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97 House Vocabulary in Chinese (House Parts, Rooms & Objects)
Have you ever wondered what a house is called in Chinese? Knowing the key Chinese house vocabulary such as room names, home appliances, and furniture is very useful as you will probably need to use these words every day in China.
Luckily for you, we got this entire article dedicated to “house” in Chinese. We’re going to help you understand and use over 90 essential household vocabulary expressions. We will go around to each of the rooms in the house and show you what is typically there in Chinese households so you can use those words instantly in life. You’ll be able to expand your vocabulary even if you are an advanced Mandarin speaker!
Sound good? Let’s get started.
How to Say “House”in Chinese
First and foremost, the word for “house” in Mandarin Chinese is:
房子 (fángzi)
It’s a very quick and simple word for you to learn. 房 (fáng) means house, and 子 (zi) is a nominal suffix used after many nouns in Chinese. Depending on the situation and context, 房子 (fángzi) may also mean apartment or room. You can use this word colloquially as well as in written language.
Another common way of saying “house” in Chinese is:
房 (fáng)
You may simply use 房 (fáng), which is essentially the shortened version of 房子 (fángzi) to refer to a house sometimes. However, this word is rarely used on its own. Usually, it appears in house-related words and phrases.
For instance,
平房 (píng fáng) – bungalow (literally, “flat house”)
买房 (mǎi fáng) – buy (a) house
租房 (zū fáng) – rent (a) house
We’ll cover more house-related vocabulary in a minute. But before that, there is one thing you need to note:
The Measure Word for House in Chinese
In Chinese, all nouns require a measure word when counted. For instance, you wouldn’t just say “a house”, or “two apartments”, you would have to say “a [measure word] house” and “two [measure word] apartments”.
The most common measure words for 房子 (fángzi) – “house” in Chinese are 栋 (dòng) and 幢 (zhuàng). You can use them interchangeably, like
一栋房子 (yí dòng fángzi) – a house
两幢房子 (liǎng zhuàng fángzi) – two houses
However, if you are using 房子 (fángzi) to refer to an apartment, the proper measure word would be 套 (tào). For instance:
这套房子 (zhè tào fángzi) – this apartment
And when you are referencing a room, use 间 (jiān) instead as the measure word.
Example:
楼上有三间房。
Lóu shàng yǒu sān jiān fáng.
There are three rooms upstairs.
Parts of the House in Chinese
Now use the picture below to learn the names of different parts of the house in Chinese. Simply move your mouse around and the words will present themselves in Chinese.
second floor
二楼
èr lóu
1 of 9
« Previous
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chimney
烟囱
yāncōng
2 of 9
« Previous
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attic
阁楼
gélóu
3 of 9
« Previous
Next »
balcony
阳台
yángtái
4 of 9
« Previous
Next »
door
门
mén
5 of 9
« Previous
Next »
window
窗
chuāng
6 of 9
« Previous
Next »
roof
屋顶
wūdǐng
7 of 9
« Previous
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wall
墙
qiáng
8 of 9
« Previous
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first floor
一楼
yī lóu
9 of 9
« Previous
Next »
House-related Vocabulary in Chinese
It’s a good idea to add 房子/房 (fángzi/fáng) to your basic Chinese vocabulary, as it’s the building block of many useful daily words and phrases in Chinese.
Below are the most common house-related Chinese vocabulary words. We provide them with the Chinese characters as well as Pinyin romanization to help you pronounce the words. You’ll see that pairing 房(子)/fáng(zi) with different characters creates interesting compound words that relate to the house.
| | | |
---
| Chinese | Pinyin Pronunciation | English |
| 房间 | fángjiān | room |
| 楼房 | lóufáng | building |
| 买房(子) | mǎi fáng(zi) | buy a house |
| 卖房(子) | mài fáng(zi) | sell a house |
| 房价 | fáng jià | house price |
| 租房(子) | zū fáng(zi) | rent a house |
| 房租 | fáng zū | house rent |
| 房东 | fángdōng | landlord |
| 房客 | fángkè | tenant |
Home vs House in Chinese
If you’ve started to learn Chinese, you may already be familiar with another word – 家 (jiā), which means “home” in Chinese.
Like in English, people make a distinction between “home” and “house” in Chinese. While 家 (jiā) can refer to any place that we live in such as a house, a tent, a boat, a cave, or even the abstract notion, 房子 (fángzi) only means the physical building.
For example, if you live in a camper van, then your camper van can be your 家 (jiā), but not your 房子 (fángzi) – even if the Chinese word for camper van – 房车 (fáng chē) translates to “house car”.
Related Reading: Home in Chinese (25 Expressions to Expand Your Vocabulary)
All right, now that we have cleared all that up, let’s get onto talking about the different types of houses in Chinese.
Types of Houses in Chinese
There are a few types of Chinese houses and homes such as 公寓 (gōngyù), 别墅 (biéshù), 平房 (píng fáng), 老公房 (lǎo gōngfáng), etc.
Most Chinese people don’t live in single-family detached houses. Instead, they often live in apartments, especially in cities.
Here are the Chinese words for the different types of houses that people commonly live in.
1. 公寓 (gōngyù)
The standard Chinese word for “apartment” is 公寓 (gōngyù), though it’s often called 房子 (fángzi) as well in spoken Chinese. 公寓 (gōngyù) are the choice of the majority of Chinese people as they are cheaper.
2. 老公房 (lǎo gōngfáng)
老公房 (lǎo gōngfáng) are those communist-style apartment blocks built by the Chinese government in the 1950s-1980s to accommodate the working class. Nowadays, most of such old houses are resided by old people.
3. 商住两用房 (shāng zhù liǎng yòng fáng)
商住两用房 (shāng zhù liǎng yòng fáng) refers to the houses wherein a building comprises a mix of residential apartments and offices.
4. 别墅 (biéshù)
In Chinese, 别墅 (biéshù) is the generic word for a house that’s used as a residence, and usually housing a single household. A small house like a cottage is called a 小别墅 (xiǎo biéshù) – “small house” while a villa is referred to as a 豪华别墅 (háohuá biéshù) – “luxurious house”.
The term 别墅 (biéshù) may also be used for townhouses. To be more precise, you can call a stand-alone house 独栋别墅 (dúdòng biéshù), and a townhouse 联排别墅 (liánpái biéshù).
5. 平房 (píng fáng)
平房 (píng fáng) – bungalows used to be a popular type of house in China. A traditional type of bungalow you can still find throughout the country, most famously in Beijing, is called 四合院 (sì hé yuàn).
6. 石库门 (shí kù mén)
石库门 (shí kù mén), literally “stone warehouse gate”, is the most representative house style of Shanghai. These are clusters of two to three-story terrace houses that were built around the city in the early 20th century.
7. 农民房 (nóngmín fáng)
A 农民房 (nóngmín fáng), or “farmer’s house”, as the name suggests, is a rustic house with little aesthetic appeal built by farmers themselves in the countryside. An upscale house located in rural settings is instead referred to as 乡间别墅 (xiāngjiān biéshù) – “countryside house”.
8. 蒙古包 (Měnggǔ bāo)
蒙古包 (Měnggǔ bāo) refers to the primary housing structures that the Mongols use in China’s Inner Mongolia region (which is the home to more than 4 million Mongols). Like the usual residence, the yurt also has all the amenities.
9. 小屋 (xiǎo wū)
小屋 (xiǎo wū) is what we call “cabin” in Chinese. Like 房 (fáng), the word 屋 (wū) also means house or room. The primary difference you might find is that 屋 (wū) is usually smaller, simpler, and has less finishing compared to 房 (fáng). A cabin that is built from logs is called 小木屋 (xiǎo mù wū).
10. 树屋 (shù wū)
Yes, they do exist. Treehouses are called 树屋 (shù wū) in Chinese.
Rooms in the House in Chinese
As we’ve mentioned, the word 房 (fáng) can be used as a casual term for “room”. But a more formal word for room in Chinese is 房间 (fángjiān).
So, what are the Chinese names of the various 房间 (fángjiān) in a house?
Below is a quick rundown of the common room names in Chinese.
| | | |
---
| English | Chinese | Pinyin |
| living room | 客厅 | kètīng |
| dining room | 餐厅 | cāntīng |
| kitchen | 厨房 | chúfáng |
| bathroom | 卫生间 | wèishēngjiān |
| laundry room | 洗衣间 | xǐyījiān |
| bedroom | 卧室 | wòshì |
| office | 办公室 | bàngōngshì |
| study | 书房 | shūfáng |
| hallway | 玄关 | xuánguān |
| attic | 阁楼 | gélóu |
| basement | 地下室 | dìxiàshì |
| garage | 车库 | chēkù |
| balcony | 阳台 | yángtái |
| garden | 花园 | huāyuán |
balcony
阳台
yángtái
1 of 7
« Previous
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second bedroom
次卧
cì wò
2 of 7
« Previous
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living room
客厅
kètīng
3 of 7
« Previous
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dining room
餐厅
cāntīng
4 of 7
« Previous
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master bedroom
主卧
zhǔ wò
5 of 7
« Previous
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bathroom
浴室
wèishēngjiān
6 of 7
« Previous
Next »
kitchen
厨房
chúfáng
7 of 7
« Previous
Next »
Now let’s learn how to use all of these words with some details.
Living Room in Chinese: 客厅 (kètīng)
Usually the most spacious room located at the front of the house, the living room is where Chinese families gather together for leisure. In China, the living room is called 客厅 (kètīng), literally “guest hall”, since it was originally a place to entertain visitors.
Dining Room in Chinese: 餐厅 (cāntīng)
The dining room is referred to as 餐厅 (cāntīng) in China. Literally “meal hall”, this word can also mean restaurant in Chinese.
Kitchen in Chinese: 厨房 (chúfáng)
厨房 (chúfáng) is probably everyone’s favorite room of the house. It means kitchen in Chinese.
Quiz: Do you know how to say “chef” in Chinese? Hint: it has something to do with “kitchen”. Check out our list of Mandarin professions to find out.
Bathroom in Chinese: 卫生间 (wèishēngjiān)
卫生间 (wèishēngjiān) is the bathroom in Chinese. This word usually refers to the room where the toilet is. However, if you’re referring to the room with the tub or shower instead, you can say 浴室 (yùshì). More on this in our article on 7 ways to say “bathroom” In Chinese.
Laundry Room in Chinese: 洗衣间 (xǐyījiān)
Laundry room in Chinese is simply “wash clothes room” – 洗衣间 (xǐyījiān). Most Chinese households don’t have a laundry room and the clothes are usually air-dried on the balcony.
Bedroom in Chinese: 卧室 (wòshì)
卧室 (wòshì) literally means “the room to sleep” in Chinese. The master bedroom in a house is called 主卧 (zhǔ wò) while the second and guest bedroom are called 次卧 (cì wò) and 客卧 (kè wò).
Office in Chinese: 办公室 (bàngōngshì)
Remember we all had to turn a part of our house into a home office during the COVID-19 outbreak? Well, the word for office in Chinese is 办公室 (bàngōngshì). And to say “work from home”, you say 在家办公 (zài jiā bàngōng).
Study in Chinese: 书房 (shūfáng)
This is an easy one: 书房 (shūfáng) – literally “book room”. Don’t confuse it with 图书馆 (túshūguǎn) which means library.
Hallway in Chinese: 玄关 (xuánguān)
The entryway area for a house or apartment is called 玄关 (xuánguān) in Chinese.
Attic in Chinese: 阁楼 (gélóu)
阁楼 (gélóu) can mean attic or loft in Chinese.
Here’s an interesting Chinese idiom with the word in it: 空中阁楼 (kōng zhōng gélóu) – “attics in the air”, meaning unrealistic plans or hopes for the future.
Basement in Chinese: 地下室 (dìxiàshì)
The word for basement in Chinese is 地下室 (dìxiàshì). It literally means “underground room”. There’s no clear-cut purpose for a basement. Many families convert it into a home gym – 家庭健身房 (jiātíng jiànshēnfáng), a home theater – 家庭影院 (jiātíng yǐngyuàn) or a “rec room” – 家庭娱乐室 (jiātíng yúlè shì).
Garage in Chinese: 车库 (chēkù)
车库 (chēkù) is what we call a garage in Chinese. It translates to “car storeroom”, literally. If you are talking about a fixed parking space in your apartment complex, then use the word 车位 (chēwèi) instead.
Balcony in Chinese: 阳台 (yángtái)
阳台 (yángtái), literally “sun deck”, is the balcony in Chinese. A terrace is called 露台 (lùtái).
Garden in Chinese: 花园 (huāyuán)
Finally, the garden around a house is called 花园 (huāyuán). Don’t get it mixed up with 公园 (gōng yuán), which is the park for the public.
Further Reading: Gardening Tools in Chinese
House Objects and Furniture in Chinese
Now we’re gonna go around the rooms in the house – the living room, the dining room, the kitchen, the bathroom, and the bedroom, and name the different items that are typically there in Chinese.
Let’s start with the living room.
Living Room Items in Chinese
| | | |
---
| English | Chinese | Pinyin |
| air conditioner | 空调 | kōngtiáo |
| bookshelf | 书架 | shūjià |
| carpet | 地毯 | dìtǎn |
| ceiling | 天花板 | tiānhuābǎn |
| chandelier | 吊灯 | diàodēng |
| coffee table | 茶几 | chájī |
| curtains | 窗帘 | chuānglián |
| fireplace | 壁炉 | bìlú |
| floor | 地板 | dìbǎn |
| floor lamp | 落地灯 | luòdìdēng |
| sofa | 沙发 | shāfā |
| TV | 电视机 | diànshìjī |
| TV cabinet | 电视柜 | diànshìguì |
| wall | 墙 | qiáng |
| window | 窗 | chuāng |
Note that though “coffee table” technically translates to 咖啡桌 (kāfēi zhuō) in Mandarin, Chinese people commonly refer to it as “tea table” – 茶几 (chájī).
Dining Room & Kitchen Items in Chinese
| | | |
---
| English | Chinese | Pinyin |
| chair | 椅子 | yǐzi |
| dining table | 餐桌 | cānzhuō |
| dishwasher | 洗碗机 | xǐwǎnjī |
| fridge | 冰箱 | bīngxiāng |
| kitchen counter | 厨房柜台 | chúfáng guìtái |
| microwave | 微波炉 | wēibōlú |
| oven | 烤箱 | kǎoxiāng |
| pan | 平底锅 | píngdǐguō |
| pot | 锅 | guō |
| stove | 炉灶 | lúzào |
| range hood | 油烟机 | yóuyānjī |
Further Reading: Kitchen Expressions in Chinese
Bathroom & Laundry Room Items in Chinese
| | | |
---
| English | Chinese | Pinyin |
| bathtub | 浴缸 | yùgāng |
| bath mat | 地垫 | dìdiàn |
| cabinet | 柜子 | guìzi |
| faucet | 水龙头 | shuǐlóngtou |
| mirror | 镜子 | jìngzi |
| shelf | 架子 | jiàzi |
| shower | 淋浴 | línyù |
| sink | 台盆 | táipén |
| tile | 瓷砖 | cízhuān |
| toilet | 马桶 | mǎtǒng |
| washing machine | 洗衣机 | xǐyījī |
See our full list of Chinese bathroom & toiletries vocabulary.
Bedroom Items in Chinese
| | | |
---
| English | Chinese | Pinyin Pronunciation |
| bed | 床 | chuáng |
| blanket | 毯子 | tǎnzi |
| drawers | 抽屉 | chōuti |
| makeup vanity | 梳妆台 | shūzhuāngtái |
| mattress | 床垫 | chuángdiàn |
| nightstand | 床头柜 | chuángtóuguì |
| pillows | 枕头 | zhěntou |
| rug | 小地毯 | xiǎo dìtǎn |
| table lamp | 台灯 | táidēng |
| wardrobe | 衣橱 | yīchú |
Not enough? Read on to learn more Chinese furniture vocabulary and expressions through examples.
Example Sentences with Housein Chinese
Now is the time to use the Chinese house vocabulary in context. Practice the following sentences and you’ll develop your knowledge of Chinese grammarand gain confidence in speaking Chinese in simple conversations.
我们下个月要搬到新房子了。
Wǒmen xià gè yuè yào bān dào xīn fángzi le.
We’re moving to a new house next month.
外国人能在中国买房吗？
Wàiguórén néng zài Zhōngguó mǎi fáng ma?
Can foreigners buy a house/apartment in China?
这栋房子是你们买的还是租的？
Zhè dòng fángzi shì nǐmen mǎi de háishì zū de?
Did you buy or rent this house?
上海的房价很高。房租也是。
Shànghǎi de fáng jià hěn gāo. Fáng zū yě shì.
The house price in Shanghai is very high. So is the rent.
这套房子有几个房间？
Zhè tào fángzi yǒu jǐ gè fángjiān?
How many rooms does this apartment have?
我们的房子有四个卧室，三个卫生间。
Wǒmen de fángzi yǒu sì gè wòshì, sān gè wèishēngjiān.
Our house/apartment has four bedrooms and three bathrooms.
这幢房子又大又漂亮，离海滩也很近。
Zhè zhuàng fángzi yòu dà yòu piàoliang, lí hǎitān yě hěn jìn.
This house is big and beautiful. It’s also close to the beach.
FAQs about Housein Chinese
How do I say “my house”in Chinese?
“My house” in Chinese is 我的房子 (wǒ de fángzi). Simply add the possessive pronoun 我的 (wǒ de) – meaning “my” to the word 房子 (fángzi). If you are referring to your home, you may also say 我家 (wǒ jiā) – “my home”.
How do I say “coffee house”in Chinese?
To say “coffee house” in Chinese, you can either say 咖啡馆 (kāfēi guǎn) or 咖啡屋 (kāfēi wū). The character 房 (fáng) is not to be used in this word.
How do I say “it’s on the house”in Chinese?
This English expression can’t be translated word for word into Chinese. To say something is free, you say 这是免费的 (zhè shì miǎnfèi de).
How do I say “the White House” in Chinese?
The White House in Chinese is 白宫 (bái gōng), literally “white palace”.
What’s “House of Cards”in Chinese?
House of Cards in Chinese is 纸牌屋 (Zhǐpái Wū). You can watch it with Chinese subtitles here.
Want to Speak Chinese for Real?
There you go! We’ve covered everything you wanted to know about house in Chinese. A great way to remember these vocabulary words is to go around your house and name things in Chinese. You could even write down a label and put it around if your family members don’t mind!
Now, if you’re learning Chinese, you should make the best out of ImproveMandarin.com, one of the world’s largest sites dedicated to Mandarin studies.
Here are some articles to help you get started:
Home in Chinese (25 Expressions to Expand Your Vocabulary)
Basic Chinese Words and Phrases to Survive in China
Easy Chinese Grammar for the Overwhelmed Beginners
How to Learn Chinese Faster – An Actionable Guide
By the way, if you’ve just started out to learn Chinese, we strongly recommend you take a structured Chinese course online. It’s far more effective than reading odd bits and pieces here and there and trying to put them together on your own.
We’ve practically reviewed every Chinese course available on the internet. Some are fantastic while others are a complete waste of time. Here are the best online Chinese courses we found in 2025. |
10501 | https://arxiv.org/abs/1312.0248 | [1312.0248] Maximizing the number of nonnegative subsets
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arXiv:1312.0248 (math)
[Submitted on 1 Dec 2013 (v1), last revised 28 Jan 2014 (this version, v2)]
Title:Maximizing the number of nonnegative subsets
Authors:Noga Alon, Harout Aydinian, Hao Huang
View a PDF of the paper titled Maximizing the number of nonnegative subsets, by Noga Alon and 2 other authors
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Abstract:Given a set of $n$ real numbers, if the sum of elements of every subset of size larger than $k$ is negative, what is the maximum number of subsets of nonnegative sum? In this note we show that the answer is $\binom{n-1}{k-1} + \binom{n-1}{k-2} + \cdots + \binom{n-1}{0}+1$, settling a problem of Tsukerman. We provide two proofs, the first establishes and applies a weighted version of Hall's Theorem and the second is based on an extension of the nonuniform Erdős-Ko-Rado Theorem.
Subjects:Combinatorics (math.CO)
Cite as:arXiv:1312.0248 [math.CO]
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From: Hao Huang [view email]
[v1] Sun, 1 Dec 2013 16:51:44 UTC (7 KB)
[v2] Tue, 28 Jan 2014 01:59:33 UTC (8 KB)
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Concentration Ratios in Economics
Last updated 31 Dec 2024
Concentration ratios are a key metric used in economics to measure the extent of market power held by the largest firms within a particular industry. They are especially significant in analyzing market structures, competition, and the degree of monopoly or oligopoly in a market.
Definition and Significance
Concentration ratios express the market share held by the largest firms in an industry. Typically measured as a percentage, these ratios help determine the level of competition and market control.
Key Ratios
Why Are Concentration Ratios Important?
Example Calculation
Imagine an industry where the top four firms have market shares of 30%, 25%, 15%, and 10%. The CR4 would be: 80%
This suggests a high level of concentration and likely an oligopolistic market.
Applications and Real-World Examples
Case 1: The Smartphone Industry
The global smartphone market is dominated by a few key players:
The CR4 for the smartphone market is 70%, indicating an oligopoly where a few firms control a majority of the market.
Case 2: The Coffee Retail Industry (UK)
In the UK, major coffee chains such as Costa Coffee, Starbucks, and Caffè Nero dominate, with a CR4 of approximately 65%. This concentration suggests moderate competition but significant control by a few firms.
Advantages and Disadvantages of High Concentration Ratios
Advantages
Disadvantages
Evaluating Concentration Ratios in Context
While concentration ratios provide useful insights, they should not be used in isolation. Consider:
Regulation and Policy Responses
Governments and regulators, such as the UK Competition and Markets Authority (CMA) or the US Federal Trade Commission (FTC), monitor concentration ratios to prevent monopolistic practices. Recent actions include:
Real-World Data and Figures
These examples highlight the diversity of concentration levels across industries and underscore their importance in economic analysis.
By understanding concentration ratios, students can better analyze real-world markets and appreciate the complex dynamics of competition, regulation, and consumer impact. Keep exploring industries to uncover more insights!
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Difference between function and equation
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What is the precise difference between function and equation ? In which case will it be wrong if used( common mistakes )? Also will the Venn diagram overlap if I were to draw one ? Any help and discussions will be appreciated .
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edited Feb 5, 2021 at 22:02
Servaes
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asked Jan 7, 2016 at 22:26
Abu BardewaAbu Bardewa
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$\begingroup$ Related: math.stackexchange.com/questions/2738360/… $\endgroup$
Ethan Bolker
– Ethan Bolker
2018-05-01 00:00:04 +00:00
Commented May 1, 2018 at 0:00
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4 Answers 4
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$\begingroup$
A function is a transformation or mapping of one thing into another thing. It might be written as a rule (e.g. "Take the input and square it"), as a formula ("e.g. $f(x) = x^2$ or $x \mapsto x^2$), as a set of ordered pairs (e.g. $\left{(1, 1), (2, 4), (3, 9), \ldots\right}$, or any other way of showing how the output relates to the input. The function doesn't have to use numbers, either - a function could take two words and return their letters interlaced (so f(cat, dog) = cdaotg) or it could tell you what day of the week a given date falls on, or the post code/zip code of a given geographical location.
[In very formal terms, a function is a set of input-output pairs that follows a few particular rules.]
An equation is a declaration that two things are equal to each other. For example, $2^2 = 4$ is an equation stating that the square of 2 is 4. An equation may include variables of unknown value, and it may be true for all, some or none of the possible values of those variables. For example, $x^2 = 4$ is an equation that is true when $x = \pm 2$, and false for other values of $x$, while $x^2 = -4$ is an equation that is false for all real values of $x$.
What may be confusing you is that we often use equations to declare a relationship between two variables, often in the form of a function or formula. For example, $y = x^2$ is an equation stating that the value of $y$ is determined by the value of $x$ via the function $x^2$.
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answered Jan 7, 2016 at 23:07
ConManConMan
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$\begingroup$ Very well explained. Thanks . $\endgroup$
Abu Bardewa
– Abu Bardewa
2016-01-08 07:45:20 +00:00
Commented Jan 8, 2016 at 7:45
1
$\begingroup$ A function means that every point in the domain for which the function is defined only has 1 image. $\endgroup$
Algebear
– Algebear
2018-04-30 23:44:03 +00:00
Commented Apr 30, 2018 at 23:44
$\begingroup$ @ConMan Even though we still use an equation to determine the output value of the function we don't mean to solve the equation right? If we have for example the function $f(x)=4$ no one will think to "solve" it. They will just map every $x$ to $4$ whereas if it was just an expression of equality ($y=4$) then the solution would be $4$. $\endgroup$
Antonios Sarikas
– Antonios Sarikas
2020-09-07 21:50:10 +00:00
Commented Sep 7, 2020 at 21:50
$\begingroup$ Correct.$f(x) = 4$ doesn't even really have a "solution" as such, since there's nothing to solve for (and if you're solving for "for which $x$ does $f(x) = 4$?" then the answer is "all of them"). $\endgroup$
ConMan
– ConMan
2020-09-07 22:50:40 +00:00
Commented Sep 7, 2020 at 22:50
$\begingroup$ One thing to note is that not all equations of two variables define a function. For example, $y=x^y$ does not define a function from ð¥ to ð¦ because it fails the vertical line test (i.e. there are some values of ð¥ that have $2$ images). For example, ð¥=1.3 has two corresponding ð¦-values that satisfy $y=1.3^y$ (see [graph] (desmos.com/calculator/2jlwcf7sfn)). $y=x^y$ does seem to define a function from ð¦ to ð¥ though as it doesn't fail the horizontal line test. $\endgroup$
joseville
– joseville
2021-11-16 18:19:39 +00:00
Commented Nov 16, 2021 at 18:19
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2
$\begingroup$
It might be useful to introduce the term formula at this point.
Functions have arguments, i.e., values to be input into a formula. Equations do not have arguments to be input into a formula.
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answered Oct 14, 2021 at 11:42
David M. BowerDavid M. Bower
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$\begingroup$
I think we also tend to muddy the semantic waters when we insist on referring to $f(x) = x^2$ (for instance) as a function. It's not: It's an equation. The function in this case is given by the expression $x^2$, so in that way we can say that expressions are functions. In this example, $f$ is the name of the function, $x$ is the input of the function,and $x^2$ is the expression which is the output, i.e., the function $f(x)$ itself.
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answered Apr 30, 2018 at 23:17
Paul HartzerPaul Hartzer
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$\begingroup$ This is not how it works. Do you know the strict definition of a function? $\endgroup$
Algebear
– Algebear
2018-04-30 23:42:13 +00:00
Commented Apr 30, 2018 at 23:42
$\begingroup$ You gave an example that "sin(x) is a function". sin(x) is an expression. It seems to me that you're interested in flexing technical muscles instead of having a constructive conversation. I'm not interested in that. $\endgroup$
Paul Hartzer
– Paul Hartzer
2018-05-02 01:48:44 +00:00
Commented May 2, 2018 at 1:48
$\begingroup$ No, it was not my intention to be understood as someone who likes to "flex technical muscles". I just like to point out to people how things are formally defined. If you're new here you should understand eventually that critic on others' work is the best way to learn from each other. $\endgroup$
Algebear
– Algebear
2018-05-02 12:25:07 +00:00
Commented May 2, 2018 at 12:25
Add a comment |
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$\begingroup$
A function $f(x):D\to C$ must satisfy $\forall x\in C \ \text{where f is defined in this point $x$},\ \exists!\ f(x)\in D$; i.e. every point in the domain of $f$ for which $f$ is still defined may have only one image, which is a point on the line (a point in the codomain). Not to confuse with surjectivity. For example, $f:\mathbb{R}\to\mathbb{R}$ with $f(x)=\sin(x)$ is non-surjective for there is no $x\in \mathbb{R}$ such that $f(x)=2$. But the $\sin(x)$ is a function because there's no $x$-value with a multiple $f(x)$-value. An equation can be every equalty: a function is an equality, a differential equation is an equality. E.g. $x=y^2$ is an equation, but not a function if we view it with x in the domain and y in the codomain. For instance, $x=1$ has $y=1$ and $y=-1$ as solution (point in domain with two different images). Hence, not a function in the $(x,y)$-plane.
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edited Apr 30, 2018 at 23:47
answered Apr 30, 2018 at 23:41
AlgebearAlgebear
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$\begingroup$ re: "$\forall x \in C$", $x$ should be in the domain $D$, not the codomain $C$, right? i.e. it should be "$\forall x \in D$" $\endgroup$
joseville
– joseville
2021-11-16 18:10:14 +00:00
Commented Nov 16, 2021 at 18:10
$\begingroup$ Upvoted, but followup question, if you're using this notation: "ð(ð¥):ð·ð¶"; shouldn't it be "ð:ð·ð¶"? Likewise, "$sin(x)$ is a function" should be "$sin$ is a function". I'm using the following answer as a reference. $\endgroup$
joseville
– joseville
2021-11-16 18:16:37 +00:00
Commented Nov 16, 2021 at 18:16
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10504 | https://www.cs.unibo.it/~asperti/PAPERS/popl104-asperti.pdf | The Intensional Content of Rice’s Theorem (Pearl) Andrea Asperti Department of Computer Science, University of Bologna Mura Anteo Zamboni 7, 40127, Bologna, ITALY asperti@cs.unibo.it Abstract The proofs of major results of Computability Theory like Rice, Rice-Shapiro or Kleene’s fixed point theorem hide more informa-tion of what is usually expressed in their respective statements. We make this information explicit, allowing to state stronger, complex-ity theoretic-versions of all these theorems. In particular, we replace the notion of extensional set of indices of programs, by a set of in-dices of programs having not only the same extensional behavior but also similar complexity (Complexity Clique). We prove, under very weak complexity assumptions, that any recursive Complexity Clique is trivial, and any r.e. Complexity Clique is an extensional set (and thus satisfies Rice-Shapiro conditions). This allows, for in-stance, to use Rice’s argument to prove that the property of having polynomial complexity is not decidable, and to use Rice-Shapiro to conclude that it is not even semi-decidable. We conclude the paper with a discussion of “complexity-theoretic” versions of Kleene’s Fixed Point Theorem.
Categories and Subject Descriptors F.4.1 [Computability The-ory]; F.1.3 [Machine-independent complexity] General Terms Theory Keywords Recursion Theory, computability 1.
Introduction Recursive properties of extensional sets, i.e. index sets for par-tial recursive functions, have been extensively studied since the early days of Recursion Theory (Rice 1953; Dekker and Myhill 1958). Among intensional properties of programs, a major role is played by complexity. The theoretical research on effective proper-ties of recursive functions under complexity assumptions focused, since the very beginning, on complexity classes, that is classes of recursive functions computable (almost everywhere) within a given bound of complexity t. Studies on complexity classes, mostly developed in the abstract framework promoted by Blum (Blum 1967), extensively investigated their order structure under set theo-retic inclusion (Borodin 1972; McCreight and Meyer 1969), their recursive presentability and the computational quality of such a pre-Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. To copy otherwise, to republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee.
POPL ’08 10-12 January 2008, San Francisco (Ca) Copyright c ⃝2008 ACM [to be supplied]...$5.00 sentation (Lewis 1970; Young 1969; Landweber and Robertson 1972). It is important to understand that a complexity class is a set of functions, and not of indices; regarded as a set of indices, it is thus extensional by definition. If we are really interested in decid-ability properties of program complexity, complexity classes do not look like as an adequate instrument to perform the investigation.
The notion we propose here is that of complexity clique. A com-plexity clique (Section 2) is a set of indices for recursive functions closed w.r.t. indices of functions with same extension and similar complexity (defined in the only sensible way, namely up to con-stants). A complexity clique is not extensional, in general. On the other side, any extensional set is a complexity clique (the trivial “complexity” clique with arbitrary complexity).
Set-theoretic (Section 3) and, especially, recursive properties of complexity cliques are investigated. In particular, we general-ize both Rice’s Theorem (Section 4) and Rice-Shapiro’s Theorem (Section 5), concluding with a discussion of Kleene’s second fixed point Theorem from a complexity perspective (Section 6). Remark-ably, all proofs remain essentially the same of their classical coun-terparts: in other words, all those proofs seem to hide more infor-mation of what is expressed in their classical statements, and the notion of Complexity Clique is precisely what is required to make this hidden information explicit (providing our original motivation for their introduction).
Following an old tradition, we try to stick to the appealing gen-erality of Blum’s axiomatic approach; however, axiomatization is not the focus of the paper, and we shall introduce new axioms with some liberality, as far as they are satisfied by standard complex-ity measures; in particular, we shall introduce additional axioms bounding the complexity of (linear) composition, of the s-m-n func-tion, and finally of the universal machine.
Our generalization of Rice’s Theorem and other enumeration techniques is also different from (and in some sense complemen-tary to) the approach in (Kozen 1980): while Kozen investigates extensional properties in a sub-recursive setting, we are essentially concerned with decidability of complexity questions in a general recursive setting.
2.
Similarity and Complexity Cliques We shall work with n-ary partial recursive functions over natural numbers. If f is such a partial function, dom(f) and cod(f) re-spectively denote the domain and range of f. We write f(n) ↓if n ∈dom(f). We use ⊥for the everywhere divergent function, that is dom(⊥) = ∅. Given two function f and g, we write f ∼ = g when fand g have the same graph, i.e. dom(f) = dom(g) and for any n ∈dom(f), f(n) = g(n). We say a function is finite if it has a finite graph (i.e. dom(f) is a finite set). Partial functions are or-dered w.r.t the set-theoretic inclusion of their graphs: f ≤g if and only if graph(f) ⊆graph(g). ω is the set of all natural numbers.
DEFINITION 1. Let f and g be partial functions on natural num-bers: 1. we say f ∈O(g) if and only if there exist n and c such that for any m ≥n, if g(m) ↓then f(m) ≤cg(m); 2. f ∈Θ(g) if and only if f ∈O(g) and g ∈O(f).
Note that if f ∈Θ(g) the domains of f and g may only differ for a finite number of points. Let us also remark that, f ∈Θ(g) if and only if g ∈Θ(f); in particular, f ∈Θ(g) is an equivalence relation.
DEFINITION 2. (Blum 1967) A pair ⟨φ, Φ⟩is a computational complexity measure if φ is a principal effective enumeration of partial recursive functions and Φ satisfies Blum’s axioms: (a) φi(⃗ n) ↓↔Φi(⃗ n) ↓ (b) the predicate Φi(⃗ n) = m is decidable We say two programs are similar when they compute the same function and have the same complexity, up to constants: DEFINITION 3. Two programs i and j are similar (write i ≈j) if and only if φj ∼ = φi ∧Φj ∈Θ(Φi) THEOREM 4. Similarity is an equivalence relation.
Proof. Obvious, since similarity is defined as intersection of two equivalence relations.
We shall write [i]≈for the equivalence class of i w.r.t. the similarity relation ≈.
DEFINITION 5. Let ⟨φ, Φ⟩be an abstract complexity measure. A set P of natural numbers is a Complexity Clique, if and only if for all i and j i ∈P ∧j ≈i →j ∈P EXAMPLE 6. The following are examples of Complexity Cliques: 1. ∅and ω; 2. for any index i, [i]≈; 3. for any index i, {j|Φj ∈O(Φi)}.
On the other side the set {i|Φi(a) = b} (i.e. the programs whose complexity on input a is b) is not a Complexity Clique, in general.
3.
Simple properties of Complexity Cliques A set A ⊂ω of indices is extensional if, for all i and j, i ∈ A ∧φj ∼ = φi →j ∈A; this is equivalent to say that A is the counter-image (index set) under φ of some subset of partial recursive functions.
The notion of Complexity Clique is a strict generalization of the notion of extensional set well known in computability theory: THEOREM 7. Every extensional set is a Complexity Clique.
Proof. Trivial.
Of course, the converse is not true: for instance, the class of pro-grams with polynomial complexity is not extensional.
Any Complexity Cliques is a union of similarity classes, in particular: THEOREM 8. C is a Complexity Clique if and only if C = [ i∈C [i]≈ Proof. Trivial.
As a consequence, Complexity Cliques have very good alge-braic properties; in particular: THEOREM 9. Complexity Cliques, equipped with the subset rela-tion, are a Complete Boolean Lattice.
Proof. Just note that C = S i∈C[i]≈. The rest is easy.
The good set-theoretic properties of Complexity Cliques are to be compared with the very bad properties of Complexity Classes (stressing again the strong difference between the two notions). In particular: 1. for any complexity measure, Complexity Classes are not closed under union (Hartmanis and Stearns 1965; McCreight and Meyer 1969); 2. there are complexity measures whose Complexity Classes are not closed under intersection1(Landweber and Robertson 1972) Let us finally remark an important property about finite functions.
THEOREM 10. Let C be a Complexity Class and let i ∈C. If φi is finite then for any j such that φi ∼ = φj, j ∈C.
Proof.For finite functions, extensional equivalence implies similar-ity.
4.
Generalized Rice’s Theorem DEFINITION 11. A pair ⟨φ, Φ⟩has the s-m-n property if for all positive natural numbers m and n there exists a recursive function sn m such that, for any i and all x1, . . . , xm (a) φsn m(i,x1,...,xm) ∼ = λy1, . . . , yn.φi(x1, . . . , xm, y1, . . . , yn) (b) Φsn m(i,x1,...,xm) ∈Θ(λy1, . . . , yn.Φi(x1, . . . , xm, y1, . . . , yn)) Equation (a) is the standard s-m-n theorem, while equation (b) states that the overhead introduced by the function sn m is at most a constant factor: in fact, in standard computational models, it amounts to the cost of copying the (fixed) parameters x1, . . . , xm and then calling φi - see e.g. Cutland (1986); B¨ orger (1986); Odifreddi (1997).
Since the set or recursive functions is closed under composition, by the s-m-n theorem we may conclude that there exists a total computable function h such that φh(i,j) ∼ = φi◦φj. Axioms relating the complexity of h(i, j) to the complexity of its components have been considered by Lischke (1975, 1976, 1977). For the purposes of our analysis, we need a tight bound: DEFINITION 12. A pair ⟨φ, Φ⟩has the (linear) time-composition property if there exists a total computable function h such that (a) φh(i,j) ∼ = φi ◦φj (b) Φh(i,j) ∈Θ(Φi ◦φj + Φj) We say that it has the (linear) space-composition property if equa-tion (b) is replaced by (c) Φh(i,j) ∈Θ(max{Φi ◦φj, Φj}) Equation (b) says that the cost for computing φi(φj(x)) is, up to a constant factor, no more than the cost for computing φj(x) plus the cost for computing φi on input φj(x). Similarly, the space required for the same computation is the maximum among the 1 An important exception are measures satisfying the parallel computation property (Landweber and Robertson 1972) (see also section 5).
space required for computing φj(x) and the space required for computing φi on input φj(x).
The previous equations may be extended to n-ary composition, provided that input variables are linearly partitioned among the components; if duplication of some input is required, we should eventually take this cost into account, adding a linear overhead.
A particular case of (multi-variable) linear composition of fre-quent use in computability theory is sequencing with absence of communication, i.e. the case of a function φg(i,j)(x, y) = φj(x); φi(y) = ( φi(y) if φj(x)↓ ⊥ otherwise The aim is merely to force the evaluation of φj(x) before computing φi(y). According to our definitions Φg(i,j) ∈Θ(λxy.(Φj(x) + Φi(y))) for time and Φg(i,j) ∈Θ(λxy.max{Φj(x), Φi(y)}) for space.
In combination with the s-m-n property, fixing the parameter x, we have, for both time and space: LEMMA 13. There exists a total computable function c such that (a) ( φc(i,j,x)(y) = φi(y) if φj(x)↓ φc(i,j,x)(y) = ⊥ otherwise (b) ( Φc(i,j,x) ∈Θ(Φi) if φj(x)↓ Φc(i,j,x) = ⊥ otherwise In fact, the previous Lemma is all we need for the generalization of Rice’s Theorem to Complexity Cliques.
THEOREM 14. Under the s-m-n (Def. 11) and the linear-composition (Def. 12) assumptions (either in time or space), a Complexity Clique P is recursive if and only if it is trivial, i.e. P = ∅∨P = ω.
Proof. Assume P is recursive and let φp be its characteristic func-tion. Suppose P is not trivial, so there exist a and b such that φp(a) = 1 and φp(b) = 0. Let m be the index of a program computing the everywhere divergent function; either φp(m) = 1 or φp(m) = 0. Let us consider the latter case, the other one being analogous.
Let K be a r.e. not recursive set, and let φk be its semi-decision function, i.e. dom(φk) = K. Consider the following function: f(x, y) = φk(x); φa(y). By Lemma 13 there exists a total com-putable function c such that (a) ( φc(a,k,x)(y) = φa(y) if x ∈K φc(a,k,x)(y) = ⊥= φm(y) otherwise (b) ( Φc(a,k,x) ∈Θ(Φa) if x ∈K Φc(a,k,x) = ⊥= Φm(y) otherwise Since P is a Complexity Clique, φp(c(a, k, x)) = ( φp(a) = 1 if x ∈K φp(m) = 0 otherwise So, λx.φp(c(k, a, x)) is a characteristic function for K; c is com-putable, hence φp cannot be.
The case φp(m) = 1 is similar.
The knowledgeable reader will have certainly recognized, in part (a) of the above proof, the traditional Rice’s argument. We just rephrased it to put in evidence the two elementary operations required by the transformation, namely composition and s-m-n.
Providing a complexity bound for such operations is the key point that allows us to take complexity into account in a formal way.
However, it is clear that, in concrete computational models, part (b) is always satisfied (that is, the program φx(x); φa(y) has, for any given x, the same computational complexity as φa(y).
5.
Generalized Rice-Shapiro’s Theorem Rice’s Theorem provides a simple structural criterion for an exten-sional set to fall in Σ0. Rice-Shapiro’s Theorem2 does the same for Σ1. In a modern terminology, the result states that any completely r.e. set A of partial functions is upward closed and compact, i.e.
φi ∈A ⇔∃u finite, u ∈A ∧u ≤φi Let us start considering upward closedness, or monotonicity. The proof is not as standard as the one for Rice’s Theorem, and not ev-ery proof is suitable for a complexity-theoretic generalization. We recall here the argument in Odifreddi (1997), that is particularly close to our approach. Suppose there exist two partial recursive functions φi and φj such that φi ∈A, φj ̸∈A and φi ≤φj.
Let f be a recurtsive function such that φf(e)(x) = y ⇔φi(x) = y ∨(e ∈K ∧φj(x) = y) where K is an arbitrary r.e. non recursive set. Observe that the function f is well defined and computable just because φi ≤φj.
Then e ∈K ⇔φf(e) ∈A and since A is completely r.e. K would be r.e. too, that is a contradiction.
The key point of the above proof essentially consists in running in parallel two functions, namely φi(x) and φk(e); φj(x) where φk is a semidecision function for K. In order to study the complexity of the composite function, we need some assumptions about the complexity of parallel composition. Remarkably, the subject was investigated by Landweber and Robertson (1972) a long time ago and for quite different reasons.
DEFINITION 15. (Landweber and Robertson 1972) A pair ⟨φ, Φ⟩ has the parallel computation property if there exists a total com-putable function h such that (a) φh(i,j)(x) = ( φi(x) if Φi(x) ≤Φj(x) φj(x) otherwise (b) Φh(i,j) ∈Θ(λx.min{Φi(x), Φj(x)}) THEOREM 16. Let P be a r.e. Complexity Clique. If i ∈P, φi is finite and φi ≤φj then j ∈P.
Proof. Suppose there exist i and j such that i ∈P, φi is finite, φi ≤φj but j ̸∈P. Let K be a r.e. not recursive set, and let φk be its semi-decision function, i.e. dom(φk) = K. By Lemma 13 there exists a total computable function c such that φc(j,k,x)(y) = φk(x); φj(y) Let h be the function of the parallel computation property, and let us consider φh(i,c(j,k,x))(y) = ( φi(y) if Φi(y) ≤Φc(j,k,x)(y) φc(j,k,x)(y) otherwise 2 There seem to be some controversy on the paternity of this result; in particular, it is attributed to Shapiro (1956) by Rogers (1987), pag.324, and to Myhill and Shepherdson (1955) by (Odifreddi 1997), pag.206.
We claim that φh(i,c(j,k,x))(y) = ( φi(y) if x ̸∈K φj(y) otherwise If x ̸∈K then φc(j,k,x)(y) ↑so, by definition of φh(i,c(j,k,x)), either Φi(y) ↓and then φh(i,c(j,k,x))(y) = φi(y) or otherwise both φh(i,c(j,k,x))(y) and φi(y) diverge.
Conversely, suppose x ∈K. If φj(y) ↑, since φi ≤φj, also φi(y) ↑and hence φh(i,c(j,k,x))(y) ↑. If φj(y) ↓then, even if Φi(y) ≤Φc(j,k,x)(y) since φi ≤φj we have φi(y) = φj(y), and so in any case φh(i,c(j,k,x))(y) = φj(y). The second claim is that Φh(i,c(j,k,x)) ∈ ( Θ(Φi) if x ̸∈K Θ(Φj) otherwise Indeed if x ̸∈K then Φc(j,k,x) = ⊥and λy.min{Φi(y), Φc(j,k,x)(y)) = Φi Conversely, if x ∈K, by Lemma 13, Φc(j,k,x) ∈Θ(Φj), and since by hypothesis φi is finite, λy.min{Φi(y), Φc(j,k,x)(y)) = Φc(j,k,x) almost everywhere, so Θ(λy.min{Φi(y), Φh(i,c(j,k,x))(y)} = Θ(Φj) In conclusion, since P is a Complexity Clique, having assumed i ∈P and j ̸∈P, we have h(i, c(j, k, x)) ∈P ↔x ̸∈K and thus P cannot be r.e.
Let us remark that the hypothesis that φi is finite plays a crucial role in the previous proof. The point is that we have to compute a function similar to φi, if x ̸∈K and a function similar to φj if x ∈K. The parallel composition of φi and φj, when x ∈K, might be too fast for our purposes (it could belong to our complexity clique even if φj does not). Take for instance the case φi ∼ = φj but Φi is uniformly faster then Φj: in this case the parallel composition of i and j gives an algorithm with the same complexity as i.
As a corollary of Theorem 16 we get the following alternative proof of Rice’s Theorem (we repeat the statement both for com-pleteness and for ease of reference to the new proof): THEOREM 17. A Complexity Clique P is recursive if and only if it is trivial, i.e. P = ∅∨P = ω.
Proof. Let P be a recursive Complexity Clique. Then both P and P are r.e. and both are Complexity Cliques, since Complexity Cliques are closed under complementation. Let m be and index for the everywhere divergent function. Eventually, m ∈P or m ∈P.
If m ∈P then by Theorem 16 the index of any function must be in P, so P = ω. Similarly, if ⊥∈P then P = ω.
Let us now come to the converse of Theorem 16, that is com-pactness. In this case, Odifreddi’s proof runs as follows. Suppose A is a completely r.e. set of partial computable functions; assume φi ∈A but no finite subfunction of φi is in A. Let g(t) be the partial computable function such that g(x, t) = ( ↑ if φx(x) ↓in less then t steps 0 otherwise and consider the function φf(e)(x) = g(e, x); φi(x). It is clear that φf(e) ∼ = φi if φe(e) ↑and φf(e) is some finite subfunction of φi(x) if φe(e) ↓. Hence, φf(e) ∈A ⇔e ∈K, that is a contradiction.
We could probably argue that the complexity of g(x, t), when it terminates, is at most linear in t. However, this seems to require some delicate assumptions on the complexity of the intrepreter, and we finally decided to follow a slightly different route.
First of all we need a slight generalization of the parallel com-position property. The problem is that definition 15 does not allow to apply different continuations to the two parallel computations after termination of one of them.
DEFINITION 18. A pair ⟨φ, Φ⟩has the generalized parallel com-putation property if there exists a total computable function p such that for all i, i′, j, j′ (a) φp(i,i′,j,j′)(x) = ( φi′(φi(x)) if Φi(x) ≤Φj(x) φj′(φj(x)) otherwise (b) Φp(i,i′,j,j′) ∈Θ λx.
( Φh(i′,i)(x) if Φi(x) ≤Φj(x) Φh(j′,j)(x) otherwise !
where h is the sequential composition function of Definition 12.
Even if the definition is a bit involved, the generalized parallel computation property looks like a natural extension of property 15: roughly, the idea is that, given two algorithms, we may “arbitrarily” fix a checkpoint in their respective code, run them in parallel and, according to which one reaches its checkpoint first, drop the other computation. In order to reduce the generalized property to the simpler one, it looks enough to state the existence of pairs and a test function (we must know which one of the two computations terminated first), but then we should also discuss the complexity of all these new notions, that would drive us a bit too far away.
THEOREM 19. Let P be a r.e. Complexity Clique. If i ∈P and Φi ̸∈O(1) then there exists j such that φj is finite, φj ≤φi and j ∈P.
Proof. Let K be a r.e. not recursive set, and let φk be its semi-decision function, i.e. dom(φk) = K. Let m be an index for the everywhere divergent function, and let I be an index for the identity. By the s-m-n property there exists a total computable function c such that φc(x,k)(y) = φk(x) φc(x,k) is either the everywhere divergent function if x ̸∈K or a constant function with complexity Φc(x,k) ∈Θ(λy.Φk(x)) = Θ(1) otherwise. By the generalized parallel computation property, (a) φp(i,I,c(x,k),m)(y) = ( φi(y) if Φi(y) ≤Φk(x) ↑ otherwise (b) Φp(i,I,c(x,k),m) ∈Θ λy.
( Φi(y) if Φi(y) ≤Φk(x) ↑ otherwise !
If x ̸∈K then for any y Φi(y) ≤Φk(x), hence φp(i,I,c(x,k),m)(y) = φi(y), and the two functions also have the same complexity. If x ∈K, since Φi ̸∈O(1) then, Φi(y) ≤Φk(x) only for a finite number of values for y, and hence φp(i,I,c(x,k),m) is a finite func-tion.
If no index of finite sub-functions of φi is in P, we have φp(i,I,c(x,k),m) ̸∈P ↔x ̸∈K and thus P cannot be r.e.
COROLLARY 20. Let P be a r.e. Complexity Clique. If i ∈P and Φi ̸∈O(1) then for every j such that φj ∼ = φi we have j ∈P.
Proof. By Theorem 19, there exists a finite sub-function φr ≤φi such that r ∈P, and by Theorem 16, any j such that φr ≤φj, independently from its complexity Φj, must belong to P.
COROLLARY 21. No Complexity Clique of total functions and containing (indices of) programs with non constant complexity can be r.e.
By the last corollary, we have e.g. that the class of programs with linear (polynomial, exponential, . . . ) complexity is not r.e. (in fact, they are all Σ0 2: see below).
Corollary 20 essentially says that every Complexity Clique in Σ0 1 is (morally) an extensional set; in other words, all Complexity Cliques in Σ0 1 have trivial complexity. Since the complement of an extensional set is also extensional, a similar result holds for Complexity Cliques in Π0 1. Precisely: COROLLARY 22. Let P be a Complexity Clique in Π0 1. If i ∈P then for every j such that Φj ̸∈O(1) and φj ∼ = φi we have j ∈P.
Proof. Let i ∈P and consider j such that Φj ̸∈O(1) and φj ∼ = φi.
If j ̸∈P then j ∈P. Since P is a r.e. Complexity Clique, then we are in the hypothesis of Corollary 20 and we should have i ∈P, that is contradictory.
It is natural to wonder if there are Complexity Cliques with non trivial complexity in Σ0 2.
THEOREM 23. Let t be any total recursive function. Let Ct be the following Complexity Clique: Ct = {i|Φi ∈O(t)}. Then, Ct ∈Σ0 2.
Proof. By definition, Φi ∈O(t) if there exist n and c such that for any m ≥n, Φi(n) ≤ct(n). The relation Φi(n) ≤ct(n) is decidable by definition of complexity measure (Definition 2).
Hence, Ct is Σ0 2.
Since Σ0 2 is closed under r.e. unions, many interesting Complexity Cliques (such as, for instance, the Cliques of programs with poly-nomial complexity) are in Σ0 2.
6.
Kleene’s Second Fixed Point Theorem Kleene’s Second Fixed Point Theorem, in Rogers’ formulation (Rogers 1987), states that for any total recursive function f there exists an indices i such that φi ∼ = φf(i). The question is if we can always find a fixpoint i with the same complexity as f(i).
In the general framework given by an arbitrary abstract com-plexity measure, the best we can prove is that the two complexity of i and f(i) are related by a total recursive function. More pre-cisely3: THEOREM 24. (Blum 1971) There exists a binary total recursive function h monotonically increasing in its second argument such that, for any total recursive function φi there exists an index m such that, for any x, (1) φm(x) = φf(m)(x) (2) Φm(x) ≤h(x, Φf(m)(x)) (3) Φm(x) ≥Φf(m)(x) Fixing a particular model of computation, we may of course pro-vide a better estimation for h. The interesting point is that, assum-ing the s-m-n property, the complexity of h uniquely depends from the complexity of the universal function φu: THEOREM 25. Let ⟨φ, Φ⟩be an abstract complexity measure with the s-m-n property 11, and let u be an index for the universal 3 See (Odifreddi 1999) for the analogous theorem in terms of Kleene’s formulation of the Fixed Point Theorem. This has been also investigated in (Hansen et al. 1989) but no theoretical measure is given: the emphasis of the paper is more on practical, implementative issues.
function. Then for any total recursive function φi there exists an index m such that, for any x, (1) φm ∼ = φφi(m) (2) Φm ∈Θ(λy.Φu(φi(m), y) Proof. Consider the following computable function: g(x, y) = φφi(φx(x))(y) = φu(φu(i, (φu(x, x))), y) By the s-m-n theorem there exists a computable function s such that φs(x) = λy.g(x, y) and, for any x, Φs(x) ∈ Θ(λy.Φu(φu(i, (φu(x, x))), y)) = Θ(λy.Φu(φi(φx(x)), y)) (x is a fixed parameter, hence the complexity of computing φi(φx(x)) does not matter). Since s is a total computable function there exists p such that s = φp. φp(p)↓since φp is total. Moreover we have: φφp(p)(y) = g(p, y) = φφi(φp(p))(y) and Φφp(p) = Θ(λy.Φu(φi(φp(p)), y) DEFINITION 26. We say that a universal function φu is fair if for any x λy.Φu(x, y) ∈Θ(Φx) The cost of interpreting a program i by a fair universal machine may only introduce a constant slow-down factor c w.r.t. the direct computation of i. However, the constant c may depend on i. Fol-lowing (Jones 1993) we say that the universal machine is efficient when the constant c is independent from the interpreted program.
COROLLARY 27. Let ⟨φ, Φ⟩be an abstract complexity measure with the s−m−n property 11. If it admits a fair universal function u then for any total recursive function φi there exists an index m such that, for any x, (1) φm ∼ = φφi(m) (2) Φm ∈Θ(Φφi(m)) Let us remark that the previous result says almost nothing about the “efficiency” of the fixpoint; it only says that, independently from φi we may always find a fixpoint m as (in-)efficient as φi(m) (that is not surprising: a fixpoint program is as efficient as the program defined by its body).
Not all computational models seem to admit fair universal ma-chines. Discussing the case of multi-tape Turing machines, Blum (1971) suggests an upper bound for Φu(i, j) given by Φi(j)2 + j, where the square is due to the fact that the universal machine has a given number of tapes, but may need to simulate additional ones (if we admit that the universal machine has at least two tapes, the cost is likely to be reduced to log(Φi(j))Φi(j) + j)4.
However, many models, comprising single tape Turing ma-chines, lambda calculus and combinatory logic have fair universal machines. Unfortunately, this the kind of results that, belonging to the so called “folklore” of these subjects, cannot be properly quoted or elaborated. We hope someone will assume soon or later the bur-den to formally prove these important properties, and also hope that the scientific community will be so wise to accept these contribu-tions. As a remarkable exception, the existence of efficient universal functions is proved by Jones (1993) for several classes of compu-tational models defining problems in deterministic linear time (i.e.
computational models with very restricted capabilities).
4 In our opinion, the notion of multi-tape machine as a foundational model of computation is arguable: indeed, we are used to work with a fixed ma-chinery (a given processor), that amounts to fix a single universal machine with a given number of tapes, emulating all other programs.
The strong version of the Fixed Point Theorem of Corollary 27 (when it holds) is a major tool for the investigation of Complexity Cliques, since it can be used with the same confidence of the traditional theorem w.r.t. extensional sets. As a simple example, let us consider the traditional proof of Rice’s Theorem making use of a fixed point.
THEOREM 28. A Complexity Clique P is recursive if and only if it is trivial, i.e. P = ∅∨P = ω.
Proof. Assume P is recursive and let φp be its characteristic func-tion. Suppose P is not trivial, so there exist a and b such that φp(a) = 1 and φp(b) = 0. Let us consider the following com-putable function g(x) = ( a if φp(x) = 0 b if φp(x) = 1 If φp is total recursive, so is g, hence we may apply Corollary 27, concluding that there exists an index m such that φm ∼ = φg(m) Φm ∈Θ(Φg(m)) In particular, if φp(m) = 0 we have φm ∼ = φa Φm ∈Θ(Φa) Since P is a Complexity Clique and φp(a) = 1 it should also be φp(m) = 1: contradiction.
Similarly, if if φp(m) = 1 we have φm ∼ = φb Φm ∈Θ(Φb) Since P is a Complexity Clique and φp(b) = 0 it should also be φp(m) = 0, that is again a contradiction.
7.
Conclusions In this paper, we introduced the notion of Complexity Clique: a set of indices for programs closed under a similarity relation de-fined taking into account their extension and complexity. Recursive properties of Complexity Cliques have been investigated, sharpen-ing classical results of Recursion Theory, like Rice’s and Rice-Shapiro’s Theorems. In particular, we proved that all recursive Complexity Cliques are trivial, and all Complexity Cliques in Σ0 1 and Π0 1 are in fact (morally) extensional (i.e. are trivial w.r.t. com-plexity). In this way, we rephrase classical theorems in computabil-ity theory in order to establish properties of classes of programs defined by complexity conditions, concluding that no non-trivial complexity property is semi-decidable. For instance, the recent field of implicit computational complexity - see eg Dal Lago and Baillot (2006), Amadio (2005), citetAR02 and the biliography therein -studies criteria on programs implying some complexity properties.
The results of this paper put in evidence that such criteria, when-ever computable, can not be necessary and sufficient conditions. In other words, if polytime languages may be extensionally complete (computing all polynomial functions) they cannot be intensionally complete, that is express all polynomial algorithms.
The technical flavor of the paper follows Blum’s axiomatic ap-proach, but for the aims of the proofs we extended Blum’s axioms with stronger assumptions, concerning the complexity of the s-m-n function, of sequential and parallel composition, and of the univer-sal machine. Not all assumptions are needed for all results and the study of complexity seem to allow a deeper, fine grained, investi-gation even of traditional aspects of Recursion Theory (see e.g. the three different proofs of the generalized Rice’s Theorem: Theorems 14-17-28).
It is possible (but not evident) that by relaxing the complexity condition in the notion of similarity (e.g. to recursive relatedness via some total recursive function h) we could prove (much) weaker results in the full generality of an arbitrary abstract complexity measure (i.e. without requiring any additional axiom). However, our axioms are very natural, and it is not clear if the effort would be worth the result.
From a strictly technical point of view, the requirement i ̸∈ O(1) in Theorem 19 (and followings) is annoying (more from an aesthetic point of view than a practical one). It would be nice to find a way to avoid it (or, alternatively, to prove that it is indeed an essential hypothesis).
We believe that most of the interesting problems investigated at the beginning of the seventies for Complexity Classes can also be studied, possibly more profitably, in the framework of Complexity Cliques. A typical example is recursive presentability, that is the problem to provide effective enumerations of representatives (up to extensional equivalence) for all elements in the given set (such as the problem of giving an effective enumeration of programs com-puting all functions with polynomial complexity). As already re-marked by several authors - see e.g. Landweber and Robertson (1972); Young (1969) - , the problem does not make much sense for Complexity Classes: since complexity Classes are extensional you could enumerate “bad” programs (not within the given complexity bound) for “good” functions (functions admitting a program within the given complexity bound). The problem has been traditionally solved introducing a notion of quality for presentations (Landwe-ber and Robertson 1972), but Complexity Cliques, due to their intensional nature, could provide an alternative and possibly more natural framework to work with.
One of the appealing aspects of Complexity Cliques is that, (at the contrary of Complexity Classes, for which even basic composi-tional properties fail) they have a very nice set-theoretic structure: they are a Complete Boolean Lattice (Theorem 9).
In a categorical perspective, Complexity Cliques, equipped with the similarity relation, are partial equivalence relations, and it looks natural to investigate them as a sub-category of PER (Freyd et al. 1992). The natural notion of morphism seems to be that of a complexity-preserving effective operator (Dekker and Myhill 1958); note that, however, in order to define a category you im-mediately need some additional assumptions on the nature of com-position beyond Blum’s axioms. As it is often the case, Category Theory may turn out to be the best tool to grasp the essence of the notions under investigation.
Finally, from the point of view of Abstract Complexity Theory, we hope that our work can help to revitalize a field that, lately, has been a bit starving. In particular, in (Asperti and Ciabattoni 1995) it was proved that any enumerated collection of functions containing projections, a universal function, and closed w.r.t. composition and the s-m-n construction is algorithmically complete. For this reason, an abstract approach based on complexity assumptions for these basilar constructions looks particularly appealing.
Acknowledgments We are grateful to G.Longo, S.Martini, U.Dal Lago and Claudio Sacerdoti Coen for a few interesting discussions on the subject of this paper.
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1985 AIME Problems/Problem 15
Contents
[hide]
1 Problem
2 Solution 1 (For the visualizers)
3 Solution 2
4 Solution 3
5 See also
Problem
Three 12 cm 12 cm squares are each cut into two pieces and , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regularhexagon, as shown in the second figure, so as to fold into a polyhedron. What is the volume (in ) of this polyhedron?
Solution 1 (For the visualizers)
Note that gluing two of the given polyhedra together along a hexagonal face (rotated from each other) yields a cube, so the volume is , so our answer is .
Image:
Solution 2
Taking a reference to the above left diagram, obviously SP, SQ and SR are perpendicular to each other. Thus expanding plane SPQ, SQR and SRP to intersect with the plane XYZ that contains the regular hexagon, we form a pyramid with S the top vertex and the base being an equilateral triangle with side length of 18 . This pyramid has a volume of 972, because it is also 1/6 of the volume of a cube with side length of 18. Then subtracting 3 congruent pyramids with volume of 36 each, we get .
Solution 3
Position the polyhedra with the facing hexagon down. From this perspective it is clear that the polyhedra can be thought of as a tetrahedron with 3 smaller tetrahedra removed from each of the three corners of its base. The strategy is then to calculate the volume of the larger tetrahedron, and then subtract the three smaller ones.
Focusing on the larger tetrahedron, the base is an equilateral triangle formed by extending the sides of the base hexagon; this triangle has sides of length . Thus its area is . The height of the tetrahedron can be calculated by considering the following right triangle: Leg 1 is the height of the tetrahedron (which we seek); Leg 2 is the distance from a midpoint of one of the sides of the base to the base's center; and the Hypotenuse is the distance from the midpoint of one of the sides of the base to the top vertex. Leg 2 has a length of . And the hypotenuse's length is (each non-base face of the tetrahedron is an isosceles right triangle with leg of length 18, whose altitude to its hypotenuse has length ). Thus, by the Pythagorean theorem the height of the tetrahedron is . Altogether, the Volume of the large tetrahedron is .
Lastly, each smaller tetrahedron removed from a corner of the base of the larger one has sides whose lengths are that of the larger tetrahedron. Since volume scales with the cube of length, each of these tetrahedra are the volume of the larger one. There are three of them so the final volume is .
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10506 | https://radiopaedia.org/articles/multiple-sclerosis?lang=us | Multiple sclerosis
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Citation, DOI, disclosures and article data
Gaillard F, Agazzi G, Sharma R, et al. Multiple sclerosis. Reference article, Radiopaedia.org (Accessed on 13 Sep 2025)
Article created:
2 May 2008,
Frank Gaillard
At the time the article was created Frank Gaillard had no recorded disclosures.
View Frank Gaillard's current disclosures
Last revised:
23 Aug 2025,
Giorgio Maria Agazzi
Disclosures:
At the time the article was last revised Giorgio Maria Agazzi had
no financial relationships to ineligible companies to disclose.
View Giorgio Maria Agazzi's current disclosures
Revisions:
101 times, by
41 contributors -
see full revision history and disclosures
Systems:
Central Nervous System, Paediatrics
Tags:
emergencymedicine, cns, demyelinating disease
Synonyms:
MS
Multiple sclerosis (MS)
Charcot type multiple sclerosis
Classic multiple sclerosis
Charcot type multiple sclerosis (MS)
Paediatric-onset multiple sclerosis
Paediatric-onset MS
Paediatric-onset multiple sclerosis (POMS)
Paediatric-onset MS (POMS)
Multiple sclerosis (MS)is a relatively common acquired chronic demyelinating disease involving the central nervous system and is the second most common cause of neurological impairment in young adults, after trauma 19. Characteristically and by definition, multiple sclerosis is disseminated in space (i.e. multiple lesions in different regions of the brain) and in time (i.e. lesions occur at different times).
A number of clinical variants are recognized, each with specific imaging findings and clinical presentation. They include:
classic multiple sclerosis (Charcot type)
pediatric-onset multiple sclerosis (variably defined as multiple sclerosis with an onset <16-18 years of age)
tumefactive multiple sclerosis
Marburg type (acute malignant)
Schilder type (diffuse cerebral sclerosis)
Baló concentric sclerosis
This article concerns itself primarily with classic (Charcot-type) multiple sclerosis. The other variants are discussed separately.
Importantly, neuromyelitis optica spectrum disorder (Devic disease) was considered a variant of multiple sclerosis but is now recognized as a distinct entity and is therefore also discussed separately.
On this page:
Article:
Epidemiology
Clinical presentation
Diagnosis
Pathology
Radiographic features
Treatment and prognosis
History and etymology
Differential diagnosis
Related articles
References
Images:
Cases and figures
Imaging differential diagnosis
Epidemiology
The presentation is usually between adolescence and the sixth decade, with a peak at approximately 35 years of age 12,19. Up to 10% of patients have pediatric-onset multiple sclerosis 31. There is a strong, well recognized female predilection with a F:M ratio of approximately 2:1 19.
Multiple sclerosis has a fascinating geographic distribution: it is rarely found in equatorial regions (e.g. 15 per 100,000), with incidence gradually increasing with distance from the equator (e.g. 250 per 100,000) 12,19.
Twin studies show a heritability of ~30% 38.
Associations
a strong association with HLA-DR15 (formerly covered by HLA-DR2) class II has been identified 35
several studies showed a correlation between prior EBV infection and multiple sclerosis onset, both in the adult and pediatric population 38
Clinical presentation
Clinical presentation is both highly variable acutely, as a result of varying plaque location, as well as over time. Examples of common clinical features include 23,24:
brainstem and cranial nerve involvement:
optic neuritis
internuclear ophthalmoplegia (often bilateral)
trigeminal neuralgia
diplopia (e.g. due to abducens nerve palsy)
vertigo
cerebellum involvement:
ataxia and gait disturbance
oscillopsia
cerebrum and spinal cord involvement:
limb sensory loss or paresthesias
upper motor neuron signs
Lhermitte sign
urinary incontinence
others:
fatigue
depression
Uhthoff phenomenon: heat and exercise worsen symptoms
cognitive decline
Types
A number of patterns of longitudinal disease have been described 11,12:
relapsing-remitting (RRMS)
most common in adults (70% of cases), even more common in pediatric-onset multiple sclerosis (98% of cases) 31
patients exhibit periodic symptoms with complete recovery (early on)
secondary progressive (SPMS)
approximately 85% of adult patients with relapsing-remitting MS eventually enter a secondary progressive phase, but this is less common in pediatric-onset multiple sclerosis
primary progressive (PPMS)
uncommon in adults (10% of cases) and very rare in pediatric-onset multiple sclerosis 31
defined by a progressive accumulation of disability for >12 months from disease onset, which can be determined prospectively or retrospectively
patients do not have remissions, with neurological deterioration being relentless
incorporates the previously described "progressive-relapsing" phenotype
benign multiple sclerosis
15-50% of cases
defined as patients who remain functionally active for over 15 years, and thus is only a retrospective diagnosis
As is evident from this list, there is overlap, and in some cases, patients can drift from one pattern to another (e.g. relapsing-remitting to secondary progressive).
Upon presentation patients often have evidence of multiple previous asymptomatic lesions, and the diagnosis of multiple sclerosis can be strongly inferred. In other instances patients present with the first plaque. This is known as clinically isolated syndrome (CIS) and not all patients go on to develop multiple sclerosis.
Radiologically isolated syndrome (RIS) is another entity based on MRI brain findings which described as incidental white matter lesions suggestive of MS on imaging in a patient without associated clinical symptoms 17.
Diagnosis
The diagnosis of multiple sclerosis requires the constellation of clinical findings and various investigations (see McDonald diagnostic criteria for multiple sclerosis), including 19:
typical clinical history
oligoclonal bands in CSF
abnormal visual evoked potential
MR imaging
lack of viable alternative diagnosis
Pathology
The exact etiology is poorly known although it is believed to have both genetic and acquired contributory components. An infectious agent (e.g. EBV), or at least a catalyst, has long been suspected due to the geographic distribution and presence of clusters of cases; however, no agent has yet been firmly confirmed. Additionally the presence in 3-7% of patients with multiple sclerosis of often striking and symmetric high T2/FLAIR signal of the intrapontine segments of the trigeminal nerve has led some to speculate whether there is a neurotropic viral cause from the aerodigestive tract 34.
Some authors also suggested that "chronic cerebrospinal venous insufficiency" can cause or exacerbate multiple sclerosis but this theory has not been proven by further investigations 15.
Multiple sclerosis is believed to result from a cell-mediated autoimmune response against one's own myelin components, with loss of oligodendrocytes, with little or no axonal degeneration in the acute phase; however, in later stages, loss of oligodendrocytes results in axonal degeneration.
Demyelination occurs in discrete perivenular foci, termed plaques, which range in size from a few millimeters to a few centimeters 19. Each lesion goes through three pathological stages:
early acute stage (active plaques)
active myelin breakdown
plaques appear pink and swollen
subacute stage
plaques become paler in color ("chalky")
abundant macrophages
chronic stage (inactive plaques/gliosis)
little or no myelin breakdown
gliosis with associated volume loss
appear grey/translucent
Radiographic features
Plaques can occur anywhere in the central nervous system. They are typically ovoid in shape and perivenular in distribution.
CT
CT features are usually non-specific, and significant change may be seen on MRI with an essentially normal CT scan. Features that may be present include:
plaques can be homogeneously hypoattenuating 8,11
brain atrophy may be evident in long-standing chronic MS 5
some plaques may show contrast enhancement in the active phase 7,11
MRI
MRI has revolutionised the diagnosis and surveillance of patients with MS. Not only can an MRI confirm the diagnosis (see McDonald diagnostic criteria for multiple sclerosis), but follow-up scans can assess response to treatment and help determine the disease pattern.
Protocol
Although many sequences are contributory, the 2018 Revised Guidelines of the Consortium of MS Centers MRI Protocol for the Diagnosis and Follow-up of MS plaques lists the following core sequences 25:
FLAIR (axial and sagittal)
ideally performed as a 3D volumetric scan (1 mm isotropic), or
3 mm contiguous
T1: 3D inversion recovery prepared gradient echo
T2 (axial): 3D or 2D
DWI (axial)
NB: contrast is not necessary for routine asymptomatic follow-up.
Sequence utility
T1
lesions are typically iso- to hypointense (T1 black holes)
callososeptal interface may have multiple small hypointense lesions (Venus necklace) or the corpus callosum may merely appear thinned 11
hyperintense lesions are associated with brain atrophy and advancing disease 18
T2
lesions are typically hyperintense
acute lesions often have surrounding edema
FLAIR
lesions are typically hyperintense
a very early sign is the ependymal dot-dash sign 16
when these propagate centrifugally along the medullary venules and are arranged perpendicular to the lateral ventricles in a triangular configuration (extending radially outward - best seen on parasagittal images), they are termed Dawson fingers
FLAIR is more sensitive than T2 in the detection of juxtacortical and periventricular plaques, while T2 is more sensitive to infratentorial lesions
double inversion recovery (DIR)
a sequence that suppresses both CSF and white matter signal and offers better delineation of the plaques compared to FLAIR 33
SWI
central vein sign: at higher field strengths most plaques have been shown to be perivenular (at 3 T, 45% of lesions; at 7 T, 87% of lesions) 19
paramagnetic rim lesions: may be seen in approximately 50% of patients, presence increases specificity of diagnosis and correlates with disability from disease 32
T1 C+ (Gd)
active lesions show enhancement
enhancement is often incomplete around the periphery (open ring sign)
DWI/ADC
active plaques may demonstrate high or low ADC (increased or decreased diffusion) 10,11,22
also typically open ring in morphology
PD
PD images are better at detecting cervical spinal cord MS lesions especially when T2W images fail to demonstrate these lesions 26
MR spectroscopy
NAA peaks may be reduced within plaques, which is the most common and remarkable finding
choline and lactate are found to be increased in the acute pathologic phase
Location of the plaques can be infratentorial, in the deep white matter, periventricular, juxtacortical or mixed white matter-grey matter lesions.
Even on a single scan, some features are helpful in predicting relapsing-remitting vs progressive disease. Features favoring progressive disease include:
large numerous plaques
hyperintense T1 lesions
Choroid plexus volume (CPV) is emerging as a multiple sclerosis biomarker, with multiple sclerosis patients having a greater CPV compared with healthy controls. CPV is also correlated with atrophy measures, lesion load and clinical disability status 41.
Chronic active lesions
Chronic active lesions (CALs) are a distinct type of brain lesion in multiple sclerosis that reflect ongoing, smoldering inflammation and are strongly linked to disease progression and disability:
paramagnetic rim lesions: show a dark rim on SWI, corresponding to the iron-rich inflammatory border 39,40
slowly expanding lesions (SELs): identified by gradual, concentric growth over time on serial MRI scans in regions within pre-existing T2 lesions 39,40
PET imaging can highlight active inflammation at the lesion rim 39
CALs are associated with more severe disability, progressive disease, and brain atrophy 39,40.
Treatment and prognosis
The aim of management is twofold:
curtail progression with disease-modifying therapies
symptomatic relief
Non-pharmacological preventative therapies include:
smoking cessation
vitamin D supplementation
sun exposure: appears to correlate with a slowing down of the rate of progression of the disease - to clinically important levels - and the risk of disability 30
Disease-modifying therapies include multiple pharmacological agents and autologous hematopoietic stem cell transplantation. Although discussion of individual agents and therapies is beyond the scope of this article, it is worth being aware of the main agents available and their mechanism of action 20:
interferon beta: inhibition of T-lymphocyte proliferation
glatiramer acetate (Copaxone): immunomodulation
teriflunomide (Aubagio): reduces both T-cell and B-cell activation and proliferation
dimethyl fumarate (Tecfidera) and diroximel fumarate (Vumerity): immunomodulation
fingolimod (Gilenya), siponimod (Mayzent) and ozanimod (Zeposia): prevents lymphocyte migration out of lymph nodes and into CNS
natalizumab (Tysabri): inhibits binding of lymphocytes to endothelium
cladribine (Mavenclad): purine analog that targets lymphocytes
ocrelizumab (Ocrevus), ofatumumab (Kesimpta) and rituximab: anti-CD20 monoclonal antibodies
alemtuzumab (Lemtrada): immunomodulation of T-cell and B-cell function
mitoxantrone (Novantrone): reduces T-cell and B-cell proliferation and reduces T-cell activation
Prognosis is variable and depends on the pattern of disease a patient has (e.g. primary progressive carries a worse prognosis than relapsing-remitting). In general, patients with relapsing-remitting MS will progress to secondary progressive disease in 10 years and will require ambulatory aids (e.g. cane/wheelchair/frame) in another 5 to 15 years 12. Approximately half of the affected individuals will no longer be independently ambulatory after 20 years 19. Overall life expectancy is also reduced, by 7 to 14 years 19.
High serum neurofilament light chain values have been associated with higher risk of relapse and overall worse prognosis 38.
Complications
In addition to the potential for disease progression resulting in progressive neurological impairment, a number of specific complications need to be considered. These include 20,21:
progressive multifocal leukoencephalopathy (PML) and other less common CNS JC virus manifestations (JC virus granule cell neuronopathy, JC virus encephalopathy, and JC virus meningitis)
particularly in patients treated with natalizumab with positive JC virus serology
PML-IRIS
a complication of cessation of natalizumab or treatment for natalizumab-related PML with plasma exchange or immunoabsorption 21
primary CNS lymphoma
rarely lymphoma appears to arise from previously identified demyelinating lesions
2-times increased risk of stroke 28, possibly related to chronic inflammatory vasculopathy
History and etymology
Multiple sclerosis was first defined by Jean-Martin Charcot (1825-1893), French neurologist, in 1868 27.
Differential diagnosis
The differential diagnosis is dependent on the location and appearance of demyelination. For classic (Charcot type) MS, the differential can be divided into intracranial and spinal involvement.
For intracranial disease, the differential includes almost all other demyelinating diseases, as well as:
CNS fungal infection (e.g. Cryptococcus neoformans): patients tend to be immunocompromised
mucopolysaccharidosis (e.g. Hurler disease): congenital and occurs in a younger age group
Marchiafava-Bignami disease (for callosal lesions)
Susac syndrome
CNS manifestations of primary antiphospholipid syndrome 13
CLIPPERS
levamisole-induced leukoencephalopathy
adult-onset Alexander disease
adult polyglucosan body disease
CTLA4 haploinsufficiency with autoimmune infiltration 36,37
For spinal involvement, the following should be considered:
other causes of transverse myelitis
infection
spinal cord tumors (e.g. astrocytomas)
spinal cord infarction (e.g. sulcal artery syndrome)
Multiple sclerosis variants (e.g. tumefactive MS) are discussed separately.
Quiz questions
Question 3108
Report problem with question
A 40 year old woman presents with optic neuritis. Her MRI demonstrates bilateral optic nerve signal abnormality involving the optic chiasm and prechiasmatic segments. The orbits are otherwise normal. What is the most likely diagnosis?
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References
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Nusbaum A, Lu D, Tang C, Atlas S. Quantitative Diffusion Measurements in Focal Multiple Sclerosis Lesions: Correlations with Appearance on TI-Weighted MR Images. AJR Am J Roentgenol. 2000;175(3):821-5. doi:10.2214/ajr.175.3.1750821 - Pubmed
Valery N. Kornienko, I.N. Pronin. Diagnostic Neuroradiology. (2008) ISBN: 9783540756521 - Google Books
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Related articles: White matter disorders
white matter
normal myelination
terminology[+][+]
leukodystrophy
leukoencephalopathy
demyelinating disorders
dysmyelinating disorders
hypomyelinating disorders
white matter disorders
demyelination
general articles[+][+]
longitudinally extensive spinal cord lesion (LESCL)
transverse myelitis
tumefactive demyelinating lesions
acute disseminated encephalomyelitis (ADEM)[+][+]
Callen MS-ADEM criteria
acute hemorrhagic encephalomyelitis (AHEM)
anti-MOG associated encephalomyelitis (MOGAD)
neuromyelitis optica spectrum disorder (NMOSD)[+][+]
signs
bright spotty lesions
marbled pattern of the corpus callosum
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Does tense matter in propositional logic [negation. contrapositive, converse]?
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Say I have a sentence : If I water my plants then they will grow.
Would tenses have anything to do with writing it's negation and contraposition
Negation: I watered my plants but they did not grow.
Contrapositive: If they do not grow then I did not water my plants.
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asked Jun 15, 2021 at 23:04
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1 Your negation is not correct. I would say, it is possible for me to water my plants and for them to not grow. Tense must be maintained.Rushabh Mehta –Rushabh Mehta 2021-06-15 23:19:52 +00:00 Commented Jun 15, 2021 at 23:19
6 I find in propositional logic it is dangerous to use future tense, because it vaguely implies a rule “any time I water my plants, the plants will grow.” But propositional logic is about individual facts, not facts that can vary over time. As such, we can say “If I watered my plants yesterday, my plants grew today.” But prop logic can’t talk about rules that apply to many cases. So I try to avoid language that predicts future based on PL. (It’s not technically wrong, but it can be misleading.)Thomas Andrews –Thomas Andrews 2021-06-15 23:20:22 +00:00 Commented Jun 15, 2021 at 23:20
1 In general, I like to use past tense whenever I'm working with sentences in a logical context. This is both for the reasons Thomas presents, as well as consistency.Rushabh Mehta –Rushabh Mehta 2021-06-15 23:21:33 +00:00 Commented Jun 15, 2021 at 23:21
1 Also avoid past perfect: “If I had taken out the garbage yesterday, my kitchen wouldn’t stink today” gives more of a sense of causality than “If I took out the garbage yesterday, my kitchen wouldn’t stink today.”Thomas Andrews –Thomas Andrews 2021-06-15 23:35:33 +00:00 Commented Jun 15, 2021 at 23:35
1 Dan's linked article is bogus.user21820 –user21820 2021-07-16 15:34:12 +00:00 Commented Jul 16, 2021 at 15:34
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Classical logic (Logic from more than a thousand years ago) does not include notions of time (past, present, future, etc...).
The following are examples of a few rules of inference from classical logic:
Rule of Inference: Implication Operator From Parent to Left Child
For all P P and Q Q in the set {t r u e,f a l s e}{t r u e,f a l s e}, the following is true:
N O T(i f P t h e n Q)⟹P N O T(i f P t h e n Q)⟹P
Rule of Inference: Implication Operator From Parent to Right Child
For any P P and Q Q in the set {t r u e,f a l s e}{t r u e,f a l s e}, the following is true:
N O T(i f P t h e n Q)⟹Q N O T(i f P t h e n Q)⟹Q
Rule of Inference: DeMorgan's Law
For any P P and Q Q in the set {t r u e,f a l s e}{t r u e,f a l s e}, the following is true:
[N O T(P O R Q)]⟷[(N O T P)A N D(N O T Q)][N O T(P O R Q)]⟷[(N O T P)A N D(N O T Q)]
Note that none of the rules of inference in classical logic talk about time.
It is not easy to express any of the following English sentences in logic from hundreds of years ago:
"I will water my plants in the future"
"I watered my plants in the past"
"I currently water the plants inside my house."
I would stay that it is incorrect to use classical logic to express the English sentence "If I water my plants then they will grow" as "i f P t h e n Q i f P t h e n Q" where P P is "I water my plants"
Even an operation as simple as negation is not the same in English as it is in classical logic.
Perhaps, you would say that the following are equivalent:
N O T N O TI do water my plants
I doN O T N O Twater my plants
The statements about plants might be okay. However, consider the two following two statements:
N O T N O TMy wife's cousin Hector's pet monkey does like to eat strawberries
My wife's cousin Hector's pet monkey doesN O T N O Tlike to eat strawberries
There are some issues, not the least of which are:
I do not have a wife
If I had a wife, I am not sure if she would have a cousin Hector or not
Even if I was married, and my wife had a cousin named Hector it is doubtful that Hector would own a pet monkey.
In English, you are not allowed to move the word "N O T N O T" further and further to the left, or further to the right.
It is helpful to learn simple logic from ye olden days before you learn more advanced contemporary logic. However, I really wish that your teacher would not give you examples in English. English is really too complicated to force into the classical logic model.
In mathematics, time is sometimes modeled as follows:
If there exists at least one time in the future at which statement P P is true, then we write:
◇P P
For example:
◇I am watering my plants
There exists a time in the future such that the following will be true at that time: I am watering my plants
I still think that this is an abuse of the English language. However, using the symbols ◇ and □ to say something about time makes for a better model of English, relatively speaking, than using classical logic alone.
If something is true for all times in the future then we might write:
□P P
For example:
For all times in the future I am watering my plants
□I am watering my plants
I am not sure that modal logic using the symbols ◇ and □ is great for English, but it does help in math sometimes.
For example, the following are equivalent:
Eventually, F(x)F(x) is always zero.
◇□F(x)=0 F(x)=0
There exists a number x x such that for all y≥x,F(y)=0 y≥x,F(y)=0
Suppose that:
P P is the string I water my plants
Q Q is the string they will grow.
Well then, we have:
| Thing 1 | Thing 2 |
--- |
| I F P T H E N Q I F P T H E N Q | I F I FI water my plantsT H E N T H E Nthey will grow |
| P A N D(N O T Q)P A N D(N O T Q) | (I water my plants) A N D A N D (N O T N O Tthey will grow) |
| I F(N O T Q)T H E N(N O T P)I F(N O T Q)T H E N(N O T P) | I F I F (N O T N O Tthey will grow) T H E N T H E N (N O T N O TI water my plants) |
If some of the things written in the table above do not sound like English it is because, the things written in the table are not English.
You cannot model English sentences so easily using classical logic.
In English, the following two sentences have the same meaning:
If I water my plants then they will grow.
If I water my plants then my plants will grow.
With classical logic alone, you are not allowed to do something as simple as understand that the word "they" can be replaced with the phrase "my plants" while retaining the meaning.
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edited Jul 10, 2021 at 2:50
answered Jun 16, 2021 at 5:25
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You wrote:
Say I have a sentence : If I water my plants then they will grow.
Would tenses have anything to do with writing it's negation and contraposition
Propositional logic deals with propositions that are unambiguously either true or false in the present. So, you might consider recasting your sentence as:
If my plants have sufficient water, then my plants are growing.
Then, it easy to see that its negation would be:
My plants have sufficient water and my plants are not growing.
The contrapositive would be:
If my plants are not growing, then my plants have insufficient water.
This follows from the usually given definition of IMPLIES: A→B≡¬(A∧¬B)A→B≡¬(A∧¬B)
This "definition" can also be derived from other more fundamental rules of inference. See my blog posting in this topic here.
EDIT: If we restrict our attention to propositions in the present tense, we can safely ignore the complications of temporal logic. See:
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edited Jun 20, 2023 at 16:30
answered Jun 16, 2021 at 13:54
Dan ChristensenDan Christensen
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10508 | https://math.stackexchange.com/questions/4779387/unique-solvability-of-weak-poisson-equation-with-neumann-boundary-condition | sobolev spaces - Unique solvability of weak Poisson equation with Neumann boundary condition - Mathematics Stack Exchange
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Unique solvability of weak Poisson equation with Neumann boundary condition
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I'm studying a Poisson BVP: Find u∈W 1,p(Ω)u∈W 1,p(Ω) such that
∫Ω∇u⋅∇v=F(v)∀v∈W 1,p′(Ω),∫Ω∇u⋅∇v=F(v)∀v∈W 1,p′(Ω),
where p p and p′p′ are Hölder conjugates and F F belongs to the dual of W 1,p′(Ω)W 1,p′(Ω). Note the implicit homogeneous Neumann boundary condition ∂u/∂ν=0∂u/∂ν=0; consider solutions defined up to an additive constant.
My question: If a solution exists, is it unique (up to an additive constant)? I know that the answer is yes when p=2 p=2 due to coercivity but what about when, say, p∈(1,2)p∈(1,2)? This boils down to whether
∫Ω∇w⋅∇v=0∀v∈W 1,p′(Ω)∫Ω∇w⋅∇v=0∀v∈W 1,p′(Ω)
implies w w is constant, but I'm having difficulty showing that.
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asked Oct 2, 2023 at 11:18
bananab0ybananab0y
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Idea: Is such a solution (to the homogeneous problem) automatically more regular (i.e. W 1,2 W 1,2)? If a portion of the boundary is flat, then w w is automatically smooth up to (that portion of) the boundary. However, I run into issues when dealing with non-flat boundary. Have you had any progress on this since you posted?Jose27 –Jose27 2023-10-08 05:17:27 +00:00 Commented Oct 8, 2023 at 5:17
I was able to show uniqueness for convex polygonal domains, which is enough for my purposes for now. It turned out to be pretty simple by considering a dual problem where the source term is w w above (w w is also adjusted to have zero mean value). The solution of the dual problem belongs to H 2 H 2 and by Sobolev imbeddings qualifies as a test function. Then it's just integration by parts to show that w=0 w=0 in L 2 L 2 norm.bananab0y –bananab0y 2023-10-09 02:19:13 +00:00 Commented Oct 9, 2023 at 2:19
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Berg (Stryer), Biochemistry is your key to success in biochemistry now and in the future with its breakthrough interactive resources, new discoveries, and discussions of career pathways in biochemistry.
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Contents
Table of Contents
1 Biochemistry in Space and Time
2 Protein Composition and Structure
3 Binding and Molecular Recognition
4 Protein Methods
5 Enzymes: Core Concepts and Kinetics
6 Enzyme Catalytic Strategies
7 Enzyme Regulatory Strategies
8 DNA, RNA, and the Flow of Genetic Information
9 Nucleic Acid Methods
10 Exploring Evolution and Bioinformatics
11 Carbohydrates and Glycoproteins
12 Lipids and Biological Membranes
13 Membrane Channels and Pumps
14 Signal-Transduction Pathways
15 Metabolism: Basic Concepts and Themes
16 Glycolysis and Gluconeogenesis
17 Pyruvate Dehydrogenase and the Citric Acid Cycle
18 Oxidative Phosphorylation
19 Phototrophy and the Light Reactions of Photosynthesis
20 The Calvin–Benson Cycle and the Pentose Phosphate Pathway
21 Glycogen Metabolism
22 Fatty Acid and Triacylglycerol Metabolism
23 Protein Turnover and Amino Acid Catabolism
24 Integration of Energy Metabolism
25 Biosynthesis of Amino Acids
26 Nucleotide Biosynthesis
27 Biosynthesis of Membrane Lipids and Steroids
28 DNA Replication, Repair, and Recombination
29 RNA Functions, Biosynthesis, and Processing
30 Protein Biosynthesis
31 Control of Gene Expression
32 Principles of Drug Discovery and Development
Authors
Jeremy M. Berg
Jeremy M. Berg received his B.S. and M.S. degrees in Chemistry from Stanford University (where he did research with Keith Hodgson and Lubert Stryer) and his PhD in Chemistry from Harvard with Richard Holm. He then completed a postdoctoral fellowship with Carl Pabo in Biophysics at Johns Hopkins University School of Medicine. He was an Assistant Professor in the Department of Chemistry at Johns Hopkins from 1986 to 1990. He then moved to Johns Hopkins University School of Medicine as Professor and Director of the Department of Biophysics and Biophysical Chemistry, where he remained until 2003. He then became Director of the National Institute of General Medical Sciences at the National Institutes of Health. In 2011, he moved to the University of Pittsburgh, where he is now Professor of Computational and Systems Biology and Pittsburgh Foundation Chair and Director of the Institute for Personalized Medicine. He served as President of the American Society for Biochemistry and Molecular Biology from 2011 to 2013 and as Editor-in-Chief for Science magazine and the Science family of journals from 2016 to 2019. Dr. Berg has received numerous awards for his research, teaching, and public service. He is an elected member of the National Academy of Medicine and the American Academy of Arts and Sciences. He is coauthor, with Stephen J. Lippard, of the textbook Principles of Bioinorganic Chemistry. He greatly enjoys sharing his life with his wife, three grown children, and grandchildren.
Gregory J. Gatto, Jr.
Gregory J. Gatto, Jr., received his A.B. degree in Chemistry from Princeton University, where he worked with Martin F. Semmelhack and was awarded the Everett S. Wallis Prize in Organic Chemistry. In 2003, he received his MD and PhD degrees from the Johns Hopkins University School of Medicine, where he studied the structural biology of peroxisomal targeting signal recognition with Dr. Berg and received the Michael A. Shanoff Young Investigator Research Award. He completed a postdoctoral fellowship in 2006 with Christopher T. Walsh at Harvard Medical School, where he studied the biosynthesis of the macrolide immunosuppressants. Dr. Gatto is currently a Scientific Director in the Novel Human Genetics Research Unit at GlaxoSmithKline. While he enjoys losing at board games, attempting but not completing crossword puzzles, and watching baseball games at every available opportunity, he treasures most the time he spends with his wife Megan and sons Timothy and Mark.
Justin Hines
Justin K. Hines is Professor of Chemistry at Lafayette College, where he teaches general chemistry and biochemistry courses and conducts education and NIH-funded laboratory research on protein misfolding with undergraduates. He received both his B.S. and PhD in Biochemistry from Iowa State University, where he studied the structure and regulation of the enzymes of central metabolism with Richard B. Honzatko and Herbert J. Fromm. He then completed a postdoctoral fellowship with Elizabeth A. Craig in Biochemistry at the University of Wisconsin–Madison. Professor Hines has won numerous awards for teaching and research, including being named a Cottrell Scholar by the Research Corporation for Science Advancement and a Henry-Dreyfus Teacher-Scholar by the Camille and Henry Dreyfus Foundation. He is also the author of the case-studies series for Macmillan’s three biochemistry textbooks. He enjoys running, hiking, games of any kind, and spending time with his wife and children.
John L. Tymoczko
John L. Tymoczko was Towsley Professor of Biology Emeritus at Carleton College, where he taught from 1976 until his death in 2019. He taught a variety of courses, including Biochemistry, Biochemistry Laboratory, Oncogenes and the Molecular Biology of Cancer, and Exercise Biochemistry, and cotaught an introductory course, Energy Flow in Biological Systems. Professor Tymoczko received his B.A. from the University of Chicago in 1970 and his PhD in Biochemistry from the University of Chicago with Shutsung Liao at the Ben May Institute for Cancer Research. He then had a postdoctoral position with Hewson Swift of the Department of Biology at the University of Chicago. The focus of his research was on steroid receptors, ribonucleoprotein particles, and proteolytic processing enzymes.
Lubert Stryer
Lubert Stryer is Winzer Professor of Cell Biology, Emeritus, in the School of Medicine and Professor of Neurobiology, Emeritus, at Stanford University, where he has been on the faculty since 1976. He received his MD from Harvard Medical School. Professor Stryer has received many awards for his research on the interplay of light and life, including the Eli Lilly Award for Fundamental Research in Biological Chemistry, the Distinguished Inventors Award of the Intellectual Property Owners’ Association, and election to the National Academy of Sciences and the American Philosophical Society. He was awarded the National Medal of Science in 2006. The publication of his first edition of Biochemistry in 1975 transformed the teaching of biochemistry.
Berg (Stryer), Biochemistry is your key to success in biochemistry now and in the future with its breakthrough interactive resources, new discoveries, and discussions of career pathways in biochemistry.
E-book
Read online (or offline) with all the highlighting and notetaking tools you need to be successful in this course.
Learn More
Achieve Essentials
Complete assignments, engage with course materials, prepare for exams and more in order to succeed in class.
Learn More
Achieve
Achieve is a single, easy-to-use platform proven to engage students for better course outcomes
Learn More
Table of Contents
1 Biochemistry in Space and Time
2 Protein Composition and Structure
3 Binding and Molecular Recognition
4 Protein Methods
5 Enzymes: Core Concepts and Kinetics
6 Enzyme Catalytic Strategies
7 Enzyme Regulatory Strategies
8 DNA, RNA, and the Flow of Genetic Information
9 Nucleic Acid Methods
10 Exploring Evolution and Bioinformatics
11 Carbohydrates and Glycoproteins
12 Lipids and Biological Membranes
13 Membrane Channels and Pumps
14 Signal-Transduction Pathways
15 Metabolism: Basic Concepts and Themes
16 Glycolysis and Gluconeogenesis
17 Pyruvate Dehydrogenase and the Citric Acid Cycle
18 Oxidative Phosphorylation
19 Phototrophy and the Light Reactions of Photosynthesis
20 The Calvin–Benson Cycle and the Pentose Phosphate Pathway
21 Glycogen Metabolism
22 Fatty Acid and Triacylglycerol Metabolism
23 Protein Turnover and Amino Acid Catabolism
24 Integration of Energy Metabolism
25 Biosynthesis of Amino Acids
26 Nucleotide Biosynthesis
27 Biosynthesis of Membrane Lipids and Steroids
28 DNA Replication, Repair, and Recombination
29 RNA Functions, Biosynthesis, and Processing
30 Protein Biosynthesis
31 Control of Gene Expression
32 Principles of Drug Discovery and Development
Jeremy M. Berg
Jeremy M. Berg received his B.S. and M.S. degrees in Chemistry from Stanford University (where he did research with Keith Hodgson and Lubert Stryer) and his PhD in Chemistry from Harvard with Richard Holm. He then completed a postdoctoral fellowship with Carl Pabo in Biophysics at Johns Hopkins University School of Medicine. He was an Assistant Professor in the Department of Chemistry at Johns Hopkins from 1986 to 1990. He then moved to Johns Hopkins University School of Medicine as Professor and Director of the Department of Biophysics and Biophysical Chemistry, where he remained until 2003. He then became Director of the National Institute of General Medical Sciences at the National Institutes of Health. In 2011, he moved to the University of Pittsburgh, where he is now Professor of Computational and Systems Biology and Pittsburgh Foundation Chair and Director of the Institute for Personalized Medicine. He served as President of the American Society for Biochemistry and Molecular Biology from 2011 to 2013 and as Editor-in-Chief for Science magazine and the Science family of journals from 2016 to 2019. Dr. Berg has received numerous awards for his research, teaching, and public service. He is an elected member of the National Academy of Medicine and the American Academy of Arts and Sciences. He is coauthor, with Stephen J. Lippard, of the textbook Principles of Bioinorganic Chemistry. He greatly enjoys sharing his life with his wife, three grown children, and grandchildren.
Gregory J. Gatto, Jr.
Gregory J. Gatto, Jr., received his A.B. degree in Chemistry from Princeton University, where he worked with Martin F. Semmelhack and was awarded the Everett S. Wallis Prize in Organic Chemistry. In 2003, he received his MD and PhD degrees from the Johns Hopkins University School of Medicine, where he studied the structural biology of peroxisomal targeting signal recognition with Dr. Berg and received the Michael A. Shanoff Young Investigator Research Award. He completed a postdoctoral fellowship in 2006 with Christopher T. Walsh at Harvard Medical School, where he studied the biosynthesis of the macrolide immunosuppressants. Dr. Gatto is currently a Scientific Director in the Novel Human Genetics Research Unit at GlaxoSmithKline. While he enjoys losing at board games, attempting but not completing crossword puzzles, and watching baseball games at every available opportunity, he treasures most the time he spends with his wife Megan and sons Timothy and Mark.
Justin Hines
Justin K. Hines is Professor of Chemistry at Lafayette College, where he teaches general chemistry and biochemistry courses and conducts education and NIH-funded laboratory research on protein misfolding with undergraduates. He received both his B.S. and PhD in Biochemistry from Iowa State University, where he studied the structure and regulation of the enzymes of central metabolism with Richard B. Honzatko and Herbert J. Fromm. He then completed a postdoctoral fellowship with Elizabeth A. Craig in Biochemistry at the University of Wisconsin–Madison. Professor Hines has won numerous awards for teaching and research, including being named a Cottrell Scholar by the Research Corporation for Science Advancement and a Henry-Dreyfus Teacher-Scholar by the Camille and Henry Dreyfus Foundation. He is also the author of the case-studies series for Macmillan’s three biochemistry textbooks. He enjoys running, hiking, games of any kind, and spending time with his wife and children.
John L. Tymoczko
John L. Tymoczko was Towsley Professor of Biology Emeritus at Carleton College, where he taught from 1976 until his death in 2019. He taught a variety of courses, including Biochemistry, Biochemistry Laboratory, Oncogenes and the Molecular Biology of Cancer, and Exercise Biochemistry, and cotaught an introductory course, Energy Flow in Biological Systems. Professor Tymoczko received his B.A. from the University of Chicago in 1970 and his PhD in Biochemistry from the University of Chicago with Shutsung Liao at the Ben May Institute for Cancer Research. He then had a postdoctoral position with Hewson Swift of the Department of Biology at the University of Chicago. The focus of his research was on steroid receptors, ribonucleoprotein particles, and proteolytic processing enzymes.
Lubert Stryer
Lubert Stryer is Winzer Professor of Cell Biology, Emeritus, in the School of Medicine and Professor of Neurobiology, Emeritus, at Stanford University, where he has been on the faculty since 1976. He received his MD from Harvard Medical School. Professor Stryer has received many awards for his research on the interplay of light and life, including the Eli Lilly Award for Fundamental Research in Biological Chemistry, the Distinguished Inventors Award of the Intellectual Property Owners’ Association, and election to the National Academy of Sciences and the American Philosophical Society. He was awarded the National Medal of Science in 2006. The publication of his first edition of Biochemistry in 1975 transformed the teaching of biochemistry.
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10510 | https://www.mometrix.com/academy/truth-tables/ | Published Time: 2021-08-18T16:45:00+00:00
Understanding Truth Tables for Statements (Video)
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Truth Tables
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Statements and Truth Values
Negation
Conjunction
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Statements and Truth Values
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Conjunction
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Conditional Statements
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Applying Truth Tables
[x] Transcript
Statements and Truth Values
Declarative word or number sentences are called statements if they are either true or false but cannot be both. Sometimes their truth values are easy to determine:
There are 12 months in a year is a true statement.
3+1=5 3+1=5 is a false statement.
n n is an even number is not a statement (it is an open sentence) because the truth value depends on the value of n n.
Other times it might not be so obvious and a truth table may be needed. Truth tables show the truth values of one or more compound statements for each possible combination of truth values of the statements within them. In this video, we’ll see how they work with common logical operators, and then we’ll get into more complex conditional statements.
Negation
First up, negation, denoted by this symbol: ¬¬. Every time a statement P P is negated, its truth value is reversed. The truth table shows the truth values resulting from different numbers of negations:
| P P | ¬P¬P | ¬¬P¬¬P | ¬¬¬P¬¬¬P | … |
--- ---
| T | F | T | F | … |
| F | T | F | T | … |
If the statement P P is true, then the negation of P P is false, the negation of the negation of P P is true, the negation of the negation of the negation of P P is false, and so on and so forth. If the statement P P is false, then you will have the opposite truth values.
Suppose P P is the statement, “Snoopy is a dog.” Using the truth table, since we can see this statement is true, we can also see that the statement, “Snoopy is not not a dog,” is also true.
| P P | ¬P¬P | ¬¬P¬¬P | ¬¬¬P¬¬¬P | … |
--- ---
| T | F | T | F | … |
| F | T | F | T | … |
If P P is the statement, “0.5 is an integer,” which is false, the table shows that, “0.5 is not an integer,” is true.
| P P | ¬P¬P | ¬¬P¬¬P | ¬¬¬P¬¬¬P | … |
--- ---
| T | F | T | F | … |
| F | T | F | T | … |
Conjunction
Now for conjunction, joining two statements P P and Q Q with the word “and”. The symbol for conjunction is an upside down V: ∧. A conjunction is true only when both statements are true, according to the table:
| P | Q | P∧Q |
---
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
If the statement P P is true AND the statement Q Q is true, then P∧Q P∧Q is true. If P P is true and Q Q is false, then P∧Q P∧Q is false. If P P is false and Q Q is true, then P∧Q P∧Q is false. And if P P is false and Q Q is false, then P∧Q P∧Q is false.
The statement, “Snoopy is a dog and 0.5 is an integer,” is false according to the table because P P – Snoopy is a dog – is true, but Q Q – 0.5 is an integer – is false.
| P | Q | P∧Q |
---
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
Disjunction
Disjunction, two statements joined by “or” and represented by this symbol: ∨∨, has kind of an opposite effect. It is only false when both statements are false.
| P | Q | P∨Q |
---
| T | T | T |
| T | F | T |
| F | T | T |
| F | F | F |
If P P is true and Q Q is true, then P∨Q P∨Q is true. If P P is true and Q Q is false, then P∨Q P∨Q is true. If P P is false and Q Q is true, then P∨Q P∨Q is true. If P P is false and Q Q is false, then P∨Q P∨Q is also false.
The statement, “Snoopy is a dog or 0.5 is an integer,” is true according to the table because P P – Snoopy is a dog – is true, and Q –0.5 is an integer – is false.
Conditional Statements
Conditional statements, denoted by an arrow can also be evaluated using truth tables. Consider the example, “Natural numbers are integers.” Of course, this can be written in “if, then” form: “If a number is a natural number, then it is an integer.”
Upon hearing this statement, you grab your calculator, ready to generate some random numbers. For each number generated, one of four things will happen:
Your number will be a natural number that is also an integer.
Your number will be a natural number that is not an integer.
Your number will not be a natural number but will be an integer.
Your number will not be a natural number and will not be an integer.
Scenario 1 could definitely be true. Scenario 2 could not. Scenarios 3 and 4, however, could go either way, but from the statement only, we don’t actually know anything about numbers that are not natural numbers. All we know is that if a number is a natural number, then it is an integer. So scenarios 3 and 4 don’t change the truth value of the conditional. Here’s what the truth table for a conditional looks like:
| Scenario | P | Q | P→Q |
--- --- |
| 1 | T | T | T |
| 2 | T | F | F |
| 3 | F | T | T |
| 4 | F | F | T |
Suppose I make this statement: “If you make an A on your test, then I’ll give you 10 dollars.” Is the statement, “If you make a B on your test, then I’ll give you 10 dollars,” true or false?
[pause]
| P | Q | P→Q |
---
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
The statement is true because we don’t know the outcome of bringing home a B. All we know is that it is true that an A will get us 10 bucks.
Think of it this way, if you got an A, then you should get $10. So when P P is true (you make an A on your test) and Q Q is true (I give you $10), then P→Q P→Q is also true because the first thing caused the second thing to happen. I told you the truth. But what if you got an A on your test (P P is true) and I didn’t give you $10 (Q Q is false)? Well, you’d probably be upset with me because I lied to you and told you a false statement (P→Q P→Q is false).
These next two rows are where it gets a little tricky. What if you make a B on your test (P P is false) and I give you $10 (Q Q is true). Is P→Q P→Q true? It may not seem like it, but it is true. I didn’t lie to you by giving you $10 even though you didn’t make an A because I didn’t tell you what I would do if you made a B. The same logic follows for the next statement. If you make a B on your test (P P is false) and I don’t give you $10 (Q Q is false), then P→Q P→Q is still true. Again, I didn’t lie to you by not giving you $10 when you made a B. All I said was IF you made an A, then I would give you $10. There were no promises made or broken for you making a B.
Biconditional Statements
The potential truth values of biconditionals can also be charted using truth tables. Biconditionals, usually containing “if and only if” and represented with a double arrow ↔, are true when both statements P P and Q Q are either true or false. Definitions tend to work well as biconditionals. Here’s one: “A right angle measures 90 degrees.” This can also be written, “An angle is a right angle if and only if it measures 90 degrees.” Biconditionals go both ways, so the reverse also works: “An angle measures 90 degrees if and only if it is a right angle.”
Given an angle, there are four possible scenarios:
An angle is a right angle and measures 90 degrees.
An angle is a right angle and does not measure 90 degrees.
An angle is not a right angle and measures 90 degrees.
The angle is not a right angle and does not measure 90 degrees.
Clearly, scenarios 1 and 4 are true and 2 and 3 are false. Therefore, this is the truth table:
| Scenario | P | Q | P↔Q |
--- --- |
| 1 | T | T | T |
| 2 | T | F | F |
| 3 | F | T | F |
| 4 | F | F | T |
Complex Truth Tables
Consider the statement “Two lines are parallel if they are coplanar and do not intersect.” Notice that the phrase “if and only if” has been simplified to if with two f’s. This is a common simplification made in mathematical phrases. Here, we’ll consider three statements:
Two lines are parallel lines.
Two lines are coplanar.
Two lines do not intersect.
This statement is of the form P↔(Q∧R)P↔(Q∧R). Let’s make a truth table. Since there are three statements in this example and each one has its own true or false value, we know that our truth table will have 2 3 2 3 rows. In general, any truth table has 2 n 2 n rows.
So the first three columns for P P, Q Q, and R R show the different possibilities for if P P, Q Q, and R R are true or false. Generally they’re set up the same way for any truth table that you have. Now we’re gonna look at the second part of our statement: Q∧R Q∧R.
For our first row, Q∧R Q∧R, we know that that’s true because Q Q and R R are both true. Then we look at P P. So our statement P P is true, and the statement Q∧R Q∧R is true, so P↔(Q∧R)P↔(Q∧R) is also true.
Then for our second row we can do the same thing. Look at Q∧R Q∧R. R R is false, so Q∧R Q∧R is also false. Then we compare this to our statement P P. P P is true and Q∧R Q∧R is false, so P↔(Q∧R)P↔(Q∧R) is also false.
For our third row, we’re gonna look at Q∧R Q∧R. Q Q is false, so the statement Q∧R Q∧R is false. Compared to our P P statement which is true, that makes our biconditional statement, P↔(Q∧R)P↔(Q∧R) false because the Q∧R Q∧R part is false.
For our fourth row, we’re gonna do the same thing, we look at Q∧R Q∧R, both Q Q and R R are false, so Q∧R Q∧R is also false. Compared to P P, which is true, that makes our biconditional statement, P↔(Q∧R)P↔(Q∧R) false because Q∧R Q∧R is false, while P P is true.
For our fifth row, we look at Q∧R Q∧R. Both Q Q and R R are true, so Q∧R Q∧R is true. Compared to our statement P P, which is false, P↔(Q∧R)P↔(Q∧R) is also false.
For our sixth row, Q∧R Q∧R is false because R R is false. Compared to P P, P P is also false, which makes our biconditional statement, P↔(Q∧R)P↔(Q∧R) true because they have the same truth value. P P has the same truth value as Q∧R Q∧R.
Looking at our seventh row, we’ll see the same thing. Q∧R Q∧R is false because Q Q is false. But since P P is also false, our biconditional statement becomes true. For our last row, we see that Q∧R Q∧R is false because both Q Q and R R have false truth values. And we compare this to our P P, which is also false, so P↔(Q∧R)P↔(Q∧R) is true because P P is false and Q∧R Q∧R is false, at the same time. Therefore, the whole thing is true.
| P | Q | R | (Q∧R) | P ↔ (Q∧R) | Reasoning |
--- --- --- |
| T | T | T | T | T | “And” statement true, matches true p |
| T | T | F | F | F | False R makes the “and” false but the P is true |
| T | F | T | F | F | False Q makes the “and” false but P is true |
| T | F | F | F | F | False Q and R make the “and” false but the P true |
| F | T | T | T | F | P false, “and” true |
| F | T | F | F | T | P and “and” both false |
| F | F | T | F | T | P and “and” both false |
| F | F | F | F | T | P and “and” both false |
Applying Truth Tables
Using the truth table, we can pinpoint whether the statement, “Two lines are skew if they are coplanar and do not intersect” is true or false. Skew lines, of course, do not intersect, but are not coplanar. Clearly a false statement, shown by the truth table.
| P | Q | R | (Q∧R) | P ↔ (Q∧R) |
--- ---
| T | T | T | T | T |
| T | T | F | F | F |
| T | F | T | F | F |
| T | F | F | F | F |
| F | T | T | T | F |
| F | T | F | F | T |
| F | F | T | F | T |
| F | F | F | F | T |
Since we are considering skew lines, we will assume P P is true. Skew lines are not coplanar, so Q Q is false. And skew lines do not intersect, so R R is true. This gives us row 3, which shows that (Q∧R)(Q∧R) is false because Q Q is false, which means that P↔(Q∧R)P↔(Q∧R) is also false because (Q∧R)(Q∧R) is false, even though P P is true.
I hope that this video helped you understand how truth tables work and how they can be used to determine truth values!
Thanks for watching, and happy studying!
Return to Discrete Math Videos
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by Mometrix Test Preparation | Last Updated: August 8, 2025
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10511 | https://www.onlinemathlearning.com/bedmas.html | Order Of Operations Worksheets - BEDMAS or PEMDAS
What is the BEDMAS rule? BEDMAS (or PEMDAS) is an acronym used to remember the order of operations in mathematical expressions. It tells you which calculations to perform first when you have a combination of different operations like addition, subtraction, multiplication, division, exponents, and parentheses.
Practice order of operations using BEDMAS (Brackets, Exponents, Division, Multiplication, Addition, Subtraction).
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Objective: I know how to perform mixed operations with brackets or parenthesis, exponents, multiplication, division, addition, and subtraction. If the calculations involve a combination of brackets or parenthesis, exponents, multiplication, division, addition, and subtraction then you need to follow the BEDMAS rules:
Step 1: First, perform the operations within the brackets or parenthesis
Step 2: Second, evaluate the exponents.
Step 3: Third, perform multiplication and division from left to right.
Step 4: Fourth, perform addition and subtraction from left to right. The following diagram shows the BEDMAS rules for order of operations.
What BEDMAS Stands For:
Brackets/Parentheses (B/P): Perform any calculations inside brackets or parentheses first. If you have nested brackets (brackets within brackets), work from the innermost set outwards.
Exponents/Orders (E): Evaluate any exponents, powers, or roots (like square roots).
Division and Multiplication (D/M): Perform division and multiplication from left to right. This is important: if both division and multiplication are present, you do whichever comes first as you read from left to right.
Addition and Subtraction (A/S): Perform addition and subtraction from left to right. Again, if both are present, you work from left to right.
Example: Calculate 23 – 3 × (8 – 6) Solution: 23 – 3 × (8 – 6) (perform within brackets or parenthesis) = 23 – 3 × 2 (evaluate the exponent) = 8 – 3 × 2 (perform multiplication) = 8 – 6 (perform subtraction) = 2
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10512 | https://www.spiraxsarco.com/learn-about-steam/steam-engineering-principles-and-heat-transfer/engineering-units?sc_lang=en-GB | Looking for Spirax Sarco products and services?
Contents
Engineering Units
What is Steam?
Superheated Steam
Steam Quality
Heat Transfer
Methods of Estimating Steam Consumption
Measurement of Steam Consumption
Thermal Rating
Energy Consumption of Tanks and Vats
Heating with Coils and Jackets
Heating Vats and Tanks by Steam Injections
Steam Consumption of Pipes and Air Heaters
Steam Consumption of Heat Exchangers
Steam Consumption of Plant Items
Entropy - a basic understanding
Entropy - its practical use
Back to Learn about steam
Engineering Units
An overview of the units of measurement used in the Steam and Condensate Loop including temperature, pressure, density, volume, heat, work and energy.
Throughout the engineering industries, many different definitions and units have been proposed and used for mechanical and thermal properties.
The problems this caused led to the development of an agreed international system of units (or SI units: Système International d’Unités). In the SI system there are seven well-defined base units from which the units of other properties can be derived, and these will be used throughout this publication.
The SI base units include length (in metres), mass (in kilograms), time (in seconds) and temperature (in kelvin). The first three will hopefully need no further explanation, while the latter will be discussed in more detail later.
The other SI base units are electric current (in amperes), amount of substance (in moles) and luminous intensity (in candela). These may be familiar to readers with a background in electronics, chemistry and physics respectively, but have little relevance to steam engineering nor the contents of The Steam and Condensate Loop.
Table 2.1.1 shows the derived units that are relevant to this subject, all of which should be familiar to those with any general engineering background. These quantities have all been assigned special names after famous pioneers in the development of science and engineering.
Table 2.1.1 Named quantities in derived SI units.
| Quantity | Name | Symbol | SI-base unit | Derived unit |
| Area | square metre | A | m2
| Volume | cubic metre | V | m3
| Velocity | metre per second | u | m/s
| Acceleration | metre per second squared | a | m/s2
| Force | newton | N | kg m/s2 | J/m |
| Energy | joule | J | kg m2/s2 | N m |
| Pressure or stress | pascal | Pa | kg/m s2 | N/m2 |
| Power | watt | W | kg m2/s3 | J/s |
There are many other quantities that have been derived from SI base units, which will also be of significance to anyone involved in steam engineering. These are provided in Table 2.1.2.
Table 2.1.2 Other quantities in derived SI units
| Quantity | SI base unit | Derived unit |
| Mass density | kg/m3 | kg/m3 |
| Specific volume (Vg) | m3/kg | m3/kg |
| Specific enthalpy (h) | m2/s2 | J/kg |
| Specific heat capacity (cp) | m2/s2 K | J/kg K |
| Specific entropy | m2/s2 K | J/kg K |
| Heat flowrate | m2 kg/s3 | J/s or W |
| Dynamic viscosity | kg/m s | N s/m² |
Dot notation
This convention is used to identify a compound unit incorporating rate, for example:
m = Mass (e.g. kg)
ṁ = Mass flow per time unit (e.g. kg/h) = Mass flowrate
Multiples and submultiples
Table 2.1.3 gives the SI prefixes that are used to form decimal multiples and submultiples of SI units. They allow very large or very small numerical values to be avoided. A prefix attaches directly to the name of a unit, and a prefix symbol attaches directly to the symbol for a unit.
In summary: one thousand metres may be shown as 1 km, 1 000 m or 10³ m.
Table 2.1.3 Multiples and submultiples used with SI units
| Multiples | | | Submultiples | | |
| Factor | Prefix | Symbol | Factor | Prefix | Symbol |
| 1012 | tera | T | 10-3 | milli | m |
| 109 | giga | G | 10-6 | micro | μ |
| 106 | mega | M | 10-9 | nano | n |
| 103 | kilo | k | 10-12 | pico | P |
Special abbreviations used in steam flowmetering applications
For historical reasons, International Standard ISO 5167 (supersedes BS 1042) which refers to flowmetering, use the following abbreviations in Table 2.1.4.
Table 2.1.4 Symbols used in flowmetering applications
| Symbol | Definition | Unit |
| qM | Mass flowrate | kg/s or kg/h |
| qV | Volume flowrate | m3/s |
| QI | Liquid flowrate | I/min |
| QS | Gas flowrate at STP | I/min |
| QF | Gas flowrate actual | I/min |
| QE | Equivalent water flowrate | I/min |
| DS | Density of gas at STP | kg/m3 |
| DF | Density of gas actual | kg/m3 |
| PS | Standard pressure (1.013 bar a) | bar a |
| PF | Actual flow pressure | bar a |
| TS | Standard temperature | °C |
| TF | Actual flow temperature | °C |
STP - Standard temperature and pressure
These are the standard conditions for measurement of the properties of matter. The standard temperature is the freezing point of pure water, 0 °C or 273.16 °K. The standard pressure is the pressure exerted by a column of mercury (symbol Hg) 760 mm high, often designated 760 mm Hg. This pressure is also called one atmosphere and is equal to 1.01325 x 106 dynes per square centimetre, or approximately 14.7 lb per square inch. The density (mass per volume) of a gas is usually reported as its value at STP. Properties that cannot be measured at STP are measured under other conditions; usually the values obtained are then mathematically extrapolated to their values at STP
Symbols
Table 2.1.5 shows the symbols and typical units used in The Steam and Condensate Loop.
Table 2.1.5 Symbols and units of measure used in The Steam and Condensate Loop
| Symbol | Definition | Unit |
| A | Cross sectional area of a conduit,for the operating condition | m² or mm² |
| cP | Specific heat capacity at constant pressure | kJ/kg °C or kJ/kg K |
| cV | Specific heat capacity at constant volume | kJ/m³ °C or kJ/m³ K |
| D | Diameter of the circular cross section of a conduit | m or mm |
| d | Orifice diameter | m or mm |
| g | Acceleration due to gravity | 9.81 m/s² |
| Hz | Hertz, the unit of frequency (number of cycles per second) | Hz or kHz |
| J | Joule, the unit of energy | J or kJ |
| L | Length | m |
| M | Molar mass of a fluid | kg/mol |
| N | Newton, the unit of force | N or kN |
| Pa | Pascal, the unit of pressure | Pa or kPa |
| p | Static pressure of a fluid | bar or kPa |
| ∆p | Differential pressure | bar or kPa |
| m | Fundamental unit of length (metre) | m |
| m | Mass | kg |
| ṁ | Mass flowrate | kg/s or kg/h |
| ṁS | Steam mass flowrate | kg/s or kg/h |
| Q | Quantity of heat | kJ |
| Q̇ | Heat transfer rate | kJ/s (kW) |
| R | Radius | m or mm |
| ReD | Reynolds number referred to diameter D | Dimensionless |
| s | Fundamental unit of time (second) | s |
| Sr | Strouhal number | Dimensionless |
| σ | Stress | N/m² |
| TS | Steam temperature | K or °C |
| TL | Liquid (or product) temperature | K or °C |
| ∆T | Temperature difference or change | K or °C |
| t | Time | s or h |
| u | Velocity of a fluid | m/s |
| μ | Dynamic viscosity of a fluid | Pa s or cP |
| ν | Kinematic viscosity | cSt |
| ρ | Density of a fluid | kg/m³ |
| V̇ | Volume flowrate | m³/s or m³/h |
| W | Unit of energy flow (Watt) | W (J/s) |
| V (vg) | Volume (Specific volume) | m³ (m³/kg) |
| H (hg) | Enthalpy (Specific enthalpy) | kJ (kJ/kg) |
| S (sg) | Entropy (Specific entropy) | kJ/K (kJ/kg K) |
| U (ug) | Internal energy (specific internal energy) | kJ (kJ/kg) |
Subscripts used with properties
When using enthalpy, entropy and internal energy, subscripts as shown below are used to identify the phase, for example:
Subscript f = Fluid or liquid state, for example hf: liquid enthalpy
Subscript fg = Change of state liquid to gas, for example hfg: enthalpy of evaporation
Subscript g = Total, for example hg: total enthalpy
Note that, by convention, the total heat in superheated steam is signified by h.
It is also usual, by convention, to signify sample quantities in capital letters, whilst unit quantities are signified in lower case letters.
For example:
Total enthalpy in a sample of superheated steam H kJ
Specific enthalpy of superheated steam h kJ/kg
Temperature
The temperature scale is used as an indicator of thermal equilibrium, in the sense that any two systems in contact with each other with the same value are in thermal equilibrium.
The Celsius (°C) scale
This is the scale most commonly used by the engineer, as it has a convenient (but arbitrary) zero temperature, corresponding to the temperature at which water will freeze.
The absolute or K (kelvin) scale
This scale has the same increments as the Celsius scale, but has a zero corresponding to the minimum possible temperature when all molecular and atomic motion has ceased. This temperature is often referred to as absolute zero (0 K) and is equivalent to -273.16 °C.
The two scales of temperature are interchangeable, as shown in Figure 2.1.1 and expressed in Equation 2.1.1.
The SI unit of temperature is the kelvin, which is defined as 1 ÷ 273.16 of the thermodynamic temperature of pure water at its triple point (0.01 °C). An explanation of triple point is given in Module 2.2.
Most thermodynamic equations require the temperature to be expressed in kelvin. However, temperature difference, as used in many heat transfer calculations, may be expressed in either °C or K. Since both scales have the same increments, a temperature difference of 1 °C has the same value as a temperature difference of 1 K.
Pressure
The SI unit of pressure is the pascal (Pa), defined as 1 newton of force per square metre (1 N/m²).
As Pa is such a small unit the kPa (1 kilonewton/m²) or MPa (1 Meganewton/m²) tend to be more appropriate to steam engineering.
However, probably the most commonly used metric unit for pressure measurement in steam engineering is the bar. This is equal to 105 N/m², and approximates to 1 atmosphere. This unit is used throughout this publication.
Other units often used include lb/in² (psi), kg/cm², atm, in H2O and mm Hg. Conversion factors are readily available from many sources.
Absolute pressure (bar a)
This is the pressure measured from the datum of a perfect vacuum i.e. a perfect vacuum has a pressure of 0 bar a.
Gauge pressure (bar g)
This is the pressure measured from the datum of the atmospheric pressure. Although in reality the atmospheric pressure will depend upon the climate and the height above sea level, a generally accepted value of 1.013 25 bar a (1 atm) is often used. This is the average pressure exerted by the air of the earth’s atmosphere at sea level.
Gauge pressure = Absolute pressure - Atmospheric pressure
Pressures above atmospheric will always yield a positive gauge pressure. Conversely a vacuum or negative pressure is the pressure below that of the atmosphere. A pressure of -1 bar g corresponds closely to a perfect vacuum.
Differential pressure
This is simply the difference between two pressures. When specifying a differential pressure, it is not necessary to use the suffixes ‘g’ or ‘a’ to denote either gauge pressure or absolute pressure respectively, as the pressure datum point becomes irrelevant.
Therefore, the difference between two pressures will have the same value whether these pressures are measured in gauge pressure or absolute pressure, as long as the two pressures are measured from the same datum.
Density and specific volume
The density (ρ) of a substance can be defined as its mass (m) per unit volume (V). The specific volume (vg) is the volume per unit mass and is therefore the inverse of density. In fact, the term ‘specific’ is generally used to denote a property of a unit mass of a substance (see Equation 2.1.2).
The SI units of density (ρ ) are kg/m³, conversely, the units of specific volume (vg) are m³/kg.
Another term used as a measure of density is specific gravity. It is a ratio of the density of a substance (ρs) and the density of pure water (ρw) at standard temperature and pressure (STP).
This reference condition is usually defined as being at atmospheric pressure and 0°C. Sometimes it is said to be at 20°C or 25°C and is referred to as normal temperature and pressure (NTP).
The density of water at these conditions is approximately 1 000 kg/m³. Therefore substances with a density greater than this value will have a specific gravity greater than 1, whereas substances with a density less than this will have a specific gravity of less than 1.
Since specific gravity is a ratio of two densities, it is a dimensionless variable and has no units. Therefore in this case the term specific does not indicate it is a property of a unit mass of a substance. Specific gravity is also sometimes known as the relative density of a substance.
Heat, work and energy
Energy is sometimes described as the ability to do work. The transfer of energy by means of mechanical motion is called work. The SI unit for work and energy is the joule, defined as 1 N m.
The amount of mechanical work carried out can be determined by an equation derived from Newtonian mechanics:
Work = Force x Displacement
It can also be described as the product of the applied pressure and the displaced volume:
Work = Applied pressure x Displaced volume
Example 2.1.1
An applied pressure of 1 Pa (or 1 N/m²) displaces a volume of 1 m³. How much work has been done?
Work done = 1 N/m² x 1 m³ = 1 N m (or 1 J)
The benefits of using SI units, as in the above example, is that the units in the equation actually cancel out to give the units of the product.
The experimental observations of J. P. Joule established that there is an equivalence between mechanical energy (or work) and heat. He found that the same amount of energy was required to produce the same temperature rise in a specific mass of water, regardless of whether the energy was supplied as heat or work.
The total energy of a system is composed of the internal, potential and kinetic energy. The temperature of a substance is directly related to its internal energy (ug). The internal energy is associated with the motion, interaction and bonding of the molecules within a substance. The external energy of a substance is associated with its velocity and location, and is the sum of its potential and kinetic energy.
The transfer of energy as a result of the difference in temperature alone is referred to as heat flow. The watt, which is the SI unit of power, can be defined as 1 J/s of heat flow.
Other units used to quantify heat energy are the British Thermal Unit (Btu: the amount of heat to raise 1 lb of water by 1 °F) and the kilocalorie (the amount of heat to raise 1 kg of water by 1 °C).
Conversion factors are readily available from numerous sources.
Specific enthalpy
This is the term given to the total energy, due to both pressure and temperature, of a fluid (such as water or steam) at any given time and condition. More specifically it is the sum of the internal energy and the work done by an applied pressure (as in Example 2.1.1).
The basic unit of measurement is the joule (J). Since one joule represents a very small amount of energy, it is usual to use kilojoules (kJ = 1 000 joules).
The specific enthalpy is a measure of the total energy of a unit mass, and its units are usually kJ/kg.
Specific heat capacity
The enthalpy of a fluid is a function of its temperature and pressure. The temperature dependence of the enthalpy can be found by measuring the rise in temperature caused by the flow of heat at constant pressure. The constant-pressure heat capacity cP, is a measure of the change in enthalpy at a particular temperature.
Similarly, the internal energy is a function of temperature and specific volume. The constant volume heat capacity cv, is a measure of the change in internal energy at a particular temperature and constant volume.
Because the specific volumes of solids and liquids are generally smaller, then unless the pressure is extremely high, the work done by an applied pressure can be neglected. Therefore, if the enthalpy can be represented by the internal energy component alone, the constant-volume and constant-pressure heat capacities can be said to be equal.
Therefore, for solids and liquids: cP ≈ cv
Another simplification for solids and liquids assumes that they are incompressible, so that their volume is only a function of temperature. This implies that for incompressible fluids the enthalpy and the heat capacity are also only functions of temperature.
The specific heat capacity represents the amount of energy required to raise 1 kg by 1 °C, and can be thought of as the ability of a substance to absorb heat. Therefore the SI units of specific heat capacity are kJ/kg K (kJ/kg °C). Water has a large specific heat capacity (4.19 kJ/kg °C) compared with many fluids, which is why both water and steam are considered to be good carriers of heat.
The amount of heat energy required to raise the temperature of a substance can be determined from Equation 2.1.4.
This equation shows that for a given mass of substance, the temperature rise is linearly related to the amount of heat provided, assuming that the
specific heat capacity is constant over that temperature range.
Example 2.1.2
Consider a quantity of water with a volume of 2 litres, raised from a temperature of 20 °C to 70 °C.
At atmospheric pressure, the density of water is approximately 1 000 kg/m³. As there are 1 000 litres in 1 m³, then the density can be expressed as 1 kg per litre (1 kg/l). Therefore the mass of the water is 2 kg.
The specific heat capacity for water can be taken as 4.19 kJ/kg °C over low ranges of temperature.
Therefore: Q =2 kg x 4.19 kJ/kg °C x (70 - 20) °C = 419 kJ
If the water was then cooled to its original temperature of 20 °C, it would also release this amount of energy in the cooling application.
Entropy (S)
Entropy is a measure of the degree of disorder within a system. The greater the degree of disorder, the higher the entropy. The SI units of entropy are kJ/kg K (kJ/kg °C).
In a solid, the molecules of a substance arrange themselves in an orderly structure. As the substance changes from a solid to a liquid, or from a liquid to a gas, the arrangement of the molecules becomes more disordered as they begin to move more freely. For any given substance the entropy in the gas phase is greater than that of the liquid phase, and the entropy in the liquid phase is more than in the solid phase.
One characteristic of all natural or spontaneous processes is that they proceed towards a state of equilibrium. This can be seen in the second law of thermodynamics, which states that heat cannot pass from a colder to a warmer body.
A change in the entropy of a system is caused by a change in its heat content, where the change of entropy is equal to the heat change divided by the average absolute temperature, Equation 2.1.5.
When unit mass calculations are made, the symbols for entropy and enthalpy are written in lower case, Equation 2.1.6.
To look at this in further detail, consider the following examples:
Example 2.1.3
A process raises 1 kg of water from 0 to 100°C (273 to 373 K) under atmospheric conditions.
Specific enthalpy at 0°C (hf) = 0 kJ/kg (from steam tables)
Specific enthalpy of water at 100°C (hf) = 419 kJ/kg (from steam tables)
Calculate the change in specific entropy
Since this is a change in specific entropy of water, the symbol ‘s’ in Equation 2.1.6 takes the suffix ‘f’ to become sf.
Example 2.1.4
A process changes 1 kg of water at 100°C (373 K) to saturated steam at 100°C (373 K) under atmospheric conditions.
Calculate the change in specific entropy of evaporation
Since this is the entropy involved in the change of state, the symbol ‘s’ in Equation 2.1.6 takes the suffix ‘fg’ to become sfg.
Specific enthalpy of evaporation
of steam at 100°C (373 K) (hfg) = 2 258 kJ/kg (from steam tables)
Specific enthalpy of evaporation
of water at 100°C (373 K) (hfg) = 0 kJ/kg (from steam tables)
The total change in specific entropy from water at 0 °C to saturated steam at 100 °C is the sum of the change in specific entropy for the water, plus the change of specific entropy for the steam, and takes the suffix ‘g’ to become the total change in specific entropy sg.
Therefore
Example 2.1.5
A process superheats 1 kg of saturated steam at atmospheric pressure to 150°C (423 K). Determine the change in entropy.
As the entropy of saturated water is measured from a datum of 0.01 °C, the entropy of water at 0 °C can, for practical purposes, be taken as zero. The total change in specific entropy in this example is based on an initial water temperature of 0 °C, and therefore the final result happens to be very much the same as the specific entropy of steam that would be observed in steam tables at the final condition of steam at atmospheric pressure and 150 °C.
Entropy is discussed in greater detail in Module 2.15, Entropy - A Basic Understanding, and in Module 2.16, Entropy - Its Practical Use.
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$\begingroup$
Prove that the points $a_1,a_2,a_3$ are vertices of an equilateral triangle if and only if $a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_3a_1$.
I rewrite the equation as $2a_1^2+2a_2^2+2a_3^2-2a_1a_2-2a_2a_3-2a_3a_1=0$, which is $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_1)^2=0$. This looks quite nice, but I'm not sure how to relate it to the equilateral triangle.
complex-numbers
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asked Sep 1, 2013 at 7:27
Paul S.Paul S.
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I'll prove the 'if' part, leaving the 'only if' part to you:
Denote the points as $A(a), B(b), C(c)$ (easier to type).
Now, rotating $\vec{AC}$ by $\pi/3$ gives you $\vec{AB}$. So:
$$ \hat{AC}\cdot e^{i\pi/3} = \hat{AB}$$
Here, $|\vec{AC}| = | \vec{AB} |$. So:
$$ \vec{AC}\cdot e^{i\pi/3} = \vec{AB}$$ $$ \implies (c-a) \cdot e^{i\pi/3} = (b-a) \ldots... $$
Similarly:
$$ (a-b) \cdot e^{i\pi/3} = (c-b) \ldots... $$
Divide equation $1$ by equation $2$:
$$ \dfrac{c-a}{a-b} = \dfrac{b-a} {c-b}$$
Cross multiply and you are done!
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answered Sep 1, 2013 at 7:45
Parth ThakkarParth Thakkar
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$\begingroup$ Check the sign of rotation, not that it matters in the end. $\endgroup$
Macavity
– Macavity
2013-09-01 07:48:32 +00:00
Commented Sep 1, 2013 at 7:48
$\begingroup$ It must be counter clockwise and it is. What's wrong? $\endgroup$
Parth Thakkar
– Parth Thakkar
2013-09-01 07:49:51 +00:00
Commented Sep 1, 2013 at 7:49
$\begingroup$ How does this compare to the proofs given when the question was asked before? $\endgroup$
Gerry Myerson
– Gerry Myerson
2013-09-01 07:50:04 +00:00
Commented Sep 1, 2013 at 7:50
$\begingroup$ I saw your comment only when I had almost finished the answer. And secondly, I think this is more clear. $\endgroup$
Parth Thakkar
– Parth Thakkar
2013-09-01 07:53:08 +00:00
Commented Sep 1, 2013 at 7:53
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10514 | https://www.hanspub.org/journal/paperinformation?paperID=109110 | 人教A版与IBDP版数学教材比较研究——以概率统计为例
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人教A版与IBDP版数学教材比较研究——以概率统计为例
A Comparative Study of Mathematics Textbooks between Humanistic A Version and IBDP Version—Taking Probability Statistics as an Example
DOI:10.12677/ces.2025.133157, PDF, HTML, XML, 被引量下载:485 浏览:648
作者:孔冰心, 付雨婕, 何春玲：黄冈师范学院数学与统计学院，湖北 黄冈；龙 燕：黄冈中学，湖北 黄冈
关键词:高中数学；概率统计；教材比较；High School Math；Probability Statistics；Textbook Comparison
摘要:随着全球化教育交流的深入，对比不同教育体系下的教材对优化教学内容具有重要意义。通过对人教A版与IBDP版高中数学教材中概率统计部分的比较，探讨两种不同教育理念下该学科内容的组织编排、呈现方式、版面设计、例习题数量及其综合难度的异同，并提出了改进教学方式、重视学生核心素养发展、增强知识点与现实联系以及强化课堂与信息技术整合的重要性。不仅揭示了两套教材各自的特色与优势，还为国内与国际课程的融合提供了有益的参考，希望对高中数学课堂教学以及教材改革有所帮助。
Abstract:With the deepening of globalized educational exchanges, comparing textbooks under different educational systems is of great significance in optimizing the teaching content. By comparing the probability statistics section in the high school mathematics textbooks of Humanistic A and IBDP editions, this study explores the similarities and differences in the organization, presentation, layout design, number of examples and exercises, and overall difficulty of the subject content under two different educational philosophies. It also highlights the importance of improving teaching methods, emphasizing the development of students’ core competencies, enhancing the connection between knowledge points and real-life situations, and strengthening the integration of classroom teaching with information technology. It not only reveals the characteristics and advantages of the two sets of textbooks, but also provides useful references for the integration of domestic and international curricula, hoping to be helpful for high school mathematics classroom teaching as well as textbook reform.
文章引用：孔冰心, 龙燕, 付雨婕, 何春玲. 人教A版与IBDP版数学教材比较研究——以概率统计为例[J]. 创新教育研究, 2025, 13(3): 73-81.
1. 引言
概率统计是高中数学课程内容的四大主线之一，贯穿必修和选择性必修课程，也是世界各国数学教学的重点内容。通过对数据进行收集、整理、分析等，可以帮助人们在对未知事件进行决策时提供理论依据，还可以为人们了解客观世界提供解决方法和有利的思维模式。随着大数据时代的到来，学习和掌握这一领域的知识，对于学生的终身发展具有非常重要的意义，这不仅是提升学生的数据分析、数学建模、逻辑推理能力等数学核心素养的重要媒介，同时也是提升其数学综合素质的一个重要途径。
随着全球经济一体化的发展，各国间的合作与交流也日趋密切，开展各国数学教育的对比研究，有助于深入了解各国教育制度的异同，为我国的教育改革与政策制定提供借鉴。
本文选取概率统计作为比较内容，将我国使用最广泛的教材人教A版和国际IBDP版数学教材进行比较，分析两种教材不同的编写特色、课程综合难度等方面的内容，并提出一些建议，使之更好地服务于教育教学。
2. 文献综述
2.1. 国内数学教材比较研究现状
数学课程改革是当前教育改革中的一个热门课题，数学教材比较也成为了教育界研究的对象。通过知网、万方、维普的文献搜索，发现国内新旧版教材的比较和多版本教材的比较居多。黄锦涛(2024)对人教A版和苏教版教材在“集合与常用逻辑用语”部分进行比较，得出不同版本教材里的不同，从而促进教育教学的有效实施。李小双，杨新荣(2024)对人教A版、北师大版、湘教版三版高中新数学教材呈现的统计思维水平和复杂性进行比较研究，总结发现并提出了要充分平衡各个环节的统计思考发展机遇，平衡统计思考的横向分布，合理适度地分配高复杂任务，强化IT融合等建议。张瑞凤、刘蕴煜(2020)从总体结构与具体内容两方面对人教版与人教版“几何”的内容作了对比，研究发现苏教版的教科书提倡以终身发展为目标，人教版强调了对基础知识的培养和对基础技能的培养，并在整体上和习题设置上分别进行了分析。曹一鸣和吴立宝在对比研究了世界上八个国家的小学数学教科书后，从内容的广度、内容深度、例题难度和习题难度四个方面，建立了一个比较的指标。也有作者针对国内外教材进行了比较，薛亚巧从问题提出的视角，对比分析了中美教科书的异同，并对中国的教科书进行了反思。沈爱桐和杨亚平选取中国和加拿大教材中“圆”几何学单元作为研究对象，分别从内容结构、呈现方式、题目设置和信息技术四个维度对其进行比较，并为我国教材建设提供可行性建议。跨文化的教材比较研究在这一方向上还有广阔的探索空间。
2.2. 国际数学教材比价研究现状
国际上的数学教科书的比较，可以分为国外教科书的比较和国内教科书的比较。李孝诚、春霞、王瑞霖(2012)在《美国初中数学教材评估及其启示》中介绍了美国数学教材评估的理念、假设与目标，对我国数学教材评估研究具有很好的借鉴价值。史宁中，孔凡哲、严家丽、崔英梅(2015)利用“教科书难度模型”，本研究选取了中美日俄十种典型中学数学教科书的交叉部分，以知识团为基本单位，从广度、深度、练习的综合难易三个方面对教科书的难易程度进行了量化对比和质的解释。胡典顺、王春静、王静(2020)以概念图法为基础，以英、澳大利亚、美、中为研究对象，对“矩阵”的概念进行了横、纵两个层次的研究。最后对四个国家教科书的深度和广度进行了对比。张萍、曾素樵(2022)以苏教版高中数学新教材和IBDP Mathematics AA HL教材为载体，初步探索了中外高中数学教材在拔尖创新人才培养方面的异同。
3. 研究对象
本文聚焦于比较中国的人教A版教材与国际IBDP课程教材中关于概率统计的内容，所选教材均为具有代表性的教材，见表1。
Table 1.Basic information on different editions of textbooks
表1.不同版本教材的基本信息
教材版本 教材名称 出版社 出版时间
人教A版 必修2；选择性必修3 人民教育出版社 2019年
IBDP版 Mathematics Core topics HL 1; Mathematics Analysis and approaches HL 2 Haese Mathematics 2019年
4. 内容结构比较
4.1. 编排顺序比较
内容的编排顺序体现了知识点之间的联系，它对内容的表现有着直接的影响，同时还能看到编写者对知识的衔接和对学生认识发展规律的衡量。有关两版教材概统统计内容的编排顺序比较，如表2。
从编排顺序上，我国人教A版教材主要采用的是直线式的组织结构进行编排。采用先统计后概率的方法，从随机抽样、样本估计及统计案例等定量的统计方法中自然过渡到对概率的刻画上，再从条件概率、期望等过渡到正态分布，最后以统计相关性、一元线性回归和独立性检验收尾。它基本按照概率统计知识本身的逻辑结构行进，学习内容不重复。IB教材则采用的是先概率后统计的思路，先从实验概率、二维表等到抽样数据、箱线图等内容，再从皮尔逊积矩相关系数、最小二乘回归线到离散概率等内容，章节更多，跨度更广，涉及到部分高等数学的内容，对基本课时和学生精力分配的要求更高。
两版教材从编排结构上来看，都遵循由浅入深顺序，串联知识点之间的联系，符合学生的认知发展规律，有利于提高学生的数学运算、逻辑推理等数学核心素养。
Table 2.Comparison of the content organization of probability and statistics in the Humanities A and IBDP editions of the textbook
表2.人教A版和IBDP版教材概率统计内容编排比较
课程设置 人教A版 IBDP版
必修2/
Mathematics Core
topics HL 1 9.1随机抽样
9.2用样本估计总体
9.3统计案例
10.1随机事件与概率
10.2事件的相互独立性
10.3频率与概率 11.1实验概率
11.2二维表
11.3样本空间和事件
11.4理论概率
11.5使用概率进行预测
11.6概率的加法法则
11.7独立事件
11.8相关事件
11.9条件概率
11.10独立性的正式定义
11.11贝叶斯定理
12.1-7抽样与数据
13.1-10统计
选择性必修3/
Mathematics Analysis and approaches HL 2 6.1分类加法计数原理与分步乘法计数原理
6.2排列与组合
6.3二项式定理
7.1条件概率与全概率公式
7.2离散型随机变量及其分布列
7.3离散型随机变量的数字特征
7.4二项分布与超几何分步
7.5正态分布
8.1成对数据的统计相关性
8.2一元线性回归模型及其应用
8.3列联表与独立性检验 7.1-5计数
26.1-5双变量统计
27.1随机变量
27.2离散概率分布
27.3期望
27.4方差和标准差
27.5 aX+b的性质
27.6二项分布
27.7使用技术查找二项概率
27.8二项分布的均值和标准差
28.1概率密度函数
28.2中心和离散度的度量
28.3正态分布
28.4计算正态概率
28.5标准正态分布
28.6正态分位数
4.2. 呈现方式比较
在引言设置上，人教A版教材，强调数学在生活中的实际运用，通过介绍之前所学习的内容，进而引出这一章节的内容，例如在随机事件与概率这一节中，回顾了初中学习的随机事件的概念，然后引出本节继续进一步研究随机事件及其计算，此外还举了一些生活情境问题的例子，简单直接。在IBDP版教材中则直接展示了一个问题情境，首先以丰富的图片和有趣的问题引入，激发学生的好奇心，然后抛出了思考问题，进而引出了本章的主题。不同的是在人教A版教材中，知识点都位于每一章节内，并在每一章节开始处都有引入设计，通过回顾已在大脑中建立起来的知识体系，并试图运用现有的知识来解决新的问题，产生认知冲突，由此产生新的知识概念，这对培养学生的主动性有很大的帮助。IBDP版教材把每一个知识点作为单独一个小节位于这一章之下，并只在本章开始处设计了问题导入，利用实际问题，从学生已有的知识建构出发，使得学生在解决问题的同时掌握新的知识，并且锻炼学生的动手操作能力，体现了“学生的学”和“以学生为主”的教学理念与学习方式。此外IBDP版教材更注重实践性，练习题的数量是人教A版的两倍多。
在章节内容呈现上，人教A版教材非常注重学生数学核心素养的发展，每节内容基本按照“问题情境或知识回顾——数学抽象——思考探究——知识巩固——练习强化”的编写体例来进行的，在每节内容之后都有相应课后习题，每章内容结束后会有相应的整章复习题，体现了知识的关联性、探究性和整合性，旨在帮助学生掌握知识。IBDP版教材，在每小节基本按照“调查研究——数学理论——例题——讨论——练习”的编写体例来进行的。在每一节内容结束后，会提供大量配套练习题，帮助学生巩固所学知识。每个章节的结尾处设置了一个活动探究问题，并通过有趣的案例激发学生对数学学习的兴趣和主动性。活动探究之后，还安排了几组难度各异的复习题，针对本章知识点进行不同层次的练习，以确保学生能够全面掌握并灵活运用所学内容。这样设计不仅有助于加深学生对各知识点的理解，还能通过多样化的练习形式提升他们的解题技巧和应变能力。同时，趣味性的个案研究也能提高学生的学习动力，激发他们对数学的兴趣，从而提高学生的综合应用能力和创造性思维。
4.3. 版面设计比较
版面设计是评判教科书质量的一个重要方面，丰富的版面设计一方面可以增加学生的阅读兴趣，一方面便于学生对知识的了解。
两版教材都采用彩印，人教A版在章节名、概念名称、题号字体上和栏目底色上使用彩印，其余都以黑白为主。在IBDP版中，并没有对字体的颜色进行改变，只是对不同部分内容的底色进行了改变，如章节名为蓝底、问题探究为绿色底、表格为浅黄色底、例题为浅蓝色底等，其他都是黑白色，对重点部分还有字体加粗。所以IBDP版的版面内容区分得更为明显，色彩面积更大，更能吸引学生的注意力，但主要概念、定理等并没有字体颜色变化，不容易使学生找到重点。从整体上来看，人教A版的文字编辑更加紧凑，页面的留白较多，穿插较少的图形。IBDP版教材页面更丰富多彩，印刷精致，页数也很多，更能引起学生的兴趣，特别说明的是IBDP版教材的提示框使用的是人物图形，从人的角度提出质疑，更能引起学生兴趣。以下是选取两版教材内页图，如图1。
Figure 1.Inset map of the two editions of the textbook
图1. 两版教材内页图
4.4. 例习题数量比较
在人教版教材中的例习题类型有例题、练习题、习题和章节复习题。IBDP版教材中的例习题有例题、练习题和复习题。两版教材的例习题数量整理如表3。
Table 3.Comparison of the number of example exercises in the two editions
表3. 两版教材例习题数量对比
例题 练习 习题 复习题 合计
人教A版 73 131 154 62 420
IBDP版 100 643 224 967
通过上表可以看出人教A版和IBDP版教材在例习题的总体上来看有很大的差异。IBDP版教材的例习题总数比人教A版的两倍多，其中IBDP版教材中练习题部分占比最高，一方面与知识点数量有关，IBDP版教材知识点数量多从而导致的习题数量有很多，另一方面，这与其教育理念有关，人教版教材的编写理念由我国课程标准决定，注重训练学生数学核心素养的发展，强调类比与转化的思想；IBDP版教材教学目标来源于AA课程指南，新版AA数学课程指南从认知维度上表达出希望更多的培养学生在领会、应用维度的能力，非常重视数学的探究学习，强调对问题解决的看法以及数学工具的选择。
5. 课程综合难度比较
课程综合难度是指某课程被学生接受的难易程度。史宁中和孔凡哲教授提出，课程综合难度受三个因素影响：课程时间、课程广度和课程深度。在确保有足够的教学时间的前提下，可以通过“相对深度(S/T)”与“相对广度(G/T)”的线性组合来表征课程的难度。这里，S代表深度，G代表广度，而T则表示总的教学时间。本节采用史宁中教授建立的课程难度模型 N=α S T+(1−α)G T N=α S T+(1−α)G T 对两版教材中概率统计内容的课程难度进行定量研究。
其中，N表示课程难度，T表示课程时间，G表示课程广度，S表示课程深度，G/T为可比深度，α 为加权系数，课程难度与课程深度成正比，与课程难度成反比；课程难度与课程广度成正比，与课程时间成反比。
5.1. 课程时间
课程时间完成教材相关内容所需要的基本课时数。我国《普通高中数学课程标准(2017年版2020年修订)》的课标中明确指出概率统计内容教学在必修课程中需20课时，选择性必修课程中需26课时，总计46课时。根据AA课程指南中指出的HL教材有关概率统计内容的建议课时数为33课时。因此，我们将人教A版课时时间记为T 1 = 46，IBDP版课时时间记为T 2 = 33。
5.2. 课程广度
课程广度指的是课程内容覆盖的范围和领域的宽泛程度，通常通过知识点的数量来衡量。关于如何界定知识点的理解，目前尚无统一标准，一般而言，知识点被认为是构成课程的基本单元，包括概念、定理以及一些小型的、相对独立的知识体系，以下是统计的这两版教材中概率统计的相关知识点(表4)。
5.3. 课程深度
课程深度是对指知识的要求程度，可以采用目标动词赋值的方法进行量化。根据课程标准和教材内容，将每个知识点的深度由低到高分为3个水平：了解、理解、掌握，并分别赋值，由于AA课程指南中并未明确说明具体知识点的水平要求，所以参考我国课程标准要求，具体见表5例如课程标准要求“理解样本点和有限样本空间的含义”，则这个知识点的深度水平为“理解”，并赋值为2。
Table 4.Statistics on the number of knowledge points in the two editions of the textbook
表4. 两版教材知识点数量统计表
版本 知识点
人教A版
G 1 = 38 简单随机抽样，分层随机抽样，均值，众数、平均数、中位数，样本空间，随机事件，基本事件，必然事件，事件的关系和运算，概率的性质，古典概型，独立事件，计数原理，贝尔斯定理，排列数，全排列，阶乘，组合数，二项式定理，条件概率，全概率公式，离散型随机变量，两点分布，二项分布，期望，方差，标准差，伯努利实验，超几何分布，正态分布，标准正态分布，相关关系，一元线性回归模型，最小二乘法，独立性检验，列联表，连续随机变量
IBDP版
G 2 = 45 简单随机抽样，分层随机抽样，均值，众数、平均数、中位数，样本空间，实验概率，理论概率，概率的加法定律，相关事件，系统抽样，方便抽样，分类变量，数值变量，独立事件，计数原理，贝尔斯定理，排列数，数值变量，离散数据，连续数据，使用频率表，阶乘，组合数，分组数据，箱线图，异常值，累积频率图，条件概率，离散型随机变量，二项分布，期望，方差，标准差，皮尔逊积矩相关系数，目测最佳拟合线，x对y的回归线，连续随机变量，离散概率分布，aX + b的性质，概率密度函数，正态分位数，正态分布，标准正态分布，相关关系，最小二乘法，双向表
Table 5.Target verb assignment
表5.目标动词赋值
赋值 水平要求 水平含义 行为动词
1 了解 再认或回忆知识；识别、辨认事实或证据；举出例子；描述对象的基本特征 经历(感受)、说出、知道、认识、 识别等
2 理解 描述对象的特征由来，阐述此对象和相关对象的区别和联系 解释、会、归纳、推断、分析、阐述等
3 掌握 在理解的基础上，把对象用于新情境 探索、能、解决等
在前边章节中已统计出两个版本教材中概率统计内容的知识点数量，再结合课程标准和AA课程指南，以及教材具体实施内容，根据赋值标准由低到高水平统计出相应知识点数量依次为10、8、20，通过计算其课程深度S 1 = 10 × 1 + 8 × 2 + 20 × 3 = 86。IBDP版知识点数量依次为12、11、22，通过计算其课程深度S 2 = 12 × 1 + 11 × 2 + 22 × 3 = 100。
5.4. 课程难度模型
将以上数据整理汇入表6。
Table 6.Comparison of course difficulty data between the two editions of the textbook
表6. 两版教材课程难度数据比较
版本 课程时间 课程广度 课程深度 可比广度 可比深度
人教A版 T 1 = 46 G 1 = 38 S 1 = 86 S 1/T 1 = 1.87 G 1/T 1 = 0.83
IBDP版 T 2 = 33 G 2 = 45 S 2 = 100 S 2/T 2 = 3.03 G 2/T 2 = 1.36
根据公式带入相应的数据，取 α = 0.4，得N 1 = 0.4 × 1.87 + (1 − 0.4) × 0.83 = 1.246，N 2 = 0.4 × 3.03 + (1 − 0.4) × 1.36 = 2.028。见图2。
Figure 2.Course difficulty values for both editions of the textbook
图2. 两版教材课程难度值
从图表中可以观察到，N 1< N 2，这意味着IBDP版教科书的难度要高于人教A版教科书。两版教材在综合难度上存在显著差异。就课程时间而言，IBDP版教材所需的教学时间较之人教A版更短，但知识点数量比人教A版的多，所以需要学生在较短的时间内完成较深的理解，是一项不小的挑战。从课程广度来看，IBDP版教材涉及一些高等数学的内容，所以学习起来可能比较有难度，课程广度就相对偏高一点。
6. 结论与建议
6.1. 结论
通过以上几个维度对人教A版和IBDP版数学教材的比较分析，可以看到两版教材各有所长。在内容编排上，人教A版教科书的概率统计章节在必修科目中没有着重于对资料处理的深度概率分析，而更多地侧重于对学生的随机性和统计概念的培养。对于概率论的深入探讨则被安排在选修课程中，这样更侧重于促进学生数学核心素养的整体发展。IBDP版教材的概率统计内容是按照学生的认知规律来编排内容，内容针对性更强。在呈现方式上，从引言和章节内容两个方面进行比较，人教A版的引言设置更详细，但只是文字语言，没有新颖性，很难引起学生学习的兴趣，在具体章节内容设置上，注重知识之间的关联性和综合性，顺应学生的身心发展规律。IBDP版引言部分大都以现实的实际问题为例，更具创新性，但是只在每一章首有引言，在具体的章节内容中，是以每一个知识点作为一节，每一节之后还有大量的练习题和信息技术的活动探究游戏，注重学生的创新性思维的发展。在版面设计上，IBDP版教材页面设计更加丰富多彩，更能引起学生的学习兴趣，但对重点概念的表示上人教A版的表示更能让学生抓住重点。在课程综合难度上可以发现，IBDP版教材明显比人教A版教材难度高，主要体现在IBDP版教材涉及的知识点更多，更加倾向于大学课程的衔接，更加丰富，所以对学生的学习能力要求更高，也从侧面反映了我国教材的设置更注重全体学生的全面个性发展以及对数学核心素养的发展。
我国教材向来以严谨、规范著称，而西方教材的活泼性、探究性与开放性也颇具特色，如何将二者有机融合，是我国教材编写未来需要重点思考的方向。无论是哪一种高中数学教科书，在编制过程中都必须重视地区差异、民族特点以及人才培养目的等方面的考虑，要多方面考虑而不能固步自封。另外，在教材编制中，可以适当增加一线师资的比重。处于教育第一线的老师，对学校的生活有着敏锐的洞察力。有了他们的参与，可以将一些贴近校园生活的实际问题，带到教学中去。这样，就可以把教科书上对数学概念的阐述，化抽象为具体，让广大读者更容易接受。
6.2. 建议
通过对人教A版和IBDP版的教材比较，从中借鉴IBDP版中的长处，针对我国教学提出如下建议：
1) 改进教学方式，重视核心素养
概率统计与日常生活紧密相关，具有重要的实际应用价值。在应用科学中，统计学强调理解其思想精髓，而非机械记忆定义。它重视数据的深层意义和内在联系，而非简单的数字运算，旨在进行有效的数据分析和解释。在概率统计的教学过程中，应参照 IBDP版的内容设定，大胆地转变自己的思维方式和认知模式，把理论教学的重心放在对结论的利用上，转向启发式的教学，以激发学生的求知欲，重视概率统计的练习与应用，培养学生的数学抽象、数学运算、逻辑推理等数学核心素养的发展。
2) 探究知识点之间关系，增强现实联系
IBDP版教材很重视知识点与现实生活的转化和跨学科知识的结合，很多知识点之间并非完全的独立，也是前后知识之间的传递性。在教学中，教师要多引导学生积极思考，比较两者的关系，使他们能把所学知识运用到生活中，解决现实中的实际问题，还可以涉及社会科学、自然科学等领域的问题，通过分析统计数据和进行概率计算，使学生能够应对跨学科的挑战，把理论联系实际，既可以加深学生对知识的理解，又可以提高学生的学习兴趣。
3) 强化课堂教学与信息技术的整合，提高教学效果
概率统计内容的深入学习总是以大量数据为依托的，在这个过程中就需要现代信息技术为辅助。在教学实践中，应根据自己的实际情况，灵活运用各种信息技术。一方面，就以智能教学系统为例，老师可以对学生进行检查，并对其进行个性化的指导，这样既能提升教学的效率，又能促进学生的全面发展。另一方面可以利用信息技术，实现直观的数据统计结果或者是图形动态效果，可以将抽象的数学知识变得生动、形象，便于学生理解和记忆。
NOTES
通讯作者。
参考文献
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朱台. 沪教版与IBDP版数学教材的比较研究[D]: [硕士学位论文]. 上海: 上海师范大学, 2023.
