the time period after measuring. Then Repeat this experiment with different masses attached to the spring. Using the equation T=2pi sqrt(m/k) Rearrange to obtain: T^2=4pi^2 m/k which is in the form y=mx+c Next plot a graph of T^2 against m. Draw a line of best fit through your data points. If you want to be more precise with this use a computer and calculate the least squares gradient (gradient of the best fit line). If you are drawing the graph, take the maximum change in T^2 with the maximum change in 1/m to reduce the percentage error of calculating the gradient. The gradient is 4pi^2/ K, so rearrange to find K. Drawing one a line of worst fit will allow you to find the uncertainty in K, by finding the difference between the best fit gradient and the worst fit gradient. The other way of doing this is to measure the length of the spring and to hang a standard mass from it and measure the extension with a meter ruler. The force on the spring will be equal to the weight of the mass. By repeating the experiment by changing the mass to a different one and measuring the extension, you can then plot a graph of the force against the extension as: F=KX K will be the gradient of the best fit line. This is because the equation is in the form: Y=mx+c Although do not hang a mass with a high value as the spring might plastically deform. Before taking another measurement check that the spring returns to its original length. Related questions How can you determine the value of a spring constant? What is the true standard value of spring constant? What happens if the spring constant increases? What is the force constant of a spring? In physics, how do you determine the spring constant? Aashay Bhogte BE (Hons.) in Chemical Engineering, Birla Institute of Technology and Science, Pilani (Graduated 2021) · 9y Related What is force constant of a spring and if a spring is cut into three pieces, what will the force constant of each? Force constant of a spring is an experimentally obtained value. "It is the amount of force exerted when a spring is elongated/compressed by unit length.” That being the technical definition. It means that if you pull a spring, say attached at one end to a wall, and you manage to pull it by about one metre, then the force u exerted on the spring is numerically equal to it's spring constant. Now for the other part of your question : Remember that for a spring, the product of its spring constant and length is always constant. Numerically: KL = constant Where, K →spring constant L →length of the spring Thi Force constant of a spring is an experimentally obtained value. "It is the amount of force exerted when a spring is elongated/compressed by unit length.” That being the technical definition. It means that if you pull a spring, say attached at one end to a wall, and you manage to pull it by about one metre, then the force u exerted on the spring is numerically equal to it's spring constant. Now for the other part of your question : Remember that for a spring, the product of its spring constant and length is always constant. Numerically: KL = constant Where, K →spring constant L →length of the spring This means for any length of the same spring, this term is going to remain constant. Take for example (your question itself) a spring with a force constant ‘k', and length ‘l'. Divide it into three EQUAL pieces. Each piece will have a length equal to l/3. Let the new spring constant be k'. As I said, KL of a spring is constant. Therefore: (KL)i=(KL)f kl = k'l/3 So: k'= 3k Thanks. If you need to find the force constant for a spring which has been divided into three unequal parts, then comment or request for an edit :) David Simpson Studied Physics & Mathematics (Graduated 2000) · Author has 1.2K answers and 1.7M answer views · 7y Related In physics, how do you determine the spring constant? As another Answerer has noted, simply hang the spring vertically and hang some weight off of it. Then add a bit to the weight. The change in weight divided by the change in the position of the mass is the spring constant. Repeating this for several weights will let you find an average, and will probably give you a better estimate. A lesser-known method is to find the spring constant from the parameters of the spring. The formula is: k=Gd48ND3 where d is the wire diameter, N is the number of (active) turns in the spring, D is the coil diameter (measured from the center of the wire), As another Answerer has noted, simply hang the spring vertically and hang some weight off of it. Then add a bit to the weight. The change in weight divided by the change in the position of the mass is the spring constant. Repeating this for several weights will let you find an average, and will probably give you a better estimate. A lesser-known method is to find the spring constant from the parameters of the spring. The formula is: k=Gd48ND3 where d is the wire diameter, N is the number of (active) turns in the spring, D is the coil diameter (measured from the center of the wire), and G is called the modulus of rigidity of the spring, and is given by G=Y2(1+ν) where Y is the Young’s modulus of the spring material, and ν is the material’s Poisson ratio. Sponsored by JetBrains Write better C++ code with less effort. Boost your efficiency with refactorings, code analysis, unit test support, and an integrated debugger. Suresh Verma M.Sc.(Physics) in Physics, Patna Science College, Patna (Graduated 1965) · Author has 1.9K answers and 4.8M answer views · 7y Related What is the real value of a spring constant? In classical physics, a spring can be seen as a device that stores specifically elastic potential energy, by straining the bonds between the atoms of an elastic material. the hook’s law holds only approximately, and only when the deformation (extension or contraction) is small compared to the spring’s overall length. Many materials have no clearly defined elastic limit, and Hooke's law can not be meaningfully applied to these materials. Moreover, for the superelastic materials, the linear relationship between force and displacement is appropriate only in the low-strain region. Hooke's law is a ma In classical physics, a spring can be seen as a device that stores specifically elastic potential energy, by straining the bonds between the atoms of an elastic material. the hook’s law holds only approximately, and only when the deformation (extension or contraction) is small compared to the spring’s overall length. Many materials have no clearly defined elastic limit, and Hooke's law can not be meaningfully applied to these materials. Moreover, for the superelastic materials, the linear relationship between force and displacement is appropriate only in the low-strain region. Hooke's law is a mathematical consequence of the fact that the potential energy of the rod is a minimum when it has its relaxed length. the force—which is the derivative of energy with respect to displacement—approximates a linear function. Force of fully compressed spring F(max) = Y. d^4 (L-n d) / {16 (1 + mu.)( D-d) ^3. n where Y – Young's modulus d – spring wire diameter L – free length of spring n – number of active windings mu = poisson ratio D – spring outer diameter we usually work with idealized spring under elastic limit and of negligible mass. see for details- Related questions How do you find the effective spring constant? How do you find the spring constant (K) from the force (F) and displacement (x)? How big and small can a spring constant get? How do you find the spring constant given the length and forces of the other 2 springs? A spring is stretched 0.50 m and the force was 30,000 N. What is the spring constant? Peter Upton BA in Physics & Mathematics, The Open University · Author has 14.4K answers and 10.7M answer views · 1y Related How do I find the centripetal force without mass? Centripetal force is not a force like gravity, magnetic forces, electrical forces , friction etc. Any of these named types of force can behave as a centripetal force. Centripetal describes the way in which another existing force is acting. It means this other force always points to a fixed point ( the centre of a circle.) So to find the centripetal force you have two options really. Look at the motion of the object subject to the centripetal force. You will be looking at circular motion formulae and these will have the mass of the object in them - like F=mv^2/r Look at the cause of the force. Lets Centripetal force is not a force like gravity, magnetic forces, electrical forces , friction etc. Any of these named types of force can behave as a centripetal force. Centripetal describes the way in which another existing force is acting. It means this other force always points to a fixed point ( the centre of a circle.) So to find the centripetal force you have two options really. Look at the motion of the object subject to the centripetal force. You will be looking at circular motion formulae and these will have the mass of the object in them - like F=mv^2/r Look at the cause of the force. Lets take an easy example - something is connected to a spring and swung in a circle in the horizontal plane. You see the spring has stretched by a certain amount (‘x). You know the stiffness of he spring ( spring constant =k) , so you lnow the size of the force F=kx. This is the size of the type of force ( which happens to be acting as a centripetal force). The spring stretched by the same amount would still exert the same sized force even if it did not always point to the centre of a circle. You could use this second method with many types of force, eg if you had two charged particles ( electron and nucleus - Bohr model of H atom) then if you know the size of the two charges and the distance between them - you can calculate the size of the force. This force just happens to supply the centripetal force that keeps the electron in orbit around the nucleus. Sponsored by OrderlyMeds Is Your GLP-1 Personalized? Find GLP-1 plans tailored to your unique body needs. Geoff Mangum Self-study 1973-2019. · Author has 3.8K answers and 3.1M answer views · 1y Related How do I find the centripetal force without mass? Centripetal force increases with mass, if radius and tangential velocity are the same. F(c) = m•v^2/r. The “force” is really whatever keeps the mass at radial distance as the mass swings around, with tangential velocity v — like a string. If the string is released or cut, the mass flies in the tangential direction at that moment dead straight at v. The “force” changes the tangential direction but not v, according to the radius and r. F = m•a a = F/m The string “accelerates” the mass by a towards the center of the circular swinging. a(centripetal) = v^2/r F(c)/m = v^/r F(c) = m•v^2/r You ask how to find Centripetal force increases with mass, if radius and tangential velocity are the same. F(c) = m•v^2/r. The “force” is really whatever keeps the mass at radial distance as the mass swings around, with tangential velocity v — like a string. If the string is released or cut, the mass flies in the tangential direction at that moment dead straight at v. The “force” changes the tangential direction but not v, according to the radius and r. F = m•a a = F/m The string “accelerates” the mass by a towards the center of the circular swinging. a(centripetal) = v^2/r F(c)/m = v^/r F(c) = m•v^2/r You ask how to find F(c) without knowing m. You can’t. F(c) requires 3 variables — m, v, r. This problem illustrates the relationship between angular and tangential velocity. Angular v (ω = Δθ / Δt or = Δrad / Δt) is the SAME at any r. But tangential velocity v(tang.) = r•ω. — not the same at different r. If a radial velocity is 1 rad/s, and the circular motion has radius = 1 m, the radial velocity of any m any r out to 1 m = the same, 1rad/s. But the tangential velocity at 1 rad/s cover 1 rad of circumference when the mass is at r = 1 m. That arc length = 1/(2•π) • circumference. The circumference is 2•r•π = 6.28 m. So the tangential velocity (linear velocity) at r = 1 is 6.28 m / 1 s = 6.28 m/s. But at r = 0.5 m, the 1-rad arc length is 2•π•0.5 = 3.14 m/s (linear velocity half as fast). The centripetal force that holds mass X swinging in a circle (string tension) is TWICE as great at r = 1 m compared to the force to hold the same mass at the same angular velocity at r = 0.5 m. The LINEAR velocity is half, so the momentum of the mass at r = 0.5 m is also half. Force “changes” momentum into the circular motion, and it takes half the force when the linear momentum is half. The fact that CHANGING the radius of mass X while keeping the angular velocity the same (1 rad/s) changes the angular velocity is because energy is conserved. Shortening the radius by half doubles the angular velocity. The skater spins fasters by tucking the arms in, which moves the mass closer in. Given the above, you need mass to calculate the centripetal force. The mass at radial velocity plus radial length determine the “acceleration” needed to change tangential to circular, and the Force then needs the mass Saleh Al-Quraishi Assistant Prof. at KFUPM, King Fahd University of Petroleum and Minerals (1981–present) · 7y Related Does spring constant depend on length? Yes it depends on length. Let us imagine that we have two identical springs (with spring constant k) of length l. If we hang a mass m on either of them, they will has the same elongation x. Applying Newton’s second law we will get; mg = kx. Now, connecting the two springs in series we will have a spring of length 2l. Let us assume its spring constant is k’. By hanging the same mass to the long spring the spring will extend by a distance 2x (each original spring will be stretched by a distance x, assuming massless springs). Applying Newton’s law we get ; mg = k’(2x). comparing the two equations Yes it depends on length. Let us imagine that we have two identical springs (with spring constant k) of length l. If we hang a mass m on either of them, they will has the same elongation x. Applying Newton’s second law we will get; mg = kx. Now, connecting the two springs in series we will have a spring of length 2l. Let us assume its spring constant is k’. By hanging the same mass to the long spring the spring will extend by a distance 2x (each original spring will be stretched by a distance x, assuming massless springs). Applying Newton’s law we get ; mg = k’(2x). comparing the two equations we conclude that k’ = ½ k. Sponsored by CDW Corporation How can Autodesk and CDW improve your design process? AutoCAD enhances 2D/3D design with automation and feedback integration, all on a collaborative platform. Aaron Chao (赵明佳) studied AP Physics in grade 12 · Author has 699 answers and 1.4M answer views · Updated 3y Related A spring has a spring constant of 20 N/m and is extended to 4m. What is the force required to extend the spring? Hooke’s law states that the force required to stretch a spring is proportional to the change in the spring’s length Formula: F=k∗Δx F is the force in N k is the spring constant in N/m Δx is the change spring length in m Here, we have that the spring constant is 20 N/m and the length changes by 4 m k=20, Δx=4 Find force required, F, by using F=k∗Δx F=20∗4 F=80 It takes exactly 80 newtons of force to extend the string by 4 m with a spring constant is 20 N/m Hooke’s law states that the force required to stretch a spring is proportional to the change in the spring’s length Formula: F=k∗Δx F is the force in N k is the spring constant in N/m Δx is the change spring length in m Here, we have that the spring constant is 20 N/m and the length changes by 4 m k=20, Δx=4 Find force required, F, by using F=k∗Δx F=20∗4 F=80 It takes exactly 80 newtons of force to extend the string by 4 m with a spring constant is 20 N/m Brad Moffat Author has 4K answers and 11.5M answer views · Updated 4y Related What is meant by the spring constant of a spring? When you pull a spring, it pulls back. The force with which it pulls depends on the material, thickness, angles, all kinds of complicated stuff. But… once the spring is built it will give a predictable force based solely on how far it’s stretched. Say I have a spring that exerts a force of 5 newtons when I stretch it 1 cm. Immediately I know that it will pull 10 newtons if I pull it 2 cm. It will pull 15 N if I pull it 3 cm, and so on . . . That 5 N/cm is called the spring constant. Well, actually, spring constant is usually expressed in terms of meters, but no one every tries to stretch it a full me When you pull a spring, it pulls back. The force with which it pulls depends on the material, thickness, angles, all kinds of complicated stuff. But… once the spring is built it will give a predictable force based solely on how far it’s stretched. Say I have a spring that exerts a force of 5 newtons when I stretch it 1 cm. Immediately I know that it will pull 10 newtons if I pull it 2 cm. It will pull 15 N if I pull it 3 cm, and so on . . . That 5 N/cm is called the spring constant. Well, actually, spring constant is usually expressed in terms of meters, but no one every tries to stretch it a full meter. This particular example the spring would have a spring constant, k, listed as 500N/m in the catalogue. If I had a stronger spring it would pull harder if I pulled it 1 cm. Say it pulls 7 N when I pull it 1 cm. Its spring constant is 700 N/m. The spring constant allows designers to predict how far a certain spring will need to stretch to produce the force they want. F=−k⋅Δx Where k is the “spring constant” and Δx is how far it is stretched. Be careful here, it is not the length of the spring, it is just the distance stretched. That’s what the Δ symbol usually indicates, “change in.” (BTW, the little negative sign is there only to indicate the direction of the force. It will always be opposite the direction of the stretch, Δx. “-” means opposite direction in vector math. If I push the spring in, it will want to push back out. If I pull it out, it will pull back in.) Gary Novosielski retired Physics teacher · Author has 17.6K answers and 11.3M answer views · 3y Related What force would be needed to extend a spring with a spring constant k=5, 0.3 m? You did not supply the units for the spring constant. If you do, the math will instantly become obvious. For example if that k is actually 5 N/m, then multiplying 5 by .3 will give you a numerical value 1.5 and multiplying N/m times m will give you N. But if you don’t know the units of k, you can’t know the units of the force. It will be 1.5, but 1.5 what? Pounds, newtons, yoctodynes? Who knows? Peter Upton BA in Physics & Mathematics, The Open University · Author has 14.4K answers and 10.7M answer views · 2y Related Can 'spring constant' be negative? I would say not. The Force is a vector and so is the displacement/ extension. The directions are included in these quantities. This means the spring constant will always be positive. I suppose an exception would be a material/object which when deformed produces a force which deforms it more. Such an object would be in unstable equilibrium and so would very rapidly exist in the more stable state of being deformed. So you would never find an object in the undeformed state, even if such an object existed. A pencil balanced on its point is a good example of unstable equilibrium. You don’t often see p I would say not. The Force is a vector and so is the displacement/ extension. The directions are included in these quantities. This means the spring constant will always be positive. I suppose an exception would be a material/object which when deformed produces a force which deforms it more. Such an object would be in unstable equilibrium and so would very rapidly exist in the more stable state of being deformed. So you would never find an object in the undeformed state, even if such an object existed. A pencil balanced on its point is a good example of unstable equilibrium. You don’t often see pencils balanced on their points! Related questions How can you determine the value of a spring constant? What is the true standard value of spring constant? What happens if the spring constant increases? What is the force constant of a spring? In physics, how do you determine the spring constant? How do you find the effective spring constant? How do you find the spring constant (K) from the force (F) and displacement (x)? How big and small can a spring constant get? How do you find the spring constant given the length and forces of the other 2 springs? A spring is stretched 0.50 m and the force was 30,000 N. What is the spring constant? Where can a spring constant be found? Does every spring has its own spring constant? What does a greater spring constant mean? What should be the value of the spring constant? If a spring is cut in half, what happens to its spring constant? 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https://math.stackexchange.com/questions/1871125/when-is-a-given-polynomial-a-square-of-another-polynomial
Skip to main content When is a given polynomial a square of another polynomial? Ask Question Asked Modified 9 years ago Viewed 435 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I meet a problem in which I hope to show a special polynomial is not a square of another polynomial. More precisely, let's consider the polynomial f(x):=1−x+2bxn−2bxn+1−b2x2n−1+2b2x2n−b2x2n+1−2bx3n−1+2bx3n−x4n−1+x4n∈k[x], where k is a field of characteristic p>0, n>2 is an integer, and b∈k with b≠0,1,−1. Indeed, in the context I meet, the field k is just the finite field with p elements, n=(pm+1)/2 with m≥1. I just try to rephrase the question in a clean and a more general way, but you may use the further assumption in your proof. I have checked for a lot of examples and I found it is always not a square of another polynomial. I believe it is true in general but I fail to give a proof for it. Please be free to give me any kind of suggestions on this problem. Although I know how to check if a concrete polynomial is a square, I have no ideas about how to systematically determine whether a general polynomial is a square or not? If you know any theory related to it, please don't hesitate to tell me. Thanks a lot for your time! abstract-algebra polynomials finite-fields Share CC BY-SA 3.0 Follow this question to receive notifications edited Jul 26, 2016 at 5:25 user26857 53.3k1414 gold badges7676 silver badges163163 bronze badges asked Jul 25, 2016 at 23:10 Joy-JoyJoy-Joy 84155 silver badges77 bronze badges 5 One way is to show that it takes on a value that is not a square of another value (i.e., suppose that f(x)=g(x)2 for every x. Then for some value a, you must have f(a)=g(a)2. But if f(a) is, for instance, a prime, then this cannot be true. – John Hughes Commented Jul 25, 2016 at 23:45 3 You are over a field. If your polynomial is a square, take its formal derivative, then the gcd of that and the original polynomial. You should be able to find reading on use of the formal derivative for this purpose. – Will Jagy Commented Jul 26, 2016 at 0:08 Examining the first and last terms, if f(x)=g(x)2, then g(x) must look like x2n−12x2n−1+⋯−12x+1 (not implying any pattern in the dots, but those must be the first and last terms). Now expand out g(x)2 using this form, and look at the coefficients of x2 and x4n−2. What does that tell you about n and b? – Greg Martin Commented Jul 26, 2016 at 0:13 1 There has been much work on polynomials whose squares (and/or powers) are "sparse", i.e. have fewer terms than the original polynomial. The results may well apply to your case. To find papers search on keywords: sparse polynomial (square OR power) Davenport, Schinzel. – Bill Dubuque Commented Jul 26, 2016 at 0:17 1 Simul-posted to MO, mathoverflow.net/questions/245085/… – without notification to either site. Please don't do that! – Gerry Myerson Commented Jul 26, 2016 at 7:28 Add a comment | 1 Answer 1 Reset to default This answer is useful 6 Save this answer. Show activity on this post. Apparently p>2. An argument specific to this polynomial relies on the following observations: x=1 is a zero with multiplicity exactly two. x=−1 is not a zero of this polynomial. f(x) is palindromic. In other words f(x)=x4nf(1/x). The claim then follows from these observations. Assume contrariwise that f(x) would be the square of another polynomial. This means that any zero γ of f(x) in some extension field K/k must have even multiplicity 2r. By item 3, unless γ=1/γ, between them γ±1 then account for 4r zeros counted with multiplicity. But γ=1/γ iff γ=±1, and items 1 and 2 show that these contribute exactly 2 zeros counted with multiplicity. Thus the number of zeros (with multiplicity) is ≡2(mod4) contradicting the fact that the degree of f(x) is a multiple of four. So we need to check those observations. It is straightforward to check that f(1)=f′(1)=0 so x=1 is a zero multiplicity ≥2. Let's denote Pk(x)=1+x+x2+⋯+xk. As Pk(x)(1−x)2=(1−xk+1)(1−x) we see that g(x):=f(x)(x−1)2=P4n−2(x)+2bxnP2n−2(x)−b2x2n−1. Therefore g(1)=4n−1+(2n−1)2b−b2=1−b2≠0, because 2n−1 is a multiple of p and b≠±1. 2. We also see that f(−1)=2(2±4b+2b2)=4(b±1)2 where the sign depends on the parity of n. 3. This is clear by observation. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jul 26, 2016 at 8:01 answered Jul 26, 2016 at 6:18 Jyrki LahtonenJyrki Lahtonen 142k3030 gold badges304304 silver badges738738 bronze badges 5 Initially I fumbled this by miscalculating that x=1 would be a simple zero. Proceed with caution. – Jyrki Lahtonen Commented Jul 26, 2016 at 8:51 It is so beautiful and it seems that you use all the assumption in such a clever way. To be honest, I noted all the observation you mentioned as above, but I failed to notice the multiplicities of its roots in the algebraic closure. Thank you so much for this clever idea. – Joy-Joy Commented Jul 26, 2016 at 15:20 I focused to much on the symmetry of this polynomial. As you may noted, if otherwise f(x) is a square of a polynomial g(x), then the coefficients of g(x) must be determined by the coefficients of terms xi in f(x) for 0≤i≤n. That is to say, only one choice of the coefficients for other terms in f(x) can make it a square. That seems impossible especially when n is big. I tried to get a contradiction from the line. But I didn't succeed even if I did some computation in this direction. – Joy-Joy Commented Jul 26, 2016 at 15:26 This argument (if/when it checks out) is so ad hoc that I'm inclined to think the root cause is hidden in the way the polynomial was constructed. Can you shed some light to that? @Joy-Joy – Jyrki Lahtonen Commented Jul 26, 2016 at 15:30 As for your curiosity about the origin of this problem, I begin with some problem on permutation polynomial. Then I meet a lot of problem about the irreducibility of polynomial over finite fields. In one quite special case I get this special polynomial after a few steps of reduction. Personally, I think this kind of polynomials are quite interesting for their own reason. As you know, I can not write all the details here. But if you are interesting, we may discuss by email in more details. Anyway, thanks again for your clever proof. – Joy-Joy Commented Jul 26, 2016 at 15:34 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 abstract-algebra polynomials finite-fields See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... 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https://www.youtube.com/watch?v=mgTyTQJBsig
Partial Sums and Telescoping Series William Lindsey 1270 subscribers 37 likes Description 3317 views Posted: 28 Feb 2017 This video is about finding the value of a series by using the limit of the partial sums. This particular series is telescoping, so it is possible to find a closed form for the partial sums. 4 comments Transcript: [Music] We're going to look at the series um when n goes from 1 to infinity of 1 / n n +1. And we would like to determine what it converges to. And that it converges. So if I write out the first couple of terms, my first term is going to be n= 1, which will be 1 over 1 2. n= 2 is going to give me 1 over 2 3. And the next one will be 1 over 3 4. and so on from here. So, I'd like to determine whether or not this converges. In order to tell if a series converges, we look at the partial sums, adding the first few terms together. So, my first partial sum is a half, just my first term. My second one is a half plus a 6th, which is my second term, plus my first term. My third partial sum will be a half plus a 6th plus a 12th. And I create a sequence of partial sums this way. If the limit of these partial sums exists, we call that the sum of the series. So generally it is not possible to tell what the limit of the partial sums is unless you can get a nice closed formula for each partial sum. A half plus a 6 is 2/3. A half plus a 6 plus a 12th will be 2/3 plus a 12th, which is going to be 3/4. If we write down the next one, S4 is going to be 3/4. That's S3 plus the next term, which will be 1 over 4 5. So 3/4 + 1 over 20 is going to be 16 20ths, which is the same as 4 fths. And so we can see a nice pattern happening here. Our nth partial sum, our s subn is going to be maybe n / n +1. In order to actually prove this, we would have to do something additional, something like induction or something a little bit easier to look at. What we're going to do instead is we're going to approach this a slightly different way. And we're going to use the idea of a telescoping series. 1 / n n plus1 looks like something that when we integrate we would use partial fractions for. So we can do the same thing here. And if the partial fractions work out nicely, our partial sums are going to be much easier to be to work with. So 1 / n n +1 should look like a n plus b / n +1. If you multiply through by your denominator, what you end up with is 1 is equal to a n + 1 plus b n. So this is equal to 1. And I'm not going to show you the steps here, but if you go through the partial fraction steps, you get your values for a and b. And you're going to end up with a is equal to 1 and b is equal to negative 1. Okay, so if I rewrite the series this way, I'm going to end up with the sum from 1 to infinity of 1 / n minus 1 / n + 1. And when a series turns out something like this, this is called telescoping because as you'll see, the partial sums have a very nice property. S1 is 1 - a half, which is just a half. But then S2 is going to be what we have here, 1 - a half plus a half minus a3. And those middle terms, the negative a half plus a half, cancel, leaving us with just 1 minus a third. S3 is going to be S2, 1 - a third, plus the next term, which is a 3 minus a 4th. So my middle terms cancel again, and I'm ending up with 1 minus a quarter. And so when it's like this, we don't have to guess at a at a form for Sn. We can see that it's going to end up as 1 - 1 / n + 1. And so if I want to know the series itself from n= 1 to infinity of 1 / n n + 1, this is the same as the limit of the partial sums. And either way that you write the partial sums, what we see is that this limit is going to be the limit of 1 / 1 - or 1 - 1 / n + 1, which is just going to be 1. So we would say that the sum from 1 to infinity of 1 / n n + 1 is equal to 1.
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https://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=8a892dbfd7e6ec03c72cd87d810327be&t
Wolfram|Alpha Widget: Truth Table Generator Truth Table Generator Enter the Statement: [Use AND, OR, NOT, XOR, NAND, NOR, and XNOR, IMPLIES and parentheses] Generate Truth Table Computing... Get this widget Build your own widget»Browse widget gallery»Learn more»Report a problem»Powered by Wolfram|Alpha Terms of use Share a link to this widget: More Embed this widget»
5091
https://openstax.org/books/algebra-1/pages/9-3-3-solve-by-completing-the-square
9.3.3 Solve by Completing the Square - Algebra 1 | OpenStax This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising purposes. Privacy Notice Customize Reject All Accept All Customize Consent Preferences We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below. The cookies that are categorized as "Necessary" are stored on your browser as they are essential for enabling the basic functionalities of the site. ...Show more For more information on how Google's third-party cookies operate and handle your data, see:Google Privacy Policy Necessary Always Active Necessary cookies are required to enable the basic features of this site, such as providing secure log-in or adjusting your consent preferences. These cookies do not store any personally identifiable data. 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Reject All Save My Preferences Accept All Loading [MathJax]/jax/input/MathML/entities/n.js Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Algebra 1 9.3.3 Solve by Completing the Square Algebra 19.3.3 Solve by Completing the Square Contents Contents Highlights Table of contents Getting Started Supporting All Learners 1 Linear Equations 2 Linear Inequalities and Systems 3 Two-Variable Statistics 4 Functions 5 Introduction to Exponential Functions 6 Working with Polynomials 7 Introduction to Quadratic Functions 8 Quadratic Equations 9 More Quadratic Equations Unit 9 Overview and Readiness 9.1 What Are Perfect Squares? 9.2 Completing the Square, Part 1 9.3 Completing the Square, Part 2 9.3.0 Lesson Overview 9.3.1 Solving Equations with Fractions 9.3.2 Using Completing the Square to Solve Equations 9.3.3 Solve by Completing the Square 9.3.4 Completing the Square with Fractions 9.3.5 Practice 9.3.6 Lesson Summary 9.4 Completing the Square, Part 3 9.5 Quadratic Equations with Irrational Solutions 9.6 The Quadratic Formula 9.7 Applying the Quadratic Formula 9.8 Deriving the Quadratic Formula 9.9 Writing Quadratics in Different Forms 9.10 Rewriting Quadratic Expressions in Vertex Form 9.11 Using Quadratic Expressions in Vertex Form to Solve Problems Project 9: Using Quadratic Equations to Model Situations and Solve Problems Unit 9 Wrap Up Research in Practice Appendix Index Search for key terms or text. Close 9.3.3 • Solve by Completing the Square Activity Here are four equations followed by worked solutions of the equations. Each solution has at least one error. Solve the equation x 2+14 x=−24 x 2+14 x=−24 x 2+14 x=−24. Then look at the following worked solution (with errors) and find and describe the error or errors in the worked solution. x 2+14 x x 2+14 x+28(x+7)2 x+7 x+7 x=−24=4=4=2=−2=−5 and x=9 Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 x 2+14 x=−24 Step 1 x 2+14 x+28=4 Step 2(x+7)2=4 Step 3 x+7=2 Step 4 x+7=−2 Step 5 x=−5 and x=9 Step 6 x 2+14 x=−24 Step 1 x 2+14 x+28=4 Step 2(x+7)2=4 Step 3 x+7=2 Step 4 x+7=−2 Step 5 x=−5 and x=9 Step 6 Compare your answer: x=−2 x=−2 x=−2 and x=−12 x=−12 x=−12 Here are four equations followed by worked solutions of the equations. Each solution has at least one error. a. Where was the error made in the worked solution above and what was the error? Compare your answer: Error: In Step 2, the number being added, 28, is not a perfect square. It should be 49. Solve the equation x 2−10 x+16=0 x 2−10 x+16=0 x 2−10 x+16=0. Then look at the following worked solution (with errors) and find and describe the error or errors in the worked solution. x 2−10 x+16 x 2−10 x+25(x−5)2 x−5 x−5 x=14=0=9=9=9=−9 and x=−4 Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 x 2−10 x+16=0 Step 1 x 2−10 x+25=9 Step 2(x−5)2=9 Step 3 x−5=9 Step 4 x−5=−9 Step 5 x=14 and x=−4 Step 6 x 2−10 x+16=0 Step 1 x 2−10 x+25=9 Step 2(x−5)2=9 Step 3 x−5=9 Step 4 x−5=−9 Step 5 x=14 and x=−4 Step 6 Compare your answer: x=8 x=8 x=8 and x=2 x=2 x=2 a. Where was the error made in the worked solution above and what was the error? Compare your answer: Errors: Step 4 should have 3 on the right side, not 9. Step 5 should have -3 on the right side, not -9. Solve the equation x 2+2.4 x=−0.8 x 2+2.4 x=−0.8 x 2+2.4 x=−0.8. Then look at the following worked solution (with errors) and find and describe the error or errors in the worked solution. x 2+2.4 x=−0.8 Step 1 x 2+2.4 x+1.44=0.64 Step 2(x+1.2)2=0.64 Step 3 x+1.2=0.8 Step 4 x 2+2.4 x=−0.8 Step 1 x 2+2.4 x+1.44=0.64 Step 2(x+1.2)2=0.64 Step 3 x+1.2=0.8 Step 4 x 2+2.4 x=−0.8 Step 1 x 2+2.4 x+1.44=0.64 Step 2(x+1.2)2=0.64 Step 3 x+1.2=0.8 Step 4 Compare your answer: x=−0.4 x=−0.4 x=−0.4 and x=−2 x=−2 x=−2 a. Where was the error made in the worked solution above and what was the error? Compare your answer: Error: In Step 4, only the positive square root is written and the negative square root is neglected. Solve the equation x 2−6 5 x+1 5=0 x 2−6 5 x+1 5=0 x 2−6 5 x+1 5=0. Then look at the following worked solution (with errors) and find and describe the error or errors in the worked solution. x 2−6 5 x+1 5 x 2−6 5 x+9 25(x−3 5)2 x−3 5 x−3 5 x=6 5=0=9 25=9 25=3 5=−3 5 and x=0 Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 x 2−6 5 x+1 5=0 Step 1 x 2−6 5 x+9 25=9 25 Step 2(x−3 5)2=9 25 Step 3 x−3 5=3 5 Step 4 x−3 5=−3 5 Step 5 x=6 5 and x=0 Step 6 x 2−6 5 x+1 5=0 Step 1 x 2−6 5 x+9 25=9 25 Step 2(x−3 5)2=9 25 Step 3 x−3 5=3 5 Step 4 x−3 5=−3 5 Step 5 x=6 5 and x=0 Step 6 Compare your answer: x=1 x=1 x=1 and x=1 5 x=1 5 x=1 5 a. Where was the error made in the worked solution above and what was the error? Compare your answer: Errors: For example: In Step 2, different amounts are added to each side. On the left, 4 25 4 25 4 25 is added. On the right, 9 25 9 25 9 25 is added. Or: While 9 25 9 25 9 25 is added on the right side, 1 5 1 5 1 5 was only subtracted from the left side. Other explanations may also be accurate. Self Check Examine the work shown. Which of the following is the error that took place when the equation was solved by completing the square? x 2−6 x−7=0 x 2−6 x−7=0 x 2−6 x=7 x 2−6 x=7 x 2−6 x+9=16 x 2−6 x+9=16 (x−3)2=16(x−3)2=16 x−3=4 x−3=4 x=7 x=7 This equation cannot be solved by completing the square. When completing the square, the wrong number was found for (b 2)2(b 2)2. When taking the square root, only the positive solution was included. The 7 should have been subtracted from each side. Additional Resources Spotting the Error When Solving Quadratics To find the error when solving quadratics, it is helpful to first solve the problem on your own. Solve the quadratic equation by completing the square and finding its zeros. Identify the error in the worked solution below. x 2+10 x−34=0 x 2+10 x=34 x 2+10 x+25=34−25(x+5)2=9 x+5=±3 x=−8,x=−2 Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 x 2+10 x−34=0 Step 1 x 2+10 x=34 Step 2 x 2+10 x+25=34−25 Step 3(x+5)2=9 Step 4 x+5=±3 Step 5 x=−8,x=−2 Step 6 x 2+10 x−34=0 Step 1 x 2+10 x=34 Step 2 x 2+10 x+25=34−25 Step 3(x+5)2=9 Step 4 x+5=±3 Step 5 x=−8,x=−2 Step 6 Now, find the error by working through the problem and comparing steps: x 2+10 x−34=0 x 2+10 x=34 x 2+10 x+25=34+25 Step 1 Step 2 Step 3 x 2+10 x−34=0 Step 1 x 2+10 x=34 Step 2 x 2+10 x+25=34+25 Step 3 x 2+10 x−34=0 Step 1 x 2+10 x=34 Step 2 x 2+10 x+25=34+25 Step 3 In Step 3, 25 was added to the left side of the equation, but 25 was SUBTRACTED from the right side of the equation. Equations must remain balanced, and 25 should be added to each side of the equation. Try it Spotting the Error When Solving Quadratics Examine the steps below and identify the error in solving the quadratic equation by completing the square: x 2+8 x+7=0 x 2+8 x=−7 x 2+8 x+16=−7+16(x+4)2=9 x+4=3 x=−1 Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 x 2+8 x+7=0 Step 1 x 2+8 x=−7 Step 2 x 2+8 x+16=−7+16 Step 3(x+4)2=9 Step 4 x+4=3 Step 5 x=−1 Step 6 x 2+8 x+7=0 Step 1 x 2+8 x=−7 Step 2 x 2+8 x+16=−7+16 Step 3(x+4)2=9 Step 4 x+4=3 Step 5 x=−1 Step 6 Here is how to find the error when completing the square: Work through the steps on your own and compare: x 2+8 x+7=0 x 2+8 x=−7 x 2+8 x+16=−7+16(x+4)2=9 x+4=±3 x=−1,x=−7 Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 x 2+8 x+7=0 Step 1 x 2+8 x=−7 Step 2 x 2+8 x+16=−7+16 Step 3(x+4)2=9 Step 4 x+4=±3 Step 5 x=−1,x=−7 Step 6 x 2+8 x+7=0 Step 1 x 2+8 x=−7 Step 2 x 2+8 x+16=−7+16 Step 3(x+4)2=9 Step 4 x+4=±3 Step 5 x=−1,x=−7 Step 6 The error occurred in Step 5 when the plus-minus sign was not included after taking the square root of each side. This led to the second possible equation not being written. PreviousNext Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis, Jay Abramson, Sharon North, Amy Baldwin, Alyssa Howell Publisher/website: OpenStax Book title: Algebra 1 Publication date: Jun 4, 2025 Location: Houston, Texas Book URL: Section URL: © Aug 26, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. Our mission is to improve educational access and learning for everyone. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. Give today and help us reach more students. 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https://www.wyzant.com/resources/answers/241513/solve_inequality_log2_4_3x_x_2_log2_2x_1_gt_1
WYZANT TUTORING Thomas L. solve inequality log2 (4+3x-x^2)-log2 (2x-1)>1 1 Expert Answer Michael J. answered • 09/18/16 Effective High School STEM Tutor & CUNY Math Peer Leader Thomas L. 09/18/16 Thomas L. 09/18/16 Michael J. 09/18/16 Venrand J. 07/23/21 Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. RELATED TOPICS RELATED QUESTIONS how do you change a^2=1.6 into a logarithm? Answers · 2 How to find the exact solution to a logarithmic equation Answers · 1 log MN=log + log Answers · 6 solve X, x^-2=9 Answers · 1 Solving Log Equation Answers · 2 RECOMMENDED TUTORS Scott S. Dayaan M. Evan W. find an online tutor Download our free app A link to the app was sent to your phone. Get to know us Learn with us Work with us Download our free app Let’s keep in touch Need more help? Learn more about how it works Tutors by Subject Tutors by Location IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Education.com 35,000 worksheets, games, and lesson plans TPT Marketplace for millions of educator-created resources Vocabulary.com Adaptive learning for English vocabulary ABCya Fun educational games for kids SpanishDictionary.com Spanish-English dictionary, translator, and learning Inglés.com Diccionario inglés-español, traductor y sitio de aprendizaje Emmersion Fast and accurate language certification
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https://en.wikipedia.org/wiki/Entropy_(order_and_disorder)
Entropy (order and disorder) - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 History 2 Overview 3 Phase change 4 Entropy-driven order 5 Adiabatic demagnetization 6 Difficulties with the term "disorder" 7 See also 8 References 9 External links Entropy (order and disorder) [x] Add languages Add 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 Add interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Interpretation of entropy as the change in arrangement of a system's particles Boltzmann's molecules (1896) shown at a "rest position" in a solid In thermodynamics, entropy is often associated with the amount of order or disorder in a thermodynamic system. This stems from Rudolf Clausius' 1862 assertion that any thermodynamic process always "admits to being reduced [reduction] to the alteration in some way or another of the arrangement of the constituent parts of the working body" and that internal work associated with these alterations is quantified energetically by a measure of "entropy" change, according to the following differential expression: ∫δ Q T≥0{\displaystyle \int !{\frac {\delta Q}{T}}\geq 0} where Q = motional energy ("heat") that is transferred reversibly to the system from the surroundings and T = the absolute temperature at which the transfer occurs. In the years to follow, Ludwig Boltzmann translated these 'alterations of arrangement' into a probabilistic view of order and disorder in gas-phase molecular systems. In the context of entropy, "perfect internal disorder" has often been regarded as describing thermodynamic equilibrium, but since the thermodynamic concept is so far from everyday thinking, the use of the term in physics and chemistry has caused much confusion and misunderstanding. In recent years, to interpret the concept of entropy, by further describing the 'alterations of arrangement', there has been a shift away from the words 'order' and 'disorder', to words such as 'spread' and 'dispersal'. History [edit] This "molecular ordering" entropy perspective traces its origins to molecular movement interpretations developed by Rudolf Clausius in the 1850s, particularly with his 1862 visual conception of molecular disgregation. Similarly, in 1859, after reading a paper on the diffusion of molecules by Clausius, Scottish physicist James Clerk Maxwell formulated the Maxwell distribution of molecular velocities, which gave the proportion of molecules having a certain velocity in a specific range. This was the first-ever statistical law in physics. In 1864, Ludwig Boltzmann, a young student in Vienna, came across Maxwell's paper and was so inspired by it that he spent much of his long and distinguished life developing the subject further. Later, Boltzmann, in efforts to develop a kinetic theory for the behavior of a gas, applied the laws of probability to Maxwell's and Clausius' molecular interpretation of entropy so as to begin to interpret entropy in terms of order and disorder. Similarly, in 1882 Hermann von Helmholtz used the word "Unordnung" (disorder) to describe entropy. Overview [edit] To highlight the fact that order and disorder are commonly understood to be measured in terms of entropy, below are current science encyclopedia and science dictionary definitions of entropy: A measure of the unavailability of a system's energy to do work; also a measure of disorder; the higher the entropy the greater the disorder. A measure of disorder; the higher the entropy the greater the disorder. In thermodynamics, a parameter representing the state of disorder of a system at the atomic, ionic, or molecular level; the greater the disorder the higher the entropy. A measure of disorder in the universe or of the unavailability of the energy in a system to do work. Entropy and disorder also have associations with equilibrium. Technically, entropy, from this perspective, is defined as a thermodynamic property which serves as a measure of how close a system is to equilibrium—that is, to perfect internal disorder. Likewise, the value of the entropy of a distribution of atoms and molecules in a thermodynamic system is a measure of the disorder in the arrangements of its particles. In a stretched out piece of rubber, for example, the arrangement of the molecules of its structure has an "ordered" distribution and has zero entropy, while the "disordered" kinky distribution of the atoms and molecules in the rubber in the non-stretched state has positive entropy. Similarly, in a gas, the order is perfect and the measure of entropy of the system has its lowest value when all the molecules are in one place, whereas when more points are occupied the gas is all the more disorderly and the measure of the entropy of the system has its largest value. In systems ecology, as another example, the entropy of a collection of items comprising a system is defined as a measure of their disorder or equivalently the relative likelihood of the instantaneous configuration of the items. Moreover, according to theoretical ecologist and chemical engineer Robert Ulanowicz, "that entropy might provide a quantification of the heretofore subjective notion of disorder has spawned innumerable scientific and philosophical narratives." In particular, many biologists have taken to speaking in terms of the entropy of an organism, or about its antonym negentropy, as a measure of the structural order within an organism. The mathematical basis with respect to the association entropy has with order and disorder began, essentially, with the famous Boltzmann formula, S=k B ln⁡W{\displaystyle S=k_{\mathrm {B} }\ln W!}, which relates entropy S to the number of possible states W in which a system can be found. As an example, consider a box that is divided into two sections. What is the probability that a certain number, or all of the particles, will be found in one section versus the other when the particles are randomly allocated to different places within the box? If you only have one particle, then that system of one particle can subsist in two states, one side of the box versus the other. If you have more than one particle, or define states as being further locational subdivisions of the box, the entropy is larger because the number of states is greater. The relationship between entropy, order, and disorder in the Boltzmann equation is so clear among physicists that according to the views of thermodynamic ecologists Sven Jorgensen and Yuri Svirezhev, "it is obvious that entropy is a measure of order or, most likely, disorder in the system." In this direction, the second law of thermodynamics, as famously enunciated by Rudolf Clausius in 1865, states that: The entropy of the universe tends to a maximum. Thus, if entropy is associated with disorder and if the entropy of the universe is headed towards maximal entropy, then many are often puzzled as to the nature of the "ordering" process and operation of evolution in relation to Clausius' most famous version of the second law, which states that the universe is headed towards maximal "disorder". In the recent 2003 book SYNC – the Emerging Science of Spontaneous Order by Steven Strogatz, for example, we find "Scientists have often been baffled by the existence of spontaneous order in the universe. The laws of thermodynamics seem to dictate the opposite, that nature should inexorably degenerate toward a state of greater disorder, greater entropy. Yet all around us we see magnificent structures—galaxies, cells, ecosystems, human beings—that have all somehow managed to assemble themselves." The common argument used to explain this is that, locally, entropy can be lowered by external action, e.g. solar heating action, and that this applies to machines, such as a refrigerator, where the entropy in the cold chamber is being reduced, to growing crystals, and to living organisms. This local increase in order is, however, only possible at the expense of an entropy increase in the surroundings; here more disorder must be created. The conditioner of this statement suffices that living systems are open systems in which both heat, mass, and or work may transfer into or out of the system. Unlike temperature, the putative entropy of a living system would drastically change if the organism were thermodynamically isolated. If an organism was in this type of "isolated" situation, its entropy would increase markedly as the once-living components of the organism decayed to an unrecognizable mass. Phase change [edit] Owing to these early developments, the typical example of entropy change Δ S is that associated with phase change. In solids, for example, which are typically ordered on the molecular scale, usually have smaller entropy than liquids, and liquids have smaller entropy than gases and colder gases have smaller entropy than hotter gases. Moreover, according to the third law of thermodynamics, at absolute zero temperature, crystalline structures are approximated to have perfect "order" and zero entropy. This correlation occurs because the numbers of different microscopic quantum energy states available to an ordered system are usually much smaller than the number of states available to a system that appears to be disordered. From his famous 1896 Lectures on Gas Theory, Boltzmann diagrams the structure of a solid body, as shown above, by postulating that each molecule in the body has a "rest position". According to Boltzmann, if it approaches a neighbor molecule it is repelled by it, but if it moves farther away there is an attraction. This, of course was a revolutionary perspective in its time; many, during these years, did not believe in the existence of either atoms or molecules (see: history of the molecule). According to these early views, and others such as those developed by William Thomson, if energy in the form of heat is added to a solid, so to make it into a liquid or a gas, a common depiction is that the ordering of the atoms and molecules becomes more random and chaotic with an increase in temperature: Thus, according to