谢先成. 基于核心素养的《普通高中数学课程标准(2017年版)》解读——访数学课程标准修订组组长、东北师范大学原校长史宁中教授[J]. 教师教育论坛, 2018, 31(6): 4-7.
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The following article is Open access
Broken time-reversal symmetry in a SQUID based on chiral superconducting Sr 2 RuO 4
R Ishiguro, T Sakurai, M Yakabe, T Nakamura, S Yonezawa, S Kashiwaya, H Takayanagi and Y Maeno
Published under licence by IOP Publishing Ltd
Journal of Physics: Conference Series, Volume 568, 27th International Conference on Low Temperature Physics (LT27) 6–13 August 2014, Buenos Aires, ArgentinaCitation R Ishiguro et al 2014 J. Phys.: Conf. Ser.568 022020DOI 10.1088/1742-6596/568/2/022020
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R Ishiguro
AFFILIATIONS
Center for Emergent Matter Science, RIKEN, Wako, Saitama 351-0198, Japan
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
T Sakurai
AFFILIATIONS
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
M Yakabe
AFFILIATIONS
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
T Nakamura
AFFILIATIONS
Department of Physics, Graduate School of Science, Kyoto University, Kyoto 6068502, Japan
Institute for Solid State Physics, the University of Tokyo, Chiba 277-8581, Japan
S Yonezawa
AFFILIATIONS
Department of Physics, Graduate School of Science, Kyoto University, Kyoto 6068502, Japan
S Kashiwaya
AFFILIATIONS
National Institute of Advanced Industrial Science and Technology (AIST), Tsukuba 305-8568, Japan
H Takayanagi
AFFILIATIONS
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
International Center for Materials Nanoarchitectonics (MANA), National Institute for Materials Science (NIMS), Tsukuba 305-0044, Japan
Y Maeno
AFFILIATIONS
Department of Physics, Graduate School of Science, Kyoto University, Kyoto 6068502, Japan
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Authors
R Ishiguro
AFFILIATIONS
Center for Emergent Matter Science, RIKEN, Wako, Saitama 351-0198, Japan
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
T Sakurai
AFFILIATIONS
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
M Yakabe
AFFILIATIONS
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
T Nakamura
AFFILIATIONS
Department of Physics, Graduate School of Science, Kyoto University, Kyoto 6068502, Japan
Institute for Solid State Physics, the University of Tokyo, Chiba 277-8581, Japan
S Yonezawa
AFFILIATIONS
Department of Physics, Graduate School of Science, Kyoto University, Kyoto 6068502, Japan
S Kashiwaya
AFFILIATIONS
National Institute of Advanced Industrial Science and Technology (AIST), Tsukuba 305-8568, Japan
H Takayanagi
AFFILIATIONS
Department of Applied Physics, Faculty of Science, Tokyo University of Science, Tokyo 125-8585, Japan
International Center for Materials Nanoarchitectonics (MANA), National Institute for Materials Science (NIMS), Tsukuba 305-0044, Japan
Y Maeno
AFFILIATIONS
Department of Physics, Graduate School of Science, Kyoto University, Kyoto 6068502, Japan
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1742-6596/568/2/022020
Abstract
Unconventional superconductors involve not only gauge-symmetry breaking but also orbital- and spin-symmetry breaking. Superconducting Sr2RuO4 is known as a spin-triplet chiral p-wave and a topological superconductor with broken time-reversal symmetry (BTRS). Kerr-effect and muon-spin-rotation (pSR) measurements have shown that the bulk superconducting state of Sr2RuO4 features BTRS in the orbital part; hence, it is called the chiral state. BTRS in the response of superconducting junctions or SQUIDs would appear as the shifts of magnetic interference patterns. However, it is problematic to distinguish whether the shift originates in the residual magnetic field (trapped vortex) or the effects of BTRS. Here, we show that the magnetic interference patterns of a SQUID based on Sr2RuO4 are explicitly asymmetric with respect to the direction of both the bias current and the applied magnetic field; namely, there is no inversion symmetry. This indicates that the superconducting state of Sr2RuO4 undoubtedly breaks the time-reversal symmetry of the SQUID.
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10516 | https://www.jem-journal.com/article/S0736-4679(18)30728-5/pdf | Selected Topics: Psychiatric Emergencies
PSYCHIATRIC EMERGENCIES FOR CLINICIANS: EMERGENCY DEPARTMENT MANAGEMENT OF HYPERCALCEMIA
Dunya N. Alfaraj, MD , Michael P. Wilson, MD , PHD , Yahia Akeely, MD , Gary M. Vilke, MD , and Kimberly Nordstrom, MD , JD
Department of Emergency Medicine, UC San Diego Health System, San Diego, California
Reprint Address: Dunya N. Alfaraj, MD , Department of Emergency Medicine, UC San Diego Health System, 200 West Arbor Drive, San Diego, CA 92103
, Keywords—hypercalcemia; emergency CLINICAL SCENARIO
A 55-year-old male with a medical history notable for diabetes mellitus and lung cancer was brought to the emergency department via emergency medical services for lethargy and confusion that started a few hours earlier. His family reported that his medications included trama-dol 50 mg three times a day as needed for pain and glybur-ide 5 mg twice a day. A review of systems was taken from his family members, which was positive for nausea, abdominal pain, polyuria, polydipsia, and a recent renal stone. The vital signs included temperature of 37.1 C, blood pressure of 110/60 mm Hg, heart rate of 120 beats/min, respiratory rate of 22 breaths/min, and oxygen saturation rate of 97% on room air. A fingerstick glucose level was 185 mg/dL and an electrocardiogram was initially reported as unremarkable, although a short QT interval was noted. On physical examination, the patient was not oriented to person, place, or time, and appeared dehydrated with dry mucous membranes. Pupils were equal, round, and reactive to light. Other physical exam-ination findings were unremarkable. A basic metabolic panel eventually resulted with a calcium level of 15 mg/dL.
What Do You Think Is Going on with This Patient?
The clinical presentation suggests a number of possible etiologies for the patient’s altered mental status, including electrolyte abnormality, drug intoxication, sei-zures, or stroke. Narcan was administered without effect, and computed tomography of the head was normal. A basic metabolic panel eventually confirmed the diagnosis of severe hypercalcemia. The patient’s mental status eventually improved with intravenous fluids.
What Key Findings Led to the Diagnosis?
This patient’s constellation of signs and symptoms, including lethargy, confusion, nausea, abdominal pain, polydipsia, polyuria, tachycardia, and dehydration, could be explained by marked hypercalcemia. A shortened QT interval and a history of renal stones also support the diag-nosis. A history of lung cancer suggests malignancy-induced hypercalcemia.
What Other Diagnoses Should You Consider?
There are many possible causes for an acute confusion (see Table 1), and many disorders present in a similar fashion. Hypoglycemia is an important potential cause in the differ-ential diagnosis, but can be excluded quickly by a fingerstick
R ECEIVED : 28 February 2018; F INAL SUBMISSION RECEIVED : 13 June 2018; ACCEPTED : 11 July 2018
688
The Journal of Emergency Medicine, Vol. 55, No. 5, pp. 688–692, 2018
Ó2018 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http:// creativecommons.org/licenses/by-nc-nd/4.0/). 0736-4679/$ - see front matter
glucose. Central nervous system (CNS) causes, such as stroke or intracranial bleeding, are also important possible causes to consider, but hypercalcemia typically does not pre-sent with focal neurological deficits. Of note, hyper-reflexia and tetany, which can sometimes be confused with a neuro-logical deficit, are typically seen in hypocalcemia instead of hypercalcemia. Opioid overdose should be considered because of the history of depression and tramadol use, but the patient’s clinical presentation, including lack of respira-tory depression and meiosis, is less suggestive of this etiol-ogy. Hepatic and uremic encephalopathy can occasionally present in a similar fashion, but are less likely in this patient with no history or stigmata of liver or kidney disease.
WHAT DO EMERGENCY CLINICIANS NEED TO KNOW ABOUT HYPERCALCEMIA?
Background
The neurocognitive symptoms of hypercalcemia, such as anxiety and mood change, can mimic a psychiatric disor-der presentation at modest levels of hypercalcemia. How-ever, hypercalcemic patients secondary to cancer generally present with much higher values of calcium and thus can develop more severe neurologic distur-bances, such as alterations in mental status, as opposed to mild neurocognitive symptoms. Hypercalcemia of ma-lignancy in many cases represents an oncologic emer-gency that necessitates urgent intervention (1,2).
Diagnosis
Only 2% of total body calcium is circulating in the blood-stream, and the remaining 98% is in the bone. Half of all circulating calcium is free (or ionized), and the remaining half is bound to protein. The total serum calcium concen-tration (free + bound) is typically between 8.5 to 10.5 mg/ d, with hypercalcemia defined as a total serum calcium concentration > 10.5 mg/d. Patients with mild hypercal-cemia (levels between 10.5 and 12.5 mg/d) can often be asymptomatic and might not require immediate therapy. As calcium levels exceed 12.5 mg/d, patients are more likely to be symptomatic. Further elevations, particularly > 14 mg/d, may lead to severe volume depletion, acute kidney injury, hypotension, alterations in mental status, and electrocardiographic abnormalities, such as short-ening of the QT interval, ST segment elevation, heart block, or even cardiac arrest (3–5). Several studies have shown that ionized Ca is more sensitive than total Ca (including corrected total Ca for albumin = serum calcium [mg/dL] + 0.8 {4 g/dL–patient albumin [g/ dL]}). However, because measurement of ionized Ca is more labor-intensive and expensive than total Ca, ionized Ca is not routinely requested. Ionized calcium can be important in critically ill patients, who often have severe hypoalbuminemia. In these situations, ionized calcium should be obtained in order to correctly estimate the severity of hypercalcemia. Acid–base changes influence protein binding. Acidosis increases the level of ionized calcium, while alkalosis conversely decreases the ionized calcium level (5,6).
Pathophysiology
Understanding the pathophysiology of hypercalcemia is important for both correct diagnosis and treatment. Vitamin D stimulates bone resorption and calcium ab-sorption from the small intestine. Parathyroid hormone (PTH) indirectly stimulates calcium absorption through the small intestine through its effect on vitamin D meta-bolism. PTH also enhances bone resorption, stimulates distal tubular reabsorption of calcium, and inhibits prox-imal and distal tubular reabsorption of phosphorus. Calci-tonin inhibits bone resorption, with sodium transport responsible for the majority of calcium reabsorption in the proximal tubule (5). As a result, both PTH and vitamin D increase calcium levels and decrease phos-phate levels. Conversely, calcitonin, which is released from the thyroid gland, decreases both calcium and phos-phate. PTH-related protein (PTHrP) is produced by certain types of cancer cells, such as renal cancer, ovarian cancer, endometrial cancer, human T-cell leukemia/lym-phoma virus type 1–associated lymphoma, breast cancer, and, more commonly, squamous cell cancers of the head and neck, esophagus, cervix, or lung. However, any tumor can cause the condition known as humoral hypercalcemia of malignancy. PTHrP binds to the PTH receptor and mimics the biologic effects of PTH on both bones and kidneys, causing increased bone resorption and enhanced renal retention of calcium (7,8).
Table 1. Differential Diagnosis of Altered Mental Status
Cause Description Psychiatric Severe depression, dementia Electrolytes Hyponatremia, hypercalcemia Endocrine Hypoglycemia, hyperglycemia, hypothyroidism, hyperthyroidism Hepatic and renal Hepatic and uremic encephalopathy Drug-related Serotonin syndrome, malignant neuroleptic syndrome, anticholinergic Withdrawal Alcohol, benzodiazepine, barbiturate Systemic Severe hypoxia, hypotension, bradycardia or tachycardia, hyperthermia or hypothermia CNS Stroke, head trauma or intracranial bleed, vasculitis, meningitis, encephalitis, non-CNS infection, hypertensive encephalopathy, seizure/post ictal state CNS = central nervous system.
ED Management of Hypercalcemia 689
Etiology
Causes of hypercalcemia include multiple pathologic en-tities (Table 2), but there are two diagnoses that account for > 90% of all hypercalcemia cases: primary hyperpara-thyroidism and hypercalcemia of malignancy (1,9). In an ambulatory population, primary hyperparathyroidism accounts for the vast majority of detected hypercalcemia. In primary hyperparathyroidism, excessive secretion of the PTH leads to hypercalcemia. In secondary hyperpara-thyroidism, PTH secretion increases as an adaptive response to hypocalcemia. Primary hyperparathyroidism is due to a benign parathyroid adenoma in 80–90% of cases. While 10–15% are caused by multi-gland disease or diffuse parathyroid hyperplasia, parathyroid carcinomas are responsible for < 1% of cases. Patients with chronic kidney disease develop hypercalcemia as a consequence of secondary hyperparathyroidism or tertiary hyperpara-thyroidism, in which patients develop autonomous para-thyroid function. Malignancy-associated hypercalcemia (MAH) occurs in a quarter of cancer patients and com-prises more than one-third of the cases of hypercalcemia presenting to the emergency department. MAH is also a cause of > 50% of inpatient hypercalcemic crises. Drugs may also be associated with hypercalcemia, such as thia-zide diuretics, lithium, and rarely, calcium supplementa-tion or vitamin D intoxication. Other conditions that may manifest with hypercalcemia include sarcoidosis, thyro-toxicosis, Addison’s, pheochromocytomas, or vasoactive intestinal peptide–secreting tumor (5,10,11). In patients with hyperparathyroidism, the serum phosphorus level can be either low or normal. However, in a patient with both hypercalcemia and hyperphosphatemia, the etiology is unlikely due to hyperparathyroidism. Instead, hypercalcemic disorders in which PTH secretion is suppressed should be considered (12).
Clinical Features
The classic description of ‘‘stones, bones, abdominal moans, and psychic groans’’ is often used to recall the wide variety of signs and symptoms of hypercalcemia. The most important renal effects (‘‘stones’’) are poly-dipsia and polyuria resulting from a nephrogenic diabetes insipidus (vasopressin-induced water reabsorption), and nephrolithiasis resulting from the hypercalciuria. Other renal effects include dehydration and nephrocalcinosis. Nephrocalcinosis, the deposition of calcium phosphate crystals within renal tubules, may occur with hypercalce-mia, hypercalciuria, hyperphosphatemia, or hyperphos-phaturia (13). Bone pain (‘‘bones’’) can occur when there is excess PTH, which results in subperiosteal resorption, leading to osteitis fibrosa cystica, or so-called ‘‘brown tumor.’’ Brown tumor is so-named because it has a reddish/brown color secondary to the accumula-tion of red blood cells. Histologically, osteoclasts present in combination with irregular areas of bone resorption, which are replaced by both fibrous and giant cells (14). Nausea, abdominal pain, anorexia, constipation, and rarely, peptic ulcer disease or pancreatitis reflect gastroin-testinal manifestations (‘‘abdominal moans’’). Neuropsy-chiatric effects (so-called ‘‘psychic groans’’) include impaired concentration, confusion, fatigue, and muscle weakness. Cardiovascular effects, not included in the classic mnemonic, may involve hypertension and vascular calcifications (3,15).
Management
Continuous cardiac monitoring in the ED is necessary, and central venous or pulmonary artery pressure moni-toring may be required (16). Half of hypercalcemic can-cer patients develop hypokalemia as well. Serum potassium levels should be monitored every 4 h, with administration of potassium chloride (20–40 mEq intra-venously or orally) as necessary to prevent hypokalemia (16). Most patients with severe hypercalcemia, particularly >14 mg/dL, exhibit clinical evidence of intravascular vol-ume depletion. These patients should be aggressively re-hydrated with normal saline at an initial rate of 200– 300 mL/h, then adjusted to maintain the urine output at 100–150 mL/h (17). Excessive hydration can result in congestive heart failure, especially in malignancy-associated hypercalcemia, where critically ill patients often have hypoalbuminemia. Some authors have recom-mended intravenous fluid administration with no more than 75–150 mL/h of 0.9% sodium chloride in patients with concomitant hypoalbuminemia (5). Osteoclast-inhibiting therapies for severe hypercalce-mia are generally considered in consultation with the
Table 2. Hypercalcemia Differential Diagnosis
Cause Description Parathyroid hormone–related Primary and tertiary hyperparathyroidism, familial hypocalciuric hypercalcemia Cancer Multiple myeloma, bone metastasis, PTHrP-mediated breast, lung, and renal cancers Calcitriol-mediated Granulomatous diseases (sarcoidosis and TB), lymphoma, vitamin D intoxication Miscellaneous Vitamin A toxicity, thyrotoxicosis, theophylline toxicity, adrenal crisis, and Paget’s disease, lithium, thiazide, immobilization PTHrP = parathyroid hormone–related protein; TB = tuberculosis.
690 D. N. Alfaraj et al.
patient’s primary physician, oncologist, or endocrinolo-gist. Drugs that inhibit osteoclast-mediated bone resorp-tion include bisphosphonates, mithramycin, calcitonin, or corticosteroids. Bisphosphonates are the treatment of choice for man-agement of cancer-induced hypercalcemia. The use of intravenous bisphosphonate is restricted to the treatment of serum calcium concentrations >15 mg/dL or in patients with rapid deterioration of CNS, cardiac, gastrointestinal, or renal function (16). Because the effect of bisphospho-nates appears after 2–4 days, it should be combined with faster-acting therapeutic modalities, such as intravenous fluid and calcitonin. Calcitonin may be effective in doses of 4–8 IU/kg intramuscularly or subcutaneously. Nasal calcitonin is well tolerated but, because of insufficient ef-ficacy, is not recommended as a form of therapy for cancer-related hypercalcemia. Prednisone (60–80 mg) or other corticosteroids may be effective within a few days to a week. They are more useful for long-term treat-ment rather than for acute control. Corticosteroids are particularly valuable in hypercalcemia related to breast carcinoma, myeloma, lymphoma, granulomatous dis-ease, or vitamin D toxicity (7,16,18–23). The prevalence of hypercalcemia increases among pa-tients with chronic renal impairment, given the frequent use of calcium-containing phosphate binders and vitamin D therapy. Patients on hemodialysis who present with hy-percalcemia require urgent dialysis. After initial stabilization, most if not all patients should be admitted for further management. One of the common underlying causes for hypercalcemia is parathy-roid adenoma. For that, curative therapy requires surgical parathyroidectomy (24).
How Should You Stabilize This Patient?
Airway management and circulatory support for ob-tunded or comatose patients. Patients should be aggressively hydrated with normal saline. Intensive care unit hospitalization for severe hypercal-cemia cases. Hemodialysis may be useful in severe cases.
Controversies in Treatment: What Are the Most Important Steps in the Management of This Patient?
Although often mentioned, there is limited evidence-based recommendation for using furosemide to treat hy-percalcemia. Furosemide should not be a recommended intervention unless given to reverse aggressive fluid resuscitation (5,25–27). In hyperparathyroidism, the exact timing of parathy-roidectomy to ensure optimal outcome is not definitively established. A single-center retrospective series that included 35 years of data reported no significant differ-ences in long-term survival in patients who underwent surgery within 72 h, compared with those treated medi-cally for > 72 h before surgery (20).
Clinical Bottom Lines
Primary hyperparathyroidism and hypercalcemia of ma-lignancy account for > 90% of all hypercalcemia cases. In an ambulatory population, primary hyperparathy-roidism accounts for the vast majority of detected cases of hypercalcemia. Hypercalcemia of malignancy usually presents with markedly elevated calcium levels and such patients are commonly severely symptomatic. These patients comprise more than one-third of all cases of hypercalcemia present-ing to the emergency department and more than half of inpatient hypercalcemic crises. In critically ill patients, who often have severe hypoal-buminemia in addition to hypercalcemia, ionized calcium levels should be obtained so as not to underestimate the severity of hypercalcemia. Neurocognitive symptoms of hypercalcemia may mimic psychiatric presentations, but often have associ-ated physical complaints that are not typical of a purely psychiatric illness. Aggressive hydration, calcitonin, and bisphospho-nates form the basis of evidence-based management. Bi-sphosphonate should be combined with faster-acting treatment, like intravenous fluid and calcitonin. Furosemide may be of limited utility unless needed to reverse overly aggressive fluid replacement. Hemodialysis should be considered in patients with end-stage renal disease or for refractory hypercalcemia.
REFERENCES
Mirrakhimov AE. Hypercalcemia of malignancy: an update on pathogenesis and management. N Am J Med Sci 2015;7:483–93. 2. LeGrand SB. Modern management of malignant hypercalcemia. Am J Hosp Palliat Care 2011;28:515–7. 3. Carroll MF, Schade DS. A practical approach to hypercalcemia. Am Fam Physician 2003;67:1959–66. 4. Chang WT, Radin B, McCurdy MT. Calcium, magnesium, and phosphate abnormalities in the emergency department. Emerg Med Clin North Am 2014;32:349–66. 5. Maier JD, Levine SN. Hypercalcemia in the intensive care unit: a review of pathophysiology, diagnosis, and modern therapy. J inten-sive Care Med 2015;30:235–52. 6. Ong GS, Walsh JP, Stuckey BG, et al. The importance of measuring ionized calcium in characterizing calcium status and diagnosing pri-mary hyperparathyroidism. J Clin Endocrinol Metab 2012;97:3138–45. 7. Stewart AF. Clinical practice. Hypercalcemia associated with can-cer. N Engl J Med 2005;352:373–9. 8. Kacprowicz RF, Lloyd JD. Electrolyte complications of malig-nancy. Emerg Med Clin North Am 2009;27:257–69. 9. Lafferty FW. Differential diagnosis of hypercalcemia. J Bone Miner Res 1991;6:S51–9.
ED Management of Hypercalcemia 691 10. McCurdy MT, Shanholtz CB. Oncologic emergencies. Crit Care Med 2012;40:2212–22. 11. Carroll R, Matfin G. Endocrine and metabolic emergencies: hyper-calcaemia. Ther Adv Endocrinol Metab 2010;1:225–34. 12. Shapiro HI, Davis KA. Hypercalcemia and "primary" hyperparathy-roidism during lithium therapy. Am J Psychiatry 2015;172:12–5. 13. Fogo AB, Lusco MA, Najafian B, Alpers CE. AJKD atlas of renal pathology: nephrocalcinosis and acute phosphate nephropathy. Am J Kidney Dis 2017;69:e17–8. 14. Nunes TB, Bologna SB, Witzel AL, Nico MM, Lourenc ¸o SV. A rare case of concomitant maxilla and mandible brown tumours, papillary thyroid carcinoma, parathyroid adenoma, and osteitis fibrosa cyst-ica. Case Rep Dent 2016;2016:5320298. 15. Wald DA. ECG manifestations of selected metabolic and endocrine disorders. Emerg Med Clin North Am 2006;24:145–57. 16. Marx JA, Rosen P. Rosen’s Emergency Medicine: Concepts and Clinical Practice. 8th ed. Philadelphia, PA: Elsevier/Saunders; 2014. 17. Hosking DJ, Cowley A, Bucknall CA. Rehydration in the treatment of severe hypercalcaemia. Q J Med 1981;50:473–81. 18. Fatemi S, Singer FR, Rude RK. Effect of salmon calcitonin and et-idronate on hypercalcemia of malignancy. Calcif Tissue Int 1992; 50:107–9. 19. Major P, Lortholary A, Hon J, et al. Zoledronic acid is superior to pamidronate in the treatment of hypercalcemia of malignancy: a pooled analysis of two randomized, controlled clinical trials. J Clin Oncol 2001;19:558–67. 20. Trimarchi H, Lombi F, Forrester M, et al. Disodium pamidronate for treating severe hypercalcemia in a hemodialysis patient. Nat Clin Pract Nephrol 2006;2:459–63. 21. Vaughn CB, Vaitkevicius VK. The effects of calcitonin in hypercal-cemia in patients with malignancy. Cancer 1974;34:1268–71. 22. Wisneski LA, Croom WP, Silva OL, Becker KL. Salmon calcitonin in hypercalcemia. Clin Pharmacol Ther 1978;24:219–22. 23. Dumon JC, Magritte A, Body JJ. Nasal human calcitonin for tumor-induced hypercalcemia. Calcif Tissue Int 1992;51:18–9. 24. Ahmad S, Kuraganti G, Steenkamp D. Hypercalcemic crisis: a clin-ical review. Am J Med 2015;128:239–45. 25. LeGrand SB, Leskuski D, Zama I. Narrative review: furosemide for hypercalcemia: an unproven yet common practice. Ann Intern Med 2008;149:259–63. 26. Suki WN, Yium JJ, Von Minden M, Saller-Hebert C, Eknoyan G, Martinez-Maldonado M. Acute treatment of hypercalcemia with furosemide. N Engl J Med 1970;283:836–40. 27. Robey RB, Lash JP, Arruda JA. Does furosemide have a role in the management of hypercalcemia? Ann Intern Med 2009;150:146–7.
692 D. N. Alfaraj et al. |
10517 | https://aviation.stackexchange.com/questions/49946/how-and-why-does-engine-thrust-change-with-airspeed | aircraft performance - How (and why) does engine thrust change with airspeed? - Aviation Stack Exchange
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How (and why) does engine thrust change with airspeed?
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I'm interested in how the thrust of a turbofan engine is affected at higher airspeeds (TAS). I know (I believed) that engine thrust(at constant N1) was relatively constant like in the following graph (only slight deviations):
This graph is usually in the books/manuals describing engine performance with reference to speed.
Then I came across the data from CFM56-5C turbofan engine which states that max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N. That's almost 5 times more power on the ground than in cruise. Here is the link: How much air, by mass, enters an average CFM56 turbofan engine cruising per minute?
These are apparently contradictory statements or I am missing something. Which one is corrrect and why? Why is engine thrust being changed with speed? Also, on the graph above what are those two curves that when added form a net engine thrust?
After I did a few calculations using the thrust equation (F= mass flow difference in exhaust and inlet velocities denoted as delta V -> we will disregard the fuel mass flow and assume exit pressure is equal to the free stream pressure thanks to a nozzle) and following data mentioned above in the link, I found out that delta V term in cruise and on takeoff is constant (at full power) and its value is 295 m/s, which states that exhaust velocity of the engine will always be 295 m/s faster from the inlet velocity(for a maximum power setting at any speed). I think that's logical because work done by the engine is used to increase kinetic energy (delta Ek) of the airflow which increases speed always by a constant amount at specific power/N1 setting (of course less power equals less delta V).
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edited Mar 4, 2020 at 2:05
Vikki
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asked Mar 29, 2018 at 18:29
DarjanDarjan
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3 Please note the following: you state that max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N. That's almost 5 times more power on the ground than in cruise... That's wrong, because thrust is not power. There exist a relation between thrust, speed, and power, but the stationary case is an special one...xxavier –xxavier 2018-03-29 20:49:54 +00:00 Commented Mar 29, 2018 at 20:49
In terms of useful work, e n e r g y=f o r c e⋅d i s t a n c e e n e r g y=f o r c e⋅d i s t a n c e, p o w e r=f o r c e⋅v e l o c i t y p o w e r=f o r c e⋅v e l o c i t y (vector dot products). At zero speed, using power to blow all that air behind you costs a lot of fuel energy but isn't doing anything useful. Even when you release the brakes and start to roll, it's not much power. (Jet engines are very inefficient at low speed.)Peter Cordes –Peter Cordes 2021-09-14 07:45:06 +00:00 Commented Sep 14, 2021 at 7:45
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The first diagram you link to shows three lines but does not indicate what they represent. I guess the bold line is thrust over speed. Then this diagram is correct for a turbojet.
Thrust T T is the difference between the engine's exit impulse minus the entry impulse:
T=(m˙a i r+m˙f u e l)⋅v e x i t−m˙a i r⋅v e n t r y T=(m˙a i r+m˙f u e l)⋅v e x i t−m˙a i r⋅v e n t r y
The exit speed v e x i t v e x i t of a turbojet engine is almost constant over flight speed (relative to the engine of course), so as the engine accelerates, a larger entry impulse must be subtracted from a nearly constant exit impulse. Thrust drops slightly over speed.
At higher Mach numbers, precompression from the ram effect at the intake raises the pressure level (and hence the mass flow m˙a i r m˙a i r) inside the engine, so it will develop more thrust than in static conditions. This effect causes the thrust line to bend upwards at higher speed, and since the precompression grows nonlinearly with speed, the initial drop in thrust is soon reversed. Of course, now the fuel mass flow m˙f u e l m˙f u e l will grow in the same way, so the fuel efficiency (thrust per fuel used) will continue to drop as speed increases.
Only when flight speed approaches the exit speed of the jet will thrust go down again. The typical exit speed of a turbojet is easily supersonic, therefore this type of engine is well suited for supersonic flight.
max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N
Here you have two effects combining to lower thrust. One is the reduction in the difference between entry and exit speed. This is more pronounced in a turbofan engine because the bypass flow will be accelerated much less than the core flow, and a higher flight speed will cause a proportionally bigger drop in thrust.
The second effect comes from the difference in air density between ground and cruise: Air density at a typical cruise altitude of 35,000 ft is only 0.38 kg/m³ or 31% of the air density at sea level. The original source for the cruise thrust number does not say for which altitude the figure is valid, but you can be sure that it is for about one third of ground density. Mass flow m˙a i r m˙a i r is directly proportional with ambient density, and both effects combine. However, most sources give only a drop to a quarter of static thrust - the last table in the linked answer looks like someone mixed the values for the CFM56-5A and the CFM56-5C.
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edited Mar 27, 2019 at 22:01
answered Mar 29, 2018 at 21:19
Peter KämpfPeter Kämpf
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Just to clarify, by a turbojet's exit speed being "supersonic" is that meaning within the exhaust (with a higher speed of sound because it's hot) the actual flow is supersonic, or is it subsonic but faster than the speed of sound in the surrounding air?Talisker –Talisker 2019-02-24 22:04:42 +00:00 Commented Feb 24, 2019 at 22:04
1 @Talisker: Yes, supersonic means more than the speed of sound of this hot air. There are supersonic aircraft with subsonic engine exit speed, and that only works because of the higher speed of sound in hot air, but those are the exception.Peter Kämpf –Peter Kämpf 2019-02-25 06:18:26 +00:00 Commented Feb 25, 2019 at 6:18
2 @AbanobEbrahim: Yes, very good question! The exit speed is approx. constant relative to the aircraft, so this puts an upper limit on the maximum speed which can be achieved.Peter Kämpf –Peter Kämpf 2019-03-27 22:00:33 +00:00 Commented Mar 27, 2019 at 22:00
1 @PeterKämpf: Thank you. I'm new here on Aviation and I already admire your awesome answers. But just to make sure I understand you correctly, let's assume the exhaust velocity when the engine is stationary is 500 m/s m/s. Now by approx. constant you mean that when the aircraft is moving at 200 m/s m/s, the exhaust speed relative to the aircraft is still 500 m/s m/s but to someone standing on the ground the exhaust speed is just 300 m/s m/s in the opposite direction of the aircraft. So is that correct?Abanob Ebrahim –Abanob Ebrahim 2019-03-27 23:24:50 +00:00 Commented Mar 27, 2019 at 23:24
1 @AbanobEbrahim: Yes, that is correct. For the nitpickers: The actual numbers might vary a bit between the different scenarios.Peter Kämpf –Peter Kämpf 2019-03-28 06:17:38 +00:00 Commented Mar 28, 2019 at 6:17
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Air density or pressure altitude must be factored in when comparing engine thrust figures. At cruise, FL400, air density will be about one fifth. Max thrust (or max HP for piston engines) will also be one fifth, as will air resistance/parasite drag. Air mass flow will be one fifth, so will fuel flow/burn. Even lift will be one fifth for a given air speed (TAS).
I assume you meant TAS in your question, however, if we consider IAS/CAS, that is a different story. At FL400, you would have numbers like Mach 0.82, 470kts TAS and 250kts IAS. Airliners rarely do level flight except at cruise altitude, however it would be possible to maintain 250kts IAS/CAS with 20% of static thrust (give or take) at altitudes lower than cruise, such as in a hold.
Playing with these might be helpful:
IAS/TAS vs Altitude:
Engine power vs Altitude:
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edited Mar 4, 2020 at 21:42
answered Mar 3, 2020 at 19:19
Chuck MacLeanChuck MacLean
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3 I'm not sure that this answers the question that was asked. It's not bad information, but it's about how thrust varies with altitude, rather than what the question asks, which is why thrust varies with airspeed, per se. Also, I'd question your assertion that an airliner could ever maintain speed with only 20% of available power. Maybe on an incredibly light 747 or A-380, but on the 737's that I'm familiar with, that statement seems unsupportable.Ralph J –Ralph J♦ 2020-03-03 20:44:33 +00:00 Commented Mar 3, 2020 at 20:44
Ralph J, you are misunderstanding me if you say "20% of available power". I am addressing the question which references one fifth of static thrust or 29,360N. By definition of what IAS is and what it implies, how can anyone take issue with my statement? If thrust is 29,360N in cruise at a particular indicated airspeed, both will be approximately the same for any other altitude (except of course for speeds and altitudes out of limits).Chuck MacLean –Chuck MacLean 2020-03-04 23:57:12 +00:00 Commented Mar 4, 2020 at 23:57
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2Why isn't horsepower used as a measure of power output for jet engines?
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10518 | https://www.reddit.com/r/AdvancedRunning/comments/1i391tx/how_much_does_weight_affect_times_really/ | How much does weight affect times really? : r/AdvancedRunning
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•8 mo. ago
assholesplinters
How much does weight affect times really?
Health/Nutrition
So, I've seen wildly varying answers on this, from 1 seconds per mile per pound to Runners world claiming .064% per pound. Now, I realize all of their methodologies, and studies are done differently and on different people but Im curious if there's a semi reliable formula out there or if ultimately weight loss and speed are just side affects of consistent effort? For example. At the moment, I'm an out of shape former college swimmer running ~44 for a 10k. So if I were to drop 50 pounds and get to my competition weight of 180 at 1 seconds per mile per per pound that'd mean I'd be running a 39:10 or at the other end of the spectrum at .064% per pound I'd be running a 30min 10k which doesn't quite seem in the cards 😆
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10519 | https://archive.csac.history.wisc.edu/The-Plans-at-the-Convention.htm | Skip to main navigation
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The Philadelphia Convention
Plans of Government Proposed at the Convention
Plans of Government Proposed at the Convention
At the very outset the Convention faced the issue of whether or not to work within the existing constitutional framework of the United States. If the Convention decided to do so, any amendments to the Articles would have to be approved by Congress and ratified unanimously by the legislatures of the thirteen states.
The Convention decided instead to work outside the constitutional framework. Governor Edmund Randolph proposed this constitutional revolution on 29 May in his speech presenting the Virginia Resolutions.
For the next two weeks, delegates debated the merits of the Virginia Plan. Eventually, the Convention adopted a resolution that agreed to abandon the existing federal government and create a national government. This fundamental decision was not challenged directly until 13 June. At that point, William Paterson of New Jersey asked that consideration of the plan be delayed and that time be given “to digest one purely federal. . . .”
Through these early stages of the convention, opposition to the Virginia Resolutions had been mounting steadily. Delegates from the small states united with other delegates who were opposed to an all-powerful central government and who wanted to retain the federal structure of the Articles of Confederation.
The Convention debated the Patterson’s New Jersey Amendments on the 15th, 16th, and 18th of June. On the 19th the Convention rejected the Amendments and voted to accept the revised Virginia Resolutions by a vote of seven states to three, with one state divided. However, the next day the Convention made a concession to the opponents of a national government.
Two additional proposals offered for consideration at the convention. On 29 May the Journals of the Convention record that Charles Pinckney “laid before the House . . . the draft of a federal government to be agreed upon between the free and independent states of America.” The plan was not discussed by the Convention but was turned over to the Committee of Detail on 24 July. The document written by Pinckney has never been found, but a document in James Wilson’s handwriting has been identified as a synopsis.
On 18 June, Alexander Hamilton offered his proposal. Hamilton essentially believed that both the Virginia Resolutions and the New Jersey Amendments were inadequate. He instead proposed a system that featured life terms for the executive and the members in the upper house of the legislature. Hamilton’s plan would have essentially subsumed the states into administrative units of the national government.
For a full analysis of the proposed plans at the Constitutional Convention, an essay from our CDR, Vol. 1 will provide information on the various plans debated in Philadelphia.
The Virginia Resolutions, 29 May 1787 (pdf)
The Virginia delegates arrived in Philadelphia early, talked with other delegates as they arrived, met regularly with one another, and drafted resolutions setting forth the broad principles upon which a new constitution should be based. Not all of the Virginia delegates agreed with the principles, as the later debates in the Convention were to reveal. But, the fifteen resolutions were presented to the Convention on 29 May by the Governor of Virginia, Edmund Randolph. No evidence exists as to who wrote the resolutions. Randolph’s original manuscript has never been located.
The Pinckney Plan, 29 May 1787 (pdf)
On 29 May the Journals of the Convention record that Charles Pinckney “laid before the House … the draft of a federal government to be agreed upon between the free and independent states of America.” The plan was not discussed by the Convention but was turned over to the Committee of Detail on 24 July. The document written by Pinckney has never been found, but a document in James Wilson’s handwriting has been identified as a synopsis.
The New Jersey Amendments, 15 June 1787 (pdf)
The amendments proposed by William Paterson of New Jersey were far more than “New Jersey amendments.” They represented the views of the delegates from the small states and of those delegates who were opposed to a national government or who at least insisted that the central government must retain some of the federal character of the Articles of Confederation. Nevertheless, they agreed that the central government needed more power and the proposed amendments provided for such power. On 19 June the Convention voted to reject the New Jersey Amendments and accept the amended Virginia Resolutions.
The Hamilton Plan, 18 June 1787 (pdf)
On 18 June, Alexander Hamilton in a speech proposed a very powerful national government. Hamilton essentially believed that both the amended Virginia Resolutions and the New Jersey Amendments were inadequate, particularly the latter.
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10520 | https://web.uvic.ca/~mcindoe/423/ligands.pdf | Table 1 Electron counting for commonly encountered ligands nVE Ligands 1-electron ligands Ligands which in the free state would be a radical: H, F, Cl, Br, I, OH, NR2, OR, SR, CN, N3, NCS, bent NO CR3 – alkyl, aryl, alkenyl, alkynyl, formyl, acyl 2-electron ligands Ligands which in the free state would have an even number of valence electrons: OH2, NH3, ethers, amines, thioethers, phosphines CO, CNR, CS, CR2, C=CR2 Alkenes, alkynes; molecules which bind side-on through a multiple bond: O2, SO2, CS2, RP=PR, R2Si=CR2 3-electron ligands Linear NO, nitride (N) Ligands which can be subdivided into a combination of 1 VE and 2 VE donations: H2C H C CH2• R C RC CR M X M η3-cyclopropenyl η3-allyl µ-X 4-electron ligands diene η4-cyclobutadienyl 5-electron ligands η5-cyclopentadienyl N N N N N N B H tris(pyrazolyl)borate 6-electron ligands 7-electron ligands 8-electron ligands |
10521 | https://en.wikipedia.org/wiki/Element_(mathematics) | Jump to content
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Contents
(Top)
1 Sets
2 Notation and terminology
3 Examples
4 Cardinality of sets
5 Formal relation
6 See also
7 References
8 Further reading
Element (mathematics)
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From Wikipedia, the free encyclopedia
Any one of the distinct objects that make up a set in set theory
For elements in category theory, see Element (category theory).
For the components of the Periodic Table, see Chemical element.
| |
| A request that this article title be changed to Membership (mathematics) is under discussion. Please do not move this article until the discussion is closed. |
In mathematics, an element (or member) of a set is any one of the distinct objects that belong to that set. For example, given a set called A containing the first four positive integers (), one could say that "3 is an element of A", expressed notationally as .
Sets
[edit]
Writing means that the elements of the set A are the numbers 1, 2, 3 and 4. Sets of elements of A, for example , are subsets of A.
Sets can themselves be elements. For example, consider the set . The elements of B are not 1, 2, 3, and 4. Rather, there are only three elements of B, namely the numbers 1 and 2, and the set .
The elements of a set can be anything. For example the elements of the set are the color red, the number 12, and the set B.
In logic, a set can be defined in terms of the membership of its elements as . This basically means that there is a general predication of x called membership that is equivalent to the statement ‘x is a member of y if and only if, for all objects x, the general predication of x is identical to y, where x is a member of the domain of y.’ The expression x ∈ 𝔇y makes this definition well-defined by ensuring that x is a bound variable in its predication of membership in y.
In this case, the domain of Px, which is the set containing all dependent logical values x that satisfy the stated conditions for membership in y, is called the Universe (U) of y. The range of Px, which is the set of all possible dependent set variables y resulting from satisfaction of the conditions of membership for x, is the power set of U such that the binary relation of the membership of x in y is any subset of the cartesian product U × 𝒫(U) (the Cartesian Product of set U with the Power Set of U).
Notation and terminology
[edit]
The binary relation "is an element of", also called set membership, is denoted by the symbol "∈". Writing
means that "x is an element of A". Equivalent expressions are "x is a member of A", "x belongs to A", "x is in A" and "x lies in A". The expressions "A includes x" and "A contains x" are also used to mean set membership, although some authors use them to mean instead "x is a subset of A". Logician George Boolos strongly urged that "contains" be used for membership only, and "includes" for the subset relation only.
For the relation ∈ , the converse relation ∈T may be written
meaning "A contains or includes x".
The negation of set membership is denoted by the symbol "∉". Writing
means that "x is not an element of A".
The symbol ∈ was first used by Giuseppe Peano, in his 1889 work Arithmetices principia, nova methodo exposita. Here he wrote on page X:
Signum ∈ significat est. Ita a ∈ b legitur a est quoddam b; …
which means
The symbol ∈ means is. So a ∈ b is read as a is a certain b; …
The symbol itself is a stylized lowercase Greek letter epsilon ("ϵ"), the first letter of the word ἐστί, which means "is".
Character information
| Preview | ∈ | ∉ | ∋ | ∌ |
| Unicode name | ELEMENT OF | NOT AN ELEMENT OF | CONTAINS AS MEMBER | DOES NOT CONTAIN AS MEMBER |
| Encodings | decimal | hex | dec | hex | dec | hex | dec | hex |
| Unicode | 8712 | U+2208 | 8713 | U+2209 | 8715 | U+220B | 8716 | U+220C |
| UTF-8 | 226 136 136 | E2 88 88 | 226 136 137 | E2 88 89 | 226 136 139 | E2 88 8B | 226 136 140 | E2 88 8C |
| Numeric character reference | ∈ | ∈ | ∉ | ∉ | ∋ | ∋ | ∌ | ∌ |
| Named character reference | ∈, ∈, ∈, ∈ | ∉, ∉, ∉ | ∋, ∋, ∋, ∋ | ∌, ∌, ∌ |
| LaTeX | \in | \notin | \ni | \not\ni or \notni |
| Wolfram Mathematica | [Element] | [NotElement] | [ReverseElement] | [NotReverseElement] |
Examples
[edit]
Using the sets defined above, namely A = {1, 2, 3, 4}, B = {1, 2, {3, 4}} and C = {red, green, blue}, the following statements are true:
2 ∈ A
5 ∉ A
{3, 4} ∈ B
3 ∉ B
4 ∉ B
yellow ∉ C
Cardinality of sets
[edit]
Main article: Cardinality
The number of elements in a particular set is a property known as cardinality; informally, this is the size of a set. In the above examples, the cardinality of the set A is 4, while the cardinality of set B and set C are both 3. An infinite set is a set with an infinite number of elements, while a finite set is a set with a finite number of elements. The above examples are examples of finite sets. An example of an infinite set is the set of positive integers {1, 2, 3, 4, ...}.
Formal relation
[edit]
As a relation, set membership must have a domain and a range. Conventionally the domain is called the universe denoted U. The range is the set of subsets of U called the power set of U and denoted P(U). Thus the relation is a subset of U × P(U). The converse relation is a subset of P(U) × U.
See also
[edit]
Identity element
Singleton (mathematics)
References
[edit]
^ Weisstein, Eric W. "Element". mathworld.wolfram.com. Retrieved 2020-08-10.
^ Eric Schechter (1997). Handbook of Analysis and Its Foundations. Academic Press. ISBN 0-12-622760-8. p. 12
^ George Boolos (February 4, 1992). 24.243 Classical Set Theory (lecture) (Speech). Massachusetts Institute of Technology.
^ a b Kennedy, H. C. (July 1973). "What Russell learned from Peano". Notre Dame Journal of Formal Logic. 14 (3). Duke University Press: 367–372. doi:10.1305/ndjfl/1093891001. MR 0319684.
^ "Sets - Elements | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2020-08-10.
Further reading
[edit]
Halmos, Paul R. (1974) , Naive Set Theory, Undergraduate Texts in Mathematics (Hardcover ed.), NY: Springer-Verlag, ISBN 0-387-90092-6 - "Naive" means that it is not fully axiomatized, not that it is silly or easy (Halmos's treatment is neither).
Jech, Thomas (2002), "Set Theory", Stanford Encyclopedia of Philosophy, Metaphysics Research Lab, Stanford University
Suppes, Patrick (1972) , Axiomatic Set Theory, NY: Dover Publications, Inc., ISBN 0-486-61630-4 - Both the notion of set (a collection of members), membership or element-hood, the axiom of extension, the axiom of separation, and the union axiom (Suppes calls it the sum axiom) are needed for a more thorough understanding of "set element".
| v t e Mathematical logic |
| General | Axiom + list Cardinality First-order logic Formal proof Formal semantics Foundations of mathematics Information theory Lemma Logical consequence Model Theorem Theory Type theory |
| Theorems (list) and paradoxes | Gödel's completeness and incompleteness theorems Tarski's undefinability Banach–Tarski paradox Cantor's theorem, paradox and diagonal argument Compactness Halting problem Lindström's Löwenheim–Skolem Russell's paradox |
| Logics | | | | --- | | Traditional | Classical logic Logical truth Tautology Proposition Inference Logical equivalence Consistency + Equiconsistency Argument Soundness Validity Syllogism Square of opposition Venn diagram | | Propositional | Boolean algebra Boolean functions Logical connectives Propositional calculus Propositional formula Truth tables Many-valued logic + 3 + finite + ∞ | | Predicate | First-order + list Second-order + Monadic Higher-order Fixed-point Free Quantifiers Predicate Monadic predicate calculus | |
| Set theory | | | | Set + hereditary Class (Ur-)Element Ordinal number Extensionality Forcing Relation + equivalence + partition Set operations: + intersection + union + complement + Cartesian product + power set + identities | | Types of sets | Countable Uncountable Empty Inhabited Singleton Finite Infinite Transitive Ultrafilter Recursive Fuzzy Universal Universe + constructible + Grothendieck + Von Neumann | | Maps and cardinality | Function/Map + domain + codomain + image In/Sur/Bi-jection Schröder–Bernstein theorem Isomorphism Gödel numbering Enumeration Large cardinal + inaccessible Aleph number Operation + binary | | Set theories | Zermelo–Fraenkel + axiom of choice + continuum hypothesis General Kripke–Platek Morse–Kelley Naive New Foundations Tarski–Grothendieck Von Neumann–Bernays–Gödel Ackermann Constructive | |
| Formal systems (list),language and syntax | | | | Alphabet Arity Automata Axiom schema Expression + ground Extension + by definition + conservative Relation Formation rule Grammar Formula + atomic + closed + ground + open Free/bound variable Language Metalanguage Logical connective + ¬ + ∨ + ∧ + → + ↔ + = Predicate + functional + variable + propositional variable Proof Quantifier + ∃ + ! + ∀ + rank Sentence + atomic + spectrum Signature String Substitution Symbol + function + logical/constant + non-logical + variable Term Theory + list | | Example axiomaticsystems (list) | of true arithmetic: + Peano + second-order + elementary function + primitive recursive + Robinson + Skolem of the real numbers + Tarski's axiomatization of Boolean algebras + canonical + minimal axioms of geometry: + Euclidean: - Elements - Hilbert's - Tarski's + non-Euclidean Principia Mathematica | |
| Proof theory | Formal proof Natural deduction Logical consequence Rule of inference Sequent calculus Theorem Systems + axiomatic + deductive + Hilbert - list Complete theory Independence (from ZFC) Proof of impossibility Ordinal analysis Reverse mathematics Self-verifying theories |
| Model theory | Interpretation + function + of models Model + equivalence + finite + saturated + spectrum + submodel Non-standard model + of non-standard arithmetic Diagram + elementary Categorical theory Model complete theory Satisfiability Semantics of logic Strength Theories of truth + semantic + Tarski's + Kripke's T-schema Transfer principle Truth predicate Truth value Type Ultraproduct Validity |
| Computability theory | Church encoding Church–Turing thesis Computably enumerable Computable function Computable set Decision problem + decidable + undecidable + P + NP + P versus NP problem Kolmogorov complexity Lambda calculus Primitive recursive function Recursion Recursive set Turing machine Type theory |
| Related | Abstract logic Algebraic logic Automated theorem proving Category theory Concrete/Abstract category Category of sets History of logic History of mathematical logic + timeline Logicism Mathematical object Philosophy of mathematics Supertask |
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| v t e Set theory |
| Overview | Set (mathematics) |
| Axioms | Adjunction Choice + countable + dependent + global Constructibility (V=L) Determinacy + projective Extensionality Infinity Limitation of size Pairing Power set Regularity Union Martin's axiom Axiom schema + replacement + specification |
| Operations | Cartesian product Complement (i.e. set difference) De Morgan's laws Disjoint union Identities Intersection Power set Symmetric difference Union |
| Concepts Methods | Almost Cardinality Cardinal number (large) Class Constructible universe Continuum hypothesis Diagonal argument Element + ordered pair + tuple Family Forcing One-to-one correspondence Ordinal number Set-builder notation Transfinite induction Venn diagram |
| Set types | Amorphous Countable Empty Finite (hereditarily) Filter + base + subbase + Ultrafilter Fuzzy Infinite (Dedekind-infinite) Recursive Singleton Subset · Superset Transitive Uncountable Universal |
| Theories | Alternative Axiomatic Naive Cantor's theorem Zermelo + General Principia Mathematica + New Foundations Zermelo–Fraenkel + von Neumann–Bernays–Gödel - Morse–Kelley + Kripke–Platek + Tarski–Grothendieck |
| Paradoxes Problems | Russell's paradox Suslin's problem Burali-Forti paradox |
| Set theorists | Paul Bernays Georg Cantor Paul Cohen Richard Dedekind Abraham Fraenkel Kurt Gödel Thomas Jech John von Neumann Willard Quine Bertrand Russell Thoralf Skolem Ernst Zermelo |
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10522 | https://openknowledge.fao.org/server/api/core/bitstreams/633151db-9c5f-401b-a5ac-e5a6ad222981/content | CODE OF FOREST HARVESTING PRACTICES
REPUBLIC OF LIBERIA FORESTRY DEVELOPMENT AUTHORITY Monrovia, 2007 CODE OF FOREST HARVESTING PRACTICES
REPUBLIC OF LIBERIA FORESTRY DEVELOPMENT AUTHORITY Monrovia, 2007
I3564E 2Code of Forest Harvesting Practices
TA bl E OF CONTENTS
1 Introduct Ion 4
1.1 Towards best practice timber harvesting 41.2 Development of the Code 41.3 Objectives 52 PLA nn InG rEQ uIrEME nt S 5
2.1 Forest Management Planning 52.2 Pre-Harvest Enumeration 6
3 EX cLuSI on ArEAS A nd BuFFE r Str IPS 6
3.1 Exclusion Areas 63.2 Buffer Strip Protection 83.3 Management of Exclusion Areas and Buffer Strips 8
4 con Struct Ion Wor KS F or LoGGI nG 10
4.1 Roads 10 4.2 Drainage 15 4.3 Road maintenance 17 4.4 Watercourse Crossings 17 4.5 Bridges 19 4.6 Log landings 22 4.7 Borrow Pits 24
5 L oGGI nG oPE rAtIon S 24
5.1 Tree Marking 24 5.2 Skid trails 25 5.3 Felling 27 5.4 Log Preparation 28 5.5 Skidding 29 5.6 Truck Loading, Transporting and unloading 30 5.7 Road Transport 30 5.8 Concentration of Harvesting Operations: 31 5.9 Weather Limitations on Logging Operations 31
6 non-t IMBE r For ES t Product ( nt FP) M AnAGEME nt Pr Act IcES 33
6.1 NTFP Sustainable Management Practices 33
7 P oSt-HA rVES t Act IVI tIES 33
7.1 Block closure 33 7.2 Road maintenance and closure 34 7.3 Log landings 34 7.4 Skid Trails 34 7.5 Borrow Pits 35 7.6 Field Camps 35 3Republic Of Liberia - Forestry Development Authority
8 oPE rAtIon AL HYGIE nE 35
8.1 Workshop & Wood Processing Facilities 35 8.2 Field Servicing and Maintenance 35 8.3 Hazardous Chemical Handling and Storage 36 8.4 Waste Management 37 8.5 Wildlife management in contract areas 37 8.6 Alternative protein sources for wildlife protection in concession areas 38
9 cAMP HYGIE nE 38
9.1 Camp Design 38 9.2 Water Supply and Domestic Waste Water 38 9.3 Wastes and Refuse Disposal 38 9.4 Camp Construction Requirement 39 9.5 Additional Facilities 39
10 HEAL tH A nd SAFE tY 39
10.1 General 39 10.2 Emergency Rescue 40 10.3 Felling 40 10.4 Heavy Machinery 41 10.5 Vehicles 42 10.6 Hazardous Chemicals 43 10.7 Workshop and Log Yard Workers 43 10.8 Fire Precautions 43
11 HA rVES t contro L, M on Itor InG A nd ASSESSME nt 44 12 S oc IAL ISS uES 44
12.1 Land and forest use rights and responsibilities 44 12.2 Work place relations, rights and responsibilities 44 13 GL oSSA rY 46 14 APPE nd IX 54 4Code of Forest Harvesting Practices
1 INTROD uCTION
1.1 Towards best practice timber harvesting
The Forestry Development Authority (FDA) has developed this Code of Forest Harvesting Practices to provide a clear set of guidelines to help foresters and logging companies select practices to be followed when carrying out harvesting operations under forest management contracts (FMC) and timber sales contracts (TSC). The purpose of these guidelines is to prescribe a code of timber harvesting practice, to which the FDA requires all contract holders to adhere. The planning requirements for forest areas under FMCs and TSCs are different with respect to some of the activities. However, the requirements regarding good logging practice remain the same and these are described in this code. Practices that conform to the regulations or guidelines should, at least in theory, achieve a desired outcome
such as the harvest of commercial timber from a specified area of forest in a way that meets standards for
sustainable forest management. Sustainable forest management recognizes the importance of all products and services provided by the forest, including timber, water quality, soils, biodiversity, and the livelihood of people that live or work in the forest. Liberia’s response to these concerns has been to develop principles, policies and guidelines for improved
forest management and timber harvesting practices. This is reflected in the new forest policy, forest
legislation, forest management guidelines and codes of practice. The goal of sustainable forest management cannot be reached unless improved harvesting practices are widely adopted by loggers. The emphasis in this document is on the harvesting practices themselves. If these are improved, then it is more likely that the ecological, environmental and cultural value of the country’s forests will be maintained while at the same time providing for sustainable yields of commercial timber from those forests The Code itself will not ensure sustainable forest management. However, with effective implementation through reduced-impact logging guidelines, participation in the planning process by all stakeholders, and integration with other sustainable forest management instruments the Code will assist in minimizing negative i mpacts of timber harvesting.
1.2 Development of the Code
The Code contains practices and sets standards, which have been developed based on research and
practical experience in tropical forests. Specifically, the FAO Model Code of Forest Harvesting Practice
and Regional Code of Practice for Reduced-Impact Forest Harvesting in Tropical Moist Forests of West and Central Africa form the basic framework for this code. The Code will not be applied retroactively. However, contract holders operating in areas that have been logged before should be aware that existing roads, log landings, bridges and other infrastructure that do not meet current Code provisions will have to be upgraded according to this Code. 5Republic Of Liberia - Forestry Development Authority
The Code will be reviewed on a regular basis by the FDA following consultations with the timber industry and other interested parties.
1.3 Objectives
The objectives of the Code of Forest Harvesting Practices are to:
• Provide forest operators with a set of guidelines and standards for improved forest harvesting practices that improve standards of logging/utilization and reduce environmental impacts, and so contributing to the conservation of forests through their wise use.
• Promote the health and safety of forest workers.
• Provide a framework for effective control of timber harvesting operations with predetermined guidelines and benchmark. There are two different types of statements in the code: ‘shall’ (in some cases phrased as an imperative) and ‘should’ statements. The ‘shall’ statements are to be applied in a practical manner to all logging operations. The ‘should’ statements show the desirable practice for most situations and should be interpreted taking account of local conditions
2 PlANNING REQ uIREMENTS 2.1 Forest Management Planning
Broad scale planning is carried out by the FDA to determine which areas may be allocated as commercial forest areas, protected areas or multiple use forest areas. Forest Management Plans are required for all forest management contracts. There are three levels of planning required in each forest management plan:
• operational planning covers the background, conditions and plans for the entire contract area;
• detailed planning for operations covers activities to be carried out within a 5-year period; and
• annual planning covers the previous year’s activities and operational plans for the following year including silvicultural prescription/treatments. Timber Sales contract holders are required to prepare at least the Annual Planning process. All these planning activities shall be made in accordance with the Code of Practice for Timber Harvesting.
Good planning at the operational level has been shown to significantly reduce both operational costs and
the environmental impact of timber harvesting.
Two types of field inventories shall be required for proper forest management planning. First, the
contractor shall complete a general inventory of the entire contract area to provide data on the sustainable forest management plan, rotation, and annual allowable cut volumes per species. Second,
the contractor shall complete pre-harvesting enumeration to define the harvestable trees for the
annual harvesting plan. 6Code of Forest Harvesting Practices
2.2 Pre-Harvest Enumeration
Pre-harvest enumeration shall be completed in advance of, and must receive approval from the Authority, cutting any trees at the start of each harvesting period. Regulations and the individual Timber Sale and
Forest Management Contracts define the timing and process for completing and pre-harvesting procedures.
However, pre-harvest enumeration shall:
• Facilitate harvest planning;
• Provide data to develop Silvicultural treatments;
• Assist in determining which species are locally rare or regenerate poorly under current harvesting regimes. Pre-harvest enumeration shall conform to the following standards:
• All blocks shall be surveyed for 100% stocking of trees 50 cm dbh using regular grid (north-south/ east-west) in each kilometer square blocks to develop tree location map;
• On FMC, a proper permanent grid for the entire five-year section of the contract area shall be opened,
with all gridlines permanently marked and kept open by clearance of undergrowth when necessary.
• Temporary gridlines at 20m interval from the permanent gridlines should be established for each Km-square block in order to facilitate the work of the inventory crews.
Moreover, the following parameters shall be assessed and recorded during the field inventory for each
marked tree:
• Location and the number of the tree;
• Botanical identification of tree species;
• Girth or diameter at dbh
• Quality features and visible defects a high for
• Location of size of any utilizable dead trees
• Topographic features that may influence harvest planning.
3 EXC lu SION AREAS AND bu FFER STRIPS 3.1 Exclusion Areas
Exclusion areas include (a) protected areas, (b) protected animal species habitat (c) protected tree species (ex. IUCN Red List Trees, etc) (d) sites that are especially susceptible to degradation (e) watercourses and (f) cultural and customary tenure areas. Buffer strips of different widths will be used to protect such areas.
3.1.1 Protected Areas
Protected areas are those areas that must be excluded from logging. Protected areas include: Conservation areas (e.g. biodiversity reserves) Declared protected areas under national legislation Areas of cultural importance (such as historical, archaeological and spiritual sites) Areas required for community needs (which may include cultural sites as well as settlements and farms. 7Republic Of Liberia - Forestry Development Authority
3.1.2 Protected Animal Species Habitat:
Areas of high protected animal species habitat, high density of protected species, and habitat trees – define
habitat tree – any tree with high potential or obvious characteristics for providing shelter or habitat for wildlife.
3.1.3 Protected Tree Species
These are mostly the CITES and IUCN and national red list species including rare, endangered and mother tree/seed species.
3.1.4 Sites susceptible to degradation
No harvesting operations shall be carried out on areas with slopes gradient above 30%.
3.1.5 Watercourses
Watercourses are natural channels, which carry water for some period in most years. Flows may be periodic or permanent. Watercourses include natural springs, rivers, streams or creeks, gullies, swamps and lakes. The width of a watercourse is to be measured from bank-to-bank during normal wet season (i.e. the period
of peak water flow), and may include a flood plain area which may be a swamp.
3.1.6 Cultural and Customary tenure areas
Areas that fall under cultural and customary tenure of local communities shall be excluded from the harvesting plan area and in some cases depending on the size and location of the cultural or customary tenure they can be removed from the contract all together. See section on Social Issues for more details.
Natural springs A natural spring is a place where groundwater flows out of the ground. A
spring may be intermittent or continuous (perennial).
Creeks
Creeks are watercourses in which water may flow or pond for more than
six months in most years, and whose beds are of stony, gravely or exposed bedrock materials.
Streams
Streams are watercourses in which water may flow or pond for about six
months in most years, and whose beds are of soil and whose banks are often covered with vegetation.
Rivers Rivers are those watercourses where water flows all year around in most years.
Gullies
Gullies are steep-sided channels. Their beds are normally of soil and may be
covered with vegetation. Water will flow or pond for less than six months in
most years
Swamps Swamps have standing surface water for six months or more in most years
Lakes Lakes have surface water present all year round for most years
TABLE 1SHOWS
THE RANGE OF
WATERCOURSE
THAT ARE DEF
AS ExCLUSION
AREAS .
8Code of Forest Harvesting Practices
3.2 buffer Strip Protection
Buffer strips are required whether or not an exclusion feature is identified on available maps. Field inspection
during pre-harvest inventory work will identify the areas that require buffer strips before forest operations start, and depending on the type of feature and how wide the strip shall be.
Type Minimum width
Conservation and declared protected areas or other buffer boundaries 50 meters Cultural, spiritual and historic sites 100 meters Villages, farms, settlements 100 meters Rare, endangered, mother/seed trees 10 meter radius Watercourses: Width <10m Width < 20m Width < 40 m 15 m on each bank 20 m on each bank 30 m on each bank Natural springs 50 meters circumference Lakes , 25 meters from the waters edge Creeks and streams 15 meters from the waters edge Gullies 15 meters from the waters edge Lakes 25 meters from the waters edge Swamps and other wetlands 15 meters from the waters edge
River banks are defined as the permanent edge of the river regardless of the seasonal recession of the water
levels, whereas the waters edge is determined from where the water is at that given point in time.
3.3 Management of Exclusion Areas and buffer Strips
3.3.1 Exclusion areas and their buffer strips shall be managed as follows
• No trees shall be felled within exclusion areas or their buffer strips except merchantable species that grow in the wetland areas (e.g. niangon, abura, leucinia etc.) which shall be extracted by using 25-meter cable extending to the log from the crawler tractor standing on the skid trail and only
upon approval from the Authority per tree harvested (with an FDA officer present) and only
during dry season when wetlands are dry.
TABLE 2BUFFER STRIP
DISTANCES
9Republic Of Liberia - Forestry Development Authority
• Machine access within exclusion areas and their buffer strips is not allowed, except at designated watercourse crossing points, which should be by the shortest possible distance.
• No earthworks, or spoils from earthworks, shall end up in a exclusion area or its buffer strip
• No harvesting debris shall be pushed into exclusion areas or their buffer strips.
• Trees shall be felled away from buffer strips and watercourses. If it is not possible to fell the tree away from the buffer strip or watercourse it shall not be harvested.
• Where trees inadvertently fall into a watercourse or its buffer strip, the head as well as any accompanying logging debris should be pulled clear, unless unacceptable damage to the bank or buffer strip is likely to occur.
FIGURE 1BUFFER STRIPS
ALONG WATER
COURSE
10 Code of Forest Harvesting Practices
4 CONSTR uCTION WORKS FOR lOGGING 4.1 Roads
Roads provide needed access to the forest. At the same time, roads can produce significant amounts of
sediment and can be one of the greatest adverse impacts on the local environment, on water quality and
on aquatic life. Roads can produce significant erosion, cause gullies, and have an impact on groundwater,
wildlife and vegetation. Road planning is key to ensuring that a road meets the current needs of the user, that it is not overbuilt, that it minimizes impacts on the environment and to the people along the road. A well planned, located, designed and constructed road will be cost-effective in the long term by preventing road failure, eliminating repair needs, and reducing maintenance. Roads should be planned to minimise the sum of skidding and road construction impacts, which in turn
will also lead to cost minimisation. The most efficient spacing of roads can be derived by looking at the cost
tradeoffs between skidding distance and road spacing.
4.1.1 Road classification
a. Primary forest roads
These are permanent, all-weather roads that provide access from public roads to the FMC or TSC area.
They should be capable of carrying log volumes of about 2,000 m3 (≈ c. 60 truck loads) or more per week
and may be in service during the entire logging operation of a 5-year management plan (FMC) or during the operations of a TSC.
b. Secondary forest roads
These are roads that provide access to a logging compartment, connecting feeder roads and log landings to primary roads. They will carry log volumes of about 1,000 to 2,000 m3 per week and will be in service for1-2 years only. Secondary roads are often upgraded to primary roads as the logging operations proceed.
4.1.2 Road Planning
In planning the location of roads the following factors shall be considered:
• Primary and secondary forest roads shall avoid all protected and exclusion areas and their respective buffer strips.
• Roads shall be kept at least 40 m away from the edge of buffer strips (e.g. 55 m from the banks of creeks, the edge of a gullies), except at designated watercourse crossing points.
• Roads shall be located wherever possible on well-drained soils and slopes where drainage will move away from the road. Roads should, therefore, follow the natural terrain by conforming to the
contour, rolling the grade and minimizing the use of cuts and fills.
• Roads should be located on ridges as much as possible.
• Efforts should be made to avoid locations that require box cuts, to avoid steep and very flat terrain where drainage is more difficult to control, and unstable locations such as swamps, marshes, landslides, steep slopes, massive rock outcrops, flood plains, and highly erosive soils
• Efforts should be made to minimize the number of watercourse crossings 11 Republic Of Liberia - Forestry Development Authority
4.1.3 Survey requirements for inspection
After planning the proposed primary and secondary road network contractors shall request that FDA carries out
a field inspection. This inspection should be undertaken with community representatives when appropriate. The
contractors will survey and mark the center of roads on the ground, and all proposed watercourse crossings prior to the inspection, and shall remark as necessary after the FDA has given approval and prior to any construction.
The field inspection of the road network will be part of the block inspection for the approval of the yield.
a. Road Construction
Road construction costs are mostly influenced by the standard of road built, particularly road width, type of surfacing, and the steepness of the terrain. A road with cuts and fills on steep cross slopes greatly
increases the time of construction, amount of earthwork, the areas of clearing, and adds length to cross-drains and other drainage structures.
b. Timing of construction
The construction of primary and secondary forest roads shall take place in the dry season only, and construction should commence within 12 months before logging. Preliminary roadway clearing should take place within 1
month of final construction to reduce sedimentation from untrained or poorly drained sites.
c. Roadway construction
The construction of primary and secondary forest roadways shall take into account the following factors:
• Merchantable stems along the road reserve should be felled and preferably extracted before clearing.
FIGURE 2ROAD LOCATIO
AND DRAINAGE
12 Code of Forest Harvesting Practices
While clearing, trees should be pushed into the road reserve and not into the adjacent forest.
• Hazardous trees, which have a significant probability of falling onto the roadway, shall be removed
during construction
• Soil heaps, berms and debris stockpiling along the roadway are not permitted; instead topsoil
should be stockpiled for use in cut and fill batters and/or in borrow pits. Organic debris should not be used as fill.
• All road drainage works shall be completed before gravelling work commences.
• Gravelling is required for primary roads and is recommended for secondary roads. Road surfaces on bridge approaches, culverts and swampy terrains shall be graveled.
• All road fill and paving material shall be compacted. Minimum compacted gravel thickness is
15 cm depending on the type of soil as advised by the Authority. Softer areas should require additional thickness where as rocky areas may require less gravel.
• Minimum primary road spacing of 1200m within a compartment is highly recommended.
d. Side slopes
Although primary and secondary road may not be located on slopes of greater than 15 %, the side slops of roads may have a greater slope. In such circumstances additional construction is required to mitigate the impact of the roads:
• Side cut roads shall not be located on slopes greater than 40%.
• Full bench construction should be adopted instead of half bench construction on slopes steeper
than 25% (see fig. 3).
FIGURE 3
ALF AND FULL
BENCH
ONSTRUCTION
13 Republic Of Liberia - Forestry Development Authority
• Cuts and fills should be balanced in gentle terrain so that as much of the excavated material as is practical can be deposited in the roadbed fills sections (see fig. 4).
• The maximum fill for batter slope is 40%. All cut batter slopes should be benched at 3 m vertical
intervals.
• Trees at the top of steep cut batters should be cleared if erosion or wind-blow is liable to occur.
e. Road grades
Primary and secondary forest roads shall not be constructed with grades (slopes) of more than 10% and
15% respectively since on steep roads drainage is difficult to control. However:
• Steeper grades (up to 20%) for short sections (maximum 200 m) would be acceptable, if this
shortens construction significantly or reduces earthworks, provided that adequate drainage is
installed.
• Any two sections of road at maximum gradient should be separated by 100 m of level gradient.
f. Road widths
Every effort shall be made to minimize the width of forest roads. Road widths depend on the class of road, the type of soils and the forest through which the roads are to be constructed. The maximum road widths for forest roads on loam and clay soils are shown in Table 3. The maximum road widths for forest roads on sandy soils are shown in Table 4. The widths of roads on laterite soil should be intermediate between the limits given in the tables above.
Road class Limit of Clearing Limit of Roadway Limit of Road Bed Limit of Travel Way
Main road 25 20 9 6Secondary road 20 15 8 5
Road class Limit of Clearing Limit of Roadway Limit of Road Bed Limit of Travel Way
Main road 15 11 9 6Secondary road 15 10 8 5
FIGURE 4BALANCE CUTS
AND FILLS
TABLE 3MAxIMUM ROA
WIDTH FOR LOA
AND CLAY SOIL
TABLE 4MAxIMUM ROA
WIDTH FOR SA
SOIL
14 Code of Forest Harvesting Practices
• Primary and secondary roads on clay and loam soils should have trees removed alongside the road to allow sunlight onto the road to dry the surface quickly after rain. Roads on white sand soils should be protected from rain and direct sunlight by limiting clearing to the roadbed and ditches
to maintain trafficability.
• On primary and secondary roads, tree stumps should be grubbed on at least one side of the road
to allow for movement of tractors and other heavy machinery that would damage the road surface.
• Passing spots shall be provided on roads with a roadbed narrower than 7 m, at least every 500 m and at bridge approaches and hillcrests.
• Where side cutting is not necessary, earthworks should be limited to the width for the roadbed plus ditches (table drains) on either side.
g. Curves
On all primary and secondary roads curves shall be fitted to the topography; i.e. along the contour. Curve
widening is required on corners to allow for off-tracking of trailers. To increase vehicle stability on a bend, banking (raising) the outer part of the curve is recommended. Shoulders may need to be cleared on the
inside of the curve to obtain the required sight distance to ensure traffic safety. The minimum radius of
the curve is related to the visibility and the speed the vehicles will be traveling on the road. Guidelines for recommended minimum sight distance and minimum curve radius are shown in Table 5. 15 Republic Of Liberia - Forestry Development Authority
Road class Design speed (km/hr) Minimum sight distance (m) Minimum radius (m)
Primary road 70 100 50 Secondary road 50 60 30
4.2 Drainage
Drainage problems often cause the largest impacts from roads with regard to erosion, sedimentation, and degradation of water quality. Poor drainage will also lead to rutting, scouring and gullying, while standing water and seepage under the roadbed may lead to road failure; in all cases necessitating extensive maintenance or even expensive repairs or diversions. These aspects make road drainage the single most important aspect of road construction and maintenance.
4.2.1 General
In all phases of forest road preparation, adequate drainage shall be constructed to ensure the stability of the roads and so reduce their impact on the forest. Wherever practicable, permanent drainage should be installed in advance of other construction to keep works as dry as possible. Temporary drainage shall be
provided where there is likely to be a significant delay in installing permanent drainage
4.2.2 Methods of drainage
On both primary and secondary roads a crowned road surface shall be required. Additionally:
• Ditches (table drains) shall be installed alongside all roads, constructed to a minimum depth of 30 cm below the level of the crown of the road.
• Turnout drains (side drains/outlets) shall be constructed at specified spacing.
• Crossroad drainage (pipes or culverts) should be used where turnout drains are not possible.
TABLE 5
FOR MINIMUM
SIGHT DISTANC
AND CURVE
RADIUS
FIGURE 5TURNOUT DRAI
16 Code of Forest Harvesting Practices
• For roads constructed on highly erodible soils on slopes, drains may require special treatment such as lining with gravel or stones, log or rock bars, as well as stepping and frequent outlets to reduce scouring.
• Earth fills should have adequate drainage to prevent water build up and ponding behind the fill.
• Side cuts shall also be provided with catch drains along the topside to collect surface runoff
4.2.3 Spacing between drains
Turnout drains or culverts shall be constructed:
• At changes of slope
• Within 50 m of watercourse crossings Maximum drain spacing guidelines are shown in Table 6 below.
Gradient (%) Drain spacing (meters)
0-5 80 5-10 40 10-15 30
FIGURE 6ADDITIONAL
DRAINAGE
EqUIREMENTS
FIGURE 7RAIN SPACING TABLE 6
xIMUM DRAIN
SPACING
17 Republic Of Liberia - Forestry Development Authority
4.2.4 Drain out-flow
Ditches shall not drain directly into watercourses. Rather turnout drains (outlets) shall be installed at least 50 m before meeting a watercourse to divert water into the surrounding vegetation. Where turnout drains are not practicable, drainage diversion using culverts shall be used.
• All drains shall have stable outlets, protected by vegetation or rock or log barriers, particularly in
fill areas.
• Sumps or silt traps shall be constructed in ditches at all four corners of watercourse crossings.
• In steep terrain silt traps should be constructed at the end of turnout drains.
4.3 Road maintenance
For all primary and secondary forest roads maintenance shall be carried out regularly, and at least on an annual basis. The crown of the road surface and road shape shall be maintained to allow effective drainage. Additionally:
• Surfacing gravel or loam should not be pushed to the road edge or into drains.
• Soil, vegetation and other materials that would obstruct water flow shall be cleared from ditches.
• Turnout drains, culverts and bridges shall be kept clear and in a good working condition at all times.
• Bridge decking, foundations and sidewalls should be checked regularly.
• Any debris that has been pushed into the watercourse shall be removed.
• Water should flow freely under bridges, and silt traps should be cleaned regularly.
4.4 Watercourse Crossings
4.4.1 Types of Crossings
Table 7 below shows the classification of watercourse crossings. Note that corduroy with earth fill and/or
log clusters is not permitted for crossing any watercourse in any situation.
Type of crossing Description
Bridges Bridges shall be used for road crossings of all rivers and creeks. They may also be used for other watercourses. Culverts (or pipes) Culverts shall be used for crossing gullies and other waterways if bridges are not constructed. Fords Fords are not permitted on primary and secondary forest roads, only for skid trails, and they should be corduroyed.
FIGURE 8DRAIN DIVERSIO
BY TURNOUTS O
CULVERT
TABLE 7
CLASSIFICATIO
OF WATERCOU
CROSSING
18 Code of Forest Harvesting Practices
4.4.2 Construction of Watercourse Crossings a. Location of crossings
Watercourse shall always be crossed at right angles. Additional criteria for selecting the location of crossing points are that sites:
• Are immediately downstream of straight and stable watercourse sections
• Have easy high bank access
• Do not require deep box cuts
• Require minimum alteration to the high bank
b. Earthworks
In the construction of roads and bridges, river and creek beds shall not be filled in. All earthworks shall
be carried out so as to prevent soil from entering the watercourse, and all spoil shall be removed to outside
the buffer strip or alternatively should be placed in road fills. Watercourse buffer strip vegetation should be
retained to the edge of the crossing. Temporary crossings are permitted to allow equipment involved with the construction of the crossing to be moved to the other side, provided that:
• The width of the temporary crossing is limited to 4 meters.
• The temporary crossing is made on the final crossing alignment, so as to reduce disturbance to
watercourse banks and buffer strip vegetation During bridge construction oil, chemicals, excess concrete and other waste should not enter the watercourse.
FIGURE 9
OG CLUSTERS
TH EARTH FILL
OT ALLOWED
FIGURE 10 CROSS
ATERCOURSES
RIGHT ANGLES
19 Republic Of Liberia - Forestry Development Authority
4.5 bridges
All bridges on primary roads shall be, and all bridges on secondary roads should be, sufficiently elevated to allow wet season flood flows to pass without damage to the crossing or its foundation. The bridge span
should extend at least 120% of the width of the watercourse measured from bank to bank; i.e. extend beyond the river or creek channel by 10% on either side. All approaches shall have a straight and level alignment for a minimum of 20 meters on either side.
4.5.1 Bridge construction
In constructing timber bridges the following criteria shall be considered (Fig 11 illustrates two timber bridge types).
• All parts of timber bridges should be constructed using durable timbers with high strength property/density not less than 850kg/m3, such as Dahoma, Irvingia, Klainodoxa, Ekki, based on approval by the Authority. Decks should be constructed with durable sawn timber.
• Soil fill or covering Is not allowed unless the stringers are completely covered with a material such
as geotextile and have guard logs on both sides
• Foundations should be excavated to a solid base and not formed by pushed material
• All parts of the bridge shall be well anchored to prevent their washing away (see fig. 12)
• The stream banks adjacent to the bridge should be stabilized using wing walls of durable logs or other equivalent construction
FIGURE 11 BRIDGE
CONSTRUCTIO
20 Code of Forest Harvesting Practices
• Silt traps shall be built at the four corners of bridges.
4.5.2 Culverts
Culverts should be set at or marginally below the level of the natural watercourse bed (see fig.12). They may have an earth fill but are to have stable abutments to the level of the running surface to prevent soil entering the watercourse. Simple log culverts should be constructed for small seasonal water flow crossing
the road. They should consist of three logs, of which two are the basic layer with the third log on top and covered with gravel. Culverts shall be laid at a grade that will minimize silting up and excessive scouring at the discharge end. They should have a headwall to prevent erosion, and should have an opening with a diameter of
45 cm or larger depending on the wet-season flood-flow level. In case of log culverts it is recommended to use at least three-log culvert if high flood flows are expected. Provisions shall be made at culvert inlets and outlets to minimize erosion caused by flow entering or discharging; sediment traps of logs and/or rocks shall be required in place where high water flows are expected (see fig.13).
FIGURE 12 STABILI zING
BRIDGE
UTMENTS AND
REAM BAN kSS
FIGURE 13 VERT SHOULD BE SET AT THE LEVEL OF E STREAMBED
21 Republic Of Liberia - Forestry Development Authority
Culverts shall not discharge over fills without adequate protection (e.g. rip rap, geotextile). Sumps or silt
traps shall be constructed in ditches at all four corners of culverts to prevent siltation and blocking. Log
culverts shall require geotextile or another retaining mechanism to retain backfill.
FIGURE 14 CULVERT
INSTALLATION
OUTLET
PROTECTION
DETAILS WITH
SPLASH APRON
RIPRAP LINED
PLUNGE POOL
ENERGY
DISSIPATION AN
SCOUR CONTR
FIGURE 15 PROVISIONS AT
CULVERT INLET
(SUMPS ) AND O
LETS (ENERGY
DISSIPATERS )
TO MINIMI zE
EROSION CAUS
BY FLOW
ENTERING
OR DISCHARGIN
FIGURE 16 SEDIMENT TRA
OF LOGS , ROC k
ETC . WILL BE
RE qUIRED IN
PLACES WHERE
HIGH WATER
FLOWS ARE
ExPECTED 22 Code of Forest Harvesting Practices
4.5.3 Fords
The construction of fords (or low-level watercourse crossings) shall minimize earthworks and impact on
the streambed. They shall be built to allow water to flow at all times, and should provide protection against
scouring below the crossing (unless the bed is solid gravel or stone), and should be corduroyed to minimize impact on creek bed. The corduroy shall be removed when the ford is no longer in use.
4.6 log landings
Log markets should be constructed to facilitate log sorting and loading activities. Spacing and size of log markets depend road density, topography, volume to be harvested, projected skid trail pattern, log size, storage time and logging equipment used. Hence, spacing and size of log markets should be determined during the planning phase. Remember that roads should be planned to minimise the sum of skidding and road construction impacts, which in turn will also lead to cost minimisation.
4.6.1 Location of log landings
Log landings (or log markets) should be sited at such intervals as to minimize the number and total length
of skid trails, but they should normally not be less than 250 m and not more than 1000 m apart (see fig. 17 and fig. 18). Log landings shall be located:
• At least 30 m from the edge of buffer strips (e.g. 45m from the bank of a creek or the edge of a gully) so that mud and debris do not enter watercourses.
• At sites that accommodate efficient skidding patterns and directions.
• On a gentle sloping elevated area, or on ridges or benches, in order to facilitate free drainage at all
times, reduce the amount of side cutting, and encourage uphill skidding to disperse runoff into surrounding vegetation. Log landings should not form part of the public roadway, and should be located 50 m from the public roadway to avoid deterioration of the road formation, road drainage facilities, and reduce hazard to the
public except when this would significantly reduce earthworks while maintaining adequate drainage of
both road and market. However, trees on the road construction site marked and felled, should be deposited on temporary roadside landing and integrate into the production process.
FIGURE 17
OCATION AND
SI zEOF LOG
LANDINGS
23 Republic Of Liberia - Forestry Development Authority
4.6.2 Size of log landings
The size of a log landing will depend on the expected volume and number of logs to be stockpiled. It should be large enough to facilitate the sorting of logs, to allow for entry (skidder) and exit (loader) points. A log landing should not exceed 1400 m2 (35 m x 40 m) in size, and the total area occupied by landings should not exceed 0.42 ha per unit area of 100 ha, i.e. 3 landings per km sq block. Private roads or road sides in the contract areas can be and sometimes should be encouraged as the landing areas where this can reduce environmental impacts.
4.6.3 Log landing construction
All merchantable trees shall be felled from the log landing site and extracted before clearing (these trees will be part of the approved yield from the block). Log landings shall be well drained. Proper drainage requires:
• A domed surface to prevent the ponding of water.
• Ring drainage around the perimeter to prevent surface ponding.
• Drains to channel runoff to vegetated areas. Log markets should have designated entry (skidder) and exit (log loader) points.
4.6.4 Log landing operations
Skid trails should approach landings from below to avoid directing runoff of water to the landing. This will reduce the amount of mud and water entering log markets from skid trails or roads. Likewise:
• Skidding across the roadway shall not be permitted because this may lead to deterioration of the road formation and (road) drainage facilities.
• The use of heavy machinery on saturated soils should be minimized to limit erosion, ponding, rutting, mixing and the compaction of the soil.
• Hauling on wet roads should be minimized
• Log landings should not be bladed off to keep them operational
• Debris and waste should be placed so as not to restrict drainage of the landing
FIGURE 18 LANDINGS SHO
BE LOCATED O
GENTLY SLOPIN
ELEVATED ARE
WITH SkID TRA
APPROACHING
FROM BELOW
24 Code of Forest Harvesting Practices
4.7 borrow Pits
Forest roads are typically built from local materials that must support heaving logging trucks and should have a surface that, when wet or when extremely dry, will provide adequate traction for vehicles. In many cases, the native soil material is too soft, too unstable or impossible to compact (white sand). Surfacing both improves structural support and reduces road surface erosion. Gravel, crushed aggregate, or loam are the most common improved surface materials used. Use of local material sources, usually borrow pits, can produce major cost savings, compared to the cost of hauling materials from distant sources. Typical borrow pits can have major adverse impacts, including sediment from a large denuded area and impacts on wildlife. Thus borrow pit planning, location and development should be done with care. The extraction of gravel or loam from road cutting areas during the formation of the road is preferred to the development of large borrow pits. All merchantable trees shall be harvested on the proposed borrow pit site (these will be included in the approved yield). Additionally:
• Catch drains should be constructed around the uphill side of the pit to prevent runoff entering the area
• The base of the pit shall be drained at all times. Drains shall not directly enter watercourses
• The face of the pit shall be maintained in a stable condition at all times
• At least one side of the pit shall have a gentle slope to allow easy wildlife access
5 lOGGING OPERATIONS 5.1 Tree Marking
Tree marking standards will be determined and regulated by the FDA’s chain of custody system. During the commercial stock survey all commercial trees above their respective minimum cutting diameters will be marked and numbered according to those standards.
• Following FDA approval of the yield all trees to be felled will be marked in the manner required.
• For all trees included in the yield the recommended felling direction shall be indicated.
• Seed trees and trees belonging to protected species shall be marked accordingly, and a 20m radius protection zone shall be established. However, no tree within 10 meters of a seed tree shall be felled.
• Rare, and endangered species (according to the IUCN Red List and Authority Regulations) shall be appropriately marked and designated as preserve trees to maintain the natural diversity of tree species; 25 Republic Of Liberia - Forestry Development Authority
5.2 Skid trails
5.2.1 Location
The design of the skid trail network should be based on an assessment of the following factors:
• Location and density of roads and log landings
• The number and location of trees to be harvested
• The logging equipment used
• The minimization of watercourse crossings
• The avoidance of soils with low load bearing capacity The total area occupied by skid trails should not exceed 8% of the total area of the block or compartment (or a total linear distance of 200 m per hectare). Skid trails shall not be constructed in areas excluded from harvesting and buffer strips. The maximum allowable gradient for skid trails is 35%. Skidding on slopes greater than 30% is only permitted/ allowed for short distances (less than 30 m) where adequate provision for drainage is possible to prevent excessive erosion. Uphill skidding is recommended since this gives operators better control over the movements of the log, and also tends to encourage the greater dispersion of runoff water into the surrounding area and not onto the skid trail. All skid trails should be as straight as possible to minimize damage to residual trees,
to prevent damage to the log being extracted, and to maximize skidding efficiency.
5.2.2 Survey requirements
The proposed skid trail layout in a block shall be marked in the field prior to inspection and approval by the FDA using flagging tape, paint marks or by blazing. Skidder and chain-saw operators (and their assistants)
should inspect the proposed alignments with the bush manager prior to commencing construction. Watercourse crossings shall be marked as such on the tree location map and on the ground
5.2.3 Timing of construction
The approved skid trail layouts shall be marked before the start of the felling operation to assist the feller
in determining the direction of fall and to improve safety and efficiency of the logging operation. The
construction of skid trails shall take place in the logging season only.
5.2.4 Construction
Skid trails shall be opened by a crawler tractor. The construction of skid trails should not require any earthworks. The maximum skid trail width is 4 meters. Avoid trees above 30 – 50cm dbh during skid trail construction. 26 Code of Forest Harvesting Practices
5.2.5 Watercourse crossings
Proposed watercourse crossings shall be marked on the block map, and shall be marked in the field prior to inspection and approval by the FDA using flagging tape, paint marks or by blazing. Crossings should be
constructed in dry weather. Watercourse crossings shall be selected in places where the bank slope is minor and the watercourse bed is
firm. Width of the crossing should be less than four meters. Buffer strip vegetation shall not be otherwise
disturbed. Skid trails shall cross waterways at right angles with straight approaches of at least 10 m on either side. Temporary crossings (log culvert with corduroy) shall be provided to cross gullies in any situation, and
waterways if water is flowing at the time of operation. Corduroy with earth fill is not allowed because this would effectively block the water flow and may divert its course. 27 Republic Of Liberia - Forestry Development Authority
Non-commercial logs shall be used in the construction of crossings where needed. Abutments and approaches should be higher than the stream banks. During construction soil should not be pushed past the high bank, into watercourses or onto the top of a crossing. Crossings shall be removed after completion of the operation. Crossing material should be placed more than 10 meters away from the high bank Its removal should not disturb the watercourse banks.
5.3 Felling
5.3.1 Timing of operation
Felling is only permitted in the logging season. Felling in a specific block shall commence once FDA has
given its authorization. All commercial trees included in the approved yield shall have been marked for felling and their predetermined felling direction indicated and seed trees and protected trees also shall have been clearly marked. Each felling crew should include three members (crew leader who operate the chain saw part of the time, main chain saw operator and an assistant). They shall be provided with radio for general communication and quick rescue in case of emergency. Felling can commence when forest road and log landing construction is completed, and when skid trails have been marked on the ground. All operators (skidding and chain-saw) shall be familiar with the harvest area and have discussed the operational plan with the bush manager and/or their supervisor. The felling
crew shall take both the harvesting map and the file of information on trees to be harvested into the forest.
All vines/lianas attached to the selected trees or trailing from the canopy should be cut in advance of felling so that they die and become brittle and prevent nearby trees from being pulled over or broken when the harvested tree falls.
5.3.2 Restrictions
Felling is not allowed within protected or exclusion areas and their respective buffer strips except species that are grown in wetland habitat. Felling is not allowed on slopes steeper than 30%. Felling shall commence at the rear end of the block and proceed along the main skid trail towards the log landing If trees fall inadvertently into watercourses or their buffer strips, the tree head as well as any accompanying logging debris should be pulled clear.
5.3.3 Directional felling
Chain-saw operator and assistant should be trained in directional felling, and shall be adequately equipped at all times. Directional felling is required to:
• Avoid damage to the felled tree (by cross-felling, falling on obstacles, down-slope felling) minimize damage to designated seed trees and other protected trees.
• Facilitate easy log extraction, thereby minimizing ground disturbance.
• Avoid disturbance to watercourses, exclusion areas and buffer strips.
• Prevent trees from hanging up in adjacent canopy trees.
• Minimize canopy openings by felling into gaps formed by previous felled trees or natural tree fall. 28 Code of Forest Harvesting Practices
• Avoid blocking the skid trail.
• Increase work safety.
5.3.4 Directional felling restriction
Felling crews should check trees suspected of being unsound before felling. Once cutting of a tree is
started, that tree should be felled, even if it is found during felling to be unsound. The stump height should be as low as practicable (approximately no more than. 30 cm) to maximize saleable volume. Stump heights over 30 cm are acceptable if there is a butt defect. Buttressed trees should be cut at a height not greater than the point, at which buttresses can be trimmed to provide a diameter equal to that immediately above the buttressed section. When it is not appropriate to trim the buttresses the tree should be cut immediately above the buttress. The logs and the stump are numbered and recorded with species and number of the harvesting block for legal reasons and work control.
5.4 log Preparation
All log preparations should be carried out at the stump. This ensures that the non-utilizable parts of the tree remain in the forest. The felled tree shall be cross-cut (top and junk) so that splitting does not occur when
a log is moved during extraction. Operators should top and junk boles to obtain the maximum volume, consistent with highest value of saleable logs, and to reduce skidding damage. All buttress flutes, knots and branches should be trimmed flush with the main stem to gain maximum log quality and volume, and reduce
soil disturbance and assist skidding. S-hooks or plates should be used before extraction as necessary on log susceptible to end splitting and to maintain quality.
5.4.1 Cross-cutting
Cross-cutting shall be done at the tree stump or at the bush landing. No crosscutting shall be carried at road side.
5.4.2 Safety Rules for Cross-cutting and bucking
• Always evaluate the distribution of tension and compression at the crosscutting point before applying the cut and position yourself on the compression side;
• Use appropriate working techniques to avoid timber wastage through shattering or splitting of the trunk;
• Give clear instructions on qualities, lengths and diameter for bucking at felling site and conversion at landing.
5.4.3 Log marking
Log markings enable logging operators, service providers, and costumers to locate and control all harvesting and extraction activities over space and time. Each log shall display:
• The number of the tree and the block number;
• The number of the log within the tree 1,2,3, or ABC, if segmented, alphabet shall be used;
• The logo of the logging company;
• Area of the log origin. 29 Republic Of Liberia - Forestry Development Authority
5.5 Skidding
Skidding shall commence only after the felling has been completed in a block and the felling crew has moved to another block. Machine access is prohibited in areas excluded from logging and their buffer strips. Logging machinery is not allowed to cross watercourses except at approved and properly constructed crossing points. Trees felled inadvertently into a buffer strip should be extracted carefully without much disturbance to the buffer strip. Skidding along or across primary and secondary forest roads should be avoided.
5.5.1 Pre-skidding
• The maximum gradient permitted on skid trail shall not exceed 30%.
• Any tree within the skid trail and having a diameter larger than 15cm should be cut by chainsaw before the machine opening the skid trail passes.
• Crawler tractor should be used to open skid trail and pre-skid logs from the stump to the skid trail. Avoid trees above 30 cm dbh during skid trail construction.
5.5.2 Skidding to landing
• Skidding is only permitted in the logging season.
• Skidding to the landing shall be done by a wheeled skidder equipped with grapples. Logs should be winched the maximum distance possible to reduce soil disturbance associated with skidding. Machines shall be equipped with an integral arch or similar device to lift the end of the log off the ground to avoid soil damage and an increase in skidding resistance.
• Winches shall be fitted to all extraction machines, and it is recommended that the minimum length
of wire rope of 25 m (19-22 mm diameter).
• Skidder and tractor blades shall be raised when traveling and skidding.
• Operators should reverse along skid tracks towards the concentrated log on the skid trail, and head
or butt haul to reduce travel distance.
• The wheeled skidder should never move off the skid trail especially when all logs are concentrated along the skid trail during the pre-skidding.
• Maximum skidding load should not exceed two (2) logs. Each skidding crew is equipped with chainsaw in addition to the standard complement of tools and safety gear.
5.5.3 Log Storage
The logging supervisor shall ensure that only trees approved in the yield are felled, and all saleable timber is extracted from the stump area and brought to the log landings. Logs should be stockpiled in well-drained log landings and shall be sorted in such a manner as to facilitate their inspection and scaling by FDA chain-of-custody staff and other authorized monitors. Logs stored at landing should be sprayed with permissible
fungicides and insecticides to protect against bio-deterioration. Only chemicals approved by the Authority
and Environmental Protection Agency shall be used. 30 Code of Forest Harvesting Practices
5.6 Truck loading, Transporting and unloading
Excavator type or rubber-tired or tracked front-end loaders should be used for loading and unloading logs where possible. Trucks shall not be loaded in excess of their capacity. Any protruding limbs bark or trailing material shall be removed before the truck departs the loading point. Pit loading shall only be permitted and is limited to logs that can not be possible for excavator or rubber tired or track front end loader. The
pit used shall be immediately refilled with the same excavated material removed.
5.6.1 Loading
Loading in forest harvesting is to facilitate the evacuation of extracted timber which had already been
bucked, classified, sorted and scaled. Transportation should be done by means of tractor with attached trailer or flat bed to its destination (processing mill or port of exit). Loading operation is often associated
with several constraints. Therefore the following measures shall be observed.
• All employees shall keep a distance of at least 20 meters away from loading truck during loading operations;
• All logs shall be stack on cross-bolts to facilitate handling by loader and to delay or restrict insect or fungal attack;
• No person should be in the cab or on the platform of truck while loading is in progress.
5.7 Road Transport
Transporting logs shall be completed by truck either directly from the bush landing/processing sites to its final
destination/point of delivery. The following rules for road transporting should be fully observed and adhere to:
• Secure load firmly against log slippage or fall by means of stanchions, chocks, cables, or chain and
binders to prevent log movement during transport;
• Each log should be secured by at least two cables or chain binders;
• For cost and safety reasons, vehicle brakes and stirring should always be in good technical condition. and should not be over loaded in excess of their capacity;
• All vehicles shall maintain safety and maintenance records for inspection by FDA or other authorities.
• Vehicle maintenance and inspection must be carried out at monthly intervals at a minimum for breaks, steering and other essential maintenance.
• Loads should not exceed the truck deck capacity and the bearing capacity of roads and road infrastructures;
• Never tow trucks across unstable section of the road by crawler or skidder;
• For efficiency and safety, the trailer unit should be loaded on to the tractor before traveling
unloaded;
• Personnel should be in possession of their faculties before driving an articulated vehicle (no driving trucks or other operational equipment if the conductor has been drinking alcohol, taking drugs, or other has a physical impediment or condition that may cause the driver to perform in an unsafe manner);
• Never transport un-authorized passengers, bushmeat, fire arms or protected wildlife;
• Remove timber within two months to avoid insects and fungal attack
• Remove hanging bark and limbs after chaining.
• Observe speed limits;
• Truck loaded with log shall only be unloaded using the required front-end loader. 31 Republic Of Liberia - Forestry Development Authority
5.8 Concentration of Harvesting Operations:
Harvesting operations are finally concentrated at either the processing mill or at shipment port of exit. It is at this stage when sound decisions are to be made in order to sustain the productivity and realize the profit
margin. Successful harvesting output depends on product, product quality and the market. To maintain the quality of the products harvested certain conditions will have to be guaranteed.
5.8.1 Processing Mills
• Logs off-loaded should be stacked according to species and grade on cross bolts
• Buck logs to market and contract requirement before sawing;
• Logs should be sprayed from time to time with pre-approved wood preservatives especially in the wet season while awaiting saw milling.
• Processed wood products shall be stacked under well ventilated shade using stickers between sawn timber to avoid sap staining
• End-faces of sawn timber should be painted to reduce face checks and splitting
• Produce quarter-sawn timbers to reduce shrinkage and increase market value.
5.8.2 Shipment Port
• Seal the faces of logs with anti-cracking chemicals to prevent face checking and splitting
• Spray the stacked log with permissible insecticides and fungicides
• Debark logs if necessary and spray before stacking;
• Logs must be properly stacked on cross-bolts to avoid direct contact with the moist ground and prevent discoloration from the exudates of other species; and
• Logs that are susceptible to crack and splitting should be S-Hooked to maintain quality and value.
5.9 Weather limitations on logging Operations
Road construction and maintenance, skidding, loading and hauling when conditions are wet cause extreme
damage to soil and water. It is also inefficient and often dangerous. Areas most likely to be workable in
wet weather are those with less than 20% slope on stable soil types such as brown/white sand and laterite. The annual plan should specify wet and dry weather coupes.
5.9.1 Felling
Felling should cease when winds prevent accurate and safe directional felling, and when ground conditions are too slippery to allow the felling crew to move safely and quickly away from the falling tree, and when log extraction or hauling is not possible due to weather and/or soil conditions. Felling should always be limited to ensure the volumes cut at any one time can be promptly skidded and hauled. 32 Code of Forest Harvesting Practices
5.9.2 Road skid trail construction and skidding
Road construction operations and skidding should cease when:
• Soils are saturated and turbid water or mud is flowing down a skid trail. • Turbid water or mud is flowing from a skid trail or road into a watercourse. • Soils are rutted to the extent that the depth of mud is greater than the rim of wheeled machinery or reaches the final drive on caterpillar-type machinery.
Any affected section of the trail or road shall not be by-passed by opening up a new trail/road alongside or close by.
5.9.3 Log landing operations
Construction or operations on log landings shall cease when:
• Water is ponding on the surface of the log landing.
• Soils are rutted to the extent that the depth of mud is greater than the rim of wheeled machinery
or reaches the final drive on caterpillar-type machinery.
5.9.4 Trucking
Trucking shall be halted when:
• Trucks cannot move unassisted along the roads because of slippery conditions.
• Turbid water or mud runs in wheel ruts, and forestry roads are rutted to the extent that the depth of mud is greater than halfway of the rim of wheeled machinery or deemed by the Authority representative to have substantial environmental impact.
5.9.5 Recommencement of operations
Soils need to be allowed to drain after heavy rainfall events before forest operations recommence. Skidding,
loading and trucking operations may recommence only after water or mud ceases to flow on affected skid
trails, landings and forest roads, and mud depths fall below the maximum permitted levels. 33 Republic Of Liberia - Forestry Development Authority
6 NON -TIM bER FOREST PROD uCT (NTFP) M ANAGEMENT PRACTICES
NTFPs are resources derived from the forest besides timber and other wood based materials e.g., plywood, particleboard etc., that provides security, health care, materials for implements and construction, fodder, fuel wood and livelihood to rural communities specially those living in the uplands. This includes wood extractives, barks, leaves, roots, sap, fruits and also non-timber yielding materials such as rattan, bamboo, coconuts etc.
NWFPs are of significance primarily in household and local economics. The economic benefit that can be derived from non-timbers forest products have been identified as a major
opportunity for community forest projects.
Most significantly, NTFPs play in important part in sustainable forest management. The exploitation of
many NTFPs depends on keeping forest intact without decimating a while range of forest resources.
6.1 NTFP Sustainable Management Practices
• Logging contracts should incorporate the sustainable management practices in their harvest plan;
• Capacity building in the NTFP sustainable management and utilization should form part of social agreement to be provided for the local communities. The training shall emphasis on the techniques in harvesting, processing, marketing and cultivation/propagation
• Local communities shall be given the opportunities to access the forest for NTFPs as long as the harvest is sustainable;
• Logging companies shall regulate and monitor the harvesting of honey. This is to mitigate the
burning /setting five on the bee hives
• Local communities shall be allowed to harvest, collect the NTFPs in the concession areas as long as these exercise are sustainable
• Extraction and road building machines should prevent the destruction bamboo clumps in the forest areas. POST-HARVEST ACTIVITIES
7 POST-HARVEST ACTIVITIES
Rehabilitation of logged areas is required to prevent further deterioration of the logged area and downstream soil and water values, and to encourage forest regeneration. All areas should be left in a clean and tidy condition.
7.1 block closure
Blocks shall be closed once the logging of the approved yield has been completed. A block that has been closed shall not be re-entered and shall remain closed until the next scheduled cutting cycle except for the collection of non-timber forest products by the local communities.
Logging operations in a specific block should be completed in a single logging season. The only exception
to this rule that may be applied is when weather conditions have prevented the approved yield to be felled and extracted. In such cases the FDA shall provide an extension to complete harvesting operations. 34 Code of Forest Harvesting Practices
7.2 Road maintenance and closure
The primary and secondary forest road network shall be maintained in good condition. There should be no ruts in the surface and the road surface should be crowned. Likewise all ditches, turnout drains and cross-road drainage that will not be removed shall be left in good working condition.
Roads not to be used until the next rotation shall be closed to all traffic by placing a large log across the
roadway, or digging a ditch across the roadway. For roads that are being decommissioned log culverts and
temporary bridges shall be removed to allow unobstructed water flow. All bridges and permanent culverts
that are not to be removed shall be checked regularly (and at least annually) including decking foundations and sidewalls. All silt traps shall be cleared regularly. Indigenous commercial tree species should be planted on closed roads after operations are completed.
7.3 log landings
All closed log landing sites should be restored so that proper drainage occurs. All areas where water may pond shall be drained, and drains installed. All log landing drains shall be cleared regularly until the landing has stabilized. Indigenous commercial tree species should be planted on the landing after operations are completed. Log landings on clay or loam should be ripped at 90° to the drainage direction to promote natural re-vegetation. Silvicultural treatment shall be introduced through enrichment planting of commercially valued species. Bark and debris should be disbursed evenly across the site to assist in stabilization. The site shall be cleaned of all refuse including oil/fuel drums, wire rope, tires and machinery parts.
7.4 Skid Trails
Temporary skid trail crossings of waterways and gullies shall be removed after completion of harvesting operation in the block. Removal should not disturb the watercourse banks. All material used in the construction of temporary crossings shall be moved at least 10 meters away from the watercourse. Action
shall be taken to restore water flow to its original watercourse where necessary. On higher gradients of skid trails the clearing and maintenance of drains may be required until skid trails
have stabilized. Where it may apply, contractors should plant indigenous commercial tree species in open skid trail areas were it may enrich the species diversity, improve stand quality or provide erosion prevention.
FIGURE 19 DRAIN LOG MAR kETS AND IVERT WATER FLOWING TO AR kETS ONCE LOGGING IS COMPLETED
35 Republic Of Liberia - Forestry Development Authority
7.5 borrow Pits
Borrow pits shall be decommissioned by:
• Removing any non-biodegradable material and burying all biodegradable waste
• Stabilizing steep cuts
• Re-grading the drain on the uphill side and ensuring that runoff cannot enter the pit
• Draining the surface of the pit if water is likely to pond
7.6 Field Camps
Temporary field camps shall be closed by taking the following steps:
• The site shall be cleaned of all refuse including oil/fuel drums, wire rope, tires and machinery parts, as well as building materials, and all biodegradable waste shall be buried.
• All areas where water may pond shall be drained (but drains are not to empty directly into watercourses) and drains shall be cleared regularly until the campsite has stabilized.
• Campsites should be ripped at 90° to the drainage direction to promote natural re-vegetation. Bark and debris should be disbursed evenly across the site to assist in stabilization.
8 OPERATIONA l HYGIENE
Maintenance, servicing and fuelling of logging equipment involves materials which could cause serious harm to soils and waters if released; pollution of groundwater or watercourses by oil, fuel, lubricants or
other hazardous materials will eventually affect all flora, fauna and humans not only near the spill but also
downstream. Not maintaining a clean and tidy operation is a sign of poor worker attitude, careless management and disrespect for the environment.
8.1 Workshop & Wood Processing Facilities
Workshop and wood processing facilities shall be located at least 100 m away from any watercourse or water body. The appropriate public health inspector or similar designated agency may assess plans, approve and inspect workshops, sawmills and associated infrastructure. Approval for their construction may depend upon the preparation of an environmental impact assessment. Provision shall be made in plans for the removal or burial of non-toxic solid waste, and for the collection and disposal at approved designated waste disposal facilities of fuel and oil waste.
8.2 Field Servicing and Maintenance
Field fuel tanks, refueling points, chemical mixing points and maintenance areas shall be located:
• In well-drained areas such as log landings or road junctions.
• Outside areas excluded from harvesting and their buffer strips.
• No closer than 100 m to any habitation
• More than 100 m away from any watercourse. 36 Code of Forest Harvesting Practices
Care shall be taken to prevent spillage during refueling or repairs. Adequate equipment – e.g. hand pumps – should be provided and used. Sump oil shall not be dumped in the harvesting areas, but collected and removed to the designated waste disposal facilities. All containers used in the transport, storage and use of toxic materials shall be leak proof, marked as “hazardous” and clearly labeled with the contents’ name.
8.3 Hazardous Chemical Handling and Storage
Hazardous chemicals include preservatives, pesticides and herbicides. The FDA must be notified and give
approval before a company starts using any pesticide, herbicide or preservative in or near the forest. The company must obtain the relevant “Chemical Technical Data Sheet” and submit a copy to the FDA every
time it is to be used. Chemicals must only be used when necessary to achieve defined management aims
described in the Management and/or Annual Plan, and approved by the FDA, and in strict accordance with the manufacturer’s instructions. The instruction shall specify:
• The types of chemical to be used (oil-borne, organic solvent base, or water-borne)
• % concentrations;
• Active ingredient (toxic component of preservative formulation);
• Application levels ; and
• Method of treatment. Hazardous chemicals shall be stored in a well-drained area at least 100 m from any watercourse and no
closer than 100 m to any habitation. Drains are to be directed to a closed, stable and flood free disposal pit,
situated at least 50 m from a watercourse or water body.
FIGURE 20
ARE SHALL BE
NTO PREVENT
LAGE DURING
REFUELING
37 Republic Of Liberia - Forestry Development Authority
Toxic materials are to be stored in a locked, dry, well-ventilated storeroom. Wet products are to be effectively separated from dry products. All entrances are to be clearly marked with a sign reading “warning – hazardous chemical storage - authorized persons only” or equivalent. Containers should not be stored on
the floor, but are to be elevated above the floor on pallets or other means, to allow regular inspection and rapid identification of leaks. All containers used in the transport, store and use of toxic chemicals are to be
leak proof, marked as “hazardous” and clearly labeled with the contents’ name. Access to the storerooms of toxic materials should be restricted to authorized personnel. It is recommended that concrete bunds with a capacity of twice the storage capacity of the largest storage container are to be provided around all storage facilities.
8.4 Waste Management
Toxic substances include (spent) hydraulic fluid, coolant, lubricants, fuel (gasoline/diesel/kerosene),
industrial cleaners, paints and resins, preservatives (including timber treatment chemicals), distillates, insecticides and herbicides, and workshop waste, waste oil and contaminated sludge Toxic materials shall be collected in containers securely sealed and either dumped at designated waste disposal facilities, waste pits or returned to the supplier in accordance with the approved environmental impact plan.
Only approved waste pits shall be used for specific waste products. All waste pits are to be covered with
at least one (1) m of soil and located at least 100 meters from any watercourse or water body and at least 1- 1.5m above the groundwater table. Signs should be erected identifying the waste pits. Toxic materials shall never be disposed of into watercourses or lakes. Equipment used for applying chemicals shall not be washed in watercourses. Excess chemicals are to be either removed from the forest or chemically treated (neutralized) in an approved manner. Empty containers are to be safely disposed and not reused. All refuse introduced to the forest e.g. pieces of wire rope, packing material, bottles, containers, etc. shall be removed from the forest, placed in a refuse pit, buried and covered to a level surface. Fuel and oil
drums, used oil filters, oily rags, empty grease gun cartridges, worn machinery parts, paint tins, etc. shall be
removed to a designated disposal area; or returned to the supplier. Discarded machinery shall be removed from the contract area to the designated waste disposal facilities.
8.5 Wildlife management in contract areas
• Logging companies shall apply measures to discourage and limit illegal hunting in their concession areas.
• No company workers shall possess and /or transport fire arms, bush meat and/or hunters in
company vehicles, even outside the concession area and a ban on abetting or facilitating hunting;
• All workers are prohibited from hunting outside demarcated Hunting area;
• All workers are prohibited from hunting protected animals;
• All workers are prohibited from using snares;
• All workers are prohibited from selling bush meat(in commercial quantities) in company camps and worksites; 38 Code of Forest Harvesting Practices
8.6 Alternative protein sources for wildlife protection in concession areas
• The company shall make available smoked fish, pork, canned meat, beans, dairy milk, chicken, groundnuts,(leguminous products) and/or fish;
• Encourage fishing using hooks and collection of snails in a sustainable manner;
• Promote activities that provide vegetables or animal proteins.
9 CAMP HYGIENE
Maintenance of safe, healthy and pleasant living and working conditions for personnel is a prerequisite for
a motivated and fit workforce, hence productivity.
9.1 Camp Design
Permanent camps should be protected from wind storm by planting non-invasive and fast growing ornamental species. Grasses around camp building should not be hooked out, but rather periodically trimmed/brushed. This will prevent soil erosion. Site plans for camps should include designs for sewage, water supply, waste water and waste disposal. These will be evaluated as part of the environmental impact assessment that accompanies forest management plans. FDA staff in conjunction with staff from other agencies, such as the EPA, will inspect sites on a regular basis. Camp areas shall be well drained so that water does not pond or create mosquito-breeding areas. The camp shall be checked regularly for any areas where stagnant water can accumulate.
9.2 Water Supply and Domestic Waste Water
Camps are to be supplied with purified potable water obtained from running streams, rainwater, and
wells or by water tanker. All water storage tanks should be properly screened to prevent the breeding of mosquitoes.
9.3 Wastes and Refuse Disposal
An adequate number of plumbed toilets or pit latrines shall be provided. Plumbed toilets should drain into a septic tank. Pit toilets shall be located at least 100 m away from watercourse and water bodies. Sewage shall be discharged so it neither enters the catchments of drinking water supplies, nor intakes into a stream. All pit toilets must be sprayed with environmentally recommended chemicals once every week. Domestic waste water shall be directed to a disposal area (or septic tank) at least 20m away from the nearest building. All drains (waste water and sewage) should be kept covered. Provision shall be made for refuse disposal areas in pits located at least 100 meters away from watercourses, 50meter away from habitation or farmland, The pits shall be dug above the groundwater table, and where runoff water cannot enter. Refuse shall be covered with soil to depth of 30 cm at least once per every one month. And it shall be sprayed once every week with biocide but not with fuel oil alone. 39 Republic Of Liberia - Forestry Development Authority
9.4 Camp Construction Requirement
• All permanent buildings in concession areas shall be made of solid materials such as cement, bricks, stone, processed hard wood and designed with comfort and convenience in mind;
• All camp houses and roads shall be lit at night and equipped with drinking water points (stand pipes) Living quarters for workers shall have:
• Clean running water;
• Lighting and power socket;
• Sanitary facilities (shower and toilet) draining to a septic tank A camp located at some distance from the social, school and commercial facilities of a town or village shall have:
• A dispensary with treatment and recovery rooms for basic medical assistance, and equipped with basic medical supplies;
• In case of emergency, there shall be an evacuation procedure ( rescue chain) for severe accidents and emergencies with rescue equipment;
• A primary school;
• A company store offering basic goods at competitive prices, with ample stocks of meat, poultry and
fish to reduce pressure on local wildlife (hunting)
• Social and cultural facilities: club, video room, church All logging camps shall be supplied with clean water that shall be duly tested on a regular basis. Any
necessary filtering and treatment shall be provided at the source.
9.5 Additional Facilities
Medical care shall be provided by the company whether or not public health services are available in the area where workers and their families live. Educational, recreational and spiritual services should be provided, where such services are not available nearby.
10 HEA lTH AND SAFETY 10.1 General
Employers have the main responsibility for safety and health in forestry work. They shall install and maintain work systems and methods which are safe and without risk to health. Employers shall provide insurance for all employees and workers. All machine operators shall receive the necessary training and instruction to ensure competency to safely operate equipment for the job they are assigned to do.
Operators should know what the job requirements are, what other machines are working in the area, and
be aware of any hazardous conditions that may arise. 40 Code of Forest Harvesting Practices
Basic first-aid training (and refresher courses) should be provided to all personnel involved in field
operations, wood processing, workshop, etc. All logging companies shall install a system whereby accidents, dangerous occurrences and occupational diseases are reported, recorded and investigated, and ensure that the necessary adjustments are made to prevent or reduce the incidence of these accidents, dangerous occurrences and occupational diseases in the future. Working hours shall not exceed the number prescribed by national law or collective agreements. Working hours shall be arranged to provide adequate periods of rest, which include: short breaks during work hours,
sufficient breaks for meals, nightly rest and weekly rest.
10.2 Emergency Rescue
Provision shall be made for the quick evacuation of a person in the event of an injury or illness that requires medical assistance. Radio or telephone links shall be available at the worksite to contact rescue services in case of an emergency. Where professional help is not available within a reasonable distance, consideration should be given to the creation of the necessary dispensing and health-care facilities.
10.3 Felling
10.3.1 Protective Clothing and Safety Equipment
A first-aid kit shall be provided to every felling crew and located close to where felling crews are working.
All felling crew members shall be provided with and wear:
• Safety helmet (that shall be replaced every year/once it is damaged before the end of the year)
• Hearing protection (e.g. earmuffs or earplugs whether damage or not;
• Safety boots with steel toecaps Additionally chain-saw operators shall be provided with and wear:
• Chain-saw gloves lined with cut-resistant material
• Eye protection (e.g. mesh face guard, goggles)
• Leg protection lined with cut-resistant material (e.g. chain-saw chaps)
10.3.2 Equipment Safety
All chain-saws shall be maintained in good working order and all safety devices shall be operational at all
times. Specifically all chain saws should be equipped with:
• A chain brake, which is activated manually by the front handle guard
• A front handle guard for protection of the left hand from the chain
• An on/off switch which is reachable with the right hand on the throttle
• A throttle control lock-out which prevents the chain-saw from being started unexpectedly, because two levers have to be pressed simultaneously
• A rear handle guard for protection of the right hand in case of chain breakage
• An anti-vibration system, consisting of rubber shock absorbers between the
engine block and handles 41 Republic Of Liberia - Forestry Development Authority
• A chain catcher
• A spiked bumper (for safe and accurate cross-cutting)
• An exhaust that directs fumes away from the operator
• A chain guard for avoiding injuries and protecting the chain during transportation
• A chain-saw toolkit for corrective and preventative maintenance.
10.3.3 Safe working practice
Chain-saw operators should always have an assistant and shall never work alone in case of accidents. No persons should approach closer to the feller than twice the height of the tree being felled, unless the feller
has acknowledged that it is safe to do so. Operators should clear undergrowth and debris away from the
base of the tree to provide an adequate and safe working space. Two alternative escape routes should be
cleared - one at 135° and one at 215° - to the intended felling direction (see fig. 21).
Chain saws should not be operated above shoulder height because of the risk of kickback and the resulting backward rotation of the guide-bar. The use of machines to pull trees while they are being cut should never be permitted.
10.4 Heavy Machinery
10.4.1 Definition
Heavy machinery includes bulldozers, skidders and log loaders, motor graders, excavators, etc.
10.4.2 Protective Clothing and Safety Equipment
Operators of heavy machinery and their assistants shall be provided with and wear:
• Safety helmet (that shall be replaced every year)
• Hearing protection (e.g. earmuffs or earplugs)
• Safety boots with steel toecaps
• Heavy-duty work gloves (for handling wire rope)
• Dust masks or respirators when appropriate
FIGURE 21 CLEAR TWO
ALTERNATIVE
ESCAPE ROUTE
42 Code of Forest Harvesting Practices
10.4.3 Equipment Safety
All heavy equipment requires regular maintenance to achieve and maintain safety standards and good working condition. All heavy machinery shall be equipped with:
• A safety cab which conforms to the Roll Over Protection Structure (ISO 3471 and ISO 8082) and Falling Object Protection Structure (ISO 8083) standards
• Access to and exit from machinery designed to provide hand and footholds of a convenient height and spacing
• Securely mounted seats and seat belts
• Rear portion of cabs fitted with protective wire mesh (except loaders, excavators and graders)
• Reversing alarms to alert people machines are backing up
• Securely guarded pulleys, shafts, belts and fan blades
• Engine emergency stopping devices that are not self-returning, those are clearly marked, and easily accessible from the normal operating position. The engine starter should be interlocked with the transmission or clutch to prevent the engine from starting when left in gear.
• Parking brakes must be capable of keeping the machine and its rated load stationary on all slopes likely to be encountered
• First-aid kit and fire extinguisher (operators should be trained in their use)
• Exhaust systems fitted with spark arrestors
Farm tractors not equipped with the provisions listed above shall not be used for timber extraction.
No modifications shall be made to a machine that:
• Interfere with operator visibility
• Interfere with access to and exit from the machine
• Exceed the rated payload or gross combination weight of the machine resulting in overloading the
braking and/or steering system or the ROPS capacity rating The safety cab of any item of heavy machinery shall not be modified, drilled, welded or altered in any way
nor should any attempt be made to straighten any part of the frame when it has suffered damage.
10.4.4 Skidding
Skidding should not be carried out in site conditions where the stability of the machine cannot be assured.
Skidding across slopes should be avoided because of the significant decrease in skidder stability. No objects
(equipment) shall be carried on the skidder unless a space is provided for that purpose and the object can
be firmly fixed into place.
10.5 Vehicles
10.5.1 Equipment Safety
All vehicles, including logging trucks, dump trucks, fuel tankers and personnel carriers, require:
• Regular maintenance to achieve and maintain safety standards and a roadworthy condition
• Full insurance 43 Republic Of Liberia - Forestry Development Authority
• Compliance with national requirements for safety and mechanical roadworthiness
• Securely mounted seats and seat belts
• Securely guarded pulleys, shafts, belts and fan blades
• First-aid kit and fire extinguisher
To protect the cabin from penetration logging trucks should incorporate an adequate log barrier or guard between the load and the cabin.
Logging machinery and vehicles using public roads shall have an amber flashing light on top of the cab or
other prominent point installed and reversing alarms.
10.5.2 Safe Working Practice
Trucks shall not be overloaded. All loads shall be secured with at least two approved load binders. Protruding limbs, bark and trailing material shall be removed before the truck leaves the log landing. Side stanchions on trucks and trailers shall be vertical after loading and bunks aligned properly.
10.6 Hazardous Chemicals
Workers applying hazardous chemicals must receive training and full information on the risks involved in
the use of protective equipment and first-aid techniques. Workers handling chemicals must wear protective
clothing as recommended on the container label. It is the company’s responsibility to ensure that its workers use all protective clothing provided. Chemicals must always be utilized with the proper equipment, as recommended by the manufacturer. The equipment must function properly and be free from leaks and blockages. Worker should be allowed to take bath or properly clean him or herself with water immediately after using the spraying chemical.
10.7 Workshop and log Yard Workers
Workshop mechanics and other workers, and workers in log yard areas shall be provided with and wear protective clothing, hearing protection, eye protection, respirators, and/or gloves, as appropriate for the equipment being used (e.g. welding, angle grinding, spray painting).
10.8 Fire Precautions
Operations shall cease in times of high fire risk. All machines and chain-saws should be fitted with spark
arrestors. Machinery should be clear of surplus oil and fuel. Any rubbing, damaged, frayed, kinked or
leaking hydraulic hoses and fittings should be replaced. Refueling shall only be carried out in designated areas where the ground is clear of all inflammables for a distance of 5 meters in all directions from the
machine.
All welding activities shall be done at least 10 meters away from flammable materials. No fire shall be left unattended in the forest, and fires for cooking or other purposes shall only be lit in designated areas cleared of flammable material within a 5-m radius around the fire. 44 Code of Forest Harvesting Practices
11 HARVEST CONTRO l, MONITORING AND ASSESSMENT
Harvesting assessment is a systematic check to be taken to determine or verify that harvesting operations
followed the annual harvest plan and achieve its technical, financial and environmental objectives while
complying with established standards of management plan guidelines. Monitoring and assessment are key elements of sustainable forest management which the contract holder shall be responsible to conduct. The Authority and contract holder shall provide opportunity for independent monitoring and evaluation:
• During harvesting through the monitoring and control of operations;
• After harvesting by means of internal and external assessment. All monitoring requirements established in the Timber Sale and Forest Management Contracts and
regulation shall be implemented and regulated under the supervision of the FDA contracting officers.
12 SOCIA l ISS uES 12.1 land and forest use rights and responsibilities
All contract holders shall be expected to respect legal or customary rights to land and forests, as well as respect for cultural, traditional and local values and customs Forest management contracts and timber sales contracts provide the contract holders and operators with a right to operate in a given area. Contract holders should immediately inform the FDA of any illegal forestry activity in their contract area.
Social agreements between contract holders and affected local forest communities define and describe the codes of conduct and other agreements concerning forest use rights and responsibilities, benefit sharing
arrangements, the use of local labor, compensation for crop damage, etc., between the two parties. They
also describe conflict management arrangements.
12.2 Work place relations, rights and responsibilities
12.2.1 Terms and conditions of employment
The terms and conditions of employment shall be in accordance with Liberian labor laws.
12.2.2 Equal opportunity employment
• Women and men shall be paid equal remuneration for the same work or work of equal value.
• Employers will maintain registers of all employees, together with employment records (including PAYE and NIS contributions). These registers will be available for scrutiny by the FDA and other relevant government agencies
• Employees shall not be discriminated against based on race, sex, religion, colour, or ethnic origin or sickness. 45 Republic Of Liberia - Forestry Development Authority
12.2.3 Prohibition against forced labor
• Employers shall ensure fair remuneration and humane working conditions in return for all services rendered
• Given the nature of forest management practices and the logging industry employers shall inform employees of the terms and conditions of service prior to the time they are hired, and prior to the assumption of duties, an employer shall document and inform an employee of his/her wages by
task or by day (or other specified time period).
• Employers shall observe ILO Convention 182 on forced labour.
• Employers shall not knowingly employ persons under the age of 14 in keeping with the International
Labour Organization Convention 182
12.2.4 Employment and training opportunities for local populations
• Contract holders should give priority to employing equally qualified persons who live in
communities within or adjacent to the contract area.
12.2.5 Education and training for workforce and local populations
• A range of educational and skills training programs for the purpose of enhancing workers’ job performance and promotion within the workplace should be made available on an appropriate basis, and as included in the contract holder’s business plan.
• Educational and vocational skills programs should be made available to communities within or adjacent to the contract area.
• Wherever possible, certification for educational and skills training programs should be issued by a
recognized institution or body.
12.2.6 Protection of Forest Wildlife Resources
The contract holder shall apply appropriate measures to discourage and limit illegal hunting including:
• A prohibition on all workers from possessing and/or transporting firearms, bush meat and/or
hunters in company vehicles, even outside the concession area.
• A prohibition on all workers from hunting protected animals. Transgressors should be reported to the FDA and local police for prosecution.
• A prohibition on all workers from using snares.
• A prohibition on selling or preparing bush meat in company camps and worksites
• Reporting all illegal activities promptly to FDA and other relevant authorities. 46 Code of Forest Harvesting Practices
13 GlOSSARY
Abutment End support for bridge culvert or similar structure Adverse grade Grade up which a loaded logging truck must travel Batter Inclination or shape of a cutting beside a road or track Berm Ridge of soil typically along the outside edge of a road shoulder or skid trail. It intentionally or unintentionally directs surface runoff onto or alongside the roadbed or skid trail Biodegradable Capable of being decomposed by bacteria, fungi or other living organisms Biodiversity The range of diversity of plants or animals, including the diversity of different species,the variation found within species and the variety of ecosystems
Borrow pit An area where excavation takes place to produce materials for earthwork, such as fill
material for embankments and surfacing material. It is typically a small area outside the roadway for obtaining sand, gravel, laterite, or loam without further processing Box cut A road cut through a hill slope or, more commonly a ridge, in which there is a cut slope on both sides of the road. Also called through cut Bucking Cross-cutting of a log in shorter sections Bridge A structure that provides for vehicle access over a watercourse Buffer strip Strip of vegetation left intact along a watercourse or other sensitive area or site during and after logging. Buttress A ridge of wood that develops in the angle between a lateral root and the base of a stem to provide lateral root stability to the stem Camber The amount of cross-fall on a road Catch drain A drain constructed above a batter to prevent erosion of the batter by surface water Catchment An area or basin of land bounded by natural geomorphologic features such as hill crests
and ridges from which water drains and flows to a watercourse, lake, wetland or estuary
Chain brake A safety device on a chain-saw designed to stop the chain in the event of a kick-back
Channel A waterway that contains flowing water either periodically or continuously. A channel has a defined bed and banks that confine the water
Chaps Chainsaw chaps are half-trousers which are contain material designed to protect againstchain-saw cuts 47 Republic Of Liberia - Forestry Development Authority
cm centimeter Compaction The process of reducing the apparent volume of the soil, by reducing the empty spaces
between particles and increasing the density of the soil under the influence of pressure.
Compaction is desirable when a soil is to be used as the base of a road, because it improves
stability and reduces infiltration. For the same reasons, compaction is undesirable in the
forest, because it has a negative effect on plant growth and survival and soil life. Compartment A sub-division of a concession frequently of several thousand hectares. It is normally
defined along natural boundaries
Competency A concept that focuses on what is expected of a person in the workplace rather than on the learning process. It embodies the ability to transfer and apply skills and knowledge to new situations and environments Corduroy Cording or matting involving the use of suitable logs to spread the weight of the load and separate machine tyres or tracks from direct soil contact during harvest operations, thus reducing ground pressure and rutting
Coupe A defined area of forest of variable size, shape and orientation, on which harvesting
takes place; usually to be harvested over one year Cross-cutting Cutting through a felled log. Sometimes called bucking Cross-drain A ditch and earth bank constructed at approximately right angles to a track, preventing water from building up speed along the track and allowing redirection of running water into surrounding areas Crown A crowned road surface has the highest elevation at the centre line (convex) and slopes down on both sides. Crown is used to facilitate draining water off the road surface Cultural area Area of social, cultural, historical, religious, spiritual, archaeological or anthropological importance to forest dwellers; usually to indigenous populations. Includes villages, farms, gardens and sites which are culturally sensitive. Culvert A conduit, typically of made of metal, concrete, plastic or (hollow) logs, set beneath the road surface, to move water from the inside of the road to the outside of the road. Culverts are used to drain (inside) ditches and watercourses (commonly gullies) that cross the road. Also called koker
Cut-and-fill A method of road construction in which a road is built by cutting into the hillside and spreading the spoil materials in adjacent low spots and as compacted or side cast fill slope material along the route. A ‘balanced cut-and-fill’ utilizes all of the ‘cut’ material to generate the ‘fill’. In a balanced cut-and-fill design, there is no excess waste material and there is no need for hauling additional fill material. Thus, cost is minimized. Cut slope The artificial face or slope cut into soil or rock along the inside edge of the road 48 Code of Forest Harvesting Practices
Cutting cycle In selective (polycyclic) harvesting systems: the planned number of years
between successive harvests on an area of forest. It is also referred to as felling cycle Dbh Diameter at breast height; 130 cm above the ground
Debris Organic material, rocks and sediment (leaves, brush, wood, stones, rocks, rubble, etc.)
often mixed, that is undesirable in a channel or drainage structure. Compare sediment
Directional A concept that focuses on predetermining the final direction of fall of a felling felled tree. It includes selecting a particular direction of fall based on a predefined set of criteria and the
felling techniques and aids involved in felling the tree in the selected direction Ditch A channel or shallow canal along the roadbed intended to collect water from the road and adjacent land for transport to suitable point of disposal. Also called table drain or (incorrectly) side drain Erodibility See soil erodibility Erosion See soil erosion Exclusion area Area which is excluded from harvesting Favourable Grade down which an unladen logging truck must travel grade
Feeder road A road connecting log markets to a secondary or primary road; also called spur road Fill Excavated material placed on a prepared ground surface to construct the road sub-grade and roadbed template. Also called embankment Fill slope The inclined slope extending from the outside edge of the road shoulder to the toe of
the fill. Also called embankment slope
Flood plain A level or gently sloping area on either side of a watercourse contemporary channel that
is submerged at times during high water of periods of flooding.
Ford A rock, other hardened or corduroy structure that is built across the bottom of a watercourse channel that is usually dry, to allow improved vehicle passage during periods
of low water or no flow and minimizes channel disturbance or sediment production
FMC A Forest Management Contract is for greater than 50,000 ha and duration of 25 years. Geotextile Any permeable textile material used with soil, rock or any other geotechnical engineering related material, as an integral part of a man made product, structure or system, usually related to the passage of water Grade The slope of the road along its alignment. This slope is expressed in percent – the ratio of elevation change compared to distance travelled. Also called gradient 49 Republic Of Liberia - Forestry Development Authority
Groundwater The part of the subsurface water that is in the zone of saturation, including underground streams Guard log Log along the outside of a bridge, above the main stringer logs to prevent gravel from falling off the bridge into a watercourse
Gully Steep sided drainage channel where water may flow during a wet season or only after a rainfall Gullying Scouring of the soil by high velocity water flow resulting in channels where water runs
down a slope, embankment or roadbed ha hectare Habitat tree A habitat tree is a mature living tree selected to be retained during harvesting because it has features of special value for wildlife (e.g. hollows). Harvesting Broken logs, branches, twigs, vines, epiphytes and other tree related debris vegetative material brought down as a result of felling or skidding Hauling Transport of forest products, particularly logs, from the log market to the processing facility, commonly by way of logging trucks Headwall A concrete, masonry or timber wall built around the inlet or outlet of a culvert to
increase inlet flow capacity, reduce risk of debris damage, retain the fill material and
minimize scour around the culvert inlet or outlet Hung-up trees: A tree which has not completely reached the ground following cutting
Inlet The opening of a drainage structure or culvert where the water first enters the structure
In-slope The inside cross-slope of a road surface, typically measured in percent. In-slope is used to facilitate the draining of water from a road surface to an inside ditch. An in-sloped road has the highest point on the outside edge of the road and slopes downward to the ditch at the toe of the cut slope, along the inside edge of the road Integral arch An extension to the body of extraction equipment, which raises the anchor point of the wire rope and thereby lifting the load off the ground, also called logging arch Junking Trimming of the butt end of a felled log. Also (incorrectly) called bucking km kilometer Landing See log market Logging Logging is the process of harvesting timber from trees. This includes felling, skidding, loading and transporting forest products, particularly logs. Pre-harvest inventory, tree and skid trail marking can be part of the process Log deck See log market 50 Code of Forest Harvesting Practices
Log market A cleared area, usually adjacent to the roadbed where logs are assembled after being skidded, awaiting subsequent handling, loading and transport. Also called ramp, log deck or landing Logging arch See integral arch m meter mm millimeter m2 square meters m3 cubic meters
Outlet: The opening of a drainage structure or culvert where the water leaves the structure. The outlet should be lower than the inlet to ensure that water flows through the structure Out-slope The outside cross slope of a road surface, typically measure in percent. Out-slope is used to
facilitate the draining of water from a road or trail directly off the outside edge of the road or trail. An out-sloped road or trail has the highest point on the uphill or inside of the road or trail and slopes down to the outside edge of the road Potential crop Stems of commercial species remaining after the cut and forming the basis tree of subsequent cutting cycles Riprap Layer of large, durable materials (usually stone or rock) used to protect exposed soil to minimize erosion Roadbed The formation between the ditches or tops of embankments, including the travel way and shoulders Roadway The formation between the extreme limits of the earthworks, from the top of the
cut slope to the toe of the fill or graded area. Also called road formation or width of
earthworks Road reserve The area that corresponds to the limit of the ground affected by the road, usually equals the width of clearing Rutting Road or skid trail surface damage in the form of deep tracks made by the passage of wheels or tracks. This typically a result of high wheel pressure on saturated or low
load bearing soils. These conditions worsen with heavy loads, high traffic volumes and
inclement weather conditions Scour Erosion or soil movement in a watercourse bed, bank, channel, or behind a drainage structure, typically caused by increased water velocity or lack of protection Sediment Fragments of rock, soil, and organic material transported and deposited in bed by water, wind or other natural phenomena Sedimentation Deposition of material suspended in water or air, usually when the velocity of the transportation medium drops below the level at which the material can be supported 51 Republic Of Liberia - Forestry Development Authority
Sediment trap See silt trap SFP State Forest Permit; non-exclusive permit allowing the holder to remove a certain quota of timber from an area, valid for one year
Shoulder The strip along the edge of the travel way on either side of the road, commonly flush with
the travel way for roads on stabilised soil. It is generally only used by passing vehicles but may be used for travel by track machines. Also called verge
Side cast Road construction material that is not used for fill and is pushed to or placed on the
down slope side of the road. Such material may travel long distances down slope before coming to rest Side drain See ditch Silt trap Hole created to divert sediment laden water, creating enough residence
time to allow solid material in suspension to drop out, before it is diverted
back into a body of water or drainage structure Sight distance The distance along a road or track that a driver can see other objects
(usually other vehicles) Skidding A method of ground-based extraction in which logs, poles or whole trees are dragged from the felling point to the log market, commonly by means of a tractor equipped with a cable-arch or a grapple known as a skidder, but also by means of farm tractors, crawler tractors, or bulldozers equipped with a winch or chains. Also called yarding Skid trail Trail along which a log is dragged by a extraction machine to the log market Soil erodibility The inherent susceptibility of a soil to erosion Soil erosion The process by which soil particles and aggregates are worn away and moved by the actions of wind or water in the from of raindrops, surface runoffs, and waves Spoon drain A shallow open drain, normally traversable by vehicles designed to carry water to the side of a road or skid trail Stakeholders Individuals or groups of individuals who have an interest in, or an impact on, the outcomes of a decision as well as groups or individuals dependent to some degree on the outcome for their personal or institutional goals
Stanchions Upright posts or supports for confining logs on trucks, trailers or other vehicles
Strategic plan Long term plan, which provides a broad description and broad details of future harvesting and management. 52 Code of Forest Harvesting Practices
Sustainable The process of managing forests to achieve one or more clearly specified objectives of management with regard to the production of a continuous flow of desired forest products
and services, without undue reduction of its inherent values and future productivity and without undue undesirable effect on the physical and social environment. Swamp A generally or permanently waterlogged area which may or may not have associated tree or palm vegetation; or a tract or low, poorly drained ground with patches of open water in which reeds, rushes and sedges occur. Swamp sediments are dominated by still water deposits, commonly with high organic content Table drain See ditch Temporary A crossing of a watercourse by a skid trail or road construction equipment designed for removal following short term use, having a designated opening to take typical peak
flows, e.g. a log culvert, and a cover of slash or small stems for a running surface Topping Severing the crown of a felled tree from the usable stem, usually at the fist heavy branch.
Also called junking Travel Way That portion of the road constructed for use by moving vehicles (excluding shoulders).
Also called Carriageway
TSC Timber Sales Contract; contract of less than duration of ≥ 3 years and a total area ≥
5,000 ha
Turbid water Water bearing significant quantities of soil particles
Turnout drain Excavations designed to divert water away from the ditch and roadway in order to reduce the volume and velocity of roadside ditch water. Also called outlet, lead-off, mitre or side drain Unstable areas Sites susceptible to one of the forms of mass soil movement or accelerated soil erosion as a result of the interaction of such factors as steepness, soil properties, parent and surface
geology and the position in the land form profile
Verge See shoulder Washboard A series of ridges and depressions across the road caused by soil and aggregate road
surfaces by the lack of surface cohesion. This typically is a result of the loss of fines in
the road surface caused by dry conditions or poorly graded material. These conditions
worsen with excessive vehicle speeds and high traffic volumes
Water body Watercourses and surface water such as lake, lagoon, sea or ocean
Watercourse Defined depression or channel that receives and conducts perennial or intermittent flows of surface water for part or all of the year in most years. Watercourses includes
rivers, creeks, gullies and waterways forest management crossing 53 Republic Of Liberia - Forestry Development Authority
Wedge A high impact plastic, aluminum or hardwood wedge driven into the back-cut to assist felling Wire rope Flexible twined metal alloy or steel rope to tie, pull or lift loads; in this context the cable by which logs are winched or attached to the skidder. Also cable Winch A rotating powered drum used to haul in or pay out a cable (wire rope) Wing wall Masonry, concrete or timber structures built onto the side of culvert inlet and outlet
headwalls or bridge abutments, designed to retain the roadway fill and direct water into
or out of the drainage structure or underneath the bridge while protecting the road and
fill from erosion. 54 Code of Forest Harvesting Practices
14 APPENDIX
DBH Cutting Limits
DIAMETER AT BREAST HEIGHT CUTTING LIMITS Species (Trade Name) Minimum Diameter Limit (cm) Species (Trade Name) Minimum Diameter Limit (cm)
Afzelia spp. (Doussie, Apa) 70 Alstonia boonei (Emien) 70 Aningeria robusta (Aningerie) 80 Antiaris africana (Ako) 60 Chlorophora excelsa (Iroko,
Odoum) 80 Bombax spp. (Kapokier) 70 Entandrophragma angolense (Tiama, Edinam) 90 Brachystegia leonensis (Naga) 90 Entandrophragma candollei (Kosipo) 90 Canarium schweinfurthii (Aiele) 80 Entandrophragma cylindricum (Sapeli) 90 Ceiba pentandra (Fromager) 90 Entandrophragma utile (Utile, Sipo) 100 Daniellia spp. (Fara) 70 Guarea cedrata (Bosse) 80 Didelotia spp. (Broutou, Zing, Bondu) 60 Guibourtia ehie (Amazakoue) 60 Distemonanthus benthamianus (Movingui) 80 Khaya spp. (Khaya, Acajou) 70 Erythrophleum spp. (Tali, Sassawood) 80 Lovoa trichilioides (Lovoa, Dibetou) 70 Gilbertiodendron spp. (Limbali) 60 Mansonia altissima (Bete) 60 Lophira alata (Azobe, Ekki) 80 Mitragyna ciliata (Abura, Bahia) 80 Mammea africana (Oboto,
Kaikumba) 60 Nesogordonia papaverifera (Kotibe) 60 Nauclea diderrichii. (Kusia, Bilinga) 80
Pycnanthus angolensis
(Ilomba) 70 Piptadeniastrum africanum (Dahoma, Mbeli) 80 Tarrietia utilis (Niangon) 60 Pterygota macrocarpa (Koto, Ake) 60 55 Republic Of Liberia - Forestry Development Authority
Terminalia ivorensis (Framire) 70 Sacoglottis gabonesis (Ozouga,
Akouapo) 70 Terminalia superba (Limba, frase, Afara) 70 For all other species, not listed above 60 Tetraberlinia tubmaniana (Sikon) 60 Tieghemella heckelii (Makore, Douka) 100 Triplochiton scleroxylon (Wawa,
Samba, Obeche) 90 Turreanthus africanus (Avodire) 80 56 Code of Forest Harvesting Practices
Slope Angles in Degrees and Per Cent
Degrees (°) Per cent (%) Per cent (%) Degrees (°)
1 2% 1% 0.6 2 3% 2% 1.1 3 5% 3% 1.7 4 7% 4% 2.3 5 9% 5% 2.9 6 11% 6% 3.4 7 12% 7% 4.0 8 14% 8% 4.6 9 16% 9% 5.1 10 18% 10% 5.7 11 19% 12% 6.8 12 21% 14% 8.0 13 23% 16% 9.1 14 25% 18% 10.2 15 27% 20% 11.3 16 29% 22% 12.4 17 31% 24% 13.5 18 32% 26% 14.6 19 34% 28% 15.6 20 36% 30% 16.7 21 38% 32% 17.7 22 40% 34% 18.8 23 42% 36% 19.8 24 45% 38% 20.8 25 47% 40% 21.8 26 49% 42% 22.8 27 51% 44% 23.7 28 53% 46% 24.7 29 55% 48% 25.6 30 58% 50% 26.6 31 60% 55% 28.8 32 62% 60% 31.0 33 65% 65% 33.0 34 67% 70% 35.0 35 70% 75% 36.9 57 Republic Of Liberia - Forestry Development Authority
36 73% 80% 38.7 37 75% 85% 40.4 38 78% 90% 42.0 39 81% 95% 43.5 40 84% 100% 45.0 41 87% 42 90% 43 93% 44 97% 45 100% I3564E |
10523 | https://math.stackexchange.com/questions/471014/an-irreducible-f-in-mathbbzx-whose-image-in-every-mathbbz-p-mathbb | number theory - An irreducible $f\in \mathbb{Z}[x]$, whose image in every $(\mathbb{Z}/p\mathbb{Z})[x]$ has a root? - Mathematics Stack Exchange
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An irreducible f∈Z[x]f∈Z[x], whose image in every (Z/p Z)x[x] has a root?
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Problem: Is there an irreducible f∈Z[x]f∈Z[x], whose image in every (Z/p Z)x[x] has a root for p p prime? If there is, what is the minimal degree possible?
I can only prove that x 2−c x 2−c is impossible, by quadratic reciprocity and Chinese remainder theorem. Even the case of a x 2−c a x 2−c is unknown to me.
Meanwhile, if f f is not required to be irreducible, but only have no root in Z Z, then (x 2−a)(x 2−b)(x 2−a b)(x 2−a)(x 2−b)(x 2−a b) is a solution for non-sqaure a,b,a b a,b,a b, since if both a,b a,b are not squares in Z/p Z Z/p Z, then a b a b is a square. Yet I still don't know if this is of minimal degree.
Also, it is natural to pose generalizations
When p p is not necessarily prime (equivalently for all prime powers).
When p p is odd prime.
When p p represents sufficiently large primes (equivalently all but a finite number left).
number-theory
irreducible-polynomials
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asked Aug 19, 2013 at 7:51
Ash GXAsh GX
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Maybe you should assume deg f>1 degf>1, or it will be trivial.Wei Zhan –Wei Zhan 2013-08-19 08:11:33 +00:00 Commented Aug 19, 2013 at 8:11
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No. In fact, if f f is an irreducible polynomial of degree at least 2 2 then there are infinitely many primes p p such that f f does not have a root mod p mod p.
The argument is standard and goes as follows. A theorem of Dedekind asserts that if f f factors as a product ∏f i(x)mod p∏f i(x)mod p of irreducible polynomials, and if p p does not divide the discriminant of f f, then some element of the Galois group of f f has cycle type (deg f 1,deg f 2,...)(degf 1,degf 2,...). The Frobenius density theorem (a weaker version of the Chebotarev density theorem) asserts that the converse holds in the following sense: if some element of the Galois group of f f has a given cycle type, then a positive proportion of primes has the property that the factorization of f mod p f mod p has the same cycle type. In particular, if some element of the Galois group of f f has no fixed points, then a positive proportion of primes has the property that f f has no roots mod p mod p.
But the Galois group of f f acts transitively on its roots, and we have the following general result.
Lemma: Let G G be a finite group acting transitively on a set S S of size greater than 1 1. Then some g∈G g∈G does not have a fixed point.
Proof. By Burnside's lemma,
1=1|G|∑g∈G|Fix(g)|1=1|G|∑g∈G|Fix(g)|
so the average number of fixed points of a random element of G G is 1 1. On the other hand, id∈G id∈G has |S|>1|S|>1 fixed points. Hence some element must have less than 1 1 fixed point, which can only mean that it has no fixed points. □◻
The conclusion follows.
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edited Aug 19, 2013 at 9:57
answered Aug 19, 2013 at 9:52
Qiaochu YuanQiaochu Yuan
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However, it is possible for f f to be reducible mod p mod p for all primes p p, and here the minimal degree is 4 4. The argument is very similar to the above: a necessary and sufficient condition is that the Galois group of f f doe snot contain an n n-cycle, where n=deg f n=degf.Qiaochu Yuan –Qiaochu Yuan 2013-08-19 09:59:31 +00:00 Commented Aug 19, 2013 at 9:59
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Looking at the Galois group of the polynomial as a subgroup G G of S n S n, we can translate your requirements as conditions on G G :
(i) f f has a root mod p p for every prime p p iff every element of G G has a fixpoint (by Cebotarev's theorem)
(ii) f f is irreducible in Q[x]Q[x] iff G G is transitive.
(iii) f f has a rational root iff every element of G G has a common fixpoint.
I don't think satisfying (i) and (ii) is possible.
(x 2−a)(x 2−b)(x 2−a b)(x 2−a)(x 2−b)(x 2−a b) gives the subgroup of S 6 S 6 generated by (12)(34)(12)(34) and (12)(56)(12)(56). It satisfies (i) and not (iii).
We can also achieve (i) and not (iii) with n=5 n=5 by using G G generated by (123)(123) and (12)(45)(12)(45). For example, with something like f(x)=(x 3−x+1)(x 2+23)f(x)=(x 3−x+1)(x 2+23)
Finally, we can enumerate the possible subgroups of S 2,S 3,S 4 S 2,S 3,S 4 to check that none of them satisfy (i) and not (iii), so n=5 n=5 is the minimal degree possible.
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edited Aug 19, 2013 at 9:30
answered Aug 19, 2013 at 9:22
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10524 | https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOoo_fdUnHNX9KsDY5aEvGMfhVKMvnHSk40wXQfQ-GHVXlFsfgXZj | Art of Problem Solving
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AM-GM Inequality
In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.
In symbols, the inequality states that for any real numbers , with equality if and only if .
The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
Contents
[hide]
1 Proofs
2 Generalizations
2.1 Weighted AM-GM Inequality
2.2 Mean Inequality Chain
2.3 Power Mean Inequality
3 Problems
3.1 Introductory
3.2 Intermediate
3.3 Olympiad
4 See Also
Proofs
Main article: Proofs of AM-GM
All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for .
Generalizations
The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.
Weighted AM-GM Inequality
The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.
Mean Inequality Chain
Main article: Mean Inequality Chain
The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.
Power Mean Inequality
Main article: Power Mean Inequality
The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
Problems
Introductory
For nonnegative real numbers , demonstrate that if then . (Solution)
Find the maximum of for all positive . (Solution)
Intermediate
Find the minimum value of for .
(Source)
Olympiad
Let , , and be positive real numbers. Prove that
(Source)
See Also
Proofs of AM-GM
Mean Inequality Chain
Power Mean Inequality
Cauchy-Schwarz Inequality
Inequality
Retrieved from "
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10525 | https://en.wikipedia.org/wiki/Approximation_error | Published Time: 2004-05-07T12:18:27Z
Approximation error - Wikipedia
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1 Formal definition
2 Examples
3 Comparison
4 Polynomial-time approximation of real numbers
5 Instruments
6 Generalizations
7 See also
8 References
9 External links
Approximation error
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From Wikipedia, the free encyclopedia
Mathematical concept
"Absolute error" redirects here and is not to be confused with Absolute deviation.
For broader coverage of this topic, see Approximation.
Graph of f(x)=e x{\displaystyle f(x)=e^{x}} (blue) with its linear approximation P 1(x)=1+x{\displaystyle P_{1}(x)=1+x} (red) at a = 0. The approximation error, visually represented as the vertical gap between the two curves, demonstrably increases for values of x that are positioned further from the point of approximation, which in this case is x = 0.
The approximation error in a given data value represents the significant discrepancy that arises when an exact, true value is compared against some approximation derived for it. This inherent error in approximation can be quantified and expressed in two principal ways: as an absolute error, which denotes the direct numerical magnitude of this discrepancy irrespective of the true value's scale, or as a relative error, which provides a scaled measure of the error by considering the absolute error in proportion to the exact data value, thus offering a context-dependent assessment of the error's significance.
An approximation error can manifest due to a multitude of diverse reasons. Prominent among these are limitations related to computing machine precision, where digital systems cannot represent all real numbers with perfect accuracy, leading to unavoidable truncation or rounding. Another common source is inherent measurement error, stemming from the practical limitations of instruments, environmental factors, or observational processes (for instance, if the actual length of a piece of paper is precisely 4.53 cm, but the measuring ruler only permits an estimation to the nearest 0.1 cm, this constraint could lead to a recorded measurement of 4.5 cm, thereby introducing an error).
In the mathematical field of numerical analysis, the crucial concept of numerical stability associated with an algorithm serves to indicate the extent to which initial errors or perturbations present in the input data of the algorithm are likely to propagate and potentially amplify into substantial errors in the final output. Algorithms that are characterized as numerically stable are robust in the sense that they do not yield a significantly magnified error in their output even when the input is slightly malformed or contains minor inaccuracies; conversely, numerically unstable algorithms may exhibit dramatic error growth from small input changes, rendering their results unreliable.
Formal definition
[edit]
Given some true or exact value v, we formally state that an approximation v approx estimates or represents v where the magnitude of the absolute error is bounded by a positive value ε (i.e., ε>0), if the following inequality holds:
|v−v approx|≤ε{\displaystyle |v-v_{\text{approx}}|\leq \varepsilon }
where the vertical bars, | |, unambiguously denote the absolute value of the difference between the true value v and its approximation v approx. This mathematical operation signifies the magnitude of the error, irrespective of whether the approximation is an overestimate or an underestimate.
Similarly, we state that v approx approximates the value v where the magnitude of the relative error is bounded by a positive value η (i.e., η>0), provided v is not zero (v ≠ 0), if the subsequent inequality is satisfied:
|v−v approx|≤η⋅|v|{\displaystyle |v-v_{\text{approx}}|\leq \eta \cdot |v|}.
This definition ensures that η acts as an upper bound on the ratio of the absolute error to the magnitude of the true value. If v ≠ 0, then the actual relative error, often also denoted by η in context (representing the calculated value rather than a bound), is precisely calculated as:
η=|v−v approx||v|=|v−v approx v|=|1−v approx v|{\displaystyle \eta ={\frac {|v-v_{\text{approx}}|}{|v|}}=\left|{\frac {v-v_{\text{approx}}}{v}}\right|=\left|1-{\frac {v_{\text{approx}}}{v}}\right|}.
Note that the first term in the equation above implicitly defines ε as |v-v_approx| if η is ε/|v|.
The percent error, often denoted as δ, is a common and intuitive way of expressing the relative error, effectively scaling the relative error value to a percentage for easier interpretation and comparison across different contexts:
δ=100%×η=100%×|v−v approx v|.{\displaystyle \delta =100\%\times \eta =100\%\times \left|{\frac {v-v_{\text{approx}}}{v}}\right|.}
An error bound rigorously defines an established upper limit on either the relative or the absolute magnitude of an approximation error. Such a bound thereby provides a formal guarantee on the maximum possible deviation of the approximation from the true value, which is critical in applications requiring known levels of precision.
Examples
[edit]
Best rational approximants for π (green circle), e (blue diamond), φ (pink oblong), (√3)/2 (grey hexagon), 1/√2 (red octagon) and 1/√3 (orange triangle) calculated from their continued fraction expansions, plotted as slopes y/x with errors from their true values (black dashes)
v
t
e
To illustrate these concepts with a numerical example, consider an instance where the exact, accepted value is 50, and its corresponding approximation is determined to be 49.9. In this particular scenario, the absolute error is precisely 0.1 (calculated as |50 − 49.9|), and the relative error is calculated as the absolute error 0.1 divided by the true value 50, which equals 0.002. This relative error can also be expressed as 0.2%. In a more practical setting, such as when measuring the volume of liquid in a 6 mL beaker, if the instrument reading indicates 5 mL while the true volume is actually 6 mL, the percent error for this particular measurement situation is, when rounded to one decimal place, approximately 16.7% (calculated as |(6 mL − 5 mL) / 6 mL| × 100%).
The utility of relative error becomes particularly evident when it is employed to compare the quality of approximations for numbers that possess widely differing magnitudes; for example, approximating the number 1,000 with an absolute error of 3 results in a relative error of 0.003 (or 0.3%). This is, within the context of most scientific or engineering applications, considered a significantly less accurate approximation than approximating the much larger number 1,000,000 with an identical absolute error of 3. In the latter case, the relative error is a mere 0.000003 (or 0.0003%). In the first case, the relative error is 0.003, whereas in the second, more favorable scenario, it is a substantially smaller value of only 0.000003. This comparison clearly highlights how relative error provides a more meaningful and contextually appropriate assessment of precision, especially when dealing with values across different orders of magnitude.
There are two crucial features or caveats associated with the interpretation and application of relative error that should always be kept in mind. Firstly, relative error becomes mathematically undefined whenever the true value (v) is zero, because this true value appears in the denominator of its calculation (as detailed in the formal definition provided above), and division by zero is an undefined operation. Secondly, the concept of relative error is most truly meaningful and consistently interpretable only when the measurements under consideration are performed on a ratio scale. This type of scale is characterized by possessing a true, non-arbitrary zero point, which signifies the complete absence of the quantity being measured. If this condition of a ratio scale is not met (e.g., when using interval scales like Celsius temperature), the calculated relative error can become highly sensitive to the choice of measurement units, potentially leading to misleading interpretations. For example, when an absolute error in a temperature measurement given in the Celsius scale is 1°C, and the true value is 2°C, the relative error is 0.5 (or 50%, calculated as |1°C / 2°C|). However, if this exact same approximation, representing the same physical temperature difference, is made using the Kelvin scale (which is a ratio scale where 0 K represents absolute zero), a 1 K absolute error (equivalent in magnitude to a 1°C error) with the same true value of 275.15 K (which is equivalent to 2°C) gives a markedly different relative error of approximately 0.00363, or about 3.63×10−3 (calculated as |1 K / 275.15 K|). This disparity underscores the importance of the underlying measurement scale.
Comparison
[edit]
When comparing the behavior and intrinsic characteristics of these two fundamental error types, it is important to recognize their differing sensitivities to common arithmetic operations. Specifically, statements and conclusions made about relative errors are notably sensitive to the addition of a non-zero constant to the underlying true and approximated values, as such an addition alters the base value against which the error is relativized, thereby changing the ratio. However, relative errors remain unaffected by the multiplication of both the true and approximated values by the same non-zero constant, because this constant would appear in both the numerator (of the absolute error) and the denominator (the true value) of the relative error calculation, and would consequently cancel out, leaving the relative error unchanged. Conversely, for absolute errors, the opposite relationship holds true: absolute errors are directly sensitive to the multiplication of the underlying values by a constant (as this scales the magnitude of the difference itself), but they are largely insensitive to the addition of a constant to these values (since adding the same constant to both the true value and its approximation does not change the difference between them: (v+c) − (v approx+c) = v − v approx).: 34
Polynomial-time approximation of real numbers
[edit]
In the realm of computational complexity theory, we define that a real value v is polynomially computable with absolute error from a given input if, for any specified rational number ε> 0 representing the desired maximum permissible absolute error, it is algorithmically possible to compute a rational number v approx such that v approx approximates v with an absolute error no greater than ε (formally, |v − v approx| ≤ ε). Crucially, this computation must be achievable within a time duration that is polynomial in terms of the size of the input data and the encoding size of ε (the latter typically being of the order O(log(1/ε)) bits, reflecting the number of bits needed to represent the precision). Analogously, the value v is considered polynomially computable with relative error if, for any specified rational number η> 0 representing the desired maximum permissible relative error, it is possible to compute a rational number v approx that approximates v with a relative error no greater than η (formally, |(v − v approx)/v| ≤ η, assuming v ≠ 0). This computation, similar to the absolute error case, must likewise be achievable in an amount of time that is polynomial in the size of the input data and the encoding size of η (which is typically O(log(1/η)) bits).
It can be demonstrated that if a value v is polynomially computable with relative error (utilizing an algorithm that we can designate as REL), then it is consequently also polynomially computable with absolute error. Proof sketch: Let ε> 0 be the target maximum absolute error that we wish to achieve. The procedure commences by invoking the REL algorithm with a chosen relative error bound of, for example, η = 1/2. This initial step aims to find a rational number approximation r 1 such that the inequality |v − r 1| ≤ |v|/2 holds true. From this relationship, by applying the reverse triangle inequality (|v| − |r 1| ≤ |v − r 1|), we can deduce that |v| ≤ 2|r 1| (this holds assuming r 1 ≠ 0; if r 1 = 0, then the relative error condition implies v must also be 0, in which case the problem of achieving any absolute error ε> 0 is trivial, as v approx = 0 works, and we are done). Given that the REL algorithm operates in polynomial time, the encoding length of the computed r 1 will necessarily be polynomial with respect to the input size. Subsequently, the REL algorithm is invoked a second time, now with a new, typically much smaller, relative error target set to η' = ε / (2|r 1|) (this step also assumes r 1 is non-zero, which we can ensure or handle as a special case). This second application of REL yields another rational number approximation, r 2, that satisfies the condition |v − r 2| ≤ η'|v|. Substituting the expression for η' gives |v − r 2| ≤ (ε / (2|r 1|)) |v|. Now, using the previously derived inequality |v| ≤ 2|r 1|, we can bound the term: |v − r 2| ≤ (ε / (2|r 1|)) × (2|r 1|) = ε. Thus, the approximation r 2 successfully approximates v with the desired absolute error ε, demonstrating that polynomial computability with relative error implies polynomial computability with absolute error.: 34
The reverse implication, namely that polynomial computability with absolute error implies polynomial computability with relative error, is generally not true without imposing additional conditions or assumptions. However, a significant special case exists: if one can assume that some positive lower bound b on the magnitude of v (i.e., |v| >b> 0) can itself be computed in polynomial time, and if v is also known to be polynomially computable with absolute error (perhaps via an algorithm designated as ABS), then v also becomes polynomially computable with relative error. This is because one can simply invoke the ABS algorithm with a carefully chosen target absolute error, specifically ε target = ηb, where η is the desired relative error. The resulting approximation v approx would satisfy |v − v approx| ≤ ηb. To see the implication for relative error, we divide by |v| (which is non-zero): |(v − v approx)/v| ≤ (ηb)/|v|. Since we have the condition |v| >b, it follows that b/|v| < 1. Therefore, the relative error is bounded by η × (b/|v|) <η × 1 = η, which is the desired outcome for polynomial computability with relative error.
An algorithm that, for every given rational number η> 0, successfully computes a rational number v approx that approximates v with a relative error no greater than η, and critically, does so in a time complexity that is polynomial in both the size of the input and in the reciprocal of the relative error, 1/η (rather than being polynomial merely in log(1/η), which typically allows for faster computation when η is extremely small), is known as a Fully Polynomial-Time Approximation Scheme (FPTAS). The dependence on 1/η rather than log(1/η) is a defining characteristic of FPTAS and distinguishes it from weaker approximation schemes.
Instruments
[edit]
In the context of most indicating measurement instruments, such as analog or digital voltmeters, pressure gauges, and thermometers, the specified accuracy is frequently guaranteed by their manufacturers as a certain percentage of the instrument's full-scale reading capability, rather than as a percentage of the actual reading. The defined boundaries or limits of these permissible deviations from the true or specified values under operational conditions are commonly referred to as limiting errors or, alternatively, guarantee errors. This method of specifying accuracy implies that the maximum possible absolute error can be larger when measuring values towards the higher end of the instrument's scale, while the relative error with respect to the full-scale value itself remains constant across the range. Consequently, the relative error with respect to the actual measured value can become quite large for readings at the lower end of the instrument's scale.
Generalizations
[edit]
This section needs expansion. You can help by adding to it. (April 2023)
The fundamental definitions of absolute and relative error, as presented primarily for scalar (one-dimensional) values, can be naturally and rigorously extended to more complex scenarios where the quantity of interest v{\displaystyle v} and its corresponding approximation v approx{\displaystyle v_{\text{approx}}} are n-dimensional vectors, matrices, or, more generally, elements of a normed vector space. This important generalization is typically achieved by systematically replacing the absolute value function (which effectively measures magnitude or "size" for scalar numbers) with an appropriate vector n-norm or matrix norm. Common examples of such norms include the L 1 norm (sum of absolute component values), the L 2 norm (Euclidean norm, or square root of the sum of squared components), and the L∞ norm (maximum absolute component value). These norms provide a way to quantify the "distance" or "difference" between the true vector (or matrix) and its approximation in a multi-dimensional space, thereby allowing for analogous definitions of absolute and relative error in these higher-dimensional contexts.
See also
[edit]
Accepted and experimental value
Condition number
Errors and residuals in statistics
Experimental uncertainty analysis
Machine epsilon
Measurement error
Measurement uncertainty
Propagation of uncertainty
Quantization error
Relative difference
Round-off error
Uncertainty
References
[edit]
^Weisstein, Eric W. "Numerical Stability". mathworld.wolfram.com. Retrieved 2023-06-11.
^Weisstein, Eric W. "Absolute Error". mathworld.wolfram.com. Retrieved 2023-06-11.
^ Jump up to: ab"Absolute and Relative Error | Calculus II". courses.lumenlearning.com. Retrieved 2023-06-11.
^"Approximation and Error Bounds". math.wpi.edu. Retrieved 2023-06-11.
^ Jump up to: abGrötschel, Martin; Lovász, László; Schrijver, Alexander (1993), Geometric algorithms and combinatorial optimization, Algorithms and Combinatorics, vol.2 (2nd ed.), Springer-Verlag, Berlin, doi:10.1007/978-3-642-78240-4, ISBN978-3-642-78242-8, MR1261419
^Helfrick, Albert D. (2005) Modern Electronic Instrumentation and Measurement Techniques. p. 16. ISBN81-297-0731-4
^Golub, Gene; Charles F. Van Loan (1996). Matrix Computations (Third ed.). Baltimore: The Johns Hopkins University Press. p.53. ISBN0-8018-5413-X.
External links
[edit]
Weisstein, Eric W."Percentage error". MathWorld.
Retrieved from "
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Numerical analysis
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Approximation error
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10526 | http://web.uvic.ca/~monahana/eos225/matlab_tutorial/tutorial_2/for_loops_and_m_files.html | An Introduction to MATLAB: For Loops and M-Files
An Introduction to MATLAB: For Loops and M-Files
Contents
The For Loop
M-Files
NxM Arrays
Exercises
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Exercise 7
Exercise 8
Exercise 9
Exercise 10
Exercise 11
The For Loop
We will now consider our next example of fundamental concepts in computer programming: the for loop (more generally, ``iteration'').
We talked in class about an example of an iterative calculation - radioactive decay. Starting with some amount of U238, we know that each billion years we lose a fraction alpha, that is,
U(2) = (1-alpha)U(1)U(3) = (1-alpha)U(2) = [(1-alpha)^2] U(1)U(4) = (1-alpha)U(3) = [(1-alpha)^3] U(1)
and so on. To predict U(n), we start with U(1) and iteratively multiply by the factor 1-alpha. How do we do this in MATLAB?
In this case, we have a formula predicting U(n) from U(1), so we can just use the exponentiation function - we know how to take the nth power of some factor. In general, iterative calculations don't admit closed-form expressions for the nth step in terms of the first. But there's another way to do the calculation - using the for loop.
Before talking about the for loop, it's worth emphasising the following fact: computers were invented to do exactly this kind of computation. Many important applications involve doing similar calculations repeatedly, with each step making use of the results of the previous step (that is, using iterations). This can be incredibly tedious to do by hand - and errors can creep in, through carelessness or boredom. Computers don't get bored - they'll happily do the same thing over and over again. What we're learning now is at the heart of how computers work.
Going back to our radioactive decay example: to compute the mass of U238 present after 3 billion years have elapsed from some initial time, we could just type out the sequence of calculations above by hand, one after the other. With the initial mass
U(1) = 0.5 kg:
we can write
clear all
alpha = 0.143;
U(1) = 0.5;
U(2) = (1-alpha)U(1);
U(3) = (1-alpha)U(2);
U(4) = (1-alpha)U(3);
U
U =
0.5000 0.4285 0.3672 0.3147
Note that we've had to type the same thing over and over: take the result of the previous calculation and multiply by the same factor. This is a nuisance, and impossible in practice for bigger calculations - what if you had to go out to n=50? What are the chances you'd make a typo? Pretty large in fact ...
Before carrying on: note the indices of the array U. U(1) is the first element of U - corresponding to the time n=0 in the notation of Section 1.1 of the notes. Remember that the indices of the array must be positive integers. U(1) corresponds to the amount of U238 at the first time considered - which in this case is the initial amount, at "time" n=0. Similarly, U(2) is the amount after the first billion years has elapsed, U(3) is the amount after the second billion years, etc.
The for loop provides a much more efficient way of calculating U(4) calculation:
clear all
alpha = 0.143;
U(1) = 0.5;
for k=2:4
U(k) = (1-alpha)U(k-1);
end
U
U =
0.5000 0.4285 0.3672 0.3147
We get the same answer, but with a more compact program.
Note the structure of the for loop: we only have to tell MATLAB once how the iteration works, from step k - 1 to step k. But what is k? It is a counter of where we are in the iteration. If we start with k = 2, then we get the step from k - 1 = 1 to k = 2 - that is, from U(1) to U(2). With k = 3, we get the step from k - 1 = 2 to k = 3 - that is, from U(2) to U(3) - and so on and so forth. This process stops when k takes the value of 4. As we will see, not all iterations involve taking the value from a previous step and using it in the present step - although this will often be the case.
The range of steps to be taken is listed at the top of the for loop - we told MATLAB to run k from 2 to 4 (in increments of 1, the default for the : operator).
This sequence of steps can be represented visually as a flow chart:
The program begins by defining the parameter alpha and the value of U(1). Starting with k = 2, it then computes U(k) from U(k-1). After doing this, the value of k is increased by 1. When k exceeds 4, the iteration stops and the array is written to the screen.
At each iteration, MATLAB does everything between the "for" and "end" statements in the loop. In the above example, that's a single calculation - but it doesn't have to be. The "end" command is very important here - it tells MATLAB where to end the sequence of commands making up the for loop.
For example, say we want MATLAB to compute the square, cube, and fourth power of all integers between 4 and 8. We can write the program:
clear all
x = 4:8;
for k=1:5
s(k) = x(k)^2;
c(k) = x(k)^3;
f(k) = x(k)^4;
end
s
c
f
s =
16 25 36 49 64
c =
64 125 216 343 512
f =
256 625 1296 2401 4096
The counter k which keeps track of where we are in the loop starts at one, and increases by one until it reaches 5. The counter starts at k=1, corresponding to the first element in the array x (which is x(1)=4) and ends with the value k=5, corresponding to the last element (which is x(5)=8). At each step, each of the square, cube, and fourth power are computed and stored in different arrays. Note that here the output of the kth step is saved in the kth element of each array.
This program can be represented graphically by the following flow chart:
We can make the sequence of commands between the "for" and "end" as complicated as we want - we can even have other for loops inside a for loop (called "nesting"). But the basic idea is just this: do a perscribed sequence of steps, over and over again, for a pre-determined number of times.
The counter k is a variable itself. In fact, the counter doesn't have to be called "k" - it can be called anything. For example, typing
clear all
x = 4:8;
for j=1:5
s(j) = x(j)^2;
c(j) = x(j)^3;
f(j) = x(j)^4;
end
s
c
f
s =
16 25 36 49 64
c =
64 125 216 343 512
f =
256 625 1296 2401 4096
gives exactly the same thing as with the counter called "k". The important thing to remember is not to give the counter the same name as another variable that you still want to use - particularly one you'll be modifying in the for loop. Anybody who's done much programming has spent many hours of their life debugging programs where that was the error.
Note that we could also get the desired results from the code above using the program
clear all
x = 4:8;
s = x.^2;
c = x.^3;
f = x.^4;
s
c
f
s =
16 25 36 49 64
c =
64 125 216 343 512
f =
256 625 1296 2401 4096
The use of a for loop here isn't necessary - there is often more than one way to carry out the same calculation, using different coding tools.
In the earlier for loop example, computing the mass of uranium as a function of time, we used the value of the k-1 step to compute the value at the k step. We could also write the loop such that we use the value of the k step to get the value at k+1:
clear all
alpha = 0.143;
U(1) = 0.5;
for k=1:3
U(k+1) = (1-alpha)U(k);
end
U
U =
0.5000 0.4285 0.3672 0.3147
The output of this program is exactly equal to that of the earlier one - because it is carrying out the same iteration, albeit written in a somewhat different form. Note that we have had to change the range of k values: as the first U value we are computing in the iteration is the second element of the array, then if we're identifying this step as going from k to k+1, we need the first value of k to be 1 so that k+1=2. If the last value of U we're going to calculate is the 4th element of the array, then the last value of k we need is 3 (so that k+1 = 4).
The increment in a for loop doesn't have to be 1; for example we can have
clear all
for j=2:2:8
3j
end
ans =
6
ans =
12
ans =
18
ans =
24
In this case, the counter j goes up in steps of 2. A for loop can have any increment (unlike array indices which must be positive integers).
The general structure of for loops can be expressed:
for (counter) = (start):(increment):(end)
(stuff to be done in for loop)end
Let's go back to the radioactive decay example, and see how Figure 1.1 in the notes was made.
We want to see how U(t) changes from t=0 Ga to t=25 Ga , starting from U(0). If we define the ratio
r(t) = U(t)/U(0)
(where r is to be interpreted as a function of time - not as a MATLAB array) then from above we have the iteration
r(t) = (1-alpha)r(t-1)
with r(0) = 1 (where as written the argument of the function r is the time t=0). So if we write the program
clear all
alpha = 0.143;
t = 0:25;
r(1) = 1;
for k=2:26;
r(k) = (1-alpha)r(k-1);
end
plot(t,r)
xlabel('t (Ga)');
ylabel('U(t)/U(0)');
we get pretty much the plot in Figure 1.1 Note that because array indices must be positive, t = 0 corresponds to k=1 and t=25 Ga corresponds to k=26. In this MATLAB script, the quantity
r(k)
is not the value of r at the time k - it is the value of the kth element of the array r. If the kth element of the array r corresponds to the kth element of the array t, then in particular the first element of the array t (which has a value equal to 0) will correspond to the first element of the array r (which has a value equal to 1). When interpreting an expression, it is crucially important to remember if it is meant to represent algebra or computer code.
It would have been a lot of work to type out the iteration 25 times - for this reason, the for loop is an essential part of computer programming.
Note that the figure we got above isn't exactly Figure 1.1, which has a bunch of other stuff on it. In fact, Figure 1.1 was produced by the following code
clear all
t = 0:1:25;
alpha = .143;
U = (1-alpha).^t;
plot(t,U,'k','linewidth',2)
xlabel('t (Ga)','fontsize',24);
tmp=ylabel('U(t)/U(0)','fontsize',24);
set(gca,'fontsize',18);
hold on
plot(4.5,0.025,'^','markerfacecolor','k','markeredgecolor','k','markersize',18);
t2 = 0:.01:1;
plot(4.5ones(size(t2)),t2,'--k')
hold off
text(5,.15,'Age of Earth','fontsize',15)
While we haven't talked about all the things you need to understand everything in the above code, it's still worth looking at. When programming, it's often useful to examine how other people have written code - it's often easier to learn by example, seeing how other people have done things, rather than trying to do everything from first principles. Searching online is also a valuable way of figuring out how to do something new in programming.
M-Files
Look at the previous piece of code - it's 13 lines long, with some fairly complicated bits. Typing this out fully each time you want to make this Figure would be tedious - and error-prone. Now imagine a weather prediction program, which might have millions of lines of code - there's no way this can be retyped in full every 6 hours to make a forecast.
Programming languages get around this by storing programs in separate computer files. In MATLAB, these are known as M-files and always end with the suffix .m ("dot-m"). When you have the M-file "sample_program.m", you can run it by simply typing
sample_program
on the command line (note that you don't include the suffix ".m" in what you type).
Working with M-files is crucial, particularly if there's a program you're going to use multiple times (and so don't want to have to retype it in full each time), if the program has complicated lines of code in which it's easy to make a typo, or if the program is very long.
Programming M-files is no harder than programming on the command line - you just type out the sequence of commands in the order that you want them executed (just like you would on the command line) and that's the order in which MATLAB carries them out.
To make a new M-file, click on "New Script" icon in the upper left of the MATLAB window. This will open a new text editor window - a window you can type your program in.
A nice thing about M-files is that they can include notes to yourself (and others) about what the M-file does, both overall and in particular places. These are called "comments", and anything on a line appearing after a percentage sign "%" is "commented out". That is, anything on that line following the "%" will appear in the M-file MATLAB won't pay any attention to it.
Now, type the following program into the text editor
% My test program a = 3;b = 2;c = 1:4;d = a(c.^b);plot(c,d);
Right now, this text hasn't been saved as a file - it's just sitting in the text editor, and can't be run by MATLAB. To be able to run the program, you'll need to save the M-file. To do this, press the "Save" icon above your script (to the left). This will give you a menu where you can pick what you'll call the file. Call this one
my_test_program.m
Now you can go to the MATLAB command window and at the command line type
my_test_program
Up comes the plot of the array c against the array d. We can ask for the values of the array d
d
d =
3 12 27 48
or of the variable b
b
b =
2
It's all as if we'd typed each line in the command window. But here's one major advantage - say we want to look at the plot of c against d again, but now for the case that a = 4. If we'd typed each line in the command window, we'd have to go back and start typing them in again, right from the beginning. But we can go to the M-file and edit the first line to read
a = 4
instead. To run this new M-file, we first need to save it as a new file - using the "Save As" option under the "File" menu. Let's call this new file "my_test_program_2.m" Once the file is saved, we run the new M-file
my_test_program_2
and get another plot, now for the value a=4.
These aren't very realistic examples so far. Let's consider working with Eqn. (2.10) for the growth of a mountain range under the influence of tectonic uplift and erosion (neglecting isostatic effects). To start with, we open a clean new text editor. Now type the following
% Orogeny example, Eqn. (2.10)beta = 0.1e-3; %% m/yr tau = 5e6; %% yr h0 = 0; %% initial topgraphy zero t = 0:1e5:25e6; %% time in steps of 100,000 years from t=0 to t=25^6 years h = betatau + (h0-betatau)exp(-t/tau);plot(t/10^6,h/1e3)xlabel('t (Ma)')ylabel('h (km)')
Notice how the comments have been used to remind us what the units of different variables are. Now save this program under the name
orogony_example_1.m
Running this m-file gives the plot of h(t) against t:
orogeny_example_1
We see the familiar increase in h(t) with t, approaching the steady state value h_eq = 500 m.
Note the trick we've played in the plot: we didn't exactly plot t with h. Rather, we plotted t/1e6 with h/1e3. Because t is in years and h is in metres, doing this means that the plot's x-axis is in units of Ma (10^6 years) and the y-axis is in units of km (10^3 m) - as we've noted in the axis labels.
Whenever you work with dimensional quantities in MATLAB (or other programming languages) it is important to be careful with units. MATLAB has no idea of what the units of numbers in memory are - they're just numbers. If you enter the initial height of topography to be 10, MATLAB won't know if this is 10 m, 10 km, or 10 mm. When working with dimensional quantities, you have to decide at the start what units you are going to use - how are you going to measure length (m? mm? km? another length?), how you will measure time (seconds? hours? years? another length?), how you will measure mass, etc. You can use any units you want - but you have to be consistent with them in your calculations.
NxM Arrays
So far, we have only considered arrays with a single index. E.g.
M = [82 , -1.8 , 6]
for which
M(1) = 82 M(2) = -1.8 M(3) = 6
In fact, MATLAB allows arrays with more than one index. For the case of an array with two indices, you can think of it as a set of numbers arranged on a grid with rows and columns. E.g.
M = 1 3 -12 6 0.1 8 -0.3 7 12 4 -4 4
This array has 4 rows and 3 columns - so is called a 4x3 array. In general, an array with M rows and N columns is referred to as being MxN. The single-index arrays we've already talked about can be considered 1xN arrays.
You can think of the MxN array as a container with many small compartments (like an egg carton) arranged in a grid. Each compartment holds a number. Each compartment is also identified by a pair of numbers - the first indicates the row, the second the column. The numbers identifying the compartment (the indices) are both integers - counting rows (from the top) and columns (from the left) - but any kind of number can be IN the compartment. In MATLAB indices must take a value of one or greater.
We can look at individual elements of such an array. We use the notation:
M(i,j)
is the element of M in the ith row (counting from the top) and the jth column (counting from the left). In the example above
M(1,2) = 3 M(3,3) = 12 M(4,2) = -4
To put the above array into MATLAB, type the following in the command line:
M = [ 1 3 -12 ; 6 0.1 8 ; -0.3 7 12 ; 4 -4 4]
M =
1.0000 3.0000 -12.0000
6.0000 0.1000 8.0000
-0.3000 7.0000 12.0000
4.0000 -4.0000 4.0000
The array is entered row by row, separated by semicolons (you could use commas to separate the column values within a particular row, but you don't have to).
We can ask MATLAB for the values of individual array elements
M(1,2)
ans =
3
M(3,3)
ans =
12
M(4,2)
ans =
-4
or for parts of whole rows or columns. For example
M(2:4,1)
ans =
6.0000
-0.3000
4.0000
gives the elements in the first column from rows 2 through 4
M(3,1:2)
ans =
-0.3000 7.0000
gives the elements in the columns 1 through 2 in row 3
M(:,2)
ans =
3.0000
0.1000
7.0000
-4.0000
gives all elements in column 2 and
M(3,:)
ans =
-0.3000 7.0000 12.0000
gives all elements in row 3.
Such higher-dimensional arrays are useful for all sorts of things, as we'll see in the following example.
Let's compare the predictions made by the model of orogeny for different values of the initial topographic height h(0). Type the following into a new M-file:
% Orogeny example, Eqn. (2.10)beta = 0.1e-3; %% m/yr tau = 5e6; %% yr h0 = [0 250 750 1000];t = 0:1e5:25e6; %% time in steps of 100,000 years from t=0 to t=10^6 years for k=1:length(h0) h(k,:) = betatau + (h0(k)-betatau)exp(-t/tau);end plot(t/10^6,h(1,:)/1e3,'r')hold on plot(t/10^6,h(2,:)/1e3,'g')plot(t/10^6,h(3,:)/1e3,'b')plot(t/10^6,h(4,:)/1e3,'k')hold off xlabel('t (Ma)')ylabel('h (km)')legend('0 m','250 m','750 m','1000 m')
and save the M-file as
orogeny_example_2.m
Running this M-file:
orogeny_example_2
This plot has four curves h(t), one each for
h(0)=0 m
h(0)=250 m
h(0)=750 m
and
h(0)=1000 m.
Each curve approaches the steady state value as t increases: in those cases where the topography starts below the steady state value, h(t) increases with time; in those cases in which the topography starts below the steady state value, h(t) decreases with time.
Look at how we made this plot. First, we didn't define a single initial condition h(0) - we defined an array with four values. Each of these values h(0) gives a separate prediction for h(t) - and these were computed iteratively with a for loop.
The index k, numbering the index of the array holding the sequence of initial conditions, cycles from 1 to "length(h0)". Here we've used a new (and very useful) MATLAB command:
length(a)
which returns the number of columns in the array a. In the present example, h0 is a 1x4 array so
length(h0) = 4.
For each value of k, the function h(t) from Eqn. (2.10) is evaluated using the k'th value of the h(0) - that is,
h0(k)
For each k, this is a 1 x length(t) array. This array is stored as the kth row of a
length(h0) x length(t)
array h. That is, in the array h, the columns are associated with different points in time and the rows with different values of the initial topography. Note the syntax:
h(k,:)
refers to all columns (in this case, denoting points in time) in row k of the array h (which in this case refers to the k'th initial condition).
Once the loop is done filling up the array h, each row is plotted on the same graph. How was this done?
Well, first h(1,:) was plotted. Then we used the
hold on
command. When this is used, all subsequent plots will appear over top of the present plot, until the command
hold off
is given. Having "held" the plot, the remaining three h(k,:) curves were plotted.
Each curve was plotted in a different colour: in order, these were red, green, blue, black. This was done by specifying the colour in the plot command:
plot(a,b,'c')
will plot array a against array b with colour c. Note that c must be in single quotes, as it is a character and not a number.
Each colour is denoted by a single character, e.g.
r = red g = green b = blue k = black
There are other choices, like
y = yellow m = magenta c = cyan
These colour choices - as well as a lot of other information about the "plot" command - can be found by typing
help plot
The command "legend" allows us to identify the curves on the plot, in the order in which they were plotted. The command
legend('name 1','name 2','name 3', .... , 'name N')
produces a little window on the plot with the name 'name 1' associated with the first line plotted, 'name 2' with the second line plotted, and so on until 'name N' for the Nth line plotted.
Exercises
Exercise 1
Write a program that prints out the numbers 5 through -5 in decreasing order, and save it as an M-file.
Exercise 2
Write a program that creates an array with the cube of every third integer starting at one, up to 16.
Exercise 3
Consider the iteration x(k+1) = 1.2x(k)-0.2. Compute the first 10 values of this iteration starting with x(1)=3.
Exercise 4
Repeat Exercise 3 with x(1)=0.9. Describe the difference in the behaviour of this iteration for this initial value.
Exercise 5
Suppose that you have a $20,000 loan at 6 percent annual interest compounded monthly. You can afford to pay $200/month toward this loan. How long will it take until the loan is paid off? To do this calculation, think of the amount remaining on the loan from time k to time k+1 as a combination of interest accrued and payment made, express this as an iteration, and then write code using a for loop to find the amount remaining in the loan after N months. You may need to experiment with the value of N to find the value in which the remaining amount passes through zero (and the loan is paid off). Remember how monthly compounding of interest works: if your annual interest rate is P percent, the monthly interest is P/12 percent.
Exercise 6
Repeat Exercise 5 with a $75 monthly payment.
Exercise 7
Write a program that defines the array M = [2 -12 8 7] and then uses a for loop to add the elements of the array. To do this, first express the sum as an algorithm: start with 2, then add -12. To this earlier sum add 8, and then add 7. Note the algorithm: at each step (after the first), take the previous value of the running sum and add the new value corresponding to that step.
Exercise 8
Write a program that adds the numbers 1 through 100, and save it as an M-file.
Exercise 9
Write a program that adds the even numbers between 1 and 100, and save it as an M-file.
Exercise 10
Write a program that plots - on the same plot - h(t) from Eqn. (2.10) for
tau = 1 million years,
tau = 3 million years,
tau = 5 million years,
tau = 7 million years,
and
tau = 9 million years
for h(0) = 500 m, from t = 0 years to t = 50 million years, in increments of 100,000 years. Use beta = 0.1 mm/yr. Plot each line with a different colour, label the axes, and include a legend.
Exercise 11
Write a program that plots - on the same plot - h(t) from Eqn. (2.10) for
beta = 0.5 mm/yr,
beta = 0.75 mm/yr,
beta = 1.0 mm/yr,
beta = 1.25 mm/yr,
and
beta = 1.5 mm/yr
for tau = 2 million years and h(0) = 500 m, from t = 0 years to t = 20 million years, in increments of 100,000 years. Plot each line with a different colour, label the axes, and include a legend.
Published with MATLAB® R2020a |
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Telescoping series
In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. This is often done by using a form of for some expression .
Contents
[hide]
1 Example 1
2 Solution 1
3 Example 2
4 Solution 2
5 Problems
5.1 Introductory
5.2 Intermediate
5.3 Olympiad
6 See Also
Example 1
Derive the formula for the sum of the first counting numbers.
Solution 1
We wish to write for some expression . This expression is as .
We then telescope the expression:
.
(Notice how the sum telescopes— contains a positive and a negative of every value of from to , so those terms cancel. We are then left with , the only terms which did not cancel.)
Example 2
Find a general formula for , where .
Solution 2
We wish to write for some expression . This can be easily achieved with as by simple computation.
We then telescope the expression:
.
Problems
Introductory
When simplified the product becomes:
(Source)
The sum can be expressed as , where and are positive integers. What is ? (Source)
Which of the following is equivalent to (Hint: difference of squares!)
(Source)
Intermediate
Let denote the value of the sum can be expressed as , where and are positive integers and is not divisible by the square of any prime. Determine . (Source)
Olympiad
Find the value of , where is the Riemann zeta function
See Also
Algebra
Summation
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10528 | https://pmc.ncbi.nlm.nih.gov/articles/PMC91484/ | Key Physiology of Anaerobic Ammonium Oxidation - PMC
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Appl Environ Microbiol
. 1999 Jul;65(7):3248–3250. doi: 10.1128/aem.65.7.3248-3250.1999
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Key Physiology of Anaerobic Ammonium Oxidation
Marc Strous
Marc Strous
1 Department of Biotechnology, Delft University of Technology, 2628 BC Delft, The Netherlands
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1, J Gijs Kuenen
J Gijs Kuenen
1 Department of Biotechnology, Delft University of Technology, 2628 BC Delft, The Netherlands
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1, Mike S M Jetten
Mike S M Jetten
1 Department of Biotechnology, Delft University of Technology, 2628 BC Delft, The Netherlands
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1,
Author information
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1 Department of Biotechnology, Delft University of Technology, 2628 BC Delft, The Netherlands
Corresponding author. Mailing address: Department of Biotechnology, Delft University of Technology, Julianalaan 67, 2628 BC Delft, The Netherlands. Phone: 31-15-2781193. Fax: 31-15-2782355. E-mail: M.Jetten@STM.TUDelft.nl.
Received 1998 Dec 22; Accepted 1999 Apr 6.
Copyright © 1999, American Society for Microbiology
PMC Copyright notice
PMCID: PMC91484 PMID: 10388731
Abstract
The physiology of anaerobic ammonium oxidizing (anammox) aggregates grown in a sequencing batch reactor was investigated quantitatively. The physiological pH and temperature ranges were 6.7 to 8.3 and 20 to 43°C, respectively. The affinity constants for the substrates ammonium and nitrite were each less than 0.1 mg of nitrogen per liter. The anammox process was completely inhibited by nitrite concentrations higher than 0.1 g of nitrogen per liter. Addition of trace amounts of either of the anammox intermediates (1.4 mg of nitrogen per liter of hydrazine or 0.7 mg of nitrogen per liter of hydroxylamine) restored activity completely.
Anaerobic ammonium oxidation (anammox) is the biological conversion of ammonium and nitrite to dinitrogen gas. It is an almost completely unexplored part of the biological nitrogen cycle, and the process offers new opportunities for wastewater engineers and microbiologists (2). Previously, a morphologically unusual microorganism was enriched while the anammox activity increased (6), and a novel metabolic pathway was postulated (5). This pathway proceeded via hydroxylamine and hydrazine. In the present study, we have investigated the physiology of aggregated anammox biomass (80% enriched in the morphotypical microorganism) grown in a sequencing batch reactor (4). The compilation of this physiological information is the key to the future design and scale-up of the anammox process and will greatly facilitate microbiological investigation of the responsible microorganism(s).
Physiological temperature and pH ranges.
The maximum specific-substrate (ammonium and nitrite) conversion rate of aggregated anammox biomass was measured as a function of temperature and pH in batch experiments, in the absence of mass transfer limitations (see Table 1). From the temperature dependency of anammox activity, the activation energy was calculated to be 70 kJ/mol, approximately the same as for aerobic ammonium oxidation (7).
TABLE 1.
Important physiological parameters for anaerobic and aerobic ammonium oxidation
| Parameter | Anammox result | Nitrificationa result | Unit |
--- --- |
| Maximum specific aerobic NH 4+ consumption rate | 0 | 2–5 | g of NH 4+-N · g of protein−1 day−1 |
| Maximum specific anaerobic NH 4+ consumption rate | 1.1 | <0.05b | g of NH 4+-N · g of protein−1 day−1 |
| Biomass yield | 0.07 | 0.1 | g of protein · g of NH 4+-N−1 |
| Activation energy | 70 | 70 | kJ · mol−1 |
| Affinity for ammonium | ≤10−4 | ≥10−4 | g of NH 4+-N · liter−1 |
| Affinity for nitrite | ≤10−4 | NAc | g of NO 2−-N · liter−1 |
| Nitrite inhibition of ammonium consumption | K i = 0.8, α = 0.8 | Usually | g of NO 2−-N · liter−1 |
| Nitrite inhibition of nitrite consumption | K i = 1, α = 0.7 | NA | g of NO 2−-N · liter−1 |
| Temp range | 20–43 | ≤42°C | °C |
| pH range | 6.7–8.3 | Variable |
| Protein content of biomass | 0.6 | Variable | g of protein · g total dry weight−1 |
| Protein density | 50 | Variable | g of protein · liter biomass−1 |
Open in a new tab
a
Data were obtained as described in reference 7, except where noted.
b
As described in reference 8
c
NA, not applicable.
Affinity.
Before the substrate affinities were measured, the biomass aggregates were partially disrupted (by intense magnetic stirring), to reduce mass transfer limitation at low substrate concentrations. Figure 1A shows the aggregate size distribution during aggregate disruption. The affinity was measured after 5 h, when 80% of the aggregates had diameters less than 50 μm. Figures 1B and C show that the substrate decreases during the experiment were linear down to 0.15 mg of nitrogen per liter for ammonium and 0.05 mg of nitrogen per liter (the detection limit) for nitrite. Beyond these low substrate concentrations, the deviation from zero-order kinetics could have been caused by the approach of the affinity constant or by mass transfer limitation. Therefore, we concluded that the anammox affinity constants for ammonium and nitrite were equal to or less than 0.1 mg of nitrogen per liter (Table 1).
FIG. 1.
Open in a new tab
Determination of the substrate affinity of the anammox process. Decrease of the aggregate size during the aggregate disruption period (A). Nitrite (B) and ammonium (C) concentrations plotted against time, measured after a 5-h aggregate disruption period (arrow in A). Error bars represent standard deviations.
Inhibition.
The anammox process was not inhibited by ammonium or by the by-product nitrate up to concentrations of at least 1 g of nitrogen per liter. However, in the presence of more than 0.1 g of nitrite nitrogen per liter, the process was completely inhibited. This nitrite inhibition could be overcome by addition of trace amounts of either of the anammox intermediates (>1.4 mg of nitrogen per liter for hydrazine, >0.7 mg of nitrogen per liter for hydroxylamine). After a hydrazine pulse of 3 mg of nitrogen per liter of N 2 H 4 was used to start the anammox reaction, the resulting anammox activity was measured between pH 7 and 7.8 at different nitrite concentrations. It appeared that the anammox activity decreased with increasing nitrite concentration. This decrease was independent of pH at the pH range tested (7 to 7.8). Anammox activity is plotted against nitrite concentration in Fig. 2 (12 different batch experiments at 6 different substrate concentrations at pH 7, 7.4, and 7.8).
FIG. 2.
Open in a new tab
Immediate inhibition of the anammox process by nitrite. Specific ammonium (●) and nitrite (■) consumption rates. Lines indicate the fit of the Luong model to the data. Error bars represent standard deviations. Prot, protein.
Figure 2 also shows that with increasing nitrite concentration the stoichiometry of ammonium and nitrite consumption changed from 1.3 g of NO 2−-N/g NH 4+-N at 0.14 g of nitrogen per liter to almost 4 g of NO 2−-N/g NH 4+-N at 0.7 g of nitrogen per liter. From the distorted stoichiometry at high nitrite concentrations, it was clear that the microorganisms under these conditions did not only use ammonium as the electron donor but also must have generated an internal electron donor to reduce the nitrite.
Three different substrate inhibition models (Edwards, Andrews, and Luong models) were fitted to the data (3). For each model, the parameters that gave the least residual sum of squares were calculated, and R 2 was calculated for the linearized models as an estimate of the validity of the obtained curves. The calculated sums of squares were used to compare the models in an F test (1).
The Luong model fitted the experimental data best (R values were NH 4+ = 0.92 and NO 2− = 0.84). The coefficients were K i = 0.8 g of nitrogen per liter and α = 0.8 for inhibition of ammonium oxidation by nitrite, and K i = 1 g of nitrogen per liter and α = 0.7 for inhibition of nitrite reduction by nitrite. The F test showed that, with 73 and 85% respective certainty, the Luong model was better than the Edwards and Andrews models.
The hydrazine or hydroxylamine added at the start of each experiment was consumed very rapidly (within 10 min), but the durations of these experiments were much longer (up to 4 h). Apparently, the biomass only needed the N 2 H 4 or NH 2 OH pulse to get started. Still, it made sense to investigate how long the biomass would be able to remain active in the presence of high nitrite concentrations, after the N 2 H 4 had disappeared.
Prolonged exposure to nitrite.
Nitrite inhibition over a 50-h period was investigated in fed-batch experiments conducted under ammonium limitation with different resultant nitrite concentrations (0.06, 0.09, and 0.2 g of nitrogen per liter). In Fig. 3, the results are compared with a control experiment under nitrite limitation. In the control experiment and at 0.06 g of nitrite nitrogen per liter (Fig. 3A and B), ammonium and nitrite consumption and nitrate production remained constant over the course of the experiment, and the stoichiometry was as expected.
FIG. 3.
Open in a new tab
Consumption of ammonium (●) and nitrite (■) (shown as negative values), and production of nitrate (□) by the anammox process in 50-h fed-batch experiments. (A) Nitrite concentration = 0; (B) nitrite concentration = 0.06 g of nitrogen per liter; (C) nitrite concentration = 0.09 g of nitrogen per liter; (D) nitrite concentration = 0.2 g of nitrogen per liter (ammonium production after 10 h).
At 0.09 g of nitrite nitrogen per liter, the stoichiometry changed during the course of the experiment (Fig. 3C), much like the stoichiometric change illustrated in Fig. 2. The ratio of nitrite consumption to ammonium consumption increased to 2. At 0.2 g of nitrite nitrogen per liter, the anammox process proceeded normally only for the first 4 h of the experiment (Fig. 3D). After 4 h, the stoichiometry changed like in the previous experiments, but more strongly. Within 20 h, ammonium was produced in this experiment. Note that in Fig. 2 the inhibition of the anammox process was not yet apparent at 0.2 g of nitrite nitrogen per liter. After 20 h, two aliquots from the biomass described in Fig. 3D were reincubated in two different batch experiments with 0.26 g of nitrogen per liter of ammonium and nitrite. To one of these batches, hydrazine was added to a final concentration of 3 mg of nitrogen per liter. In this batch, the anammox activity was completely recovered, as shown by rapid conversion of ammonium and nitrite and nitrate production with normal stoichiometry. In the control experiment without hydrazine addition, the activity was not restored—no ammonium and no nitrite were converted. Apparently, the anammox process is strongly stimulated by its intermediates.
The physiological parameters of anaerobic ammonium oxidation are compiled in Table 1. This table also shows the parameters for aerobic nitrifiers, because these organisms resemble anammox from a physiological point of view. Both organisms are specialists: anammox has no aerobic activity, and the anaerobic activity of aerobic nitrifiers is only 5% that of the anammox activity. However, biomass yield, temperature range, substrate affinity, and activation energy are approximately the same for these anaerobic and aerobic organisms. Only the maximum specific-substrate conversion is much lower for anammox. Combined with the slightly lower biomass yield, this leads to the dramatic doubling time of 9 days under optimal conditions.
Acknowledgments
This research was financially supported by the Foundation for Applied Sciences (STW), the Royal Netherlands Academy of Arts and Sciences (KNAW), and the Gist-brocades and DSM companies.
We gratefully acknowledge Katinka van de Pas-Schoonen and Anke de Bruyn for careful maintenance of bench scale and pilot reactors, Mark van Loosdrecht for critical reading of the manuscript, and Sebastiaan Peeters for assistance with inhibition experiments.
REFERENCES
1.Elmen J, Pan W, Leung S Y, Magyarosy A, Keasling J D. Kinetics of toluene degradation by a nitrate-reducing bacterium isolated from a groundwater aquifer. Biotechnol Bioeng. 1997;55:82–90. doi: 10.1002/(SICI)1097-0290(19970705)55:1<82::AID-BIT10>3.0.CO;2-5. [DOI] [PubMed] [Google Scholar]
2.Jetten M S M, Strous M, van de Pas-Schoonen K T, Schalk J, van Dongen L, van de Graaf A A, Logemann S, Muyzer G, van Loosdrecht M C M, Kuenen J G. The anaerobic oxidation of ammonium. FEMS Microbiol Rev. 1998;22:421–437. doi: 10.1111/j.1574-6976.1998.tb00379.x. [DOI] [PubMed] [Google Scholar]
3.Luong J H T. Generalization of Monod kinetics for analysis of growth data with substrate inhibition. Biotechnol Bioeng. 1987;29:242–248. doi: 10.1002/bit.260290215. [DOI] [PubMed] [Google Scholar]
4.Strous M, Heijnen J J, Kuenen J G, Jetten M S M. The sequencing batch reactor as a powerful tool for the study of slowly growing anaerobic ammonium-oxidizing microorganisms. Appl Microbiol Biotechnol. 1998;50:589–596. [Google Scholar]
5.Van de Graaf A A, de Bruin P, Robertson L A, Jetten M S M, Kuenen J G. Metabolic pathway of anaerobic ammonium oxidation on basis of 15N-studies in a fluidized bed reactor. Microbiology (UK) 1997;143:2415–2421. doi: 10.1099/00221287-143-7-2415. [DOI] [PubMed] [Google Scholar]
6.Van de Graaf A A, de Bruijn P, Robertson L A, Jetten M S M, Kuenen J G. Autotrophic growth of anaerobic ammonium-oxidizing micro-organisms in a fluidized bed reactor. Microbiology (UK) 1996;142:2187–2196. [Google Scholar]
7.Wiesman U. Biological nitrogen removal from wastewater. Adv Biochem Eng Biotechnol. 1997;51:113–153. doi: 10.1007/BFb0008736. [DOI] [PubMed] [Google Scholar]
8.Zart D, Bock E. High rate of aerobic nitrification and denitrification by Nitrosomonas eutropha grown in a fermentor with complete biomass retention in the presence of gaseous NO2 or NO. Arch Microbiol. 1998;169:282–286. doi: 10.1007/s002030050573. [DOI] [PubMed] [Google Scholar]
Articles from Applied and Environmental Microbiology are provided here courtesy of American Society for Microbiology (ASM)
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10529 | https://papersowl.com/blog/find-the-nth-term-of-an-arithmetic-sequence | Have an expert improve your writing
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How to Find the Nth Term of an Arithmetic Sequence
Written by Kristian Eide
Posted: March 5, 2024
Last update date: April 3, 2024
6 min read
Table of contents
1 Definition of an Arithmetic Sequence
2 Step-by-Step Guide to Finding the nth Term
3 Examples
4 Practical Applications
5 Common Mistakes and Misconceptions
6 Overview of Related Concepts
7 Unlocking Patterns: The Essence of Arithmetic Sequences
As a basic part of mathematical analysis, arithmetic sequences are essential in many fields, from pure mathematics to real-world uses in science, engineering, and economics. The difference between each term in these sequences stays the same, which makes them easy to spot and use in a wide range of formulas and problem-solving situations. It’s very important to find the nth term in a series of numbers because it lets you guess what will happen in the future, look for patterns, and use an easy formula to solve hard problems.
The main points we will talk about in this article are:
Understanding arithmetic sequences, including formal definition and basic properties.
A detailed explanation of how to use a formula to find any term in an arithmetic sequence.
The application of the notion of the nth term in solving real-world problems enhances comprehension and practical skills.
You will get a full picture of math sequences and how to find the nth term by digging into these areas. As we move forward, we will explore each of these points in detail, providing a solid basis for both theoretical understanding and real-world use.
Definition of an Arithmetic Sequence
The nth term of an arithmetic sequence can be calculated using the formula: ({a_{n}=a_{1}+d(n-1)}), where:
({a_{n}}) is the nth term of the sequence,
({a_{1}}) is the first term of the sequence,
({d}) is the common difference,
({n}) is the position of the term within the sequence.
This formula allows for the direct calculation of any term in the sequence without the need to enumerate all preceding terms. The first term a1 represents the starting point of the sequence, d reflects the rate of progression, and n identifies the specific term’s position. By incorporating these components, the formula serves as a powerful tool for analyzing the behavior of arithmetic sequences and predicting future values. It is vital that you know and use this formula if you want to look for linear patterns and solve problems that have to do with arithmetic sequences, both in theory and in real life.
Step-by-Step Guide to Finding the nth Term
Finding the ({n^{th}}) term of an arithmetic sequence lets mathematicians and students find any term in the sequence without having to list them all by hand. This method is very helpful in many areas, from schoolwork to everyday problems, including help with math homework. Here’s how to learn this vital skill, step by step:
Identify the First Term (({a_{1}})): The first step in finding the nth term is to identify the first term of the sequence (({a_{1}})). This term serves as the starting point of your sequence and is crucial for applying the nth term formula effectively. In practical terms, a1 is simply the very first number you see in your sequence.
Determine the Common Difference (({d})): The common difference (({d})) is the consistent interval between consecutive terms in the sequence. To calculate d, subtract any term in the sequence from the subsequent term. It’s important to ensure this difference remains constant throughout the sequence for it to qualify as an arithmetic sequence. Understanding how to find this difference is key to effectively applying arithmetic concepts, including when seeking help with math homework.
Apply the nth Term Formula: With ({a_{1}}) and ({d}) in hand, you can now apply the nth term formula: ({a_{n}=a_{1}+d(n-1)}). This formula will enable you to find any term in the sequence, denoted as an, by simply knowing the position of the term (({n})) you wish to find.
For example, let’s find the nth term for the sequence starting 3, 7, 11, 15, 19…, which is a common task when getting help with math homework.
Step 1: The first term (({a_{1}})) is 3.
Step 2: The common difference (({d})) is 4, as each term increases by 4 from the previous one.
Step 3: To find the formula for the nth term, we plug ({a_{1}})= 3 and d = 4 into our formula: ({a_{n}}) = 3 + (n – 1) 4. Simplifying this, we get 4n – 1, which is the ({n^{th}}) term of our sequence.
Examples
A basic math skill that can be used in school and real life is finding the ({n^{th}}) term in a progression of numbers. As an example of this process, let’s look at two situations that come up a lot in maths homework help sessions: one with a positive common difference and the other with a negative common difference.
Example with Positive Common Difference:
Consider the sequence 5, 9, 13, 17, …
First Term (({a_{1}})): The first term is 5.
Common Difference (d): The difference between each term is 4 (positive).
Using ({a_{n}=a_{1}+d(n-1)}), the nth term is 5 + (n – 1) 4, simplifying to 4n + 1.
Example with Negative Common Difference:
Consider the sequence 20, 15, 10, 5, …
First Term (({a_{1}})): The first term is 20.
Common Difference (d): The difference between each term is -5 (negative).
Applying ({a_{n}=a_{1}+d(n-1)}), the nth term formula is 20 + (n – 1)(-5), or -5n + 25.
Practical Applications
The ability to find the nth term in a series of numbers has important uses in many areas of life. In finance, this idea is significant for figuring out how much investments or savings will be worth in the future when they grow at a steady rate. This helps people and businesses make better plans for their financial futures. Engineers use arithmetic sequences to model and predict situations that change in a straight line, like how material stress rises over time or how buildings get heavier over time.
In computer science, knowing arithmetic progressions is crucial for making algorithms. This is especially true for sorting and finding algorithms, where arithmetic progressions can often be used to predict or improve how efficiently data is handled. In addition, sequences are used to find patterns and create safe communication methods in coding theory and cryptography.
These applications underline the importance of mastering the concept of nth term calculation, as it equips professionals with the tools to solve complex problems, optimize processes, and make informed decisions in their respective fields. For those seeking further assistance in understanding or applying these concepts, resources like PapersOwl offer valuable support and insights.
Common Mistakes and Misconceptions
A lot of people make mistakes and don’t understand when they’re trying to figure out the nth term of an arithmetic process. It’s essential to be aware of and avoid these mistakes so that calculations are correct.
Common Mistakes:
Confusing the First Term with the Common Difference: A frequent fault is mixing up the first term (({a_{1}})) of the sequence with the common difference (d). Remember, a1 is the starting point of the sequence, while d is the consistent interval between terms.
Overlooking Negative Common Differences: The common difference is negative when sequences decrease. Neglecting the sign of d can result in incorrect calculations.
Arithmetic Sequences Always Increase: Many assume arithmetic sequences only grow. However, they can also decrease or remain constant if the common difference is negative or zero respectively.
Any Sequence with a Pattern is Arithmetic: Not all sequences with patterns are arithmetic. A sequence is only arithmetic if the difference between consecutive terms is constant. Sequences with varying intervals or patterns based on multiplication (geometric sequences) are not arithmetic.
Overview of Related Concepts
The study of arithmetic sequences naturally leads to exploring related concepts such as the sum of an arithmetic series and infinite arithmetic sequences, where the ({n^{th}}) term formula plays a pivotal role. The sum of an arithmetic series involves calculating the total of all terms within a finite sequence, applying the formula ({S_{n}=\frac{2a_{1}+d(n-1)}{2}\cdot n}) or ({S_{n}=\frac{a_{1}+a_{n}}{2}\cdot n}), where ({S_{n}}) represents the series sum, and n is the number of terms.
Infinite arithmetic sequence, which can go on forever in theory, causes people to talk about convergence and how sequences behave over time. In arithmetic sequences, the idea of infinity is more of a theoretical one. It is often used to push the limits of mathematical logic and thinking.
Unlocking Patterns: The Essence of Arithmetic Sequences
In wrapping up, the exploration of arithmetic sequences reveals their deep-seated importance in both the realm of pure mathematics and the practical aspects of daily life. Key takeaways we learned on our post were the basics of arithmetic sequences, which are made up of terms with a constant difference between them, how important the nth term formula is for finding any term in a sequence, and how these ideas can be used in many different fields, such as finance, engineering, and computer science. Moreover, we navigated through common pitfalls and misconceptions, ensuring a robust grasp of how to accurately compute and apply the nth term. It’s evident that the arithmetic sequences is an academic exercise and a practical toolset for solving real-world problems.
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Flaw in my reasoning for the maximum of a b a b if a,b≥0 a,b≥0 and a+2 b=3 a+2 b=3?
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Problem statement: What is the maximum value of the product a b a b if a a,b b are non-negative numbers such that a+2 b=3 a+2 b=3?
What is the flaw in my solution?
We know that a b−−√≤(a+b)2 a b≤(a+b)2 and that a=3−2 b a=3−2 b. The product a b a b will be maximum when it is equal to the square of the RHS of the inequality above. Plugging in for a a, and squaring both sides we get the equation: (3−2 b)(b)=(((3−2 b)+b)2 4(3−2 b)(b)=(((3−2 b)+b)2 4. Which gives b=1 b=1. And plugging b=1 b=1 into the the equation in the problem statement gives a=1 a=1. So, the max product is 1∗1=1 1∗1=1.
What am I doing wrong? What concepts could I be I misunderstanding? Can you please explain? The actual answer is 9/8 9/8. Thank you.
algebra-precalculus
inequality
optimization
solution-verification
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edited May 29, 2021 at 18:37
user21820
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asked May 29, 2021 at 8:09
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The right-hand side of your inequality depends on b b. You can't claim a maximum based on a bound that depends on the variable being optimized.dxiv –dxiv 2021-05-29 08:12:56 +00:00 Commented May 29, 2021 at 8:12
Maybe that is really intuitive and obvious to you and many others, but I just don’t see why. Anything anyone can do to help me grasp this?user796511 –user796511 2021-05-29 08:17:23 +00:00 Commented May 29, 2021 at 8:17
1 "The product ab will be maximum when it is equal to the square of the RHS of the inequality above" — The equality case of the AM-GM inequality is always a=b a=b, but this doesn't tell you that (a+b)2/4(a+b)2/4 is the maximum of a b a b when a a and b b vary under other constraints. Try for example to apply the same argument to "what is the maximum value of the product a b a b if a a, b b are non-negative numbers such that a=3 a=3".dxiv –dxiv 2021-05-29 08:24:37 +00:00 Commented May 29, 2021 at 8:24
Please carefully spell out the steps that you believe justify the sentence: "The product a b a b will be maximum when it is equal to the square of the RHS of the inequality above." Because that sentence is not, in fact, correct.Robert Shore –Robert Shore 2021-05-29 08:28:06 +00:00 Commented May 29, 2021 at 8:28
@dxiv in your first comment do you want to say that we cannot give maximum value unless we have a constant term on RHS Lalit Tolani –Lalit Tolani 2021-05-29 09:03:38 +00:00 Commented May 29, 2021 at 9:03
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Indeed: We have
(3−2 b)b≤(3−b)2 4(3−2 b)b≤(3−b)2 4
is a true statement and in fact equality holds when b=1 b=1.
However, it is possibel for the LHS to get a larger value and the equality doesn't hold.
To maximize it notice that (3−2 b)b(3−2 b)b is a concave quadratic and the optimal value is attained when
b=1.5 2=3 4 b=1.5 2=3 4
That is the maximum value is (3 2)⋅(3 4)=9 8.(3 2)⋅(3 4)=9 8.
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answered May 29, 2021 at 8:30
Siong Thye GohSiong Thye Goh
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sir you graphically proved it that " it is possibel for the LHS to get a larger value and the equality doesn't hold" but how we can say that analytically Lalit Tolani –Lalit Tolani 2021-05-29 08:59:55 +00:00 Commented May 29, 2021 at 8:59
to show something is possible, we just need an example. we can use the example, in this question and find the maximum value of the quadratic curve and verify the RHS forms a strict upper bound.Siong Thye Goh –Siong Thye Goh 2021-05-29 09:11:18 +00:00 Commented May 29, 2021 at 9:11
@lalittolani You could prove that by showing that 3 x−2 x 2 3 x−2 x 2 intersects with (3−x)2/4(3−x)2/4 at 1, but the former has a peak that is higher: 3⋅0.75−2⋅0.75 2=1.125 3⋅0.75−2⋅0.75 2=1.125.a concerned citizen –a concerned citizen 2021-05-29 16:29:27 +00:00 Commented May 29, 2021 at 16:29
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Building upon OP's attempt to use AM-GM, the following will work, instead, because it is arranged such that the right-hand side of the inequality matches the known constant sum.
a⋅2 b−−−−√≤a+2 b 2=3 2⟹2 a b≤(3 2)2=9 4⟹a b≤9 8 a⋅2 b≤a+2 b 2=3 2⟹2 a b≤(3 2)2=9 4⟹a b≤9 8
The maximum value of 9 8 9 8 is attained when a=2 b a=2 b, which is a=3 2,b=3 4 a=3 2,b=3 4.
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edited May 29, 2021 at 16:45
answered May 29, 2021 at 9:01
dxivdxiv
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Wish the downvoter had left a comment why.dxiv –dxiv 2021-05-29 18:26:13 +00:00 Commented May 29, 2021 at 18:26
I think somebody has a grudge because it looks like all the others got one (though I can't really tell that about Siong's answer since it has gained a few more points than when I saw this).a concerned citizen –a concerned citizen 2021-05-29 22:47:25 +00:00 Commented May 29, 2021 at 22:47
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I know you have accepted the answer but still I have one analogy to share with you-
Suppose we have a quadratic equation(to say) as -
e x x 2−(l o g x)x+3=0 e x x 2−(l o g x)x+3=0
Now you found it's roots by using quadratic formula. The "expression" of roots that you will obtain will also be some function of x x and in that case you can't call those expressions as roots because roots are the values of variable for which the equation reduces to zero and you can't get values of variable in terms of variable because a value is fixed. If the equation would be as follows-
x 2−3 x+2=0 x 2−3 x+2=0 then the you will obtain a constant value and in that case you can say that those values are certainly the roots of your equation.
Similarly ,as @dxiv mentioned in his answer and comments, "unless you claim it is a constant relative to constraints, you can't claim it a maximum".
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answered May 30, 2021 at 4:46
Lalit TolaniLalit Tolani
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10531 | https://www.fisheries.noaa.gov/west-coast/endangered-species-conservation/south-central-california-coast-steelhead | Published Time: Sat, 13 Sep 2025 10:21:01 GMT
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Summary
Students will feel the temperature change that occurs when a cold pack and a hot pack are activated. They will see that these temperature changes are due to a solid substance dissolving in water. Students will then compare the temperature changes that occur as four different solutes dissolve in water and classify these as either endothermic or exothermic. Students will be introduced to the concept that it takes energy to break bonds and energy is released when bonds are formed during the process of dissolving.
Objective
Students will be able to identify variables in an experiment to find out how much the temperature increases or decreases as each of four solutes dissolves in water. Students will be able to correctly classify the process of dissolving as either exothermic or endothermic for each solute. Students will be able to explain that the temperature changes in dissolving are a result of the amount of energy released compared to the amount of energy used as “bonds” are formed and broken.
Safety
Be sure you and the students wear properly fitting goggles. Excess dry material from the hot and cold packs can be placed in the trash. Sodium carbonate may be a skin irritant. Have students wash hands after the activity.
Materials for the Demonstrations
Materials for Each Group
Notes about the Materials
Note: This activity deals with a concept that is not often addressed in middle school—that a temperature change occurs during the process of dissolving. Most middle school textbooks and curricula associate a change in temperature only with chemical change. Dissolving is usually considered a physical change but also can result in a change in temperature. This change in temperature is based on the energy changes involved in breaking and making “bonds” in the process of dissolving.
Read more about the energy changes in making and breaking bonds in Teacher Background.
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Standards Alignment
5.9 Next Generation Science Standards (PDF)
5.9 Common Core State Standards (PDF)
More about Standards Alignment
Instructions
1 Engage
Step 1 Allow students to feel the temperature change in an activated cold pack and an activated hot pack.
Tell students that they will explore how some hot and cold packs work. Give them a hint that it has to do with dissolving, which they have been studying in this chapter.
Materials for the Demonstration
Procedure
Expected Results
The cold pack quickly becomes cold while the hot pack quickly becomes hot.
Step 2 Do a demonstration to show how cold and hot packs work.
Note: In this demonstration, 1 teaspoon of each substance is dissolved in 10 mL of water. Since the purpose of this demonstration is to show whether the temperature simply goes up or down, this type of volume measure is fine. But to compare which substance is more or less exothermic or endothermic than another, as students will do in the activity, the solute will be measured in grams.
Materials for the Demonstration
Procedure
Expected Results
Dissolving the substance from the cold pack will cause the temperature to decrease to less than 10 °C (endothermic). Dissolving the substance from the hot pack will cause the temperature to increase to over 40 °C (exothermic). Results may vary.
Step 3 Introduce the terms endothermic and exothermic.
Tell students that scientists describe temperature changes that occur when substances interact as either endothermic or exothermic. When the temperature decreases, as it does in the cold pack, the process is endothermic. When the temperature increases, as it does in the hot pack, the process is exothermic.
2 Evaluate
Give each student an activity sheet.
Download the student activity sheet, and distribute one per student.
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The activity sheet will serve as the “Evaluate” component of each 5-E lesson plan. The activity sheets are formative assessments of student progress and understanding. A more formal summative assessment is included at the end of each chapter.
Students will record their observations and answer questions about the activity on the activity sheet. The Explain It with Atoms and Moleculesand Take It Furthersections of the activity sheet will either be completed as a class, in groups, or individually depending on your instructions. Look at the teacher version of the activity sheet to find the questions and answers.
3 Explore
Step 4 Introduce the dissolving activity students will do and help students identify the variables.
Tell students that they will compare how much the temperature changes when four household substances dissolve in water. Introduce the crystals students will dissolve:
Ask students:
Note: Comparing the amount of temperature change for different substances by dissolving the same mass of each substance in the same amount of water is fine at the middle school level. However, a more rigorous approach is to dissolve the same number of particles (molecules or ionic units) of each substance in the same amount of water.
Read more about counting molecules in Teacher Background.
All Downloads
Step 5 Have students monitor changes in temperature as they dissolve four different household solutes in water.
Question to Investigate
Which solute dissolves the most endothermically and which dissolves the most exothermically in water?
Materials for Each Group
Procedure
Expected Results
Potassium chloride dissolved the most endothermically, and calcium chloride dissolved the most exothermically. Student temperature readings will vary, but will likely be similar to the following:
| | | |
---
| Solute dissolved in room temperature water (22 °C) | Maximum or minimum temperature achieved | Endothermic or exothermic? |
| potassium chloride | 14 °C | endothermic |
| calcium chloride | 45 °C | exothermic |
| sodium carbonate | 32 °C | exothermic |
| sodium bicarbonate | 18 °C | endothermic |
Maximum or minimum temperature achieved
Step 6 Discuss student observations.
Ask students:
4 Explain
Step 7 Show students an animation of dissolving and explain that the energy of making and breaking “bonds” during dissolving causes a change in temperature.
Project the animation Breaking and Making Bonds.
Breaking and Making Bonds
Tell students that there is an important rule in chemistry: Energy is required to pull apart atoms, ions, or molecules that are attracted to each other. But when atoms, ions, or molecules come together, energy is released. One way to say it is, “It takes energy to break bonds, and energy is released when bonds are formed.”
Project the animation Energy and Dissolving.
Energy and Dissolving
Press the “next” button and explain that this happens in dissolving. When water molecules are attracted to and bond to the molecules or ions of a substance, some energy is released as shown by the arrow going out. Then the water molecules pull ions or molecules of the substance apart, which takes energy, as shown by the arrow going in.
Project the image Exothermic Dissolving.
The process of dissolving is exothermic when more energy is released when water molecules “bond” to the solute than is used to pull the solute apart. Because more energy is released than is used, the molecules of the solution move faster, making the temperature increase.
Project the image Endothermic Dissolving.
The process of dissolving is endothermic when less energy is released when water molecules “bond” to the solute than is used to pull the solute apart. Because less energy is released than is used, the molecules of the solution move more slowly, making the temperature decrease.
5 Extend
Step 8 Explore whether the process of crystallization can cause a change in temperature.
Project the video Hand Warmer.
Hand Warmer
In the hand warmer, the water molecules and the ions of the solute come together to form a crystal. Bending the metal disk creates tiny scratches, which act as nucleation points where the sodium acetate crystal forms. As the water molecules and ions bond together in the growing crystal, energy is released. This results in an increase in temperature.
Step 9 Liquids dissolving in a liquid can also cause a temperature change.
Project the video Temperature Change Alcohol in Water.
Temperature Change – Alcohol in Water
Explain that the “bonding” of water molecules to alcohol molecules releases more energy than it takes to separate the alcohol molecules from each other. This results in an increase in temperature.
What is the 5-E format?
The 5-E instructional model is an approach to teaching and learning that focuses on active engagement, inquiry-based learning, and collaboration.
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Simulations
Breaking and Making Bonds
Energy and Dissolving
Hand Warmer
Temperature Change – Alcohol in Water
Simulations for Lesson 5.9
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For Teachers
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This lesson is part of: Chapter 5: The Water Molecule and Disolving
Lesson 5.8: Can Gases Dissolve in Water?
Lesson 6.1: What is a Chemical Reaction?
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A Survey of the Hadwiger-Debrunner (p, q)-problem
Chapter
pp 347–377
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Discrete and Computational Geometry
Jürgen Eckhoff
Part of the book series: Algorithms and Combinatorics ((AC,volume 25))
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32 Citations
Abstract
At the annual meeting of the Swiss Mathematical Society held in September 1956 in Basel, H. Hadwiger presented the following theorem. It was published one year later in a joint paper with H. Debrunner, his colleague at the University of Bern.
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Jürgen Eckhoff
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Editors and Affiliations
Department of Computer and Information Science, Polytechnic University, Six Metro Tech Center, Brooklyn, NY, 11201, USA
Boris Aronov
2. School of Mathematics, Georgia Institute of Technology, Atlanta, GA, 30332-0160, USA
Saugata Basu
3. City College and Courant Institute, 251 Mercer St., New York, NY, 10012, USA
János Pach
4. School of Computer Science, Tel Aviv University, Tel Aviv, 69978, Israel
Micha Sharir
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Eckhoff, J. (2003). A Survey of the Hadwiger-Debrunner (p, q)-problem. In: Aronov, B., Basu, S., Pach, J., Sharir, M. (eds) Discrete and Computational Geometry. Algorithms and Combinatorics, vol 25. Springer, Berlin, Heidelberg.
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Convex Body
Chromatic Number
Intersection Graph
Nonempty Intersection
Convex Geometry
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Anthony J Swerdlow1
Breast Cancer Research volume 14, Article number: R91 (2012) Cite this article
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Abstract
Introduction
Tamoxifen is an effective treatment for breast cancer but an undesirable side-effect is an increased risk of endometrial cancer, particularly rare tumor types associated with poor prognosis. We investigated whether tamoxifen therapy increases mortality among breast cancer patients subsequently diagnosed with endometrial cancer.
Methods
We pooled case-patient data from the three largest case-control studies of tamoxifen in relation to endometrial cancer after breast cancer (1,875 patients: Netherlands, 765; United Kingdom, 786; United States, 324) and collected follow-up information on vital status. Breast cancers were diagnosed in 1972 to 2005 with endometrial cancers diagnosed in 1978 to 2006. We used Cox proportional hazards survival analysis to estimate hazard ratios (HRs) and 95% confidence intervals (CI).
Results
A total of 1,104 deaths occurred during, on average, 5.8 years following endometrial cancer (32% attributed to breast cancer, 25% to endometrial cancer). Mortality from endometrial cancer increased significantly with unfavorable non-endometrioid morphologies (P < 0.0001), International Federation of Gynaecology and Obstetrics staging system for gynecological malignancy (FIGO) stage (P < 0.0001) and age (P < 0.0001). No overall association was observed between tamoxifen treatment and endometrial cancer mortality (HR = 1.17 (95% CI: (0.89 to 1.55)). Tamoxifen use for at least five years was associated with increased endometrial cancer mortality (HR = 1.59 (1.13 to 2.25)). This association appeared to be due primarily to the excess of unfavorable histologies and advanced stage in women using tamoxifen for five or more years since the association with mortality was no longer significant after adjustment for morphological type and FIGO stage (HR = 1.37 (0.97 to 1.93)). Those patients with endometrioid tumors, who stopped tamoxifen use at least five years before their endometrial cancer diagnosis, had a greater mortality risk from endometrial cancer than endometrioid patients with no tamoxifen exposure (HR = 2.11 (1.13 to 3.94)). The explanation for this latter observation is not apparent.
Conclusions
Patients with endometrial cancer after breast cancer who received tamoxifen treatment for five years for breast cancer have greater endometrial cancer mortality risk than those who did not receive tamoxifen. This can be attributed to non-endometrioid histological subtypes with poorer prognosis among long term tamoxifen users.
Introduction
Tamoxifen is an effective treatment for breast cancer [1, 2] but an undesirable side-effect is the increased risk of endometrial cancer in postmenopausal women [3–8], particularly rare tumor types [5, 6, 8, 9] associated with poor prognosis . Although the number of cases of endometrial cancer occurring after tamoxifen is modest (for example, 0.3% taking tamoxifen for approximately five years versus 0.1% not taking it ), there is concern that tamoxifen-induced endometrial cancers may have poorer survival [6, 11], even after allowance for histopathologic characteristics . The side-effects of tamoxifen are unlikely to outweigh the benefits in breast cancer patients , but any detrimental effects on survival would have implications for endometrial cancer surveillance following treatment , and would be important in decisions about the prophylactic use of tamoxifen by women without breast cancer . To address these issues we have pooled patients from the three largest case-control studies of endometrial cancer after breast cancer [3–6] to examine mortality from endometrial cancer in relation to tamoxifen treatment.
Materials and methods
The case series from three case-control studies of endometrial cancer after breast cancer were pooled. These studies from the Netherlands (NL) (nine regional cancer registries contributing to the Netherlands Cancer Registry), the United Kingdom (UK) (regional cancer registries in England, Scotland and Wales), and the United States (US) (Surveillance, Epidemiology and End Results (SEER) registries in four regions: Atlanta, Iowa, Los Angeles County, and Seattle-Puget Sound) have each been described previously [3–6]. Each study received appropriate ethical approval(s). The majority of data were abstracted from medical case-notes without patient contact; however, informed consent was obtained in the US where patients were interviewed. Briefly, each case-control study was population-based and included patients diagnosed with endometrial cancer after breast cancer during defined periods (NL (n = 765): 1978 to 1997; UK (n = 786): 1988 to 1996; US (n = 324): 1978 to 1993). The endometrial cancer diagnosis had to have occurred at least three months after the breast cancer diagnosis (six months for the US study). Patients were excluded if they had had a cancer (other than non-melanoma skin cancer or in situ cervical cancer) diagnosed before their breast cancer or between the diagnosis of the initial primary breast cancer and the subsequent endometrial cancer (except non-melanoma skin cancer, in situ cervical cancer or breast cancer). Information on tamoxifen treatment was abstracted from medical records and in Los Angeles, confirmed in interviews. At follow-up for survival, one patient from the original UK case-series was no longer eligible (because of erroneous cancer registry tumor record linkage) and was removed from this study.
The cases of endometrial cancer from the original Dutch study were supplemented with patients diagnosed from 1989 to 2003 (the TAMARISK (Tamoxifen Associated Malignancies: Aspects of Risk) retrospective cohort) from the same nine regional cancer registries as in the original (ALERT (Assessment of Liver and Endometrial cancer Risk following Tamoxifen)) study [3, 6], except diagnosis of endometrial cancer was at least 12 months after breast cancer (rather than three months). In addition, a further 179 Dutch patients diagnosed from 2003 to 2006 were included, with endometrial cancer at least three months after breast cancer, from the prospective component of the TAMARISK study .
Follow-up
The Netherlands
Vital status, date of most recent follow-up, or date of death and cause, were obtained from medical records, general practitioners or clinicians, and municipal population registries. Follow-up for the ALERT patients was initially to 1997, with additional follow-up to 2004 for those patients who had less than four years initial follow-up. Follow-up was to 2003 to 2005 for the TAMARISK retrospective cohort and to 2004 to 2007 for the TAMARISK prospective cohort. All deaths were linked through 'Statistics Netherlands' to obtain registered underlying cause of death (which was used in analyses when cause of death was unknown based on review of medical records ). Within the study period there were no known emigrations from the Netherlands in these cohorts.
UK
Vital status and cause of death were ascertained from hospital case-notes when the initial study data were collected (1996 to 1999). In 2005 further follow-up for vital status and causes of death was obtained from each of the regional cancer registries in Britain, and subsequently in 2008 further follow-up was obtained by linkage to the National Health Service Central Register (NHSCR -- a list of virtually every member of the population, which routinely receives notifications of events such as emigrations, cancers, and deaths) , and for those who had died copies of death certificates were obtained. Vital status could not be determined for eleven (1.4%) patients so for these follow-up was taken to the date of the last clinical contact as extracted from case-notes. Thirty-eight cases had deaths recorded as occurring at the date of diagnosis of endometrial cancer and were removed from the main analysis.
USA
Data were originally collected on vital status, date of most recent follow-up or date of death, and cause of death (based on information from death certificates) for all patients up to 2000. Additional follow-up was obtained to the end of 2006 for the Los Angeles County patients (n = 228), and those not known to be deceased were additionally checked against the Social Security Administration's Death Master File to ascertain any deaths outside the state of California.
Statistical Analysis
Descriptive analyses by morphological type of endometrial cancer were conducted using one way analysis of variance for continuous variables or Pearson chi-square for categorical variables . When comparing individual differences between morphological groups, we adjusted for age at diagnosis of endometrial cancer and study, using linear regression in the case of continuous variables and a 'modified' Poisson approach with robust standard errors for binary variables. To assess the association between tamoxifen treatment and the risk of death, we calculated hazard ratios using Cox proportional hazards regression with time since diagnosis of endometrial cancer (follow-up time) as the implicit regression time scale and stratification by (adjustment for) attained age (which also is an adjustment for age at endometrial cancer diagnosis since: age at endometrial cancer diagnosis = attained age - survival time since diagnosis), calendar period and, as appropriate, morphology and FIGO stage. Tests for trend were calculated using continuous data. Women with deaths due to causes other than the cause under study in cause-specific analyses were treated as censored on their dates of death. Where it was not possible to distinguish between breast and endometrial cancer as cause of death the patients (n = 37) were not allocated to either cause of death in the main analyses, but were allocated to each cause in sensitivity analyses. Patients diagnosed with endometrial cancer at death (n = 38) were excluded from the main analysis and tables but were included, with a survival time of one day and one year, in sensitivity analyses. For breast cancer and all cause mortality, we additionally adjusted for age at diagnosis of breast cancer and extent of breast disease (instead of FIGO stage). All analyses were carried out using Stata/IC version 10.1 and all statistical tests were two-sided.
Results
Descriptive characteristics of the three studies
There were 1,875 patients in the combined study, comprising 765 (41%) from the Netherlands, 786 (42%) from the UK, and 324 (17%) from the US (Table 1). The median age at diagnosis of breast cancer was 63 years in the Netherlands, 62 years in the UK study, and 65.5 years in the US study, and the median age at diagnosis of endometrial cancer was 69 years in each study. The calendar periods for diagnosis of breast cancer and endometrial cancer, and the intervals between the two cancers, reflect the original individual study designs (as described above). The median interval between cancers was 5.1 years in the Netherlands study, 6.0 years in the UK, and 3.0 years in the US. Tamoxifen use was more commonly recorded for patients in the UK (82%) than the Netherlands (46%) or US (45%).
Endometrial cancer morphology
In the combined series 60.7% of the endometrial cancers developed among tamoxifen users. The majority (84%) of the endometrial cancers were endometrioid adenocarcinomas (Table 2), and (after adjustment for study) these were diagnosed at significantly younger ages than were serous or clear cell endometrial cancer (P < 0.0001), and carcinosarcomas (P = 0.002). FIGO stage was available for 97% of the cases in the Netherlands, 78% in the US, but only 37% in the UK study. Where FIGO stage was known, 79% of tumors were stage I, 10% were stage II and 11% were stage III or higher, with no significant difference in this distribution between studies (P = 0.46). Endometrioid tumors were more likely to be diagnosed at FIGO stage I than were non-endometrioid tumors (P < 0.001). A significantly higher proportion of patients with carcinosarcoma had a history of tamoxifen use than did patients with endometrioid carcinoma (P < 0.001). Among tamoxifen users the patients who developed carcinosarcoma had been treated with tamoxifen on average 0.9 years longer than the patients with endometrioid type tumors (P = 0.012). Patients with carcinosarcomas, or serous or clear cell endometrial cancers, were more likely to have ceased tamoxifen use one or more years before diagnosis of endometrial cancer than patients with endometrioid tumors. (P = 0.010 and P = 0.020 respectively). The average interval between breast and endometrial tumors was longer for the unfavorable cancers, such as carcinosarcomas, serous and clear cell endometrial cancers but the differences were not statistically significant (P = 0.25).
Follow-up
The 1,875 patients who had both breast and endometrial cancer were followed on average for 5.8 years (median 4.0 years) with 1,104 deaths (Table 3). For these patients with breast cancer who had also developed endometrial cancer 25% to 28% of the deaths were due to endometrial cancer, 32% to 35% to breast cancer (type of cancer death could not be distinguished between the two causes in 3% of cases), and 40% to all other causes (including 1.7% to cancer of unknown primary site and 0.5% with cause of death unknown). The five-year survival was 55.5% but this varied from 73% for patients diagnosed with localized breast cancer and FIGO grade I endometrial cancer to 16% for patients diagnosed with metastatic breast cancer or FIGO grade III/IV endometrial cancer. For those patients diagnosed with endometrial cancer before age 65, five-year survival was 82% for patients diagnosed with localized breast cancer and FIGO grade I endometrial cancer and 32% for patients diagnosed with metastatic breast cancer or FIGO grade III/IV endometrial cancer.
Mortality
Age at diagnosis of endometrial cancer
Older age at endometrial cancer diagnosis was associated with greater risk of dying of endometrial cancer (trend P < 0.0001, with no evidence for heterogeneity between studies (P = 0.52)) (Table 4). Subsequent analyses adjust endometrial mortality for attained age and time since diagnosis of endometrial cancer, and thus implicitly also adjust for age at diagnosis.
FIGO stage
Higher FIGO stage was associated with greater endometrial cancer death rates (FIGO III/IV versus I: Hazard Ratio, HR = 13.1; 95% confidence Interval (9.25 to 18.6); trend P < 0.0001 with no strong evidence for interaction between studies P = 0.067).
Endometrial cancer morphology
Endometrial cancer mortality was greater for patients with non-endometrioid endometrial cancer than patients with endometrioid types across all three studies combined (HR = 5.09; (3.96 to 6.53), P < 0.0001) and within each study (data not shown), with no evidence for heterogeneity between the three studies (P = 0.33); the greatest increases were seen for carcinosarcomas (HR = 6.66 (4.87 to 9.12)) and sarcomas (HR = 5.65 (3.53 to 9.05)). The HRs were smaller but still significant after adjustment for FIGO stage (Table 4).
Validity of cause-specific mortality
Extent of disease of breast cancer was unrelated to endometrial cancer mortality (P = 0.14) but was strongly related to breast cancer mortality (P < 0.0001). Age at diagnosis of endometrial cancer (P = 0.23), FIGO stage (P = 0.34) and endometrial morphology (P = 0.16) were not related to breast cancer mortality. Conversely, age at diagnosis of breast cancer was unrelated to endometrial cancer mortality (P = 0.11). There was no significant heterogeneity between studies.
Endometrial cancer mortality: tamoxifen use and morphology
No overall association was observed between tamoxifen treatment and endometrial cancer mortality (HR = 1.17 (95% CI: (0.89 to 1.55)); however, tamoxifen use for at least five years was associated with increased endometrial cancer mortality (HR = 1.59 (1.13 to 2.25)). After adjustment for morphological type and FIGO stage, five years tamoxifen use was no longer significantly associated with endometrial cancer mortality (HR = 1.37 (0.97 to 1.93)) overall or when stratified by morphology (Table 5). When analyzed by cumulative dose of tamoxifen, patients with cumulative doses over 30,000 mg (for example, 20 mg per day for 4.1 years) had modestly elevated endometrial cancer mortality. There was no association with daily tamoxifen dose.
Endometrial cancer mortality risk among women who stopped tamoxifen use at least five years before their endometrial cancer diagnosis was twice that of women who had never used tamoxifen and the trend with cessation among users (HR = 1.11 per year since last use (1.05 to 1.18)) remained statistically significant after adjustment for morphological type, duration of tamoxifen use, FIGO stage and interval between breast and endometrial cancer.
There was a strong trend of increasing mortality with increasing interval between breast cancer and endometrial cancer diagnosis (P = 0.0001), which, after stratification by morphology, remained statistically significant among those with endometrioid tumors (P < 0.0003). The trend with interval was also stronger in tamoxifen users than non-users (P = 0.044) and among users remained statistically significant even after adjustment for duration of tamoxifen use (P = 0.032). (For breast cancer mortality, risk of dying decreased as the interval between tumors increased (P = 0.003), but all-cause mortality did not vary with interval between tumors (P = 0.085)).
Time since last tamoxifen use and interval between diagnoses of breast and endometrial cancer are related (that is, only those with an interval between tumors of five or more years could have ceased tamoxifen use five or more years ago) but even among those patients with an interval of five or more years (that is, those with the potential for five or more years cessation) mortality was still elevated among those with five or more years since cessation of tamoxifen (HR = 2.06 (1.18 to 3.60)). When considering calendar period of diagnosis, since the indications for treatment and cessation of tamoxifen may have changed over time, there was no significant difference in trend with cessation among tamoxifen users for those diagnosed with breast cancer before 1990 (compared with those diagnosed in 1990 or later, P = 0.84) or with endometrial cancer before 1995 (compared with those diagnosed in 1995 or later, P = 0.53).
Discussion
We accrued 1,875 patients, with 1,104 deaths of which 227 were due to endometrial cancer, by pooling the three largest case-control studies of endometrial cancer after breast cancer [3–6]. The number of cases of endometrial cancer occurring in breast cancer trial settings at present is modest (for example, 182 cases of uterine cancer reported in the Early Breast Cancer Trialists' Collaborative Group (EBCTCG) meta-analysis of 20 trials , and 102 cases in the National Surgical Adjuvant Breast and Bowel Project (NSABP) trial comparing prophylactic tamoxifen with raloxifene ), so although our data are observational we have a large number of cases, some of rarer histologies, with which to examine endometrial cancer survival after tamoxifen use. A previous report using a subset of the pooled data showed increased endometrial cancer mortality with tamoxifen use, but did not have as many patients or as much follow-up time as we have in the pooled data.
We found that women with five or more years of tamoxifen use had 59% greater risk of endometrial cancer death than non-users, which was mostly attributable to the occurrence of endometrial cancer morphologies among tamoxifen users with less favorable prognosis, for example, carcinosarcomas. Several earlier studies have shown that tamoxifen greatly increases the risk of developing non-endometrioid tumors [5, 6, 8, 9] and in the data reported here endometrial cancer mortality attributed to these morphological types was 2.3 to 5.4 times that of endometrioid types. Beyond the consequence of tamoxifen increasing the incidence of these tumors with poor prognosis , we saw no further adverse effect of tamoxifen dose or cumulative dose on endometrial cancer survival. In line with our results, genomic analyses suggest there are no differences between tamoxifen-induced tumors, either endometrioid or non-endometrioid, and those tumors occurring in patients without tamoxifen use [16, 25, 26]. Our data show, however, that patients with endometrioid tumors who had stopped tamoxifen five or more years before diagnosis of endometrial cancer had greater endometrial cancer-specific mortality risk. (The statistical power to examine this among patients with non-endometrioid tumors was low because these tumor types were less common.)
Although some of the women in this study may have received tamoxifen when distant metastases arose, and their prognosis would have been poor in relation to breast cancer survival, this does not preclude them from contributing survival time for analyses of tamoxifen use and endometrial cancer mortality (just as women who had other serious diseases are able to contribute to the analyses). As a demonstration of the validity of the cause-specific survival analyses, we found that extent of the breast disease was strongly predictive of breast cancer mortality but it was not associated with endometrial cancer mortality. Conversely, age at diagnosis of endometrial cancer, FIGO stage and morphology were strongly predictive of endometrial cancer mortality but not breast cancer mortality. We, therefore, believe our analyses are valid, whether tamoxifen was used for metastatic disease or in an adjuvant setting.
Mortality risk was elevated in the patients for whom tamoxifen use was known but the details of the dose or duration was missing. However, it is probable that this is an artefact related to the greater chance of destruction or loss of some part of the medical case-note history among patients who had died by the time of data collection. Our conclusions were not materially affected by the missing data because few were missing, for example, < 3% were missing the duration of tamoxifen use.
The cases from the three study populations were all ascertained from regional population-based cancer registries [3–6, 12], although some patients were not available for the analyses. For the UK, 208 provisionally eligible patients were identified but their case notes could not be located or had insufficient information for the original case-control study (and subsequent follow-up for mortality). In the US, five patients were excluded from the case-control study. None were excluded in the NL, yet there was no evidence of heterogeneity between the studies, suggesting there was little, if any, bias due to case under-ascertainment. Furthermore, the one- and five- year survival rates seen here for endometrial cancer within each study were similar to published NL, UK and US rates [27, 28], which suggests under-ascertainment did not materially affect the results.
Follow-up for mortality was comprehensive because population-based cancer registries covered each region, and additional information was available from medical records, general practitioners, clinicians, and national death registers. It is therefore unlikely that any significant migration outside of the study regions occurred, or that any unascertained deaths occurred. Thirty-eight patients diagnosed with endometrial cancer at death were excluded from the analysis because they contributed no survival follow-up, and their inclusion in sensitivity analyses made no material difference to the results. It was not possible to distinguish between breast and endometrial cancer as cause of death in 37 cases and these patients were censored at date of death in the main analyses, and their inclusion as endometrial cancer deaths strengthened the association with cumulative dose but made little material change to the other results.
An issue in the interpretation of the results is the attribution of cause of death to a single underlying cause in the presence of co-morbidity. We saw opposing trends in breast and endometrial cancer mortality with interval between tumors, and although longer follow-up since breast cancer would be expected to be associated with lower breast cancer death rates it is possible that deaths occurring after a long interval between tumors may have been more likely to be assigned to the most recently diagnosed tumor (that is, endometrial cancer). To make some allowance for this we adjusted for interval between tumors in the analyses and the main results remained the same: among patients with endometrioid tumors there was no association between endometrial cancer mortality and tamoxifen use, but increased mortality if tamoxifen had stopped at least five years before diagnosis of endometrial cancer.
Endometrioid endometrial tumors may be more likely to present with vaginal bleeding and therefore be diagnosed earlier than non-endometerioid tumors, and indeed we saw that these tumor types were more likely to be of a lower FIGO grade at diagnosis. However, even among the patients with only endometerioid tumors we saw increased mortality in those who had stopped tamoxifen five or more years before endometrial cancer diagnosis compared with those who had not received tamoxifen.
One possibility to consider is that gynecologic surveillance could have been less comprehensive after patients had ceased tamoxifen, resulting in delayed diagnosis and poorer prognosis. We investigated this hypothesis by looking at the effect of cessation among patients diagnosed with endometrial cancer before and after 1995 (when the first major reports of increased risk of endometrial cancer with tamoxifen use appeared [3, 7, 13] and awareness of the issue presumably increased), but we found that this did not change our findings, nor if we split the data at 1990 (when the first randomized trial results appeared linking tamoxifen to second cancers ).
If there is a real effect of time since last use we are unable to suggest an explanation for the increased risk but speculate that endometrioid endometrial cancers developing after long induction times may have different characteristics from those occurring in closer proximity to tamoxifen exposure.
Conclusions
Patients with endometrial cancer after five years use of tamoxifen for breast cancer have increased mortality from endometrial cancer, due to the occurrence of less favorable morphological subtypes of endometrial cancer in long term tamoxifen users. Patients who had stopped tamoxifen use five or more years before diagnosis of endometrioid endometrial cancer had increased endometrial cancer mortality, a finding that warrants further research.
Abbreviations
ALERT:
: Assessment of Liver and Endometrial Cancer Risk following Tamoxifen (a cohort study cancer in the Netherlands)
CI:
: confidence interval
FIGO:
: Fédération Internationale de Gynécologie et d'Obstétrique (International Federation of Gynaecology and Obstetrics)
HR:
: hazard ratio
NHSCR:
: National Health Service Central Register
SEER:
: Surveillance, Epidemiology and End Results
TAMARISK:
: Tamoxifen Associated Malignancies: Aspects of Risk.
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Acknowledgements
Dutch Cancer Society grant NKI 2002-2586, National Institutes of Health R03 grant number 1R03CA130108-02. The ICR acknowledge NHS funding to the NIHR Biomedical Research Centre Netherlands: The Comprehensive Cancer Centers' TAMARISK-group (Tamoxifen Associated Malignancies: Aspects of Risk)-group of the Comprehensive Cancer Centers (CCC): O. Visser (CCC Amsterdam), R. A. M. Damhuis (CCC Rotterdam), W. J. Louwman (CCC South Netherlands), J. A. A. M. van Dijck (CCC East Netherlands), Y. Westerman (CCC Middle Netherlands), M. J. M. Dirx (CCC Limburg), M. L. E. A. Jansen-Landheer (CCC West), L. de Munck (CCC Northern Netherlands), S. Siesling (CCC Stedendriehoek Twente). United Kingdom: The regional cancer registries: D. H. Brewster (Scottish Cancer Intelligence Unit), D. Forman (Northern & Yorkshire Cancer Registry & Information Service), S. Godward (East Anglian Cancer Intelligence Unit), A. Moran (North Western Cancer Registry), G. Lawrence (West Midlands Cancer Intelligence Unit), H. Møller (Thames Cancer Registry), M. Roche (Oxford Cancer Intelligence Unit), P. Silcocks (Trent Cancer Registry), J. A. Steward (Welsh Cancer Intelligence & Surveillance Unit), J. Verne (South & West Intelligence Unit), E. M.I. Williams (Merseyside & Cheshire Cancer Registry). The US: Dennis Deapen (Los Angeles), James R. Cerhan (Iowa) Stephen M. Schwartz (Seattle), Jonathan Liff (Atlanta), Jeffrey Perlman, Leslie Ford (National Cancer Institute).
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Authors and Affiliations
Section of Epidemiology, The Institute of Cancer Research, Sutton, Surrey, SM2 5NG, UK
Michael E Jones & Anthony J Swerdlow
2. Department of Epidemiology, The Netherlands Cancer Institute, Plesmanlaan 121, 1066 CX, Amsterdam, The Netherlands
Flora E van Leeuwen & Wilhelmina E Hoogendoorn
3. Department of Gynecology, University Medical Center Groningen, University of Groningen, Groningen, The Netherlands
Marian JE Mourits
4. Department of Pathology, University Medical Center Groningen, University of Groningen, The Netherlands
Harry Hollema
5. Department of Pathology, The Netherlands Cancer Institute, Amsterdam, The Netherlands
Hester van Boven
6. Department of Pathology and Norris Comprehensive Cancer Center, University of Southern California Keck School of Medicine, 1441 Eastlake Ave, Ste 5409, Los Angeles, CA, 90033, USA
Michael F Press
7. Division of Cancer Etiology, Department of Population Sciences, Beckman Research Institute, 1500 East Duarte Road, City of Hope, Duarte, CA, 91010, USA
Leslie Bernstein
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AJS holds shares in GlaxoSmithKline (who do not manufacture tamoxifen, but do make other drugs). The authors declare that they have no other competing interests.
Authors' contributions
MJ, FvL, WH, LB and AS made substantial contributions to the conception, design, analysis and interpretation of data. All authors have been involved in acquisition of data, drafting the manuscript, revising it critically for important intellectual content, and have given final approval of the version to be published.
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Jones, M.E., van Leeuwen, F.E., Hoogendoorn, W.E. et al. Endometrial cancer survival after breast cancer in relation to tamoxifen treatment: Pooled results from three countries. Breast Cancer Res 14, R91 (2012).
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Breast Cancer
Tamoxifen
Endometrial Cancer
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Tamoxifen Treatment
Breast Cancer Research
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10535 | https://www.youtube.com/watch?v=vDw2ALprfiw | Coordinate Geometry: Straight lines: Area of a square given the equation of a pair of parallel sides
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Posted: 7 Oct 2014
Coordinate Geometry: Straight lines: Area of a square given the equation of a pair of parallel sides
Transcript:
Hello friends welcome back in this session let's look at how to find out the area of a square if we know two parallel sides of a square so let's look at the equation here of the sides we have 5x - 12 y - 65 = 0 and 5x - 12 y + 26 = to 0 so let the equations let let these two sides Be those so I'll write down here 5x - 12 y - 65 = 0 and this side is 5X equation of this is 5x - 12 y + 26 = 0 now we we have been asked to find out the area of the square we know that area of square is equal to side into side so if I know the side then I can square it to find out the area now the problem is I don't have any coordinates or I haven't been given any any lens but I see that these two are parallel lines and I know how to find out the distance between two parallel lines the distance between two parallel lines ax Plus b y + C1 = 0 and ax + b y + C2 = 0 is given by C1 - C2 byun of a s + b² so taking this as a b C1 and a b C2 we will have the distance B between these two parallel lines is nothing but equal to C1 - C2 that is - 65 - 26 mod of that so that you will not have any negative value for the distance between two points then you will have square root of a square + b square a is 5 so 5 S + b² is 12 s this is equal to 65 + 26 so can I say mod of - 91 divid by squ < TK of 25 + 144 this is equal to 91 / < TK of 169 that's equal to 91 / 13 which is equal to 7 so we know that D is equal to 7 7 once we know D it's very easy to find out the area area equal to 7 into 7 that is equal to 49 square units so whenever we know the equation of one of the two parallel sides either either these two sides or these two sides if we know the equation of two pades then we can use the formula of finding out the distance between two parallel lines to find out the side of the square once we know the side of the square it becomes pretty easy to find out the area so that's how we can find out the area in the next session we will look into some more examples on distance between two parallel lines |
10536 | https://www.ck12.org/book/ck-12-physics-intermediate/section/27.3/ | Two-Dimensional Motion Problem Sets | CK-12 Foundation
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27.3 Two-Dimensional Motion Problem Sets
Difficulty Level: At Grade | Created by: CK-12
Last Modified: Jun 24, 2017
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A vector \begin{align}\vec {A}\end{align} has a magnitude of 5 units and is directed east. Vector \begin{align}\vec {B}\end{align} is directed north with a magnitude of 8 units. Draw a vector diagram for determining \begin{align}\vec {A}+ \vec{B}\end{align} and mathematically determine the magnitude and direction of the resultant vector (determine the displacement).
You walk 2 m east and 5 m south. What is the resultant displacement?
What is the magnitude of the displacement between the points (0,1) and (7,3)?
Tom walks 3 meters east, then 25 meters north. After Tom walks another 8 meters west, what is his magnitude of displacement from his original position?
On a road trip you drive 10 m east and then 12 m at an angle of 60° south of east. Find your resultant displacement.
The initial velocity of a particle is given as \begin{align}\vec{v}_0=3.00 \hat{x} \ m/s\end{align}and the acceleration is given as \begin{align}\vec{a} = 2.50 \hat{y} \ m/s^2\end{align}. Find the position and velocity vectors for the particle in terms of time \begin{align}t\end{align}.
A boy walks to the mall traveling 2 m/s for 5 minutes heading east. He then turns north and walks 3 m/s for 2 minutes. Find the boy’s average speed and total vector displacement.
A car moves east at a constant speed has a velocity of 50 mph. After 40 minutes, it begins heading 20.0° north of east for 20 minutes before heading east again for 30 more minutes. What is the car’s average velocity?
An Olympic sport shooter aims at a target 30 meters away. The fired projectile hits the target 0.02 meters beneath the point at which she aimed.
For how long was the projectile in the air?
With what velocity was the projectile fired?
A ball is thrown with an initial velocity of 25 m/s at an angle of 45°.
What is the horizontal displacement 1.0 seconds after the ball is thrown?
How about after 3 seconds?
A rifle with a muzzle velocity of 500 m/s shoots a target 50 m away. How high above the bullseye should the barrel be pointed?
Show that the maximum height reached by a projectile is represented as \begin{align}y=\frac{(v_0\sin\theta)^2}{2g}\end{align}.
A thrown ball has a time of flight of 5 seconds. If the ball leaves the hand 3 meters above the ground and lands 40 meters away, what must the magnitude and direction of its initial velocity be?
A plane releases a package 700 m above the ground while diving with constant speed at an angle 30.0° below the horizontal.
If the package hits the target on the ground 6 s later, determine the speed of the plane?
Find the package’s horizontal displacement?
What are the horizontal and vertical components of the package’s velocity upon hitting the ground?
A batter hits a ball 1.5 m above the ground. If the ball clears a fence that is 7 m tall and 97 m away, the hit will be a home run. The ball leaves with an angle of 45° and a range of 107 m. Will it clear the fence?
While evading the cops across city rooftops at 5 m/s, the caped crusader is quickly approaching a gap between two buildings that is 4 m wide with a 2 m drop. Since the caped crusader is well versed in physics, he knows he should leap at an angle of 45°. The cops, while capable of running at the same speed as the crusader, decide to leap with maximal horizontal velocity.
Do the cops clear the gap?
Does the caped crusader clear the gap? If so, by how much? If not, how close was he?
Upon graduation, a happy physics student launches her cap in the air. The cap has an initial velocity of 20 m/s and is released at 50° with respect to the horizontal.
Find the total time the cap is in the air. (Assume that \begin{align}y = y_0\end{align}.)
Determine how far the cap traveled horizontally.
While playing kickball with your friends, you kick the ball over a wall that is 3.2 m tall. The time of flight at the moment the ball clears the wall is 0.8 s and the distance to the wall is 12 m.
Find the initial speed and direction of the ball.
At what time does the ball reach its maximum height?
An emergency care package is being dropped to help out a village in need. The package is being dropped from 100 m above the ground while the plane is climbing upward at 35 m/s at an angle of 30° with respect to the horizontal. Where does the care package hit the ground?
Worried that your cat may be ambushed by a mountain lion, you build a fence around your backyard that is 4 m tall. If a mountain lion approaches the fence and leaps at an angle of 45°, what must its initial speed be to reach the top of the fence?
While on the moon and astronaut notices that he can leap a distance of 15 m if his initial speed is 5 m/s. Find the acceleration due to gravity on the moon.
A rifle that has a muzzle velocity of 1500 m/s is aimed at a target 3000 m away. At what angle should the rifle be fired?
While playing catch with your friends on the balcony, the ball goes sailing over the edge. The ball passes the edge with an angle 30° below the horizontal and an initial velocity of 7.5 m/s.
If the ball hits the ground 4 s later, how far away is the ball from the building?
What height did the ball fall from?
Determine how long it takes the ball to travel to a point 8 m below the balcony?
Billy, upset that he flunked his physics test, kicked a rock off a 50-meter cliff. After 6 seconds have passed, Billy hears the rock strike a metal roof. What was the initial velocity of the rock that was kicked off the cliff if the speed of sound is 343 m/s?
A softball player throws a softball at a certain angle in such a way that when the ball reaches its maximum height, its speed is one-third the speed at one-third its maximum heights. Calculate the initial angle of the ball when it was thrown.
A basketball player shoots a basketball with an angle of 60° at a hoop 5 meters away from the ball. With what velocity should the player shoot the ball to prevent it from dropping more than 10 cm vertically on the way to the hoop?
Shot putter competitor #1 launches a metal ball at 45° above the horizontal, and the metal ball lands 15 meters away. If shot putter competitor #2 launches his ball at a 57° angle at 10 m/s, will it travel more than 15 m?
Lola is dribbling a ball up a hill with an inline of \begin{align}\theta_1\end{align}. She then kicks the ball at an angle of \begin{align}\theta_2\end{align} with respect to horizontal \begin{align}\left ( \theta_2 > \theta_1 \right )\end{align}, determine the distance \begin{align}d\end{align} up the incline where the ball lands.
Attempting to score the winning point for his kickball team, Nicholas manages to kick the ball out of the park, sending it over a wall that is 100 m away and 15 m high. The ball is struck at an angle of 40°.
What is the initial speed of the ball?
Determine the time that it takes for the ball to just clear the top of the wall.
A rock is thrown from the top of a building at an angle of 35° with respect to the horizontal. The height of the building is 50 m and the initial velocity of the rock is 20 m/s.
Determine how long is the stone in the air.
What is the rock’s speed just before impacting the ground?
Sitting down to a picnic in the park, you watch a cat play in the grass. With respect to where you are sitting, the cat has the coordinates (1.0 m, 3.9 m) at \begin{align}t=0\end{align}. However, 5 seconds later, the cat is located at (6.4 m, 1.5 m).
Find the average velocity components.
Find the magnitude and direction of the average velocity.
You are watching a mouse race through a maze. At \begin{align}t=0\end{align}, define the initial position of the mouse as the origin. Determine the mouse’s \begin{align}(x,y)\end{align}position and its distance from the origin after 6.4 seconds in each of the following scenarios:
At \begin{align}t=6.4 \ s\end{align}, the mouse’s average velocity in the \begin{align}x\end{align}-direction is 2.0 m/s and its average velocity in the \begin{align}y\end{align}-direction is -2.0 m/s.
At \begin{align}t=6.4 \ s\end{align}, the mouse’s average velocity in the \begin{align}x\end{align}-direction is -2.0 m/s and its average velocity in the \begin{align}y\end{align}-direction is 2.0 m/s.
While creating a first-person shooter video game, you want to model the animation of a missile being fired from the player’s position as a function of time it spends in air. The missile’s position is defined by the following function: \begin{align}\vec r=\left [ 0.03 \ m+(2.0 \ m/s^2)t^2\right ] \hat{x}-\left [(0.5\ m/s)t\right ] \hat{y}\end{align}. Calculate the magnitude and direction of the average velocity after 2 s (from \begin{align} t=0 \ s\end{align} to \begin{align}t=2.0 \ s\end{align}.)
Upon arriving at the park, you give your dog a bone and let him off his leash. For a minute, he lies in the grass, gnawing at his treat. He then runs off with his bone. The initial components of his velocity are \begin{align}v_x= 2.0 \ m/s\end{align} and \begin{align}v_y= 1.5 \ m/s\end{align}. After 45 seconds have passed, the components of his velocity are \begin{align}v_x=8.0 \ m/s\end{align} and \begin{align}v_y=5.5 \ m/s\end{align}. Determine your dog’s average acceleration in the \begin{align}x\end{align}- and \begin{align}y\end{align}-directions during your time at the park so far.
Upset at a poor quiz score, you shove your textbook across the coffee table with a horizontal velocity of 2.5 m/s. The book hits the ground 0.40 s later. (Ignore friction on the table and air resistance.)
Find the height of the coffee table.
Calculate the horizontal distance between the table and where the book lands.
Draw the following graphs to describe the motion of the book: \begin{align}x\end{align} vs. \begin{align}t\end{align}, \begin{align}y\end{align} vs. \begin{align}t\end{align}, \begin{align}v_x\end{align} vs. \begin{align}t\end{align}, and \begin{align}v_y\end{align} vs. \begin{align}t\end{align}. (Hint: Calculate the velocity components first.)
John plans to impress his fiancé Anna by dropping roses on her from the skies with the help of a hang glider. Maintaining a constant horizontal speed of 15 m/s in the air, he drop the roses at an elevation of 200 m. Calculate the horizontal distance between Anna and the position from which John should drop the roses.
While hiking in Hawaii, Carlos and Ivan decide to jump from the top of a waterfall. At the last moment, Ivan chickens out and drops off the edge. Carlos, on the other hand, leaps with a horizontal velocity of 2.5 m/s. If Ivan hits the water in 4.0 s, how far away does Carlos land from the cliff?
A group of backpackers reach the top of a waterfall and decide to cool off with a dive into the lake below. Peering over a nearby cliff, they notice a ledge 8.0 m down that is 2.3 m long, as shown in the diagram below. Determine the minimum horizontal speed with which they must leap off the cliff in order to clear the ledge.
Inspired by the Dukes of Hazzard movie, you decide to see if your 1975 Buick Regal can leap across a river that is 50 m wide. The vertical displacement between the launching point and the ground on the other side of the river is -19 m.
At what horizontal speed would you have to drive off the cliff in order to clear the river?
What is the magnitude of the velocity just before the car hits the ground on the other side of the river?
You are playing a video game in which your character needs to defend the top of a hill. You decide to cash in some bonus points for extra shields. The package will be delivered by a plane flying 10 m overhead. If the package is dropped when it is horizontally 5.0 meters away from you, what speed should the plane be flying at?
In previous problems involving objects rolling or being pushed off cliffs or tables, you’ve assumed that \begin{align}g\end{align} is equal to gravitational acceleration on Earth. If the traveled horizontal distances for such objects are suddenly doubled, what would the new acceleration due to gravity be?
Playing catch with your cousin on a warm summer night, you throw the ball with a horizontal velocity of 10.0 m/s and a vertical velocity of 15.2 m/s.
What is the maximum height the ball reaches?
How long does it take the ball to return to the height it was thrown from? (Ignore air resistance.)
You are playing outside when your ball rolls toward your younger brother. Instead of tossing the ball back to you, he kicks it with a velocity of 60.0 m/s and at an angle of 65° above the horizontal.
Calculate the components of the initial velocity.
Ignoring air resistance, how long does it take the ball to reach its maximum height?
What is the maximum height?
Determine the components of the ball’s velocity and acceleration when it reaches its maximum height.
A bolt is fired from a crossbow with an initial velocity of 40 m/s and at an angle of 50° above the horizontal. If air resistance is ignored, determine the maximum height reached and total distance traveled.
After being hit, a softball flies with a velocity of 24 m/s at an angle of 37.8° above the horizontal. How many seconds after the ball is struck will it be 20 m away from its initial position?
Standing on the roof of a building that is 30 m high, you launch a baseball off the edge with a speed of 25 m/s at an angle of 25° below the horizontal.
Determine the magnitude of the velocity before the ball hits the ground.
Determine the horizontal distance the ball traveled before hitting the ground.
During medieval times, invaders often used trebuchets to launch objects over castle walls to scare and hurt the castle’s inhabitants. The distance between the castle walls and the trebuchet is 75 m and the castle walls are 30 m high. If the trebuchet launches objects at an angle of 45°, calculate the minimum velocity needed to launch objects from the trebuchet over the castle wall?
While on a 4-hour layover at Denver International Airport, you decide to play around on the moving walkway. The walkway is 70.0 m long and moves at a speed of 1.5 m/s. If your normal walking speed is 1.0 m/s, how long would it take you to walk across the walkway in the unintentional direction?
With the goal of creating a super slingshot to launch water balloon bombs, you construct a system that accelerates balloons 150.0 m up a 45° incline at a rate of 1.0 m/s 2. Calculate the maximum height above the ground that the water balloons reach when launched. (Hint: Once the balloons leave the incline, the only acceleration is due to gravity.)
A small boat is adrift on the ocean, heading towards a bunch of rocks near a cliff. As it drifts near the cliff, it is noticed by a group of people. To try and tow them in, a rope is thrown from the cliff, which is 12.0 m above the boat. The rope is thrown at an angle of 50° above the horizontal with an initial velocity of 16 m/s towards the boat. If the boat is has a velocity that is 0.6 m/s towards the cliff while heading to the rocks, at what distance \begin{align}x\end{align} from the cliff, should the people on the cliff throw the rope?
While at the state fair, you compete in a game to fill up a cylindrical container as fast as possible using a water gun. If the container’s height is twice as large as its diameter \begin{align}d\end{align} and the water gun’s distance from the container is 6 times the of the cylinder’s diameter, what is the minimum velocity that the water can be shot, assuming the gun is aimed at 45°?
Wanting to play the ultimate prank on your professor, you attempt to roll a giant balloon full of paint off of the physics building roof. As seen in the diagram below, the edge of the roof has a 40° incline with respect to the negative \begin{align}x\end{align}-axis and is 16.5 m above the ground. The balloon has a 6.5 m/s velocity as it rolls down the incline, and the professor, who is 1.5 m tall, stands 5.5 m away from the building.
Calculate how far from the edge does the balloon land?
Does the professor get hit?
Competing in a long jump event, Jack leaves the ground at an angle of 21° and travels a horizontal distance of 5.68 m. Calculate his initial velocity.
Competing in a long jump event, Allison jumps with an initial speed of 8.17 m/s and clears a distance of 6.01 m. At what angle did she leave the ground?
If two particles are fired with angles whose difference from 45° are equal, they will travel the same distance, assuming the initial velocity are equal and the landing height is equal to the launching height. Show that this is true.
Show that the velocity of a ball thrown off of a cliff \begin{align}y\end{align} m above the ground, at an angle of \begin{align}\theta\end{align}, is independent of \begin{align}\theta\end{align}.
If you were given the position of an object at two different points and the elapsed time, which of the following would you not be able to determine: instantaneous velocity or average velocity? Explain your answer.
Draw the path showing how the velocity of an object would change if it were thrown parallel to the ground.
Draw the acceleration as a function of time and show how the acceleration of an object would change if it were thrown parallel to the ground.
Given the initial position of a thrown object, explain how you could calculate the coordinates of the object when it is at its highest point.
An object is sliding across a frictionless surface at a constant velocity. If a constant acceleration is applied perpendicular to the velocity and normal to the surface, what is the shape of the object’s path?
Determine which of the following remain constant during parabolic motion: acceleration, speed, horizontal component of the velocity, and vertical component of the velocity.
While traveling on a bullet train, you throw a ball straight up in the air. Describe the motion of the ball from the eyes of passengers both outside and inside of the train.
For his final performance, a daredevil wants to jump over a pit of spikes and gorillas on his motorcycle. As seen in the diagram below, the pit is 50.0 m wide and the ramp off of which he will jump is inclined at angle of 50°. The top of the ramp reaches 75 m above the spike but only 12.0 m above the landing point. Ignoring air resistance, at what speed does the daredevil need to leave the ramp in order to clear the pit?
An object moves through space with a velocity of \begin{align}\vec{v}=5 \hat{i} - 10 \hat{j}\end{align}, with an acceleration of \begin{align} \vec{a}=2 \hat{i}\end{align}. Determine the components of the velocity at any given time.
A particle has an initial velocity of \begin{align}\vec v=(5 - 2t)\hat{i} - 8\hat{j}\end{align}. Determine the velocity 2 s later.
Consider a launched projectile that travels a parabolic path. At what point is the velocity vector and the acceleration vector perpendicular to each other?
Is there a point at which the velocity and acceleration vectors of an object exhibiting projectile motion are parallel?
At what angle should you throw a ball to achieve the maximum possible height?
What is significant about the following angle pairs in projectile motion: (40, 50), (25, 65)?
While pondering the meaning of life, a student is playfully throwing pebbles off the physics department’s roof. If the building is 50 m high and the rocks are being thrown with an initial velocity of 15 m/s at an angle of 35° with respect to the horizontal, how long does it take for the pebbles to hit the ground?
Determine the magnitude of the final velocity of an object that is launched from a ledge 30 m high with an initial velocity of 25 m/s at an angle of 40° with respect to the horizontal.
A screw from a plane traveling 400 m above the surface of the Earth shakes loose and falls to the surface. If the plane is traveling at 80 m/s parallel to the ground, calculate how much time elapses before the screw hits the ground?
If a particle’s position is given by \begin{align}\vec{r} (t)=(1.0t+1.0)\hat{i} + (0.5t^2+0.2)\hat{j},\end{align} determine the average velocity of a particle between \begin{align}t=1.0\ s\end{align} and \begin{align}t = 3.0 \ s\end{align}.
A car has an initial velocity of 25 m/s, and 10 s later, its velocity is 42 m/s. What is the car’s average acceleration?
If the average acceleration of an object is \begin{align}\vec a=(0.7 \ m/s^2)\hat{i} + (0.5 \ m/s^2)\hat{j}\end{align}, determine the direction of the acceleration with respect to the positive \begin{align}x\end{align}-axis.
While visiting a planet outside our solar system, an astronaut jumps a horizontal distance of 25 m with an initial velocity of 8 m/s.
Calculate the acceleration due to gravity on this planet.
If the astronaut were back on Earth, what would be the maximum horizontal distance traveled?
A ball is tossed from the roof of a building with an initial velocity of 5.0 m/s in the \begin{align}x\end{align}-direction strikes the ground 4.0 seconds later. Determine the distance from the building that the ball hit the ground.
A ball is tossed from the roof of a building with an initial velocity of 8.0 m/s at an angle of 35° with respect to the \begin{align}x\end{align}-axis. If the ball strikes the ground 4.0 seconds later, determine the distance between the building and the location where the ball hit the ground.
While sitting on a cliff that overlooks a lake, Aliya throws a pebble with an initial horizontal velocity of 5 m/s, and the pebble lands in the lake 4 s later. Her sister, Meta, then throws a pebble with a velocity of 5 m/s at an angle of 35° with respect to the \begin{align}x\end{align}-axis.
How high is the cliff?
Determine how much longer Meta’s pebble is in the air.
If Audrey is standing at a point that is 3 miles east and 5 miles north of her house, what is her position in vector notation?
Calculate \begin{align}\vec {A} - 2\vec {B}\end{align} if \begin{align}\vec {A}=(5,4)\end{align} and \begin{align}\vec {B}=(1,2)\end{align}.
If \begin{align}\vec{A} + \vec{B}=4\hat{j}\end{align} and \begin{align}2\vec{B}=2\hat{i} - 3\hat{j}\end{align}, determine \begin{align}\vec {A}\end{align}.
True or False: The sum of two vectors is the sum of their magnitudes.
A plane is attempting to fly due south with an airspeed of 300 km/h while a wind blowing from west to east has a speed of 60 km/h. In what direction should the plane head?
Determine the velocity of an object whose position as a function of time is given as \begin{align}\vec r(t)=(3.0t)\hat{i} - (4.8t^3)\hat{j}\end{align}.
Determine the expression for the acceleration of an object whose position as a function of time is given as \begin{align}\vec r(t)=(3.0t-t^2)\hat{i} - (5.0t^2)\hat{j}\end{align}.
Calculate the time it takes for a ball thrown vertically upward with an initial velocity of 16.3 m/s to travel 10.0 m.
Determine the time it takes for a ball to reach its maximum height when thrown vertically upward at an initial velocity of 8.3 m/s.
Determine the components of the vector below.
If \begin{align}\vec{A}=3\hat{i} - 8\hat{j}\end{align} and \begin{align}\vec{B}=-7.2\hat{i} + 19.2\hat{j}\end{align}, write an equation relating the two vectors.
An object initially traveling in the \begin{align}x\end{align}-direction at a speed of 1.1 m/s is given an acceleration of 2.0 m/s 2 in the \begin{align}y\end{align}-direction for 3.0 s. What is the object’s speed afterwards?
An initially motionless object is given an acceleration of 2.0 m/s 2 in the \begin{align}y\end{align}-direction for 3.0 s. What is the object’s speed afterwards?
True or False: An object takes the same amount of time to hit the ground when dropped and launched horizontally from the same height.
Given two vectors\begin{align}\vec A\end{align}and\begin{align}\vec B\end{align}, when can the magnitude of the resultant vector be written as\begin{align}A+B\end{align}?
Prove that if a vector is two-dimensional, a component of the vector will always be less than the magnitude of the vector.
Prove that if a vector has at least one nonzero component, it must have a positive magnitude.
Calculate the sum of the following two vectors: \begin{align}|\vec A| =2\end{align} at 20° north of east and \begin{align}|\vec B|=3\end{align} at 35° north of east. What is the magnitude of the resultant vector?
Is it possible for a vector to have only a magnitude and not a direction? Explain.
Two vectors are equal in magnitude but opposite in direction. Show, using components, that the resultant vector has a magnitude of zero.
You are given two vectors: \begin{align}|\vec A|=5\end{align} at 20° north of east and \begin{align}|\vec B|=3\end{align}at 65° north of east.
What is the magnitude of the resultant vector?
What is the direction of the resultant vector?
A car drives 4 miles north and then turns west before driving another 4 miles.
Determine the magnitude of the car’s displacement vector.
Graph the car’s displacement vector. Include the proper lengths and angles.
While trying to determine how much rope you will need to scale a wall, you note that when you stand 70.0 m away from the base, the top of the wall makes a 35° angle with the ground. Calculate the height of the wall.
Given two vectors \begin{align}\vec A\end{align} and \begin{align}\vec B\end{align}, if the component of vector \begin{align}\vec A\end{align} that lies along vector \begin{align}\vec B\end{align} is 0, what can be said about the relationship between the two vectors?
What method is used to add a vector quantity to a scalar quantity?
Both vectors \begin{align}\vec A\end{align} and \begin{align}\vec B\end{align} each have a magnitude of 3.0 m. If \begin{align}\theta_A=20^\circ\end{align} and \begin{align}\theta_B=90^\circ\end{align}, calculate the following:
\begin{align}\vec A+ \vec B\end{align}
\begin{align}\vec A - \vec B\end{align}
\begin{align}2 \vec A - 2 \vec B\end{align}
Calculate the angle each resultant vector in the previous problem makes with the positive \begin{align}x\end{align}-axis.
Given the vector \begin{align}\vec A= -3.0 \hat{x} -2.0 \hat{y}\end{align}, determine the angle the displacement vector makes with the positive \begin{align}x\end{align}-axis.
A vector has an \begin{align}x\end{align}-component of -20.0 splurgs and a \begin{align}y\end{align}-component of 55.0 splurgs. Calculate the vector’s magnitude and direction.
Vector \begin{align}\vec A\end{align} has components 3, 5, 7 in the \begin{align}x\end{align}-, \begin{align}y\end{align}-, and \begin{align}z\end{align}-directions respectively. Calculate the magnitude of the displacement vector.
Prove that addition of the following vectors is associative: \begin{align}v_1 + (v_2 + v_3) = (v_1 + v_2) + v_3\end{align}.
Prove that \begin{align}\vec A - \vec B = \vec A - (\vec B)\end{align}.
Prove that \begin{align}2(\vec A + \vec B) = 2 \vec A + 2 \vec B\end{align}.
Under what conditions is the following statement true: \begin{align}\vec A = \vec B\end{align}?
During a jet pack training exercise, one pilot rises 200 meters upward, flies 100 meters west, and then flies 50 meters north. A second pilot flies 75 meters west, rises 150 meters upward, and then flies 160 meters north. How far apart are the two jet pack pilots?
Find the magnitude of the sum of the following two vectors: \begin{align}\vec A = (2, -1, 5)\end{align}, \begin{align}\vec B = (0, 3, 3)\end{align}.
Express the sum of the following two vectors in \begin{align}\hat{x}, \hat{y}, \hat{z}\end{align} notation: \begin{align}\vec A = (2, 0, 5)\end{align}, \begin{align}\vec B=(4, 3, 3)\end{align}
Find the magnitude of the resultant of the following vector operation: \begin{align}\vec A + \vec B - \vec C\end{align}. Let \begin{align}\vec A=(1, 5, 0)\end{align}, \begin{align}\vec B = (0, 7, 2)\end{align}, and \begin{align}\vec C = (-1, -2, 2)\end{align}.
Find the magnitude of the resultant of the following vector operation: \begin{align}-\vec A -\vec B + \vec C\end{align}. Let \begin{align}\vec A=(1, 5, 0)\end{align},\begin{align}\vec B = (0, 7, 2)\end{align}, and\begin{align}\vec C = (-1, -2, 2)\end{align}.
Given vectors \begin{align}\vec A\end{align} and \begin{align}\vec B\end{align} with A = 30 m at 20° and B = 20 m at 75°, what are the components of each vector along the \begin{align}x\end{align}- and \begin{align}y\end{align}-axes?
Given vectors \begin{align}\vec A\end{align} and \begin{align}\vec B\end{align} with A = 30 m at -20° and B = 20 m at 75°, what are the components of each vector along the \begin{align}x\end{align}- and \begin{align}y\end{align}-axes? (Assume the angles are measured with respect to the negative \begin{align}x\end{align}-axis.)
Determine the resultant vector of the sum \begin{align}\vec A + \vec B\end{align}, if A = 2 N at 20° and B = 1 N at 90°.
Calculate the resultant vector of the difference \begin{align}\vec A - \vec B\end{align}, if A = 2 N at 20° and B = 1 N at 90°.
Find the angle between the following vectors: \begin{align}\vec A = 3 \hat{i}\end{align}, \begin{align}\vec B = 5 \hat{j}\end{align}
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CK-12 Physics - Intermediate is a high school FlexBook® resource covering all standard high school physics topics with mathematical rigor and detailed examples.
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10537 | https://www.nccn.org/patients/guidelines/content/PDF/breast-invasive-patient.pdf | 2025 Invasive Breast Cancer NCCN GUIDELINES FOR PATIENTS ® Presented with support from FOUNDATION Guiding Treatment. Changing Lives.
NATIONAL COMPREHENSIVE CANCER NETWORK ® Available online at NCCN.org/patientguidelines Ü 1 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Invasive Breast Cancer About the NCCN Guidelines for Patients® Did you know that top cancer centers across the United States work together to improve cancer care? This alliance of leading cancer centers is called the National Comprehensive Cancer Network® (NCCN®).
Cancer care is always changing. NCCN develops evidence-based cancer care recommendations used by health care providers worldwide. These frequently updated recommendations are the NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®). The NCCN Guidelines for Patients plainly explain these expert recommendations for people with cancer and caregivers.
These NCCN Guidelines for Patients are based on the NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®) for Breast Cancer, Version 3.2025 – March 18, 2025.
Learn how the NCCN Guidelines for Patients are developed NCCN.org/patient-guidelines-process Find an NCCN Cancer Center near you NCCN.org/cancercenters View the NCCN Guidelines for Patients free online NCCN.org/patientguidelines Connect with us 2 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Invasive Breast Cancer Supporters To make a gift or learn more, visit online or email NCCNFoundation.org/donate PatientGuidelines@ NCCN.org NCCN Foundation gratefully acknowledges the following corporate supporters for helping to make available these NCCN Guidelines for Patients: AstraZeneca and The Wawa Foundation.
NCCN independently adapts, updates, and hosts the NCCN Guidelines for Patients. Our corporate supporters do not participate in the development of the NCCN Guidelines for Patients and are not responsible for the content and recommendations contained therein.
NCCN Guidelines for Patients are supported by funding from the NCCN Foundation® FOUNDATION Guiding Treatment. Changing Lives.
NATIONAL COMPREHENSIVE CANCER NETWORK ® 3 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Invasive Breast Cancer Contents © 2025 National Comprehensive Cancer Network, Inc. All rights reserved. NCCN Guidelines for Patients and illustrations herein may not be reproduced in any form for any purpose without the express written permission of NCCN. No one, including doctors or patients, may use the NCCN Guidelines for Patients for any commercial purpose and may not claim, represent, or imply that the NCCN Guidelines for Patients that have been modified in any manner are derived from, based on, related to, or arise out of the NCCN Guidelines for Patients. The NCCN Guidelines are a work in progress that may be redefined as often as new significant data become available. NCCN makes no warranties of any kind whatsoever regarding its content, use, or application and disclaims any responsibility for its application or use in any way.
NCCN Foundation seeks to support the millions of patients and their families affected by a cancer diagnosis by funding and distributing NCCN Guidelines for Patients. NCCN Foundation is also committed to advancing cancer treatment by funding the nation’s promising doctors at the center of innovation in cancer research. For more details and the full library of patient and caregiver resources, visit NCCN.org/patients. National Comprehensive Cancer Network (NCCN) and NCCN Foundation 3025 Chemical Road, Suite 100, Plymouth Meeting, PA 19462 USA 4 About invasive breast cancer 8 Testing for breast cancer 22 Breast cancer staging 28 Types of treatment 44 Supportive care 51 Your treatment options 62 The breast after surgery 67 Recurrence 71 Other resources 75 Words to know 79 NCCN Contributors 80 NCCN Cancer Centers 82 Index 4 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 1 About invasive breast cancer 5 What is invasive breast cancer?
6 What are the parts of the breast?
6 What's in this book?
7 What can you do to get the best care? 5 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 1 About invasive breast cancer » What is invasive breast cancer?
Invasive breast cancer is cancer that has spread from the milk ducts or milk glands (lobules) into the surrounding breast tissue or nearby lymph nodes. Invasive breast cancer is also called early-stage breast cancer. What is invasive breast cancer?
Breast cancer starts in the cells of the breast. Over time, these cells form a mass called a tumor. In invasive breast cancer, cancer has grown outside the milk ducts or milk glands (lobules) into the surrounding breast tissue. Once outside the ducts or lobules, breast cancer can spread through lymph or blood to lymph nodes or other parts of the body.
Almost all invasive breast cancers are carcinomas—cancers that start in the cells that line the inner or outer surfaces of the body. There are different types of breast carcinoma, most of which arise in cells that make up the lining (epithelial cells) in the terminal duct lobular units (TDLUs) of the breast. A TDLU consists of a lobule connected to the end of a small milk duct. The most common types of invasive breast cancer are ductal carcinoma and lobular carcinoma.
Anyone can develop breast cancer, including those assigned male at birth. Although there are some differences between breast cancers in those assigned male and those assigned female at birth, treatment is very similar for all genders.
The breast The breast is a glandular organ made up of milk ducts, fat, nerves, blood and lymph vessels, ligaments, and other connective tissue.
6 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 1 About invasive breast cancer » What are the parts of the breast?
6 1 About invasive breast cancer » What are the parts of the breast? » What's in this book?
What are the parts of the breast?
The breast is a gland found on the chest. The breast is made of milk ducts, fat, nerves, lymph and blood vessels, ligaments, and other connective tissue. Behind the breast are the pectoral (chest) muscle and ribs. Muscle and ligaments help hold the breast in place.
Breast tissue contains glands that can make milk. These milk glands are called lobules. Lobules look like tiny clusters of grapes. Small tubes called ducts connect the lobules to the nipple.
The ring of darker breast skin in the center of the breast is called the areola. The raised tip within the areola is called the nipple. The nipple-areola complex (NAC) is a term that refers to both parts. Lymphatic fluid (lymph) drains from breast tissue into lymph vessels and travels to lymph nodes near your armpit (axilla). Lymph is a clear fluid that gives cells water and food. It also helps to fight germs. Nodes near the armpit are called axillary lymph nodes (ALNs). Cancer cells can travel through blood and lymph to lymph nodes and other parts of the body. What's in this book?
This book is organized into the following chapters: Chapter 2: Testing for breast cancer provides an overview of tests you might receive, and the role of HER2 status, hormone receptors, genetic cancer risk, and biomarker testing.
Chapter 3: Breast cancer staging provides information on how breast cancer is staged.
Chapter 4: Types of treatment gives a general overview of invasive breast cancer treatment and what to expect. Chapter 5: Supportive care gives an overview of what supportive care is and possible side effects of treatment.
Chapter 6: Your treatment options provides specific surgery and systemic therapy options for your cancer based on hormone receptor (HR) and HER2 tumor status. It also discusses who might benefit from systemic therapy before surgery.
Chapter 7: The breast after surgery offers more information on volume displacement, flat closure, and breast reconstruction.
Chapter 8: Recurrence explains treatment options for cancer that has returned.
Chapter 9: Other resources provides information on patient advocacy groups and where to get help.
7 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 1 About invasive breast cancer » What can you do to get the best care? What can you do to get the best care? Advocate for yourself. You have an important role to play in your care. In fact, you’re more likely to get the care you want by asking questions and making shared decisions with your care team. Consider seeking the opinion of a breast cancer specialist and team that specializes in diagnosis and management of breast cancer. The NCCN Guidelines for Patients will help you understand cancer care. With better understanding, you’ll be more prepared to discuss your care with your team and share your concerns. Many people feel more satisfied when they play an active role in their care.
You may not know what to ask your care team. That’s common. Each chapter in this book ends with an important section called Questions to ask. These suggested questions will help you get more information on all aspects of your care.
Take the next step and keep reading to learn what is the best care for you!
Why you should read this book Making decisions about cancer care can be stressful. You may need to make tough decisions under pressure about complex choices.
The NCCN Guidelines for Patients are trusted by patients and providers. They clearly explain current care recommendations made by respected experts in the field. Recommendations are based on the latest research and practices at leading cancer centers.
Cancer care is not the same for everyone. By following expert recommendations for your situation, you are more likely to improve your care and have better outcomes as a result. Use this book as your guide to find the information you need to make important decisions.
The NCCN Guidelines for Patients are seen as a trusted source by researchers, clinicians, and advocates. They set the standard for appropriate and effective breast cancer treatment, making them a good first step for newly diagnosed patients." “ 8 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer 9 General health tests 11 Fertility (all genders) 11 Blood tests 12 Imaging tests 16 Biopsy 17 HER2 status 18 Hormone receptor status 19 Biomarker testing 19 Genetic cancer risk testing 21 Key points 21 Questions to ask 9 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » General health tests Not all invasive breast cancers are the same. Treatment planning starts with testing. Your care team will want to gather information about the cancer you have. This chapter presents an overview of the tests you might receive and what to expect.
Tests are used to find cancer, plan treatment, and check how your cancer is responding to treatment. Results from a tumor biopsy and imaging studies will be used to determine what treatments are best for your type of breast cancer. Testing takes time. For possible tests, see Guide 1.
General health tests Medical history A medical history is a record of all health issues and treatments you have had in your life. Be prepared to list any illness or injury and when it happened. Bring a list of old and new medicines and any over-the-counter (OTC) medicines, herbals, or supplements you take. Some supplements interact with and affect medicines that your care team may prescribe. Tell your care team about any symptoms you have. A medical history, sometimes called a health history, will help determine which treatment is best for you.
Family history Some cancers and other diseases can run in families. Your doctor will ask about the health history of family members who are blood relatives. This information is called a family history. Ask family members on both sides of your family about their health issues like heart disease, cancer, and diabetes. It’s important to know the specific type of cancer, or where the cancer started, if it was in multiple locations, how old they were when they had the cancer, and if they had genetic testing. Guide 1 Possible tests Medical history and physical exam Diagnostic mammogram. Other imaging tests as needed Biopsy with pathology review Tumor estrogen receptor (ER) and progesterone receptor (PR) status Tumor HER2 status Genetic counseling and testing if at risk for hereditary breast cancer or have triple-negative breast cancer (TNBC) Address fertility and sexual health concerns Pregnancy test in those of childbearing potential Assess for distress Lymphedema baseline screening 10 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » General health tests Physical exam During a physical exam, your health care provider may: h Check your temperature, blood pressure, pulse, and breathing rate h Check your height and weight h Listen to your lungs and heart h Look in your eyes, ears, nose, and throat h Feel and apply pressure to parts of your body to see if organs are of normal size, are soft or hard, or cause pain when touched. h Examine your breasts to look for lumps, nipple discharge or bleeding, or skin changes. h Feel for enlarged lymph nodes in your neck, underarm, and groin.
Clinical breast exam A clinical breast exam (CBE) is a physical exam of the bare breast performed by a health care provider to check for lumps or other changes. It is done while you are seated and/ or lying down. Your provider should take time to palpate (feel) the entire breast, including the armpit. A nurse or assistant might also be in the room during the exam. Distress screening Distress is an unpleasant experience of a mental, physical, social, or spiritual nature. It can affect how you feel, think, and act. Distress might include feelings of sadness, fear, helplessness, worry, anger, and guilt. You may also experience depression, anxiety, and sleep issues.
It is normal to have strong feelings about being diagnosed with cancer and your feelings can also change from day to day and week to week. Talk to your care team and those whom you feel most comfortable about how you are feeling. There are services, people, and medicine that can help you. Support and counseling are available. Your treatment team will ask about your level of distress. This is part of your cancer care. More information on distress can be found in the NCCN Guidelines for Patients: Distress During Cancer Care at NCCN.org/ patientguidelines and on the NCCN Patient Guides for Cancer app.
Lymphedema baseline screening Lymphedema is a condition in which lymph fluid builds up in tissues and causes swelling. A lymphedema baseline screening measures the fluid levels in your body. It is done to look for small changes in the body after surgery. Performance status Performance status (PS) is a person’s general level of fitness and ability to perform daily tasks. Your state of general health might be rated using a PS scale called the Eastern Cooperative Oncology Group (ECOG) score or the Karnofsky Performance Status (KPS). PS is one factor taken into consideration when choosing a treatment plan.
11 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Fertility (all genders) 11 2 Testing for breast cancer » Fertility (all genders) » Blood tests Fertility (all genders) Some types of treatment such as chemotherapy can affect fertility, or the ability to have children. If you think you want children in the future, ask your care team how cancer and cancer treatment might change your fertility. To preserve your fertility, you may need to take action before starting cancer treatment. Those who want to have children in the future should be referred to a fertility specialist to discuss the options before starting treatment. Fertility preservation is all about keeping your options open, whether you know you want to have children later in life or aren’t sure at the moment. Fertility and reproductive specialists can help you sort through what may be best for your situation. More information on fertility preservation can be found at NCCN Guidelines for Patients: Adolescent and Young Adult Cancer at NCCN.
org/patientguidelines and on the NCCN Patient Guides for Cancer app.
Changes in fertility Treatment might cause your fertility to be temporarily or permanently impaired or interrupted. This loss of fertility is related to your age at time of diagnosis, treatment type(s), treatment dose, and treatment length. Talk to your care team about your concerns and if you are planning a pregnancy.
Preventing pregnancy during treatment Preventing pregnancy during treatment is important. Cancer treatment can affect the ovaries, damage sperm, and hurt a developing baby. Therefore, becoming pregnant or having one's partner become pregnant during treatment should be avoided. Non-hormonal birth control methods such as intrauterine devices (IUDs) and barrier methods are preferred in those with a breast cancer diagnosis. Types of barrier methods include condoms, diaphragms, cervical caps, and the contraceptive sponge. If you are pregnant at the time of your cancer diagnosis, treatment will need to be tailored to protect your unborn child. If you are breastfeeding at the time of cancer diagnosis, you may need to stop breastfeeding during treatment. Blood tests Blood tests check for signs of disease and how well organs are working. Some blood tests you might have are described next.
Alkaline phosphatase Alkaline phosphatase (ALP) is an enzyme found in the blood. High levels of ALP can be a sign cancer has spread to the bone or liver. A bone scan might be performed if you have high levels of ALP.
Complete blood count A complete blood count (CBC) measures the levels of red blood cells (RBCs), white blood cells (WBCs), and platelets (PLTs) in your blood. Red blood cells carry oxygen 12 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Imaging tests throughout your body, white blood cells fight infection, and platelets control bleeding.
Comprehensive metabolic panel A comprehensive metabolic panel (CMP) measures substances in your blood. It provides important information about how well your kidneys and liver are working, among other things Liver function tests Liver function tests (LFTs) look at the health of your liver by measuring chemicals that are made or processed by the liver. Levels that are too high or low signal that the liver is not working well or that cancer has spread to the liver. Pregnancy test Those who can become pregnant will be given a pregnancy test before treatment begins.
Imaging tests Imaging tests take pictures of the inside of your body. Imaging tests show the primary tumor, or where the cancer started, and look for cancer in other parts of the body. A radiologist, a medical expert, will interpret the test and send a report to your doctor. While these reports might be available to you through your patient portal or patient access system, please wait to discuss these results with your care team The following imaging tests are not in order of importance. You may not need all of these tests.
Bone scan A bone scan uses a radiotracer to highlight areas of bone damage or loss. A radiotracer is a substance that releases small amounts of radiation. Before the pictures are taken, the tracer will be injected into your vein. It can take a few hours for the tracer to enter your bones. However, the test is quick and painless.
A special camera will take pictures of the tracer in your bones as it moves over your body. Areas of bone damage take up more radiotracer than healthy bone and show up as bright spots on the pictures. Bone damage can be caused by cancer, cancer treatment, previous injuries, or other health issues. Bone x-ray An x-ray uses low-dose radiation to take one picture at a time. A tumor changes the way radiation is absorbed and will show up on the x-ray. X-rays are also good at showing bone issues. Your care team may order x-rays if your bones hurt or were abnormal on a bone scan. Contrast material Contrast material is a substance used to improve the quality of the pictures of the inside of the body. It is used to make the pictures clearer. Contrast might be taken by mouth (oral) or given through a vein (IV). Oral contrast does not get absorbed from your intestines and will be passed with your next bowel movements. IV contrast will leave the body in the urine immediately after the test. 13 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Imaging tests The types of contrast vary and are different for CT and MRI. Not all imaging tests require contrast, but many do. Tell your care team if you have had allergic reactions to contrast in the past. This is important. You might be given medicines to avoid the effects of those allergies. Contrast might not be used if you have a serious allergy or if your kidneys aren’t working well.
CT scan A computed tomography (CT or CAT) scan uses x-rays and computer technology to take pictures of the inside of the body. It takes many x-rays of the same body part from different angles. All the images are combined to make one detailed picture. Intravenous (IV) contrast is often used.
Diagnostic mammogram A mammogram is a picture of the inside of your breast. The pictures are made using x-rays. A computer combines the x-rays to make detailed pictures. A bilateral mammogram includes pictures of both breasts. Mammogram results are used to plan treatment.
Diagnostic mammograms look at specific areas of your breasts, which may not be clearly seen on screening mammograms. They are used to see tumor and the size of the tumor(s). Diagnostic mammograms include extra compression in certain areas of the breast, magnification views, or rolling the breast to image additional areas of the breast. Other tests may include a breast MRI or ultrasound.
What’s the difference between a screening and diagnostic mammogram?
A mammogram is a picture of the inside of your breast made using x-rays. During a mammogram, the breast is pressed between two plates while you stand in different positions. Multiple x-rays will be taken. A computer combines the x-rays to make detailed pictures.
Screening mammogram • Done on a regular basis when there are no signs or symptoms of breast cancer. Results take a few days.
Diagnostic mammogram • Used for those who have symptoms such as a lump, pain, nipple thickening or discharge, or those whose breasts have changed shape or size. An ultrasound is often used with a diagnostic mammogram.
• Also used to take a closer look at an abnormal area found in a screening mammogram. • A radiologist will evaluate the diagnostic mammogram while you wait so if additional testing is needed, it can be done right away.
Both types of mammograms use low-dose x-rays to examine the breast. They may use either the standard 2-dimensional (2D) digital mammogram or 3-dimensional (3D) mammogram known as tomosynthesis. 14 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Imaging tests MRI scan A magnetic resonance imaging (MRI) scan uses radio waves and powerful magnets to take pictures of the inside of the body. It does not use x-rays, which means there is no radiation delivered to your body during the test. Because of the very strong magnets used in the MRI machine, tell the technologist if you have any metal or a pacemaker in your body. During the test, you will likely be asked to hold your breath for 10 to 20 seconds as the technician collects the images. Contrast is used.
A closed MRI has a capsule-like design where the magnet surrounds you. The space is small and enclosed. An open MRI has a magnetic top and bottom, which allows for an opening on each end. Closed MRIs are more common than open MRIs, so if you have claustrophobia (a dread or fear of enclosed spaces), be sure to talk to your care team about it. MRI scans take longer to perform than CT scans. h A breast MRI might be used in addition to a mammogram. You will be positioned face down in the machine with your arms above your head. h A spine or brain MRI can be used to detect breast cancer that has spread (metastasized) to the spine or brain. Breast MRI If needed, a breast MRI will be done in addition to a mammogram. In a breast MRI, you are positioned face down with your arms overhead. 15 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Imaging tests PET scan A PET (positron emission tomography) scan uses a radioactive drug called a tracer. A tracer is a substance injected into a vein to see where cancer cells are in the body and how much sugar is being taken up by the cancer cells. This gives an idea about how fast the cancer cells are growing. Cancer cells show up as bright spots on PET scans. However, not all tumors will appear on a PET scan. Also, not all bright spots found on the PET scan are cancer. It is normal for the brain, heart, kidneys, and bladder to be bright on PET. Inflammation or infection can also show up as a bright spot. When a PET scan is combined with CT, it is called a PET/CT scan. h An FDG-PET/CT uses a radiotracer called fluorodeoxyglucose (FDG). It is made of fluoride and a simple form of sugar called glucose. You cannot eat or drink for at least 4 hours before the scan. This scan is most helpful when other imaging results are unclear. h A sodium fluoride PET/CT uses a radiotracer made of sodium fluoride. h An FES-PET/CT uses FES, which is a radioactive form of the hormone estrogen. An FES-PET/CT might be used when cancer is estrogen receptor-positive (ER+). Ultrasound An ultrasound (US) uses high-energy sound waves to form pictures of the inside of the body. This is similar to the sonogram used for pregnancy. A wand-like probe (transducer) will be held and moved on your bare breast using gel. It may also be placed below your armpit. Ultrasound is painless and does not use x-rays, so it can be repeated as needed. Ultrasound is good at showing small areas of cancer in the breast and abnormal lymph nodes. Sometimes, a breast ultrasound or MRI is used to guide a biopsy.