Boltzmann, owing to increases in thermal motion, whenever heat is added to a working substance, the rest position of molecules will be pushed apart, the body will expand, and this will create more molar-disordered distributions and arrangements of molecules. These disordered arrangements, subsequently, correlate, via probability arguments, to an increase in the measure of entropy. Entropy-driven order [edit] Entropy has been historically, e.g. by Clausius and Helmholtz, associated with disorder. However, in common speech, order is used to describe organization, structural regularity, or form, like that found in a crystal compared with a gas. This commonplace notion of order is described quantitatively by Landau theory. In Landau theory, the development of order in the everyday sense coincides with the change in the value of a mathematical quantity, a so-called order parameter. An example of an order parameter for crystallization is "bond orientational order" describing the development of preferred directions (the crystallographic axes) in space. For many systems, phases with more structural (e.g. crystalline) order exhibit less entropy than fluid phases under the same thermodynamic conditions. In these cases, labeling phases as ordered or disordered according to the relative amount of entropy (per the Clausius/Helmholtz notion of order/disorder) or via the existence of structural regularity (per the Landau notion of order/disorder) produces matching labels. However, there is a broad class of systems that manifest entropy-driven order, in which phases with organization or structural regularity, e.g. crystals, have higher entropy than structurally disordered (e.g. fluid) phases under the same thermodynamic conditions. In these systems phases that would be labeled as disordered by virtue of their higher entropy (in the sense of Clausius or Helmholtz) are ordered in both the everyday sense and in Landau theory. Under suitable thermodynamic conditions, entropy has been predicted or discovered to induce systems to form ordered liquid-crystals, crystals, and quasicrystals. In many systems, directional entropic forces drive this behavior. More recently, it has been shown it is possible to precisely engineer particles for target ordered structures. Adiabatic demagnetization [edit] In the quest for ultra-cold temperatures, a temperature lowering technique called adiabatic demagnetization is used, where atomic entropy considerations are utilized which can be described in order-disorder terms. In this process, a sample of solid such as chrome-alum salt, whose molecules are equivalent to tiny magnets, is inside an insulated enclosure cooled to a low temperature, typically 2 or 4 kelvins, with a strong magnetic field being applied to the container using a powerful external magnet, so that the tiny molecular magnets are aligned forming a well-ordered "initial" state at that low temperature. This magnetic alignment means that the magnetic energy of each molecule is minimal. The external magnetic field is then reduced, a removal that is considered to be closely reversible. Following this reduction, the atomic magnets then assume random less-ordered orientations, owing to thermal agitations, in the "final" state: Entropy "order"/"disorder" considerations in the process of adiabatic demagnetization The "disorder" and hence the entropy associated with the change in the atomic alignments has clearly increased. In terms of energy flow, the movement from a magnetically aligned state requires energy from the thermal motion of the molecules, converting thermal energy into magnetic energy. Yet, according to the second law of thermodynamics, because no heat can enter or leave the container, due to its adiabatic insulation, the system should exhibit no change in entropy, i.e. Δ S = 0. The increase in disorder, however, associated with the randomizing directions of the atomic magnets represents an entropy increase? To compensate for this, the disorder (entropy) associated with the temperature of the specimen must decrease by the same amount. The temperature thus falls as a result of this process of thermal energy being converted into magnetic energy. If the magnetic field is then increased, the temperature rises and the magnetic salt has to be cooled again using a cold material such as liquid helium. Difficulties with the term "disorder" [edit] In recent years the long-standing use of term "disorder" to discuss entropy has met with some criticism. Critics of the terminology state that entropy is not a measure of 'disorder' or 'chaos', but rather a measure of energy's diffusion or dispersal to more microstates. Shannon's use of the term 'entropy' in information theory refers to the most compressed, or least dispersed, amount of code needed to encompass the content of a signal. See also [edit] Entropy Entropy production Entropy rate History of entropy Entropy of mixing Entropy (information theory) Entropy (computing) Entropy (energy dispersal) Second law of thermodynamics Entropy (statistical thermodynamics) Entropy (classical thermodynamics) References [edit] ^Mechanical Theory of Heat – Nine Memoirs on the development of concept of "Entropy" by Rudolf Clausius [1850–1865] ^Mahon, Basil (2003). The Man Who Changed Everything – the Life of James Clerk Maxwell. Hoboken, NJ: Wiley. ISBN0-470-86171-1. ^Anderson, Greg (2005). Thermodynamics of Natural Systems. Cambridge University Press. ISBN0-521-84772-9. ^Oxford Dictionary of Science, 2005 ^Oxford Dictionary of Chemistry, 2004 ^Barnes & Noble's Essential Dictionary of Science, 2004 ^Gribbin's Encyclopedia of Particle Physics, 2000 ^Landsberg, P.T. (1984). "Is Equilibrium always an Entropy Maximum?" J. Stat. Physics 35: 159–69. ^ Jump up to: abcMicrosoft Encarta 2006. © 1993–2005 Microsoft Corporation. All rights reserved. ^ Jump up to: abGreven, Andreas; Keller, Gerhard; Warnercke, Gerald (2003). Entropy – Princeton Series in Applied Mathematics. Princeton University Press. ISBN0-691-11338-6. ^ Jump up to: abcdUlanowicz, Robert, E. (2000). Growth and Development – Ecosystems Phenomenology. toExcel Press. ISBN0-595-00145-9.{{cite book}}: CS1 maint: multiple names: authors list (link) ^Kubat, L.; Zeman, J. (1975). Entropy and Information in Science and Philosophy. Elsevier. ^ Jump up to: abJorgensen, Sven E; Svirezhev, Yuri M (2004). Towards a Thermodynamic Theory for Ecological Systems. Elsevier. ISBN0-08-044167-X. ^Strogatz, Steven (2003). the Emerging Science of Spontaneous Order. Theia. ISBN0-7868-6844-9. ^Brooks, Daniel, R.; Wiley, E.O. (1988). Entropy as Evolution – Towards a Unified Theory of Biology. University of Chicago Press. ISBN0-226-07574-5.{{cite book}}: CS1 maint: multiple names: authors list (link) ^Cercignani, Carlo (1998). Ludwig Boltzmann: The Man Who Trusted Atoms. Oxford University Press. ISBN978-0-19-850154-1. ^Boltzmann, Ludwig (1896). Lectures on Gas Theory. Dover (reprint). ISBN0-486-68455-5.{{cite book}}: ISBN / Date incompatibility (help) ^van Anders, Greg; Klotsa, Daphne; Ahmed, N. Khalid; Engel, Michael; Glotzer, Sharon C. (2014). "Understanding shape entropy through local dense packing". Proc Natl Acad Sci USA. 111 (45): E4812 –E4821. arXiv:1309.1187. Bibcode:2014PNAS..111E4812V. doi:10.1073/pnas.1418159111. PMC4234574. PMID25344532. ^Onsager, Lars (1949). "The effects of shape on the interaction of colloidal particles". Annals of the New York Academy of Sciences. 51 (4): 627. Bibcode:1949NYASA..51..627O. doi:10.1111/j.1749-6632.1949.tb27296.x. S2CID84562683. ^Haji-Akbari, Amir; Engel, Michael; Keys, Aaron S.; Zheng, Xiaoyu; Petschek, Rolfe G.; Palffy-Muhoray, Peter; Glotzer, Sharon C. (2009). "Disordered, quasicrystalline and crystalline phases of densely packed tetrahedra". Nature. 462 (7274): 773–777. arXiv:1012.5138. Bibcode:2009Natur.462..773H. doi:10.1038/nature08641. PMID20010683. S2CID4412674. ^Damasceno, Pablo F.; Engel, Michael; Glotzer, Sharon C. (2012). "Predictive Self-Assembly of Polyhedra into Complex Structures". Science. 337 (6093): 453–457. arXiv:1202.2177. Bibcode:2012Sci...337..453D. doi:10.1126/science.1220869. PMID22837525. S2CID7177740. ^Geng, Yina; van Anders, Greg; Dodd, Paul M.; Dshemuchadse, Julia; Glotzer, Sharon C. (2019). "Engineering Entropy for the Inverse Design of Colloidal Crystals from Hard Shapes". Science Advances. 5 (7): eeaw0514. arXiv:1712.02471. Bibcode:2019SciA....5..514G. doi:10.1126/sciadv.aaw0514. PMC6611692. PMID31281885. ^ Jump up to: abcHalliday, David; Resnick, Robert (1988). Fundamentals of Physics, Extended 3rd ed. Wiley. ISBN0-471-81995-6. ^ Jump up to: abcNASA – How does an Adiabatic Demagnetization Refrigerator Work? ^Denbigh K. (1981). The Principles of Chemical Equilibrium: With Applications in Chemistry and Chemical Engineering. London: Cambridge University Press. pp.55–56. ^Jaynes, E.T. (1989). Clearing up mysteries—the original goal, in Maximum Entropy and Bayesian Methods, J. Skilling, Editor, Kluwer Academic Publishers, Dordrecht, pp. 1–27, page 24. ^Grandy, Walter T. Jr. (2008). Entropy and the Time Evolution of Macroscopic Systems. Oxford University Press. pp.55–58. ISBN978-0-19-954617-6. ^Frank L. Lambert, 2002, "Disorder—A Cracked Crutch for Supporting Entropy Discussions," Journal of Chemical Education 79: 187. ^Carson, E. M., and Watson, J. R., (Department of Educational and Professional Studies, King's College, London), 2002, "Undergraduate students' understandings of entropy and Gibbs Free energy," University Chemistry Education – 2002 Papers, Royal Society of Chemistry. ^Sozbilir, Mustafa, PhD studies: Turkey, A Study of Undergraduates' Understandings of Key Chemical Ideas in Thermodynamics, Ph.D. Thesis, Department of Educational Studies, The University of York, 2001. ^Shannon, C.E. (1945). A mathematical theory of cryptography, Memorandum for file, MM-45-110-98, 135 pages, page 20; found in File 24 at page 203 in Claude Elwood Shannon: Miscellaneous Writings edited by N.J.A. Sloane, and Aaron D. Wyner (revision of 2013), Mathematical Sciences Research Center, AT&T Bell Laboratories, Murray Hill, NJ; previously partly published by IEEE Press. ^Gray, R.M. (2010). Entropy and Information Theory, Springer, New York NY, 2nd edition, p. 296. ^Mark Nelson (24 August 2006). "The Hutter Prize". Archived from the original on 2018-03-01. Retrieved 2008-11-27. External links [edit] Lambert, F. L. Entropy Sites — A Guide Lambert, F. L. Shuffled Cards, Messy Desks, and Disorderly Dorm Rooms – Examples of Entropy Increase? Nonsense! Journal of Chemical Education Retrieved from " Categories: Thermodynamic entropy State functions Hidden categories: CS1 maint: multiple names: authors list CS1 errors: ISBN date Articles with short description Short description is different from Wikidata This page was last edited on 11 September 2025, at 23:36(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. 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https://www.blackhc.net/blog/2021/binomial-stirling/
Stirling's Approximation for Binomial Coefficients An example of using information-theoretic quantities for outcomes In MacKay (2003) on page 2, the following straightforward approximation for a binomial coefficient is introduced: [\begin{equation} \log \binom{N}{r} \simeq(N-r) \log \frac{N}{N-r}+r \log \frac{N}{r}. \end{equation}] The derivation in the book is short but not very intuitive although it feels like it should be. Information theory would be the likely candidate to provide intuitions. But information-theoretic quantities like entropies do not apply to fixed observations, only random variables, or do they? We will derive Stirling’s approximation for binomial coefficients using a neat notation for information-theoretical quantities . It allows quantifying information-theoretic quantities for more than just random variables. At the same time, we will see that this might allow using other tools from probability theory to estimate the approximation error. This might be interesting beyond this toy example. ( \require{newcommand} \require{mathtools} \newcommand{ifblank}{#3} \DeclareMathOperator{\argmax}{arg\,max} \DeclareMathOperator{\argmin}{arg\,min} \newcommand{\defeq}{\vcentcolon=} \DeclareMathOperator{\opExpectation}{\mathbb{E}} % Expectation over #1 of #2 \DeclarePairedDelimitersXPP{\E}{\opExpectation_{#1}}{\lbrack}{\rbrack}{}{ % \ifblank{#2}{\:\cdot\:}{#2} } %\newcommand{\E}{\opExpectation_{#1} \left [ \ifblank{#2}{\:\cdot\:}{#2} \right ]} % Expectation over #1 without brackets \newcommand{\simpleE}{\opExpectation_{#1}} % {\ifblank{#2}{\left [ \:\cdot\: \right ]}{#2}}} % can be useful to refer to this outside \Set \newcommand\MidSymbol[]{ % \nonscript\:#1 %\allowbreak \nonscript\: \mathopen{} } % Helper commands for specifying conditionals etc % (from mathtools documentation p 28) % Make sure it exists \newcommand\given{\MidSymbol[\vert]} %% Shannon's Information Content \DeclareMathOperator{\opInformationContent}{h} \DeclarePairedDelimitersXPP{\ICof}{\opInformationContent}{(}{)}{}{ % \ifblank{#1}{\:\cdot\:}{#1} } %% Entropy \DeclareMathOperator{\opEntropy}{H} \DeclarePairedDelimitersXPP{\Hof}{\opEntropy}{[}{]}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#1}{\:\cdot\:}{#1} } \DeclarePairedDelimitersXPP{\xHof}{\opEntropy}{(}{)}{}{ % \ifblank{#1}{\:\cdot\:}{#1} } \DeclareMathOperator{\opMI}{I} \DeclarePairedDelimitersXPP{\MIof}{\opMI}{[}{]}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#1}{\:\cdot\:}{#1} } \DeclareMathOperator{\opSuprise}{S} \DeclarePairedDelimitersXPP{\Sof}{\opSuprise}{[}{]}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#1}{\:\cdot\:}{#1} } %% Cross-Entropy \DeclarePairedDelimitersXPP{\CrossEntropy}{\opEntropy}{(}{)}{}{ % \ifblank{#1#2}{\:\cdot\: \MidSymbol[\delimsize\Vert] \:\cdot\:}{#1 \MidSymbol[\delimsize\Vert] #2} } %\newcommand{\CrossEntropy}{\E{#1}{-\log(#2)}} %% Kullback-Leibler Divergence \DeclareMathOperator{\opKale}{D_\mathrm{KL}} \DeclarePairedDelimitersXPP{\Kale}{\opKale}{(}{)}{}{ % \ifblank{#1#2}{\:\cdot\: \MidSymbol[\delimsize\Vert] \:\cdot\:}{#1 \MidSymbol[\delimsize\Vert] #2} } %% Probabilities \DeclareMathOperator{\opp}{p} \DeclarePairedDelimitersXPP{\pof}{\opp}{(}{)}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#1}{\:\cdot\:}{#1} } \DeclarePairedDelimitersXPP{\pcof}{\opp_{#1}}{(}{)}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#2}{\:\cdot\:}{#2} } \DeclarePairedDelimitersXPP{\hpcof}{\hat{\opp}_{#1}}{(}{)}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#2}{\:\cdot\:}{#2} } \DeclareMathOperator{\opq}{q} \DeclarePairedDelimitersXPP{\qof}{\opq}{(}{)}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#1}{\:\cdot\:}{#1} } %% Variational Entropy (Cross-Entropy) \DeclarePairedDelimitersXPP{\varHof}{\opEntropy_{\ifblank{#1}{\:\cdot\:}{#1}}}{[}{]}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#2}{\:\cdot\:}{#2} } \DeclarePairedDelimitersXPP{\xvarHof}{\opEntropy_{\ifblank{#1}{\:\cdot\:}{#1}}}{(}{)}{}{ % \renewcommand\given{\MidSymbol[\delimsize\vert]} \ifblank{#2}{\:\cdot\:}{#2} } \Hof{B, r} = \Hof{B \given r} ) Notation A quick revision of the notation in “A Practical & Unified Notation for Information-Theoretic Quantities in ML ”. It allows for information quantities between (untied) random variables and tied random variables with observed outcomes. All quantities, which we will use below, can be consistently defined as follows: (For random variables (R) and (B) with outcomes (r) and (b), respectively, we shorten “(R=r)” and “(B=b)” to “(r)” and “(b)” in expressions.) (H[r] := h(p(r)) := -\log p(r),) (H[b,r] := h(p(b,r)) = -\log p(b, r),) (H[b \mid r ] := h(p(b \mid r)) = -\log p(b \mid r),) (H[B,r] := \mathbb{E}_{p(b|r)} H[b, r] = -\mathbb{E}_{p(b|r)} \log p(b, r),) and (H[B \mid r] := \mathbb{E}_{p(b|r)} H[b \mid r] = -\mathbb{E}_{p(b|r)} \log p(b \mid r).) (\Hof{r}) denotes the number of nats to encode event (R=r), given probability distribution (\pof{r}). Similarly, (\Hof{B, r}) denotes the number of nats to encode both (B) and (R) when (R=r) unbeknownst to encoding scheme. The main idea behind this notation is that it is consistent with taking expectations over the quantities to obtain regular entropies and so on. Most rules for information-theoretic quantities also directly apply to quantities that mix random variables and outcomes. More information and intuitions are found in the paper. Figure 1. The relationship between between the information quantities. (B) is the joint of the Bernoulli random variables, (R) is the number of successes in (B) with observed outcome (r) (and thus follows a Binomial distribution). The arrow below (\Hof{r}) symbolizes that we minimize (\Hof{r}) by optimizing the success probability (\rho) to close the gap between (\simpleE{\pof{b \given r}}{\Hof{b}}) and (\Hof{B \given r}). This follows the diagrams in MacKay (2003) , e.g. Figure 8.1. We will start by introducing a simple probabilistic model that will allow us to formulate the approximation as an upper-bound. We will then show how the upper-bound follows from dropping one term and then pick the success probability to tighten the bound as much as possible. This sounds more complicated than it is but is a fun way to show some tools that are often employed in mathematics. We will relate the quantities we examine to information theory (as a means of quantifying optimal communication) by asking: what message does the quantity represent? This will hopefully provide additional intuitions to the reader. Setup. Let (B_1, \ldots, B_N) be (N) Bernoulli random variables with success probability (\rho), and let (B) be the joint of these random variables. Further, let (R) be the random variable that counts the number of successes in (B). (R) follows a Binomial distribution with success probability (\rho) and (N) trials. Main Idea. For a given outcome (r) of (R), we have: [\begin{equation} \Hof{B, r} = \Hof{B \given r} + \Hof{r} \ge \Hof{B \given r}, \end{equation}] as (\Hof{\cdot}) is non-negative for discrete random variables. We will examine this inequality to obtain the approximation. Note that (\Hof{B \given r}) is the additional number of bits needed to encode (B) when the number of successes is already known. Similarly, (\Hof{B, r}) is the number of bits needed to encode both (B) and (R) under the circumstance that (R=r). Determining (\Hof{B, r}). (R) is fully determined by (B), and thus we have (\Hof{B, R}=\Hof{B}) and hence also This also follows immediately from (\Hof{R \given B} = 0 \implies \forall r: \Hof{r \given B} = 0). : [\begin{equation} \Hof{B, r} = \simpleE{\pof{b \given r}}{\Hof{b}}. \end{equation}] (\simpleE{\pof{b \given r}}{\Hof{b}}) is the expected number of bits needed to transmit the outcome (b) of (B) when (r) is given. When we encode (b), we do not know (r) upfront, so we need to transmit (N) Bernoulli outcomes. Hence, we need to transmit (r) successes and (N-r) failures. Given the success probability (\rho), the optimal message length for this is: [\begin{align} &\simpleE{\pof{b \given r}}{\Hof{b}} = r \, \ICof{\rho} + (N-r)\, \ICof{1-\rho} \ & \quad = - r \log \rho - (N - r) \log (1-\rho). \end{align}] All this is visualized in Figure 1. Alternative Argument. We can also look at the terms (\Hof{B \given r} + \Hof{r}) separately. We have [\begin{equation} \Hof{r} = -\log \pof{r} = - \log \left ( \binom{N}{r} \, \rho^r \,(1-\rho)^{N-r}\right ), \end{equation}] and [\begin{equation} \Hof{B \given r} = -\simpleE{\pof{b \given r}} \log \pof{b \given r} = \log \binom{N}{r}. \end{equation}] The former follows from (R) being binomially distributed. For the latter, we observe that we need to encode (B) while knowing (r) already. Given (r), (\pof{b \given r} = \text{const}) for all valid (b). There are (\binom{N}{r}) possible (b) for fixed (r). Hence, we can simply create a table with all possible configurations with (r) successes. There are (\binom{N}{r}) many. We then encode the index into this table. Each configuration with (r) successes has an equal probability, so we have a uniform discrete distribution with entropy (\log \binom{N}{r}) and obtain the same result. Determining (\rho). With this, we are almost done. We already have [\begin{align} \Hof{B \given r} + \Hof{r} &= -r \log \rho - (N - r) \log (1-\rho) \notag \ &\ge \log \binom{N}{r} = \Hof{B \given r}. \end{align}] How do we make this inequality as tight as possible? We need to minimize the gap (\Hof{r}) which creates the inequality in the first place, and (\Hof{r}=-\log \pof{r}) is minimized exactly when (\pof{r}) becomes maximal. Hence, we choose the success probability (\rho) to do so: the maximum likelihood solution (\argmax_p \pof{r \given \rho}) is (\rho = \frac{r}{N}). The Binomial distribution of (R) then has its mode, mean, and median at (r). Altogether, after substituting (\rho = \frac{r}{N}) and rearranging, we obtain the wanted approximation: [\begin{align} \log \binom{N}{r} &\le -r \log \rho - (N - r) \log (1-\rho) \ & = r \log \frac{N}{r} + (N - r) \log \frac{N}{N-r}. \end{align}] Approximation Error (\Hof{r}). The approximation error is just (\Hof{r}). We can easily upper-bound it with (\Hof{r} \le \log N): First, (\Hof{R} \le \log N) as the uniform distribution with entropy (\log N) is the maximum entropy distribution in this case (discrete random variable with finite support). Second, (\Hof{R}) is the expectation over different (\Hof{R=r'}). Now given the (\rho) we have chosen, (\pof{r}) is maximal (and (\Hof{r}) is minimal). Hence, (\Hof{r} \le \log N) by contradiction: otherwise, we would have (\log N < \Hof{r} \le \Hof{R}), yet (\Hof{R} \le \log N). Empirical Evaluation In Figure 2, below we show an empirical evaluation of our approximation in comparison to the true (\log) binomial coefficients for (N) up to (10^5). We chose (r \approx \tfrac{N}{2}). (a) exact (\log) binomials vs our approximation (b) approximation error vs estimated error (c) relative approximation error vs estimated error Figure 2. Comparison between exact (\log) binomial and our approximation (r \log \frac{N}{r} + (N - r) \log \frac{N}{N-r}). Qualitatively the approxmation error quickly seems to vanish in (a). While the absolute error is slowly increasing in (b), we see that the relative error quickly vanishes in (c). The estimated approximation error (\log N) is not tight but still a good predictor. More details can be found in this Google Colab. Conclusion We have deduced an approximation of Stirling’s approximation for binomial coefficients and upper-bounded the approximation error with new intuitions. We have done so by using simple probability theory and information theory. Acknowledgements We would like to thank Yarin Gal, Tim Rudner, Ravid Shwartz-Ziv, Clare Lyle, Joost van Amersfoort, as well as the members of OATML in general for their feedback. More from OATML For more blog posts by OATML in Oxford, check out our group’s blog Follow me on Twitter @blackhc