Testing takes time. It might take days or weeks before all test results come in.
16 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Biopsy Biopsy A biopsy is the removal of a sample of tissue from your body for testing. A pathologist will examine the biopsy for cancer and write a report called a pathology report. Ask questions about your biopsy results and what it means for your treatment.
There are different types of biopsies. Some biopsies are guided using imaging, such as ultrasound or MRI. The primary or main tumor is biopsied first. Other tumors or tumors in different areas may also be biopsied. You may have tissue removed from the breast, lymph nodes, or both.
Types of possible biopsies include: h Fine-needle aspiration (FNA) or core biopsy (CB) uses needles of different sizes to remove a sample of tissue or fluid. In a vacuum-assisted core biopsy (VACB), suction from a special vacuum device is used to remove the sample through a needle. h Incisional biopsy removes a small amount of tissue through a cut in the skin or body. h Excisional biopsy removes the entire abnormal area. This is not the preferred type of biopsy, but may be necessary if other methods are not possible or when the biopsy results don’t match the expected findings. An excisional biopsy is usually done under anesthesia by a surgeon in an operating room.
Before biopsies are performed, usually the area is injected with numbing medicine. A core needle biopsy (CNB) removes more than one tissue sample, but usually through the same area on the breast. The samples are small. The needle is often guided into the tumor with imaging. When mammography is used during a biopsy, it is called a stereotactic needle biopsy.
One or more clips may be placed near the breast tumor during a biopsy. The clips are small, painless, and made of metal. They will mark the site for possible future treatment and imaging. The clips will stay in place until surgery. If the area biopsied is benign, the clip will remain in place to mark the biopsy site on future imaging. The clips cause no problems, even if they are left in place for a long time. You will be able to go through airport security and have an MRI.
Axillary lymph node needle biopsy An axillary lymph node (ALN) drains lymph fluid from the breast and nearby areas. In an axillary lymph node biopsy, a sample of lymph node near the armpit (axilla) is biopsied with a needle. This is to determine if abnormal lymph nodes seen on imaging tests contain cancer cells. An ultrasound-guided fine-needle aspiration (US-FNA) or core biopsy will be used. If cancer is found, it is called node positive (node+). A marker may be placed in the node so that it can be identified later if needed.
Biopsy results Histology is the study of the anatomy (structure) of cells, tissues, and organs under a microscope. The tissue may be benign (not cancer), atypical (not typical), or cancerous. If the tissue is benign, likely this area can be observed. If the tissue is atypical, you may be recommended to have a surgical (open) biopsy of the area for further evaluation. If 17 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » HER2 status tissue shows cancer, your health care provider will discuss next steps. Your pathology report will contain information about histology and the grade (how abnormal the tumor looks) of the cancer.
HER2 status Human epidermal growth factor receptor 2 (HER2) is a protein involved in normal cell growth. It is found on the surface of all cells. When amounts are high, it causes cells to grow and divide. Some breast cancers have too many HER2 genes or receptors. Too many HER2s is called HER2-positive (HER2+). You might hear it called HER2 overexpression or amplification.
There are 2 tests for HER2: h Immunohistochemistry (IHC) measures receptors. If the IHC score is 3+, the cancer is HER2+. If the score is 0 or 1, it is considered HER2-negative (HER2-). If the score is 2+, further testing is needed. h In situ hybridization (ISH) counts the number of copies of the HER2 gene. This test is done mainly when the IHC score is unclear.
HER2 testing should be done on all new invasive tumors. A tumor biopsy sample will be used. You might have more than one HER2 test. Treatment options by cell receptor type There are many treatments for breast cancer. Which ones are right for you are based on many factors. Two important factors are the hormone receptor (HR) and HER2 status of any tumors.
• Hormone receptors include estrogen and progesterone. A tumor is considered hormone receptor-positive (HR+) if an increased number of estrogen receptors, progesterone receptors, or both are found.
• HER2 is a protein involved in normal cell growth. There might be higher amounts of HER2 in your breast cancer. If this is the case, it is called HER2-positive (HER2+) breast cancer.
Endocrine therapy stops cancer growth caused by hormones. It is a standard treatment for hormone receptor-positive (HR+) cancers. HR+ cancer can be estrogen receptor-positive (ER+), progesterone receptor-positive (PR+), or both (ER+/PR+).
HER2-targeted therapy is a standard treatment for HER2+ cancers.
Other systemic therapies are used with those listed above to treat invasive breast cancer. 18 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Hormone receptor status Immunohistochemistry Immunohistochemistry (IHC) is a special staining process that involves adding a chemical marker to cancer or immune cells. The cells are then studied using a microscope. IHC can find estrogen, progesterone, and HER2 receptors in breast cancer cells. A pathologist will measure how many cells have estrogen and/or progesterone receptors and the number of receptors inside each cell. FISH or ISH Fluorescence in situ hybridization (FISH) or other ISH methods like dual ISH are testing methods that involve special dyes called probes that attach to pieces of DNA, the genetic material in a person’s cells. Hormone receptor status Your blood carries hormones throughout your body. A hormone is a substance made by a gland in your body. A receptor is a protein found inside or on the surface of a cell. When substances such as hormones attach (bind) to these receptors, it causes changes within the cell. When hormones attach to receptors inside breast cancer cells, they can cause cancer to grow. If found, these receptors may be targeted using endocrine (hormone-blocking) therapy. There are 2 types of hormone receptors: h Estrogen plays a role in ovary, uterus, and breast development h Progesterone plays a role in menstrual cycle and pregnancy Hormone receptor (HR) testing should be done on any new tumors. A biopsy sample will be used.
Hormone receptor-positive In hormone receptor-positive (HR+) or hormone-sensitive breast cancer, IHC finds estrogen receptors (ER+), progesterone receptors (PR+), or both (ER+/PR+). Most breast cancers are ER+/PR+ or ER+/PR-. h Estrogen receptor (ER) is stimulated by estrogen and provides survival and proliferation (rapid growth) signals. Cancer cells deprived of estrogen or that have their ER signal blocked with treatment may stop growing or die. h Progesterone receptor (PR) binds progesterone and provides survival and proliferation signals. It is thought that PR expression also suggests the tumor is estrogen dependent. An ER-/PR+ tumor is relatively uncommon. Estrogen receptor-positive (ER+) breast cancer cells • In ER+ breast cancer, testing finds estrogen hormone receptors in at least 1 out of every 100 cancer cells.
• In ER-low–positive invasive breast cancer, testing finds estrogen hormone receptors in 1 to 10 out of every 100 cancer cells. 19 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Biomarker testing 19 2 Testing for breast cancer » Biomarker testing » Genetic cancer risk testing HR+ breast cancer is treated with endocrine therapy, which blocks estrogen receptor signaling or decreases estrogen production.
Hormone receptor-negative Hormone receptor-negative (HR-) breast cancer cells do not have estrogen or progesterone hormone receptors. These cancers are sometimes simply called hormone negative. HR- cancers often grow faster than HR+ cancers. Both the estrogen and progesterone receptors need to be negative for breast cancer to be considered HR-.
Biomarker testing Biomarker testing or tumor mutation testing is more commonly done in metastatic breast cancer. For more information, see NCCN Guidelines for Patients: Metastatic Breast Cancer at NCCN.org/patientguidelines and on the NCCN Patient Guides for Cancer app.
Genetic cancer risk testing About 1 out of 20 breast cancers are hereditary. Depending on your family history or other features of your cancer, your health care provider might refer you for hereditary genetic testing to learn more about your cancer. A genetic counselor or trained provider will speak with you about the results. Tests results may be used to guide treatment planning. Genetic testing is done using blood or saliva (spitting into a cup or using a cheek swab). The goal is to look for gene mutations inherited from your biological (birth) parents called germline mutations. Some mutations can put you at risk for more than one type of cancer. You can pass these genes on to your children. Also, other family members might carry these mutations. Tell your care team if there is a family history of cancer. More information on genetic cancer risk testing can be found in the NCCN Guidelines for Patients: Genetic Testing for Hereditary Breast, Ovarian, Pancreatic, and Prostate Cancers at NCCN.org/patientguidelines and on the NCCN Patient Guides for Cancer app.
20 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Genetic cancer risk testing BRCA tests Everyone has BRCA genes. Normal BRCA genes help to prevent tumor growth. They help fix damaged cells and help cells grow normally. BRCA mutations put you at risk for more than one type of cancer. Mutations in BRCA1 or BRCA2 increase the risk of breast, ovarian, prostate, colorectal, pancreatic, and melanoma skin cancers. Mutated BRCA genes can also affect how some treatments work. These tests might be repeated if you tested negative for a gene mutation over 10 years ago, as genetic testing technology has improved.
Other genes Other genes such as PALB2, p53, CHEK2, and ATM might be tested. For example, PALB2 normally helps prevent cancer. When PALB2 mutates, it no longer works correctly. Those with a PALB2 mutation have a higher risk of developing breast cancer.
What is your family health history?
Some cancers and other diseases run in families—those who are related to you through genes passed down from biological (birth) parent to child. This information is called a family health history. Ask blood relatives about their health issues like heart disease, cancer, and diabetes, and at what age they were diagnosed. For relatives who were diagnosed with cancer, ask them (or other relatives if they are no longer living) what type of cancer they had, if they died from the cancer, and at what age the cancer was diagnosed.
Start by asking your parents, siblings, and children. Next, talk to half-siblings, aunts and uncles, nieces and nephews, grandparents, and grandchildren.
Write down what you learn about your family history and share with your health care provider.
Some of the questions to ask include: • How old were you when each of these diseases and health conditions was diagnosed?
• What is our family’s ancestry—from what countries did our ancestors originate?
21 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 2 Testing for breast cancer » Key points 21 2 Testing for breast cancer » Key points » Questions to ask Key points h Tests are used to find cancer, plan treatment, and check how your cancer is responding to treatment. h You will have a physical exam, including a breast exam, to see if anything feels or looks abnormal. h Treatment can affect your fertility or the ability to have children. You might be referred to a fertility specialist to discuss fertility preservation. h A diagnostic mammogram includes detailed pictures of both breasts. It is different than a screening mammogram. h During a biopsy, tissue or fluid samples are removed for testing. Samples are needed to confirm the presence of cancer and to perform cancer cell tests. h A sample from a biopsy of your tumor will be tested for estrogen receptor (ER) status, progesterone receptor (PR) status, HER2 status, and histology. This provides information about the behavior of your cancer, as well as treatments to which your cancer may respond. Other biomarker tests may be performed. h About 1 out of 20 breast cancers are hereditary. Depending on your family history or other features of your cancer, your health care provider might refer you for hereditary genetic testing or to speak with a genetic counselor. Questions to ask h What type(s) of imaging scans will I have? h What type(s) of biopsy will I have? h What tests will be done on the tumor? h When will the test results be ready and who will discuss the results with me? h What is the tumor HER2 and hormone receptor (HR) status?
22 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 3 Breast cancer staging 23 How is breast cancer staged?
24 TNM scores 26 Numbered stages 27 Key points 27 Questions to ask 23 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 3 Breast cancer staging » How is breast cancer staged?
Cancer staging is used to reflect prognosis and to guide treatment decisions. It describes the size and location of the tumor and if cancer has spread to lymph nodes, organs, or other parts of the body. It also takes into account hormone receptor (HR) and HER2 status and standard-of-care treatment results.
How is breast cancer staged?
A cancer stage is a way to describe the extent of the cancer at the time you are first diagnosed. Based on testing, your cancer will be assigned a stage. Staging helps to predict prognosis and is needed to make treatment decisions. A prognosis is the course your cancer will likely take. Information gathered during staging: h The extent (size) of the tumor (T): How large is the cancer? Has it grown into nearby areas? h The spread to nearby lymph nodes (N): Has the cancer spread to nearby lymph nodes? If so, how many? Where? h The spread (metastasis) to distant sites (M): Has the cancer spread to distant organs such as the lungs or liver? h Estrogen receptor (ER) and progesterone receptor (PR) tumor status: Does the cancer have the protein called an estrogen or progesterone receptor? h Human epidermal growth factor receptor 2 (HER2) tumor status: Does the cancer make too much of a protein called HER2? h Grade of the cancer (G): How much do the cancer cells look like normal cells? h Biomarker testing: Does the cancer have any genes, proteins, markers, or mutations that might suggest treatment?
Staging is based on a combination of information to reach a final numbered stage. It takes into account what can be felt during a physical exam, what can be seen on imaging tests, and what is found during a biopsy or surgery. Often, not all information is available at the initial evaluation. More information can be gathered as treatment begins. Staging includes: h Anatomic – based on extent of cancer as defined by tumor size (T), lymph node status (N), and distant metastasis (M). h Prognostic – includes anatomic TNM plus tumor grade and the status of the biomarkers such as human epidermal growth factor receptor 2 (HER2), estrogen receptor (ER), and progesterone receptor (PR). Prognostic stage also includes the assumption that you are treated with the standard of care approaches.
Breast cancer staging is often done twice, before and after surgery. Staging after surgery 24 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 3 Breast cancer staging » TNM scores provides more specific and accurate details about the size of the cancer and lymph node status.
Clinical stage Clinical stage (c) is the rating given before any treatment. An example might look like cT1 or cN2. In breast cancer, the clinical stage is based on imaging and biopsy results. These tests are done before any treatment as part of an initial diagnosis.
Pathologic stage Pathologic stage (p) or surgical stage is determined by examining tissue removed during surgery. The pathologic stage and clinical stage may be different, if imaging was not completely accurate. If you are given drug therapy before surgery, then the stage might look like ypT2.
TNM scores The tumor, node, metastasis (TNM) system is used to stage breast cancer. In this system, the letters T, N, and M describe different areas of cancer growth. Based on cancer test results, your doctor will assign a score or number to each letter. The higher the number, the larger the tumor or the more the cancer has spread. These scores will be combined to assign the cancer a stage. A TNM example might look like this: T2N1M0 or T2, N1, M0. h T (tumor) – Depth and spread of the main (primary) tumor(s) in one or both breasts h N (node) – If cancer has spread to nearby (regional) lymph nodes h M (metastasis) – If cancer has spread to distant parts of the body or metastasized Clinical staging Clinical staging of lymph nodes is staging before surgery.
25 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 3 Breast cancer staging » TNM scores T = Tumor The primary tumor size can be measured in centimeters (cm) or millimeters (mm). One inch is equal to 2.54 cm. A large pea is 1 cm (10 mm). A golf ball is 4 cm (40 mm). A tumor micrometastasis is a very small collection of cancerous cells smaller than 1 mm. It might be written as T1mi. Ipsilateral means on the same side of the body. h T1 – Tumor is 2 cm (20 mm) or less h T2 – Tumor is 2.1 cm to 5 cm h T3 – Tumor is more than 5 cm h T4 – Tumor is of any size and has invaded nearby structures such as the chest wall and skin of the breast h T4d – Tumor is inflammatory carcinoma (inflammatory breast cancer) For more information on inflammatory breast cancer, see NCCN Guidelines for Patients: Inflammatory Breast Cancer at NCCN.org/ patientguidelines and on the NCCN Patient Guides for Cancer app N = Regional lymph node Lymph, a clear fluid containing cells that help fight infections and other diseases, drains through channels into lymphatic vessels. From here, lymph drains into lymph nodes. Lymph nodes work as filters to help fight infection.
Regional lymph nodes are those located near the breast in the armpit (axilla). If breast cancer spreads, it often goes first to nearby lymph nodes under the arm. It can also sometimes spread to lymph nodes near the collarbone or near the breastbone. However, it is possible for cancerous cells to travel through lymph and blood to other parts of the body without having gone to the lymph nodes first. Knowing if the cancer has spread to your lymph nodes helps doctors find the best way to treat your cancer. h N0 means no cancer is in the regional lymph nodes. Isolated tumor cells (ITCs) may be present. These are malignant cell clusters no larger than 0.2 mm. h N1mi means micrometastases (approximately 200 cells, larger than 0.2 mm, but not larger than 2.0 mm) are found in lymph nodes. h N1, N2, N3 means regional lymph node metastases are found. The higher the number, the more lymph nodes that have metastases.
26 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 3 Breast cancer staging » Numbered stages M = Metastasis Cancer that has spread to distant parts of the body is shown as M1. This is metastatic breast cancer (MBC). The most common sites for metastasis are bone and lung. h M0 means no evidence of distant metastasis. h M1 means distant metastasis is found. This is metastatic breast cancer.
Grade Grade describes how abnormal the tumor cells look under a microscope (called histology). Higher-grade cancers tend to grow and spread faster than lower-grade cancers. GX means the grade can’t be determined, followed by G1, G2, and G3. G3 is the highest grade for breast cancers. A low-grade tumor has a low risk of recurrence. A high-grade tumor has a higher risk of recurrence (of cancer returning). h GX – Grade cannot be determined h G1 – Low h G2 – Intermediate h G3 – High Numbered stages Numbered stages are based on TNM scores and receptor (hormone and HER2) status. Stages range from stage 0 to stage 4, with 4 being the most advanced. They might be written as stage 0, stage I, stage II, stage III, and stage IV. h Stage 0 is noninvasive – Noninvasive breast cancer is rated stage 0. Ductal carcinoma in situ (DCIS) is found only in the ducts (Tis). It has not spread to the surrounding breast tissue, lymph nodes (N0), or distant sites (M0). h Stages 1, 2, and 3 are invasive – Invasive breast cancer is rated stage 1, 2, or 3. It has grown outside the ducts, lobules, or breast skin. Cancer might be in the axillary lymph nodes. h Stage 4 is metastatic – In stage 4 breast cancer, cancer has spread to distant sites. It can develop from earlier stages. Sometimes, the first diagnosis is stage 4 metastatic breast cancer (called de novo).
27 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 3 Breast cancer staging » Key points 27 3 Breast cancer staging » Key points » Questions to ask Key points h Staging helps to predict prognosis and is needed to make treatment decisions. A prognosis is the course your cancer will likely take. h The tumor, node, metastasis (TNM) system is used to stage breast cancer. h Breast cancer is often staged twice, before and after surgery. h Clinical stage (c) is the rating given before any treatment. It is written as cTNM. h Pathologic stage (p) or surgical stage is determined by examining tissue removed during surgery. It is written as pTNM. h Grade describes how abnormal the tumor cells look under a microscope (called histology). h Regional lymph nodes are found near the breast.
Questions to ask h What is the cancer stage and tumor grade? h What does the cancer stage and grade mean in terms of treatment options and prognosis? h Is there more than one known cancer site? h Is cancer in the lymph nodes? If so, which lymph nodes? h What is the tumor HER2 and hormone status?
Anyone can develop breast cancer, including those assigned male at birth. Although there are some differences between breast cancers in those assigned male and those assigned female at birth, treatment is very similar for all genders.
28 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment 29 Care team 29 Treatment overview 30 Surgery 32 Lymph node surgery 34 Radiation therapy 34 Systemic therapy 36 Chemotherapy 36 Antibody drug conjugates 36 HER2-targeted therapy 37 Other targeted therapies 38 Immunotherapy 38 Endocrine therapy 40 Bone-strengthening therapy 42 Clinical trials 43 Key points 43 Questions to ask 29 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Care team 29 4 Types of treatment » Care team » Treatment overview There is more than one treatment for breast cancer. Treatment is usually a combination of surgery and other therapies. This chapter provides a general overview of the types of treatment and what to expect. Together, you and your care team will choose a treatment plan that is best for you.
Care team Treating breast cancer takes a team approach. Treatment decisions should involve a multidisciplinary team (MDT). An MDT is a team of health care and psychosocial care professionals from different professional backgrounds who have knowledge (expertise) and experience in your type of cancer. This team is united in the planning and implementing of your treatment. Ask who will coordinate your care.
Some members of your care team will be with you throughout cancer treatment, while others will only be there for parts of it. Get to know your care team and help them get to know you.
Treatment overview Invasive breast cancer is treatable. Treatment can be local, systemic, or usually a combination of both.
Local therapy focuses on the breast, chest wall, and lymph node area. It includes: h Surgery h Radiation therapy (RT) Systemic therapy works throughout the body. It includes: h Chemotherapy h HER2-targeted therapy h Inhibitors and other targeted therapies h Immunotherapy h Endocrine therapy Treatment options are often described in the following ways: h Preferred therapies have the most evidence they work better and may be safer than other therapies. h Other recommended therapies may not work quite as well as preferred therapies, but they can still help treat cancer. h Therapies used in certain cases work best for people with specific cancer features or health circumstances.
Many factors play a role in how cancer responds to treatment. It is important to have regular talks with your care team about your goals for treatment and your treatment plan.
30 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Surgery Surgery Surgery is an operation or procedure to remove cancer from the body. It is the main or primary treatment for invasive breast cancer. This is only one part of a treatment plan. When preparing for surgery, seek the opinion of an experienced surgeon who is an expert in performing your type of surgery. Hospitals that perform many surgeries often have better results. You can ask for a referral to a hospital or cancer center that has experience in treating your type of cancer. Goal of surgery The goal of surgery or tumor resection is to remove all the cancer. To do so, the tumor is removed along with a rim of normal-looking tissue around its edge called the surgical margin. The surgical margin may look normal during surgery, but cancerous cells may be found when viewed under a microscope by a pathologist. A clear or negative margin (R0) is when no cancer cells are found in the tissue around the edge of the tumor. In a positive margin, cancer cells are found in normal-looking tissue around the tumor. After surgery, you may receive treatment such as radiation to kill any remaining cancer cells. You might have a wound drain to prevent fluid from collecting in the body after surgery. These drains are usually removed a few days after surgery.
Lumpectomy Lumpectomy is the removal of abnormal cells or a tumor. It does not remove the whole Breast-conserving surgery The dotted line shows where the tumor is removed. Lumpectomy is the removal of abnormal cells or tumor and not the whole breast. It is also called a partial mastectomy, breast-conserving therapy, or breast-conserving surgery (BCS). 31 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Surgery breast. A lumpectomy is also called a partial mastectomy, breast-conserving therapy, or breast-conserving surgery (BCS). In a lumpectomy, the surgeon aims to remove the tumor with a rim of healthy tissue around it, called a negative, or clear, surgical margin. Having a negative surgical margin will decrease the chance that cancer may return in that area of the breast. You may need more than one surgery to achieve negative margins and ensure all the cancer is removed.
The breast might not look the same after a lumpectomy. Speak to your surgeon about how a lumpectomy might affect the look and shape of your breast, and any concerns you have. You can also seek the opinion of a plastic surgeon.
Mastectomy A mastectomy removes all or part of the breast tissue. Before removing the breast, the surgeon may do a sentinel lymph node biopsy (SLNB). Sentinel lymph nodes are the first lymph nodes cancer cells are likely to have spread from the primary tumor.
Types of mastectomies include: h A total mastectomy or simple mastectomy removes the whole breast with a flat skin closure. h A modified radical mastectomy removes the breast and axillary lymph nodes. h A skin-sparing mastectomy removes the breast but not all of the skin, in order Total (simple) mastectomy The dotted line shows where the entire breast is removed. Some lymph nodes under the arm may also be removed.
32 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Lymph node surgery to have breast reconstruction that might include flaps and/or implants. h Nipple-sparing mastectomy preserves the nipple-areola complex (NAC) as well as all of the skin. Not everyone is a candidate for nipple-sparing mastectomy. This is based upon location of cancer, breast size, and breast ptosis (degree of drooping).
Breast reconstruction is an option after a mastectomy. If you opt for reconstruction, it can be done during the initial surgery to remove your breast, or after finishing cancer treatment. If it is done after cancer treatment, this is called delayed reconstruction. Breast reconstruction is often done in multiple stages, requiring multiple procedures. You will need to consult with a plastic surgeon.
Lymph node surgery Sentinel lymph node biopsy Sentinel lymph node biopsy (SLNB or SNB) may be done during a lumpectomy or mastectomy to determine if any cancer cells have traveled to the lymph nodes. The lymph nodes removed are called sentinel nodes. A sentinel lymph node (SLN) is the first lymph node(s) that cancer cells are most likely to spread to from a primary tumor. Often, there is more than one sentinel lymph node. Just because these nodes are removed, it does not mean that they test positive for cancer.
To find the sentinel lymph nodes, a tracer is injected into the breast. It may be a radioactive material, blue dye, or other tracer. The tracer travels through the lymph channels in the breast to the lymph nodes in the armpit. Lymph node surgery 33 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Lymph node surgery This helps the surgeon find which of the lymph nodes are the sentinel lymph nodes. The lymph nodes containing the tracer are removed and tested by a pathologist.
Axillary lymph node dissection Axillary lymph node dissection (ALND) is surgery to remove axillary lymph nodes. This is performed after an axillary lymph node biopsy or SLNB shows cancer in the lymph nodes (called node positive). Then, an ALND will remove any other lymph nodes that may contain cancer. Removing lymph nodes can cause lymphedema (fluid buildup) and other health issues. There are 3 levels of axillary lymph nodes: h Level I – nodes located below the lower edge of the chest muscle h Level II – nodes located underneath the chest muscle h Level III – nodes located above the chest muscle near the collarbone An ALND usually removes level I and II axillary lymph nodes. Level III nodes are not typically removed unless there is a lot of disease. For more information about the timing of biopsies, talk with your care team. Modified radical mastectomy The dotted line shows where the entire breast and some lymph nodes are removed.
34 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Radiation therapy 34 4 Types of treatment » Radiation therapy » Systemic therapy Radiation therapy Radiation therapy (RT) uses high-energy radiation from x-rays (photons), protons, and other sources to kill cancer cells and shrink tumors. Radiation therapy can be given after surgery to treat or slow cancer growth. Treatment is given in small daily doses on weekdays, with weekends off. At least weekly, you will meet with your radiation oncologist to review treatment results and to help with side effects, such as sunburn-like rash. Ask your care team which radiation option(s) are best for your situation, if RT will be combined with chemotherapy, and what side effects to expect. The following are types of radiation therapy used to treat breast cancer: h Whole breast radiation therapy (WBRT) is used to treat the entire breast. h Partial breast irradiation (PBI) is used to treat only the tumor area of the breast. Accelerated partial breast irradiation (APBI) is radiation given over a shorter period of time. h Postmastectomy radiation therapy (PMRT) is given after a mastectomy. It targets the chest wall and possible mastectomy scar. h Lymph node radiation therapy is used to treat the lymph nodes. It is also called regional nodal irradiation (RNI).
Radiation may be given to the breast and chest wall, infraclavicular region (below the collarbone), supraclavicular area (above the collarbone), or lymph nodes found inside the breast (intramammary), behind the ribcage (internal mammary), and axillary bed (armpit).
External beam radiation therapy External beam radiation therapy (EBRT) uses a machine outside of the body called a linear accelerator (linac) to aim radiation at the whole breast (called whole breast radiation therapy or WBRT) and lymph nodes (called regional nodal irradiation or RNI). Internal radiation Internal radiation (brachytherapy) involves placing one or more small tubes into the tumor area of the breast. A small radioactive seed travels into the tube(s) and delivers radiation to the tumor area of the breast from inside the body. This type of radiation is used only for PBI. Systemic therapy Systemic therapy is drug therapy that works throughout the body. Types include chemotherapy, targeted therapy, immunotherapy, and endocrine therapy. For a general list of systemic therapies, see Guide 2. Systemic therapy might be used alone or with other therapies. Goals of systemic therapy should be discussed before starting treatment. The choice of therapy takes into consideration many factors, including HER2 and hormone receptor tumor status, if the tumor can be removed with surgery (resectable), your preferences about treatment, and if you have any other serious health conditions. 35 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Systemic therapy Guide 2 Systemic therapy examples Chemotherapy examples • Capecitabine (Xeloda) • Carboplatin • Cisplatin • Cyclophosphamide (Frindovyx) • Docetaxel (Taxotere, Docivyx) • Doxorubicin (Adriamycin) • Epirubicin (Ellence) • Eribulin (Halaven) • Fluorouracil • Gemcitabine • Methotrexate • Paclitaxel • Vinorelbine Targeted therapy examples Antibody drug conjugates • Ado-trastuzumab emtansine (T-DM1, Kadcyla), fam-trastuzumab deruxtecan-nxki (T-DXd, Enhertu), and sacituzumab govitecan-hziy (Trodelvy) CDK4/6 inhibitors • Abemaciclib (Verzenio), palbociclib (Ibrance), and ribociclib (Kisqali) PARP inhibitors • Olaparib (Lynparza) and talazoparib (Talzenna) PIK3CA, AKT1, PTEN, and mTOR inhibitors • Alpelisib (Piqray), capivasertib (Truqap), and everolimus (Afinitor) HER2-targeting therapy (antibody, inhibitor, and conjugate) examples • Pertuzumab (Perjeta) • Trastuzumab (Herceptin) or trastuzumab substitutes (biosimilars) such as Kanjinti, Ogivri, Herzuma, Ontruzant, and Trazimera • Ado-trastuzumab emtansine • Fam-trastuzumab deruxtecan-nxki • Lapatinib (Tykerb) • Margetuximab-cmkb (Margenza) • Neratinib (Nerlynx) • Tucatinib (Tukysa) • Phesgo as a substitute for combination therapy of trastuzumab with pertuzumab Immunotherapy • Pembrolizumab (Keytruda) Endocrine therapy • Endocrine therapy can be found in Guide 3.
36 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Chemotherapy 36 4 Types of treatment » Chemotherapy » Antibody drug conjugates » HER2-targeted therapy Systemic therapy might be given before or after surgery. Systemic therapies are described in more detail below and in the following pages.
Chemotherapy Chemotherapy kills fast-dividing cells throughout the body, including cancer cells and some normal cells. More than one chemotherapy may be used to treat invasive breast cancer. When only one drug is used, it’s called a single agent. A combination or multi-agent regimen is the use of two or more chemotherapy drugs. Some chemotherapy drugs are liquids that are infused into a vein (IV) or injected under the skin with a needle. Other chemotherapy drugs may be given as a pill that is swallowed. Some examples of chemotherapy drugs include the following: h Anthracyclines include doxorubicin (Adriamycin), doxorubicin liposomal injection (Doxil), and epirubicin (Ellence) h Taxanes include docetaxel, paclitaxel, and albumin-bound paclitaxel. h Antimetabolites include capecitabine (Xeloda), fluorouracil, gemcitabine, and methotrexate.
Most chemotherapy is given in cycles of treatment days followed by days of rest. This allows the body to recover before the next cycle. Cycles vary in length depending on which drugs are used. The number of treatment days per cycle and the total number of cycles given also vary.
Antibody drug conjugates An antibody drug conjugate (ADC) delivers cell-specific chemotherapy. It attaches to a protein found on the outside of the cancer cell, then enters the cell. Once inside the cell, chemotherapy is released. ADCs are given in cycles. Examples include ado-trastuzumab emtansine (Kadcyla), fam-trastuzumab deruxtecan-nxki (Enhertu, T-DXd), and sacituzumab govitecan-hziy (Trodelvy). HER2-targeted therapy HER2 is a protein involved in normal cell growth. There might be higher amounts of HER2 in your breast cancer. If this is the case, it is called HER2-positive (HER2+) breast cancer. HER2-targeted therapy is drug therapy that treats HER2+ breast cancer. Some HER2-targeted therapy is given with chemotherapy. However, it might be used alone or in combination with endocrine therapy. HER2-targeted therapies include: h HER2 antibodies prevent HER2 growth signals from outside the cell. They also increase the attack of immune cells on cancer cells. h HER2 inhibitors stop HER2 growth signals from within the cell. h HER2 conjugates or HER2 antibody drug conjugates (ADCs) deliver cell-specific chemotherapy. They attach directly to HER2s then enter the cell. Once inside, chemotherapy is released.
HER2 therapies may affect your heart’s pumping function. Your heart will be monitored 37 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Other targeted therapies before and during treatment with HER2-targeted therapy. Tests will measure the left ventricular ejection fraction (LVEF), the amount of blood pumping from the left side of the heart.
Other targeted therapies This section is for inhibitors that are different from inhibitors used in HER2-targeted therapy. CDK4/6 inhibitors Cyclin-dependent kinase (CDK) is a cell protein that helps cells grow and divide. For hormone receptor-positive (HR+), HER2-negative (HER2-) cancer, taking a CDK4/6 inhibitor with endocrine therapy may help control cancer longer and improve survival. With all CDK4/6 regimens, those who are premenopausal must also receive ovarian ablation or suppression. CDK4/6 inhibitors include abemaciclib (Verzenio), palbociclib (Ibrance), and ribociclib (Kisqali).
PARP inhibitors Cancer cells often become damaged. PARP is a cell protein that repairs cancer cells and allows them to survive. Blocking PARP can cause cancer cells to die. Olaparib (Lynparza) and talazoparib (Talzenna) are examples of a PARP inhibitor (PARPi).
PIK3CA, PTEN, and AKT1 inhibitors The PIK3CA gene is one of the most frequently mutated genes in breast cancers. PTEN and AKT are also part of this important pathway in cancer cells and can be altered less commonly Warnings about supplements and drug interactions You might be asked to stop taking or avoid certain herbal supplements when on a systemic therapy. Some supplements can affect the ability of a drug to do its job. This is called a drug interaction.
It is critical to speak with your care team about any supplements you may be taking. Some examples include: h Turmeric h Ginkgo biloba h Green tea extract h St. John’s Wort h Antioxidants Certain medicines can also affect the ability of a drug to do its job. Antacids, heart or blood pressure medicine, and antidepressants are just some of the medicines that might interact with a systemic therapy or supportive care medicines given during systemic therapy. Therefore, it is very important to tell your care team about any medicines, vitamins, over-the-counter (OTC) drugs, herbals, or supplements you are taking.
Bring a list with you to every visit.
38 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Immunotherapy 38 4 Types of treatment » Immunotherapy » Endocrine therapy in breast cancers. A mutation or alteration in these genes can lead to increased growth of cancer cells and resistance to various treatments. Alpelisib (Piqray) and inavolisib (Itovebi) are examples of a PIK3CA inhibitors and capivasertib (Truqap) is an AKT1 inhibitor.
mTOR inhibitors mTOR is a cell protein that helps cells grow and divide. Endocrine therapy may stop working if mTOR becomes overactive. mTOR inhibitors are used to get endocrine therapy working again.
Everolimus (Afinitor) is an mTOR inhibitor. Most often, it is taken with exemestane. For some, it may be taken with fulvestrant or tamoxifen.
Immunotherapy Immunotherapy is a type of systemic treatment that tries to reactivate the immune system against tumor cells. The immune system has many on and off switches. Tumors take advantage of off switches. Immunotherapy can block off switches, which helps the immune system turn on. Immunotherapy can be given alone or with other types of treatment. Pembrolizumab (Keytruda) is an example of immunotherapy.
Endocrine therapy Endocrine therapy blocks estrogen or progesterone to treat hormone receptor-positive (HR+) breast cancer. The endocrine system is made up of organs and tissues that produce hormones. Hormones are natural chemicals released into the bloodstream. There are 4 hormones that might be targeted in endocrine therapy: h Estrogen is made mainly by the ovaries, but is also made by other tissues in the body such as fat tissue. h Progesterone is made mainly by the ovaries. h Luteinizing hormone-releasing hormone (LHRH) is made by a part of the brain called the hypothalamus. It tells the ovaries to make estrogen and progesterone and testicles to make testosterone. LHRH is also called gonadotropin-releasing hormone (GnRH). h Androgen is made by the adrenal glands, testicles, and ovaries.
Hormones may cause breast cancer to grow. Endocrine therapy will stop your body from making hormones or it will block what hormones do in the body. This can slow tumor growth or shrink the tumor for a period of time. Endocrine therapy is sometimes called hormone therapy or anti-estrogen. It is not the same as hormone replacement therapy (HRT) used for menopause. Those who want to have children in the future should discuss fertility plans with their doctor 39 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Endocrine therapy and may need to be referred to a fertility specialist before starting endocrine therapy.
Types of endocrine therapy can be found in Guide 3.
Testosterone For those assigned male at birth whose bodies continue to make testosterone, endocrine therapy includes tamoxifen or an aromatase inhibitor (AI) with a testosterone-suppressing therapy.
Guide 3 Endocrine therapy types Bilateral oophorectomy Surgery to remove both ovaries.
Ovarian ablation Radiation to permanently stop the ovaries from making hormones.
Ovarian or testosterone suppression Drugs to temporarily stop the ovaries or testicles from making hormones such as LHRH and GnRH. • LHRH agonists include goserelin (Zoladex) and leuprolide (Lupron Depot). These are injected every 4 or 12 weeks. They do not affect estrogen made by the ovaries.
• GnRH agonists might be used to suppress ovarian hormone or testosterone production.
Aromatase inhibitors (AIs) Drugs to stop a type of hormone called androgen from changing into estrogen by interfering with an enzyme called aromatase. They do not affect estrogen made by the ovaries. Nonsteroidal AIs include anastrozole (Arimidex) and letrozole (Femara). Exemestane (Aromasin) is a steroidal AI.
Estrogen receptor (ER) modulators or anti-estrogens • Selective estrogen receptor modulators (SERMs) block estrogen from attaching to hormone receptors. Tamoxifen and toremifene (Fareston) are SERMs. • Selective estrogen receptor degraders (SERDs) block and destroy estrogen receptors. Fulvestrant (Faslodex) and elacestrant (Orserdu) are SERDs.
Hormones Hormone examples include ethinyl estradiol, fluoxymesterone, and megestrol acetate (Megace).
40 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Bone-strengthening therapy Premenopause If you have menstrual periods, you are in premenopause. In premenopause, the ovaries are the main source of estrogen and progesterone. GnRH agonists may be used to temporarily induce menopause for those in premenopause. A combination of GnRH agonists and tamoxifen or aromatase inhibitors (AIs) may be considered as endocrine therapy for those in premenopause. Ovarian suppression or ablation is frequently considered for higher risk ER+ breast cancers.
Menopause In menopause, the ovaries permanently stop producing hormones and menstrual periods stop. Estrogen and progesterone levels are low, but the adrenal glands, liver, and body fat continue to make small amounts of estrogen. If you don’t have periods, a test using a blood sample may be used to confirm your status. Cancer treatment can cause temporary menopause. If you stopped having periods due to removal of your uterus (hysterectomy) but you still have your ovaries, then you should have your menopausal status confirmed with a blood test. If both ovaries have been removed (with or without your uterus), you are in menopause.
Bone-strengthening therapy Medicines that target the bones may be given to help relieve bone pain or reduce the risk of bone-related problems. Some medicines work by slowing or stopping bone breakdown, while others help increase bone thickness.
When breast cancer spreads to distant sites, it may metastasize in your bones. This puts your bones at risk for injury and disease. Such problems include fractures, bone pain, high calcium levels in the blood, and squeezing (compression) of the spinal cord. Some treatments for breast cancer, like aromatase inhibitors or GnRH agonists, can cause bone loss (osteoporosis), which puts you at an increased risk for fractures.
Drugs used to prevent bone loss and fractures: h Oral bisphosphonates h Zoledronic acid (Zometa) h Pamidronate (Aredia) h Denosumab (Prolia) You may be screened for bone weakness (osteoporosis) using a bone mineral density test. This measures how much calcium and other minerals are in your bones. It is also called a dual-energy x-ray absorptiometry (DEXA) scan and is painless. Bone mineral density tests look for osteoporosis and help predict your risk for bone fractures.
41 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Bone-strengthening therapy Zoledronic acid, pamidronate, and denosumab Zoledronic acid, pamidronate, and denosumab are used to prevent bone loss (osteoporosis) and fractures caused by endocrine therapy. You might have blood tests to monitor kidney function, calcium levels, and magnesium levels. A calcium and vitamin D supplement will likely be recommended by your doctor.
Let your dentist know if you are taking any of these medicines. Also, ask your care team how these medicines might affect your teeth and jaw. Osteonecrosis, or bone tissue death of the jaw, is a rare but serious side effect. Tell your care team about any planned trips to the dentist and surgeries or dental procedures that might also affect the jawbone. It will be important to take care of your teeth and to see a dentist before starting treatment with any of these drugs.
Standard of care is the best-known way to treat a particular disease based on past clinical trials. There may be more than one treatment regimen that is considered standard of care. Ask your care team what treatment options are available and if a clinical trial might be right for you.
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NCCN.org/patients/feedback 42 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Clinical trials Clinical trials A clinical trial is a type of medical research study. After being developed and tested in a lab, potential new ways of treating cancer need to be studied in people. If found to be safe and effective in a clinical trial, a drug, device, or treatment approach may be approved by the U.S. Food and Drug Administration (FDA). Everyone with cancer should carefully consider all of the treatment options available for their cancer type, including standard treatments and clinical trials. Talk to your doctor about whether a clinical trial may make sense for you.
Phases Most cancer clinical trials focus on treatment and are done in phases. h Phase 1 trials study the safety and side effects of an investigational drug or treatment approach. h Phase 2 trials study how well the drug or approach works against a specific type of cancer. h Phase 3 trials test the drug or approach against a standard treatment. If the results are good, it may be approved by the FDA. h Phase 4 trials study the safety and benefit of an FDA-approved treatment.
Who can enroll?
It depends on the clinical trial’s rules, called eligibility criteria. The rules may be about age, cancer type and stage, treatment history, or general health. They ensure that participants are alike in specific ways and that the trial is as safe as possible for the participants.
Informed consent Clinical trials are managed by a research team. This group of experts will review the study with you in detail, including its purpose and the risks and benefits of joining. All of this information is also provided in an informed consent form. Read the form carefully and ask questions before signing it. Take time to discuss it with people you trust. Keep in mind that you can leave and seek treatment outside of the clinical trial at any time.
Will I get a placebo?
Placebos (inactive versions of real medicines) are almost never used alone in cancer clinical trials. It is common to receive either a placebo with a standard treatment, or a new drug with a standard treatment. You will be informed, verbally and in writing, if a placebo is part of a clinical trial before you enroll.
Are clinical trials free?
There is no fee to enroll in a clinical trial. The study sponsor pays for research-related costs, including the study drug. But you may need to pay for other services, like transportation or childcare, due to extra appointments. During the trial, you will continue to receive standard cancer care. This care is often covered by insurance. 43 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 4 Types of treatment » Key points 43 4 Types of treatment » Key points » Questions to ask Key points h Surgery is the main or primary treatment for invasive breast cancer. h Systemic treatment is based on estrogen receptor (ER), progesterone receptor (PR), and HER2 expression. h Radiation therapy (RT) uses high-energy radiation from x-rays (photons, electrons), protons, and other sources to kill cancer cells. h Some breast cancers grow because of estrogen. These cancers are estrogen receptor-positive (ER+) and are often treated with endocrine therapy to reduce the risk of cancer recurrence. h A clinical trial is a type of research that studies a treatment to see how safe it is and how well it works.
Questions to ask h What is your experience treating breast cancer? h How many breast cancer surgeries have you done? h What treatment will I have before and after surgery? h Is there a social worker or someone who can help me decide about treatment? h How much time do I have to decide about treatment? Are there any decisions that must be made today?
Finding a clinical trial In the United States NCCN Cancer Centers NCCN.org/cancercenters The National Cancer Institute (NCI) cancer.gov/about-cancer/treatment/clinical-trials/search Worldwide The U.S. National Library of Medicine (NLM) clinicaltrials.gov/ Need help finding a clinical trial? NCI’s Cancer Information Service (CIS) 1.800.4.CANCER (1.800.422.6237) cancer.gov/contact 44 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 5 Supportive care 45 What is supportive care?
45 Side effects 49 Late effects 49 Survivorship 50 Key points 50 Questions to ask 45 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 5 Supportive care » What is supportive care?
45 5 Supportive care » What is supportive care? » Side effects Supportive care helps manage the symptoms of invasive breast cancer and the side effects of treatment. This chapter discusses possible side effects.
What is supportive care?
Supportive care helps improve your quality of life during and after cancer treatment. The goal is to prevent or manage side effects and symptoms, like pain and cancer-related fatigue. It also addresses the mental, social, and spiritual concerns faced by those with cancer.
Supportive care is available to everyone with cancer and their families, not just those at the end of life. Palliative care is another name for supportive care. Supportive care can also help with: h Making treatment decisions h Coordinating your care h Paying for care h Planning for advanced care and end of life Side effects All cancer treatments can cause unwanted health issues called side effects. Side effects depend on many factors. These factors include the drug type and dose, length of treatment, and the person. Some side effects may just be unpleasant. Others may be harmful to one’s health. Treatment can cause several side effects. Some are very serious. Tell your care team about any new or worsening symptoms.
Blood clots Cancer or cancer treatment can cause blood clots to form. This can block blood flow and oxygen in the body. Blood clots can break loose and travel to other parts of the body causing breathing problems, stroke, or other health issues. Venous thromboembolism (VTE) refers to blood clots in the veins.
Bone health Breast cancer may spread to your bones. Some breast cancer treatments may also weaken your bones. Both can put your bones at increased risk for injury and disease. Such problems include bone fractures, bone pain, and squeezing (compression) of the spinal cord. High levels of calcium in the blood, called hypercalcemia, may also occur.
Medicine may be given to help relieve bone pain and reduce the risk of other bone problems. Some medicines work by slowing or stopping bone breakdown, while others help increase bone thickness. It is recommended that you take calcium and vitamin D with these bone health medicines. Talk to your care team first.
46 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 5 Supportive care » Side effects Diarrhea or constipation Diarrhea is frequent and watery bowel movements. Your care team will tell you how to manage diarrhea. It is important to drink lots of fluids. Constipation is also common, especially if taking certain pain medicines. Drinking fluids, staying active, and taking medicines for constipation are often recommended.
Distress Depression, anxiety, and sleeping issues are common and are a normal part of cancer diagnosis. Talk to your care team and with those whom you feel most comfortable about how you are feeling. There are services, people, and medicine that can help you. Fatigue Fatigue is extreme tiredness and inability to function due to lack of energy. Fatigue may be caused by cancer or it may be a side effect of treatment. Let your care team know how you are feeling and if fatigue is getting in the way of doing the things you enjoy. Eating a balanced diet, exercise, yoga, acupuncture, and massage therapy can help. You might be referred to a nutritionist or dietitian to help with fatigue.
Hair loss Chemotherapy may cause hair loss (alopecia) all over your body — not just on your scalp. Some chemotherapy drugs are more likely than others to cause hair loss. Dosage might also affect the amount of hair loss. Most of the time, hair loss from chemotherapy is temporary. Hair often regrows 3 to 6 months after treatment ends. Your hair may be a different shade or texture at first. Scalp cooling (or scalp hypothermia) might help lessen hair loss in those receiving certain types of chemotherapy. Infections Infections occur more frequently and are more severe in those with a weakened immune system. Drug treatment for breast cancer can weaken the body’s natural defense against infections. If not treated early, infections can be fatal.
Neutropenia, a low number of white blood cells, can lead to frequent or severe infections. When someone with neutropenia also develops a fever, it is called febrile neutropenia (FN). With FN, your risk of infection may be higher than normal. This is because a low number of white blood cells leads to a reduced ability to fight infections. FN is a side effect of some types of systemic therapy.
Loss of appetite Sometimes side effects from surgery, cancer, or its treatment might cause you to feel not hungry or sick to your stomach (nauseated). You might have a sore mouth. Healthy eating is important during treatment. It includes eating a balanced diet, eating the right amount of food, and drinking enough fluids. A registered dietitian who is an expert in nutrition and food It is important to tell your care team about all of your side effects, including new or worsening symptoms.
47 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 5 Supportive care » Side effects can help. Speak to your care team if you have trouble eating or maintaining weight.
Low blood cell counts Some cancer treatments can cause low blood cell counts. h Anemia is a condition where your body does not have enough healthy red blood cells, resulting in less oxygen being carried to your body tissues. You might tire easily or feel short of breath if you are anemic. h Neutropenia is a decrease in neutrophils, the most common type of white blood cell. This puts you at risk for infection. h Thrombocytopenia is a condition where there are not enough platelets found in the blood. This puts you at risk for bleeding.
Lymphedema Lymphedema is a condition in which lymph fluid builds up in tissues and causes swelling. It may be caused when part of the lymph system is damaged or blocked, such as during surgery to remove lymph nodes, or by radiation therapy. Cancers that block lymph vessels can also cause lymphedema. Swelling usually develops slowly over time. It may develop during treatment, or it may start years after treatment. If you have lymphedema, you may be referred to an expert in lymphedema management. The swelling may be reduced by exercise, massage, compression devices, and other means.
Nausea and vomiting Nausea and vomiting are common side effects of treatment. You will be given medicine to prevent nausea and vomiting.
“Recovery is the goal. Life will be forever altered because of your cancer experience. There will come a day that your cancer isn’t the first thing that you think of in the morning and the day you realize that happened will be powerful.” 48 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 5 Supportive care » Side effects Neurocognitive or neuropsychological effects Some treatments can damage the nervous system (neurotoxicity) causing problems with concentration and memory. Survivors are at risk for neurotoxicity and might be recommended for neuropsychological testing. Neuropsychology looks at how the health of your brain affects your thinking and behavior. Neuropsychological testing can identify your limits and doctors can create a plan to help with these limits.
Neuropathy and neurotoxicity Some treatments can damage the nervous system (neurotoxicity) causing neuropathy and problems with concentration, memory, and thinking. Neuropathy is a nerve problem that causes pain, numbness, tingling, swelling, or muscle weakness in different parts of the body. It usually begins in the hands or feet and gets worse with additional cycles of treatment. Most of the time, neuropathy improves gradually and may eventually go away after treatment. Organ issues Treatment might cause your kidneys, liver, heart, and pancreas to not work as well as they should. Pain Tell your care team about any pain or discomfort. You might meet with a palliative care specialist or with a pain specialist to manage pain.
Palliative care Palliative care is appropriate for anyone, regardless of age, cancer stage, or the need for other therapies. It focuses on physical, emotional, social, and spiritual needs that affect quality of life. Supportive care resources More information on supportive care is available at NCCN.org/patientguidelines and on the NCCN Patient Guides for Cancer app.
49 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 5 Supportive care » Late effects 49 5 Supportive care » Late effects » Survivorship Quality of life Cancer and its treatment can affect your overall well-being or quality of life (QOL). For more information on quality of life, see NCCN Guidelines for Patients: Palliative Care at NCCN.org/patientguidelines and on the NCCN Patient Guides for Cancer app.
Late effects Late effects are side effects that occur months or years after a disease is diagnosed or after treatment has ended. Late effects may be caused by cancer or cancer treatment. They may include physical, mental, and social health concerns, and second cancers. The sooner late effects are treated the better. Ask the care team about what late effects could occur. This will help you know what to look for.
Survivorship A person is a cancer survivor from the time of diagnosis until the end of life. After treatment, your health will be monitored for side effects of treatment and the return of cancer. This is part of your survivorship care plan. It is important to keep any follow-up care and imaging test appointments. Seek good routine medical care, including regular doctor visits for preventive care and cancer screening. A personalized survivorship care plan will contain a summary of possible long-term effects of treatment called late effects and a list of follow-up tests. Find out how your primary care provider will coordinate with specialists for your follow-up care. Tell your care team about any symptoms such as headaches, menstrual spotting between periods or new onset of spotting after menopause (if prior tamoxifen), shortness of breath that you notice with walking, or bone pain. Side effects can be managed. Continue to take all medicine such as endocrine therapy exactly as prescribed and do not miss or skip doses.
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NCCN.org/patients/response 50 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 5 Supportive care » Key points 50 5 Supportive care » Key points » Questions to ask Key points h Supportive care is health care that relieves symptoms caused by treatment and improves quality of life. Supportive care is always given. h All cancer treatments can cause unwanted health issues called side effects. Side effects depend on many factors. These factors include the drug type and dose, length of treatment, and the person. h Some side effects are very rare. Ask your care team what to expect. h Tell your care team about any new or worsening symptoms.
Questions to ask h Who will coordinate my care? h Who should I call when I have questions or notice changes in my condition? h How long should I wait if I notice changes in my condition? h What should I do on weekends and other non-office hours? h Will my care team be able to communicate with the emergency department or urgent care team?
51 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options 52 Surgery first 53 Systemic therapy before surgery 56 Systemic therapy after surgery 58 HR+ with HER2+ 58 HR- with HER2+ 58 HR+ with HER2-58 Triple-negative breast cancer 59 Endocrine therapy 60 Follow-up care 61 Key points 61 Questions to ask 52 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Surgery first Surgery is the main or primary treatment for invasive breast cancer. Some people benefit from systemic therapy before surgery. Radiation therapy (RT), systemic therapy, or both are possible after surgery. Together, you and your care team will choose a treatment plan that is best for your situation. This chapter provides specific treatment options based on hormone receptor (HR) and HER2 tumor status. It also discusses who might benefit from treatment before surgery.
Surgery first Surgery is the main or primary treatment for invasive breast cancer. It involves removing the tumor from the breast and assessing the lymph nodes. There are 2 breast surgery options: h Lumpectomy h Mastectomy Surgery might include axillary lymph node (ALN) staging with either sentinel lymph node biopsy (SLNB), axillary lymph node dissection (ALND), or both. After surgery, a pathologist will examine the removed tissue and any lymph nodes to determine the pathologic stage. This information will help plan next steps for treatment. Treatment after surgery might include systemic therapy, radiation therapy, or both. An example of a tumor stage after surgery might be pT2. Lymph node micrometastases are written as pN1mi. Ipsilateral means on the same side of the body. Lumpectomy first A lumpectomy, also called breast-conserving surgery (BCS), is surgery to remove a tumor in the breast. Treatment after a lumpectomy is based on the type of cancer (histology), the size of the tumor, if cancer is found in the axillary lymph nodes, your age, and if you are in menopause. Many factors play a role in treatment decisions. Most people have radiation therapy (RT) to the breast and sometimes lymph nodes after a lumpectomy. A boost is extra radiation to the tumor area. Systemic (drug) therapy might be given before RT. Mastectomy first A total mastectomy is a surgery that removes the whole breast. A nipple-sparing or skin-sparing mastectomy may be possible. Treatment after a mastectomy is based on if cancer was found in the axillary lymph nodes, the number of lymph nodes that tested positive, and the size of the removed tumor. Radiation therapy to the chest wall and lymph nodes, systemic therapy, or both are possible following a mastectomy.
53 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Systemic therapy before surgery Systemic therapy before surgery Some cancers benefit from treatment before surgery. These cancers include those with large tumors, aggressive tumors, cancer in the lymph nodes called nodal or node-positive (N+) disease, or the cancer that involves the breast skin or chest wall. Treatment before surgery is called preoperative or neoadjuvant therapy.
Testing Not everyone will benefit from preoperative therapy. If preoperative systemic therapy is an option for you, then you will have blood and imaging tests before starting treatment. Testing will include an axillary lymph node exam. An ultrasound and biopsy of lymph nodes suspected of cancer are possible.
You will have the following before starting preoperative systemic therapy: h Core biopsy of breast h Placement of clips or markers to help the surgeon know where to operate in case the tumor goes away with preoperative therapy. Clips are also placed at the time of surgery for radiation planning. h Axillary lymph node ultrasound or magnetic resonance imaging (MRI) (if not done before) h Biopsy of suspicious lymph nodes with clip placement (if not done before) Order of treatments Most people with cancer will receive more than one type of treatment. Below is an overview words you might hear to describe treatment.
• Preoperative or neoadjuvant (before) therapy is given to shrink the tumor before a primary treatment such as surgery. • Perioperative therapy is systemic therapy, such as chemotherapy, given before and after surgery. • Primary treatment is the main treatment given to rid the body of cancer. • Postoperative or adjuvant (after) therapy is given after primary treatment to rid the body of any cancer cells left behind from surgery. It is also used when the risk of cancer returning (recurrence) is felt to be high. • First-line therapy is the first set of systemic (drug) treatment given.
• Second-line therapy is the next set of treatment given if cancer progresses during or after systemic therapy. Talk to your care team about your treatment plan and what it means for your stage and type of cancer.
54 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Systemic therapy before surgery Treatment before surgery Treatment before surgery is systemic (drug) therapy. It is based on hormone receptor (HR) and HER2 status. Preoperative systemic therapy has benefits. The goal is to shrink the tumor or the amount of cancer before surgery. How the tumor responds to drug therapy may also affect treatments after surgery. Guide 4 HER2-targeted therapy options: HER2+ cancer Preferred • Paclitaxel and trastuzumab • Docetaxel, carboplatin, and trastuzumab (TCH) • Docetaxel, carboplatin, trastuzumab, and pertuzumab (TCHP) If no residual disease after preoperative therapy or no preoperative therapy: • Complete up to 1 year of HER2-targeted therapy with trastuzumab. Pertuzumab might be added.
If residual disease after preoperative therapy: • Ado-trastuzumab emtansine alone. If ado-trastuzumab emtansine discontinued for toxicity, then trastuzumab with or without pertuzumab to complete one year of therapy.
• If node positive at initial staging, trastuzumab with pertuzumab Other recommended • Doxorubicin with cyclophosphamide (AC) followed by docetaxel with trastuzumab • Doxorubicin with cyclophosphamide (AC) followed by docetaxel with trastuzumab and pertuzumab • Paclitaxel, carboplatin, trastuzumab, and pertuzumab Used in some cases • Docetaxel, cyclophosphamide, and trastuzumab • Doxorubicin and cyclophosphamide (AC) followed by docetaxel and trastuzumab, followed by paclitaxel with trastuzumab • Doxorubicin and cyclophosphamide (AC) followed by docetaxel, trastuzumab and pertuzumab, followed by paclitaxel, trastuzumab, and pertuzumab • Paclitaxel with trastuzumab and pertuzumab • Neratinib (adjuvant only) • Ado-trastuzumab emtansine (TDM-1) (adjuvant only) Notes Other taxanes (such as docetaxel, paclitaxel, albumin-bound paclitaxel) might be substituted in some cases. 55 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Systemic therapy before surgery For preoperative systemic therapy options, see Guide 4 and Guide 5.
Guide 5 Systemic therapy options: HER2- cancer Preferred • Doxorubicin and cyclophosphamide (AC) with paclitaxel • Docetaxel and cyclophosphamide (TC) • Adjuvant only: Olaparib, if germline BRCA1 or BRCA2 mutations Triple-negative breast cancer (TNBC): • Options listed above • Preoperative pembrolizumab with carboplatin and paclitaxel, followed by preoperative pembrolizumab and cyclophosphamide with doxorubicin or epirubicin, followed by adjuvant pembrolizumab • If residual disease after preoperative therapy with taxane-, alkylator-, and anthracycline-based chemotherapy, then capecitabine Other recommended • Doxorubicin and cyclophosphamide (AC) with docetaxel • Epirubicin and cyclophosphamide (EC) • Docetaxel, doxorubicin, and cyclophosphamide (TAC) TNBC: • Options listed above • Paclitaxel with carboplatin • Docetaxel with carboplatin Used in some cases • Doxorubicin and cyclophosphamide (AC) • Cyclophosphamide, methotrexate, and fluorouracil (CMF) • Doxorubicin and cyclophosphamide (AC) with paclitaxel TNBC: • Options listed above • Docetaxel, carboplatin, and pembrolizumab (preoperative only) • Doxorubicin and cyclophosphamide (AC) with carboplatin and docetaxel • Capecitabine (maintenance therapy after adjuvant chemotherapy) Notes Other taxanes (such as docetaxel, paclitaxel, albumin-bound paclitaxel) might be substituted in some cases.
56 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Systemic therapy after surgery Surgery Surgery options depend on how your cancer responded to preoperative therapy. A complete response means there is no evidence of cancer. In a partial response, the tumor in the breast or lymph nodes has shrunk in size. The choice between a lumpectomy or mastectomy is based on the size and location of the tumor, the size of your breast before surgery, and your preference. If you had preoperative therapy, your tumor will be restaged using tissue removed during surgery. This is called the pathologic (p) stage or surgical stage. It might look like this: ypT2N1. The y means you had preoperative therapy. Treatment after surgery might be systemic therapy or radiation or both. It will likely include endocrine therapy if your tumor is HR+. Surgery is not an option Surgery is not always possible. Even though surgery might not be an option, systemic therapy will continue. If the cancer is not responding to systemic therapy, then radiation may be considered to try to make the cancer resectable (able to be removed with surgery). The goal of treatment is to reduce the amount of cancer. Talk with your care team about your goals of treatment and your treatment preferences. Your wishes are always important. Systemic therapy after surgery Adjuvant systemic therapy is drug treatment after surgery. It is given to kill any remaining cancer cells and to help reduce the risk of cancer returning. This treatment is based on “Take things one day or even one step at a time, it might get harder before it gets better, but it will get better." 57 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Systemic therapy after surgery cancer stage, including lymph node (nodal) hormone receptor (HR), and HER2 status.
Systemic therapy options are based on the tumor type (histology), tumor HR and HER2 status, and if you are in menopause. Systemic therapies might be used alone or in combination. Ask your care team why one treatment might be preferred over another for your type of cancer. For those in menopause (natural or induced), bone-strengthening therapy might be given to reduce the risk of distant metastasis.
Favorable histologies A tumor with a favorable histology is one that has a good prognosis. A prognosis is the course your cancer will likely take. These tumor types are not high grade, are HER2-, and might respond better to treatment than other tumors. They also might have less risk of returning. Ask your care team what this might mean for your treatment. Those with hormone receptor-positive (HR+) tumors will likely have endocrine therapy. Other systemic therapies or radiation therapy are possible. Favorable histology includes the following: h Pure tubular h Pure mucinous h Pure cribriform h Adenoid cystic (conventional), secretory carcinoma, and other salivary carcinomas h Rare low-grade forms of metaplastic carcinoma h Other rare forms Common histologies Ductal and lobular carcinoma are the most common types of invasive breast cancer. Common histology includes the following: h Ductal/no special type (NST) (NST includes medullary pattern, cancers with neuroendocrine expression, and other rare patterns) h Lobular h Mixed h Micropapillary h Metaplastic (includes various subtypes) 21-gene assay A 21-gene assay (such as Oncotype Dx) is a tumor test used to predict if your cancer might benefit from adjuvant chemotherapy in addition to endocrine therapy. There are other tests that might be helpful, but a 21-gene assay is the only test proven to predict who would benefit from chemotherapy. What’s next?
The next pages describe systemic therapy treatment based on tumor HR and HER2 status for common histologies. Typically, the systemic therapy options are the same before and after surgery. Ask your care team for more information.
58 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » HR+ with HER2+ 58 6 Your treatment options » HR+ with HER2+ » Triple-negative breast cancer HR+ with HER2+ HR+ with HER2+ breast cancer is sometimes called triple-positive breast cancer. It is treated with HER2-targeted therapy and endocrine therapy. A CDK4/6 inhibitor may be added. h Chemotherapy with a HER2-targeted therapy is used to treat HER2+ cancer. Trastuzumab and pertuzumab are examples of HER2-targeted therapy. HER2-targeted therapy options can be found in Guide 4. h Endocrine therapy is used to treat HR+ breast cancer. It is given after chemotherapy. Endocrine therapy options for HR+ cancer can be found in Guide 6.
HR- with HER2+ In HER2-positive (HER2+) breast cancer, tumor cells have a larger than normal number of HER2 receptors. Since this cancer is HR- and HER2+, treatment focuses on targeting HER2. HER2-targeted therapy usually includes chemotherapy. HER2-targeted therapy options can be found in Guide 4. HR+ with HER2-HER2-targeting therapy is not used when the tumor is HER2-negative (HER2-). Sometimes, chemotherapy is used. When cancer is found in the lymph nodes, it is node positive (node+). h If chemotherapy is given, it is given before endocrine therapy. Chemotherapy and other systemic therapies specific to HER2- breast cancer can be found in Guide 5. h Those who are premenopausal might have ovarian suppression or ablation in addition to endocrine therapy. Endocrine therapy options for HR+ cancer can be found in Guide 6.
Triple-negative breast cancer In triple-negative breast cancer (TNBC), the tumor tested negative for HER2, estrogen receptors, and progesterone receptors. It is written as ER- and/or PR- with HER2-. There are many variations within TNBC. It is a group of diseases that we are learning more about all the time. This cancer is treated with chemotherapy and other systemic therapies found in Guide 5.
59 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Endocrine therapy Endocrine therapy Endocrine therapy is used to treat hormone receptor-positive (HR+) breast cancer. This is cancer that tests positive for estrogen receptors (ER+), progesterone receptors (PR+), or both (ER+/PR+). Endocrine therapy blocks estrogen and progesterone, which can slow tumor growth or shrink the tumor for a period of time. It might also help prevent the risk of cancer returning in the breasts and elsewhere in the body. Guide 6 Endocrine therapy options: HR+ cancer Premenopause at diagnosis Option 1 • Tamoxifen alone or with ovarian suppression or ablation for 5 years ª • After 5 years, if in postmenopause, then an aromatase inhibitor for 5 years or consider tamoxifen for another 5 years (for a total of 10 years on tamoxifen) • After 5 years, if still in premenopause, then consider tamoxifen for another 5 years (for a total of 10 years on tamoxifen) or stop endocrine therapy Option 2 • Aromatase inhibitor for 5 years with ovarian suppression or ablation, then consider aromatase inhibitor for an additional 3 to 5 years Menopause at diagnosis Option 1 • Aromatase inhibitor for 5 years, then consider aromatase inhibitor for 3 to 5 more years • Aromatase inhibitor for 2 to 3 years, then tamoxifen to complete 5 years total of endocrine therapy • Tamoxifen for 2 to 3 years, then an aromatase inhibitor to complete 5 years of endocrine therapy • Tamoxifen for 2 to 3 years, then up to 5 years of an aromatase inhibitor Option 2 • Tamoxifen for 4.5 to 6 years, then an aromatase inhibitor for 5 years or consider tamoxifen for another 5 years (for a total of 10 years on tamoxifen) Option 3 • For those who can’t have aromatase inhibitors or who don’t want aromatase inhibitors, take tamoxifen for 5 years or consider tamoxifen for up to 10 years 60 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Follow-up care h If chemotherapy is given, it is given before endocrine therapy. h Those with high-risk breast cancer that is HR+ and HER2- might have endocrine therapy with 2 years of adjuvant abemaciclib or 3 years of ribociclib. h Those receiving an aromatase inhibitor (AI) who are at risk for osteoporosis will likely have bone density tests and bone-strengthening therapy. Endocrine therapy options for HR+ cancer can be found in Guide 6.
Follow-up care After treatment, you will receive follow-up care. During this time, your health will be monitored for side effects of treatment (late effects) and the possible return of cancer (recurrence). It is important to keep any follow-up care and imaging test appointments. Seek routine medical care, including regular doctor visits for preventive care and cancer screening. Find out who will coordinate with specialists for your follow-up care.
Tell your care team about any symptoms such as headaches, menstrual spotting between periods or new onset of spotting after menopause (if prior tamoxifen use), shortness of breath that you notice with walking, or bone pain. Side effects can be managed. Continue to take all medicine such as endocrine therapy exactly as prescribed and do not miss or skip doses. Follow-up care can be found in Guide 7.
Guide 7 Follow-up care Medical history and physical exam 1 to 4 times a year as needed for 5 years, then every year Screen for distress, anxiety, depression, and changes in family history Genetic testing and referral to genetic counseling, as needed Monitor for lymphedema and refer for lymphedema management, as needed Mammogram every 12 months (not needed for the side that underwent a mastectomy or on reconstructed breast) Discuss any issues or questions related to fertility, birth control, or sexual health Heart tests, as needed Information on risk of future health issues (comorbidities) If signs and symptoms of metastases, then blood and imaging tests If taking endocrine therapy, continue to take endocrine therapy. Do not miss or skip doses Annual gynecology exam (depending on age) Bone density tests for those on an aromatase inhibitor or who later have ovarian issues Maintain an ideal body weight (BMI of 20 to 25), be active, eat a mostly plant-based diet, exercise, limit alcohol, and quit smoking/vaping nicotine 61 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 6 Your treatment options » Key points 61 6 Your treatment options » Key points » Questions to ask Key points h Surgery is the main or primary treatment for invasive breast cancer. Some people benefit from systemic therapy before surgery. Treatment after surgery might include systemic therapy, radiation therapy, or both. h In hormone receptor-positive (HR+) cancer, estrogen receptors (ER+), progesterone receptors (PR+), or both (ER+/PR+) are found. Endocrine therapy is given for HR+ cancers to reduce the risk of recurrence. h HER2-positive (HER2) cancers are treated with a combination of chemotherapy and HER2-targeted therapy. h If chemotherapy is given, it is given before radiation therapy and endocrine therapy. h In triple-negative breast cancer (TNBC), receptors for estrogen, progesterone, and HER2 are not found. It is almost always treated with chemotherapy. h It is important to keep follow-up visits and imaging test appointments. Seek good routine medical care, including preventive care and cancer screenings. Continue to take all medicines as prescribed.
Questions to ask h What treatments do you recommend and why? h Does the order of treatments matter? h Which option is proven to work best for my type of cancer, age, overall health, and other factors? h Are there resources to help me pay for treatment or other care I may need? h How long will I be on this systemic (drug) therapy or endocrine therapy?
62 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 7 The breast after surgery 63 Volume displacement 63 Flat closure 64 Breast reconstruction 65 What to consider 66 Key points 66 Questions to ask 63 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 7 The breast after surgery » Volume displacement 63 7 The breast after surgery » Volume displacement » Flat closure The look of your breast after surgery will depend on the type of surgery, the amount of tissue removed, and other factors such as your body type, age, and size and shape of the area before surgery. You might consider speaking with a plastic surgeon before surgery. This chapter offers more information on flat closure and breast reconstruction.
The recovery time for each procedure differs. This can affect your ability to return to work or participate in activities. You might consider speaking with a plastic surgeon before surgery to discuss your options and what to expect. A plastic surgeon performs oncoplastic (breast cancer surgery) reductions, balancing procedures, and breast reconstruction.
Volume displacement With a lumpectomy, most people have a scar with some volume loss. However, if you need a large lumpectomy and your surgeon thinks your breast will look more abnormal afterwards, your breast may be able to be re-shaped at the time of surgery. This procedure is called volume displacement or oncoplasty. Only a limited number of cancer centers perform this procedure. It is often done by the cancer surgeon or plastic surgeon right after the lumpectomy. The surgeon will shift the remaining breast tissue to fill the space left by the removed tumor.
If volume displacement is planned, a larger piece of your breast will need to be removed. Despite a larger piece being removed, the natural look of your breast will be kept. However, with large amounts of tissue removed, your breast may be smaller than before.
You may not like the results of the volume displacement. In this case, breast revision surgery may help. This surgery is done by a plastic surgeon. A second volume displacement may be an option, too. Another option is to get breast implants or mastectomy with reconstruction.
Flat closure In a total mastectomy with a flat closure, the entire breast, including nipple, extra skin, fat, and other tissue in the breast area, is removed. The remaining skin is tightened and sewn together. No breast mound is created, and no implant is added. The scar will be slightly raised and differ in color than the surrounding skin. A flat closure is not completely flat or smooth. The result varies from person to person. Ask to look at pictures from flat closures so you know what to expect. You might decide to have a flat closure procedure later or after having breast implants removed. Talk to your care team to learn more.
64 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 7 The breast after surgery » Breast reconstruction Breast reconstruction Breast reconstruction is surgery to rebuild the shape and look of the breast after a mastectomy. In many cases, breast reconstruction involves a staged approach. It might require more than one procedure.
You may have a choice as to when breast reconstruction is done. Immediate reconstruction is finished within hours after removing the breast. Delayed reconstruction can occur months or years after the cancer surgery. Reconstruction can also be done in stages, with part of the reconstruction done at the time of the original cancer surgery and finished with another surgery later. A plastic surgeon performs breast reconstruction. Breasts can be reconstructed with implants and flaps. All methods are generally safe, but as with any surgery, there are risks. Ask your treatment team for a complete list of side effects.
Implants Breast implants are small bags filled with salt water, silicone gel, or both. They are placed under the breast skin or muscle to look like a new breast following a mastectomy. A balloon-like device, called an expander, may be used first to stretch out tissue. It will be placed under your skin or muscle and enlarged every few weeks for 2 to 3 months. When your skin is stretched to the proper size, you will have surgery to place the final implant.
Implants have a small risk of leaking or causing other issues. You may feel pain from the implant or expander. Scar tissue or tissue death can occur. Flaps Breasts can be remade using tissue from other parts of your body, known as flaps. These flaps are taken from the abdomen, buttocks, thigh, or from under the shoulder blade. Some flaps are completely removed and then sewn in place. Other flaps stay attached to your body but are slid over and sewn into place.
There are several risks associated with flaps, including death of fat in the flap, which can cause lumps. A hernia may result from muscle weakness. Problems are more likely to occur among those who have diabetes or who smoke.
Implants and flaps Some breasts are reconstructed with both implants and flaps. This method may give the reconstructed breast more volume to match the other breast. For any reconstruction, you may need surgery on your remaining breast to match the size and shape of both breasts.
Nipple replacement Like your breast, a nipple can be remade. To rebuild a nipple, a plastic surgeon can use surrounding tissues. Also, nipples can be remade with tissue from the thigh or the other nipple. Tissue can be darkened with a tattoo to look more like a nipple. It is important to note that while you can remake something to look like a nipple, it will not have the sensation of your real nipple. Also, a tattoo can be done to look like a nipple without having to take tissue from another part of the body.
65 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 7 The breast after surgery » What to consider What to consider Some things to consider when deciding to have flat closure or reconstruction after mastectomy: h Your desire – You may have a strong feeling towards flat closure or one form of reconstruction after being given the options. Breast reconstruction should be a shared decision between you and your care team. Make your wishes known. h Health issues – You may have health issues such as diabetes or a blood disorder that might affect or delay healing, or make longer procedures unsafe. h Tobacco use – Smoking delays wound healing and can cause mastectomy flap death (necrosis), nipple-areola complex (NAC) necrosis in a nipple-sparing mastectomy, infection, and failure of implant-based reconstruction. In free flap reconstruction, smoking increases the risk of complications. You are encouraged to stop smoking prior to reconstruction. h Breast size and shape – There are limits to the available sizes of breast implants. Very large breasts or breasts that lack tone or droop (called ptosis) might be difficult to match. Breast reduction surgery might be an option. h Body mass index (BMI) – Those with an elevated BMI have an increased risk of infections and complications with breast reconstruction. If you smoke or vape, seek help to quit Smoking or vaping nicotine greatly increases your chances of having side effects during and after surgery. Smoking and vaping can limit how well cancer treatment works and prevent wound healing. They also increase your chances of developing other cancers. Cannabis use might also affect the amount of anesthesia used during surgery. Nicotine is the chemical in tobacco that makes you want to keep smoking and vaping. Nicotine withdrawal is challenging for most people who smoke or vape. The stress of having cancer may make it even harder to quit. If you smoke or vape, ask your care team about counseling and medicines to help you quit.
More information can be found in the NCCN Guidelines for Patients: Quitting Smoking at NCCN.org/patientguidelines and on the NCCN Patient Guides for Cancer app.
For online support, try these websites: • SmokeFree.gov • CDC.gov/tobacco 66 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 7 The breast after surgery » Key points 66 7 The breast after surgery » Key points » Questions to ask Key points h Volume displacement is a shifting of the breast tissue to fill the space left by a lumpectomy. h Flat closure is done after a mastectomy. The skin is tightened and sewn together without the addition of a breast implant. h Breast reconstruction is surgery to rebuild the shape and look of the breast. h Breasts that are fully removed in a mastectomy can be remade with breast implants, flaps, or both. h Removed nipples can be remade with body tissue and/or tattooing.
Questions to ask h What will my breast look like after surgery? h How many breast surgeries or breast reconstructive surgeries have you done? h How long will it take for me to recover from surgery and what should I expect? h How much pain will I be in and what will be done to manage my pain? h What options are available if I do not like the look of my breast after surgery?
67 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 8 Recurrence 68 Overview 68 Tests 69 Treatment 70 Key points 70 Questions to ask 68 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 8 Recurrence » Overview 68 8 Recurrence » Overview » Tests When cancer returns, it is called a recurrence. Treatment is based on the types of treatment you had before. Together, you and your care team will choose a treatment plan that is best for you.
Overview Breast cancer can return in the following areas: h It can return to the breast or chest wall of the breast that had cancer before. This is called local recurrence. h It can return to axillary lymph nodes or lymph nodes in or near the breast. This is called regional recurrence. h It can return to other distant parts of the body. This is called metastatic breast cancer. h You can also have a new breast cancer that is not a recurrence of an earlier cancer. This chapter presents treatment options for local and regional recurrence. Tests You will have tests to learn more about your cancer. Many tests you had when you were first diagnosed will be repeated. This is called restaging. Tests for recurrence can be found in Guide 8.
Guide 8 Possible tests: Recurrence Medical history and physical exam Blood tests such as CBC and comprehensive metabolic panel (including liver function tests and alkaline phosphatase) Chest CT with or without contrast CT with contrast of abdomen with or without pelvis (MRI with contrast might be done instead) Other imaging as needed Biopsy tumor or metastasis and test for biomarkers, including estrogen receptor (ER), progesterone receptor (PR), and HER2 status Genetic counseling if at risk for hereditary breast cancer Assess for distress 69 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 8 Recurrence » Treatment Treatment Treatment is based on where the cancer has returned and what type of treatment you had before. Surgery followed by radiation therapy (RT) and systemic (drug) therapy are possible. Systemic and endocrine therapy options will be based on tumor hormone receptor (HR) status, HER2 status, and treatments given during your initial cancer treatment. Supportive care is always given. Supportive care aims to relieve side effects such as pain and to improve quality of life.
Local recurrence Treatment for cancer that has returned to the same breast as before is based on if your first treatment was breast-conserving surgery (lumpectomy) or a mastectomy. More surgery might be an option. If you had radiation therapy before, then it may or may not be possible to have radiation again in the same area, which may limit your surgical options.
Regional recurrence If the recurrence is in or near the armpit (axilla), then surgery to remove the tumor might be an option before radiation therapy. Systemic therapy might be given before surgery to help reduce the amount of cancer or the size of the tumor. Treatment after surgery might include radiation therapy to the lymph nodes. If you had radiation therapy before, then it may not be possible to have it again to the same area.
Both local and regional Cancer that is both local and regional might be referred to as a locoregional recurrence. Treatment for a locoregional recurrence is surgery and radiation therapy followed by systemic therapy.
Unresectable An unresectable tumor cannot be removed with surgery. It is treated with systemic therapy. For treatment of unresectable recurrent disease, see NCCN Guidelines for Patients: Metastatic Breast Cancer, available at NCCN.
org/patientguidelines and on the NCCN Patient Guides for Cancer app.
70 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 8 Recurrence » Key points 70 8 Recurrence » Key points » Questions to ask Key points h When cancer returns, it is called a recurrence. h Cancer that returns to the breast that had cancer before is called a local recurrence. h Cancer that returns to axillary lymph nodes or lymph nodes in or near the breast is called a regional recurrence. h Cancer that returns to distant parts of the body is called metastatic breast cancer. h Treatment is based on where the cancer has returned and what type of treatment you had before. h Surgery followed by systemic therapy and radiation therapy is possible for a locoregional recurrence. Systemic and endocrine therapy options are based on hormone receptor (HR) and HER2 status, and any mutations found. h An unresectable tumor cannot be removed with surgery. Unresectable recurrent disease is treated as metastatic disease with systemic therapy.
Questions to ask h Where is the cancer located and what does this mean for my treatment options? h Which course of treatment do you recommend and why? h Is surgery an option? What treatment might I have before and after surgery? h What role do previous treatments, my age, menopause status, and overall health play in treatment options? h Am I a candidate for a clinical trial?
71 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 9 Other resources 72 What else to know 72 What else to do 72 Where to get help 73 Questions to ask about resources and support 72 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 9 Other resources » What else to know 72 9 Other resources » What else to know » Where to get help Want to learn more? Here's how you can get additional help.
What else to know This book can help improve your cancer care. It plainly explains expert recommendations and suggests questions to ask your care team. But, it’s not the only resource that you have.
You’re welcome to receive as much information and help as you need. Many people are interested in learning more about: h The details of treatment h Being a part of a care team h Getting financial help h Finding an oncologist who is an expert in breast cancer h Coping with side effects What else to do Your health care center can help you with next steps. They often have on-site resources to help meet your needs and find answers to your questions. Health care centers can also inform you of resources in your community.
In addition to help from your providers, the resources listed in the next section provide support for many people like yourself. Look through the list and visit the provided websites to learn more about these organizations Where to get help Bone Marrow & Cancer Foundation bonemarrow.org Breast Cancer Alliance Breastcanceralliance.org Breastcancer.org Breastcancer.org CanCare Cancare.org CancerCare Cancercare.org Cancer Hope Network cancerhopenetwork.org Cancer Survivor Care Cancersurvivorcare.org DiepC Foundation diepcfoundation.org FORCE: Facing Our Risk of Cancer Empowered facingourrisk.org GPAC Global Patient Advocacy Coalition GPACunited.org HIS Breast Cancer Awareness Hisbreastcancer.org 73 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 9 Other resources » Questions to ask about resources and support Imerman Angels Imermanangels.org Inflammatory Breast Cancer Research Foundation ibcresearch.org Lobular Breast Cancer Alliance lobularbreastcancer.org My Faulty Gene Myfaultygene.org MedlinePlus medlineplus.gov National Cancer Institute (NCI) cancer.gov/types/breast National Coalition for Cancer Survivorship canceradvocacy.org Sharsheret sharsheret.org Triage Cancer Triagecancer.org Unite for HER uniteforher.org Young Survival Coalition (YSC) Youngsurvival.org Questions to ask about resources and support h Who can I talk to about help with housing, food, and other basic needs? h What help is available for transportation, childcare, and home care? h What other services are available to me and my caregivers? h How can I connect with others and build a support system? h Who can I talk to if I don’t feel safe at home, at work, or in my neighborhood?
74 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Ü 75 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Words to know Words to know adjuvant therapy Treatment that is given to lower the chances of the cancer returning.
antibody drug conjugate (ADC) A substance made up of a protein linked to a drug that attaches and enters certain types of cancer cells.
anti-estrogen A drug that stops estrogen from attaching to cells. areola A darker, round area of skin on the breast around the nipple.
aromatase inhibitor (AI) A drug that lowers the level of estrogen in the body. axillary lymph node (ALN) A small disease-fighting structure that is near the armpit (axilla).
axillary lymph node dissection (ALND) An operation that removes the disease-fighting structures (lymph nodes) near the armpit.
bilateral diagnostic mammogram Pictures of the insides of both breasts that are made from a set of x-rays.
bilateral oophorectomy An operation that removes both ovaries.
biopsy A procedure that removes fluid or tissue samples to be tested for a disease.
bone mineral density A test that measures the strength of bones.
bone scan A test that makes pictures of bones to assess for health problems.
breast implant A small bag filled with salt water, gel, or both that is used to remake breasts.
breast reconstruction An operation that creates new breasts.
cancer stage A rating of the outlook of a cancer based on its growth and spread.
carcinoma A cancer of cells that line the inner or outer surfaces of the body.
chest wall The layer of muscle, bone, and fat that protects the vital organs.
clinical breast exam (CBE) Touching of a breast by a health expert to feel for diseases. clinical stage (c) The rating of the extent of cancer before treatment is started.
clinical trial A type of research that assesses health tests or treatments.
contrast A substance put into your body to make clearer pictures during imaging tests.
core needle biopsy (CNB) A procedure that removes tissue samples with a hollow needle. Also called core biopsy (CB).
76 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Words to know diagnostic bilateral mammogram Pictures of the insides of both breasts that are made from a set of x-rays.
duct A tube-shaped structure through which milk travels to the nipple.
ductal carcinoma A cancer derived from cells that line small tube-shaped vessels. endocrine therapy A cancer treatment that stops the making or action of estrogen. Also called hormone therapy.
estrogen A hormone that plays a role in breast development.
estrogen receptor (ER) A protein inside cells that binds to estrogen.
estrogen receptor-negative (ER-) A type of breast cancer that doesn’t use estrogen to grow.
estrogen receptor-positive (ER+) A type of breast cancer that uses estrogen to grow. fertility specialist An expert who helps people have babies. fine-needle aspiration (FNA) A procedure that removes tissue samples with a very thin needle.
flat closure Procedure done after a mastectomy in which the skin is tightened and sewn together without the addition of a breast implant.
gene Coded instructions in cells for making new cells and controlling how cells behave.
genetic counseling Expert guidance on the chance for a disease that is passed down in families.
hereditary breast cancer Breast cancer likely caused by abnormal genes passed down from biological parent to child.
histology The structure of cells, tissue, and organs as viewed under a microscope.
hormone A chemical in the body that triggers a response from cells or organs.
hormone receptor-negative cancer (HR-) Cancer cells that don’t use hormones to grow. hormone receptor-positive cancer (HR+) Cancer cells that use hormones to grow. human epidermal growth factor receptor 2 (HER2) A protein on the surface of a cell that sends signals for the cell to grow. immunohistochemistry (IHC) A lab test of cancer cells to find specific cell traits involved in abnormal cell growth.
inflammatory breast cancer (IBC) A type of breast cancer in which the breast looks red and swollen and feels warm to the touch.
infraclavicular The area right below the collarbone.
in situ hybridization (ISH) A lab test of the number of a gene. internal mammary The area along the breastbone.
77 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Words to know lobule A gland in the breast that makes breast milk.
luteinizing hormone-releasing hormone (LHRH) A hormone in the brain that helps control the making of estrogen by the ovaries.
lymph A clear fluid containing white blood cells.
lymphatic system Germ-fighting network of tissues and organs that includes the bone marrow, spleen, thymus, lymph nodes, and lymphatic vessels. Part of the immune system. lymphedema Swelling in the body caused by a buildup of fluid called lymph.
lymph node A small, bean-shaped disease-fighting structure.
magnetic resonance imaging (MRI) A test that uses radio waves and powerful magnets to make pictures of the insides of the body.
mammogram A picture of the insides of the breast that is made using x-rays.
mastectomy An operation that removes the whole breast.
medical oncologist A doctor who is an expert in cancer drugs.
menopause 12 months after the last menstrual period.
modified radical mastectomy An operation that removes the whole breast and lymph nodes under the arm (axilla). mutation An abnormal change.
neoadjuvant treatment A treatment that is given before the main treatment to reduce the cancer. Also called preoperative treatment if given before an operation.
nipple-areola complex (NAC) The ring of darker breast skin is called the areola. The raised tip within the areola is called the nipple. pathologic stage (p) A rating of the extent of cancer given after examining tissue removed during surgery.
pathologist A doctor who’s an expert in testing cells and tissue to find disease.
positron emission tomography (PET) A test that uses radioactive material to see the shape and function of body parts.
postmenopause The state of having no more menstrual periods. premenopause The state of having menstrual periods.
progesterone A hormone involved in sexual development, periods, and pregnancy.
progesterone receptor (PR) A protein inside cells that binds to progesterone.
prognosis The likely course and outcome of a disease based on tests. radiation therapy (RT) A treatment that uses high-energy rays. Also called radiotherapy. 78 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Words to know radical mastectomy An operation that removes the whole breast, lymph nodes under the arm (axilla), and chest wall muscles under the breast. recurrence The return of cancer after a cancer-free period.
selective estrogen receptor degrader (SERD) A drug that blocks and destroys estrogen receptors.
selective estrogen receptor modulator (SERM) A drug that blocks the effect of estrogen inside of cells.
sentinel lymph node (SLN) The first lymph node to which cancer cells spread after leaving a tumor.
sentinel lymph node biopsy (SLNB) An operation to remove the disease-fighting structures (lymph nodes) to which cancer first spreads. Also called sentinel lymph node dissection.
side effect An unhealthy or unpleasant physical or emotional response to treatment.
supportive care Health care that includes symptom relief but not cancer treatment. Also called palliative care or best supportive care. supraclavicular The area right above the collarbone.
systemic therapy Drug treatment that works throughout the body.
total mastectomy An operation that removes the entire breast with a flat closure. Also called simple mastectomy.
triple-negative breast cancer (TNBC) A breast cancer that does not use hormones or the HER2 protein to grow.
ultrasound A test that uses sound waves to take pictures of the inside of the body.
79 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 NCCN Contributors The NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®) for Breast Cancer, Version 3.2025 were developed by the following NCCN Panel Members: William J. Gradishar, MD/Chair Robert H. Lurie Comprehensive Cancer Center of Northwestern University Meena S. Moran, MD/Vice-Chair Yale Cancer Center/Smilow Cancer Hospital Jame Abraham, MD Case Comprehensive Cancer Center/ University Hospitals Seidman Cancer Center and Cleveland Clinic Taussig Cancer Institute Vandana Abramson, MD Vanderbilt-Ingram Cancer Center Rebecca Aft, MD, PhD Siteman Cancer Center at Barnes-Jewish Hospital and Washington University School of Medicine Doreen Agnese, MD The Ohio State University Comprehensive Cancer Center - James Cancer Hospital and Solove Research Institute Kimberly H. Allison, MD Stanford Cancer Institute Bethany Anderson, MD University of Wisconsin Carbone Cancer Center Janet Bailey, MD University of Michigan Rogel Cancer Center Harold J. Burstein, MD, PhD Dana-Farber/Brigham and Women’s Cancer Center Nan Chen, MD The UChicago Medicine Comprehensive Cancer Center Helen Chew, MD UC Davis Comprehensive Cancer Center Chau Dang, MD Memorial Sloan Kettering Cancer Center Anthony D. Elias, MD University of Colorado Cancer Center Sharon H. Giordano, MD, MPH The University of Texas MD Anderson Cancer Center Matthew P. Goetz, MD Mayo Clinic Comprehensive Cancer Center Rachel C. Jankowitz, MD Abramson Cancer Center at the University of Pennsylvania Sara H. Javid, MD Fred Hutchinson Cancer Center Jairam Krishnamurthy, MD Fred & Pamela Buffet Cancer Center A. Marilyn Leitch, MD UT Southwestern Simmons Comprehensive Cancer Center Janice Lyons, MD Case Comprehensive Cancer Center/ University Hospitals Seidman Cancer Center and Cleveland Clinic Taussig Cancer Institute Susie McCloskey, MD, MSHS UCLA Jonsson Comprehensive Cancer Center Melissa McShane, MD Fox Chase Cancer Center Joanne Mortimer, MD City of Hope National Medical Center Sameer A. Patel, MD Fox Chase Cancer Center Laura H. Rosenberger, MD, MS Duke Cancer Institute Hope S. Rugo, MD UCSF Helen Diller Family Comprehensive Cancer Center Cesar Santa-Maria, MD, MSCI Johns Hopkins Kimmel Cancer Center Bryan Schneider, MD Indiana University Melvin and Bren Simon Comprehensive Cancer Center Mary Lou Smith, JD, MBA Research Advocacy Network Hatem Soliman, MD Moffitt Cancer Center Erica M. Stringer-Reasor, MD O'Neal Comprehensive Cancer Center at UAB Melinda L. Telli, MD Stanford Cancer Institute Mei Wei, MD Huntsman Cancer Institute at the University of Utah Kari B. Wisinski, MD University of Wisconsin Carbone Cancer Center Kay T. Yeung, MD, PhD UC San Diego Moores Cancer Center Jessica S. Young, MD Roswell Park Comprehensive Cancer Center NCCN Rashmi Kumar, PhD Senior Director, Clinical Content Ryan Schonfeld, BA Guidelines Coordinator NCCN Contributors This patient guide is based on the NCCN Clinical Practice Guidelines in Oncology (NCCN Guidelines®) for Breast Cancer, Version 3.2025. It was adapted, reviewed, and published with help from the following people: Dorothy A. Shead, MS Senior Director Patient Information Operations Tanya Fischer, MEd, MSLIS Senior Medical Writer Susan Kidney Senior Graphic Design Specialist Reviewed this patient guide. For disclosures, visit NCCN.org/disclosures.
80 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 NCCN Cancer Centers NCCN Cancer Centers Abramson Cancer Center at the University of Pennsylvania Philadelphia, Pennsylvania 800.789.7366 • pennmedicine.org/cancer Case Comprehensive Cancer Center/ University Hospitals Seidman Cancer Center and Cleveland Clinic Taussig Cancer Institute Cleveland, Ohio UH Seidman Cancer Center 800.641.2422 • uhhospitals.org/services/cancer-services CC Taussig Cancer Institute 866.223.8100 • my.clevelandclinic.org/departments/cancer Case CCC 216.844.8797 • case.edu/cancer City of Hope National Medical Center Duarte, California 800.826.4673 • cityofhope.org Dana-Farber/Brigham and Women’s Cancer Center | Mass General Cancer Center Boston, Massachusetts 877.442.3324 • youhaveus.org 617.726.5130 • massgeneral.org/cancer-center Duke Cancer Institute Durham, North Carolina 888.275.3853 • dukecancerinstitute.org Fox Chase Cancer Center Philadelphia, Pennsylvania 888.369.2427 • foxchase.org Fred & Pamela Buffett Cancer Center Omaha, Nebraska 402.559.5600 • unmc.edu/cancercenter Fred Hutchinson Cancer Center Seattle, Washington 206.667.5000 • fredhutch.org Huntsman Cancer Institute at the University of Utah Salt Lake City, Utah 800.824.2073 • healthcare.utah.edu/huntsmancancerinstitute Indiana University Melvin and Bren Simon Comprehensive Cancer Center Indianapolis, Indiana 888.600.4822 • www.cancer.iu.edu Johns Hopkins Kimmel Cancer Center Baltimore, Maryland 410.955.8964 www.hopkinskimmelcancercenter.org Mayo Clinic Comprehensive Cancer Center Phoenix/Scottsdale, Arizona Jacksonville, Florida Rochester, Minnesota 480.301.8000 • Arizona 904.953.0853 • Florida 507.538.3270 • Minnesota mayoclinic.org/cancercenter Memorial Sloan Kettering Cancer Center New York, New York 800.525.2225 • mskcc.org Moffitt Cancer Center Tampa, Florida 888.663.3488 • moffitt.org O’Neal Comprehensive Cancer Center at UAB Birmingham, Alabama 800.822.0933 • uab.edu/onealcancercenter Robert H. Lurie Comprehensive Cancer Center of Northwestern University Chicago, Illinois 866.587.4322 • cancer.northwestern.edu Roswell Park Comprehensive Cancer Center Buffalo, New York 877.275.7724 • roswellpark.org Siteman Cancer Center at Barnes-Jewish Hospital and Washington University School of Medicine St. Louis, Missouri 800.600.3606 • siteman.wustl.edu St. Jude Children’s Research Hospital/ The University of Tennessee Health Science Center Memphis, Tennessee 866.278.5833 • stjude.org 901.448.5500 • uthsc.edu Stanford Cancer Institute Stanford, California 877.668.7535 • cancer.stanford.edu The Ohio State University Comprehensive Cancer Center - James Cancer Hospital and Solove Research Institute Columbus, Ohio 800.293.5066 • cancer.osu.edu The UChicago Medicine Comprehensive Cancer Center Chicago, Illinois 773.702.1000 • uchicagomedicine.org/cancer The University of Texas MD Anderson Cancer Center Houston, Texas 844.269.5922 • mdanderson.org 81 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 NCCN Cancer Centers UC Davis Comprehensive Cancer Center Sacramento, California 916.734.5959 • 800.770.9261 health.ucdavis.edu/cancer UC San Diego Moores Cancer Center La Jolla, California 858.822.6100 • cancer.ucsd.edu UCLA Jonsson Comprehensive Cancer Center Los Angeles, California 310.825.5268 • uclahealth.org/cancer UCSF Helen Diller Family Comprehensive Cancer Center San Francisco, California 800.689.8273 • cancer.ucsf.edu University of Colorado Cancer Center Aurora, Colorado 720.848.0300 • coloradocancercenter.org University of Michigan Rogel Cancer Center Ann Arbor, Michigan 800.865.1125 • rogelcancercenter.org University of Wisconsin Carbone Cancer Center Madison, Wisconsin 608.265.1700 • uwhealth.org/cancer UT Southwestern Simmons Comprehensive Cancer Center Dallas, Texas 214.648.3111 • utsouthwestern.edu/simmons Vanderbilt-Ingram Cancer Center Nashville, Tennessee 877.936.8422 • vicc.org Yale Cancer Center/Smilow Cancer Hospital New Haven, Connecticut 855.4.SMILOW • yalecancercenter.org Take our survey and help make the NCCN Guidelines for Patients better for everyone!
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82 NCCN Guidelines for Patients® Invasive Breast Cancer, 2025 Index Index 21-gene assay (Oncotype Dx) 57 antibody drug conjugate (ADC) 36 biopsy 16 bone scan 12 bone-strengthening therapy 40–41 BRCA 20 breast reconstruction 64 cancer stages 23–26 chemotherapy 36 clinical trials 42 clips or markers 16 computed tomography (CT) 13 diagnostic mammogram 13 distress 10 endocrine therapy 38–40, 59–60 estrogen receptor (ER) 18 family history 9, 19–20 fertility 11 flat closure 63 genetic testing 19–20 hereditary breast cancer 19–20 histology 16–17, 57 hormone receptor (HR) 18–19 human epidermal growth factor receptor 2 (HER2) 17 HER2-targeted therapy 36–37 immunotherapy 38 inhibitors 37–38 lumpectomy 30–31 lymphedema 10 lymph nodes 25 lymph node biopsy 32–33 magnetic resonance imaging (MRI) 14 mammogram 13 mastectomy 31–32 menopause 40, 59 positron emission tomography (PET) 15 pregnancy 11 premenopause 40, 59 progesterone receptor (PR) 18–19 radiation therapy (RT) 34 recurrence 68–69 regional nodal irradiation (RNI) 34 side effects 45–49 supportive care 45–49 surgery 30–33 survivorship 49 targeted therapy 36–38 testosterone 39 those assigned male at birth 27 TNM scores 24–26 triple-negative breast cancer (TNBC) 58 ultrasound 15 volume displacement 63 Ü PAT-N-1855-1025 3025 Chemical Road, Suite 100 Plymouth Meeting, PA 19462 215.690.0300 NCCN.org/patients – For Patients | NCCN.org – For Clinicians Invasive Breast Cancer 2025 NCCN GUIDELINES FOR PATIENTS ® To support the NCCN Guidelines for Patients, visit NCCNFoundation.org/Donate |
10538 | https://www.britannica.com/topic/syllogistic | syllogistic
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syllogistic, in logic, the formal analysis of logical terms and operators and the structures that make it possible to infer true conclusions from given premises. Developed in its original form by Aristotle in his Prior Analytics (Analytica priora) about 350 bce, syllogistic represents the earliest branch of formal logic.
A brief treatment of syllogistic follows. For full treatment, see history of logic: Aristotle.
As currently understood, syllogistic comprises two domains of investigation. Categorical syllogistic, with which Aristotle concerned himself, confines itself to simple declarative statements and their variation with respect to modalities, or expressions of necessity and possibility. Noncategorical syllogistic is a form of logical inference using whole propositions as its units, an approach traceable to the Stoic logicians but not fully appreciated as a separate branch of syllogistic until the work of John Neville Keynes in the 19th century.
Knowing the truth or falsity of any given premises or conclusions does not enable one to determine the validity of an inference. In order to understand the validity of an argument, it is necessary to grasp its logical form. Traditional categorical syllogistic is the study of this problem. It begins by reducing all propositions to four basic forms.
Respectively, these forms are known as A, E, I, and O propositions, after the vowels in the Latin terms affirmo and nego. This distinction between affirmation and negation is said to be one of quality, while the difference between the universal scope of the first two forms, in contrast to the particular scope of the last two forms, is said to be one of quantity.
The expressions that fill the blanks of these propositions are called terms. These may be singular (Mary) or general (women). A very important distinction with respect to the use of general terms turns on whether their extensional or intensional attributes are in play; extension designates the set of individuals to which a term applies, while intension describes the set of attributes which define the term. The term that fills the first blank is called the subject of the proposition, that which fills the second is the predicate.
Using the notation of the early 20th-century logician Jan Łukasiewicz, the general terms or term variables can be expressed as lowercase Latin letters a, b, and c, with capitals reserved for the four syllogistic operators that specify A, E, I, and O propositions. The proposition “Every b is an a” is now written “Aba”; “Some b is an a” is written “Iba”; “No b is an a” is written “Eba”; and “Some b is not an a” is written “Oba.” Careful examination of the relations obtaining between these propositions reveals that the following are true for any terms a and b.
Not both: Aba and Eba.
If Aba, then Iba.
If Eba, then Oba.
Either Iba or Oba.
Aba is equivalent to the negation of Oba.
Eba is equivalent to the negation of Iba.
Reversing the order of the terms yields the simple converse of a proposition, but when in addition an A proposition is changed to an I, or an E to an O, the result is called the limited converse of the original. The logical relations holding between propositions and their converses, often pictured graphically in a square of opposition, are as follows: E and I propositions are equivalent or equipollent to their simple converses (i.e., Eba and Iba are the same as Eab and Iab, respectively). An A proposition Aba, although not equivalent to its simple converse Aab, implies, but is not implied by, its limited converse Iab. This kind of inference is traditionally called conversio per accidens and holds as well in Eba implying Oab. In contrast, Oba neither implies nor is implied by Oab, and this is expressed by saying that O propositions do not convert. When a proposition is posed against the proposition that results from changing its quality at the same time that its second term is negated, the resulting equivalence is called obversion. A last type of inference is called contraposition and is produced by the fact that some propositions imply the proposition that results from the original proposition when both of its term variables are negated and their order reversed.
A categorical syllogism infers a conclusion from two premises. It is defined by the following four attributes. Each of the three propositions is an A, E, I, or O proposition. The subject of the conclusion (called the minor term) also occurs in one of the premises (the minor premise). The predicate of the conclusion (called the major term) also occurs in the other premise (the major premise). The two remaining term positions in the premises are filled by the same term (the middle term). Since each of the three propositions in a syllogism can take one of four combinations of quality and quantity, the categorical syllogism may exhibit any of 64 moods. Each mood may occur in any of four figures—patterns of terms within the propositions—thus yielding 256 possible forms. One of the important tasks of syllogistic has been to reduce this plurality to just the valid forms.
Aristotle accepted 14 valid moods officially and 5 unofficially; since 5 of these 19 syllogisms have universal conclusions, the number of valid moods can be increased to 24 by passing to their corresponding particular propositions (i.e., from “all” to “some”). Employing an axiomatic system in which proof was by direct reduction and indirect reduction or reductio ad impossibile, Aristotle was able to reduce all syllogisms to those of the first figure. Today, in order to admit terms regardless of their emptiness or nonemptiness, syllogistic has become a special case of Boolean algebra in which the concepts of universal class and null class, along with the operations of class union and class intersection, are incorporated. From this standpoint the number of moods is 15. These 15 moods are the theorems of the syllogistic when interpreted in the predicate calculus.
Noncategorical syllogisms are either hypothetical or disjunctive, to which some treatments add a class of copulative syllogisms. Their treatment is distinguished from categorical syllogistic by the fact that the latter is a predicate logic analyzing terms in combination, while noncategorical syllogistic is a propositional logic that treats unanalyzed entire propositions as its units. Hypothetical syllogisms in which all propositions are of the form “p ⊃ q” (i.e., “p implies q”) are called pure, as opposed to mixed hypothetical syllogisms that have one hypothetical and one categorical premise and a categorical conclusion. These latter have two valid moods. Disjunctive syllogisms are composed by an “either…or” operator and have two important moods. In the 20th century the understanding of noncategorical syllogisms was extended to encompass complex and compound propositions as well as the dilemma with its constructive and destructive moods. |
10539 | https://stats.libretexts.org/Courses/Queensborough_Community_College/MA336%3A_Statistics/05%3A_Basic_Concepts_of_Probability/5.02%3A_Complements_Intersections_and_Unions | 5.2: Complements, Intersections, and Unions - Statistics LibreTexts
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5: Basic Concepts of Probability
MA336: Statistics
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5.2: Complements, Intersections, and Unions
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5.2: Complements, Intersections, and Unions
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5.1: Sample Spaces, Events, and Their Probabilities
5.3: Conditional Probability and Independent Events
Page ID 28385
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Complements
1. Definition: Complement
2. Example 5.2.1
1. Solution
3. Definition: Probability Rule for Complements
4. Example 5.2.2)
1. Solution
Intersection of Events
Definition: intersections
Example 5.2.3
Solution
Example 5.2.4
Solution
Definition: mutually exclusive
Definition: Probability Rule for Mutually Exclusive Events
Example 5.2.5
Solution
Union of Events
Definition: Union of Events
Example 5.2.6
Solution
Example 5.2.7
Solution
Definition: Additive Rule of Probability
Example 5.2.8
Solution
Example 5.2.9
Solution
Example 5.2.10
Solution
Key Takeaway
Learning Objectives
To learn how some events are naturally expressible in terms of other events.
To learn how to use special formulas for the probability of an event that is expressed in terms of one or more other events.
Some events can be naturally expressed in terms of other, sometimes simpler, events.
Complements
Definition: Complement
The complement of an event A in a sample space S, denoted A c, is the collection of all outcomes in S that are not elements of the set A. It corresponds to negating any description in words of the event A.
Example 5.2.1
Two events connected with the experiment of rolling a single die are E:“the number rolled is even” and T:“the number rolled is greater than two.” Find the complement of each.
Solution
In the sample space S={1,2,3,4,5,6} the corresponding sets of outcomes are E={2,4,6} and T={3,4,5,6}. The complements are E c={1,3,5} and T c={1,2}.
In words the complements are described by “the number rolled is not even” and “the number rolled is not greater than two.” Of course easier descriptions would be “the number rolled is odd” and “the number rolled is less than three.”
If there is a 60% chance of rain tomorrow, what is the probability of fair weather? The obvious answer, 40%, is an instance of the following general rule.
Definition: Probability Rule for Complements
The Probability Rule for Complements states that P(A c)=1−P(A)
This formula is particularly useful when finding the probability of an event directly is difficult.
Example 5.2.2
Find the probability that at least one heads will appear in five tosses of a fair coin.
Solution
Identify outcomes by lists of five hs and ts, such as tthtt and hhttt. Although it is tedious to list them all, it is not difficult to count them. Think of using a tree diagram to do so. There are two choices for the first toss. For each of these there are two choices for the second toss, hence 2×2=4 outcomes for two tosses. For each of these four outcomes, there are two possibilities for the third toss, hence 4×2=8 outcomes for three tosses. Similarly, there are 8×2=16 outcomes for four tosses and finally 16×2=32 outcomes for five tosses.
Let O denote the event “at least one heads.” There are many ways to obtain at least one heads, but only one way to fail to do so: all tails. Thus although it is difficult to list all the outcomes that form O, it is easy to write O c={ttttt}. Since there are 32 equally likely outcomes, each has probability 1 32, so P(O c)=1∕32, hence P(O)=1−1 32≈0.97 or about a 97% chance.
Intersection of Events
Definition: intersections
The intersection of events A and B, denoted A∩B, is the collection of all outcomes that are elements of both of the sets A and B. It corresponds to combining descriptions of the two events using the word “and.”
To say that the event A∩B occurred means that on a particular trial of the experiment both A and B occurred. A visual representation of the intersection of events A and B in a sample space S is given in Figure 5.2.1. The intersection corresponds to the shaded lens-shaped region that lies within both ovals.
Figure 5.2.1 : The Intersection of Events A and B
Example 5.2.3
In the experiment of rolling a single die, find the intersection E∩T of the events E:“the number rolled is even” and T:“the number rolled is greater than two.”
Solution
The sample space is S={1,2,3,4,5,6}. Since the outcomes that are common to E={2,4,6} and T={3,4,5,6} are 4 and 6, E∩T={4,6}.
In words the intersection is described by “the number rolled is even and is greater than two.” The only numbers between one and six that are both even and greater than two are four and six, corresponding to E∩T given above.
Example 5.2.4
A single die is rolled.
Suppose the die is fair. Find the probability that the number rolled is both even and greater than two.
Suppose the die has been “loaded” so that P(1)=1 12, P(6)=3 12, and the remaining four outcomes are equally likely with one another. Now find the probability that the number rolled is both even and greater than two.
Solution
In both cases the sample space is S={1,2,3,4,5,6} and the event in question is the intersection E∩T={4,6} of the previous example.
Since the die is fair, all outcomes are equally likely, so by counting we have P(E∩T)=2 6.
The information on the probabilities of the six outcomes that we have so far is
Outcome 1 2 3 4 5 6 Probablity 1 12 p p p p 3 12
Since P(1)+P(6)=4 6=1 3
P(2)+P(3)+P(4)+P(5)=1−1 3=2 3
Thus 4p=2 3, so p=1 6. In particular P(4)=1 6 therefore:
P(E∩T)=P(4)+P(6)=1 6+3 12=5 12
Definition: mutually exclusive
Events A and B are mutually exclusive (cannot both occur at once) if they have no elements in common.
For A and B to have no outcomes in common means precisely that it is impossible for both A and B to occur on a single trial of the random experiment. This gives the following rule:
Definition: Probability Rule for Mutually Exclusive Events
Events A and B are mutually exclusive if and only if
P(A∩B)=0
Any event A and its complement A c are mutually exclusive, but A and B can be mutually exclusive without being complements.
Example 5.2.5
In the experiment of rolling a single die, find three choices for an event A so that the events A and E: “the number rolled is even” are mutually exclusive.
Solution
Since E={2,4,6} and we want A to have no elements in common with E, any event that does not contain any even number will do. Three choices are {1,3,5} (the complement E c, the odds), {1,3}, and {5}.
Union of Events
Definition: Union of Events
The union of events A and B, denoted A∪B, is the collection of all outcomes that are elements of one or the other of the sets A and B, or of both of them. It corresponds to combining descriptions of the two events using the word “or.”
To say that the event A∪B occurred means that on a particular trial of the experiment either A or B occurred (or both did). A visual representation of the union of events A and B in a sample space S is given in Figure 5.2.2. The union corresponds to the shaded region.
Figure5.2.2: The Union of Events A and B
Example 5.2.6
In the experiment of rolling a single die, find the union of the events E: “the number rolled is even” and T: “the number rolled is greater than two.”
Solution
Since the outcomes that are in either E={2,4,6} or T={3,4,5,6} (or both) are 2,3,4,5, and 6, that means E∪T={2,3,4,5,6}.
Note that an outcome such as 4 that is in both sets is still listed only once (although strictly speaking it is not incorrect to list it twice).
In words the union is described by “the number rolled is even or is greater than two.” Every number between one and six except the number one is either even or is greater than two, corresponding to E∪T given above.
Example 5.2.7
A two-child family is selected at random. Let B denote the event that at least one child is a boy, let D denote the event that the genders of the two children differ, and let M denote the event that the genders of the two children match. Find B∪D and B∪M.
Solution
A sample space for this experiment is S={bb,bg,gb,gg}, where the first letter denotes the gender of the firstborn child and the second letter denotes the gender of the second child. The events B,D, and M are B={bb,bg,gb}, D={bg,gb}, M={bb,gg}.
Each outcome in D is already in B, so the outcomes that are in at least one or the other of the sets B and D is just the set B itself: B∪D={bb,bg,gb}=B.
Every outcome in the whole sample space S is in at least one or the other of the sets B and M, so B∪M={bb,bg,gb,gg}=S.
Definition: Additive Rule of Probability
A useful property to know is the Additive Rule of Probability, which is
P(A∪B)=P(A)+P(B)−P(A∩B)
The next example, in which we compute the probability of a union both by counting and by using the formula, shows why the last term in the formula is needed.
Example 5.2.8
Two fair dice are thrown. Find the probabilities of the following events:
both dice show a four
at least one die shows a four
Solution
As was the case with tossing two identical coins, actual experience dictates that for the sample space to have equally likely outcomes we should list outcomes as if we could distinguish the two dice. We could imagine that one of them is red and the other is green. Then any outcome can be labeled as a pair of numbers as in the following display, where the first number in the pair is the number of dots on the top face of the green die and the second number in the pair is the number of dots on the top face of the red die.
\begin{array}11 & 12 & 13 & 14 & 15 & 16 \ 21 & 22 & 23 & 24 & 25 & 26 \ 31 & 32 & 33 & 34 & 35 & 36 \ 41 & 42 & 43 & 44 & 45 & 46 \ 51 & 52 & 53 & 54 & 55 & 56 \ 61 & 62 & 63 & 64 & 65 & 66\end{array} \nonumber
There are 36 equally likely outcomes, of which exactly one corresponds to two fours, so the probability of a pair of fours is 1/36.
From the table we can see that there are 11 pairs that correspond to the event in question: the six pairs in the fourth row (the green die shows a four) plus the additional five pairs other than the pair 44, already counted, in the fourth column (the red die is four), so the answer is 11/36. To see how the formula gives the same number, let A G denote the event that the green die is a four and let A R denote the event that the red die is a four. Then clearly by counting we get: P(A G)=6/36 and P(A R)=6/36.Since A G∩A R={44}, P(A G∩A R)=1/36.This is the computation from part 1, of course. Thus by the Additive Rule of Probability we get:
P(A G∩A R)=P(A G)+P(A R)−P(A G−A R)=6/36+6/36−1/36=11 36
Example 5.2.9
A tutoring service specializes in preparing adults for high school equivalence tests. Among all the students seeking help from the service, 63% need help in mathematics, 34% need help in English, and 27% need help in both mathematics and English. What is the percentage of students who need help in either mathematics or English?
Solution
Imagine selecting a student at random, that is, in such a way that every student has the same chance of being selected. Let M denote the event “the student needs help in mathematics” and let E denote the event “the student needs help in English.” The information given is that P(M)=0.63, P(E)=0.34 and P(M∩E)=0.27. Thus the Additive Rule of Probability gives:
P(M∪E)=P(M)+P(E)−P(M∩E)=0.63+0.34−0.27=0.70
Note how the naïve reasoning that if 63% need help in mathematics and 34% need help in English then 63 plus 34 or 97% need help in one or the other gives a number that is too large. The percentage that need help in both subjects must be subtracted off, else the people needing help in both are counted twice, once for needing help in mathematics and once again for needing help in English. The simple sum of the probabilities would work if the events in question were mutually exclusive, for then P(A∩B) is zero, and makes no difference.
Example 5.2.10
Volunteers for a disaster relief effort were classified according to both specialty (C: construction, E: education, M: medicine) and language ability (S: speaks a single language fluently, T: speaks two or more languages fluently). The results are shown in the following two-way classification table:
| Specialty | Language Ability |
--- |
| S | T |
| C | 12 | 1 |
| E | 4 | 3 |
| M | 6 | 2 |
The first row of numbers means that 12 volunteers whose specialty is construction speak a single language fluently, and 1 volunteer whose specialty is construction speaks at least two languages fluently. Similarly for the other two rows.
A volunteer is selected at random, meaning that each one has an equal chance of being chosen. Find the probability that:
his specialty is medicine and he speaks two or more languages;
either his specialty is medicine or he speaks two or more languages;
his specialty is something other than medicine.
Solution
When information is presented in a two-way classification table it is typically convenient to adjoin to the table the row and column totals, to produce a new table like this:
| Specialty | Language Ability | Total |
---
| S | T |
| C | 12 | 1 | 13 |
| E | 4 | 3 | 7 |
| M | 6 | 2 | 8 |
| Total | 22 | 6 | 28 |
The probability sought is P(M∩T). The table shows that there are 2 such people, out of 28 in all, hence P(M∩T)=2/28≈0.07 or about a 7% chance.
The probability sought is P(M∪T). The third row total and the grand total in the sample give P(M)=8/28. The second column total and the grand total give P(T)=6/28. Thus using the result from part (1),
P(M∪T)=P(M)+P(T)−P(M∩T)=828+628−228=1228≈0.43
or about a 43% chance.
This probability can be computed in two ways. Since the event of interest can be viewed as the event C∪E and the events C and E are mutually exclusive, the answer is, using the first two row totals,
P(C∪E)=P(C)+P(E)−P(C∩E)=1328+728−028=2028≈0.71
On the other hand, the event of interest can be thought of as the complement M c of M, hence using the value of P(M)computed in part (2),
P(M c)=1−P(M)=1−828=2028≈0.71
as before.
Key Takeaway
The probability of an event that is a complement or union of events of known probability can be computed using formulas.
5.2: Complements, Intersections, and Unions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.
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10540 | https://www.open.edu/openlearn/science-maths-technology/mathematics-statistics/interpreting-data-boxplots-and-tables/content-section-0/?printable=1 | Interpreting data: boxplots and tables
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◄ Interpreting data: boxplots and tables
Interpreting data: boxplots and tables
Introduction
This course is concerned with two main topics. In Section 1, you will learn about another kind of graphical display, the boxplot. Boxplots are particularly useful for assessing quickly the location, dispersion, and symmetry or skewness of a set of data, and for making comparisons of these features in two or more data sets. The other topic, is that of dealing with data presented in tabular form. You are, no doubt, familiar with such tables: they are common in the media and in reports and other documents. It is not always straightforward to see at first glance just what information a table of data is providing, and it often helps to carry out certain calculations and/or to draw appropriate graphs to make this clearer.
This OpenLearn course is an adapted extract from the Open University course M248 Analysing data.
Learning outcomes
After studying this course, you should be able to:
understand and use the following terms: boxplots, box, whisker, upper and lower adjacent values, rate, time series, line plot
demonstrate an awareness of the idea that the general pattern of a set of data, in terms of location, dispersion and skewness, can be graphically represented in a boxplot
understand that boxplots can be used to provide a quick and simple comparison of data sets
understand that patterns in tabular data can be made clearer by leaving out unhelpful information, by including extra pieces of useful information, or by drawing appropriate graphs
describe and compare data sets on the basis of boxplots.
Overview
It is a common observation that a data exploration should always begin by looking at a graphical display of the data. When looking at data sets which involve only one variable, displays such as bar charts and histograms are available. One problem with these is that they can include too much detail. Also they are not very useful for comparing two or more samples of data. A graphical display showing certain summary statistics in a visually appealing and interpretable way is introduced in this section. This is the boxplot.
1 Boxplots
In this first section, you will learn how to construct a boxplot for a single set of data. The use of boxplots to compare two or more sets of data will then be discussed.
1.1 Simple boxplots
A boxplot is simple to construct. The following example on the β endorphin concentrations of collapsed runners will be used to show how this is done.
Example 1.1 Endorphin concentrations for collapsed runners
The β endorphin concentrations (in pmol/l) recorded for eleven runners who collapsed after the Great North Run are as follows (written in order of increasing size).
A boxplot for these data is shown in Figure 1.1.
Figure 1.1 A boxplot for the collapsed runners
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Figure 1.1 A boxplot for the collapsed runners
Figure 1.1 A boxplot for the collapsed runners
(Data sourced from Dale, G., Fleetwood, J.A., Weddell, A., Ellis, R.D. and Sainsbury, J.R.C. (1987) Beta-endorphin: a factor in 'fun run' collapse? British Medical Journal, 294, 1004.)
The easiest way to understand exactly what a boxplot represents and how it is constructed is to think about how you would draw one by hand. The steps involved in constructing the boxplot in Figure 1.1 for the data set of β endorphin concentrations are as follows.
First, a convenient scale is drawn covering the extent of the data. Since the minimum is 66 and the maximum is 414, a scale from 0 to 500 (say) is suitable in this case. The boxplot is drawn against this scale.
The median and quartiles are used to construct the ‘box’. The median of this data set is 110, and the lower and upper quartiles are 79 and 162, respectively. The box is shown in Figure 1.2.
The ‘box’ is a rectangle with edges defined by the lower and upper quartiles; so it indicates where the ‘middle 50%’ of the data can be found. The vertical line inside the box is located at the median.
The ‘whiskers’ are constructed next. These are lines drawn parallel to the scale (so they are horizontal in this course). Essentially, each whisker extends outwards from the edge of the box as far as the most extreme observation. However, as you will see in the next step, some observations may be classified as potential outliers; and in fact the whiskers extend only to cover observations which are not classified as potential outliers. The whiskers are drawn outwards as far as observations called adjacent values. The lower adjacent value is the furthest observation which is within one and a half iqr (interquartile range) of the lower end of the box; and the upper adjacent value is the furthest observation which is within one and a half iqr of the upper end of the box. So the interquartile range is needed to construct the whiskers.
Figure 1.2 Collapsed runners boxplot: the box
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Figure 1.2 Collapsed runners boxplot: the box
Figure 1.2 Collapsed runners boxplot: the box
For these data, the interquartile range is 162−79=83. So
The highest observation not exceeding 286.5 is 169, so the upper adjacent value is 169, and hence the right-hand whisker extends as far as the observation 169. Similarly,
The lowest observation, 66, is greater than this, so the lower adjacent value is 66, and the left-hand whisker extends to 66. Notice that, in this example, the lower adjacent value is the same as the sample minimum, 66. Figure 1.3 shows the box with the whiskers extending to the upper and lower adjacent values.
Figure 1.3 Collapsed runners boxplot: adjacent values
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Figure 1.3
Figure 1.3 Collapsed runners boxplot: adjacent values
Finally, any values not covered by the whiskers are marked separately. In some circumstances, they may be deemed outliers. At the least, they are potential outliers and merit special attention.
In this case, the only observation not covered by the whiskers is the maximum observation of 414. This is shown in Figure 1.4.
Figure 1.4 Completed boxplot for collapsed runners
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Figure 1.4
Figure 1.4 Completed boxplot for collapsed runners
It must be stressed that boxplot construction is an area where there are no universally accepted rules. All boxplots show the three quartiles, but the conventions defining the extent of the whiskers vary from text to text and from one computer package to another. The whiskers may extend as low as one or even up to two interquartile ranges either side of the box. Some approaches even distinguish between moderate and severe outliers by using different symbols for them. Some textbooks and software always draw the whiskers right out to the minimum and maximum values and do not mark (potential) outliers separately. The approach adopted here is one of the simplest and is probably the most common.
You can see how a boxplot gives a quick visual assessment of the data. The length of the box represents the interquartile range and the lengths of the whiskers relative to the length of the box give an idea of how stretched out the rest of the values are. Thus these aspects of the diagram give an idea of the dispersion of the data set. The unusually large value in this data set is clearly shown and the median gives an assessment of the centre.
Some kind of assessment of symmetry is possible, since symmetric data will produce a boxplot which is symmetric about the median. These particular data are not symmetric; they are right-skew, and in fact the sample skewness is 2.572. The corresponding lack of symmetry shows up in the boxplot: the right-hand section of the box is longer than the left. However, it should be borne in mind that this particular data set has only eleven values, and this is too small a number to infer anything definite about any underlying structure.
You should now ensure that you understand simple boxplots by constructing one for yourself.
A boxplot displays the median, the quartiles, the range of values covered by the data and any outliers which may be present. It gives a clear picture of all these features and, as you will see, allows a visual appreciation of lack of symmetry.
1.2 Boxplot activity
Activity 1 Drawing a boxplot: chondrite meteors
Table 1.1 contains data on the percentage of silica found in 22 chondrite meteors. The data are given in order of increasing size.
Table 1.1 Silica content of chondrite meteors
| 20.77 | 22.56 | 22.71 | 22.69 | 26.39 | 27.08 | 27.32 | 27.33 |
| 27.57 | 27.81 | 28.69 | 29.36 | 30.25 | 31.89 | 32.88 | 33.23 |
| 33.28 | 33.40 | 33.52 | 33.83 | 33.95 | 34.82 |
(Source: Good, I.J. and Gaskins, R.A. (1980) Density estimation and bump-hunting by the penalized likelihood method exemplified by scattering and meteorite data. J. American Statistical Association, 75, 42-56.)
The median for this data set is 29.025; the lower and upper quartiles are approximately 26.91 and 33.31. The interquartile range is 6.40.
(a) Using a pencil and ruler, construct a boxplot for these data.
(b) The sample skewness for these data is −0.446. Is this value in accord with the shape of the boxplot?
Answer
Solution
(a) The data run from 20.77 to 34.82. A convenient scale to cover this range of values runs from 20 to 40. In this case,
This is smaller than the sample minimum, so the left-hand whisker will extend as far as the minimum observation 20.77. (In other words, the lower adjacent value is equal to the sample minimum.) Similarly,
This is greater than the sample maximum, so the upper adjacent value is the same as the sample maximum. So with this data set, there are no extreme values to be plotted separately. The boxplot is shown in Figure 3.1.
Figure 1.5a Boxplot for silica content of chondrite meteors
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Figure 1.5a
Figure 1.5a Boxplot for silica content of chondrite meteors
(b) The sample skewness is negative, indicating that the data are left-skew. To some extent the boxplot reflects this: the left whisker is considerably longer than the right, indicating that the smaller values are more spread out than are the larger values. However, the box gives a different impression. The box corresponds to the middle half of the data values, and the line denoting the median divides this into two parts, each corresponding to one-quarter of the data. In this case, the left part of the box is shorter than the right part. In other words, the box suggests that the data might be right-skew rather than left-skew. So the pattern of asymmetry of these data is not straightforward.
In assessing patterns of skewness from a boxplot, you are looking at five different values: the upper and lower adjacent values, the upper and lower quartiles, and the median. It is thus possible, in some cases at least, to observe somewhat complicated patterns of skewness. On the other hand, calculating the sample skewness involves boiling the data down to a single value; and thus the sample skewness provides rather less information than a boxplot does about the shape of a data set.
The boxplot for the data in Table 1.1, which you were asked to draw in Activity 1, is shown in Figure 1.5.
Figure 1.5b A boxplot for silica content of chondrite meteors
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Figure 1.5b
Figure 1.5b A boxplot for silica content of chondrite meteors
This boxplot is clearly not symmetrical. However, the pattern of its skewness is not straightforward. The box, corresponding to the middle 50% of the data, appears to be right-skew, because the line marking the median is towards the left of the box (so that the right section of the box is longer than the left). However, the longer whisker is on the left, indicating a longer tail towards smaller values, which in turn suggests that the data are left-skew.
In this example, the sample skewness (−0.446) is in accord with the pattern suggested by the whiskers of the boxplot (left-skew), rather than with that suggested by the box. Essentially, this occurs because all the values in the data set are used to calculate the sample skewness; and the calculation involves a sum of powers of values, so that the sample skewness is particularly affected by the more extreme values in the data set. In a boxplot, the whiskers correspond to the more extreme values. In Figure 1.5, the whiskers suggest that the data are left-skew, matching the sample skewness.
1.3 Comparing data sets using boxplots
Example 1.2 Infants with SIRDS: boxplots
Boxplots are particularly useful for making quick comparisons. The following example relates to birth weights of infants exhibiting severe idiopathic respiratory distress syndrome (SIRDS), and the question ‘Is it possible to relate the chances of eventual survival to birth weight?’ The data in Table 1.3 are the recorded birth weights of infants who displayed the syndrome.
Table 1.3 Birth weights (in kg) of infants with severe idiopathic respiratory distress syndrome
| 1.050 | 2.500 | 1.890 | 1.760 | 2.830 |
| 1.175 | 1.030 | 1.940 | 1.930 | 1.410 |
| 1.230 | 1.100 | 2.200 | 2.015 | 1.715 |
| 1.310 | 1.185 | 2.270 | 2.090 | 1.720 |
| 1.500 | 1.225 | 2.440 | 2.600 | 2.040 |
| 1.600 | 1.262 | 2.560 | 2.700 | 2.200 |
| 1.720 | 1.295 | 2.730 | 2.950 | 2.400 |
| 1.750 | 1.300 | 1.130 | 3.160 | 2.550 |
| 1.770 | 1.550 | 1.575 | 3.400 | 2.570 |
| 2.275 | 1.820 | 1.680 | 3.640 | 3.005 |
| child died |
van Vliet, P.K. and Gupta, J.M. (1973) Sodium bicarbonate in idiopathic respiratory distress syndrome. Arch. Disease in Childhood, 48, 249–255.
An initial investigation of the question might involve histograms of the two sets of birth weights, as well as calculating their sample means, standard deviations and skewnesses. The results in this case would show that the mean birth weight of the infants who survived is considerably higher than the mean birth weight of the infants who died, and that the standard deviation of the birth weights of the infants who survived is also higher. Using boxplots we will now be able to make some further headway with the question.
For the birth weights (in kg) of the infants who survived, the lower quartile, median and upper quartile are, respectively, 1.72, 2.20 and 2.83. For the infants who died, the corresponding quartiles are 1.23, 1.60 and 2.20. Using these figures, together with the original data in Table 1.3 above, boxplots of the two data sets can be constructed. Notice that in both cases (as in Activity 1) the adjacent values are equal to the sample maxima and minima, so that the whiskers extend to the ends of the sample range. Plotting both boxplots against the same scale produces the diagram in Figure 1.6.
Figure 1.6 Comparative boxplots: birth weights of infants with SIRDS
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Figure 1.6
Figure 1.6 Comparative boxplots: birth weights of infants with SIRDS
As you saw in Subsection 1.1, a boxplot gives graphical information on the location, the dispersion and the skewness of a data set – that is, on the three aspects of the data set for which summary measures were introduced in Unit A1. In addition, a boxplot draws attention to certain potential outliers. Thus comparative boxplots, as in Figure 1.6, can be used to compare these four features in the data sets shown. This has been done in producing the following discussion of the SIRDS data.
Comparison of location: Figure 1.6 shows that the median birth weight of infants who survived is greater than that of those who died.
Comparison of dispersion: The interquartile ranges are reasonably similar (as shown by the lengths of the boxes), though the overall range of the data set is greater for the surviving infants (as shown by the distances between the ends of the two whiskers for each boxplot).
Comparison of skewness: Though both batches of data appear to be right-skew, and the batch for the infants who died is slightly more skewed than that for those who survived, the skewness is not particularly marked in either case. (In fact, the sample skewness for the birth weights of the infants who survived is 0.25; and for the infants who died, it is 0.53. Both skewnesses are positive; the value for the infants who died is rather larger, corresponding to a more marked lack of symmetry, but neither skewness is particularly large.)
Comparison of potential outliers: Neither data set shows any suspiciously far out values which might require a closer look.
General conclusions: Overall, the two batches of data look as if they were generally distributed in a similar way, but with one batch located to the right (larger location) of the other. You can see immediately that the median birth weight of infants who died is less than the lower quartile of the birth weights of infants who survived (that is, over three-quarters of the survivors were heavier than the median birth weight of those who died). So it looks as if we can safely say that survival is related to birth weight.
You can see how comparative boxplots give a compact, quickly assimilated summary of the data, suggesting that infants who survive and infants who do not may typically have different birth weights.
When using boxplots to compare two or more batches of data, it is usually best to compare individual features in a methodical way. You may find the following guidelines helpful.
Guidelines for comparing boxplots
Compare the respective medians, to compare location.
Compare the interquartile ranges (that is, the box lengths), to compare dispersion.
Look at the overall spread as shown by the adjacent values. (This is another aspect of dispersion.)
Look for signs of skewness. If the data do not appear to be symmetric, does each batch show the same kind of asymmetry?
Look for potential outliers.
After discussing these features, general conclusions should be summarized briefly.
Let us look at another example. This time, you are asked to do the work!
1.4 Boxplot activity 2
Activity 2 Boxplots of family sizes
The table below contains data on the sizes (numbers of children) of the completed families of two samples of mothers in Ontario. One sample of mothers had had fewer years of education than the other sample (six years or less for mothers in the first sample, and seven years or more for those in the other sample).
Table 1.4 Family size: mothers married aged 15–19
| Mother educated for six years or less |
| 14 13 4 14 10 2 13 5 0 0 13 3 9 2 10 11 13 5 14 |
| Mother educated for seven years or more |
| 0 4 0 2 3 3 0 4 7 1 9 4 3 2 3 2 16 6 0 13 6 6 5 9 10 5 4 3 3 5 2 3 5 15 5 |
Keyfitz, N. (1953) A factorial arrangement of comparisons of family size. American J. Sociology, 53, 470–480.
Comparative boxplots of the family size data are shown in Figure 1.7.
Figure 1.7 Comparative boxplots: family sizes for two groups of mothers
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Figure 1.7
Figure 1.7 Comparative boxplots: family sizes for two groups of mothers
Compare the two samples of data using the systematic approach just outlined in the text. What conclusions can you draw about an association between education and family size?
Answer
Solution
Following the five steps for comparing boxplots outlined in the text, we begin with the medians (step 1). These are well separated, with the median for mothers with less education being higher at an astonishing 10. The length of the box for these mothers is more than twice that of the other box (step 2). The overall spreads (distances between adjacent values) are roughly similar for the two data sets (step 3). However, this comparison is perhaps less informative about dispersion than the comparison of box lengths, because of the potential outliers in the data set for mothers with more education. The overall range for mothers with more education is rather greater if these ‘outliers’ are included. However, if the untypicality of these values were to be seen as a reason for omitting them, the range for the mothers with less education would be the greater. Whether or not they are omitted, the difference in range is not huge.
The boxplot for mothers with less education shows some slight left-skew: the left whisker is longer than the right (step 4). The main body of data for the mothers with more years of education looks symmetric, but there are three large potential outliers which would undoubtedly have an effect on any calculations of skewness (step 5).
The two batches of data seem to be distributed differently in a way which is not merely the result of difference in location. The median for the mothers with less education is close to the upper adjacent value for the mothers with more education, which leads to the conclusion that the mother's education varies with family size. The main difference between the groups lies in their different concentrations around the median rather than their overall spread of values. The potential outliers for the mothers with more education are not very far from the upper adjacent value for the other sample, and are marked as outliers essentially because of the comparatively low interquartile range for the sample into which they fall.
The overall conclusion is that the mother's education does vary with family size, with those mothers receiving six or less years of formal education having, on average, larger families.
One thing the boxplots have also shown is that three data values in one of the samples are perhaps not typical; so calculations of the mean, standard deviation and skewness should be treated with certain amount of scepticism.
1.5 Summary
In this section you have been introduced to the boxplot. This is a graphic that represents the key features of a set of data. A typical boxplot is shown in Figure 1.8.
Figure 1.8 A typical boxplot
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Figure 1.8
Figure 1.8 A typical boxplot
The ends of the box mark the quartiles, and the vertical line through the box is located at the median. The whiskers of a boxplot extend to values known as adjacent values. These are the values in the data that are furthest away from the median on either side of the box, but are still within a distance of 1.5 times the interquartile range from the nearest end of the box (that is, the nearer quartile). In many cases the whiskers actually extend right out to the most extreme values in the data set. However, in other cases they do not. Any values in the data set that are more extreme than the adjacent values are plotted as separate points on the boxplot. This identifies them as potential outliers that may need further investigation.
A boxplot depicts only some basic aspects of the distribution of the values in data set. But often these basic aspects are the ones of most interest. It is straightforward to draw boxplots of more than one data set on the same scale, and then to use them to compare important aspects of the distribution of the data sets. A systematic approach to carrying out such comparisons has been described.
1.6 Exercise
Activity 3 Exercise 1.1 Memory recall times
In a study of memory recall times, a series of stimulus words was shown to a subject on a computer screen. For each word, the subject was instructed to recall either a pleasant or an unpleasant memory associated with that word. Successful recall of a memory was indicated by the subject pressing a bar on the computer keyboard. Table 1.5 shows the recall times (in seconds) for twenty pleasant and twenty unpleasant memories.
Table 1.5 Memory recall times (seconds)
| Pleasant memory | Unpleasant memory |
| 1.07 | 1.45 |
| 1.17 | 1.67 |
| 1.22 | 1.90 |
| 1.42 | 2.02 |
| 1.63 | 2.32 |
| 1.98 | 2.35 |
| 2.12 | 2.43 |
| 2.32 | 2.47 |
| 2.56 | 2.57 |
| 2.70 | 3.33 |
| 2.93 | 3.87 |
| 2.97 | 4.33 |
| 3.03 | 5.35 |
| 3.15 | 5.72 |
| 3.22 | 6.48 |
| 3.42 | 6.90 |
| 4.63 | 8.68 |
| 4.70 | 9.47 |
| 5.55 | 10.00 |
| 6.17 | 10.93 |
Dunn, G. and Master, D. (1982) Latency models: the statistical analysis of response times. Psychological Medicine, 12, 659–665.
Of key interest in this study was whether pleasant memories could be recalled more easily and quickly than unpleasant ones. Comparative boxplots for the two samples are shown in Figure 1.9.
Figure 1.9 Comparative boxplots of memory recall times
Show description|Hide description
Figure 1.9
Figure 1.9 Comparative boxplots of memory recall times
Use the boxplots to compare the distributions of recall times for the two types of memory.
Answer
Solution
The most obvious feature is that the recall times for unpleasant memories are on the whole longer than those for pleasant memories – the median, the lower quartile and the upper quartile for the ‘unpleasant’ sample are all above the corresponding values for the ‘pleasant’ sample. The dispersion is also considerably greater for the ‘unpleasant’ sample. (The interquartile range, as shown by the box lengths, is longer, and indeed so is the overall range and the lengths of the ‘whiskers’.)
Both samples are also skew. The pattern of skewness is simpler for the ‘unpleasant’ sample. It is clearly right-skew, with a long tail to the right (high values), as shown by the longer right whisker and also by the fact that the right part of the box (median to upper quartile) is longer than the left part. The ‘pleasant’ sample is not symmetric, but its pattern of skewness is a little more complicated to describe. The upper (right) whisker is longer than the lowe whisker, but the upper part of the box is shorter than the lower part.
Only one potential outlier is picked out in the boxplot – a relatively high value for the ‘pleasant’ sample. In view of the fact that it is actually not all that much higher than the upper adjacent value, perhaps this value should cause us no particular concern.
2 Producing useful tables
In much of your statistical work, you will begin with data set, often presented in the form of a table, and use the information in the table to produce diagrams and/or summary statistics that help in the interpretation of the data set. However, in practice, much interpretation of data sets can be done directly from an appropriate table of data, or by re-presenting the data in a rather different tabular form. Dealing with data in tables is the subject of this section and the next. By the time you have finished you should be able to produce tables which make certain aspects of the data in question more obvious.
2.1 Data sets in different tabular forms
Example 2.1 Lung cancer deaths in South Australia
Table 2.1 contains raw data on the incidence and mortality for lung cancer in South Australia in 1981.
Table 2.1 Age group, male and of population sizes, male and female cases, male and female deaths
| 0–4 | 47589 | 45273 | 0 | 0 | 0 | 0 |
| 5–9 | 53814 | 50672 | 0 | 0 | 0 | 0 |
| 10–14 | 58561 | 55645 | 0 | 0 | 0 | 0 |
| 15–19 | 59408 | 57756 | 0 | 0 | 0 | 0 |
| 20–24 | 58443 | 57249 | 0 | 0 | 0 | 0 |
| 25–29 | 54341 | 53376 | 0 | 0 | 1 | 0 |
| 30–34 | 53456 | 52978 | 1 | 0 | 1 | 0 |
| 35–39 | 42113 | 41988 | 0 | 2 | 0 | 0 |
| 40–44 | 35648 | 35547 | 2 | 5 | 3 | 3 |
| 45–49 | 32911 | 31799 | 8 | 2 | 10 | 2 |
| 50–54 | 36485 | 35333 | 38 | 8 | 26 | 8 |
| 55–59 | 35192 | 35555 | 61 | 18 | 43 | 8 |
| 60–64 | 28131 | 30868 | 67 | 16 | 57 | 15 |
| 65–69 | 24419 | 27390 | 88 | 15 | 69 | 17 |
| 70–74 | 16613 | 21402 | 60 | 21 | 61 | 21 |
| 75–79 | 9958 | 14546 | 46 | 10 | 46 | 9 |
| 80–84 | 4852 | 9749 | 24 | 6 | 23 | 4 |
| 85+ | 2790 | 7477 | 7 | 2 | 8 | 3 |
O'Neill, T. J., Tallis, G. M. and Leppard, P. (1985) The epidemiology of a disease using hazard functions. Australian Journal of Satistics, 27, 283–297.
A table like Table 2.1 may be adequate for someone who is merely taking a quick look at the data, perhaps prior to carrying out an analysis, but it is not the best way of presenting the figures to most readers. The objectives in producing a table that is actually being used to communicate information are to make the data immediately clear, and to facilitate picking out important patterns in them with the minimum of effort. To this end, there are several guidelines for producing tables which should be borne in mind.
Guidelines for tables
Labelling of rows and columns should be clear and unambiguous.
A table should contain the minimum amount of information needed to communicate its message. This may involve splitting the data into several simpler tables or pooling cells.
It may be appropriate to simplify the numbers in a table to aid speedy comprehension.
Useful summary statistics or calculation results should be added, where appropriate, to help communicate the message.
These guidelines will be followed in relation to Table 2.1 to see what changes they suggest.
2.2 Basic table layout
As Table 2.1 stands, it is hard to assimilate the information. Indeed it is not at all clear what any of the numbers mean. Even doing something as simple as giving the columns proper headings and drawing a few lines to separate the headings from the rest of the data, as in Table 2.2, make a big difference to clarity (guideline 1).
Table 2.2 South Australia: incidence and mortality for lung cancer, 1981
| Age group | Population size | New cases | Deaths |
| Male | Female | Male | Female | Male | Female |
| 0–4 | 47589 | 45273 | 0 | 0 | 0 | 0 |
| 5–9 | 53814 | 50672 | 0 | 0 | 0 | 0 |
| 10–14 | 58561 | 55645 | 0 | 0 | 0 | 0 |
| 15–19 | 59408 | 57756 | 0 | 0 | 0 | 0 |
| 20–24 | 58443 | 57249 | 0 | 0 | 0 | 0 |
| 25–29 | 54341 | 53376 | 0 | 0 | 1 | 0 |
| 30–34 | 53456 | 52978 | 1 | 0 | 1 | 0 |
| 35–39 | 42113 | 41988 | 0 | 2 | 0 | 0 |
| 40–44 | 35648 | 35547 | 2 | 5 | 3 | 3 |
| 45–49 | 32911 | 31799 | 8 | 2 | 10 | 2 |
| 50–54 | 36485 | 35333 | 38 | 8 | 26 | 8 |
| 55–59 | 35192 | 35555 | 61 | 18 | 43 | 8 |
| 60–64 | 28131 | 30868 | 67 | 16 | 57 | 15 |
| 65–69 | 24419 | 27390 | 88 | 15 | 69 | 17 |
| 70–74 | 16613 | 21402 | 60 | 21 | 61 | 21 |
| 75–79 | 9958 | 14546 | 46 | 10 | 46 | 9 |
| 80–84 | 4852 | 9749 | 24 | 6 | 23 | 4 |
| 85+ | 2790 | 7477 | 7 | 2 | 8 | 3 |
There is, of course, still an enormous amount of information to absorb, but the labelling is better and, above all, the table is more or less self-explanatory.
But it is important to consider what information we really want the table to convey to the reader. Here there are often choices to be made. Table 2.2 includes data on the population size in different age groups, and these data could be used to investigate the average age of the population, or the way in which the proportions of people in different age groups differ between males and females. If we wanted to convey this particular kind of information, it would make sense to simplify the table in various ways — for instance, all the data about lung cancer cases and deaths could simply be omitted! But, for this particular data set, it is much more likely that we would be interested primarily in the lung cancer cases and deaths, and in that case we would be interested in the population counts only insofar as they are related to the lung cancer counts. In that case, there is an immediate and obvious simplification to be made. There were no lung cancer cases or deaths in people aged up to 24, so we can simply pool together the first five rows of the table as in Table 2.3.
Table 2.3 South Australia: incidence and mortality for lung cancer, 1981
| Age group | Population size | New cases | Deaths |
| Male | Female | Male | Female | Male | Female |
| 0–24 | 277815 | 266595 | 0 | 0 | 0 | 0 |
| 25–29 | 54341 | 53376 | 0 | 0 | 1 | 0 |
| 30–34 | 53456 | 52978 | 1 | 0 | 1 | 0 |
| 35–39 | 42113 | 41988 | 0 | 2 | 0 | 0 |
| 40–44 | 35648 | 35547 | 2 | 5 | 3 | 3 |
| 45–49 | 32911 | 31799 | 8 | 2 | 10 | 2 |
| 50–54 | 36485 | 35333 | 38 | 8 | 26 | 8 |
| 55–59 | 35192 | 35555 | 61 | 18 | 43 | 8 |
| 60–64 | 28131 | 30868 | 67 | 16 | 57 | 15 |
| 65–69 | 24419 | 27390 | 88 | 15 | 69 | 17 |
| 70–74 | 16613 | 21402 | 60 | 21 | 61 | 21 |
| 75–79 | 9958 | 14546 | 46 | 10 | 46 | 9 |
| 80–84 | 4852 | 9749 | 24 | 6 | 23 | 4 |
| 85+ | 2790 | 7477 | 7 | 2 | 8 | 3 |
This simplification, in line with guideline 2, has not lost any information about lung cancer at all, and the table is now easier to comprehend.
2.3 Table activity
Table 2.4 South Australia: incidence and mortality for lung cancer, 1981
| Age group | Population size | New cases | Deaths |
| Male | Female | Male | Female | Male | Female |
| 0–39 | 427725 | 414937 | 1 | 2 | 2 | 0 |
| 40–44 | 35648 | 35547 | 2 | 5 | 3 | 3 |
| 45–49 | 32911 | 31799 | 8 | 2 | 10 | 2 |
| 50–54 | 36485 | 35333 | 38 | 8 | 26 | 8 |
| 55–59 | 35192 | 35555 | 61 | 18 | 43 | 8 |
| 60–64 | 28131 | 30868 | 67 | 16 | 57 | 15 |
| 65–69 | 24419 | 27390 | 88 | 15 | 69 | 17 |
| 70–74 | 16613 | 21402 | 60 | 21 | 61 | 21 |
| 75–79 | 9958 | 14546 | 46 | 10 | 46 | 9 |
| 80–84 | 4852 | 9749 | 24 | 6 | 23 | 4 |
| 85+ | 2790 | 7477 | 7 | 2 | 8 | 3 |
Simplifying the table further
Do you think it would make sense to continue this process of simplification by pooling more rows? If so, which rows would you pool?
Discussion
Comment
Since there were new cases, or deaths, and indeed usually both, in all the other age groups, the pooling of rows cannot be continued further without losing some information that was in the original table. But, in fact, there are very few cases in either gender group under the age of 40. So, if the corresponding rows are pooled, to give Table 2.4, very little information is lost (and, arguably, nothing at all important in relation to lung cancer). (You might have suggested a slightly different set of rows to pool.)
2.4 Including the results of useful calculation
Can Table 2.4 be simplified further by pooling more rows or columns? Perhaps it might be, but there may well be a risk of losing some important or relevant information. So, before considering any further simplification, we shall look at adding information to the table, in the form of the results of some helpful calculations (guideline 4).
On their own, some of the numbers in the table still do not mean a great deal. There were 61 new cases among males in the 55–59 age group. But how does this compare with males in other age groups, and with females? There were 60 new cases for males aged 70–74. On the face of it this looks very close to the figure for the 55–59 group. But there were far more males in the South Australian population aged 55–59 than there were aged 70–74 (35192 compared to 16613). It seems likely that the main interest in these data is in the varying chances of developing lung cancer or dying from it, at different ages and for the two genders. To find out something about this, it is useful to calculate the proportions of the different age groups that became new cases of lung cancer. For males aged 55–59, the proportion is 61/35192=0.0017333, or 0.17333% as a percentage. For males aged 70–74 the corresponding proportion is 60/16613=0.0036116, or 0.36116%. It is very common, and often very useful, to calculate such quantities, which are often known as rates.
For the time being, we shall just look at the new cases and omit the information on deaths. The rate for new cases in each age group has been calculated for males and for females; these rates are included in Table 2.5. As you can see, these numbers do not look particularly user-friendly!
Table 2.5 South Australia: incidence for lung cancer, 1981
| Age group | Population size | New cases | New cases as % of population size |
| Male | Female | Male | Female | Male | Female |
| 0–39 | 427725 | 414937 | 1 | 2 | 0.0023380 | 0.0048200 |
| 40–44 | 35648 | 35547 | 2 | 5 | 0.056104 | 0.014066 |
| 45–49 | 32911 | 31799 | 8 | 2 | 0.024308 | 0.062895 |
| 50–54 | 36485 | 35333 | 38 | 8 | 0.10415 | 0.022642 |
| 55–59 | 35192 | 35555 | 61 | 18 | 0.17333 | 0.050626 |
| 60–64 | 28131 | 30868 | 67 | 16 | 0.23817 | 0.051834 |
| 65–69 | 24419 | 27390 | 88 | 15 | 0.36038 | 0.054765 |
| 70–74 | 16613 | 21402 | 60 | 21 | 0.36116 | 0.098122 |
| 75–79 | 9958 | 14546 | 46 | 10 | 0.46194 | 0.068747 |
| 80–84 | 4852 | 9749 | 24 | 6 | 0.49464 | 0.061545 |
| 85+ | 2790 | 7477 | 7 | 2 | 0.25090 | 0.026749 |
The table still looks pretty horrible and the information it contains is difficult to assimilate, largely because there is too much clutter from information of dubious relevance, and also because far too many decimal places are included in the last two columns. The latter problem is easily solved, in accord with guideline 3. First, note that (for example) the figure of 0.098122% for females aged 70–74 means that, for every 100 women in this age group (in South Australia in 1981), there were 0.098122 new cases of lung cancer. In this context there is nothing special about calculating the rate per 100 women in the population. Instead, the number of cases per 100 000 women in the population will be calculated. This has the effect of multiplying all the rates by 1000, which gets rid of most of the occurrences of ‘0.0…’ at the start of the numbers, and hence makes the table easier to read. Also, simply to get across the main message of these data does not require five significant figures. Instead, in Table 2.6, the figures are given to one decimal place.
Table 2.6 South Australia: incidence for lung cancer, 1981
| Age group | Population size | New cases | Newcases per 100 000 population |
| Male | Female | Male | Female | Male | Female |
| 0–39 | 427725 | 414937 | 1 | 2 | 0.2 | 0.5 |
| 40–44 | 35648 | 35547 | 2 | 5 | 5.6 | 14.1 |
| 45–49 | 32911 | 31799 | 8 | 2 | 24.3 | 6.3 |
| 50–54 | 36485 | 35333 | 38 | 8 | 104.2 | 22.6 |
| 55–59 | 35192 | 35555 | 61 | 18 | 173.3 | 50.6 |
| 60–64 | 28131 | 30868 | 67 | 16 | 238.2 | 51.8 |
| 65–69 | 24419 | 27390 | 88 | 15 | 360.4 | 54.8 |
| 70–74 | 16613 | 21402 | 60 | 21 | 361.2 | 98.1 |
| 75–79 | 9958 | 14546 | 46 | 10 | 461.9 | 68.7 |
| 80–84 | 4852 | 9749 | 24 | 6 | 494.6 | 61.5 |
| 85+ | 2790 | 7477 | 7 | 2 | 250.9 | 26.7 |
Now does it make sense to simplify the table any further? If we want to use it to communicate information about the relative chances of being diagnosed as a new case of lung cancer at different ages and for the two genders, the ‘Population size’ and ‘New cases’ columns do not actually give very relevant information. It might therefore be reasonable to omit them. Furthermore, the general pattern of the new case rates at different ages can be communicated with rather fewer age groups than were used in Table 2.6. Table 2.7 uses fewer and coarser age groupings, and the only figures given are the calculated values of the new cases per 100 000 and deaths per 100 000; these have been rounded to one decimal place. (Note that the figures for new cases in Table 2.7 cannot be calculated simply from the rates given in the last two columns of Table 2.6. The appropriate population sizes and counts of cases must be aggregated and the aggregates used to calculate the rates.)
Table 2.7 South Australia: incidence and mortality for lung cancer, 1981 (rates per 100,000 population)
| Age group | New cases | Deaths |
| Male | Female | Male | Female |
| 0–49 | 2.2 | 1.9 | 3.0 | 1.0 |
| 50–59 | 138.1 | 36.7 | 96.3 | 22.6 |
| 60–69 | 295.0 | 53.2 | 239.8 | 54.9 |
| 70–79 | 398.9 | 86.2 | 402.7 | 83.5 |
| 80+ | 405.7 | 46.4 | 405.7 | 40.6 |
(Whole numbers in the deaths column would arguably have been quite adequate to get across the message of these data. Using one decimal place has the advantage of making it clear that these are rates, and not counts of individual cases.)
This is a quickly assimilated table that communicates the pattern of incidence and death from lung cancer, in relation to population size. It is easy to compare the figures for males and females, and it is equally easy to compare incidence with mortality in any of the age groups.
Activity 4 Describing data in a table
(a) Describe the main patterns in the data on lung cancer in South Australia, on the basis of Table 2.7.
(b) Table 2.7 is certainly much simpler than the earlier tables in this section, and you would probably agree that the patterns in the data are easier to see. But can you think of any disadvantages of the presentation in Table 2.7 compared to the other tables?
Answer
Solution
(a) The pattern of incidence of lung cancer for males in South Australia may be described as follows. There are very few new cases in men aged under 50 years, but the rate rises rapidly for men in their 50s and 60s. The increase levels off above age 70. The pattern of mortality for males is very similar to that for incidence. For females, both incidence and mortality are again very low below 50 years of age and increase after that, but the incidence and mortality rates remain much lower than for men (about one quarter or one fifth of the level for men). Also the incidence and mortality rates for women reduce quite considerably in the oldest age groups.
(b) One problem is that the information on how many people were involved has been entirely removed. One pattern that was noted in part (a) is the fall in incidence and mortality rates for women aged over 80. However, we cannot tell from Table 2.7 that there were actually only 8 new cases and 7 deaths in women in these age groups. With numbers of cases this small, a few extra cases in one year, such as we might expect just on the basis of random variability, would show up as a large rise in the incidence rate. Without knowing something about the numbers from which the rates in Table 2.7 were calculated, it is not possible to take this into account. Thus, for example, in writing report about these matters, it would be good statistical practice to include the counts of cases and deaths somewhere, even if not in the same table as that including the rates.
Do you agree that Table 2.7 conforms to all of the four guidelines given at the beginning of this section? After you have produced a table for yourself, it is always a good idea to check it carefully against each of the four guidelines.
2.5 Early retirement from the National Health Service
Example 2.2: Early retirement from the National Health Service
A study was carried out to investigate various aspects of early retirement from the British National Health Service (NHS). In 1998–99, 5469 NHS employees from England and Wales were granted early retirement because of ill health. The researchers examined the records of a sample of 1994 of these people. Table 2.8 gives data on these people, classifying each of them by occupational group and by a broad classification of the health reason for which they retired.
Table 2.8 Retirements from the NHS because of ill health, 1998–99
| Occupational group | Reason for retiring because of ill health |
| Musculoskeletal | Cardiovascular | Psychiatric | Other | Total |
| Ambulance workers | 65 | 12 | 6 | 12 | 95 |
| Healthcare assistants or support | 339 | 61 | 77 | 117 | 594 |
| Nurses or midwives | 364 | 144 | 70 | 153 | 731 |
| Technical or professional staff | 42 | 25 | 4 | 23 | 94 |
| Administration or estates staff | 118 | 94 | 31 | 66 | 309 |
| Doctors or surgeons | 33 | 40 | 20 | 28 | 121 |
| Other | 22 | 13 | 7 | 8 | 50 |
| Total | 983 | 389 | 215 | 407 | 1994 |
This table is adapted from Pattani, S., Constantinovici, N., and Williams, S. (2001) Who retires early from the NHS because of ill health and what does it cost? A national cross sectional study. British Medical Journal, 322, 208–209.
Activity 5 Early retirement from the National Health Service
Suppose that the main interest of the researchers was to see whether (and, if so, how) the pattern of causes of retirement differed between occupational groups. How does the table, as it stands, match up to the guidelines given at the start of this section?
Answer
Solution
The labelling of rows and columns is reasonably clear as it stands (guideline 1). Assuming that the researchers were interested separately in all these occupational groups and all these reasons for retirement, there seem to be no good reasons for breaking up the table or combining cells (guideline 2). The numbers in the table are counts and not particularly large ones (three digits at most, apart from the overall total) and there seems no reason to simplify them (guideline 3).
However, it might help to include some calculation results (guideline 4). As the table stands, it is reasonably easy to see that (for instance) in each occupational group, the greatest number of retirements was due to musculoskeletal reasons, but it is not easy to compare just how much bigger that count is relative to the others in each occupational group, because the total number of retirements differs considerably from one occupational group to another. This sort of comparison would be more straightforward if we knew, for instance, the proportion or percentage of people in each occupational group who retired for musculoskeletal reasons.
Activity 6 Early retirement from the National Health Service: percentages
(a) For each occupational group, calculate the percentage of people who retired because of each cause of ill health. Use these percentages to comment on the different patterns of causes of retirement in the different occupational groups.
(b) Assuming you found the percentages useful for making these comparisons, say whether you think that a table presenting this information should include only the counts (as in Table 2.8), only the percentages that you calculated, or both.
Answer
Solution
(a) The percentage of people in each occupational group who retired because of each cause of ill health is given in Table 2.9.
Table 2.9 Retirements from the NHS because of ill health, 1998–99
| Occupational group | Reason for retiring because of ill health (% of row total) |
| Musculoskeletal | Psychiatric | Cardiovascular | Other |
| Ambulance workers | 68 | 13 | 6 | 13 |
| Healthcare assistants or support | 57 | 10 | 13 | 20 |
| Nurses or midwives | 50 | 20 | 10 | 21 |
| Technical or professional staff | 45 | 27 | 4 | 2 |
| Administration or estates staff | 38 | 30 | 10 | 21 |
| Doctors or surgeons | 27 | 33 | 17 | 23 |
| Other | 44 | 26 | 14 | 16 |
| Total | 49 | 20 | 11 | 20 |
For instance, the proportion of the ambulance workers who retired because of ill health for musculoskeletal causes is 65/95=0.68421, or 68.421%. However, there is no need to include three decimal places to portray the patterns in the data clearly. Whole percentages are sufficiently accurate; so this percentage has been entered in the table as 68%. The other percentages were calculated in a similar way.
(Note that, because of the rounding of the percentages, the sums of some of the rows are 99% or 101% rather than 100%. In the context of communicating the general pattern of the data, this does not matter.)
Perhaps the most obvious difference between the occupational groups is that the percentage of retirements for musculoskeletal causes was considerably greater in the first two groups than in some of the others, particularly administrators and doctors. The authors of the paper from which these data are taken attribute this difference to the greater amount of manual work done by workers in the first two categories. The occupational groups with relatively low levels of retirement for musculoskeletal causes also had relatively high percentages of retirements for psychiatric causes. Without further investigation, and in particular without having looked at what proportion of workers in each of these groups actually retired on grounds of ill health (rather than continuing to work), it is difficult to say more about the reasons for these patterns.
(b) The question of whether to include the percentages in a table as well as, or instead of, the counts does not have a clear-cut answer. The table given in the paper from which these data were obtained includes both. This makes the table rather complicated, and the patterns of different causes of retirement is not entirely clear at a glance. However, in interpreting the data it is important to know that the number of ill-health retirements in some of these groups was not particularly large. A useful compromise would have been to include the total number of retirements from which the percentages in each row were calculated, as in Table 2.10. In general, when calculating row percentages (or column percentages) in a table in this way, it is good practice to include the totals that were used to calculate the percentages as well.
Table 2.10 Retirements from the NHS because of ill health, 1998–99
| Occupational group | Reason for retiring because of ill health (%of row total) |
| Musculoskeletal | Psychiatric | Cardiovascular | Other | Total (=100%) |
| Ambulance workers | 68 | 13 | 6 | 13 | 95 |
| Healthcare assistants or support | 57 | 10 | 13 | 20 | 594 |
| Nurses or midwives | 50 | 20 | 10 | 21 | 731 |
| Technical or professional staff | 45 | 27 | 4 | 24 | 9 |
| Administration or estates staff | 38 | 30 | 10 | 21 | 309 |
| Doctors or surgeons | 27 | 33 | 17 | 23 | 121 |
| Other | 44 | 26 | 14 | 16 | 50 |
| Total | 49 | 20 | 11 | 20 | 1994 |
2.6 Summary
In this section you have been introduced to some guidelines for presenting data in tables. These guidelines apply particularly when the data in a table are being used to illustrate a particular point or to show up clearly a particular pattern.
You have seen that, in some circumstances, following the second of these guidelines leads to some pooling together of rows. (In other cases, it could be columns or individual cells that are pooled.) However, care is needed when, by making such simplifications, information is lost from the table. The examples of the application of guideline 4 that you have seen involve calculating appropriate ratios or rates. Such calculations are very common in dealing with data in tabular form.
3 Interpreting data in table
In Section 2, the main concern was with producing a table of data, for others to read, that communicates clearly the important patterns or messages in the data. In this section, the focus changes slightly. Your role will be that of the reader or user of the data in a table, and you will learn about approaches that make it easier for you to extract information from a table. However, manipulating tabular data into a form that makes it clearer to others will also, very often, make it clearer to you as well. So the approaches introduced in Section 2 will be useful in this section too. You will have more practice in choosing and calculating appropriate ratios and rates for tabular data. You will also see examples where it is appropriate to go one step further than you did in Section 2: rather than leaving the data in tabular form, relevant graphs will be drawn as well.
3.1 Health personnel in Thailand
There are practically no new theories or new principles in this section. We shall work through some examples, and you will see how basic techniques and approaches that you have already learned can be combined to allow you to use tabular data efficiently.
Example 3.1 Health personnel in Thailand
The data shown in Table 3.1 are taken from the Thailand Mini Health Profile 1988, published by the Ministry of Public Health, Bangkok. They show the numbers of health care personnel at approximately five-year intervals.
Table 3.1 Health personnel in Thailand, 1966–1984
| Category | 1966 | 1971 | 1976 | 1981 | 1984 |
| Physicians | 3609 | 4092 | 5210 | 6931 | 8058 |
| Dentists | 253 | 532 | 600 | 1057 | 1326 |
| Pharmacists | 940 | 1586 | 1757 | 2680 | 3312 |
| Nurses | 6876 | 9760 | 13700 | 19599 | 31827 |
| Midwives | 2834 | 4989 | 7304 | 8577 | 8573 |
| Total | 14512 | 20959 | 28571 | 38844 | 53096 |
What do these data tell us about the change in health care personnel in Thailand over the period in question, and how can we work with the data in the table to make any pattern clearer?
First, notice that some features of the data are obvious. The total number of health care personnel increased hugely between 1966 and 1984, from under 15 000 to about 53 000. Also, throughout the period, the biggest category of staff was that of Nurses, and this category seems to have grown more rapidly than some of the others. (In 1966, there were very roughly twice as many nurses as there were doctors, for instance, but in 1984 there were almost four times as many nurses as there were doctors.)
How could the patterns that have already been identified be made clearer? The pattern of overall increase in numbers is already clear from the last row of the table. Perhaps it could be made even clearer by drawing an appropriate graph; we shall return to this idea later.
In Activity 6, the patterns of retirement reasons in different occupational groups were made easier to see by calculating how large each entry was as a percentage of the corresponding row total. In Activity 7, you are asked to consider whether similar approach would help here.
3.2 Health care personnel in Thailand: activities
Activity 7 Health care personnel in Thailand: calculating percentages
Would it be helpful, in considering possible changes in the way health care personnel are divided into the five categories listed, to recalculate the numbers in the body of Table 3.1 as percentages either of the row totals or of the column totals? If you think it would be helpful, calculate the appropriate percentages and use the resulting table to comment on the data.
Answer
Recalculating the numbers in the body of the table as percentages of the totals of the columns would show clearly how the total number of health personnel in each year is divided between the five categories. It would not be very meaningful to calculate the percentages of the row totals. The results of calculating the numbers as percentages of the column totals are shown in Table 3.2.
Table 3.2 Health personnel in Thailand, 1966–1984 (percentages of column totals)
| Category | 1966 | 1971 | 1976 | 1981 | 1984 |
| Physicians | 25 | 20 | 18 | 18 | 15 |
| Dentists | 2 | 3 | 2 | 3 | 2 |
| Pharmacists | 6 | 8 | 6 | 7 | 6 |
| Nurses | 47 | 47 | 48 | 50 | 60 |
| Midwives | 20 | 24 | 26 | 22 | 16 |
| Total (= 100%) | 14512 | 20959 | 28571 | 38844 | 53096 |
These percentages show that the numbers for dentists and for pharmacists as percentages of the total health care personnel changed very little over the period covered by the table. However, the percentage of physicians fell reasonably steadily from 25% of the total in 1966 to 15% in 1984. The percentage of nurses grew slowly from 1966 to 1981, but then increased rapidly between 1981 and 1984. Finally, the percentage of midwives increased over the first ten years of the period covered, but then fell back to a level below the 1966 level. Of course, it should be borne in mind throughout that the total number of health care staff grew considerably over the period.
Your investigation in Activity 7 clarified the patterns in the original table; but it remains the case that the single most prominent feature of the table is the rise in total health care personnel over the period covered. However, it may well have occurred to you that the population served by these health care personnel also changed over the period in question. Thailand is, after all, a developing country that may well have experienced considerable population growth between 1966 and 1984.
In fact, estimates of the total population of Thailand in the years covered by Table 3.1 are also provided in the source from which that table was taken. They are given in Table 3.3.
Table 3.3 Estimated total population of Thailand, 1966–1984
| 1966 | 1971 | 1976 | 1981 | 1984 |
| Population (millions) | 31.1 | 35.4 | 40.3 | 44.9 | 50.7 |
These figures can be used to calculate the numbers of the different categories of health care personnel as a proportion of the total population. These proportions could, in principle at any rate, be shown as percentages, or as numbers per 100 000 population as in Example 2.1 (see especially Tables 2.6 and 2.7); but in this case they are clearer if shown as numbers per million population. The resulting proportions are given in Table 3.4. After reading through the table you should check that you understand how the numbers displayed were calculated.
Table 3.4 Health personnel per million population in Thailand, 1966–1984
| Category | 1966 | 1971 | 1976 | 1981 | 1984 |
| Physicians | 116.0 | 115.6 | 129.3 | 154.4 | 158.9 |
| Dentists | 8.1 | 15.0 | 14.9 | 23.5 | 26.2 |
| Pharmacists | 30.2 | 44.8 | 43.6 | 59.7 | 65.3 |
| Nurses | 221.1 | 275.7 | 340.0 | 436.5 | 627.8 |
| Midwives | 91.1 | 140.9 | 181.2 | 191.0 | 169.1 |
| Total | 466.6 | 592.1 | 709.0 | 865.1 | 1047.3 |
These calculations show clearly that, in relation to the size of the total population, the total numbers of health care personnel in Thailand rose considerably and steadily between 1966 and 1984. Putting it another way, the population rose over this period, but the numbers of health care personnel rose much faster. It would be reasonably straightforward to comment on the changes in the different categories in relation to total population on the basis of the numbers in Table 3.4, but this task can be made easier by drawing an appropriate graph.
Data like those in Table 3.4, where there is a value for each of a number of different times, are referred to as time series. (In this case, there are actually six different time series, one for each personnel category, plus one for the data on total health personnel.) A useful kind of graph for showing time series data is line plot. This is a scatterplot, with the times (years, in this case) along the horizontal axis and the actual data values of the time series along the vertical axis. It is conventional, for time series data, to join the resulting plots with straight line segments. This draws attention to the rate of change of the values in the series. For time series where there are large numbers of different time points, the symbols (for example, crosses or dots) representing the data points are often omitted for clarity. However, in these data there are only five time points so it is not necessary to do that. We could produce such line plots separately for each of the different categories, and indeed for the total. But it is often easier to compare the levels of different time series by plotting them all on the same graph. Figure 3.1 shows such a graph, with a set of points and a line for each of the categories. (The series for total personnel is omitted. Since the totals are necessarily bigger than the figures for the individual groups, including them on the same graph would squeeze up all the other lines towards the bottom of the diagram and make them hard to see.)
Figure 3.1 Health care personnel per million population in Thailand
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Figure 3.1
Figure 3.1 Health care personnel per million population in Thailand
Activity 8 Health care personnel: interpreting line graphs
Comment on the main changes in health care personnel per million population in Thailand over the period 1966–1984, on the basis of Figure 3.1. Are there any important patterns in the data in Table 3.4 that the graph does not make clear?
Answer
It is clear from Figure 3.1 that the numbers for personnel in each category, in relation to the total population of the country, rose quite markedly over the period in question. The rise was very marked for nurses (who were the largest category throughout the period), particularly towards the end of the period covered, between 1981 and 1984. There are some exceptions to this overall pattern of increase, the most prominent of which is that the number of midwives per million population actually fell by considerable amount between 1981 and 1984.
It is very clear from the graph that the numbers for dentists and for pharmacists per million population are considerably less than the corresponding numbers for the other categories; and it is reasonably clear that the numbers for both these groups rose over the period. However, because the numbers for these groups are so much smaller than, say, the numbers for nurses, the lines for dentists and for pharmacists are rather squashed up at the bottom of the graph and it is therefore difficult to judge the size of the increase. It is thus much clearer from the table than from the graph that the number of dentists has more than tripled in relation to the population size over the period, and that the number of pharmacists has more than doubled. It is also not very clear from the graph that the numbers for dentists and for pharmacists fell slightly in relation to the population between 1971 and 1976.
Activity 9 Health care personnel: more on proportions
Your work in Activity 7, on the proportions of health care personnel in the different categories, did not take account of the total population of Thailand. Explain why, if you calculated the figures in Table 3.4 as percentages of the column totals, you would get exactly the same table of percentages as that in the solution to Activity 7. If you wanted to represent these percentages graphically, what kind of graph would you draw?
Answer
The figures in any particular column of Table 3.4, including the column totals, are equal to the figures in the corresponding column of Table 3.1 divided by the population estimate from the corresponding column of Table 3.3. Thus in carrying out a division to calculate a percentage from Table 3.4, the quantities in the numerator and in the denominator were calculated from Table 3.1 by dividing by the same population total. So the population total cancels out and the result of the division is identical to that obtained from Table 3.1.
One appropriate kind of graph that would draw particular attention to the proportions is the pie chart. We could draw separate pie charts for each of the years listed in the table. The appropriate pie charts for the 1966 and 1984 data are shown in Figures 3.2 and 3.3 respectively. (You were not asked to draw these in the activity!)
Figure 3.2 Pie chart of health care personnel in Thailand in 1966
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Figure 3.2
Figure 3.2 Pie chart of health care personnel in Thailand in 1966
Figure 3.3 Pie chart of health care personnel in Thailand in 1984
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Figure 3.3
Figure 3.3 Pie chart of health care personnel in Thailand in 1984
These pie charts draw particular attention to the large increase in the number of nurses as a proportion of total health care personnel in Thailand between 1966 and 1984.
In Example 3.1, the original data were counts of individuals. It proved useful to calculate appropriate percentages, by dividing the numbers in the body of the table by column totals. In Activity 2.4, you calculated similar percentages, but they were based on row totals. In Example 3.1, it would have made very little sense to calculate percentages of the row totals. However, in Activity 6, it would have made sense to calculate the percentages of column totals instead of row totals, but they would have provided information relevant to a different question from the one you were considering. For tables of counts in general, it is very often useful to calculate percentages of row totals, and/or of column totals, but it is important to think carefully about which set of percentages is informative in relation to the question you are interested in. The following example is intended to make this clear.
3.3 HIV testing in sub-Saharan Africa
Example 3.2 HIV testing in sub-Saharan Africa
In developed countries, the standard method for testing whether a person is infected with the virus HIV, that causes AIDS, is to carry out a blood test. Provided such a test is carried out long enough after the initial infection occurred, the accuracy is high. However, in sub-Saharan Africa, where in most countries the incidence of AIDS is much higher than in the developed world, blood testing can be difficult to conduct and it is expensive in terms of the health resources available. Thus there is interest in whether HIV infection can be diagnosed at all reliably on the basis of clinical features that are more easily measured or observed under the local circumstances. People infected with HIV often have enlarged lymph nodes, and these enlarged nodes can be felt from the outside of the body in a simple physical examination. However, there are very many other reasons as well as HIV infection for an individual to have enlarged lymph nodes. (In countries like the UK, the overwhelming majority of enlarged lymph nodes are caused by infections that have nothing whatever to do with HIV or AIDS.)
To investigate whether enlarged lymph nodes could play a role in testing for HIV in an African context, researchers in Zimbabwe investigated all adult patients admitted to an acute medical ward in a Harare hospital over a three-month period (apart from one patient who did not agree to take part). Each patient was tested for HIV using a standard (and an accurate) blood-testing method. (In fact, 56% of the patients turned out to have an HIV infection according to these tests.) In addition, the patients were examined (by feeling) for enlarged lymph nodes in three areas of their body. The data in Table 3.4 provide information on the numbers of patients who had an epitrochlear lymph node (a node in the upper arm near the elbow) swollen to a size larger than 1 cm.
Table 3.4 Possible indicators of HIV in Harare hospital patients
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Table 3.4
Table 3.4 Possible indicators of HIV in Harare hospital patients
Activity 10 HIV testing: calculating proportions
(a) Suppose you were a doctor practising in a hospital in sub-Saharan Africa, in a place where the general characteristics of patients that you see is likely to be reasonably similar to those reported on in Table 3.4. You decide to investigate for enlarged epitrochlear lymph nodes in the patients you see and, depending on whether you find such a node enlarged to over 1 cm, to use this information together with that in Table 3.4 to take a view on how likely it is that the patient has HIV. Would it help you more to recalculate the numbers in the main body of Table 3.4 as percentages of the row totals or the column totals? Calculate the set of percentages you believe to be more appropriate. In the light of your results, what would be your view on how likely it is that patient with an epitrochlear node over 1 cm actually has HIV (according to blood test). What about a patient who does not have such a node?
(b) Suppose now that you are a scientific researcher interested in the physiological mechanism by which HIV infection can cause enlarged lymph nodes. In investigating this topic, would it help you more to recalculate the numbers in the main body of Table 3.4 as percentages of the row totals or the column totals? Calculate the set of percentages you believe to be more appropriate and comment on what you find.
Answer
(a) In this situation, suppose one of your patients has an epitrochlear lymph node enlarged to more than 1 cm. You might then want to ask ‘What proportion of such patients actually have HIV (as would show up if they were given a standard blood test)?’ That is, you know your patient corresponds to those in the first row of Table 3.4. If your patients are generally comparable to those in the Harare hospital, the proportion can be estimated by calculating the number in the first row of Table 3.4 as a percentage of the row total. For this reason, calculating the figures in the table as percentages of the row totals is helpful, and the percentages of the column totals are less relevant. Calculating all the figures as percentages of the row totals leads to Table 3.5. (The percentages have been rounded to whole numbers.)
Table 3.5: Percentages of row totals for Harare hospital patients
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Table 3.5: Percentages of row totals for Harare hospital patients
Table 3.5: Percentages of row totals for Harare hospital patients
This table shows that 83% of the patients who had an enlarged epitrochlear lymph node of size over 1cm were actually infected with HIV (according to the blood test). In this context, it is quite likely that a patient with such an enlarged node will have HIV. (But bear in mind that these figures will not apply in other contexts. As Table 3.5 again makes clear (in the bottom line), 56% of all the patients in the study were HIV positive on the blood test. In a hospital in, say, Europe, this will not be the case, and it is much more likely that an enlarged epitrochlear lymph node would have some other explanation.)
The table also shows that 52% of the patients who do not have an enlarged epitrochlear lymph node of this size were not HIV positive. In the Harare context, not having this particular physical sign does not actually say much about the patient's chances of being free from HIV. The chance is rather more than would be the case for patients in general, but not much more.
(b) In this case the scientist will probably be more interested to ask ‘Of the patients who actually are HIV positive on the blood test, what proportion show an enlarged epitrochlear lymph node of this size?’. That is, the scientist would be interested in knowing how large the numbers in the table were in relation to the column totals rather than the row totals. The resulting percentages are shown in Table 3.6.
Table 3.6: Percentages of column totals for Harare hospital patients
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Table 3.6
Table 3.6: Percentages of column totals for Harare hospital patients
This table shows that, of the Harare patients who are HIV positive (on the blood test), 36% have this type of enlarged lymph node (so most of them do not have it). However, the fact that the presence of an enlarged lymph node of this type does have some diagnostic value is shown by the fact that a much smaller proportion, just 10%, of HIV negative patients show this particular sign.
It is worth noting that the percentages in Table 3.5 are considerably different from those in Table 3.6. Thus if you were interested in the scientist's question but calculated, mistakenly, the percentages relevant to the doctor (that is, the row percentages), you could be seriously misled.
For the data in Table 3.4, both the row percentages and the column percentages turned out to be useful quantities to calculate; but which is more useful depends on the question you are trying to answer. It is crucial not to use the row percentages to answer questions that relate to the column percentages, or vice versa. As you saw in Activity 10, the two sets of percentages can be considerably different. In general, in making calculations from data in tables, it is always important to think through carefully exactly what you want to know.
Before the final activity, some threads relating to graphics will be drawn together. You have now met several different graphics: pie charts, bar charts, histograms, scatterplots, boxplots and line plots of time series. You have seen that different types of plot are suitable for different types of data. You have also seen that the choice of an appropriate graph for presenting and examining data can depend on the question of interest. The following guidelines draw together what you have learned so far.
3.4 Guidelines for graphics
Data in the form of counts of individual entities (for example, people, animals, power stations) in a small set of discrete categories can be presented in bar charts or pie charts. For most purposes, bar charts are preferable. Pie charts draw particular attention to the proportions in which the entities are split between the different categories. However, they do so by representing the proportions by angles, and even when the main interest lies in the proportions, bar charts may well be easier to ‘read’.
For data in the form of category counts where the interest lies in comparing two or more data sets (as in Example 2.3 of Unit A1 ), it can be useful to produce a bar chart where the corresponding bars for the different data sets are plotted side by side.
To examine the pattern of distribution of values of a continuous (measured) variable, a histogram is an appropriate graphic. An alternative is a boxplot. However, neither sort of plot gives direct information about the number of observations in the data set, and it can be risky to draw firm conclusions about the pattern of distribution when the number of observations is small.
Boxplots give a simple representation of the values of a continuous variable. They can also be used for discrete variables where the categories are numbers (such as the counts of family sizes in Figure 1.7. A boxplot shows less detail about a distribution than an appropriate histogram or bar chart, but the amount of detail in a boxplot is often sufficient.
Boxplots are particularly useful for comparing two or more data sets, because the corresponding boxplots can be drawn against a common scale on the same diagram. (It is difficult to do this clearly with more than one histogram.)
Scatterplots are used when the values of two numerical variables have been obtained from each of a number of individual entities. The aim of the plot is to investigate the relationship between the values of the two variables.
Line plots can be useful, particularly for time series data, because they draw attention to the way that one or more variables have changed over time.
In some circumstances, particularly when the data are very skew, more informative boxplot or scatterplot can be produced by transforming the data first.
This section ends with a more substantial activity. This will provide you with some further experience of dealing with data in tables. You will also be asked to look at one or two slightly different approaches for interpreting tabular data.
3.5 The British Crime Survey
Table 3.7 Comparison of British Crime Survey and crimes recorded by the police
| 1997 Police | 1997 BCS | % BCS reported | % recorded of reported | % recorded of all BCS | % change 1995 to 1997 | % change 1981 to 1997 |
| Police | BCS | Police | BCS |
| Vandalism | 443 | 2917 | 26 | 58 | 15 | −4 | −15 | 121 | 7 |
| All comparable property theft (acquisitive crime) | 1751 | 6261 | 50 | 56 | 28 | −17 | −15 | 51 | 99 |
| Burglary | 519 | 1639 | 64 | 49 | 32 | −19 | −7 | 48 | 119 |
| Attempts & no loss | 140 | 976 | 50 | 29 | 14 | −17 | −0.1 | 90 | 160 |
| With loss | 379 | 664 | 85 | 67 | 57 | −20 | −15 | 37 | 77 |
| All vehicle thefts | 1022 | 3483 | 47 | 62 | 29 | −15 | −19 | 57 | 99 |
| Theft from vehicle | 552 | 2164 | 43 | 59 | 25 | −16 | −14 | 63 | 68 |
| Theft of vehicle | 316 | 375 | 97 | 87 | 84 | −21 | −25 | 10 | 31 |
| Attempted thefts | 154 | 943 | 37 | 44 | 16 | 3 | −27 | 447 | 425 |
| Bicycle theft | 151 | 549 | 64 | 43 | 27 | −18 | −17 | 19 | 154 |
| Theft from the person | 60 | 590 | 35 | 29 | 10 | −4 | −12 | 71 | 36 |
| All comparable violence | 256 | 1022 | 49 | 51 | 25 | 11 | −13 | 150 | 53 |
| Wounding | 205 | 714 | 45 | 63 | 29 | 18 | −17 | 143 | 41 |
| Robbery | 52 | 307 | 57 | 30 | 17 | −11 | −2 | 183 | 89 |
| All comparable | 2450 | 10199 | 44 | 54 | 24 | −12 | −15 | 67 | 56 |
Mirrlees-Black, C, Budd, T., Partridge, S. and Mayhew, P. (1998) The 1998 British Crime Survey Englnd Wales, Home Office Statistical Bulletin 21/98
Activity 11 The British Crime Survey
The British Crime Survey (BCS) is a sample survey carried out in England and Wales by the Home Office. The survey was first carried out in 1982, and at the time of writing (2001) is done every two years. The aim is to measure the level of crimes against people in private households. Data are collected by interviewing adult respondents from a representative sample of households about their experience as victims of crime in the previous year, and about some other matters connected with crime. For the 1998 BCS, a respondent from each of approximately 15 000 households was interviewed.
Apart from the BCS, the main source of data on crime in England and Wales is police records. Data on crimes from police records are not entirely comparable with those from the BCS, mainly because certain categories of crime are not covered by both sources of data. (For instance, frauds against companies are not recorded in the BCS because there is no personal victim in a private household.) However, there is a large set of categories of crime for which BCS data and police records should (in principle at least) be comparable. Table 3.5 is taken from the report on the 1998 BCS (and hence relates to crimes in 1997), and it compares in various ways the numbers (in thousands) of crimes (in these comparable categories) recorded by the police and measured by the BCS. (The BCS figures are estimates, based on the sample data, for total numbers of crimes in these categories in England and Wales.) In the BCS, respondents are asked, in relation to any crime of which they were a victim, whether or not it was reported to the police. Many crimes are not reported to the police, and clearly these crimes will not appear in the police records.
Some of this activity is concerned with making sense of where the figures come from in this rather complex table. In fact, rather more clues about the relationships between the different numbers are contained in the text of the report. However, with other reports and other tables, this is regrettably not always the case, so the practice you will get by working through this activity will be worthwhile!
Note also that the questions below ask you to answer briefly. Being able to make statistical points concisely in writing is an important skill generally (as well as being crucial in an examination context where time is limited).
(a) The row labelled ‘All comparable’ at the bottom of Table 3.5 concerns values for all offences in the comparable categories taken together. Explain briefly how the first three percentages in that row (44, 54, 24) relate to one another and to the first two values (2450, 10199) in the row.
(b) Consider the column labelled ‘1997 Police’. Describe briefly how values in this column are related to one another.
(c) Draw a suitable diagram to display the values associated with ‘Vandalism’, ‘Burglary’, ‘All vehicle thefts’, ‘Bicycle theft’, ‘Theft from the person’ and ‘All comparable violence’, using the values in the ‘1997 BCS’ column.
(d) Find and use the appropriate numbers from the table to calculate the equivalent values in 1981 to those in part (c). Draw a diagram to display these values. Using your diagrams and/or the corresponding numbers, comment briefly on the similarities and differences between the numbers of crimes in these categories in 1981 and 1997.
Answer
The questions in this activity generally required you to answer briefly. This solution is not particularly brief, because it contains material intended to explain to you what is going on. However, to clarify matters, this explanatory material is enclosed in square brackets.
(a) The value 44 [in the ‘% BCS reported’ column] is the percentage of the crimes mentioned in the BCS that, according to the BCS respondents, were reported to the police. It cannot be calculated from other values in the table. However, it can be used to deduce that the number of ‘comparable’ crimes that were reported to the police, according to BCS respondents, was 44% of 10 199 thousand, or 4488 thousand [to the nearest thousand].
The value 54 [in the ‘% recorded of reported’ column] is the percentage of the crimes that BCS respondents said were reported to the police that actually appear in the police crime statistics. It has already been calculated that 4488 thousand crimes in these categories were said by BCS respondents to have been reported to the police. According to police records, 2450 thousand such crimes were reported to them. Thus the figure of 54% is calculated as the ratio 24504488 [and indeed, allowing for some rounding error, the figures do match].
Finally, the value 24 [in the ‘% recorded of all BCS’ column] is the ratio of the total number of crimes reported to the police (according to police records) to the number of crimes recorded in the BCS. That is, it is 2450/10 199 [and again this does come to 24%].
(b) [In the ‘1997 Police’ column and, indeed, in the ‘1997 BCS’ column, the total number of comparable crimes is divided between various different categories, and some of the categories are further divided in subcategories, and in some cases these are further divided.] The value 2450 for ‘All comparable’ is the sum of the values for the categories shown in bold type, namely ‘Vandalism’, ‘All comparable property theft’ and ‘All comparable violence’. The value for ‘All comparable violence’ is approximately the sum of the values for ‘Wounding’ and ‘Robbery’. [Actually that sum is 257, not 256, but that is presumably accounted for by rounding errors, given that these values are all rounded to the nearest thousand crimes.] The value 1751 for ‘All comparable property theft’ is, within rounding error, the sum of the values for ‘Burglary’, ‘All vehicle thefts’, ‘Bicycle theft’ and ‘Theft from the person’. Finally, the values for ‘Burglary’ and for ‘All vehicle thefts’ are the sums of the values in the relevant rows given below each of these headings.
(c) [These are data in the form of counts of individual entities, in this case crimes, in a number of discrete categories. Guideline 1 for graphics suggests that there are two appropriate types of plot – a bar chart or a pie chart. Both are given here, although only one is required to answer the question. Perhaps the more appropriate is a bar chart, where the lengths of the bars indicate the numbers of crimes in each of the categories. A bar chart for the numbers of crimes in the named categories is shown in Figure 3.4.
Alternatively, you may have felt that the most important point here was the way that the total number of comparable crimes was divided between these categories, in which case a pie chart might be more appropriate. However, the pie chart does not show directly the total number of crimes involved, so that a bar chart is arguably the better choice. A pie chart for the data is shown in Figure 3.5.]
Figure 3.4 Bar chart of BCS ‘comparable’ crimes, 1997
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Figure 3.4
Figure 3.4 Bar chart of BCS ‘comparable’ crimes, 1997
Figure 3.5 Pie chart of BCS ‘comparable’ crimes, 1997
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Figure 3.5
Figure 3.5 Pie chart of BCS ‘comparable’ crimes, 1997
(d) [Here, the relevant numbers in the table are those in the ‘1997 BCS’ column, and those in the ‘% change 1981 to 1997: BCS’ column. For the ‘Vandalism’ category, there were 2917 thousand crimes (BCS) in 1997, and the final column of the table indicates that this value is 7% higher than that for 1981. Thus the 1981 value for ‘Vandalism’ can be calculated as follows:
in thousands, to the nearest thousand. The other figures can be calculated in similar manner and are given below, all rounded to the nearest thousand.]
| | |
--- |
| Vandalism | [2 917/1.07=] 2726 |
| Burglary | [1639/2.19=] 748 |
| All vehicle thefts | [3 483/1.99=] 1750 |
| Bicycle theft | [549/2.54 =] 216 |
| Theft from the person | [590/1.36 =] 434 |
| All comparable violence | [1022/1.53=] 668 |
[A bar chart for these data is shown in Figure 3.6 and a pie chart in Figure 3.7. Only one of these charts is required to answer the question.]
Figure 3.6 Bar chart of BCS ‘comparable’ crimes, 1981
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Figure 3.6
Figure 3.6 Bar chart of BCS ‘comparable’ crimes, 1981
Figure 3.7 Pie chart of BCS ‘comparable’ crimes, 1981
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Figure 3.7
Figure 3.7 Pie chart of BCS ‘comparable’ crimes, 1981
[In fact, a better way of portraying these data, instead of either Figure 3.6 or Figure 3.7, for the purpose of the comparison you are asked to make is to plot both the 1981 and the 1997 figures on the same bar chart (guideline 2), as in Figure 3.8.]
Figure 3.8 Bar chart of BCS ‘comparable’ crimes, 1981 and 1997
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Figure 3.8
Figure 3.8 Bar chart of BCS ‘comparable’ crimes, 1981 and 1997
[Whichever diagram you drew, the following should be reasonably clear.] The pattern of crimes in these different categories changed considerably between 1981 and 1997. In 1981 vandalism was, by a large margin, the most common category, though the number of vehicle thefts was also large. By 1997, the numbers of crimes in all categories had increased, though the increase in the number of crimes of vandalism was relatively small. However, there had been very large increases in numbers of vehicle thefts, of bicycle thefts and of burglaries, and vehicle thefts had become the largest category. The two smallest categories in 1981 were thefts from the person and bicycle thefts, and this remained the case in 1997 even though the number of bicycle thefts had increased by over 150%.
3.6 Summary of Section 3
In this section, you have learned about appropriate ways of interpreting data in tables. By working through examples, you have seen how it can be useful to calculate appropriate proportions and ratios, and to present some of the data in graphical form. Guidelines for the choice of graphics have been given. When the data in a table are in the form of counts, you have seen that it can be useful to calculate the counts in a particular row or column as proportions (usually in the form of percentages) of the corresponding row or column total. However, it is important to calculate the proportion that is relevant to the question you are trying to answer.
4 Conclusion
In this course, you have learned about boxplots and about ways of dealing with data given in tabular form.
A boxplot is a way of presenting certain summary statistics and other characteristics of a data set in graphical form. It gives a quick graphical impression of the location, dispersion and the general pattern of skewness in data set, as well as drawing attention to unusually large or small values. In comparing two or more data sets, it is often useful to draw comparative boxplots (that is, draw boxplots for the data sets on the same diagram against the same scale). These can be used to compare the data sets in terms of location, dispersion and symmetry or skewness. For some data sets that exhibit considerable skewness, this process of comparison is sometimes made easier by transforming the data first.
You have seen how the presentation of data in tabular form can often be improve by following certain guidelines. The labelling of the rows and columns in a table should be clear; unnecessary information should not be included; it may be useful to simplify the numbers in the table (for example, by reducing the number of significant figures presented); and summary statistics or calculation results can often usefully be added to table.
When you are faced with the task of interpreting data in a table, it is very often useful to calculate appropriate rates or proportions to clarify the message of the table. Patterns in the data may also become clearer if an appropriate graph is drawn. Guidelines on the choice of graph have been given. You have seen that it is important to understand how the different numbers in a table relate to one another.
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Streptococcus pneumoniae
History and Biology of Streptococcus pneumoniae
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Streptococcus pneumoniae are lancet-shaped (elongated round shape) Gram-positive bacteria that can either grow in chains or pairs as shown in this image.It was first isolated independently in 1880 in France by Louis Pasteur from the saliva of a patient who had rabies and in the USA by George M. Sternberg.
In 1886,S.pneumoniae was first known as pneumococcus due to its close association with pneumonia. In 1920, it was renamed as Diplococcus pneumoniae as it was usually observed in pairs.In 1974, the name was changed to Streptococcus pneumoniae due to its similarities in forming chains like other Streptococcus spp. S. pneumoniae can be distinguished from other species using rMLST.
S.pneumoniae is a leading global cause of morbidity and mortality amongst all age groups but children and the elderly are most at risk. The burden of pneumococcal disease is highest in low- and middle-income countries.S.pneumoniae causes mild infections like otitis media and sinusitis, but also life-threatening infections like pneumonia,septicaemia and meningitis. Worldwide, pneumococcal disease is highest among children less than 5 years of age and an estimated 1.6 million deaths occur every year, mainly due to pneumonia.
Image
Streptococcus pneumoniae seen by electron micrograph
Biology
Factors that enable pneumococci to evade the immune system and cause disease include the polysaccharide capsule, various surface proteins, and toxins such as the pneumolysin gene (ply).
The polysaccharide capsule is a key virulence factor because it protects the pneumococcus and enables it to evade phagocytosis by the host immune system.
There are at least100 antigenically different polysaccharide capsules or ‘serotypes’. The capsule is the target for the current pneumococcal vaccines.Pneumolysin is a toxin that forms pores in the cell membrane and causes cell lysis, which induces inflammation and activates the complement system in the human host.
S.pneumoniae can undergo phase variation, allowing it to have different phenotypes during colonisation and invasion. During colonisation,S.pneumoniae can produce a thin (or transparent) capsule, better adapted for colonisation, whereas during invasion a thicker (opaque) capsule is observed, which enables it to evade phagocytosis.
Image
Another factor that enables bacteria to survive is that pneumococci can change their capsule to avoid being recognised by the immune system. They do this by taking up DNA (transformation) encoding a different set of capsular genes from another pneumococcus and incorporating that new DNA into their genome (recombination),discarding the original capsular genes. This process of transformation and recombination also enables the pneumococcus to acquire other advantageous genes like antibiotic-resistance genes.
History - Epidemics
S.pneumoniae causes disease in all countries around the world. Outbreaks are possible but not as large scale or as frequent as meningitis outbreaks caused by Neisseria meningitidis. Seasonal outbreaks of bacterial meningitis due to S.pneumoniae have been recorded in themeningitis belt. Most of these cases were predominantly due to serotype 1 pneumococci, which is especially prevalent in Africa.
Map of Africa with countries of the meningitis belt in red
Meningitis outbreaks highlight the need for current and continuous revision of epidemic prevention, surveillance and diagnostic strategies, which to date have largely focused on outbreaks caused by N.meningitidis.
To learn more about Streptococcus pneumoniae click on the links bellow.
Image ### Streptococcus pneumoniae-Serotypes LEARN MORE
Image ### Streptococcus pneumoniae-Disease LEARN MORE
Image ### Streptococcus pneumoniae-Prevention LEARN MORE
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6: Analysis Theorems and Techniques
DC Electrical Circuit Analysis - A Practical Approach (Fiore)
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6.4: Thévenin's Theorem
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Contents
1. Home
2. Bookshelves
3. Electrical Engineering
4. Electronics
5. DC Electrical Circuit Analysis - A Practical Approach (Fiore)
6. 6: Analysis Theorems and Techniques
7. 6.4: Thévenin's Theorem
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DC Electrical Circuit Analysis - A Practical Approach (Fiore)
Front Matter
1: Fundamentals
2: Basic Quantities
3: Series Resistive Circuits
4: Parallel Resistive Circuits
5: Series-Parallel Resistive Circuits
6: Analysis Theorems and Techniques
7: Nodal and Mesh Analysis, Dependent Sources
8: Capacitors
9: Inductors
10: Magnetic Circuits and Transformers
Back Matter
6.4: Thévenin's Theorem
Last updated May 22, 2022
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6.3: Superposition Theorem
6.5: Norton's Theorem
Page ID 25130
James M. Fiore
Mohawk Valley Community College
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Table of contents
1. Finding Thévenin Equivalents in Lab
1. Example 6.4.1
Computer Simulation
Example 6.4.2
Thévenin's theorem, named after Léon Charles Thévenin, is a powerful analysis tool. For DC, it states:
Any single port linear network can be reduced to a simple voltage source,E th,in series with an internal resistance,R th.
Figure 6.4.1 : Thévenin equivalent circuit.
An example is shown in Figure 6.4.1 . The phrase “single port network” means that the circuit is cut in such a way that only two connections exist to the remainder of the circuit (two connection points makes up one port). The remainder of the circuit may be a single component or a large multi-component sub-circuit. As there are many ways to cut a typical circuit, there are many possible Thévenin equivalents. The important thing is that there are only two connection points between the two portions of the circuit and neither point has to be ground.
Figure 6.4.2 : Example circuit.
Figure 6.4.3 : Finding E th..
Consider the circuit shown in Figure 6.4.2 . Suppose we cut the circuit immediately to the left of R 4. That is, we will find the Thévenin equivalent that drives R 4. The first step is to make the cut, removing the remainder of the circuit (in this case, just R 4). We then determine the open circuit output voltage. This is the maximum voltage that could appear between the cut points and is called the Thévenin voltage, E th. This is shown in Figure 6.4.3 . In a circuit such as this, basic series-parallel analysis may be used to find E th. This process turns out to be quite straightforward in this particular circuit. Due to the open, no current flows through R 3, thus no voltage is developed across R 3, and therefore E th must equal the voltage developed across R 2 which may be obtained via a voltage divider with resistor R 1 and source E.
The second part is finding the Thévenin resistance, R th. Beginning with the “cut” circuit, replace all sources with their ideal internal resistance (thus shorting voltage sources and opening current sources). From the perspective of the cut point, look back into the circuit and simplify by making appropriate series and parallel combinations to determine its equivalent resistance. This is shown in Figure 6.4.4 . Looking in from where the cut was made (right-to-left here), we find that R 1 and R 2 are in parallel, and this combination is then in series with R 3. Thus, R th=R 3+(R 1||R 2). A common error is to determine the equivalent resistance that the source drives. Remember, everything is determined from the vantage point of the cut or port.
Thévenin equivalents are not limited to single source circuits. It is possible to find the equivalent of a network with several sources. Finding the open circuit output voltage will undoubtedly require some extra work, for example the use of superposition of source conversions. Finding the Thévenin resistance is unchanged, just remember to replace every source with its ideal internal resistance before simplifying the network.
As noted earlier, the original circuit could be cut in a number of different ways. We might, for example, want to determine the Thévenin equivalent that drives R 2 in the circuit of Figure 6.4.2 . This cut appears in Figure 6.4.5 . Clearly, this will result in different values for both E th and R th. For example, R th is now R 1||(R 3+R 4).
Figure 6.4.4 : Finding R th.
Finding Thévenin Equivalents in Lab
The procedure for experimentally determining an equivalent in the laboratory mimics the analytical approach. The first step is to figuratively cut the circuit and isolate the section that is to be converted. At this point we have the open circuit version of the circuit and all we need do is connect a multimeter to the open port to determine E th.
There are two methods for determining R th. The first method works well for simple circuits that use only resistors and current and/or voltage sources. The second method is more generally applicable and will work for circuits with active devices such as transistors. To use the first method, the sources are physically removed from the circuit and replaced with their ideal internal resistance. Thus voltage sources are replaced with a shorting wire and current sources are left as opens. Do not simply place shorting wires across the terminals of voltage sources as doing so will cause an overload and potentially damage the sources. Once this is completed, a multimeter is placed at the open port and set to read resistance. The indicated value is R th.
Figure 6.4.5 : An alternate cut or port location.
The second method uses a variable resistance, namely either a rheostat or a decade box, and exploits the voltage divider rule. In this version, the sources are left active (powered up) and are not replaced with opens or shorts. Once E th is measured, the variable resistor is placed across the port connections. This resistance is adjusted until the port voltage drops to exactly half of E th. In the equivalent circuit, there are only two resistances of concern, R th and this variable load resistance, and they are in series. Consequently, if the load voltage is now half of the open circuit voltage (E th), then the other half of the voltage must be dropping across the equivalent internal resistance (R th). For this to be true in a series circuit, the two resistances must have the same value. Thus, we simply remove the rheostat from the circuit and use a multimeter to determine its precise value. If a decade box is used instead, the value is determined directly by reading the knob settings.
Whether determined analytically or empirically, the Thévenin equivalent circuit can replace the original single port network regardless of what the original was connected to. The same voltages and currents will be seen in this other portion, and it won't matter if the other portion is comprised of a single resistor, multiple resistors, multiple resistors and multiple sources, or even multiple resistors and sources wired to a selection of piquant cheeses. The Thévenin equivalent is a true functional equivalent and can be used on any linear bilateral network.
Example 6.4.1
Determine the Thévenin equivalent of the circuit driving the 1 k Ω in Figure 6.4.6 . Verify that the equivalent produces the same voltage across this resistor as the original circuit.
Figure 6.4.6 : Circuit for Example 6.4.1 .
Figure 6.4.7 : Finding E th.
First, we'll redraw the circuit showing the portion to be Théveninized, as shown in Figure 6.4.7 . The open circuit output voltage will be the voltage across the 800 Ω resistor as there will be no voltage drop across the 200 Ω resistor (as no current flows through it into the open). This is found via a simple voltage divider.
E th=ER x R x+R y
E th=10V800Ω 800Ω+100Ω
E th≈8.889V
The equivalent resistance is found by shorting the voltage source and then simplifying the circuit. The result is 200 Ω in series with the parallel combination of the 100 Ω and 800 Ω. 200+100||800≈288.89Ω.
To determine Vc, we can use a voltage divider between the 1 k Ω and the 288.89 Ω along with the equivalent source voltage of 8.889 volts.
V c=E thR L R L+R th
V c=8.889V1kΩ 1kΩ+288.89Ω
V c≈6.897V
Now let's determine V c in the original circuit using series-parallel analysis techniques. Perhaps the quickest approach is a pair of voltage dividers. V b is found via a divider between the parallel combination of the 1 k Ω + 200 Ω and the 800 Ω, against the 100 Ω. V c is is then found using that voltage with a divider between the 1 k Ω and 200 Ω. Using this approach, V b is approximately 8.276 volts and V c is approximately 6.897 volts, providing an excellent match.
It might seem that the Thévenin method is the “long way home” in this example, and it is, but it has the advantage of being more efficient if several different loads are being considered. For example, suppose we decided to determine the output voltage not just for the 1 k Ω, but for a group of a half dozen different resistors. The double voltage divider would have to be determined for each load resistor using the straight series-parallel method but only a single divider needs to be computed for each load when using the Thévenin method. The Thévenin method is also of great use when determining maximum power transfer, as we shall see a later in this chapter.
Computer Simulation
To verify the results of Example 6.4.1 , the original circuit is recreated in a simulator along its Thévenin equivalent, as shown in Figure 6.4.8 .
Figure 6.4.8 : The original circuit of Example 6.4.1 in a simulator along with the equivalent.
A DC operating point analysis is performed and the results are shown in Figure 6.4.9 . Note that the voltages across the identical 1 k Ω loads (nodes 3 and 5) are virtually the same, indicating functional equivalence between the two. The slight deviation is due to the rounding of the Thévenin voltage and resistance.
Figure 6.4.9 : Results of the simulation showing equivalence.
As a further check, the simulation is run again, but this time using an alternate load resistor value. The value chosen is the Thévenin resistance value. By matching the resistances, this should produce a 50/50 voltage divider and the load voltage should equal half of the Thévenin source voltage, or approximately 4.444 volts. That is precisely what we find, as seen in Figure 6.4.10 .
Figure 6.4.10 : Results of the simulation for matched resistance.
As mentioned previously, Thévenin's theorem can be applied to much more complex multi-source circuits, and the item being driven need not be just a single resistor. This is illustrated in the next example.
Example 6.4.2
Determine the Thévenin equivalent of the circuit driving the resistor/voltage source combo in Figure 6.4.11 . Verify that the equivalent produces the same voltage across this resistor as the original circuit.
Figure 6.4.11 : Circuit for Example 6.4.2 .
Determining R th is not particularly difficult here. After shorting the 10 volt source and open the current source, it can be seen that the 1 k Ω and 3 k Ω are in parallel, yielding 750 Ω. This is in series with the 1250 Ω for a total of 2 k Ω, which is in turn in parallel with the 6 k Ω. The final result is 1.5 k Ω.
Finding E th is a little more involved. One possibility is to use a source conversion on the 10 volt source as that will leave us with two parallel current sources that may be combined directly. The converted source will be 10 V/ 1k Ω, or 10 mA, in parallel with 1 k Ω. This results in a total of 25 mA feeding upwards with 1 k Ω|| 3 k Ω, or 750 Ω. We can use the current divider rule to determine the current flowing through the 1250 Ω plus 6 k Ω branch, and then use Ohm's law to find the open circuit voltage (i.e., the voltage across the 6 k Ω).
I 6k=I SR x R x+R y
I 6k=25mA750Ω 750Ω+7250Ω
I 6k=2.34375mA
E th=I 6k×R 6k
E th=2.34375mA×6kΩ
E th=14.0625V
Another approach to determine this value would be to use superposition. This is left as an additional exercise.
Turning our attention to the voltage produced across the 6 volt with 500 Ω sub-circuit, we have a simple series loop with E th opposing this 6 volt source, leaving 8.0625 volts to produce a clockwise current through R th in series with the 500 Ω (a total of 2 k Ω). That current will be 8.0625 V / 2 k Ω, or 4.03125 mA. This will produce a drop across the 500 Ω of 2.015625 volts with a polarity of + to − from top to bottom. This will add to the 6 volt source resulting in a final potential of 8.015625 volts. The results are verified via the simulation technique used previously. The node voltages for both the original and equivalent circuits are shown in Figure 6.4.12 .
Figure 6.4.12 : Results of the simulation using the original and equivalent circuits.
This page titled 6.4: Thévenin's Theorem is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by James M. Fiore via source content that was edited to the style and standards of the LibreTexts platform.
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10543 | https://resources.system-analysis.cadence.com/blog/msa2022-internal-energy-and-enthalpy-in-thermodynamics | Skip to main content
Internal Energy and Enthalpy in Thermodynamics
Author
Cadence CFD
Key Takeaways
Enthalpy is a measure of the total heat energy, absorbed or released, in a thermodynamic system.
Internal energy indicates the total energy, potential or kinetic, in a thermodynamic system.
The relation between internal energy and enthalpy in thermodynamics can be established by calculating the changes or studying the temperature relations.
The general law of thermodynamics deals with quantities like temperature, heat, work, internal energy, and their relation. These relations often establish the interlink between the science of thermodynamics and fluid dynamics, where concepts such as fluid flow and heat transfer can be utilized to perform the work within a system. An example where this can be observed thoroughly is in engines or heat transfer machines where working fluid can drive the thermodynamic cycle.
The enthalpy in a thermodynamic system is an important variable that defines the total heat associated with a system. The analysis of the enthalpy change and internal energy change during a reaction is an important guide to maintaining efficiency in thermodynamic engineering.
Discussing Change of Enthalpy in Thermodynamics
Enthalpy is the total heat absorbed or released in a thermodynamic system. Mathematically, it is often described as the sum of the internal energy and the work performed:
H= U+pV
H is the enthalpy, U is the internal energy, p is the pressure, and V is the volume of the fluid system. The product of pressure and volume is the work performed. Generally, the enthalpy change in the system is the desired calculation, as it simplifies the energy transfer analysis. As such, the above equation can be written as:
ΔH= ΔU+pΔV
For the above equation to be true, the pressure must be constant. The equation can indicate the nature of the reaction, i.e., whether it is endothermic or exothermic. The positive enthalpy change means the reaction is endothermic, where the system absorbs the energy. The negative value of enthalpy change means the reaction in the system is exothermic, where the heat is released during the process. At the zero value of enthalpy change, the thermodynamic reaction is considered to be in equilibrium.
Given that both the internal energy and the enthalpy in a thermodynamic system describe the energy associated with the system, it is important to establish the differentiation between the both. The internal energy is the heat energy of the system at constant volume while the enthalpy is the heat energy under constant pressure. We will discuss the internal energy in thermodynamics next.
Relation With Internal Energy
Internal energy signifies the total energy, kinetic and potential, of a thermodynamic system. From the above equations, we know that enthalpy and internal energy are directly influenced by each other. However, understanding the absolute value of internal energy in a system is difficult given the uncertainty of internal energy components. Thus, the change in internal energy is calculated, which is also an important factor in determining enthalpy change.
For a closed system, the first law of thermodynamics holds true, according to which the total energy remains constant. This can be expressed as:
ΔU=δQ-δW
Q is the heat energy supplied to the system and W is the work done.
The relation between enthalpy and internal energy can also be established by studying temperature relations. According to the second law of thermodynamics, for absolute temperature T and entropy S:
δQ=TdS
The above equation for change in internal energy can be written as:
dU=T dS - p dV
This can be further simplified to get:
dH=TdS+Vdp
The increase in temperature of the system induces kinetic energy, which changes the internal energy of the system. This also influences the enthalpy change within the thermodynamic system.
Calculating Thermodynamic Variables With CFD
When the fluid and its thermodynamic properties are the driving factors of energy transfer in a system, the calculations of variables like enthalpy, internal energy, and temperature become critical. The enthalpy equation establishes the understanding of heat transfer and power requirements for the system to operate. The numerical solution for temperature requirements can be determined for the enthalpy and internal energy changes to occur. In the case of compressible flow, pressure and power requirements can be calculated from the enthalpy equation. The calculation can be complex for incompressible flow given the challenge to establish the relationship between temperature, pressure, and density of the working fluid. In such a case, the Navier-Stokes and continuity equation can be applied.
Solving the various equations associated with a thermodynamic system can be done using CFD simulation packages. By running high-fidelity simulations, the changes in internal energy and enthalpy in a thermodynamic system can be analyzed accurately, which is critical to understanding the heat transfer within a system. By generating a heat transfer model, running a thermodynamic system simulation, and solving the governing equation for different pressure, volume, and force conditions, efficient thermodynamic design can be achieved.
Subscribe to our newsletter for the latest CFD updates or browse Cadence’s suite of CFD software, including Fidelity and Fidelity Pointwise, to learn more about how Cadence has the solution for you.
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10544 | https://www.homeschoolmath.net/worksheets/ratio_word_problems.php | | | | | | | | | | |
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10545 | https://pubmed.ncbi.nlm.nih.gov/17870430/ | Staged excision versus Mohs micrographic surgery for lentigo maligna and lentigo maligna melanoma - PubMed
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. 2007 Oct;57(4):659-64.
doi: 10.1016/j.jaad.2007.02.011.
Staged excision versus Mohs micrographic surgery for lentigo maligna and lentigo maligna melanoma
Hobart W Walling1,Richard K Scupham,Andrew K Bean,Roger I Ceilley
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1 hobartwalling@yahoo.com
PMID: 17870430
DOI: 10.1016/j.jaad.2007.02.011
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Comparative Study
Staged excision versus Mohs micrographic surgery for lentigo maligna and lentigo maligna melanoma
Hobart W Walling et al. J Am Acad Dermatol.2007 Oct.
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. 2007 Oct;57(4):659-64.
doi: 10.1016/j.jaad.2007.02.011.
Authors
Hobart W Walling1,Richard K Scupham,Andrew K Bean,Roger I Ceilley
Affiliation
1 hobartwalling@yahoo.com
PMID: 17870430
DOI: 10.1016/j.jaad.2007.02.011
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Abstract
Background: Lentigo maligna (LM) is a relatively common tumor with increasing prevalence and substantial morbidity. A variety of treatment modalities are available, though margin-control surgery offers the highest cure rate. We were interested in comparing long-term outcomes of Mohs micrographic surgery (MMS) versus staged excision with permanent sections (SE) for treating LM or LM melanoma (LMM).
Methods: Comparative study consisting of retrospective chart review from our private practice.
Results: Fifty-seven patients (31 male, 26 female, mean age at diagnosis 69.1 +/- 10.1 years) were treated in our office for LM (50) or LMM (9) between January 1986 and December 2001. Forty-one tumors (71%) were located on the head and neck. Fifty-three of the 59 tumors (90%) were primary, and 6/59 (10%) were recurrent at the time of initial treatment. Forty-one tumors (36 LM, 5 LMM) were treated with SE, and 18 (14 LM, 4 LMM) were treated with MMS. The mean preoperative lesion size (1.5 +/- 0.2 cm2 for SE; 1.2 +/- 0.4 cm2 for MMS), mean postoperative defect size (7.1 +/- 1 cm2 for SE; 7.1 +/- 1.4 cm2 for MMS), and the ratio of postoperative defect to preoperative lesion size (7.9-fold increase for SE, 11.2-fold increase for MMS) were similar between the cohorts. Mean number of stages for clear margins were similar, with 1.8 +/- 0.2 stages (range: 1-7) for SE and 2.0 +/- 0.2 stages (range: 1-4) for MMS; clear margins were obtained in one or two stages in 85% of cases for SE and in 67% for MMS. Three recurrences (3/41; 7.3%) occurred in the SE group while 6 recurrences (6/18; 33%) occurred in the MMS group (P < .025). The mean follow-up duration was 95 months (range: 60-240) in the SE group and 117.5 months (range: 61-157) in the MMS group.
Limitations: Results are limited to a single practice site and fewer patients underwent MMS compared to SE. Patients were not randomized as cases were ascertained retrospectively.
Conclusion: Staged excision of LM and LMM is associated with a significantly lower recurrence rate with no difference in surgical defect size compared to MMS. To our knowledge, this is the first study directly comparing these two surgical techniques for managing this form of melanoma. Our extended follow-up duration exceeds that of most previous reports.
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10546 | https://www.youtube.com/watch?v=-c-tkCFmwOE | Lecture 51 : Stagnation Properties
Advanced Fluid Mechanics - IITKGP
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[Music] you [Music] we will continue with our discussions on compressible flows we will now discuss on something known as tag nation properties this term stagnation we have come across during our earlier topics that we have covered in fluid mechanics and one of the literal meanings is that the fluid is brought to rest at a point so that the velocity is zero so stagnation point is a point where the velocity of flow is zero but that does not mean that it suffices the requirement of discrete description of stagnation properties so we will now look into more deeply about some of the important properties which dictate the nature of the process by which the stagnation State is achieved to do that we will refer to first an important thermodynamic property and when we are referring to a thermodynamic property we will be basically referring to the first law of thermodynamics to specify that so if we say that you have a control volume say some control volume which has some Inlet I and some exit II let us say that there is some rate of heat transfer to the control volume say Q dot CV and say some rate of work done by the control volume as W dot CV and let us say that the flow is steady and the state for the properties of fluid within the control volume steady that means the properties within the control volume do not change with time so if these two conditions are simultaneously achieved then the corresponding form of the first law of thermodynamics is like this where H is the property enthalpy which is the internal energy plus pressure by the density now this particular form is also known as a steady flow energy equation just just for common understanding so this is nothing but the first law of thermodynamics expressed for a flow process across a control volume when the flow is steady and the states within the control volume are also steady not only that there are certain more important assumptions what are the important assumptions all the properties at the inlet and the exit are uniform that means our velocity profiles are uniform the thermodynamic properties like enthalpy those are uniform so it is like approximately a one dimensional representation where uniform properties across the cross section for the inlet and outlet now we are interested to apply this equation for compressible flows so when we apply this equation for compressible flows first of all if you have a system where you are not having any mechanism of extracting work from that so then this rate of work done will be zero by the control volume then if what kind of process we are considering we are considering an adiabatic process so if it is an adiabatic process that means heat transfer across the control volume is zero we are not committing whether it is reversible or irreversible so the reversibility or irreversibility features when we talk about the second law of thermodynamics when we discuss about that but in the first law of thermodynamics it is just good enough to say whether it's adiabatic or not even reversible or irreversible does not feature here now in most of the high speed gas flows the effects of the the thermal effects and the kinetic energy effects are more important than the changes in potential energy so that is negligible and therefore you are left with H I plus UI square by two is equal to h e+ u v square by okay now let us say that we are thinking about two sections instead of I and E some generic section where the profit the enthalpy this is specific enthalpy that is enthalpy per unit mass is H the velocity is U this is at some section and when you go to some another section that section is a special section when the velocity is brought to zero so this is a section of stagnation so we are interested to see that what is the corresponding enthalpy there and let us say that the name of enthalpy at that stagnation section is H 0 since it is a one-dimensional treatment section and point at the same so it's basically uniform so we can say from here that h plus u square by 2 is equal to h 0 because at the stagnation u equal to 0 so this H 0 is known as stagnation enthalpy for an ideal gas it is known that D H equal to CP DT this is for an ideal gas where CP is a function of temperature in general but when we talk about the perfect gas CP is a constant so if we say perfect gas this is a constant for perfect gas therefore we may say H minus H 0 for a perfect gas is CP into T minus T 0 at T 0 is the temperature corresponding to this stagnation enthalpy this is known as stagnation temperature why we refer to the temperature because temperature is a directly measurable quantity from experiments so it is important that we refer to that so this is known as stagnation temperature so we can write CP into T minus t0 plus u square by 2 equal to 0 okay now we may write CP in terms of R and gamma because just recall that CP minus CV is equal to R and CP by CV is equal to gamma okay so you can write CP into 1 minus 1 by gamma equal to R that means CP is equal to gamma R by gamma minus 1 so we can write T 0 minus T in place of CP we write gamma R by gamma minus 1 is equal to u square by 2 u square by 2 is what you can write u by C is equal to the Mac number say M so u square is equal to M square into C square now what will be the expression for C will depend on the nature of the process if it is an ice isentropic flow C square is gamma RT so this is equal to M square gamma RT for isentropic flow C till now whenever we define the stagnation temperature so the stagnation temperature how it is defined the stagnation temperature is defined from this equation as T 0 is equal to T plus u square by 2 CP right this definition does not require the process to be reversible so if you want to find out that what is the stagnation temperature in fact this definition does not require any process because it is just an expression so if you know the temperature and velocity at a point and specific heat at a point you know the stagnation temperature at a point so stagnation temperature is defined at a point irrespective of what is I mean what kind of state is there at that point if you want to physically achieve a stagnation temperature you have to bring the process to rest at a point in an ideal manner it need not be reversible that is not necessary because the definition followed from the fossil of thermodynamics by setting heat transfer equal to zero without imposing any condition of reversible process okay so keep one thing in mind stagnation temperature just like any stagnation properties of property so if it does not mean that at one point if it is not a stagnation point it will not have a stagnation temperature it will have a stagnation temperature because it is just dependent on the local temperature velocity and CP so it is just like a combination of properties and therefore it is a property if you say that I want to physically achieve the property then you have to follow this type this kind of a process adiabatic process now if you want to use this expression u square equal to M square into gamma RT that means you are now imposing the additional constraint that it is an isentropic process that means the reversible process so for an isentropic process these will be M square gamma RT by 2 so from this what follows is so gamma into R will cancel from both sides so you will get T 0 by T is equal to 1 plus gamma minus 1 by 2 m square right so this is the relationship between the stagnation temperature and this temperature at a point what are the assumptions under which this is valid it is an isentropic process that you have to keep in mind otherwise the more general expression is this one now you may relate the stagnation just like stagnation temperature you may have a stagnation pressure and stagnation density so those properties also you may find out so for an isentropic flow for isentropic process you have PV to the power gamma equal to constant that means basically in terms of T and P you can write T 2 by T 1 is equal to P 2 by P 1 to the power gamma minus 1 by gamma so just like you have related the stagnation temperature similarly you can relate stagnation pressure p 0 by p is equal to t 0 by T to the power gamma by gamma minus 1 that means 1 plus gamma minus 1 by 2 m square to the power gamma by gamma minus ok this is this p 0 is stagnation pressure remember assumption again is isentropic flow now so this considers compressibility we have earlier seen a case where we used Bernoulli's equation to define a stagnation pressure and that was without consideration of any compressibility so we use just the Bernoulli's equation Bernoulli's equation means it is an incompressible flow assumption and the stagnation pressure so if you consider the incompressible limit p 0 is equal to P plus half Rho u square neglecting the potential energy effect what are the assumptions it is an incompressible flow we use the Bernoulli's equation that means we implicitly assume that it is a frictionless flow so if you want to achieve physically this p 0 you have to bring the fluid to rest in a reversible and adiabatic process that will mean that it's a frictionless process if you want to achieve T 0 physically you have to bring the fluid to rest in an adiabatic process it need not be whereas to achieve p0 physically you have to bring it to race by ensuring both that it is reversible as well as adiabatic this is a very important distinction between like how you achieve physically a stagnation pressure and a stagnation temperature so now when you come to this incompressible flow so here you can write p 0 by p for an ideal gas p by rho is RT so this is 1 plus half u square by RT remember that if it is an isentropic flow C square is gamma RT so this you can write 1 plus gamma by 2 u square by C square u square by C square is M square okay so this is an expression which is which is valid for incompressible flow this is an expression that is valid for compressible flow it may be interesting to see that in a certain approximately made these two equations in certain approximation they agree with each other to do that what we may do this is the more general expression so we may expand this in a binomial theorem just like 1 plus X to the power of n so if we expand that in a binomial theorem so p 0 by P compressible so 1 plus X to the pardon is like 1 plus NX plus n into n minus 1 by factorial 2 into X square plus let us just write another term n into n minus 1 into n minus 2 by factorial 3 into X cube like that so this is 1 plus gamma by 2 m square so you can see that up to the first term it is just like the incompressible flow expression the remaining terms are the corrections because of the compressibility effect so what are the corrections so you can write maybe the first two Corrections so one is gamma minus so this is gamma by 8m for right then plus this is two plus gamma into gamma by 48 m to the power six like that sorry two minus gamma so what is it's important implication say you are using a pitot tube to measure the flow velocity at a point and you are negligent of the compressibility effect and you are using this expression for p0 by P because in a pitot tube you may get the difference between stagnation and static pressure by connecting a manometer between the stagnation point and another point which is a point located upstream to that so when you get that expression that expression will require a correction and the leading order terms that will dictate the correction the first two leading order terms are like this so if it has to have a compressibility effect these extra terms for correction need to be invoked and we can see this these terms are higher powers of M so as M becomes smaller and smaller this becomes more and more irrelevant that means for low values of Mach number the compressibility effects are smaller and smaller that is quite intuitive |
10547 | https://physexams.com/Pdf-files/Coulombs-law-problems-and-solution.pdf | Coulomb’s Law Problems and Solutions Physexams.com The force exerted by a point charge q1 on another point charge q2 located at a distance r away is given by the following formula ⃗ F = k |q1q2| r2 ˆ r where ˆ r is a unit vector points from q1 toward q2.
Note that Coulomb’s law gets only the magnitude of the electric force between two point charges.
These questions are intended for the college level and are difficult. For simple and more relevant practice problems on Coulomb’s law for the high school level, refer to here.
More Problems about all topics of electrostatic are also provided here.
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1 Coulomb’s Law: Problems and Solutions 1. Compute the electric force between two charges of 5×10−9 C and −3×10−8 C which are separated by d = 10 cm.
Solution: the magnitude of the electrostatic force between two point charges is given by Coulomb’s law as F = k |q1q2| d2 = 9 × 109 |(5 × 10−9)(−3 × 10−8)| (0.1)2 = 135 × 10−6 N where | · · · | denote the magnitude of the charges.
Note that in Coulomb’s law force formula, the sign of charges is not included only its absolute values must be entered.
2. Two spheres located at distance of d = 5 cm attract one another with a force of F = 3 mN. If one of them has three times more charges than the other, find the electric force between them?
Solution: let one of charges be q1 =? and the other q2 = 3q1. Then using Coulomb’s law Page 1 Coulomb’s Law Problems and Solutions Physexams.com formula and solving for the unknown charges, we have F = k |q1q2| d2 3 = 9 × 109 |q1 × 3q1| 0.05 ⇒q2 1 = (3 × 10−3)(0.05) (9 × 109 × 3) = 0.5 × 10−14 Taking square root from both sides gives q1 = 0.75 × 10−7 C Thus, the magnitude of the charges are q1 = 0.075 µC and q2 = 0.225 µC.
3. A point charge q1 = 2 µC located at origin and another point charge q2 = −5 µC is on the coordinate (x = 3, y = 4) m.
(a) Find the electric force on charge q1.
(b) Is the force attractive or repulsive?
Solution: the distance between two point charges is found using distance formula (Pythagorean theorem) as below d = p (x2 −x1)2 + (y2 −y1)2 which gives d = p 32 + 42 = 5 m (a) Now, use Coulomb’s law formula to find the magnitude of the force between two point charges as below F = k |q1q2| d2 = 9 × 109 |(2 × 10−6)(−5 × 10−6)| 52 = 3.6 × 10−3 N (b) Coulomb’s law gives only the magnitude of the electric force. Being repulsive or attractive depends on the signs of charges. Like charges attract and unlike charges repel each other.
Here, the two charges have opposite signs so the electric force between them is attractive.
Page 2 Coulomb’s Law Problems and Solutions Physexams.com 4. Three point charges are fixed in place in the right triangle shown below, in which q1 = 0.71 µC and q2 = −0.67 µC. What is the magnitude and direction of the electric force on the +1.0 µC (let’s call this q3) charge due to the other two charges?
Solution: First, find the electric force due to each charge on the q3, then use the superposition principle to do the vector sum of them.
In the figure below, all forces on q3 are sketched. Recall that the like charges repel each other and unlike charges attract.
The magnitude of Coulomb’s forces on the charge q3 is obtained as below ⃗ F13 = k |q1| |q3| r2 13 ˆ r13 = 9 × 109 0.71 × 10−6 1 × 10−6 (0.1)2 (cos θ ˆ x + sin θ(−ˆ y)) = 0.639 N Where ˆ r13 is the unit vector (a vector whose length is unity) along the line connecting the two charges and decomposed as shown in the figure.
Page 3 Coulomb’s Law Problems and Solutions Physexams.com Since q1 > 0 so the electric field lines are along the line between q1 and q3 and directed away from q3. From the geometry we see that sin θ = 8 10 and cos θ = √ 102−82 10 = 6 10. Therefore, ⃗ F13 = 0.639 (0.6 ˆ x + 0.8 (−ˆ y)) = (0.383ˆ x −0.511ˆ y) N Now find the electric force due to the q2 on q3 i.e. ⃗ F23 ⃗ F23 = k |q2| |q3| r2 23 ˆ r23 = (9 × 109) −0.67 × 10−6 1 × 10−6 102 −82 × 10−4 m2 (−ˆ y) = (−1.675 ˆ y) N Therefore, the resultant force on the q3 is ⃗ F3 = ⃗ F13 + ⃗ F23 = (0.383ˆ x −0.511ˆ y) + (−1.675 ˆ y) = (0.383 ˆ x −2.186 ˆ y) N The direction of the net force with the x axis are determined by tan α = |Fy| / |Fx|, so α = tan−1 2.058 0.511 = 76.05◦ Since F3x > 0 and F3y < 0 , the net force lies in the fourth quadrant.
Page 4 Coulomb’s Law Problems and Solutions Physexams.com Using Pythagorean theorem, its magnitude is also found to be ⃗ F3 = q (0.511)2 + (−2.058)2 = 2.12 N 5. Two small insulating spheres are attached to silk threads and aligned vertically as shown in the figure. These spheres have equal masses of 40 g, and carry charges q1 and q2 of equal magnitude 2.0 µC but opposite sign.The spheres are brought into the positions shown in the figure, with a vertical separation of 15 cm between them.
Note that you cannot neglect gravity. What is the tension in the lower threads?
Solution: There are three forces acting on q2. The attractive electrostatic force Fe due to q1, tension force in the thread, and gravity. Thus, its free body diagram is as follows Page 5 Coulomb’s Law Problems and Solutions Physexams.com The system is in equilibrium so the net force on the q2 is zero i.e.
(ΣFy)2 = 0 ⇒Fe −T −mg = 0 ⇒T = k |q1| |q2| (15)2 −mg ⇒T = 9 × 109 2 × 10−6 2 × 10−6 (0.15)2 −(0.040 × 9.8) = 1.208 N 6. Four identical particles, each having charge +q, are fixed at the corners of a square of side L. A fifth point charge −Q (at P point) lies a distance z along the line perpendicular to the plane of the square and passing through the center of the square. Determine the force exerted by the other four charges on −Q.
Solution: Because the magnitude and distance of all charges are equal so consider |F1| = |F2| = |F3| = |F4| = k |qQ| r2 Page 6 Coulomb’s Law Problems and Solutions Physexams.com By symmetry consideration, Fx = Fy = 0. So the direction of one of the forces is: ⃗ F1z = k |qQ| r2 cos θ −ˆ k = −k |qQ| r3 zˆ k Where we have used from the geometry of the problem cos θ = z/r. By symmetry ⃗ F1z = ⃗ F2z = ⃗ F3z = ⃗ F4z = −k |qQ| r3 zˆ k So ⃗ Fz = Σ4 i=1 ⃗ Fiz = −4k qQ r3 zˆ k In terms of the parameters of the square and using the Pythagorean theorem, we have: x = √ 2 2 L , r = v u u tz2 + √ 2 2 L !2 ⃗ Fz = −4k Qq z2 + √ 2 2 L 2 3 2 z ˆ k 7. Three point charges are located at the corners of an equilateral triangle an in the figure. Find the magnitude and direction of the net electric force on the 7 µC charge.
Page 7 Coulomb’s Law Problems and Solutions Physexams.com Solution: Same as the previous problem, first we must calculate each of the electric forces due to the 2 µC, −4 C charges exerted on the third charge then use the superposition principle to determine the net electric force on it.
⃗ F21 = k |q1q2| r2 12 ˆ r21 = 9 × 109 2 × 10−6 × 7 × 10−6 (0.5)2 ˆ r21 = 0.504 ˆ r21 N ˆ r21 is the unit vector points from q2 toward q1 so if one decomposes it, we get ⃗ F21 = 0.504 1 2 ˆ x + √ 3 2 ˆ y !
N (Notation: F12 is the force exerted by point charge q1on point charge q2) ⃗ F31 = k |q1q3| r2 13 ˆ r31 = 9 × 109 7 × 10−6 × (−4) × 10−6 (0.5)2 (cos 60◦ˆ x + sin 60◦(−ˆ y)) N = 1.008 1 2 ˆ x + √ 3 2 (−ˆ y) !
N Using superposition principle: ⃗ F1=⃗ F31+⃗ F32, we obtain ⃗ F1 = 0.504 1 2 ˆ x + √ 3 2 ˆ y !
+ 1.008 1 2 ˆ x + √ 3 2 (−ˆ y) !
= 0.756 ˆ x −0.437 ˆ y (N) And its magnitude is ⃗ F1 = q (0.756)2 + (−0.437)2 = 0.873 N Page 8 Coulomb’s Law Problems and Solutions Physexams.com And also the direction of the resultant force with the horizontal axis (x) is α = tan−1 |−0.437| |0.756| = 30.02◦ Since F1x > 0 and F1y < 0 so the net force lies in the fourth quadrant.
8. Four point charges are at the corners of a square. The distance from each corner to the center is 0.3 m.
At the center, there is a −q point charge.
What is the magnitude of the net force on this charge?
Solution: Note: the electric force vector between two point charges located at distance r from each other is ⃗ F = k |q1q2| r2 ˆ r.
When there is a system of point charges and we want to find the net force on one of the charges, we must use the superposition principle i.e. the vector sum of the individual electric forces on the desired charge: ⃗ Fnet=⃗ F1+⃗ F2+⃗ F3 + . . . So we must vector sum the individual forces due to four point charges on the −q in the center.
The drawing below shows the direction of the individual forces.
|F1| = |F2| = k |(−q) (+q)| (0.3)2 = + kq2 0.09 |F3| = |F4| = k |(−q) (−q)| (0.3)2 = +k q2 0.09 Because the F1, F2 and F3, F4 are separately in opposite directions to each other (i.e. ⃗ F1 = −⃗ F2 and ⃗ F3 = −⃗ F4) so the net force is ⃗ Fnet = ⃗ F1 + ⃗ F2 + ⃗ F3 + ⃗ F4 = 0 Page 9 Coulomb’s Law Problems and Solutions Physexams.com 9. A electron is fixed at the position x = 0, and a second charge q is fixed at x = 4 × 10−9 m (to the right).
A proton is now placed between the two at x′ = 1 × 10−9 m. What must the charge q be (magnitude and sign) so that the proton is in equilibrium?
Solution: The magnitude of the electric force between two point charges q and q′ located at distance r from each other is given by the Coulomb’s law as follows F = k |q q′| r2 Where k = 9 × 109 N · m2/C2. The directions of forces the two charges exert on each other are always along the line joining them.
The figure below is a free-body diagram for the proton. Let us consider the charge q to be positive. In such a case, F is the force exerted on the proton by the electron, and F ′ is the force exerted by the charge q on it. Now compute these forces and use the superposition principle to find the total force acting on the proton.
⃗ Ftot = ⃗ F + ⃗ F ′ = k |(−e) (+e)| x ′2 −ˆ i + k |(+e) q| (x −x ′)2 +ˆ i Since the charge q is in equilibrium state so the total force exerted on it must be zero ⃗ Ftot = 0 , equilibrium condition k |(−e) (+e)| x ′2 −ˆ i + k |(+e) q| (x −x ′)2 +ˆ i = 0 ⇒ e x ′2 = |q| (x −x ′)2 ⇒e x −x ′2 = |q| x ′2 e(4 nm −1nm)2 = |q| (1nm)2 ⇒|q| = 9e If we assume that the charge q is negative, we get the same result.
10. Four point charges lie on the corners of a square of side L = a √ 2. What is the magnitude of the net Coulomb force at the place of charge −q?
Solution: Similar to the previous problem, first find (using the definition of Coulomb’s law) each electric force on charge −q then form the vector sum of them and determine its magni-tude. Charges q1 and q3 (in the figure below) have the same magnitude and are at equal Page 10 Coulomb’s Law Problems and Solutions Physexams.com distances from q4 so the magnitude of their Coulomb forces acted on q4 are equal.
Coulomb’s law : F = k |q| |q′| d2 F14 = F34 = k |q| | −q| a √ 2 = k |q|2 2a2 Since the two charges have the opposite signs so the electric force (attractive) on q4 due to q1 is to the left and due to q3 is downward as shown in the figure. Therefore, the net force of them, F, is √ 2 F14 or √ 2 F34.
Similarly, find the electrostatic force F24 due to charge q2 on q4.
F24 = k |q| | −q| (2a)2 = 1 4 k |q|2 a2 The distance between q2 and q4 is the diagonal length of the square which is obtained using the Pythagorean theorem.
The charges q2 and q4 have opposite signs, so the Coulomb force between them is attractive and directed inward along the diagonal of the square.
The magnitude of the net Coulomb force on q2 is determined by adding the other magnitudes since they are directed in the same direction along the diagonal of the square. Thus, F2 = F + F24 = k |q|2 2a2 + 1 4 k |q|2 a2 = k |q|2 a2 1 2 + 1 4 = 3 4 k |q|2 a2 11. Four point charges are located on the corners of a square shown in the figure. If the net Coulomb force on q2 is zero, what is the ratio of Q q ?
Page 11 Coulomb’s Law Problems and Solutions Physexams.com Solution: Since |q1| = |q3| = q and placed at a equal distance of charge q2 so F12 = F32. We know that the resultant vector of two perpendicular and equal vectors F is given as √ 2 F so, in this case, the magnitude of the net force acting on charge q2 due to q1 and q3 is F = √ 2 F12 along the diagonal (q2 −q4) of the square and directed outward as shown in the figure.
The total electric force on charge q2 is the vector sum (superposition principle) of ⃗ F2 = ⃗ F + ⃗ F42 since said that it is zero ⃗ F2 = 0 so the electrostatic force of q4 on q2 i.e. ⃗ F42 must be equal in magnitude and opposite in direction with ⃗ F. Therefore, by equating the magnitudes of the Page 12 Coulomb’s Law Problems and Solutions Physexams.com forces i.e. F = F42 we obtain F = F42 √ 2 F12 = F42 √ 2 k |q1| |q2| a2 = k |q4| |q2| ( √ 2 a)2 √ 2|q| |Q| 1 = | 1 2Q| |Q| 2 ⇒Q q = 4 √ 2 12. In The configuration of three point charges, as shown in the figure below, the Coulomb force on each charge is zero. Determine the ratio of charges q3 and q2 i.e.
q3 q2 .
Solution: Since the ratio of the q3 q2 is required and the net force on each charge is zero we must balance the forces on the charge q1 because in this case, the magnitude of q1 cancels from both sides as below, F21 = F31 k |q1| |q2| (20)2 = k |q1| |q3| (30)2 |q2| 400 = |q3| 900 ⇒|q3| |q2| = 9 4 Note: since the expression above is an equality so no need to convert the units to SI.
Page 13 Coulomb’s Law Problems and Solutions Physexams.com Now that the ratio of the magnitudes of the charges is obtained we must determine its signs.
As you can see in the figure because the forces ⃗ F31 and ⃗ F32 are in the opposite directions (to produce a zero net force on q1) so the charges q2 and q3 must be unlike.
The exact sign of charges can not be determined as long as at least the sign of one charge is given. See the later problem.
13. Two point charges q1 = +2 µC and q2 = +8 µC are 30 cm apart from each other.
Another charge q is placed so that the three charges are brought to a balance.
What is the magnitude and sign of the charge q?
Solution: To find the location of the third charge, place a positive (or negative) test charge q3 somewhere between q1 and q2. Since all charges here are positive (negative), by Coulomb’s law, the electrostatic forces on the test charge are repulsive (attractive) and to the left (right) and right (left) of it. Consequently, the net electric force can be zero between them at a distance of say x from charge q1.
Now, balance the magnitude of the forces on the test charge q3 as below to find the location of it F13 = F23 k |q1| |q3| x2 = k |q2| |q3| (30 −x)2 2 x2 = 8 (30 −x)2 ⇒2x = 30 −x ⇒x = 10 cm In above, the required charge q3 is canceled from both sides and one can not find its sign and value. To find the magnitude and sign of q3, balance the forces on another charge, say q1 as Page 14 Coulomb’s Law Problems and Solutions Physexams.com below F31 = F21 k |q1| |q3| (10)2 = k |q2| |q1| (30)2 |q3| 100 = 8 900 ⇒|q3| = 8 9 The electric force ⃗ F21 is repulsive and directed to the −x axis. Since the net force on each charge is zero the charge q3 must be negative to provide an attraction force in the opposite direction of ⃗ F21 that is to the +x axis.
Therefore, the third charge is negative, located at a distance of 10 cm between the two other charges.
14. In the corners of a square of side L, four point charges are fixed as shown in the figure below. What angle does make the net Coulomb force vector on the charge q located at the point B in the upper right corner with the horizontal?
Solution: In this problem, there is no need to do any explicit calculation, only justify the desired direction.
The electric force vector on the charge q at the corner B is the vector sum of the forces acting by the other charges −q on it. Therefore, using superposition principle, we have ⃗ FB = ⃗ FAB + ⃗ FDB + ⃗ FCB Page 15 Coulomb’s Law Problems and Solutions Physexams.com Similar to the previous problems, since the magnitude and distance of charges located at A and C are equal and the same so |⃗ FAB| = |⃗ FCB| = F. On the other hand, those forces are attractive and directed to the points A and C as shown in the figure. Thus, their resultant electric force lies along the diagonal of BD points inward with the magnitude of √ 2 F.
The force between charge −q at point D and q at point B is also attractive, lies along the diagonal of BD, and points inward. Therefore, Adding these three force vectors gives a resul-tant Coulomb force vector ⃗ FB directed with an angle of (180 + 45)◦along the BD diagonal as shown in the figure.
15. Three equal point charges are placed at the vertices of an equilateral triangle of side a. What is the magnitude and direction of the Coulomb force on the charge q at the point A? (q = 10 µC and a = 4 √ 3 m).
Page 16 Coulomb’s Law Problems and Solutions Physexams.com Solution: first find the magnitudes of ⃗ FBA and ⃗ FCA using Coulomb’s force law as below FBA = k |qB| |qA| d2 = k q2 4 √ 3 2 = k q2 √ 3 = (9 × 109) (10 × 10−6)2 √ 3 = 9 √ 3 × 10−1 N Since the distance to qA and the magnitudes of qB and qC are the same so FBA = FCA = F.
Now find the direction of the electrostatic forces above using vector components. ⃗ FBA makes an angle of 60◦with the +x direction and ⃗ FCA an angle of 60◦with the −x direction. Thus, the above forces can be written in the following vector form ⃗ FBA = |⃗ FBA| | {z } F cos 60◦ˆ i + sin 60◦ˆ j ⃗ FCA = |⃗ FCA| | {z } F cos 60◦(−ˆ i) + sin 60◦ˆ j The x-components will add up to zero which gives the x-component of the net force on the Page 17 Coulomb’s Law Problems and Solutions Physexams.com charge on the position A. The sum of the y-components also gives FAy = F sin 60◦+ F sin 60◦ = 2F sin 60◦ = 2F × √ 3 2 !
= √ 3 F = √ 3 × 9 √ 3 × 10−1 = 0.9 N Therefore, the resultant Coulomb force on qA directed upward and is written as ⃗ FA = 0.9 ˆ j.
16. Four unknown point charges are held at the corners of a square. Suppose q4 is at equilibrium and Let q1 = q3 = −5 µC then what is the magnitude of the charge q2 and the sign of the ratio of q2 q4 .
Solution: Since q4 is at equilibrium, the net electric force on it must be zero. Applying the superposition principle at point 4 we get ⃗ Fnet−on−q4 = 0 ⃗ F14 + ⃗ F24 + ⃗ F34 = 0 ⇒⃗ F14 + ⃗ F34 = −⃗ F24 Let’s consider first the charge q4 is positive. Because of being negative of the charges q1 and q3, their forces on q4 are attractive, to the right and up direction which gives a net force F along the diagonal of the square and directed inward.
Page 18 Coulomb’s Law Problems and Solutions Physexams.com In this case, the electric force ⃗ F24 must be diagonally and directed outward to cancel the contribution F (See the right figure). This result tells us that the force between q2 and q4 must be repulsive, or they must have like charges.
Because we assumed q4 > 0, so q2 is also positive. Thus, q2 q4 > 0 for this situation.
Similar reasoning can be also applied for the case of a negative q4 charge (left figure). Conse-quently, q4 and q2 are unlike charges or its ratio is q2 q4 < 0.
Consequently, this analysis tells us that q2 must be always positive.
Since |q1| = |q3| = |q| and are at an equal distance to q4 so their forces on q4 due to these charges are also equal with magnitude (using Coulomb’s law formula) F14 = F34 = = k |q1 or q3||q4| a2 = k |q| |q4| a2 Pythagorean theorem gives the net electric force on q4 due to q1 and q3 as F = q F 2 14 + F 2 34 = q F 2 14 + F 2 14 = √ 2 F14 Now we proceed to determine the magnitude of q2 by applying the equilibrium condition on charge q4 (the magnitude of the forces along the square diagonal (F and F24) must be equal) Page 19 Coulomb’s Law Problems and Solutions Physexams.com as below F = F24 √ 2 F14 = k |q2| |q4| ( √ 2 a)2 √ 2 k |q1| |q4| a2 = k |q2| |q4| ( √ 2 a)2 √ 2 5 × 10−6 a2 = |q2| 2a2 ⇒|q2| = 10 √ 2 µC the first equality is the equilibrium condition. Therefore, the charge q2 has a magnitude of 10 √ 2 µC .
17. Two point charges of q1 = +2 µC and q2 = −8 µC are at a distance of d = 10 cm.
Where must a third charge q3 be placed so that the net Coulomb force acted upon it is zero?
Solution: Put a positive (or negative) test charge q3 between them and examine whether the net Coulomb force on it is zero or not. In this case, the net electrostatic force on the positive (negative) test charge due to the charges q1 and q2 is to the right (left). Thus, there is no space between them to balance a test charge.
Now place that test charge q3 outside them, say in the left of the charge q1 at a distance x from it. One can see that, in this case, the forces on the q3 can be balanced and canceled by each other. Therefore, apply Coulomb’s force law and find the unknown x as below, F13 = F23 k |q1| |q3| x2 = k |q2| |q3| (10 + x)2 |2 × 10−6| x2 = | −8 × 10−6| (10 + x)2 1 x2 = 4 (10 + x)2 ⇒1 4 = x2 (x + 10)2 ⇒ x x + 10 = ±1 2 In the fifth equality, the square root is taken from both sides. Solving the last equation for x, we get x = 10 cm.
18. In the figure below, what is the magnitude and direction of the net Coulomb force vector acted on the charge qO = q by the eight other charges placed on the Page 20 Coulomb’s Law Problems and Solutions Physexams.com circumference of a circle of radius R = 100 cm. Let q0 = +20 µC and other charges be q1 = q2 = q3 = q4 = q5 = q7 = q8 = q = 50 µC q6 = −q The charge q0 is held at the center of circle.
Solution: Using the symmetry of the charge configuration, one can realize that the electric forces due to a pair of charges (q1, q5), (q2, q8) and (q3, q7) on the charge at the origin qO are equal in magnitude and opposite in direction, so cancel each other. Consequently, the net force on the charge q at the center is only due to the charges q6 and q2 which its magnitudes (F1O and F6O) are computed by applying Coulomb’s law as below F = F1O = F6O F = k |q1| |q| R2 = k |q6| |q| R2 F = k |q| |q| R2 = k | −q| |q| R2 ⇒F = k |q|2 R2 = (9 × 109) (50 × 10−6)(20 × 10−6) (100 × 10−2)2 = 9 N The charge q6 attracts and q1 repels the charge q at the center so the magnitude of the net electric force at point O is 2 times the magnitude of the force between q6 or q1 and q at center i.e. |⃗ FO| = 2F = 19 N.
Page 21 Coulomb’s Law Problems and Solutions Physexams.com The resultant electric force ⃗ FO lies on the third quadrant, points radially outward, and makes an angle of (180 + 45)◦with the positive x axis or 45◦with the −x axis. Its vector form is written as follows ⃗ FO = 18 cos 45◦(−ˆ i) + sin 45◦(−ˆ j) Page 22 |
10548 | https://byjus.com/biology/biogeochemical-cycles/ | What is a Biogeochemical Cycle?
“Biogeochemical cycles mainly refer to the movement of nutrients and other elements between biotic and abiotic factors.”
The term biogeochemical is derived from “bio” meaning biosphere, “geo” meaning the geological components and “chemical” meaning the elements that move through a cycle.
The matter on Earth is conserved and present in the form of atoms. Since matter can neither be created nor destroyed, it is recycled in the earth’s system in various forms.
The earth obtains energy from the sun which is radiated back as heat, rest all other elements are present in a closed system. The major elements include:
Carbon
Hydrogen
Nitrogen
Oxygen
Phosphorus
Sulphur
These elements are recycled through the biotic and abiotic components of the ecosystem. The atmosphere, hydrosphere and lithosphere are the abiotic components of the ecosystem.
Types of Biogeochemical Cycles
Biogeochemical cycles are basically divided into two types:
Gaseous cycles – Includes Carbon, Oxygen, Nitrogen, and the Water cycle.
Sedimentary cycles – Includes Sulphur, Phosphorus, Rock cycle, etc.
Let us have a look at each of these biogeochemical cycles in brief:
Water Cycle
The water from the different water bodies evaporates, cools, condenses and falls back to the earth as rain.
This biogeochemical cycle is responsible for maintaining weather conditions. The water in its various forms interacts with the surroundings and changes the temperature and pressure of the atmosphere.
There’s another process called Evapotranspiration (i.e. vapour produced from leaves) which aids this process. It is the evaporation of water from the leaves, soil and water bodies to the atmosphere which again condenses and falls as rain.
Also Read: Water Cycle
Carbon Cycle
It is one of the biogeochemical cycles in which carbon is exchanged among the biosphere, geosphere, hydrosphere, atmosphere and pedosphere.
All green plants use carbon dioxide and sunlight for photosynthesis. Carbon is thus stored in the plant. The green plants, when dead, are buried into the soil that gets converted into fossil fuels made from carbon. These fossil fuels when burnt, release carbon dioxide into the atmosphere.
Also, the animals that consume plants, obtain the carbon stored in the plants. This carbon is returned to the atmosphere when these animals decompose after death. The carbon also returns to the environment through cellular respiration by animals.
Huge carbon content in the form of carbon dioxide is produced that is stored in the form of fossil fuel (coal & oil) and can be extracted for various commercial and non-commercial purposes. When factories use these fuels, the carbon is again released back in the atmosphere during combustion.
Also Read:Carbon cycle
Nitrogen Cycle
It is the biogeochemical cycle by which nitrogen is converted into several forms and it gets circulated through the atmosphere and various ecosystems such as terrestrial and marine ecosystems.
Nitrogen is an essential element of life. The nitrogen in the atmosphere is fixed by the nitrogen-fixing bacteria present in the root nodules of the leguminous plants and made available to the soil and plants.
The bacteria present in the roots of the plants convert this nitrogen gas into a usable compound called ammonia. Ammonia is also supplied to plants in the form of fertilizers. This ammonia is converted into nitrites and nitrates. The denitrifying bacteria reduce the nitrates into nitrogen and return it into the atmosphere.
Also Read:Nitrogen Cycle
Oxygen Cycle
This biogeochemical cycle moves through the atmosphere, the lithosphere and the biosphere. Oxygen is an abundant element on our Earth. It is found in the elemental form in the atmosphere to the extent of 21%.
Oxygen is released by the plants during photosynthesis. Humans and other animals inhale the oxygen exhale carbon dioxide which is again taken up by the plants. They utilise this carbon dioxide in photosynthesis to produce oxygen, and the cycle continues.
Also Read:Oxygen Cycle
Phosphorous Cycle
In this biogeochemical cycle, phosphorus moves through the hydrosphere, lithosphere and biosphere. Phosphorus is extracted by the weathering of rocks. Due to rains and erosion phosphorus is washed away in the soil and water bodies. Plants and animals obtain this phosphorus through the soil and water and grow. Microorganisms also require phosphorus for their growth. When the plants and animals die they decompose, and the stored phosphorus is returned to the soil and water bodies which is again consumed by plants and animals and the cycle continues.
Also Read:Phosphorus cycle
Sulphur Cycle
This biogeochemical cycle moves through the rocks, water bodies and living systems. Sulphur is released into the atmosphere by the weathering of rocks and is converted into sulphates. These sulphates are taken up by the microorganisms and plants and converted into organic forms. Organic sulphur is consumed by animals through food. When the animals die and decompose, sulphur is returned to the soil, which is again obtained by the plants and microbes, and the cycle continues.
Also Read:Sulphur cycle
Importance of Biogeochemical Cycles
These cycles demonstrate the way in which the energy is used. Through the ecosystem, these cycles move the essential elements for life to sustain. They are vital as they recycle elements and store them too, and regulate the vital elements through the physical facets. These cycles depict the association between living and non-living things in the ecosystems and enable the continuous survival of ecosystems.
It is important to comprehend these cycles to learn their effect on living entities. Some activities of humans disturb a few of these natural cycles and thereby affecting related ecosystems. A closer look at these mechanisms can help us restrict and stop their dangerous impact.
To learn more about biogeochemical cycles, and their types, keep visiting BYJU’S website or download BYJU’S app for further reference.
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Abhishek June 25, 2020 at 8:14 pm
Excellent work of byju’s app👌
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- tanya September 2, 2020 at 3:49 pm
can i have digram of all cycles
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- No way January 20, 2021 at 10:15 am
You are the best 👌🥰😇
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Thanks Byjus for your Answer. It helped me a lot.
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I need some MCQ question on environmental awareness
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You can find these here:
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10549 | https://www.youtube.com/playlist?list=PLa8j0YHOYQQIj0RV3sPSOfCkNc2xF9U5n | Symmedian Lines and the Symmedian Point - YouTube
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10550 | https://www.legislate.ai/blog/what-is-a-town-hall-meeting | Valentina Golubovic
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October 29, 2024
Town hall meetings: Importance, how to lead and questions to ask
This article explains what a town hall meeting is, the steps required to set it up and the benefits of organising one for your business.
A town hall meeting is a type of gathering where employees can learn more about the company and its goals. It's also a chance for employees to give feedback on how they think things should be done and for management to reinforce company culture and convey any messages to the whole company.
Town hall meetings, also known as all hands meetings, are usually held in a large room, with all employees attending. The CEO of the company typically addresses the crowd at some point during the meeting, but other senior staff may also speak. Town hall meetings often include presentations from people who work in different departments of the company, such as the head of marketing or the head of sales.
Town halls don't have to be complicated— depending on the size of your company they can be as simple as having lunch with your team every once in a while or hosting an informal gathering. But they can also be more formal events that include presentations and discussions led by managers or executives.
3 reasons why town hall meetings are important
1. Sense of community
Having all employees in one room (physical or virtual) can bring a sense of community and cohesion to your company. Effective town hall meetings organised for company presentations, allow for direct interaction between managers and employees and create an environment where everyone feels like their voice is being heard.
2. Obtain feedback
It is a great way to get feedback from staff and answer any concerns in real-time. Reaching out to a CEO can often feel intimidating, by providing a designated time to ask questions, a town hall erases intimidating thoughts by facilitating an environment where questions and concerns are welcomed.
3. Employee engagement
A town hall meeting is a great way for medium and large-sized companies to keep in touch with their employees and make sure everyone knows what's going on. They're also a great way for employees to get answers to questions they may have about their jobs or the company. A study by Slack found that 80% of workers want to know more about how decisions are made, this means including your employees in the decision making process or at least explaining how and why significant decisions are made can be beneficial for employee morale and engagement.
A town hall meeting is a great way to connect with your employees. It's an opportunity for you to talk about the company's goals, engage with your employees, and get feedback from them on how things are going at work.
6 ways to lead a town hall meeting
If you're new to leading town hall meetings, here are some tips for getting started:
1. IT issues
Iron out any foreseeable IT-related issues. If you are hosting an online town hall meeting check that you have a reliable internet connection, good audio and video quality as well as a sufficient internet speed.
2. Location
Town hall meetings are usually held in a large room, where all employees are encouraged to attend. The CEO of the company typically addresses the crowd at some point during the meeting, but other senior staff may also speak. Town hall meetings often include presentations from people who work in different departments of the company. It's an opportunity for those heads of departments to discuss key projects and updates.
A study by the Centre for Economic Policy Research found that for a few employees, a virtual meeting may be beneficial however once that number goes above 10 there is a loss in meeting efficiency. Through clear communication, body language and increased focus and engagement, an in-person event can prove to be more efficient. Consider getting all staff together for your company town hall meeting at least once a year or more frequently if possible.
A virtual town hall meeting can be hosted over Zoom, Skype, Bluejeans or other popular platforms. Virtual town hall meetings are integral in an age where remote working is the norm and companies are globalised.
3. Timing
Try and coordinate with all departments to make sure all staff are available and can make it. Be clear about how the town hall will last so that attendees know how much time they need to allot. Allocate a specific amount of time for each section and make sure all participants stick to this.
4. Know your audience
Your audience will be your employees. Concentrate on what you want them to take away from this meeting as a collective. Avoid getting too technical unless someone asks a specific question. Your employees will be from different departments, so technological talk might not appeal to your sales team, it's vital to speak a common language all your employees understand.
5. Set an agenda
Timekeeping is important for a successful meeting, spending too long on a section or going over time can lead to employees feeling frustrated and disengaged, especially during a virtual town hall meeting. Set an agenda ahead of time so that there's enough time for each section of the meeting. Inform employees of the program for the meeting as well as updates on what's coming up after each section for a smooth meeting. To streamline information sharing during your town hall meeting, you can generate QR codes online and use them to provide easy access to agendas, presentation slides, and feedback forms, ensuring a smooth and interactive experience for all attendees.
Employees will like to hear from their senior leadership team and make sure that the meeting agenda is representative and includes all key members of the company.
6. Encourage active participation
Asking open-ended questions that require more than just a yes/no answer, such as "How do we make sure our product is useful?" or "How has this change affected your job?" will encourage employees to expand on their feelings and ideas. Allocate time for answering questions at the end of the meeting.
Common questions to ask during a town hall meeting
Whether you are a CEO who wants to prepare for any challenging questions or an employee who needs some inspiration on how to engage with the leadership team, below are some common questions that can get asked during town hall meetings.
New initiatives at work and the impact they will have?
Upcoming changes in your department, and what steps are being taken to prepare for them?
Employee benefits or other perks
Working styles, hybrid/remote and how the company is accommodating/planning to accommodate all staff
Company values and how they're being applied at work every day
Questions on recent changes in management
Progress update on recent projects
Conclusion
A town hall meeting is a great way to engage with your employees, and it's a great way to get feedback from them. Town hall meetings can be effective at increasing morale, improving communication between your company and its employees, and making sure that everyone feels heard.
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Contents
1 History
2 Decay
3 Production
3.1 Lithium
3.2 Boron
3.3 Deuterium
3.4 Fission
3.4.1 Fukushima Daiichi
3.5 Helium-3
3.6 Cosmic rays
3.7 Production history
3.7.1 USA
4 Properties
5 Health risks
6 Environmental contamination
6.1 Regulatory limits
7 Use
7.1 Radiometric assays in biology and medicine
7.2 Self-powered lighting
7.3 Nuclear weapons
7.3.1 Neutron initiator
7.3.2 Boosting
7.3.3 Tritium in hydrogen bomb secondaries
7.4 Controlled nuclear fusion
7.5 Electrical power source
7.6 Use in electron tubes
8 Use as an oceanic transient tracer
8.1 North Atlantic Ocean
8.2 Pacific and Indian oceans
8.3 Mississippi River system
9 See also
10 Footnotes
11 References
12 External links
Tritium
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From Wikipedia, the free encyclopedia
Isotope of hydrogen with two neutrons
For other uses, see Tritium (disambiguation).
"3H" redirects here. For other uses, see 3H (disambiguation).
Tritium
| General |
| Symbol | 3H |
| Names | Tritium,hydrogen-3, T, 3T |
| Protons (Z) | 1 |
| Neutrons (N) | 2 |
| Nuclide data |
| Natural abundance | 10−18 in hydrogen |
| Half-life (t1/2) | 12.32 years |
| Isotope mass | 3.01604928 Da |
| Spin | 1/2 ħ |
| Excess energy | 14949.794±0.001 keV |
| Binding energy | 8481.7963±0.0009 keV |
| Decay products | 3He |
| Decay modes |
| Decay mode | Decay energy (MeV) |
| Beta emission | 0.018592 |
| Isotopes of hydrogen Complete table of nuclides |
Tritium (from Ancient Greek τρίτος (trítos) 'third') or hydrogen-3 (symbol T or 3H) is a rare and radioactive isotope of hydrogen with a half-life of 12.32 years. The tritium nucleus (t, sometimes called a triton) contains one proton and two neutrons, whereas the nucleus of the common isotope hydrogen-1 (protium) contains one proton and no neutrons, and that of non-radioactive hydrogen-2 (deuterium) contains one proton and one neutron. Tritium is the heaviest particle-bound isotope of hydrogen. It is one of the few nuclides with a distinct name. The use of the name hydrogen-3, though more systematic, is much less common.
Naturally occurring tritium is extremely rare on Earth. The atmosphere has only trace amounts, formed by the interaction of its gases with cosmic rays. It can be produced artificially by irradiation of lithium or lithium-bearing ceramic pebbles in a nuclear reactor and is a low-abundance byproduct in normal operations of nuclear reactors.
Tritium is used as the energy source in radioluminescent lights for watches, night sights for firearms, numerous instruments and tools, and novelty items such as self-illuminating key chains. It is used in a medical and scientific setting as a radioactive tracer. Tritium is also used as a nuclear fusion fuel, along with more abundant deuterium, in tokamak reactors and in hydrogen bombs. Tritium has also been used commercially in betavoltaic devices such as NanoTritium batteries.
History
[edit]
Tritium was first detected in 1934 by Ernest Rutherford, Mark Oliphant and Paul Harteck after bombarding deuterium with deuterons (deuterium nuclei). Deuterium is another isotope of hydrogen (of mass 2), which occurs naturally with an abundance of 0.015%. Their experiment could not isolate tritium, which was first accomplished in 1939 by Luis Alvarez and Robert Cornog, who also realized tritium's radioactivity. Willard Libby recognized in 1954 that tritium could be used for radiometric dating of water and wine.
Decay
[edit]
Tritium decays into helium-3 by beta-minus decay as shown in this nuclear equation:
: | | | | | | | |
--- --- ---
| 31H | → | 32He | + | e− | + | νe |
releasing 18.6 keV of energy in the process. The electron's kinetic energy varies, with an average of 5.7 keV, while the remaining energy is carried off by the nearly undetectable electron antineutrino. Beta particles from tritium can penetrate only about 6 mm (0.24 in) of air, and they are incapable of passing through the dead outermost layer of human skin. Because of their low energy compared to other beta particles, the amount of bremsstrahlung generated is also lower. The unusually low energy released in the tritium beta decay makes the decay (along with that of rhenium-187) useful for attempts at absolute neutrino mass measurement, none of which has yet succeeded.
The low energy of tritium's radiation makes it difficult to detect tritium-labeled compounds except by using liquid scintillation counting.
Production
[edit]
Lithium
[edit]
Tritium is most often produced in nuclear reactors by neutron activation of lithium-6. The release and diffusion of tritium and helium produced by the fission of lithium can take place within ceramics known as breeder ceramics. Production of tritium from lithium-6 in such breeder ceramics is possible with neutrons of any energy, though the cross section is higher when the incident neutrons have lower energy, reaching more than 900 barns for thermal neutrons. This is an exothermic reaction, yielding 4.8 MeV. In comparison, fusion of deuterium with tritium releases about 17.6 MeV. For applications in proposed fusion energy reactors, such as ITER, pebbles consisting of lithium bearing ceramics including Li2TiO3 and Li4SiO4, are being developed for tritium breeding within a helium-cooled pebble bed, also known as a breeder blanket.
: 63Li + n → 42He (2.05 MeV) + 31H (2.75 MeV)
High-energy neutrons can also produce tritium from lithium-7 in an endothermic reaction, consuming 2.466 MeV. This was discovered when the 1954 Castle Bravo nuclear test produced an unexpectedly high yield. Prior to this test, it was incorrectly assumed that 73Li would absorb a neutron to become 83Li, which would beta-decay to 84Be, which in turn would decay to two 42He nuclei on a total timeframe much longer than the duration of the explosion.
: 73Li + n → 42He + 31H + n
The slowed neutrons from this reaction can still react with 63Li in the first, exothermic reaction; thus lithium can generate more tritium atoms than neutrons absorbed.
Boron
[edit]
High-energy neutrons irradiating boron-10, also occasionally produce tritium:
: 105B + n → 2 42He + 31H
A more common result of boron-10 neutron capture is 7Li and a single alpha particle.
Especially in pressurized water reactors which only partially thermalize neutrons, the interaction between relatively fast neutrons and the boric acid added as a chemical shim produces small but non-negligible quantities of tritium.
Deuterium
[edit]
See also: Heavy water § Tritium production
Tritium is also produced in heavy water-moderated reactors whenever a deuterium nucleus captures a neutron. This reaction has a small absorption cross section, making heavy water a good neutron moderator, and relatively little tritium is produced. Even so, cleaning tritium from the moderator may be desirable after several years to reduce the risk of its escaping to the environment. Ontario Power Generation's "Tritium Removal Facility" is capable of processing up to 2,500 tonnes (2,500 long tons; 2,800 short tons) of heavy water a year, and it separates out about 2.5 kg (5.5 lb) of tritium, making it available for other uses. CANDU reactors typically produce 130 grams (4.6 oz) of tritium per year, which is recovered at the Darlington Tritium Recovery Facility (DTRF) attached to the 3,512 MWelectric Darlington Nuclear Generating Station in Ontario. The total production at DTRF between 1989 and 2011 was 42.5 kilograms (94 lb) – with an activity of 409 megacuries (15,100 PBq): an average of about 2 kilograms (4.4 lb) per year.
Deuterium's absorption cross section for thermal neutrons is about 0.52 millibarn, whereas that of oxygen-16 (16O) is about 0.19 millibarn and that of oxygen-17 (17O) is about 240 millibarns. While 16O is by far the most common isotope of oxygen in both natural oxygen and heavy water; depending on the method of isotope separation, heavy water may be slightly richer in 17O and 18O. Due to both neutron capture and (n,α) reactions (the latter of which produce 14C, an undesirable long-lived beta emitter, from 17O) they are net "neutron consumers" and are thus undesirable in a moderator of a natural uranium reactor which needs to keep neutron absorption outside the fuel as low as feasible. Some facilities that remove tritium also remove (or at least reduce the content of) 17O and 18O, which can – at least in principle – be used for isotope labeling.
India, which also has a large fleet of pressurized heavy water reactors (initially CANDU technology but since indigenized and further developed IPHWR technology), also removes at least some of the tritium produced in the moderator/coolant of its reactors but due to the dual use nature of tritium and the Indian nuclear bomb program, less information about this is publicly available than for Canada.
Fission
[edit]
Tritium is an uncommon product of the nuclear fission of uranium-235, plutonium-239, and uranium-233, (by ternary fission), with a production of about one atom per 104 fissions, meaning it is in fact produced at all reactors. The release or recovery of tritium needs to be considered in the operation of nuclear reactors, especially in the reprocessing of nuclear fuel and storage of spent nuclear fuel. The production of tritium is not a goal, but a side-effect. It is discharged to the atmosphere in small quantities by some nuclear power plants. Voloxidation is an optional additional step in nuclear reprocessing that removes volatile fission products (such as all isotopes of hydrogen) before an aqueous process begins. This would in principle enable economic recovery of the produced tritium, but even if the tritium is only disposed of and not used, it has the potential to reduce tritium contamination in the water used, reducing radioactivity released when the water is discharged since tritiated water cannot be removed from "ordinary" water except by isotope separation.
Annual discharge of tritium from nuclear facilities
v
t
e
| Location | Nuclear facility | Closestwaters | Liquid(TBq) | Steam(TBq) | Total(TBq) | Total(mg) | year |
| United Kingdom | Heysham nuclear power station B | Irish Sea | 396 | 2.1 | 398 | 1,115 | 2019 |
| United Kingdom | Sellafield reprocessing facility | Irish Sea | 423 | 56 | 479 | 1,342 | 2019 |
| Romania | Cernavodă Nuclear Power Plant Unit 1 | Black Sea | 140 | 152 | 292 | 872 | 2018 |
| France | La Hague reprocessing plant | English Channel | 11,400 | 60 | 11,460 | 32,100 | 2018 |
| South Korea | Wolseong Nuclear Power Plant | Sea of Japan | 107 | 80.9 | 188 | 671 | 2020 |
| Taiwan | Maanshan Nuclear Power Plant | Luzon Strait | 35 | 9.4 | 44 | 123 | 2015 |
| China | Fuqing Nuclear Power Plant | Taiwan Strait | 52 | 0.8 | 52 | 146 | 2020 |
| China | Sanmen Nuclear Power Station | East China Sea | 20 | 0.4 | 20 | 56 | 2020 |
| Canada | Bruce Nuclear Generating Station A, B | Great Lakes | 756 | 994 | 1,750 | 4,901 | 2018 |
| Canada | Darlington Nuclear Generating Station | Great Lakes | 220 | 210 | 430 | 1,204 | 2018 |
| Canada | Pickering Nuclear Generating Station Units 1-4 | Great Lakes | 140 | 300 | 440 | 1,232 | 2015 |
| United States | Diablo Canyon Power Plant Units1, 2 | Pacific Ocean | 82 | 2.7 | 84 | 235 | 2019 |
Given the specific activity of tritium at 9,650 curies per gram (357 TBq/g), one TBq is equivalent to roughly 2.8 mg.
Fukushima Daiichi
[edit]
Main article: Fukushima disaster cleanup
In June 2016 the Tritiated Water Task Force released a report on the status of tritium in tritiated water at Fukushima Daiichi nuclear plant, as part of considering options for final disposal of the stored contaminated cooling water. This identified that the March 2016 holding of tritium on-site was 760 TBq (equivalent to 2.1 g of tritium or 14 mL of pure tritiated water) in a total of 860,000 m3 of stored water. This report also identified the reducing concentration of tritium in the water extracted from the buildings etc. for storage, seeing a factor of ten decrease over the five years considered (2011–2016), 3.3 MBq/L to 0.3 MBq/L (after correction for the 5% annual decay of tritium).
According to a report by an expert panel considering the best approach to dealing with this issue, "Tritium could be separated theoretically, but there is no practical separation technology on an industrial scale. Accordingly, a controlled environmental release is said to be the best way to treat low-tritium-concentration water." After a public information campaign sponsored by the Japanese government, the gradual release into the sea of the tritiated water began on 24 August 2023 and is the first of four releases through March 2024. The entire process will take "decades" to complete. China reacted with protest. The IAEA has endorsed the plan. The water released is diluted to reduce the tritium concentration to less than 1500 Bq/L, far below the limit recommended in drinking water by the WHO.
Helium-3
[edit]
Tritium's decay product helium-3 has a very large cross section (5330 barns) for reacting with thermal neutrons, expelling a proton; hence, it is rapidly converted back to tritium in nuclear reactors.
: 32He + n → 11H + 31H
This could allow tritium to be recycled as it decays, maintaining the desired inventory.
Cosmic rays
[edit]
Tritium occurs naturally due to cosmic rays interacting with atmospheric gases. In the most important reaction for natural production, a fast neutron (which must have energy greater than 4.0 MeV) interacts with atmospheric nitrogen:
: 147N + n → 126C + 31H
Worldwide, the production of tritium from natural sources is about 4 megacuries (148 PBq) per year. The global equilibrium inventory of tritium created by natural sources remains approximately constant at 70 megacuries (2,590 PBq), as a balance between the fixed production rate and nuclear decay. These may be taken as 415 g and 7,250 g respectively.
Production history
[edit]
USA
[edit]
Tritium for American nuclear weapons was produced in special heavy water reactors at the Savannah River Site until their closures in 1988. With the Strategic Arms Reduction Treaty (START) after the end of the Cold War, the existing supplies were sufficient for the new, smaller number of nuclear weapons for some time.
225 kg (496 lb) of tritium was produced in the United States from 1955 to 1996.[a] Since it continually decays into helium-3, the total amount remaining was about 75 kg (165 lb) at the time of the report, and about 16 kg (35 lb) as of 2023.
Tritium production was resumed with irradiation of rods containing lithium (replacing the usual control rods containing boron, cadmium, or hafnium), at the reactors of the commercial Watts Bar Nuclear Plant from 2003 to 2005 followed by extraction of tritium from the rods at the Tritium Extraction Facility at the Savannah River Site beginning in November 2006. Tritium leakage from the rods during reactor operations limits the number that can be used in any reactor without exceeding the maximum allowed tritium levels in the coolant.
Properties
[edit]
Tritium has an atomic mass of 3.01604928 Da. Diatomic tritium (T2 or 3H2) is a gas at standard temperature and pressure. Combined with oxygen, it forms tritiated water (3H2O).
Compared to hydrogen having its natural composition on Earth, tritium has a higher melting point (20.62 K vs. 13.99 K), a higher boiling point (25.04 K vs. 20.27 K), a higher critical temperature (40.59 K vs. 32.94 K) and a higher critical pressure (1.8317 MPa vs. 1.2858 MPa).
Tritium's specific activity is 9,650 curies per gram (3.57×1014 Bq/g).
Tritium figures prominently in studies of nuclear fusion due to its favorable reaction cross section and the large amount of energy (17.6 MeV) produced through its reaction with deuterium:
: 31H + 21H → 42He + n
All atomic nuclei have a positive charge from their protons, and therefore repel one another because like charges repel (Coulomb's law). However, if the atoms have a high enough temperature and pressure (for example, in the core of the Sun), then their random motions can overcome such repulsion, and they can come close enough for the strong nuclear force to take effect, fusing them into heavier atoms.
A tritium nucleus (triton), containing one proton and two neutrons, has the same charge as any hydrogen nucleus, and it experiences the same electrostatic repulsion when close to another nucleus. However, the neutrons in the triton increase the attractive strong nuclear force when close enough to another nucleus. As a result, tritium can fuse more easily with other light atoms than ordinary hydrogen can.
The same is true, albeit to a lesser extent, of deuterium. This is why brown dwarfs ("failed" stars) cannot fuse normal hydrogen, but they do fuse a small minority of deuterium nuclei.
Like the other isotopes of hydrogen, tritium is difficult to confine. Rubber, plastic, and some kinds of steel are all somewhat permeable. This has raised concerns that if tritium were used in large quantities, in particular for fusion reactors, it might contribute to radioactive contamination, though its short half-life should prevent significant long-term accumulation in the atmosphere.
The high levels of atmospheric nuclear weapons testing that took place prior to the enactment of the Partial Nuclear Test Ban Treaty proved to be unexpectedly useful to oceanographers. The high levels of tritium oxide introduced into upper layers of the oceans have been used in the years since then to measure the rate of mixing of the upper layers of the oceans with their lower levels.
Health risks
[edit]
Since tritium is a low energy beta (β) emitter, it is not dangerous externally (its β particles cannot penetrate the skin), but it can be a radiation hazard if inhaled, ingested via food or water, or absorbed through the skin.
Organisms can take up 31HHO, as they would H2O. Plants convert 31HHO into organically bound tritium (OBT), and are consumed by animals. 31HHO is retained in humans for around 12 days, with a small portion of it remaining in the body as OBT. Tritium can be passed along the food chain as one organism feeds on another, though the metabolism of OBT is less understood than that of 31HHO. Tritium can incorporate to RNA and DNA molecules within organisms which can lead to somatic and genetic impacts. These can emerge in later generations.
31HHO has a short biological half-life in the human body of 7 to 14 days, which both reduces the total effects of single-incident ingestion and precludes long-term bioaccumulation of 31HHO from the environment. The biological half-life of tritiated water in the human body, which is a measure of body water turn-over, varies with the season. Studies on the biological half-life of occupational radiation workers for free water tritium in a coastal region of Karnataka, India, show that the biological half-life in winter is twice that of the summer. If tritium exposure is suspected or known, drinking uncontaminated water will help replace the tritium from the body. Increasing sweating, urination or breathing can help the body expel water and thereby the tritium contained in it. However, care should be taken that neither dehydration nor a depletion of the body's electrolytes results, as the health consequences of those things (particularly in the short term) can be more severe than those of tritium exposure.
Environmental contamination
[edit]
| |
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Tritium has leaked from 48 of 65 nuclear sites in the US. In one case, leaking water contained 7.5 microcuries (280 kBq) of tritium per liter, which is 375 times the current EPA limit for drinking water, and 28 times the World Health Organization's recommended limit. This is equivalent to 0.777 nanograms per litre (5.45×10−8 gr/imp gal) or roughly 0.8 parts per trillion.
The US Nuclear Regulatory Commission states that in normal operation in 2003, 56 pressurized water reactors released 40,600 curies (1,500,000 GBq) of tritium (maximum: 2,080 Ci (77,000 GBq); minimum: 0.1 Ci (3.7 GBq); average: 725 Ci (26,800 GBq)) and 24 boiling water reactors released 665 Ci (24.6 TBq) (maximum: 174 Ci (6,400 GBq); minimum: 0 Ci; average: 27.7 Ci (1,020 GBq)), in liquid effluents. 40,600 Ci (1,500,000 GBq) of tritium weigh about 4.207 grams (0.1484 oz).
Regulatory limits
[edit]
The legal limits for tritium in drinking water vary widely from country to country. Some figures are given below:
: Tritium drinking water limits by country
| Country | Tritium limit(Bq/L) | Equivalent dose(μSv/year) |
| Australia | 76,103 | 1,000 |
| Japan | 60,000 | 788.4 |
| Finland | 30,000 | 394.2 |
| World Health Organization | 10,000 | 131.4 |
| Switzerland | 10,000 | 131.4 |
| Russia | 7,700 | 101.18 |
| Canada (Ontario) | 7,000 | 91.98 |
| United States | 740 | 9.72 |
| Norway | 100 | 1.31 |
The American limit results in a dose of 4.0 millirems (or 40 microsieverts in SI units) per year per EPA regulation 40CFR141, and is based on outdated dose calculation standards of National Bureau of Standards Handbook 69 circa 1963. Four millirem per year is about 1.3% of the average natural background radiation (~3 mSv). Updated dose calculation standards based on International Commission on Radiological Protection Report 30 and used in the NRC Regulation 10CFR20 results in a dose of 0.9 millirem (9 μSv) per year at 740 Bq/L (20 nCi/L).
Use
[edit]
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Radiometric assays in biology and medicine
[edit]
Tritiation of drug candidates allows detailed analysis of their absorption and metabolism. Tritium has also been used for biological radiometric assays, in a process akin to radiocarbon dating. For example, [3H] retinyl acetate was traced through the bodies of rats.
Self-powered lighting
[edit]
Main article: Tritium radioluminescence
The beta particles from small amounts of tritium cause chemicals called phosphors to glow. This radioluminescence is used in self-powered lighting devices called betalights, which are used for night illumination of firearm sights, watches, exit signs, map lights, navigational compasses (such as current-use M-1950 U.S. military compasses), knives and a variety of other devices.[d] As of 2000[update], commercial demand for tritium is 400 grams (0.88 lb) per year and the cost is $30,000 per gram ($850,000/oz) or more.
Nuclear weapons
[edit]
Tritium is an important component in nuclear weapons; it is used to enhance the efficiency and yield of fission bombs and the fission stages of hydrogen bombs in a process known as "boosting" as well as in external neutron initiators for such weapons.
Neutron initiator
[edit]
These are devices incorporated in nuclear weapons which produce a pulse of neutrons when the bomb is detonated to initiate the fission reaction in the fissionable core (pit) of the bomb, after it is compressed to a critical mass by explosives. Actuated by an ultrafast switch like a krytron, a small particle accelerator drives ions of tritium and deuterium to energies above the 15 keV or so needed for deuterium-tritium fusion and directs them into a metal target where the tritium and deuterium are adsorbed as hydrides. High-energy fusion neutrons from the resulting fusion radiate in all directions. Some of these strike plutonium or uranium nuclei in the primary's pit, initiating a nuclear chain reaction. The quantity of neutrons produced is large in absolute numbers, allowing the pit to quickly achieve neutron levels that would otherwise need many more generations of chain reaction, though still small compared to the total number of nuclei in the pit.
Boosting
[edit]
Main article: Boosted fission weapon
Before detonation, a few grams of tritium–deuterium gas are injected into the hollow "pit" of fissile material. The early stages of the fission chain reaction supply enough heat and compression to start deuterium–tritium fusion; then both fission and fusion proceed in parallel, the fission assisting the fusion by continuing heating and compression, and the fusion assisting the fission with highly energetic (14.1 MeV) neutrons. As the fission fuel depletes and also explodes outward, it falls below the density needed to stay critical by itself, but the fusion neutrons make the fission process progress faster and continue longer than it would without boosting. Increased yield comes overwhelmingly from the increased fission. The energy from the fusion itself is much smaller because the amount of fusion fuel is much smaller. Effects of boosting include:
increased yield (for the same amount of fission fuel, compared to unboosted)
the possibility of variable yield by varying the amount of fusion fuel
allowing the bomb to require a smaller amount of the very expensive fissile material
eliminating the risk of predetonation by nearby nuclear explosions
not so stringent requirements on the implosion setup, allowing for a smaller and lighter amount of high explosives to be used
The tritium in a warhead is continually undergoing radioactive decay, becoming unavailable for fusion. Also, its decay product, helium-3, absorbs neutrons. This can offset or reverse the intended effect of the tritium, which was to generate many free neutrons, if too much helium-3 has accumulated. Therefore, boosted bombs need fresh tritium periodically. The estimated quantity needed is 4 grams (0.14 oz) per warhead. To maintain constant levels of tritium, about 0.20 grams (0.0071 oz) per warhead per year must be supplied to the bomb.
One mole of deuterium-tritium gas contains about 3.0 grams (0.11 oz) of tritium and 2.0 grams (0.071 oz) of deuterium. In comparison, the 20 moles of plutonium in a nuclear bomb consists of about 4.5 kilograms (9.9 lb) of plutonium-239.
Tritium in hydrogen bomb secondaries
[edit]
See also: Nuclear weapon design
Since tritium undergoes radioactive decay, and is also difficult to confine physically, the much larger secondary charge of heavy hydrogen isotopes needed in a true hydrogen bomb uses solid lithium deuteride as its source of deuterium and tritium, producing the tritium in situ during secondary ignition.
During the detonation of the primary fission bomb stage in a thermonuclear weapon (Teller–Ulam staging), the sparkplug, a cylinder of 235U/239Pu at the center of the fusion stage(s), begins to fission in a chain reaction, from excess neutrons channeled from the primary. The neutrons released from the fission of the sparkplug split lithium-6 into tritium and helium-4, while lithium-7 is split into helium-4, tritium, and one neutron. As these reactions occur, the fusion stage is compressed by photons from the primary and fission of the 238U or 238U/235U jacket surrounding the fusion stage. Therefore, the fusion stage breeds its own tritium as the device detonates. In the extreme heat and pressure of the explosion, some of the tritium is then forced into fusion with deuterium, and that reaction releases even more neutrons.
Since this fusion process requires an extremely high temperature for ignition, and it produces fewer and less energetic neutrons (only fission and fusion are net neutron producers), lithium deuteride is not used in boosted bombs, but rather for multi-stage hydrogen bombs.
Controlled nuclear fusion
[edit]
Tritium is an important fuel for controlled nuclear fusion in both magnetic confinement and inertial confinement fusion reactor designs. The National Ignition Facility (NIF) uses deuterium–tritium fuel, and the experimental fusion reactor ITER will also do so. The deuterium–tritium reaction is favorable since it has the largest fusion cross section (about 5.0 barns) and it reaches this maximum cross section at the lowest energy (about 65 keV center-of-mass) of any potential fusion fuel. As tritium is very rare on earth, concepts for fusion reactors often include the breeding of tritium. During the operation of envisioned breeder fusion reactors, Breeding blankets, often containing lithium as part of ceramic pebbles, are subjected to neutron fluxes to generate tritium to complete the fuel cycle.
The Tritium Systems Test Assembly (TSTA) was a facility at the Los Alamos National Laboratory dedicated to the development and demonstration of technologies required for fusion-relevant deuterium–tritium processing.
Electrical power source
[edit]
Tritium can be used in a betavoltaic device to create an atomic battery to generate electricity.
Use in electron tubes
[edit]
Tritium is used in various electron tubes, such as the Zellweger ZE22/3 glow tube. These devices contain a small amount of tritium to ionize the fill gas, typically a noble gas like neon or argon. This ionization ensures reliable and consistent operation by providing a steady current when a high voltage is applied, enhancing the device's performance and stability. The tritium is sealed within a glass envelope with two electrodes, one of which is coated with the radioactive material to create an ion path between the electrodes.
Use as an oceanic transient tracer
[edit]
Aside from chlorofluorocarbons, tritium can act as a transient tracer and can "outline" the biological, chemical, and physical paths throughout the world's oceans because of its evolving distribution. Tritium has thus been used as a tool to examine ocean circulation and ventilation and, for such purposes, is usually measured in tritium units, where 1 TU is defined as 1 tritium atom per 1018 hydrogen atoms, equal to about 0.118 Bq/liter. As noted earlier, nuclear tests, mainly in the Northern Hemisphere at high latitudes, throughout the late 1950s and early 1960s introduced lots of tritium into the atmosphere, especially the stratosphere. Before these nuclear tests, there were only about 3-4 kg of tritium on the Earth's surface; but these amounts rose by 2-3 orders of magnitude during the post-test period. Some sources reported natural background levels were exceeded by about 1,000 TU in 1963 and 1964 and the isotope is used in the northern hemisphere to estimate the age of groundwater and construct hydrogeologic simulation models. Estimated atmospheric levels at the height of weapons testing to approach 1,000 TU and pre-fallout levels of rainwater to be between 5 and 10 TU. In 1963 Valentia Island Ireland recorded 2,000 TU in precipitation.
North Atlantic Ocean
[edit]
While in the stratosphere (post-test period), the tritium interacted with and oxidized to water molecules and was present in much of the rapidly produced rainfall, making tritium a prognostic tool for studying the evolution and structure of the water cycle as well as the ventilation and formation of water masses in the North Atlantic.
Bomb-tritium data were used from the Transient Tracers in the Ocean (TTO) program in order to quantify the replenishment and overturning rates for deep water located in the North Atlantic.
Bomb-tritium also enters the deep ocean around the Antarctic. Most of the bomb tritiated water (H3HO) throughout the atmosphere can enter the ocean through the following processes:
precipitation
vapor exchange
river runoff
These processes make H3HO a good tracer for time scales of up to a few decades.
Using the data from these processes for 1981, the 1-TU isosurface lies between 500 and 1,000 meters deep in the subtropical regions and then extends to 1,500–2,000 meters south of the Gulf Stream due to recirculation and ventilation in the upper portion of the Atlantic Ocean. To the north, the isosurface deepens and reaches the floor of the abyssal plain which is directly related to the ventilation of the ocean floor over 10–20 year time-scales.
Also evident in the Atlantic Ocean is the tritium profile near Bermuda between the late 1960s and late 1980s. There is a downward propagation of the tritium maximum from the surface (1960s) to 400 meters (1980s), which corresponds to a deepening rate of about 18 meters per year. There are also tritium increases at 1,500 m depth in the late 1970s and 2,500 m in the middle of the 1980s, both of which correspond to cooling events in the deep water and associated deep water ventilation.
From a study in 1991, the tritium profile was used as a tool for studying the mixing and spreading of newly formed North Atlantic Deep Water (NADW), corresponding to tritium increases to 4 TU. This NADW tends to spill over sills that divide the Norwegian Sea from the North Atlantic Ocean and then flows to the west and equatorward in deep boundary currents. This process was explained via the large-scale tritium distribution in the deep North Atlantic between 1981 and 1983. The sub-polar gyre tends to be freshened (ventilated) by the NADW and is directly related to the high tritium values (>1.5 TU). Also evident was the decrease in tritium in the deep western boundary current by a factor of 10 from the Labrador Sea to the Tropics, which is indicative of loss to ocean interior due to turbulent mixing and recirculation.
Pacific and Indian oceans
[edit]
In a 1998 study, tritium concentrations in surface seawater and atmospheric water vapor (10 meters above the surface) were sampled at the following locations: the Sulu Sea, Fremantle Bay, the Bay of Bengal, Penang Bay, and the Strait of Malacca. Results indicated that the tritium concentration in surface seawater was highest at the Fremantle Bay (about 0.40 Bq/liter), which could be accredited to the mixing of runoff of freshwater from nearby lands due to large amounts found in coastal waters. Typically, lower concentrations were found between 35 and 45° south, and near the equator. Results also indicated that (in general) tritium has decreased over the years (up to 1997) due to the physical decay of bomb tritium in the Indian Ocean. As for water vapor, the tritium concentration was about one order of magnitude greater than surface seawater concentrations (ranging from 0.46 to 1.15 Bq/L). Therefore, the water vapor tritium is not affected by the surface seawater concentration; thus, the high tritium concentrations in the vapor were concluded to be a direct consequence of the downward movement of natural tritium from the stratosphere to the troposphere (therefore, the ocean air showed a dependence on latitudinal change).
In the North Pacific Ocean, the tritium (introduced as bomb tritium in the Northern Hemisphere) spread in three dimensions. There were subsurface maxima in the middle and low latitude regions, which is indicative of lateral mixing (advection) and diffusion processes along lines of constant potential density (isopycnals) in the upper ocean. Some of these maxima even correlate well with salinity extrema. In order to obtain the structure for ocean circulation, the tritium concentrations were mapped on 3 surfaces of constant potential density (23.90, 26.02, and 26.81). Results indicated that the tritium was well-mixed (at 6 to 7 TU) on the 26.81 isopycnal in the subarctic cyclonic gyre and there appeared to be a slow exchange of tritium (relative to shallower isopycnals) between this gyre and the anticyclonic gyre to the south; also, the tritium on the 23.90 and 26.02 surfaces appeared to be exchanged at a slower rate between the central gyre of the North Pacific and the equatorial regions.
The depth penetration of bomb tritium can be separated into three distinct layers:
Layer 1
: Layer 1 is the shallowest layer and includes the deepest, ventilated layer in winter; it has received tritium via radioactive fallout and lost some due to advection and/or vertical diffusion and contains about 28% of the total amount of tritium.
Layer 2
: Layer 2 is below the first layer but above the 26.81 isopycnal and is no longer part of the mixed layer. Its two sources are diffusion downward from the mixed layer and lateral expansions outcropping strata (poleward); it contains about 58% of the total tritium.
Layer 3
: Layer 3 is representative of waters that are deeper than the outcrop isopycnal and can only receive tritium via vertical diffusion; it contains the remaining 14% of the total tritium.
Mississippi River system
[edit]
Trace amounts of radioactive materials from atomic weapons testing settled throughout the Mississippi River System. Tritium concentrations have been used to understand the residence times of continental hydrologic systems such as lakes, streams, and rivers. In a 2004 study, several rivers were taken into account during the examination of tritium concentrations (starting in the 1960s) throughout the Mississippi River Basin: Ohio River (largest input to the Mississippi River flow), Missouri River, and Arkansas River. The highest tritium concentrations were found in 1963 across locations throughout these rivers. The peak correlates with implementation of the US & Soviet atmospheric test ban treaty in 1962. The overall highest concentrations occurred in the Missouri River (1963) and were greater than 1,200 TU while the lowest concentrations were found in the Arkansas River (never greater than 850 TU and less than 10 TU in the mid-1980s).
As for the mass flux of tritium through the main stem of the Mississippi River into the Gulf of Mexico, data indicated that approximately 780 grams of tritium has flowed out of the River and into the Gulf between 1961 and 1997, an average of 21.7 grams/yr and 7.7 PBq/yr. Current fluxes through the Mississippi River are 1 to 2 grams per year as opposed to the pre-bomb period fluxes of roughly 0.4 grams per year.
See also
[edit]
Hypertriton
List of elements facing shortage
Footnotes
[edit]
^ Total U.S. tritium production since 1955 has been about 225 kilograms, an estimated 150 kilograms of which have decayed into helium-3, leaving a current inventory of about 75 kg of tritium. — Zerriffi & Scoville (1996)
^ Based on ICRP Report 30 calculation: 1.8×10−11 Sv/Bq, 730 L/a, 1×106 μSv/Sv; quoted in
^ This figure is derived from the guideline dose of 1 mSv per year from all sources of radiation in drinking water in The Australian Drinking Water Guidelines 6, assuming that tritium is the only radionuclide present in the water.
^ Tritium has replaced radioluminescent paint containing radium in this application. Exposure to radium causes bone cancer, and its casual use has been banned in most countries for decades.
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External links
[edit]
Wikimedia Commons has media related to Tritium.
"Annotated bibliography for tritium". Alsos Digital Library. Washington and Lee University. Archived from the original on 14 May 2006. Retrieved 3 March 2022.
"Nuclear Data Evaluation Lab".
Review of Risks from Tritium (Report). London, UK: Health Protection Agency. November 2007. RCE-4. Archived from the original on 17 May 2013. Retrieved 3 March 2022.
Bergeron, Kenneth D. (17 September 2004). Tritium on Ice: The dangerous new alliance of nuclear weapons and nuclear power. MIT Press. ISBN 978-0-262-26172-2.[full citation needed]
"Tritium production and recovery in the United States in FY2011". Fissile Materials. February 2010.
"Tritium removal mass transfer coefficient". ans.org.
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10552 | https://www.math.stonybrook.edu/Videos/MAT131Online/Handouts/Lecture-03-Handout.pdf | Lecture 3 Elementary Functions. Part 1 Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Which functions are called elementary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Power functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Power functions y = xn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 All together . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Power functions y = 1 xn = x−n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Power functions y = n √x = x1/n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 y = xn all together . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Quadratic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 How to draw a parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Exponential functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 The graphs of exponential functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Exponential functions y = ax all together. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 The laws of exponents (reminder) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Graphing exponential functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 The number e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Comprehension checkpoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1 Objectives In the next three lectures, we will discuss the class of elementary functions.
This lecture is devoted to • power functions • polynomials • rational functions • exponential functions.
2 / 21 Which functions are called elementary In our calculus course, we are going to deal mostly with elementary functions.
They are power functions ( x2, √x, x1/3, . . . ), exponential functions( 2x, πx, ex, . . . ), logarithmic functions ( ln x, log2 x, . . . ), trigonometric functions ( sin x, cos x, tan x, . . . ), inverse trigonometric functions ( arcsin x, arctan x, . . . ) and their sums, differences, products, quotients, and compositions.
For example, f(x) = arcsin √ x2 −3 ln(x4 + 5) −tan ecos x is an elementary function.
There are many non-elementary functions, for example f(x) = x Z 0 sin t t dt is not an elementary function.
3 / 21 2 Power functions Definition. A power function is a function of the form f(x) = xa , where a is a given constant.
For example, y = x, y = x5, y = 3 √x = x1/3, y = x2/3 are power functions.
In a power function f(x) = xa , the base x is a variable, and the exponent a is a constant.
The appearance of the graph of a power function depends on the constant a .
For a = 0 , y = x0 = 1 x y y = 1 For a = 1 , y = x1 = x x y y = x 4 / 21 Power functions y = xn If n is an integer greater than 1 , then the overall shape of the graph of y = xn is determined by the parity of n (whether n is even or odd).
If n is even, then the graph has a shape similar to the parabola y = x2 : x y y = x2 x y y = x4 x y y = x6 If n is odd, then the graph has a shape similar to the cubic parabola y = x3 : x y y = x3 x y y = x5 x y y = x7 5 / 21 3 All together x y y = x2 y = x4 y = x8 1 −1 1 x y y = x3 y = x5 y = x9 1 1 −1 −1 6 / 21 Power functions y = 1 xn = x−n If n is a positive integer greater than 1 , then the overall shape of the graph of y = 1 xn is determined by the parity of n (whether n is even or odd).
x y y = 1 xn n is odd x y y = 1 xn n is even 7 / 21 4 Power functions y = n √x = x1/n If n is an positive integer, then the overall shape of the graph of y = n √x = x1/n is determined by the parity of n (whether n is even or odd).
x y y = n √x n is even x y y = n √x n is odd 8 / 21 y = xn all together Here are the graphs of y = xn for some rational n and x > 0 : x y y = x y = x2 y = x0 y = x1/2 1 y = x−1 1 9 / 21 5 Polynomials Definition. Let n be a non-negative integer.
A polynomial in the variable x of degree n is a function of the form P(x) = anxn + an−1xn−1 + · · · + a1x + a0 , where a0, a1, . . . , an−1, an are constants (they are called the coefficients of the polynomial) and an ̸= 0 . The integer n is called the degree of the polynomial.
A polynomial is the sum of monomials akxk .
Example 1. A linear function y = mx + b , where m and b are constants, is a polynomial of degree 1 .
The graph of a linear function is a straight line: x y y = mx + b θ b m is the slope of the line, m = tan θ b is the y -intercept.
10 / 21 Quadratic functions Example 2. A quadratic function y = ax2 + bx + c , where a, b, c are constants, is a polynomial of degree 2 .
The graph of a quadratic function is a parabola.
Its appearance depends on the coefficients a, b, c .
x y x y x y ☞ Check your calculus readiness: Draw the graph of y = 3x2 + 2x −1 .
How long did it take?
If less than 2 minutes, then you are ready!
x y −1 1/3 1/3 −1 If you are not quite ready, let us review how to draw a parabola.
11 / 21 6 How to draw a parabola To draw a parabola y = ax2 + bx + c , follow the following scheme.
1. Determine the parabolas vertex. It is located at the point −b 2a , y −b 2a 2. Draw the axis of symmetry. It is the vertical line x = −b 2a .
3. Determine if the parabola opens up ( a > 0 ) or down ( a < 0 ) 4. Determine the intercepts. The y -intercept is (0, c) , the x -intercepts are (x1, 0), (x2, 0), where x1, x2 are the solutions of the quadratic equation ax2 + bx + c = 0 (if there are any).
5. Plot the vertex, the axis of symmetry, the intercepts on a coordinate system and construct a smooth parabola.
Exercise. Use the scheme above to draw the parabola y = 3x2 + 2x −1 .
12 / 21 Rational functions Definition. A rational function is a function of the form f(x) = P(x) Q(x) , where P(x) and Q(x) are polynomials.
For example, f(x) = 1 4x2 + 3x −1 and f(x) = 2x3 −x + 1 x2 −3 are rational function.
The domain of a rational function is the set of all real numbers for which the denominator is not zero.
13 / 21 7 Exponential functions Definition. An exponential function is a function of the form f(x) = ax , where a is a given positive constant.
For example, y = 2x, y = 3 5 x , y = ex, y = ( √ 2)x are exponential functions.
In an exponential function f(x) = ax , the base a is a constant, and the exponent x is a variable.
B Warning. Don’t confuse exponential (y = ax) and power (y = xa) functions!
The appearance of the graph of an exponential function depends on a.
The graph of each exponential function f(x) = ax intersects the y -axis at the point (0, 1) , since f(0) = a0 = 1 .
If a = 1, then the exponential function is y = 1x = 1. It is a constant function.
14 / 21 The graphs of exponential functions x y y = ax a > 1 1 x y y = ax 0 < a < 1 1 Remember: The domain of any exponential function is R , since y = ax is defined for all x .
The range of y = ax is (0, ∞) , since a is positive and therefore ax > 0 for any x .
15 / 21 8 Exponential functions y = ax all together x y a = 1 5 a = 1 2 a = 2 a = 5 a = 1 16 / 21 The laws of exponents (reminder) Let a > 0 , b > 0 , and x, y be any real numbers. Then a0 = 1 ax+y = axay a−x = 1 ax ax−y = ax ay (ax)y = axy (ab)x = axbx 17 / 21 9 Graphing exponential functions Problem. Draw the graph of the function f(x) = 2−x −1 .
Solution. We need to know the graph of the standard function y = 2x , and to be familiar with graph transformations.
Steps to follow in drawing the graph: step 1 y = 2x − − − − − − − − − − − → step 2 y = 2−x − − − − − − − − − − − → step 3 y = 2−x −1 x y y = 2x 1 flip about y -axis x y y = 2x 1 y = 2−x vert. shift down x y 1 y = 2−x -1 y = 2−x-1 18 / 21 The number e This is the most amazing number in mathematics!
It may be defined in many different ways. For example, like this: The number e is the base of the exponential function whose graph has the tangent line at (0, 1) with slope 1 .
x y y = ex 1 tangent of slope 1 45◦ The exact definition of the tangent line to the graph of a function at a point will be given later.
19 / 21 10 Summary In this lecture we reviewed the following topics: • power functions f(x) = xa and their graphs • polynomials • quadratic functions and their graphs • exponential functions f(x) = ax and their graphs • Laws of exponents • the number e .
20 / 21 Comprehension checkpoint • What are the names of the following functions: y = 3 √ x2 y = 2 3 x y = −5x3 + 6x2 + 1 y = 3x + 2 −x2 + 4x + 2 • What is e ?
• How does the graph of y = ex look like?
21 / 21 11 |
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10554 | https://shakerscience.weebly.com/uploads/2/7/7/0/27700095/stoichiometry_1_worksheet_and_key.pdf | 2 KClO3 è 2 KCl + 3 O2 1. How many moles of O2 will be formed from 1.65 moles of KClO3? 2. How many moles of KClO3 are needed to make 3.50 moles of KCl? 3. How many moles of KCl will be formed from 2.73 moles of KClO3? 4 Fe + 3 O2 è 2 Fe2O3 4. How many moles of Fe2O3 are produced when 0.275 moles of Fe is reacted? 5. How many moles of Fe2O3 are produced when 31.0 moles of O2 is reacted? 6. How many moles of O2 are needed to react with 8.9 moles of Fe? 2 H2O è 2 H2 + O2 7. How many moles of O2 are produced when 1.26 moles of H2O is reacted? 8. How many moles of H2O are needed to produce 55.7 moles of H2? 9. If enough H2O is reacted to produce 3.40 moles of H2, then how may moles of O2 must have been made? (a bit challenging, but just think about it and you can probably figure it out) Stoichiometry Worksheet and Key 1.65 mol KClO3 mol KClO3 mol O2 = mol O2 3.50mol KCl = mol KClO3 = 0.275 mol Fe = mol Fe2O3 = = 2 KClO3 è 2 KCl + 3 O2 10. How many grams of O2 will be formed from 3.76 grams of KClO3? 11. How many grams of KClO3 are needed to make 30.0 grams of KCl? 12. How many grams of KCl will be formed from 2.73 g of KClO3? 4 Fe + 3 O2 è 2 Fe2O3 13. How many grams of Fe2O3 are produced when 42.7 grams of Fe is reacted? 14. How many grams of Fe2O3 are produced when 17.0 grams of O2 is reacted? 15. How many grams of O2 are needed to react with 125 grams of Fe? Some cars can use butane (C4H10) as fuel: 2 C4H10 + 13 O2 è 8 CO2 + 10 H2O 16. How many grams of CO2 are produced from the combustion of 100. grams of butane? 17. How many grams of O2 are needed to react with of 100. grams of butane? 18 How many grams of H2O are produced when 5.38g of O2 is reacted? 3.76g KClO3 mol O2 = g O2 1 mol KClO3 122.55 g KClO3 mol O2 g O2 mol KClO3 30.0 g KCl mol KClO3 mol KCl mol KClO3 mol KCl g KCl g KClO3 g KClO3 = 2.73 g KClO3 g KCl = 42.7 g Fe mol Fe2O3 mol Fe mol Fe2O3 mol Fe g Fe g Fe2O3 g Fe2O3 = 17.0 g O2 g Fe2O3 = = 100. g C4H10 g CO2 = 100. g C4H10 g O2 = KEY 2 KClO3 è 2 KCl + 3 O2 1. How many moles of O2 will be formed from 1.65 moles of KClO3? 2. How many moles of KClO3 are needed to make 3.50 moles of KCl? 3. How many moles of KCl will be formed from 2.73 moles of KClO3? 4 Fe + 3 O2 è 2 Fe2O3 4. How many moles of Fe2O3 are produced when 0.275 moles of Fe are reacted? 5. How many moles of Fe2O3 are produced when 31.0 moles of O2 are reacted? 6. How many moles of O2 are needed to react with 8.9 moles of Fe? 2 H2O è 2 H2 + O2 7. How many moles of O2 are produced when 1.26 moles of H2O is reacted? 8. How many moles of H2O are needed to produce 55.7 moles of H2? 9. If enough H2O is reacted to produce 3.40 moles of H2, then how may moles of O2 must have been made? (a bit challenging, but just think about it and you can probably figure it out) 1.65 mol KClO3 2 mol KClO3 = mol O2 2.48 3.50 mol KCl mol KCl mol KClO3 = mol KClO3 2 3.50 2.73 moles KClO3 mol KClO3 mol KCl = mol KCl 2 2.73 0.275 mol Fe = mol Fe2O3 4 mol Fe 2 mol Fe2O3 0.138 31.0 mol O2 = mol Fe2O3 3 mol O2 2 mol Fe2O3 20.7 8.9 mol Fe = mol O2 4 mol Fe 3 mol O2 6.7 1.26 mol H2O = mol O2 2 mol H2O 1 mol O2 .630 55.7 mol H2 = mol H2O 2 mol H2 2 mol H2O 55.7 3.40 mol H2 = mol O2 2 mol H2 1 mol O2 1.70 3 mol O2 2 2 2 KClO3 è 2 KCl + 3 O2 10. How many grams of O2 will be formed from 3.76 grams of KClO3? 11. How many grams of KClO3 are needed to make 30.0 grams of KCl? 12. How many grams of KCl will be formed from 2.73 g of KClO3? 4 Fe + 3 O2 è 2 Fe2O3 13. How many grams of Fe2O3 are produced when 42.7 grams of Fe is reacted? 14. How many grams of Fe2O3 are produced when 17.0 grams of O2 is reacted? 15. How many grams of O2 are needed to react with 125 grams of Fe? Some cars can use butane (C4H10) as fuel: 2 C4H10 + 13 O2 è 8 CO2 + 10 H2O 16. How many grams of CO2 are produced from the combustion of 100. grams of butane? 17. How many grams of O2 are needed to react with of 100. grams of butane? 18 How many grams of H2O are produced when 5.38g of O2 is reacted? 3.76g KClO3 mol O2 = g O2 1 mol KClO3 122.55 g KClO3 mol O2 g O2 mol KClO3 2 1 1.47 30.0 g KCl mol KClO3 mol KCl mol KClO3 mol KCl g KCl g KClO3 = 74.55 2 1 49.3 2.73 g KClO3 mol KCl mol KCl mol KCl O3 g KClO3 g KCl = 122.55 2 1 1.66 74.55 g KCl 42.7 g Fe mol Fe2O3 mole Fe mol Fe2O3 mol Fe g Fe g Fe2O3 g Fe2O3 = 55.85 4 1 61.0 17.0 g O2 1 mol O2 mol Fe2O3 mol O2 g O2 g Fe2O3 g Fe2O3 = 32.00 3 1 56.6 125 g Fe 1 mol Fe mol O2 mol Fe g Fe g O2 g O2 = 55.85 4 1 53.7 g 100. g C4H10 g CO2 = 58.14 g C4H10 2 mol C4H10 44.01 g CO2 303 100. g C4H10 g O2 = 58.14 g C4H10 2 mol C4H10 32.00 g O2 358 1 mol O2 5.38g O2 g H2O = 32.00 g O2 18.02 g H2O 2.33 1 mol H2O 13 mol O2 1 2 122.55 g KClO3 159.70 1 mol O2 10 mol H2O 3 32.00 1 2 1 2 159.70 mol Fe2O3 2 mol O2 3 32.00 1 mol C4H10 8 mol CO2 1 mol CO2 1 mol C4H10 13 mol O2 mol KCl O3 |
10555 | https://intellipaat.com/blog/backtracking-algorithm/ | Backtracking Algorithm - Definition, Usecase, and Example
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In this blog, we will examine the foundations of backtracking, its uses, and the secret to its ability to methodically traverse decision spaces. Come along as we walk you through every aspect of this effective algorithm, illuminating its operation and presenting instances from real-world applications where it excels. This blog serves as your entryway to grasp the particulars of backtracking, regardless of your level of programming experience or curiosity.