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https://epubs.siam.org/doi/10.1137/15M1011032
Skip to main content Next article The “Runs” Theorem Authors: Hideo Bannai, Tomohiro I, Shunsuke Inenaga, Yuto Nakashima, Masayuki Takeda, and Kazuya TsurutaAuthors Info & Affiliations Get Access Tools Add to favorites Download Citations Track Citations Abstract We give a new characterization of maximal repetitions (or runs) in strings based on Lyndon words. The characterization leads to a proof of what was known as the “runs” conjecture [R. M. Kolpakov and G. Kucherov, Proceedings of the IEEE Symposium on Foundations of Computer Science (FOCS), IEEE Computer Society, Los Alamitos, CA, 1999, pp. 596--604]), which states that the maximum number of runs 𝜌⁡(𝑛) in a string of length 𝑛 is less than 𝑛. The proof is remarkably simple, considering the numerous endeavors to tackle this problem in the last 15 years, and significantly improves our understanding of how runs can occur in strings. In addition, we obtain an upper bound of 3n for the maximum sum of exponents 𝜎⁡(𝑛) of runs in a string of length 𝑛, improving on the best known bound of 4.1n by Crochemore et al. [J. Discrete Algorithms, 14 (2012), pp. 29--36], as well as other improved bounds on related problems. The characterization also gives rise to a new, conceptually simple linear-time algorithm for computing all the runs in a string. A notable characteristic of our algorithm is that, unlike all existing linear-time algorithms, it does not utilize the Lempel--Ziv factorization of the string. We also establish a relationship between runs and nodes of the Lyndon tree, which gives a simple optimal solution to the 2-period query problem that was recently solved by Kociumaka et al. [Proceedings of the Twenty-Sixth Annual ACM-SIAM Symposium on Discrete Algorithms, (SODA) 2015, San Diego, CA, SIAM, Philadelphia, 2015, pp. 532--551]. Keywords maximal repetitions Lyndon words combinatorics on words string algorithms MSC codes 68R15 68W32 Get full access to this article View all available purchase options and get full access to this article. Get Access References H. Bannai, T. I, S. Inenaga, Y. Nakashima, M. Takeda, and K. Tsuruta, A new characterization of maximal repetitions by Lyndon trees, in Proceedings of the Annual ACM-SIAM Symposium on Discrete Algorithms (SODA), SIAM, Philadelphia, 2015, pp. 562--571. Google Scholar H. Barcelo, On the action of the symmetric group on the free Lie algebra and the partition lattice, J. Combin. Theory, Ser. A, 55 (1990), pp. 93--129. Web of Science Google Scholar P. Baturo, M. Piatkowski, and W. Rytter, The maximal number of runs in standard Sturmian words, Electron. J. Combin., 20 (2013), P13. Crossref Web of Science Google Scholar M. A. Bender and M. Farach-Colton, The LCA problem revisited, in Proc. LATIN 2000, Springer, Berlin, 2000, pp. 88--94. Google Scholar D. 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Crossref Web of Science Google Scholar Information & Authors Information Published In SIAM Journal on Computing Volume 46 • Issue 5 • January 2017 Pages: 1501 - 1514 DOI: 10.1137/15M1011032 ISSN (online): 1095-7111 Copyright © 2017, Society for Industrial and Applied Mathematics. History Submitted: 4 March 2015 Accepted: 19 July 2017 Published online: 21 September 2017 Keywords maximal repetitions Lyndon words combinatorics on words string algorithms MSC codes 68R15 68W32 Authors Affiliations Hideo Bannai View all articles by this author Tomohiro I View all articles by this author Shunsuke Inenaga View all articles by this author Yuto Nakashima View all articles by this author Masayuki Takeda View all articles by this author Kazuya Tsuruta View all articles by this author Funding Information Japan Society for the Promotion of Science : 25280086, 16H02783, 26280003, 25240003 Metrics & Citations Metrics Metrics Downloads Citations No data available. 306 58 Total 6 Months 12 Months Total number of downloads Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Simply select your manager software from the list below and click Download. 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Copying failed. Share with email Email a colleague Share on social media FacebookX (formerly Twitter)LinkedInemail Get Access Get Access Purchase Save for later Item saved, go to cart Article Pay-Per-View $40.00 Add to cart Article Pay-Per-View Checkout Access via your Institution Questions about how to access this content? Contact SIAM at service@siam.org. References References 1. H. Bannai, T. I, S. Inenaga, Y. Nakashima, M. Takeda, and K. Tsuruta, A new characterization of maximal repetitions by Lyndon trees, in Proceedings of the Annual ACM-SIAM Symposium on Discrete Algorithms (SODA), SIAM, Philadelphia, 2015, pp. 562--571. Google Scholar 2. H. Barcelo, On the action of the symmetric group on the free Lie algebra and the partition lattice, J. Combin. Theory, Ser. A, 55 (1990), pp. 93--129. Crossref Web of Science Google Scholar 3. P. Baturo, M. Piatkowski, and W. Rytter, The maximal number of runs in standard Sturmian words, Electron. J. Combin., 20 (2013), P13. Crossref Web of Science Google Scholar 4. M. A. Bender and M. Farach-Colton, The LCA problem revisited, in Proc. LATIN 2000, Springer, Berlin, 2000, pp. 88--94. Google Scholar 5. D. Breslauer, Efficient String Algorithmics, Ph.D. thesis, Columbia University, New York, 1992. Google Scholar 6. S. Brlek, J.-O. Lachaud, X. Provençal, and C. Reutenauer, Lyndon + Christoffel = digitally convex, Pattern Recognit., 42 (2009), pp. 2239--2246. Crossref Web of Science Google Scholar 7. M. Chemillier, Periodic musical sequences and Lyndon words, Soft Comput., 8 (2004), pp. 611--616. Crossref Web of Science Google Scholar 8. G. Chen, S. J. Puglisi, and W. F. Smyth, Fast and practical algorithms for computing all the runs in a string, in Combinatorial Pattern Matching, Springer, Berlin, 2007, pp. 307--315. Google Scholar 9. K. T. Chen, R. H. Fox, and R. C. Lyndon, Free differential calculus, IV. The quotient groups of the lower central series, Ann. of Math. (2), 68 (1958), pp. 81--95. Crossref Web of Science Google Scholar 10. M. Crochemore and L. Ilie, Computing longest previous factor in linear time and applications, Inform. Process. Lett., 106 (2008), pp. 75--80. Crossref Web of Science Google Scholar 11. M. Crochemore and L. Ilie, Maximal repetitions in strings, J. Comput. System Sci., 74 (2008), pp. 796--807. Crossref Web of Science Google Scholar 12. M. Crochemore, L. Ilie, and W. Rytter, Repetitions in strings: Algorithms and combinatorics, Theoret. Comput. Sci., 410 (2009), pp. 5227--5235. Crossref Web of Science Google Scholar 13. M. Crochemore, L. Ilie, and L. Tinta, The “runs” conjecture, Theoret. Comput. Sci., 412 (2011), pp. 2931--2941. Crossref Web of Science Google Scholar 14. M. Crochemore, C. Iliopoulos, M. Kubica, J. Radoszewski, W. Rytter, and T. Waleń, The maximal number of cubic runs in a word, J. Comput. System Sci., 78 (2012), pp. 1828--1836. Crossref Web of Science Google Scholar 15. M. Crochemore, C. Iliopoulos, M. Kubica, J. Radoszewski, W. Rytter, and T. Waleń, Extracting powers and periods in a word from its runs structure, Theoret. Comput. Sci., 521 (2014), pp. 29--41. Crossref Web of Science Google Scholar 16. M. Crochemore, M. Kubica, J. Radoszewski, W. Rytter, and T. Waleń, On the maximal sum of exponents of runs in a string, J. Discrete Algorithms, 14 (2012), pp. 29--36. Crossref Google Scholar 17. O. Delgrange and E. Rivals, STAR: An algorithm to search for tandem approximate repeats, Bioinformatics, 20 (2004), pp. 2812--2820. Crossref Web of Science Google Scholar 18. A. Deza and F. Franek, A 𝑑-step approach to the maximum number of distinct squares and runs in strings, Discrete Appl. Math., 163 (2014), pp. 268--274. Crossref Web of Science Google Scholar 19. A. Deza and F. Franek, 𝑑-Step Method and the Number of Runs, Technical report AdvOL2015/01, Advanced Optimization Laboratory, Department of Computing and Software, McMaster University, Hamilton, Canada, 2015. Google Scholar 20. Y. Dieudonné, F. Levé, F. Petit, and V. Villain, Deterministic geoleader election in disoriented anonymous systems, Theoret. Comput. Sci., 506 (2013), pp. 43--54. Crossref Web of Science Google Scholar 21. Y. Dieudonné and F. Petit, Circle formation of weak robots and Lyndon words, Inform. Process. Lett., 101 (2007), pp. 156--162. Crossref Web of Science Google Scholar 22. J.-P. Duval, Factorizing words over an ordered alphabet, J. Algorithms, 4 (1983), pp. 363--381. Crossref Web of Science Google Scholar 23. N. J. Fine and H. S. Wilf, Uniqueness theorems for periodic functions, Proc. Amer. Math. Soc., 16 (1965), pp. 109--114. Crossref Web of Science Google Scholar 24. J. Fischer and V. Heun, Theoretical and practical improvements on the RMQ-problem, with applications to LCA and LCE, in Combinatorial Pattern Matching, Elsevier, Amsterdam, 2006, pp. 36--48. Google Scholar 25. F. Franek and Q. Yang, An asymptotic lower bound for the maximal number of runs in a string, Internat. J. Found. 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Lorentz, An 𝑂⁡(𝑛⁢log⁡𝑛) algorithm for finding all repetitions in a string, J. Algorithms, 5 (1984), pp. 422--432. Crossref Web of Science Google Scholar 42. U. Manber and G. Myers, Suffix arrays: A new method for on-line string searches, SIAM J. Comput., 22 (1993), pp. 935--948. Abstract Web of Science Google Scholar 43. W. Matsubara, K. Kusano, A. Ishino, H. Bannai, and A. Shinohara, New lower bounds for the maximum number of runs in a string, in Prague Stringology Conference, World Scientific, Singapore, 2008, pp. 140--145. Google Scholar 44. M. Mucha, Lyndon words and short superstrings, in Proceedings of the Annual ACM-SIAM Symposium on Discrete Algorithms (SODA) '13, SIAM, Philadelphia, 2013, pp. 958--972. Google Scholar 45. S. J. Puglisi, J. Simpson, and W. F. Smyth, How many runs can a string contain?, Theoret. Comput. Sci., 401 (2006), pp. 165--171. Crossref Web of Science Google Scholar 46. C. Reutenauer, Free Lie Algebras, Oxford University Press, Oxford, 1993. Google Scholar 47. W. Rytter, The number of runs in a string: Improved analysis of the linear upper bound, in Proceedings of the Symposium on Theoretical Aspects of Computer Science (STACS) 2006, Lecture Notes in Comput. Sci. 3884, Springer, Berlin, 2006, pp. 184--195. Google Scholar 48. J. Simpson, Modified Padovan words and the maximum number of, Australas. J. Combin., 46 (2010), pp. 129--145. Google Scholar 49. W. F. Smyth, Repetitive perhaps, but certainly not boring, Theoret. Comput. Sci., 249 (2000), pp. 343--355. Crossref Web of Science Google Scholar 50. W. F. Smyth, Computing regularities in strings: A survey, European J. Combinatorics, 34 (2013), pp. 3--14. Crossref Web of Science Google Scholar 51. W. F. Smyth, Large-scale detection of repetitions, R. Soc. Phil. Trans. Lond. Ser. A Math. Phys. Eng. Sci., 372 (2014), 20130138. Crossref Google Scholar 52. J. Vuillemin, A unifying look at data structures, Comm. ACM, 23 (1980), pp. 229--239. 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5096
https://en.wikipedia.org/wiki/Kapandji_score
Kapandji score - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Scoring 2 References Kapandji score [x] Add languages Add 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 Add interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia The Kapandji score is a tool useful for assessing the opposition of the thumb, based on where on their hand the patient is able to touch with the tip of their thumb. Scoring [edit] | Score | Location achieved | --- | | 1 | Radial side of the proximal phalanx of the index finger | | 2 | Radial side of the middle phalanx of the index finger | | 3 | Tip of the index finger | | 4 | Tip of the middle finger | | 5 | Tip of the ring finger | | 6 | Tip of the little finger | | 7 | Distal interphalangeal joint crease of the little finger | | 8 | Proximal interphalangeal joint crease of the little finger | | 9 | Metacarpophalangeal joint crease of the little finger | | 10 | Distal palmar crease | References [edit] ^Kapandji A (1986). "Clinical test of apposition and counter-apposition of the thumb". Ann Chir Main. 5 (1): 67–73. doi:10.1016/s0753-9053(86)80053-9. PMID3963909. This article about orthopedic surgery is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Orthopedic clinical prediction rules Musculoskeletal examination Orthopedics stubs Hidden category: All stub articles This page was last edited on 22 October 2022, at 16:34(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Kapandji score Add languagesAdd topic
5097
https://chemistrynotes.com/blogs/chemistry-help/entropy-changes-and-gibbs-free-energy
Entropy Changes in Chemical Reactions, and Gibbs Free Energy Search 0 0 Home General Chemistry Notes +- FULL COURSE (Sections 1-20) 1 - Foundations of Chemistry 2 - Atoms, Molecules, and Ions 3 - Chemical Quantities and Stoichiometry 4 - Types of Chemical Reactions and Solution Stoichiometry 5 - Gases 6 - Thermochemistry 7 - Quantum Mechanical View of the Atom, and Periodicity 8 - Chemical Bonding 9 - Covalent Bonding and Molecular Orbitals 10 - Liquids, Solids, and Intermolecular Forces 11 - Solutions and Their Properties 1ST SEMESTER (Sections 1-11) 12 - Chemical Kinetics 13 - Chemical Equilibrium 14 - Acids and Bases 15 - Applications of Acid-Base Equilibria 16 - Spontaneity, Entropy, and Free Energy 17 - Electrochemistry 18 - Transition Metals and Coordination Chemistry 19 - The Nucleus and Nuclear Chemistry 20 - An Introduction to Organic Chemistry 2ND SEMESTER (Sections 12-20) Organic Chemistry Notes +- FULL COURSE (Sections 1-23) 1 - Structural, Bonding, Molecular Properties 2 - The Nature of Organic Compounds: Alkanes and Cycloalkanes 3 - Stereochemistry of Alkanes and Cycloalkanes: 3-D Structures of Molecules 4 - The Study of Organic Reactions: An Overview 5 - Structure and Reactivity of Alkenes 6 - Reactions and Synthesis of Alkenes 7 - Alkynes 8 - Stereochemistry of Organic Compounds 9 - Alkyl Halides (R-X) 10 - Substitution (SN2, SN1) and Elimination (E2, E1) Reactions 11 - Mass Spectrometry and IR Spectroscopy 12 - NMR Spectroscopy 1ST SEMESTER (Sections 1-12) 13 - Conjugated Systems and UV Spectroscopy 14 - Aromatic Compounds and Aromaticity 15 - Electrophilic Aromatic Substitution (E.A.S.) 16 - Reactions and Synthesis of Alcohols 17 - Ethers, Epoxides, and Sulfides 18 - Aldehydes and Ketones: Nucleophilic Addition Reactions 19 - Carboxylic Acids 20 - Carboxylic Acid Derivatives and Nucleophilic Acyl Substitution Reactions 21 - Carbonyl Alpha-Substitution Reactions 22 - Carbonyl Condensation Reactions 23 - Amines 2ND SEMESTER (Sections 13-23) Full Semester +- 1st Semester General Chemistry Notes 2nd Semester General Chemistry Notes 1st Semester Organic Chemistry Notes 2nd Semester Organic Chemistry Notes Entire Year +- Entire Year General Chemistry Notes Entire Year Organic Chemistry Notes Check Out the Video Notes! 🖥️ Search Home General Chemistry Notes FULL COURSE (Sections 1-20) 1 - Foundations of Chemistry 2 - Atoms, Molecules, and Ions 3 - Chemical Quantities and Stoichiometry 4 - Types of Chemical Reactions and Solution Stoichiometry 5 - Gases 6 - Thermochemistry 7 - Quantum Mechanical View of the Atom, and Periodicity 8 - Chemical Bonding 9 - Covalent Bonding and Molecular Orbitals 10 - Liquids, Solids, and Intermolecular Forces 11 - Solutions and Their Properties 1ST SEMESTER (Sections 1-11) 12 - Chemical Kinetics 13 - Chemical Equilibrium 14 - Acids and Bases 15 - Applications of Acid-Base Equilibria 16 - Spontaneity, Entropy, and Free Energy 17 - Electrochemistry 18 - Transition Metals and Coordination Chemistry 19 - The Nucleus and Nuclear Chemistry 20 - An Introduction to Organic Chemistry 2ND SEMESTER (Sections 12-20) Organic Chemistry Notes FULL COURSE (Sections 1-23) 1 - Structural, Bonding, Molecular Properties 2 - The Nature of Organic Compounds: Alkanes and Cycloalkanes 3 - Stereochemistry of Alkanes and Cycloalkanes: 3-D Structures of Molecules 4 - The Study of Organic Reactions: An Overview 5 - Structure and Reactivity of Alkenes 6 - Reactions and Synthesis of Alkenes 7 - Alkynes 8 - Stereochemistry of Organic Compounds 9 - Alkyl Halides (R-X) 10 - Substitution (SN2, SN1) and Elimination (E2, E1) Reactions 11 - Mass Spectrometry and IR Spectroscopy 12 - NMR Spectroscopy 1ST SEMESTER (Sections 1-12) 13 - Conjugated Systems and UV Spectroscopy 14 - Aromatic Compounds and Aromaticity 15 - Electrophilic Aromatic Substitution (E.A.S.) 16 - Reactions and Synthesis of Alcohols 17 - Ethers, Epoxides, and Sulfides 18 - Aldehydes and Ketones: Nucleophilic Addition Reactions 19 - Carboxylic Acids 20 - Carboxylic Acid Derivatives and Nucleophilic Acyl Substitution Reactions 21 - Carbonyl Alpha-Substitution Reactions 22 - Carbonyl Condensation Reactions 23 - Amines 2ND SEMESTER (Sections 13-23) Full Semester 1st Semester General Chemistry Notes 2nd Semester General Chemistry Notes 1st Semester Organic Chemistry Notes 2nd Semester Organic Chemistry Notes Entire Year Entire Year General Chemistry Notes Entire Year Organic Chemistry Notes Check Out the Video Notes! 🖥️ S16E2 - Entropy Changes in Chemical Reactions, and Gibbs Free Energy •SECTION 16 - Spontaneity Entropy and Free Energy Second Law of Thermodynamics 2nd Law of Thermodynamics = the entropy (S) of the universe is constantly increasing. In any spontaneous process, the entropy of the universe increases. The change in entropy of the universe is given the symbol, ΔS univ ➞ if ΔS univ> 0 , the process is spontaneous. ➞ if ΔS univ< 0 , the reverse process is spontaneous. ➞ if ΔS univ= 0 , the process has no tendency to occur (it's at equilibrium). ➞ Sometimes ΔS sys and ΔS surr have opposite signs, and their magnitudes determine whether or not the overall process is spontaneous (ΔS surr> 0 ). Gibbs Free Energy, G Gibbs Free Energy, G = a thermodynamic function that not only is related to spontaneity, but also deals with the temperature dependence of spontaneity: ΔG =ΔH - TΔS ➞ if ΔG<0 , the process is spontaneous. ➞ if ΔG> 0 , the process is not spontaneous. ➞ To calculate ΔG, you must compare and consider the ΔH and ΔS values, and you must consider the Kelvin temperature (T). ex: At what temperatures is the following process spontaneous at 1atm? Br 2(l)➞Br 2(g) ΔH° = 31.0 kJ/mol, and ΔS° = 93.0 J/K.mol Well, ΔG must be negative inΔG° =ΔH° - TΔS° (° symbol = 25°C, 1atm conditions)... ... So we can find the threshold temperature (T) by setting ΔG° = 0 NOTE - When T = 333K, the free-energy change (ΔG) = 0 and the liquid and gas phases of Br 2 coexist at 1atm. -- so, the normal boiling point (bp) = 333K What is Entropy? Entropy Changes in Chemical Reactions -- It's most important to focus on the amount of gas molecules. ex: Does entropy increase or decrease in the following reactions? i.N 2(g)+3H 2(g)➞2NH 3(g) 4 gas molecules "goes to" 2 gas molecules, and fewer gas molecules means fewer possible configurations. fewer gas molecules = less disorder =entropy decreases,ΔS < 0 ii. 4NH 3(g)+5O 2(g)➞4NO(g)+6H 2 O(g) entropy increases, ΔS > 0, because 9 gas molecules becomes 10 gas molecules. Standard Entropy Values Standard Entropy Values (S°) are listed in a lengthy table or appendix in the back of your textbook. You can find one here as well: Table of Standard Entropies. ➞ notice that indeed S° increases for a substance in going from solid to a liquid to a gas. ➞ S° has units of " J / K.mol " ➞To calculate ΔS° for a reaction: ΔS°rxn = (sum of all entropies for products) - (sum of all entropies of reactants) , or... ex: Calculate ΔS°for this reaction: Al 2 O 3(s)+3H 2(g)➞2Al(s)+3H 2 O(g) First, we will need a Table of Standard Entropy Values (S°) from the back of our textbook, or use the link I gave you earlier in this article (above). This gives us... Question: In the sample problem above, if 3 gas molecules (reactants) are converted to 3 gas molecules (products), how come the ΔS° value is large instead of nearly 0 J/K ?... Answer: The large ΔS° value is because H 2 O is more complex than H 2, and thus has a higher standard entropy value (S°). ➞ Why? Because H 2 O's molecular structure, being more sophisticated, has a greater number of vibrations/rotations/etc. which increase its randomness and disorder (entropy). In my next post covering SECTION 16 - Spontaneity, Entropy, and Free Energy, We'll talk extensively about Spontaneous Reactions and the Gibbs Free Energy Change,ΔG° Want FULL ACCESS to ALL 143 Chemistry Videos? Our blog only gives you the first 3 videos from each of 20 Sections (Chapters). 60 Blog Videos vs. All 143 Videos Inside. Click the blue button below to access the FULL COURSE VIDEO NOTES :-) GET ALL 143 VIDEOS 〉 Share Share on FacebookTweet Tweet on TwitterPin it Pin on Pinterest Categories SECTION 1 - Foundations of Chemistry SECTION 10 - Liquids Solids and Intermolecular Forces SECTION 11 - Solutions and Their Properties SECTION 12 - Chemical Kinetics SECTION 13 - Chemical Equilibrium SECTION 14 - Acids and Bases SECTION 15 - Applications of Acid-Base Equilibria SECTION 16 - Spontaneity Entropy and Free Energy SECTION 17 - Electrochemistry SECTION 18 - Transition Metals and Coordination Chemistry SECTION 19 - The Nucleus and Nuclear Chemistry SECTION 2 - Atoms Molecules and Ions SECTION 20 - Intro to Organic Chemistry SECTION 3 - Chemical Quantities and Stoichiometry SECTION 4 - Types of Chemical Reactions and Solution Stoichiometry SECTION 5 - Gases SECTION 6 - Thermochemistry SECTION 7 - Quantum-Mechanical View of the Atom and Periodicity SECTION 8 - Chemical Bonding SECTION 9 - Covalent Bonding and Molecular Orbitals General Chemistry Notes Individual General Sections 1st Semester General (Sections 1-11) 2nd Semester General (Sections 12-20) Full Course General (Sections 1-20) Organic Chemistry Notes Individual Organic Sections 1st Semester Organic (Sections 1-12) 2nd Semester Organic (Sections 13-23) Full Course Organic (Sections 1-23) Video Notes My Blog and YouTube Channel only give you the first 3 videos from each of 20 Sections. 60 Blog Videos vs. ALL 143 Videos Inside. 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5098