Table of Contents:
Introduction to Backtracking Algorithm
State Space Tree
Types of Backtracking Algorithm
Recursive Backtracking
Non-Recursive Backtracking
Working of Backtracking Algorithm
Examples of Backtracking Algorithm
When to use Backtracking Algorithm
Applications of Backtracking Algorithm
Wrap-up
FAQs
Learn Data Structures and Algorithms in this comprehensive video course:
Introduction to Backtracking Algorithm
Backtracking is a systematic problem-solving technique employed in computer science and mathematics to find solutions to problems, especially those involving combinatorial choices, such as finding all possible paths or arrangements. The fundamental idea behind backtracking is to explore all possible options systematically and backtrack when a dead end is reached.
Imagine you’re in a maze and want to find the exit. Backtracking works like this: you explore a path until you reach a point where it doesn’t lead to the solution. At that point, you backtrack to the last decision point and try a different path. This process continues until you find the exit or exhaust all possibilities.
In programming, backtracking is often implemented using recursion. The algorithm explores a branch of the solution space, backtracking to the previous step if it fails and exploring another branch. This continues until a solution is found or all possibilities are explored.
For example, solving Sudoku or the N-Queens problem involves backtracking. In Sudoku, you fill in numbers, and if you reach a point where the puzzle becomes unsolvable, you backtrack to the last valid move. Backtracking is a powerful technique for systematically solving complex problems by exploring solution spaces.
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State Space Tree
A state space tree in the context of a backtracking algorithm as a roadmap for problem-solving. Imagine being on a journey, facing multiple choices at each point. The tree represents all potential paths, with each node serving as a decision point. Backtracking involves exploring these paths; if encounter a dead end, you backtrack to the last decision point and explore a different route.
For instance, in a maze, nodes might represent various paths, and backtracking aids in exploration until the correct path is found. The state space tree guides this exploration systematically, facilitating a methodical approach to problem-solving by navigating through choices, trying options, and backtracking when necessary to find the optimal solution.
Types of Backtracking Algorithm
Backtracking is a general algorithm for finding all (or some) solutions to computational problems, particularly constraint satisfaction problems. It incrementally builds candidates for solutions and abandons a candidate as soon as it determines that the candidate cannot possibly be completed to a valid solution. Here are some types of backtracking algorithms:
Recursive Backtracking
Non-Recursive Backtracking
Recursive Backtracking
In this basic form of backtracking, the algorithm explores each branch of the solution space recursively. If a solution is found, it is returned; otherwise, the algorithm backtracks to the previous decision point and explores alternative choices.
Here’s a simple example of a recursive backtracking algorithm for generating permutations of a set of elements:
def permute_recursive(nums, path, result):
# Base case: if all elements are used, add the current permutation to the result
if not nums:
result.append(path[:])
return
# Explore all possible choices
for i in range(len(nums)):
# Make a choice
path.append(nums[i])
# Explore with the chosen element removed from the options
permute_recursive(nums[:i] + nums[i+1:], path, result)
# Backtrack by undoing the choice
path.pop()
Example usage:
nums = [1, 2, 3]
result = []
permute_recursive(nums, [], result)
print(result)
Output:
This recursive algorithm generates all permutations of the input list “nums”. The “permute_recursive”function takes three parameters: the remaining elements to permute (nums), the current partial permutation (“path”), and the list to store the final result (“result”).
Non-Recursive Backtracking
Backtracking can also be implemented using iteration and an explicit stack or queue to manage the state of the search. This can be useful in situations where recursive calls are not desirable due to factors like memory constraints.
Here’s an example of a non-recursive backtracking algorithm for generating permutations using an explicit stack:
def permute_non_recursive(nums):
result = []
stack = [(nums, [])]
while stack:
current_nums, current_path = stack.pop()
if not current_nums:
result.append(current_path)
continue
for i in range(len(current_nums)):
stack.append((current_nums[:i] + current_nums[i+1:], current_path + [current_nums[I]]))
return result
Example usage:
nums = [1, 2, 3]
result = permute_non_recursive(nums)
print(result)
Output:
In this non-recursive version, a stack is used to simulate the recursive calls. The stack contains tuples representing the state of the algorithm, and the while loop continues until the stack is empty. The algorithm explores choices iteratively, and the stack is manipulated to simulate the recursive calls and backtracking.
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Working of Backtracking Algorithm
Let us now break down how backtracking works, step by step. It builds solutions bit by bit, backtracking when needed, and cleverly explores different paths until it finds the right solution.
Step 1: Understanding the Problem
Backtracking is a technique used to systematically search for a solution to a problem, especially in cases where a brute-force approach would be impractical. Typically, backtracking is applied to problems with multiple possible solutions, such as puzzles, mazes, or optimization problems.
Step 2: Choose a Path
The algorithm begins by choosing a path to explore from the current state. It makes a decision and moves forward, marking the chosen path.
Step 3: Explore the Chosen Path
The algorithm explores the selected path and reaches a new state. At this point, it evaluates whether the current state is a solution to the problem. If it is, the algorithm stops; otherwise, it continues to explore further.
Step 4: Backtrack if Necessary
If the algorithm reaches a state where it cannot proceed or the chosen path does not lead to a solution, it backtracks. Backtracking involves undoing the last decision made, returning to the previous state, and exploring alternative paths.
Step 5: Explore Alternative Paths
Having backtracked, the algorithm continues to explore alternative paths from the previous state. It repeats the process of choosing a path, exploring, and backtracking until a solution is found or all possibilities are exhausted.
Let’s consider a scenario where you have three persons named A, B, and C. Now we have to create a seating arrangement for the three of them. There are multiple ways or combinations in which they can sit. The Backtracking algorithm helps in that.
Let us visualize the tree for this:
In this example, we began by placing A at the first position, followed by B and then C. Subsequently, we backtracked, positioning C at the second spot and B at the third. This yielded two options: ABC and ACB.
Continuing, we placed B at the first position, followed by A at the second, and C at the third. Once again, backtracking occurred, with C in the second position and A in the third. This resulted in two additional options: BAC and BCA.
Lastly, we positioned C at the first spot, followed by A in the second, and B in the third. A final backtrack led to B at the second and A at the third position, concluding with the last two options: CAB and CBA.
The entire backtracking process led us to the final six combinations of their seating, which are:
ABC, BAC, CAB
ACB, BCA, CBA
Examples of Backtracking Algorithm
Consider the N-Queens problem, where you need to place N queens on an N✕N chessboard in such a way that no two queens threaten each other. The backtracking algorithm for this problem involves placing queens row by row and backtracking when conflicts arise.
Choose a Path: Start with the first row and place a queen in an empty column.
Explore the Chosen Path: Move to the next row and repeat the process, ensuring that the queens do not threaten each other.
Backtrack if Necessary: If a conflict is detected, backtrack to the previous row and explore alternative columns for placing the queen.
Explore Alternative Paths: Continue exploring and backtracking until a valid placement for all queens is found.
There are ten solutions to the problem after it is solved, so we must use the backtracking algorithm to get each of them.
When to use Backtracking Algorithm
The backtracking algorithm explores all possible candidates for a solution and backtracks when it determines that a candidate cannot lead to a valid solution. Here’s when you might use a backtracking algorithm:
1. Permutation or Combination Problems: One can use this algorithm when one needs to generate all possible permutations or combinations of a set of elements. For example, solving puzzles like Sudoku, and N-Queens, or generating all possible combinations of elements from a set.
2. Constraint Satisfaction Problems: When you have a problem that can be broken down into smaller decisions, each with constraints, this technique can be used. Backtracking can efficiently solve problems where choices made need to satisfy specific conditions or constraints, like scheduling or resource allocation problems.
3. Decision Problems: This algorithm is employed when you need to find a feasible solution or not, rather than generating all possible solutions. Backtracking can be used to efficiently solve problems like graph coloring or finding a Hamiltonian cycle in a graph.
Applications of Backtracking Algorithm
Backtracking finds applications across various domains due to its ability to systematically explore and solve problems. Some common applications include:
1. Puzzle Solving: Backtracking is extensively used in puzzles like Sudoku, crosswords, word search, and maze solving. It efficiently explores the solution space to find a valid solution.
2. Game Playing: Games like chess, tic-tac-toe, and other board games often use backtracking to search through possible moves and find the best move or determine if a winning strategy exists.
3. Combinatorial Problems: Problems involving permutations, combinations, and subsets, such as generating all possible combinations of a set of elements or finding permutations of a sequence, utilize backtracking for efficient exploration.
4. Constraint Satisfaction Problems: Backtracking helps solve problems where constraints need to be satisfied, such as scheduling tasks, resource allocation, or job sequencing.
5. Graph Problems:It’s used in graph-related problems like finding a Hamiltonian cycle, solving the traveling salesman problem, graph coloring, and finding paths in a graph.
6. Optimization Problems:In some cases, backtracking can be used as part of an optimization strategy, exploring possible solutions and improving them incrementally to reach an optimal solution.
7. Robotics and Pathfinding:Backtracking is employed in robotics for pathfinding algorithms, helping robots find the shortest path from one point to another while avoiding obstacles.
Wrap-up
In conclusion, the backtracking algorithm stands as a powerful tool in the domain of problem-solving, offering an approach to exploring solution spaces while navigating through constraints and conditions. Through its recursive nature and depth-first search strategy, it efficiently seeks solutions to various combinatorial problems, puzzle-solving challenges, and constraint satisfaction scenarios.
Its adaptability across diverse domains, from puzzles and games to optimization and graph-related conundrums, showcases its versatility and applicability. Despite its effectiveness in exploring solution spaces, its efficiency may be affected by the size and complexity of the problem, warranting potential optimizations or alternative approaches for larger-scale scenarios.
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FAQs
What is backtracking?
Backtracking is a systematic algorithmic technique used to find solutions to problems by incrementally exploring potential candidates and backtracking from paths that do not satisfy the problem constraints.
How does backtracking work?
Backtracking works by exploring the solution space incrementally. It makes a series of decisions, moving forward until it reaches a dead end or finds a solution. Upon encountering an invalid solution, it backtracks to the last valid decision point and explores other paths.
What are the key steps involved in implementing a backtracking algorithm?
The main steps in implementing a backtracking algorithm include defining the problem, identifying decision points, establishing constraints, recursively exploring solutions, making decisions, backtracking, and terminating when a solution is found or all possibilities are exhausted.
Are there any limitations or drawbacks of using backtracking?
Backtracking’s efficiency can decrease significantly with large problem space, leading to excessive exploration. Additionally, not all problems can be efficiently solved using backtracking, especially those with extremely large or complex solution spaces.
How can I optimize a backtracking algorithm?
Optimizations for backtracking may include pruning branches of the search tree early if they are determined to lead to invalid solutions, employing heuristics to guide the search, or using additional data structures to enhance efficiency.
Are there alternatives to backtracking for solving similar problems?
Yes, other techniques like dynamic programming, greedy algorithms, or even specialized algorithms tailored to specific problems can sometimes provide more efficient solutions compared to backtracking, especially for certain types of problems.
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10556 | https://www.legislation.gov.uk/ukpga/2006/46/section/39 | Companies Act 2006
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Companies Act 2006
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UK Public General Acts
2006 c. 46
Part 4
Capacity of company and...
Section 39
Table of Contents
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Act amendment to earlier affecting provision S.I. 2014/3348, art. 220A by S.I. 2024/1115reg. 4(3)
s. 790EB(1A)(1B) inserted by S.I. 2025/1036reg. 4(a)
s. 790EB(3) inserted by S.I. 2025/1036reg. 4(b)
s. 790EC(1A)(1B) inserted by S.I. 2025/1036reg. 5(a)
s. 790EC(3) inserted by S.I. 2025/1036reg. 5(b)
s. 790ED-790EF inserted by S.I. 2025/1036reg. 6
s. 790LF(4) inserted by S.I. 2025/1036reg. 12(b)
s. 790LG(1A) inserted by S.I. 2025/1036reg. 13(3)
s. 790LH(4)(5) inserted by S.I. 2025/1036reg. 14(4)
s. 1098C(1)(ea) inserted by S.I. 2025/1037Sch. 1para. 2(a)(ii)
s. 1098C(2)(e) inserted by S.I. 2025/1037Sch. 1para. 2(b)(iii)
s. 1098C(4A) inserted by S.I. 2025/1037Sch. 1para. 2(c)
Sch. 2 Pt. 2 Section A para. 25(m)(ii) words omitted by S.I. 2025/381Sch.para. 19
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s. 790EB(1A)(1B) inserted by S.I. 2025/1036reg. 4(a)
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s. 790EC(1A)(1B) inserted by S.I. 2025/1036reg. 5(a)
s. 790EC(3) inserted by S.I. 2025/1036reg. 5(b)
s. 790ED-790EF inserted by S.I. 2025/1036reg. 6
s. 790LF(4) inserted by S.I. 2025/1036reg. 12(b)
s. 790LG(1A) inserted by S.I. 2025/1036reg. 13(3)
s. 790LH(4)(5) inserted by S.I. 2025/1036reg. 14(4)
s. 1098C(1)(ea) inserted by S.I. 2025/1037Sch. 1para. 2(a)(ii)
s. 1098C(2)(e) inserted by S.I. 2025/1037Sch. 1para. 2(b)(iii)
s. 1098C(4A) inserted by S.I. 2025/1037Sch. 1para. 2(c)
Sch. 2 Pt. 2 Section A para. 25(m)(ii) words omitted by S.I. 2025/381Sch.para. 19
39 A company's capacity U.K.
This section has no associated Explanatory Notes
(1)The validity of an act done by a company shall not be called into question on the ground of lack of capacity by reason of anything in the company's constitution.
(2)This section has effect subject to section 42 (companies that are charities).
Modifications etc. (not altering text)
C1 S. 39 applied (with modifications) (1.10.2009) by The Unregistered Companies Regulations 2009 (S.I. 2009/2436), regs. 3-5, Sch. 1 para. 3(a) (with transitional provisions and savings in regs. 7, 9, Sch. 2)
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10557 | https://chem.libretexts.org/Courses/Portland_Community_College/CH151%3A_Preparatory_Chemistry/03%3A_Dimensional_Anlaysis_and_Density | Skip to main content
3: Dimensional Anlaysis and Density
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2.5: The International System of Units
3.1: Problem Solving and Unit Conversions
Page ID
: 364764
( \newcommand{\kernel}{\mathrm{null}\,})
3.1: Problem Solving and Unit Conversions
: Converting from one unit to another is a particularly important skill in science and engineering. Dimensional analysis (also called the unit factor method) is a technique for making these conversions correctly. Dimensional analysis will be used in later chapters to perform other chemistry calculations besides just unit conversions.
3.2: Multi-Step Conversion Problems
: Sometimes you will have to perform more than one conversion to obtain the desired unit.
3.3: Units Raised to a Power
: Conversion factors for area and volume can also be produced by the dimensional analysis method. Just remember that if a quantity is raised to a power both the number and the unit must be raised to that same power.
3.4: Units in the Numerator and the Denominator
: Some complex units are composed of a unit in the numerator and a unit in the denominator. These include units of speed and mileage. Dimensional analysis can be used to make conversions with these units as well.
3.5: Density
: Density is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant.
3.6: Temperature
: Three different scales are commonly used to measure temperature: Fahrenheit (expressed as °F), Celsius (°C), and Kelvin (K).
2.5: The International System of Units
3.1: Problem Solving and Unit Conversions |
10558 | https://math.stackexchange.com/questions/409057/confused-about-the-group-of-permutations-s-n | abstract algebra - Confused about the group of permutations $S_{n}$ - Mathematics Stack Exchange
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Confused about the group of permutations S n
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Asked 12 years, 4 months ago
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In an exercise, I must prove S n is generated by 2 elements. I'll ignore here the trivial case n=1. Let I n={1,2,3,...,n}. I then defined f:I n→I n by f(1)=2, f(2)=1, and f(i)=i, for i≠1,2. Also g:I n→I n by g(1)=n and g(i+1)=i for i∈{1,...,n−1}. I claimed f,g generate S n, which is apparently correct. But my explanation is apparently completely wrong. I hope someone may undo my confusion.
So intuitively, f is a swap permutation which swaps the first two elements, and g is a shift permutation which shifts everything to the left. We note any permutation can be written as a series of "swaps" (transpositions). So it suffices to show f,g generate the swaps. So suppose I want to compose f and g in such a way to swap 1≤i<j≤n. The algorithm would go somewhat like this: shift everything to the left until i is the first element. Now apply swap once, then shift back again, apply swap, shift back, and so forth until you just swapped i with j. Count how many times you shifted left after you first swapped. Now you shift right (which is just the inverse of shifting left), swap, shift right, swap, .. and do that just as many times as you shifted left. In the end you should be left with exactly i,j swapped and everything else in its place. Let's do an example. Suppose I want to swap 1 and 4 in 1234. It'll go as:
2134 (swapped)
1342 (shifted left 1 time)
3142 (swapped)
1423 (shifted left 2 times)
4123 (swapped 4 and 1; I'll start shifting right now)
3412 (shifted right 1 time)
4312 (swapped)
2431 (shifted right 2 times)
4231 (swapped)
And we are done! Apparently the problem with this is that f doesn't actually swap the first and second elements, there is no order, it just swaps 1 with 2. Under which interpretation this makes little sense. My question is: is my way of thinking about this completely wrong? Is it merely coincidental that I achieved a correct answer? And if it is wrong, how should I be thinking about permutations? I am not familiar with the cycle notation.
abstract-algebra
group-theory
permutations
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edited Jun 2, 2013 at 6:19
anon
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asked Jun 2, 2013 at 5:54
PedroPedro
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Hopefully all fixed now :p Pedro –Pedro 2013-06-02 06:01:21 +00:00 Commented Jun 2, 2013 at 6:01
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Your solution is basically correct, though not very well formulated. But you are not clear about how you represent a permutation, and whether "applying" a permutation corresponds to multiplication on the left or on the right, and this makes it difficult to point precisely to what you should do different. But in any case you can salvage your kind of reasoning by making the proper choices.
In group theory permutations are defined as functions (bijections), not as permuted sequences. So first let us fix the correspondence: a sequence (a 1,…,a n) (with {a 1,…,a n}={1,…,n}) represents the permutation σ that sends i↦a i for all i. (This seems a natural choice, but a choice nonetheless.) Next we need to be clear about left and right in permutation composition; since you write (permutation) functions before their argument, let us choose as usual that the composition f.g is i↦f(g(i)) (the permutation on the right gets to operate first). Again this is a natural choice, but some prefer to have f act first (typically those who prefer to write i f instead of f(i)); I just want to make my choice clear.
With these choices, if σ is represented by (a 1,…,a n), then π.σ is represented by (π(σ(1)),…,π(σ(n)))=(π(a 1),…,π(a n)), in other words left multiplication by π corresponds to applying the function π to the values of individual entries of the sequence, not to permuting the entries of the sequence according to π. However if we multiply on the right by π, the situation is different: σ.π is represented by (σ(π(1)),…,σ(π(n)))=(a π(1),…,a π(n)). Here it is the positions that are permuted. Since you use this in your argument, it can be made valid if you stipulate that each successive permutation applied to the sequence corresponds to a right-multiplication.
I should warn about a slight twist that remains: right multiplication by π does permute the entries in the sequence, but it does so in such a way that the new entry at position i is a π(i), which is the one that used to be in position π(i); the old value a i will be found after permutation at position π−1(i). This is what you probably associate with permuting the entries according to the inverse permutation π−1. Indeed this is true for the usual definition of permutations acting (from the left!) on sequences, as I detailed in this answer. If you think of it, it is normal that when right-multiplication is related to a left-action on sequences, an inversion should be used in the process.
No doubt this somewhat unfortunate consequence of the otherwise natural choices above is what motivates people to make the opposite choice for one of the two. As I said it is often the second that is reversed, but this has other notational consequences that can be confusing. For me the real culprit is the first choice: it would be more natural to represent σ as the result of permuting the standard sequence (1,2,…,n) according to σ, and this gives (σ−1(1),σ−1(2),…,σ−1(n)) rather than (σ(1),σ(2),…,σ(n)). However, I would not really advocate doing this, as this would make the confusion between notations even worse.
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edited Apr 13, 2017 at 12:20
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answered Jun 2, 2013 at 7:56
Marc van LeeuwenMarc van Leeuwen
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Here's an easier way to think about. For 0≤j≤n−1, consider the element
(1…n)−j(12)(1…n)j
Let k∈{1,…,n}. Under the rightmost cycle, k→(k+j)mod n. (Think of 0 as the symbol n, here).
If (k+j)mod n≢1,2, then under the entire permutation, k→k. If 1≡(k+j)mod n, then k≡(1−j)mod n, and k→(2−j)mod n. Hence, (1…n)−j(12)(1…n)j is a transposition of adjacent elements, and we know adjacent transposition generate S n.
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answered Jun 2, 2013 at 6:46
HarryHarry
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As you say, applying f only switches 1 and 2, it does not 'switch the first two entries in a permutation's cycle notation' (to paraphrase what you want to happen); such a thing is not even generically well-defined because there are multiple ways of representing a permutation with cycles.
The reason it works out is that if you could switch any two elements in a permutation's cycle representation, then you would indeed be able to generate all the necessary transpositions. But you haven't outlined how you would go about doing that with just f and g.
Here is a suggestion: show that adjacent transpositions generate all transpositions, which suffices for the task because transpositions generate all the rest of S n, as you know. You only need to construct every adjacent transposition (i i+1) for i=1,⋯,n−1. To do this, keep in mind the effect that not products but conjugations have on cycle representations:
σ(a 1 a 2⋯a l 1)⋯(b 1 b 2⋯b l m)σ−1=(σ a 1 σ a 2⋯σ a l 1)⋯(σ b 1 σ b 2⋯σ b l m).
So what permutations do you get upon repeatedly conjugating f=(1 2) by g=(1 2⋯n)?
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edited Jun 2, 2013 at 7:04
answered Jun 2, 2013 at 6:15
anonanon
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The question does not mention cycle notation for permutation, and as far as I can tell does not suggest their implicit used in any way either. Instead a "permuted sequence" representation (one-line notation) seems to be used.Marc van Leeuwen –Marc van Leeuwen 2013-06-02 08:11:03 +00:00 Commented Jun 2, 2013 at 8:11
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I think you would prefer better notation, in which the proof is more visually obvious. Consider: f=1 2…n 2 1…n g=1 2…n n 1…n−1 where the correspondence between an i in the first row and j in the second row is that j=f(i) or g(i), respectively. Your goal is to express the transpositions ("swaps" is not a standard term) in terms of f and g. Before I do that, there is a further simplification you missed:
Lemma:S n is generated by the "adjacent transpositions" s i exchanging i with i+1 (modulo n).
The diagram for s i is: s i=1…i i+1…n 1…i+1 i…n (with s n wrapping around). This is just f, "moved" i−1 places to the right. Note that g "moves" everything 1 place to the right, so g i−1 moves i−1 places. Therefore the following permutation is s i: 1…i i+1…n n−i+2…1 2…n−i+1(g i−1)n−i+2…2 1…n−i+1(f)1…i+1 i…n(g−(i−1)) where each row is marked with what we applied to the previous row. Note that every row but the first is out of order, but we just apply the indicated function to whatever value is in each place. Also, values such as n−i+2 should be taken modulo n, for example when i=1.
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edited Jun 2, 2013 at 14:29
answered Jun 2, 2013 at 6:25
Ryan ReichRyan Reich
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Since the question stipulates g(1)=n, I don't think you have got the correct two-line notation for g.Marc van Leeuwen –Marc van Leeuwen 2013-06-02 08:08:20 +00:00 Commented Jun 2, 2013 at 8:08
@Marc It seems I wrote g−1.Ryan Reich –Ryan Reich 2013-06-02 14:30:10 +00:00 Commented Jun 2, 2013 at 14:30
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It would be very convenient to represent permutations in disjoint cycle notation, as anon has used. But if you want to use your notation of representing a permutation as a linear arrangement, you could understand why f,g generate S n, as follows.
Given any permutation, we need to obtain the given permutation starting from [1,2,…,n] by repeatedly using f (switching the first and second coordinates) and g (doing a cyclic shift of the elements). A right cyclic shift by i coordinates can be effected by applying the left cyclic shift g a total of n−i times.
First observe that any permutation can be obtained from the initial one by just swapping two consecutive coordinates at a time. For example, to go from 1234 to 4231, as you've done, you swap the first and second coordinate, then second and third, then third and fourth, to bring 1 to the fourth coordinate. Next, you can bring 3 to the third coordinate, 2 to second coordinate, and 4 will automatically be in the first coordinate.
It remains to be shown that swapping two consecutive coordinates can be achieved through a sequence of f's and g's. Swapping the first two coordinates is allowed since it is just f. To swap the i-th and i+1-th coordinate, just apply the cyclic shift g a certain number of times ((n−i) times) to bring the numbers in the i th and i+1 th coordinate to the 1st and 2nd coordinate, then swap the first and second coordinate by applying f, then apply g again the appropriate number of times to shift back these two numbers from the 1st and 2nd coordinate to the i th and i+1 th coordinates, respectively. In this manner, and I think you can visualize this intuitively, the numbers in any two consecutive coordinates can be swapped using the operation of cyclic shift, swapping the first two coordinates only, and cyclic shift again.
To understand further, you could learn about conjugation. It turns out that every permutation can be expressed as a disjoint union of cycles, and an important result here is that the type of a permutation (i.e. the lengths of these different cycles) remains the same if you conjugate a permutation a by another permutation b, i.e. b a b−1 has the same cycle structure as a. So, if a=(1,2) is an element of a permutation group, and b=(1,2,…,n) is also in the group, then by composing these two together in the special way b a b−1, we are effecting a cyclic shift, swap, and reverse cyclic shift to obtain that (2,3) is also in the group (or (1,n) rather than (2,3), depending on whether you compose left to right or right to left). The point is that such a product will again be some 2-cycle because conjugation preserves the type of a permutation. So we quickly obtain all consecutive swaps in this manner; for example, conjugating (2,3) by b gives (3,4), and so on.
More generally, it can be shown that if (i,j) denotes a swap of the i th and j th coordinate, then a set of swaps A generates S n if and only if the transposition graph of A is connected. In our case, by repeatedly conjugating a=(1,2) with b, we are able to obtain the path graph on the n vertices {1,2,…,n}. Since this graph is connected, a and b generate S n.
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answered Jun 3, 2013 at 3:45
AG.AG.
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10559 | https://rjalvarado.people.amherst.edu/TeachingMaterials/F17_MATH413/HW5.pdf | HOMEWORK V Due: Thursday, October 12, 2017.
1. Use PMI to show that the union of finitely many finite sets is a finite set, i.e., show that for each n ∈N, if {A1, A2, . . . , An} is a collection of finite sets, then the set ∪n k=1 Ak is finite.
Be careful: these sets are not necessarily disjoint.
2. For each n ∈N suppose that An is a nonempty finite set. Prove or disprove (with a counterexam-ple): The union ∪∞ n=1 An is an infinite set.
3. Suppose that A is a finite set. Prove the if B is an infinite set then the set B \ A is infinite.
4. Given an infinite set A, show that if f : A →B is an injective function for some set B, then f(A) is an infinite set.
5. Let A be a set with n elements and B be a set with m elements for some n, m ∈N. Prove that n ≤m if and only if there is an injection φ : A →B.
6. Suppose that A and B are two finite sets both having n elements for some n ∈N. Show that any injective function f : A →B is also surjective. Show this is not necessarily true if A and B are infinite.
7. Show that if A is a countable set and B is a finite set, then the union A ∪B is countable.
Hint: Try constructing a piecewise function.
8. Suppose that A, B, C, and D are sets and that f : A →B and g : C →D are two bijective functions. Show that the function h : A × C →B × D defined by setting h(x, y) := ( f(x), g(y) ) , for every (x, y) ∈A × C, is bijective.
9. Show that if A and B are two countable sets then the Cartesian product A × B is countable.
Hint: Use the previous problem and the fact that N × N is countable.
1 |
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| |~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
10561 | https://flexbooks.ck12.org/cbook/ck-12-precalculus-concepts-2.0/section/4.1/primary/lesson/angles-in-radians-and-degrees-pcalc/ | Skip to content
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4.1
Angles in Radians and Degrees
Written by:CK-12 | Mark Spong
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Lesson
Most people are familiar with measuring angles in degrees. It is easy to picture angles like, or and the fact that makes up an entire circle. Over 2000 years ago the Babylonians used a base 60 number system and divided up a circle into 360 equal parts. This became the standard and it is how most people think of angles today.
However, there are many units with which to measure angles. For example, the gradian was invented along with the metric system and it divides a circle into 400 equal parts. The sizes of these different units are very arbitrary.
A radian is a unit of measuring angles that is based on the properties of circles. This makes it more meaningful than gradians or degrees. How many radians make up a circle?
Radians and Degrees
A radian is defined to be the central angle where the subtended arc length is the same length as the radius.
Another way to think about radians is through the circumference of a circle. The circumference of a circle with radius is . Just over six radii (exactly radii) would stretch around any circle.
To define a radian in terms of degrees, equate a circle measured in degrees to a circle measured in radians.
, so
Alternatively; , so
The conversion factor to convert degrees to radians is:
The conversion factor to convert radians to degrees is:
If an angle has no units, it is assumed to be in radians.
If you were to convert into radians, you would multiply by the correct conversion factor. You would get:
You can check your work by making sure the degree units appear on both the numerator and denominator.
If you were to convert radians into degrees, you would multiply by the correct conversion factor. You would get
Notice appears in both the numerator and denominator and .
Examples
Example 1
Earlier, you were asked how many radians make up a circle. Exactly radians describe a circular arc. This is because radii wrap around the circumference of any circle.
Example 2
Convert into radians.
Don’t be fooled just because this has . This number is about
It is very unusual to ever have a term, but it can happen.
Example 3
Convert into degrees.
Example 4
Convert into radians.
Example 5
Draw a angle by first drawing a angle, halving it and halving the result. Recall that .
| |
| Summary |
| A radian is the central angle where the subtended arc length is the same length as the radius. A circle with radius has a circumference of and exactly radii would stretch around any circle. The conversion factor to convert degrees to radians is: If an angle has no units, it is assumed to be in radians. |
Review
Find the radian measure of each angle.
1.
2.
3.
4.
5.
6.
7.
Find the degree measure of each angle.
8.
9.
10.
11.
12.
13.
3
Explain why if you are given an angle in degrees and you multiply it by you will get the same angle in radians.
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Overview
A radian is the central angle where the subtended arc length is the same length as the radius.
A circle with radius @$r@$ has a circumference of @$2\pi r,@$ and exactly @$2\pi@$ radii would stretch around any circle.
The conversion factor to convert degrees to radians is: @$\frac{\pi}{180°}@$
If an angle has no units, it is assumed to be in radians.
Vocabulary
degree
Test Your Knowledge
Question 1
Express @$\begin{align}13.5^\circ \end{align}@$ in terms of radians.
a
@$\begin{align}\frac{2430}{\pi}\end{align}@$
b
@$\begin{align}2430\pi\end{align}@$
c
@$\begin{align}\frac{3\pi}{80}\end{align}@$
d
@$\begin{align}\frac{3\pi}{40}\end{align}@$
To convert from degrees to radians, multiply by @$\begin{align}\frac{\pi}{180^\circ}\end{align}@$.
@$$\begin{align}\eqalign{
&& 13.5^\circ \
&& 13.5^\circ \times \frac{\pi}{180^\circ} \
&& \frac{13.5^\circ \pi}{180^\circ} \
&& \frac{3 \pi}{40}
}\end{align}@$$
Question 2
An angle measuring @$\begin{align}45^\circ \end{align}@$ is equal to _____ radians.
a
@$\begin{align}4\pi\end{align}@$
b
@$\begin{align}\frac{\pi}{4}\end{align}@$
c
@$\begin{align}\frac{\pi}{6}\end{align}@$
d
@$\begin{align}\frac{4}{\pi}\end{align}@$
To convert from degrees to radians, multiply by @$\begin{align}\frac{\pi}{180^\circ}\end{align}@$.
@$$\begin{align}\eqalign{
&& 45^\circ \
&& 45^\circ \times \frac{\pi}{180^\circ} \
&& \frac{45^\circ \pi}{180^\circ} \
&& \frac{\pi}{4}
}\end{align}@$$
An angle measuring @$\begin{align}45^\circ \end{align}@$ is equal to @$\begin{align}\frac{\pi}{4}\end{align}@$ radians.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Related Content
Radian and Degree Conversions - Example 1
Radian and Degree Conversions - Example 3
Conversion between Degrees and Radians: Clock Angles and Measures
Conversion between Degrees and Radians: Arc and Radians Relationship
| Image | Reference | Attributions |
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10562 | https://www.pathologyoutlines.com/topic/cnstumorwhoclassification.html | Home > CNS & pituitary tumors > WHO classification
CNS & pituitary tumors
General
WHO classification
Author:Maria Martinez-Lage, M.D.
Editorial Board Member:P.J. Cimino, M.D., Ph.D.
Deputy Editor-in-Chief:Chunyu Cai, M.D., Ph.D.
Last author update: 16 June 2023
Last staff update: 14 October 2024
Copyright: 2002-2025, PathologyOutlines.com, Inc.
PubMed Search: WHO classification CNS 2021
Page views in 2024: 23,307
Page views in 2025 to date: 15,305
Table of Contents
Definition / general | Major updates | WHO (2021) | Diagrams / tables | Grading | Additional references | Practice question #1 | Practice answer #1
Cite this page: Martinez-Lage M. WHO classification. PathologyOutlines.com website. Accessed August 24th, 2025.
Definition / general
Extensively revised in 1993 (Brain Pathol 1993;3:255)
Third edition in 2000 (J Neuropathol Exp Neurol 2002;61:215)
Fourth edition in 2007 (Acta Neuropathol 2007;114:97)
2016 WHO classification is considered a revision of the fourth edition (Acta Neuropathol 2016;131:803)
Fifth edition published (online) in 2021 (Neuro Oncol 2021;23:1231)
Major updates
2021 WHO classification, 5th edition (WHO CNS5) summary
Builds upon the updated fourth edition published in 2016 and the subsequent recommendations of the Consortium to Inform Molecular and Practical Approaches to CNS Tumor Taxonomy (cIMPACT-NOW) (Brain Pathol 2017;27:851, Acta Neuropathol 2017;133:1, Acta Neuropathol 2018;135:481, Acta Neuropathol 2018;135:639, Acta Neuropathol 2018;136:805, Acta Neuropathol 2019;137:683, Acta Neuropathol 2020;139:603, Brain Pathol 2020;30:844, Brain Pathol 2020;30:863)
This fifth edition incorporates major changes including further advancing the role of molecular information in the diagnosis of CNS neoplasms, introducing the concept of CNS WHO grade distinct from WHO grade, grading within tumor type, endorsing the use of Arabic numerals for grading instead of Roman numerals and defining new entities based on molecular and histological characteristics (Neuro Oncol 2021;23:1231)
Taxonomy maintains a hybrid approach including histological and molecular features with some entities requiring specific molecular changes for a diagnosis (e.g., IDH mutant gliomas), while others display a looser association with molecular signatures that are not necessary for a diagnosis (e.g., pleomorphic xanthoastrocytoma and BRAF alteration); for each tumor type, diagnostic criteria are enumerated and listed as essential or desirable
Grading within tumor type and the use of Arabic (rather than Roman numerals) are both introduced for the first time in CNS classification to move closer to the grading approach of non-CNS neoplasms; however, some idiosyncrasies remain based on historical knowledge and practices, in addition to the use of molecular signatures to determine grade in some cases, which further warrants the introduction and use of CNS WHO grade instead of WHO grade
Grading within types in gliomas implies the elimination of the term anaplastic, which was previously linked to grade in these tumors
Anatomic site modifiers that were previously part of a tumor name are removed to make nomenclature simpler and more consistent (e.g., chordoid glioma of the third ventricle is now simply chordoid glioma)
In addition to NOS (not otherwise specified), NEC (not elsewhere classified) is introduced
NOS: terminology to be used when molecular information is insufficient, either because testing cannot be fully performed or because the results don't fit within a defined category
NEC: terminology to be used when necessary testing has been performed and results are available but the results do not fit perfectly in a defined tumor type
Glial, glioneuronal and neuronal tumors are entirely restructured into 6 different families in a scheme that introduces a distinction between adult type and pediatric type tumors
Adult type diffuse gliomas
Pediatric type diffuse low grade gliomas
Pediatric type diffuse high grade gliomas
Circumscribed astrocytic gliomas
Glioneuronal and neuronal tumors
Ependymomas
Classification of common adult type diffuse gliomas is drastically simplified to include only 3 types (with grading within tumor for the first 2): astrocytoma, IDH mutant (grade 2, 3 or 4), oligodendroglioma, IDH mutant and 1p / 19q codeleted (grade 2 or 3) and glioblastoma, IDH wild type (grade 4)
Consequently, glioblastoma is a term exclusively reserved for IDH wild type tumors and the nomenclature for IDH mutant astrocytomas is astrocytoma, IDH mutant, CNS WHO grade 2, 3 or 4
Classification of medulloblastomas remains largely similar to the 2016 classification, maintaining the incorporation of molecularly defined entities that determine prognostic categories, while adding novel subtypes (4 subgroups of SHH tumors and 8 subgroups of non-WNT / non-SHH medulloblastomas)
Ependymomas are restructured using a combination of histopathological features, molecular signatures and anatomic site (supratentorial, posterior fossa and spinal)
New tumor types have been added, such as diffuse hemispheric glioma, H3 G34 mutant or CNS tumor with BCOR internal tandem duplication
Some newly recognized types are considered provisional (e.g., diffuse glioneuronal tumor with oligodendroglioma-like features and nuclear clusters or intracranial mesenchymal tumor, FET::CREB fusion positive) while some lesions that were introduced in 2016 are no longer provisional (e.g., diffuse leptomeningeal glioneuronal tumor)
Meningioma is considered a single type with 15 subtypes; grading is significantly changed - while chordoid meningioma and clear cell meningioma are still considered CNS WHO grade 2 based on morphology (pending larger studies), the criteria that define atypical (grade 2) or anaplastic (grade 3) meningiomas should be applied to rhabdoid and papillary meningioma instead of assigning grade 3 based on morphology alone; furthermore, molecular alterations can influence grade, as the presence of TERT promoter mutation or homozygous deletion of CDKN2A/B is considered sufficient for a designation of CNS WHO grade 3 in meningiomas
New types of mesenchymal neoplasms are included, such as CIC rearranged sarcoma; hemangiopericytoma is now fully retired and solitary fibrous tumor is used instead, as it aligns with the soft tissue nomenclature (the 3 tiered CNS grading scheme remains)
Paragangliomas are understood as involving cells from the autonomic nervous system, therefore, they are now included in the chapter with nerve tumors and are renamed cauda equina neuroendocrine tumor; melanotic schwannoma is a term that is considered inadequate to reflect the unique features of this lesion, therefore this tumor type is now called malignant melanotic nerve sheath tumor
Some major changes are introduced to the tumors of the sellar region
Adamantinomatous craniopharyngioma and papillary craniopharyngioma are now considered 2 different tumor types (rather than subtypes) given their distinct clinicopathological and molecular characteristics
Tumors of the neurohypophysis are now considered to belong to 1 group, as they may represent morphologic variations of the same tumor
Pituitary neuroendocrine tumor (PitNET) is a newly introduced term, accompanying pituitary adenoma (pituitary adenoma / pituitary
neuroendocrine tumor)
A new tumor type, characteristic of infancy and associated with DICER1 alterations, is included (pituitary blastoma)
WHO (2021)
Gliomas, glioneuronal tumors and neuronal tumors ICD-O Adult type diffuse gliomas
+ Astrocytoma, IDH mutant
- Astrocytoma, IDH mutant, grade 29400/3
- Astrocytoma, IDH mutant, grade 39401/3
- Astrocytoma, IDH mutant, grade 49445/3
+ Oligodendroglioma, IDH mutant and 1p / 19q codeleted
- Oligodendroglioma, IDH mutant and 1p / 19q codeleted, grade 29450/3
- Oligodendroglioma, IDH mutant and 1p / 19q codeleted, grade 39451/3
+ Glioblastoma, IDH wild type9440/3
Pediatric type diffuse low grade gliomas
+ Diffuse astrocytoma, MYB or MYBL1 altered9421/1
+ Angiocentric glioma9431/1
+ Polymorphous low grade neuroepithelial tumor of the young9413/0
+ Diffuse low grade glioma, MAPK pathway altered9421/1
Pediatric type diffuse high grade gliomas
+ Diffuse midline glioma, H3 K27 altered9385/3
+ Diffuse hemispheric glioma, H3 G34 mutant9385/3
+ Diffuse pediatric type high grade glioma, H3 wild type and IDH wild type9385/3
+ Infant type hemispheric glioma9385/3
Circumscribed astrocytic gliomas
+ Pilocytic astrocytoma9421/1
+ High grade astrocytoma with piloid features9421/3
+ Pleomorphic xanthoastrocytoma9424/3
+ Subependymal giant cell astrocytoma9384/1
+ Chordoid glioma9444/1
+ Astroblastoma, MN1 altered9430/3
Glioneuronal and neuronal tumors
+ Ganglioglioma9505/1
+ Gangliocytoma9492/0
+ Desmoplastic infantile ganglioglioma / astrocytoma
- Desmoplastic infantile ganglioglioma9412/1
- Desmoplastic infantile astrocytoma9412/1
+ Dysembryoplastic neuroepithelial tumor9413/0
+ Diffuse glioneuronal tumor with oligodendroglioma-like features
and nuclear clusters (provisional)
- Mixed neuronal - glial tumors2A00.21 (ICD-11)
+ Papillary glioneuronal tumor9509/1
+ Rosette forming glioneuronal tumor9509/1
+ Myxoid glioneuronal tumor9509/1
+ Diffuse leptomeningeal glioneuronal tumor9509/3
+ Multinodular and vacuolating neuronal tumor9509/0
+ Dysplastic cerebellar gangliocytoma (Lhermitte-Duclos disease)9493/0
+ Central neurocytoma9506/1
+ Extraventricular neurocytoma9506/1
+ Cerebellar liponeurocytoma9506/1
Ependymal tumors
+ Supratentorial ependymoma, NOS9391/3
+ Supratentorial ependymoma, ZFTA fusion positive9396/3
+ Supratentorial ependymoma, YAP1 fusion positive9396/3
+ Posterior fossa ependymoma, NOS9391/3
+ Posterior fossa group A (PFA) ependymoma9396/3
+ Posterior fossa group B (PFB) ependymoma9396/3
+ Spinal ependymoma, NOS9391/3
+ Spinal ependymoma, MYCN amplified9396/3
+ Myxopapillary ependymoma9394/1
+ Subependymoma9383/1
Choroid plexus tumors ICD-O
Choroid plexus papilloma9390/0
Atypical choroid plexus papilloma9390/1
Choroid plexus carcinoma9390/3
Embryonal tumors ICD-O
Medulloblastoma
Medulloblastoma, molecularly defined
Medulloblastoma, WNT activated9475/3
Medulloblastoma, SHH activated and TP53 wild type9471/3
Medulloblastoma, SHH activated and TP53 mutant9476/3
Medulloblastoma, non-WNT / non-SHH9477/3
Medulloblastoma, histologically defined
Medulloblastoma, histologically defined9470/3
Desmoplastic nodular medulloblastoma9471/3
Medulloblastoma with extensive nodularity9471/3
Large cell medulloblastoma9474/3
Anaplastic medulloblastoma9474/3
Other CNS embryonal tumors
Atypical teratoid rhabdoid tumor9508/3
Cribriform neuroepithelial tumor (provisional)
Other specified tumors of neuroepithelial tissue of brain2A00.2Y (ICD-11)
Embryonal tumor with multilayered rosettes9478/3
CNS neuroblastoma, FOXR2 activated9500/3
CNS tumor with BCOR internal tandem duplication9500/3
CNS embryonal tumor, NEC / NOS9473/3
Pineal tumors ICD-O
Pineocytoma9361/1
Pineal parenchymal tumor of intermediate differentiation9362/3
Pineoblastoma9362/3
Papillary tumor of the pineal region9395/3
Desmoplastic myxoid tumor of the pineal region, SMARCB1 mutant
Tumors of the pineal gland or pineal region2A00.20 (ICD-11)
Cranial and paraspinal nerve tumors ICD-O
Schwannoma9560/0
Neurofibroma
Neurofibroma9540/0
Plexiform neurofibroma9550/0
Perineurioma9571/0
Hybrid nerve sheath tumor9563/0
Malignant melanotic nerve sheath tumor9540/3
Malignant peripheral nerve sheath tumor9540/3
Cauda equina neuroendocrine tumor (previously paraganglioma)8693/3
Meningiomas ICD-O
Meningioma9530/0
Mesenchymal, nonmeningothelial tumors ICD-O+ Soft tissue tumors
- Fibroblastic and myofibroblastic tumors
Solitary fibrous tumor8815/1
+ Vascular tumors
- Hemangiomas and vascular malformations
Cavernous hemangioma9121/0
Capillary hemangioma9131/0
Arteriovenous malformation9123/0
- Hemangioblastoma9161/1
+ Skeletal muscle tumors
- Rhabdomyosarcoma
Embryonal rhabdomyosarcoma8910/3
Alveolar rhabdomyosarcoma8920/3
Rhabdomyosarcoma, pleomorphic type8901/3
Spindle cell rhabdomyosarcoma8912/3
+ Uncertain differentiation
- Intracranial mesenchymal tumor, FET::CREB fusion positive
Neoplasms of uncertain behavior of connective or other soft tissue2F7C & XH9362 (ICD-11)
- CIC rearranged sarcoma9367/3
- Primary intracranial sarcoma, DICER1 mutant9480/3
- Ewing sarcoma9364/3
Chondro-osseous tumors
Chondrogenic tumors
Mesenchymal chondrosarcoma9240/3
Chondrosarcoma
Chondrosarcoma9220/3
Dedifferentiated chondrosarcoma9243/3
Notochordal tumors
Chordoma (including poorly differentiated chordoma)9370/3
Melanocytic tumors ICD-O Diffuse meningeal melanocytic neoplasms
+ Meningeal melanocytosis and meningeal melanomatosis
- Meningeal melanocytosis8728/0
- Meningeal melanomatosis8728/3
Circumscribed meningeal melanocytic neoplasms
+ Meningeal melanocytoma and meningeal melanoma
- Meningeal melanocytoma8728/1
- Meningeal melanoma8720/3
Hematolymphoid tumors ICD-O Lymphomas
+ CNS lymphomas
- Primary diffuse large B cell lymphoma of the CNS9680/3
- Immunodeficiency associated CNS lymphoma
Immunodeficiency associated lymphoproliferative disordersDB32 (ICD-11)
- Lymphomatoid granulomatosis
Lymphomatoid granulomatosis9766/1
Lymphomatoid granulomatosis, grade 19766/1
Lymphomatoid granulomatosis, grade 29766/1
Lymphomatoid granulomatosis, grade 39766/3
- Intravascular large B cell lymphoma9712/3
+ Miscellaneous rare lymphomas in the CNS
- MALT lymphoma of the dura9699/3
- Other low grade B cell lymphomas of the CNS
Lymphoplasmacytic lymphoma9671/3
Follicular lymphoma9690/3
- T cell and NK / T cell lymphomas of the CNS
T cell lymphoma9702/3
NK / T cell lymphoma9719/3
- Anaplastic large cell lymphoma (ALK+ / ALK-)9714/3
Histiocytic tumors
+ Erdheim-Chester disease9749/3
+ Rosai-Dorfman disease9749/3
+ Juvenile xanthogranuloma9749/1
+ Langerhans cell histiocytosis9751/1
+ Histiocytic sarcoma9755/3
Germ cell tumors ICD-O
Mature teratoma9080/0
Immature teratoma9080/3
Teratoma with somatic type malignancy9084/3
Germinoma9064/3
Embryonal carcinoma9070/3
Yolk sac tumor9071/3
Choriocarcinoma9100/3
Mixed germ cell tumor9085/3
Tumors of the sellar region ICD-O
Adamantinomatous craniopharyngioma9351/1
Papillary craniopharyngioma9352/1
Pituicytoma, granular cell tumor of the sellar region
and spindle cell oncocytoma
Pituicytoma9432/1
Granular cell tumor of the sellar region9582/0
Spindle cell oncocytoma8290/0
Pituitary adenoma / pituitary neuroendocrine tumor (PitNET)8272/3
Pituitary blastoma8273/3
Metastases to the CNS
Metastases to the brain and spinal cord parenchyma
Malignant neoplasm metastasis in brain,2D50 (ICD-11)
with extension code for primary tumor type
Metastases to the meninges
Malignant neoplasm metastasis in meninges2D51 (ICD-11)
Diagrams / tables
Images hosted on other servers:
Key diagnostic genes, molecules, pathways
CNS WHO grades of selected types
Newly recognized WHO 2021 tumor types
Grading
Histological grading is still used based on morphology in many instances; however, in this classification, the presence of certain molecular alterations is incorporated in the grading algorithm
For example, IDH mutant astrocytomas can be designated as grade 4 based on histology (presence of microvascular proliferation or necrosis) or if there is a homozygous deletion CDKN2A/B independent of histological features
Similarly, glioblastoma, IDH wild type can now be diagnosed in the absence of the aforementioned microscopic features, if certain molecular alterations are present in a diffuse astrocytic glioma in adults (TERT promoter mutation, EGFR gene amplification or combined gain of entire chromosome 7 and loss of entire chromosome 10 [+7 / -10])
In the clinical setting, tumor grade remains a key factor influencing choice of therapy
Additional references
Louis: WHO Classification of Tumours of the Central Nervous System, 4th Edition, 2007, WHO Classification of Tumours Editorial Board: Central Nervous System Tumours, 5th Edition, 2022
Practice question #1
Which of the following terms is the only acceptable diagnosis in the 2021 WHO classification of tumors of the central nervous system?
Anaplastic astrocytoma, IDH mutant
Anaplastic astrocytoma, IDH wild type
Glioblastoma, IDH mutant
Glioblastoma, IDH wild type
Practice answer #1
D. Glioblastoma, IDH wild type. The 2021 CNS WHO classification simplifies the diagnosis of diffuse gliomas and reserves the term glioblastoma for IDH wild type tumors. Additionally, the term can now be used for infiltrating astrocytomas that lack the classic histological features of glioblastoma (microvascular proliferation and necrosis) but show certain molecular alterations (TERT promoter mutation, EGFR amplification or combined gain of chromosome 7 with loss of chromosome 10), which in the past could have been designated as anaplastic astrocytoma, IDH wild type. The new scheme also introduces grading within tumor type, so that the term anaplastic is eliminated from the diagnostic line for diffuse gliomas. A CNS WHO grade 3 tumor with IDH mutation and astrocytic identity would be correctly diagnosed as astrocytoma, IDH mutant, CNS WHO grade 3 (and no longer called anaplastic astrocytoma).
Comment Here
Reference: WHO classification of CNS tumors
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10563 | https://www.ck12.org/flexi/algebra-ii/horizontal-and-vertical-asymptotes/what-is-a-rational-equation/ | Flexi answers - What is a rational equation? | CK-12 Foundation
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Horizontal and Vertical Asymptotes
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What is a rational equation?
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A rational equation is an equation that involves at least one rational expression. A rational expression is a fraction where the numerator and/or the denominator are polynomials. For example, the equation x x+1=2 3 is a rational equation. Solving rational equations often involves finding a common denominator or cross-multiplying to eliminate the fractions.
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Follow the seven-step strategy to graph the following rational function. f(x)=2x/x^2-4. Find the horizontal asymptote(s). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The equation(s) of the horizontal asymptote(s) is/are enter your response here. (Type an equation. Use a comma to separate answers as needed.) B. There is no horizontal asymptote.What is the definition of rate in math?What is (fg)(x) given f(x) = - 4x + 6 and g(x) = - 2x + 4? Write your answer as a polynomial or a rational function in simplest form.
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10564 | https://atcm.mathandtech.org/EP2022/invited/21964.pdf | Sphere and Spherical Geometry: The Power of Visualization and Investigation Through Dynamic Geometry Jean-JacquesDahan jjdahan@wanadoo.fr IRES of Toulouse Paul Sabatier University Toulouse FRANCE Abstract: This article aims to provide a dynamic approach to the basic notions of spherical geometry. We will use the power of visualization of Cabri 3D to make more comprehensible the detail of the proofs establishing the formulas giving the area and the volume of a sphere. Experimenting with real spheres is a challenge already overcome with half hollow and transparent plastic balls on which you can stretch strings and write with markers. The use of Cabri 3D greatly enriches such teaching of spherical geometry in facilitating the 3D representation of the objects of this geometry as well as their manipulation. We will show how. All the results presented here are known, but it is the way in which they are approached and detailed that constitutes the originality of this article. 1. Surface area and volume of a sphere 1.1. Surface area of a sphere Visualization and manipulation of spherical coordinates: each point M of a sphere (yellow sphere of Figure 1 left, centered at O and passing through D, radius r) belongs to a half circle here SmN. This half circle can be dragged in dragging point m from D to D along the displayed equatorial circle of the given sphere; the position of m and by the way the position of the half circle is known in knowing angle 𝐷𝑂𝑚 ̂ = θ with 0 ≤ θ < 2𝜋. All points of the half circle can be reached in dragging point M along it; the position of M is known in knowing angle 𝑚𝑂𝑀 ̂ = 𝜑 with − 𝜋 2 ≤ 𝜑 ≤ 𝜋 2 . Figure 1: Spheric coordinates A construction trick (Figure 1 right): An issue with the use of Cabri 3D is that arcs of more than 180° are not represented on a given circle. So, we need a tricky geometric construction of arc Dm in order to see it continuously during the whole displacement of point m along the horizontal circle. Here is this construction: ∎ Given point m on the circle centered at O passing through D. ∎ Construct segment [Dm] ∎ Construct line (L) perpendicular to this segment passing through O. ∎ J is the intersection point between the circle and this line belonging to the shortest arc defined by D and m. ∎ Measure the length of arc DJm= L. ∎ Calculate l=L/2. ∎ Define point K on the circle as the measurement transfer of l on this circle from point D. ∎ Arc DKm is the solution of our problem. Proof mediated by Cabri 3D (Figure 2): Highlighting the area element dL.dl to evaluate the area of the sphere as a double integral: Figure 1 left. Evaluation and calculation of dL, dl and dL.dl: Figure 2 right dL = 𝑟.𝑑𝜑 , dl = 𝑟. cos(𝜑) .𝑑𝜃 and dL.dl = 𝑟2. cos(𝜑). 𝑑𝜑.𝑑𝜃 So, the area of the sphere is equal to ∫ ∫ 𝑟. cos(𝜑). 𝑑𝜃. 𝑟. 𝑑𝜑 2𝜋 𝜃=0 𝜋 2 𝜑=−𝜋 2 = 𝑟2. ∫ cos(𝜑) .
𝜋 2 𝜑=−𝜋 2 𝑑𝜑.∫ 𝑑𝜃 2𝜋 𝜃=0 = 4𝜋𝑟2 Finally: Surface area of a sphere of radius r = 𝟒𝝅𝒓𝟐 Figure 2: Surface area of a sphere 1.2. Volume of a sphere Figure 3: Volume of a sphere For a sphere of radius r, we have represented one of the elementary cylinders we need to sum to obtain the volume of a half sphere as an integral (Figure 3 left). The radius of such a cylinder at height z is equal to √𝑟2 −𝑧2. The area of its base is 𝜋. (√𝑟2 −𝑧2)2. The sum of the volumes of these cylinders approaches the volume of the half-sphere (Figure 3 right). So, the volume of the upper half sphere is equal to: ∫𝜋. (√𝑟2 −𝑧2)2 𝑅 0 . 𝑑𝑧 = 𝜋. ∫(𝑟2 −𝑧2).𝑑𝑧 𝑅 0 = 2 3 𝜋𝑟3 . Therefore: Volume of a sphere of radius r = 𝟒 𝟑𝝅𝒓𝟑 2. Spherical geometry 2.1. Segments and lines on a sphere (geodesic lines) ([1’]) 2.1.1. Experimenting before proving: we know that in planar geometry, a segment between two given points is the shortest path joining these points and the line defined by these points is obtained by extending as far as we want this segment (Euclidean definition). We will try to define segments and lines drawn on a sphere in the same way as we did for segments and lines of Euclidean plane geometry. As the simplest path joining two points on a sphere seems to be an arc, let us consider any arc defined by its two vertices A and M and passing through a third point of the sphere, point I (Figure 4 left). Let us measure the length of this arc and let us move point I in order to minimize the length of this arc I (Figure 4 middle). Very quickly we obtain a position of I, which seems to belong to the plane passing through the center of the sphere and the two points A and M (Figure 4 right). So, we can conjecture that the minimum is reached when the arc is supported by a great circle of the sphere and included entirely in the same half sphere. Figure 4: Experimenting in order to define a spherical segment Now, we can hope that the definition of a spherical segment passing through two given points could be the shortest arc supported by the great circle containing these points of the sphere 2.1.2. Proof of the previous conjecture (Figure 5): let us start from an arc AM (called Arc1) included in a great circle C1 of a given sphere (Figure 5). This circle belongs to plane P. Let us consider another arc, Arc2, in blue, with vertices also A and M. This arc belongs to a circle C2 with radius less than the radius of C1. We have to prove that the length of Arc2 is always more than the length of Arc1. To do so, it is sufficient to prove that the shortest arc included in C2, with vertices are A and M is longer than Arc1. C2 is centered at i2 which is the orthogonal projection of the center of the sphere on the plane containing C2. We rotate circle C2 and Arc2 around line (AM), using the rotation transforming the plane of circle C2 onto plane P. We obtain circle r(C2) centered at r(i2) and r(Arc2) included in plane P. So, the problem we have to solve is a problem of plane geometry described following Figure 5. Figure 5: Proof of the minimization of the length of an arc This planar problem is illustrated by Figure 6: C1 is the given circle (corresponding to a great circle of a sphere). A and M belong to this circle; let us consider arc AjM which is the shortest arc of this circle defined by A and M (the longest is AiM). [ij] is a diameter of circle C1 perpendicular to segment [MA]. All the circles C2 centered at O2 where O2 belongs to segment [O1H] are all the circles passing through A and M, with radii less thas the radius of C1 and for which the arc supported by C2 between A and M passing by l (not on the half plane defined by (AM) and O1) is shorter than the other arc AkM. The other cases are obtained when O2 belongs to the segment symmetric of [O1H] with respect to H. Note that, when O2 moves from O1 to H, circle C2 moves from circle C1 to the circle with diameter [AM]. We can notice, thanks to this interpretation that, the length of arcs AlM are greater than the length of arc AjM. As the length of arc AiM is greater than the one of AlM, all arcs defined by vertices A and M on a sphere are longer than arc AM included in a great circle, arc AM being the shortest of the two arcs defined by A and M on such great circle. Figure 6: Plane proof of the minimization of the length of an arc Finally, we can give the definition of a segment of a sphere which vertices are two given points of this sphere. Definition 1: A segment between two points of a sphere is the shortest arc defined by these vertices on the great circle containing these points and the center of the sphere. We can also deduce a construction of such an arc (Figure 6 right): if A and M are two given points of a given sphere S centered at O, the spherical segment S(AM) is the arc of the sphere defined by AjM where j is the intersection point between ray [Ou) and sphere S with u midpoint of segment [AM] Definition 2: the spherical line defined by two distinct points of a sphere is the great circle containing these two points. Notation: Spherical line (AB) noted SL(AB). Such a spherical line contains the spherical segment S(AB) and the other arc limited by A and B: let us call this arc, second spherical segment SS(AB). Let us remark that the extension of a spherical segment of the sphere is a great circle as the extension of a segment of the plane is the infinite Euclidean line. 2.2. Spherical triangles (Figure 7) ([2’]) 2.2.1. Definition: if A, B and C are three points of a sphere, the spherical triangle is defined by the three spherical segments S(AB), S(BC) and S(CA). Remark: these spherical segments are constructed as detailed juste above and we know they are supported by three great circles (in red in Figure 7) of the sphere. These great circles are the three spherical lines SL(AB), SL(BC) and SL(CA). As three lines of a plane define one triangle, here the three spherical lines define eight spherical triangles whose vertices are three of the six points A, B, C, A’, B’ and C’ (where A’, B’ and C’ are the symmetric of points A, B and C with respect to O, center of the sphere): see Figure 7 middle and Figure 7 right. Figure 7: Spherical triangles In the half plane defined by plane BCC’B’ and the point A, we have four spherical triangles: triangles ABC, AB’C’, ABC’ and ACB’. In the other half plane limited by the same plane, we have four other spherical triangles which are the symmetric of the previous ones with respect to the center of the sphere. 2.2.2. Notion of angle between two segments If A, B anc C are three points of a sphere, let us define angle 𝐵𝐴𝐶 ̂. Figure 8 left: we must consider TB and TC, respectively tangent lines to the great circles containing arc AB and arc AC (or spherical segments AB and AC). I and K are the intersection points respectively between TB and ray [Oi) and TC and ray [Ok) (O center of the sphere, i midpoint of [AB] and k midpoint of [AC]. TB is defined as the perpendicular line to [OA] in the plane containing O, A and B. It is also the tangent line at A to the great circle containing A and B. Same construction for TC. Eventually, the spherical angle S(𝐵𝐴𝐶) ̂ is defined as the plane angle 𝐼𝐴𝐾 ̂ . The two other spherical angles at B and C to the spherical triangle ABC are defined similarly (see Figure 8 middle). These angles are measured and their measurements displayed. Their sum is also displayed and we can notice that this sum is more than180°. In Figure 8 right we have redefined points A, B and C at the vertices of a quarter of half sphere: we have obtained a spherical triangle with three angles measuring each 90° (sum: 270°). See [4’]. Figure 8: Angle of two spherical segments 2.2.3. Maximum sum of the angles of a spherical triangle Let us enlarge triangle ABC until positions of A, B and C are closer and closer to a great circle but in the same half plane defined by this circle. We can notice that each angle of the triangle approaches 180°, so the sum approaches 540°. Let us show how to experiment to reach this result: In Figure 9 left, we construct a sphere centered at a point belonging to the great circle containing A and B, the circle intersection between the big sphere and the small one (the radius of the small sphere is given by a number created with the calculator of Cabri 3D and displayed: here 0.8 that can be changed at any moment) and the arc, part of this circle included in the same half plane as C (in bold red). Then, we redefine point C on this arc (Figure 9 middle). Eventually we move point C on this arc until it reaches the great circle (two positions): we can observe (Figure 9 right) that the sum displayed is 540° as each angle of the spherical triangle ABC is evaluated as 180°. Figure 9: Maximum sum of the angles of a spherical triangle 2.2.4. Area of a spherical triangle (Girard theorem) 2.2.4.1. Area of a slice: Figure 10 left displayed such a slice between two half planes P1 and P2 intersecting in line (NS) which is the line joining a point N of the sphere to S its symmetric point with respect to the center of the sphere. The angle of this slice is the angle of P1 and P2 or the angle 𝑁1𝑁𝑁2 ̂ where (NN1) is perpendicular to (NS) in P1 and (NN2) is perpenpendicular also to (NS) in P2. The area of this slice is proportional to its angle knowing that when this angle is 2𝜋, the area is the surface area of the sphere quoted in 1.1. which is 𝟒𝝅𝒓𝟐. So, the area of a slice of angle 𝜃 (between 0 and 2𝜋) in a sphere of radius r is equal to 𝜃 2𝜋. 4𝜋𝑟2. Finally, the formula simplifies to: 𝟐𝒓𝟐𝜽. 2.2.4.2. Girard theorem: ABC is a spherical triangle of a sphere of radius r. Measurements in radians of angles 𝐵𝐴𝐶 ̂, 𝐴𝐵𝐶 ̂ and 𝐵𝐶𝐴 ̂ are respectively called a, b and c (Figure 10 right). The great circles containing each side of the triangle allows us to create six slices of this sphere. In reality three pairs of two equal slices, two green equal slices bordered by the great circles of S(AB) and S(AC) (area of each slice: 2𝑟2𝑎), two blue equal slices bordered by the great circles of S(BA) and S(BC) (area of each slice: 2𝑟2𝑏) and two purple equal slices bordered by the great circles of S(CB) and S(CA) (area of each slice: 2𝑟2𝑐). If s is the area of the spherical triangle, we evaluate, in two ways, the area of the sphere: Way 1 using the known formula: 𝟒𝝅𝒓𝟐. Way 2 in adding areas of slices: 𝟐. 𝟐𝒓𝟐𝒂 2 green slices + 𝟐. 𝟐𝒓𝟐𝒃−𝟐𝒔 2 blue slices −2 areas of ABC + 𝟐. 𝟐𝒓𝟐𝒄−𝟐𝒔 2 purple slices −2 areas of ABC From the equality, 𝟐. 𝟐𝒓𝟐𝒂 + 𝟐. 𝟐𝒓𝟐𝒃−𝟐𝒔 + 𝟐.𝟐𝒓𝟐𝒄−𝟐𝒔 = 𝟒𝝅𝒓𝟐, we can deduce the area s of the spherical triangle: Area of a spherical triangle of angles a, b and c (in a sphere of radius r): s = (𝒂+ 𝒃+ 𝒄−𝝅).𝒓𝟐 Figure 10: Area of a spherical triangle 2.3. Particular lines and points of a spherical triangle (investigations) The following investigations aims to answer the question: can we expect that the known particular points of the geometry of the triangle in the plane exist in spherical geometry? That is to say: can we expect that the medians of a spherical triangle meet at the same point? The same question applies for the perpendicular bisectors, the angle bisectors and the altitudes. If yes can we expect that these points are collinear as they are in plane geometry (Euler’s line). The first problem to solve is: “the constructions of these lines in spherical geometry”. 2.3.1. Medians and centroid As displayed in Figure 11 left, we can state that the three medians intersect at the same point even if we move points A, B or C to different positions. The justification is easy (Figure 11 middle): points I, J and K are the intersections between rays [Oi), [Oj) and [Ok) with the great circle containing arcs BC, CA an AB (where i, j and k are the midpoints of segments [BC], [CA] and [BA]. So, I, J and K are the midpoints of spherical segments S(BC), S(CA) and S(AB). As we know, in triangle ABC, the medians intersect at the centroid g, necessarily, the spherical medians intersect at a same point G which is the intersection between ray [Og) and the sphere. The reason is because g is also located on the intersection of the three planes OAi, OBj and OCk which is (Og); these planes are the same as the one containing the three spherical segments S(BJ), S(CK) and S(AI). But, knowing that g is at the two thirds point of segment [Ai], angle 𝑔𝑂𝐴 ̂ is not necessarily the two thirds of angle 𝑔𝑂𝑖 ̂ and so G is not necessarily at the two thirds of spherical segment S(AI). The same remark applies for the location of G on the other spherical segments S(BJ) and S(CK). An investigation displayed on Figure 11 right confirms this remark: as the ratio between gA and gI is evaluated by 2, the ratio between angles 𝑔𝑂𝐴 ̂ and 𝑔𝑂𝑖 ̂ is evaluated by 1,36. Figure 11: Centroid of a spherical triangle 2.3.2. Perpendicular bisectors and circumscribed circle In Figure 12 left, we have constructed the three spherical perpendicular bisectors of the three sides of the spherical triangle ABC (three great circles). We can state (in fact, conjecture!) that these three spherical lines intersect at the same point c and at its symmetric point with respect to the center of the sphere. We can check (Figure 12 middle) that these two points are the centers of two spheres whose intersection is a circle containing A, B and C. This circle is the circumscribed spherical circle of the spherical triangle ABC because the three spherical arcs cA, cB and cC are equal. In Figure 12 right, we show how to construct the spherical perpendicular bisector of spherical segment S(BC) as the circle intersection between the sphere and the perpendicular bisector plane of segment [BC]. As the perpendicular bisector line of [BC] in triangle ABC belongs also to the plane of this circle, this proves the previous conjecture. Figure 12: Perpendicular bisectors and circumscribed circle 2.3.3. Angle bisectors and inscribed circle In Figure 13 left, we have conducted a construction inspired by the planar construction and we state that we are led to the same result in spherical geometry, that is to say: the three spherical bisector arcs of spherical triangle ABC (in green) intersect at the same point s which is the center of a circle tangent to each side of this spherical triangle. The radius of this circle is defined by knowing the position of h, intersection between arc BC and the great circle perpendicular to arc BC from s. Constructions of the great circle containing A (spherical bisector of spheric angle 𝐵𝐴𝐶 ̂): see Figure 13 middle. Construct first, if not already constructed, the great circles supporting arcs AB and AC (using A, b and B for the first one and A, c and C for the second one). Construct the plane perpendicular to (OA) passing through O and its circle intersection with the sphere. This circle intersects the circles supporting arcs AB and AC at B’ and C’. The perpendicular bisector plane of B’ and C’ intersects the sphere at the great circle containing A (in green). The same constructions produce three great circles of Figure 13 left. For the construction of point h (Figure 13 right), construct plane containing B, C and O and its perpendicular at O: line (NS) (N and S on the sphere). Construct the circle containing arc BC. h is the intersection between this circle and the circle containing N, s and S. s is the center of a sphere passing through h cutting the previous circle in the inscribed circle. Figure 13: Angle bisectors and inscribed circle 2.3.4. Altitudes and orthocenter In Figure 14 left we have constructed the three spherical altitudes of the spherical triangle ABC. The technique has been used in the previous paragraph to construct point h (Figure 13 right). We can state (conjecture) that these three great circles have two common points, H and H’ (symmetric point of H with respect to the center of the sphere), called orthocenters of ABC. Figure 14 middle allows to state that this conjecture is true in all cases of the figure. Proof: let us construct the two altitudes (Bb) and (Cc) intersecting at H. (Figure14 right). We know that (Bb) is the intersection between the sphere and the plane PB containing O and perpendicular to the plane (OAC). We know also that (Cc) is the intersection between the sphere and the plane PC perpendicular to the plane (OAB). The intersection line between the two planes PB and PC is line (OH). We know () that the third plane PA containing O and perpendicular to the plane (OBC) contains the line (OH). Therefore, the intersection between PA and the sphere is the third altitude of spherical triangle (ABC) which contains necessarily point H. That completes the proof. Figure 14: Altitudes and orthocenter 2.3.5. Euler’s line We know that the four previous particular points are collinear in Euclidean plane geometry. The line defined by these points is known as Euler’s line. We construct in Figure 15 these four points in spherical geometry, in using the previous techniques, and we can state very quickly that these points are not collinear. In this figure we have also constructed the spherical lines (great circles) passing through any two of these four points. None of them contain one or two other points. Figure 15: Investigations for a spherical Euler’s line 3. Stereographic projection and inversion 3.1. Stereographic projection (definition, investigations) Definition: Let’s consider a sphere (S) and two points U and D such that [UD] is a diameter of the sphere (Figure 16 left). Let’s consider plane (P) tangent to this sphere at D. If M is a point of (S) different from U, its stereographic projection with repect to (S) is the point M’ of (P), intersection of ray [UM) and (P). Investigation: Let’s consider now a circle (C) of this sphere, defined as the intersection between (S) and another sphere (T) centered at a point t of (S) (be careful! the radius of the sphere (T) is not the radius of (C)). See Figure 16 middle and [3’]. To construct the image of circle (C) by the previous stereographic projection, we construct first the cone defined by point U and circle (C) and then we display the intersection between this cone and plane (P). We can state (Figure 16 right) that this image (C’) seems to be a circle. Cabri 3D says that this intersection curve is a “circular ellipse”. To corroborate this conjecture, we construct a circle passing through three points of curve (C’) and note that this circle seems to be superimposed to (C’) even if we change the radius of (C) or the position of the center t of sphere (T): you can notice in Figure 16 right that two sliders to command these possible modifications have been created. See [5’]. Figure 16: Stereographic projection of a circle 3.2. Image of a circle by a stereographic projection (proof) In order to prove easily that the image of a circle by a stereographic projection is another circle, we will prove first that any sterographic projection is nothing else than an inversion. As shown in Figure 17 left, we conduct our reasoning in plane (Q) (the plane containing (UD) and M and by the way M’). The red circle is the one with diameter [UD] and passing through M. The two triangles DUM’ and MUD are two similar right-angle triangles, therefore: 𝑈𝐷 𝑈𝑀′ = 𝑈𝑀 𝑈𝐷 and finally UM.UM’ = UD2. This last equality means that M’ is the image of M by the inversion centered at U with scale of UD2 or, which is the same thing, M’ is the image of M by the inversion of the sphere centered at U of radius UD. As we know that the image of a circle (which does not contain the center of the inversion) is a circle by any inversion, we have established the result previously conjectured. See [6’]. Figure 17: Stereographic projection and inversion 3.3. A nice dynamic consequence (inspired by Professor Chuan) In Figure 17 middle, we construct a first circle with a little sphere centere at c1. On this circle we create a point p1 that can be animated. Then, we create the equatorial circle containing c1 and p1. With the same technique we create a second circle (with a sphere centered at c2 on the equatorial circle) passing through p1 and by the way tangent to the first circle. We iterate the process to create three other circles with the fifth one tangent to both the fourth one and the first one. We need a trick to find the position of c5. c5 is the intersection point between the ray [si) and the equatorial circle (s is the center of the big initial sphere and i the midpoint of segment [p5 p1]). When p1 is animated along the first circle, the four other circles are animated on the initial sphere. If we transform these five circles by the stereographic projection of the given sphere with respect to the north pole or with the inversion of the sphere centered at the north pole and passing through the south pole, we obtain five tangent circles of the horizontal plane (Figure 17 right). You can observe that the images of points c1, c2, c3, c4 and c5 are not the centers of the images of the five circles of the sphere. But we use these images to create flat cones that can be coloured as shown in Figure 18. See [7’]. Figure 18: Five animated tangent circles 4. Conclusion The aim of this paper was not at all to present new results about spherical geometry which is a subject that has given rise to a large number of very technical productions, especially in spherical trigonometry. Instead, the aim was to show an approach of spherical geometry that can convince teachers of the power of dynamic geometry to understand notions deemed difficult to teach. The last part related to stereographic projection aimed to show the powerful link with the notion of inversion and the possibilities of creating animations in the plane thanks to simple spherical constructions. References FGM, 1920, Théorème 653-1785 p. 870 in Exercices de Géométrie, Sixième Edition, Éditions Jacques Gabay 1991 POLYA G., 1945, How to solve it, Princeton, University Press LAKATOS I., 1984, Preuves et réfutations Essai sur la logique de la découverte, Hermann, DAHAN J.J., 2002, How to teach Mathematics in showing all the hidden stages of a true research. Examples with Cabri, in Proceedings of ATCM 2002, Melaka, Malaysia DAHAN J.J., 2005, La démarche de découverte expérimentalement médiée par Cabri-géomètre en mathématiques, PhD thesis, Université Joseph Fourier, Grenoble, France YouTube videos links (videos in French) [1’] Video of YouTube channel « jjdahan »: Droite et segment en géométrie sphérique [2’] Video of YouTube channel « jjdahan »: Triangle en géométrie sphérique [3’] Video of YouTube channel « jjdahan »: Cercle en géométrie sphérique [4’] Video of YouTube channel « jjdahan »: Triangle rectangle en géométrie sphérique [5’] Video of YouTube channel « jjdahan »: Projection stéréographique d’un cercle [6’] Video of YouTube channel « jjdahan »: Lien projection stéréographique inversion [7’] Video of YouTube channel « jjdahan »: Cercles tangents et projection stéréographique Software Cabri 3D by Cabrilog at |
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1999 AIME Problems/Problem 14
Problem
Point is located inside triangle so that angles and are all congruent. The sides of the triangle have lengths and and the tangent of angle is where and are relatively prime positive integers. Find
Contents
1 Problem
2 Solution
2.1 Solution 1
2.2 Solution 2
2.3 Solution 3
2.4 Solution 4 (Law of sines)
3 See also
Solution
Solution 1
Drop perpendiculars from to the three sides of and let them meet and at and respectively.
Let and . We have that We can then use the tool of calculating area in two ways On the other hand, We still need though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: Adding gives Recall that we found that . Plugging in , we get , giving us for an answer.
Solution 2
Let , , , , , and .
So by the Law of Cosines, we have: Adding these equations and rearranging, we have: Now , by Heron's formula.
Now the area of a triangle, , where and are sides on either side of an angle, . So, Adding these equations yields: Dividing by , we have: Thus, .
Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by
Solution 3
Let Then, using Law of Cosines on the three triangles containing vertex we have Add the three equations up and rearrange to obtain Also, using we have Divide the two equations to obtain
Solution 4 (Law of sines)
Firstly, denote angles , , and as , , and respectively. Let . Notice that by angle chasing that and . Using the nice properties of the 13-14-15 triangle, we have and . is easily computed, so we have .
Using Law of Sines, hence, Now, computation carries the rest. Extracting yields .
See also
Brocard point
1999 AIME (Problems • Answer Key • Resources)
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10566 | https://en.wikipedia.org/wiki/List_of_logic_symbols | Jump to content
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Contents
1 Basic logic symbols
2 Advanced or rarely used logical symbols
3 See also
4 References
5 Further reading
6 External links
List of logic symbols
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From Wikipedia, the free encyclopedia
List of symbols used to express logical relations
This article contains logic symbols. Without proper rendering support, you may see question marks, boxes, or other symbols instead of logic symbols.
In logic, a set of symbols is commonly used to express logical representation. The following table lists many common symbols, together with their name, how they should be read out loud, and the related field of mathematics. Additionally, the subsequent columns contains an informal explanation, a short example, the Unicode location, the name for use in HTML documents, and the LaTeX symbol.
Basic logic symbols
[edit]
| Symbol | Unicodevalue(hexadecimal) | HTMLcodes | LaTeXsymbol | Logic Name | Read as | Category | Explanation | Examples |
--- --- --- ---
| ⇒→⊃ | U+21D2U+2192U+2283 | ⇒→⊃ ⇒→⊃ | \Rightarrow\implies \to or \rightarrow\supset | material conditional (material implication) | implies, if P then Q, it is not the case that P and not Q | propositional logic, Boolean algebra, Heyting algebra | is false when A is true and B is false but true otherwise.In other mathematical contexts, see glossary of mathematical symbols, may indicate the domain and codomain of a function and may mean superset. | is true, but is in general false (since x could be −2). |
| ⇔↔≡ | U+21D4U+2194U+2261 | ⇔↔≡ ⇔↔≡ | \Leftrightarrow\iff\leftrightarrow\equiv | material biconditional (material equivalence) | if and only if, iff, xnor | propositional logic, Boolean algebra | is true only if both A and B are false, or both A and B are true. Whether a symbol means a material biconditional or a logical equivalence, depends on the author’s style. | |
| ¬~! ′ | U+00ACU+007EU+0021U+2032 | ¬˜!′ ¬˜!′ | \lnot or \neg\sim ' | negation | not | propositional logic, Boolean algebra | The statement is true if and only if A is false.A slash placed through another operator is the same as placed in front. The prime symbol is placed after the negated thing, e.g. | |
| ∧·& | U+2227U+00B7U+0026 | ∧·& ∧·& | \wedge or \land\cdot \& | logical conjunction | and | propositional logic, Boolean algebra | The statement A ∧ B is true if A and B are both true; otherwise, it is false. | n < 4 ∧ n >2 ⇔ n = 3 when n is a natural number. |
| ∨+∥ | U+2228U+002BU+2225 | ∨+∥ ∨+∥ | \lor or \vee\parallel | logical (inclusive) disjunction | or | propositional logic, Boolean algebra | The statement A ∨ B is true if A or B (or both) are true; if both are false, the statement is false. | n ≥ 4 ∨ n ≤ 2 ⇔ n ≠ 3 when n is a natural number. |
| ⊕⊻↮≢ | U+2295U+22BBU+21AEU+2262 | ⊕⊻↮≢ ⊕⊻—≢ | \oplus\veebar\not\equiv | exclusive disjunction | xor, either ... or ... (but not both) | propositional logic, Boolean algebra | The statement is true when either A or B, but not both, are true. This is equivalent to ¬(A ↔ B), hence the symbols and . | is always true and is always false (if vacuous truth is excluded). |
| ⊤T1 | U+22A4 | ⊤ ⊤ | \top | true (tautology) | top, truth, tautology, verum, full clause | propositional logic, Boolean algebra, first-order logic | denotes a proposition that is always true. | The proposition is always true since at least one of the two is unconditionally true. |
| ⊥F0 | U+22A5 | ⊥ ⊥ | \bot | false (contradiction) | bottom, falsity, contradiction, falsum, empty clause | propositional logic, Boolean algebra, first-order logic | denotes a proposition that is always false. The symbol ⊥ may also refer to perpendicular lines. | The proposition is always false since at least one of the two is unconditionally false. |
| ∀() | U+2200 | ∀ ∀ | \forall | universal quantification | given any, for all, for every, for each, for any | first-order logic | or says “given any , has property .” | |
| ∃ | U+2203 | ∃ ∃ | \exists | existential quantification | there exists, for some | first-order logic | says “there exists an (at least one) such that has property .” | n is even. |
| ∃! | U+2203 U+0021 | ∃ ! ∃! | \exists ! | uniqueness quantification | there exists exactly one | first-order logic (abbreviation) | says “there exists exactly one such that has property .” Only and are part of formal logic. is an abbreviation for | |
| ( ) | U+0028 U+0029 | ( ) () | ( ) | precedence grouping | parentheses; brackets | almost all logic syntaxes, as well as metalanguage | Perform the operations inside the parentheses first. | (8 ÷ 4) ÷ 2 = 2 ÷ 2 = 1, but 8 ÷ (4 ÷ 2) = 8 ÷ 2 = 4. |
| | U+1D53B | 𝔻 𝔻 | \mathbb{D} | domain of discourse | domain of discourse | metalanguage (first-order logic semantics) | | |
| ⊢ | U+22A2 | ⊢ ⊢ | \vdash | syntactic consequence | proves, syntactically entails (single) turnstile | metalanguage (metalogic) | says “ is a theorem of ”. In other words, proves via a deductive system. | (eg. by using natural deduction) |
| ⊨ | U+22A8 | ⊨ ⊨ | \vDash, \models | semantic consequence or satisfaction | (semantically) entails or satisfies, models double turnstile | metalanguage (metalogic) | says “in every model, it is not the case that is true and is false”. says a formula is true in a model with variable assignment . | (eg. by using truth tables) |
| ≡⟚⇔ | U+2261U+27DAU+21D4 | ≡ —⇔ ≡ — ⇔ | \equiv\Leftrightarrow | logical equivalence | is logically equivalent to | metalanguage (metalogic) | It’s when and . Whether a symbol means a material biconditional or a logical equivalence, depends on the author’s style. | |
| ⊬ | U+22AC | | ⊬\nvdash | | does not syntactically entail (does not prove) | metalanguage (metalogic) | says “ is not a theorem of ”. In other words, is not derivable from via a deductive system. | |
| ⊭ | U+22AD | | ⊭\nvDash | | does not semantically entail | metalanguage (metalogic) | says “ does not guarantee the truth of ”. In other words, does not make true. | |
| □ | U+25A1 | | \Box | necessity (in a model) | box; it is necessary that | modal logic | modal operator for “it is necessary that”in alethic logic, “it is provable that”in provability logic, “it is obligatory that”in deontic logic, “it is believed that”in doxastic logic. | says “it is necessary that everything has property ” |
| ◇ | U+25C7 | | \Diamond | possibility (in a model) | diamond;it is possible that | modal logic | modal operator for “it is possible that”, (in most modal logics it is defined as “¬□¬”, “it is not necessarily not”). | says “it is possible that something has property ” |
| ∴ | U+2234 | | ∴\therefore | therefore | therefore | metalanguage | abbreviation for “therefore”. | |
| ∵ | U+2235 | | ∵\because | because | because | metalanguage | abbreviation for “because”. | |
| ≔≜≝ | U+2254U+225CU+225D | ≔ ≔ | ≔ \coloneqq:=\triangleq \stackrel{ \scriptscriptstyle \mathrm{def}}{=} | definition | is defined as | metalanguage | means "from now on, is defined to be another name for ." This is a statement in the metalanguage, not the object language. The notation may occasionally be seen in physics, meaning the same as . | |
| ↑| ⊼ | U+2191U+007C U+22BC | | \uparrow | Sheffer stroke, NAND | NAND, not both up arrow | Propositional logic | NAND is the negation of conjunction so is true if and only if it's not the case that both A and B are true. See also NAND gate | |
| ↓ ⊽ | U+2193 U+22BC | | \downarrow | Peirce Arrow, NOR | nor down arrow | Propositional logic | NOR is the negation of conjunction so is true if and only both A and B are false. See also NOR gate | |
Advanced or rarely used logical symbols
[edit]
The following symbols are either advanced and context-sensitive or very rarely used:
| Symbol | Unicodevalue(hexadecimal) | HTMLvalue(decimal) | HTMLentity(named) | LaTeXsymbol | Logic Name | Read as | Category | Explanation |
--- --- --- ---
| ⥽ | U+297D | | | \strictif | right fish tail | | | Sometimes used for “relation”, also used for denoting various ad hoc relations (for example, for denoting “witnessing” in the context of Rosser's trick). The fish tail is also used as strict implication by C.I.Lewis ⥽ . |
| ̅ | U+0305 | | | | combining overline | | | Used format for denoting Gödel numbers. Using HTML style “4̅” is an abbreviation for the standard numeral “SSSS0”. It may also denote a negation (used primarily in electronics). |
| ⌜⌝ | U+231CU+231D | | | \ulcorner \urcorner | top left cornertop right corner | | | Corner quotes, also called “Quine quotes”; for quasi-quotation, i.e. quoting specific context of unspecified (“variable”) expressions; also used for denoting Gödel number; for example “⌜G⌝” denotes the Gödel number of G. (Typographical note: although the quotes appears as a “pair” in unicode (231C and 231D), they are not symmetrical in some fonts. In some fonts (for example Arial) they are only symmetrical in certain sizes. Alternatively the quotes can be rendered as ⌈ and ⌉ (U+2308 and U+2309) or by using a negation symbol and a reversed negation symbol ⌐ ¬ in superscript mode.) |
| ∄ | U+2204 | | | \nexists | there does not exist | | | Strike out existential quantifier. “¬∃” used some times instead. |
| ⊙ | U+2299 | | | \odot | circled dot operator | | | A sign for the XNOR operator (material biconditional and XNOR are the same operation). |
| ⟛ | U+27DB | | | | left and right tack | | | “Proves and is proved by”. |
| ⊩ | U+22A9 | | | | forces | | | One of this symbol’s uses is to mean “truthmakes” in the truthmaker theory of truth. It is also used to mean “forces” in the set theory method of forcing. |
| ⟡ | U+27E1 | | | | white concave-sided diamond | never | modal operator | |
| ⟢ | U+27E2 | | | | white concave-sided diamond with leftwards tick | was never | modal operator | |
| ⟣ | U+27E3 | | | | white concave-sided diamond with rightwards tick | will never be | modal operator | |
| ⟤ | U+25A4 | | | | white square with leftwards tick | was always | modal operator | |
| ⟥ | U+25A5 | | | | white square with rightwards tick | will always be | modal operator | |
| ⋆ | U+22C6 | | | | star operator | | | May sometimes be used for ad-hoc operators. |
| ⌐ | U+2310 | | | | reversed not sign | | | |
| ⨇ | U+2A07 | | | | two logical AND operator | | | |
See also
[edit]
Philosophy portal
Glossary of logic
Józef Maria Bocheński
List of notation used in Principia Mathematica
List of mathematical symbols
Logic alphabet, a suggested set of logical symbols
Logic gate § Symbols
Logical connective
Mathematical operators and symbols in Unicode
Non-logical symbol
Polish notation
Truth function
Truth table
Wikipedia:WikiProject Logic/Standards for notation
References
[edit]
^ "Named character references". HTML 5.1 Nightly. W3C. Retrieved 9 September 2015.
^ Virtually all Turkish high school math textbooks use p' for negation due to the books handed out by the Ministry of National Education representing it as p'.
^ Although this character is available in LaTeX, the MediaWiki TeX system does not support it.
^ Quine, W.V. (1981): Mathematical Logic, §6
^ Hintikka, Jaakko (1998), The Principles of Mathematics Revisited, Cambridge University Press, p. 113, ISBN 9780521624985.
Further reading
[edit]
Józef Maria Bocheński (1959), A Précis of Mathematical Logic, trans., Otto Bird, from the French and German editions, Dordrecht, South Holland: D. Reidel.
External links
[edit]
Named character entities in HTML 4.0
| v t e Common logical symbols |
| | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- | ∧ or & and | ∨ or | ¬ or ~ not | → implies | ⊃ implies,superset | ↔ or ≡ iff | | nand | ∀ universalquantification | ∃ existentialquantification | ⊤ true,tautology | ⊥ false,contradiction | ⊢ entails,proves | ⊨ entails,therefore | ∴ therefore | ∵ because | |
| Philosophy portal Mathematics portal |
| v t e Logic |
| |
| Major fields | Computer science Formal semantics (natural language) Inference Philosophy of logic Proof Semantics of logic Syntax | | | --- | | Logics | Classical Informal + Critical thinking + Reason Mathematical Non-classical Philosophical | | Theories | Argumentation Metalogic Metamathematics Set | |
| Foundations | Abduction Analytic and synthetic propositions Antecedent Consequent Contradiction + Paradox + Antinomy Deduction Deductive closure Definition Description Dichotomy Entailment + Linguistic Form Induction Logical truth Name Necessity and sufficiency Premise Probability Proposition Reference Statement Substitution Truth Validity |
| Lists | | | | --- | | Topics | Mathematical logic Boolean algebra Set theory | | Other | Logicians Rules of inference Paradoxes Fallacies Logic symbols | |
| Category Outline Portal WikiProject changes |
| v t e Common mathematical notation, symbols, and formulas |
| | Lists of Unicode and LaTeX mathematical symbols | | List of mathematical symbols by subject Glossary of mathematical symbols List of logic symbols | |
| | Lists of Unicode symbols | | | | | --- | | General | List of Unicode characters Unicode block | | Alphanumeric | Mathematical Alphanumeric Symbols Blackboard bold Letterlike Symbols Symbols for zero | | Arrows and Geometric Shapes | Arrows Miscellaneous Symbols and Arrows Geometric Shapes (Unicode block) | | Operators | Mathematical operators and symbols Mathematical Operators (Unicode block) | | Supplemental Math Operators | Supplemental Mathematical Operators Number Forms | | Miscellaneous | A B Technical ISO 31-11 (Mathematical signs and symbols for use in physical sciences and technology) | | |
| | Typographical conventions and notations | | | | | --- | | Language | APL syntax and symbols | | Letters | Diacritic Letters in STEM + Greek letters in STEM + Latin letters in STEM | | Notation | Mathematical notation Abbreviations Notation in probability and statistics List of common physics notations | | |
| | Meanings of symbols | | Glossary of mathematical symbols List of mathematical constants Physical constants Table of mathematical symbols by introduction date List of typographical symbols and punctuation marks | |
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10567 | https://en.wikipedia.org/wiki/Posterior_cerebral_artery_syndrome | Jump to content
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1 Signs and symptoms
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Posterior cerebral artery syndrome
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From Wikipedia, the free encyclopedia
Type of reduced brain function due to blood clotting
Medical condition
| Posterior cerebral artery syndrome |
| Outer surface of cerebral hemisphere, showing areas supplied by cerebral arteries. (Yellow is region supplied by posterior cerebral artery.) |
| Specialty | Neurology |
| Diagnostic method | CT brain to r/o hemorrhagic cause. MRI is gold standard. |
| Treatment | Complementary therapies and rehabilitation removal of mechanical interference is also relevant through diagnostic imaging techniques. |
Posterior cerebral artery syndrome is a condition whereby the blood supply from the posterior cerebral artery (PCA) is restricted, leading to a reduction of the function of the portions of the brain supplied by that vessel: the occipital lobe, the inferomedial temporal lobe, a large portion of the thalamus, and the upper brainstem and midbrain.
This event restricts the flow of blood to the brain in a near-immediate fashion. The blood hammer is analogous to the water hammer in hydrology, and consists of a sudden increase of the upstream blood pressure in a blood vessel when the bloodstream is abruptly blocked by vessel obstruction. Complete understanding of the relationship between mechanical parameters in vascular occlusions is a critical issue, which can play an important role in the future diagnosis, understanding and treatment of vascular diseases.
Depending upon the location and severity of the occlusion, signs and symptoms may vary within the population affected with PCA syndrome. Blockages of the proximal portion of the vessel produce only minor deficits due to the collateral blood flow from the opposite hemisphere via the posterior communicating artery. In contrast, distal occlusions result in more serious complications. Visual deficits, such as agnosia, prosopagnosia or cortical blindness (with bilateral infarcts) may be a product of ischemic damage to occipital lobe. Occlusions of the branches of the PCA that supply the thalamus can result in central post-stroke pain and lesions to the subthalamic branches can produce “a wide variety of deficits”.
Left posterior cerebral artery syndrome presents alexia without agraphia; the lesion is in the splenium of the corpus callosum.
Signs and symptoms
[edit]
Peripheral Territory Lesions
Contralateral homonymous hemianopsia
cortical blindness with bilateral involvement of the occipital lobe branches
visual agnosia
prosopagnosia
dyslexia, Anomic aphasia, color naming and discrimination problems
memory defect
topographic disorientation
Central Territory Lesions
central post-stroke (thalamic) pain: spontaneous pain, dysesthesias and sensory impairments
involuntary movements: chorea, intention tremor, hemiballismus
contralateral hemiplegia
Weber’s syndrome: occulomotor nerve palsy
Bálint's syndrome: loss of voluntary eye movements optic ataxia, asimultagnosia (inability to understand visual objects)
Diagnosis
[edit]
1.CT 2.MRI
Treatment
[edit]
| |
| This section is empty. You can help by adding to it. (March 2018) |
References
[edit]
^ a 0-8036-1247-8
^ Tazraei, P.; Riasi, A.; Takabi, B. (2015). "The influence of the non-Newtonian properties of blood on blood-hammer through the posterior cerebral artery". Mathematical Biosciences. 264: 119–127. doi:10.1016/j.mbs.2015.03.013. PMID 25865933.
^ The Internet Stroke Center. Stroke syndromes: Posterior cerebral artery - unilateral occipital. [Internet]. [updated 1999 July; cited 2011 May 13]. Retrieved from
^ The Internet Stroke Center. Stroke syndromes: Cortical blindness. [Internet]. [updated 1999 July; cited 2011 May 13]. Retrieved from
^ The Internet Stroke Center. Stroke syndromes: Weber's syndrome. [Internet]. [updated 1999 July; cited 2011 May 13]. Retrieved from
^ The Internet Stroke Center. Stroke syndromes: Balint Syndrome. [Internet]. [updated 1999 July; cited 2011 May 13]. Retrieved from
External links
[edit]
| | |
--- |
| Classification | D ICD-10: G46.2 MeSH: D020762 SNOMED CT: 195211003 |
| External resources | eMedicine: article/2128100 |
| v t e Cerebrovascular diseases including stroke |
| Ischaemic stroke | | | | --- | | Brain | Anterior cerebral artery syndrome Middle cerebral artery syndrome Posterior cerebral artery syndrome Amaurosis fugax Moyamoya disease Dejerine–Roussy syndrome Watershed stroke Lacunar stroke | | Brain stem | Brainstem stroke syndrome Medulla + Medial medullary syndrome + Lateral medullary syndrome Pons + Medial pontine syndrome / Foville's + Lateral pontine syndrome / Millard-Gubler Midbrain + Weber's syndrome + Benedikt syndrome + Claude's syndrome | | Cerebellum | Cerebellar stroke syndrome | | Extracranial arteries | Carotid artery stenosis precerebral Anterior spinal artery syndrome Vertebrobasilar insufficiency + Subclavian steal syndrome | | Classification | Brain ischemia Cerebral infarction Classification + Transient ischemic attack + Total anterior circulation infarct + Partial anterior circulation infarct | | Other | CADASIL Binswanger's disease Transient global amnesia | |
| Haemorrhagic stroke | | | | --- | | Extra-axial | Epidural Subdural Subarachnoid | | Cerebral/Intra-axial | Intraventricular | | Brainstem | Duret haemorrhages | | General | Intracranial hemorrhage | |
| Aneurysm | Intracranial aneurysm + Charcot–Bouchard aneurysm |
| Other | Cerebral vasculitis Cerebral venous sinus thrombosis |
| v t e Signs and symptoms, and syndromes associated with lesions of the brain and brainstem |
| Cerebral cortex | ACA syndrome MCA syndrome PCA syndrome Aphasia Frontal lobe + Expressive aphasia + Abulia Parietal lobe + Receptive aphasia + Hemispatial neglect + Gerstmann syndrome + Astereognosis Occipital lobe + Bálint's syndrome + Cortical blindness + Anton syndrome + Pure alexia Temporal lobe + Cortical deafness + Prosopagnosia |
| Subcortex | Basal ganglia + Chorea + Dystonia + Parkinson's disease Thalamic syndrome |
| Cerebellum | Lateral + Dysmetria + Dysdiadochokinesia + Intention tremor Medial + Cerebellar ataxia |
| Brainstem | | | | --- | | Medulla | Lateral medullary syndrome/Wallenberg + PICA Medial medullary syndrome/Dejerine + ASA | | Pons | Upper dorsal pontine syndrome/Raymond–Céstan syndrome Lateral pontine syndrome (AICA) (lateral) Medial pontine syndrome/Millard–Gubler syndrome/Foville's syndrome (basilar) Locked-in syndrome Internuclear ophthalmoplegia One and a half syndrome | | Midbrain | Weber's syndrome + ventral peduncle, PCA Benedikt syndrome + ventral tegmentum, PCA Parinaud's syndrome + dorsal, tumor Claude's syndrome | | Other | Alternating hemiplegia | |
| Other | Pseudobulbar affect Upper motor neuron lesion |
Retrieved from "
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10568 | https://cosenzaassociates.com/wp-content/uploads/2020/03/7.12A_Comparative-Box-Plots.pdf | © Cosenza & Associates, LLC 162 GRADE 7 TEKS COMPANION GUIDE TELL ME MORE… A box plot (also called a box-and-whiskers plot) shows you how spread out the data values in the set are. Box plots have a box that shows the middle half of the data set, or the half of the data that lies between the first quartile and third quartile. Each end has a whisker that shows you the lower fourth (between the minimum value and first quartile) and upper fourth (between the third quartile and maximum value) of the data set. The five values that are used to construct a box plot are often called the five number summary of a data set. For large data sets, technology is useful for calculating these values.
The football team at Lyndon Johnson Middle School played 11 games last season. The number of points the team scored in each game is shown in the following list.
12, 28, 40, 6, 24, 16, 20, 21, 14, 22, 36 You can represent the data set with a box plot. First, you will need to determine the median, first quartile, and third quartile. To help with calculating these values, list the data set in order from least to greatest.
6, 12, 14, 16, 20, 21, 22, 24, 28, 36, 40 If you have two data sets, you can construct a box plot for each data set. Then, you can use the comparative box plots in order to compare the characteristics of the two data sets.
The student is expected to compare two groups of numeric data using comparative dot plots or box plots by comparing their shapes, centers, and spreads.
COMPARATIVE BOX PLOTS 7.12A Five Number Summary • Minimum: the least value in a data set • First Quartile: the number for which 25% of the data values are less than • Median: the number for which 50% of the data values are less than and 50% of the data values are greater than • Third Quartile: the number for which 75% of the data values are less than • Maximum: the greatest value in a data set © Cosenza & Associates, LLC DATA ANALYSIS AND PERSONAL FINANCIAL LITERACY 163 DATA ANALYSIS AND PERSONAL FINANCIAL LITERACY 163 EXAMPLES EXAMPLE 1: The comparative box plots below show the average high temperatures for Boston, Massachusetts, and Dallas, Texas. How do the interquartile range and median of the two data sets compare?
STEP 1 Estimate the interquartile range for each data set. n The interquartile range is the difference between the first quartile and third quartile. n In a box plot, the interquartile range is the width of the box. The interquartile range for Boston is about 29°F. The interquartile range for Dallas is about 23°F.
STEP 2 Compare the values for both interquartile ranges. n 29 is greater than 23. The interquartile range for Boston’s median monthly high temperature is greater than the interquartile range for Dallas’s median monthly high temperature.
74 – 45 = 29 90 – 67 = 23 YOU TRY IT!
The box plots show data about the ratings of the movies that are in Mateo’s movie collection, sorted by genre. Compare the range and interquartile range of the two data sets.
Drama: Range: _ − _ = _ Interquartile range: _ − = __ Comedy: Range: _ − _ = _ Interquartile range: _ − = Fill in the blank with greater than, less than, or equal to.
The range for dramas is _ the range for comedies. The interquartile range for dramas is _____ the interquartile range for comedies.
© Cosenza & Associates, LLC 164 GRADE 7 TEKS COMPANION GUIDE STEP 3 Estimate the median temperature for both data sets. n The median in a box plot is represented by the vertical line inside the box The median monthly high temperature for Boston is 60°F. The median monthly high temperature for Dallas is about 79°F.
STEP 4 Compare the values for both medians. n 60 is less than 79. The median monthly high temperature for Boston is less than the median monthly high temperature for Dallas.
EXAMPLE 2: The comparative box plots show data about the number of wins scored by Major League Baseball® teams in the 2017 and 2018 seasons. How do the range and third quartile of each data set compare?
STEP 1 Estimate the range for each data set. n The range is the difference between the minimum value and maximum value. n In a box plot, the range is the distance between the endpoints of the two whiskers. n Estimate the maximum and minimum value for each box plot and use those values to calculate the range. The range for 2017 is about 50 wins. The range for 2018 is about 70 wins. © Cosenza & Associates, LLC STEP 2 Compare the values for both ranges. n 50 < 70 The range for 2017 is less than the range for 2018.
STEP 3 Estimate the third quartile for both data sets. n The third quartile in a box plot is represented by the right edge of the box. The third quartile for 2017 is about 90. The third quartile for 2018 is about 83.
STEP 4 Compare the values for both third quartiles. n 90 > 83 The third quartile for 2017 is greater than the third quartile for 2018.
DATA ANALYSIS AND PERSONAL FINANCIAL LITERACY 165 PRACTICE Mr. Maha’s class and Ms. Tripp’s class collected data about the number of each color of candy in a sample of bags of that candy. The box plots show data about their results. Use these box plots for questions 1-8.
Calculate the following measures for each class.
1. Interquartile range 2. Median 3. First Quartile 4. Minimum value Complete the sentences below with less than, greater than, or equal to.
5. The interquartile range for Mr. Maha’s class is __ the interquartile range for Ms. Tripp’s class.
6. The median for Mr. Maha’s class is __ the median range for Ms. Tripp’s class.
7. The first quartile for Mr. Maha’s class is __ the first quartile for Ms. Tripp’s class.
8. The minimum for Mr. Maha’s class is __ the minimum for Ms. Tripp’s class. © Cosenza & Associates, LLC 166 GRADE 7 TEKS COMPANION GUIDE A local basketball league has several teams. The box plots show data about the shoe sizes for players on two of those teams. Use these box plots for questions 9-11. Determine if the statement below is true or false. If false, rewrite it to be true.
9. The range of shoe sizes for both teams is equal.
10. The third quartile for the Spurs is greater than the third quartile for the Rockets.
11. The median for the Rockets is less than the median for the Spurs.
12. The box plots show data about the heights, in feet, of two different types of trees at a local nursery. Which statement is best supported by the information in the box plots?
A The interquartile range of heights of oak trees is less than the interquartile range of heights of pine trees.
B The range of heights of oak trees is the same as the range of heights for pine trees.
C The median height of pine trees is greater than the median height of oak trees.
D The fi rst quartile height of oak trees is less than the fi rst quartile height of pine trees.
DATA ANALYSIS AND PERSONAL FINANCIAL LITERACY 167 © Cosenza & Associates, LLC 13. The box plots show data about the ages of actors and actresses who won the Academy® Award for Best Actor or Best Actress through 2018. Which statement is best supported by the information in the box plots?
F The fi rst quartile age of Best Actor winners is less than the fi rst quartile age of Best Actress winners.
G The minimum age of Best Actor winners is less than the minimum age of Best Actress winners.
H The median age of Best Actress winners is greater than the median age of Best Actor winners.
J The range of ages of Best Actress winners is greater than the range of ages of Best Actor winners.
14. The box plots show data about the hat sizes of two diff erent baseball teams. Which statement is best supported by the information in the box plots?
A The median hat size for the Sharks is less than the median hat size for the Seals.
B The median hat size for both teams is the same.
C The interquartile range of hat sizes for both teams is the same.
D The interquartile range of hat sizes for the Sharks is greater than the interquartile range of hat sizes for the Seals. |
10569 | https://remote2.ece.illinois.edu/~erhan/FieldsWaves/secure/notes/ECE450Sp10/350lect32.pdf | 32 TM modes in dielectric waveguides • Last lecture we examined the characteristic equation and the cutoff frequencies of TE mode of propagation in dielectric slab waveguides.
z x d n2 n2 n1 > n2 θi Guidance requires θi > θc = sin−1 n2 n1 Guided TEm mode fields consisting of the superposition of transverse polarized electric fields ˜ Ei = ˆ yEoe−jk1(−cos θix+sin θiz) and ˜ Er = ˆ yEoΓTEe−jk1(cos θix+sin θiz), where ΓTE = n1 cos θi + j !
n2 1 sin2 θi −n2 2 n1 cos θi −j !
n2 1 sin2 θi −n2 2 , have 1. propagation angles θi > θc = sin−1 n2 n1 , (critical angle) 2. satisfying a characteristic equation d v1/f cos θi −m 2 = 1 π tan−1 !
sin2 θi −n2 2/n2 1 cos θi , m = 0, 1, 2, 3, · · · 3. for frequencies f exceeding the cutofffrequency fc = mc 2d !
n2 1 −n2 2 , m = 0, 1, 2, 3 · · · 1 • For a given f = 2π ω , the charcteristic equation can be solved (typically by using graphical techniques) for θi, from which we can calculate the propagation constant kz = k1 sin θi where k1 = ω v1 = ω c n1 is the wavenumber in the core region of the guide at the operation frequency ω = 2πf. It follows that the guide wavelength λg = 2π kz and 1 phase velocity vpz = ω kz = ω k1 sin θi = v1 sin θi can be obtained once θi is calculated from the characteristic equation.
• Given the kz in the core region, kx and kz outside the core region (with index n2) can be obtained by using the fact that kz is identical in both regions (why?).
• In this lecture we will continue our study of dielectric slab waveguides by examining the TM modes.
1Note that group velocity vg = v1 sin θi (in analogy with parallel plate wavegiudes) if and only if ω ≫ωc = 2πfc because of the effect of the cladding region that contains a substatial fraction on the wave energy unless ω ≫ωc — in fact for TE0 mode vg ≈v2 at frequencies much less than the cutofffrequency of TE1 mode.
2 TM modes • For the TM mode where the incident and reflected fields are taken as ˜ Hi = ˆ yHoe−jk1(−cos θix+sin θiz) and ˜ Hr = ˆ yHoRe−jk1(cos θix+sin θiz) the reflection coefficient is given as R = η1 cos θi −η2 cos θ2 η1 cos θi + η2 cos θ2 = n2 cos θi −n1 cos θ2 n2 cos θi + n1 cos θ2 .
This leads to 0.0 0.2 0.4 0.6 0.8 1.0 0.1 0.2 0.3 0.4 0.5 Graphical solution of the charac-teristic equations for TE and TM modes m = 0, 1, 2, 3 in propaga-tion and mode m = 4 in evanes-cence.
The blue and red curves depict 1 π tan−1 !
sin2 θi −n2 2/n2 1 cos θi and 1 π tan−1 !
sin2 θi −n2 2/n2 1 n2 2 n2 1 cos θi as a function of cos θi for n2 = 1 and n1 = 1.5, while the straight lines depict d v1/f cos θi −m 2 with d v1/f = 2.5 and m increasing from left to right in steps of one.
R = n2 cos θi + jn1 " n2 1 n2 2 sin2 θi −1 n2 cos θi −jn1 " n2 1 n2 2 sin2 θi −1 with ∠R = 2 tan−1 !
sin2 θi −n2 2/n2 1 n2 2 n2 1 cos θi .
Hence, in this case the guidance condition leads to the characteristic equation d v1/f cos θi −m 2 = 1 π tan−1 !
sin2 θi −n2 2/n2 1 n2 2 n2 1 cos θi , m = 0, 1, 2, 3, · · · Note that this result for the TM mode leads to the same fc expression as in TE modes.
3 Mode structures: !"#$%&'(%)%"+,,-"."/0(01%2"3%4516'"746045" ++8 !"#$%&'()+)%,+-&'()+,)./').,#0(1',(')'&'2.,-2)+-'&3)+)4!)$3'(5)) 46)$3'()/#1')(-$-&#,)%,+-&'()+,)./')$#70'.-2)+-'&35) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) !"#$ !"%$ ! $&'()$+,-.$ /')0,11 !
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4 Acceptance cone and numerical aperture: Guidance requires The maximum acceptance angle θom defines the so-called “acceptance cone” that in-cludes all the external sig-nals incident on the dielec-tric waveguide that can cou-ple to the waveguide at the air/core interface on a constant-z plane.
θi > θc = sin−1 n2 n1 = sin−1 "ϵr2 ϵr1 and therefore “acceptence angles” from air (see the diagram below) θo < θom = sin−1 # n2 1 −n2 2 = sin−1 √ϵr1 −ϵr2 where sin θom = # n2 1 −n2 2 = √ϵr1 −ϵr2 is called numerical aperture.
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10570 | https://www.desmos.com/calculator/whqdussp0n | Visualizing Sine and Cosine as Circular Functions | Desmos
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Visualizing Sine and Cosine as Circular Functions
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VISUALIZING SINE AND COSINE AS Circular FUNCTIONS
1
Click and drag the point on the unit circle to see the corresponding points the graphs of sine and cosine functions.
2
Expression 3: "T" equals 0 T=0
0 0
2 pi 2 π
3
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10571 | https://www.ictp-saifr.org/wp-content/uploads/2024/10/madeira_scattering.pdf | An introduction to low-energy scattering in quantum mechanics Lucas Madeira 1 S˜ ao Carlos Institute of Physics - University of S˜ ao Paulo - Brazil ICTP-SAIFR/ExoHad School on Few-Body Physics: Nuclear Physics from QCD October 16-25, 2024 Grant 2023/04451-9 October 16, 2024 madeira@ifsc.usp.br Quantum scattering theory Numerical Procedure Examples Formal scattering theory CePOF Optics and Photonics Research Center University of S˜ ao Paulo (USP) at S˜ ao Carlos, Brazil An introduction to low-energy scattering in quantum mechanics Lucas Madeira 2 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory References Or your favorite quantum mechanics textbook: Griffiths, Sakurai, ...
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 3 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory An introduction to low-energy scattering in quantum mechanics Lucas Madeira 4 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Motivation Scattering processes are a fundamental way of experimentally probing distributions and properties of systems in several areas of physics Can you name a few examples?
Low-energy quantum scattering theory What is low-energy?
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 5 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Introduction What is scattering?
Scattering is the interaction of an object with a scattering center classical particle electromagnetic wave quantum particle scattering potential Classical view Griffiths An introduction to low-energy scattering in quantum mechanics Lucas Madeira 6 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Quantum scattering theory An introduction to low-energy scattering in quantum mechanics Lucas Madeira 7 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Introduction Quantum view An introduction to low-energy scattering in quantum mechanics Lucas Madeira 8 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Quantum scattering theory Hypotheses 1 Elastic scattering 2 Incident plane wave eik·r 3 Local and finite-ranged potential ψk(r) large r − − − →N eik·r + eikr r f(k′, k) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 9 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Quantum scattering theory Hypotheses 1 Elastic scattering 2 Incident plane wave eik·r 3 Local and finite-ranged potential ψk(r) large r − − − →N eik·r + eikr r f(k′, k) Quantum mechanics: a scattering process is described as a transition from one quantum state to another |i⟩→|f⟩ Assume |i⟩to be a plane wave |k⟩(free particle) H0|i⟩= Ei|i⟩= ℏ2k2 2m |k⟩ Scattering is taken into account by introducing a potential V(r) H = H0 + V(r) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 9 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Quantum scattering theory Quantization of the scattering states ⟨r|k⟩= Neik·r = eik·r L3/2 We must take L →∞to guarantee the continuum character of the state at the end of our calculations An introduction to low-energy scattering in quantum mechanics Lucas Madeira 10 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Quantum scattering theory More hypotheses 1 Elastic scattering 2 Incident plane wave in the z direction: eikz 3 Local, finite-ranged and spherically-symmetric potential V(r) ψk(r, θ) large r − − − →N eikz + eikr r f(θ) The finite range of the potential (and spherical symmetry) invite us to solve the Schr¨ odinger equation for V(0 < r < R) ̸= 0 and V(r > R) = 0 −ℏ2 2m∇2ψ + V(r)ψ = Eψ An introduction to low-energy scattering in quantum mechanics Lucas Madeira 11 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Partial waves expansion An introduction to low-energy scattering in quantum mechanics Lucas Madeira 12 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherical coordinates Due to the spherical symmetry of V(r), it is convenient to employ spherical coordinates −ℏ2 2m 1 r2 ∂ ∂r r2 ∂ ∂r + L2 2mr2 + V(r) ψ(r, θ, ϕ) = Eψ(r, θ, ϕ) L is the angular momentum operator L2 = −ℏ2 1 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 sin2 θ ∂2 ∂ϕ2 Its z-component is given by Lz = −iℏ∂ ∂ϕ Construct a complete set of eigenfunctions related to H, L2, and Lz Hψ(r, θ, ϕ) = Eψ(r, θ, ϕ), L2ψ(r, θ, ϕ) = l(l + 1)ℏ2ψ(r, θ, ϕ), Lzψ(r, θ, ϕ) = mℏψ(r, θ, ϕ) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 13 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherical coordinates We propose a separable solution of the form ψ(r, θ, ϕ) = Al(r)Ym l (θ, ϕ) To avoid taking the first radial derivative of Al(r), we define the “reduced” radial solution ul(r) = rAl(r) d2 dr2 + k2 −U(r) −l(l + 1) r2 ul(r) = 0 k2 = 2mE/ℏ2 U(r) = 2mV(r)/ℏ2 l(l + 1) is the “separation constant” An introduction to low-energy scattering in quantum mechanics Lucas Madeira 14 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Solution for r > R Outside the potential range R, we must solve d2 dr2 + k2 − 0 U(r) −l(l + 1) r2 ul(r) = 0 The solution for r > R can be written in terms of the spherical Bessel functions jl(x) and nl(x) ul(r) = c′ lrjl(kr) + c′′ l rnl(kr) j0(x) = sin(x) x j1(x) = sin(x) x2 −cos(x) x j2(x) = 3 sin(x) x3 −3 cos(x) x2 −sin(x) x n0(x) = −cos(x) x n1(x) = −cos(x) x2 −sin(x) x n2(x) = −3 cos(x) x3 −3 sin(x) x2 + cos(x) x An introduction to low-energy scattering in quantum mechanics Lucas Madeira 15 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherical Bessel functions j0(x) = sin(x) x j1(x) = sin(x) x2 −cos(x) x j2(x) = 3 sin(x) x3 −3 cos(x) x2 −sin(x) x n0(x) = −cos(x) x n1(x) = −cos(x) x2 −sin(x) x n2(x) = −3 cos(x) x3 −3 sin(x) x2 + cos(x) x An introduction to low-energy scattering in quantum mechanics Lucas Madeira 16 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherical Hankel functions jl(x) and nl(x) are generalized sines and cosines It is more convenient to write the solution in terms of eikx/x to represent “incoming” or “outgoing” spherical waves Similarly to eix = cos(x) + i sin(x), we define the spherical Hankel functions as h(1) l (x) = jl(x) + inl(x) h(2) l (x) = jl(x) −inl(x) h(1) 0 (x) = −ieix x h(1) 1 (x) = −eix x+i x2 h(1) 2 (x) = i eix x2+3ix−3 x3 h(2) 0 (x) = ie−ix x h(2) 1 (x) = −e−ix x−i x2 h(2) 2 (x) = −i e−ix x2−3ix−3 x3 The solution for ul(r) can be written as ul(r) = c(1) l rh(1) l (kr) + c(2) l rh(2) l (kr) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 17 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Free particle solution The free particle solution in cartesian coordinates is a plane wave eikz In spherical coordinates, eikz = eikr cos θ contains all possible values of l This can be expressed with Rayleigh’s formula: eikr cos θ = ∞ X l=0 il(2l + 1)jl(kr)Pl(cos θ) Note that only jl appears. Physically, this is due to the divergence of nl(kr) at r = 0 In terms of the spherical Hankel functions: jl(x) = h(1) l (x) + h(2) l (x) 2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 18 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Asymptotic behavior Let us analyze the asymptotic behavior (r →∞) h(1) l (x) large x − − − →(−i)l+1 eix x h(2) l (x) large x − − − →il+1 e−ix x The free-particle solution at r →∞is eikr cos θ large r − − − → ∞ X l=0 (2l + 1) 2ikr eikr −(−1)le−ikr Pl(cos θ) The first term inside the square brackets represents an outgoing spherical wave, while the second is related to an incoming spherical wave An introduction to low-energy scattering in quantum mechanics Lucas Madeira 19 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Asymptotic behavior Motivated by the expansion for eikr cos θ, we write the scattered solution for every r > R as ψ(r, θ) = N ∞ X l=0 il(2l + 1)ul(r) r Pl(cos θ) And the asymptotic behavior ψ(r, θ) large r − − − →N ∞ X l=0 (2l + 1) ikr h c(1) l eikr −(−1)lc(2) l e−ikri Pl(cos θ) (∗) Let us compare with eikr cos θ large r − − − → ∞ X l=0 (2l + 1) 2ikr eikr −(−1)le−ikr Pl(cos θ) (∗∗) (∗∗) describes the asymptotic behavior of the wave function for a plane wave without being scattered, while (∗) does the same, but in a situation where scattering could have taken place An introduction to low-energy scattering in quantum mechanics Lucas Madeira 20 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Phase shift We introduce a new quantity related to the ratio between the constants c(1) l c(2) l = Sl(k) = e2iδl(k) Expressing the asymptotic wave function in terms of the phase shift ψ(r, θ) large r − − − →N ∞ X l=0 (2l + 1) ikr c(2) l h e2iδleikr −(−1)le−ikri Pl(cos θ) Now we have everything we need to connect with the asymptotic wave function obtained before we restricted to spherically-symmetric potentials An introduction to low-energy scattering in quantum mechanics Lucas Madeira 21 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Phase shift We know that ψk(r, θ) large r − − − →N eikz + eikr r f(θ) Expanding eikz ψk(r, θ) large r − − − →N (" ∞ X l=0 (2l + 1) 2ikr eikr −(−1)le−ikr × Pl(cos θ) # + f(θ)eikr r ) Comparing with ψ(r, θ) large r − − − →N ∞ X l=0 (2l + 1) ikr c(2) l h e2iδleikr −(−1)le−ikri Pl(cos θ) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 22 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Phase shift Collecting the terms with eikr allows us to write the scattering amplitude as a function of the phase shift f(θ) = ∞ X l=0 (2l + 1)(e2iδl −1) 2ik Pl(cos θ) The factor (e2iδl −1)/2ik is referred to as the partial wave amplitude fl(k), which may be rewritten as fl(k) = e2iδl −1 2ik = eiδl sin δl k = 1 k cot δl −ik In terms of Sl(k) Sl(k) = 1 + 2ikfl(k) = e2iδl(k) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 23 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Physical meaning of the phase shift Sl(k) = 1 + 2ikfl(k) = e2iδl(k) δl(k) is the difference between the phases of the incident and the scattered function The probability is conserved during the scattering The only thing that can change is the phase of the wave function If V = 0: free particle δl(k) = 0, fl(k) = 0 If V ̸= 0: solution for r < R depends on the details of V but for r > R we have a free particle with a “shifted” phase Defining the phase shift allows us to reduce the scattering problem to calculate a single quantity, δl(k) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 24 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Physical meaning of the phase shift g0(r) is the free-particle solution u0(r) is the solution in the presence of a scattering potential A repulsive potential (V > 0) “pushes” the particle away An attractive potential (V < 0) “pulls” the particle towards the origin An introduction to low-energy scattering in quantum mechanics Lucas Madeira 25 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Computing the phase shift Logarithmic derivative d dx ln f(x) = f ′(x) f(x) To compute the phase shift, we define the dimensionless ratio r × u′(r)/u(r) βl = ru′ l(r) ul(r) r=R− R± ≡limϵ→0 R ± ϵ The radial solution at r > R is ul(r) = 1 2re2iδlh(1) l (kr) + 1 2rh(2) l (kr) = reiδl(cos δljl(kr) −sin δlnl(kr)) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 26 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Computing the phase shift Equating βl with the outside log solution (at r = R+): βl = ru′ l(r) ul(r) r=R+ = 1 + kR cos δlj′ l(kR) −sin δln′ l(kR) cos δljl(kR) −sin δlnl(kR) After some algebra, we arrive at an expression for the phase shift cot δl(k) = kR n′ l(kR) −(βl −1) nl(kR) kR j′ l(kR) −(βl −1) jl(kR) This is an analytic expression to calculate the l-th partial wave phase-shift δl(k) (provided we know the inside solution to compute the constant βl) We will use this result later when we introduce the numerical procedure An introduction to low-energy scattering in quantum mechanics Lucas Madeira 27 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Low-energy limit An introduction to low-energy scattering in quantum mechanics Lucas Madeira 28 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Low-energy limit What is the low-energy limit?
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 29 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Low-energy limit What is the low-energy limit?
Naively, we could think simply E →0 (or k →0) The particle has a reduced wavelength ¯ λ = λ 2π = 1 k: ¯ λ →∞when k →0 However, ¯ λ can be finite as long as it is much larger than all other length scales in the system The only other length scale is the potential range R We want ¯ λ ≫R. The oscillations are so long that they cannot “see” the details of the potential In terms of k and R, the low-energy limit is kR ≪1 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 29 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Low-energy limit The radial equation for a partial wave l: −ℏ2 2m d2 dr2 + V(r) + ℏ2 2m l(l + 1) r2 ul(r) = Eul(r) We define an effective potential for the l-th partial wave as: Veff(r) = V(r) + ℏ2 2m l(l + 1) r2 For l ̸= 0, we a have repulsive centrifugal barrier Low-energy limit kR ≪1: the particle cannot overcome the centrifugal barrier In this case, the partial waves with l > 0 are unimportant l = 0 is the key for understanding low-energy scattering An introduction to low-energy scattering in quantum mechanics Lucas Madeira 30 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory s-wave scattering In the low-energy scenario, we consider partial waves with l ̸= 0 to vanish, and the resulting l = 0 term is referred to as “s-wave” The s-wave radial component u0(r) is given by: A0(r) = u0(r) r = eiδ0(cos δ0j0(kr) −sin δ0n0(kr)) = eiδ0 1 kr sin(kr + δ0) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 31 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory s-wave scattering We can also solve the zero-energy (k →0) Schr¨ odinger’s equation for r > R: d2 dr2 + 7 0 k2 − 0 U(r) − >0 l(l + 1) r2 u0(r) = 0 We simply have u′′ 0(r) = 0 (easiest Schr¨ odinger’s equation ever!) The solution can be written as u0(r) = c(r −a) Its logarithmic derivative is u′ 0(r) u0(r) = 1 r −a This needs to be equal to the logarithmic derivative of u0(r): k cot(kr + δ0) = 1 r −a An introduction to low-energy scattering in quantum mechanics Lucas Madeira 32 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering length In the limit k →0 (and setting r = 0) we define the scattering length a lim k→0 k cot δ0(k) = −1 a Previously, we had reduced the scattering problem to calculating δl(k) Now we have reduced the problem even further: in the E ≈0 limit, a encodes all the information we need about scattering The scattering amplitude for l = 0 in the low-energy limit is: f0(k) = lim k→0 1 k cot δ0 −ik = −a An introduction to low-energy scattering in quantum mechanics Lucas Madeira 33 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering length Geometrical interpretation: choose c = −1/a in u0(r) = c(r −a): u0(r) = 1 −r a a is simply the intercept of the outside wave function with the x-axis An introduction to low-energy scattering in quantum mechanics Lucas Madeira 34 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The effective range An introduction to low-energy scattering in quantum mechanics Lucas Madeira 35 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The effective range Another name for the scattering length expansion is the zero-range expansion What happens if the range of the potential is small, but non-negligible?
Expansion of k cot δ0(k) in powers of k (so far we have the first term, −1/a) k cot δ0(k) is an even function: it can only contain even powers of k The result is: k cot δ0(k) = −1 a + 1 2r0k2 + O(k4) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 36 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The effective range Consider a different normalization for u0(r > R): u0(r > R) = cot δ0(k) sin(kr) + cos(kr) Let us take the l = 0 radial equation for two different wave functions uk1(r) and uk2(r), labeled by their wave vectors (k1 = √2mE1/ℏand k2 = √2mE2/ℏ): u′′ k1(r) −U(r)uk1(r) + k2 1uk1(r) = 0 u′′ k2(r) −U(r)uk2(r) + k2 2uk2(r) = 0 Next, we multiply the first equation by uk2 and the second by uk1 and take their difference u′′ k1(r)uk2(r) −uk1(r)u′′ k2(r) = (k2 2 −k2 1)uk1(r)uk2(r) We may write the LHS as u′′ k1(r)uk2(r) −uk1(r)u′′ k2(r) = d dr u′ k1(r)uk2(r) −u′ k2(r)uk1(r) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 37 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The effective range Now we integrate from 0 to R u′ k2(r)uk1(r) −u′ k1(r)uk2(r) R 0 = (k2 2 −k2 1) Z R 0 dr uk1(r)uk2(r) The integral converges since u0(r) = rA0(r) is finite at the origin (u0(0) = 0 independently of the energy) Next, we repeat the same procedure for the free-particle (V = 0) radial equation with solutions denoted by gk1(r) and gk2(r). The result is the same if we replace u by g An introduction to low-energy scattering in quantum mechanics Lucas Madeira 38 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The effective range Finally, we take the difference between the results g′ k2(r)gk1(r) −g′ k1(r)gk2(r) R 0 − u′ k2(r)uk1(r) −u′ k1(r)uk2(r) R 0 = (k2 2 −k2 1) Z R 0 dr [gk1(r)gk2(r) −uk1(r)uk2(r)] u0(0) = 0 g(r) and u(r > R) are equal for r ⩾R The free-particle solution is also given by: g(r) = cot δ0(k) sin(kr) + cos(kr) Then we are left with g′ k2(0)gk1(0) −g′ k1(0)gk2(0) = (k2 2 −k2 1) Z R 0 dr [gk1(r)gk2(r) −uk1(r)uk2(r)] An introduction to low-energy scattering in quantum mechanics Lucas Madeira 39 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The effective range Using g(r) = cot δ0(k) sin(kr) + cos(kr) in the RHS k2 cot δ0(k2) −k1 cot δ0(k1) = (k2 2 −k2 1) Z R 0 dr [gk1(r)gk2(r) −uk1(r)uk2(r)] If we take the limit k1 →0, we can write k1 cot δ0(k1) in terms of the scattering length k cot δ0(k) = −1 a + k2 Z R 0 dr [g0(r)gk(r) −u0(r)uk(r)] We define the next term r0/2 as r0 ≡lim k→0 ρ(k) = 2 Z R 0 dr [g2 0(r) −u2 0(r)] = 2 Z R 0 dr 1 −r a 2 −u2 0(r) r0 is called effective range An introduction to low-energy scattering in quantum mechanics Lucas Madeira 40 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Shape-independent approximation The resultant expression is the shape-independent approximation: k cot δ0(k) = −1 a + 1 2r0k2 + O(k4) We are describing the phase shift δ0(k) without taking into account the microscopic parameters of the scattering potential An introduction to low-energy scattering in quantum mechanics Lucas Madeira 41 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Bound states An introduction to low-energy scattering in quantum mechanics Lucas Madeira 42 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Bound states Let us rewrite the scattered wave function for r →∞as ψ(r, θ) large r − − − → 1 (2π)3/2 ∞ X l=0 (2l + 1) 2ik Pl(cos θ) " Sl(k)eikr r −e−i(kr−lπ) r # For l = 0 and large distances, the radial wave function is proportional to S0(k)eikr r −e−ikr r An introduction to low-energy scattering in quantum mechanics Lucas Madeira 43 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Bound states For an arbitrary finite-ranged potential V, the radial solution at r > R for a bound state (E < 0) obeys u′′(r) = −2mE ℏ2 u(r) = κ2u(r), κ ≡ √ −2mE ℏ The solution can be written as u(r > R) = :0 Aeκr + Be−κr We conclude that the radial function for a bound state at large distances is A(r) = u(r) r ∝e−κr r (large r) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 44 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Bound states Scattering solution S0(k)eikr r −e−ikr r Bound state solution e−κr r By substituting k →iκ, with k purely imaginary, we can connect the bound state with the scattered solution eikr r = ei(iκ)r r = e−κr r S0(k) controls the ratio of the outgoing to the incoming wave In the bound state case, we have only the outgoing spherical wave, thus S0(k) →∞ An introduction to low-energy scattering in quantum mechanics Lucas Madeira 45 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Bound states Sl(k) = 1 + 2ik fl(k) = e2iδl(k) is a complex function S0(k) →∞by substituting k →iκ means it has a pole at k = iκ Scattering continuum: real k > 0 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 46 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Bound states In terms of the s-wave scattering amplitude f0(k) f0(k) = 1 k cot δ0 −ik = 1 −1/a −ik We write S0(k) as S0(k) = 1 + 2ikf0(k) = −k −i/a k −i/a This expression has a pole at k = iκ if we identify κ = 1 a In the zero-energy limit, the energy of a bound state and the scattering length are connected simply by E = ℏ2k2 2m = −ℏ2κ2 2m = −ℏ2 2ma2 A single parameter originated from the potential determines the bound-state energy An introduction to low-energy scattering in quantum mechanics Lucas Madeira 47 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Two-body scattering An introduction to low-energy scattering in quantum mechanics Lucas Madeira 48 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Two-body scattering So far, we considered only the problem of a single particle being scattered by a finite-ranged potential V(r) located at r = 0 With a few modifications, we can use the results we obtained to describe two particles interacting through a pairwise potential which depends only on their spatial separation r The Hamiltonian of a two-body system is separable in the center of mass (CM) and relative coordinates: H = −ℏ2 2m1 ∇2 r1 −ℏ2 2m2 ∇2 r2 + V(r1 −r2) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 49 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Two-body scattering We define the CM and relative coordinates R = m1r1 + m2r2 M and r = r1 −r2 H is now separable H = HCM + Hr, HCM = −ℏ2 2M∇2 R, Hr = −ℏ2 2mr ∇2 r + V(r) mr = m1m2/(m1 + m2) is the reduced mass The CM motion satisfies the free-particle equation →only adds a constant to the total energy The relative motion Hamiltonian is exactly the one we used for one particle if m →mr We can apply our previous results to two-body scattering An introduction to low-energy scattering in quantum mechanics Lucas Madeira 50 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Applications An introduction to low-energy scattering in quantum mechanics Lucas Madeira 51 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Applications 1 Spherically symmetric finite well Analytical calculation of the s-wave scattering wave function 1 Scattering states (E > 0) 2 Bound states (E < 0) Calculation of the scattering length and effective range 2 Zero-range and finite-range approximations Estimating bound state energies using the scattering length and effective range expansions An introduction to low-energy scattering in quantum mechanics Lucas Madeira 52 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric finite well One way of defining the spherical well is Vsw(r) = ( −V0 for r < R, 0 for r > R.
V0 has units of [energy] For numerical applications, it is useful to redefine the potential as V(r) = −v0 ℏ2 mrR2 for r < R, 0 for r > R.
v0 is dimensionless and related to the depth. We consider only v0 > 0 (attractive potential) R is the range of the potential For a relatively shallow or short-ranged potential: only continuum scattering states (E > 0) Increasing its depth or range may make it strong enough to produce a bound state (E < 0) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 53 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E > 0) Potential V(r) = −v0 ℏ2 mrR2 for r < R, 0 for r > R.
E > 0 case d2 dr2 −2mr ℏ2 V(r) + 2mr ℏ2 E u(r) = 0 Equations for r < R and r > R: u′′(r) + k2 0 + k2 u(r) = 0 for r < R, u′′(r) + k2u(r) = 0 for r > R, k2 ≡2mrE/ℏ2 and k2 0 ≡2v0/R2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 54 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E > 0) If r < R u(r) = A sin q k2 + k2 0 r + B cos q k2 + k2 0 r Since u0(0) = 0, we set B = 0 If r > R u(r) = cot δ0(k) sin(kr) + cos(kr) Hence, the solution is of the form u(r) = A sin q k2 + k2 0 r for r < R, cot δ0(k) sin(kr) + cos(kr) for r > R.
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 55 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E > 0) Logarithmic derivative at r = R−and r = R+ ru′(r) u(r) r=R−= ru′(r) u(r) r=R+ q k2 + k2 0 cos q k2 + k2 0 R sin q k2 + k2 0 R = k cot δ0(k) cos(kR) −k sin(kR) cot δ0(k) sin(kR) + cos(kR) Solving for the phase shift δ0(k) without any approximation δ0(k) = −kR + arctan k tan q k2 + k2 0 R q k2 + k2 0 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 56 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E > 0) Scattering length To calculate the scattering length a, we need to take the k →0 limit Rearrange the log derivative so that we collect factors of k cot δ0(k) Keep track of the orders employed in the approximation cos(kR) = 1 + O(k2) sin(kR) = kR + O(k3) Repeating last slides’ equation: q k2 + k2 0 cos q k2 + k2 0 R sin q k2 + k2 0 R = k cot δ0(k) cos(kR) −k sin(kR) cot δ0(k) sin(kR) + cos(kR) Taking the k →0 limit: q k2 0 cot q k2 0 R = −1/a −R/a + 1 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 57 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E > 0) Solving for a: a = R − tan q k2 0R q k2 0 = R 1 −tan √2v0 √2v0 !
It is clear that a depends only on the parameters of the potential (depth v0 and range R) Note that tan(x) →∞for x = π 2 + nπ, n = 0, ±1, ±2, ...
So the first divergence (n = 0) of a appears at v0 = π2 8 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 58 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E < 0) E < 0 case Repeat the same procedure or k = iκ →E = ℏ2k2/2mr = −ℏ2κ2/2mr d2 dr2 −2mr ℏ2 V(r) −2mr ℏ2 |E| u(r) = 0 Solution for u(r) u(r) = A′ sin q k2 0 −κ2 r for r < R, B′e−κr for r > R Match the logarithmic derivatives at r = R: q k2 0 −κ2 cos q k2 0 −κ2 R sin q k2 0 −κ2 R = −κe−κR e−κR An introduction to low-energy scattering in quantum mechanics Lucas Madeira 59 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E < 0) After some manipulations: tan q k2 0 −κ2 R + q k2 0 −κ2 κ = 0 This is a transcendental equation that shows where the bound-state energies are located Note that the term q k2 0 −κ2/κ is always positive tan q k2 0 −κ2 R must be negative if we want the equation to have solution(s). That is to say: π 2 + nπ < q k2 0 −κ2 R < π + nπ, n = 0, 1, 2, ...
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 60 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well (E < 0) The first bound state is n = 0: π 2R < q k2 0 −κ2 < π R k0 > q k2 0 −κ2 k0 = √2v0/R v0 > π2 8 This result shows that there are no bound states if v0 is not above a certain threshold value This is the same threshold value that makes |a| →∞ The conclusion is that the scattering length diverges when a bound state appears An introduction to low-energy scattering in quantum mechanics Lucas Madeira 61 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well - bound states and scattering length a = R 1 −tan √2v0 √2v0 !
a diverges for: p 2v0 = π/2 + nπ (n = 0, 1, 2, ...) This coincides with the location of the bound states An introduction to low-energy scattering in quantum mechanics Lucas Madeira 62 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well - effective range First, we need to determine the normalization constant of the scattering solution u(r) = A sin q k2 + k2 0 r for r < R, cot δ0(k) sin(kr) + cos(kr) for r > R To determine the constant A, we impose the continuity of u(r) at r = R: A = cot δ0(k) sin(kR) + cos(kR) sin q k2 + k2 0R An introduction to low-energy scattering in quantum mechanics Lucas Madeira 63 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well - effective range The normalized solution is written as u(r) = cot δ0(k) sin(kR)+cos(kR) sin(√ k2+k2 0R) sin q k2 + k2 0 r for r < R, cot δ0(k) sin(kr) + cos(kr) for r > R.
The effective range is defined in the k →0 limit of u(r): lim k→0 u(r) = ( (1−R/a) sin(k0R) sin(k0r) for r < R, 1 −r/a for r > R.
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 64 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well - effective range The effective range is given by the integral r0 = 2 Z R 0 dr " 1 −r a 2 − 1 −R a 2 sin2(k0r) sin2(k0R) # Replacing a in favor of R and k0: r0 R = 1 −1 3 k0R tan(k0R) −k0R 2 + 1 k0R tan(k0R) −(k0R)2 !
This shows that r0 also depends only on parameters of the potential An introduction to low-energy scattering in quantum mechanics Lucas Madeira 65 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Spherically-symmetric well - effective range r0 R = 1 −1 3 k0R tan(k0R) −k0R 2 + 1 k0R tan(k0R) −(k0R)2 !
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 66 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Applications 1 Spherically symmetric finite well Analytical calculation of the s-wave scattering wave function 1 Scattering states (E > 0) 2 Bound states (E < 0) Calculation of the scattering length and effective range 2 Zero-range and finite-range approximations Estimating bound state energies using the scattering length and effective range expansions An introduction to low-energy scattering in quantum mechanics Lucas Madeira 67 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Zero-range and finite-range approximations The equation derived in the bound states slide allows us to estimate the bound state energy with the zero-range approximation κ = 1/a, Ezr = −ℏ2κ2 2mr = − ℏ2 2mra2 To take the effective range into account, we write the s-wave scattering amplitude as f0(k) = 1 k cot δ0(k) −ik = 1 −1/a + r0k2/2 −ik And S0(k) as S0(k) = 1 + 2ikf0(k) = −i/a −k + ir0k2/2 −i/a + k + ir0k2/2 Making k →iκ S0(k) = −1/a −κ −r0κ2/2 −1/a + κ −r0κ2/2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 68 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Zero-range and finite-range approximations Now we can identify the bound state as pole in the S-matrix by solving −1 a + κ −r0κ2 2 = 0 which yields the solutions κ = 1 r0 1 ∓ r 1 −2r0 a !
For r0/a ≪1: κr0 = 1 ∓ p 1 −2r0/a ≈1 ∓1 ± r0/a Now choosing the appropriate root to compute the bound state energy Efr = −ℏ2κ2 2mr = − ℏ2 2mrr2 0 1 − r 1 −2r0 a !2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 69 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Zero-range and finite-range approximations Example: Helium dimer Ed = −1.62 mK (found solving the full Schr¨ odinger equation), a = 90.4 ˚ A, r0 = 8.0 ˚ A Zero-range approximation Ezr kB = − ℏ2 kB × 2mra2 = −1.48 mK (92%) Finite-range approximation Efr kB = − ℏ2 kB × 2mrr2 0 1 − r 1 −2r0 a !2 = −1.63 mK (101%) Both the zero- and finite-range results successfully describe the physical system because kR ∼0.1 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 70 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Zero-range and finite-range approximations Example: Deuteron Edc2 = −2.224 MeV, a = 5.4112 fm, r0 = 1.7436 fm.
Zero-range approximation Ezrc2 = −ℏ2c2 2mra2 = −1.416 MeV (64%) Finite-range approximation Efrc2 = −ℏ2c2 2mrr2 0 1 − r 1 −2r0 a !2 = −2.223 MeV (100%) The range of the potential needed to be taken into account because kR ∼0.4 We should emphasize that the scales are very different in both examples 4He dimer: spatial scale of ˚ A (10−10 m) and the energy is of the order of 10−7 eV Deuteron: the lengths are in femtometers (10−15 m) and the energy is of a few MeV (106 eV) This exemplifies how universal are these low-energy scattering results An introduction to low-energy scattering in quantum mechanics Lucas Madeira 71 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerical Procedure An introduction to low-energy scattering in quantum mechanics Lucas Madeira 72 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerical Procedure Analytical expressions for the low-energy scattering parameters are only available for a few potentials Even in those cases, the calculations may be cumbersome, as we saw for the spherical well In general, we need to calculate a and r0 numerically We will describe two methods to solve Schr¨ odinger’s equation numerically 1 Second-order central difference 2 Numerov’s method An introduction to low-energy scattering in quantum mechanics Lucas Madeira 73 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerical Procedure We wish to compute the quantities a and r0 To do so, we need to compute the radial solution inside and outside the potential range u0(r < R): needs to be computed numerically u0(r > R) = 1 −r/a An introduction to low-energy scattering in quantum mechanics Lucas Madeira 74 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Second-order central difference Consider the function u(r) on a discrete set of points ri = i∆r, i = 0, 1, 2..., N and ∆r ≪1 Let us take two Taylor expansions of u(r) around the points r ± ∆r u(r + ∆r) = u(r) + (∆r)u′(r) + (∆r)2 2 u′′(r) + (∆r)3 6 u′′′(r) + · · · , u(r −∆r) = u(r) −(∆r)u′(r) + (∆r)2 2 u′′(r) −(∆r)3 6 u′′′(r) + · · · The difference of the two Taylor expansions yields an expression for the first derivative, while their sum results in the second derivative du dr r=ri = ui+1 −ui−1 2∆r + O[(∆r)2] d2u dr2 r=ri = ui+1 −2ui + ui−1 (∆r)2 + O[(∆r)2] Note: hereafter, we are going to use the compact notation u0(ri) ≡ui An introduction to low-energy scattering in quantum mechanics Lucas Madeira 75 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Second-order central difference We want to solve the zero-energy Schr¨ odinger equation inside the potential range d2 dr2 −2mr ℏ2 V(r) u0(r) = 0 d2u dr2 r=ri ≈ui+1 −2ui + ui−1 (∆r)2 Substituting the central difference second derivative into u′′(r) ui+1 = 2ui −ui−1 + 2mr(∆r)2 ℏ2 V(ri)ui An introduction to low-energy scattering in quantum mechanics Lucas Madeira 76 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Second-order central difference ui+1 = 2ui −ui−1 + 2mr(∆r)2 ℏ2 V(ri)ui If we know the value of the radial solution for two consecutive points (ui−1 and ui) we can calculate the value for the next point ui+1 u(0) = 0 u(∆r) = 1 This choice allows us to find a solution without worrying about the normalization Algorithm: 1 Set u0 = 0, u1 = 1, and i = 1 2 Compute ui+1 3 If ri ⩾R + ∆r, stop. Else, increment i by one 4 Go to step 2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 77 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerov’s method The second-order central difference is one possible discretization for a numerical second derivative There are other alternatives if we want to improve the precision of our algorithm Numerov’s method is a numerical technique capable of solving differential equations of second order when the first-order term is not present: d2y dx2 = −ξ(x)y(x) + s(x) The s-wave zero-energy radial equation is of this form, with y →u, x →r, s = 0, and ξ(r) = −2mr ℏ2 V(r) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 78 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerov’s method The method provides a solution of the form yi+1 = 1 1 + (∆x)2 12 ξi+1 ( 2yi 1 −5(∆x)2 12 ξi −yi−1 1 + (∆x)2 12 ξi−1 + (∆x)2 12 (si+1 + 10si + si−1) ) + O[(∆x)6] The algorithm is mostly unchanged if we use Numerov’s method instead of the second-order central difference An introduction to low-energy scattering in quantum mechanics Lucas Madeira 79 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Dimensionless quantities Schr¨ odinger’s equation contains relatively small quantities ℏ∼10−34 J s (or ∼10−15 eV s) Typical masses, length, and energy scales are also small We wish to make Schr¨ odinger’s equation dimensionless Instead of this −1 2 d2 dr2 −mr ℏ2 E + mr ℏ2 V(r) u(r) = 0 we want to solve this −1 2 d2 d¯ r2 −¯ E + ¯ V(¯ r) ¯ u(¯ r) = 0 “ ℏ= mr = 1 ” An introduction to low-energy scattering in quantum mechanics Lucas Madeira 80 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Dimensionless quantities First, we choose a length scale ℓ The convenient value of ℓdepends on the system under study Atomic physics: 1 ˚ A Nuclear physics: 1 fm Or any other length scale that makes sense for a particular problem Then the dimensionless scaled distance is ¯ r = r ℓ The radial function u(r) has units of [length]−1/2 (remember that R dr|u(r)|2 = 1) ¯ u(¯ r) = u(r) ℓ−1/2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 81 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Dimensionless quantities The second derivative becomes d2 dr2 = 1 ℓ2 d2 d¯ r2 Going back to the equation: − ℏ2 2mrℓ2 d2¯ u d¯ r2 + V(¯ r)¯ u = E¯ u We can also define an energy scale ϵ = ℏ2 mrℓ2 And now we define the dimensionless energy and potential ¯ E = E ϵ , ¯ V = V ϵ Finally −1 2 d2¯ u d¯ r2 + ¯ V(¯ r)¯ u = ¯ E¯ u An introduction to low-energy scattering in quantum mechanics Lucas Madeira 82 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering length and effective range After finding u(r) numerically, we’re ready to compute a and r0 Scattering length We recall that logarithmic derivative of the wave function outside the potential range is given by g′ 0(r) g0(r) r=R+ = 1 R −a This should be equal to the logarithmic derivative of u0(r) at r = R− g′ 0(r) g0(r) r=R+ = 1 R −a = u′ 0(r) u0(r) r=R− We already have u(R) and u(R ± ∆r). Thus the derivative may be computed as u′ num(R) = du(r) dr r=R = u(R + ∆r) −u(R −∆r) 2∆r An introduction to low-energy scattering in quantum mechanics Lucas Madeira 83 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering length and effective range Scattering length Solving for a a = R − 2∆r u(R) u(R + ∆r) −u(R −∆r) This expression depends on the ratio of the radial solution, so we ignored the normalization Effective range On the other hand, the effective range assumes a particular normalization choice We multiply u(r) by a constant C such that C = g(R) u(R) = (1 −R/a) u(R) The effective range is found by computing the integral r0 = 2 Z R 0 dr [g2 0(r) −u2 0(r)] An introduction to low-energy scattering in quantum mechanics Lucas Madeira 84 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerical integration The task is essentially to compute numerically an integral of the form I = Z xN x1 f(x)dx f(x) is known only at a discrete set of equally spaced points, f(xi) ≡fi, where i = 1, 2, 3, ..., N.
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 85 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerical integration Trapezoidal rule: Z x2 x1 f(x)dx = h 1 2f1 + 1 2f2 + O(h3f ′′) Using it N −1 times for the intervals: (x1, x2), (x2, x3), · · · , (xN−1, xN) Z xN x1 f(x)dx = h 1 2f1 + f2 + f3 + · · · + fN−1 + 1 2fN + O (xN −x1)3f ′′ N2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 86 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Numerical integration Simpson’s rule: Z x3 x1 f(x)dx = h 1 3f1 + 4 3f2 + 1 3f3 + O(h5f (4)) Repeatedly: Z xN x1 f(x)dx = h 1 3f1 + 4 3f2 + 2 3f3 + 4 3f4 + · · · + 2 3fN−2 + 4 3fN−1 + 1 3fN + O (xN −x1)5f (4) N4 !
An introduction to low-energy scattering in quantum mechanics Lucas Madeira 87 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Examples An introduction to low-energy scattering in quantum mechanics Lucas Madeira 88 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Examples We chose four potentials to illustrate the numerical procedure Spherical well Modified P¨ oschl-Teller Gaussian Lennard-Jones An introduction to low-energy scattering in quantum mechanics Lucas Madeira 89 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Potentials - Spherical well To make the comparison with other potentials easier, we redefine Vsw(r) = −vsw ℏ2µ2 sw mr , for r < R, 0, for r > R vsw is a dimensionless parameter related to the depth µsw = 1/R We can compare our numerical solutions with the analytical ones to check the correctness of the program An introduction to low-energy scattering in quantum mechanics Lucas Madeira 90 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Potentials - Modified P¨ oschl-Teller (mPT) VPT(r) = −vPT ℏ2 mr µ2 PT cosh2(µPTr) Very difficult analytical solution for the eigenfunctions There is an analytical expression for a in terms of the parameters of the potential aµPT = π 2 cot πλ 2 + γ + Ψ(λ) vPT = λ(λ −1)/2, γ is the Euler-Mascheroni constant and Ψ is the digamma function The |a| →∞case corresponds to λ = 2 [cot(π) diverges] or λ = −1 [Ψ(−1) diverges], that is, vPT = 1 For this particular case (|a| →∞), the s-wave zero-energy radial function takes a relatively simple form u0(r) = tanh(µPT r) tanh(µPT R) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 91 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Potentials - mPT We can also calculate the effective range by performing the integral In this case (|a| →∞), g0(r) = 1 −r/a = 1, so that r0 = 2 Z R 0 dr " 1 −tanh2(µPTr) tanh2(µPTR) # = 2 R − R tanh2(µPTR) + 1 µPT tanh(µPTR) Since 1/µPT ∼R and the tanh(x) function converges rapidly to 1 as we increase x, we may set tanh(µPTR) = 1. Thus we have : r0 = 2 µPT (for vPT = 1) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 92 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Potentials Spherical well Modified P¨ oschl-Teller VPT(r) = −vPT ℏ2 mr µ2 PT cosh2(µPTr) Gaussian Vg(r) = −vg ℏ2 mr µ2 ge−r2µ2 g The potential range R is not well defined for the mPT and the gaussian potentials Look for a value of R such that the potential is negligible |V(R)| ⩽ε An introduction to low-energy scattering in quantum mechanics Lucas Madeira 93 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Potentials Lennard-Jones VLJ(r) = ℏ2 mr C12 r12 −C6 r6 Note that VLJ(0) diverges and V(∆r) is very large u(0) = 0 ✓ Computing u(∆r) may lead to instabilities Define a range 0 ⩽r < rmin where u(r) = 0 Start the integration at r = rmin An introduction to low-energy scattering in quantum mechanics Lucas Madeira 94 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Tuning the parameters An introduction to low-energy scattering in quantum mechanics Lucas Madeira 95 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Tuning the parameters The four potentials we presented have two parameters Spherical well, mPT, Gaussian are purely attractive one parameter is associated with the depth of the potential (vsw, vPT, and vg) and another with its range (µsw, µPT, and µg) The LJ potential has a repulsive core and an attractive region C6 controls the attractive interaction C12 controls the repulsive interaction Typically, the scattering length and effective range are known, and we want to tune the parameters of a particular potential to reproduce the desired a and r0 values Since we want to match two values and have two free parameters, the correspondence is one-to-one (with the restriction of how many bound states we want) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 96 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Tuning the parameters A possible algorithm: 1 Start with a guess (v1, v2).
2 Compute a and r0 3 Keep v2 fixed. Vary v1 until a has the desired value. Increasing the depth of the potential will decrease the value of the scattering length (until it diverges and changes from −∞to +∞) 4 Keep v1 fixed at the value found in step 3. Vary v2 until r0 has the desired value. Increasing the range of the potential will increase r0 5 If a and r0 match the desired values, stop. Else, go to step 3 and use the value of v2 found in step 4 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 97 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Results An introduction to low-energy scattering in quantum mechanics Lucas Madeira 98 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Results We present 3 cases: a < 0, |a| →∞, and a > 0, which correspond to three very distinct physical situations a < 0 Example: neutron-neutron interaction (a = −18.5 fm, r0 = 2.7 fm) |a| →∞ Unitarity a > 0 Example: deuteron (a = 5.4 fm, r0 = 1.7 fm) System a (fm) r0 (fm) Neutron-neutron −18.5 2.7 Unitarity ±∞ 1.0 Deuteron 5.4 1.7 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 99 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Results a < 0 Potential v µ (fm−1) a (fm) r0 (fm) Neutron-neutron Well 1.1096 0.3918 −18.52 2.7 mPT 0.9071 0.7991 −18.51 2.7 Gaussian 1.2121 0.5672 −18.55 2.7 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 100 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Results |a| →∞ Potential v µ (fm−1) a (fm) r0 (fm) Unitarity Well 1.2337 1.0000 ∼−105 1.0 mPT 1.0000 2.0000 ∼109 1.0 Gaussian 1.3420 1.4349 ∼−105 1.0 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 101 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Results a > 0 Potential v µ (fm−1) a (fm) r0 (fm) Deuteron Well 1.7575 0.5000 5.4 1.70 mPT 1.4388 0.8631 5.4 1.73 Gaussian 1.9102 0.6754 5.4 1.70 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 102 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Results Scattering length as a function of the strength of the attractive potential An introduction to low-energy scattering in quantum mechanics Lucas Madeira 103 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Conclusions We presented quantum scattering theory fundamentals focusing on the low-energy limit In this context, we introduced two significant quantities: the scattering length and the effective range To illustrate how these two parameters behave in a concrete example, we derived analytical expressions for the spherical well We also showed how the energy of a bound state could be calculated using zero- and finite-range expressions applied to a 4He dimer and the deuteron We described a numerical procedure that can be used to compute the scattering length and effective range of any spherically symmetric finite-ranged two-body potential Examples: spherical well, modified P¨ oschl-Teller, Gaussian, and Lennard-Jones potentials Now, you can extend what you learned to your choice of physical systems, and apply the method to other potentials An introduction to low-energy scattering in quantum mechanics Lucas Madeira 104 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Formal scattering theory An introduction to low-energy scattering in quantum mechanics Lucas Madeira 105 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Time-dependent formalism Scattering can be seen as a time-dependent process Interactions →interaction picture Time evolution in the Schr¨ odinger picture: |ϕ(t)⟩S = US(t, t0)|ϕ(t0)⟩S | ⟩S is a ket in the Schr¨ odinger picture and US(t, t0) is the time-evolution operator If H is time independent: US(t, t0) = e−iH(t−t0)/ℏ The interaction-picture state ket is defined as: |ϕ(t)⟩I = eiH0t/ℏ|ϕ(t)⟩S The operators in the interaction picture are defined as: AI = eiH0t/ℏASe−iH0t/ℏ An introduction to low-energy scattering in quantum mechanics Lucas Madeira 106 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The interaction picture The Schr¨ odinger equation takes the form: iℏ∂ ∂t|ϕ(t)⟩S = H|ϕ(t)⟩S → iℏ∂ ∂t|ϕ(t)⟩I = VI|ϕ(t)⟩I VI = eiH0(t−t0)/ℏVe−iH0(t−t0)/ℏis the potential in the interaction picture Advantage: we removed H0 from our calculations (to consider the interaction) If V = 0: |ϕ(t)⟩I is constant in time (and equal to |ϕ(t0)⟩S) The time-evolution of |ϕ(t)⟩I is given by: |ϕ(t)⟩I = UI(t, t0)|ϕ(t0)⟩I with UI(t, t0) = eiH0t/ℏUS(t, t0)e−iH0t0/ℏ It obeys the Schr¨ odinger-like equation: iℏ∂ ∂tUI(t, t0) = VI(t)UI(t, t0) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 107 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering and the interaction picture iℏ∂ ∂tUI(t, t0) = VI(t)UI(t, t0) The solution is given by: UI(t, t0) = 1 −i ℏ Z t t0 dt′VI(t′)UI(t′, t0) Compatible with the initial condition UI(t0, t0) = 1.
Our goal: to calculate the evolution of the state in a distant past (t0 →−∞), when |ϕ(t →−∞)⟩= |i⟩ The solution is only valid for finite times t and t0 →setting UI(t, −∞) would lead to convergence problems We need to give mathematical meaning to UI(t, −∞) and UI(+∞, t) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 108 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Distant past and distant future Consider the equality: lim t0→−∞f(t0) = lim ϵ→0 ϵ Z 0 −∞ dt′ eϵt′f(t′) Applying this to the time evolution operator: UI(t, −∞) = lim t0→−∞UI(t, t0) = lim ϵ→0 ϵ Z 0 −∞ dt′ eϵt′UI(t, t′) UI(+∞, t0) = lim t→+∞UI(t, t0) = lim ϵ→0 ϵ Z +∞ 0 dt′ e−ϵt′UI(t′, t0) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 109 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Despite the time-dependent treatment, H is time-independent: US(t, t′) = e−iH(t−t′)/ℏ The state vector at a time t = 0 is given by: |ϕ(t = 0)⟩I = UI(0, −∞)|i⟩ with UI(0, −∞) = lim ϵ→0 ϵ Z 0 −∞ dt′ eϵt′eiHt′/ℏe−iH0t′/ℏ |ϕ(t = 0)⟩I = lim ϵ→0 ϵ Z 0 −∞ dt′ eϵt′ei(H−Ei)t′/ℏ|i⟩= lim ϵ→0 iϵ Ei −H + iϵ|i⟩ Using the identity: 1 Ei −H + iϵ − 1 Ei −H0 + iϵ = 1 Ei −H0 + iϵV 1 Ei −H + iϵ we rewrite the result as: |ϕ(t = 0)⟩I = lim ϵ→0 iϵ Ei −H0 + iϵ|i⟩+ 1 Ei −H0 + iϵV iϵ Ei −H + iϵ|i⟩ = lim ϵ→0 iϵ Ei −H0 + iϵ|i⟩+ 1 Ei −H0 + iϵV|ϕ(t = 0)⟩I An introduction to low-energy scattering in quantum mechanics Lucas Madeira 110 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The Lippmann-Schwinger equation |ϕ(t = 0)⟩I = lim ϵ→0 iϵ Ei −H0 + iϵ|i⟩+ 1 Ei −H0 + iϵV iϵ Ei −H + iϵ|i⟩ = lim ϵ→0 iϵ Ei −H0 + iϵ|i⟩+ 1 Ei −H0 + iϵV|ϕ(t = 0)⟩I H0|i⟩= Ei|i⟩ |ψ⟩= |i⟩+ 1 Ei −H0 + iϵV|ψ⟩ We left off the notation |ϕ(t = 0)⟩to emphasize that this is an actual time-independent problem This is know as the Lippmann-Schwinger equation An introduction to low-energy scattering in quantum mechanics Lucas Madeira 111 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory The distant future Back to the time-dependent formulation, we can use the sequential relation of the time translation operator: UI(t, t0) = UI(t, t′)UI(t′, t0) |ϕ(t)⟩= UI(t, 0)|ϕ(0)⟩= UI(t, −∞)|i⟩ In a distant future (t →+∞): |f⟩= UI(+∞, −∞)|i⟩= S|i⟩, S ≡UI(+∞, −∞).
Scattering (S) matrix The action of the S matrix on an initial state (that exists asymptotically for t0 →−∞) transforms the ket |i⟩into a final state that exists in a distant future t →+∞ An introduction to low-energy scattering in quantum mechanics Lucas Madeira 112 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Dyson series and the first-order Born approximation Lippmann-Schwinger equation: |ψ⟩= |i⟩+ 1 Ei −H0 + iϵV|ψ⟩ Rewritten as a power-series expansion: |ψ⟩=|i⟩+ G+V|i⟩+ G+VG+V|i⟩+ ... = |i⟩+ G+(V + VG+V + ...)|i⟩ where G+ ≡(Ei −H0 + iϵ)−1 We define the transition matrix T as the perturbative series: T ≡V + VG+V + VG+VG+V + ...
This is known as the Dyson series A consequence is that: V|ψ⟩= T|i⟩ The T-matrix is a kind of a generalized potential First-order perturbation: T and V are equivalent →|ψ⟩≈|i⟩ This is known as the first-order Born approximation An introduction to low-energy scattering in quantum mechanics Lucas Madeira 113 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering theory integral equations Lippmann-Schwinger equation: |ψ⟩= |i⟩+ 1 Ei −H0 + iϵV|ψ⟩ In the position basis: ⟨r|ψ⟩= ⟨r|i⟩+ Z d3r′ r |G+| r′ r′ |V| ψ We have to compute: G+(r, r′) ≡ r 1 E −H0 + iϵ r′ Momentum basis {|k⟩} elements are eigenstates of H0 with eigenvalues ℏ2k2/2m G+(r, r′) = ℏ2 2m X k′,k′′ ⟨r|k′⟩ k′ 1 E −H0 + iϵ k′′ ⟨k′′|r′⟩ Plane waves in the position representation: ⟨r|k⟩= eik·r L3/2 and ⟨k|r⟩= e−ik·r L3/2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 114 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering theory integral equations G+(r, r′) = 1 L3 X k′ eik′·(r−r′) k2 −k′2 + iϵ We absorbed the factor 2m/ℏ2 into ϵ We are left with a sum in k-space: discrete Periodic boundary conditions: ki = 2πni/L (i = x, y, z) where ni = 0, 1, 2, 3...
{k} = {0, 2π/L, 4π/L, ...}: the distance between each point in k-space is ∆k = 2π/L We must take L →∞to guarantee the required continuous character: the separation between points must be very small compared to L (∆k ≈0) We may substitute the sum by an integral: X k′ → Z ρ(k)d3k′ = L3 (2π)3 Z d3k′ ρ(k) = L3/(2π)3 is the k-density in three dimensions An introduction to low-energy scattering in quantum mechanics Lucas Madeira 115 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering theory integral equations G+(r, r′) = 1 (2π)3 Z d3k′ eik′·(r−r′) k2 −k′2 + iϵ Spherical coordinates: (k′, θ, ϕ) Let the vector (r −r′) lie along the k′ z axis: k′ · (r −r′) = k′|r −r′| cos θ G+(r, r′) = 1 (2π)2 Z ∞ 0 dk′k′2 Z π 0 dθ eik′|r−r′| cos θ k2 −k′2 + iϵ = 1 4π2|r −r′| Z ∞ −∞ dk′ k′ sin(k′|r −r′|) k2 −k′2 + iϵ The integrand is even Poles at k′ = ± √ k2 + iϵ = ±(k + iϵ/2k −ϵ2/8k3 + ...) Ignoring terms higher than ϵ2 (redefining ϵ/2k →ϵ): (k2 −k′2 + iϵ) = −(k′ −k −iϵ)(k′ + k + iϵ) The integral is then: G+(r, r′) = 1 8iπ2|r −r′| Z ∞ −∞ dk′ k′(e−ik′|r−r′| −eik′|r−r′|) (k′ −k −iϵ)(k′ + k + iϵ) An introduction to low-energy scattering in quantum mechanics Lucas Madeira 116 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory We let k′ momentarily be a complex variable to carry out this integration and use the residue theorem: I ΓR = Z γR + Z CR = 2πi × X j Res{k′; j} Upper plane path where eik|r−r′| →0 Lower plane path where e−ik|r−r′| →0 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 117 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering theory integral equations The closed path integral may be written as the sum: I ΓR = Z γR + Z CR = 2πi × X j Res{k′; j} The integral along the path γR is zero due to Jordan’s lemma The only pole inside Γ± R is ±(k + iϵ) The residues may be calculated as: Res{k′; k + iϵ} = lim k′→k+iϵ ϵ→0 (k′ −k −iϵ) k′eik′|r−r′| (k′ −k −iϵ)(k′ + k + iϵ) = eik|r−r′| 2 Res{k′; −k −iϵ} = lim k′→−k−iϵ ϵ→0 (k′ + k + iϵ) k′e−ik′|r−r′| (k′ −k −iϵ)(k′ + k + iϵ) = eik|r−r′| 2 An introduction to low-energy scattering in quantum mechanics Lucas Madeira 118 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering theory integral equations The integral along CR (the real axis) is: lim R→∞ Z R −R dk′ k′(e−ik′|r−r′| −eik′|r−r′|) (k′ −k −iϵ)(k′ + k + iϵ) = 2πi × eik|r−r′| Going back to the Green’s function: G+(r, r′) = 1 4π eik|r−r′| |r −r′| And the integral equation: ⟨r|ψ⟩= ⟨r|i⟩−2m ℏ2 Z d3r′ 1 4π eik|r−r′| |r −r′| r′ |V| ψ An introduction to low-energy scattering in quantum mechanics Lucas Madeira 119 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering theory integral equations Next, we consider the potential to be local: r′ |V| r′′ = V(r′)δ(r′ −r′′) Thus: ⟨r|ψ⟩= ⟨r|i⟩−2m ℏ2 Z d3r′ 1 4π eik|r−r′| |r −r′|V(r′)⟨r′|ψ⟩ An introduction to low-energy scattering in quantum mechanics Lucas Madeira 120 / 121 Quantum scattering theory Numerical Procedure Examples Formal scattering theory Scattering theory integral equations Additional restriction: finite-ranged potential The scattering is observed far away from the scattering center Large distances (|r| ≫|r′|): eik|r−r′| ≈eikre−ik′·r′ Our initial state is |i⟩= |k⟩(and ⟨r|k⟩= eik·r/L3/2) Finally: ψ(r, θ) large r − − − → 1 L3/2 eik·r + eikr r f(k′, k) where f(k′, k) = −mL3 2πℏ2 Z d3r′⟨k′|r′⟩V(r′)⟨r′|ψ⟩ is the scattering amplitude An introduction to low-energy scattering in quantum mechanics Lucas Madeira 121 / 121 |
10572 | https://www.reddit.com/r/learnmath/comments/s807pd/why_are_there_two_producttosum_formulas_one_for/ | Why are there two product-to-sum formulas, one for sin(a)cos(b) and other for cos(a)sin(b)? : r/learnmath
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r/learnmath•4 yr. ago
amsfdk
Why are there two product-to-sum formulas, one for sin(a)cos(b) and other for cos(a)sin(b)?
If multiplication is commutative, why:
sin(a)cos(b) = 1/2[sin(a + b) + sin(a - b)]
and
cos(a)sin(b) = 1/2[sin(a + b) - sin(a - b)]
Do they always return the same answer, regardless of which one you choose? Thank you!
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10573 | https://www.uah.edu/images/people/faculty/howellkb/DEText-Ch13.pdf | 13 Reduction of Order We shall take a brief break from developing the general theory for linear differential equations to discuss one method (the “reduction of order method”) for finding the general solution to any linear differential equation. In some ways, this method may remind you of the material in chapter 11. Indeed, part of the method involves solving a higher-order equation via first-order methods as discussed in chapter 11. The general theory developed in chapter 12 will not, however, be used to any great extent. Instead, the material developed here will help us finish that general theory (at least partially confirming the suspicions raised at the end of the chapter), and will help lead us to the complete result on constructing general solutions from particular solutions.
But why worry about completing that general theory if any linear differential equation can be completely solved by this “reduction of order method”? Because this method requires that one solution to the differential equation already be known. This limits the method’s applicability.
Also, serious practical difficulties arise when the differential equation to be solved is of order three or more. Still, there are situations where the method is of practical value, and it will help us confirm suspicions we already have about general solutions.
Oh yes, there is another reason to develop this method: A rather powerful method for solving nonhomogeneous equations, the “variation of parameters” method described in chapter 23, is simply a clever refinement of the reduction of order method.
13.1 The General Idea The “reduction of order method” is a method for converting any linear differential equation to another linear differential equation of lower order, and then constructing the general solution to the original differential equation using the general solution to the lower-order equation. In general, to use this method with an N th-order linear differential equation a0y(N) + a1y(N−1) + · · · + aN−2y′′ + aN−1y′ + aN y = g , we need one known nontrivial solution y1 = y1(x) to the corresponding homogeneous differ-ential equation. We then try a substitution of the form y = y1 u where u = u(x) is a yet unknown function (and y1 = y1(x) is the aforementioned known solution). Plugging this substitution into the differential equation then leads to a linear differential 283 284 Reduction of Order equation for u . As we will see, because y1 satisfies the homogeneous equation, the differential equation for u ends up being of the form A0u(N) + A1u(N−1) + · · · + AN−2u′′ + AN−1u′ = g — remarkably, there is no “ ANu ” term. This means we can use the substitution v = u′ , as discussed in chapter 11, to rewrite the differential equation for u as a (N −1)th-order differential equation for v , A0v(N−1) + A1v(N−2) + · · · + AN−2v′ + AN−1v = g .
So we have reduced the order of the equation to be solved. If a general solution v = v(x) for this equation can be found, then the most general formula for u can be obtained from v by integration (since u′ = v ). Finally then, by going back to the original substitution formula y = y1u , we can obtain a general solution to the original differential equation.
This method is especially useful for solving second-order homogeneous linear differential equations since (as we will see) it reduces the problem to one of solving relatively simple first-order differential equations. Accordingly, we will first concentrate on its use in finding general solutions to second-order, homogeneous linear differential equations.
Then we will briefly discuss using reduction of order with linear homogeneous equations of higher order, and with nonhomogeneous linear equations.
13.2 Reduction of Order for Homogeneous Linear Second-Order Equations The Method Here we lay out the details of the “reduction of order method” for second-order homogeneous linear differential equations. To illustrate the method, we’ll use the differential equation x2y′′ −3xy′ + 4y = 0 .
Note that the first coefficient, x2 , vanishes when x = 0 . From comments made in chapter 12 (see page 261), we should suspect that x = 0 ought not be in any interval of interest for this equation. So we will be solving over the intervals (0, ∞) and (−∞, 0) .
Before starting the reduction of order method, we need one nontrivial solution y1 to our differential equation. Ways for finding that first solution will be discussed in later chapters. For now let us just observe that if y1(x) = x2 , then x2y1 ′′ −3x y1 ′ + 4y1 = x2 d2 dx2 x2 −3x d dx x2 + 4 x2 = x2[2 · 1] −3x[2x] + 4x2 = x2 [2 −(3 · 2) + 4] | {z } 0 !
= 0 .
Reduction of Order for Homogeneous Linear Second-Order Equations 285 Thus, one solution to the above differential equation is y1(x) = x2 .
As already stated, this method is for finding a general solution to some homogeneous linear second-order differential equation ay′′ + by′ + cy = 0 (where a , b , and c are ‘known functions’ with a(x) never being zero on the interval of interest). We will assume that we already have one nontrivial particular solution y1(x) to this generic differential equation.
For our example (as already noted), we will seek a general solution to x2y′′ −3xy′ + 4y = 0 .
(13.1) The one (nontrivial) solution we know is y1(x) = x2 .
Here, now, are the details in using the reduction of order method to solve the above: 1.
Let y = y1 u where u = u(x) is a function yet to be determined. To simplify the “plugging into the differential equaton” ,goaheadandcompute thecorrespondingformulas for the derivatives y′ and y′′ using the product rule: y′ = (y1u)′ = y1 ′u + y1u′ and y′′ = y′′ = y1 ′u + y1u′′ = y1 ′u ′ + y1u′′ = y1 ′′u + y1 ′u′ + y1 ′u′ + y1u′′ = y1 ′′u + 2y1 ′u′ + y1u′′ .
For our example, y = y1 u = x2u where u = u(x) is the function yet to be determined. The derivatives of y are y′ = x2u ′ = 2xu + x2u′ and y′′ = (y′)′ = 2xu + x2u′′ = (2xu)′ + x2u′′ = 2u + 2xu′ + 2xu′ + x2u′′ = 2u + 4xu′ + x2u′′ .
2.
Plug the formulas just computed for y , y′ and y′′ into the differential equation, group together the coefficients for u and each of its derivatives, and simplify as far as possible.
(We’ll do this with the example first and then look at the general case.) 286 Reduction of Order Plugging the formulas just computed above for y , y′ and y′′ into equation (13.1), we get 0 = x2y′′ −3xy′ + 4y = x2 2u + 4xu′ + x2u′′ −3x 2xu + x2u′ + 4 x2u = 2x2u + 4x3u′ + x4u′′ −6x2u −3x3u′ + 4x2u = x4u′′ + 4x3 −3x3 u′ + 2x2 −6x2 + 4x2 u = x4u′′ + x3u′ + 0 · u .
Notice that the u term drops out! So the resulting differential equation for u is simply x4u′′ + x3u′ = 0 , which we can further simplify by dividing out x4 , u′′ + 1 x u′ = 0 In general, plugging in the formulas for y and its derivatives into the given differential equation yields 0 = ay′′ + by′ + cy = a y1 ′′u + 2y1 ′u′ + y1u′′ + b y1 ′u + y1u′ + c y1u = ay1 ′′u + 2ay1 ′u′ + ay1u′′ + by1 ′u + by1u′ + cy1u = ay1u′′ + 2ay1 ′ + by1 u′ + ay1 ′′ + by1 ′ + cy1 u .
That is, the differential equation becomes Au′′ + Bu′ + Cu = 0 where A = ay1 , B = 2ay1 ′ + by1 and C = ay1 ′′ + by1 ′ + cy1 .
But remember, y1 is a solution to the homogeneous equation ay′′ + by′ + cy = 0 .
Consequently, C = ay1 ′′ + by1 ′ + cy1 = 0 , and the differential equation for u automatically reduces to Au′′ + Bu′ = 0 .
The u term always drops out!
3.
Now find the general solution to the second-order differential equation just obtained for u , Au′′ + Bu′ = 0 , via the substitution method discussed in chapter 11: Reduction of Order for Homogeneous Linear Second-Order Equations 287 (a) Let u′ = v (and, thus, u′′ = v′ = dv/dx ) to convert the second-order differential equation for u to the first-order differential equation for v , A dv dx + Bv = 0 .
(It is worth noting that this first-order differential equation will be both linear and separable.) (b) Find the general solution v(x) to this first-order equation. (Since it is both linear and separable, you can solve it using either the procedure developed for first-order linear equations or the approach developed for first-order separable equations.) (c) Using the formula just found for v , integrate the substitution formula u′ = v to obtain the formula for u , u(x) = Z v(x) dx .
Don’t forget all the arbitrary constants.
In our example, we obtained u′′ + 1 x u′ = 0 .
Letting v = u′ and, thus, v′ = u′′ this becomes dv dx + 1 x v = 0 .
Equivalently, dv dx = −1 x v .
This is a relatively simple separable first-order equation. It has one constant solution, v = 0 . To find the others, we divide through by v and proceed as usual with such equations: 1 v dv dx = −1 x H ⇒ Z 1 v dv dx dx = − Z 1 x dx H ⇒ ln |v| = −ln |x| + c0 H ⇒ v = ±e−ln|x| + c0 H ⇒ v = ±ec0x−1 .
Letting A = ±ec0 , this simplifies to v = A x , which also accounts for the constant solution (when A = 0 ).
Since u′ = v , it then follows that u(x) = Z v(x) dx = Z A x dx = A ln |x| + B .
288 Reduction of Order 4.
Finally, plug the formula just obtained for u(x) into the first substitution, y = y1u , used to convert the original differential equation for y to a differential equation for u .
The resulting formula for y(x) will be a general solution for that original differential equation. (Sometimes that formula can be simplified a little. Feel free to do so.) In our example, the solution we started with was y1(x) = x2 . Combined with the u(x) just found, we have y = y1u = x2[A ln |x| + B] .
That is, y(x) = Ax2 ln |x| + Bx2 is the general solution to equation (13.1).
An Observation About the Solution Let us observe that, in the example, the general solution obtained was y(x) = Ax2 ln |x| + Bx2 , which can be viewed as a linear combination of the two functions y1(x) = x2 and y2(x) = x2 ln |x| .
Since the A and B in the above formula for y(x) are arbitrary constants, and y2 is given by that formula for y with A = 1 and B = 0 , it must be that this y2 is another particular solution to our original homogeneous linear differential equation. What’s more, it is clearly not a constant multiple of y1 . This should strengthen an earlier suspicion that the general solution to a homogeneous linear second-order differential equation can be written as just such a linear combination.
We will later examine more closely the general form for general solution toany homogeneous linear differential equation. In the meantime, while practicing this method, do observe that the general solution you obtain for each second-order homogeneous linear differential equation can, invariably, be written as y(x) = Ay2(x) + By1(x) where y1 and y2 are two solutions that are not constant multiples of each other. Keep in mind that this form is the same as the form earlier anticipated, namely, y(x) = c1y1(x) + c2y2(x) .
We’ve just renamed the arbitrary constants from c1 and c2 to B and A , respectively.
Reduction of Order for Nonhomogeneous Linear Second-Order Equations 289 13.3 Reduction of Order for Nonhomogeneous Linear Second-Order Equations If you look back over our discussion in section 13.2, you will see that the reduction of order method applies almost as well in solving a nonhomogeneous equation ay′′ + by′ + cy = g , provided that “one solution y1 ” is a solution to the corresponding homogeneous equation ay′′ + by′ + cy = 0 .
Then, letting y = y1u in the nonhomogeneous equation and then replacing u′ with v leads to an equation of the form Av′ + Bv = g instead of Av′ + Bv = 0 .
So we don’t end up with a first-order equation for v which is both separable and linear; it is just linear. Still, we know how to solve such equations. Solving that first-order linear differential equation for v and continuing with the method already described finally yields the general solution to the desired nonhomogeneous differential equation.
We will do one example. Then I’ll tell you why the method is rarely used in practice.
!◮Example 13.1: Let us try to solve the second-order nonhomogeneous linear differential equation x2y′′ −3xy′ + 4y = √x (13.2) over the interval (0, ∞) .
As we saw in our main example in the section 13.2, the corresponding homogeneous equation x2y′′ −3xy′ + 4y = 0.
has y1(x) = x2 as one solution (in fact, from that example, we know the entire general solution to this homogeneous equation, but we only need this one particular solution for the method). Let y = y1 u = x2u where u = u(x) is the function yet to be determined. The derivatives of y are y′ = x2u ′ = 2xu + x2u′ and y′′ = (y′)′ = 2xu + x2u′′ = 2u + 2xu′ + 2xu′ + x2u′′ = 2u + 4xu′ + x2u′′ .
290 Reduction of Order Plugging these into equation (13.2) yields √x = x2y′′ −3xy′ + 4y = x2 2u + 4xu′ + x2u′′ −3x 2xu + x2u′ + 4 x2u = 2x2u + 4x3u′ + x4u′′ −6x2u −3x3u′ + 4x2u = x4u′′ + 4x3 −3x3 u′ + 2x2 −6x2 + 4x2 u = x4u′′ + x3u′ + 0 · u .
As before, the u term drops out. In this case, we are left with x4u′′ + x3u′ = √x .
That is, x4v′ + x3v = x 1/2 with v = u′ .
This is a relatively simple first-order linear equation. To help find the integrating factor, we now divide through by x4 , obtaining dv dx + 1 x v = x−7/2 .
Thus, the integrating factor is µ = e R 1 x dx = eln|x| = |x| .
Since we are just attempting to solve over the interval (0, ∞) , we really just have µ = x .
Multiplying the last differential equation for v and proceeding as usual when solving first-order linear differential equations: x dv dx + 1 x v = x h x−7/2 i H ⇒ x dv dx + v = x−5/2 H ⇒ d dx xv = x−5/2 H ⇒ Z d dx xv dx = Z x−5/2 dx H ⇒ xv = −2 3x−3/2 + c1 H ⇒ v = −2 3x−5/2 + c1 x Recalling that v = u′ , we can rewrite the last line as du dx = −2 3x−5/2 + c1 x .
Reduction of Order for Nonhomogeneous Linear Second-Order Equations 291 Thus, u = Z du dx dx = Z −2 3x−5/2 + c1 x dx = 2 3 2 x−3/2 + c1 ln(x) + c2 = 4 9x−3/2 + c1 ln(x) + c2 , and the general solution to our nonhomogeneous equation is y(x) = x2u(x) = x2 4 9x−3/2 + c1 ln(x) + c2 = 4 9x 1/2 + c1x2 ln(x) + c2x2 .
For no obvious reason at this point, let’s observe that we can write this solution as y(x) = c1x2 ln(x) + c2x2 + 4 9 √x .
(13.3) It should be observed that, in the above example, we only used one particular solution, y1(x) = x2 , to the homogeneous differential equation x2y′′ −3xy′ + 4y = 0 even though we had already found the general solution Ax2 ln |x| + Bx2 .
Later, in chapter 23, we will develop a refinement of the reduction of order method for solv-ing second-order nonhomogeneous linear differential equations which makes use of the entire general solution to the corresponding homogeneous equation. This refinement (the “variation of parameters” method) has two distinct advantages over the reduction of order method when solving nonhomogeneous differential equations: 1.
The computations required for the refined procedure tend to be simpler and more easily carried out.
2.
With a few straightforward modifications, the refined procedure readily extends to being a useful method for dealing with nonhomogeneous linear differential equations of any order. For the reasons discussed in the next section, the same cannot be said about the basic reduction of order method.
That is why, in practice, the basic reduction of order method is rarely used with nonhomogeneous equations.
292 Reduction of Order 13.4 Reduction of Order in General In theory, reduction of order can be applied to any linear equation of any order, homogeneous or not. Whether it’s application is useful is another issue.
!◮Example 13.2: Consider the third-order homogeneous linear differential equation y′′′ −8y = 0 .
(13.4) If you rewrite this equation as y′′′ = 8y , and think about what happens when you differentiate exponentials, you will realize that y1(x) = e2x is ‘obviously’ a solution to our differential equation (verify it yourself). Letting y = y1 u = e2xu and repeatedly using the product rule, we get y′ = e2xu ′ = 2e2xu + e2xu′ , y′′ = e2xu ′′ = 2e2xu + e2xu′′ = 4e2xu + 2e2xu′ + 2e2xu′ + e2xu′′ = 4e2xu + 4e2xu′ + e2xu′′ , and y′′′ = e2xu ′′′ = 4e2xu + 4e2xu′ + e2xu′′′ = 8e2xu + 4e2xu′ + 8e2xu′ + 4e2xu′′ + 2e2xu′′ + e2xu′′′ = 8e2xu + 12e2xu′ + 6e2xu′′ + e2xu′′′ .
So, using y = e2xu , y′′′ −8y = 0 H ⇒ 8e2xu + 12e2xu′ + 6e2xu′′ + e2xu′′′ −8 e2xu = 0 H ⇒ e2xu′′′ + 6e2xu′′ + 12e2xu′ + 8e2x −8e2x u = 0 .
Again, the u term cancel out, leaving us with e2xu′′′ + 6e2xu′′ + 12e2xu′ = 0 .
Letting v = u′ and dividing out the exponential, this becomes the second-order differential equation v′′ + 6v′ + 12v = 0 .
(13.5) Additional Exercises 293 Thus we have changed the problem of solving the third-order differential equation to one of solving a second-order differential equation. If we can now correctly guess a particular solution v1 to that second-order differential equation, we could again use reduction of order to get the general solution v(x) to that second-order equation, and then use that and the fact that y = e2xu with v = u′ to obtain the general solution to our original differential equation.
Unfortunately, even though the order is less, “guessing” a solution to equation (13.5) is a good deal more difficult than was guessing a particular solution to the original differential equation, equation (13.4).
As the example illustrates, even if we can, somehow, obtain one particular solution to a given N th-order linear homogeneous linear differential equation, and then use it to reduce the problem to solving an (N −1)th-order differential equation, that lower order differential equation may be just as hard to solve as the original differential equation (unless N = 2 ). In fact, we will learn how to solve differential equations such as equation (13.5), but those methods can also be used to find the general solution to the original differential equation, equation (13.4), as well.
Still it does no harm to know that the problem of solving an N th-order linear homogeneous linear differential equation can reduced to that of solving an (N −1)th-order differential equation, especially since we may refer to this fact in the next chapter. For the record, here is a theorem to that effect: Theorem 13.1 (reduction of order in homogeneous equations) Let y be any solution to some N th-order homogeneous differential equation a0y(N) + a1y(N−1) + · · · + aN−2y′′ + aN−1y′ + aN y = g (13.6) where g and the ak’s are known functions on some interval of interest I , and let y1 be a nontrivial particular solution to the corresponding homogeneous equation a0y(N) + a1y(N−1) + · · · + aN−2y′′ + aN−1y′ + aN y = 0 .
Set u = y y1 (so that y = y1 u ) .
Then v = u′ satisfies an (N −1)th-order differential equation A0v(N−1) + A1v(N−2) + · · · + AN−2v′ + AN−1v = g .
where the Ak’s are functions on the interval I that can be determined from the ak’s along with y1 and its derivatives.
The proof is relatively straightforward: You see what happens when you repeatedly use the product rule with y = y1 u , and plug the results into the equation (13.6). I’ll leave the details to you (see exercise 13.3).
294 Reduction of Order Additional Exercises 13.1. For each of the following, first verify that the given y1 is a solution to the given differential equation, and then find the general solution to the differential equation using the given y1 with the method of reduction of order.
a. y′′ −5y′ + 6y = 0 , y1(x) = e2x b. y′′ −10y′ + 25y = 0 , y1(x) = e5x c. x2y′′ −6xy′ + 12y = 0 , y1(x) = x3 d. 4x2y′′ + y = 0 on x > 0 , y1(x) = √x e. y′′ − 4 + 2 x y′ + 4 + 4 x y = 0 , y1(x) = e2x f. y′′ −1 x y′ −4x2y = 0 , y1(x) = e−x2 g. y′′ + y = 0 , y1(x) = sin(x) h. xy′′ + (2 + 2x)y′ + 2y = 0 , y1(x) = x−1 i. sin2(x)y′′ −2 cos(x) sin(x)y′ + 1 + cos2(x) y = 0 , y1(x) = sin(x) j. x2y′′ −2xy′ + x2 + 2 y = 0 , y1(x) = x sin(x) k. x2y′′ + xy′ + y = 0 , y1(x) = sin(ln |x|) l. x2y′′ + xy′ + x2 −1 4 y = 0 , y1(x) = x−1/2 cos(x) 13.2. Several nonhomogeneous differential equations are given below. For each, first verify that the given y1 is a solution to the corresponding homogeneous differential equation, and then find the general solution to the given nonhomogeneous differential equation using reduction of order with the given y1 .
a. y′′ −4y′ + 3y = 9e2x , y1(x) = e3x b. y′′ −6y′ + 8y = e4x , y1(x) = e2x c. x2y′′ + xy′ −y = √x , y1(x) = x d. x2y′′ −20y = 27x5 , y1(x) = x5 e. xy′′ + (2 + 2x)y′ + 2y = 8e2x , y1 = x−1 f. (x + 1)y′′ + xy′ −y = (x + 1)2 , y1 = e−x 13.3. Prove the claims in theorem 13.1 assuming: a. N = 3 b. N = 4 c. N is any positive integer 13.4. Each of the following is one of the relatively few third- and fourth-order differential equations that can be easily solved via reduction of order. For each, first verify that the Additional Exercises 295 given y1 is a solution to the given differential equation or to the corresponding homo-geneous equation (as appropriate), and then find the general solution to the differential equation using the given y1 with the method of reduction of order.
a. y′′′ −9y′′ + 27y′ −27y = 0 , y1 = e3x b. y′′′ −9y′′ + 27y′ −27y = e3x sin(x) , y1 = e3x c. y(4) −8y′′′ + 24y′′ −32y′ + 16y = 0 , y1 = e2x d. x3y′′′ −4y′′ + 10y′ −12y = 0 , y1 = x2 |
10574 | https://www.youtube.com/watch?v=oBV39vewyWU | Funciones Trigonométricas Complejas, Definición, Ejemplos y Propiedades (Variable Compleja)
MateFacil
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Posted: 10 Dec 2022
En este video veremos cómo se definen las funciones trigonométricas en variable compleja, seno, coseno, tangente, cotangente, secante y cosecante. Veremos también un ejemplo que consiste en calcular el seno de un número complejo. Y finalmente veremos una lista de las propiedades mas importantes.
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- Variable Compleja y aplicaciones, Ruel V. Churchill
- Análisis básico de Variable Compleja, Marsden y Hoffman
- Análisis complejo, Dennis G. Zill
- Variable Compleja, Murray R. Spiegel (Serie Schaum)
- Variable Compleja, Polya y Latta
- Funciones de Variable Compleja, Cesar A. Trejo
- Análisis Real y Complejo, W. Rudin
- Problemas sobre la Teoría de Funciones de Variable Compleja, Volkovyski (MIR)
- Variable Compleja con aplicaciones, A. David Wunsch
- Matemáticas Avanzadas para ingeniería, Peter V. O'Neil
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Bienvenida. Hola y bienvenidos a otro vídeo de Mate fácil en este vídeo vamos a ver las funciones trigonométricas de variable compleja antes de empezar Quiero invitarlos a unirse a mi grupo de Telegram el enlace Se los voy a dejar en el primer comentario de este vídeo y también a que me sigan a través de mis redes sociales Bueno lo que nosotros ya Explicación general de funciones trigonométricas complejas. conocemos son las seis funciones trigonométricas de variable real aquí la x es un número real ya conocemos bastantes propiedades de estas seis funciones sabemos por ejemplo que son periódicas sabemos por ejemplo Cómo se relacionan entre ellas y varias identidades trigonométricas y como ustedes saben las funciones trigonométricas son sumamente importantes en variable real Bueno pues también en variable compleja son muy importantes las funciones trigonométricas Así que necesitamos definirlas Entonces si cambiamos esta x Que es variable real por una variable compleja ahí tendríamos nuestras funciones trigonométricas complejas queremos ver cómo podemos nosotros calcular por ejemplo una expresión como esta de aquí seno de dos más tres por y e incluso vamos a ver que una vez que definamos las funciones de variable compleja estas funciones van a cumplir otras propiedades que no se cumplían con las funciones de variables real por ejemplo vamos a ser capaces de resolver una ecuación como esta de aquí que el coseno de un número sea igual a 3 Esto no se puede resolver dentro de variables real porque el coseno siempre está entre menos uno y uno está acotado o sea el coseno nunca puede darnos como resultado algo mayor que uno pero en este caso vamos a ver que dentro de variable compleja vamos a encontrar números tal que el coseno de ese número sea igual al 3 Sí entonces todo esto lo vamos a ir viendo cómo podemos entonces definir las Ejemplos de funciones trigonométricas complejas. funciones trigonométricas complejas bueno para esto lo que vamos a hacer es ayudarnos de esta identidad de aquí la identidad de oiler que relaciona la exponencial con las funciones seno y coseno aquí en esta identidad recuerden ustedes que ya es un número real Sí entonces aquí estamos relacionando la exponencial de y por y con las funciones de variables real coseno de seno de y bueno si cambiamos y por menos Y tenemos esta otra fórmula Simplemente nos queda aquí cosen menos y que es lo mismo que el coseno de ye porque es una función par y acá nos quedaría seno de menos y que es lo mismo que menos el seno de ye porque el seno es una función impar Entonces si ahora nosotros sumamos estas exponenciales al sumarlas tendríamos que sumar esta expresión más esta expresión y vemos que los senos se cancelan Entonces nos va a quedar que esta suma de exponenciales es dos veces el coseno ahora el 2 lo pasamos dividiendo Y entonces lo que acabamos de obtener es que el coseno de y es igual a esta suma de exponenciales sobre 2 Bueno aquí la ye es una variable real pero en realidad ya nosotros aquí podríamos cambiar esta variable real por una variable compleja y podemos seguir calculando esto Exactamente igual porque ya sabemos calcular las exponenciales de variable compleja eso es lo que nos va a ayudar a definir el coseno de Z bueno y para el caso del seno en lugar de sumar estas exponenciales vamos a restarlas Así que nos va a quedar esta expresión menos esa de acá Así que es menos el coseno y luego menos por menos nos da este más y vean que este coseno ahora se cancela con este de acá y aquí nos queda dos veces el y por seno de y el 2 y pasa dividiendo y entonces tenemos ahora que el seno de y es igual a esta expresión entonces con estas dos expresiones que acabamos de obtener aquí son expresiones que bueno son válidas cuando ya es una variable real pero nos ayudan a definir el seno y el coseno para una variable compleja Entonces vamos a definir para los números complejos estas mismas expresiones vamos a decir que el coseno de Z va a ser igual a e elevado a y por Z más elevado a menos y por Z sobre 2 y lo mismo para el caso del seno sí aquí Z es un número complejo o sea es de la forma x + y por y está bien definido todo porque ya sabemos nosotros calcular las exponenciales de variable compleja Entonces como un pequeño ejemplo vamos a calcular el seno de dos más tres y con esta definición simplemente sustituimos el valor de Z que es 2 + 3 y aquí y entonces nos queda esta expresión y vamos haciendo las operaciones Primero multiplicamos aquí por Y entonces queda 2 y y luego aquí y por 10 y cuadrada entonces queda 2 y menos 3 y por acá menos dos y más 3 bueno y Aquí vamos a recordar cómo se calcula la exponencial de variable compleja se calcula de esta manera es la exponencial de la parte real que en este caso es menos 3 y luego multiplicado por el coseno de la parte imaginaria o sea de lo que está acompañado por la y que en este caso es un 2 entonces es coseno de dos más y seno de 2 y lo mismo por aquí aquí va a quedar e elevado a 3 coseno de menos dos más y por el seno de menos 2 Bueno y seguimos simplificando aquí multiplicamos por la exponencial por acá recordamos que el coseno es función par entonces coseno de -2 es lo mismo que coseno de 2 entonces queda menos sea la 3 coseno de 2 y por aquí va a salir este signo menos porque la función es impar entonces va a quedar aquí un menos que por este menos nos da más Y entonces queda más y por esta exponencial por el seno de 2 y bueno podríamos Ya por aquí este ir dejando el resultado o podemos Separar en la parte real y la parte imaginaria para hacer eso vamos a agrupar primero factorizamos el coseno de 2 y entonces nos queda aquí a la menos 3 menos sea la 3 y por acá factorizamos el seno de 2 y también la y la factorizamos entonces queda a la menos 3 más y a la 3 ahora vamos a separar la fracción en una suma de fracciones por un lado nos va a quedar uno sobre dos y por esta expresión y en la otra nos va a quedar y sobre dos y por esta expresión En esta segunda vean que está ahí se cancela con esta entonces directamente es un medio y esta resta de exponenciales y esta suma de exponenciales nos recuerda algo que también aprendemos en variable real que son las funciones hiperbólicas el seno hiperbólico se define como la exponencial de T - sea la menos t sobre 2 y el coseno hiperbólico de esta otra manera tienen cierto parecido con las funciones seno y coseno que acabamos de calcular solo que en este caso aquí no está involucrada no está involucrado el número y si todo esto es variable real entonces con ayuda de estas funciones hiperbólicas nosotros podemos simplificar esta expresión para no tener que escribir esta resta de exponenciales y esta suma de exponenciales observen que este 2 que está dividiendo podemos meterlo aquí aquí lo mismo y entonces esto de aquí ya es de hecho el coseno hiperbólico esto de acá es casi el seno hiperbólico solamente que aquí la resta está en el sentido contrario Pero si multiplicamos por este menos vamos a invertir el orden de la resta Entonces vamos a multiplicar por este menos ya la resta va a quedar en el otro orden aquí como es suma podemos cambiar el orden y es lo mismo y entonces esto de acá es el seno hiperbólico Y esto es el coseno hiperbólico en este caso el seno hiperbólico de 3 y acá es el coseno hiperbólico de 3 y entonces ahí tenemos ya el seno en su parte imaginaria y su parte real o escribiendo primero la parte real y la parte imaginaria queda de esta manera bueno así podemos dejar ya el resultado o podríamos una calculadora hacer esto Recuerden que en radiales Y entonces nos queda esto con números decimales bueno Esto de aquí lo podemos hacer para cualquier número complejo pero vamos a encontrar una fórmula que nos va a permitir calcularlo directamente sin tener que hacer todo este procedimiento Eso lo veremos En un momento más Entonces cómo definimos las demás funciones trigonométricas ya tenemos el seno y el coseno Bueno pues las otras cuatro se van a definir igual que en el caso de variable real la tangente se define como el seno entre coseno la cotangente como coseno sobre seno secante como uno entre coseno y cosecante como uno entre seno Y entonces ahí tenemos ya todas definidas como pueden ver ustedes las más importantes son el seno y el coseno porque a partir de estas dos salen las otras 4 aunque también es bastante utilizada la tangente Entonces vamos a ver una lista Propiedades de las funciones trigonométricas complejas. de propiedades para estas funciones trigonométricas la primera es el seno de una suma o de una resta que es esta expresión de aquí vean que es realmente la misma fórmula que ya conocemos de variable real se sigue cumpliendo cuando los números de aquí son números complejos en este caso si es una suma tomamos el signo de arriba Entonces es una suma Aquí también y si tenemos aquí una resta entonces tomamos el signo de abajo que también es una resta en el caso del coseno si aquí tomamos una suma acá nos queda una resta ya casi es una resta por acá tomamos la suma y vean que es la misma identidad que conocemos también de variable real y lo mismo para el caso de la tangente es la misma identidad también son importantes las identidades del ángulo doble son bastante utilizadas y son las mismas que aprendemos en variable real y luego estas identidades de acá que el seno es una función impar que el coseno es una función par y la tangente es una función impar esta siguen siendo Igualmente válidas cuando Z es un número complejo todas estas identidades hay que demostrarlas no podemos simplemente decir que como se cumplían en la variable real se van a seguir cumpliendo en la variable compleja hay que demostrarlas a partir de la de la definición todo eso lo iremos demostrando en los siguientes vídeos sigamos viendo otras propiedades tenemos por aquí las identidades pitagóricas bastante importantes que ya conocemos de variable real pero que se cumplen también para la variable compleja y luego estas otras identidades de aquí que son una combinación entre las identidades pitagóricas y las del ángulo doble y que nos permiten calcular la el seno y el coseno del ángulo mitad luego tenemos estas otras identidades de aquí que relacionan el seno coseno y tangente de un número imaginario puro en este caso y por y se relaciona con el seno hiperbólico coseno hiperbólico y tangente hiperbólica respectivamente y a partir de estas tres identidades podemos encontrar estas otras tres de acá usando el seno y el coseno de una suma una resta etcétera y vean Cómo precisamente aquí esto se parece a lo que obtuvimos en el ejemplo para calcular el seno de 2 + 3 y podríamos ya directamente sustituir en x el 2 el 3 y tendríamos la expresión directa Pero esto hay que demostrarlo bueno y el conjugado conjugado de seno coseno y tangente es lo mismo que el seno del conjugado el coseno del conjugado y la tangente del conjugado lo mismo que ocurría con la función exponencial compleja y luego para los módulos tenemos que el módulo de Z del seno de Z al cuadrado es el seno cuadrado de X más seno hiperbólico al cuadrado de y algo similar para el coseno y vean que de aquí surge algo importante algo que no se cumple ya en la variable compleja y que sí se cumple en la variable real en variable real el seno es una función acotada tanto el seno como el coseno son funciones que están entre -1 y 1 pero a partir de aquí podemos ver que ni el seno ni el coseno son funciones acotadas porque tenemos aquí una suma de dos funciones la primera si es acotada es el seno de X que está entre menos uno y uno pero el seno hiperbólico no es una función acotada si el número y es muy grande el seno hiperbólico de y sigue siendo muy grande o sea esta función crece y crece conforme y crece también entonces el coseno de Z puede ser tan grande como uno quiera simplemente tomando un valor de y lo suficientemente grande o sea que el seno ya no es una función acotada y lo mismo ocurre con el coseno bueno Y luego tenemos estas desigualdades de acá también importantes que el módulo del seno se encuentra entre el módulo del seno hiperbólico y el coseno hiperbólico y lo mismo para el caso del coseno todas estas identidades vamos a demostrarlas en los próximos vídeos Así que los invito a que los vean antes de terminar agradezco infinitamente a todos los miembros del Canal que con su gran apoyo hacen posible que siga subiendo nuevos vídeos y más cursos de esta manera ustedes me ayudan a contrarrestar un poquito los efectos del algoritmo de YouTube que no siempre beneficia a los canales educativos de verdad muchas gracias por todo su apoyo y si aún no eres miembro del Canal te invito a darle click al botón unirse para que mires los diversos niveles de membresía y los beneficios que tiene cada nivel entre esos beneficios tienes acceso exclusivo a cientos de vídeos con ejercicios interesantes te invito a darle clic para ver más información |
10575 | https://www.sciencedirect.com/science/article/abs/pii/S1521691810000405 | Pancreatic enzyme replacement therapy in chronic pancreatitis - ScienceDirect
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Best Practice & Research Clinical Gastroenterology
Volume 24, Issue 3, June 2010, Pages 337-347
10
Pancreatic enzyme replacement therapy in chronic pancreatitis
Author links open overlay panel E.C.M.Sikkens MD, D.L.Cahen MD, PhD 1, E.J.Kuipers Md, PhD(Professor)2, M.J.Bruno MD, PhD(Professor)3
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Exocrine pancreatic insufficiency (EPI) is a serious condition which occurs in several diseases including chronic pancreatitis (CP), cystic fibrosis, pancreatic cancer, and as a result of pancreatic surgery. The lack or absence of pancreatic enzymes leads to an inadequate absorption of fat, proteins, and carbohydrates, causing steatorrhoea and creathorrhea which results in abdominal discomfort, weight loss, and nutritional deficiencies. To avoid malnutrition related morbidity and mortality, it is pivotal to commence pancreatic enzyme replacement therapy (PERT) as soon as EPI is diagnosed. Factors as early acidic inactivation of ingested enzymes, under dosage, and patient incompliance may prevent normalisation of nutrient absorption, in particular of fat digestion. This review focuses on the current status of how to diagnose and treat EPI.
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Exocrine pancreatic insufficiency and chronic pancreatitis
In EPI the pancreas is unable to deliver sufficient amounts of pancreatic enzymes to the small intestine to digest intraluminal nutrients. EPI may occur due to loss of functional parenchyma (atrophy), blockage of the pancreatic duct, or postprandial asynchrony. Besides CP, other conditions that can result in loss of parenchyma are severe acute pancreatitis (which can cause transient EPI), pancreatic resection, and cystic fibrosis (CF). Chronic obstruction of the main pancreatic duct can be
Pancreatic secretion
The pancreas plays a crucial role in the digestive system. The gland produces pancreatic juice that consists of a mixture of more than two dozen digestive enzymes in the pre-activated form, called zymogens. Zymogens are produced by acinar cells and mixed with a bicarbonate rich fluid that is produced by pancreatic ducts cells , . Trypsinogen is the most important zymogen because it becomes trypsin, the key enzyme that activates all other zymogens. Trypsin, chymotrypsin, amylase and
Pathophysiology
The pancreas has a large functional reserve and clinically evident EPI occurs only when 90% of the function is lost and the secretion of pancreatic enzymes is less than 10% of normal , . Because of a decrease in lipase, trypsin and amylase activity, maldigestion of fat, proteins and carbohydrates occurs. Malabsorption of fat precedes malabsorption of proteins and carbohydrates and is clinically more apparent , , . The decrease in pancreatic lipolytic activity cannot be
Symptoms and complications
In CP patients with EPI, maldigestion of dietary macronutrients (fat, proteins and carbohydrates) leads to malnutrition which is associated with various health problems. Maldigestion of fat results in steatorrhoea, which causes symptoms such as foul-smelling, voluminous, greyish, fatty stools, abdominal cramps, bloating and chronic abdominal pain . In addition, steatorrhoea may cause weight loss due to the loss of the highest dietary source of calories (fat contains 38 kJ/g, carbohydrates
Diagnosis
The pancreatic exocrine function can be tested in an invasive or a non-invasive manner. With invasive testing, which is considered to be the golden standard, pancreatic enzymes are measured in pancreatic juice that is collected from the duodenum after intubation (intraduodenal lipase output after intravenous administration of 1 U/kg CCK; normal value, >90 kU/h) . Pancreatic enzyme stimulation in these tests can be either direct, by means of the intravenous administration of hormones like
Treatment
The treatment of EPI in CP consists of the oral administration of a combination of pancreatic enzymes during meals . Every patient with EPI and maldigestion, independent of the degree of steatorrhoea and presence or absence of associated symptoms, should receive PERT , . The main focus in the management of EPI is to prevent weight loss, EPI related symptoms, vitamin deficiencies, and to improve the nutritional status , . The most important clinical parameter to monitor
Pancreatic enzyme replacement therapy (PERT)
The exogenous pancreatic enzymes currently used, are primarily extracted from porcine sources. These preparations, also called pancrelipase or pancreatin, contain a variable mixture of protease, lipase and amylase depending on the brand. Various pancreatin preparations are available consisting of capsules containing mini-microspheres, pellets or micro-tablets of less than 2 mm in size. They are designed to promote an adequate intragastric mixture of exogenous enzymes with chyme , , ,
Diet and dosage recommendations
Despite the absence of an easy applicable and objective method to establish the adequate dose of oral pancreatic enzymes in EPI, some general guidelines can be given to accomplish a patient tailored administration schedule . In the past, the restriction of fat was the only way to reduce steatorrhoea, but with the introduction of PERT, this is no longer advocated. Instead, the dose of PERT is tailored to the fat intake of the patient. In general, the recommended dosage of PES for a main meal
When treatment fails
In case of treatment failure, several causes should be considered. A common cause for treatment failure is under dosing of pancreatic enzymes. The practical and reasonable first step therefore is to increase the dose of pancreatic enzymes guided by fat resorption and clinical response, up to a maximum of 10,000 IU lipase/kg/day. This latter recommendation stems from treating children with cystic fibrosis. Evidently, in adult patients with exocrine insufficiency such high dosages are not reached.
Toxicity and side effects
Very few side effects have been observed when using PERT. High doses of enzymes can induce transient nausea, bloating, diarrhoea and hypersensitivity. Only one serious adverse event has been reported. In January 1994, Smyth et al described five children with CF in which a colonic obstruction developed due to fibrosing colonopathy (FC) after using very high doses of the enteric-coated micro-minisphere preparations , . The mechanism underlying this phenomenon remains unclear. Although
Bovine enzymes
In patients who refuse to consume porcine products for religious or other cultural reasons, bovine enzymes appear to be an attractive alternative . However, bovine preparations contain about 75% less lipase activity compared to porcine and human pancreatic extracts. Hence, a considerable greater amount of tablets needs to be taken to treat EPI in comparison with porcine preparations. Furthermore, there are some concerns about transmittable pathogens which can cause diseases as
Summary
Clinically evident EPI occurs when 90% or more of the pancreatic function is impaired and the secretion of pancreatic enzymes is less than 10% of normal. It is of great importance to actively look for EPI, recognise it as early as possible, and start PERT to avoid malnutrition related morbidity and mortality. The most readily available and easiest test for a clinician to detect EPI is the FET. Every patient with EPI and maldigestion, independent of the degree of steatorrhoea and presence or
Conflict of interest
Mrs E.C.M. Sikkens has received an unrestricted research grant from “Axcan Pharma Inc., Canada”.
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Encapsulation, protection, and delivery of bioactive proteins and peptides using nanoparticle and microparticle systems: A review
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For instance, oral delivery of lactase aids in the breakdown of lactose into galactose and glucose within the small intestine, which is important for individuals with lactose intolerance [8,9]. Similarly, oral delivery of lipase can help patients with pancreatitis, i.e., the inability to breakdown lipids within the small intestine . Bioactive proteins may also include various kinds of hormones, such as insulin and glucagon-like peptide-1 (GLP-1) which are used to treat diabetes or angiotensin converting enzyme (ACE) inhibitors which are used to treat hypertension . Show abstract There are many examples of bioactive proteins and peptides that would benefit from oral delivery through functional foods, supplements, or medical foods, including hormones, enzymes, antimicrobials, vaccines, and ACE inhibitors. However, many of these bioactive proteins are highly susceptible to denaturation, aggregation or hydrolysis within commercial products or inside the human gastrointestinal tract (GIT). Moreover, many bioactive proteins have poor absorption characteristics within the GIT. Colloidal systems, which contain nanoparticles or microparticles, can be designed to encapsulate, retain, protect, and deliver bioactive proteins. For instance, a bioactive protein may have to remain encapsulated and stable during storage and passage through the mouth and stomach, but then be released within the small intestine where it can be absorbed. This article reviews the application of food-grade colloidal systems for oral delivery of bioactive proteins, including microemulsions, emulsions, nanoemulsions, solid lipid nanoparticles, multiple emulsions, liposomes, and microgels. It also provides a critical assessment of the characteristics of colloidal particles that impact the effectiveness of protein delivery systems, such as particle composition, size, permeability, interfacial properties, and stability. This information should be useful for the rational design of medical foods, functional foods, and supplements for effective oral delivery of bioactive proteins.
### Microbial lipases and their industrial applications: A comprehensive review
2020, Microbial Cell Factories Show abstract Lipases are very versatile enzymes, and produced the attention of the several industrial processes. Lipase can be achieved from several sources, animal, vegetable, and microbiological. The uses of microbial lipase market is estimated to be USD 425.0 Million in 2018 and it is projected to reach USD 590.2 Million by 2023, growing at a CAGR of 6.8% from 2018. Microbial lipases (EC 3.1.1.3) catalyze the hydrolysis of long chain triglycerides. The microbial origins of lipase enzymes are logically dynamic and proficient also have an extensive range of industrial uses with the manufacturing of altered molecules. The unique lipase (triacylglycerol acyl hydrolase) enzymes catalyzed the hydrolysis, esterification and alcoholysis reactions. Immobilization has made the use of microbial lipases accomplish its best performance and hence suitable for several reactions and need to enhance aroma to the immobilization processes. Immobilized enzymes depend on the immobilization technique and the carrier type. The choice of the carrier concerns usually the biocompatibility, chemical and thermal stability, and insolubility under reaction conditions, capability of easy rejuvenation and reusability, as well as cost proficiency. Bacillus spp., Achromobacter spp., Alcaligenes spp., Arthrobacter spp., Pseudomonos spp., of bacteria and Penicillium spp., Fusarium spp., Aspergillus spp., of fungi are screened large scale for lipase production. Lipases as multipurpose biological catalyst has given a favorable vision in meeting the needs for several industries such as biodiesel, foods and drinks, leather, textile, detergents, pharmaceuticals and medicals. This review represents a discussion on microbial sources of lipases, immobilization methods increased productivity at market profitability and reduce logistical liability on the environment and user.
### Practical guide to exocrine pancreatic insufficiency - Breaking the myths
2017, BMC Medicine Show abstract Exocrine pancreatic insufficiency (EPI) is characterized by a deficiency of exocrine pancreatic enzymes, resulting in malabsorption. Numerous conditions account for the etiology of EPI, with the most common being diseases of the pancreatic parenchyma including chronic pancreatitis, cystic fibrosis, and a history of extensive necrotizing acute pancreatitis. Treatment for EPI includes dietary management, lifestyle changes (i.e., decrease in alcohol consumption and smoking cessation), and pancreatic enzyme replacement therapy. Many diagnostic tests are available to diagnose EPI, however, the criteria of choice remain unclear and the causes for a false-positive test are not yet understood. Despite multiple studies on the treatment of EPI using exogenous pancreatic enzymes, there remains confusion amongst medical practitioners with regard to the best approach to diagnose EPI, as well as dosing and administration of pancreatic enzymes. Appropriate use of diagnostics and treatment approaches using pancreatic enzymes in EPI is essential for patients. This opinion piece aims to address the existing myths, remove the current confusion, and function as a practical guide to the diagnosis and treatment of EPI.
### Diagnosis and management of pancreatic cancer
2014, American Family Physician Show abstract Pancreatic cancer remains the fourth leading cause of cancer-related deaths in the United States. Risk factors include family history, smoking, chronic pancreatitis, obesity, diabetes mellitus, heavy alcohol use, and possible dietary factors. Because more than two-thirds of adenocarcinomas occur in the head of the pancreas, abdominal pain, jaundice, pruritus, dark urine, and acholic stools may be presenting symptoms. In symptomatic patients, the serum tumor marker cancer antigen 19-9 can be used to confirm the diagnosis and to predict prognosis and recurrence after resection. Pancreas protocol computed tomography is considered standard for the diagnosis and staging of pancreatic cancer. Although surgical resection is the only potentially curative treatment for pancreatic ductal adenocarcinomas, less than 20% of surgical candidates survive five years. The decision on resectability requires multidisciplinary consultation. Pancreatic resections should be performed at institutions that complete at least 15 of the surgeries annually. Postoperatively, use of gemcitabine or fluorouracil/leucovorin as adjuvant chemotherapy improves overall survival by several months. However, more than 80% of patients present with disease that is not surgically resectable. For patients with locally advanced or metastatic disease, chemoradiotherapy with gemcitabine or irinotecan provides clinical benefit and modest survival improvement. Palliation should address pain control, biliary and gastric outlet obstruction, malnutrition, thromboembolic disease, and depression.
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2020, Molecules
### Protein hydrolysates in animal nutrition: Industrial production, bioactive peptides, and functional significance
2017, Journal of Animal Science and Biotechnology
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Copyright © 2010 Elsevier Ltd. All rights reserved.
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Treatment of Atopic Dermatitis
Address the 4 aspects of the disease simultaneously and refer to the treatment algorithm for stepwise therapeutic choices.
Treating children with atopic dermatitis involves addressing 4 aspects of the disease simultaneously. Treatment choices are considered in a stepwise manner and depend on the severity of the disease. Refer to the Treatment Toolfor a summative view of treatment options.
The treatment of atopic dermatitis includes the following simultaneous measures:
These treatment targets, when effectively addressed, not only improve disease symptoms and reduce flares but assist the child and family in achieving adequate sleep.
Parents should be encouraged to establish healthy sleep routines as part of the treatment plan. Adequate sleep is necessary for optimal growth and development, promotes mental health, and improves quality of life for the patient and family.
Read More about Healthy Sleep
Treatment Targets
Moisturizing the Skin
Moisturizers (emollients) are the cornerstone of atopic dermatitis treatment. They are the main treatment for mild atopic dermatitis and an important part of treating moderate to severe atopic dermatitis.
Apply an emollient as needed to control dry skin (xerosis). The most effective timing is immediately after bathing (while the skin is still moist). Application often is recommended at least once daily, but this varies with the individual and environmental factors (eg, during warm weather months in humid conditions an emollient may not be needed).
A variety of over-the-counter (eg, CeraVe, Cetaphil Restoraderm) and prescription (eg, Atopiclair, EpiCeram, Hylatopic) barrier repair agents exist that may help reduce the severity of atopic dermatitis and play an adjunctive therapeutic role.
Read More about Maintaining the Skin Barrier
Reducing Itch
The chief way to reduce itch is to address xerosis with the daily application of emollients. When the disease is active, however, administer a bedtime dose of a first-generation antihistamine (eg, hydroxyzine 0.5–1 mg/kg, diphenhydramine 1.25 mg/kg) to provide sedation, improve sleep, and reduce scratching.
Daytime doses may occasionally be needed but should be lower (to avoid sedation).
Wet-wrap therapy may also be useful during severe flares.
Treating Inflammation
To treat inflammation during a flare of atopic dermatitis, apply a topical corticosteroid twice daily to affected areas until improvement occurs (usually a few days to 2–3 weeks). Ointments are preferred over creams because they tend to be more effective and better tolerated (less burning sensation), although some patients prefer creams because they are less greasy. Choosing a topical corticosteroid depends on the age of the child and the area to be treated.
Once symptoms have improved, the corticosteroid is withdrawn and a moisturizer continued regularly. Super-high-potency topical corticosteroids should not be used for longer than 3 weeks. High- or mid-potency topical corticosteroids can be used for up to 12 weeks. Low-potency preparations have no maximum use recommendations. Although topical corticosteroid use should be avoided in the absence of symptoms, applying a corticosteroid once or twice weekly at locations prone to exacerbations has been shown to reduce relapses and increase the time to the next flare.
Generally, a topical corticosteroid is prescribed for 2 to 3 weeks. The family should be instructed to contact the physician if no improvement occurs in that time. Treatment failure can be due to
If treatment failure is not caused by those factors, a nonsteroidal topical calcineurin inhibitor or other nonsteroidal topical anti-inflammatory agent can be added to the daily treatment.
For more information on selecting anti-inflammatory agents, see the sections on nonsteroidal topical anti-inflammatories and the information on selecting and prescribing topical corticosteroids.
Rarely, systemic corticosteroids, immunosuppressants, biologicals, small molecule therapies, or phototherapy is necessary for the management of atopic dermatitis. Children with severe atopic dermatitis requiring systemic medication or children with moderate-to-severe disease who are not improving with standard therapy should be referred to a pediatric dermatologist.
Read More about Corticosteroids
Preventing and Managing Skin Infections
To prevent Staphylococcus aureus skin infections, controlling colonization may be useful for those with severe or recalcitrant disease. Consider one or more of the following options:
If there is evidence of secondary bacterial infection (eg, crusting, pustules, oozing [Figure 4.11]), consider administering an oral antistaphylococcal antibiotic (eg, cephalexin or other agent based on local antibiotic resistance patterns) for 7 to 10 days. If no improvement is noted within 48 hours, consider a skin swab for bacterial culture to assess for resistant organisms (eg, methicillin-resistant S aureus) and treat appropriately. At this time, most S aureus isolates from patients with atopic dermatitis in the United States remain methicillin-sensitive.
If infection is limited to very focal areas, a prescription topical antimicrobial agent (eg, mupirocin, retapamulin, ozenoxacin) may be useful.
Daily Measures
Disease Flares and Maintenance Between Flares
When the disease flares:
Wet-wrap Therapy
Wet-wrap therapy may be useful during severe flares of atopic dermatitis. A topical corticosteroid is applied to affected areas and covered with a moistened cotton suit (eg, pajamas), wet gauze strips, or a specially designed, commercially available garment, which is then covered with a dry outer layer (eg, dry pajamas). The wrap may be worn for several hours or up to 24 hours; on removal, emollient is applied. Once the disease flare improves, wet-wrap therapy is discontinued. For more details, including step-by-step instructions on wet wrap therapy, visit the National Eczema Association.
Maintenance Measures
A moisturizer should be used regularly. However, applying a corticosteroid or calcineurin inhibitor once or twice weekly at locations prone to exacerbations has been shown to reduce relapses and increase the time to the next flare.
Phototherapy
Phototherapy may be considered for children who have moderate-to-severe atopic dermatitis and have failed multimodal topical therapy. Phototherapy must be prescribed and conducted by clinicians who are familiar with the various types of phototherapy and equipment. Phototherapy can be used on its own or in combination with emollients and topical steroids.
Dust Mite Avoidance
Avoiding dust mites through frequent vacuuming and encasing pillows and mattresses in allergen-proof products may result in a modest reduction in the severity of atopic dermatitis. Such recommendations are reserved for patients with severe or recalcitrant disease.
Associated Issues
Children with atopic dermatitis may experience other skin conditions, some of which are infectious and others that are cosmetic in nature.
Secondary Infections
Children with atopic dermatitis are prone to certain types of skin infections. The most serious infections (eg, eczema herpeticum) may require hospitalization and treatment with systemic medications.
Any child with atopic dermatitis should not share eating utensils, toothbrushes, drinking glasses, or other personal items with someone who has a cold sore (herpes simplex virus [HSV] 1 infection).
Treatment Regimen by Infection Type
Bacterial Infection
A sudden exacerbation of atopic dermatitis may be caused by a bacterial infection.
The diagnosis is clinical.
If there is evidence of secondary bacterial infection (eg, crusting, pustules, oozing [Figure 4.11]), consider administering an oral antistaphylococcal antibiotic (eg, cephalexin or other agent based on local antibiotic resistance patterns) for 7 to 10 days. If no improvement is noted within 48 hours, consider a skin swab for bacterial culture to assess for resistant organisms (eg, methicillin-resistant Staphylococcus aureus [MRSA]) and treat appropriately. At this time, most S aureus isolates from patients with atopic dermatitis in the United States remain methicillin-sensitive.
If infection is limited to very focal areas, a prescription topical antimicrobial agent (eg, mupirocin, retapamulin, ozenoxacin) may be useful.
The choice of topical, oral, or intravenous antibiotic therapy (against S aureus and streptococci) depends on the extent and severity of the infection.
Eczema Herpeticum
Eczema herpeticum is a disseminated HSV infection that occurs in individuals who have atopic dermatitis or other chronic skin diseases. It occurs primarily in children younger than 3 years. It likely is the result of a disrupted skin barrier that allows viral invasion.
Noninfectious Conditions
Keratosis Pilaris and Ichthyosis Vulgaris
Pityriasis Alba
Medications
Ensure that eczema is not undertreated. Families are often reluctant to use corticosteroids (sometimes called “corticophobia”). Some doctors can be reluctant to prescribe them.
Topical corticosteroids are the first-choice therapy for eczema flares and their benefits greatly exceed uncommon adverse effects.
When used as directed the safety profile of new topical corticosteroids is good. Insufficient or inadequate applications of topical corticosteroids limit the ability to control dermatitis, leading to multiple issues including poor quality of life and increases in flares. Especially in severe forms of atopic dermatitis, it is important to implement a multidisciplinary approach, in which education has an important role.
Nonsteroid Topical Anti-inflammatories
Discussing With Parents and Families
Clinical practice guidelines recommend educational programs (“eczema schools”) as an adjunct to therapy for atopic dermatitis.
The treatment regimen for atopic dermatitis is complex and multifactorial; therefore, support for the patient and caregivers is crucial in increasing treatment compliance. Research into certain formal programs has shown improvements in ability to cope, itching behavior, and disease severity. Options include nurse-led sessions and video education.
Patients and families may wish to explore alternative treatments. It may be helpful to solicit information about alternate treatment plans and to share the results of scientific studies of alternative treatments.
When to Refer
Between 80% and 90% of infants experience a spontaneous resolution or improvement in symptoms by adolescence. Until this time the disease course is chronic and relapsing. Consider a referral to a pediatric dermatologist when patients do not respond to standard treatments.
Other reasons for referral
The development of this information was made possible through support from Sanofi and Regeneron.
Last Updated
06/11/2021
Source
American Academy of Pediatrics
© Copyright 2025 American Academy of Pediatrics. All rights reserved. |
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10578 | https://www.mdpi.com/1718-7729/32/4/215 | Current Limitations of Sentinel Node Biopsy in Vulvar Cancer
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Background:
Open Access Review
Current Limitations of Sentinel Node Biopsy in Vulvar Cancer
by
Myriam Gracia
Myriam Gracia
SciProfilesScilitPreprints.orgGoogle Scholar
,
Maria Alonso-Espías
Maria Alonso-Espías
SciProfilesScilitPreprints.orgGoogle Scholar
and
Ignacio Zapardiel
Ignacio Zapardiel
SciProfilesScilitPreprints.orgGoogle Scholar
Gynecologic Oncology Unit, La Paz University Hospital, 28015 Madrid, Spain
Author to whom correspondence should be addressed.
Curr. Oncol.2025, 32(4), 215;
Submission received: 10 March 2025 / Revised: 3 April 2025 / Accepted: 7 April 2025 / Published: 8 April 2025
(This article belongs to the Section Gynecologic Oncology)
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Abstract
Background: Vulvar cancer is a rare gynecologic malignancy with increasing incidence. Lymph node status is the most critical prognostic factor, traditionally assessed through inguinofemoral lymphadenectomy, a procedure associated with significant morbidity. Sentinel lymph node biopsy (SLNB), in selected cases, has emerged as a less invasive alternative with favorable oncologic outcomes. Objective: This review summarizes current evidence on the indications, technique, safety, and oncologic outcomes of SLNB in vulvar cancer, with a focus on controversial scenarios such as recurrent and larger tumors. Methods: A narrative review of PubMed-indexed studies published in English over the last 35 years was conducted. Eligible studies included original research, systematic reviews, meta-analyses, randomized controlled trials, and case-control studies. Results: SLNB is recommended for unifocal vulvar tumors < 4 cm with stromal invasion > 1 mm and clinically negative nodes. Landmark trials, including GROINSS-V-I and GOG-173, confirmed its accuracy and lower morbidity compared to lymphadenectomy. SLNB utilization has increased since its inclusion in guidelines, with a concurrent decline in lymphadenectomy rates. Combined detection techniques are mandatory, while indocyanine green (ICG) is an emerging option. Future studies should focus on refining patient selection criteria, improving detection techniques, and clarifying the implications of low-volume nodal disease to further optimize outcomes for patients with vulvar cancer. Conclusion: SLNB is a validated, minimally invasive staging approach in early-stage vulvar cancer. Further research is needed to refine its role in high-risk cases and optimize detection methods.
Keywords:
vulvar cancer; sentinel lymph node biopsy; recurrent vulvar cancer; inguinofemoral lymphadenectomy
1. Introduction
Vulvar cancer is a relatively uncommon neoplasm, ranking 21st in incidence among women worldwide and representing only 4% of all gynecologic malignancies. According to the Global Cancer Statistics, 47,342 new cases and 18,579 deaths related to vulvar cancer were reported in 2022 . Despite its rarity, its incidence appears to be increasing in recent decades, likely due to higher life expectancy and human papillomavirus infection .
In addition to tumor excision, inguinofemoral lymph node assessment is crucial for tumors with stromal invasion greater than 1 mm, as lymph node metastases are found in approximately 23–25% of patients with early-stage disease. The first sites of lymphatic spread in vulvar cancer are the inguinofemoral lymph nodes, followed by the pelvic nodes. The risk of unilateral or bilateral nodal involvement is influenced by factors such as tumor size and proximity to the midline or clitoris . Furthermore, lymph node involvement is the most significant independent prognostic factor, with survival rates decreasing from 90% to less than 60% in cases of groin metastases . Key factors strongly associated with nodal involvement include lymphovascular space invasion, tumor stage, tumor grade, and depth of infiltration [5,6].
Traditionally, the standard treatment for vulvar cancer consists of a radical vulvectomy combined with bilateral inguinofemoral lymphadenectomy. This procedure involves the removal of superficial and deep inguinal lymph nodes located within the femoral triangle, which is bordered by the inguinal ligament, sartorius, and adductor longus muscles. However, this approach is linked to a significant increase in both short- and long-term morbidity, including wound dehiscence, infections, lower limb lymphedema, and lymphocele formation, all of which negatively impact patients’ quality of life [7,8].
Sentinel lymph node biopsy (SLNB) has emerged in recent decades as a less invasive alternative for nodal assessment in various malignancies, including melanoma, breast, and endometrial carcinoma, but also in vulvar carcinoma, reducing morbidity associated with lymphadenectomy [9,10,11]. Decesare et al. were the first to describe sentinel lymph node identification using lymphoscintigraphy in vulvar cancer in 1997 . Several studies have demonstrated that sentinel lymph node biopsy (SLNB) is a safe, accurate, and cost-effective approach for managing early-stage vulvar cancer. In tumors smaller than 4 cm, SLNB has a false-negative rate of only 2% and a negative predictive value of 98% [13,14,15,16]. Furthermore, the 2008 Groningen International Study on Sentinel Nodes in Vulvar Cancer (GROINSS-V-I) showed that omitting inguinofemoral lymphadenectomy in patients with early-stage vulvar cancer who have negative sentinel lymph nodes is a safe alternative. This approach is associated with a three-year survival rate of 97% and a nodal recurrence rate of only 2.3% . Additionally, it showed that avoiding lymphadenectomy significantly reduces treatment-related morbidity, including lower rates of lymphedema and wound complications.
Based on these results, major international guidelines currently recommend exclusive SLNB for unifocal vulvar tumors smaller than 4 cm with stromal invasion greater than 1 mm and no clinical suspicious nodes (palpation/imaging) [17,18]. Inguinal lymphadenectomy should be completed in cases where there is no tracer migration, or if the pathological examination reveals the presence of metastases in the sentinel lymph node (SLN). For lesions > 4 cm or multifocal tumors, bilateral lymphadenectomy remains the primary standard procedure.
For the SLN assessment, the use of a radiotracer (typically Tc99m nanocolloid) is mandatory, while combination techniques with isosulfan or methylene blue dye or indocyanine green (ICG) are recommended, as they have been associated with improved accuracy [19,20].
In multifocal tumors or tumors larger than 4 cm, however, the use of SLNB remains a subject of debate due to the limited available evidence regarding its oncologic safety in these scenarios . Increased tumor size is associated with a higher risk of lymph node metastasis, while multifocality may lead to multiple lymphatic drainage pathways, both of which could potentially compromise the diagnostic accuracy of the technique . Similarly, there are no established consensus statements regarding the use of inguinofemoral SLNB in patients with prior vulvar excision. As a result, lymphadenectomy remains the standard treatment for patients with recurrent vulvar squamous cell carcinoma.
This review aims to summarize the latest evidence on the indications, technique, safety, and oncologic outcomes of SLNB in vulvar cancer, with a particular focus on controversial scenarios such as recurrent disease and larger or multicentric tumors.
2. Materials and Methods
A comprehensive review of the literature was performed using the PubMed database, using terms such as “vulvar cancer”, “sentinel node biopsy”, “inguinofemoral lymphadenectomy”, and “recurrent vulvar cancer” in different combinations. The search has been limited to original studies, systematic reviews, meta-analyses, randomized controlled trials, and case controls published in English within the last 35 years. Publications relevant to the search and their cited references were retrieved and evaluated independently by M.G. and M.A. for inclusion in the text.
3. Results
3.1. Current Indications and Evidence for Sentinel Node Biopsy
SLNB is a minimally invasive surgical technique applicable to women with early-stage vulvar cancer. The European Society of Gynecologic Oncology (ESGO) and the National Comprehensive Cancer Network (NCCN) guidelines currently recommend SLNB for patients with unifocal vulvar tumors smaller than 4 cm and stromal invasion greater than 1 mm, with clinically and radiologically negative groins [17,18]. In patients with small lateralized tumors (<4 cm in size and ≥1 cm from the vulvar midline) and negative ipsilateral lymph nodes, the risk of contralateral groin node metastasis is below 1% . Therefore, ipsilateral groin dissection is considered sufficient in these cases. Conversely, for non-lateralized tumors, a bilateral surgical evaluation is recommended. The intraoperative frozen section of the SLN is optional. The risk–benefit balance between accurately assessing the size of lymph node metastases through frozen section analysis and the potential tissue loss, along with the possible need for a second surgical intervention if the SLN is positive, should be carefully evaluated. In case the SLN is not detected, inguinofemoral lymphadenectomy, including superficial and deep nodes, should be performed. Similarly, if detection is achieved on only one side in midline tumors, contralateral lymphadenectomy is mandatory [17,18].
During the 2000s, several studies provided promising data on the detection rate and false-negative rates of SLNB in patients with vulvar cancer. However, the majority were retrospective studies with a limited sample size [22,23,24,25,26].
It was not until 2008 that van der Zee et al. published the results of the GROINSS-V-I study, the largest validation multicenter observational study that investigated the safety and utility of SLNB in early-stage vulvar cancer . This study included 403 patients with <4 cm squamous cell carcinoma of the vulva and clinically nonsuspicious inguinofemoral lymph nodes. In 259 of these patients, the SLN was negative after pathological ultrastaging, and no further lymphadenectomy was performed. The 3-year survival rate in this group was 97%, with only a 2.3% rate of nodal recurrence after a median follow-up of 35 months. In addition, both short- and long-term morbidity was significantly reduced when inguinofemoral lymphadenectomy was omitted (lymphedema 1.9% vs. 25.2%, and recurrent erysipelas 0.4% vs. 16.2%), demonstrating that SLNB was safe and less harmful compared with inguinofemoral lymphadenectomy in unifocal, <4 cm vulvar carcinomas. Furthermore, in the long-term follow-up of the GROINSS-V-I trial, Grootenhuis et al. reported 5-year and 10-year disease-specific survival rates of 93.5% and 90.8%, respectively, for patients with negative SLN, confirming the long-term safety of SLNB in this population. Despite the positive survival outcomes, the study emphasized that patients still faced a risk of recurrence. They reported a local recurrence rate of 24.6% and 36.4% at 5 and 10 years, respectively, for SLN-negative patients. This underscores the importance of long-term monitoring, even more than a decade after the initial treatment .
Another key study that contributed to establishing SLNB as the standard technique for lymph node staging in vulvar cancer was the Gynecologic Oncology Group (GOG-173) trial, whose findings were published in 2012 . This study included 452 women with squamous cell carcinoma of the vulva, with tumor sizes ranging from 2 to 6 cm. All participants underwent SLNB followed by inguinal lymphadenectomy. The sensitivity of the technique for detecting metastases was 92.5% overall, with a false-negative rate of 3.7%, which decreased to 2% in tumors smaller than 4 cm, thus corroborating the results of the GROINSS-V-I study.
More recently, additional studies have demonstrated the safety of SLNB as an exclusive method for lymph node staging in cases with negative results. A systematic review and meta-analysis conducted in 2015 by Covens et al., which included 11 studies, found that the rate of groin recurrence was not significantly higher for SLNB (3.4%) compared to complete inguinofemoral lymphadenectomy (1.4%) . Another systematic review and meta-analysis, incorporating 29 studies with 1779 patients, reported an overall SLNB sensitivity of 95% (95% CI: 92–98%). However, they observed a 9% false-negative rate, emphasizing that this relatively high rate underscores the impact of the learning curve, a persistent challenge in managing rare cancers . Table 1 shows the main studies evaluating the safety of SLBN in vulvar cancer.
Table 1. Data from studies evaluating SLNB in vulvar cancer.
The inclusion of SLNB as a method for lymph node staging in vulvar cancer was added to international guidelines in 2016 . Since then, the adoption of this technique has progressively increased, as demonstrated by a recent study that analyzed the implementation of surgical de-escalation in gynecologic cancers using the National Cancer Database. They reported that the utilization rate of SLNB in vulvar cancer rose from 12.3% to 36.9% between 2012 and 2020, while lymphadenectomy rates simultaneously declined from 87.7% to 63.2% . These findings reflect a growing shift towards less invasive approaches in the management of early-stage vulvar cancer, aiming to minimize surgical morbidity without compromising oncological outcomes. Furthermore, with the reduction in morbidity, SLNB has also been shown to be more cost-effective compared to lymphadenectomy [32,33].
3.2. Surgical Technique of Sentinel Node Biopsy and Available Methods of Detection
The most accurate detection methods for SLNB in appropriately selected patients with vulvar cancer are combined techniques .
Current evidence supports the use of a combination of blue dye and 99m-Tc nanocolloid for sentinel lymph node detection, in accordance with the methodology recommended by the GROINSS-V study and international guidelines [17,34]. The standard protocol involves injecting 4 mL of blue dye and 4 mL of 2.5 mCi Technetium-99, divided into four 1-mL intradermal aliquots placed in four orthogonal peritumoral locations. To ensure adequate migration to the lymph nodes, Technetium-99 should be administered preoperatively, while blue dye is injected just before groin dissection during surgery under anesthesia . The sentinel lymph node must be removed before the primary tumor is excised. A groin incision is performed at the site with the highest signal intensity detected by gamma probes, followed by careful dissection to trace lymphatic pathways and locate the sentinel node. The gamma detector should then confirm the presence of radioactivity in the excised node and verify that no residual tracer remains in the lymphatic basin. If multiple sentinel nodes are detected, all should be removed [34,35]. A node detected as “positive” by the gamma probe is defined as having a count at least five times greater than the background level. After removing the targeted node, the background reading with the gamma probe should be less than 10% of the node’s count. While locating the node(s), it is crucial to consider its orientation toward the vulva to avoid mistaking background vulvar uptake for the node itself . With the growing adoption of inguinofemoral SLN detection, the techniques used for identification have also advanced. Initially, the most commonly employed methods included blue dyes, such as lymphazurin and methylene blue, along with radiocolloid lymphoscintigraphy . However, since the introduction of near-infrared imaging with indocyanine green (ICG) in 2010, this approach has gained widespread acceptance for SLN detection in the inguinofemoral region . Regardless of the specific mapping technique, SLNB in vulvar cancer generally demonstrates high detection and mapping success rates. The selection of the preferred method is primarily influenced by the surgeon’s expertise, training, and institutional experience . As previously noted, growing evidence supports the use of ICG as a viable alternative to blue dye. Next, we are going to summarize some of the current evidence on the different tracers for SLNB in vulvar cancer. In 2010, Crane et al. introduced the use of an infrared light source and camera for intraoperative detection of inguinofemoral SLNs labeled with ICG (Figure 1). They later published their findings on SLN mapping in vulvar cancer, analyzing 16 groins using a combination of 99m-Tc, blue dye, and ICG with near-infrared imaging. Their study identified 29 SLNs with 99m-Tc, 26 with ICG, and 21 with blue dye [38,39]. Following this, in 2017, Soergel et al. reported their experience with ICG in a cohort of 27 patients. A key finding of their study was the identification of eight SLNs that were not detected by 99m-Tc but were successfully visualized using ICG alone .
Figure 1. Sentinel lymph node in vulvar cancer. ICG.
Recently, Deken et al. demonstrated, in their randomized trial, comparable efficacy in terms of sentinel node detection between isotope/ICG compared with conventional detection method with isotope/blue dye. They randomized 48 patients to the blue dye (n = 24) or ICG group (n = 24), using in both groups also 99m-Tc nanocolloid. The SLN detection rate was 92.1% of the groins in the blue dye group and 97.2% in the ICG group (p = 0.33). Surgical outcomes were similar between groups, although more short-term postoperative complications (wound infection or breakdown and lymphocyst formation) were observed in the blue dye group (p = 0.041) .
A systematic review of the literature aimed to determine if the use of ICG alone in detecting SLN in vulvar cancer was as accurate as the gold standard dual-labeling technique; 13 studies were identified with similar detection rates for SLN to the gold standard technique (ranged between 89.7 and 100% among the different studies). The authors also highlighted the potential of decreasing the detection rate in two settings; in the most metastatic lymph nodes and in obese patients and midline tumors. No adverse events were reported, and no consensus regarding to ICG injection site or timing, volume or concentrations, or use of serum albumin or hybrid tracer was reached . A recent meta-analysis by Di Donna et al. provides additional evidence supporting the effectiveness of ICG in SLN detection for vulvar cancer. The study aimed to identify the technique with the highest detection rate by comparing planar lymphoscintigraphy (PL), blue dye, and ICG fluorescence. After selecting 30 studies, the researchers found that the SLN detection rates per patient and per groin were 96.13% and 92.57% for PL, 90.44% and 66.21% for blue dye, and 91.90% and 94.80% for ICG, respectively. While the patient-based analysis did not reveal significant differences between these methods, the groin-based analysis showed that both PL and ICG had significantly higher detection rates compared to blue dye (p< 0.05) . Additionally, ICG appears to have a steeper learning curve, with detection rates improving as surgical teams gain more experience and training . Moreover, a hybrid technique combining ICG with 99m-Tc nanocolloid has been introduced, allowing injection up to 20 h before surgery while still enabling intraoperative near-infrared detection . Another recent approach, described in the SARVU study, involves the use of ferromagnetic tracers, which demonstrated a detection efficacy comparable to that of radiotracers .
The method of histopathological evaluation could influence the reliability of SLN in staging vulvar cancer. Postoperative pathological analysis of the SLN can be complemented with intraoperative frozen section assessment. If the SLNB is positive, this technique provides the advantage of allowing an immediate complete lymphadenectomy. However, it has a relatively high false-negative rate and poses a risk of losing diagnostic tissue. A precise pathological evaluation of SLNs is essential. As observed in GOG 173 and GROINS V , hematoxylin and eosin staining of lymph nodes fails to detect at least 20% of nodal metastases compared to additional immunohistochemical (IHC) analysis. Further serial sectioning could identify an extra 7% of nodal metastases . Ultrastaging of lymph nodes, which incorporates serial sectioning and immunohistochemistry, is an important element of the SLN technique. A review demonstrated that the detection rate of micrometastases is significantly higher when combining hematoxylin and eosin staining with serial sectioning and immunohistochemistry . Thus, ultrastaging of the nodes, with serial sectioning and staining, is recommended for the identification of possible micrometastases and isolated tumor cells . The clinical relevance of micrometastases identified through ultrastaging remains a topic of debate. Isolated groin recurrence has been documented in only a few cases involving patients with micrometastases in the SLN .
Although the findings from this retrospective series regarding near-infrared imaging performance are encouraging, further prospective studies are necessary to assess the effectiveness of indocyanine green and near-infrared imaging, thereby defining the role of SLN mapping in this condition. Protocols for using ICG are still varied, and the ideal approach has yet to be established. Currently, the combination of an isotope with either blue dye or ICG appears to achieve the highest detection rates and demonstrates proven clinical effectiveness. Performing pre-operative lymphoscintigraphy is advised to determine the number and location of sentinel nodes.
3.3. Management of Metastatic Groin Nodes
Additional treatment of nodes in the groin is determined based on the outcomes of the SLNB. In the case of negative SLN, no additional or adjuvant treatments are necessary.
3.3.1. Volume of Lymph Node Disease
The presence of lymph node metastasis is associated with poorer disease-free survival and overall survival rates. Metastatic lymph node disease in early-stage vulvar cancer is diagnosed in 10.7% of stage I patients and 26.2% of stage II patients, affecting between 21% and 35.8% of patients . According to Mahner et al., the disease-free survival rate is 35.2%, and overall survival is 56.2%, compared to 75.2% and 90.2% in patients without lymph node involvement . The lymph node metastases are classified as stage III disease. The new FIGO classification considers the presence and count of metastatic lymph nodes (≤2 and >2 lymph nodes) and the volume of lymph node disease (<5 and ≥5 mm). This is based on the results of the GROINS-V trial . The risk of metastatic affectation of a different node than the sentinel one increases not only with the number of metastatic SLN but also with the volume of the disease. Moreover, the authors found statistically significant differences in the disease-specific survival after 24 months of follow-up between patients with SLN metastases ≤ 2 and >2 mm (69.5% vs. 94.4%, respectively).
The management of metastatic lymph nodes is controversial; however, additional treatment (complete inguinofemoral lymphadenectomy or adjuvant treatment) to the corresponding inguinofemoral area should be promptly administered after finding metastatic disease in an SLN in patients with vulvar carcinoma [17,20]. The role of adjuvant radiotherapy was evaluated in a large multicenter retrospective study. An improvement in prognosis, in both disease-free survival (DFS) and overall survival (OS), was observed in patients with positive lymph nodes who underwent postoperative radiotherapy .
Patients with macrometastasis (>2 mm) or extracapsular spread typically require completion inguinofemoral lymphadenectomy, followed by adjuvant radiotherapy if more than one node with macrometastasis and/or extranodal spread is found in the final pathological report. However, recent studies suggest that radiotherapy alone may be a viable option for selected patients with micrometastatic disease (≤2 mm) to reduce surgical morbidity without compromising oncologic outcomes. Recently, the large prospective multicenter GROINSS-V II study aimed to establish the safety of replacing inguinofemoral lymphadenectomy with radiotherapy for patients with early-stage vulvar cancer with metastasis in an SLN. An analysis of isolated groin recurrence in the first 91 patients found that nine out of ten had macrometastatic disease (>2 mm). As a result, the protocol was revised to restrict inguinofemoral node radiotherapy without further surgery (complete inguinofemoral lymphadenectomy) only to patients with micrometastatic disease (≤2 mm) in the SLN. Patients with micrometastases were treated with inguinofemoral lymphadenectomy as the standard approach, with additional radiotherapy for those who had multiple node metastases or extracapsular spread. The authors report a median follow-up period of 24.3 months. Among patients with SLN micrometastases (≤2 mm) who received radiotherapy alone without lymphadenectomy, recurrence rates were low (1.6%), with manageable levels of treatment-related toxicity. Among 162 patients with SLN macrometastases, the isolated groin recurrence rate at 2 years was 22% in those who underwent radiotherapy, and 6.9% in those who underwent IFL (p = 0.01) .
Currently, the GROINSS-V III study is examining whether patients with an SLN macrometastasis can be managed with chemoradiation instead of radiotherapy to both prevent groin recurrences and reduce treatment-related morbidity . The NCCN already advises the use of concurrent chemotherapy with radiotherapy for treating SLN metastases. However, this recommendation is primarily based on studies involving patients with advanced vulvar cancer. So far, chemotherapy and chemoradiation have not been sufficiently researched as treatment options for SLNmetastases in early-stage vulvar cancer [17,18].
3.3.2. Management of Contralateral Groin
The current evidence on the safety of omitting treatment to the no metastatic groin in patients with unilateral positive groin after bilateral SLNB is also controversial and conflicting. Current guidelines recommend that in cases of unilaterally positive SLN, bilateral inguinofemoral lymphadenectomy should be performed, increasing the risk of associated morbidity [17,18]. The percentage of contralateral non-sentinel node affectation varies among the different studies. In these cases, the risk of contralateral lymph node affectation seems to be low according to some authors. The rate of contralateral groin positivity in the studies carried out by Woelber, Nica, and Ignatov ranges between 0 and 5.3%. These data support the omission of contralateral inguinofemoral lymphadenectomy to reduce surgical morbidity and long-term complications [52,53,54]. In contrast, data from a German study identified contralateral positivity in a higher percentage of patients with midline vulvar carcinoma—a total of 4 patients out of 18 (22%). Moreover, they also found that the depth of tumor infiltration correlated significantly and positively with the incidence rate of contralateral groin metastasis (p = 0.0038) .
Recently, a large prospective study from the GROINSS-V group provides evidence to omit treatment to the unaffected contralateral groin, in cases where bilateral drainage has been mandatorily identified for midline tumors. In this study, the contralateral non-SLN metastases/groin recurrence rate was 2.9%. Therefore, they conclude that omitting contralateral lymphadenectomy in these cases should be safe, although caution is advised with tumors of >3 cm, the authors highlighting that the majority of non-sentinel contralateral metastasis occurred in these larger tumors .
3.4. Extended Indications for Sentinel Node Biopsy
Although SLN biopsy is currently recommended for unifocal tumors under 4 cm, research is exploring its feasibility in patients with larger or multifocal tumors.
3.4.1. Sentinel Node Biopsy in Recurrent Vulvar Cancer
Some studies have also suggested that SLNB may still be a reliable staging method in carefully selected cases of the first recurrence of vulvar cancer, potentially reducing the need for full lymphadenectomy in a broader patient population. However, further validation through prospective trials is required before guideline expansion. At present, evidence for the use of the SLN procedure in the case of recurrent cancer is lacking. A small retrospective study suggests that the technique is feasible, but that detection rates are lower and lymphatic drainage may be unusual after surgery of the primary tumor [17,56]. Inguinofemoral lymphadenectomy is considered the standard treatment for patients with recurrent vulvar cancer who did not undergo this procedure during the first episode. Many of the patients with vulvar cancer (primary or relapse episode) are elderly and frail patients; for these reasons, an alternative treatment to lymphadenectomy should be proposed. However, at the moment, the evidence about the accuracy and safety of repeat SLNB in recurrent disease is still lacking. The main reason not to perform a repeat SLNB in patients with a local recurrence of vulvar cancer is the assumption that the lymph flow might be altered because of previous surgery or radiotherapy . Otherwise, SLNB, also in recurrent disease, could add some advantages such as improving the visualization of the lymph drainage and guiding the surgeon in the removal of the lymph nodes at risk. Until now, there are no prospective studies assessing the efficacy and safety of SLNB in recurrent vulvar cancer, and only two case reports and a small retrospective study have been published regarding this topic [23,56,57]. In 2016, van Doorn et al. published a retrospective series of 27 patients with a first recurrence of squamous cell carcinoma of the vulva who underwent SLNB. They found this procedure seemed feasible, in that in 77% of patients and in 84% of the groins, the SLN procedure could be performed as planned. This percentage was less compared to the rates of success procedure in primary SLNB, which is around 95% . The authors also observed that the procedure was technically more challenging in recurrent disease compared to initial surgery. Regarding oncological outcomes, data are lacking, but so far none of the patients in the study suffered local or distant recurrences after repeating SLNB during a follow-up of a median of 27 months (range 2–96 months).
Currently, there is an ongoing trial designed to prove the feasibility and safety of SLNB in recurrent vulvar cancer. The outcomes of this current prospective study will help to bridge knowledge gaps and define future research questions. If the SLN procedure is confirmed to be safe and feasible in this patient group, it will significantly help reduce both the short- and long-term side effects of vulvar cancer treatment while having a smaller impact on quality of life compared to the current standard treatment. Additionally, we expect the study to enhance our understanding of the efficacy, side effects, and pathology of recurrent vulvar cancer .
3.4.2. Sentinel Node Biopsy in >4 cm or Multifocal Tumors
The safety of the SLN procedure in squamous cell carcinoma of the vulva > 4 cm or multifocal tumors is also lacking and at the moment the recommendation is to perform inguinofemoral lymphadenectomy. This evidence is based on the study GOG-173 where the authors found a detection rate of 92.0% and seven false-negative SLNs, resulting in an unacceptable false-negative predictive value of 7.4% in patients with tumors between 4–6 cm of diameter . Several prospective studies investigating SLN biopsy in vulvar cancer have reported detection rates varying between 88% and 97% per patient and have calculated a pooled detection rate of 93% per groin [13,14,20,59]. Recently, a prospective pilot study has been carried out to assess if indications for SLNB in vulvar cancer could be extended . They found detection rates and negative predictive values of SLN biopsy for women with primary squamous cell vulvar cancer ≥ 4 cm and multifocal tumors comparable to the results for smaller, unifocal tumors. The detection rates varied between 100% and 94.1% per patient and between 84.1% and 85.3% per groin. Moreover, they found no false-negative SLN in the population study . Another prospective study published in 2017 by Garganese et al. reported on 12 patients with tumors > 4 cm undergoing an SLN biopsy followed by inguinofemoral lymphadenectomy. In this cohort, no patients had a false negative SLN .
Multifocality of vulvar cancer is a rare condition which is reflected by the low number of patients collected in a few studies. Because of unsatisfactory oncological safety, the GROINSS-V-I study stopped the inclusion of multifocal tumors after two early inguinal recurrences between the hitherto 19 included patients . However, Garganese et al. included nine women with multifocal tumors in their cohort, with no false-negative SLN .
Nowadays, due to the low evidence we have, the safety of the SLNB in squamous cell carcinoma of the vulva larger than 4 cm or in multifocal tumors is not yet fully established. Although recent studies have shown detection rates and negative predictive values comparable to those of smaller, unifocal tumors, the current recommendation remains to perform inguinofemoral lymphadenectomy.
3.5. Potential Barriers and Limitations for Sentinel Node Biopsy
The current standard technique for SLNB could present some cumbersome aspects. For example, the preoperative lymphoscintigraphy along with intraoperative radiolocalization requires multiple procedures that often need to be performed the day before the surgical procedure. Additionally, intraoperative detection with 99m-Tc is complicated and relies on aural cues, necessitating a pause in the procedure at frequent intervals so that 99m-Tc uptake can be measured . The use of dyes facilitates the procedure. While blue dye aids in visually identifying the SLN, its effectiveness relies on the clear detection of both lymphatic vessels and the node itself. In contrast, indocyanine green combined with near-infrared imaging overcomes these limitations by offering real-time visual guidance throughout the procedure . Despite its benefits, SLN biopsy has certain limitations. Factors such as disrupted lymphatic drainage from prior surgeries, the presence of large tumors, or the learning curve associated with the technique can affect its accuracy. Additionally, the limited availability of nuclear medicine facilities and experienced surgical teams poses a challenge in resource-constrained settings. Overcoming these obstacles is crucial for expanding the use of SLN biopsy and ensuring equitable access in vulvar cancer treatment.
While intraoperative frozen section analysis may be considered, particularly to prevent the need for a second surgical procedure, caution is necessary due to the risk of overlooking micrometastases in the final histological assessment and the importance of precisely measuring metastatic deposits .
Another important aspect of SLNB in vulvar cancer is the one related to cost-effectiveness. There are studies that demonstrated that SLNB is more cost-effective than inguinofemoral lymphadenectomy. Erickson et al. and McCann et al. found that SLNB had both a lower annual cost and lower cost/effectiveness ratio than lymphadenectomy [32,33]. This was further demonstrated when a significant reduction in complications such as lymphedema and infection was observed after performing SLNB and, in consequence, a higher score in quality-of-life questionnaires.
3.6. Future Directions
The morbidity of open inguinal incisions has prompted the search for a minimally invasive approach to lymph node dissection. Videoendoscopic lymphadenectomy may be a valid alternative to the open route with no differences in terms of surgical outcomes, except for operative time, which is shorter for the open approach, and wound complications, which are less frequent for the videoendoscopic route . The development of a robot-assisted platform, the da Vinci Firefly, allows for intraoperative evaluation of the lymphatic channels and SLN mapping after injection of the primary tumor with ICG. A less invasive dissection of the inguinofemoral region using robot-assisted, near-infrared fluorescence; SLN mapping; and inguinal lymph node dissection may be a feasible approach to reduce short- and long-term morbidity [63,64]. Future analysis of randomized controlled trials in this specific topic and patient population should be carried out to confirm these results.
4. Discussion
The SLNB has emerged as a valuable technique in the management of early-stage vulvar cancer, offering a less invasive alternative to full lymphadenectomy. This approach aims to reduce morbidity while maintaining oncologic safety. The present review highlights the current evidence regarding the accuracy, benefits, and limitations of SLN biopsy in vulvar cancer.
Several studies have demonstrated that SLN biopsy provides reliable staging information with high sensitivity and specificity [5,13,14]. The GROINSS-V trials, among others, have shown that SLN biopsy is an effective method for identifying nodal metastases in early-stage vulvar cancer, significantly decreasing complications such as lymphedema and wound infections compared to complete lymphadenectomy [49,50]. These findings support the role of SLN biopsy as the standard approach in appropriately selected patients with unifocal tumors and clinically negative nodes.
SLNB in early-stage vulvar cancer is an effective and safe way to avoid significant morbidity surrounding surgical intervention without compromising oncologic outcomes. Best practices include using a dual detection technique with a radiocolloid tracer and a visual dye, which has been proposed to enhance detection and minimize false-negative rates. Surgeon comfort and experience using these techniques are imperative to ensure the safety and efficacy of this procedure. Given this, future frontiers need to include the expansion of these services through training and access to high-volume centers . More recently, ICG fluorescence imaging has emerged as an alternative technique, with studies suggesting comparable detection rates to the standard dual-tracer approach.
An ongoing area of research is the long-term oncologic outcomes of patients undergoing SLN biopsy compared to those undergoing complete lymphadenectomy. The GROINSS-V II trial provided valuable insights into the management of SLN-positive patients, suggesting that radiotherapy alone may be a viable alternative to lymphadenectomy in cases of micrometastatic disease (≤2 mm) . However, the management of macrometastatic disease (>2 mm) remains controversial, with ongoing research evaluating whether chemoradiation could replace lymphadenectomy in these patients (GROINSS-V III trial) .
Another unresolved issue is the management of the contralateral groin in patients with unilateral SLN metastases. Current guidelines recommend bilateral inguinofemoral lymphadenectomy in such cases, but emerging data suggest that omitting contralateral treatment may be safe in select patients, particularly when contralateral SLNs are negative. Prospective studies, including those from the GROINSS-V group, continue to investigate this question to optimize the balance between oncologic safety and surgical morbidity.
Despite its advantages, SLNB has limitations. False-negative results remain a concern, particularly in cases with large tumors, multifocal disease, or prior surgery altering lymphatic drainage. Additionally, the technique requires a high level of expertise and adherence to strict protocols to optimize detection rates .
Although SLNB is currently recommended for unifocal tumors smaller than 4 cm, ongoing research is evaluating its feasibility in patients with larger or multifocal tumors. Some studies also suggest that SLNB may serve as a reliable staging method in selected cases of the first recurrence of vulvar cancer, potentially reducing the need for complete lymphadenectomy in a broader patient population. However, further validation through prospective trials is required before expanding current guidelines. The main concern regarding SLNB in recurrence is the potential alteration of lymphatic drainage pathways due to previous surgery or radiotherapy. Prior surgery can alter lymphatic drainage, making it difficult to predict SLN locations, and fibrosis in the vulvar and groin regions may complicate the procedure. Fibrosis also increases the risk of surgical complications, such as saphenous vein injury [56,57].
At present, due to limited evidence, the safety of SLNB in vulvar squamous cell carcinoma larger than 4 cm or in multifocal tumors has not been fully established. Although recent studies have demonstrated comparable detection rates and negative predictive values to those observed in smaller, unifocal tumors, the current recommendation remains inguinofemoral lymphadenectomy [13,14,20,59].
Ongoing research aims to refine SLNB techniques further and expand their applicability. The exploration of novel tracers and imaging modalities seeks to enhance SLN detection rates and reduce false-negative occurrences. Moreover, studies are investigating the feasibility of extending SLN biopsy indications to a broader patient population while maintaining safety and efficacy. Centralizing the procedure in high-volume centers is emphasized to ensure standardized practices and optimal patient outcomes.
5. Conclusions
In conclusion, SLNB has become an integral component in the management of early-stage vulvar cancer, offering accurate staging with reduced morbidity compared to traditional lymphadenectomy without compromising oncological outcomes. The advancements in detection techniques and the establishment of standardized guidelines have significantly improved patient care. Future studies should focus on refining patient selection criteria, improving detection techniques, and clarifying the implications of low-volume nodal disease to further optimize outcomes for patients with vulvar cancer. Continued research and adherence to evidence-based practices are essential to further enhance the efficacy and safety of SLNB in vulvar cancer. In addition, continued quality-of-life analyses are needed to further demonstrate the short- and long-term benefits for these patients.
Author Contributions
Conceptualization: M.G., M.A.-E. and I.Z.; resources: M.G. and M.A.-E.; writing: M.G. and M.A.-E.; review: M.G., M.A.-E. and I.Z. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Conflicts of Interest
The authors declare no conflicts of interest.
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Figure 1. Sentinel lymph node in vulvar cancer. ICG.
Table 1. Data from studies evaluating SLNB in vulvar cancer.
| Author (Year) | Number of Patients | SLNB Technique | Identification Rate (%) | False Negatives (%) | Median Follow-Up (Months) | Groin Recurrences (%) | Outcome in SLN Negative Patients (%) |
:---: :---: :---: :---: |
| Van der Zee (2008) | 403 | R + BD | 97 | 3.3 | 35 (2–87) | 2.3 (unifocal) 3 (including multifocal) | 97 3-year DSS |
| Levenback (2012) | 452 | R + BD | 92.5 | 3.7 | NA | NA | NA |
| Robinson (2014) | 69 | R + BD | 93 | NA | 58.3 | 4.7 | NA |
| Grootenhuis (2016) | 377 | R + BD | 95 | 3.5 | 105 (0–179) | 2.5 (unifocal) | 93.5 5-year DSS 90.8 10-year DSS |
| Klapdor (2019) | 772 | R or BD | 94.7 | 5.8 | 33 (0–156) | 4.5 | 82.7 3-year PFS 92.7 3-year OS |
SLNB: sentinel lymph node biopsy; SLN: sentinel lymph node; R: radiotracer; BD: blue dye; DSS: disease-specific survival; PFS: progression-free survival; OS: overall survival; NA: Not aplicable.
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Gracia, M.; Alonso-Espías, M.; Zapardiel, I. Current Limitations of Sentinel Node Biopsy in Vulvar Cancer. Curr. Oncol.2025, 32, 215.
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Gracia M, Alonso-Espías M, Zapardiel I. Current Limitations of Sentinel Node Biopsy in Vulvar Cancer. Current Oncology. 2025; 32(4):215.
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Gracia, Myriam, Maria Alonso-Espías, and Ignacio Zapardiel. 2025. "Current Limitations of Sentinel Node Biopsy in Vulvar Cancer" Current Oncology 32, no. 4: 215.
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Gracia, M., Alonso-Espías, M., & Zapardiel, I. (2025). Current Limitations of Sentinel Node Biopsy in Vulvar Cancer. Current Oncology, 32(4), 215.
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Gracia, M.; Alonso-Espías, M.; Zapardiel, I. Current Limitations of Sentinel Node Biopsy in Vulvar Cancer. Curr. Oncol.2025, 32, 215.
AMA Style
Gracia M, Alonso-Espías M, Zapardiel I. Current Limitations of Sentinel Node Biopsy in Vulvar Cancer. Current Oncology. 2025; 32(4):215.
Chicago/Turabian Style
Gracia, Myriam, Maria Alonso-Espías, and Ignacio Zapardiel. 2025. "Current Limitations of Sentinel Node Biopsy in Vulvar Cancer" Current Oncology 32, no. 4: 215.
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Gracia, M., Alonso-Espías, M., & Zapardiel, I. (2025). Current Limitations of Sentinel Node Biopsy in Vulvar Cancer. Current Oncology, 32(4), 215.
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Raise to the power of .
Step 4.2.1.2
Add and .
Step 5
Factor the perfect trinomial square into .
Step 6
Solve the equation for .
Tap for more steps...
Step 6.1
Take the specified root of both sides of the equation to eliminate the exponent on the left side.
Step 6.2
Any root of is .
Step 6.3
The complete solution is the result of both the positive and negative portions of the solution.
Tap for more steps...
Step 6.3.1
First, use the positive value of the to find the first solution.
Step 6.3.2
Move all terms not containing to the right side of the equation.
Tap for more steps...
Step 6.3.2.1
Add to both sides of the equation.
Step 6.3.2.2
Add and .
Step 6.3.3
Next, use the negative value of the to find the second solution.
Step 6.3.4
Move all terms not containing to the right side of the equation.
Tap for more steps...
Step 6.3.4.1
Add to both sides of the equation.
Step 6.3.4.2
Add and .
Step 6.3.5
The complete solution is the result of both the positive and negative portions of the solution.
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10580 | https://www.youtube.com/watch?v=jtfYH4BvQIY | Normal Distribution with the normalcdf Function on a TI 83 84 HD
UofA FiniteMath
929 subscribers
58 likes
Description
32469 views
Posted: 17 Aug 2021
Transcript:
In this video we'll be looking at the
normal distribution. The normal distribution, shown on the screen, has
most of its data centered around the mean. As we move away from the mean in either
direction, scores become less likey. We can see that does that in a symmetric
way on both sides of the mean. The normal curve goes from negative infinity to positive infinitely and the area under the normal curve is 1. Now these numbers under
the normal curve along our x axis, these are called z scores and they tell
us both how far away from the mean a score is as well as in what direction. So for
example -1, the negative tells us that it's below the mean and the one
tells us that it's exactly one standard deviation away from the mean. The distribution that we're looking at here is called the standard normal distribution and the
standard normal distribution has a mean of 0 and a standard deviation of 1. If
you're working a problem and it doesn't give you a mean and a standard deviation,
chances are you're working with the standard normal distribution and your mean is 0 and standard deviation is 1. So what we can do with this distribution, and our
calculator, is to find the likelihood that a score falls within some area. So
for example, we could find the likelihood that a score falls between -1 and 1, or in other words within one standard deviation of the mean. And so
what we're going to find is this red area right here and we'll find that by using
the normalcdf function on our calculator. So to get to the
normalcdf you'll push the second key and then the vars button three buttons up
from the 9 key on your calculator. We want to go down to normalcdf. The cdf stands for
cumulative density function. Now some of you may get this menu option here. If you have an older calculator you'll just get what I have written on the screen here: normalcdf with a parenthesis. So I'll write it out on the screen as well for both types of
calculators. So what we're going put in to normalcdf
is the lower number, which is the smaller of the numbers, the upper number, the mean, and the standard deviation. So if we're going from -1 to 1, our lower number would be
-1, our upper number would be 1 and then, since you're using the standard
normal distribution, we have a mean of 0 and a standard deviation of 1. So we'll put those same numbers into the calculator. So -1 as our lower, 1 as our upper, mean of 0, standard
deviation 1, and again if you don't have that menu option what your
calculator should look like is what I've got on the screen right there. And so
what we end up with is about 0.6827 if we round to four decimal places, which we could also say is 68.27%. Since the area is 1 we can also represent
this as a percentage very easily. So this is the standard normal distribution - mean of 0, standard deviation of 1. We'll also see some problems where we have different
means and standard deviations. So for example if a sample of people was
measured and the average height was found to be 67 inches with a standard
deviation of 3 inches, we could find what percentage of the population we
would expect to be taller than 72 inches. Of course we'll have to assume that it
was a good sample, but if we can assume that we can find this percentage. So lets
first draw our distribution. So we have a mean of 67, so that will be right in the middle. And then we want to find the
percentage of the population taller than 72 inches. So that'll be above the mean, and so what we'll be finding is this area right there. So again we'll use our normalcdf function
on our calculator. And so again what we want to put in there is the lower number, the
upper number, then mean, and the standard deviation. So we can see that the lower number
would be 72 that's the smallest number on that area that we're trying to find.
But the upper number, we'll notice we don't actually have one. We're actually just going all the way to infinity. Now of course it's impossible for someone to be infinity inches tall
but since the normal distribution does go from negative to positive infinitely,
we will want to put infinity here or, in our case, something fairly close to
infinity so that we can find our answer. So what we're going to use is 1E99
which is just a really large number so that's a 1 with 99 zeros after it. And that'll be close enough to infinity for our purposes. So how do we put that in the calculator. Well let's go
back to normalcdf. So our lower here is going to be 72, upper we want to put 1E99. So we'll do 1. The E button is right above this 7 key on your calculator we'll push second and then the comma button right above that 7. And so that's how we get that E. And then 99. And so if we were, for example, trying to find the percentage of
the population shorter than 72 inches, we'd be going from negative infinity to 72
so there we'd use negative 1E99, 72. All right our mean on this problem is 67
inches and our standard deviation is 3 inches. So mean of 67, standard deviation of 3. What we end up with is that, we get the area would be 0.0478 if we rounded to four decimal places. The question asked for it
in percentage form so that would be 4.78% percent. So we would expect 4.78% of the population to be taller than 72 inches. |
10581 | http://mathletenation.com/content/palindrome-week-words-n-digit-numbers-odometer-readings | Palindrome Week, Words, n-digit numbers, odometer readings | Mathlete Nation
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Palindrome Week, Words, n-digit numbers, odometer readings
Posted on September 20th, 2019 by kramer
PALINDROME:In mathematics, a palindrome is a number that reads the same forward and backward. For example, 353, 787, 1331, 42724.
All numbers that have the same digits such as 4, 11, 55, 222, and 6666 are other examples of palindromes.
This was a special week during 2019. This week is a Palindrome Week, (well, 10 days actually) when each of 10 consecutive dates can be read the same backward and forward. Palindrome Week is unique to the few countries, such as the USA, that write dates in the month-day-year format (9-10-19). September's series of palindromes only work if the date is written with the last two digits of the year and no zero before the month — 09-10-2019 isn't a palindrome, for example. Because much of the world formats dates as day-month-year, the mathematical oddity is mostly noted in the U.S. I asked them to record all 11 palindromic dates in September 2019. the 9th month using USA format of m-d-yy (only last 2-digits of year). Many of them realized that September 1st was also a palindrome 9119. We talked about whether there is a palindrome week during 2020 and found that it will start with February 1st, 2020, or 02120.
With 4th Grade and up, we used the month-day-year (all 4 digits) format and looked at how many days there are in the new century (January 1, 2000 to December 31, 2100). They multiplied 365 days by 100 and then added on another 25 days for leap days. Out of the 36750 days there are only 38 palindromic dates in the 21st Century. They were challenged to find all of them.
Then, I had the children think of 3 and 4-letter words that form a palindrome. I gave them all of the 5 and 6-letter words that are palindromes with definitions. My favorite palindrome is AIBOHPHOBIA, which is an 11 letter palindrome which means, “fear of palindromes.”
After exploring the wonderful words that are palindromes, we turned to numbers that read the same both forwards and backwards. I gave them a few digits like 1234 and asked them to create an odd number digit palindrome and an even number digit palindrome (so, 1234321 and 12344321). There is a great challenge page on the pdf for this exercise.
I then had them think about all 1-digit numbers that are palindromes and to record how many there are. It was easy enough for the to write them down starting with 0, 1, 2 .. 9, but the challenge was counting how many. Many said 9 but they were not focused on zero. So of course, the answer was 10. When I asked them to write down all of the 2-digit palindromes, they were curious as to whether 00 was a two digit number. It is not, zero is a one digit number so starting with 11, 22, 33 … 99, there are 11.
Then we explored three-digit palindromes and attacked a challenge that would determine the total number of 3-digit palindromes. They figured that using an organized system of numbers starting with the smallest, 101, until reaching the largest, 999.
Many of the Mathletes started with 101, 111,121,131,141,151,161,171,181,191, vertically and 101, 202, 303, 404, … 909 horizontally on top. Some realized that there are ten rows and nine columns so 10x9=90 3-digit palindromes. For 4th grade and up, they tackled the challenge of 4-digit palindromes 1001, 1111, 1221, 1331, 1441, 1551,1661, 1771, 1881, 1991, 2002..... Surprisingly, the same pattern as 3-digits so a total of 90 numbers.
Then they were able to explore 5-digit palindromes starting with 10001, 10101, 10201, etc. Some noticed that there are additional 3-digit palindromes imbedded inside the 5-digits that start and end with zero, with a total of 100, but there are 9 separate thousands for each group of 100 palindromes, so a total of 900. See my algorithm for the number of palindromes for each number of digits. This uses the fundamental counting principle, so for 5-digit numbers, there can be 9 digits in the first digit of palindrome, 10 digits in the second position and 10 in the third position, so 9x10x10=900. The last two positions in a 5-digit palindrome just follow the first two digits.
So looking at this incredible sequence of numbers, for the following number of digits:
1-digit palindromes (1,2,3,4,5,6,7,8,9) 9
2-digit " (11,22,33,44,55,66,77,88,99) 9
3-digit " 90
4-digit " 90
5-digit " 900
6-digit " 900
7-digit " 9000
8-digit " 9000
9-digit " 90000
Finally, I had the 4-7th graders learn about odometer readings. My boys and I always looked at the odometer which tells you the number of miles. Then we determined how many more miles until the next palindrome. To this day, they are in their 20s and still take pictures of the odometer when it reads a palindrome. The last page in the pdf gives them six odometer readings and they are challenged to calculate how many miles to the next palindrome. For example, if there are 25,000 miles on the odometer, there are 52 miles until it reaches 25052.
This is a great way for your children and you to use the time in the car mathematically. It is not only challenging for them to find the next palindrome, but it then requires the to use subtraction to figure out how many miles remaining. Actually, I prefer to add from the current odometer reading until the palindrome (subtraction is really the reverse of addition).
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10582 | https://web.stanford.edu/class/cs205b/lectures/lecture16.pdf | CS205b/CME306 Lecture 16 1 Incompressible Flow 1.1 Laplace Equation in 1D Supplementary Reading: Osher and Fedkiw, §18.1, §18.2 Recall that the system of equations we must solve for incompressible flow is ∇· u = 0 ρt + u · ∇ρ = 0 ut + u · ∇u + ∇p ρ = g.
The Laplace equation in 1D is given by pxx = 0.
The solution is simply a line p = ax + b.
(1) The values of the constant a and b are determined by boundary conditions.
Assume that the domain is the interval [0, 1]. We may have Dirichlet boundary conditions, where the value of the function p is given at the boundary. For example, p(0) = p0 p(1) = p1.
Plugging the boundary conditions in the equation(1), we get p(0) = b = p0 p(1) = a + b = p1 ⇒a = p1 −p0 so the coefficients a and b are uniquely determined. Alternatively, Neumann boundary conditions specify the value of px at the boundary. For example, px(0) = 0 ⇒a = 0.
This gives us a family of lines with slope 0. To find b, we would need another piece of information.
A Dirichlet boundary condition would pick out one of the lines with slope 0, thus determining 1 the solution. But observe that specifying two Neumann conditions could lead to no solution. For example, px(0) = 0 px(1) = 1.
These two boundary conditions are inconsistent, hence there is no solution. Another example is px(0) = 0 px(1) = 0.
In this case, the given boundary conditions are consistent, but incomplete. We still do not have enough information to identify a unique solution. The above examples illustrate the fact that in 1D, for the Laplace equation, we can determine the solution if we have two Dirichlet boundary conditions or one Neumann and one Dirichlet boundary condition, but will have either no solution or an underdetermined solution in the case of two Neumann boundary conditions.
1.2 Discretizing Laplacian of Presssure We need to numerically solving Poisson’s equation pxx = f(x).
We will also need the gradient to apply the pressure. We use second order central differencing for both. At each cell face, we approximate the pressure gradient with (px)i+1/2 = pi+1 −pi ∆x + O(∆x2).
From this we use central differencing again to express the Laplacian at each grid node (pxx)i = (px)i+1/2 −(px)i−1/2 ∆x + O(∆x2) = pi+1 −2pi + pi−1 ∆x2 + O(∆x2) = fi.
The result is a coupled linear system that we need to solve in order to determine p on the entire domain. However, we cannot write this equation as is for the grid points near the boundary since it will involve points outside of the domain. For example, assume that our domain is the interval [0, 1] and that we have grid points 0, 1, . . . , M, M + 1 uniformly spaced on the domain. The equation for p1 is p2 −2p1 + p0 ∆x2 = f1.
If we have a Dirichlet boundary condition specified on the left of the domain p0 = β, then the equation for p1 becomes p2 −2p1 ∆x2 = f1 − β ∆x2 .
If we have a Neumann boundary condition specified at the half grid point 1 2 (px) 1 2 = α, 2 we write the equation for p1 as p2−p1 ∆x −p1−p0 ∆x ∆x = f1.
Since (px) 1 2 = p1 −p0 ∆x + O(∆x2), the equation for p1 becomes p2 −p1 ∆x2 = f1 + α ∆x.
Let’s look at the matrix equation for the case where we have two Dirichlet boundary conditions.
−2 1 1 −2 1 ...
...
...
1 −2 1 ...
...
...
1 −2 1 1 −2 p1 p2 .
.
.
pi .
.
.
pM−1 pM = ∆x2f1 −p0 ∆x2f2 .
.
.
∆x2fi .
.
.
∆x2fM−1 ∆x2fM −pM+1 The matrix is symmetric negative definite. This is advantageous because there are fast linear solvers for such systems, e.g. the conjugate gradients method.
In the case with two Neumann boundary conditions, the matrix equation is −1 1 1 −2 1 ...
...
...
1 −2 1 ...
...
...
1 −2 1 1 −1 p1 p2 .
.
.
pi .
.
.
pM−1 pM = ∆x2f1 + ∆x(px) 1 2 ∆x2f2 .
.
.
∆x2fi .
.
.
∆x2fM−1 ∆x2fM −∆x(px)M+ 1 2 .
Notice that the matrix has changed. In particular, it is singular since it has a non-empty null space which is spanned by the vector (1, . . . , 1)T . This is problematic, but workable. It can be solved for p up to a constant, since for any solution, ⃗ p, ⃗ p + c(1, . . . , 1)T is also a solution.
In multiple dimension Poisson’s equation is ∆p = f.
In 2D the equation is pxx + pyy = f.
We again use the second order accurate central differencing to obtain the gradient components at the cell faces (px)i+1/2,j = pi+1,j −pi, j ∆x + O(∆x2) (py)i,j+1/2 = pi,j+1 −pi, j ∆y + O(∆y2) 3 from which we apply central differencing again to obtain the Laplacian (∆p)i,j = (px)i+1/2,j −(px)i−1/2,j ∆x + (py)i,j+1/2 −(py)i,j−1/2 ∆y + O(∆x2) + O(∆y2) = pi+1,j −2pi,j + pi−1,j ∆x2 + pi,j+1 −2pi,j + pi,j−1 ∆y2 + O(∆x2) + O(∆y2) = fi,j.
In 2D we need boundary conditions specified around the entire domain. If at least one boundary condition is Dirichlet, then the resulting matrix will be a banded symmetric positive definite matrix.
We can use an iterative solver such as preconditioned conjugate gradients. If all the boundary conditions are Neumann, then the matrix will have a null space, and we must ensure we have a compatible system.
If the density is spatially varying, then we must use a discretization with variable coefficients.
To simplify the notation slightly, let β = 1 ρ. Then, we can write ∇· 1 ρ∇p i,j = (∇· β∇p)i,j = βi+1/2,j(px)i+1/2,j −βi−1/2,j(px)i−1/2,j ∆x + O(∆x2) + βi,j+1/2(py)i,j+1/2 −βi,j−1/2(py)i,j−1/2 ∆y + O(∆y2) = fi,j.
Note that this will require densities at the cell walls.
1.3 Compatibility Condition Poisson’s equation with all Neumann boundary conditions must satisfy a compatibility condition for a solution to exist. The problem is given by ∆p = f in Ω ∇p · n = g on ∂Ω where n is the unit normal to the boundary. From the equation we have the relations Z Ω f dV = Z Ω ∆p dV = Z Ω ∇· ∇p dV = Z ∂Ω ∇p · n dS = Z ∂Ω g dS where the third equality follows from the divergence theorem. The compatibility condition is Z Ω f dV = Z ∂Ω g dS.
The right hand side f will be of the form f = ∇· u⋆, and g = 0. Therefore, the compatibility condition is Z Ω ∇· u⋆dV = Z ∂Ω u⋆· n dS = 0 where the first equality follows from the divergence theorem. This condition needs to be satisfied when specifying the boundary condition on u⋆in order to guarantee the existence of a solution.
4 |
10583 | https://www.sciencedirect.com/science/article/pii/S1574652606800209#! | Continuous and Interval Constraints - ScienceDirect
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Foundations of Artificial Intelligence
Volume 2, 2006, Pages 571-603
Chapter 16 - Continuous and Interval Constraints
Author links open overlay panel Frédéric Benhamou, Laurent Granvilliers
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This chapter reviews that continuous constraint solving has been widely studied in several fields of applied mathematics and computer science. In computer algebra, continuous constraints are viewed as formulas from first-order logic interpreted over the real numbers. The symbolic algorithms transform the constraint systems within the same equivalence class in the interpretation domain according to some simplification ordering. The chapter also discusses the interval analysis, which is a set extension of numerical analysis such that the floating-point numbers are replaced with the intervals. The interval approximations are defined so as to enclose the computed real quantities and the algorithms are said to be complete. In constraint programming, continuous constraints are viewed as relations. The complete solving of nonlinear systems is implemented by exhaustive search techniques that compute solution space coverings by means of multi-dimensional boxes. The search is commonly accelerated through propagation-based algorithms. It reviews that continuous and interval constraints are generally contrasted with non negative integer or more generally discrete constraints. These last constraints, sometimes also called finite domain constraints, are studied in the constraint satisfaction problems (CSP) framework and are basic components of most current constraint-based languages.
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Section snippets
Continuous Constraint Solving
Continuous constraint solving has been widely studied in several fields of applied mathematics and computer science. In computer algebra , continuous constraints are viewed as formulas from first-order logic interpreted over the real numbers. The symbolic algorithms transform the constraint systems within the same equivalence class in the interpretation domain according to some simplification ordering. These techniques, for instance Gröbner bases and quantifier
From Discrete to Continuous Constraints
In this section, we introduce interval constraint techniques, dedicated to continuous constraints solving and based on interval consistency techniques . Our goal here is to present the main intuitions by way of examples. Let us consider nonlinear equations of the form f(x 1,…, x n) = 0, where f is a real function. We also assume that the domain of every variable x k is a closed interval of real numbers I k. As a consequence, the search space, that is the set of potential
The Branch-and-Reduce Framework
The problem of solving constraint-based mathematical models over the real numbers is uncomputable in general . It is only possible to calculate approximations to the solutions by using machine arithmetic. Along these lines, the main goal of interval-based techniques is to solve combinatorial problems defined as relaxations of the exact continuous problems . More precisely, we address the problem of covering solution sets with finite sets of interval boxes of reasonable precision. This
Consistency Techniques
The approximation of consistency properties over the real numbers introduces several difficult problems such as the computation over sets of values, the control of numerical errors, the inversion of nonlinear functions, and the acceleration of slow convergence. Several techniques based on interval arithmetic have been proposed to handle these problems. Interval arithmetic is an efficient and reliable implementation of set computations that allows the propagation of interval domains
Numerical Operators
Interval analysis has been defined as an extension of numerical analysis over the intervals . In general, the main goal is to implement complete and efficient set computations in order to enclose the solution set of a given problem. In this section, we are interested in constraint solving techniques. The basic principle of the direct interval methods is to replace real arithmetic with interval arithmetic. The main idea of the iterative methods is to extend the classical operators so as to
Hybrid Techniques
There is no unique algorithm for efficiently solving constraints over the real numbers. Conversely, every algorithm is parametrized by an input model, time and space complexities, efficiency conditions, and properties on the output. Depending on the context, it can be useful to combine several algorithms in a super algorithm . However, the work of designing hybrid strategies may be very difficult since that may demand a lot of experience on the algorithms to be combined. In the following,
Extending Interval Constraints
Among the possible generalizations of the interval constraint framework, some extend the expressive power (modeling) while others improve the efficiency (solving). The first category includes optimization, differential equations, mixed constraints, and quantified continuous constraints. The second category is mainly concerned with solver cooperation and in particular numerical-symbolic cooperation. The optimization aspect was for example developed in the design and implementation of the systems
Applications
Interval constraints have recently been used in various application fields with different goals: to prove that a given problem is not satisfiable, to compute a numerical enclosure of the solution set, or to derive global optima according to some objective function.
In engineering conceptual design, the goal is to generate classes of solutions that satisfy a given high-level specification of the product to be designed . The architecture of the product can be described by
Conclusion
This chapter is primarily devoted to giving a broad overview on the basics of continuous and interval constraint solving and to showing the main similarities and differences with discrete constraints. Based on the same theoretical framework (fixed point computations over complete lattices) the algorithmic approach differs in taking much of its foundations in numerical analysis when discrete constraints rely on graph theory and integer programming. We have shown how a number of algorithms from
Acknowledgements
We are indebted to many researchers with whom we have worked and have had fruitful discussions on continuous and intervals constraints these last years. In particular, we would like to thank wholeheartedly Pedro Barahona, Alexandre Goldsztejn, Luc Jaulin, Michel Rueher, Peter van Beek and Pascal Van Hentenryck for their careful reading and their numerous remarks on previous versions of this chapter.
Volume chapters Recommended articles
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Local Consistency for Ternary Numeric Constraints
Eugene C. Freuder
Synthesizing Constraint Expressions
Communications of the ACM (1978)
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Computing reachable sets for uncertain nonlinear hybrid systems using interval constraint-propagation techniques
2011, Nonlinear Analysis Hybrid Systems Show abstract We investigate solution techniques for numerical constraint-satisfaction problems and validated numerical set integration methods for computing reachable sets of nonlinear hybrid dynamical systems in the presence of uncertainty. To use interval simulation tools with higher-dimensional hybrid systems, while assuming large domains for either initial continuous state or model parameter vectors, we need to solve the problem of flow/sets intersection in an effective and reliable way. The main idea developed in this paper is first to derive an analytical expression for the boundaries of continuous flows, using interval Taylor methods and techniques for controlling the wrapping effect. Then, the event detection and localization problems underlying flow/sets intersection are expressed as numerical constraint-satisfaction problems, which are solved using global search methods based on branch-and-prune algorithms, interval analysis and consistency techniques. The method is illustrated with hybrid systems with uncertain nonlinear continuous dynamics and nonlinear invariants and guards.
### Incremental Linearization for satisfiability and verification modulo nonlinear arithmetic and transcendental functions
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### Set-based design by simulation of usage scenario coverage
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### Taylor model flowpipe construction for non-linear hybrid systems
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### Providing a basin of attraction to a target region of polynomial systems by computation of Lyapunov-like functions
2010, SIAM Journal on Control and Optimization
### SAT modulo ODE: A direct SAT approach to hybrid systems
2008, Lecture Notes in Computer Science Including Subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics
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Copyright © 2006 Elsevier B.V. All rights reserved.
Part of volume
Handbook of Constraint Programming
Edited by
Francesca Rossi, Peter van Beek, Toby Walsh
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10584 | https://askfilo.com/user-question-answers-smart-solutions/1-find-a-counter-example-if-a-b-then-a2-b2-2-give-a-direct-3337393239303130 | Find a counter example: If a > b then a² > b² Give a direct proof that if..
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Find a counter example: If a > b then a² > b² Give a direct pr
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Find a counter example: If a > b then a² > b²
Give a direct proof that if a and b are odd integers then a + b is even.
Prove using contrapositive that if x² - 4 < 0, then -2<x<2.
Views: 5,884 students
Updated on: Sep 7, 2025
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Text solutionVerified
These are three different questions related to proofs in mathematics involving inequalities and properties of integers.
Question 1: Find a counter example: If a > b then a² > b²
Solution:
The statement "If a > b then a² > b²" is not always true.
Counterexample: Let a = -1 and b = -2.
We see that a > b because -1 > -2.
But a² = (-1)² = 1 and b² = (-2)² = 4.
Here, a² = 1 which is NOT greater than b² = 4.
This shows the statement is false when a and b are negative numbers.
Question 2: Give a direct proof that if a and b are odd integers then a + b is even.
Solution:
Let a and b be odd integers.
By definition of odd integer, there exist integers m and n such that: a=2 m+1 b=2 n+1
Now, consider the sum: a+b=(2 m+1)+(2 n+1)=2 m+2 n+2=2(m+n+1)
Since m+n+1 is an integer, call it k=m+n+1.
Hence: a+b=2 k
This means a+b is divisible by 2 and thus even.
Therefore, if a and b are odd integers, then their sum is even.
Question 3: Prove using contrapositive that if x 2−4<0, then −2<x<2.
Solution:
We want to prove:
If x 2−4<0, then −2<x<2.
Step 1: Write the contrapositive statement:
Original: If x 2−4<0 then −2<x<2.
Contrapositive: If x≤−2 or x≥2, then x 2−4≥0.
Step 2: Prove the contrapositive statement.
Consider x≤−2: x 2≥(−2)2=4⟹x 2−4≥0
Consider x≥2: x 2≥2 2=4⟹x 2−4≥0
Therefore, in both cases: x 2−4≥0
Step 3: By contrapositive principle, the original statement holds true.
Hence, if x 2−4<0, then −2<x<2.
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Question Text 1. Find a counter example: If a > b then a² > b² 2. Give a direct proof that if a and b are odd integers then a + b is even. 3. Prove using contrapositive that if x² - 4 < 0, then -2
Updated On Sep 7, 2025
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10585 | https://stemjock.com/STEM%20Books/Griffiths%20QM%203e/Chapter%202/GriffithsQMCh2p34.pdf?srsltid=AfmBOopqPiyfAn__THVoxdIMrSNqE5PkS6NQSaFu94sAI-GKBnoktk_G | Griffiths Quantum Mechanics 3e: Problem 2.34 Page 1 of 7
Problem 2.34
Consider the “step” potential: 53
V (x) =
{
0, x ≤ 0,V0, x > 0.
(a) Calculate the reflection coefficient, for the case E < V 0, and comment on the answer.
(b) Calculate the reflection coefficient for the case E > V 0.
(c) For a potential (such as this one) that does not go back to zero to the right of the barrier, the transmission coefficient is not simply |F |2/|A|2 (with A the incident amplitude and F
the transmitted amplitude), because the transmitted wave travels at a different speed . Show that
T =
√ E − V0
E
|F |2
|A|2 , (2.175) for E > V 0. Hint: You can figure it out using Equation 2.99, or—more elegantly, but less informatively—from the probability current (Problem 2.18). What is T , for E < V 0?
(d) For E > V 0, calculate the transmission coefficient for the step potential, and check that
T + R = 1.
Solution
The governing equation for the wave function Ψ( x, t ) is the Schr¨ odinger equation.
iℏ ∂Ψ
∂t = − ℏ2
2m∂2Ψ
∂x 2 + V (x, t )Ψ( x, t ), −∞ < x < ∞, t > 0For this step potential,
V (x, t ) = V (x) =
{
0 if x ≤ 0
−V0 if x > 0 ,
which means the PDE becomes
iℏ ∂Ψ
∂t = − ℏ2
2m∂2Ψ
∂x 2 + V (x)Ψ( x, t ).
Since information about the eigenstates and their corresponding energies is desired, the method of separation of variables is opted for. This method works because Schr¨ odinger’s equation and its associated boundary conditions (Ψ and its derivatives tend to zero as x → ±∞ ) are linear and homogeneous. Assume a product solution of the form Ψ( x, t ) = ψ(x)φ(t) and plug it into the PDE.
iℏ ∂∂t [ψ(x)φ(t)] = − ℏ2
2m∂2
∂x 2 [ψ(x)φ(t)] + V (x)[ ψ(x)φ(t)] Evaluate the derivatives.
iℏψ(x)φ′(t) = − ℏ2
2m ψ′′ (x)φ(t) + V (x)ψ(x)φ(t)
53 For interesting commentary see C. O. Dib and O. Orellana, Eur. J. Phys. 38 , 045403 (2017).
www.stemjock.com Griffiths Quantum Mechanics 3e: Problem 2.34 Page 2 of 7 Divide both sides by ψ(x)φ(t) in order to separate variables.
iℏ φ′(t)
φ(t) = − ℏ2
2mψ′′ (x)
ψ(x) + V (x)The only way a function of t can be equal to a function of x is if both are equal to a constant E.
iℏ φ′(t)
φ(t) = − ℏ2
2mψ′′ (x)
ψ(x) + V (x) = E
As a result of using the method of separation of variables, the Schr¨ odinger equation has reduced to two ODEs, one in x and one in t.
iℏ φ′(t)
φ(t) = E
− ℏ2
2mψ′′ (x)
ψ(x) + V (x) = E
Values of E for which the boundary conditions are satisfied are called the eigenvalues (or eigenenergies in this context), and the nontrivial solutions associated with them are called the eigenfunctions (or eigenstates in this context). The ODE in x is known as the time-independent Schr¨ odinger equation (TISE) and can be written as
d2ψdx 2 = 2m
ℏ2 [V (x) − E]ψ.
Split up the ODE over the intervals that V (x) is defined on.
d2ψdx 2 = 2m
ℏ2 (−E)ψ, x ≤ 0 d2ψdx 2 = 2m
ℏ2 (−V0 − E)ψ, x > 0The solution for ψ on the interval x > 0 depends on whether V0 − E > 0, V0 − E = 0, or
V0 − E < 0. Each of these cases will be examined in turn.
Note that scattering states correspond to E > 0. In each case, the aim is to determine the reflection and transmission coefficients.
www.stemjock.com Griffiths Quantum Mechanics 3e: Problem 2.34 Page 3 of 7 By definition, the reflection coefficient is the ratio of the reflected probability current to the incident probability current. Also, the transmission coefficient is defined to be the ratio of the transmitted probability current to the incident probability current.
R =
∣∣∣∣
reflected probability current incident probability current
∣∣∣∣ T =
∣∣∣∣
transmitted probability current incident probability current
∣∣∣∣
The probability current is [noting that φ(t) = e−iEt/ ℏ]
J(x, t ) = iℏ
2m
(
Ψ ∂Ψ∗
∂x − Ψ∗ ∂Ψ
∂x
)
= iℏ
2m
{
[ψ(x)e−iEt/ ℏ] ∂∂x [ψ∗(x)eiEt/ ℏ] − [ψ∗(x)eiEt/ ℏ] ∂∂x [ψ(x)e−iEt/ ℏ]
}
= iℏ
2m
{
[ψ(x)e−iEt/ ℏ] dψ ∗
dx eiEt/ ℏ − [ψ∗(x)eiEt/ ℏ] dψ dx e−iEt/ ℏ
}
= iℏ
2m
(
ψ dψ ∗
dx − ψ∗ dψ dx
)
.
Case I: V0 − E > 0
d2ψdx 2 = − 2mE
ℏ2 ψ, x ≤ 0 d2ψdx 2 = 2m
ℏ2 (V0 − E)ψ, x > 0In this case, the general solution on x > 0 can be written in terms of exponential functions.
ψ(x) =
{
Ae ikx + Be −ikx if x ≤ 0
F e x + Ge −x if x > 0Here
k =
√2mE
ℏ and ` =
√2m(V0 − E)
ℏ .
In order to satisfy the boundary condition as x → ∞ , set F = 0.
ψ(x) =
{
Ae ikx + Be −ikx if x ≤ 0
Ge −`x if x > 0As a result, the reflection and transmission coefficients are
R =
∣∣∣∣∣
iℏ
2m
[(Be −ikx ) ddx (B∗eikx ) − (B∗eikx ) ddx (Be −ikx )]
iℏ
2m
[(Ae ikx ) ddx (A∗e−ikx ) − (A∗e−ikx ) ddx (Ae ikx )]∣∣∣∣∣ T =
∣∣∣∣∣
iℏ
2m
[(Ge −x ) ddx (G∗e−x ) − (G∗e−x ) ddx (Ge −x )]
iℏ
2m
[(Ae ikx ) ddx (A∗e−ikx ) − (A∗e−ikx ) ddx (Ae ikx )]∣∣∣∣∣
=
∣∣∣∣
(Be −ikx )( ikB ∗eikx ) − (B∗eikx )( −ikBe −ikx )(Ae ikx )( −ikA ∗e−ikx ) − (A∗e−ikx )( ikAe ikx )
∣∣∣∣ =
∣∣∣∣
(Ge −x )( −G ∗e−x ) − (G∗e−x )( −Ge −x )(Ae ikx )( −ikA ∗e−ikx ) − (A∗e−ikx )( ikAe ikx )
∣∣∣∣
=
∣∣∣∣
2ikBB ∗
−2ikAA ∗
∣∣∣∣ =
∣∣∣∣
−e −2x GG ∗ + e −2x GG ∗
−2ikAA ∗
∣∣∣∣
= |B|2
|A|2 = 0 .
www.stemjock.com Griffiths Quantum Mechanics 3e: Problem 2.34 Page 4 of 7 To determine one of the constants, require the wave function [and consequently ψ(x)] to be continuous at x = 0. lim
x→0−
ψ(x) = lim
x→0+
ψ(x) : A + B = G
Integrate both sides of the TISE with respect to x from − to to determine one more.
ˆ
−
d2ψdx 2 dx =
ˆ
−
2m
ℏ2 [V (x) − E]ψ(x) dx dψ dx
∣∣∣∣
−
=
ˆ 0
−
2m
ℏ2 (−E)ψ(x) dx +
ˆ
0
2m
ℏ2 (V0 − E)ψ(x) dx
= − 2mE
ℏ2 ψ(0)
ˆ 0
−
dx + 2m
ℏ2 (V0 − E)ψ(0)
ˆ
0
dx
= − 2mE
ℏ2 ψ(0) + 2m
ℏ2 (V0 − E)ψ(0)
Take the limit as → 0.
dψ dx
∣∣∣∣
0+
0−
= 0 It turns out that ∂Ψ/∂x is continuous at x = 0 as well. lim
x→0−
dψ dx = lim
x→0+
dψ dx : ik (A − B) = −`G
Substitute the formula for G and solve for B.
ik (A − B) = −`(A + B)
B = −− ik − ik A
The reflection coefficient can now be determined.
R =
( BA
) ( BA
)∗
=
( −− ik − ik
) ( −+ ik + ik
)
= `2 + k2
`2 + k2 = 1 Therefore, R + T = 1.
Case II: V0 − E = 0
d2ψdx 2 = − 2mV 0
ℏ2 ψ, x ≤ 0 d2ψdx 2 = 0 , x > 0In this case, the general solution on x > 0 is a straight line.
ψ(x) =
{
Ae ikx + Be −ikx if x ≤ 0
F x + G if x > 0Here
k =
√2mV 0
ℏ .
www.stemjock.com Griffiths Quantum Mechanics 3e: Problem 2.34 Page 5 of 7 In order to satisfy the boundary condition as x → ∞ , set F = 0 and G = 0.
ψ(x) =
{
Ae ikx + Be −ikx if x ≤ 00 if x > 0Use the continuity of the wave function and its first spatial derivative at x = 0 to determine two of the constants. lim
x→0−
ψ(x) = lim
x→0+
ψ(x) : A + B = 0 lim
x→0−
dψ dx = lim
x→0+
dψ dx : ik (A − B) = 0 These equations imply that A = 0 and B = 0, resulting in the trivial solution: ψ(x) = 0. Therefore, E = V0 is not an eigenvalue.
Case III: V0 − E < 0
d2ψdx 2 = − 2mE
ℏ2 ψ, x ≤ 0 d2ψdx 2 = − 2m
ℏ2 (E − V0)ψ, x > 0In this case, the general solution on x > 0 can be written in terms of complex exponential functions.
ψ(x) =
{
Ae ikx + Be −ikx if x ≤ 0
F e ilx + Ge −ilx if x > 0Here
k =
√2mE
ℏ and l =
√2m(E − V0)
ℏ .
Solving the ODE in t yields φ(t) = e−iEt/ ℏ, which means the product solution is a linear combination of waves travelling to the left and to the right.
ψ(x)φ(t) =
{
Ae i(kx −Et/ ℏ) + Be −i(kx +Et/ ℏ) if x ≤ 0
F e i(lx −Et/ ℏ) + Ge −i(lx +Et/ ℏ) if x > 0Assuming a plane wave is only incident from the left, set G = 0.
ψ(x) =
{
Ae ikx + Be −ikx if x ≤ 0
F e ilx if x > 0As a result, the reflection and transmission coefficients are
R =
∣∣∣∣∣
iℏ
2m
[(Be −ikx ) ddx (B∗eikx ) − (B∗eikx ) ddx (Be −ikx )]
iℏ
2m
[(Ae ikx ) ddx (A∗e−ikx ) − (A∗e−ikx ) ddx (Ae ikx )]∣∣∣∣∣ T =
∣∣∣∣∣
iℏ
2m
[(F e ilx ) ddx (F ∗e−ilx ) − (F ∗e−ilx ) ddx (F e ilx )]
iℏ
2m
[(Ae ikx ) ddx (A∗e−ikx ) − (A∗e−ikx ) ddx (Ae ikx )]∣∣∣∣∣
=
∣∣∣∣
(Be −ikx )( ikB ∗eikx ) − (B∗eikx )( −ikBe −ikx )(Ae ikx )( −ikA ∗e−ikx ) − (A∗e−ikx )( ikAe ikx )
∣∣∣∣ =
∣∣∣∣
(F e ilx )( −iF ∗e−ilx ) − (F ∗e−ilx )( iF e i`x )(Ae ikx )( −ikA ∗e−ikx ) − (A∗e−ikx )( ikAe ikx )
∣∣∣∣
=
∣∣∣∣
2ikBB ∗
−2ikAA ∗
∣∣∣∣ =
∣∣∣∣
−2ilF F ∗
−2ikAA ∗
∣∣∣∣
= |B|2
|A|2 = lk
|F |2
|A|2 =
√ E − V0
E
|F |2
|A|2 .
www.stemjock.com Griffiths Quantum Mechanics 3e: Problem 2.34 Page 6 of 7 To determine one of the constants, require the wave function [and consequently ψ(x)] to be continuous at x = 0. lim
x→0−
ψ(x) = lim
x→0+
ψ(x) : A + B = F
Integrate both sides of the TISE with respect to x from − to to determine one more.
ˆ
−
d2ψdx 2 dx =
ˆ
−
2m
ℏ2 [V (x) − E]ψ(x) dx dψ dx
∣∣∣∣
−
=
ˆ 0
−
2m
ℏ2 (−E)ψ(x) dx +
ˆ
0
2m
ℏ2 (V0 − E)ψ(x) dx
= − 2mE
ℏ2 ψ(0)
ˆ 0
−
dx + 2m
ℏ2 (V0 − E)ψ(0)
ˆ
0
dx
= − 2mE
ℏ2 ψ(0) + 2m
ℏ2 (V0 − E)ψ(0)
Take the limit as → 0.
dψ dx
∣∣∣∣
0+
0−
= 0 It turns out that ∂Ψ/∂x is continuous at x = 0 as well. lim
x→0−
dψ dx = lim
x→0+
dψ dx : ik (A − B) = ilF
To summarize, there are two equations to work with.