https://probstatsdata.com/RBT.html
Chapter 9 Rank Based Tests | Probability, Statistics, and Data Type to search Probability, Statistics and Data: A Fresh Approach Using R Welcome Preface Software Installation 1 Data in R 1.1 Arithmetic and variable assignment 1.2 Help 1.3 Vectors 1.4 Indexing vectors 1.5 Data types 1.5.1 Missing data 1.6 Data frames 1.7 Reading data from files 1.8 Packages 1.9 Errors and warnings 1.10 Useful idioms Vignette: Data science communities Vignette: An R Markdown primer R Markdown files Document content Code chunks Onward Exercises 2 Probability 2.1 Probability basics 2.2 Simulations 2.2.1 Simulation with sample 2.2.2 Using simulation to estimate probabilities 2.2.3 Using replicate to repeat experiments 2.3 Conditional probability and independence 2.3.1 Independent events 2.3.2 Simulating conditional probability 2.3.3 Bayes’ Rule and conditioning 2.4 Counting arguments Vignette: Negative surveys Exercises 3 Discrete Random Variables 3.1 Probability mass functions 3.2 Expected value 3.3 Binomial and geometric random variables 3.3.1 Binomial 3.3.2 Geometric 3.4 Functions of a random variable 3.5 Variance, standard deviation, and independence 3.5.1 Independent random variables 3.6 Poisson, negative binomial, and hypergeometric 3.6.1 Poisson 3.6.2 Negative binomial 3.6.3 Hypergeometric Vignette: Loops in R Exercises 4 Continuous Random Variables 4.1 Probability density functions 4.2 Expected value 4.3 Variance and standard deviation 4.4 Normal random variables 4.4.1 Computations with normal random variables 4.4.2 Normal approximation to the binomial 4.5 Uniform and exponential random variables 4.5.1 Uniform random variables 4.5.2 Exponential random variables 4.6 Summary Exercises 5 Simulation of Random Variables 5.1 Estimating probabilities 5.2 Estimating discrete distributions 5.3 Estimating continuous distributions 5.4 Central Limit Theorem 5.5 Sampling distributions 5.5.1 The χ 2 χ 2 distribution 5.5.2 The t t distribution 5.5.3 The F F distribution 5.5.4 Summary 5.6 Point estimators 5.6.1 Properties of point estimators 5.6.2 Variance of unbiased estimators 5.6.3 MSE and bias-variance decomposition Vignette: Stein’s paradox Exercises 6 Data Manipulation 6.1 Data frames and tibbles 6.2 dplyr verbs 6.3 dplyr pipelines 6.3.1 Group by and summarize 6.4 The power of dplyr 6.5 Working with character strings 6.6 Structure of data 6.6.1 Tidy data: pivoting 6.6.2 Using join to merge data frames 6.7 The apply family Vignette: dplyr murder mystery Vignette: Data and gender Exercises 7 Data Visualization with ggplot 7.1 ggplot fundamentals 7.1.1 The grammar of graphics 7.1.2 Basic plot creation 7.1.3 Structured data 7.2 Visualizing a single variable 7.2.1 Histograms 7.2.2 Barplots 7.2.3 Density plots 7.2.4 Boxplots 7.2.5 QQ plots 7.3 Visualizing two or more variables 7.3.1 Scatterplots 7.3.2 Line graphs and smoothing 7.3.3 Faceting 7.4 Customizing 7.4.1 Color 7.4.2 Labels and themes 7.4.3 Text annotations 7.4.4 Highlighting Vignette: Choropleth maps Vignette: COVID-19 State-by-state comparisons Exercises 8 Inference on the Mean 8.1 Sampling distribution of the sample mean 8.2 Confidence intervals for the mean 8.3 Hypothesis tests of the mean 8.4 One-sided confidence intervals and hypothesis tests 8.5 Assessing robustness via simulation 8.5.1 Symmetric, light tails 8.5.2 Skew 8.5.3 Heavy tails and outliers 8.5.4 Independence 8.5.5 Summary 8.6 Two sample hypothesis tests 8.6.1 Two independent samples t t-test 8.6.2 Paired two sample t t-test 8.7 Type II errors and power 8.7.1 Effect size 8.8 Resampling 8.8.1 Bootstrapping 8.8.2 Permutation tests Exercises 9 Rank Based Tests 9.1 One sample Wilcoxon signed rank test 9.2 Two sample Wilcoxon tests 9.2.1 Paired two sample test 9.2.2 Independent two sample test 9.2.3 Ordinal data 9.3 Power and sample size 9.3.1 Sample size 9.4 Effect size and consistency 9.4.1 Effect size 9.4.2 Consistency 9.5 Summary Vignette: ROC curves and the Wilcoxon rank sum statistic Exercises 10 Tabular Data 10.1 Tables and plots 10.2 Inference on a proportion 10.2.1 Exact binomial test 10.2.2 One sample test of proportions 10.3χ 2 χ 2 tests 10.3.1χ 2 χ 2 test for given probabilities 10.4χ 2 χ 2 goodness of fit 10.4.1 Simulations 10.5χ 2 χ 2 tests on cross tables 10.5.1χ 2 χ 2 test of independence 10.5.2χ 2 χ 2 test of homogeneity 10.5.3 Two sample test for equality of proportions 10.6 Exact and Monte Carlo methods Vignette: Tables Exercises 11 Simple Linear Regression 11.1 Least squares regression line 11.2 Correlation 11.3 Geometry of regression 11.4 Residual analysis 11.4.1 Linearity 11.4.2 Heteroscedasticity 11.4.3 Normality 11.4.4 Outliers and leverage 11.4.5 Independence 11.5 Inference 11.5.1 The summary command 11.5.2 Confidence intervals for parameters 11.5.3 Prediction intervals for response 11.5.4 Confidence intervals for response 11.6 Simulations for simple linear regression 11.6.1 Residuals 11.6.2 Prediction intervals 11.7 Cross validation 11.8 Bias-variance tradeoff Vignette: Simple logistic regression Exercises 12 Analysis of Variance and Comparison of Multiple Groups 12.1 ANOVA 12.1.1 Groups and means 12.1.2 The ANOVA model 12.1.3 Simulations 12.2 The ANOVA test 12.2.1 Example: THC mice 12.2.2 Example: Humanization 12.3 Unequal variance 12.3.1 The oneway.test 12.3.2 Error simulations 12.4 Pairwise t t-tests 12.4.1 FWER and p p-value adjustments Vignette: Reproducibility Exercises 13 Multiple Regression 13.1 Two explanatory variables Call Residuals Coefficients Residual standard error Multiple R-squared F statistic 13.2 Categorical variables 13.2.1 Equal slopes model 13.2.2 Interaction terms 13.2.3 Cross validation 13.3 Variable selection Vignette: External data formats Exercises Published with bookdown Facebook Twitter LinkedIn Weibo Instapaper A A Serif Sans White Sepia Night Probability, Statistics, and Data Probability, Statistics, and Data: A Fresh Approach Using R Chapter 9 Rank Based Tests In a 2013 paper, D.K. Milton et al.68 studied the impact of wearing a surgical facemask on exhaled aerosol droplets for patients with influenza. Each subject performed two 30-minute trials, exhaling into a collection device (Figure 9.1). The subjects were tested once while wearing a facemask and once without, and the researchers counted the number of copies of the influenza virus in their fine particle droplets. Figure 9.1: Inlet cone for the human exhaled breath air sampler used to measure influenza virus, from Milton et al. Image credit: Milton et al.Image Credit Milton et al.This is an open-access article distributed under the terms of the Creative Commons Attribution License. Virus counts are in the variables mask_fine and no_mask_fine within the data set fosdata::masks. The difference no_mask_fine - mask_fine will be positive if the subject exhaled less influenza virus while wearing a mask. masks <- fosdata::masks %>% mutate(virus_difference = no_mask_fine - mask_fine) summary(masks$virus_difference) ``` Min. 1st Qu. Median Mean 3rd Qu. Max. -531 2 75 3809 417 102348 ``` The virus_difference variable has two large positive outliers (26422 and 102348) that make visualization challenging. Figure 9.2 shows a dotplot69 of the variable with those two values removed. Figure 9.2: Each dot represents one patient and gives the difference in fine particle virus count between their masked and maskless trials. A positive value means the exhaled virus count was larger when not wearing a mask. Two large positive points are omitted. Even without the largest values, it is obvious from the plot that most patients shed fewer virus particles while wearing a mask. The seven patients who shed more particles while wearing a mask had smaller differences in viral count than the 28 who shed fewer particles while wearing a mask. However, the presence of outliers (the two shown in the figure and the two even larger ones not shown) renders the t t-test powerless to detect a difference between masked and unmasked patients. A paired two-sample t t-test on this data results in a p p-value of 0.19, not significant. Wilcoxon rank based tests take the data and sort it from smallest to largest, replacing the actual measurements with their ranks. In this example, the ranks range from 1 to 37, for the 37 subjects of the study. Using ranks instead of the actual data values makes the test resistant to outliers while still accounting for the preponderance of positive values and the fact that those values tend to be larger. wilcox.test(masks$mask_fine, masks$no_mask_fine, paired = TRUE) ``` Wilcoxon signed rank test with continuity correction data: masks$mask_fine and masks$no_mask_fine V = 88.5, p-value = 0.000214 alternative hypothesis: true location shift is not equal to 0 ``` We interpret this result as saying that the median difference in viral count is positive, with p=0.0002 p=0.0002. The low p p-value suggests that masks reduce exhaled viral particles. In this chapter we discuss rank based tests that can be used, and are still effective, on a population with outliers. Rank based tests may also be used when the measured quantity is ordinal rather than numeric. 9.1 One sample Wilcoxon signed rank test The Wilcoxon signed rank test tests H 0:m=m 0 H 0:m=m 0 versus H a:m≠m 0 H a:m≠m 0, where m m is the median of the underlying population. We will assume that the data is centered symmetrically around the median (so that the median is also the mean). There can be outliers, bimodality or any kind of tails. Let’s look at how the test works by hand by examining a simple, made up data set. Suppose you wish to test H 0:m=6 v s H a:m≠6 H 0:m=6 v s H a:m≠6 You collect the data x 1=15,x 2=7,x 3=3,x 4=10 x 1=15,x 2=7,x 3=3,x 4=10 and x 5=13 x 5=13. The test works as follows: Compute y i=x i−m 0 y i=x i−m 0 for each i i. Here m 0=6 m 0=6 and we get y 1=9,y 2=1,y 3=−3,y 4=4 y 1=9,y 2=1,y 3=−3,y 4=4 and y 5=7 y 5=7. Let R i R i be the rank of the absolute value of y i y i. That is, R 1=5,R 2=1,R 3=2,R 4=3,R 5=4 R 1=5,R 2=1,R 3=2,R 4=3,R 5=4 since |y 1||y 1| is largest, |y 2||y 2| is smallest, etc. Let r i r i be the signed rank of R i R i; i.e., r i=R i×s i g n(y i)r i=R i×s i g n(y i), so r 1=5,r 2=1,r 3=−2,r 4=3,r 5=4 r 1=5,r 2=1,r 3=−2,r 4=3,r 5=4. Add all of the positive ranks. We get r 1+r 2+r 4+r 5=13 r 1+r 2+r 4+r 5=13. That is the test statistic for this test, and it is traditionally called V V. Compute the p p-value for the test, which is the probability that we get a test statistic V V which is this, or more, extreme relative to the expected value under the assumption of H 0 H 0. In order to perform the last step, we need to understand the sampling distribution of V V, under the assumption of the null hypothesis, H 0 H 0. The ranks of our five data points will always be the numbers 1, 2, 3, 4, and 5. When H 0 H 0 is true, each data point is equally likely to be positive or negative, and its corresponding rank will be included in the sum in item 4 above half of the time. So, the expected value is E(V)=1 2⋅1+1 2⋅2+1 2⋅3+1 2⋅4+1 2⋅5=15 2=7.5 E(V)=1 2⋅1+1 2⋅2+1 2⋅3+1 2⋅4+1 2⋅5=15 2=7.5 In our example, V=13 V=13, which is 5.5 away from the expected value of 7.5. For the probability distribution of V V, Table 9.1 lists all 2 5=32 2 5=32 possibilities of how the ranks could be signed. Since we have assumed that the distribution is centered about its mean, each of the possibilities is equally likely. Therefore, we can compute the proportion of rows in the table that lead to a test statistic at least 5.5 away from the expected value of 7.5. Table 9.1: All possible sums of positive signed ranks.| r1 | r2 | r3 | r4 | r5 | Sum of positive ranks | Far from 7.5? | ---: ---: :---: | -1 | -2 | -3 | -4 | -5 | 0 | Yes | | -1 | -2 | -3 | -4 | 5 | 5 | | | -1 | -2 | -3 | 4 | -5 | 4 | | | -1 | -2 | -3 | 4 | 5 | 9 | | | -1 | -2 | 3 | -4 | -5 | 3 | | | -1 | -2 | 3 | -4 | 5 | 8 | | | -1 | -2 | 3 | 4 | -5 | 7 | | | -1 | -2 | 3 | 4 | 5 | 12 | | | -1 | 2 | -3 | -4 | -5 | 2 | Yes | | -1 | 2 | -3 | -4 | 5 | 7 | | | -1 | 2 | -3 | 4 | -5 | 6 | | | -1 | 2 | -3 | 4 | 5 | 11 | | | -1 | 2 | 3 | -4 | -5 | 5 | | | -1 | 2 | 3 | -4 | 5 | 10 | | | -1 | 2 | 3 | 4 | -5 | 9 | | | -1 | 2 | 3 | 4 | 5 | 14 | Yes | | 1 | -2 | -3 | -4 | -5 | 1 | Yes | | 1 | -2 | -3 | -4 | 5 | 6 | | | 1 | -2 | -3 | 4 | -5 | 5 | | | 1 | -2 | -3 | 4 | 5 | 10 | | | 1 | -2 | 3 | -4 | -5 | 4 | | | 1 | -2 | 3 | -4 | 5 | 9 | | | 1 | -2 | 3 | 4 | -5 | 8 | | | 1 | -2 | 3 | 4 | 5 | 13 | Yes | | 1 | 2 | -3 | -4 | -5 | 3 | | | 1 | 2 | -3 | -4 | 5 | 8 | | | 1 | 2 | -3 | 4 | -5 | 7 | | | 1 | 2 | -3 | 4 | 5 | 12 | | | 1 | 2 | 3 | -4 | -5 | 6 | | | 1 | 2 | 3 | -4 | 5 | 11 | | | 1 | 2 | 3 | 4 | -5 | 10 | | | 1 | 2 | 3 | 4 | 5 | 15 | Yes | The last column in Table 9.1 is “Yes” if the sum of positive ranks is greater than or equal to 13 or less than or equal to 2. As you can see, we have that 6 of the 32 possibilities are at least as far away from the test statistic V=13 V=13 as our data, so the p p-value would be 6 32=.1875 6 32=.1875. Let’s check it with the built-in R command. wilcox.test(c(15, 7, 3, 10, 13), mu = 6) ``` Wilcoxon signed rank exact test data: c(15, 7, 3, 10, 13) V = 13, p-value = 0.1875 alternative hypothesis: true location is not equal to 6 ``` We see that our test statistic V=13 V=13 and the p p-value is 0.1875, just as we calculated. In general, if you have n n data points, the expected value of V V is E(V)=n(n+1)4 E(V)=n(n+1)4 (Exercise 9.3). To deal with ties, give each data point the average rank of the tied values. The sampling distribution of the test statistic V V under the null hypothesis is a built-in R distribution with root signrank. As usual, this function has prefixes d, p, q, and r, which correspond to the pmf, cdf, quantile function, and random generator, respectively. So, in the above example, we also could have computed the p p-value as the probability that V V is in {0,1,2,13,14,15}{0,1,2,13,14,15} as sum(dsignrank(c(0:2, 13:15), 4)) ## 0.1875 When the sample size n n is large, the sampling distribution of the test statistic V V is approximately normal with mean n(n+1)4 n(n+1)4 and variance n(n+1)(2 n+1)24 n(n+1)(2 n+1)24. Figure 9.3 shows the distribution of V V with n=10 n=10 with a superimposed normal curve that has mean 27.5 and standard deviation 9.81. Figure 9.3: pmf of V V when n=10 n=10 with superimposed normal distribution. Example 9.1 Consider the normtemp data set in the fosdata package. Recall that this data is the gender, body temperature and heart rate of 130 patients. We are interested in whether the mean or median body temperature is 98.6. A t t-test looks like this: normtemp <- fosdata::normtemp t.test(normtemp$temp, mu = 98.6) ``` One Sample t-test data: normtemp$temp t = -5.4548, df = 129, p-value = 2.411e-07 alternative hypothesis: true mean is not equal to 98.6 95 percent confidence interval: 98.12200 98.37646 sample estimates: mean of x 98.24923 ``` resulting in a p p-value of 2×10−7 2×10−7. If we instead perform the Wilcoxon signed rank test, we get wilcox.test(normtemp$temp, mu = 98.6) ``` Wilcoxon signed rank test with continuity correction data: normtemp$temp V = 1774.5, p-value = 1.174e-06 alternative hypothesis: true location is not equal to 98.6 ``` The p p-value still leads to the same conclusion at most reasonable α α levels, but it is about 5 times as large as the t t-test. Both the one sample t t-test and the Wilcoxon signed rank test are estimating the probability of obtaining data “this unlikely” or “more unlikely,” given that the null hypothesis is true. However, the two tests are summarizing what it means to be “this unlikely” in different ways. We will explore the pros and cons of the Wilcoxon signed rank test versus the one sample t t-test in Sections 9.3 and 9.4. 9.2 Two sample Wilcoxon tests In this section, we will explain how to use wilcox.test to compare two populations. We will split the discussion into the cases when the data is paired and when it is independent. 9.2.1 Paired two sample test In many studies, observations are naturally paired. Each observation consists of two measurements coming from two different populations. For example, the two measurements might be test scores before and after studying, opinions of a married couple, or pairs of agricultural fields matched for similar soil and weather characteristics. Any setting from Section 8.6 where a paired t t-test is appropriate may also be studied with a paired Wilcoxon test. The paired two sample Wilcoxon test has the following hypotheses and assumptions: H 0 H 0 : The two populations have the same distribution. H a H a : The two populations have different distributions. Assumptions 9.1 (for paired Wilcoxon test) 1. Subtracting measured values is possible and meaningful. We will need to assume that a difference of 2 is less than a difference of 3, for example. 2. The only dependence is between pairs of data points, one from each population. Unpaired observations are independent. Under these assumptions, the differences in the paired values in the two populations will be a symmetric distribution under the null hypothesis. Therefore, assuming the null hypothesis is true, we can use the Wilcoxon signed rank test on the differences of the values between the two populations. Example 9.2 The flint data set in the fosdata package gives the results of tap water lead testing during the Flint, Michigan water crisis in 2015.70 In the study, households filled three sample collection bottles from their faucets. One bottle was filled at “first draw,” after the faucet had been off for 6 hours. The second and third bottles were filled after the faucet had been running for 45 seconds, and after it had been running for 2 minutes. All bottles were sent to a lab at Virginia Tech for lead testing. Does the length of time that a faucet is left on influence the measurement of lead in the water coming from the faucets? We restrict to two populations: the amount of lead in the water on first draw (Pb1) and the amount of lead in the water two minutes later (Pb3). This is paired data. It would be wrong to assume independence between these two measurements within one household in this experiment. A faucet that has high levels of lead at first draw is also likely to have higher levels of lead two minutes later. A careful look at the data reveals two houses that have notes saying they were tested twice. It is unreasonable to expect those pairs of data points to be independent of each other, so we remove the observations that correspond to those houses. This is a reasonable thing to do, especially if the houses weren’t tested twice based on the outcome of the first test. flint <- fosdata::flint flint <- flint %>% filter(Notes == "") The null hypothesis is that the distribution of Pb1 (lead draw in first sample) and Pb3 (lead draw after 2 minutes) are the same. Under the null hypothesis, the difference between values in Pb1 and Pb3 will be symmetric, so the assumptions of the Wilcoxon signed rank test are met on the differences. We start by providing a histogram of the differences. flint %>% ggplot(aes(x = Pb1 - Pb3)) + geom_histogram() + labs(title = "Difference in lead levels: first draw and after 2 minutes") The histogram indicates that the difference in lead levels is positive and smallish, with a heavy right tail and outliers both positive and negative. The heavy tail and outliers make this data unsuitable for a t t-test. We perform the paired two sample Wilcoxon test as follows: wilcox.test(flint$Pb1, flint$Pb3, paired = TRUE) ``` Wilcoxon signed rank test with continuity correction data: flint$Pb1 and flint$Pb3 V = 32931, p-value < 2.2e-16 alternative hypothesis: true location shift is not equal to 0 ``` The p p-value is very low, so we reject H 0 H 0 and conclude that running the water does make a significant difference to the lead level. 9.2.2 Independent two sample test This section uses wilcox.test to compare independent samples from two populations, referred to as the Wilcoxon rank sum test. We no longer need to assume that the population distributions are symmetric, but we must assume that all of the observations are independent. The null hypothesis and alternative hypotheses can be stated as follows. H 0 H 0: the distribution of population one is the same as the distribution of population two. H a H a: the distribution of population one is different than the distribution of population two. To perform the test, we choose a random sample of size m m from population one and independently choose a random sample of size n n from population two. For each data point x x in group one, count the number of data points in group two that are less than x x. The total number of “wins” for group one (counting ties as 0.5) is the test statistic W W. Under the null hypothesis, the expected value of the test statistic is E[W]=m n 2,E[W]=m n 2, which is exactly half of the possible pairings between the two groups. Let’s look at a simple example for concreteness. Suppose that we obtain the following data, which has two observations in group one and three observations in group two. | group | value | --- | | 1 | 5 | | 1 | 11 | | 2 | 0 | | 2 | 3 | | 2 | 10 | The group one value x=5 x=5 is bigger than two observations in group two, and the group one value x=11 x=11 is larger than three observations in group two. So, the value of the test statistic is W=2+3=5 W=2+3=5. The expected value of the test statistic is 2×3/2=3 2×3/2=3. The p p-value of the test is the probability (under the null) of obtaining a test statistic either 5 or larger, or obtaining one that is 1 or smaller, i.e., P(|W−3|≥2)P(|W−3|≥2). To compute this probability, we imagine that we have sorted the 5 values from smallest to largest. Under the null hypothesis, each possible arrangement of group one and group two within the sorted values would be equally likely. There are (5 3)=10(5 3)=10 possible permutations of the values, shown as the ten rows of Table 9.2. Table 9.2: Possible permutations of three 2’s and two 1’s, with associated test statistic. | V1 | V2 | V3 | V4 | V5 | W | --- --- --- | | 2 | 2 | 2 | 1 | 1 | 6 | | 2 | 2 | 1 | 2 | 1 | 5 | | 2 | 1 | 2 | 2 | 1 | 4 | | 1 | 2 | 2 | 2 | 1 | 3 | | 2 | 2 | 1 | 1 | 2 | 4 | | 2 | 1 | 2 | 1 | 2 | 3 | | 1 | 2 | 2 | 1 | 2 | 2 | | 2 | 1 | 1 | 2 | 2 | 2 | | 1 | 2 | 1 | 2 | 2 | 1 | | 1 | 1 | 2 | 2 | 2 | 0 | Under the assumption that the distribution of population one is the same as the distribution of population two, each outcome is equally likely. There are four rows of Table 9.2 with |W−3|≥2|W−3|≥2, so the p p-value is P(|W−3|≥2)=4/10 P(|W−3|≥2)=4/10. The sampling distribution of the test statistic W W is in R with root name wilcox, where as always, prefixes d, p, q, and r correspond to the pmf, cdf, quantile function, and random generator, respectively. For example, if we wish to compute P(|W−3|≥2)P(|W−3|≥2) when the two sample sizes are 2 and 3 as above, we could do the following. sum(dwilcox(c(0, 1, 5, 6), m = 2, n = 3)) ## 0.4 Figure 9.4 shows the distribution of the test statistic W W with group sizes m=5 m=5 and n=10 n=10. The sampling distribution of the test statistic W W is approximately normal when the groups are even modestly large. Figure 9.4: pmf of W W when m=5 m=5, n=10 n=10, with superimposed normal distribution. The function wilcox.test computes the test statistic W W and the p p-values in one simple step. wilcox.test(c(5, 11), c(0, 3, 10)) ``` Wilcoxon rank sum exact test data: c(5, 11) and c(0, 3, 10) W = 5, p-value = 0.4 alternative hypothesis: true location shift is not equal to 0 ``` Example 9.3 Hillier et al.71 studied whether academic papers on climate change got more or fewer citations based on their narrative style. They assessed various aspects of the narrative style of the papers and also counted the number of citations. The data set climate in the fosdata package summarizes their data for articles published in three well-respected journals. Let’s load the data and do some exploring. climate <- fosdata::climate For the purposes of this example, we are going to look at whether the distribution of articles with an appeal in the abstract are associated with different citation rates than those without an appeal. An appeal being “Yes” indicates that the author made an explicit appeal to the reader or a clear call for action. The normalized_citation variable gives the number of citations per year since publication. Let’s restrict to those two variables to clean things up a bit. climate <- climate %>% select(binary_appeal, normalized_citations, abstract_number) climate <- climate %>% mutate(climate, binary_appeal = factor(binary_appeal, levels = 0:1, labels = c("No", "Yes")) ) ggplot(climate, aes(x = binary_appeal, y = normalized_citations)) + geom_boxplot(outlier.size = -1) + geom_jitter(height = 0, width = 0.2) It’s a bit hard to see what is going on. Let’s compute some summary statistics. climate %>% group_by(binary_appeal) %>% summarize( mean = mean(normalized_citations), sd = sd(normalized_citations), skew = e1071::skewness(normalized_citations), N = n() ) ``` # A tibble: 2 × 5 binary_appeal mean sd skew N 1 No 15.5 19.6 4.54 213 2 Yes 16.7 18.4 4.17 519 ``` One thing that we notice is that there is quite a bit of skewness in these variables. The authors took the log of the citation variable, which lowered the skewness and made the data look more normal. A t t-test without transforming the data would be problematic even with 200-500 samples because of the level of skewness of the data, as well as the outliers. Instead, we will use a Wilcoxon rank sum test with hypotheses: H 0 H 0: the distribution of papers with an appeal is the same as the distribution of papers without an appeal. H a H a: the distribution of papers with an appeal is different than the distribution of papers without an appeal. wilcox.test(normalized_citations ~ binary_appeal, data = climate) ``` Wilcoxon rank sum test with continuity correction data: normalized_citations by binary_appeal W = 47608, p-value = 0.003182 alternative hypothesis: true location shift is not equal to 0 ``` Note the use of the formula notation ~ in wilcox.test; this is similar to its use in t.test, where we are telling R how we want to group the data. We reject the null hypothesis that the two distributions are the same at the α=.05 α=.05 level. Recall that the expected value of the test statistic is 213 519 / 2 ## 55273.5 The test statistic that we obtained is 47608, which is lower than the expected value. This indicates that the “no appeal” papers are cited less frequently than would be expected under the null hypothesis. 9.2.3 Ordinal data Another common use of wilcox.test is when the data is ordinal rather than numeric. Definition 9.4 Data is ordinal when, given data points p 1 p 1 and p 2 p 2, we can determine whether p 1 p 1 is bigger than p 2 p 2, less than p 2 p 2, or equal to p 2 p 2. For example, airplane passengers may fly first class, business class, or coach. The seat type of an airline passenger is ordinal data because the three types of seats are ordered by quality. Ordinal data is different from numeric data because there is no natural way to assign numbers to ordinal data. It is quite common to assign numbers to ordinal categories, but that doesn’t necessarily mean that it is a correct thing to do. Example 9.5 Surveys commonly ask for responses in the form of Likert-scale data. The Likert scale is an ordinal five-point scale with choices reflecting agreement with a statement. Often, those options are given numeric values, as in this example: Circle the response that most accurately reflects your opinion | Question | Strongly Agree | Agree | Neutral | Disagree | Strongly Disagree | :---: :---: :---: | | Pride and Prejudice | | | | | | | is a great book | 1 | 2 | 3 | 4 | 5 | | | | | | | | | wilcox.test is a | | | | | | | useful R command | 1 | 2 | 3 | 4 | 5 | Although the options are listed as numbers, it does not necessarily make sense to treat the responses numerically. For example, if one respondent Strongly Disagrees that wilcox.test is useful, and another Strongly Agrees, is the mean of those responses Neutral? What is the mean of Strongly Agree and Agree? If the responses are numeric, what units do they have? So, it may not (and often does not) make sense to treat the responses to the survey as numeric data, even though it is encoded as numbers. For compelling examples of how numbering Likert scales can lead you astray, see the somewhat whimsical booklet, “Do not use averages with Likert scales.”72 What we can do with Likert scale data is to order the responses in terms of amount of agreement. It is clear that Strongly Agree > Agree > Disagree > Strongly Disagree. Neutral is omitted because Neutral responses open a whole new can of worms. For example, we don’t know whether someone who put Neutral for Question (1) did so because they have never read Pride and Prejudice, or because they read it and don’t have an opinion on whether it is a great book. In one case, the response might better be treated as missing data, while in the other it would be natural to include the result between Agree and Disagree. Example 9.6 Consider the MovieLens data set fosdata::movies from Chapter 6. Each movie review has a star rating, which is a subjective opinion of how much enjoyment the viewer had. The number of stars that a movie receives can definitely be ordered from 5 stars being the best, down to 1/2 star being the worst. It is not as clear that the number of stars should be treated numerically. To perform inference with the star ratings, we prefer the Wilcoxon test, which depends only on the ordinal nature of the data. Let’s compare the movies Toy Story and Toy Story 2. Recall that the data that we have in our MovieLens data set is a random sample from all users. Is there sufficient evidence to suggest that there is a difference in the ratings of Toy Story and Toy Story 2 in the full data set? The movies data set consists of observations pulled from 610 distinct users. If one user rates both Toy Story and Toy Story 2, those two ratings are not independent of each other. So, let’s recast the question as follows. Among those people who have only rated one or the other movie, is there evidence of a difference in the ratings of Toy Story and Toy Story 2 in the full data set? We should also decide whether we want to do a one-sided or a two-sided test. Some people have strong opinions about which Toy Story movie is better, but that is not sufficient evidence or reason to do a one-sided test. So, let X X be the rating of Toy Story for a randomly sampled person in the large data set who only rated one of the two movies. Let Y Y be the rating of Toy Story 2 for a randomly sampled person who only rated one of the two movies. Our null and alternative hypotheses are: H 0 H 0: X X and Y Y have the same distribution. H a H a: X X and Y Y do not have the same distribution. After loading the data, we create a data frame that only has the ratings of the first two Toy Story movies (by excluding Toy Story 3 which came out years later). toy_story <- fosdata::movies %>% filter(stringr::str_detect(title, "Toy Story [^3]")) Next, we filter out only those ratings from people who rated just one of the two movies. toy_story_once <- toy_story %>% group_by(userId) %>% mutate(N = n()) %>% filter(N == 1) Finally, we perform the Wilcoxon test: wilcox.test(rating ~ title, data = toy_story_once) ``` Wilcoxon rank sum test with continuity correction data: rating by title W = 1060.5, p-value = 0.9448 alternative hypothesis: true location shift is not equal to 0 ``` There is not a statistically significant difference (p=0.94 p=0.94) between the ratings of Toy Story and Toy Story 2. We do not reject the null hypothesis. 9.3 Power and sample size In this section, we compare the power of a t t-test with that of a one sample Wilcoxon test. We do so with three populations: normal, heavy tails, and normal with an outlier. We begin by comparing the power of Wilcoxon rank sum test with t t-test when the underlying data is normal. We assume we are testing H 0:μ=1 H 0:μ=1 versus H a:μ≠1 H a:μ≠1 at the α=.05 α=.05 level, when the underlying population is truly normal with mean 0 and standard deviation 1 with a sample size of 10. Let’s first estimate the percentage of time that a t t-test correctly rejects the null-hypothesis. mean(replicate(10000, t.test(rnorm(10, 0, 1), mu = 1)$p.value < .05)) ## 0.8011 We see that we correctly reject H 0 H 0 80.1% of the time in this simulation. Let’s see what happens with the Wilcoxon test. mean(replicate(10000, wilcox.test(rnorm(10, 0, 1), mu = 1)$p.value < .05)) ## 0.7864 Here, we see that we correctly reject H 0 H 0 78.6% of the time in this simulation. If you repeat the simulations, you will see that indeed a t t-test correctly rejects H 0 H 0 more often than the Wilcoxon test does, on average. However, if there is a departure from normality in the underlying data, Wilcoxon can outperform t.test. Let’s repeat the above simulations with a heavy tailed distribution. We sample from a t t distribution with 3 degrees of freedom. mean(replicate(10000, t.test(rt(10, 3), mu = 1)$p.value < .05)) ## 0.5378 mean(replicate(10000, wilcox.test(rt(10, 3), mu = 1)$p.value < .05)) ## 0.544 Here, we see that Wilcoxon outperforms t t, but not by a tremendous amount. Finally, we look at the third example, which is normal with a single outlier. Our model for an outlier will be that we multiply one of the values by 100. This does not change the mean of the distribution (since it is still zero), but it will often introduce relatively large outliers. We can simulate data of this type as follows: dat <- rnorm(10, 0, 1) dat <- dat 100 dat ``` -0.07719331 1.84950460 -0.08484561 0.09717147 -0.59400957 -0.60122548 0.55263885 -1.04205402 -0.09278696 -57.35763723 ``` If you run the above code multiple times, you will see that the last value usually does not appear to come from a standard normal distribution, though sometimes it can appear to do so. We again estimate the power of a t t-test and a one sample Wilcoxon test on this type of data, again with H 0:μ=1 H 0:μ=1. Since the simulations are a bit more complicated, we write out the simulations in a longer format. p_vals <- replicate(10000, { dat <- rnorm(10, 0, 1) dat <- dat 100 t.test(dat, mu = 1)$p.value }) mean(p_vals < .05) ## 0.033 p_vals <- replicate(10000, { dat <- rnorm(10, 0, 1) dat <- dat 100 wilcox.test(dat, mu = 1)$p.value }) mean(p_vals < .05) ## 0.4275 Now we see a substantial difference in the power of the two tests, with the one sample Wilcoxon test being much more powerful. Alert 9.1 If your data is symmetric with outliers, the one sample Wilcoxon test will be much more powerful than the one sample t t-test. To summarize the results of this section: If your population is normal, then a t t-test will be a little bit more powerful than a one sample Wilcoxon test. If your population has moderately heavy tails, then a one sample Wilcoxon test will likely be a little bit more powerful than a t t-test. If your population has outliers, you should use a one sample Wilcoxon test rather than a t t-test. 9.3.1 Sample size When performing rank-based tests, we recommend using simulation to estimate power and sample sizes. With power computations, we will set the significance level α α we wish to test, and we will need some estimate of the effect size we wish to detect. This can be challenging to do well. We illustrate the technique with an example. Example 9.7 Suppose we are designing an experiment in which students are randomly assigned to either a traditional classroom or a flipped classroom. At the end of the semester, students will be given a statement and asked whether they Strongly Agree, Agree, Disagree or Strongly Disagree with the statement. We wish to determine whether there is a statistically significant difference in the responses between those in the traditional classroom and those in the flipped classroom. For an experiment with significance level α=0.05 α=0.05 and power of 80%, what number of students would need to be assigned to each group? To perform the simulation, we need to estimate the specific distribution of responses in the two groups under the alternate hypothesis. The power computation will tell us how large a sample we need to take to be able to detect the difference between our estimated distributions. Without knowledge of what sort of effect size matters to educational decisions, this is impossible. So imagine we read literature about educational studies, and decide that a reasonable effect would be probability distributions given in the table below. | | Traditional | Flipped | --- | Strongly Agree | 0.4 | 0.6 | | Agree | 0.3 | 0.3 | | Disagree | 0.1 | 0.1 | | Strongly Disagree | 0.2 | 0.0 | We recommend estimating the power for various sample sizes and plotting a smoothed version of the resulting curve. When using this technique, we don’t need to estimate each probability as accurately as we would if we were only doing it once. ``` p_trad <- c(.4, .3, .1, .2) p_flip <- c(.6, .3, .1, 0) sample_sizes <- seq(10, 150, length.out = 20) powers <- sapply(sample_sizes, function(x) { tt <- replicate(500, { d1 <- sample(1:4, x, T, prob = p_trad) d2 <- sample(1:4, x, T, prob = p_flip) suppressWarnings(wilcox.test(d1, d2))$p.value }) mean(tt < .05) }) for_plot <- data.frame( sample_size = sample_sizes, power = powers ) ``` ggplot(for_plot, aes(x = sample_size, y = power)) + geom_smooth(se = F) + geom_point() We read off of the plot that in order to have a power of about 0.8, we would need about 58 samples in each group. 