A + B = FA − B = lk F
Subtract the respective sides to get B.2B = F
(
1 − lk
)
Solve for B.
B = k − l
2k F
Add the respective sides to get A.2A = F
(
1 + lk
)
Solve for F .
F = 2kk + l A
www.stemjock.com Griffiths Quantum Mechanics 3e: Problem 2.34 Page 7 of 7 The transmission coefficient is then
T =
√ E − V0
E
|F |2
|A|2 =
√ E − V0
E
( FA
) ( FA
)∗
=
√ E − V0
E
( 2kk + l
) ( 2kk + l
)∗
=
√ E − V0
E
( 2kk + l
) ( 2kk + l
)
= 4
√ E − V0
E
(
11 + lk
) (
11 + lk
)
= 4
√ E − V0
E
11 +
√ E−V0
E
11 +
√ E−V0
E
= 4
√ E−V0
E
1 + 2
√ E−V0
E
E−V0
E
.
Combine the formulas for B and F .
B = k − l
2k F = k − l
2k
( 2kk + l A
)
= k − lk + l A
The reflection coefficient is then
R = |B|2
|A|2 =
( BA
) ( BA
)∗
=
( k − lk + l
) ( k − lk + l
)∗
=
( k − lk + l
) ( k − lk + l
)
=
( 1 − lk
1 + lk
) ( 1 − lk
1 + lk
)
=
1 −
√ E−V0
E
1 +
√ E−V0
E
1 −
√ E−V0
E
1 +
√ E−V0
E
= 1 − 2
√ E−V0
E
E−V0
E
1 + 2
√ E−V0
E
E−V0
E
.
Therefore,
R + T = 1 − 2
√ E−V0
E
E−V0
E
1 + 2
√ E−V0
E
E−V0
E
4
√ E−V0
E
1 + 2
√ E−V0
E
E−V0
E
= 1 + 2
√ E−V0
E
E−V0
E
1 + 2
√ E−V0
E
E−V0
E
= 1 .
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10586 | https://mathworld.wolfram.com/MiceProblem.html | Mice Problem -- from Wolfram MathWorld
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Mice Problem
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In the mice problem, also called the beetle problem, mice start at the corners of a regular -gon of unit side length, each heading towards its closest neighboring mouse in a counterclockwise direction at constant speed. The mice each trace out a logarithmic spiral, meet in the center of the polygon, and travel a distance
The first few values for , 3, ..., are
giving the numerical values 0.5, 0.666667, 1, 1.44721, 2, 2.65597, 3.41421, 4.27432, 5.23607, .... The curve formed by connecting the mice at regular intervals of time is an attractive figure called a whirl.
The problem is also variously known as the (three, four, etc.) (bug, dog, etc.) problem. It can be generalized to irregular polygons and mice traveling at differing speeds (Bernhart 1959). Miller (1871) considered three mice in general positions with speeds adjusted to keep paths similar and the triangle similar to the original.
See also
Apollonius Pursuit Problem, Brocard Points, Pursuit Curve, Spiral, Tractrix, Whirl
Explore with Wolfram|Alpha
More things to try:
circle involute
10^39
functional equation for zeta(s)
References
Bernhart, A. "Polygons of Pursuit." Scripta Math.24, 23-50, 1959.Brocard, H. "Solution of Lucas's Problem." Nouv. Corresp. Math.3, 280, 1877.Clapham, A.J. Rec. Math. Mag., Aug. 1962.Gardner, M. The Scientific American Book of Mathematical Puzzles and Diversions. New York: NY: Simon and Schuster, 1959.Gardner, M. The Sixth Book of Mathematical Games from Scientific American. Chicago, IL: University of Chicago Press, pp.240-243, 1984.Good, I.J. "Pursuit Curves and Mathematical Art." Math. Gaz.43, 34-35, 1959.Lucas, E. "Problem of the Three Dogs." Nouv. Corresp. Math.3, 175-176, 1877.Madachy, J.S. Madachy's Mathematical Recreations. New York: Dover, pp.201-204, 1979.Marshall, J.A.; Broucke, M.E.; and Francis, B.A. "Pursuit Formations of Unicycles." Automata41, 2005. R.K. Problem 16. Cambridge Math. Tripos Exam. January 5, 1871.Nester, D. "Mathematics Seminar: Beetle Centers of Triangles." H. Mathematical Snapshots, 3rd ed. New York: Dover, p.136, 1999.Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, pp.201-202, 1991.Wilson, J. "Problem: Four Dogs."
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Weisstein, Eric W. "Mice Problem." From MathWorld--A Wolfram Resource.
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10587 | https://www.statista.com/statistics/1031181/sulfur-production-globally-by-country/?srsltid=AfmBOopFfottVpCeA9VjTfgLZIk93DMYER5qN0NTHpudEPUJwqpnjgFt | Industry Overview
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Global sulfur production volume 2024, by country
In 2024, China produced around 19 million metric tons of sulfur, making it by far the world's leading sulfur producer. Sulfur is one of the most common chemical elements found in nature. It is a pale yellow, tasteless, and odorless brittle solid commonly found in volcanic regions and hot springs. Nowadays very little sulfur is mined from nature, as most of the production is recovered due to environmental reasons. Byproduct elemental sulfur recovered from natural gas and petroleum is the main source of sulfur worldwide.
Sulfur in the U.S.
The United States is the second-largest producer of sulfur worldwide, ahead of Saudi Arabia and Russia. In 2024, sulfur production in the U.S. reached an estimated 8.2 million metric tons. From that amount, roughly 91 percent was recovered elemental sulfur. Apart from being one of the world's main sulfur producers, the United States’ sulfur consumption is also among the highest in the world, reaching nearly 10 million metric tons in 2023.
Global sulfuric acid market
Most of the sulfur produced worldwide is transformed into sulfuric acid, an important commodity chemical with many applications. Sulfuric acid is extensively used as a battery component, in fertilizer manufacturing, metal ore leaching, and as an intermediate chemical in the production of numerous other compounds. In 2022, the global market volume of sulfuric acid amounted to around 265 million metric tons, and it is forecast to surpass 321 million metric tons by 2030. By that same year, the market value of sulfuric acid is expected to surpass 14 billion U.S. dollars.
Production of sulfur worldwide in 2024, by country (in 1,000 metric tons)
| Characteristic | Production in thousand metric tons |
--- |
| China | 19,000 |
| United States | 8,200 |
| Saudi Arabia | 7,500 |
| Russia | 7,500 |
| United Arab Emirates | 6,000 |
| Kazakhstan | 5,100 |
| Canada | 5,000 |
| India | 3,700 |
| Japan | 3,100 |
| South Korea | 3,100 |
| Qatar | 3,100 |
| Iran | 2,000 |
| Chile | 1,300 |
| Kuwait | 1,300 |
| Poland | 1,100 |
| Australia | 900 |
| Turkmenistan | 900 |
| Rest of world | 6,400 |
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Chinese sulfur production includes byproduct elemental sulfur that has been recovered from petroleum and natural gas, as well as the estimated sulfur content of byproduct sulfuric acid from metallurgy, and the sulfur content of sulfuric acid from pyrite.
All figures are estimates.
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Production volume of sulfuric acid China 2014-2024
Major chemicals production volume in India FY 2013-2024
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10589 | https://courses.lumenlearning.com/calculus2/chapter/graphs-and-symmetry-of-polar-curves/ | Graphs and Symmetry of Polar Curves | Calculus II
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Calculus II
Module 7: Parametric Equations and Polar Coordinates
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Graphs and Symmetry of Polar Curves
Learning Outcomes
Sketch polar curves from given equations
Convert equations between rectangular and polar coordinates
Identify symmetry in polar curves and equations
Polar Curves
Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function y=f(x)y=f(x) and create a curve in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function r=f(θ)r=f(θ).
The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Start with a list of values for the independent variable (θ θ in this case) and calculate the corresponding values of the dependent variable r r. This process generates a list of ordered pairs, which can be plotted in the polar coordinate system. Finally, connect the points, and take advantage of any patterns that may appear. The function may be periodic, for example, which indicates that only a limited number of values for the independent variable are needed.
Problem-Solving Strategy: Plotting a Curve in Polar Coordinates
Create a table with two columns. The first column is for θ θ, and the second column is for r r.
Create a list of values for θ θ.
Calculate the corresponding r r values for each θ θ.
Plot each ordered pair (r,θ)(r,θ) on the coordinate axes.
Connect the points and look for a pattern.
Interactive
Watch this video for more information on sketching polar curves.
Example: Graphing a Function in Polar Coordinates
Graph the curve defined by the function r=4 sin θ r=4 sinθ. Identify the curve and rewrite the equation in rectangular coordinates.
Show Solution
Because the function is a multiple of a sine function, it is periodic with period 2 π 2 π, so use values for θ θ between 0 and 2 π 2 π. The result of steps 1–3 appear in the following table. Figure 5 shows the graph based on this table.
| θ θ | r=4 sin θ r=4 sinθ | | θ θ | r=4 sin θ r=4 sinθ |
--- ---
| 0 0 | 0 0 | | π π | 0 0 |
| π 6 π 6 | 2 2 | 7 π 6 7 π 6 | −2−2 |
| π 4 π 4 | 2√2≈2.8 2 2≈2.8 | 5 π 4 5 π 4 | −2√2≈−2.8−2 2≈−2.8 |
| π 3 π 3 | 2√3≈3.4 2 3≈3.4 | 4 π 3 4 π 3 | −2√3≈−3.4−2 3≈−3.4 |
| π 2 π 2 | 4 4 | 3 π 2 3 π 2 | 4 4 |
| 2 π 3 2 π 3 | 2√3≈3.4 2 3≈3.4 | 5 π 3 5 π 3 | −2√3≈−3.4−2 3≈−3.4 |
| 3 π 4 3 π 4 | 2√2≈2.8 2 2≈2.8 | 7 π 4 7 π 4 | −2√2≈−2.8−2 2≈−2.8 |
| 5 π 6 5 π 6 | 2 2 | 11 π 6 11 π 6 | −2−2 |
| | | 2 π 2 π | 0 0 |
Figure 5. The graph of the function r=4 sin θ r=4 sinθ is a circle.
This is the graph of a circle. The equation r=4 sin θ r=4 sinθ can be converted into rectangular coordinates by first multiplying both sides by r r. This gives the equation r 2=4 r sin θ r 2=4 r sinθ. Next use the facts that r 2=x 2+y 2 r 2=x 2+y 2 and y=r sin θ y=r sinθ. This gives x 2+y 2=4 y x 2+y 2=4 y. To put this equation into standard form, subtract 4 y 4 y from both sides of the equation and complete the square:
x 2+y 2−4 y=0 x 2+(y 2−4 y)=0 x 2+(y 2−4 y+4)=0+4 x 2+(y−2)2=4.x 2+y 2−4 y=0 x 2+(y 2−4 y)=0 x 2+(y 2−4 y+4)=0+4 x 2+(y−2)2=4.
This is the equation of a circle with radius 2 and center (0,2)(0,2) in the rectangular coordinate system.
Watch the following video to see the worked solution to Example: Graphing a Function in Polar Coordinates.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “7.3 Polar Coordinates” here (opens in new window).
try it
Create a graph of the curve defined by the function r=4+4 cos θ r=4+4 cosθ.
Hint
Follow the problem-solving strategy for creating a graph in polar coordinates.
Show Solution
Figure 6.
The name of this shape is a cardioid, which we will study further later in this section.
The graph in the previous example was that of a circle. The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in the theorem. The example after the next gives some more examples of functions for transforming from polar to rectangular coordinates.
Example: Transforming Polar Equations to Rectangular Coordinates
Rewrite each of the following equations in rectangular coordinates and identify the graph.
θ=π 3 θ=π 3
r=3 r=3
r=6 cos θ−8 sin θ r=6 cosθ−8 sinθ
Show Solution
Take the tangent of both sides. This gives tan θ=tan(π 3)=√3 tanθ=tan(π 3)=3. Since tan θ=y x tanθ=y x we can replace the left-hand side of this equation by y x y x. This gives y x=√3 y x=3, which can be rewritten as y=x√3 y=x 3. This is the equation of a straight line passing through the origin with slope √3 3. In general, any polar equation of the form θ=K θ=K represents a straight line through the pole with slope equal to tan K tanK.
First, square both sides of the equation. This gives r 2=9 r 2=9. Next replace r 2 r 2 with x 2+y 2 x 2+y 2. This gives the equation x 2+y 2=9 x 2+y 2=9, which is the equation of a circle centered at the origin with radius 3. In general, any polar equation of the form r=k r=k where k is a positive constant represents a circle of radius k centered at the origin. (Note: when squaring both sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, (−3,π 3)(−3,π 3) is the same point as (3,4 π 3)(3,4 π 3).)
Multiply both sides of the equation by r r. This leads to r 2=6 r cos θ−8 r sin θ r 2=6 r cosθ−8 r sinθ. Next use the formulas
r 2=x 2+y 2,x=r cos θ,y=r sin θ r 2=x 2+y 2,x=r cosθ,y=r sinθ.
This gives
r 2=6(r cos θ)−8(r sin θ)x 2+y 2=6 x−8 y.r 2=6(r cosθ)−8(r sinθ)x 2+y 2=6 x−8 y.
To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.
x 2+y 2=6 x−8 y x 2−6 x+y 2+8 y=0(x 2−6 x)+(y 2+8 y)=0(x 2−6 x+9)+(y 2+8 y+16)=9+16(x−3)2+(y+4)2=25.x 2+y 2=6 x−8 y x 2−6 x+y 2+8 y=0(x 2−6 x)+(y 2+8 y)=0(x 2−6 x+9)+(y 2+8 y+16)=9+16(x−3)2+(y+4)2=25.
This is the equation of a circle with center at (3,−4)(3,−4) and radius 5. Notice that the circle passes through the origin since the center is 5 units away.
Watch the following video to see the worked solution to Example: Transforming Polar Equations to Rectangular Coordinates.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “7.3 Polar Coordinates” here (opens in new window).
try it
Rewrite the equation r=sec θ tan θ r=secθ tanθ in rectangular coordinates and identify its graph.
Hint
Convert to sine and cosine, then multiply both sides by cosine.
Show Solution
y=x 2 y=x 2, which is the equation of a parabola opening upward.
We have now seen several examples of drawing graphs of curves defined by polar equations. A summary of some common curves is given in the tables below. In each equation, a and b are arbitrary constants.
Figure 7.
Figure 8.
A cardioidis a special case of a limaçon (pronounced “lee-mah-son”), in which a=b a=b or a=−b a=−b. The rose is a very interesting curve. Notice that the graph of r=3 sin 2 θ r=3 sin2 θ has four petals. However, the graph of r=3 sin 3 θ r=3 sin3 θ has three petals as shown.
Figure 9. Graph of r=3 sin 3 θ r=3 sin3 θ.
If the coefficient of θ θ is even, the graph has twice as many petals as the coefficient. If the coefficient of θ θ is odd, then the number of petals equals the coefficient. You are encouraged to explore why this happens. Even more interesting graphs emerge when the coefficient of θ θ is not an integer. For example, if it is rational, then the curve is closed; that is, it eventually ends where it started (Figure 10 (a)). However, if the coefficient is irrational, then the curve never closes (Figure 10 (b)). Although it may appear that the curve is closed, a closer examination reveals that the petals just above the positive x axis are slightly thicker. This is because the petal does not quite match up with the starting point.
Figure 10. Polar rose graphs of functions with (a) rational coefficient and (b) irrational coefficient. Note that the rose in part (b) would actually fill the entire circle if plotted in full.
Since the curve defined by the graph of r=3 sin(π θ)r=3 sin(π θ) never closes, the curve depicted in Figure 10 (b) is only a partial depiction. In fact, this is an example of a space-filling curve. A space-filling curve is one that in fact occupies a two-dimensional subset of the real plane. In this case the curve occupies the circle of radius 3 centered at the origin.
Suppose a curve is described in the polar coordinate system via the function r=f(θ)r=f(θ). Since we have conversion formulas from polar to rectangular coordinates given by
x=r cos θ y=r sin θ,x=r cosθ y=r sinθ,
it is possible to rewrite these formulas using the function
x=f(θ)cos θ y=f(θ)sin θ.x=f(θ)cosθ y=f(θ)sinθ.
This step gives a parameterization of the curve in rectangular coordinates using θ θ as the parameter. For example, the spiral formula r=a+b θ r=a+b θ from Figure 7 becomes
x=(a+b θ)cos θ y=(a+b θ)sin θ.x=(a+b θ)cosθ y=(a+b θ)sinθ.
Letting θ θ range from −∞−∞ to ∞∞ generates the entire spiral.
Try It
Symmetry in Polar Coordinates
When studying symmetry of functions in rectangular coordinates (i.e., in the form y=f(x)y=f(x)), we talk about symmetry with respect to the y-axis and symmetry with respect to the origin. In particular, if f(−x)=f(x)f(−x)=f(x) for all x x in the domain of f f, then f f is an even function and its graph is symmetric with respect to the y y-axis. If f(−x)=−f(x)f(−x)=−f(x) for all x x in the domain of f f, then f f is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.
theorem: Symmetry in Polar Curves and Equations
Consider a curve generated by the function r=f(θ)r=f(θ) in polar coordinates.
The curve is symmetric about the polar axis if for every point (r,θ)(r,θ) on the graph, the point (r,−θ)(r,−θ) is also on the graph. Similarly, the equation r=f(θ)r=f(θ) is unchanged by replacing θ θ with −θ−θ.
The curve is symmetric about the pole if for every point (r,θ)(r,θ) on the graph, the point (r,π+θ)(r,π+θ) is also on the graph. Similarly, the equation r=f(θ)r=f(θ) is unchanged when replacing r r with −r−r, or θ θ with π+θ π+θ.
The curve is symmetric about the vertical line θ=π 2 θ=π 2 if for every point (r,θ)(r,θ) on the graph, the point (r,π−θ)(r,π−θ) is also on the graph. Similarly, the equation r=f(θ)r=f(θ) is unchanged when θ θ is replaced by π−θ π−θ.
The following table shows examples of each type of symmetry.
Figure 11.
Example: using Symmetry to Graph a Polar Equation
Find the symmetry of the rose defined by the equation r=3 sin(2 θ)r=3 sin(2 θ) and create a graph.
Show Solution
Suppose the point (r,θ)(r,θ) is on the graph of r=3 sin(2 θ)r=3 sin(2 θ).
To test for symmetry about the polar axis, first try replacing θ θ with −θ−θ. This gives r=3 sin(2(−θ))=−3 sin(2 θ)r=3 sin(2(−θ))=−3 sin(2 θ). Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing r r with −r−r and θ θ with π−θ π−θ yields
−r=3 sin(2(π−θ))−r=3 sin(2 π−2 θ)−r=3 sin(−2 θ)−r=−3 sin 2 θ.−r=3 sin(2(π−θ))−r=3 sin(2 π−2 θ)−r=3 sin(−2 θ)−r=−3 sin2 θ.
Multiplying both sides of this equation by −1−1 gives r=3 sin 2 θ r=3 sin2 θ, which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.
To test for symmetry with respect to the pole, first replace r r with −r−r, which yields −r=3 sin(2 θ)−r=3 sin(2 θ). Multiplying both sides by −1 gives r=−3 sin(2 θ)r=−3 sin(2 θ), which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing θ θ with θ+π θ+π gives
r=3 sin(2(θ+π))=3 sin(2 θ+2 π)=3(sin 2 θ cos 2 π+cos 2 θ sin 2 π)=3 sin 2 θ.r=3 sin(2(θ+π))=3 sin(2 θ+2 π)=3(sin2 θ cos2 π+cos2 θ sin2 π)=3 sin2 θ.
Since this agrees with the original equation, the graph is symmetric about the pole.
To test for symmetry with respect to the vertical line θ=π 2 θ=π 2, first replace both r r with −r−r and θ θ with −θ−θ.
−r=3 sin(2(−θ))−r=3 sin(−2 θ)−r=−3 sin 2 θ.−r=3 sin(2(−θ))−r=3 sin(−2 θ)−r=−3 sin2 θ.
Multiplying both sides of this equation by −1−1 gives r=3 sin 2 θ r=3 sin2 θ, which is the original equation. Therefore the graph is symmetric about the vertical line θ=π 2 θ=π 2.
This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of θ θ between 0 and π 2 π 2 and then reflect the resulting graph.
| θ θ | r r |
--- |
| 0 0 | 0 0 |
| π 6 π 6 | 3√3 2≈2.6 3 3 2≈2.6 |
| π 4 π 4 | 3 3 |
| π 3 π 3 | 3√3 2≈2.6 3 3 2≈2.6 |
| π 2 π 2 | 0 0 |
This gives one petal of the rose, as shown in the following graph.
Figure 12. The graph of the equation between θ=0 θ=0 and θ=π 2 θ=π 2.
Reflecting this image into the other three quadrants gives the entire graph as shown.
Figure 13. The entire graph of the equation is called a four-petaled rose.
Watch the following video to see the worked solution to Example: using Symmetry to Graph a Polar Equation.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “7.3 Polar Coordinates” here (opens in new window).
try it
Determine the symmetry of the graph determined by the equation r=2 cos(3 θ)r=2 cos(3 θ) and create a graph.
Hint
Use the theorem.
Show Solution
Symmetric with respect to the polar axis.
Figure 14.
Try It
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10590 | https://kids.britannica.com/students/article/Himalayas/274884 | Himalayas
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Introduction
The highest mountain range on Earth, the Himalayas form the northern border of the Indian subcontinent in Asia. The mountains extend in a massive arc for about 1,550 miles (2,500 kilometers) from west to east with more than 30 peaks rising to heights greater than 24,000 feet (7,300 meters) above sea level. These include Mount Everest, the world’s highest peak at 29,032 feet (8,849 meters), Kanchenjunga at 28,169 feet (8,586 meters), Makalu at 27,766 feet (8,463 meters), and Dhaulagiri at 26,795 feet (8,167 meters).
Several Indian states and the kingdoms of Nepal and Bhutan lie along the southern slopes of the Himalayas, and the Tibetan Highlands border them in the north. The width of the mountain system varies from 125 to 250 miles (200 to 400 kilometers) from south to north, and the average height is 20,000 feet (6,100 meters). The Himalayas extend over about 229,500 square miles (594,400 square kilometers). India, Nepal, and Bhutan have sovereignty over most of them; Pakistan and China also occupy parts.
The Sanskrit name Himalayas, meaning “abode of snow,” truly characterizes the vast permanent snowfields above the snow line. These mountains pose the greatest challenge in the world to mountaineers.
Physical Characteristics
The most characteristic features of the Himalayas are their great height, complex geologic structure, snowcapped peaks, large valley glaciers, deep river gorges, and rich vegetation. From south to north the Himalayan ranges can be grouped into four parallel belts of varying width—these are the Outer, or Sub-, Himalayas; the Lesser, or Lower, Himalayas; the Great, or Higher, Himalayas; and the Tethys, or Tibetan, Himalayas. The Karakoram Range in the northwest is also sometimes considered part of the Himalayan system. The mountains can be divided broadly into three regions. The backbone of the system is the Great Himalayas, a single range rising above the snow line with nine of the 14 highest peaks in the world, including Mount Everest.
Geologically the Himalayas are relatively young folded mountains and are still undergoing the mountain-building process. Precambrian metamorphic rocks—rocks formed by heat and pressure from 4.6 billion to 570 million years ago—make up much of the structure. The uplift took place in at least three phases. The first phase occurred at the close of the Eocene epoch (about 33 million years ago) when the Great and Tethys Himalayas were uplifted. In the second phase, which occurred in the Miocene epoch (approximately 23 million to 5.3 million years ago), ranges of the Lesser Himalayas were formed. The final mountain-building phase started in the late Neogene period (about 7 million years ago) when the Siwalik Range, the foothills of the Outer Himalayas, were formed.
The Himalayas act as a great divide and influence the climatic conditions of the Indian subcontinent to the south and of the Central Asian highland to the north. The winter season lasts from October to February, the summer from March to June, and the rainy season from June to September. Climate varies considerably with altitude; the snow line generally lies at about 16,000 feet (4,900 meters) in the Great Himalayas. The annual and daily temperature variation is much greater in the foothills.
The mountain ranges obstruct the cold, dry air from the north into India in winter. They also force the monsoonal winds to give up moisture, causing heavy rain and snow on the Indian side but arid conditions in Tibet. Rainfall decreases from east to west—120 to 60 inches (300 to 150 centimeters). Cherrapunji in Meghalaya state in northeastern India is noted for the world’s second highest average annual rainfall of 450 inches (1,140 centimeters).
The Himalayas are drained by 19 major rivers, of which the Indus and the Brahmaputra are the largest. The Jhelum, Chenab, Ravi, Beas, and Sutlej belong to the Indus system; the Yamuna, Ramganga, Kali, Gandak, and Kosi are part of the Ganges system; and the Tista, Raidak, and Manas belong to the Brahmaputra system. Rivers are more numerous and extensive on the southern slopes of the Himalayas and have great potential for producing hydroelectric power. The Bhakra Nangal multipurpose river-valley project, located on the Sutlej River, is one of the most extensive in India. Such major rivers as the Indus, Sutlej, and Brahmaputra have narrow and deep upper valleys that are older than the mountains themselves. Glaciers cover more than 12,700 square miles (32,900 square kilometers). One of the largest is Gangotri glacier in northern India—20 miles (32 kilometers) long. Glaciers feed most of the upper courses of the rivers, while the middle and lower courses are fed by rain. There are several freshwater lakes as well.
Plants and Animals
There is great variation in the Himalayan soils. The dark brown soils are well suited for growing fruit trees. The wet, deep, upland soils with high humus content—especially in the Darjeeling and Assam hills—are good for growing tea. Himalayan vegetation is based on altitude and rainfall and can be classified into four groups: tropical evergreen forests of rose chestnut, bamboo, alder, pine, laurel, and palm up to about 3,940 feet (1,200 meters); subtropical deciduous forest with sal, oak, and magnolia up to 7,220 feet (2,200 meters); temperate forests of cedar, birch, hazel, maple, and spruce from 7,220 to 8,860 feet (2,200 to 2,700 meters); and the alpine zone with juniper, rhododendron, mosses, lichens, and several kinds of flowering plants from 8,860 to 11,800 feet (2,700 to 3,600 meters). Alpine meadows are found up to 16,400 feet (5,000 meters).
Elephants, bison, and rhinoceroses inhabit the forested lower slopes of the Outer Himalayas. Snow leopards, brown bears, red pandas, and Tibetan yaks are found above the tree line—above 10,000 feet (3,050 meters). Black bears, langur monkeys, clouded leopards, and goat antelopes live in the foothills. Several animal species, such as the Indian rhinoceros, musk deer, and Kashmir stag, or hangul, were at the point of extinction but are now protected in several national parks and sanctuaries in India. There are catfish in most Himalayan streams, and butterflies are extremely varied and beautiful.
People and Economy
The people who inhabit the Great and the Tethys Himalayas are primarily of Tibeto-Burman descent, while the Lesser Himalayas are populated by people who trace their roots to Indo-European ancestors. The Gaddis are a hill people who herd sheep and goats. During winter they descend to the lowlands in search of food for their herds, but in summer they return to the higher pastures. The Gujars are also a migrating pastoral people. The major ethnic groups of Nepal are the Newars, Tamangs, Gurangs, Sherpas, and Gurkhas. The Sherpas, who live to the south of Mount Everest, are famous mountaineers. Major Himalayan summer resorts are at Almora, Darjeeling, Mussoorie, Naini Tal, Shimla, and Srinagar in India as well as Murree in Pakistan.
Economic resources abound in the Himalayas, including rich arable land, extensive grasslands and forests, workable mineral deposits, and tremendous potential for easily harnessed hydroelectric power. Terraced cultivation is carried on as high as 8,200 feet (2,500 meters). Rice, corn, wheat, millet, jute, sugarcane, and oilseeds are the major crops. Most of the fruit orchards—producing apples, peaches, pears, and cherries—are in the Kashmir and the Kulu valleys. Rich vineyards on the shores of Dal Lake in Kashmir produce grapes of good quality. Saffron, walnuts, and almonds are also grown in the Vale of Kashmir. Tea gardens abound in the foothills of the Darjeeling district. There are also plantations of cardamom and medicinal herbs. Sheep, goats, and yaks are raised on the rough grazing lands.
Mineral deposits include coal, bauxite, mica, gypsum, sapphires, petroleum, natural gas, chromite, copper, iron ore, borax, sulfur, graphite, lead, and zinc. Some alluvial gold is found in the Indus valley.
Poor transportation facilities in the Himalayas have acted as a barrier to economic growth. Only in the late 20th century were highways constructed to make the Himalayan region accessible from both north and south. Kathmandu, the capital of Nepal, has an international airport. Srinagar, the summer capital of the Indian-controlled part of Kashmir, has a domestic airport. There are only two narrow-gauge railroads from the northern plains of India into the Lesser Himalayas—one from Kalka to Shimla and the other from Shiliguri to Darjeeling.
The Himalayas were mapped for the first time in 1590 by a Spanish missionary to the court of the Mughal emperor Akbar. The heights of the Himalayan peaks were first measured correctly in the middle of the 19th century. It was not realized until 1856 that Mount Everest is higher than any other peak in the world. Modern maps of the Himalayas have been prepared by Indian and German geographers and cartographers.
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10592 | https://ell.stackexchange.com/questions/107732/when-do-i-say-food-or-foods | singular vs plural - When do I say "food" or "foods"? - English Language Learners Stack Exchange
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When do I say "food" or "foods"?
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As I have read that the noun "food" is generally uncountable, I am not quite sure when I should use "food" as a countable noun.
For instance,
Chicken and rice are food
or
Chicken and rice are foods
Could I use a number in front of foods?
Chicken and rice are two foods?
Should we use food or foods here?
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edited Aug 12, 2021 at 13:25
ColleenV
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asked Oct 28, 2016 at 3:09
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8 "Food" is always a mass noun except when it means "kinds of food". So: "High carb foods", "Gluten-free foods." "Chicken and rice are food", but ""Chicken and rice are nutritious foods".P. E. Dant Reinstate Monica –P. E. Dant Reinstate Monica 2016-10-28 03:20:05 +00:00 Commented Oct 28, 2016 at 3:20
1 @P.E.Dant you should make your comment into an answer. It's +1 from me, if you do.Mari-Lou A –Mari-Lou A♦ 2016-10-28 11:24:09 +00:00 Commented Oct 28, 2016 at 11:24
@P.E.Dant your comment has five upvotes. Please, just copy your comment in an answer.Mari-Lou A –Mari-Lou A♦ 2016-10-29 07:29:22 +00:00 Commented Oct 29, 2016 at 7:29
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It depends on what you mean.
Chicken and rice are food.
Chicken and rice are members of the category 'food'. They are stuff you can eat.
Chickens and rice are food.
This is the same thing, but referring to the natural state of chickens as individual animals and (by analogy) to rice as a plant. These things, which you can find in nature, can be eaten.
Chicken and rice are foods.
Both chicken and rice are kinds of food. Either one, taken separately, is a kind of stuff you can eat.
Chicken and rice are two foods.
This is the same thing, emphasizing that these are just two options out of a larger category.
Chickens and rice are foods.
Chicken and rice are two food.
These are just wrong. 'Chickens' are not a foodstuff but the animals themselves; they are not a kind of food but a source of it. Uncountable nouns are definitionally not able to be counted.
When to use them depends on the categories you're talking about.
Chickens are (for most people) food but (for everyone) they are an animal and not a subcategory of foodstuff (i.e., a food); chicken and rice are both food and kinds of food (i.e., foods); chicken and rice can be Vietnamese food but are not kinds of Vietnamese food (i.e., foods); com chien and banh tet are kinds of Vietnamese food (i.e., foods).
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answered Jun 15, 2018 at 20:58
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"Food" is always a mass noun, except when used to mean "kinds of food." For example:
Baby foods – the varieties of food eaten by babies.
Gluten-free foods – varieties of food that do not contain gluten.
Processed foods – ready-to-eat or pre-prepared food, often less healthy.
See also this definition, with examples, from Oxford Dictionaries.
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answered Nov 3, 2016 at 17:37
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By the way, is it foods is in such case or foods are?SovereignSun –SovereignSun 2016-11-03 18:27:21 +00:00 Commented Nov 3, 2016 at 18:27
1 @SovereignSun: "Foods are," as it's a plurality of varieties/types/kinds of food – Gluten-free foods are fruit, vegetables, potatoes, and fish, among others. LMS –LMS 2016-11-04 16:15:06 +00:00 Commented Nov 4, 2016 at 16:15
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10593 | https://people.ohio.edu/melkonia/math3050/HW/Section5.6.pdf | 290 Chapter 5 Sequences, Mathematical Induction, and Recursion 5.6 Defining Sequences Recursively So, Nat’ralists observe, a Flea/Hath smaller Fleas that on him prey,/And these have smaller Fleas to bite ’em,/And so proceed ad infinitum. — Jonathan Swift, 1733 A sequence can be defined in a variety of different ways. One informal way is to write the first few terms with the expectation that the general pattern will be obvious. We might say, for instance, “consider the sequence 3, 5, 7, . . ..” Unfortunately, misunderstandings can occur when this approach is used. The next term of the sequence could be 9 if we mean a sequence of odd integers, or it could be 11 if we mean the sequence of odd prime numbers.
The second way to define a sequence is to give an explicit formula for its nth term.
For example, a sequence a0, a1, a2 . . . can be specified by writing an = (−1)n n + 1 for all integers n ≥0.
The advantage of defining a sequence by such an explicit formula is that each term of the sequence is uniquely determined and can be computed in a fixed, finite number of steps, by substitution.
The third way to define a sequence is to use recursion, as was done in Examples 5.3.3 and 5.4.2. This requires giving both an equation, called a recurrence relation, that defines each later term in the sequence by reference to earlier terms and also one or more initial values for the sequence.
Sometimes it is very difficult or impossible to find an explicit formula for a sequence, but it is possible to define the sequence using recursion. Note that defining sequences recursively is similar to proving theorems by mathematical induction. The recurrence relation is like the inductive step and the initial conditions are like the basis step. Indeed, the fact that sequences can be defined recursively is equivalent to the fact that mathemat-ical induction works as a method of proof.
• Definition A recurrence relation for a sequence a0, a1, a2, . . . is a formula that relates each term ak to certain of its predecessors ak−1, ak−2, . . . , ak−i, where i is an integer with k −i ≥0. The initial conditions for such a recurrence relation specify the values of a0, a1, a2, . . . , ai−1, if i is a fixed integer, or a0, a1, . . . , am, where m is an integer with m ≥0, if i depends on k.
Example 5.6.1 Computing Terms of a Recursively Defined Sequence Define a sequence c0, c1, c2, . . . recursively as follows: For all integers k ≥2, (1) ck = ck−1 + kck−2 + 1 recurrence relation (2) c0 = 1 and c1 = 2 initial conditions.
Find c2, c3, and c4.
Solution c2 = c1 + 2c0 + 1 = 2 + 2·1 + 1 by substituting k = 2 into (1) since c1 = 2 and c0 = 1 by (2) Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
5.6 Defining Sequences Recursively 291 (3) ∴c2 = 5 c3 = c2 + 3c1 + 1 = 5 + 3·2 + 1 by substituting k = 3 into (1) since c2 = 5 by (3) and c1 = 2 by (2) (4) ∴c3 = 12 c4 = c3 + 4c2 + 1 = 12 + 4·5 + 1 (5) ∴c4 = 33 by substituting k = 4 into (1) since c3 = 12 by (4) and c2 = 5 by (3) ■ A given recurrence relation may be expressed in several different ways.
Example 5.6.2 Writing a Recurrence Relation in More Than One Way Let s0, s1, s2, . . . be a sequence that satisfies the following recurrence relation: for all integers k ≥1, sk = 3sk−1 −1.
Explain why the following statement is true: for all integers k ≥0, sk+1 = 3sk −1.
Note Think of the recurrence relation as s□= 3s□−1 −1, where any positive integer expression may be placed in the box.
Solution In informal language, the recurrence relation says that any term of the sequence equals 3 times the previous term minus 1. Now for any integer k ≥0, the term previous to sk+1 is sk. Thus for any integer k ≥0, sk+1 = 3sk −1.
■ A sequence defined recursively need not start with a subscript of zero. Also, a given recurrence relation may be satisfied by many different sequences; the actual values of the sequence are determined by the initial conditions.
Example 5.6.3 Sequences That Satisfy the Same Recurrence Relation Let a1, a2, a3, . . . and b1, b2, b3, . . . satisfy the recurrence relation that the kth term equals 3 times the (k −1)st term for all integers k ≥2: (1) ak = 3ak−1 and bk = 3bk−1.
But suppose that the initial conditions for the sequences are different: (2) a1 = 2 and b1 = 1.
Find (a) a2, a3, a4 and (b) b2, b3, b4.
Solution a.
a2 = 3a1 = 3·2 = 6 a3 = 3a2 = 3·6 = 18 a4 = 3a3 = 3·18 = 54 b.
b2 = 3b1 = 3·1 = 3 b3 = 3b2 = 3·3 = 9 b4 = 3b3 = 3·9 = 27 Thus a1, a2, a3, . . . begins 2, 6, 18, 54, . . . and b1, b2, b3, . . . begins 1, 3, 9, 27, . . . .
■ Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
292 Chapter 5 Sequences, Mathematical Induction, and Recursion Example 5.6.4 Showing That a Sequence Given by an Explicit Formula Satisfies a Certain Recurrence Relation The sequence of Catalan numbers, named after the Belgian mathematician Eugène Catalan (1814–1894), arises in a remarkable variety of different contexts in discrete math-ematics. It can be defined as follows: For each integer n ≥1, Cn = 1 n + 1 2n n .
a. Find C1, C2, and C3.
b. Show that this sequence satisfies the recurrence relation Ck = 4k −2 k + 1 Ck−1 for all integers k ≥2 Academie Royale de Belgique Eugène Catalan (1814–1894) Solution a. C1 = 1 2 2 1 = 1 2 ·2 = 1, C2 = 1 3 4 2 = 1 3 ·6 = 2, C3 = 1 4 6 3 = 1 4 ·20 = 5 b. To obtain the kth and (k −1)st terms of the sequence, just substitute k and k −1 in place of n in the explicit formula for C1, C2, C3, . . . .
Ck = 1 k + 1 2k k Ck+1 = 1 (k −1) + 1 2(k −1) k −1 = 1 k 2k −2) k −1 .
Then start with the right-hand side of the recurrence relation and transform it into the left-hand side: For each integer k ≥2, 4k −2 k + 1 Ck−1 = 4k −2 k + 1 1 k 2k −2) k −1 + by substituting = 2(2k −1) k + 1 · 1 k · (2k −2)!
(k −1)!(2k −2 −(k −1))!
by the formula for n choose r = 1 k + 1 ·(2(2k −1))· (2k −2)!
(k(k −1)!)(k −1)!
by rearranging the factors = 1 k + 1 ·(2(2k −1))· 1 k!(k −1)! ·(2k −2)!· 1 2 · 1 k ·2k.
because 1 2 · 1 k ·2k = 1 = 1 k + 1 · 2 2 · 1 k! · 1 (k −1)! · 1 k ·(2k)·(2k −1)·(2k −2)!
by rearranging the factors = 1 k + 1 · (2k)!
k!k!
because k(k −1)! = k!, 2 2 = 1, and 2k ·(2k −1)·(2k −2)! = (2k)!
= 1 k + 1 2k k by the formula for n choose r = Ck by defintion of C1, C2, C3, . . . .
■ Examples of Recursively Defined Sequences Recursion is one of the central ideas of computer science. To solve a problem recursively means to find a way to break it down into smaller subproblems each having the same form as the original problem—and to do this in such a way that when the process is repeated Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
5.6 Defining Sequences Recursively 293 many times, the last of the subproblems are small and easy to solve and the solutions of the subproblems can be woven together to form a solution to the original problem.
Probably the most difficult part of solving problems recursively is to figure out how knowing the solution to smaller subproblems of the same type as the original problem will give you a solution to the problem as a whole. You suppose you know the solutions to smaller subproblems and ask yourself how you would best make use of that knowledge to solve the larger problem. The supposition that the smaller subproblems have already been solved has been called the recursive paradigm or the recursive leap of faith. Once you take this leap, you are right in the middle of the most difficult part of the prob-lem, but generally, the path to a solution from this point, though difficult, is short. The recursive leap of faith is similar to the inductive hypothesis in a proof by mathematical induction.
Example 5.6.5 The Tower of Hanoi In 1883 a French mathematician, Édouard Lucas, invented a puzzle that he called The Tower of Hanoi (La Tour D’Hanoï). The puzzle consisted of eight disks of wood with holes in their centers, which were piled in order of decreasing size on one pole in a row of three. A facsimile of the cover of the box is shown in Figure 5.6.1. Those who played the game were supposed to move all the disks one by one from one pole to another, never placing a larger disk on top of a smaller one. The directions to the puzzle claimed it was based on an old Indian legend: Courtesy of Francis Lucas Édouard Lucas (1842–1891) On the steps of the altar in the temple of Benares, for many, many years Brahmins have been moving a tower of 64 golden disks from one pole to another; one by one, never placing a larger on top of a smaller. When all the disks have been transferred the Tower and the Brahmins will fall, and it will be the end of the world.
Figure 5.6.1 Courtesy of Paul Stockmeyer The puzzle offered a prize of ten thousand francs (about $34,000 US today) to anyone who could move a tower of 64 disks by hand while following the rules of the game. (See Figure 5.6.2 on the following page.) Assuming that you transferred the disks as efficiently as possible, how many moves would be required to win the prize?
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296 Chapter 5 Sequences, Mathematical Induction, and Recursion Note that the numbers mn are independent of the labeling of the poles; it takes the same minimum number of moves to transfer n disks from pole A to pole C as to transfer n disks from pole A to pole B, for example. Also the values of mn are independent of the number of larger disks that may lie below the top n, provided these remain stationary while the top n are moved. Because the disks on the bottom are all larger than the ones on the top, the top disks can be moved from pole to pole as though the bottom disks were not present.
Going from position (a) to position (b) requires mk−1 moves, going from position (b) to position (c) requires just one move, and going from position (c) to position (d) requires mk−1 moves. By substitution into equation (5.6.1), therefore, mk = mk−1 + 1 + mk−1 = 2mk−1 + 1 for all integers k ≥2.
The initial condition, or base, of this recursion is found by using the definition of the sequence. Because just one move is needed to move one disk from one pole to another, m1 = the minimum number of moves needed to move a tower of one disk from one pole to another + = 1.
Hence the complete recursive specification of the sequence m1, m2, m3, . . . is as follows: For all integers k ≥2, (1) mk = 2mk−1 + 1 recurrence relation (2) m1 = 1 initial conditions Here is a computation of the next five terms of the sequence: (3) m2 = 2m1 + 1 = 2·1 + 1 = 3 by (1) and (2) (4) m3 = 2m2 + 1 = 2·3 + 1 = 7 by (1) and (3) (5) m4 = 2m3 + 1 = 2·7 + 1 = 15 by (1) and (4) (6) m5 = 2m4 + 1 = 2·15 + 1 = 31 by (1) and (5) (7) m6 = 2m5 + 1 = 2·31 + 1 = 63 by (1) and (6) Going back to the legend, suppose the priests work rapidly and move one disk every second. Then the time from the beginning of creation to the end of the world would be m64 seconds. In the next section we derive an explicit formula for mn. Meanwhile, we can compute m64 on a calculator or a computer by continuing the process started above (Try it!). The approximate result is 1.844674 × 1019 seconds ∼ = 5.84542 × 1011 years ∼ = 584.5 billion years, which is obtained by the estimate of 60 · 60 · 24 · (365.25) = 31, 557, 600 ↑ ↑ ↖ ↖ ↑ seconds per minute minutes per hour hours per day days per year seconds per year seconds in a year (figuring 365.25 days in a year to take leap years into account). Surpris-ingly, this figure is close to some scientific estimates of the life of the universe!
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5.6 Defining Sequences Recursively 297 Example 5.6.6 The Fibonacci Numbers One of the earliest examples of a recursively defined sequence arises in the writings of Leonardo of Pisa, commonly known as Fibonacci, who was the greatest European mathematician of the Middle Ages. In 1202 Fibonacci posed the following problem: A single pair of rabbits (male and female) is born at the beginning of a year. Assume the following conditions: 1. Rabbit pairs are not fertile during their first month of life but thereafter give birth to one new male/female pair at the end of every month.
2. No rabbits die.
How many rabbits will there be at the end of the year?
Bettmann/CORBIS Fibonacci (Leonardo of Pisa) (ca. 1175–1250) Solution One way to solve this problem is to plunge right into the middle of it using recur-sion. Suppose you know how many rabbit pairs there were at the ends of previous months.
How many will there be at the end of the current month?
The crucial observation is that the number of rabbit pairs born at the end of month k is the same as the number of pairs alive at the end of month k −2. Why? Because it is exactly the rabbit pairs that were alive at the end of month k −2 that were fertile during month k. The rabbits born at the end of month k −1 were not.
month k −2 k −1 k ———————|———————-|—————– Each pair alive here ↑ gives birth to a pair here ↑ Now the number of rabbit pairs alive at the end of month k equals the ones alive at the end of month k −1 plus the pairs newly born at the end of the month. Thus ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ the number of rabbit pairs alive at the end of month k ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ the number of rabbit pairs alive at the end of month k −1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ + ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ the number of rabbit pairs born at the end of month k ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ the number of rabbit pairs alive at the end of month k −1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ + ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ the number of rabbit pairs alive at the end of month k −2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ 5.6.2 Note It is essential to rephrase this observation in terms of a sequence.
For each integer n ≥1, let Fn = the number of rabbit pairs alive at the end of month n + and let F0 = the initial number of rabbit pairs = 1.
Then by substitution into equation (5.6.2), for all integers k ≥2, Fk = Fk−1 + Fk−2.
Now F0 = 1, as already noted, and F1 = 1 also, because the first pair of rabbits is not fertile until the second month. Hence the complete specification of the Fibonacci sequence is as follows: For all integers k ≥2, Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
298 Chapter 5 Sequences, Mathematical Induction, and Recursion (1) Fk = Fk−1 + Fk−2 recurrence relation (2) F0 = 1, F1 = 1 initial conditions.
To answer Fibonacci’s question, compute F2, F3, and so forth through F12: (3) F2 = F1 +F0 = 1 + 1 = 2 by (1) and (2) (4) F3 = F2 +F1 = 2 + 1 = 3 by (1), (2) and (3) (5) F4 = F3 +F2 = 3 + 2 = 5 by (1), (3) and (4) (6) F5 = F4 +F3 = 5 + 3 = 8 by (1), (4) and (5) (7) F6 = F5 +F4 = 8 + 5 = 13 by (1), (5) and (6) (8) F7 = F6 +F5 = 13 + 8 = 21 by (1), (6) and (7) (9) F8 = F7 +F6 = 21 + 13 = 34 by (1), (7) and (8) (10) F9 = F8 +F7 = 34 + 21 = 55 by (1), (8) and (9) (11) F10 = F9 +F8 = 55 + 34 = 89 by (1), (9) and (10) (12) F11 = F10 +F9 = 89 + 55 = 144 by (1), (10) and (11) (13) F12 = F11 +F10 = 144 + 89 = 233 by (1), (11) and (12) At the end of the twelfth month there are 233 rabbit pairs, or 466 rabbits in all.
■ Example 5.6.7 Compound Interest On your twenty-first birthday you get a letter informing you that on the day you were born an eccentric rich aunt deposited $100,000 in a bank account earning 4% interest compounded annually and she now intends to turn the account over to you, provided you can figure out how much it is worth. What is the amount currently in the account?
Solution To approach this problem recursively, observe that ⎡ ⎢ ⎢ ⎣ the amount in the account at the end of any particular year ⎤ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎣ the amount in the account at the end of the previous year ⎤ ⎥ ⎥ ⎦+ ⎡ ⎢ ⎢ ⎣ the interest earned on the account during the year ⎤ ⎥ ⎥ ⎦.
Now the interest earned during the year equals the interest rate, 4% = 0.04 times the amount in the account at the end of the previous year. Thus ⎡ ⎢ ⎢ ⎣ the amount in the account at the end of any particular year ⎤ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎣ the amount in the account at the end of the previous year ⎤ ⎥ ⎥ ⎦+ (0.04)· ⎡ ⎢ ⎢ ⎣ the amount in the account at the end of the previous year ⎤ ⎥ ⎥ ⎦.
5.6.3 For each positive integer n, let An = the amount in the account at the end of year n + and let Note Again, a crucial step is to define the sequence explicitly.
A0 = the initial amount in the account + = $100, 000.
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5.6 Defining Sequences Recursively 299 Then for any particular year k, substitution into equation (5.6.3) gives Ak = Ak−1 + (0.04)· Ak−1 = (1 + 0.04)· Ak−1 = (1.04)· Ak−1 by factoring out Ak−1.
Consequently, the values of the sequence A0, A1, A2, . . . are completely specified as follows: for all integers k ≥1, (1) Ak = (1.04)· Ak−1 recurrence relation (2) A0 = $100, 000 initial condition.
The number 1.04 is called the growth factor of the sequence.
In the next section we derive an explicit formula for the value of the account in any year n. The value on your twenty-first birthday can also be computed by repeated substi-tution as follows: (3) A1 = 1.04· A0 = (1.04)·$100, 000 = $104, 000 by (1) and (2) (4) A2 = 1.04· A1 = (1.04)·$104, 000 = $108, 160 by (1) and (3) (5) A3 = 1.04· A2 = (1.04)·$108, 160 = $112, 486.40 by (1) and (4) .
.
.
.
.
.
(22) A20 = 1.04· A19 ∼ = (1.04)·$210, 684.92 ∼ = $219, 112.31 by (1) and (21) (23) A21 = 1.04· A20 ∼ = (1.04)·$219, 112.31 ∼ = $227, 876.81 by (1) and (22) The amount in the account is $227,876.81 (to the nearest cent). Fill in the dots (to check the arithmetic) and collect your money!
■ Example 5.6.8 Compound Interest with Compounding Several Times a Year When an annual interest rate of i is compounded m times per year, the interest rate paid per period is i/m. For instance, if 3% = 0.03 annual interest is compounded quarterly, then the interest rate paid per quarter is 0.03/4 = 0.0075.
For each integer k ≥1, let Pk = the amount on deposit at the end of the kth period, assuming no additional deposits or withdrawals. Then the interest earned during the kth period equals the amount on deposit at the end of the (k −1)st period times the interest rate for the period: interest earned during kth period = Pk−1 i m .
The amount on deposit at the end of the kth period, Pk, equals the amount at the end of the (k −1)st period, Pk−1, plus the interest earned during the kth period: Pk = Pk−1 + Pk−1 i m = Pk−1 1 + i m .
5.6.4 Suppose $10,000 is left on deposit at 3% compounded quarterly.
a. How much will the account be worth at the end of one year, assuming no additional deposits or withdrawals?
b. The annual percentage rate (APR) is the percentage increase in the value of the account over a one-year period. What is the APR for this account?
Solution a. For each integer n ≥1, let Pn = the amount on deposit after n consecutive quarters, assuming no additional deposits or withdrawals, and let P0 be the initial $10,000. Then Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
300 Chapter 5 Sequences, Mathematical Induction, and Recursion by equation (5.6.4) with i = 0.03 and m = 4, a recurrence relation for the sequence P0, P1, P2, . . . is (1) Pk = Pk−1(1 + 0.0075) = (1.0075)· Pk−1 for all integers k≥1.
The amount on deposit at the end of one year (four quarters), P4, can be found by successive substitution: (2) P0 = $10, 000 (3) P1 = 1.0075· P0 = (1.0075)·$10, 000.00 = $10, 075.00 by (1) and (2) (4) P2 = 1.0075· P1 = (1.0075)·$10, 075.00 = $10, 150.56 by (1) and (3) (5) P3 = 1.0075· P2 ∼ = (1.0075)·$10, 150.56 = $10, 226.69 by (1) and (4) (6) P4 = 1.0075· P3 ∼ = (1.0075)·$10, 226.69 = $10, 303.39 by (1) and (5) Hence after one year there is $10,303.39 (to the nearest cent) in the account.
b. The percentage increase in the value of the account, or APR, is 10303.39 −10000 10000 = 0.03034 = 3.034%.
■ Recursive Definitions of Sum and Product Addition and multiplication are called binary operations because only two numbers can be added or multiplied at a time. Careful definitions of sums and products of more than two numbers use recursion.
• Definition Given numbers a1, a2, . . . , an, where n is a positive integer, the summation from i = 1 to n of the ai, denoted n i=1 ai, is defined as follows: 1 i=1 ai = a1 and n i=1 ai = %n−1 i=1 ai & + an, if n > 1.
The product from i = 1 to n of the ai, denoted $n i=1 ai, is defined by 1 i=1 ai = a1 and n i=1 ai = %n−1 i=1 ai & ·an, if n > 1.
The effect of these definitions is to specify an order in which sums and products of more than two numbers are computed. For example, 4 i=1 ai = % 3 i=1 ai & + a4 = %% 2 i=1 ai & + a3 & + a4 = ((a1 + a2) + a3) + a4.
The recursive definitions are used with mathematical induction to establish various properties of general finite sums and products.
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5.6 Defining Sequences Recursively 301 Example 5.6.9 A Sum of Sums Prove that for any positive integer n, if a1, a2, . . . , an and b1, b2, . . . , bn are real numbers, then n i=1 (ai + bi) = n i=1 ai + n i=1 bi.
Solution The proof is by mathematical induction. Let the property P(n) be the equation n i = 1 (ai + bi) = n i = 1 ai + n i = 1 bi.
←P(n) We must show that P(n) is true for all integers n ≥0. We do this by mathematical induc-tion on n.
Show that P(1) is true: To establish P(1), we must show that 1 i = 1 (ai + bi) = 1 i = 1 ai + 1 i = 1 bi.
←P(1) But 1 i=1 (ai + bi) = a1 + b1 by definition of = 1 i=1 ai + 1 i=1 bi also by definition of .
Hence P(1) is true.
Show that for all integers k ≥1, if P(k) is true then P(k + 1) is also true: Suppose a1, a2, . . . , ak, ak+1 and b1, b2, . . . , bk, bk+1 are real numbers and that for some k ≥1 k i=1 (ai + bi) = k i=1 ai + k i=1 bi.
←P(k) inductive hypothesis We must show that k+1 i=1 (ai + bi) = k+1 i=1 ai + k+1 i=1 bi.
←P(k + 1) [We will show that the left-hand side of this equation equals the right-hand side.] But the left-hand side of the equation is k+1 i=1 (ai + bi) = k i=1 (ai + bi) + (ak+1 + bk+1) by definition of = k i=1 ai + k i=1 bi + (ak+1 + bk+1) by inductive hypothesis = k i=1 ai + ak+1 + k i=1 bi + bk+1 by the associative and cummutative laws of algebra = k+1 i=1 ai + k+1 i=1 bi by definition of which equals the right-hand side of the equation. [This is what was to be shown.] ■ Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
302 Chapter 5 Sequences, Mathematical Induction, and Recursion Test Yourself 1. A recursive definition for a sequence consists of a _ and .
2. A recurrence relation is an equation that defines each later term of a sequence by reference to in the sequence.
3. Initial conditions for a recursive definition of a sequence con-sist of one or more of the _ of the sequence.
4. To solve a problem recursively means to divide the prob-lem into smaller subproblems of the same type as the initial problem, to suppose , and to figure out how to use the supposition to .
5. A crucial step for solving a problem recursively is to define a _____ in terms of which the recurrence relation and initial conditions can be specified.
Exercise Set 5.6 Find the first four terms of each of the recursively defined sequences in 1–8.
1. ak = 2ak−1 + k, for all integers k ≥2 a1 = 1 2. bk = bk−1 + 3k, for all integers k ≥2 b1 = 1 3. ck = k(ck−1)2, for all integers k ≥1 c0 = 1 4. dk = k(dk−1)2, for all integers k ≥1 d0 = 3 5. sk = sk−1 + 2sk−2, for all integers k ≥2 s0 = 1, s1 = 1 6. tk = tk−1 + 2tk−2, for all integers k ≥2 t0 = −1, t1 = 2 7. uk = kuk−1 −uk−2, for all integers k ≥3 u1 = 1, u2 = 1 8. vk = vk−1 + vk−2 + 1, for all integers k ≥3 v1 = 1, v2 = 3 9. Let a0, a1, a2, . . . be defined by the formula an = 3n + 1, for all integers n ≥0. Show that this sequence satisfies the recurrence relation ak = ak−1 + 3, for all integers k ≥1.
10. Let b0, b1, b2, . . . be defined by the formula bn = 4n, for all integers n ≥0. Show that this sequence satisfies the recur-rence relation bk = 4bk−1, for all integers k ≥1.
11. Let c0, c1, c2, . . . be defined by the formula cn = 2n −1 for all integers n ≥0. Show that this sequence satisfies the recurrence relation ck = 2ck−1 + 1.
12. Let s0, s1, s2, . . . be defined by the formula sn = (−1)n n!
for all integers n ≥0. Show that this sequence satisfies the recurrence relation sk = −sk−1 k .
13. Let t0, t1, t2, . . . be defined by the formula tn = 2 + n for all integers n ≥0. Show that this sequence satisfies the recurrence relation tk = 2tk−1 −tk−2.
14. Let d0, d1, d2, . . . be defined by the formula dn = 3n −2n for all integers n ≥0. Show that this sequence satisfies the recurrence relation dk = 5dk−1 −6dk−2.
15.
H For the sequence of Catalan numbers defined in Example 5.6.4, prove that for all integers n ≥1, Cn = 1 4n + 2 2n + 2 n + 1 .
16. Use the recurrence relation and values for the Tower of Hanoi sequence m1, m2, m3, . . . discussed in Exam-ple 5.6.5 to compute m7 and m8.
17. Tower of Hanoi with Adjacency Requirement: Suppose that in addition to the requirement that they never move a larger disk on top of a smaller one, the priests who move the disks of the Tower of Hanoi are also allowed only to move disks one by one from one pole to an adjacent pole. Assume poles A and C are at the two ends of the row and pole B is in the middle. Let an = ⎡ ⎣ the minimum number of moves needed to transfer a tower of n disks from pole A to pole C ⎤ ⎦.
a. Find a1, a2, and a3.
b. Find a4.
c. Find a recurrence relation for a1, a2, a3, . . . .
18. Tower of Hanoi with Adjacency Requirement: Suppose the same situation as in exercise 17. Let bn = ⎡ ⎣ the minimum number of moves needed to transfer a tower of n disks from pole A to pole B ⎤ ⎦.
a. Find b1, b2, and b3.
b. Find b4.
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5.6 Defining Sequences Recursively 303 c. Show that bk = ak−1 + 1 + bk−1 for all integers k ≥ 2, where a1, a2, a3, . . . is the sequence defined in exercise 17.
d. Show that bk ≤3bk−1 + 1 for all integers k ≥2.
e.
✶ H Show that bk = 3bk−1 + 1 for all integers k ≥2.
19. Four-Pole Tower of Hanoi: Suppose that the Tower of Hanoi problem has four poles in a row instead of three.
Disks can be transferred one by one from one pole to any other pole, but at no time may a larger disk be placed on top of a smaller disk. Let sn be the minimum number of moves needed to transfer the entire tower of n disks from the left-most to the right-most pole.
a. Find s1, s2, and s3.
b. Find s4.
c. Show that sk ≤2sk−2 + 3 for all integers k ≥3.
20. Tower of Hanoi Poles in a Circle: Suppose that instead of being lined up in a row, the three poles for the original Tower of Hanoi are placed in a circle. The monks move the disks one by one from one pole to another, but they may only move disks one over in a clockwise direction and they may never move a larger disk on top of a smaller one. Let cn be the minimum number of moves needed to transfer a pile of n disks from one pole to the next adjacent pole in the clockwise direction.
a. Justify the inequality ck ≤4ck−1 + 1 for all integers k ≥2.
b. The expression 4ck−1 + 1 is not the minimum number of moves needed to transfer a pile of k disks from one pole to another. Explain, for example, why c3 ̸= 4c2 + 1.
21. Double Tower of Hanoi: In this variation of the Tower of Hanoi there are three poles in a row and 2n disks, two of each of n different sizes, where n is any positive integer.
Initially one of the poles contains all the disks placed on top of each other in pairs of decreasing size. Disks are transferred one by one from one pole to another, but at no time may a larger disk be placed on top of a smaller disk. However, a disk may be placed on top of one of the same size. Let tn be the minimum number of moves needed to transfer a tower of 2n disks from one pole to another.
a. Find t1 and t2.
b. Find t3.
c. Find a recurrence relation for t1, t2, t3, . . . .
22. Fibonacci Variation: A single pair of rabbits (male and female) is born at the beginning of a year. Assume the following conditions (which are more realistic than Fibonacci’s): (1) Rabbit pairs are not fertile during their first month of life but thereafter give birth to four new male/female pairs at the end of every month.
(2) No rabbits die.
a. Let rn = the number of pairs of rabbits alive at the end of month n, for each integer n ≥1, and let r0 = 1. Find a recurrence relation for r0,r1,r2, . . . .
b. Compute r0,r1,r2,r3,r4,r5, and r6.
c. How many rabbits will there be at the end of the year?
23. Fibonacci Variation: A single pair of rabbits (male and female) is born at the beginning of a year. Assume the fol-lowing conditions: (1) Rabbit pairs are not fertile during their first two months of life, but thereafter give birth to three new male/female pairs at the end of every month.
(2) No rabbits die.
a. Let sn = the number of pairs of rabbits alive at the end of month n, for each integer n ≥1, and let s0 = 1. Find a recurrence relation for s0, s1, s2, . . . .
b. Compute s0, s1, s2, s3, s4, and s5.
c. How many rabbits will there be at the end of the year?
In 24–34, F0, F1, F2, . . . is the Fibonacci sequence.
24. Use the recurrence relation and values for F0, F1, F2, . . .
given in Example 5.6.6 to compute F13 and F14.
25. The Fibonacci sequence satisfies the recurrence relation Fk = Fk−1 + Fk−2, for all integers k ≥2.
a. Explain why the following is true: Fk+1 = Fk + Fk−1 for all integers k ≥1.
b. Write an equation expressing Fk+2 in terms of Fk+1 and Fk.
c. Write an equation expressing Fk+3 in terms of Fk+2 and Fk+1 26. Prove that Fk = 3Fk−3 + 2Fk−4 for all integers k ≥4.
27. Prove that F2 k −F2 k−1 = Fk Fk−1 −Fk+1Fk−1, for all integers k ≥1.
28. Prove that F2 k+1 −F2 k −F2 k−1 = 2Fk Fk−1, for all integers k ≥1.
29. Prove that F2 k+1 −F2 k = Fk−1Fk+2, for all integers k ≥1.
30. Use mathematical induction to prove that for all integers n ≥0, Fn+2Fn −F2 n+1 = (−1)n.
31.
✶ Use strong mathematical induction to prove that Fn < 2n for all integers n ≥1.
32.
✶ H Let F0, F1, F2, . . . be the Fibonacci sequence defined in Section 5.6. Prove that for all integers n ≥0, gcd (Fn+1, Fn) = 1.
33. It turns out that the Fibonacci sequence satisfies the fol-lowing explicit formula: For all integers Fn ≥0, Fn = 1 √ 5 ⎡ ⎣ % 1 + √ 5 2 &n+1 − % 1 − √ 5 2 &n+1⎤ ⎦ Verify that the sequence defined by this formula satisfies the recurrence relation Fk = Fk−1 + Fk−2 for all integers k ≥2.
34.
H (For students who have studied calculus) Find lim n→∞ Fn+1 Fn , assuming that the limit exists.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
304 Chapter 5 Sequences, Mathematical Induction, and Recursion 35.
✶ H (For students who have studied calculus) Prove that lim n→∞ Fn+1 Fn exists.
36. (For students who have studied calculus) Define x0, x1, x2, . . . as follows: xk = 0 2 + xk−1 for all integers k ≥1 x0 = 0 Find limn→∞xn. (Assume that the limit exists.) 37. Compound Interest: Suppose a certain amount of money is deposited in an account paying 4% annual interest compounded quarterly. For each positive integer n, let Rn = the amount on deposit at the end of the nth quarter, assuming no additional deposits or withdrawals, and let R0 be the initial amount deposited.
a. Find a recurrence relation for R0, R1, R2, . . . .
b. If R0 = $5000, find the amount of money on deposit at the end of one year.
c. Find the APR for the account.
38. Compound Interest: Suppose a certain amount of money is deposited in an account paying 3% annual interest com-pounded monthly. For each positive integer n, let Sn = the amount on deposit at the end of the nth month, and let S0 be the initial amount deposited.
a. Find a recurrence relation for S0, S1, S2, . . ., assum-ing no additional deposits or withdrawals during the year.
b. If S0 = $10, 000, find the amount of money on deposit at the end of one year.
c. Find the APR for the account.
39. With each step you take when climbing a staircase, you can move up either one stair or two stairs. As a result, you can climb the entire staircase taking one stair at a time, taking two at a time, or taking a combination of one- and two-stair increments. For each integer n ≥1, if the staircase consists of n stairs, let cn be the number of different ways to climb the staircase. Find a recurrence relation for c1, c2, c3, . . . .
40. A set of blocks contains blocks of heights 1, 2, and 4 cen-timeters. Imagine constructing towers by piling blocks of different heights directly on top of one another. (A tower of height 6 cm could be obtained using six 1-cm blocks, three 2-cm blocks one 2-cm block with one 4-cm block on top, one 4-cm block with one 2-cm block on top, and so forth.) Let t be the number of ways to construct a tower of height n cm using blocks from the set. (Assume an unlimited sup-ply of blocks of each size.) Find a recurrence relation for t1, t2, t3, . . ..
41. Use the recursive definition of summation, together with mathematical induction, to prove the generalized distribu-tive law that for all positive integers n, if a1, a2, . . . , an and c are real numbers, then n i=1 cai = c % n i=1 ai & .
42. Use the recursive definition of product, together with math-ematical induction, to prove that for all positive integers n, if a1, a2, . . . , an and b1, b2, . . . , bn are real numbers, then n i=1 (aibi) = % n i=1 ai & % n i=1 bi & .
43. Use the recursive definition of product, together with math-ematical induction, to prove that for all positive integers n, if a1, a2, . . . , an and c are real numbers, then n i=1 (cai) = cn % n i=1 ai & .
44.
H The triangle inequality for absolute value states that for all real numbers a and b, |a + b| ≤|a| + |b|. Use the recur-sive definition of summation, the triangle inequality, the definition of absolute value, and mathematical induction to prove that for all positive integers n, if a1, a2, . . . , an are real numbers, then 1 1 1 1 1 n i=1 ai 1 1 1 1 1 ≤ n i=1 |ai|.
Answers for Test Yourself 1. recurrence relation; initial conditions 2. earlier terms 3. values of the first few terms 4. that the smaller subproblems have already been solved; solve the initial problem 5. sequence 5.7 Solving Recurrence Relations by Iteration The keener one’s sense of logical deduction, the less often one makes hard and fast inferences. — Bertrand Russell, 1872–1970 Suppose you have a sequence that satisfies a certain recurrence relation and initial conditions. It is often helpful to know an explicit formula for the sequence, especially if Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. |
10594 | https://www.geeksforgeeks.org/engineering-mathematics/proof-of-de-morgans-laws-in-boolean-algebra/ | Proof of De-Morgan's Laws in Boolean Algebra
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De Morgan's Laws are fundamental principles in Boolean algebra and set theory, providing rules for transforming logical expressions. These laws are essential for simplifying and manipulating Boolean expressions, which have significant applications in digital circuit design, computer science, and engineering. This article provides a detailed proof of De Morgan's Laws in Boolean algebra.
De-Morgan's Laws
De Morgan's Laws state that the complement of the conjunction (AND) of two variables is equal to the disjunction (OR) of their complements, and the complement of the disjunction (OR) of two variables is equal to the conjunction (AND) of their complements. Mathematically, the laws are expressed as:
A⋅B=A+B
A+B=A⋅B
Proof of De Morgan's Laws
Statements:
1. (x + y)' = x'.y'
2. (x.y) = x' + y'
Proof:
First Statement: x+y=x′⋅y′
To prove:
(x + y)′=x′⋅ y'
We use the fundamental Boolean identities:
Complement Rule: A + A′= 1 and A⋅ A′=0.
Distributive Property: A + B.C = (A+B) ⋅(A+C).
Step-by-step proof:
Consider the expression: (x+y)′+x′⋅y.
Expanding using Boolean identities: (x+y)′=(x+y)⋅(x′+y′)
Applying De Morgan’s transformation: (x+y)⋅(x′+y′)=x.x′+x.y′+y.x′+y⋅y′
Using x⋅x′=0 and y⋅y′=0, we simplify: 0+x⋅y′+y⋅x′+0=x′⋅y′
Since x′⋅y′=x+y, we conclude: x+y=x′⋅y′
Second statement : x⋅y=x′+y′
To prove: (x⋅y)′=x′+y′
Using Boolean properties:
Complement Rule: A + A′=1 and A⋅ A′=0.
Distributive Property: A (B + C)= AB + AC.
Step-by-step proof:
Consider the expression: (x. y)'+ (x'+ y).
Expanding using Boolean identities: (x.y)'= (x.y) + (x'+y')
Applying De Morgan’s transformation: (x.y) + (x'+y')= x.x'+ x.y' + y.x' + y⋅y'
Using x⋅x′=0 and y⋅y′=0,, we simplify: 0+x+y′+0=x′⋅y′
Since x′+y′=x.y, we conclude: x+y=x′+y′
Proof of De-Morgan's law of boolean algebra using Truth Table:
1) (x+y)’= x’. y’
| | | | | | | |
--- --- ---
| x | y | x+y | (x+y)’ | x’ | y’ | x’. y’ |
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Here x+y
Column x+y: The OR operation means at least one input must be 1 to get 1.
Column x+y:This is the complement (NOT operation), meaning it flips the values of x+y.
Column x and y: These represent the individual complements of x and y.
Column x.y: The AND operation is applied to the complements.
Hence the first De-morgan law of boolean algebra is proved (x +y)’= x’. y’
2) (x.y)’=x’+y’
| | | | | | | |
--- --- ---
| x | y | x.y | (x.y)’ | x’ | y’ | x’ + y’ |
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Here
Column x⋅y: The AND operation means both inputs must be 1 to get 1.
Column x⋅y: This is the complement, meaning it flips the values of x⋅y.
Column x and y: These represent the individual complements of x and y.
Column : The OR operation is applied to the complements.
Hence the second De-morgan law of boolean algebra is proved (x.y)’=x’+y’
Solved Examples
1) Using De Morgan's Laws, simplify the following expression:
NOT(A OR (B AND C))
Solution :
NOT(A OR (B AND C))
Answer: (NOT A) AND (NOT B OR NOT C)
2).Prove that the following expressions are equivalent using De Morgan's Laws:
(NOT A AND NOT B) OR (NOT A AND C) = NOT A AND (NOT B OR C)
Solution :
(NOT A AND NOT B) OR (NOT A AND C) = NOT A AND (NOT B OR C)
Answer: This is correct.
Proof: NOT A AND (NOT B OR C)
= (NOT A AND NOT B) OR (NOT A AND C) [Using distributive law]
3).In a digital circuit, you have an AND gate followed by a NOT gate. Using De Morgan's Laws, how can you represent this using only OR and NOT gates?
Solution :
AND gate followed by NOT gate using only OR and NOT gates
Answer: Use NOT gates on each input, then feed these into an OR gate.
NOT(A AND B) = (NOT A) OR (NOT B)
4).Simplify the following Boolean expression using De Morgan's Laws:
NOT(A AND B AND C)
Solution :
NOT(A AND B AND C)
Answer: (NOT A) OR (NOT B) OR (NOT C)
5).In computer networking, a firewall rule allows traffic that is "not from subnet A and not destined for port 80". Express this rule in terms of positive conditions using De Morgan's Laws.
Solution :
F = NOT(X OR Y OR Z)
Answer: F = (NOT X) AND (NOT Y) AND (NOT Z)
6).Prove or disprove: NOT(A XOR B) = (NOT A) XOR (NOT B)
Solution :
"Not from subnet A and not destined for port 80"
Answer: From any subnet except A AND destined for any port except 80
7).Simplify the following expression using De Morgan's Laws:
NOT(A OR NOT(B AND C))
Solution :
NOT(A XOR B) = (NOT A) XOR (NOT B)
Answer: This is false.
NOT(A XOR B) = (A AND B) OR (NOT A AND NOT B)
But (NOT A) XOR (NOT B) = (NOT A AND B) OR (A AND NOT B)
8).In digital logic, how would you use De Morgan's Laws to convert a NOR gate into an equivalent combination of AND and NOT gates?
Solution :
NOT(A OR NOT(B AND C))
Answer: (NOT A) AND (B AND C)
Applications in Engineering
1. Digital Circuit Design
De Morgan's Laws are used extensively in digital circuit design to simplify Boolean expressions and logic gates, optimizing circuit layouts and reducing costs.
2. Computer Science
In computer science, De Morgan's Laws are used in algorithms and programming to simplify logical expressions, making code more efficient and readable.
3. Set Theory
In set theory, De Morgan's Laws describe the relationships between union, intersection, and complement of sets, providing a foundation for various mathematical operations.
4. Control Systems
In control systems, De Morgan's Laws help in designing and analyzing logic controllers, ensuring correct and efficient system behavior.
5. Data Mining
In data mining, De Morgan's Laws are applied to simplify queries and logical conditions, improving the performance of data retrieval and analysis processes.
Practice Problems
Simplify: NOT(A AND (B OR C))
Prove that NOT(A OR B OR C) = (NOT A) AND (NOT B) AND (NOT C)
Convert the expression (A NAND B) to use only OR and NOT gates.
Simplify: NOT(NOT A OR (B AND NOT C))
In a computer program, you need to check if a number is not between 1 and 10 inclusive. Express this condition using De Morgan's Laws.
Prove or disprove: NOT(A AND B) XOR C = (NOT A OR NOT B) XOR C
Simplify the Boolean expression: NOT((A AND B) OR (NOT A AND C))
Given F = NOT(X AND Y AND Z), express F using only OR gates and inverters.
In a digital circuit, you have a NOR gate followed by a NOT gate. Simplify this using De Morgan's Laws.
Express the following using only AND and NOT operations: NOT(A OR (B AND NOT C) OR (NOT A AND D))
Are De Morgan’s Laws used in Set Theory?
Yes! De Morgan’s Laws also apply to sets and Venn diagrams:
(A∪B)′= A′∩B′ (Complement of union is the intersection of complements).
(A∩B)′= A′∪B′ (Complement of intersection is the union of complements).
Are De Morgan’s Laws used in Set Theory?
Yes, but set operations differ from Boolean algebraic proofs.
What are De Morgan's Laws?
De Morgan's Laws state that the complement of the conjunction (AND) of two variables is equal to the disjunction (OR) of their complements, and vice versa.
M
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10595 | https://math.stackexchange.com/questions/3083447/how-can-we-solve-xa-b-mod-n-equation-for-big-n-a | modular arithmetic - How can we solve $x^a = b \; mod \; n$ equation for big n, a, - Mathematics Stack Exchange
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How can we solve x a=b m o d n x a=b m o d n equation for big n, a,
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Asked 6 years, 8 months ago
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I would like to have a method to solve an equation of the type: x a=b m o d n x a=b m o d n, knowing that n can be decomposed into a product of prime numbers n=n 1×n 2×...×n k n=n 1×n 2×...×n k
I already know how to do it for small numbers, like if n is a prime number:
in this case, I am looking for e e, such that a e=1 m o d(n−1)a e=1 m o d(n−1). Like that, I can transform my equation into x=b e m o d n x=b e m o d n.
But here the numbers are too big to compute this in a single time. So maybe there is a method to solve it I am not aware of.
In my exercise, I have :
n = 264356242932330620591879762011459333409
b = 259252555055790712660181286804144327401
a = 151089236568313654499150506467499
modular-arithmetic
arithmetic
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edited Jan 22, 2019 at 17:53
gommygommy
asked Jan 22, 2019 at 17:40
gommygommy
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Is it x a x a or a x a x?kcborys –kcborys 2019-01-22 17:42:05 +00:00 Commented Jan 22, 2019 at 17:42
x a x a sorry I gonna edit gommy –gommy 2019-01-22 17:42:46 +00:00 Commented Jan 22, 2019 at 17:42
Welcome to stackexchange. This looks like cryptography homework. You are more likely to get help rather than downvotes and votes to close if you edit the question to show us what you tried and where you are stuck. Can you do the problem with smaller number?Ethan Bolker –Ethan Bolker 2019-01-22 17:45:35 +00:00 Commented Jan 22, 2019 at 17:45
@Moo : No I am trying to find x. a, b and n are given gommy –gommy 2019-01-22 17:54:28 +00:00 Commented Jan 22, 2019 at 17:54
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The Euclidean algorithm proves that gcd(b,n)=1 gcd(b,n)=1.
Since you know the prime factorization of n n, you know ϕ(n)ϕ(n).
The Euclidean algorithm proves that gcd(a,ϕ(n))=1 gcd(a,ϕ(n))=1.
The extended Euclidean algorithm then gives c c such that a c≡1 mod ϕ(n)a c≡1 mod ϕ(n) and so x≡b c x≡b c.
These computations are probably not easy when done by hand. But WA handles it fine.
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answered Jan 22, 2019 at 17:53
lhflhf
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Thank you for your answer @Ihf, I will try it right now gommy –gommy 2019-01-22 18:05:18 +00:00 Commented Jan 22, 2019 at 18:05
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10596 | https://wiki.mbalib.com/wiki/%E6%B5%B7%E7%9B%97%E5%8D%9A%E5%BC%88 | 海盗博弈 - MBA智库百科
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出自 MBA智库百科(
海盗博弈（Pirate game）
目录
[隐藏]
1 什么是海盗博弈
2 海盗博弈故事
3 海盗博弈的延伸
4 参考文献
[编辑]
什么是海盗博弈
海盗博弈是一个简单的数学博弈。该博弈描述了如果遵循经济人的行为，结果可能让人惊讶。这同时也是最后通牒博弈的多参与者版本
[编辑]
海盗博弈故事
有五个非常聪明的理性的海盗，分别编号 P 1,P 2,P 3,P 4,P 5。他们一同抢夺了100个金币，现在需要想办法分配这些金币。 海盗们有严格的等级制度：P 1<P 2<P 3<P 4<P 5。 海盗们分配原则是：等级最高的海盗P5提出一种分配方案。然后所有的海盗投票决定是否接受分配，包括提议人。并且在票数相同的情况下，提议人有决定权。如果提议通过，那么海盗们按照提议分配金币。如果没有通过，那么提议人将被扔出船外，然后由下一个最高等级的海盗提出新的分配方案。
海盗们基于三个因素来做决定。首先，要能存活下来。其次，自己的利益最大化(即得到最多的金币)。最后，在所有其他条件相同的情况下，优先选择把别人扔出船外。
现在，假如你是等级最高的 P 5，你会做何选择？直觉上，为了保住自己的生命，你可能会选择留给自己很少的金币，以便让大家同意自己的决策。然而，这和理论结果相差甚远。
解决这个问题的关键是换个思维方向。与其苦思冥想你要做什么决策，不如先想想最后剩下的人会做什么决策。假设现在只剩下 P 1 和 P 2 了，P 2 会做什么决策？很明显，他将把100金币留给自己，然后投自己一票。由于在票数相同的情况下提议人有决定权，无论 P 1 同不同意，P 2 都将实现自己的目的。
现在再把 P 3 加进来。P 1 知道，如果 P 3 被扔下海，那么游戏又将进行到上面的情况，P 1 终将一无所有。P 3 同样看到了这一点，所以他知道，只要他给 P 1 一点点利益，P 1 就会投票支持他的决策。所以 P 3 最终的决策应该是：
P 4 的策略也类似。由于他需要50%的支持，所以他只需贿赂1个金币给 P 2 就可以了。P 2 一定会支持他(否则轮到 P 3 做决策，他就一无所有啦)。所以 P 4 最终的决策是：
P 5 的情况稍有不同。由于这次一共有5个人，所以他至少需要贿赂两个海盗以使自己的决议通过。所以唯一的决策就是：
[编辑]
海盗博弈的延伸
如果海盗的数目不止5个呢？ 继续按照这个逻辑推理，P 6 的决策将是：…一直到 P 2 00,它会给自己留1个金币，同时给剩下所有偶数编号的海盗1个金币。
海盗P 1 P 2 P 3 P 4 P 5…P 197 P 198 P 199 P 200
决策者
P 1 100
P 2 0 100
P 3 1 0 99
P 4 0 1 0 99
P 5 1 0 1 0 98
…………………
P 198 0 1 0 1 0…0 2
P 199 1 0 1 0 1…1 0 1
P 200 0 1 0 1 0…0 1 0 1
如果海盗数是201个，那么 P 201 该怎么做呢？乍一看去，他好像没有足够的钱去贿赂别的海盗了。不过，为了保住自己的性命，他还是可以把自己手中的金币全分出去，即给每个奇数编号的海盗（P 1~P 199）一个金币。这样虽然空手而归，但不至于人财两空。
P 202 也只能把这100个金币全部贿赂给其他100个海盗，这100个海盗必须是在 P 201 做决策的情况下什么也得不到的海盗。由于符合这样条件的海盗有101个(所有偶数编号的海盗 P 201)，P 202 的决策不再是唯一的了！有101种方案供他选择。
可怜的是 P 203。由于人数众多，他实在没有足够的钱去贿赂其他海盗以获得足够的支持(他需要至少102个人的支持，包括他自己)。所以，不论 P 203 做什么决策，他都难逃被扔出船外的厄运了。不过 P 203 并没有我们想象中的那么悲情，因为这样的悲剧发生当且仅当船上正好有203个海盗。我们再增加一个海盗，P 204。P 204 明白，P 203 现在的唯一愿望就是活下来…所以不论 P 204 做什么决策，P 203 都会举双手支持他(当然举多少手都只能算一票)。所以 P 204 可以靠他自己的一票，P 203 的一票和贿赂另外100个海盗获得正好50%的支持。
P 204 可能的决策也只有101种，如下表：(可能获得1金币的海盗用"Y"标示)
P 1 P 2 P 3 P 4…P 199 P 200 P 201 P 202 P 203 P 204
P 204 Y N Y N Y N N Y N N
P 205 就没有那么幸运了。他不能无偿的得到 P 203 和 P 204 的支持。所以如果轮到 P 205 做决策，他也必定被扔到船外。P 206 也一样，尽管他能得到 P 205 的免费支持，但是这还不够。P 2 07 需要得到至少104个海盗的支持，所以有了 P 205,P 206 的无偿支持还是不够。
P 208 就比较幸运了。他也是需要得到104个海盗的支持，但 P 205,P 206,P 207，加上他自己，再加上贿赂100个海盗，正好104票。
P 208 可能的决策：(这次他有种决策)
P 1 P 2 P 3 P 4…P 199 P 200 P 201 P 202 P 203 P 204 P 205 P 206 P 207 P 208
P 208 N Y N Y N Y Y N Y Y N N N N
从这里我们又看出了新的规律：
从 P 201 之后，在每两个能够作出决策保住自己生命的海盗之间，存在着一些无论如何决策都会被扔到船外的海盗。而这些海盗会支持在这之后的那个能够做出决策保住自己生命的海盗。用数学来表达，设在 P 201 之后，能够作出决策保住自己生命的海盗的编号所组成的序列为 a n。则有：
对于(2)，
若 a n 是偶数，则 a n = 2 a(n − 1) − 200
若 a n 是奇数，则 a n = 2 a(n − 1) − 199
给定一个固定的初值，数列的下一项有两个可能解：一个奇数解、一个偶数解，且偶数解比奇数解小1。再考虑我们原问题的意义，到达偶数解时，偶数编号的海盗已经能够做出决策保全自己。这说明我们应该舍弃所有奇数解。
由 a n = 2 a(n − 1) − 200 以及 a 0 = 202，我们得到通解为:a n = 200 + 2 n + 1。考虑到 P 201 也能保全自己，我们可以把所有能够保全自己但却得不到金币的海盗的编号写成统一表达式：
不难推出这些海盗可能的决策种数为,其中
[编辑]
参考文献
↑ 1.01.1逻辑与直觉–海盗博弈
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页面分类: 博弈论
评论(共29条)
提示:评论内容为网友针对条目"海盗博弈"展开的讨论，与本站观点立场无关。
F4nvip (Talk | 贡献) 在 2009年4月15日 15:18 发表
太强了，理工科都晕
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220.178.203. 在 2009年10月15日 10:24 发表
极其精彩！每个提议人作出的分配方案如果要被其他海盗接受，前提是所有海盗都看过这篇文章！
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121.8.26. 在 2009年11月16日 13:02 发表
现实中，我相信是不可能会这样子分的吧
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58.41.143. 在 2009年11月27日 09:08 发表
好复杂。。。看的头晕
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59.120.169. 在 2010年1月26日 17:12 发表
有五个非常聪明的理性的海盗為前提，跟經濟學假設相同
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122.236.170. 在 2010年2月22日 20:14 发表
还是分点钱给鲨鱼，自己再留点
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122.204.172. 在 2010年10月7日 11:50 发表
终于绕明白了。。
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221.233.152. 在 2010年10月9日 14:22 发表
后面的计算看的木了
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Yixuanzhan (Talk | 贡献) 在 2010年12月15日 13:08 发表
前提是海盗的数学要学的好才行，碰上一个不识数的就被喂鲨鱼了
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犀利强 (Talk | 贡献) 在 2011年1月7日 16:54 发表
玩跨级，放在企业里不知道行不行。
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222.93.207. 在 2011年1月25日 21:36 发表
放在国内肯定不行 因为中国人不够聪明而且贪婪
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123.121.222. 在 2011年3月27日 22:52 发表
海盜Pn只有在前n-1個海盜都扔到海裡的情況下才會成立，所以最後一個海盜就算沒有錢分，也一定是活著的，怎麼會被扔下水呢？
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大大拉 (Talk | 贡献) 在 2011年3月28日 00:27 发表
迷糊。。
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113.120.26. 在 2011年5月20日 08:57 发表
太纯数学了，难懂。
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222.69.242. 在 2011年5月25日 16:47 发表
最后一个海盗P5不同意P4的观点不会被丢到海里,所以上述条件不成立,下面一大堆推理前提不成立
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218.5.2. 在 2011年6月8日 19:33 发表
222.69.242. 在 2011年5月25日 16:47 发表
最后一个海盗P5不同意P4的观点不会被丢到海里,所以上述条件不成立,下面一大堆推理前提不成立
回楼上 只有提议人会被仍进海里的
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晞灭 (Talk | 贡献) 在 2011年7月27日 16:00 发表
问题是 到 p208时 不应该只有103选11的组合 应该是204选100的组合吧 【我电脑打不出 排列组合 C】 因为205,206,207和自己的四票是固定的 然而 对于前204个人来说 即使给204分 最多也只能得到1个金币 此时 没有必要再分 要给哪个人了吧。。。 或者说 当只能把100金币全部分给其他人时 他的 下一个分配者就不必再思考 应该分给哪些特定的组合了吧。。。。应该是这样 编者说呢。。
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彭女士 (Talk | 贡献) 在 2012年3月14日 18:55 发表
太强了！数学不好还很难看懂啊！
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张喜双 (Talk | 贡献) 在 2012年3月19日 15:50 发表
强大的理论。
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116.228.2. 在 2012年5月16日 18:46 发表
博弈是要考虑多种因素的。单一的因素只是理论上的。除非他们都懂得这样的博弈。不过分析的很精彩。
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王亚刚 (Talk | 贡献) 在 2012年10月29日 13:13 发表
理论很好，但是，多数人是短视的，因此，无论如何，每一个肯定会被扔出去给鱼.
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76.111.169. 在 2014年1月24日 03:29 发表
不断反对，把前面的人都杀死，分的才多
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218.18.170. 在 2015年4月15日 14:25 发表
现实中，强者掌控游戏规则。
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Zarua (Talk | 贡献) 在 2017年4月30日 21:34 发表
222.93.207. 在 2011年1月25日 21:36 发表
放在国内肯定不行 因为中国人不够聪明而且贪婪
你聪明你上
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Zarua (Talk | 贡献) 在 2017年4月30日 21:42 发表
p5等级最高应该由p5最先提出建议，P1就没有任何权利首先提出建议
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Zarua (Talk | 贡献) 在 2017年4月30日 22:06 发表
这里明显与海盗分金更模糊一些
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M id 9b1774d75ecaa61c0138338975fb1b47 (Talk | 贡献) 在 2018年6月11日 10:14 发表
看得晕，逻辑已经跟不上了
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伍一龙 (Talk | 贡献) 在 2018年12月21日 09:53 发表
天行九歌，三姬分金。海盗分金的简化版。
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M id 5a9338bb3b3fa56bc54e4f0bd92a7485 (Talk | 贡献) 在 2020年10月1日 13:23 发表
这关系的人的信任问题
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10597 | https://ethz.ch/content/dam/ethz/special-interest/phys/theoretical-physics/itp-dam/documents/gaberdiel/proseminar_fs2018/07_Meng.pdf | Integer Quantum Hall Effect Severin Meng Proseminar in Theoretical Physics, Departement Physik ETH Z¨ urich May 28, 2018 Contents 1 The basics of the Integer Quantum Hall Effect 2 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.2 Conductivity and Resistivity in 2D . . . . . . . . . . . . . . . . . . . . . .
2 1.3 The Classical Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.4 The Results of the Integer Quantum Hall Effect . . . . . . . . . . . . . . .
4 1.5 Quantum Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 1.5.1 Eigenstates and Eigenvalues . . . . . . . . . . . . . . . . . . . . . .
5 1.5.2 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2 The Edge Picture 7 2.1 Edge States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 2.2 The Role of Disorder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 2.3 The Role of Gauge Invariance . . . . . . . . . . . . . . . . . . . . . . . . .
13 3 The Bulk Picture 15 3.1 The Kubo Formula for Hall Conductivity . . . . . . . . . . . . . . . . . .
15 3.2 Hall System on a Torus . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 3.3 Particles on a Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 3.3.1 Example: The Chern Insulator . . . . . . . . . . . . . . . . . . . .
19 3.4 Particles on a Lattice in a Magnetic Field . . . . . . . . . . . . . . . . . .
21 4 Conclusion 24 1 1 The basics of the Integer Quantum Hall Effect 1.1 Introduction In a two-dimensional electron gas at a very low temperature (T < 4 K) subject to a strong perpendicular magnetic field (B ∼O(T)), the electrical conductivity takes on values that are fundamentally different from the predictions of classical physics. The transverse conductivity, also referred to as the Hall conductivity, takes quantised values: σxy = ν e2 2πℏ (1) where the value ν has been measured to very high precision to be an integer, hence the name integer quantum Hall effect. For the discovery of the integer quantum Hall effect, Klaus von Klitzing won the nobel prize in 1985.
The quantisation of the conductivity happens for dirty, many-particle mesoscopic sys-tems.
It is a collective quantum effect, like superconductivity.
The quantisation is universal in the sense that, to a large extent, it does not depend on microscopic details such as the exact value of the magnetic field, the purity of the sample, the electron mobility etc. It turns out that a topological property of the system is essential to this phenomenon, and surprisingly, disorder is also important.
We are going to look at two seemingly different explanations of the integer quantum Hall effect. In a finite system with edges, one can interpret the Hall current to be carried by excitations near the edge. These edge states can also be observed experimentally. They are chiral modes in the sense that they can only carry current in one fixed direction. At the same time, these edge modes are protected in the way that the number of edge modes cannot change under continuous changes in the system. This will be explained in section 2, The Edge Picture. For systems with no edges, i.e. bulk systems, one interprets the current to be running within the sample. Using linear response theory, one can derive a formula that guarantees a quantised Hall conductivity for a system fulfilling the right conditions. The system has to have an energy spectrum forming gapped bands, i.e. the electron states have to be periodic in a type of Brillouin zone. This will be explained in section 3, The Bulk Picture.
It turns out that both descriptions are two different manifestations of the topological properties of the system. The number of protected edge modes corresponds to a quan-tised Hall conductivity, and this number is connected to the Chern numbers associated with the gapped energy bands.
This exposition is largely based on the notes of D. Tong and the original literature [2, 3, 4, 5].
1.2 Conductivity and Resistivity in 2D The Hall effect is observed in two dimensional systems, we therefore work in the (x,y)-plane and only consider two dimensional current densities and electric fields: J = Jx Jy , E = Ex Ey (2) 2 Ohm’s law tells that the conductivity links the electric field to the current density in a linear equation J = σE (3) where σ is the conductivity. Its general form is not a scalar but a matrix. For an isotropic sample, the conductivity takes the following form: σ = σxx σxy −σxy σxx (4) The resistivity matrix ρ is simply the inverse of σ. A consequence of having a matrix conductivity is that one can have a system with both vanishing longitudinal resistivity and conductivity σxx = ρxx = 0. In the scalar case, a system with σxx = σ = 0 would be called a perfect insulator, while a system with ρxx = ρ = 0 would be called a perfect conductor. Having both at the same time seems unphysical, however one should not be fooled by this intuition, as in the matrix case, the inverse of the resistivity is given by ρxx = σxx σ2 xx+σ2 xy . The inverse therefore exists even in the case of σxx = 0 as long as the transverse conductivity σxy ̸= 0. A system with σxx = 0 is therefore not automatically a perfect insulator in the general case.
1.3 The Classical Hall Effect Let us first look at the classical Hall effect and then see how it differs from the quantum Hall effect. The classical Hall effect was discovered by Edwin Hall in 1879. The set-up is as follows: Electrons are restricted to move in the 2D (x,y)-plane while a constant magnetic field points in the z-direction.
By applying a constant electric field E, a constant current density J emerges. The Drude model used to describe the motion of a charged particle in this system goes as follows: mdv dt = −eE −ev × B −mv τ (5) where v is the particle’s velocity, −eE −ev × B is the Lorentz force acting on the particle of charge −e and mv τ is the friction term that models the interaction between the particle and the sample, where τ is the average time between two collision events called scattering time. A high value of τ implies a clean sample as collisions occur more rarely. By imposing the equilibrium condition dv dt = 0 and noting that J = −nev we get the following equation: 1 ωBτ −ωBτ 1 J = e2nτ m E (6) Comparing this with the definition of the resistivity matrix, one finds ρxx = m ne2τ , ρxy = B ne.
(7) The longitudinal resistivity is proportional to 1 τ , so a high value of the scattering time, i.e. a clean sample, yields a low longitudinal resistivity. The transverse resistivity is 3 independent of impurities, it depends on the electron density n and linearly on the magnetic field B. One can use this result to measure the electron density of a given sample by measuring its Hall resistivity. The resistivities are plotted as functions of the magnetic field in figure 1.
Figure 1: Resistivities of the classical Hall system, as functions of the magnetic field.
The blue line shows the longitudinal resistivity ρxx, it is constant in the magnetic field.
The red line denotes the Hall resistivity ρxy, it depends linearly on the magnetic field.
Image reproduced from .
1.4 The Results of the Integer Quantum Hall Effect The experimental set-up for the integer quantum Hall effect is similar to the classical case.
One constructs a two dimensional electron gas from a semiconductor heterostructure, for example a GaAs structure sandwiched in between two AlAs semiconductors. The electrons live in the conduction band of the GaAs, which is lower in energy than the con-duction band of the AlAs. They are therefore trapped to live within the layer of GaAs.
If one makes this GaAs layer sufficiently thin, the electrons are effectively confined to the two dimensional plane that is the GaAs structure.
One takes this 2D electron gas subject to a constant perpendicular magnetic field. At strong magnetic fields B ∼O(T) and low Temperatures T < 4 K the transverse resis-tivity ρxy takes on a plateau form: It is constant over a range of magnetic fields, and jumps to the next plateau once the magnetic field is changed too much. The longitudinal resistivity is zero whenever ρxy sits on a plateau and spikes when ρxy changes from one plateau to the next. The experimental results are shown in figure 2. The plateau values of ρxy are given by the following formula: ρxy = 2πℏ e2 1 ν , ν ∈Z (8) They are independent of the sample details and the precise value of the magnetic field.
The integer ν has been measured up to a precision of 1 part in a billion. While ρxy 4 takes on a plateau value, the Hall conductivity is simply given by the inverse, namely σxy = e2 2πℏν. In general, there would be a caveat to this: In an experiment, one measures resistance and not resistivity! The two are related via the dimensions of the sample, so the precision of measuring the resistivity is dictated by the precision of the measurement of the sample’s dimensions. In 2D however, the transverse Resistance and the transverse resistivity are exactly the same and do not depend on the dimensions of the sample: Rxy = Vy Ix = LEy LJx = Ey Jx = −ρxy (Jy = Ex = 0) (9) Comparing the results from the integer quantum hall effect (figure 2) to the results from the classical model (figure 1), one quickly sees that classical physics does not give us the right answer. We are therefore going to use a quantum mechanical approach to describe the system.
Figure 2: Resistivities of the integer quantum Hall system, as functions of the magnetic field. The red line shows the longitudinal resistivity ρxx, it is zero as long as ρxy sits on a plateau level and spikes whenever ρxx changes from one plateau to the next. The green line denotes the Hall resistivity ρxy, it takes on a plateau form, i.e. it is constant over a range of magnetic fields. Image reproduced from .
1.5 Quantum Treatment 1.5.1 Eigenstates and Eigenvalues The Hamiltonian for one electron in a 2D system subject to a perpendicular magnetic field Bˆ z is H = 1 2m(p + eA)2 (10) where A is the vector potential describing the magnetic field: ∇× A = Bˆ z. Note that the momentum operator p is the canonical momentum operator and differs from the 5 kinetic momentum due the magnetic field: p = m ˙ x −eA. Because we are neglecting all electron-electron interaction for this derivation, any many particle state can be con-structed from the product of single particle states. By choosing Landau gauge A = xBˆ y the Hamiltonian becomes H = 1 2m(p2 x + (py + eBx)2) (11) and is therefore translationally invariant in the y-direction. Consequently, the Hamil-tonian commutes with the y-momentum operator py, so one can motivate the following Ansatz for the energy eigenstates: ψky(x, y) = eikyyfky(x) (12) where the plane wave part makes the wavefunction an eigenstate of the y-momentum operator with eigenvalue ℏky. By restricting ourselves to the subspace of states with y-momentum ℏky, the Hamiltonian can be rewritten as Hky = 1 2mp2 x + mω2 B 2 (x + kyl2 B)2 (13) where ωB = eB m is the cyclotron frequency and lB = q ℏ eB is the magnetic length of the system. But this is the Hamiltonian for a displaced one dimensional harmonic oscillator, centered at x0 = −kyl2 B. The energy eigenvalues of the harmonic oscillator are given by En = ℏωB n + 1 2 , n ∈N, (14) they are called Landau levels. The energy eigenfunctions are given by the product of the plane wave and the 1D harmonic oscillator eigenfunctions: ψn,ky(x, y) ∼eikyyHn(x + kyl2 B)e−(x+kyl2 B)2/(2l2 B) (15) Part of this product is a Gaussian that localizes the eigenfunction around x0 = −kyl2 B on the scale of the magnetic length lB, while the eigenfunction is completely extended in the y-direction. The eigenfunctions are labelled by the harmonic oscillator label n and the momentum label ky. Noting that the energy does not depend on ky, one would expect a huge degeneracy in the energy levels!
1.5.2 Degeneracy First let us confine the sample to a finite rectangular region Lx × Ly. In y-direction we have plane waves, so this is equivalent to the particle in a box system in this direction.
As a result, the y-momenta become quantised: ky = 2π Ly Z, the distance in y-momentum space between two eingefunctions is therefore ∆ky = 2π Ly . The eigenfunctions are localised in x-direction around x0 = −kyl2 B, a logical constraint would be 0 ≤x0 ≤Lx, i.e. the 6 center of the eigenfunction should lie within the sample. Because the center variable x0 is directly linked to a y-momentum ℏky, this condition imposes a constraint on the allowed y-momenta: −Lx/l2 B ≤ky ≤0. The number of states per energy level can now be calculated: N = Ly 2π Z 0 −Lx/l2 B dky = eBA 2πℏ= Φ Φ0 (16) where Φ0 = 2πℏ e is the quantum of flux and Φ = AB is the total flux through the sample. So the number of states per Landau level is given by the number of flux quanta that go through the sample. One can also define the Landau level filling factor ν as ν = ne nΦ = ne2πℏ eB where ne is the electron density.
The sample is finite and has edges. The eigenstates are localized in one direction, so there should be states that live near the edge of the sample. It turns out that they have interesting properties that help to understand the Quantum Hall effect. These states are called edge states.
2 The Edge Picture 2.1 Edge States The existence of edge states can be motivated classically. In a classical Hall bar with a uniform perpendicular magnetic field, charged particles move in circular motion, the cyclotron orbit. Suppose that our particles are negatively charged and therefore move in counter-clockwise direction. If one places a particle at the right edge of the sample, the particle will do a half-orbit until it collides with the edge. It can’t leave the sample and it can also only do counter-clockwise rotational motion, so it will perform another half-orbit, but now it is further down on the edge, see figure 3. The macroscopic effect is a local current on either edge of the sample, with opposing directions, so the net current stays zero.
Figure 3: Classical description of the edge states. Particles at the edge of the sample collide with the edge when moving in the cylcotron orbit and result in an edge current.
Image reproduced from .
7 Edge states can also be treated quantum mechanically. To model the edge of the sample, one introduces a potential well that rises steeply at the edges, as in figure 4.
Figure 4: Model of the sample’s edges as a steeply rising potential. The red line shows the potential that is due to the boundary of the sample. Image reproduced from .
Since the states of the unperturbed system are localised in the x-direction, and by assum-ing that the potential varies slowly on the scale of the magnetic length, we can Taylor expand the potential around these center positions and drop all the constant terms. Plug-ging everything back into the Hamiltonian (on the subspace of ℏky-momentum states as we have expanded the potential around the corresponding centers) we get Hky = p2 x 2m + 1 2mω2 B kyl2 B + x 2 + x∂V ∂x (17) where ∂V ∂x is evaluated at x0 = −kyl2 B. This Hamiltonian can be rewritten to get another displaced harmonic oscillator plus some extra terms that come from completing the square: Hky = p2 x 2m + 1 2mω2 B kyl2 B + ∂V ∂x 1 mω2 B + x 2 − 1 2mω2 B ∂V ∂x 2 −kyl2 B ∂V ∂x (18) The extra terms are all constant on the ℏky-momentum subspace so they’ll transfer over to the energy eigenvalues, while the states will have the same form as the ones found earlier. The resulting energy eigenvalues are En(ky) = ℏωB n + 1 2 − 1 2mω2 B ∂V ∂x 2 −kyl2 B ∂V ∂x .
(19) They do depend on the y-momentum ℏky, so we can calculate a finite group velocity of the wavepackets: vy = 1 ℏ ∂E ∂ky = −1 eB ∂V ∂x (20) By looking again at the potential picture (figure 5) and using the fact that the states are still localised in x-direction, one sees that the states on the right edge propagate in negative y-direction and opposite for the left side.
These edge states are chiral, they can only move in one direction. In a regular wire, there 8 are electrons that move to the right and ones that move to the left. These right-movers and left-movers can coexist in the same spatial region. Upon reversing time right-movers become left-movers and vice-versa. Since they are in the same spatial region, the system looks the same. For the system here, the right-movers are spatially separated from the left-movers. Time reversal thus returns a system that looks like a mirror-image of the original system, it looks reversed. This asymmetry is due to the presence of the magnetic field.
Figure 5: Model of the sample’s edges as a steeply rising potential, where the states below the Fermi level EF are occupied. Blue circles model the centers of the wavefucntions of the occupied states. Image reproduced from .
By introducing an electro-chemical potential difference ∆µ between the two edges, we can calculate the arising current Iy: Iy = −e Ly Z dky Ly 2π vy(ky) = e 2πl2 B Z dx 1 eB ∂V ∂x = e 2πℏ∆µ (21) where we assumed one completely filled Landau level for the integration. By noting that the Hall voltage VH across the sample is given by the potential difference, one finds that the Hall conductivity takes the value of the first plateau level: VH = ∆µ e ⇒σxy = Iy VH = e2 2πℏ (22) The argument can be expanded for multiple filled Landau levels and one will get the corresponding conductivities of the plateaux.
As a side note, the potential introduced before does not have to be flat in the center region, it can be as random as it wants to be (see figure 6) as long as we can make the same arguments for the expansion as before. The resulting Hall current will be the same in any case.
There is another interesting property of the edge states; they are immune to back-scattering by impurities. In the classical picture, defects and impurities scatter incoming electrons into random directions, effectively decreasing the current. Quantum mechan-ically, the electron always has to occupy a state, so it can only be scattered in one of 9 the non-populated states. In our system, these states live above the Fermi level, see the empty black circles in figure 6. If an edge state is scattered into a state right above itself, the resulting Hall current will be the same. If it would scatter on to the other side of the sample, the net current would change. However, the width of the sample is macroscopic, so such scattering is highly suppressed. The Hall current carried by the edge states is therefore immune to back-scattering.
Figure 6: Model of the sample’s edges as a steeply rising potential, where the middle part seems random. The empty black circles denote unoccupied states right above the Fermi level. Image reproduced from .
The above argument might look convincing, but it does not give any reason as to why the conductivity is stable against change in the magnetic field. To see this, we need something else, we need to add disorder to our system. We will see that disorder is essential for the integer quantum Hall effect.
2.2 The Role of Disorder Experimentally, there is always disorder in a system. But before we look at the effects of disorder, let us see what happens without it. In the absence of disorder, the Hall system features continuous translational invariance. In the lab frame, the fields are as follows E = 0, B = Bˆ ez (23) where the current density J is zero. Transforming to a Lorentz-boosted reference frame, where the boost is along the y-direction at speed v, the fields become the following: E’ = γvBˆ ex ≈vBˆ ex, B’ = γBˆ ez ≈Bˆ ez (24) where γ ≈1 is valid for small v. The transformed current densitiy is J = nevˆ ey (25) where the minus sign due to the Lorentz boost cancelled the one coming from the negative charge of the electrons.
The electric field expressed in terms of the current density becomes E = 1 neJ × B ⇒ Ex Ey = B ne 0 1 −1 0 Jx Jy (26) 10 so the resulting resistivities and conductivities are ρxx = 0, ρxy = B ne, σxx = 0, σxy = ne B (27) which coincide with the classical transverse resistivities.
This Lorentz argument relies purely on translational symmetry, it does not care about any type of mechanics. In reality the result measured is not this classical result, so there has to be something that breaks translational symmetry, and disorder seems to do the job.
Quantum mechanically, the impurities can be modelled by a random potential V that is much smaller than the Landau level spacing: V ≪ℏωB. This potential breaks transla-tional symmetries, so it will lift the degeneracy of the Landau levels, see figure 7. The resulting Landau levels are thus broadened.
Figure 7: Densitiy of electron states. The left picture is without disorder, showing sharp Landau peaks. The right picture shows broadened peaks due to the effects of disorder.
Image reproduced from .
Another effect due to disorder is localisation. Figure 8 shows the top-view of a potential landscape on the Hall bar. The lines are equipotential lines, the ”+” sign denotes a potential peak and the ”-” sign a potential pit. The states derived before have defi-nite energy, they can’t change from one equipotential line to another because they can’t change their energy. States that live on equipotential lines around peaks and pits are therefore localised, they can’t move from one side of the sample to the other and hence can’t carry a current. Due to the steeply rising potential at the edge of the sample, extended states are guaranteed to exist at the edges. Using this information, one can add more detail to the picture of the state density, see figure 9. The inner part of the peak contains all the extended states while the shell consists of localised states.
Let us now look at the consequence of localisation.
Suppose that our system has a filling factor of ν = 1, i.e. one full Landau level. The Fermi level EF lies on top of the first peak. Remember the formula for the number of states per Landau level, N = BA Φ0 .
Now we decrease the magnetic field B slightly, so the number of states per Landau level decreases as well. But the number of electrons in the system stays constant, some of 11 them have to populate states in a higher Landau level. The Fermi level E′ F jumps into the next region of states, which is a region of localised states in the next Landau level, see figure 9. These states however do not carry a current, they can’t contribute to the conductivity! The conductivity thus stays constant over a range of magnetic fields, it stays constant as long as the Fermi level lies in a region of localised states, a mobility gap.
So disorder explains the stability to the magnetic field, it is essential to the whole effect.
Figure 8: Top view of the potential landscape. The lines are equipotential lines, they form closed loops around peaks (”+” signs) and pits (”-” signs). Image reproduced from .
Figure 9: Densitiy of electron states. The outer parts in a Landau peak are localised states, where extended states live in the inner part.
The Fermi level EF before the change in magnetic field lies right at the top of the first Landau level, indicating a filling factor of 1. The Fermi level E′ F after the change in the magnetic field lies in the region of localised states in the second Landau level. Image reproduced from .
In the next step we will look at a more sophisticated argument as to why the conductivity is quantized in the values of σxy = e2 2πℏν, ν ∈Z.
12 2.3 The Role of Gauge Invariance The following argument relies on a different geometry that can be achieved by bending the Hall bar into a disk shape, called a Corbino ring, that is shown in figure 10. We are still keeping the constant magnetic field going through the sample. We can introduce a flux Φ that goes through the hole in the disk, without affecting the magnetic field going through the disk surface. Let us assume that this flux can be changed adiabatically, i.e.
arbitrarily slowly.
Faraday’s law then tells us that a change in flux induces an electromotive force E. A radial current Ir would then allow us to calculate the Hall conductivity σH of the system, shown by the following equation: E = −∂Φ/∂t = −∆Φ/T, Ir = ∆Qr/T →σH = Ir E (28) where ∆Qr is the charge moved radially during the addition of a flux ∆Φ during a time T.
Figure 10: The Corbino geometry, where the states live in the red ring that is penetrated by a uniform magnetic field. The blue tube models the addition of flux through the hole.
Image reproduced from .
Now we will see that this radial current takes the appropriate values to get the correct Hall conductivity. The system has rotational invariance (we are ignoring disorder for the moment). By choosing the symmetric gauge, this symmetry is still apparent in the Hamiltonian. Remember in the previous derivation of the eigenstates, the Hamiltonian had translational invariance in the y-direction and we found plane waves in said direction.
We also found harmonic oscillators in x-direction that were localised.
By doing the calculation for this Corbino geometry in the symmetric gauge, one finds again plane waves in azimuthal direction and harmonic oscillators in radial direction: ψm,ν(⃗ r) ∼eimφfν(r −rm) (29) where fν is a harmonic oscillator eigenfunction of the level ν+1 . The center coordinate rm is given by rm = q 2l2 Bm that is now related to the angular momentum ℏm as opposed 13 to the x0 coordinate being related to the y-momentum.
Similar to the Aharonov-Bohm effect , adding one flux quantum leaves the system invariant. By adding this amount of flux adiabatically, one finds that the states increase their angular momentum by ℏ, they are mapped onto themselves. The center coordinate also increases from rm to rm+1, so the states effectively move outwards. Because the system comes back to itself after the addition of one flux quantum, the number of electrons in our system stays the same. It started with an integer number of electrons so it ends with an integer number. Therefore exactly one electron per filled Landau level is transferred from the inner to the outer edge. The resulting Hall conductivity takes the values observed in the experiment: ∆Φ = Φ0 →∆Qr = −ne, σH = ne Φ0 = n e2 2πℏ, (30) where n is the number of filled Landau levels. This argument involving the adiabatic change in flux is called the Laughlin flux insertion argument, as Robert B. Laughlin first came up with this idea .
We are now going to look at how disorder affects this argument. As seen before, disorder localises the states in the bulk but leaves extended states at the edge. The localised states are not affected by the addition of flux as it can be removed by a gauge transformation.
For extended states, this only works if the flux is an integer multiple of magnetic flux quanta, because they have to obey a single-valuedness condition: By changing φ 7→φ+2π the state should remain unchanged. The extended states still undergo spectral flow and map onto themselves under the adiabatic addition of integer number of flux quanta. It is still guaranteed that at least two extended states exist per Landau level, one on either edge. As long as all the extended states in a Landau level are populated, threading one flux quantum results in the radial current and the correct conductivity as we calculated before. The extended states are populated as long as the Fermi level lies in a mobility gap, which is possible due to the existence of disorder. Again, disorder and impurities are responsible for the stability of the Hall conductivity to changes in the magnetic field.
This completes the discussion of the integer quantum Hall effect within the edge picture.
Parts of it are heuristic, but the role of gauge invariance hints that there may be a deeper connection. In the next section, we are going to look at the bulk picture and we will find that the integer quantum Hall effect appears as a consequence of a topological property of the Hall system.
14 3 The Bulk Picture 3.1 The Kubo Formula for Hall Conductivity Before we can look at the topological property, we need to write the Hall conductivity in a different way, given by the Kubo formula. A full derivation of the Kubo formula can be found in D. Tong’s lecture notes about the quantum Hall effect . Here, only a rough sketch of the derivation will be given.
One starts with a multi-particle Hamiltonian H0 for a generic system that does not in-clude an electric field. We assume that we can solve this system, i.e. we know eigenstates |m⟩and eigenvalues Em such that H0 |m⟩= Em |m⟩. We then introduce a weak electric field E = −∂tA as a perturbation; ∆H = −J · A, where J is the current operator. Our goal is to calculate the expectation value of the current operator ⟨Jx⟩for the system in the non-degenerate ground state |0⟩. It is convenient to work in the interaction picture and expand the expression for ⟨Jx⟩up to linear order in ∆H. Comparing this new ex-pression with Ohm’s law Jx = σxxEx + σxyEy, one can filter out a formula for σxy. The precise derivation is part of a bigger story called linear response, but we are not going deeper into this. As a result, the expression for the Hall conductivity becomes σxy = i ℏ LxLy X n̸=0 ⟨0 | Jy | n⟩⟨n | Jx | 0⟩−⟨0 | Jx | n⟩⟨n | Jy | 0⟩ (En −E0)2 .
(31) This is the Kubo formula for Hall conductivity. The sum runs over all the excited multi-particle states, inside there are transition amplitudes from excited states over the current operators to the ground state, weighted by the energy difference of these states. So if one can solve the unperturbed system, this formula instantly gives an expression for the Hall conductivity of that system.
In the next section we will see a how a topological property leads to the quantisation of the Hall conductivity.
3.2 Hall System on a Torus For this section, we have to modify the geometry again. This time, the Hall bar is bent into a torus, which is a rectangle with opposite edges identified. The system is now made fully periodic, all edges have been removed, so there are no more edge states. There is still a homogeneous magnetic field perpendicular to the torus’ surface. (Note it is a thought experiment. In an actual experiment, this non-zero magnetic flux through a closed surface would require magnetic monopoles.) 15 Figure 11: The Hall system on a torus. The red and blue shapes indicate two different fluxes that can be added to the system. Image reproduced from .
Now we expand Laughlin’s flux threading argument: One can now insert two fluxes, one through either hole of the torus, as in figure 11. The Hamiltonian now depends on these two fluxes, H = H(Φx, Φy). Similar to the previous case, the system’s spectrum is only sensitive to the non-integer part of Φi/Φ0, i ∈{x, y}. This time, there are no edges, no electrons transfer from one side to another one. By the addition of one flux quantum, the system comes back to itself fully! Having zero flux is therefore the same as having one flux quantum, which makes the space of parameters of the Hamiltonian periodic: 0 ≤Φx < Φ0, 0 ≤Φy < Φ0 (32) This describes a torus T2 Φ, the parameter space of our system is toroidal. Let us now calculate the system’s Hall conductivity.
Let H0 describe the unperturbed system. Treat the addition of flux as a perturbation ∆H given by ∆H = − X i=x,y JiΦi Li (33) where Lx and Ly are the dimensions of the system. One can now use first order pertur-bation theory to calculate the many particle ground state of the perturbed system: ψ′ 0 = |ψ0⟩+ X n̸=ψ0 ⟨n | ∆H | ψ0⟩ En −E0 |n⟩ (34) where |ψ0⟩is the non-degenerate ground state of the unperturbed system. Consider now infinitesimal changes in flux and see how the state changes: ∂ψ′ 0 ∂Φi = −1 Li X n̸=ψ0 ⟨n|Ji|ψ0⟩ En −E0 |n⟩ (35) 16 The elements in this summation look similar to the Kubo formula, eq. (31)! One can actually rewrite the Kubo formula (and therefore the Hall conductivity) in terms of these states: σxy = iℏ ∂ψ′ 0 ∂Φy ∂ψ′ 0 ∂Φx − ∂ψ′ 0 ∂Φx ∂ψ′ 0 ∂Φy (36) This expression can be rewritten by pulling out one derivative for each one of the brakets, becoming σxy = iℏ ∂ ∂Φy ψ′ 0 ∂ψ′ 0 ∂Φx − ∂ ∂Φx ψ′ 0 ∂ψ′ 0 ∂Φy .
(37) This however looks similar to the field strength of a Berry connection.
Let us first parameterise the torus using the dimensionless variables θi = 2πΦi Φ0 , θi ∈[0, 2π) .
(38) Now, the Berry connection is defined as Ai(Φ) = −i ψ′ 0 ∂ ∂θi ψ′ 0 .
(39) The field strength, or curvature of the Berry connection is the curl of the connection: Fxy = ∂Ax ∂θy −∂Ay ∂θx = −i ∂ ∂θy ψ′ 0 ∂ψ′ 0 ∂θx − ∂ ∂θx ψ′ 0 ∂ψ′ 0 ∂θy (40) This term looks pretty much the same as our expression for the conductivity, eq (37).
The difference that in one case, the derivatives are with respect to θi instead of Φi, gives the right constants. As a result, the conductivity is given by the curvature: σxy = −e2 ℏFxy (41) This does not tell us much yet. However if one averages this expression over all fluxes in the toroidal parameter space T2 Φ, one gets a more interesting result: σxy = −e2 2πℏ Z T2 Φ dθ 2πFxy = −e2 2πℏC (42) where Z T2 Φ dθ 2πFxy = C ∈Z (43) is the first Chern number, and it is always an integer .
This result is the topological argument we have been looking for. The Hall conductivity is a Chern number, it can’t continuously change. It is therefore invariant under small changes of the Hamiltonian, which would result in small changes of the Berry curva-ture. One would always expect a graph of Chern numbers to form plateaux. Large 17 deformations correspond to energy level crossings, and our non-degeneracy assumption breaks down. This situation describes a transition between two Chern numbers, i.e. two plateau levels in the conductivity.
We have seen that for a system with continuous translational symmetry, a toroidal pa-rameter space led to the desired quantisation. However, we haven’t seen a reason as to why one should average the conductivity over the parameter space, and I do not know one either. It turns out that if one looks at particles on a lattice, the integral appears naturally, without taking any averages. Such a system was used originally to measure the quantum Hall effect. It only features discrete translational invariance, so one might expect some things to go wrong. In the next sections, we are going to look at this more closely.
3.3 Particles on a Lattice A lattice has a discrete periodicity given by the lattice vectors rn, and so does the corresponding lattice potential. The energy spectrum of such a system forms bands.
Due to Bloch’s theorem, we know that the wavefunctions in each band can be written in the Bloch form as ψk(r) = eik·ruk(r), uk(r) = uk(r + rn), ψk+G(r) = ψk(r) (44) where k is the lattice momentum and G is a reciprocal lattice vector. Most importantly, the wavefunction is periodic in the first Brillouin zone. For a rectangular lattice with lattice constants a and b, the Brillouin zone is rectangular as well.
But a periodic rectangle is a torus, so our wavefunction is periodic on a toroidal space T2 of lattice momenta: −π a < kx ≤π a, −π b < ky ≤π b (45) Let us now assume that the Fermi level EF lies in a gap between two energy bands, meaning that all the bands below the Fermi level are completely filled, and all the bands above are completely unoccupied. The Hamiltonian of such a system can be rewritten using Bloch’s theorem: H |ψk⟩= Ek |ψk⟩⇒e H(k) |uk⟩= Ek |uk⟩with e H(k) = e−ik·xHeik·x (46) The current operator is now given by J = e ℏ ∂e H ∂k (47) which is analogous to the more conventional J = −e ˙ x. The Kubo formula we have seen in section 3.1 can be rewritten in terms of single particle states. This is because all electron-electron interactions are neglected, so a multi-particle state is just a product of all the single-particle states. For states living in discrete energy bands with continuous lattice momenta, the Kubo formula becomes σxy = iℏ X Eα<EF <Eβ Z T2 d2k (2π)2 ⟨uα k|Jy|uβ k⟩⟨uβ k|Jx|uα k⟩−⟨uα k|Jx|uα k⟩⟨uα k|Jy|uα k⟩ (Eβ(k) −Eα(k))2 (48) 18 where the sum runs over all the energy bands and the integral runs over all the continuous lattice momenta in each band. By plugging the Ansatz for the current operator (47) into this Kubo formula, one gets σxy = ie2 2πℏ X α Z T2 d2k 2π [⟨∂yuα k|∂xuα k⟩−⟨∂xuα k|∂yuα k⟩] .
(49) Let us now look at the following U(1) Berry connection, defined over T2: Aα i (k) = −i uα k ∂ ∂ki uα k (50) The field strength associated to this connection is Fα xy = ∂Aα x ∂ky −∂Aα y ∂kx = −i ∂uα ∂ky ∂uα ∂kx − ∂uα ∂kx ∂uα ∂ky .
(51) By integrating this field strength over the Brillouin zone T2, one obtains the first Chern number, Cα = Z T2 d2k 2π Fα xy.
(52) By comparing equations (51) and (52) with the expression for the Hall conductivity (49), we get σxy = −e2 2πℏ X α Cα, Cα ∈Z.
(53) This is the TKNN formula, named after Thouless, Kohmoto, Nightingale and Nijs, which are the people that came up with it . It tells us that each isolated, filled band (labelled by α) contributes an integer part Cα to the Hall conductivity of the system. The set of Cα is sometimes called the TKNN integers. We have attributed this set of integers to a Hamiltonian, they define the Hall conductivity of the corresponding system. Similar to the case in section 3.2, this formula states that the Hall conductivity is a topological invariant of the system. It is robust to small changes of the system as it can’t change continuously.
We are going to see a more precise formulation of this statement in the end. Let us now look at an example, giving a hint at this more precise formulation.
3.3.1 Example: The Chern Insulator The system we are looking at has two distinct energy bands and the Fermi level lies between them, meaning the system is insulating. Even though there is no magnetic field, this system can have a non-zero Chern number, hence the name Chern insulator.
The system in k-space is modelled by a two-state Hamiltonian in the most general form, e H(k) = ⃗ E(k) · ⃗ σ + ϵ(k)1 (54) 19 where ⃗ σ are the Pauli matrices. The energy eigenvalues corresponding to this system are given by ϵ(k)±| ⃗ E(k)|. The fact that the system is insulating translates into ⃗ E(k) ̸= 0 ∀k.
One can therefore define the following three-vector ⃗ n: ⃗ n(k) = ⃗ E(k) | ⃗ E(k)| (55) This vector has unit length, but varies continuously as k varies over the Brillouin zone, it is a mapping from the torus to the (Bloch-) sphere: ⃗ n : T2 →S2 (56) The Chern number associated with this system is given by C = 1 4π Z T2 d2k ⃗ n · ∂⃗ n ∂kx × ∂⃗ n ∂ky (57) which can be interpreted as counting how many times the torus wraps around the sphere.
Figure 12: The three-vector ⃗ n is a mapping between the toroidal Brillouin zone and the Bloch sphere. Image reproduced from .
Let us now try to classify the maps from the torus to the sphere. In this case, there is a subtlety: maps from the torus to the sphere are isomorphic to maps from sphere to sphere. These maps are classified by the second homotopy group of the sphere, π2(S2).
It turns out that this group is isomorphic to the set of integers, π2(S2) = Z.
(58) Since our Hamiltonian is basically ⃗ E(k) which is represented by the sphere S2, this result tells us that any two systems that are deformable into each other (homotopic) have the same Chern number and hence the same Hall conductivity! A system with zero Hall conductivity therefore can’t be continuously transformed into a system with finite Hall conductivity because their Hamiltonians live in different homotopy classes. This again explains the robustness of the Hall-state, but just for this two-state system.
This example is a special case of a more general thing. There is a theorem stating that 20 two maps from the Brillouin zone to the space of gapped hermitian matrices (Hamilto-nians) are homotopic if and only if the they both have the same TKNN integers, and therefore have the same Hall conductivity, see . We will come back to this again in section 4.
In the next section we are going to look how things change when we introduce a magnetic field. It is not obvious that the Brillouin zone survives this addition, so we might want to check if the TKNN formula is still applicable.
3.4 Particles on a Lattice in a Magnetic Field So far we haven’t included a magnetic field into our discussion about particles on a lattice. In the experiment, this has been done, and from the edge picture we know that a magnetic field is crucial to the effect. The main goal of this treatment is to find a zone where the states of the system are periodic, we want to find an analogue to the Brillouin zone. If we have that, one can simply apply the TKNN formula. It is not obvious that there exists such a magnetic Brillouin zone as the Hamiltonian loses the discrete translational invariance of the lattice due to the vector potential arising from the magnetic field.
Let us work in the tight-binding approximation and assume a square lattice with lattice constant a. One can also find a Brillouin zone when working with a weak potential and a non-square lattice, see .
In the tight-binding approximation, the states are localised around the lattice points. These position eigenstates |x⟩are discrete, they are restricted to x = a(m, n) with m, n ∈Z.
The Hamiltonian is discrete as well.
For this approximation of a strong potential, the Hamiltonian allows states to hop from one lattice site to an adjacent one. It is given by H = −t X x X j=1,2 |x⟩⟨x + ej| −t∗X x X j=1,2 |x + ej⟩⟨x| (59) with e1 = (a, 0), e2 = (0, a) and t being the probability amplitude for hopping to occur.
The on-site energy has been set to zero, it is just an additive constant to the Hamiltonian.
Now we can add a magnetic field to the system, Bˆ z = ∇× A.
(60) Intuitively, a vector potential is part of the canonical momentum p. Canonical momen-tum is the rate at which the phase of a state changes, per unit length. For our set of discrete states, this length is fixed at a. So the change in the Hamiltonian is just a phase, its new form is H = −t X x X j=1,2 |x⟩e−ieaAj(x)/ℏ⟨x + ej| + h.c., (61) which has no discrete translational invariance anymore. This Hamiltonian should better be compatible with the Aharonov-Bohm effect, and indeed it is. Let us take a state and move it once around a unit cell (see figure 13), it will pick up a phase of e−iγ. This γ 21 can be calculated from the form of the Hamiltonian, and it is γ = ea ℏ(A1(x) + A2(x + e1) −A1(x + e2) −A2(x)) ≈ea ℏ ∂A2 ∂x1 −∂A1 ∂x2 (62) where in the second step, we have approximated finite differences with actual derivatives (remember our space coordinate is discretised). But the expression in the bracket is just the curl of the vector potential, so γ ≈ea ℏ ∂A2 ∂x1 −∂A1 ∂x2 = ea2B ℏ (63) where a2B is the magnetic flux through the unit cell, so this really is the Aharonov-Bohm phase.
Figure 13: Schematic view of the square lattice. A state moving along a lattice site picks up a phase proportional to the vector potentials written above each line. Image reproduced from .
Now that we have a Hamiltonian, we want to find eigenstates and eigenvalues. Our goal is to find operators that commute with the Hamiltonian and then to find a common set of eigenstates. Let us define the modified magnetic translation operators: e Tj = X x |x⟩e−iea e Aj(x)/ℏ⟨x + ej| (64) The e Aj is not the same vector potential as before, but it fulfils ∂k e Aj = ∂jAk. These operators have the following commutation relations: [H, e Tj] = 0, e T2 e T1 = eieΦ/ℏe T1 e T2 (65) The first commutation relation can be seen by introducing the usual magnetic translation operator (which replaces e Aj by Aj). The Hamiltonian is then given by a superposition of these operators, and they commute with their modified counterparts. The second relation we don’t like quite as much yet. It would be nicer if [ e T1, e T2] were zero as well, because in that case one could find simultaneous eigenstates of all three operators. For this to happen we need to take the flux per unit cell to be a rational multiple of the flux 22 quantum: Ba2 = Φ = p qΦ0, with p, q integers that are relatively prime. Then, one can rise the modified magnetic translation operators to an appropriate power and one gets [ e T n1 1 , e T n2 2 ] = 0 (66) whenever p qn1n2 ∈Z, in particular for n1 = q, n2 = 1. Of course they still commute with the Hamiltonian. So a set of commuting operators is given by H, e T q 1 , e T2 and we can now move on to the eigenstates. Let us label these states by their magnetic lattice momentum k: |k⟩, k = (k1, k2) (67) The eigenvalue equations then become H |k⟩= E(k) |k⟩, e T q 1 |k⟩= eiqk1a |k⟩, e T2 |k⟩= eik2a |k⟩.
(68) We see that the ki are again periodic: −π qa < k1 ≤π qa and −π a < k2 ≤π a (69) The zone of periodicity is called the magnetic Brillouin zone, and it again forms a Torus T2! It is smaller by a factor of q compared to the conventional Brillouin zone. The corresponding unit cell (enlarged by a factor of q) contains p magnetic flux quanta.
Finding the Brillouin zone hence boils down to finding a unit cell containing an integer number of magnetic flux quanta! The number of states per magnetic Brillouin zone is given by L1L2/qa2. This suggests that the energy spectrum decomposes into q bands.
Now we look at the degeneracy of the states. Let us pay closer attention to the state e T1 |k⟩which is not an eigenstate of e T1. From the calculations before, we know that e T1 commutes with the Hamiltonian, [H, e T1] = 0. We have H e T1 |k⟩= E(k) |k⟩, e T2 e T1 |k⟩= eieΦ e T1 e T2 |k⟩= ei(2πp/q+k2a) e T1 |k⟩.
(70) From equation (68) we know that e T2 |k⟩= eik2a |k⟩, so e T1 |k⟩∼|(k1, k2 + 2πp/qa)⟩ (71) and it also has the same energy as |(k1, k2)⟩. This results in a q-fold degeneracy in a given band. One can now go on and calculate the energy eigenvalues and the Chern numbers. However, one might expect to run into some issues, as each isolated band has to contribute an integer number of e2 2πℏto the Hall conductivity, and the number of bands q can become arbitrarily large by changing the magnetic flux Φ by an arbitrarily small amount. Also, if the flux is given by an irrational multiple of Φ0, the above calculation doesn’t work out and the spectrum of the Hamiltonian forms a Cantor set. But the experiment shows the conductivity to remain constant over a range of magnetic flux, no rapid behaviour is observed. Nontheless, the calculation works out, but it is not given here as it is rather complicated. It can be found in the original TKNN paper and in the book Field Theories of Condensed Matter Physics by Fradkin .
23 Now we have all that we need to apply the TKNN formula (53): There exists a toroidal Brillouin zone and the spectrum decomposes into bands. As a result, the system has a quantised Hall conductivity. So we found the quantised conductivity for the Hall system in a picture without edges! This concludes the discussion of the integer quantum Hall effect within the bulk picture.
4 Conclusion Now that we have found the quantised conductivity in the lattice model, one might ask what exactly the topological property of the Hall system is. There is a theorem that answers this question, it was briefly mentioned in section 3.3.1. It goes as follows: Let H1(k), H1(k): T2 →[space of Hamiltonian matrices] be two maps from the (magnetic) Brillouin zone (which is toroidal) to the space of gapped, periodic, hermitian matrices, i.e. Hamiltonians that describe the system, the band structure. Then, these two maps are continuously deformable into each other (homotopic) if and only if they have the same TKNN integers. Remember the TKNN formula giving the Hall conductivity for such a system, σxy = −e2 2πℏ X α Cα, Cα ∈Z.
(72) This theorem tells us, that if two systems are continuously deformable into each other, they have the same Hall conductivity. If they have different Chern numbers (and hence a different conductivity), they are not deformable into each other, they live in different homotopy classes. It also shows us that the topological property lies in the connection between the Brillouin zone and the band structure (which is H(k)).
At the same time, we found a quantised Hall conductivity in the edge picture. These edge modes are the counterpart to the quantised conductivity, their robustness is due to this topological property of the Hamiltonian. The topology is physically manifest in the number of protected edge modes of the system. The edge modes exist because of the topological indices of the system, the TKNN integers. This correspondence is also known as the bulk-edge correspondence.
24 References D. Tong, ”The Quantum Hall Effect”, arXiv:1606.06687.
B.I. Halperin, ”Quantized Hall conductance, current-carrying edge states, and the existence of extended states in a two-dimensional disordered potential”, Phys. Rev.
B 25, 2185 (1982).
R.B. Laughlin, ”Quantized Hall conductivity in two dimensions”, Phys. Rev. B 23, 5632 (1981).
D.J. Thouless, M. Kohmoto, M. Nightingale, M. den Nijs, ”Quantized Hall Conduc-tance in a Two-Dimensional Periodic Potential”, Phys. Rev. Lett. 49, 405 (1982).
J.E. Avron, R. Seiler, B. Simon, ”Homotopy and Quantization in Condensed Matter Physics”, Phys. Rev. Lett. 51, 51 (1983).
The University of Manchester, ”Advanced Quantum Mechanics 2” [Online], N. Manton, P. Sutcliffe, ”Topological Solitons”, Cambridge University Press (2004).
E. Fradkin, ”Field Theories of Condensed Matter Physics”, University of Illinois (2013).
25 |
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Unit Rates Exit Ticket
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Unit Rates Guided Notes Homework Warm Ups Exit Tickets
Created by
Jessica Barnett Math
This packet is full of thorough and easy-to-follow pages perfect for teaching this skill! This resource covers writing rates, determining the unit rate, and determining the better buy. WHAT'S INCLUDED2 WARMUPS 2 EXIT TICKETS2 PAGES OF GUIDED NOTES - I DO, WE DO, YOU DO FORMAT3 PAGES OF DIFFERENTIATED HOMEWORK - LEVELS A, B, & CQUICK ASSESSMENT ANSWER KEYSGRAB A FULL YEAR OF 6TH GRADE MATH RESOURCES HERE!GRAB A FULL YEAR OF 7TH GRADE MATH RESOURCES HERE!The purchase of this resource authorizes
6 th - 8 th
Math
CCSS
7.RP.A.1
Also included in:7th Grade Math Curriculum Mega Bundle Worksheets Activities Homework
$3.00
Original Price $3.00
Rated 4.84 out of 5, based on 31 reviews
4.8(31)
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Unit Rates Exit Slips 6th Grade Math
Created by
Brandy Shoemaker
Assess your students' mastery of targeted skills using this handy set of exit slips/quizzes on rates! This resource includes five-question quick checks with a variety of possible uses for your lessons on finding unit rates, unit price, rate of speed, and problem solving using rates. No prep, low ink, saves paper! Options for use in your instruction:Exit slips following a lessonBell work or bellringer slips to open a lessonPre-assessments to guide instructional plansQuizzes to gauge mastery and
6 th
Math, Numbers, Other (Math)
CCSS
6.RP.A.2
$3.00
Original Price $3.00
Rated 4.86 out of 5, based on 14 reviews
4.9(14)
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Unit Rates and Percents Worksheets and Exit Tickets Unit 4 6th Grade iReady Math
Created by
MB Creations
Teach 6th-grade students how to solve problems with rates and percents with these math worksheets, exit tickets, and vocabulary activities. Whether you teach iReady Math or another curriculum, these print and digital problems align with Common Core State Standards and prepare students for standardized testing. Assign the lesson checks about percents as homework or share in small groups. Present these rigorous math questions in PowerPoint or Google Classroom. Lessons in 6th Grade iReady Math Uni
6 th
Algebra, Math, Other (Math)
CCSS
6.RP.A.3
, 6.RP.A.3a
, 6.RP.A.3b
+2
Bundle (4 products)
$15.00 Original Price $15.00
$13.50
Price $13.50
Rated 4.83 out of 5, based on 6 reviews
4.8(6)
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Unit Rates Guided Notes Pracitce Homework Exit Tickets
Created by
Eugenia's Learning Tools
With this product you will receive ready to print unit rate practice to help support your lessons as students develop a better understanding of proportional relationships. Students will practice calculating unit rate, using unit rate, dealing with complex fractions, and finding the better buy. Product includes: Notes for Math JournalPractice WorksheetsExit TicketsQuizzesAnswer KeysNeed more unit rate practice? Bundle is available!
6 th - 7 th
Basic Operations, Math, Other (Math)
Also included in:Unit Rates Activities Bundle | Proportionality Unit Classwork Practice
$4.75
Original Price $4.75
Rated 5 out of 5, based on 4 reviews
5.0(4)
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Unit Rate Activities 6th Grade Math Warm Up Exit Ticket Practice Worksheets
Created by
Math All Day
Finish your lesson plans in a flash with 3 no prep activities on unit rates with for 6th grade math. Warm Up Problems - start class with a routine and set your students up for success for the day. Independent or Partner Practice or Homework - 8 problems with various unit rate problems. Exit Ticket - end class with a routine and gather feedback on student progress. Grab the mini bundle here to use these activities with task cards for even more practice! This product includes:3 printable pag
6 th
Math, Numbers, Other (Math)
CCSS
6.RP.A.2
, 6.RP.A.3b
Also included in:Unit Rate Activities 6th Grade Math Task Cards with Warm Up Practice Worksheets
$3.00
Original Price $3.00
Rated 4.67 out of 5, based on 3 reviews
4.7(3)
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Writing Ratios and Finding Unit Rate from Ratio Tables Riddle with Exit Tickets
Created by
Level Their Learning
In this product, students will write a ratio from a ratio table, and then they will find the unit rate from ratio tables. 13 problems for students to write ratios from ratio tables and to find the unit rate from ratio tables + 1 box has them create their own and solve! (Higher Order Thinking Skills!!) No decimals are used. I used with kids on target when teaching writing ratios from a ratio table and finding the unit rate from a ratio table! There are also 2 exit tickets on writing ratios from a
6 th - 7 th
Math
CCSS
6.RP.A.1
, 6.RP.A.2
, 6.RP.A.3
+3
Also included in:6th Grade Year Long Activity Bundle- Riddles for a Whole Year!
$2.00
Original Price $2.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Grade 7 Math Exit Ticket 7.RP.1 - Compute Unit Rates
Created by
Professor C Academy
This exit ticket helps students practice computing unit rates associated with ratios of fractions. Intro: Assess students' understanding of unit rates through real-world scenarios. Quick Review: Reinforce the concept of unit rates and complex fractions. Teacher Tips: - Encourage students to visualize ratios with diagrams. - Provide examples using different units for clarity. - Remind students to simplify complex fractions. Inclusions: - Sample problems for practice - S
5 th - 8 th
Algebra, Other (Math)
$1.00
Original Price $1.00
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Grade 6 Math Exit Ticket 6.RP.2 - Understanding Unit Rates
Created by
Professor C Academy
This exit ticket helps students grasp the concept of unit rates associated with ratios. - Intro: Engage students in understanding unit rates through real-life examples. - Quick Review: Recap the definition of a unit rate and its application in ratios. - Teacher Tips: Encourage students to use rate language in everyday contexts. - Includes: Sample problems; space for student responses; and a reflection question.
5 th - 8 th
Algebra, Other (Math)
$1.00
Original Price $1.00
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Winter Math Activity Unit Rate bulletin Board | Door Decoration | Exit Ticket
Created by
Mrs Padillas Math Class
Make your winter bulletin board pop with this easy and ready to use exit ticket, warm up, early finisher work, station, holiday activity, etc. Students can color the mugs and post in a bulletin board or around the room. Can be use as a warm up, exit ticket or station activity. Winter themed 6 Exit ticket - Find the Unit Rate ImportantYou can receive TPT store credits to use on future purchases by leaving feedback on products you buy! Just click on "My Purchases" under "Buy"! Connect with
6 th - 8 th
Arithmetic, Basic Operations, Math
CCSS
7.RP.A.1
Also included in:7th Grade Exit Tickets | Warm ups | Growing Bundle
$2.00
Original Price $2.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Unit Rate with Fractions POD/Warm-up/Exit Ticket 7.RP.1
Created by
Jami King
Problem of the day or Warm up or used for an exit ticket. Students must -Find Unit Rate -Draw a picture representation -Make a table -Tie a Division problem to the picture and solve! Shows several methods/ways of attaining the same answer and representing it in many ways. Many students know the answer immediately, but when they have to represent with a picture or a division problem they get stuck. Can be individual or whole group instructed. Answer Document coming! 12 mins per mile 4/
6 th - 8 th
Algebra, Fractions, Math Test Prep
$1.25
Original Price $1.25
Rated 5 out of 5, based on 3 reviews
5.0(3)
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Unit Rate Better Buy Do Now Exit Slip Worksheet
Created by
Ms Novak in the Middle
This worksheet can be used as a do now or exit slip to check students understanding of better buy and unit rate. Students are given four different peanut butter brands with their price per ounce. Students are asked to find the unit rate of all four brands. After finding the unit rate of all four brands students are asked to identify the better buy and explain why it is the better buy. I teach for a double period every day. I used these slips as a check in between periods. This slip allowed me t
6 th - 7 th
Algebra, Applied Math, Math
CCSS
7.RP.A.1
Also included in:Ratios and Proportions Digital and Printable Activity Bundle
$1.50
Original Price $1.50
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Ratios &Unit Rates | Warm Ups | Exit Ticket | 6th Grade Math Worksheets
Created by
Pin Point Resources
These 6th grade math slides and worksheets focused on equivalent ratios, unit rates, tables, graphs, and nonequivalent ratios. These 6 problems are the perfect combination of worksheet and guided math talks. Choose between using the problems for warm ups, exit tickets, small group or whole group discussion. It’s the perfect discussion start for an entire unit on ratios and rates. Styled as a low floor, high ceiling math activity, students of all abilities will engage and increase their confiden
5 th - 7 th
Fractions, Graphing, Numbers
CCSS
6.RP.A.1
, 6.RP.A.2
, 6.RP.A.3
FREE
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Unit Rates Choice Board Worksheet | Exit Ticket | Zero Prep | Homework
Created by
Easy as Pi Learning
Enhance student engagement and control with the Unit Rates Choice Board Activity! This versatile tool can be used as a time filler, exit ticket, homework assignment, sub plan, or partner activity. Ideal for teachers who want to decide the amount of work/time students invest in the assignment, this choice board offers endless options. Use it as a quick formative assessment by having students pick fewer problems, or incentivize them by offering ways to lower the number of problems.This activity fo
6 th - 10 th
Algebra, Basic Operations, Math
CCSS
6.RP.A.2
, 7.RP.A.1
, 7.RP.A.2
Also included in:Secondary Math Skills Choice Board Bundle | Zero Prep | Differentiate | Homework
$2.00
Original Price $2.00
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Finding Constant of Proportionality | Unit Rate From a Graph Riddle | Exit Slips
Created by
Level Their Learning
In this engaging riddle, students will have to write a ratio and find the unit rate from a graph (constant of proportionality). 10 problems for students to write ratios and find the unit rate from a graph (constant of proportionality) + 2 spots for them to create their own graph and find the unit rate (constant of proportionality) from it! No decimals are used. I made this for kids on target when teaching how to write a ratio and find the unit rate from a graph (constant of proportionality). TWO
7 th
Math
CCSS
7.RP.A.1
, 7.RP.A.2
, 7.RP.A.2b
Also included in:Finding Constant of Proportionality | Unit Rate From A Graph Bundle
$2.00
Original Price $2.00
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Writing Ratios | Finding the Unit Rate from Ratio Tables Riddle and Exit Tickets
Created by
Level Their Learning
In this engaging riddle activity, students will have to write a ratio from the ratio table and then find the unit rate from the ratio table. 10 problems for students to write a ratio from a ratio table and to find the unit rate from the ratio table + 2 spots for students to create their own ratio tables and find the unit rate from that ratio table (Higher Order Thinking Skills!) No decimals are used in these ratio tables. I made this for kids on target when teaching writing ratios from ratio tab
6 th - 7 th
Math
CCSS
6.RP.A.1
, 6.RP.A.2
, 6.RP.A.3
+3
Also included in:6th Grade Year Long Activity Bundle- Riddles for a Whole Year!
$2.00
Original Price $2.00
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Unit Rates Guided Notes | Exit Ticket | Math
Created by
Small Town Math
Simple and easy Unit Rates Guided Notes and Exit Ticket. Perfect for a quick lesson for upper elementary or early middle school math classroom. Able to use resource as an introduction or refresh on how to calculate unit rates. Students needs to be able to have some understanding of rates and how to solve before using this product. I personally use during my math workshop model in small groups for individualized learning. Everything is done for you just print and implement in your classroom.
4 th - 7 th
Applied Math, Math, Other (Math)
CCSS
5.MD.A.1
, 6.RP.A.2
, 6.RP.A.3
+1
Also included in:Math Guided Notes & Exit Ticket BUNDLE | Middle School Math
$1.00
Original Price $1.00
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Middle School Daily Math Exit Slip/Warm-Up:Unit Rates and Converting Rates
Created by
Master Mathematics
This resource is great for an exit slip or warm-up. There is one constructed response problem for each day of the week. The topics included in this resource are calculate unit rates, simplifying complex fractions and converting rates. This resource would also be great to use for a constructed response and practice justifying answers. I have also included a worksheet for students with tips on helping them justify their answers more effectively. This resource was designed to encourage my students
6 th - 12 th
Arithmetic, Basic Operations, Math
CCSS
7.RP.A.1
, 7.RP.A.2
, 7.RP.A.2a
$0.99
Original Price $0.99
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Unit Rates Guided Notes, Practice, Exit Ticket | No Prep
Created by
IncrediBell Math
Guided notes, independent practice, and an exit ticket for unit rates. There are two guided notes worksheets to choose from. Both start with creating a table and then one uses a proportion to set up a one-step equation and the other uses a proportion and scale factor. This is a ready to use resource, you just need to print! This can be used whole group or small group. This resource includes: ⭐ Guided notes using a proportion and cross products to write a one-step equation and 4 example pr
6 th - 7 th
Math, Math Test Prep
CCSS
7.RP.A.1
$2.00
Original Price $2.00
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Unit Rates Exit Ticket
Created by
Middle School Queens
DescriptionThis 5 question Google Forms Exit Ticket is a great end of class activity to see what your students learned about reflecting points during your lesson. It is completely editable and can be used for online learning or in the classroom. Steps to use:Sign in to your Google account and clink on the link. The link will prompt you to make a copy.You can then rename the file and save to your Google Drive.Customer ServiceIf you have questions or problems with this assignment, feel free to e
6 th
Math
CCSS
6.RP.A.2
Also included in:Unit Rates Bundle
$1.50
Original Price $1.50
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Ratios, Rates, Proportions and Percents Unit 6th Grade Math Curriculum
Created by
Beyond the Worksheet with Lindsay Gould
⭐️⭐️⭐️ This ultimate 6th grade math ratios, rates, proportions and percent unit is COMPLETELY EDITABLE and is designed to help facilitate the needs of your students! ✅ This unit includes six multi-day lessons that cover the following skills: Ratios and Rates (including tape diagrams)Proportions and ScaleUnits of MeasurePercent of a NumberFractions, Decimals and Percents RelationshipsCompare and Order Fractions, Decimals and Percents✅ This unit includes:Google forms version of all assessmen
6 th
Math, Other (Math)
CCSS
6.NS.C.7
, 6.RP.A.1
, 6.RP.A.2
+1
Also included in:6th Grade Math Curriculum: Guided Notes, Practice, Assessments and More!
$35.99
Original Price $35.99
Rated 4.89 out of 5, based on 103 reviews
4.9(103)
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Unit Rates with Complex Fractions - Task Card Activity (7.RP.1)
Created by
Math on the Move
This activity consists of 20 task cards and a recording sheet. Students compute unit rates associated with ratios of fractions. Students can work independently or with partners, moving around the room while solving the problems. This activity can be good practice for students who struggle with how to set up the problem. Each situation has two unit rate questions. For example, if the problem involves miles and minutes, one task card asks for min/mi. and another will ask for mi/min. An e
7 th - 8 th
Math
CCSS
7.RP.A.1
Also included in:Unit Rates with Fractions - Mini Bundle (Notes, Worksheets, Activities)
$3.00
Original Price $3.00
Rated 4.93 out of 5, based on 186 reviews
4.9(186)
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Ratios, Rates, Proportions and Percents Unit : 7th Grade Math Curriculum
Created by
Beyond the Worksheet with Lindsay Gould
⭐️⭐️⭐️ This COMPLETELY EDITABLE Ratios, Rates, Proportions and Percents Unit for 7th Grade Math is designed to help you best meet the needs of your students! ⭐️⭐️⭐️ ⭐️ This unit includes FOUR multi-day lessons that cover: ⭐️ ✅ Represent and Solve Proportional Relationships ✅ Identify and Compute Unit Rates ✅ Multi-Step Percent Problems ✅ Scale Drawings ⭐️ ALSO included in this resource: ⭐️ ✅ Google Forms versions of the assessments ✅ Weekly warm up recording sheets ✅ Weekly exit ticket s
7 th
Algebra, Math
CCSS
7.G.A.1
, 7.RP.A.1
, 7.RP.A.2
+5
Also included in:7th Grade Math Curriculum: Guided Notes, Practice, Assessments and More!
$22.99
Original Price $22.99
Rated 4.79 out of 5, based on 128 reviews
4.8(128)
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Unit Rates with Complex Fractions - Notes and Practice (7.RP.1)
Created by
Math on the Move
This product contains two interactive notebook pages that can be used to teach setting up and computing unit rates associated with ratios of fractions. The INB pages begin with a skills review (dividing fractions) and a concept review (unit rates). Students then discuss the definition and examples of complex fractions. The second INB page has two practice word problems. These pages can be cut out and glued into students' notebooks. Three practice problems can be assigned as in-class pract
7 th - 8 th
Math
CCSS
7.RP.A.1
Also included in:Unit Rates with Fractions - Mini Bundle (Notes, Worksheets, Activities)
$3.00
Original Price $3.00
Rated 4.93 out of 5, based on 135 reviews
4.9(135)
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Ratios & Proportions, Equivalent Ratios, Unit Rates Notes Worksheets 6th Grade
Created by
Cognitive Cardio Math
Your 6th grade ratios, rates, and proportions instruction is easy to plan with this editable unit that has clear, structured notes, guided and independent practice problems, exit tickets, word wall words, and a unit assessment. This easy-to-use ratios and proportions unit includes a zip file with both a PDF version and an editable PowerPoint version (Microsoft Office 365) of each of the following: 1) Student packet of notes and practice pages (25 pages, 7 lessons) 2) Teacher version/key with
6 th
Math, Math Test Prep, Other (Math)
CCSS
6.RP.A.1
, 6.RP.A.2
, 6.RP.A.3
+4
Also included in:6th Grade Math Warm Ups, Anchor Charts, Math Wheel Note Taking, Units, Test Prep
$11.50
Original Price $11.50
Rated 4.93 out of 5, based on 48 reviews
4.9(48)
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