9.4 Effect size and consistency 9.4.1 Effect size When communicating statistical results to others, it may be useful to include a common language effect size. The goal of a common language effect size statistic is to present results in a way that is easy to understand without advanced statistical training. We recommend the following for the two sample Wilcoxon rank sum test. Definition 9.8 Given two populations, Vargha and Delaney’s A A is the probability that a sample from one population will be larger than a sample from the other population plus one-half the probability that they will be equal. Intuitively, Vargha and Delaney’s A A is the probability that a sample from one population will be larger than a sample from the other population, where we are assuming that ties are broken 50-50. The effect size A A is directly related to the test statistic for the two sample Wilcoxon rank sum test. The Wilcoxon test statistic W W counts the number of times that samples from population one are larger than those in population two, plus one half the number of ties. To compute A A, divide W W by the number of comparisons that are made. Example 9.9 Let’s make sense of the difference between Sense and Sensibility (1995) and The Sixth Sense (1999), the two most popular movies with “sense” in the title. The corresponding movieId’s for those movies are 17 and 2762. Which movie is preferred, is there a significant difference, and what is the effect size? To maintain independence of the samples, restrict to those raters who only rated one of the two movies: sensible_movies <- fosdata::movies %>% filter(movieId %in% c(17, 2762)) %>% group_by(userId) %>% mutate(N = n()) %>% filter(N == 1) Now test whether the distribution of ratings are the same for these two sensible movies: wilcox.test(rating ~ title, data = sensible_movies) ``` Wilcoxon rank sum test with continuity correction data: rating by title W = 3089.5, p-value = 0.65 alternative hypothesis: true location shift is not equal to 0 ``` The p p-value is 0.65, so there is not a significant difference in the ratings. For the common language effect size A A, we need to know the number of comparisons. How many ratings of each sensible movie were there? table(sensible_movies$title) ``` Sense and Sensibility (1995) Sixth Sense, The (1999) 42 154 ``` Since each pair of ratings is compared, there are 42×154=6468 42×154=6468 comparisons. Then A=W/6468=3089.5/6468≈47.8%A=W/6468=3089.5/6468≈47.8%. Since Sense and Sensibility comes first in the ordering of titles, Sense and Sensibility is preferred by about 47.8% of raters, while The Sixth Sense is preferred by about 53.2% of raters. (This is only formally true if we imagine that for ties we flip a coin.) Without using the W W statistic, we can compute A A directly: s_and_s_ratings <- filter(sensible_movies, movieId == 17) %>% pull(rating) t_s_s_ratings <- filter(sensible_movies, movieId == 2762) %>% pull(rating) sum(sapply(s_and_s_ratings, function(x) { sum(x > t_s_s_ratings) + 1 / 2 sum(x == t_s_s_ratings) })) / (length(s_and_s_ratings) length(t_s_s_ratings)) ## 0.4776592 There is a package, effsize, that has a function to compute Vargha and Delaney’s A A from the data with no manipulation: effsize::VD.A(rating ~ title, data = sensible_movies) ``` Vargha and Delaney A A estimate: 0.4776592 (negligible) ``` It’s still 47.8% in favor of Sense and Sensibility, described as a “negligible” effect. You should take the adjectives assigned to effect sizes by the effsize package with a grain of salt, as effect sizes are domain dependent. As with Cohen’s d d, what is considered a negligible effect in one field might be considered a large effect in a different field. 9.4.2 Consistency In this section, we discuss the consistency of the one sample t.test and the two independent sample wilcox.test. Definition 9.10 A hypothesis test is consistent if the probability of rejecting the null hypothesis given that the null hypothesis is false converges to 1 as the sample size goes to infinity. Consistency is a desirable property for a hypothesis test because it means that a sufficiently large sample size will always have the power to detect a false H 0 H 0. We will not provide proofs of consistency, but instead examine it via simulations. Example 9.11 Consider the t.test with one population, where we are testing H 0:μ=0 H 0:μ=0 versus H a:μ≠0 H a:μ≠0. We assume that the underlying population has mean μ a≠0 μ a≠0, so H 0 H 0 is false. Also assume the population is normally distributed with standard deviation σ=1 σ=1, and choose significance level α=0.01 α=0.01. We start by computing the percentage of times H 0 H 0 will be rejected when μ a=1 μ a=1 and the sample size is n=20 n=20. sigma <- 1 nsize <- 20 mua <- 1 pvals <- replicate(10000, { dat <- rnorm(nsize, mua, sigma) t.test(dat, mu = 0)$p.value }) mean(pvals < .01) ## 0.9338 We see that with 20 samples, we already reject H 0 H 0 about 93% of the time. By repeating this simulation with various values of n n and μ a μ a, we create the plot in Figure 9.5. It appears that the probability of rejecting H 0 H 0 is getting larger as the sample size increases, and it also appears to be converging to 1. We would need to take more samples to see this for μ a=0.02 μ a=0.02, but 0.02 is a very small difference between the null and alternative hypothesis. Figure 9.5: Each curve shows the probability that a t t-test rejects H 0 H 0 as sample size grows. The different curves correspond to different population means μ a μ a. As μ a μ a gets closer to the hypothesized μ=0 μ=0, larger samples are needed to detect the difference. The story for the Wilcoxon rank sum test is a bit more involved. We begin with an example where the null hypothesis is false, but we see that the test is not consistent. Example 9.12 Let population 1 consist of numbers uniformly distributed on [−1,1][−1,1]. Let population 2 consist of numbers uniformly distributed on [−2,−1][−2,−1] with probability 1/2 1/2, and numbers uniformly distributed on [1,2][1,2] with probability 1/2 1/2, as shown. Fixing α=0.05 α=0.05, let’s see what percentage of times, on average, the Wilcoxon test rejects H 0 H 0 with a sample size of n=100 n=100. N <- 100 sim_data <- replicate(10000, { d1 <- runif(N, -1, 1) d2 <- sample(c(-1, 1), N, replace = TRUE) runif(N, 1, 2) wilcox.test(d1, d2)$p.value }) mean(sim_data < .05) ## 0.0889 We see that the test rejects H 0 H 0 more than with probability α=.05 α=.05, but the test still has very low power. In other words, it does not do a good job of distinguishing population 1 from population 2. Repeating the computation with different sample sizes n n produces Table 9.3. Table 9.3: Power of Wilcoxon test. | n | power | --- | | 10 | 0.105 | | 20 | 0.098 | | 50 | 0.125 | | 100 | 0.092 | | 200 | 0.107 | | 500 | 0.101 | | 1000 | 0.106 | From Table 9.3 we see that as n n grows, the power of the test is not approaching 1. It can be shown that when n→∞n→∞, the power converges to about 0.1095 0.1095. With these populations, the Wilcoxon rank sum test is not consistent. However, we do have the following theorem: Theorem 9.1 Suppose X X and Y Y are random variables with finite variance and P(X>Y)≠P(XY)≠P(X<Y). Then the Wilcoxon rank sum test of X X against Y Y is consistent. That is, as the number of samples goes to infinity, the power converges to 1. The condition in Theorem 9.1 is equivalent to a statement about Vargha and Delaney’s A A. If A≠0.5 A≠0.5, then the two sample Wilcoxon test is consistent. Alert 9.2 We recommend using the Wilcoxon rank sum test only when we suspect, or wish to detect, a difference in populations that is of the type P(X>Y)≠P(XY)≠P(X<Y). In this case, the test is consistent. To finish off this section, we provide an example of two populations such that P(X>Y)≠P(XY)≠P(X<Y), and give evidence with simulation that the Wilcoxon rank sum test is consistent. Example 9.13 Let X X and Y Y be random variables, with X∼Unif(1,2)X∼Unif(1,2) and Y Y taking values 0 and 3 with probabilities p<1/2 p<1/2 and 1−p 1−p. Then P(X>Y)=p P(X>Y)=p and P(X<Y)=1−p P(X<Y)=1−p. Fix significance level α=0.01 α=0.01. Here is a single computation when p=0.4 p=0.4 and the sample size n=100 n=100. p <- 0.4 N <- 100 pvals <- replicate(1000, { x <- runif(N, 1, 2) y <- sample(c(0, 3), size = N, prob = c(p, 1 - p), replace = TRUE) suppressWarnings(wilcox.test(x, y))$p.value }) mean(pvals < 0.01) ## 0.416 The probability of correctly rejecting H 0 H 0 in this case is 0.416. Repeating the simulation for different values of p p and n n produces the plot shown in Figure 9.6. From the figure, it appears that the percentage of times H 0 H 0 is rejected converges to 1 as the sample size goes to infinity, so the test is consistent. Values of A A closer to 1/2 1/2 require larger sample sizes. Figure 9.6: Each curve shows the probability that a Wilcoxon test rejects H 0 H 0 as sample size grows. The curves correspond to fixed population values of A=P(X<Y)A=P(X<Y). As A A gets closer to 1/2, larger samples are needed to detect the difference between X X and Y Y. 9.5 Summary In this chapter, we provided an alternative to t.test. In the one sample case, we can use the Wilcoxon signed rank test for a random sample from a symmetric population. The Wilcoxon signed rank test is preferred over t.test when either the data is ordinal and means cannot be taken, or when in the presence of outliers. For a population that is normally distributed, use t.test because it has slightly larger power. When the population distribution is skewed or unknown (but with no outliers), use t.test with the caveats regarding sample size presented in Chapter 8. In the independent two sample case, we can use the Wilcoxon rank sum test as an alternative to t.test. Use the Wilcoxon rank sum test when the data is ordinal, when there may be outliers, or when the sample sizes are small enough that the Central Limit Theorem may not apply. Recall that the sample size required for the Central Limit Theorem to apply depends on the distribution. Use t.test when the populations are normal or when the sample size is large enough that the Central Limit Theorem applies. If the data is ordinal, but can be transferred meaningfully to a numeric scale, then you can use t.test on the transferred data. However, it can be difficult to re-interpret your result in the original scale when you do this. For the two sample Wilcoxon rank sum test, the null hypothesis is that the populations are the same and the alternative is that they are different. The power of a Wilcoxon rank sum test goes to 1 as the sample size goes to infinity in the case that P(X>Y)≠P(XY)≠P(XY)≠P(XY)≠P(XY)P(X>Y). If the populations appear to be different for reasons other than P(X>Y)≠P(XY)≠P(X<Y), such as having different variances, then the Wilcoxon rank sum test will typically have poor power, and should not be used. For a more advanced theoretical treatment of this topic, we recommend the text by Lehmann.73 In the paired two sample case, we can use the Wilcoxon signed rank test on the differences as an alternative to a t t-test on the differences. For the Wilcoxon signed rank test to be valid, the differences in values must be meaningful, and in particular, must satisfy the conditions of the Wilcoxon signed rank test. Use Wilcoxon signed rank when the data is ordinal with meaningful differences or when there are outliers. Vignette: ROC curves and the Wilcoxon rank sum statistic Suppose you wish to classify an object into one of two groups. It would be helpful if there were a variable X X and a value x 0 x 0 such that whenever X<x 0 X<x 0 we would classify the object into group 1, and whenever X≥x 0 X≥x 0 we would classify the object into group 2. A receiver operating characteristic (ROC) curve is a graphical measurement of how well a variable discriminates between two alternatives. As an example, let’s consider palmerpenguins::penguins and suppose that we are trying to distinguish between Adelie and chinstrap penguins, based solely on bill length. penguins <- palmerpenguins::penguins %>% filter(species %in% c("Adelie", "Chinstrap")) %>% mutate(species = droplevels(species)) ggplot( penguins, aes(x = bill_length_mm, y = flipper_length_mm, color = species) ) + geom_point() + geom_vline(xintercept = c(40.9, 46), linetype = "dashed") Based on this graph, we can see that there is not a value of bill length that completely separates the two variables. Whatever x 0 x 0 we pick so that we assign any penguin with bill length less than x 0 x 0 to be Adelie, we will be misclassifying some of the penguins. The ROC curve is computed as follows. Start at the smallest bill length and imagine that is your split value; all penguins with bill length less than that are classified as Adelie. The y y-coordinate is the percentage of penguins correctly classified as Adelie with that split value, and the x x-coordinate is the percentage of penguins incorrectly classified as Adelie with that split value. An ideal ROC curve would look like this: This means that there is at least one x 0 x 0 such that all of the objects in group 1 have values less than x 0 x 0, while all of the objects in group 2 have values at least x 0 x 0. Let’s look at the ROC curve for the penguins using the ROCR package. penguins <- penguins %>% select(species, bill_length_mm) penguins <- penguins[complete.cases(penguins), ] pred <- ROCR::prediction(penguins$bill_length_mm, penguins$species) perf <- ROCR::performance(pred, "tpr", "fpr") plot(perf, colorize = TRUE, main = "ROC curve for two penguin species") Note that the curve looks similar to the idealized version, except that it cuts off the corner at the top left. The curve there is green, which corresponds to bill lengths of around 45 mm. Those values of bills are exactly where we are unsure how the penguin should be classified. The area underneath an ideal ROC curve is 1. For the penguin ROC curve, a little bit of area is missing in the upper left corner. The area under the ROC curve is abbreviated AUC and measures the separation between the two groups. It is used as a performance measure for machine learning algorithms. Proposition 9.1 The area under the ROC curve (AUC) is equal to Vargha and Delaney’s A A. Proposition 9.1 allows another interpretation of the AUC: it measures the probability that a randomly selected member of group 2 has a higher value of the variable under question than a randomly selected member of group 1. We will not prove Proposition 9.1, but the following code verifies that three ways of computing the AUC are the same in this context. penguins <- penguins %>% mutate(species = factor(species, levels = c("Chinstrap", "Adelie"))) wilcox.test(bill_length_mm ~ species, data = penguins)$statistic / (151 68) ``` W 0.9901636 ``` effsize::VD.A(bill_length_mm ~ species, data = penguins) ``` Vargha and Delaney A A estimate: 0.9901636 (large) ``` perf <- ROCR::performance(pred, "auc") perf@y.values ## 0.9901636 Exercises Exercises 9.1 – 9.6 require material through Section 9.1. 9.1 Suppose you wish to test H 0:μ=2 H 0:μ=2 versus H a:μ≠2 H a:μ≠2 using a Wilcoxon signed rank test. If you collect 30 samples, what is the expected value of the test statistic V V under the null hypothesis? 9.2 Suppose you are testing H 0:μ=0 H 0:μ=0 versus H a:μ≠0 H a:μ≠0. You collect 40 data points and compute V=601 V=601. Use the built-in R function dsignrank or psignrank to compute the p p-value associated with the test. 9.3 Suppose x 1,…,x n x 1,…,x n is a random sample from a symmetric distribution with mean 0. Show that the expected value of the Wilcoxon test statistic V V is E(V)=n(n+1)4 E(V)=n(n+1)4. Find lim x n→∞E[V]lim x n→∞E[V], which is the scenario of a single, arbitrarily large outlier. 9.4 Suppose you wish to test H 0:μ=3 H 0:μ=3 versus H a:μ≠3 H a:μ≠3. You collect the data points x 1=−4 x 1=−4, x 2=0 x 2=0, x 3=2 x 3=2 and x 4=8 x 4=8. Go through all of the steps of a Wilcoxon signed rank test and determine a p p-value. Check your answer using R. 9.5 In this example, we explore what happens when applying the Wilcoxon signed rank test to skewed data. Suppose you take a random sample of size 20 from an exponential random variable with rate 1. The mean of the distribution is 1, and the median is log(2)log⁡(2). All tests below are to be conducted at the α=.05 α=.05 level. Estimate the effective type I error when testing H 0:μ=1 H 0:μ=1 versus H a:μ≠1 H a:μ≠1 in this setting. Estimate the effective type I error when testing H 0:m=log(2)H 0:m=log⁡(2) versus H a:m≠log(2)H a:m≠log⁡(2) in this setting. Even though the test is not working correctly for either of those values, there is a value that makes the test work approximately correctly. The value is called a pseudo-median, and for exponential random variables with rate 1 it is approximately 0.84. Confirm that the effective type I error of testing H 0:m=.84 H 0:m=.84 versus H a:m≠.84 H a:m≠.84 is approximately 0.05. (It is, based on our simulations, slightly larger than .05.) 9.6 Consider the weight_estimate data set in the fosdata package. Children, and some adults, were asked to estimate the weight of an object while watching a professional actor pick it up. For the purposes of this problem, consider only the mean200 variable. Plot a boxplot of mean200. Does it appear to be reasonable symmetric? Conduct a Wilcoxon signed rank test of H 0:μ=200 H 0:μ=200 versus H a:μ≠200 H a:μ≠200 at the α=.05 α=.05 level and interpret. Exercises 9.7 – 9.16 require material through Section 9.2. 9.7 Suppose you wish to perform a two sample Wilcoxon rank sum test. You collect data x 1=2,x 2=6 x 1=2,x 2=6 and y 1=4,y 2=9 y 1=4,y 2=9 and y 3=10 y 3=10. Compute the test statistic and p p-value by hand, and check your answer using R. 9.8 On episode 2 of the Nicolas Cage documentary, “The History of Swear Words,” six comedians took part in an experiment. Each was asked to submerge their hand in a bucket of ice water for as long as they could stand it. Four of the comedians were allowed to swear out loud while performing the experiment, the other two were forbidden from swearing. The two clean subjects spent 53 and 58 seconds in the water, while the four swearing comedians spent 69, 78, 87, and 140 seconds in the water. Test for a difference between the two groups using a Wilcoxon rank sum test and report your p p-value. Repeat your analysis with a t t-test. What happens if you change to a one-sided hypothesis? Were any of your analyses significant at the α=0.05 α=0.05 level? What conclusions would you draw from this experiment? 9.9 The data set fosdata::bechdel is described in Example 7.4. Make a boxplot comparing the distributions of budget_2013 for movies that pass and movies that fail the binary Bechdel test. Perform a Wilcoxon rank sum test to compare the budgets of movies that pass the Bechdel test with movies that fail. Report your conclusion with a p p-value. Explain why a t t-test would be inappropriate to use with this data. 9.10 In 1876, Charles Darwin studied the growth of pairs of corn seedlings, one produced by cross-fertilization and the other produced by self-fertilization. These seedlings were grown in pairs, with one of each type planted under the same conditions in the same pot. The final heights of the plants are in the cross and self variables of the data ZeaMays in the HistData package. Make a boxplot to compare the distribution of heights of cross- and self-fertilized plants. Why would a rank based test be preferable to a t t-test for this data? Carry out an appropriate rank based test to compare the two groups and state your conclusion with a p p-value. 9.11 In an experiment, 37 volunteers with influenza or influenza symptoms were observed, each with and without a face mask. Experimenters measured the number of fine influenza particles in both settings. The data is available in fosdata::masks. Make a plot or plots to confirm that the mask_fine and no_mask_fine variables are highly skewed and inappropriate for a t t-test. Does a t t-test detect a difference in particles for the mask and no mask conditions? Does a Wilcoxon test detect a difference in particles for the mask and no mask conditions? Transform both variables by x→log(1+x)x→log⁡(1+x) and check that they would now be reasonable to use in a t t-test. Does a t t-test detect a difference in particles for the mask and no mask conditions, after the log(1+x)log⁡(1+x) transformation? Overall, does it appear that wearing a mask reduces the number of fine influenza particles? 9.12 This exercise explores how the Wilcoxon test changes when data values are transformed. Suppose you wish to test H 0:μ=0 H 0:μ=0 versus H a:μ≠0 H a:μ≠0 using a Wilcoxon test You collect data x 1=−1,x 2=2,x 3=−3,x 4=−4 x 1=−1,x 2=2,x 3=−3,x 4=−4 and x 5=5 x 5=5. What is the p p-value? Now, suppose you multiply everything by 2: H 0:μ=0 H 0:μ=0 versus H a:μ≠0 H a:μ≠0 and your data is x 1=−2,x 2=4,x 3=−6,x 4=−8 x 1=−2,x 2=4,x 3=−6,x 4=−8 and x 5=10 x 5=10. What happens to the p p-value? (Try to answer this without using R. Check your answer using R, if you must.) Now, suppose you square the magnitudes of everything. H 0:μ=0 H 0:μ=0 versus H a:μ≠0 H a:μ≠0, and your data is x 1=−1,x 2=4,x 3=−9,x 4=−16 x 1=−1,x 2=4,x 3=−9,x 4=−16 and x 5=25 x 5=25. What happens to the p p-value? Compare your answers to those that you got when you did the same exercise for t t-tests in Exercise 7.8. 9.13 Consider the data in ex0221 in the Sleuth3 package. This data gives the humerus length of 24 adult male sparrows that perished in a winter storm, as well as the humerus length of 35 adult male sparrows that survived. The question under consideration is whether those that perished did so because of some physical characteristic; measured in this case by humerus length. Create a boxplot of humerus length for the sparrows that survived and those that perished. Note that there is one rather extreme outlier in the Perished group. Use a Wilcoxon rank sum test to test whether the median humerus length is the same in both groups. Is there evidence to conclude that the humerus length is different in the two groups? 9.14 Consider the data in ex0428 in the Sleuth3 package. This gives the plant heights in inches of plants of the same age, one of which was grown from a seed from a cross-fertilized flower, and the other of which was grown from a seed from a self-fertilized flower. Is there a significant difference in the height of the flowers at the α=.05 α=.05 level? Create a boxplot of the height of the flowers for each type of fertilization. Comment on any abnormalities apparent. (Hint: this data is in “wide” format rather than “long” format, and you may want to use pivot_longer to convert it.) 9.15 The data set malaria from the ISwR package contains 100 observations of children in Ghana. (Note: this is not the same as the malaria data set from the fosdata package!) The data records each child’s age, levels of a particular antibody, and whether or not they have malaria symptoms. State and carry out a hypothesis test that the antibody levels (ab) differ between the groups with and without malaria symptoms. Use the Wilcoxon rank sum test. Inspect a boxplot or histogram of the ab variable by mal. Would you use this variable in a t t-test? Explain. Inspect a boxplot or histogram of log(ab)log⁡(ab) by mal. Would you use this variable in a t t-test? Explain. Carry out a t t-test that the log of antibody levels differ between malaria groups and compare your results to the Wilcoxon test. 9.16 The data set flint from the fosdata package contains data on lead in tap water from Flint, Michigan in 2015. Is there a difference in the mean lead level after running water for 45 seconds (Pb2) and for 2 minutes (Pb3)? Explain why it is very likely that each observation of Pb2 is dependent on the observation of Pb3. There are two houses that are sampled twice. What does this imply about the independence of Pb2 and Pb3between observations? Since this only affects 4 of the 271 observations, we will ignore this. Plot the data. Would it be appropriate to use a t t-test directly on this data? If appropriate, test using a Wilcoxon signed rank test at the α=.01 α=.01 level. State your conclusions, including either a point estimate or a confidence interval for the pseudo-median (see Exercise 9.5 on the difference in lead levels). Exercises 9.17 – 9.21 require material through Section 9.3. 9.17 Compare the effective power of t.test versus wilcox.test in the case of testing H 0:μ=0 H 0:μ=0 versus H a:μ≠0 H a:μ≠0 when the underlying population is uniform on the interval [−0.5,1][−0.5,1], and the sample size is 30. 9.18 This problem explores the effective type I error rate for a one sample Wilcoxon and t t-tests. Choose a sample of 21 values where X 1,…,X 20 X 1,…,X 20 are iid normal with mean 0 and sd 1, and X 21 X 21 is 10 or -10 with equal probability. Test H 0:m=0 H 0:m=0 versus H a:m≠0 H a:m≠0 at the α=.05 α=.05 level. How often does the Wilcoxon test reject H 0 H 0? Compare with the effective type I error rate for a t t-test of the same data. Which test is performing closer to how it was designed? 9.19 How well can hypothesis tests detect a small change? Suppose the population is normal with mean μ=0.1 μ=0.1 and standard deviation σ=1 σ=1. We test the hypothesis H 0:μ=0 H 0:μ=0 versus H a:μ≠0 H a:μ≠0. When n=100 n=100, what percent of the time does a t t-test correctly reject H 0 H 0? When n=100 n=100, what percent of the time does a Wilcoxon test correctly reject H 0 H 0? Repeat parts (a) and (b) with n=1000 n=1000. The assumptions for both tests are satisfied, since the population is normal. Which test would you recommend? 9.20 In this problem, we estimate the probability of a type II error. Suppose you wish to test H 0:μ=1 H 0:μ=1 versus H a:μ≠1 H a:μ≠1 at the α=.05 α=.05 level. Suppose the true underlying population is t t with 3 degrees of freedom, and you take a sample of size 20. What is the true mean of the underlying population? What type of error would be possible to make in this context, type I or type II? In the problems below, if the error is impossible to make in this context, the probability would be zero. Approximate the probability of a type I error if you use a t t-test. Approximate the probability of a type I error if you use a Wilcoxon test. Approximate the probability of a type II error if you use a t t-test. Approximate the probability of a type II error if you use a Wilcoxon test. Note that t t random variables with small degrees of freedom have heavy tails and contain data points that look like outliers. Does it appear that Wilcoxon or t t-test is more powerful with this type of population? 9.21 Suppose that you are planning an experiment, and you wish to use a two sample Wilcoxon test on your data. You wish your experiment to have power of 80% when the test is performed at the α=.05 α=.05 level and when population one is normal with mean 0 and standard deviation 1, and population 2 is normal with mean 0.4 and standard deviation 1. Assume that you will take equal sample sizes from each population. Create a power versus sample size plot for 5 to 10 sample sizes between 20 and 200, and indicate how many samples from each population you would recommend. Exercises 9.22 – 9.27 require material through Section 9.4. 9.22 Consider the MovieLens data fosdata::movies. Is there sufficient evidence at the α=.05 α=.05 level to conclude that among people who have only seen one of the movies, those people have a preference between Sleepless in Seattle and While You Were Sleeping? Report Vargha and Delaney’s A A effect size for this data, and interpret. 9.23 Consider the sharks data in the fosdata package.74 Participants either watched a video or listened to an audio documentary about sharks while different types of music played. Participants were then asked to rate sharks on various dimensions such as gracefulness and viciousness. For this problem, consider only those participants who watched a video and heard either ominous or uplifting music. Is there a difference in those participants’ responses to how vicious sharks are? Test at the α=.05 α=.05 level. If there is a difference, provide the common language effect size and explain. 9.24 Consider the plastics data set in the fosdata package. Snow was filtered in both the Arctic and Europe to find microfibers, which are microplastics that are in fibrous shape. Perform a two sample Wilcoxon rank sum test to determine whether the length of the microfibers is the same in the Arctic as in Europe. The authors in the paper75 reported a test statistic of 13723, which is not what R reports. Confirm that the authors used the following method to obtain their test statistic. They computed the test statistic W W from wilcox.test and then added the total number of comparisons between lengths of fibers from the Arctic with other lengths of fibers from the Arctic. Provide an effect size in terms of Vargha and Delaney’s A A and interpret. 9.25 Provide an estimate for the effect size in Exercise 9.16 and interpret. 9.26 Use simulation to show that if population 1 is standard normal, and population 2 is normal with mean 1 and standard deviation 1, then the Wilcoxon rank sum test is consistent at the α=.05 α=.05 level. 9.27 Suppose population 1 is standard normal, and population 2 is normal with mean 0 and standard deviation 5. Use simulation to show that P(X>Y)=.5 P(X>Y)=.5, where X X is a random sample from population 1 and Y Y is a random sample from population 2. Use sample sizes of n=10,100,1000 n=10,100,1000 and 5000 for each population, and estimate the probability that the null hypothesis will be rejected at the α=.05 α=.05 level. Use 1000 replications rather than 10,000 so that it runs in a reasonable amount of time. Does the Wilcoxon rank sum test appear to be consistent in this context? Donald K Milton et al., “Influenza Virus Aerosols in Human Exhaled Breath: Particle Size, Culturability, and Effect of Surgical Masks,”PLOS Pathogens 9, no. 3 (March 2013): 1–7, A dotplot is similar to a histogram, with each data point shown as an individual dot. This one was made with geom_dotplot.↩︎ Rebekah Martin et al., “Lead Results from Tap Water Sampling in Flint, MI,” 2015, Ann Hillier, Ryan P Kelly, and Terrie Klinger, “Narrative Style Influences Citation Frequency in Climate Change Science,”PLOS One 11, no. 12 (2016): e0167983.↩︎ Dwight Barry, “Do Not Use Averages with Likert Scale Data,” 2017, E L Lehmann, Nonparametrics: Statistical Methods Based on Ranks, Holden-Day Series in Probability and Statistics (Holden-Day, Inc., San Francisco, Calif.; McGraw-Hill International Book Co., New York-Düsseldorf, 1975).↩︎ Andrew P Nosal et al., “The Effect of Background Music in Shark Documentaries on Viewers’ Perceptions of Sharks,”PLOS One 11, no. 8 (August 2016): 1–15, Melanie Bergmann et al., “White and Wonderful? Microplastics Prevail in Snow from the Alps to the Arctic,”Science Advances 5, no. 8 (2019): eaax1157,
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(CON) Edition: 9th ISBN13: 9781118921876 ISBN10: 1118921879 Format: Loose-leaf Pub. Date: 2015-03-02 Publisher(s): John Wiley & Sons Inc Other versions by this Author This Item Qualifies for Free Shipping! Excludes marketplace orders. List Price: ~~$99.91~~ Rent Book Select for Price Add to Cart There was a problem. Please try again later. Buy New Arriving Soon. Will ship when available. $95.15 Add to Cart Used Book We're Sorry Sold Out eBook We're Sorry Not Available Buy from our Marketplace starting at $86.59 How Marketplace Works: This item is offered by an independent seller and not shipped from our warehouse Item details like edition and cover design may differ from our description; see seller's comments before ordering. Sellers much confirm and ship within two business days; otherwise, the order will be cancelled and refunded. Marketplace purchases cannot be returned to eCampus.com. Contact the seller directly for inquiries; if no response within two days, contact customer service. Additional shipping costs apply to Marketplace purchases. Review shipping costs at checkout. Currently unavailable Summary Fox & McDonald’s Introduction to Fluid Mechanics 9 th Editionhas been one of the most widely adopted textbooks in the field. This highly-regarded text continues to provide readers with a balanced and comprehensive approach to mastering critical concepts, incorporating a proven problem-solving methodology that helps readers develop an orderly plan to finding the right solution and relating results to expected physical behavior. The ninth edition features a wealth of example problems integrated throughout the text as well as a variety of new end of chapter problems. Author Biography Philip J. Pritchard is the author of Fox and McDonald's Introduction to Fluid Mechanics, 9th Edition, published by Wiley. John W. Mitchell is the author of Fox and McDonald's Introduction to Fluid Mechanics, 9th Edition, published by Wiley. Table of Contents CHAPTER 1 INTRODUCTION 1 1.1 Introduction to Fluid Mechanics 2 Note to Students 2 Scope of Fluid Mechanics 3 Definition of a Fluid 3 1.2 Basic Equations 4 1.3 Methods of Analysis 5 System and Control Volume 6 Differential versus Integral Approach 7 Methods of Description 7 1.4 Dimensions and Units 9 Systems of Dimensions 9 Systems of Units 10 Preferred Systems of Units 11 Dimensional Consistency and “Engineering” Equations 11 1.5 Analysis of Experimental Error 13 1.6 Summary 14 Problems 14 CHAPTER 2 FUNDAMENTAL CONCEPTS 17 2.1 Fluid as a Continuum 18 2.2 Velocity Field 19 One-, Two-, and Three-Dimensional Flows 20 Timelines, Pathlines, Streaklines, and Streamlines 21 2.3 Stress Field 25 2.4 Viscosity 27 Newtonian Fluid 28 Non-Newtonian Fluids 30 2.5 Surface Tension 31 2.6 Description and Classification of Fluid Motions 34 Viscous and Inviscid Flows 34 Laminar and Turbulent Flows 36 Compressible and Incompressible Flows 37 Internal and External Flows 38 2.7 Summary and Useful Equations 39 References 40 Problems 40 CHAPTER 3 FLUID STATICS 47 3.1 The Basic Equation of Fluid Statics 48 3.2 The Standard Atmosphere 51 3.3 Pressure Variation in a Static Fluid 52 Incompressible Liquids: Manometers 52 Gases 57 3.4 Hydrostatic Force on Submerged Surfaces 59 Hydrostatic Force on a Plane Submerged Surface 59 Hydrostatic Force on a Curved Submerged Surface 66 3.5 Buoyancy and Stability 69 3.6 Fluids in Rigid-Body Motion (www.wiley.com/college/pritchard) 72 3.7 Summary and Useful Equations 72 References 73 Problems 73 CHAPTER 4 BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME 82 4.1 Basic Laws for a System 84 Conservation of Mass 84 Newton’s Second Law 84 The Angular-Momentum Principle 84 The First Law of Thermodynamics 85 The Second Law of Thermodynamics 85 4.2 Relation of System Derivatives to the Control Volume Formulation 85 Derivation 86 Physical Interpretation 88 4.3 Conservation of Mass 89 Special Cases 90 4.4 Momentum Equation for Inertial Control Volume 94 Differential Control Volume Analysis 105 Control Volume Moving with Constant Velocity 109 4.5 Momentum Equation for Control Volume with Rectilinear Acceleration 111 4.6 Momentum Equation for Control Volume with Arbitrary Acceleration (on the Web) 117 4.7 The Angular-Momentum Principle 117 Equation for Fixed Control Volume 117 4.8 The First and Second Laws of Thermodynamics 121 Rate of Work Done by a Control Volume 122 Control Volume Equation 123 4.9 Summary and Useful Equations 128 Problems 129 CHAPTER 5 INTRODUCTION TO DIFFERENTIAL ANALYSIS OF FLUID MOTION 144 5.1 Conservation of Mass 145 Rectangular Coordinate System 145 Cylindrical Coordinate System 149 5.2 Stream Function for Two-Dimensional Incompressible Flow 151 5.3 Motion of a Fluid Particle (Kinematics) 153 Fluid Translation: Acceleration of a Fluid Particle in a Velocity Field 154 Fluid Rotation 160 Fluid Deformation 163 5.4 Momentum Equation 167 Forces Acting on a Fluid Particle 167 Differential Momentum Equation 168 Newtonian Fluid: Navier–Stokes Equations 168 5.5 Introduction to Computational Fluid Dynamics 176 The Need for CFD 176 Applications of CFD 177 Some Basic CFD/Numerical Methods Using a Spreadsheet 178 The Strategy of CFD 182 Discretization Using the Finite-Difference Method 183 Assembly of Discrete System and Application of Boundary Conditions 184 Solution of Discrete System 185 Grid Convergence 185 Dealing with Nonlinearity 186 Direct and Iterative Solvers 187 Iterative Convergence 188 Concluding Remarks 189 5.6 Summary and Useful Equations 190 References 192 Problems 192 CHAPTER 6 INCOMPRESSIBLE INVISCID FLOW 198 6.1 Momentum Equation for Frictionless Flow: Euler’s Equation 199 6.2 Bernoulli Equation: Integration of Euler’s Equation Along a Streamline for Steady Flow 202 Derivation Using Streamline Coordinates 202 Derivation Using Rectangular Coordinates 203 Static, Stagnation, and Dynamic Pressures 205 Applications 207 Cautions on Use of the Bernoulli Equation 212 6.3 The Bernoulli Equation Interpreted as an Energy Equation 213 6.4 Energy Grade Line and Hydraulic Grade Line 217 6.5 Unsteady Bernoulli Equation: Integration of Euler’s Equation Along a Streamline (on the Web) 219 6.6 Irrotational Flow 219 Bernoulli Equation Applied to Irrotational Flow 219 Velocity Potential 220 Stream Function and Velocity Potential for Two-Dimensional, Irrotational, Incompressible Flow: Laplace’s Equation 221 Elementary Plane Flows 223 Superposition of Elementary Plane Flows 225 6.7 Summary and Useful Equations 234 References 235 Problems 236 CHAPTER 7 DIMENSIONAL ANALYSIS AND SIMILITUDE 244 7.1 Nondimensionalizing the Basic Differential Equations 245 7.2 Nature of Dimensional Analysis 246 7.3 Buckingham Pi Theorem 248 7.4 Significant Dimensionless Groups in Fluid Mechanics 254 7.5 Flow Similarity and Model Studies 256 Incomplete Similarity 258 Scaling with Multiple Dependent Parameters 263 Comments on Model Testing 266 7.6 Summary and Useful Equations 267 References 268 Problems 268 CHAPTER 8 INTERNAL INCOMPRESSIBLE VISCOUS FLOW 275 8.1 Internal Flow Characteristics 276 Laminar versus Turbulent Flow 276 The Entrance Region 277 PART A. FULLY DEVELOPED LAMINAR FLOW 277 8.2 Fully Developed Laminar Flow between Infinite Parallel Plates 277 Both Plates Stationary 278 Upper Plate Moving with Constant Speed, U 283 8.3 Fully Developed Laminar Flow in a Pipe 288 PART B. FLOW IN PIPES AND DUCTS 292 8.4 Shear Stress Distribution in Fully Developed Pipe Flow 293 8.5 Turbulent Velocity Profiles in Fully Developed Pipe Flow 294 8.6 Energy Considerations in Pipe Flow 297 Kinetic Energy Coefficient 298 Head Loss 298 8.7 Calculation of Head Loss 299 Major Losses: Friction Factor 299 Minor Losses 303 Pumps, Fans, and Blowers in Fluid Systems 308 Noncircular Ducts 309 8.8 Solution of Pipe Flow Problems 309 Single-Path Systems 310 Multiple-Path Systems 322 PART C. FLOW MEASUREMENT 326 8.9 Restriction Flow Meters for Internal Flows 326 The Orifice Plate 329 The Flow Nozzle 330 The Venturi 332 The Laminar Flow Element 332 Linear Flow Meters 335 Traversing Methods 336 8.10 Summary and Useful Equations 337 References 340 Problems 341 CHAPTER 9 EXTERNAL INCOMPRESSIBLE VISCOUS FLOW 353 PART A. BOUNDARY LAYERS 355 9.1 The Boundary-Layer Concept 355 9.2 Laminar Flat-Plate Boundary Layer: Exact Solution (www.wiley.com/college/pritchard) 359 9.3 Momentum Integral Equation 359 9.4 Use of the Momentum Integral Equation for Flow with Zero Pressure Gradient 363 Laminar Flow 364 Turbulent Flow 368 Summary of Results for Boundary-Layer Flow with Zero Pressure Gradient 371 9.5 Pressure Gradients in Boundary-Layer Flow 371 PART B. FLUID FLOW ABOUT IMMERSED BODIES 374 9.6 Drag 374 Pure Friction Drag: Flow over a Flat Plate Parallel to the Flow 375 Pure Pressure Drag: Flow over a Flat Plate Normal to the Flow 378 Friction and Pressure Drag: Flow over a Sphere and Cylinder 378 Streamlining 384 9.7 Lift 386 9.8 Summary and Useful Equations 400 References 402 Problems 403 CHAPTER 10 FLUID MACHINERY 412 10.1 Introduction and Classification of Fluid Machines 413 Machines for Doing Work on a Fluid 413 Machines for Extracting Work (Power) from a Fluid 415 Scope of Coverage 417 10.2 Turbomachinery Analysis 417 The Angular-Momentum Principle: The Euler Turbomachine Equation 417 Velocity Diagrams 419 Performance—Hydraulic Power 422 Dimensional Analysis and Specific Speed 423 10.3 Pumps, Fans, and Blowers 428 Application of Euler Turbomachine Equation to Centrifugal Pumps 428 Application of the Euler Equation to Axial Flow Pumps and Fans 429 Performance Characteristics 432 Similarity Rules 437 Cavitation and Net Positive Suction Head 441 Pump Selection: Applications to Fluid Systems 444 Blowers and Fans 455 10.4 Positive Displacement Pumps 461 10.5 Hydraulic Turbines 464 Hydraulic Turbine Theory 464 Performance Characteristics for Hydraulic Turbines 466 Sizing Hydraulic Turbines for Fluid Systems 470 10.6 Propellers and Wind-Power Machines 474 Propellers 474 Wind-Power Machines 482 10.7 Compressible Flow Turbomachines 490 Application of the Energy Equation to a Compressible Flow Machine 490 Compressors 491 Compressible-Flow Turbines 495 10.8 Summary and Useful Equations 495 References 497 Problems 499 CHAPTER 11 FLOW IN OPEN CHANNELS 507 11.1 Basic Concepts and Definitions 509 Simplifying Assumptions 509 Channel Geometry 511 Speed of Surface Waves and the Froude Number 512 11.2 Energy Equation for Open-Channel Flows 516 Specific Energy 518 Critical Depth: Minimum Specific Energy 521 11.3 Localized Effect of Area Change(Frictionless Flow) 524 Flow over a Bump 524 11.4 The Hydraulic Jump 528 Depth Increase across a Hydraulic Jump 531 Head Loss across a Hydraulic Jump 532 11.5 Steady Uniform Flow 534 The Manning Equation for Uniform Flow 536 Energy Equation for Uniform Flow 541 Optimum Channel Cross Section 543 11.6 Flow with Gradually Varying Depth 544 Calculation of Surface Profiles 545 11.7 Discharge Measurement Using Weirs 548 Suppressed Rectangular Weir 548 Contracted Rectangular Weirs 549 Triangular Weir 549 Broad-Crested Weir 550 11.8 Summary and Useful Equations 551 References 552 Problems 553 CHAPTER 12 INTRODUCTION TO COMPRESSIBLE FLOW 556 12.1 Review of Thermodynamics 557 12.2 Propagation of Sound Waves 563 Speed of Sound 563 Types of Flow—The Mach Cone 567 12.3 Reference State: Local Isentropic Stagnation Properties 570 Local Isentropic Stagnation Properties for the Flow of an Ideal Gas 571 12.4 Critical Conditions 577 12.5 Basic Equations for One-Dimensional Compressible Flow 577 Continuity Equation 577 Momentum Equation 578 First Law of Thermodynamics 578 Second Law of Thermodynamics 579 Equation of State 579 12.6 Isentropic Flow of an Ideal Gas: Area Variation 580 Subsonic Flow, M Supersonic Flow, M >1 583 Sonic Flow, M =1 583 Reference Stagnation and Critical Conditions for Isentropic Flow of an Ideal Gas 584 Isentropic Flow in a Converging Nozzle 589 Isentropic Flow in a Converging-Diverging Nozzle 593 12.7 Normal Shocks 598 Basic Equations for a Normal Shock 599 Normal-Shock Flow Functions for One-Dimensional Flow of an Ideal Gas 601 12.8 Supersonic Channel Flow with Shocks 605 12.8 Supersonic Channel Flow with Shocks 12.9 Flow in a Constant-Area Duct with Friction (www.wiley.com/college/pritchard) 607 12.10 Frictionless Flow in a Constant-Area Duct with Heat Exchange (www.wiley.com/college/pritchard) 607 12.11 Oblique Shocks and Expansion Waves(www.wiley.com/college/pritchard) 607 12.12 Summary and Useful Equations 607 References 610 Problems 610 APPENDIX A FLUID PROPERTY DATA 615 APPENDIX B VIDEOS FOR FLUID MECHANICS 627 APPENDIX C SELECTED PERFORMANCE CURVES FOR PUMPS AND FANS 629 APPENDIX D FLOW FUNCTIONS FOR COMPUTATION OF COMPRESSIBLE FLOW 640 APPENDIX E ANALYSIS OF EXPERIMENTAL UNCERTAINTY 643 APPENDIX F ADDITIONAL COMPRESSIBLE FLOW FUNCTIONS (WWW.WILEY.COM/ COLLEGE/PRITCHARD ONLINE) WF-1 APPENDIX G A BRIEF REVIEW OF MICROSOFT EXCEL (WWW.WILEY.COM/COLLEGE/ PRITCHARD) WG-1 Answers to Selected Problems 649 Index 656 We are currently experiencing difficulties. 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